Download to read offline
![F 4
( ) =
F <-This does not work
F x
( ) x
2
4
−
:=
1.6 GRAPHICS
It has been demonstrated that Mathcad can be used as a calculator and as an equation solver.
It can also be used as a versatile visualization tool that supports a full set of plot types, an
animation facility, as well as simple image processing.
1.7 GRAPHING OF FUNCTIONS AND PLOTTING OF DATA
To create a plot in Mathcad, type the expression you would like to plot, say t
9
. Then click the
graphing button on your palette bar (See Figure 1.2 ) to bring up the graphing palette and click
the upper left button for an XY plot. Click outside the graph or press [Enter] to see the result:
4
− 2
− 0 2 4
1
− 10
5
×
5
− 10
4
×
0
5 10
4
×
1 10
5
×
t
9
t
Parametric plots can also be done easily . .
1
− 0.5
− 0 0.5 1
1
−
0.5
−
0
0.5
1
sin 5 t
⋅
( )
cos 7 t
⋅
( )
Chapter 1: Basics of Mathcad 5](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-15-320.jpg)
![Several mathcad palettes can be found on the palette strip at the top of the window that give
access to Mathcad's mathematical operators. A click in your worksheet will place a red
crosshair cursor. Math operators can now be placed in the worksheet via these palettes. To
access Mathcad's built-in functions, go to the Insert menu as shown in Figure 1.5, and
select Function , or click on the Insert Function button on the toolbar
.
EXAMPLES
In the following examples, the number of decimal places required may be put in by choosing
Result from the Format menu and then selecting Number Format.
The red crosshair
Use the Arithmetic Palette (See Figure 1.1)
, and type = to obtain the answer. The basic
operations are +, -, *, and / which are
available on the keyboard .
2.645 10
8
⋅
30007
5
5
6
+
1.0426928428 10
7
−
×
= <--
<- Examples of built-in functions (See
Figure 1.5)
log 1768.985
( ) sin
4
79
π
⋅
⎛
⎜
⎝
⎞
⎟
⎠
⋅ 0.514
=
14.95 5.7i
( )
+
[ ]
3
e
6 5i
−
+ 1.999 10
3
× 4.024i 10
3
×
+
= <-- Complex numbers.
Use of units- Choose Unit from
the Insert menu (See Figure 1.5)
<--- .
7600 km
⋅
1 hr
⋅
2.11 10
3
× m s
1
−
⋅
=
Use of := by typing a colon
character or using the
Arithmetic Palette
a 8
:= a
3 a
4
+ 513.414
= <----
f x
( )
sin x
( )
a
x
6
+
:= f 37
( ) 0.045
−
= <-- Use of a defined function
Chapter 1: Basics of Mathcad 11](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-21-320.jpg)
![Controlling Calculation and The Status Line
The calculation mode, whether manual or automatic, is a property saved in your worksheet and
template files. Mathcad starts in the automatic mode and all calculations and results are updated
automatically. The word " Auto " can be seen in the message line at the bottom of the Mathcad
window. This line provides status alerts, tips, keyboard shortcuts, and other helpful information
along with the calculation status of the worksheet . Here, "auto" refers to the automatic
mode , in which Mathcad automatically recalculates any math expressions when changes in the
worksheet are made. When "WAIT" appears on the status line, the cursor changes to a flashing
lightbulb, indicating that Mathcad is still completing computations. Besides giving the page
number of the current worksheet, the message line will also indicate whether the Caps Lock or
the Num Lock key is depressed on your keyboard. In manual mode, Mathcad does not
compute or show results until recalculation is specifically requested. However, while in manual
mode, Mathcad does keep track of pending computations . Once a change is made that requires
recalculation, the word " Calc " appears on the status line to indicate to the user that the results
being displayed are not correct and that recalculation is necessary to ensure accuracy. The
screen can then be updated by going to the Tools menu and choosing Calculate and then
Calculate Now . Alternatively, click = on the Standard toolbar or press [F9]. To force
Mathcad to recalculate all equations in the worksheet, go to the Tools menu and choose
Calculate and then Calculate Worksheet .
1.13 MATHCAD REGIONS
Mathcad allows you to enter equations, text and graphs as separate objects anywhere in the
worksheet and each equation, text paragraph or plot is considered a region. A region can be
selected by clicking in math or text in your worksheet, after which it is indicated by a thin
rectangle around it. Moving the cursor to one of the edges of the region, will change it to a
small hand with which the region can be moved to anywhere in the worksheet. While clicking
in the math region will bring blue selecting lines under the material selected, clicking in a text
region will bring black boxes to each corner and the middle of each line. With these boxes
text regions can be resized as needed. To add a border around a region or regions, select the
region(s), then right-click and choose Properties from the menu. Then click on the Display tab
and check the box next to " Show Border ".
A math region looks like: A text region looks like:
x 1535.56
:= What is shown on the left was
created by typing
x:1535.56
Chapter 1: Basics of Mathcad 15](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-25-320.jpg)
![f x
( ) x 6
+
( ) x
3
5
−
( )
⋅
:=
Type
f(x):x+6[Spacebar]*(x^3[Spacebar]-5)
Example 2
To create
f x
( ) x 765 x
6 4
−
( )
⋅
+
:=
Type
f(x):x+765*(x^6-4)
The exponent operator is called a sticky operator because unless you get out by pressing
[Spacebar] , your keystrokes will "stick" to the exponent . This stickiness applies to
exponents, square roots, subscripts, and division.
Example 3
To create the expression
x
7
675
+
780
Type
x^7[Spacebar] <-- Puts x 7 in blue editing lines
+675[Spacebar] <-- Puts the whole equation in the editing lines
/780[Enter] <-- Expression is complete
Example 4
To create
22 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-32-320.jpg)
![x
7 675
780
+
do the same as above but without pressing the spacebar after +675
Example 5
To create
x
17
t
7
450
type
x^17/t^7[Spacebar][Spacebar][Spacebar]/450
Editing Expressions
The equation editor functions very much like a text editor and goes from left to right. Most
problems dealing with editing equations stem from working with operators. Although Mathcad
automatically inserts parentheses wherever necessary, the Mathcad user must put in
parentheses himself in accordance with his own judgement to give clarity to expressions .
When expressions become complicated, it is definitely preferable to work with smaller and
more manageable subexpressions within them. The reader is referred to Chapter 4 of the
Mathcad 14 User's Guide for a detailed discussion on Editing Expressions. This can be
accessed by choosing Tutorials from the Help menu.
1.19 DEFINING RANGE VARIABLES
To assign a range of values to a variable x going from 0 to 30, say, with an increment
of 3, type
x:0,3;30
The result on the screen will be
x 0 3
, 30
..
:=
To evaluate a function f(x) for x= 0 to 7 with an increment of 1, type
x:0;7 f(x):x^5[Spacebar]-6*x^3[Spacebar]+9
Chapter 1: Basics of Mathcad 23](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-33-320.jpg)
![To create a vector V , click in a blank space, Choose Matrix from the Insert menu (See
Figure 1.6 ) or click on the button inside the Vector and Matrix Palette . Then, fill in the
appropriate number of rows and columns, click on Insert and finally fill in the placeholders with
given values. To move from placeholder to placeholder inside the vector, use [Tab] or click on
the appropriate placeholder to select it
V
7
8
9
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
To access a particular element of a vector, use the subscript operator , which can be
created by typing a left square bracket ( [ ), or by using the Xn button in the Matrix
Palette .
In Mathcad , by default the first element has the index 0. Type
V[0=
to see on the screen
V0 7
=
The next two elements will have indices 1 and 2 Thus typing V[1= and V[2= will
produce on the screen
V1 8
= V2 9
=
In order to obtain all the elements of the vector at the same time, the index can be defined
as a range variable as shown below::
i 0 2
..
:= giving Vi
7
8
9
=
<--- Type V[i=
The elements of a vector can also be used as the arguments of a function as shown below.
Define a function f(x) as.
f x
( ) sin 2 x
⋅
( ) cos 3 x
⋅
( )
+
:=
Let a vector V be a three-element vector as given below
26 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-36-320.jpg)
![v
1.3
2.6
−
7.9
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
Then, typing i:0;2 and then f(v[i)= will show the following on the screen
i 0 2
..
:= f vi
( )
-0.21
0.937
0.046
=
The Vector and Matrix Palette contains most of the common matrix and vector operators.
These are listed below.
Operation Keystroke
Palette
Button Display
Dot product [Shift]8 v w
⋅ Displays like scalar
multiplication.
Cross product [Ctrl]8 v w
×
Determinant | M
Column [Ctrl]6 M
2
〈 〉
Returns the 2nd
column of M.
There is also a wide range of built-in functions in Mathcad that return information about the
size of an array and its elements. For a given matrix [M], various pieces of information can
be generated as shown below.
M
9
4
12
10
5
13
11
7
14
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
Chapter 1: Basics of Mathcad 27](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-37-320.jpg)
![Number of columns: cols M
( ) 3
= Number of rows: rows M
( ) 3
=
Largest value in matrix: max M
( ) 14
= Determinant of matrix: M 3
=
Eigenvalues of a matrix: eigenvals M
( )
28.768
0.176
−
0.592
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
1.21 CREATING GRAPHS
To create an x-y plot, choose Graph/ X-Y Plot from the Insert Menu as shown in Figure
1.7. Press [Enter] and fill in the placeholders on the x and y axes. Alternatively, type an
expression depending on one variable such as: sin(x) + cos (3x) and click on the X-Y Plot
button on the Graph toolbar . The resulting plot will be
x 10
− 9.9
−
, 10
..
:=
10
− 5
− 0 5 10
2
−
1
−
1
2
sin x
( ) cos 3 x
⋅
( )
+
x
28 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-38-320.jpg)
![Using Range Variables to Plot a Function
The range of values being plotted on the x-axis must, in general, be specified. If this range is
not prescibed by you, Mathcad will choose a default range for the dependent variable. For
example, let us plot
f x
( ) x
3
8 x
2
⋅
+ 9 x
⋅
+ 14
+
:=
The graph of this function over the range x= 0 to x= 6 can be accomplished as follows.
x 0 1
, 6
..
:= f x
( ) x
3
8 x
2
⋅
+ 9 x
⋅
+ 14
+
:=
Now create the plot by clicking in the worksheet window. Type @ to create the x-y plot, type
x in the middle placeholder on the horizontal axis and type f(x) in the middle placeholder
on the vertical axis. Press [Enter] . The following graph should then appear on the screen.
0 2 4 6
0
200
400
600
f x
( )
x
The plot generated above does not seem to be very smooth. In order to obtain a smoother
trace, change the definition of x to x 0 0.1
, 6
..
:= . The smaller step enables Mathcad to
calculate more points. This will make the plot a lot smoother (see graph below ) because now
there are more points or dots being connected together.
Formatting of an x-y plot can be accomplished as follows. Double-click on it or choose Graph
from the Format menu to open up a dialog box. This dialog box will allow several options in
terms of grid lines, legends, trace types, markers, colors, axis limits etc.
30 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-40-320.jpg)
![x 0 0.1
, 6
..
:=
0 2 4 6
200
400
600
f x
( )
x
Plotting a Vector of Data Points
Let us say that we need to plot a vector of data points called Temp , which is the rising
outside temperature on different days of a week in a summer month. This vector which will
have 7 rows and 1 column can be created using the Matrix command on the Insert menu.
Temp
78
82
84
86
88
90
92
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
This data can be graphed by making the horizontal axis an index variable or a vector with
the same number of elements as the vector Temp . Define the index i as:
i 0 6
..
:=
Create your plot by typing [@] and typing in
Temp[i
Chapter 1: Basics of Mathcad 31](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-41-320.jpg)
![ε xmidi 1
+ xmidi
−
= (3.1)
(3.2)
εrel
xmidi 1
+ xmidi
−
xmidi 1
+
100
⋅
= percent
where xmid i+1 and xmid i are the midpoints in the current and previous iterations. While,
in general, the relative error should not be greater than 5 %, an error of 0.01 % is the
largest that is tolerable for some classes of problems, that require immense precision.
The true error , ε True , is an indicator of the real accuracy of a solution and can be
evaluated only if the true solution, x True , is known. It is defined as
εTrue
xTrue xmidi
−
xTrue
100
⋅
= (3.3)
Calculation of the true error clearly requires knowledge of the true solution, which, in general,
will not be known to us. Therefore, the quantity ε rel may have to be mostly used to
determine the error associated with a solution process.
Advantages and Disadvantages of Bisection
While Bisection is a simple, robust technique for finding one root in a given interval, when
the root is known to exist and it works even for non-analytic functions, its convergence
process is generally slow, making it a somewhat inefficient procedure.
Sometimes, a singularity may be identified as if it were a root, since the method does not
distinguish between roots and singularities, at which the function would go to infinity.
Therefore, as the method proceeds, a check must be made to see if the absolute value of
[f(xend)-f(xstart)], in fact, converges to zero. If this quantity diverges, the method is chasing
a singularity rather than a root .
When there are multiple roots, Bisection is not a desirable technique to use, since the
function may not change signs at points on either side of the roots. Therefore, a graph of
the function must first be drawn before proceeding to do the calculations.
Example 3.1
Obtain a root of f(x) in the range of 4 < x < 20
f(x)= ( 750.5 / x ) [ 1 - exp( -0.15245x) ] - 40
Chapter 3: Roots of Equations 55](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-65-320.jpg)
![x
0
0
1
2
3
4
5
6
7
8
9
10
0
-1
-0.8774
-0.9381
-0.9095
-0.9233
-0.9167
-0.9199
-0.9184
-0.9191
-0.9188
=
<--- Convergence occurs here ---> y x10
( ) 5.0333
− 10
4
−
×
=
3.9 MULTIPLE ROOTS AND DIFFICULTIES IN COMPUTATION
When the given function makes contact with the x- axis and has a slope of zero, multiple roots
can occur as indicated in Figure 3.5. These roots are difficult to compute by the methods
discussed in this chapter for the following reasons.
1. In the case of the Bisection method, the function does not change sign at the root.
2. In the case of Newton-Raphson and Secant methods, the derivative at the
multiple root is zero.
Figure 3.5. Case of multiple roots
Because the Newton-Raphson and Secant methods in their orginal forms are methods that
resort to linear convergence, they cannot be employed to generate multiple roots. The following
modification to the original Newton-Raphson equation has been suggested [ 4 ].
80 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-90-320.jpg)
![xi 1
+ xi m
f xi
( )
fprime xi
( )
⋅
−
= (3.13)
where m is the multiplicity of the root ( for example, m = 2 for a double root ). However, this
may not be a very practical route since it assumes prior knowledge about the multiplicity of
a root.
Another approach, also suggested [ 4 ] involves defining a new function u(x)
u x
( )
f x
( )
fprime x
( )
= (3.14)
which has the same roots as the original given function f(x), and then resorting to the
following alternative form of the Newton-Raphson formula
xi 1
+ xi
u xi
( )
uprime xi
( )
−
= (3.15)
where uprime(x) represents the derivative of the new function u(x). Substitution of Equation
(3.14) into the Newton-Raphson equation, which is now Equation (3.15), yields the desired
formula
xi 1
+ xi
f xi
( ) fprime xi
( )
⋅
fprime xi
( )2
f xi
( ) fdoubleprime xi
( )
⋅
−
−
= (3.16)
where fprime and fdoubleprime represent first and second derivatives respectively.
The modified formula for the Secant Method that will be suitable for the generation of multiple
roots can be obtained by simply substituting u(x) for f(x) in the Secant method formula giving
xi 1
+ xi
u xi
( ) xi 1
− xi
−
( )
⋅
u xi 1
−
( ) u xi
( )
−
−
= (3.17)
Example 3.9
The function
f(x)= x3 - 4 x 2 + 5 x - 2 = (x-1) (x-1) (x-2)
Chapter 3: Roots of Equations 81](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-91-320.jpg)
![(11) x 1.8429
= y 3.0579
= x
10.5 1.2 x
⋅ y
⋅
−
1.1
:= x 1.8433
= y
58.2 y
−
3.2 x
⋅
:= y 3.0575
=
Notice that after ten iterations convergence has taken place yielding the solution as : x =1.843
and y = 3.058. This example illustrates a very serious disadvantage associated with the
successive substitution method which is the dependence of convergence on the format in which
the equations are put and utilized in the iteration process. Also , even in situations where a
converged solution can be attained, initial estimates that are fairly close to the true solution
must be resorted to, because, otherwise, divergence may occur and a solution may never be
obtained. The results of the iterative process are summarized in the following table.
----------------------------------------------------------------------------------------------------------------------------
Iteration # x y
------------------------------------------------------------------------------------------------------------------------------
0 init=1.0 3.00
1 2.505 2.624
2 1.541 3.357
3 1.975 2.946
4 1.788 3.107
5 1.866 3.037
6 1.833 3.066
7 1.847 3.054
8 1.841 3.059
9 1.844 3.057
10 1.843 3.058
11 1.843 3.058
----------------------------------------------------------------------------------------------------------------------------------
3.11 SOLVING SYSTEMS OF EQUATIONS USING MATHCAD'S Given
AND Find FUNCTIONS
Mathcad allows you to solve a system of 50 simultaneous equations in 50 unknowns. The
procedure is as follows:
1. Provide initial guesses for all unknowns. These initial guesses give Mathcad a place to
start searching for the solution.
2. Type the word Given. This tells Mathcad that what follows is a system of equations. Given
can be typed in any combination of upper and lower case letters, and in any font. However, it
should not be typed in a text region.
3. Then type the equations and inequalities in any order below the word Given . Make sure
you separate the left and right sides of an equation by an " = ". Press [Ctrl]= to type " = ".
You can of course separate the left and right sides of an inequality with the appropriate symbol.
4. a:= Find(x,y.....) returns the solution to the system of equations.
86 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-96-320.jpg)
![Example 3.11.
Using the Given and Find functions, solve
1.5 x
2
⋅ 2 y
2
⋅
+ 6.5
= . x y
+ 2.75
=
Provide initial guesses: x 1
:= y 1
:=
Given
1.5 x
2
⋅ 2 y
2
⋅
+ 6.5
=
x y
+ 2.75
=
a Find x y
,
( )
:=
a
1.5
1.25
⎛
⎜
⎝
⎞
⎟
⎠
= <--- Solution
Alternatively, do the following sequence to obtain xval and yval as the solution
Given 1.5 x
2
⋅ 2 y
2
⋅
+ 6.5
= x y
+ 2.75
=
xval
yval
⎛
⎜
⎝
⎞
⎟
⎠
Find x y
,
( )
:= xval 1.5
= yval 1.25
= <--Solution
3.12. APPLICATIONS IN ROOT-FINDING
The following are examples of situations where roots of nonlinear equations need to be
evaluated.
3.12.1 Maximum Design Load for a Column
A steel pipe with an outside diameter of 10.70 inches and a wall thickness of 0.375 inches is
to serve as a pin-ended column that is 22 feet long . It supports an eccentric load at an
eccentricity of e = 1.5 inches from the column center-line. The maximum load that the
column can carry with a factor of safety of 2.75 is to be determined. The modulus of
elasticity E for steel is 30 x 10 6 psi and its yield strength, σyld , is 40,000 psi .
The equation for the stress level σ max resulting from the application of the maximum load
P max is [ 3,18 ]
Chapter 3: Roots of Equations 87](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-97-320.jpg)
![e 1.50
:= L 264.
:= c
do
2
:= c 5.35
=
I
π
64
do
4
di
4
−
( )
⋅
:= ( in 4 ) A
π
4
do
2
di
2
−
( )
⋅
:= (in 2 )
A 12.1639
= I 162.3057
= K
I
A
:= K 3.6528
= (in )
With a factor of safety of 2.75, the maximum stress level that the column can have is
σmax
σyld
FS
:= σmax 1.4545 10
4
×
= psi
The corresponding maximum allowable load may be determined from Equation (3.18) using
the root function as follows.
g Pmax
( )
Pmax
E A
⋅
L
2 K
⋅
⋅
:=
f Pmax
( ) σmax
Pmax
A
1
e c
⋅
K
2
sec g Pmax
( )
( )
⋅
+
⎛
⎜
⎝
⎞
⎟
⎠
⋅
−
:=
Pmax 125000.
:= (lbs) <--Guess value for the root
a root f Pmax
( ) Pmax
,
( )
:=
a 1.0221 10
5
×
= (lbs) <-- Maximum allowable load column can carry
3.12.2. Natural Frequencies of Vibration of a Uniform Beam
For a uniform beam that is clamped at one end and free at the other, a shown in Figure 3.7 ,
the natural frequencies of vibration [20] are given by the equation
1 cos βprime
( ) cosh βprime
( )
⋅
+ 0
= (3.19)
Chapter 3: Roots of Equations 89](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-99-320.jpg)
![β
a
L
:= β 0.0938
=
ω
EI β
4
⋅
ρ
:= ω 27.7106
= (rad/sec) <-- First or fundamental
natural frequency
βprime 5
:= <-- Guess value for next root
f βprime
( ) 1 cos βprime
( ) cosh βprime
( )
⋅
+
:=
a root f βprime
( ) βprime
,
( )
:=
a 4.6941
=
β
a
L
:= β 0.2347
=
ω
EI β
4
⋅
ρ
:= ω 173.6595
= (rad/sec) <-- Second natural frequency
Similarly by making appropriate guesses, higher natural frequencies can be determined
using the root function. Some of these are listed below.
Third natural frequency: 486.3 rad/sec
Fourth natural frequency: 952.8 rad/sec
Fifth natural frequency: 1575.1 rad/sec
Sixth natural frequency: 2353 rad/sec
Seventh natural frequency: 3286 rad/sec.
3.12.3. Solving the Characteristic Equation in Control Systems Engineering
A characteristic equation is an algebraic equation that is formulated from the differential
equation or equations of a control system [14] . Its solution, which often requires evaluation
of the roots of a polynomial of degree higher than two, is crucial in determining system
stability and assessing system transient response in terms of its time constant, natural
frequencies , damping qualities etc. An application involving the use of Mathcad' s polyroots
function in determining the roots of a characteristic polynomial of a control system is
presented below.
Chapter 3: Roots of Equations 91](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-101-320.jpg)
![<-- Negative
real parts
of root
indicate a
stable
system
K 100
:= v
100
20
12
1
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
:= r polyroots v
( )
:= r
11.0084
−
0.4958
− 2.9729i
+
0.4958
− 2.9729i
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
<-- Positive
real
parts of root
indicate an
unstable
system
K 400
:= v
400
20
12
1
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
:= r polyroots v
( )
:= r
12.8628
−
0.4314 5.5598i
+
0.4314 5.5598i
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
3.12.4. Horizontal Tension in a Uniform Cable
Flexible cables are often used in suspension bridges, transmission lines, telephone lines ,
mooring lines and many other applications. A cable hanging under its own weight takes the
shape of a catenary in which the deflection y along the span is given by the following
relationship [ 8 ]
y
H
w
cosh
w x
⋅
H
⎛
⎜
⎝
⎞
⎟
⎠
1
−
⎛
⎜
⎝
⎞
⎟
⎠
⋅
= (3.22)
where x is the horizontal distance along the span a shown in Figure 3.9 , H is the horizontal
component of tension in the cable and w is the weight per unit length of the cable.
Figure 3.9. Catenary cable
Chapter 3: Roots of Equations 93](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-103-320.jpg)
![3.24. The dynamic response of an underdamped second-order system, which is a
common mathematical model in control system analysis, to a unit step function is given
by [ 14 ]
c t
( ) 1
1
1 ζ
2
−
e
ζ
− ωn
⋅ t
⋅
⋅ cos ωn 1 ζ
2
−
⋅ t
⋅ ϕ
−
⎛
⎝
⎞
⎠
⎡
⎣
⎤
⎦
⋅
−
=
where ζ is the system damping ratio, ω n is the undamped natural frequency, and φ is
ϕ atan
ζ
1 ζ
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
A precise analytical relationship between the damping ratio and the rise time which is the time
required for the response to go from 10 % of the final value to 90 % ( that is, from 0.1 to 0.9 in
this case ) cannot be determined . However, using Mathcad's root function, the normalized
rise time which is ω n times the rise time, can be found for a specified damping ratio ζ by
solving for the values of ω n t that yield c(t) = 0.9 and c(t)= 0.1 . Subtracting these two values
yields the normalized rise time for the value of ζ considered . Using a range of damping ratios
from 0.1, 0.2, 0.3... .. 0.9 , construct a plot depicting normalized rise time as a function of
the damping ratio .
3.25. The frequency equation for the free undamped longitudinal vibration of a slender bar
of length L and mass m with one end fixed and with the other end carrying a mass M is
given by [ 21 ]
β tan β
( )
⋅ α
=
where
β ω
L
c0
⋅
=
c0
E
ρ
=
Figure P 3.25.
100 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-110-320.jpg)
![α
m
M
=
in which E is Young's modulus for the material of the bar, ρ is the mass per unit volume and
ω is the natural frequency .
For a steel bar with mass ratio α = 0.70, E = 30 x 10 6 psi , L = 98 in., and ρ = 7.35x 10 -4
lb-sec 2 / in 4 , determine the first three natural frequencies.
3.26. For a uniform beam of length L with one end fixed and the other simply supported, the
natural frequencies of transverse vibration are given by the relationship [ 21 ]
tan k L
⋅
( ) tanh k L
⋅
( )
=
where
k
2
ω
ρ A
⋅
E I
⋅
⋅
=
Figure P 3.26
in which ω is the natural frequency , ρ is the mass density of the beam material, A is the
area of cross section of the beam , E is Young's modulus, and I is the moment of inertia of
the beam cross section.
Determine the first three natural frequencies of a 36 in. long steel beam whose area of
cross section A is 12.50 in 2 , moment of inertia I is 12.75 in 4, and ρ = 7.35x 10 -4 lb-sec 2 /
in 4 . E for steel is 30 x 10 6 psi.
Chapter 3: Roots of Equations 101](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-111-320.jpg)
![For the above case, C12 1
−
( )
3
31
( )
⋅
:= , that is, C12 31
−
=
Adjoint Matrix:
The adjoint of a square matrix is the transpose of the matrix of its cofactors.
Inverse of a Matrix:
The inverse A -1 of a matrix A must satisfy the relationship
A A -1 = I , (4.3)
where I is the Identity or the Unit matrix which is a matrix with "1" along the diagonal and
zero elsewhere. The inverse A -1 is calculated as follows
A -1 = [adjoint (A)] / Det (A) , (4.4)
where Det (A) is the determinant of the given matrix, which must be non-zero for the inverse to
exist.
Example 4.1
Determine the inverse of the following matrix A
A
1
1
1
2
3
0
3
2
4
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
The determinant of this matrix is A 1
−
=
The minors of A are
M11
3
0
2
4
⎛
⎜
⎝
⎞
⎟
⎠
:= M12
1
1
2
4
⎛
⎜
⎝
⎞
⎟
⎠
:= M13
1
1
3
0
⎛
⎜
⎝
⎞
⎟
⎠
:=
M21
2
0
3
4
⎛
⎜
⎝
⎞
⎟
⎠
:= M22
1
1
3
4
⎛
⎜
⎝
⎞
⎟
⎠
:= M23
1
1
2
0
⎛
⎜
⎝
⎞
⎟
⎠
:=
104 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-114-320.jpg)
![Given the two matrices A and B:
A
1
1
7
0
3
8
2
2
9
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= B
3
1
1
2
2
5
1
3
9
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
Addition:
A B
+
4
2
8
2
5
13
3
5
18
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Subtraction:
A B
−
2
−
0
6
2
−
1
3
1
1
−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Multiplication:
<--Use the asterisk ( * ) to multiply two matrices
A B
⋅
5
8
38
12
18
75
19
28
112
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
<-- Order in which multiplication is done is important.
Note that [A] [B] is not equal to [B][A]
B A
⋅
12
24
69
14
30
87
19
33
93
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Scalar subtraction:
A 5
−
4
−
4
−
2
5
−
2
−
3
3
−
3
−
4
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Chapter 4: Matrices and Linear Algebra 107](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-117-320.jpg)
![Scalar multiplication: ( Use the asterisk "*" )
A 5
⋅
5
5
35
0
15
40
10
10
45
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Matrix inverse: Use : ^ -1
A
1
−
0.733
−
0.333
−
0.867
1.067
−
0.333
0.533
0.4
0
0.2
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Determinant of A : A 15
−
=
Transpose of A: Use [Ctrl]1
A
T
1
0
2
1
3
2
7
8
9
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Matrix subscripts:
Use " [ " to put in subscripts. Note that, in this case, subscripts will go from 0 to 2.
because, in Mathcad, vector and matrix elements are ordinarily numbered starting with row
zero and column zero. To make " one " the " ORIGIN " , go to the Tools menu, choose
Worksheet Options , as shown in Figure 4.2 , and change the array origin from 0 to 1.
4.3 SOLUTION OF LINEAR ALGEBRAIC EQUATIONS BY USING
THE INVERSE
For a system of linear algebraic equations that can be written in matrix form as
[A]{X} = {B} (4.5)
where the matrix [A] and the column vector {B} contain known constants, the inverse of the matrix
[A] can be utilized in the determination of the solution vector {X} as follows
{X} = [A] -1 {B} (4.6)
108 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-118-320.jpg)
![Example 4.2
Solve: 2x + 4 y + z = -11
-x + 3 y -2 z = -16
2x - 3y + 5 z = 21
Putting the given equations in the form [A] {X}= {B}, where
A
2
1
−
2
4
3
3
−
1
2
−
5
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= B
11
−
16
−
21
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
The inverse of [A] can be easily obtained as
A
1
−
0.474
0.053
0.158
−
1.211
−
0.421
0.737
0.579
−
0.158
0.526
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
and [A] multiplied by [A]-1 will yield the identity matrix as shown
A A
1
−
⋅
1
0
0
0
1
0
0
0
1
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
The solution can then be generated by performing the matrix operation
X A
1
−
B
⋅
:=
yielding
X
2
4
−
1
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
4.4 SOLUTION OF LINEAR ALGEBRAIC EQUATIONS BY
CRAMER'S RULE
As an alternative to the above procedure, the solution {X} to the matrix equation
(Equation 4.5)
110 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-120-320.jpg)
![[A] {X} = {B}
can be computed using Cramer's rule as shown below
Letting
A
A11
A21
A31
A12
A22
A32
A13
A23
A33
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= B
B1
B2
B3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
the solution {X} can be computed using
X1
B1
B2
B3
A12
A22
A32
A13
A23
A33
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
A11
A21
A31
A12
A22
A32
A13
A23
A33
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= X2
A11
A21
A31
B1
B2
B3
A13
A23
A33
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
A11
A21
A31
A12
A22
A32
A13
A23
A33
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
(4.7)
X3
A11
A21
A31
A12
A22
A32
B1
B2
B3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
A11
A21
A31
A12
A22
A32
A13
A23
A33
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
For Example 4.2,
A
2
1
−
2
4
3
3
−
1
2
−
5
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= B
11
−
16
−
21
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= D A
:= D 19
=
Chapter 4: Matrices and Linear Algebra 111](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-121-320.jpg)
![,
x
B1
B2
B3
A1 2
,
A2 2
,
A3 2
,
A1 3
,
A2 3
,
A3 3
,
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
D
:= y
A1 1
,
A2 1
,
A3 1
,
B1
B2
B3
A1 3
,
A2 3
,
A3 3
,
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
D
:= z
A1 1
,
A2 1
,
A3 1
,
A1 2
,
A2 2
,
A3 2
,
B1
B2
B3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
D
:=
where the matrix elements have been specified in the format required by Mathcad. These
calculations result in
x 2
= y 4
−
= z 1
=
4.5 SOLUTION OF LINEAR ALGEBRAIC EQUATIONS USING THE
FUNCTION lsolve
The lsolve function in Mathcad can be used to solve a system of linear algebraic equations .
Given:
[A] {X}= {B},
the function lsolve (A,B) will return the solution vector {X}, provided the matrix [A] is not
singular or nearly singular. For Example 4.2,
A
2
1
−
2
4
3
3
−
1
2
−
5
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= B
11
−
16
−
21
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
X lsolve A B
,
( )
:= gives X
2
4
−
1
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Example 4.3
Application of Matrices to Electrical Circuit Analysis.
Formulate a set of linear equations that represent the relationship between voltage, current and
resistances for the circuit shown in Figure 4.3. Solve the simultaneous equations generated using
(1) the inverse (2) Cramer's Rule and (3) the Mathcad function lsolve
112 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-122-320.jpg)
![Figure 4.3. Electrical circuit example
The governing equations are:
i1 + i2 = i3
7 i1 + 3 i3 = 6
7 i1 - 5 i2 = 6
These must be put in the form: [A] {X} = {B} and solved for {X}
Using the inverse
A
1
7
7
1
0
5
−
1
−
3
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= B
0
6
6
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
<--These are the currents i1, i2 and
i3
X A
1
−
B
⋅
:= X
0.6761
0.2535
−
0.4225
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Using Cramer's Rule
L
0
6
6
1
0
5
−
1
−
3
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= M
1
7
7
0
6
6
1
−
3
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= N
1
7
7
1
0
5
−
0
6
6
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
i1
L
A
:= i2
M
A
:= i3
N
A
:= i1 0.676
= i2 0.254
−
= i3 0.4225
=
Chapter 4: Matrices and Linear Algebra 113](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-123-320.jpg)
![Using Mathcad Function lsolve
X lsolve A B
,
( )
:= X
0.676
0.254
−
0.423
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= <--These are the currents i1, i2 and i3
4.6 THE EIGENVALUE PROBLEM
Eigenvalue problems occur in a variety of physical situations. Some examples of eigenvalue
problems are given below.
1. Determination of natural frequencies and mode shapes of oscillating systems.
2. Computation of principal stresses and principal directions
3. Computation of principal moments of inertia and principal axes.
4. Buckling of structures.
5. Oscillations of electrical networks.
The eigenvalue problem can be mathematically stated as
[A] {X} = λ {X} (4.8)
where [A] is a known square matrix, {X} is a column vector of unknowns, and λ is an unknown
scalar quantity. The solution of the eigenvalue problem involves the determination of vectors
{X} and associated constants λ satisfying equation (4.8). These vectors {X} are termed
eigenvectors. With each eigenvector is associated a constant λ which is called an
eigenvalue.
Equation (4.8) can be written in the form
[A- λ I ] {X} = {0} (4.9)
A trivial solution to the above is
{X} = 0
which is not of interest to us. The condition for the existence of a non-trivial solution is
det ( A- λ I ) = 0 (4.10)
114 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-124-320.jpg)
![Expansion of the above determinant produces a polynomial in λ. For instance, if [A] is a 3x 3
matrix, then Equation (4.10) would give a cubic in λ. This polynomial is called the
characteristic polynomial and its roots are called eigenvalues. Associated with each
eigenvalue is a solution vector {X} called the eigenvector.
It can be shown that the adjoint of matrix (A- λiI) contains columns, each of which is the ith
eigenvector {X} i associated with the eigenvalue λ i multiplied by an arbitrary constant.
Thus, to determine an eigenvector corresponding to an eigenvalue λi , compute the adjoint of
(A- λiI) and choose any column of it as the eigenvector.
Example 4.4
Compute the eigenvalues and eigenvectors of
17
45
6
−
16
−
⎛
⎜
⎝
⎞
⎟
⎠
A
17
45
6
−
16
−
⎛
⎜
⎝
⎞
⎟
⎠
:=
The [A- λI] matrix can be written as
A λ I
⋅
−
17 λ
−
45
6
−
16
− λ
−
⎛
⎜
⎝
⎞
⎟
⎠
=
A non-trivial solution is obtained by setting the determinant of [ A - λ I ] equal to zero
17 λ
−
45
6
−
16
− λ
−
⎛
⎜
⎝
⎞
⎟
⎠
0
=
which gives λ 2 - λ - 2 = 0, or, λ 1 = 2, λ 2 = -1. The adjoint of the matrix (A- λ I ), in
this case, is
16
− λ
−
45
−
6
17 λ
−
⎛
⎜
⎝
⎞
⎟
⎠
For λ 1 = 2, the corresponding eigenvector is the first or second column of the above
adjoint matrix with λ = λ 1 which gives
Chapter 4: Matrices and Linear Algebra 115](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-125-320.jpg)
![differential equations of motion can be seen to be
2 m d2x1 /dt2 + 3 k x1 - k x2 = 0
m d2x2 /dt2 +k ( x2 - x1 ) = 0
(4.11)
Expressing the above equations in matrix form as
..
(4.12)
[M] {x} + [K ] {x} = {0}
where [M] is the mass or inertia matrix , and [K] is the stiffness matrix, and assuming that
{x} = {A}sin ωt (4.13)
we obtain a set of linear homogeneous algebraic equations that can be written in matrix form as
(4.14)
([K]- ω2 [M ] ) {A} = 0
For a non-trivial solution, the determinant | K- ω2 M | must be zero. The problem can also be
posed slightly differently in the familiar form
|A-λI| = 0 (4.15)
where the λ 's ( λ = ω2) are the eigenvalues of the matrix [A] , which, in this case, is
[A] = [M] -1 [K] (4.16)
To find the natural frequencies and mode shapes in this case, we need to determine the
eigenvalues and eigenvectors of the matrix [A]. The square roots of the eigenvalues will then
give the natural frequencies and the corresponding eigenvectors will be the mode shapes.
Letting m = 1, and k =1 in this problem for convenience
m 1
:= k 1
:=
Then, the mass and stiffness matrices
M
2 m
⋅
0
0
m
⎛
⎜
⎝
⎞
⎟
⎠
:= K
3 k
⋅
k
−
k
−
k
⎛
⎜
⎝
⎞
⎟
⎠
:=
118 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-128-320.jpg)
![give
M
2
0
0
1
⎛
⎜
⎝
⎞
⎟
⎠
= K
3
1
−
1
−
1
⎛
⎜
⎝
⎞
⎟
⎠
=
The [A] matrix, which is called the dynamic matrix in the language of vibrations is ,
A M
1
−
K
⋅
:= which gives A
1.5
1
−
0.5
−
1
⎛
⎜
⎝
⎞
⎟
⎠
=
The eigenvalues of [A] are obtained from setting the determinant of [A-λI] equal to zero, as
follows
A λ I
⋅
− 0
=
This yields
1.5 λ
−
( ) 1 λ
−
( )
⋅ 0.5
− 0
= , or, λ1 = 0.5 , and λ2 = 2
The adjoint of the matrix [A-λI] is
1 λ
−
( )
1
0.5
1.5 λ
−
( )
⎡
⎢
⎣
⎤
⎥
⎦
In Vibration Engineering, eigenvectors are termed as natural modes or mode shapes, and
these are simply the configurations that the system takes when vibrating at a natural
frequency. Because the system considered here is a two-degree-of-freedom system, there
will be two natural frequencies and correspondingly two modes. While the natural
frequencies are simply the square roots of the eigenvalues, the associated eigenvectors or
modeshapes can be determined by putting λ = λ1(for the first mode ) and λ = λ2 (for the
second mode ) in either column of the adjoint of [A- λ I]
This procedure generates the following mode shapes
FirstMode
0.5
1
⎛
⎜
⎝
⎞
⎟
⎠
:= and SecondMode
1
1
⎛
⎜
⎝
⎞
⎟
⎠
:=
The determination of eigenvalues and eigenvectors using Mathcad is done below.
Chapter 4: Matrices and Linear Algebra 119](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-129-320.jpg)
![Figure 4.6. Mathematical model of three-story building
The equations of motion of the mathematical model resorted to can be shown to be
m1
2
t
x1
d
d
2
⋅ k1 x1
⋅
+ k2 x2 x1
−
( )
⋅
− 0
=
m2
2
t
x2
d
d
2
⋅ k2 x2 x1
−
( )
⋅
+ k3 x3 x2
−
( )
⋅
− 0
= (4.17)
m3
2
t
x3
d
d
2
⋅ k3 x3 x2
−
( )
⋅
+ 0
=
These can be put in the following matrix format
[M] d2 {x} /dt2 + {K] { x } = { 0} (4.18)
where
122 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-132-320.jpg)
![[M] =
m1
0
0
0
m2
0
0
0
m3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
(4.19)
[K] =
k1 k2
+
( )
k2
−
0
k2
−
k2 k3
+
( )
k3
−
0
k3
−
k3
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
Letting m1 = 2x 10 6 kg , m2 = 14x 10 5 kg , m3 = 7x10 5 kg; and k1 = 12x10 8 N/ m,
k2= 8x 10 8 N/m , k3 = 4x 10 8 N/ m , the [A] matrix can be computed, using Mathcad ,
as follows
m1 2 10
6
⋅
:= m2 14 10
5
⋅
:= m3 7 10
5
⋅
:= k1 12 10
8
⋅
:= k2 8 10
8
⋅
:= k3 4 10
8
⋅
:=
M
m1
0
0
0
m2
0
0
0
m3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= K
k1 k2
+
( )
k2
−
0
k2
−
k2 k3
+
( )
k3
−
0
k3
−
k3
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
:=
M
2 10
6
⋅
0
0
0
1.4 10
6
⋅
0
0
0
0.7 10
6
⋅
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
:= K
2 10
9
×
8
− 10
8
×
0
8
− 10
8
×
1.2 10
9
×
4
− 10
8
×
0
4
− 10
8
×
4 10
8
×
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
=
A M
1
−
K
⋅
:= A
1 10
3
×
571.429
−
0
400
−
857.143
571.429
−
0
285.714
−
571.429
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
The eigenvalues and the eigenvectors are
lambda eigenvals A
( )
:= lambda
1.495 10
3
×
761.106
172.146
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Chapter 4: Matrices and Linear Algebra 123](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-133-320.jpg)
![4.9 APPLICATION OF THE EIGENVALUE PROBLEM TO STRESS
ANALYSIS- DETERMINATION OF PRINCIPAL STRESSES AND
PRINCIPAL DIRECTIONS
If [σ ] is the state of stress at a point in a material given by the stress components σxx ,
σyy, σzz, σxy , σyz and σzx , the principal stresses are the eigenvalues λ of the stress matrix
and the principal directions, which are the x, y and z components of the unit vectors along
the normals to the principal planes , are the eigenvectors {η} of the stress matrix. Thus ,
the eigenvalue problem in this case is stated as
[σ] - λ [ I ]) { η } = {0} (4.20)
where the eigenvalues λ and the eigenvectors η are to be determined.
Example 4.6
The two-dimensional state of stress in a mechanical element, which is shown in Figure 4.8, is
given by the components:
σ xx = 80 MPa, σyy = 0 , and σ xy = 50 MPa.
Determine the principal stresses and their associated directions.
Figure 4.8. Two-dimensional state of stress
Because the stress matrix is symmetrical, σ yx must be equal to σ xy. Thus, the matrix of
stresses is
Chapter 4: Matrices and Linear Algebra 125](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-135-320.jpg)
![considered as eigenvectors. The following examples will illustrate this situation.
Example 4.8
For the given matrix [A]
A
0
1
−
1
1
−
0
1
1
1
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
the eigenvalues and eigenvectors are found using Mathcad as
<-- double root at λ = 1
lambda eigenvals A
( )
:= lambda
1
2
−
1
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
X eigenvecs A
( )
:= X
0.816
0.408
−
0.408
0.577
−
0.577
−
0.577
0.226
−
0.793
0.566
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
It is clear that the first and third eigenvectors corresponding to the repeated root λ = 1 are
arbitrary, since they are both different , while the one corresponding to λ = - 2 is
1
−
1
−
1
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Example 4.9
The following problem is, clearly, a case of repeated roots with a double root at λ = 1
Chapter 4: Matrices and Linear Algebra 129](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-139-320.jpg)
![A
1
0
0
1
−
1
0
0
1
1
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
The eigenvalues and eigenvectors are
lambda eigenvals A
( )
:= lambda
1
1
1
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
X eigenvecs A
( )
:= X
1
0
0
1
0
0
0.218
−
0.436
−
0.873
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
Note that the first and second eigenvectors corresponding to the repeated (double) root at λ
= 1 are, in fact , arbitrary since they have two components that are zero, while the one
corresponding to λ = -1 is the third one which is
1
−
2
−
4
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Example 4.10
For the following matrix [A]
A
14
−
0
1
1
2
0
0
0
2
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
the eigenvalues and eigenvectors are
130 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-140-320.jpg)
![ah + bj = -p and ch + d j = -q (4.24)
When these two simultaneous equations are solved for h and j , we generate new
improved values for x and y, say x i+1 and y i+1 , where , x i + 1 = x i + h and y i+1 = yi +
j . These new improved values replace the old values x i and y i in the next step in an effort
to obtain still better solutions. This procedure is repeated until convergence occurs. The
following example will clarify the procedure.
Example 4.11
Determine the solution to: f(x,y) = x2 + 3 y 2 - 54 = 0 , g(x, y ) = -4 x 2 + 3x y - 2 y + 15 = 0
Iteration Number 1
f x y
,
( ) x
2
3 y
2
⋅
+ 54
−
:= g x y
,
( ) 4
− x
2
⋅ 3 x
⋅ y
⋅
+ 2 y
⋅
− 15
+
:=
partfx x y
,
( ) 2 x
⋅
:= partfy x y
,
( ) 6 y
⋅
:= partgx x y
,
( ) 8
− x
⋅ 3 y
⋅
+
:= partgy x y
,
( ) 3 x
⋅ 2
−
:=
where the above are partial derivatives of f(x,y) and g(x,y) with respect to x and y.
xold 1.5
:= yold 2.0
:= <-- Assumed starting values of solution
p f xold yold
,
( )
:= q g xold yold
,
( )
:=
a partfx xold yold
,
( )
:= b partfy xold yold
,
( )
:= c partgx xold yold
,
( )
:=
d partgy xold yold
,
( )
:=
a 3
= b 12
= c 6
−
= d 2.5
= p 39.75
−
= q 11
=
Now solve the equations obtained from the linear Taylor series approximations, namely
a h + b j = -p and c h + d j = -q
in the form: [A] {H} = {P} using lsolve as follows
A
a
c
b
d
⎛
⎜
⎝
⎞
⎟
⎠
:= P
p
−
q
−
⎛
⎜
⎝
⎞
⎟
⎠
:= H lsolve A P
,
( )
:= H
2.91
2.585
⎛
⎜
⎝
⎞
⎟
⎠
=
h H1
:= j H2
:= h 2.91
= j 2.585
=
xnew xold h
+
:= ynew yold j
+
:= xnew 4.41
= ynew 4.585
=
132 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-142-320.jpg)
![a6
y6 a1
− a2 x6 x1
−
( )
⋅
− a3 x6 x1
−
( )
⋅ x6 x2
−
( )
⋅
− a4 x6 x1
−
( )
⋅ x6 x2
−
( )
⋅ x6 x3
−
( )
⋅
−
1
−
( ) a5
⋅ x6 x1
−
( ) x6 x2
−
( )
⋅ x6 x3
−
( )
⋅ x6 x4
−
( )
⋅
⎡
⎣ ⎤
⎦
⋅
+
...
x6 x1
−
( ) x6 x2
−
( )
⋅ x6 x3
−
( )
⋅ x6 x4
−
( )
⋅ x6 x5
−
( )
⋅
=
(5.11)
Example 5.2 -
Given three data points: X: 100 200 400
Y: 975 1575 2054
Determine the Gregory -Newton Polynomial f(x) and evaluate f(251)
The form of this polynomial will be :
y = f(x) = a1 + a2(x-X1 + a3(x-X1)(x-X2)
with a1 = Y1
a2 = (Y2-Y1) / (X2-X1)
a3 = [ Y3- a1 -a2(X3-X1)] /[ (X3-X1)(X3-X2)]
The data is put in in vector form as follows.
X
100
200
400
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= Y
975
1575
2054
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:=
and the coefficients a1, a2 and a3 are computed using
a1 Y1
:= a2
Y2 Y1
−
X2 X1
−
:= a3
Y3 a1
− a2 X3 X1
−
( )
⋅
−
X3 X1
−
( ) X3 X2
−
( )
⋅
:=
146 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-156-320.jpg)
![Δf = f(x+Δx)- f(x) (5.12)
The second finite difference is
Δ 2 f = Δ[Δf] =[f(x+2Δx)-f(x+Δx)]-[f(x+Δx)- f(x)] (5.13)
Similarly, the (n-1)th finite difference will be
Δn f = Δ[Δ n-1 f ] (5.14)
The finite differences as derived above are arranged in a table as shown in Table 5.1 which
can be used to generate an interpolating polynomial, as will be demonstrated later.
TABLE 5.1
Table of finite differences
--------------------------------------------------------------------------------------------------------------------------------
x f x
( ) Δf Δ
2
f Δ
3
f Δ
n
f
-------------------------------------------------------------------------------------------------------------------------------
x f x
( )
Δf x
( )
x Δx
+ f x Δx
+
( ) Δ
2
f x
( )
Δf x Δx
+
( ) Δ
3
f x
( )
x 2 Δx
⋅
+ f x 2 Δx
⋅
+
( ) Δ
2
f x Δx
+
( )
Δf x 2 Δx
⋅
+
( )
x 3 Δx
⋅
+ f x 3 Δx
⋅
+
( )
------------------------------------------------------------------------------------------------------------------------------------
150 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-160-320.jpg)
![fpspline 415
( ) 962.582
= fcspline 415
( ) 955.728
=
5.9. APPLICATIONS IN NUMERICAL INTERPOLATION
In the examples that follow, data from various practical areas such as design of
machinery, noise control and vibration analysis, is utilized in interpolation procedure s to
generate function values at points other than the given data points.
5.9.1. Stress-Strain Data for Titanium
The stress versus strain data given in the table below is obtained from a tensile test of
annealed A-40 titanium [ 18 ]. A cubic spline can be developed using Mathcad as follows
and the stresses interpolated for any values of strain in the range provided.
The cubic spline interpolation generated together with the given data points is presented in
Figure 5.12. Notice that the spline generated passes through all the given data points as is
expected of an interpolating polynomial.
Strain (in/in) Stress (kpsi)
<--Given stress versus strain data
σ
0
5
10
15
20
25
30
34.9
39.9
44.9
49.9
60.9
73.3
78.2
86.2
94.5
98.7
118.2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
ε
0
0.00030
0.00060
0.00090
0.00120
0.00175
0.00220
0.00285
0.00349
0.00469
0.00698
0.01610
0.04807
0.07676
0.14364
0.26677
0.34470
0.72202
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Vε ε
:= Vσ σ
:=
Vs cspline Vε Vσ
,
( )
:=
168 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-178-320.jpg)
![fspline ε
( ) interp Vs Vε
, Vσ
, ε
,
( )
:= <-- Generates a cubic spline function for given data
fspline 0.004
( ) 42.699
=
<-- interpolated values of stress at strains of 0.004 and 0.65
fspline 0.65
( ) 114.503
=
i 1 18
..
:= ε 0 0.00010
, 0.722
..
:=
1 .10
4
1 .10
3
0.01 0.1 1
1
10
100
1 .10
3
fspline ε
( )
Vσi
ε Vεi
,
Figure 5.12. Mathcad cubic spline for given stress-strain data
5.9.2. Notch Sensitivity of Aluminum
The fatigue stress concentration factor K f is used with the nominal stress σ 0 to compute
the maximum resulting stress σ max at a discontinuity in a machine part via the following
relationship [18] .
σmax Kf σ0
⋅
= (5.31)
The fatigue stress concentration factor K f , which is, in fact , just a reduced value of the
geometrical stress concentration factor K t, is a function of the sensitivity of the material of
which the part is made to notches ( known as notch sensitivity q ). These quantities are
connected by the following relationship .
Chapter 5: Numerical Interpolation 169](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-179-320.jpg)
![fspline 0.13
( ) 0.787
= fspline 0.15
( ) 0.813
= f 0.15
( ) 0.814
= fspline 0.15
( ) 0.813
=
r 0.015 0.016
, 0.16
..
:= i 1 7
..
:=
0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16
0.2
0.4
0.6
0.8
Given data points
Interpolation by undet coeff method
Mathcad cubic spline interpolation
Given data points
Interpolation by undet coeff method
Mathcad cubic spline interpolation
Mathcad interpolation/ notch sensitivity
Notch radius, r, inches
Notch
sensitivity,
q
Vqi
f r
( )
fspline r
( )
Vri r
,
Figure 5.13. Interpolation of notch sensitivity data
5.9.3 Speech Interference Level
In the field of Noise and Vibration Control, it often becomes necessary to determine the
effect of steady background noise on speech communication in a given setting. The
preferred speech interference level (PSIL) was established in an effort to study this effect
under the constraint that speech sounds would not be allowed to be reflected back to the
listener [ 2, 10 ]. For effective communication at a given voice level , the maximum
distance that there can be between the speaker and the listener is a function of the
preferred speech interference level existing at the location. The following data which is
provided for communication at the level of a normal male voice can be utilized in an
interpolation scheme to determine the maximum PSIL permitted for a given distance
between the speaker and the listener.
Dist (ft) 0 2 4 6 8 10 12
PSIL(dB) 83 62.5 56 52.5 50 47.5 46.5
172 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-182-320.jpg)
![0 2 4 6 8 10 12
40
50
60
70
80
90
Given data points
Gregory-Newton interpolation
Mathcad's interp/cspline
Given data points
Gregory-Newton interpolation
Mathcad's interp/cspline
Interpolation- speech communication
Distance from speaker to listener (ft)
Preferred
speech
interference
level
(
dB
Vyi
f x
( )
fspline x
( )
Vxi x
,
Figure 5.14. Interpolation of speech communication data
5.9.4. Load-Deflection Data for Elastomeric Mounts
Elastomeric mounts are employed when small electrical and mechanical devices have to be
isolated from high forcing frequencies [ 9, 10 ]. They are especially useful in the protection of
delicate electronic instruments . Mount performance characteristics in the form of load
versus static deflection data are provided by manufacturers of vibration control products to
enable a designer to select an appropriate isolator for a given application.
The following data is available for an elastomeric mount.
Deflection (cm): 0 0 0.02 0.04 0.06 0.08 0.10
Load (N): 5 11 17.5 23.5 28 30.5
There are six data points prescribed. Thus, if a Lagrangian interpolating polynomial is to be
generated, it will be of order 5, having the form
L D
( )
1
6
i
wi D
( ) . Li
⋅
( )
∑
=
=
where
Chapter 5: Numerical Interpolation 175](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-185-320.jpg)
![5.7. The amplitude of vibration of an automobile in the vertical direction, after passing over
a road bump is found to be as follows.
Time, t (seconds) 0 0.65 1.30 1.95
Amplitude, A (mm) 5.25 2.50 0.80 0.40
Using the Gregory-Newton method, compute the amplitude for t = 1.80 seconds.
5.8. The Charpy test [ 18 ] provides material toughness data under dynamic conditions.
In this test, which is helpful in comparing several materials and in determining
low-temperature brittleness and impact strength, the specimen is struck by a pendulum
released from a fixed height and the energy absorbed by the specimen, termed the impact
value , computed from the height of the swing after fracture. The following table gives the
impact value, V, as a function of temperature, T, for a certain material. Using the
Gregory-Newton method, derive an interpolating polynomial and estimate V for T = -25 deg.
F and - 75 deg. F
T (deg F) -200 -150 -100 -50 0
V (ft-lbs) 0 1.5 3 10 35
5.9. The data in the following table are the distances traveled by a motorcycle, which was
stationary at time t = 0, for selected times.
time, t (seconds) 0 2 4 6 8 10
Distance, D (ft) 0 10 50 150 330 610
Generate a finite-difference table and obtain an interpolating polynomial using Newton's
method. Estimate the distance D traveled at t = 7 seconds.
5.10. For the given data, generate both a finite-difference table and an interpolating
polynomial.
Angle, θ, (degrees) Cosine (θ)
0 1.00000
10 0.98481
20 0.93969
30 0.86603
40 0.76604
50 0.64279
60 0.50000
Evaluate the error of the estimated value of cos (25 degrees). The true value is given to be
0.90631.
Chapter 5: Numerical Interpolation 179](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-189-320.jpg)
![5.11. The following data gives the notch sensitivity , q , of a steel as a functionof the
notch radius , r [ 18 ]. Compute a finite-difference table and derive an interpolating
polynomial by Newton's method. Estimate q for r = 0.11 inches.
q 0.45 0.65 0.75 0.78 0.785
r ( in) 0.01 0.05 0.09 0.13 0.17
5.12. The deflection, δ, of a structure under loading is measured at a point for four different
values of the applied force, F. The measured data is as follows.
F (kn) 0.50 1.50 2.0 2.50
δ (cm) 3.0 5.2 7.3 10.5
Using Lagrangian interpolation , compute the deflection, δ, for F = 2.2 kN.
5.13. When a vibration problem is solved with nonlinearities included, the natural
frequencies of vibration become dependent on the amplitudes of vibration [ 9,16,20,21 ].
The following table relates the nonlinear natural frequency / linear natural frequency ratio,
r, of a rotating blade vibrating in the plane of rotation to its vibration amplitude , A.
A (in) 10 8.5 5 0
r (%) 99.9325 99.9575 99.9921 100
Derive a Lagrange interpolating polynomial and compute r for A = 4.25 in .
5.14. The behavior of a mooring line employed to control the excursions of a floating
ocean structure resembles that of a nonlinear spring with tension-displacement
characteristics which depend upon its length, weight, elastic properties, anchor holding
capacities and water depth [ 22 ]. In the table below, the horizontal component of mooring
line tension , H, is given as a function of the horizontal distance, X, between the ends of the
line.
X (ft) 440 470 485 492.5 495 498
H (Kips)0 5 10 42 150 440
Using Lagrangian interpolation, generate an interpolating polynomial and compute H for
X= 487 ft.
180 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-190-320.jpg)
![5.15. The following table gives the pressure (P) versus temperature (T) relationship in the
liquid-vapor region for water [ 17 ]
T(deg Kelvin) 403.15 453.15 503.15 553.15 633.15
P (MPa) 0.2701 1.0021 2.795 6.412 18.651
Using Lagrangian interpolation, generate an interpolating polynomial and estimate P for T=
525 deg. K.
5.16. While the value of the endurance limit, S end , of a material is based on its
tensile strength, S ult , it is also dependent on the condition of its surface[19 ]. The following
data relates to a machined , unnotched specimen subjected to reversed bending.
S ult ( kpsi): 60 80 100 140 180 220
S end (kpsi): 22 30 38 50 60 64
Develop linear and cubic splines using linterp and interp for the above data. Also ,
graph the functions generated with proper labels and a title.
Chapter 5: Numerical Interpolation 181](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-191-320.jpg)
![which can be solved for the C's using algebraic or matrix methods or by accessing the
Mathcad function lsolve. The above can be expressed as
[M]{C} = {A}, (6.11)
where
(6.12)
M
n
1
n
i
xi
( )2
∑
=
1
n
i
xi
( )2
∑
=
1
n
i
xi
( )4
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
= , A
1
n
i
yi
∑
=
1
n
i
xi
( )2
yi
⋅
⎡
⎣
⎤
⎦
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
Example 6.2
A data set is given below .
x y
0.5 0.51
1.0 2.35
1.5 7.54
2.0 13.23
2.5 17.65
3.0 24.21
3.5 28.94
4.0 37.63
4.5 58.32
5.0 63.21
Determine the quadratic function that will best fit the data given
Put in the given data :
i 1 2
, 10
..
:= x1 0.5
:= xi 1
+ xi 0.5
+
:= n 10
:=
y1 0.51
:= y2 2.35
:= y3 7.54
:= y4 13.23
:= y5 17.65
:=
y6 24.21
:= y7 28.94
:= y8 37.63
:= y9 58.32
:= y10 63.21
:=
In this problem, the [M} and {A} matrices are
Chapter 6: Curve-Fitting 189](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-199-320.jpg)
![x
0.18
0.32
0.65
1.12
1.65
2.34
2.52
2.85
3.12
3.61
3.92
4.51
4.72
5.25
6.75
7.85
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:= y
4.5921
5.2345
6.2321
6.5467
7.2976
7.7654
7.7237
8.2543
7.9765
8.4328
9.6322
8.8764
9.0276
9.2568
8.6432
10.2137
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Yi ln yi
( )
:= Xi ln xi
( )
:=
Equation 6.17 is: [D] { K} = {P}, where
P
1
n
i
Yi
( )
∑
=
1
n
i
Yi
( ) Xi
⋅
⎡
⎣ ⎤
⎦
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
D
n
1
n
i
Xi
∑
=
1
n
i
Xi
∑
=
1
n
i
Xi
( )2
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
and {K} is the vector of unknown coefficients C and B
The constants C and B can be obtained using lsolve as shown below, and then the coefficients
A and B of the power function can be calculated
K lsolve D P
,
( )
:= K
1.883
0.2
⎛
⎜
⎝
⎞
⎟
⎠
= <------ This means: C= 1.883 and B = 0.2
C K1
:= B K2
:= C 1.883
= B 0.2
=
Chapter 6: Curve-Fitting 193](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-203-320.jpg)
![f xi
( )
1
m
j
Cj fj xi
( )
⋅
⎡
⎣ ⎤
⎦
∑
=
= (6.28)
In this case, the sum of the squares of the deviations from the true values y i will be:
s y1 f x1
( )
−
( )2
y2 f x2
( )
−
( )2
+ ......................
+ yn f xn
( )
−
( )2
+
=
(6.29)
For S to be a minimum , the partial derivatives of S with respect to C1, C2 ....... C m
must be zero and the second partials must be positive. These requirements lead to the following
matrix equation in the unknowns, C 1, C2 ...... C m
[M] {C} = {L} (6.30)
where
C
C1
C2
Cm
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:= , L
1
n
i
f1 xi
( )
⎡
⎣ ⎤
⎦ yi
( )
⋅
⎡
⎣ ⎤
⎦
∑
=
1
n
i
f2 xi
( )
⎡
⎣ ⎤
⎦ yi
( )
⋅
⎡
⎣ ⎤
⎦
∑
=
1
n
i
fm xi
( )
⎡
⎣ ⎤
⎦ yi
( )
⋅
⎡
⎣ ⎤
⎦
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:= , and
200 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-210-320.jpg)
![yanal x
( ) C1 f1 x
( )
⋅ C2 f2 x
( )
⋅
+
:= <--- Curve-fit generated by analysis
yanal 1
( ) 6.281
= yanal 2
( ) 8.733
=
<---y-values generated by curve-fit
yanal 3
( ) 15.386
=
z 1 1.1
, 5
..
:= i 1 5
..
:=
1 2 3 4 5
0
20
40
60
Results of Analysis
Given data points
Results of Analysis
Given data points
Analysis - Generated Fit/ Given Data
X- Values
Y-
Values
yanal z
( )
yi
z xi
,
Figure 6.7. Curve-fit with a linear combination of known functions
6.8 CURVE-FITTING WITH POLYNOMIALS
If an mth order polynomial seems to be a good fit to the data points given, the procedure used
in fitting a linear combination of known functions can still be resorted to. In this case,
f 1 ( x) = 1, f 2 (x) = x , f 3 ( x ) = x 2, f 4 ( x ) = x 3 , ............. ,
f m+1 (x ) = x m
(6.31)
f(x) = C1 f 1 (x ) + C 2 f 2 ( x ) +....................... C m+1 f m+1 ( x ) (6.32)
in which the C 's must be solved for using
[M] {C} = {L} (6.33)
Chapter 6: Curve-Fitting 203](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-213-320.jpg)
![where
C
C1
C2
Cm 1
+
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
M
n
1
n
i
xi
∑
=
1
n
i
xi
( )2
∑
=
1
n
i
xi
( )m
∑
=
1
n
i
xi
∑
=
1
n
i
xi
( )2
∑
=
1
n
i
xi
( )m
∑
=
1
n
i
xi
( )m 1
+
∑
=
1
n
i
xi
( )m 2
+
∑
=
1
n
i
xi
( )2 m
⋅
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:= L
1
n
i
yi
∑
=
1
n
i
xi yi
⋅
( )
∑
=
1
n
i
xi
( )m
yi
⋅
⎡
⎣
⎤
⎦
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
,
where the ( x i, y i ) are the "n" given data points.
Example 6.7.
Given the data points
x: 0 1 2 3 4 5
y: 0 8.47 17.48 19.57 14.69 11.23
Obtain a curve-fit of the form y x
( ) f x
( )
= C1 C2 x
⋅
+ C3 x
2
⋅
+
=
Here, n = 6 and m = 2 and the matrix equation to be solved is
[M]{C} = {L}
204 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-214-320.jpg)
![where [M] is a known 3x3 matrix and {L} is a known column vector as shown below. The
The column vector {C} , which involves the undetermined coefficients, is
C1
C2
C3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
The given data is
n 6
:= x1 0
:= x2 1
:= x3 2
:= x4 3
:= x5 4
:= x6 5
:=
y1 0
:= y2 8.47
:= y3 17.48
:= y4 19.57
:= y5 14.69
:= y6 11.23
:=
The matrices M and L are
L
1
n
i
yi
∑
=
1
n
i
xi yi
⋅
( )
∑
=
1
n
i
xi
( )2
yi
⋅
⎡
⎣
⎤
⎦
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
M
n
1
n
i
xi
∑
=
1
n
i
xi
( )2
∑
=
1
n
i
xi
∑
=
1
n
i
xi
( )2
∑
=
1
n
i
xi
( )3
∑
=
1
n
i
xi
( )2
∑
=
1
n
i
xi
( )3
∑
=
1
n
i
xi
( )4
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
The unknown constants are determined below using lsolve
C lsolve M L
,
( )
:= C
0.444
−
12.484
2.057
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
C1 C1
:= C2 C2
:= C3 C3
:=
C1 0.444
−
= C2 12.484
= C3 2.057
−
=
The polynomial fitting the given data , then, is
Chapter 6: Curve-Fitting 205](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-215-320.jpg)
![0 1 2 3 4 5 6 7
0
2000
4000
Given data points
Mathcad's pwrfit
Given data points
Mathcad's pwrfit
vy
f x
( )
vx x
,
Mathcad's expfit
The function expfit(vx,vy, [vg]) returns a vector containing the coefficients a, b and c for an
exponential curve of the form A exp (b x ) + C that best fits the real data values represented by
vectors vx and vy. The use of this function requires that there be at least three data points and the
vx values must all be greater than or equal to zero. If a fit has be done on data that has negative
x-values, then the data must be shifted to the positive axis. . Otherwise, a curve-fit with erroneous
values may be generated. The vector vg is a three-element vector of real guess values for the
parameters A, b and C, but it is optional, and does not have to be used. However, if it is not
used, then expfit generates a guess that fits a line to the logs of vy. If an exponential fit is
desired that is different from the above form, then use genfit .
For the data points represented by the vectors vx and vy as given below, and with the use of the
vector vg of guess values shown for the coefficients a, b and c ,.
vg
3
2
10
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
vx
0
1
2
3
4
5
6
7
8
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:= vy
21
24.892
31.31
41.89
59.334
88.095
135.513
213.693
342.589
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
the curve-fit returned by Mathcad's expfit is defined by
Chapter 6: Curve-Fitting 219](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-229-320.jpg)
![Example 6.17
Fit the data points :
x: 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
y: 0.59 0.83 0.90 0.95 1.02 1.43 1.75 3.15
with a function of the form : y = C1 f 1 (x ) + C2 f 2 ( x) +C 3 f 3(x) + C4 f4 (x) where
f 1 (x ) = 1 , f 2 ( x ) = x2 , f 3 ( x) = sin (2x) , f 4 ( x) = exp( 0.95x )
In this case, we have the following matrix equation
[M] {C} = {L} ,
where {C} is the column vector of unknown coefficients C1, C2, C3 and C4 , and [M] and {L} are
known matrices as generated below.
The given data is
N 8
:=
x1 0.1
:= x2 0.2
:= x3 0.3
:= x4 0.4
:= x5 0.5
:= x6 0.6
:= x7 0.7
:= x8 0.8
:=
y1 0.59
:= y2 0.83
:= y3 0.90
:= y4 0.95
:= y5 1.02
:= y6 1.43
:= y7 1.75
:= y8 3.15
:=
i 1 2
, 8
..
:= f1 x
( ) 1
:= f2 x
( ) x
2
:= <--- The subscipts 1 and 2 are literal
subscripts , and do not denote
vector components. These are
done by including a period in the variable
name . Whatever follows the period then
becomes the subscript. For
example , "vel init" is created by typing
a period after "vel" and then typing "init"
f3 x
( ) sin 2 x
⋅
( )
:=
f4 x
( ) exp 0.95 x
⋅
( )
:=
In this case, the M and L matrices are
230 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-240-320.jpg)
![6.10 APPLICATIONS IN CURVE-FITTING
In the world of engineering, science, and business, it is very common to see a function in
the form of data prescribed at many points, and in addition, there may also be
considerable scatter in the available data . The methods discussed in the earlier sections
of this chapter will now be used to generate appropriate curve-fits to actual data from
several practical applications.
6.10.1 Fatigue Failure Curve for Loading in the Finite Life Range
The fatigue failure curve of a material provides useful information pertinent to the design of
machinery for finite life [ 19 ]. It is essentially a relationship between the completely
reversed applied stress, S, and fatigue life, L , which is measured in terms of the number
of stress reversals to failure.
The following data pertains to a steel with an endurance limit of 40 kpsi and an ultimate
strength of 90 kpsi. It is required to curve-fit the data with a suitable function.
L ( million cycles) Stress (kpsi)
vL
0.02
0.03
0.04
0.05
0.075
0.1
0.2
0.4
0.6
0.8
1.0
1.1
1.2
1.3
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:= vS
80
70
65
61
57
55
50
45
42
40.5
40.1
40
39.9
39.85
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
From the plot of the given data shown in Figure 6.20, it appears that a power function would
furnish a reasonable curve-fit. This has the form
S(L)= A*LB ,
236 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-246-320.jpg)
![where A and B are undetermined coefficients to be evaluated on a least squares basis.
i 1 14
..
:=
0 0.2 0.4 0.6 0.8 1 1.2
40
60
80
Given Data
Given Data
Given Data Points
Given L- values (million cycles )
Given
S-
values
(kpsi)
vSi
vLi
Figure 6.20. Fatigue life data
Using S(L) = A L B , Sstar= C+ B . Lstar , where Lstar= ln (L) , C = ln (A) and
Sstar= ln (S), the matrix equation [D]{X} = {P} is obtained , where {X} is the column vector
containing the unknowns C and B.
n 14
:= Sstar ln vS
( )
:= Lstar ln vL
( )
:= D X
⋅ P
=
D
n
1
n
i
Lstar i
∑
=
1
n
i
Lstar i
∑
=
1
n
i
Lstar i
( )2
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
P
1
n
i
Sstari
∑
=
1
n
i
Sstari Lstar i
⋅
( )
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
The unknown constants C and B can be computed using lsolve as shown below
K lsolve D P
,
( )
:= K
3.684
0.155
−
⎛
⎜
⎝
⎞
⎟
⎠
= <------Thus, C= 3.684 and B = -0.155
C K1
:= B K2
:= C 3.684
= B 0.155
−
= A e
C
:= A 39.808
=
Chapter 6: Curve-Fitting 237](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-247-320.jpg)
![6.10.2. Temperature Response of an Object Placed in a Hot Stream of Air
The following data pertains to the temperatures of a solid steel sphere suspended in a hot stream
of air measured at several instants of time [ 17 ]. It is required to curve-fit the data with a suitable
function.
t= Time (secs) T=Temp (deg C) t=Time(Secs) T=Temp (deg C)
0.0 14 500 51
100 28 600 54
200 37.5 700 55
300 44 800 57.8
400 48 900 59.1
The given data is put in as follows and plotted in Figure 6.22.
n 10
:= t1 0
:= t2 100
:= t3 200
:= t4 300
:= t5 400
:= t6 500
:= t7 600
:=
T1 14
:= T2 28
:= T3 37.5
:= T4 44
:= T5 48
:= T6 51
:= T7 54
:=
t8 700
:= t9 800
:= t10 900
:= T8 55
:= T9 57.8
:= T10 59.1
:=
i 1 10
..
:=
0 200 400 600 800
0
20
40
60
Given data points
Given data points
Ti
ti
Figure 6.22. Given temperature versus time data
A glance at the plot of Figure 6.22 suggests that a function of the following form would provide a
reasonable fit to the given data
f t
( ) A B e
t
−
τ
⋅
+
=
where f(t) is a linear combination of functions in which A and B are constants to be determined
on a least squares basis. From the data plot, it appears that the temperature profile attains
steady state in approximately 1200 seconds, which , when equated to four time constants, will
yield a time constant, τ, of 300 seconds. Thus, the coefficient of t in the exponential term can
be selected as (-1/300).
Chapter 6: Curve-Fitting 239](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-249-320.jpg)
![In this case, the matrix equation to be solved is
[M] {C} = {A}
in which {M] and {A} contain the functions f1 t
( ) 1
:= f2 t
( ) e
1
−
300
t
⋅
:=
and
M
1
n
i
f1 ti
( )2
∑
=
1
n
i
f1 ti
( ) f2 ti
( )
⋅
( )
∑
=
1
n
i
f1 ti
( ) f2 ti
( )
⋅
( )
∑
=
1
n
i
f2 ti
( )2
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:= , C
C1
C2
⎛
⎜
⎝
⎞
⎟
⎠
= , A
1
n
i
f1 ti
( ) Ti
⋅
( )
∑
=
1
n
i
f2 ti
( ) Ti
⋅
( )
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
In the above, note that n = 10 and m =2, and the unknown constants are C1 and C2 which are
determined below using lsolve.
C lsolve M A
,
( )
:= A C1
:= B C2
:=
C
60.477
45.967
−
⎛
⎜
⎝
⎞
⎟
⎠
=
The function fitting the given data , then, is
f t
( ) A B e
1
−
300
t
⋅
⋅
+
:=
which is shown in Figure 6.23 .
f 0
( ) 14.511
= f 200
( ) 36.877
= f 800
( ) 57.283
= f 900
( ) 58.189
=
Mathcad's function linfit can also be used to evaluate the undetermined coefficients A and B in
the regression function employed which is a linear combination of the functions selected for the
curve-fit. This can be done as shown in the following steps.
vt t
:= vT T
:= F t
( )
1
e
1
−
300
t
⋅
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
:= S linfit vt vT
, F
,
( )
:=
240 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-250-320.jpg)
![6.10.3. The Effect of Operating Temperature on the Strength of a Mechanical
Element
When a mechanical element is subjected to reversed stresses at temperatures below room
temperature, there is a strong possibility of the occurrence of brittle fracture . On the other
hand, when the operating temperatures are higher than room temperature, the yield strength
of the material drops off very rapidly with increase in temperature, and yielding can take place
. The temperature-corrected value of tensile strength, then, to be used in design calculations
is obtained by multiplying the tensile strength at room temperature by a factor K d which is
a function of the temperature of the operating environment.
The following data was collected from tests done on carbon and alloy steels and shows the
effect of operating temperature on tensile strength [ 18 ]. An appropriate curve-fit to the
given data is sought.
Temp., deg F. K d Temp.,deg F. Kd
0 0.975 600 0.963
100 1.008 700 0.927
200 1.020 800 0.872
300 1.024 900 0.797
400 1.018 1000 0.698
500 0.995 1100 0.567
The given data points, which are plotted in Figure 6.24 below, are put in as follows.
n 12
:= T1 0
:= T2 100
:= T3 200
:= T4 300
:= T5 400
:= T6 500
:=
T7 600
:= T8 700
:= T9 800
:= T10 900
:= T11 1000
:= T12 1100
:=
Kd1 0.975
:= Kd2 1.008
:= Kd3 1.020
:= Kd4 1.024
:= Kd5 1.018
:= Kd6 0.995
:=
Kd7 0.963
:= Kd8 0.927
:= Kd9 0.872
:= Kd10 0.797
:= Kd11 0.698
:= Kd12 0.567
:=
242 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-252-320.jpg)
![0 200 400 600 800 1000
0.5
1
Given data points
Given data points
Given data points
Temperature, deg. F
Factor
Kd
Kdi
Ti
i 1 12
..
:=
Figure 6.24. Given Kd versus temperature data
We will fit the given data with a second-order polynomial of the following form.
Kd(T) = f(T) = C1 + C2 T + C3T 2
Here, n = 12 and m = 2 and the matrix equation to be solved is:
[M]{C} = {A},
where [M] is a known 3x3 matrix and {A} is a known column vector as shown below. The
column vector {C} , which involves the undetermined coefficients, is
C
C1
C2
C3
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
The [M] and {A} matrices are
Chapter 6: Curve-Fitting 243](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-253-320.jpg)
![0 200 400 600 800 1000
0.4
0.6
0.8
1
1.2
Given data points
Analysis-generated function
Mathcad regress
Mathcad loess
Given data points
Analysis-generated function
Mathcad regress
Mathcad loess
Curve-fits generated
Temperature, deg F
Multiplying
factor
Kd
vKdi
f T
( )
freg T
( )
floess T
( )
vTi T
,
Figure 6.25. Effect of operating temperature on the strength of a mechanical element
6.10.4. Drop-Testing of Packaged Articles
An interesting parameter that often warrants investigation is the height from which a package can
be dropped before it suffers any accountable damage, which can be assessed from the maximum
acceleration imparted to the package at the end of the drop [ 20 ]. This is a question that arises
when packages must be properly cushioned before shipment to another location. The maximum
acceleration to gravity ratio, a , has been found to be a function of the ratio, h , of twice the
distance dropped to the static deflection of the package.
The following a versus h data is provided , using which a reasonable curve-fit is to be
generated .
h a h a
________________________
1 1.4 9 3.25
2 1.8 10 3.3
3 2 11 3.55
4 2.25 12 3.65
5 2.65 13 3.75
6 2.8 14 3.85
7 2.9 15 4.2
8 3.01 16 4.3
_______________________________
Chapter 6: Curve-Fitting 245](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-255-320.jpg)
![These data points are plotted in Figure 6.26.
n 16
:= i 1 2
, 16
..
:= h1 1
:= hi 1
+ hi 1
+
:=
a1 1.4
:= a2 1.8
:= a3 2.
:= a4 2.25
:= a5 2.65
:= a6 2.8
:= a7 2.9
:= a8 3.01
:=
a9 3.25
:= a10 3.3
:= a11 3.55
:= a12 3.65
:= a13 3.75
:= a14 3.85
:= a15 4.2
:= a16 4.3
:=
We will generate a curve-fit of the form:
a = f(h) = C1 + C2 h + C3 h 2 + C 4 h 3
Here, n = 16 and m = 3 and the matrix equation to be solved is:
[M]{C} = {A},
0 2 4 6 8 10 12 14 16
1
2
3
4
5
Given data points
Given data points
Given data points
(2h/ static deflection ) ratio
(max
acceleration/
gravity)
ratio
ai
hi
Figure 6.26. Given acceleration ratio versus height ratio data
246 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-256-320.jpg)
![where [M] is a known 4x4 matrix and {A} is a known column vector as shown below. The
column vector {C} , which involves the undetermined coefficients, is
C
C1
C2
C3
C4
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
=
M
n
1
n
i
hi
∑
=
1
n
i
hi
( )2
∑
=
1
n
i
hi
( )3
∑
=
1
n
i
hi
∑
=
1
n
i
hi
( )2
∑
=
1
n
i
hi
( )3
∑
=
1
n
i
hi
( )4
∑
=
1
n
i
hi
( )2
∑
=
1
n
i
hi
( )3
∑
=
1
n
i
hi
( )4
∑
=
1
n
i
hi
( )5
∑
=
1
n
i
hi
( )3
∑
=
1
n
i
hi
( )4
∑
=
1
n
i
hi
( )5
∑
=
1
n
i
hi
( )6
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
A
1
n
i
ai
∑
=
1
n
i
hi ai
⋅
( )
∑
=
1
n
i
hi
( )2
ai
⋅
⎡
⎣
⎤
⎦
∑
=
1
n
i
hi
( )3
ai
⋅
⎡
⎣
⎤
⎦
∑
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
:=
C lsolve M A
,
( )
:= C
0.997
0.433
0.029
−
9.472 10
4
−
×
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
= C1 C1
:= C2 C2
:= C3 C3
:= C4 C4
:=
C1 0.997
= C2 0.433
= C3 0.029
−
= C4 9.472 10
4
−
×
=
f h
( ) C1 C2 h
⋅
+ C3 h
2
⋅
+ C4 h
3
⋅
+
:= f 15
( ) 4.114
= f 9
( ) 3.219
=
The Mathcad regress and loess functions also can be utilized as shown below to generate
curve-fits to the given data.
Chapter 6: Curve-Fitting 247](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-257-320.jpg)
![Strain,e, ( %) Stress,S, (Kpsi)
10 63
20 63.6
40 62
60 60
80 58
100 56
120 52
140 48
150 47
Using an approximating polynomial of the form
S C1 C2 e
⋅
+ C3 e
2
⋅
+
=
obtain a least squares fit to the given data. Determine S(105%)
6. 16. The following data relating the distance traveled by an object to time was obtained from a
test in an experimental test track . The distance traveled, S, is in meters and the time , t, is in
seconds.
Time, t (secs) 0.1 0.5 1.5 3.0
Distance, S (m) 0.26 1.56 7.41 30.90
Generate a curve-fit of the form S(t) = A + B t + C t 2. Compute S for t= 2 seconds.
6.17. The pressure drag , D P, on an object can be reduced in comparison with the total
drag , DT, by streamlining it , that is, making its length, L, in the direction of flow larger with
respect to its maximum thickness or diameter, D . [ 17 ] . Obtain an appropriate curve-fit to
the following data which is provided to demonstrate the effect of streamlining on the pressure
drag of a body of symmetrical airfoil cross section.
L
D
1 2 3 4 5 6 7 8 9 10 11
( )
=
DP
DT
1.25 0.65 0.42 0.25 0.20 0.10 0.08 0.06 0.04 0.02 0.01
( )
=
Determine D P / DT for L / D = 5.5
6.18. The following data , in which the time, t, is given in seconds provides the
252 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-262-320.jpg)
![response c ( t ) of a control system to a step input.
t 0 0.75 1.50 2 3 4 5 6
( )
:=
c t
( ) 0 1.30 1.40 1.10 0.85 1.02 1.01 1.009
( )
:=
Obtain a least-squares fit of the form.
c(t)= A+ B e -t sin (2.828 t ) + C e -t cos ( 2.828 t )
and determine c at t = 3.5 seconds.
6.19. The Brinell hardness number, Bhn , for steel [ 18 ] is given as a function of the
tempering temperature, T, in the following table
T, deg F. 400 600 800 1000 1200
Bhn 540 500 445 375 310
Obtain an appropriate curve-fit to the given data and determine the Brinell hardness
number for T = 850 deg F.
6.20. The average strain rate used in obtaining stress/ strain data for a material
specimen is about 0.001 in / (in-sec ). However, when the strain rate, SR, is
increased, as it happens, under conditions of impact loading, the strength of the
material also increases. [ 18 ] The following data shows the influence of the
increase in strain rate on the yield strength, SYLD, of a steel specimen
SR 10
6
−
10
5
−
10
4
−
10
2
−
1 10
2
10
3
( )
= in/(in-sec)
SYLD 30 28 30 37 45 65 78
( )
= kpsi
Obtain an appropriate least-squares fit to the given data and determine SYLD
corresponding to an SR of 50.
6.21. The endurance limit of a steel specimen is, typically, a function of the condition
of its surface [ 19 ]. For an unnotched, ground specimen in reversed bending,
the endurance limit, S end , is related to its tensile strength, S ult , as given in the
following table of data.
Sult 60 80 100 120 140 160 180 200 220
( )
= kpsi
Send 25 35 45 55 65 72 80 85 90
( )
= kpsi
Chapter 6: Curve-Fitting 253](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-263-320.jpg)
![Obtain a reasonable curve-fit to the given data. Determine S end corresponding to S ult = 135
kpsi .
6.22. Expected voice levels required for speech communication with various back-
ground noise levels and separations between the speaker and the listener [ 10 ]
are given in the following table of data .
Obtain appropriate curve-fits to the given data in the form of separate curves for
(a) Peak shouting (b) Shouting (c) Very loud voice (d) Raised voice and (e)
Normal voice.
Determine the voice levels required in the following situations.
(1) Speaker listener distance = 2.5 m , background noise level = 68 dBA
(2) Speaker listener distance = 4.5 m , background noise level = 86 dBA
(3) Speaker listener distance = 8.5 m , background noise level = 45 dBA
(4) Speaker listener distance = 6.5 m , background noise level = 65 dBA
BACKGROUND NOISE LEVELS (dBA)
Speaker
/listener
distance
( m )
Normal
Voice
Raised
Voice
Very
Loud
Voice
Shouting Peak
Shouting
0.25
0.5
1.0
2
3
4
5
6
7
8
9
9.5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
90
80
72
63
57
54.5
52
50
49.5
49
48
47.5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
62
60
57
56.5
55
54
53.5
52.5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
95
88
78
72
68
66
64
62.5
61.5
60
58
57.5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
100
92
84
77
74
72
68
67.5
66
65
64.5
64
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
118
112
103
96
92.5
90
88
86.5
86
85
84.5
84
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
254 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-264-320.jpg)
![ffourprime
1
∆x
4
1.853 4 1.723
⋅
− 6 1.456
⋅
+ 4 1.212
⋅
− 0.703
+
( )
⋅
:= ffourprime 0.448
−
=
7.3. INTERPOLATING POLYNOMIAL METHOD
In the interpolating polynomial method, an interpolating polynomial passing through given
data points is determined by any of the methods discussed in Chapter 5. Then, the
polynomial generated can be differentiated as is done in traditional calculus. If the polynomial
derived , for instance, is
f x
( ) a0 a1 x
⋅
+ a2 x
2
⋅
+ a3 x
3
⋅
+ ............
+ an x
n
⋅
+
= (7.10)
then, its derivative can be obtained as
x
f x
( )
d
d
a1 2 a2
⋅ x
⋅
+ 3 a3
⋅ x
2
⋅
+ ..............
+ n an
⋅ x
n 1
−
⋅
+
= (7.11)
Example 7.2
Given the following data :
x 2 3 4 5 6
y 0.80354 1.23591 1.62312 1.89765 1.96789
Use the Method of Undetermined Coefficients to fit an interpolating polynomial to the given
data and estimate: yprime(4.5), yprime(2.7), ydblprime(4.5) and ydblprime(2.7).
Since 5 data points are given, we can generate a 4 th order polynomial.
Let y(x) = f(x)= a + bx + cx2 + dx3 + e x4
Substituting the given values of (x,y) in the assumed polynomial above yields the following
matrix equation:
[M]{A}= {V}
where
Chapter 7: Numerical Differentiation 259](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-269-320.jpg)
![Determine the velocity and acceleration of the train as functions of time. Estimate the
velocity and acceleration of the train at t = 32 seconds
We first use the method of undetermined coefficients to fit an interpolating polynomial to
the given data. Since six data points are given, a fifth order polynomial of the following
form will be generated.
s(t) = f(t)= a + bt + ct2 + dt3 + e t4 + f t 5
Substituting the given values of (t,s) in the assumed polynomial above yields the following
matrix equation:
[M]{A} = {V}
where
V
0
75
175
360
480
600
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
M
1
1
1
1
1
1
0
10
20
30
35
40
0
2
10
2
20
2
30
2
35
2
40
2
0
3
10
3
20
3
30
3
35
3
40
3
0
4
10
4
20
4
30
4
35
4
40
4
0
5
10
5
20
5
30
5
35
5
40
5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
Using the Mathcad function lsolve, the unknown coefficients in the column vector {A} can be found
A lsolve M V
,
( )
:=
A
0
9.41429
0.36131
−
0.01667
7.7381 10
5
−
×
4.52381
− 10
6
−
×
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Chapter 7: Numerical Differentiation 263](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-273-320.jpg)
![As done in the preceding problem, we first use the method of undetermined coefficients to fit an
interpolating polynomial to the given data. Since six data points are prescribed , a fifth order
polynomial of the following form can be generated.
x(t) = f(t)= a + bt + ct2 + dt3 + e t4 + f t 5
Substituting the given values of (t,x) in the assumed polynomial above yields the following matrix
equation:
[M]{A}= {V}
where
M
1
1
1
1
1
1
0
0.5
1
1.5
2
2.5
0
2
0.5
2
1
2
1.5
2
2
2
2.5
2
0
3
0.5
3
1
3
1.5
3
2
3
2.5
3
0
4
0.5
4
1
4
1.5
4
2
4
2.5
4
0
5
0.5
5
1
5
1.5
5
2
5
2.5
5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:= V
0
0.23
0
0.075
−
0.025
0
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
A lsolve M V
,
( )
:= A
0
1.7375
3.69917
−
2.65167
0.76333
−
0.07333
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
The coefficients a, b, c, d. e and f of the assumed polynomial are
a A1
:= a 0
= c A3
:= c 3.69917
−
= d A4
:= d 2.65167
=
b A2
:= b 1.737
= e A5
:= e 0.76333
−
= f A6
:= f 0.073
=
Chapter 7: Numerical Differentiation 267](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-277-320.jpg)
![As shown below , the damping, c , and stiffness, k , furnished by the shock absorber
can be computed using the pertinent formulas available in textbooks in Vibrations and
Controls [ 6, 9, 14, 16, 20, 21].
Damping factor x undamped natural frequency = 1/ system time constant :
ζωn
1
τ
:= ζωn 1.6
=
Since ζωn and ωd are known , the damping factor ζ can be calculated.
ωd ωn 1 ζ
2
−
⋅
=
ωd 2
π
τd
⋅
:= ωd 3.142
= rad/sec ζ
ζωn
2
ωd
2
ζωn
2
+
:= ζ 0.454
=
The undamped natural frequency of the system along with the stiffness and damping coefficients
of the shock absorber can now be computed using
ωn
ζωn
ζ
:= ωn 3.526
= rad/sec
The stiffness and damping coefficients of the shock absorber can now be determined.
m 200
:= kg
k m ωn
2
⋅
:= k 2.486 10
3
×
= N/m
c 2 m
⋅ ωn
⋅ ζ
⋅
:= c 640
= N-sec/ m
The shock absorber spring and damper restoring forces which can now be calculated as functions
of time using
270 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-280-320.jpg)
![The procedure for computing the integral of f(x) over a prescribed range of x= a to x= b by
the Romberg method is as follows.
1. For a range of integration from x =a to x= b, a shown in Figure 8.8 (a) ,obtain
the first trapezoidal estimate I 0,1 of the integral of the given function f(x) by the trapezoidal
rule as
(8.19)
I0 1
,
1
2
h
⋅ f a
( )
( f b
( )
+
[ ]
⋅
=
where h b a
−
= (8.20)
2. Divide the given range into two segments as shown in Figure 8.8 (b) and obtain
the second trapezoidal estimate I 1,1 as
I1 1
,
h
2
y1 2 y2
⋅
+ y3
+
( )
⋅
= (8.21)
where
h
b a
−
2
=
y1 f a
( )
= <--Value of function at the beginning of the range of
integration
(8.22)
y2 f a h
+
( )
= <--Value of function at the middle of the range of
integration
y3 f a 2 h
⋅
+
( )
= <--Value of function at the end of the range of
integration
3 Now divide the given range further into four segments and obtain the third
trapezoidal estimate I 2,1 with h = one-half of the previous h = (b-a)/4 .
4. Obtain a fourth trapezoidal estimate I 3,1with eight segments and h = (b-a)/8
5. Obtain a fifth trapezoidal estimate I 4,1 with sixteen segments and h = (b-a)/16
6. Using the Richardson extrapolation formula which is
Chapter 8: Numerical Integration 289](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-299-320.jpg)
![Ii j
, Ii j 1
−
,
1
4
j 1
−
1
−
Ii j 1
−
, Ii 1
− j 1
−
,
−
( )
⋅
+
= (8.23)
compute the other unknown components of the matrix [I] , namely ,
I1 2
, I2 2
, I2 3
, I3 2
, I3 3
, I3 4
,
I4 2
, I4 3
, I4 4
, I4 5
, ............
and fill in the Romberg table which is shown in Table 8.1. Keep filling in the table
until the integral values converge , i.e,
I3 3
, I3 2
,
= or I4 3
, I4 2
,
= or I4 4
, I4 3
,
=
whichever occurs earlier.
TABLE 8.1
The Romberg table
R
I0 1
,
I1 1
,
I2 1
,
I3 1
,
I4 1
,
I5 1
,
I1 2
,
I2 2
,
I3 2
,
I4 2
,
I5 2
,
I2 3
,
I3 3
,
I4 3
,
I5 3
,
I3 4
,
I4 4
,
I5 4
,
I4 5
,
I5 5
, I5 6
,
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
:=
The procedure of this method should be clear from the examples that follow.
290 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-300-320.jpg)
![8.6 APPLICATIONS IN NUMERICAL INTEGRATION
Applications of numerical integration can be encountered in situations dealing with location of
centroids, determination of moments of inertia of cross sectional shapes and computation of
periods of oscillations in nonlinear vibration problems. Some examples are dealt with below.
8.6.1 Centroid of a Rod Bent into the Shape of a Parabola
The length of an element dL of a rod bent into the shape of the parabola shown in Figure 8.8
can be expressed in terms of the differentials dx and dy as [ 8 ]
dL dx
2
dy
2
+
= or 1
y
x
d
d
⎛
⎜
⎝
⎞
⎟
⎠
2
+ dy
⋅ (8.24)
Since x 2 y
2
⋅
= ,
y
x
d
d
4 y
×
= and dL 1 16 y
2
⋅
+ dy
⋅
=
y x
( )
x
2
:= x 0 0.001
, 0.5
..
:=
0 0.1 0.2 0.3 0.4
0
0.2
0.4
bent shape of rod
bent shape of rod
Rod of parabolic shape: x = 2 y .y
x , meters
y,
meters
y x
( )
x
Figure 8.8. Rod bent into a parabolic shape
The length of the rod , then, can be written as
L L
1
⌠
⎮
⎮
⌡
d
= or L
0
0.5
y
1 16 y
2
⋅
+
⌠
⎮
⌡
d
:= L 0.739
= meters
Chapter 8: Numerical Integration 301](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-311-320.jpg)
![The location of the centroid ( x CG , y CG ) can then be calculated using
xCG
L
x
⌠
⎮
⎮
⌡
d
L
= yCG
L
y
⌠
⎮
⎮
⌡
d
L
=
or,
xCG
1
L 0
0.5
y
2 y
2
⋅ 1 16 y
2
⋅
+
⋅
⌠
⎮
⌡
d
⋅
:= yCG
1
L 0
0.5
y
y 1 16 y
2
⋅
+
⋅
⌠
⎮
⌡
d
⋅
:=
giving
xCG 0.205
= m yCG 0.287
= m
8.6.2. Moment of Inertia of a Beam of Semi-Elliptic Cross Section
For the semi-elliptic beam cross section shown in Figure 8.9 , its moments of inertia about
the x and y axes can be calculated using [ 8 ]
Ixx A
y
2
⌠
⎮
⎮
⌡
d
= Iyy A
x
2
⌠
⎮
⎮
⌡
d
= (8.25)
where
dA 2 x
⋅ dy
⋅
= with x y
( ) 1
y
2
b
2
− a
⋅
= and a 4
:= in b 2
:= in
dA y dx
⋅
= with y x
( ) 1
x
2
a
2
− b
⋅
:= x 4
− 3.99
−
, 4.
..
:=
4 2 0 2 4
0
2
Semi-elliptic cross section
y x
( )
x
The moments of inertia , I xx and I yy,
can now be computed as follows,using
Equations (8.25)
Ixx 2
a
b
⋅
0
b
y
y
2
b
2
y
2
−
⋅
⌠
⎮
⌡
d
⋅
:=
Figure 8.9. Beam cross section
Iyy
b
a a
−
a
x
x
2
a
2
x
2
−
⋅
⌠
⎮
⌡
d
⋅
:= giving Ixx 12.566
= in 4 and Iyy 50.266
= in 4
302 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-312-320.jpg)
![8.6.3. Launch of a Projectile
A projectile is launched with an initial vertical velocity vy init from a height y 0 above the ground.
Neglecting air resistance, the relationship between its vertical velocity v y at any time t and the
height y above the ground that it attains on the way up is given by [ 8 ]
vy vyinit
2
2 G
⋅ y0 y
−
( )
⋅
+
= (8.26)
where G is the acceleration due to gravity ( 32.2 ft/ sec 2 or 9.81 m/ sec 2 ) . Equating the
above quantity to dy/dt , and then integrating with respect to time from 0 to t seconds, the
following expression can be derived for the time t taken by the projectile to attain the height
y above the ground
t
y0
y
y
1
vyinit
2
2 G
⋅ y0 y
−
( )
⋅
+
⌠
⎮
⎮
⎮
⌡
d
=
(8.27)
The time taken to reach the maximum height , y max , is then the above integral with
y max as the upper limit of integration , where
ymax
vyinit
2
2 G
⋅
y0
+
=
(8.28)
For an initial velocity of 55 ft/ sec at an angle of 65 degrees from a height of 3.5 ft above ground, the
time taken to attain the maximum height is calculated below using Equations (8.27) and (8.28).
vinit 55
:= ft/sec θ 65 deg
⋅
:= sin θ
( ) 0.906
= G 32.2
:= ft/ sec 2
vyinit vinit sin θ
( )
⋅
:= vyinit 49.847
= ft/sec y0 3.5
:= ft
ymax
vyinit
2
2 G
⋅
y0
+
:= ymax 42.083
= ft
tmaxheight
y0
ymax
y
1
vyinit
2
2 G
⋅ y0 y
−
( )
⋅
+
⌠
⎮
⎮
⎮
⌡
d
:= tmaxheight 1.548
= sec
Chapter 8: Numerical Integration 303](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-313-320.jpg)
![8.6.4. Large Oscillations of a Simple Pendulum
A simple pendulum, which is a mass attached to one end of a light rigid rod, is shown in Figure
8.10.
Figure 8.10. Vibration of a simple pendulum
The large-amplitude motion of this pendulum is governed by the following nonlinear differential
equation [ 6,9,16,20 ].
2
t
θ t
( )
d
d
2 G
L
sin θ t
( )
( )
⋅
+ 0
= (8.29)
where L is the length of the rod and G is the acceleration due to gravity . However, if the
amplitudes of vibration are considered small, the above differential equation assumes
its linear form
2
t
θ t
( )
d
d
2 G
L
θ t
( )
⋅
+ 0
= (8.30)
and the motion, then, has a natural period
τlinear 2 π
⋅
L
G
⋅
:= seconds (8.31)
For the nonlinear problem , it can be shown that the period of one oscillation, which is now a
function of the amplitude of vibration , is given by the integral
304 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-314-320.jpg)
![T
H1
H2
h
A
Q k h
0.5
⋅
−
⌠
⎮
⎮
⎮
⌡
d
=
where H1 and H2 are the initial and final heights of the water surface above the base of the
tank , A is the cross sectional area of the tank , Q is the inflow rate, and k is a constant
that depends on the coefficient of discharge and the area of the orifice. Determine how long
it will take for the water level to fall 10 ft in a tank 50 ft long by 30 ft wide with vertical sides,
when water enters the tank at the rate of 6 cu. ft / sec and the initial level of the water
surface in the tank is 16 ft above the bottom. Take k to be 3.
8.16. The motion of bodies moving with moderate speed in fluids such as
water or air is resisted by a damping force that is proportional to the square of the speed. In
order to estimate the resonant amplitude of the motion, the equivalent linear viscous
damping, C eq , is needed which is found by equating the energy dissipated by the viscous
damping in one cycle to that of the nonviscous damping [ 20 ]. The equivalent viscous
damping, in this case , can be shown to be
Ceq
2 a
⋅ ω
⋅ X
⋅
π 0
π
θ
sin θ
( )
3
⌠
⎮
⌡
d
⋅
=
where a is a proportionality constant that depends on the body geometry and the
properties of the fluid medium , ω is the frequency of excitation and X is the
amplitude of the resulting vibration. Determine C eq if a = 0.85 N-sec 2 / m , ω = 550
rads/ sec and X = 16 mm.
8.17. The moment of inertia of a sphere of radius R about a diametral axis is given by
the integral [ 8 ]
Idia
1
2
ρ
⋅ π
⋅
R
−
R
x
R
(
2
x
2
−
⎡
⎣ ⎤
⎦
2
⌠
⎮
⎮
⌡
d
⋅
=
where ρ is the mass density of the material of the sphere. Estimate I dia for a 1 m
diameter sphere made of steel with ρ = 7.8 Mg / m 3
8.18. The shear stress distribution in a beam cross section subjected to a shear force
V is given by [ 18 ]
Chapter 8: Numerical Integration 309](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-319-320.jpg)
![Example 9.7.
Using the Runge-Kutta fourth-order method, obtain a solution to
dx/dt = f(x,y,t) = xy2 + t; dy/dt= g(x,y,t) = ty + x2
for t = 0 to t= 0.5 seconds. The initial conditions are given as: x(0)= -1; y (0) = 1
Use a time increment of h = 0.1 second.
Note that both x and y are functions of the independent variable " t".
Thus, in this case,
dx/dt = f(x,y,t ) = xy2 + t ,
and dy/dt = g(x,y,t) = ty + x2
Version 1 of the Runge-Kutta Method which utilizes Simpson's one-third rule, will then resort
to the following sequential equations to compute x and y at t1, where t1 = t0 + h., and h is
the increment or step size
k1 = h f(x0, y0, t0) ; m1= h g(x0,y0, t0) ;
tmp = t0 + 0.5 h ; xmp= x0+ 0.5 k1; ymp= y0+0.5 m1
k2= h f(xmp, ymp, tmp) ; m2 = h g( xmp, ymp, tmp )
xmp* = x0 + 0.5 k2 ; ymp* = y0+ 0.5 m2
k3 = h f ( xmp*, ymp*, tmp ) ; m3 = h g ( xmp*, ymp*, tmp )
x1 = x0 + k3 ; y1 = y0 + m3 ; t1 = t0 + h
k4 = h f ( x1, y1, t1 ) ; m4 = h g ( x1, y1, t1 )
x1= x0 + (1/6) [ k1 +2 k2 + 2 k3 + k4 ] ; y1 = y0 + (1/6 ) [ m1 + 2 m2 + 2 m3 + m4 ]
In order to compute the solution (x1,y1) at t1, we start with known values x0, y0 and t0 as
shown above. Similarly, to compute the solution (x2, y2) at t2, the above procedure needs to
be repeated with ( x1, y1 , t1) replacing (x0, y0,t0) and (x2,y2,t2) replacing (x1,y1,t1).
Define the functions f(x,y,t) and g(x,y,t), the step size and the initial conditions:.
342 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-352-320.jpg)
![t x y
Note: The first column contains the
t-values, the second column contains
the x-values, and the third contains the
y- values
Value
0
5 10
3
−
×
0.03
0.155
0.374387
0.5
1
−
1.005025
−
1.030966
−
1.190367
−
1.800212
−
3.026613
−
1
1.005038
1.031388
1.197505
1.741152
2.524711
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
<---
It is noted that the Runge-Kutta- Version 1 solution agrees reasonably well with the
Mathcad solutions.
9.7 SOLUTION OF HIGHER-ORDER ORDINARY DIFFERENTIAL
EQUATIONS
In many areas of engineering and physics, the solution of higher-order differential equations is
often required. Examples can be encountered in vibration engineering, control systems
engineering , electrical circuit analysis etc.
The general form of a second-order differential equation is
d2y/dx2 = f ( x, y, dy/dx ) (9.27)
A more special form often encountered in vibrations, control systems, and electrical circuit
analysis is
a d2y/dx2 + b dy/dx + c y = q(x) (9.28)
with initial conditions y(0) and dy/dx(0) specified.
Although any of the methods discussed in sections 9.2-9.5 can certainly be used to obtain
solutions to Equation 9.28, it must first be converted to a system of first-order differential
equations as follows.
dy/dx = z = f ( x,y, z)
dz/dx = [q(x) - b z - c y] / a = g (x,y, z )
(9.29)
Chapter 9: Numerical Solution of Ordinary Differential Equations 349](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-359-320.jpg)
![with the initial conditions now becoming y(0) and z(0) . Thus, if a solution to the second-order
differential equation, which is now represented by the two first-order equations of Equations
9.29 , is to be generated by Version 1 of the fourth-order Runge-Kutta method, the
calculations shown in the following procedure must be resorted to.
k1 = h f (y0, z0, x0) ; m1= h g (y0,z0, x0) ;
xmp = x0 + 0.5 h ; ymp= y0+ 0.5 k1; zmp= z0+0.5 m1
k2= h f (ymp, zmp, xmp) ; m2 = h g ( ymp, zmp, xmp )
ymp* = y0 + 0.5 k2 ; zmp* = z0+ 0.5 m2
k3 = h f ( ymp*, zmp*, xmp ) ; m3 = h g ( ymp*, zmp*, xmp )
y1 = y0 + k3 ; z1 = z0 + m3 ; x1 = x0 + h
k4 = h f ( y1, z1, x1 ) ; m4 = h g ( y1, z1, x1 )
y1= y0 + (1/6) [ k1 +2 k2 + 2 k3 + k4 ] ; z1 = z0 + (1/6 ) [ m1 + 2 m2 + 2 m3 + m4 ]
(9.30)
where h is the increment or step size. Note that in order to generate the solution
(y1, z1), we started with known values ( x0, y0, z0 ). Similarly, to obtain ( y2, z2) , the above
procedure must be repeated with (x1,y1,z1) replacing ( x0, y0,z0) and (x2, y2,z2) replacing
(x1,y1,z1).
Example 9.8
Using the Runge-Kutta fourth-order
method, obtain a solution to
m d2y/dt2 +c dy/dt + k y = F(t) ;
m= 1kg, c= 3 N-sec/m, k = 9 N/m , F(t)= 9 N.
for t = 0 to t= 3 seconds. The initial conditions
are given as: y(0)=0; dy/dt (0) = 0 .
Use a time increment of 0.5 seconds.
350 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-360-320.jpg)
![The above differential equation represents the equation of motion of a linear
spring-mass-damper system subjected to a step forcing function as shown in the above figure.
As mentioned earlier , application of the fourth order Runge- Kutta Method to the solution of a
second-order differential equation is quite similar to that of a first order differential equation ,
except that the following two first-order equations must now be considered.
dy/dt = z ;
and m dz/dt + c z +k y = F(t)
Note that, now, both y and z are functions of the independent variable " t".
Thus, in this case,
dy/dt = f( y,z, t ) = z ,
and dz/dt = g(y,z,t) =[F(t) - ( k y + c z)] /m
Version 1 of the R-K procedure will then resort to the following equations to compute y and z at
t1:
k1 = h f(y0, z0, t0) ; m1= h g(y0,z0, t0) ;
tmp = t0 + 0.5 h ; ymp= y0+ 0.5 k1; zmp= z0+0.5 m1
k2= h f(ymp, zmp, tmp) ; m2 = h g( ymp, zmp, tmp )
ymp* = y0 + 0.5 k2 ; zmp* = z0+ 0.5 m2
k3 = h f ( ymp*, zmp*, tmp ) ; m3 = h g ( ymp*, zmp*, tmp )
y1 = y0 + k3 ; z1 = z0 + m3 ; t1 = t0 + h
k4 = h f ( y1, z1, t1 ) ; m4 = h g ( y1, z1, t1 )
y1= y0 + (1/6) [ k1 +2 k2 + 2 k3 + k4 ] ; z1 = z0 + (1/6 ) [ m1 + 2 m2 + 2 m3 + m4 ]
Chapter 9: Numerical Solution of Ordinary Differential Equations 351](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-361-320.jpg)
![solution process. Upon completion of the solution process, a check must be made to
determine if the condition prescribed at the point x = L is satisfied or not. If not, a second
guess must be made and the procedure redone. If the second guess still does not satisfy the
prescribed end condition, a third guess has to be estimated. This can be accomplished by
noting the difference between the given and the computed conditions at x = L and resorting to
a linear interpolation process that utilizes the results of the first two trials. This iteration
process must be repeated until the prescribed end condition is , in fact, satisfied. In this
method, it is important for the analyst to be skillful at estimating the unprescribed initial
condition properly at the starting point in order to maintain efficiency and avoid divergence.
Because the method is analogous to aiming properly and shooting at a given object, it has
been appropriately labelled as the shooting method. When higher-order differential equations
must be solved involving several boundary conditions, this method has proven to be quite
uncouth and laborious and, thus, inefficient.
Although any of the methods discussed in this chapter can be employed in the solution of a
boundary-value problem by the shooting method. the following example will illustrate its
application with Mathcad's Rkadapt and rkadapt functions.
Example 9.9
Solve: d2y/ dx2 = M/EI = (30 x - 1.5 x^2)/ EI , where EI = 121,500 kips-ft 2
The boundary conditions prescribed are:
x = 0, y = 0 and x= 20, y =0
This differential equation represents the relationship between the bending moment M and
the deflection "y" at a distance "x" along the span from the left hand end of a 20 foot long
simply supported beam loaded with a uniformly distributed load of 3 kips/ ft. as shown in
Figure 9.17. The slope and deflection at any point " x " can be obtained by solving the above
differential equation.
Consider the two equations:
dy/dx = z ; and
(9.32)
dz/dx = M/EI = (30 x - 1.5 x^2 )/ EI
Note that, now, both y and z are functions of the dependent variable " x". Thus, in this
case,
dy/dx = f ( x, y, z ) = z ,
(9.33)
and dz/dx = g( x, y, z ) =[30 x - 1.5 x ^2 ] / EI
As said earlier, the shooting method is a trial- and- error method used in the solution of
boundary-value problems. Although two boundary conditions are prescribed here, there is
only one initial condition ( x= 0, y = 0 ) provided to begin the solution process. The initial
Chapter 9: Numerical Solution of Ordinary Differential Equations 359](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-369-320.jpg)
![New value of y1 to be tried = [second value of the first derivative y1 assumed times the
first deflection value at the end computed - first value of the first derivative y1 assumed
times the second deflection value at the end computed ] / [ first deflection value at the end
computed - the second deflection value at the end computed ] . That is,
y1
Value20 2
, Value14 1
,
⋅ Value10 2
, Value24 1
,
⋅
−
Value14 1
, Value24 1
,
−
:=
y1 8.23045
− 10
3
−
×
=
Then, the solution with the new guess is
Value3 rkadapt y 0
, 20
, 0.001
, D
, 1000
, 0.00001
,
( )
:=
which yields
Note the negligible
deflection
computed at the end with
the current assumed value
of the slope (derivative) at
x= 0
<----
Value3
0
0.2
1.2
6.2
20
0
1.645763
− 10
3
−
×
9.807565
− 10
3
−
×
0.042741
−
1.922074
− 10
15
−
×
8.230453
− 10
3
−
×
8.225547
− 10
3
−
×
8.059786
− 10
3
−
×
4.465547
− 10
3
−
×
8.230453 10
3
−
×
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Now , using Rkadapt , the solution can be evaluated at 100 points between 0 and 20 with the
above initial slope value put in which led to the computation of the true displacement (y= 0) at the
end x= 20 Ft. The following commands will then generate the solution matrix [ Z ] which will
contain 3 columns- the "x" values, the corresponding "y" values and the "dy/dx " values.
Z Rkadapt y 0
, 20
, 100
, D
,
( )
:=
i 0 rows Z
( ) 1
−
..
:=
The results of the above Mathcad computation are plotted in Figures 9.18 and 9.19.
362 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-372-320.jpg)
![0 100
0.2
0.1
0
Slope along beam span
Distance from left end
Slope
Z
2
〈 〉
( )i
Z
0
〈 〉
( )i
0 50 100 150
20
10
0
Deflection along beam span
Distance from left end, inches
Deflection
,
inches
Z
1
〈 〉
( )i
Z
0
〈 〉
( )i
Figure 9.23. Beam deflection curve Figure 9.24. Beam slope along span
9.9.3. Temperature Response of a Solid Steel Ball Placed in a Hot Stream Of Air
When a body which is initially at a uniform temperature T 0 is suddenly placed in a hot
fluid, the rate at which the temperature T(t) of the body changes with time is given by the
differential equation [ 17 ]
ρ c
⋅ Lc
⋅
t
T t
( )
d
d
⋅ h Tinf T t
( )
−
( )
⋅
= (9.38)
where ρ is the mass density of the body, c is its specific heat at constant pressure, T inf is
the temperature of the surrounding fluid, h is the average convection heat transfer coefficient,
and L c is the characteristic length. For a solid steel sphere of diameter 2 cm initially at 10 deg
C., and placed in a stream of hot air at temperature, T inf = 75 deg C. , with h =25 W/ m 2 . K,
we have
d 0.02
:= m ρ 7854
:= kg/ m 3
c 434
:= J/ kg . K h 25
:= W/ m 2 . K
Tinf 75
:= deg. K T0 10
:= deg. K V π
d
3
6
⋅
:= A π d
2
⋅
:=
Lc
V
A
:= Lc 3.333333 10
3
−
×
= m
and the Mathcad statements needed to generate the temperature response of the steel
ball to the hot stream of air are
366 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-376-320.jpg)
![T0 10
:= D t T
,
( )
h
ρ c
⋅ Lc
⋅
Tinf T
−
( )
⋅
:= Z Rkadapt T 0.
, 2000
, 100
, D
,
( )
:=
i 0 rows Z
( ) 1
−
..
:=
0 500 1000 1500 2000
0
50
100
Temp response of steel ball
Time, seconds
Temperature,
deg.
C.
Z
1
〈 〉
( )i
Z
0
〈 〉
( )i
Figure 9.25. Temperature response of steel ball
The Mathcad solution generated, which is plotted in Figure 9.25 shows how the steel ball attains
its steady-state temperature of 75 deg C. in a time t equal to 4τ, where τ is the system time
constant which, in this case, is ρ c
⋅
Lc
h
⋅ = 454.5 seconds.
9.9.4. Nonlinear Vibration of a Simple Pendulum
The natural frequency of large oscillation of a simple pendulum, which is a mass attached
to one end of a light rigid rod, has been generated in Section 6 of Chapter 8 as a function
of the maximum amplitude of the pendulum. The nonlinear behavior of the pendulum is
governed by the differential equation [ 6, 9, 16, 20 ]
2
t
θ t
( )
d
d
2 G
L
sin θ t
( )
( )
⋅
+ 0
= (9.39)
where L is the length of the pendulum and G is the acceleration due to gravity .
Chapter 9: Numerical Solution of Ordinary Differential Equations 367](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-377-320.jpg)
![9.9.6. Nonlinear Vibration of a Damped System with a Hardening Spring.
Consider the following nonlinear equation known as Duffing's equation [ 6, 16, 20, 21 ],
which is the differential equation of motion of a spring-mass-damper system with a
hardening spring
M
2
t
x
d
d
2
⋅ c
t
x
d
d
⋅
+ k x
⋅
+ μ x
3
⋅
+ F t
( )
= (9.42)
where the last term on the right hand side of the equation is due to the effect of the hardening
spring in which the spring stiffness increases as the load on it increases.
The term F(t) on the right hand side of the equation is the exciting force that the system
is subjected to. We will utilize Mathcad to obtain solutions to Equation (9.42) for m
= 1.2 kg, c= 0.6 N-sec/m , k= 1.2 N/ m and F(t) = (a) a step function and then (b) a
ramp function for values of μ ranging from 0.01 to 1.0. The results will then be compared with
those for μ =0
(a) Step input with μ =0, 0.01, 0.1, 1.0:
M 1.2
:= k 1.2
:= c 0.6
:=
xinit 0
:= vinit 0
:= <-- Specify initial displacement and velocity
Express initial conditions as a vector y , the first
component being the initial displacement and the
second being the initial velocity.
y
xinit
vinit
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:= <----
F t
( ) Φ t
( )
:= This creates a unit step function (Φ(t)= unit
step or Heaviside function )
<----
0 0.2 0.4 0.6 0.8
0.998
0.999
1
1.001
Unit step input
Time, seconds
F(t),
Newtons
F t
( )
t
t 0 0.1
, 1.0
..
:=
Chapter 9: Numerical Solution of Ordinary Differential Equations 373](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-383-320.jpg)
![0 5 10 15 20
0
5
10
15
20
mu= 0.01
mu= 0.1
mu= 1.0
mu = 0
mu= 0.01
mu= 0.1
mu= 1.0
mu = 0
Reponse to ramp input
Time, seconds
Displacement
,
m
xlowmui
xmidmui
xhighmui
xzeromui
ti
Figure 9.31. Response of system with hardening spring to a unit ramp input
9.9.7. Temperature Distribution in the Wall of a Pipe Carrying a Hot Fluid
The differential equation relating temperature in the wall of a metal pipe to radial distance is
[ 17 ]
r
2
r
T r
( )
d
d
2
⋅
t
T r
( )
d
d
+ 0
= (9.43)
where T(r) is the temperature in the pipe wall at a radial distance r from the pipe centerline. The
inner radius of the pipe is r i and the outer radius is r o . The temperature of the fluid inside the
pipe is T f while the temperature of the outer wall of the pipe is T o. Although two boundary
conditions are prescribed here, namely the temperatures at the inner and outer walls of the pipe,
there is only one initial condition available, that is , T(r i )= T f to begin the solution process.
The initial temperature gradient, namely, dT/dr , is not known. Thus, in order to resort to the
Shooting Method, this initial temperature gradient must be assumed so that the process of
generating a solution may be initiated. The solution T at ro must then be determined with these
initial values of T and dT/dr ( at r i ). If the solution obtained does not match up with the given
condition at r = r o , the assumed initial condition must be revised and a new solution T(r o )
determined. This process must be repeated until the solution obtained at r o agrees with the
prescribed condition . As discussed in Section 9.7, Equation (9.43) must first be converted to
the following two first- order differential equations before a solution can be obtained.
r
T
d
d
0
= ,
r
z
d
d
1
−
r
z
⋅
= (9.44)
378 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-388-320.jpg)
![Then, the solution with the new initgrad is
initgrad 640
−
= T
650
initgrad
⎛
⎜
⎝
⎞
⎟
⎠
:=
D r T
,
( )
initgrad
1
−
r
initgrad
⋅
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
Value3 rkadapt T ri
, ro
, 0.001
, D
, 1000
, 0.00001
,
( )
:=
which yields
Note that the temperature
generated at the outer wall now
matches the prescribed
temperature (10 deg. C. ).
<----
Value3
1
1.01
1.06
1.31
2
650
643.6
611.6
451.6
10
640
−
633.632
−
602.708
−
467.183
−
196.382
−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Now , using Rkadapt , the solution can be evaluated at 100 points between r= 1 and r =
2 cm with the above initial dT/dr value put in which led to the computation of the true
temperature at the outer wall . The following commands will then generate the solution
matrix [ Z ] which will contain 3 columns- the "r" values, the corresponding "T" values
and the "dT/dr " values.
initgrad 640
−
= Z Rkadapt T ri
, ro
, 100
, D
,
( )
:=
D r T
,
( )
initgrad
1
−
r
initgrad
⋅
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
i 0 rows Z
( ) 1
−
..
:=
The results of the above Mathcad computation are plotted in Figures 9.32 and 9.33.
Chapter 9: Numerical Solution of Ordinary Differential Equations 381](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-391-320.jpg)
![9.9.9. The Effect of Damping on the Step Response of a Second-Order Control System
The general differential equation governing the behavior of a second-order control system
subjected to a unit step input is [ 14 ]
2
t
y t
( )
d
d
2
2 ζ
⋅ ωn
⋅
t
y t
( )
d
d
⎛
⎜
⎝
⎞
⎟
⎠
+ ωn
2
y t
( )
⋅
+ ωn
2
= (9.47)
where y(t) is the system response to the unit step input, ζ is the system damping ratio,
and ω n is the natural frequency of the undamped system. We want to investigate the dynamic
response of the system to different levels of damping obtained by varying ζ. If ωn = 5 rad/sec
and ζ is varied from 0.1 to1.4, the following Mathcad statements will be needed to generate the
rseponses .
ωn 5
:= y
0
0
⎛
⎜
⎝
⎞
⎟
⎠
:=
ζ1 0.1
:= D1 t y
,
( )
y1
ωn
2
2 ζ1
⋅ ωn
⋅ y1
⋅
− ωn
2
y0
⋅
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:= Z1 Rkadapt y 0
, 10
, 1000
, D1
,
( )
:=
ζ2 0.2
:= D2 t y
,
( )
y1
ωn
2
2 ζ2
⋅ ωn
⋅ y1
⋅
− ωn
2
y0
⋅
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:= Z2 Rkadapt y 0
, 10
, 1000
, D2
,
( )
:=
ζ3 0.6
:= D3 t y
,
( )
y1
ωn
2
2 ζ3
⋅ ωn
⋅ y1
⋅
− ωn
2
y0
⋅
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
Z3 Rkadapt y 0
, 10
, 1000
, D3
,
( )
:=
ζ4 0.8
:= D4 t y
,
( )
y1
ωn
2
2 ζ4
⋅ ωn
⋅ y1
⋅
− ωn
2
y0
⋅
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:= Z4 Rkadapt y 0
, 10
, 1000
, D4
,
( )
:=
384 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD](https://image.slidesharecdn.com/anintroductiontonumericalmethodsusingmathcadmathcadrelease14-230805170449-0ea94cc2/85/AN-INTRODUCTION-TO-NUMERICAL-METHODS-USING-MATHCAD-Mathcad-Release-14-394-320.jpg)
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AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD Mathcad Release 14
This document provides an introduction and overview to the textbook "An Introduction to Numerical Methods Using Mathcad" by Khyruddin Akbar Ansari. The textbook is designed as a guide for teaching numerical methods using Mathcad software. It covers topics such as root-finding, matrices, interpolation, curve-fitting, differentiation, integration, and solving differential equations. The author has many years of experience in engineering education and research.
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AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD Mathcad Release 14
- 1. AN INTRODUCTION TONUMERICAL METHODS USING MATHCAD Mathcad Release 14 Khyruddin Akbar Ansari, Ph.D., P.E. Professor of Mechanical Engineering School of Engineering and Applied Science Gonzaga University Spokane, Washington
- 2. Copyright 2007 KhyruddinAkbar Ansari. All rights reserved No part of this book may be used or reproduced or distributed in any form , or by any means or stored in a database or retrieval system, without the prior written permission of the author . Mathcad screenshots are reprinted by permission of Parametric Technology Corporation (PTC). Mathcad is a registered trademark of PTC, 140 Kendrick Street, Needham, MA 02494. All rights reserved
- 3. Dedicated to mydear wife, Fatima
- 4. ABOUT THE AUTHOR KhyruddinAkbar Ansari, who has been a professor of mechanical engineering at Gonzaga University in Spokane, Washington, since 1986, received his bachelor’s degree in electrical engineering from Osmania University, Hyderabad, India in 1964, his master’s degree in mechanical engineering from the University of California at Berkeley in 1965 and his Ph.D. degree in engineering mechanics from the University of Texas at Arlington in 1972. He did his doctoral thesis work in the area of nonlinear vibrations of rotating blades under the guidance of the late Professor Nils O. Myklestad. Professor Ansari has many years of varied industrial , teaching and research experience, having been associated in the past with such organizations as Bell Helicopter, Westinghouse Electric , Bell Aerospace, Brown and Root, King Fahd’s University of Petroleum and Minerals and Battelle Pacific Northwest Laboratories . At Gonzaga University, Professor Ansari is currently involved in the teaching of courses in the areas ofengineeringmechanics,numerical methods,advancedengineeringmathematics,vibrations, system dynamics and controls , and machine design. His research, which has been varied, has resulted in several published articles and contributions to books including chapters in the Encyclopedia of Fluid Mechanics and Developments in Offshore Engineering. Professor Ansari is a member of the American Society of Mechanical Engineers and the American Society for Engineering Education. He is also a registered professional engineer.
- 5. TABLE OF CONTENTS Prefacev 1. Basics of Mathcad. 1 1.1 Introduction 1 1.2 The Mathcad Screen 1 1.3 Exact Answers 3 1.4 Variables, Functions and Live math. 3 1.5 Feedback 4 1.6 Graphics 5 1.7 Graphing of Functions and Plotting of Data 5 1.8 Animations 7 1.9 The Mathcad Tutorials 7 1.10 Advantages of Mathcad 7 1.11 Computations in Mathcad 10 1.12 The Mathcad Window, Toolbars and Palettes 13 1.13 Mathcad Regions 15 1.14 Entering Math and Text 16 1.15 Mathcad Worksheets,Templates and Styles 17 1.16 Defining Variables 18 1.17 Defining Functions in Mathcad 19 1.18 Building and Editing Mathematical Expressions 21 1.19 Defining Range Variables 23 1.20 Defining Vectors and Matrices 24 1.21 Creating Graphs 28 1.22 Formatting Math, Text and Results 35 1.23 Using Units 39 2. Introduction to Numerical Methods. 43 2.1 The Use of Numerical Methods in Science and Engineering 43 2.2 Comparison of Numerical Methods with Analytical Methods 43 2.3 Sources of Numerical Errors and their Computation 44 2.4 Taylor Series Expansion 44 Problems 49 3. Roots of Equations. 53 3.1 Introduction 53 3.2 Methods Available 53 3.3 Bisection Method 53 3.4. The Regula Falsi or the False Position Method 58 3.5 Newton-Raphson Method 65 3.6 Use of Mathcad’s root and polyroots Functions 71 Table of Contents: i
- 6. 3.7 Secant Method72 3.8. Method of Successive Substitution 78 3.9 Multiple Roots and Difficulties in Computation 80 3.10. Solution of Systems of Nonlinear Equations 84 3.11 Solving Systems of Equations using Mathcad's Given and Find Functions 86 3.12 Applications in Root-Finding 87 3.12.1 Maximum Design Load for a Column 87 3.12.2 Natural frequencies of Vibration of a Uniform Beam 89 3.12.3 Solving the Characteristic Equation in Control Systems Engineering 91 3.12.4 Horizontal Tension in a Uniform Cable 93 Problems 96 4. Matrices and Linear Algebra. 103 4.1 Basic Matrix Operations 103 4.2 Use of Mathcad in Performing Matrix Operations 105 4.3 Solution of Linear Algebraic Equations by Using the Inverse 108 4.4 Solution of Linear Algebraic Equations by Cramer’s Rule 110 4.5 Solution of Linear Algebraic Equations Using the Function lsolve 112 4.6 The Eigenvalue Problem 114 4.7 Solving the Eigenvalue Problem with Mathcad 116 4.8 Application of the Eigenvalue Problem to Vibration Engineering 117 4.9 Application of the Eigenvalue Problem to Stress Analysis- Determination of Principal Stresses and Principal Directions 125 4.10 Repeated Roots in the Determinantal Equation 128 4.11 Solution of Nonlinear Simultaneous Equations 131 Problems 134 5. Numerical Interpolation. 141 5.1 Linear Interpolation 141 5.2 The Method of Undetermined Coefficients 141 5.3 The Gregory-Newton Interpolating Polynomial 144 5.4 Interpolation Using Finite Differences 149 5.5 Newton’s Method Utilizing Finite Differences 151 5.6 The Lagrange Interpolating Polynomial 155 5.7 Interpolation Using Linear, Quadratic and Cubic Splines 158 5.8 Interpolation with Mathcad 159 5.9 Applications in Numerical Interpolation 168 5.9.1 Stress-Strain data for Titanium 168 5.9.2 Notch Sensitivity of Aluminum 169 5.9.3 Speech Interference Level 172 5.9.4 Load-Deflection Data for Elastomeric Mounts 175 Problems 177 ii AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 7. 6. Curve-Fitting. 183 6.1The Need to Fit a Function to Measured Data 183 6.2 The Method of Least Squares. 184 6.3 Straight Line Regression 185 6.4 Curve-Fitting with a Quadratic Function 188 6.5 Curve-Fit with a Power Function 191 6.6 Curve-Fitting with an Exponential Function 194 6.7 Curve-Fitting with a Linear Combination of Known Functions 199 6.8 Curve-Fitting with Polynomials. 203 6.9 Use of Mathcad's Regression Functions for Curve-Fitting 207 6.9.1 Linear Regression with Mathcad 207 6.9.2 Nonlinear Regression with Mathcad 209 6.9.3 Use of the Function linfit 211 6.9.4 Use of the Function genfit 213 6.9.5 Use of the Mathcad Functions logfit, lnfit, pwrfit and expfit 215 6.9.6 More Examples with Mathcad 220 6.10 Applications in Curve-Fitting 236 6.10.1 Fatigue Failure Curve for Loading in the Finite Life Range 236 6.10.2 Temperature Response of an Object Placed in a Hot Stream of Air 239 6.10.3 The Effect of Operating Temperature on the Strength of a Mechanical Element 242 6.10.4 Drop-Testing of Packaged Articles 245 Problems 248 7. Numerical Differentiation 255 7.1 Introduction to Numerical Differentiation and the Use of the Mathcad Derivative Operators 255 7.2 Method of Finite Differences 255 7.3 Interpolating Polynomial Method 259 7.4 Applications in Numerical Differentiation 262 7.4.1 Determination of Velocities and Accelerations from Given Displacement Data 262 7.4.2 Determination of Shock Absorber Parameters, and Damper and Spring Restoring Forces from Given Vehicle Displacement Data 266 Problems 271 8. Numerical Integration 277 8.1 Introduction to Numerical Integration and the Use of the Mathcad Integral Operator 277 8.2 The Interpolating Polynomial Method 279 8.3 Trapezoidal Rule 280 8.4 Simpson’s Rules 283 Table of Contents: iii
- 8. 8.4.1 Simpson’s One-ThirdRule 283 8.4.2 Simpson’s Three-Eighth Rule 286 8.4.3 Limitations of Simpson’s Rules 287 8.5 Romberg Integration 288 8.6 Applications in Numerical Integration 301 8.6.1 Centroid of a Rod Bent into the Shape of A Parabola 301 8.6.2 Moment of Inertia of a Beam of Semi-Elliptic Cross Section 302 8.6.3 Launch of a Projectile 303 8.6.4 Large Oscillations of a Simple Pendulum 304 Problems 306 9. Numerical Solution of Ordinary Differential Equations. 311 9.1 Introduction 311 9.2 Taylor Series Method 312 9.3 Euler’s Method 317 9.4 Modified Euler’s Method 323 9.5 Runge- Kutta Methods 329 9.5.1 Fourth-Order Runge-Kutta Method 329 9.5.2 Mathcad Solutions to a First-Order Differential Equation 335 9.6 Systems of Ordinary Differential Equations 341 9.7 Solution of Higher-Order Ordinary Differential Equations 349 9.8 Boundary-Value Problems and the Shooting Method 358 9.9 Applications in Numerical Solution of Ordinary Differential Equations 363 9.9.1 Response of an Electric R-L Circuit to a Unit-Step Voltage Input 363 9.9.2 Deflection Curve of a Cantilevered Beam with a Uniformly Distributed Load 364 9.9.3 Temperature Response of a Solid Steel Ball Placed in a Hot Stream of Air 366 9.9.4 Nonlinear Vibration of a Simple Pendulum 367 9.9.5 Transient Vibration of a Spring-Mass-Damper System Excited by a Pulse Function 370 9.9.6 Nonlinear Vibration of a Damped System with a Hardening Spring 373 9.9.7 Temperature Distribution in the Wall of a Pipe Carrying a Hot Fluid 378 9.9.8 Response of an R-L Circuit with a Nonlinear Resistor 382 9.9.9 The Effect of Damping on the Step Response of a Second-Order Control System 384 Problems 386 Bibliography 399 Index 401 iv AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 9. PREFACE This book, whichis designed to be used in a first numerical methods course in a computer science, mathematics, science , engineering or engineering technology curriculum, has been developed from notes that I generated over several years in the process of teaching a one- semester / three-credit sophomore level course in computer methods at Gonzaga University. It should also provide a reliable source of reference material to practicing engineers and scientists and in other junior and senior-level courses such as machine design, vibrations, system dynamics and controls, where Mathcad can be effectively utilized as a software tool in problem solving. A principal goal of this book , then, is to furnish the background needed to create Mathcad documents for the generation of solutions to a variety of problems. Specific applications involving root-finding, interpolation, curve-fitting, matrices, derivatives, integrals and differential equations are discussed and the broad applicability of Mathcad demonstrated. The material contained herein should be easily grasped by students with a background in calculus, elementary differential equations and some linear algebra, and when utilized in a mathematics, science, engineering, computer science or engineering technology course sequence, it should give a good basic coverage of numerical methods while simultaneously demonstrating the general applicability of Mathcad to problem- solving and solution documentation. The first chapter discusses Mathcad basics while the second gives a general introduction to numerical methods in science and engineering and presents the computation of numerical errors along with the Taylor series as a basis of approximation in numerical analysis. Chapter 3 offers a coverage of the popular methods of finding roots of equations such as Bisection, Regula-Falsi, Newton-Raphson, Secant and Iteration . Solutions to systems of nonlinear algebraic equations obtained by the iteration method are also presented. The applicability of root-finding to practical problems such as those that occur in column design, vibration analysis and control systems engineering is demonstrated. In the fourth chapter, matrices and linear algebra are dealt with along with the solution of eigenvalue problems. Applications to the field of vibration engineering and stress analysis are also discussed. Chapters 5 and 6 focus on numerical interpolation and curve- fitting respectively. The common techniques of interpolation and the functions generally resorted to for curve-fitting of data are covered quite thoroughly. The applicability of these to the analysis of scientific and engineering data such as stress-strain, load-deflection and fatigue failure is presented. Chapters 7 and 8 address numerical differentiation and integration. The concept of finite differences is introduced and the common methods of differentiation and integration , such as the interpolating polynomial method, trapezoidal and Simpson’s rules and Romberg integration are covered. Applications of numerical differentiation and integration such as determination of velocities and accelerations from given displacement data and computation of the moment of inertia of a cross section are included. The last chapter, which is the ninth one, deals with coverage of numericalsolutionofordinarydifferentialequations,andalthough it does not discuss each and every technique that is available, it does focus on the popular ones such as Taylor series method, Euler, modified Euler and Runge-Kutta methods. A variety of practical applications of numerical procedures are included in this chapter, ranging from the response of an electric circuit to an input Preface: v
- 10. voltage to theeffect of damping on the response of a control system. Partial differential equations is a topic that is typically dealt with in detail only in graduate level courses and , therefore, I did not think it very appropriate to introduce it in this book which is really aimed at the needs of undergraduate students. Mathcad documents are generated and examples showing the applicability and use of Mathcad are presented throughout the book. Wherever appropriate, the use of Mathcad functions offering shortcuts and alternatives to otherwise long and tedious numerical solutions is also demonstrated. At the end of every chapter is included a set of problems to be solved covering the material presented. A solutions manual that provides solutions to these exercises can be made available to instructors. Although Mathcad is a very powerful and versatile tool that combines text, graphics and equations in a single worksheet, several figures and sketches in the book had to be done using AutoCad. The help provided by Carlos G. Alfaro earlier in generating these figures and Ryan Lambert later in making some additions and modifications to them is gratefully acknowledged. I am also thankful to those of my students who have so kindly made useful suggestions and pointed out errors and mistakes in the manuscript. Their input and feedback have certainly been helpful. Last but not least, I am indebted to my wife, Fatima , for her patience, support and understanding during the preparation of this book. Without her encouragement, its completion would still be wishful thinking. Users of this book are welcome to e-mail any comments and suggestions to me at ansari@gonzaga.edu , and, of course, I will be pleased to address these in subsequent versions / editions. Khyruddin . Akbar Ansari October 2007 vi: AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 11. C H AP T E R 1 BASICS OF MATHCAD 1.1 INTRODUCTION Mathcad 14 is the first global release of Mathcad presented by Parametric Technology Corporation (PTC) with many significant enhancements since it acquired Mathsoft in April 2006. This is a very powerful software that offers a wide variety of tools in the solution of problems requiring the use of numerical analysis. New ways of using Mathcad are constantly being found leading to overwhelming possibilities. The sections that follow give an overview of the features in Mathcad 14 and will help the novice to get introduced to its problem-solving capabilities. Mathcad 14 has a number of improvements and added capabilities designed to increase the productivity of the user as well as Web connectivity. There is a wealth of information that can be found by clicking on "Help" and going to "Tutorials" , where you can get an overview and a quick tour of Mathcad , see the new features available in Mathcad 14, learn the fundamental skills for working in Mathcad, and obtain on-line help if needed. By clicking on "Help", the Mathcad Quicksheets can be accessed, which are ready-made recipes for performing a variety of tasks in Mathcad. E-books and Reference Tables, user-contributed files, samples, animations and graphs along with other example files for using and extending Mathcad can also be accessed. By choosing "User Forums" under the "Help" menu, questions can be posted for other Mathcad users. The Mathcad web site can always be accessed for updates, technical support information and user files , and Mathcad documents can now be saved on the Web or in Microsoft Word using Rich Text Format (RTF). 1.2 THE MATHCAD SCREEN Mathcad has a full screen numerical and symbolic calculator, which is essentially a blank workspace that you see when you open up a new Mathcad document. To use Mathcad's calculating capabilities, type the expression you want to evaluate and then type an equals sign to get the result: 435 57 24 + 437.375 = As shown in Figure 1.1. the palette buttons open up palettes that include a host of math operations . An expression such as the following can be assembled quite easily. sin π 8 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ln 76.5 ( ) e 95.2 ⋅ 1.952 10 40 × = Chapter 1: Basics of Mathcad 1
- 12. Figure 1.1 TheMath palettes 2 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 13. Matrix arithmetic, calculusoperations such as differentiation and integration, and essentially all of the mathematical operations normally required in engineering and scientific analysis can be easily done in Mathcad. Some examples are given below. 2 7 x e 2x sin 4 x ⋅ ( ) ⋅ ⌠ ⎮ ⌡ d 2.641 10 5 × = 7 9i + 13 7i − 0.128 0.761i + = 5 4 7 2 8 3 7 9 6 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 − 0.168 − 0.312 − 0.352 0.072 − 0.152 8 10 3 − × 0.304 0.136 0.256 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 1.3 EXACT ANSWERS If exact answers are needed in terms of fractions and radicals instead of decimals, the "symbolic equals sign" (-->) on the fourth math palette can be used to obtain exact answers. Some examples are given below. 546 59 17 + 9341 17 → sin π 16 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ln 75.6 ( ) e 90.5 ⋅ 9.0751625112586372168e37 → 5 4 9 2 8 3 3 9 5 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 − 13 7 61 7 60 7 − 1 7 − 2 7 − 3 7 6 7 − 33 7 − 32 7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ → 2 7 x e 2 x ⋅ sin 4 x ⋅ ( ) ⋅ ⌠ ⎮ ⌡ d cos 8 ( ) e 4 ⋅ 5 cos 28 ( ) e 14 ⋅ 5 − e 4 sin 8 ( ) ⋅ 10 − e 14 sin 28 ( ) ⋅ 10 + → 1.4 VARIABLES, FUNCTIONS AND LIVE MATH When there is a need to evaluate an expression for several different inputs, the desired variables and functions can be defined and calculated like built-in constants and functions. This computational feature in Mathcad is called live math. Chapter 1: Basics of Mathcad 3
- 14. For example, ifthe area of a circle for various values of the radius is to be calculated, define a radius variable r and give it a value. r 8 := The := symbol can be inserted by typing a colon. Evaluate the area as π r 2 ⋅ 201.062 = The numerical value of r can now be changed to some other number . Click anywhere outside the definition to recalculate the area . . This is live math . If this area computation is to be usable in other parts of the worksheet, an area function must be defined. area r ( ) π r 2 ⋅ := Use this function as needed to calculate an area . Length units for the radius can also be assigned as shown.. area 60 cm ⋅ ( ) 1.131 m 2 = Live math works for symbolic answers as well. Put in a function , obtain the derivative of the function, and then change the function to some other function. Click outside the definition to see how Mathcad redoes the derivative symbolically. 1.5 FEEDBACK There is continuous feedback in Mathcad. As you edit math expressions, the status line will provide advice and information . When Mathcad cannot understand something, it will color the expression red. Clicking on this red expression will give an error message indicating what is wrong. To see an example, click on the expression below. p q + = p In the above, variables p and q were not defined. Therefore, Mathcad tells us so. Also, Mathcad calculates from top to bottom and left to right. p 5 := q 9 := <--This works ---> p q + 14 = 4 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 15. F 4 ( )= F <-This does not work F x ( ) x 2 4 − := 1.6 GRAPHICS It has been demonstrated that Mathcad can be used as a calculator and as an equation solver. It can also be used as a versatile visualization tool that supports a full set of plot types, an animation facility, as well as simple image processing. 1.7 GRAPHING OF FUNCTIONS AND PLOTTING OF DATA To create a plot in Mathcad, type the expression you would like to plot, say t 9 . Then click the graphing button on your palette bar (See Figure 1.2 ) to bring up the graphing palette and click the upper left button for an XY plot. Click outside the graph or press [Enter] to see the result: 4 − 2 − 0 2 4 1 − 10 5 × 5 − 10 4 × 0 5 10 4 × 1 10 5 × t 9 t Parametric plots can also be done easily . . 1 − 0.5 − 0 0.5 1 1 − 0.5 − 0 0.5 1 sin 5 t ⋅ ( ) cos 7 t ⋅ ( ) Chapter 1: Basics of Mathcad 5
- 16. Figure 1.2 Thegraph palette 6 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 17. . . .and like all Mathcad calculations, plots are also live, which means that changes can be easily made. 1.8 ANIMATIONS Mathcad can create and play short animation clips by using the built-in variable FRAME and the Animation dialog box. Choose Animation, and then Record from the Tools menu to bring up this dialog box Plots as well as numerical results can be animated, and the animation played back at different speeds. 1.9 THE MATHCAD TUTURIALS The Mathcad tutorials offer a step-by-step guide to building and editing expressions, creating and formatting graphs, using Mathcad's wordprocessing capability to enter and format text, working with units and utilizing Mathcad's built-in functions and operators. For a Mathcad novice, this is a good place to begin. The Mathcad tutorials can be accessed as indicated in Figure 1.3 by choosing Help and then Tutorials . 1.10 ADVANTAGES OF MATHCAD Mathcad offers a great way to work with equations, numbers, text, and graphs. Unlike other programs, Mathcad uses the language of mathematics and does mathematics in a very natural way. It works like a scratchpad and pencil combination . Mathcad's on-screen interface is a blank worksheet on which equations, graph data or functions can be entered and combined with text -- anywhere on the page. In Mathcad, an equation looks exactly as in a textbook : x b − b 2 4 a ⋅ c ⋅ − + 2 a ⋅ = Mathcad's equations and graphs are live, which means that any data, variable, or equation can be changed with Mathcad doing an instant recalculation and redrawing of the graphs. Mathcad can solve a wide range of problems, numerically as well as symbolically, and equations and data can be evaluated with 2D and 3D plots. Mathcad Quicksheets are full-featured worksheets that offer ready-made Mathcad templates that can be customized to perform a wide range of mathematical tasks from solving equations to graphing and calculus. They also demonstrate how some of Mathcad's special features like programming and animation can be used. Choose Help and then QuickSheets to use this capability. Chapter 1: Basics of Mathcad 7
- 18. Figure 1.3 Accessingthe Mathcad tutorials 8 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 19. Mathcad's Electronic Bookshave a wealth of mathematical knowledge and reference material -- all live - which can be dragged and dropped into your worksheets. In order to do this, choose Help and then E-books . Mathcad has proven to be an extremely remarkable , effective and powerful tool for the solution of problems. A wide range of problems can indeed be explored, formulated, analyzed and optimized, after which the best solution can be documented and presented. By choosing Help and then User Forums, users can connect with and share their work with colleagues and other professionals around the world , and collaboration is easy . A host of forums can be accessed as shown in Figure 1.4. Figure 1.4 Accessing the Mathcad Collaboratory Chapter 1: Basics of Mathcad 9
- 20. 1.11 COMPUTATIONS INMATHCAD Figure 1.5 The Insert/ Function buttons in Mathcad 10 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 21. Several mathcad palettescan be found on the palette strip at the top of the window that give access to Mathcad's mathematical operators. A click in your worksheet will place a red crosshair cursor. Math operators can now be placed in the worksheet via these palettes. To access Mathcad's built-in functions, go to the Insert menu as shown in Figure 1.5, and select Function , or click on the Insert Function button on the toolbar . EXAMPLES In the following examples, the number of decimal places required may be put in by choosing Result from the Format menu and then selecting Number Format. The red crosshair Use the Arithmetic Palette (See Figure 1.1) , and type = to obtain the answer. The basic operations are +, -, *, and / which are available on the keyboard . 2.645 10 8 ⋅ 30007 5 5 6 + 1.0426928428 10 7 − × = <-- <- Examples of built-in functions (See Figure 1.5) log 1768.985 ( ) sin 4 79 π ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 0.514 = 14.95 5.7i ( ) + [ ] 3 e 6 5i − + 1.999 10 3 × 4.024i 10 3 × + = <-- Complex numbers. Use of units- Choose Unit from the Insert menu (See Figure 1.5) <--- . 7600 km ⋅ 1 hr ⋅ 2.11 10 3 × m s 1 − ⋅ = Use of := by typing a colon character or using the Arithmetic Palette a 8 := a 3 a 4 + 513.414 = <---- f x ( ) sin x ( ) a x 6 + := f 37 ( ) 0.045 − = <-- Use of a defined function Chapter 1: Basics of Mathcad 11
- 22. Use of therange operator in the Arithmetic Palette. You can also use a semicolon (;). This will evaluate any function of z for z values from zero to 1 in increments of 0.2 z 0 .2 , 1 .. := <-- z 0 0.2 0.4 0.6 0.8 1 = f z ( ) 0 0.025 0.048 0.07 0.088 0.103 = cos f z ( ) ( ) z ⋅ 0 0.2 0.4 0.599 0.797 0.995 = <-- Create these tables by typing z=, f(z)=, etc. x 0 0.2 , 20 .. := <-- Define a range of values for the plot. 0 5 10 15 20 0.2 − 0.1 − 0 0.1 0.2 f x ( ) cos x ( ) f x ( ) x Use the X-Y Plot button in the Graph Palette, and type expressions to be plotted in the middle placeholders on the x and y axes. Several expressions separated by a comma (,) can be entered in these placeholders. <-- 0 100 n 1 n! cos 2 n ⋅ ( ) ⋅ ∑ = 2.28758499 − = Use the Calculus Palette (See Figure 1.1) to do sums, derivatives and integrals. Click in your worksheet to position the red crosshair, click on the proper button in the palette, then fill in each placeholder as appropriate . <-- 0 25 x 1 1 x 3 e x ⋅ + ⌠ ⎮ ⎮ ⎮ ⌡ d 0.842 = x 5 := x e x sin x ( ) ⋅ ( ) d d 100.218 − = 12 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 23. Matrices can beentered using the Vector and Matrix palette or by going to the Insert menu and choosing Matrix . A 3 1 1 7 5 10 1 6 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := <-- A 1 − 0.427 0.034 − 0.043 − 0.034 0.043 − 0.197 0.316 − 0.145 0.068 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = <-- Type A ^-1= to obtain the inverse A 117 − = <-- Use the Vector and Matrix Palette to obtain the determinant. t 2.5 := root t 4 cos 3 t ⋅ ( ) − t , ( ) 0.502 = <-- Find the root of an expression using the root function, starting with the guess of t =2.5 1.12 THE MATHCAD WINDOW, TOOLBARS AND PALETTES The Mathcad Window and the Main Menu This does math, graphics and symbolic algebra besides handling the editing and management of a worksheet. The array of computational and formatting capabilities available can be seen by clicking on each of the menus. The Math Toolbar This is below the main menu. It opens up palettes of math operators as described below. Chapter 1: Basics of Mathcad 13
- 24. Calculator toolbar Booleantoolbar Graph toolbar Programming toolbar Matrix toolbar Greek toolbar Evaluation toolbar Symbolic toolbar Calculus toolbar Just click on one of these as appropriate and start inserting math symbols into the worksheet. The Standard Toolbar This provides shortcuts for many common tasks from opening and saving files, cutting and deleting to spell checking and bringing up lists of built-in functions and units. Hover over each button to see tooltips with a brief description. The Formatting Toolbar This formats your text and math at the click of a button. In case the Format Bar does not show, go to the View Menu, select Toolbars and then Formatting. The Resources Window and E- books If you seek examples, want information that can be utilized in your Mathcad worksheets or wish to access web information from within Mathcad, go to the Help menu, and then open Tutorials, QuickSheets , Reference Tables or E-books. Tutorials includes Getting Started Primers, Migration Guide and Features In-Depth, while Quicksheets are live examples showing the use of Mathcad functions, graphs and programming features. Information on physical constants, chemical and physical data and mathematical formulas in Mathcad format can be found in Reference tables . Mathcad E-books can be accessed by opening E-books. These E-books have the advantage that all equations are live in them and you can change the values of variables, constants etc. to test and evaluate different results. A number of Mathcad E-books and articles are also available in the Resources section on http://www.ptc.com/go/mathsoft/mathcad/. 14 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 25. Controlling Calculation andThe Status Line The calculation mode, whether manual or automatic, is a property saved in your worksheet and template files. Mathcad starts in the automatic mode and all calculations and results are updated automatically. The word " Auto " can be seen in the message line at the bottom of the Mathcad window. This line provides status alerts, tips, keyboard shortcuts, and other helpful information along with the calculation status of the worksheet . Here, "auto" refers to the automatic mode , in which Mathcad automatically recalculates any math expressions when changes in the worksheet are made. When "WAIT" appears on the status line, the cursor changes to a flashing lightbulb, indicating that Mathcad is still completing computations. Besides giving the page number of the current worksheet, the message line will also indicate whether the Caps Lock or the Num Lock key is depressed on your keyboard. In manual mode, Mathcad does not compute or show results until recalculation is specifically requested. However, while in manual mode, Mathcad does keep track of pending computations . Once a change is made that requires recalculation, the word " Calc " appears on the status line to indicate to the user that the results being displayed are not correct and that recalculation is necessary to ensure accuracy. The screen can then be updated by going to the Tools menu and choosing Calculate and then Calculate Now . Alternatively, click = on the Standard toolbar or press [F9]. To force Mathcad to recalculate all equations in the worksheet, go to the Tools menu and choose Calculate and then Calculate Worksheet . 1.13 MATHCAD REGIONS Mathcad allows you to enter equations, text and graphs as separate objects anywhere in the worksheet and each equation, text paragraph or plot is considered a region. A region can be selected by clicking in math or text in your worksheet, after which it is indicated by a thin rectangle around it. Moving the cursor to one of the edges of the region, will change it to a small hand with which the region can be moved to anywhere in the worksheet. While clicking in the math region will bring blue selecting lines under the material selected, clicking in a text region will bring black boxes to each corner and the middle of each line. With these boxes text regions can be resized as needed. To add a border around a region or regions, select the region(s), then right-click and choose Properties from the menu. Then click on the Display tab and check the box next to " Show Border ". A math region looks like: A text region looks like: x 1535.56 := What is shown on the left was created by typing x:1535.56 Chapter 1: Basics of Mathcad 15
- 26. Moving Regions inthe Worksheet This can be accomplished by clicking in a region and drag-selecting it to put it in a dashed selection box and then moving the cursor to the edge where it shows up as a small black hand. Now, use the mouse to move the region as necessary by holding down the left mouse button and dragging it. Once the regions are positioned in the desired manner, the mouse button may be released . Then, an empty part of the screen may be clicked on to deselect the regions. Alternatively, choose Regions from the View menu, which will highlight the boundaries of the region against a contrasting background. Cut and Paste can also be used to move regions. Once regions are inserted in the worksheet, they can be aligned horizontally or vertically by going to the Format menu and choosing Align Regions. How Mathcad Reads Your Document Mathcad reads your document from left to right and top to bottom . For example, if the value of a function , ln(y) is desired for y= 67, either of the following setups will work. (a) y 67 := ln y ( ) 4.205 = (b) y 67 := ln y ( ) 4.205 = Deleting Selected Regions Selected regions in dashed line boxes can be deleted by choosing Cut from the Edit menu. Copying and Pasting Selected Regions Selected regions can also be copied and put into any place within a document or into another document by choosing the Copy and Paste commands from the Edit menu . 1.14 ENTERING MATH AND TEXT Entering Math Math can be entered by clicking the mouse in a blank space to see a red crosshair cursor. Type 1+ which will get you 16 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 27. on the screen.The little black box delimited by blue editing lines that you see is called a placeholder . Continue typing. Type 6 in the placeholder and press the equals key (=) to see the result. 1 6 + 7 = The Arithmetic Palette can be seen by clicking on the Math Palette. This is shown below Math Palette button Operation Keystroke Button Addition + Subtraction - Multiplication * Division / Powers ^ Entering Text A text region can be created by clicking in a blank space to see the red crosshair cursor. Then choose Text Region from the Insert menu or type the double quote (") . The crosshair transforms to an insertion point with a black text box around it. 1.15 MATHCAD WORKSHEETS, TEMPLATES AND STYLES Creating and Using Templates Mathcad styles and templates are similar to those of any word-processing software. When you create a new worksheet in Mathcad, you can start with Mathcad's default choices, or you Chapter 1: Basics of Mathcad 17
- 28. can use atemplate containing customized formats. Several predefined templates are available each with a variety of styles, and if necessary, new templates can also be created. Any Mathcad worksheet can be saved as a new template. To create a new worksheet based on a template, choose File/New , and then select a template from the list, or browse to your own template directory. In the placeholders, substitute your own text or bitmaps in any of the placeholders, or use the built-in styles. You can also revise them if needed. When you create a worksheet based on a template, all of the formatting information, and any math, text and image regions are copied into the new worksheet. By creating a new template or revising another template , a customized format can be generated. Text styles and number formats can be created, number formats, fonts and sizes can be set, bitmaps can be added , and also, page numbers, filenames and dates can be inserted . Thus, by using templates, you maintain consistency across multiple worksheets through definition of math styles, text styles, printing margins, numerical result formats, units etc. To save a template, choose Save As from the File menu and use the file extension .mct. The settings, styles and bitmaps saved will be available for the next file you may want to create, leading to greater consistency in your files. Using Styles The use of Text Styles allows you to create a consistent appearance in your worksheets. Styles in each template are available by choosing Style from the Format menu. Any specific style with a defined font, size, etc. can be applied to a text region. To create or modify a style, again choose Style from the Format menu . To save your styles for use in new files, you must make a template file. Math Styles can be used to assign specific fonts, font sizes etc. to mathematical expressions. There are predefined ones but additional styles can be defined and applied. This can be accomplished by going to the Format menu and choosing Equation. 1.16 DEFINING VARIABLES There are times when the value of a number may need to be changed several times in a document . Such a number is termed a variable . Let Apples be, then, a variable . Type Apples:50 to see on the screen Apples 50 := By doing the above, we have given the value of 50 to the variable Apples . Similarly, another variable Unitprice can be defined 18 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 29. Unitprice 0.35 := A thirdvariable can also be defined as Totprice Apples Unitprice ⋅ := Type Totprice= to see on the screen Totprice 17.5 = Now, if new values need to be given to Apples and to Unitprice, change the numbers above as necessary and the result which is Totprice will also change accordingly as shown below. Apples 75 := Unitprice 0.40 := Totprice Apples Unitprice ⋅ := Totprice 30 = 1.17 DEFINING FUNCTIONS IN MATHCAD The value of a function depends on the values assigned to its arguments. For example, to define a function f(x), type f(x):x^7 to see on the screen f x ( ) x 7 := Putting in a numerical value of 6 as the argument of f(x) will give Chapter 1: Basics of Mathcad 19
- 30. f 6 ( )2.799 10 5 × = This can also be accomplished by doing x 6 := f x ( ) 2.799 10 5 × = If a range of values for the argument is given x 1 6 .. := f(x) will yield a table of numbers as shown below x 1 2 3 4 5 6 = f x ( ) 1 128 3 2.187·10 4 1.638·10 4 7.813·10 5 2.799·10 = In addition to defining a customized function, Mathcad's library of built-in functions can also be utilized. To do this, select Function from the Insert menu, or click on the function button on the toolbar The Insert Function dialog box, shown below, will allow insertion of a function name directly into the placeholder Built-in functions can also be inserted directly from the keyboard. Some examples are given below. Trigonometry and Logarithms ln 56 ( ) 4.025 = cos 50 deg ⋅ ( ) 0.455 2 ⋅ = asinh 0.95 ( ) 0.846 = 20 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 31. Matrix Functions identity 6 () 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = cols identity 6 ( ) ( ) 6 = 1.18 BUILDING AND EDITING MATHEMATICAL EXPRESSIONS Building Mathematical Expressions In Mathcad, various parts of an expression are assembled by observing the rules of precedence and some additional procedures that enable entering denominators, exponents and expressions inside of radicals. The following examples will illustrate the process. Example 1 To create Chapter 1: Basics of Mathcad 21
- 32. f x ( )x 6 + ( ) x 3 5 − ( ) ⋅ := Type f(x):x+6[Spacebar]*(x^3[Spacebar]-5) Example 2 To create f x ( ) x 765 x 6 4 − ( ) ⋅ + := Type f(x):x+765*(x^6-4) The exponent operator is called a sticky operator because unless you get out by pressing [Spacebar] , your keystrokes will "stick" to the exponent . This stickiness applies to exponents, square roots, subscripts, and division. Example 3 To create the expression x 7 675 + 780 Type x^7[Spacebar] <-- Puts x 7 in blue editing lines +675[Spacebar] <-- Puts the whole equation in the editing lines /780[Enter] <-- Expression is complete Example 4 To create 22 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 33. x 7 675 780 + do thesame as above but without pressing the spacebar after +675 Example 5 To create x 17 t 7 450 type x^17/t^7[Spacebar][Spacebar][Spacebar]/450 Editing Expressions The equation editor functions very much like a text editor and goes from left to right. Most problems dealing with editing equations stem from working with operators. Although Mathcad automatically inserts parentheses wherever necessary, the Mathcad user must put in parentheses himself in accordance with his own judgement to give clarity to expressions . When expressions become complicated, it is definitely preferable to work with smaller and more manageable subexpressions within them. The reader is referred to Chapter 4 of the Mathcad 14 User's Guide for a detailed discussion on Editing Expressions. This can be accessed by choosing Tutorials from the Help menu. 1.19 DEFINING RANGE VARIABLES To assign a range of values to a variable x going from 0 to 30, say, with an increment of 3, type x:0,3;30 The result on the screen will be x 0 3 , 30 .. := To evaluate a function f(x) for x= 0 to 7 with an increment of 1, type x:0;7 f(x):x^5[Spacebar]-6*x^3[Spacebar]+9 Chapter 1: Basics of Mathcad 23
- 34. to see onthe screen x 0 7 .. := f x ( ) x 5 6 x 3 ⋅ − 9 + := Now, typing x= and f(x)= will give the following output table on the screen x 0 1 2 3 4 5 6 7 = f x ( ) 9 4 -7 90 649 3 2.384·10 3 6.489·10 4 1.476·10 = If the range is to have increments other than 1, which is the default value, the next value in the range must be entered. The following will provide a range of x values going from 2 to 4 with increments of 0.5 and the corresponding f(x) values. x 2 2.5 , 4 .. := x 2 2.5 3 3.5 4 = f x ( ) -7 12.906 90 276.969 649 = 1.20 DEFINING VECTORS AND MATRICES To create a vector or a matrix in Mathcad, the dimensions of the array must be chosen and the placeholders must be filled in. 24 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 35. Figure 1.6 Insertingmatrices in Mathcad Chapter 1: Basics of Mathcad 25
- 36. To create avector V , click in a blank space, Choose Matrix from the Insert menu (See Figure 1.6 ) or click on the button inside the Vector and Matrix Palette . Then, fill in the appropriate number of rows and columns, click on Insert and finally fill in the placeholders with given values. To move from placeholder to placeholder inside the vector, use [Tab] or click on the appropriate placeholder to select it V 7 8 9 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := To access a particular element of a vector, use the subscript operator , which can be created by typing a left square bracket ( [ ), or by using the Xn button in the Matrix Palette . In Mathcad , by default the first element has the index 0. Type V[0= to see on the screen V0 7 = The next two elements will have indices 1 and 2 Thus typing V[1= and V[2= will produce on the screen V1 8 = V2 9 = In order to obtain all the elements of the vector at the same time, the index can be defined as a range variable as shown below:: i 0 2 .. := giving Vi 7 8 9 = <--- Type V[i= The elements of a vector can also be used as the arguments of a function as shown below. Define a function f(x) as. f x ( ) sin 2 x ⋅ ( ) cos 3 x ⋅ ( ) + := Let a vector V be a three-element vector as given below 26 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 37. v 1.3 2.6 − 7.9 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Then, typing i:0;2and then f(v[i)= will show the following on the screen i 0 2 .. := f vi ( ) -0.21 0.937 0.046 = The Vector and Matrix Palette contains most of the common matrix and vector operators. These are listed below. Operation Keystroke Palette Button Display Dot product [Shift]8 v w ⋅ Displays like scalar multiplication. Cross product [Ctrl]8 v w × Determinant | M Column [Ctrl]6 M 2 〈 〉 Returns the 2nd column of M. There is also a wide range of built-in functions in Mathcad that return information about the size of an array and its elements. For a given matrix [M], various pieces of information can be generated as shown below. M 9 4 12 10 5 13 11 7 14 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Chapter 1: Basics of Mathcad 27
- 38. Number of columns:cols M ( ) 3 = Number of rows: rows M ( ) 3 = Largest value in matrix: max M ( ) 14 = Determinant of matrix: M 3 = Eigenvalues of a matrix: eigenvals M ( ) 28.768 0.176 − 0.592 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 1.21 CREATING GRAPHS To create an x-y plot, choose Graph/ X-Y Plot from the Insert Menu as shown in Figure 1.7. Press [Enter] and fill in the placeholders on the x and y axes. Alternatively, type an expression depending on one variable such as: sin(x) + cos (3x) and click on the X-Y Plot button on the Graph toolbar . The resulting plot will be x 10 − 9.9 − , 10 .. := 10 − 5 − 0 5 10 2 − 1 − 1 2 sin x ( ) cos 3 x ⋅ ( ) + x 28 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 39. Figure 1.7 Creatinggraphs in Mathcad X-Y Plot Button Chapter 1: Basics of Mathcad 29
- 40. Using Range Variablesto Plot a Function The range of values being plotted on the x-axis must, in general, be specified. If this range is not prescibed by you, Mathcad will choose a default range for the dependent variable. For example, let us plot f x ( ) x 3 8 x 2 ⋅ + 9 x ⋅ + 14 + := The graph of this function over the range x= 0 to x= 6 can be accomplished as follows. x 0 1 , 6 .. := f x ( ) x 3 8 x 2 ⋅ + 9 x ⋅ + 14 + := Now create the plot by clicking in the worksheet window. Type @ to create the x-y plot, type x in the middle placeholder on the horizontal axis and type f(x) in the middle placeholder on the vertical axis. Press [Enter] . The following graph should then appear on the screen. 0 2 4 6 0 200 400 600 f x ( ) x The plot generated above does not seem to be very smooth. In order to obtain a smoother trace, change the definition of x to x 0 0.1 , 6 .. := . The smaller step enables Mathcad to calculate more points. This will make the plot a lot smoother (see graph below ) because now there are more points or dots being connected together. Formatting of an x-y plot can be accomplished as follows. Double-click on it or choose Graph from the Format menu to open up a dialog box. This dialog box will allow several options in terms of grid lines, legends, trace types, markers, colors, axis limits etc. 30 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 41. x 0 0.1 ,6 .. := 0 2 4 6 200 400 600 f x ( ) x Plotting a Vector of Data Points Let us say that we need to plot a vector of data points called Temp , which is the rising outside temperature on different days of a week in a summer month. This vector which will have 7 rows and 1 column can be created using the Matrix command on the Insert menu. Temp 78 82 84 86 88 90 92 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := This data can be graphed by making the horizontal axis an index variable or a vector with the same number of elements as the vector Temp . Define the index i as: i 0 6 .. := Create your plot by typing [@] and typing in Temp[i Chapter 1: Basics of Mathcad 31
- 42. In the placeholderon the y axis, and i in the placeholder on the x axis. The resulting plot should be 0 2 4 6 75 80 85 90 95 Tempi i Notice here that box symbols have been used on a dashed line. Alternatively, two vectors of equal size can be plotted against each other. Thus, the second vector which will be called day is defined as day 1 2 3 4 5 6 7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Temp 78 82 84 86 88 90 92 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Now create the plot by typing @ , as was done previously, and type day in the placeholder on the x-axis and Temp in the placeholder on the y-axis The resulting plot should look like this 32 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 43. 0 5 10 70 85 100 Notethat the symbol type used here is O's , and the grid lines have been turned off on both axes. Plotting a Function of Vector Elements Sometimes, a function may need to be plotted over points that are not evenly spaced . As an example, let us define a function f x ( ) 3 x 3 ⋅ 5 x 2 ⋅ + 9 x ⋅ + 12 + := to be plotted over the range x as shown: -2.0, 0,3.50,10, 15. In order to do this, x has to be defined as a vector of the given numbers in the prescribed range. This can be done using the Matrices dialog box with the placeholders properly filled in as shown. x 2.0 − 0 3.5 10 15 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := The index variable must be defined as i 0 4 .. := Now, type @ and fill in the middle placeholders on the x and y axes with xi and f xi ( ). The resulting graph should be Chapter 1: Basics of Mathcad 33
- 44. 5 − 0 510 15 5 − 10 3 × 5 10 3 × 1 10 4 × 1.5 10 4 × f xi ( ) xi When plotting a function of a range variable , notice that the horizontal axis can be any set of prescribed numbers and does not have to be in even increments Plotting Several Functions on One Graph Plotting multiple functions over a single domain on a single graph can be done by typing these functions on the y- axis one after the other but separated by commas. For example, the following should illustrate how the two functions sin(x) and cos(2x) can be plotted on the same graph. x 0 0.1 , 6 .. := 0 2 4 6 1 − 0.5 − 0.5 1 sin x ( ) cos 2x ( ) x 34 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 45. 1.22 FORMATTING MATH,TEXT AND RESULTS Using the Format Bar in the Mathcad window, the appearance of the text as well as the math can be altered quite easily, and different fonts, sizes and styles can be readily selected. If the Format Bar is hidden, ensure that the Formatting bar on the View/Toolbars menu is checked. Formatting Text Using the drop-down lists in the Format Bar, different fonts as well as point sizes can be selected . Appropriate buttons must be clicked to generate special effects like boldface, italics, etc. Several options become available upon choosing Text from the Format menu . (See Figure 1.8). Text Styles and Templates Text styles provide consistency in the appearance of worksheets and enable the application of text formatting to the text regions. Available text styles depend on the template used to create a worksheet. To examine the different templates and text styles provided in Mathcad, choose New from the File menu and modify them or create new ones. Formatting Math In Mathcad different font tags can be applied to variables and to constants. To make changes in the font, click on a variable or constant in a math region and use the Format Bar. Alternatively, you can choose Equation from the Format menu. Highlighting of equations can also be done in Mathcad. To do this, click on an equation and choose Properties/Highlight Region from the Format menu in the main worksheet window. Below is shown an example of a highlighted equation. S 1 n i ai xi ⋅ ( ) ∑ = = In Mathcad, you can specify how answers can be displayed in terms of the desired decimal places, precision etc. Also, fonts and their sizes can be controlled and highlighting of equations can be done as desired. Chapter 1: Basics of Mathcad 35
- 46. Figure 1.8 Formattingtext in Mathcad Formatting Results and Graphs To control the format of a result, double-click on the result , or click on the result and choose Result from the Format menu . In the Number Format dialog box that shows up, change the Exponential Threshold , and Number of decimal places as necessary and click on OK . This will set the format only for this particular result. (See Figure 1.9 ). However, if this needs to be done for the entire worksheet, click on a blank part of the worksheet, and do the above . 36 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 47. Figure 1.9 Formattingresults in Mathcad Chapter 1: Basics of Mathcad 37
- 48. The reformatting ofgraphs can be easily done using a dialog box generated by double- clicking on the plot, or by going to Insert/Graph/X-Y Plot and clicking on the Traces tab shown in Figure 1.10. Under the " Legend label " column, type the desired name of the trace . Symbols, line types, colors, trace types, etc. can all be controlled by clicking on the appropriate options. Finally, preview the changes and click OK to finalize them . An example is shown at the end of this section. Figure 1.10 Formatting graphs in Mathcad 38 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 49. f x ( )sin x ( ) := g x ( ) cos 3 x ⋅ ( ) := h x ( ) f x ( ) g x ( ) + := x 10 − 9.9 − , 10 .. := 10 − 5 − 0 5 10 2 − 1 − 0 1 2 sin(x) cos(3x) sin(x)+cos(3x) Graphing Functions and labelling Plots x- values y-values f x ( ) g x ( ) h x ( ) x 1.23 USING UNITS Although units are not required to be put into Mathcad equations, a great feature within Mathcad is its ability to track standard units during calculations and to automatically perform conversions. The program will also flag incorrect and inconsistent dimensional calculations, and mixing and matching of units can be done as desired.The default system in Mathcad is the SI unit system To define a variable in terms of the built-in unit kilometers , for example, just multiply the given number by km. Type R:4*km to see on the screen R 4 km ⋅ := Chapter 1: Basics of Mathcad 39
- 50. Figure 1.11 Insertingunits in Mathcad 40 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 51. The equation fora circular area of radius R now will automatically generate the result in terms of the appropriate units A π R 2 ⋅ := A 5.027 10 7 × m 2 = Note that the above result is automatically shown in terms of the base units of the default unit system, which, in this case, is SI. To change this result to feet , for example, click in the name of the unit that you want to replace, then drag-select it and type in the new desired unit in its place. Finally click outside the equation to see the new result. Alternatively, click in the placeholder next to the unit name , go to Insert/Unit (See Figure 1.11) , click on the appropriate Dimension name, then double-click on the unit name desired , and finally click outside the equation to see the desired result. R 1.312 10 4 × ft ⋅ = To see a list of buit-in units in Mathcad, go to the Insert menu . The following equivalents can be easily generated using the procedures described above. Work 14000. J ⋅ := Work 1.4 10 4 × m 2 kg s 2 − ⋅ ⋅ = Work 13.269 BTU ⋅ = Power 2000 kW ⋅ := Power 2.682 10 3 × hp ⋅ = Force 1500lbf := Force 6.672 10 3 × N ⋅ = A calculation involving mixing of units is shown below. Although each term in the calculation is expressed in a different unit of length, the result is calculated in the base SI unit for length, namely, meters 6 ft ⋅ 30 in ⋅ + 1.4 m ⋅ + 0.8 yd ⋅ + 4.722 m = The following calculation, obviously, cannot be completed and Mathcad indicates the same. 7 ft ⋅ 6 sec ⋅ + = 6 sec ⋅ Click on the expression to see the error message,which is: This value has units: Time, but must have units: Length Chapter 1: Basics of Mathcad 41
- 52. To change theunit system of the worksheet from SI to U.S. units, select Worksheet Options from the Tools menu. Then, click the Unit System tab, and select U.S. This option gives you results in "English System" measures. In the following calculation, a mass is multiplied by an acceleration to give force in proper units. The result should be in Newtons or an equivalent force unit. mass 5 kg ⋅ := acc 20 m sec 2 ⋅ := Force mass acc ⋅ := Force 100 m kg s 2 − ⋅ ⋅ = To display units, choose Result from the Format menu. Then, go to Unit Display and check Format Units and Simplify units when possible, and results will display as shown below. g 9.807 s 2 − ⋅ m ⋅ := 60 N ⋅ 20 ⋅ m ⋅ 1.2 10 3 × J = 8 kg ⋅ 6 ⋅ m sec 2 ⋅ 48 N = 42 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 53. C H AP T E R 2 INTRODUCTION TO NUMERICAL METHODS 2.1 THE USE OF NUMERICAL METHODS IN SCIENCE AND ENGINEERING Analysis of problems in engineering and the physical sciences typically involves four steps as follows. (1) Development of a suitable mathematical model that realistically represents a given physical system. (2) Derivation of the system governing equations using physical laws such as Newton's laws of motion, conservation of energy, the laws governing electrical circuits etc. (3) Solution of the governing equations, and (4) Interpretation of the results. Because real world problems are generally quite complex with the generation of closed-form analytical solutions becoming impossible in many situations, there exists, most definitely, a need for the proper utilization of computer-based techniques in the solution of practical problems. The advancement of computer technology has made the effective use of numerical methods and computer-based techniques very feasible, and thus, solutions can now be obtained much faster than ever before and with much better than acceptable accuracy. However, there are advantages as well as disadvantages associated with any numerical procedure that is resorted to , and these must be kept in mind when using it. 2.2 COMPARISON OF NUMERICAL METHODS WITH ANALYTICAL METHODS While an analytical solution will be exact if it exists, a numerical method , on the other hand, will generally require iterations to generate a solution, which is only an approximation and which certainly cannot be considered exact by any means. A disadvantage associated with analytical solution techniques is that they are generally applicable only to very special cases of problems. Numerical solutions, on the contrary, will solve complex situations as well. While numerical techniques have several advantages including easy programming on a computer and the convenience with which they handle complex problems, the initial estimate of the solution along with the many number of iterations that are sometimes required to generate a solution can be looked upon as disadvantages. Chapter 2: Introduction to Numerical Methods 43
- 54. 2.3 SOURCES OFNUMERICAL ERRORS AND THEIR COMPUTATION It is indeed possible for miscalculations to creep into a numerical solution because of various sources of error. These include inaccurate mathematical modeling, wrong programming, wrong input, rounding off of numbers and truncation of an infinite series. Round-off error is the general name given to inaccuracies that affect the calculation scene when a finite number of digits are assigned to represent an actual number. In a long sequence of calculations, this round-off error can accumulate, then propagate through the process of calculation and finally grow very rapidly to a significant number. A truncation error results when an infinite series is approximated by a finite number of terms, and, typically, upper bounds are placed on the size of this error. The true error is defined as the difference between the computed value and the true value of a number. ETrue XComp XTrue − = (2.1) while the relative true error is the error relative to the true value (2.2) er XComp XTrue − XTrue = Expressed as a percentage, the relative true error is written as er XComp XTrue − XTrue 100 ⋅ = (%) (2.3) 2.4 TAYLOR SERIES EXPANSION The Taylor series is considered as a basis of approximation in numerical analysis. If the value of a function of x is provided at " x 0", then the Taylor series provides a means of evaluating the function at " x 0 + h", where " x 0" is the starting value of the independent variable and " h " is the difference between the starting value and the new value at which the function is to be approximated f x0 h + ( ) f x0 ( ) h x f x0 ( ) ⎡ ⎣ ⎤ ⎦ d d ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ + h 2 2! 2 x f x0 ( ) d d 2 ⋅ + h 3 3! 3 x f x0 ( ) d d 3 ⋅ + ..................... + = (2.4) This equation can be used for generating various orders of approximations as shown below. The order of approximation is defined by the highest derivative included in the series. For example, If only terms up to the second derivative are retained in the series, the result is a second order approximation. 44 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 55. First order approximation: fx0 h + ( ) f x0 ( ) h x f x0 ( ) d d ⋅ + = (2.5) Second order approximation: f x0 h + ( ) f x0 ( ) h x f x0 ( ) d d ⋅ 2 x f x0 ( ) d d 2 ⋅ + = (2.6) Third order approximation: f x0 h + ( ) f x0 ( ) h x f x0 ( ) d d ⋅ + h 2 2! 2 x f x0 ( ) d d 2 ⋅ + h 3 3! 3 x f x0 ( ) d d 3 ⋅ + ........................... + = (2.7) It is to noted that the significance of the higher order terms in the Taylor series increases with the nonlinearity of the function involved as well as the difference between the " starting x" value and the "x" value at which the function is to be approximated. Thus , the fewer the terms that are included in the series, the larger will be the error associated with the computation of the function value. If the function is linear , however, only terms up to the first derivative term need to be included. Example 2.1 Using the Taylor series expansion for f(x) = - 0.15 x4 - 0.17 x 3 - 0.25 x 2 -0.25 x + 1.25 determine the zeroth, first, second, third, fourth and fifth order approximations of f(x0 + h ) where x0 = 0 and h = 1,2, 3, 4,5 and compare these with the exact solutions. h=1.0: Put in the function and generate its derivatives: f x ( ) 0.15 − x 4 ⋅ 0.17 x 3 ⋅ − 0.25 x 2 ⋅ − 0.25x − 1.25 + := x0 0 := h 1. := fprime x ( ) 0.60 − x 3 ⋅ 0.51 x 2 ⋅ − 0.50 x ⋅ − 0.25 − := <--Generate derivatives f2prime x ( ) 1.8 − x 2 ⋅ 1.02 x ⋅ − 0.50 − := f3prime x ( ) 3.6 − x ⋅ 1.02 − := f4prime x ( ) 3.6 − := f5prime x ( ) 0. := Chapter 2: Introduction to Numerical Methods 45
- 56. term1 f x0 () := term2 h fprime x0 ( ) ⋅ := term3 h 2 2 f2prime x0 ( ) ⋅ := term4 h 3 6 f3prime x0 ( ) ⋅ := term5 h 4 24 f4prime x0 ( ) ⋅ := term6 h 5 120 f5prime x0 ( ) ⋅ := ftaylor0 term1 := <---- one-term or zero-order approximation ftaylor1 term1 term2 + := <---- first order approximation with two terms ftaylor2 term1 term2 + term3 + := <---second order approximation with 3 terms ftaylor3 term1 term2 + term3 + term4 + := <---third order approximation with 4 terms ftaylor4 term1 term2 + term3 + term4 + term5 + := <---fourth order approximation with 5 terms fifth order approximation with 6 terms ftaylor5 term1 term2 + term3 + term4 + term5 + term6 + := <----- x x0 h + := x 1 = These are the zero- fifth order approximations of the given function f(x) using the Taylor series. ftaylor0 1.25 = ftaylor1 1 = ftaylor2 0.75 = <-- ftaylor3 0.58 = ftaylor4 0.43 = ftaylor5 0.43 = f 1 ( ) 0.43 = <---EXACT ANSWER USING FUNCTION GIVEN. err0 f x ( ) ftaylor0 − := err0 0.82 − = err1 f x ( ) ftaylor1 − := err1 0.57 − = err2 f x ( ) ftaylor2 − := err2 0.32 − = These are errors ( differences between exact values and approximations ) for the above zero - fifth order approximations. <-- err3 f x ( ) ftaylor3 − := err3 0.15 − = err4 f x ( ) ftaylor4 − := err4 0 = err5 f x ( ) ftaylor5 − := err5 0 = Similarly, by using h= 2, 3, 4, 5 , the zeroth- fifth order approximations for f(2), f(3), f(4), f(5) and the associated errors can be determined. These are given in Tables 2.1 and 2.2 Plots of the various Taylor series approximations of the given function and associated errors are generated below and are presented in Figs. 2.1 and 2.2 46 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 57. x0 0 := x0 0.01 , 5 .. := ftaylor0 x ( ) f x0 ( ) := <-- zeroth-order approximation ftaylor1 x ( ) ftaylor0 x ( ) x x0 − ( ) fprime x0 ( ) ⋅ + := <---first-order approximation ftaylor2 x ( ) ftaylor1 x ( ) x x0 − ( ) 2 2 f2prime x0 ( ) ⋅ + := <--second-order approximation ftaylor3 x ( ) ftaylor2 x ( ) x x0 − ( ) 3 6 f3prime x0 ( ) ⋅ + := <---third-order approximation ftaylor4 x ( ) ftaylor3 x ( ) x x0 − ( ) 4 24 f4prime x0 ( ) ⋅ + := <---fourth-order approximation ftaylor5 x ( ) ftaylor4 x ( ) x x0 − ( ) 5 120 f5prime x0 ( ) ⋅ + := <---fifth- order approximation Errors generated with the various approximations are as follows Zero order approximation: err0 x ( ) f x ( ) ftaylor0 x ( ) − := First order approximation: err1 x ( ) f x ( ) ftaylor1 x ( ) − := Second order approximation: err2 x ( ) f x ( ) ftaylor2 x ( ) − := Third order approximation: err3 x ( ) f x ( ) ftaylor3 x ( ) − := Fourth order approximation: err4 x ( ) f x ( ) ftaylor4 x ( ) − := Fifth order approximation: err5 x ( ) f x ( ) ftaylor5 x ( ) − := The various approximations generated by the above calculations and the associated errors are compared in the Table 2.1. Chapter 2: Introduction to Numerical Methods 47
- 58. 0 1 23 4 5 150 100 50 0 50 Zeroth order approx Third order approx Fifth order approx/ given function Zeroth order approx Third order approx Fifth order approx/ given function Taylor series approx of given function x- value Function approximated by Taylor series ftaylor0 x ( ) ftaylor3 x ( ) ftaylor5 x ( ) x Figure 2.1. Taylor series approximation of given function x 0 0.01 , 5 .. := 0 1 2 3 4 5 150 100 50 0 50 Zero order approx Third order approx Fifth order approx Zero order approx Third order approx Fifth order approx Errors gen due to Taylor-series approx x- value errors as function of x err0 x ( ) err3 x ( ) err5 x ( ) x Figure 2.2. Errors associated with the various Taylor series approximations 48 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 59. The various approximationsgenerated by the above calculations and the associated errors are compared in the following tables. Table 2.1 h 1 5 .. := Various orders of approximation generated by Taylor series approach versus true values of given function x 0 1 , 5 .. := zeroth order first order second order third order fourth order fifth order True Value x 0 1 2 3 4 5 = ftaylor0 x ( ) 1.25 1.25 1.25 1.25 1.25 1.25 = ftaylor1 x ( ) 1.25 1 0.75 0.5 0.25 0 =ftaylor2 x ( ) 1.25 0.75 -0.25 -1.75 -3.75 -6.25 = ftaylor3 x ( ) 1.25 0.58 -1.61 -6.34 -14.63 -27.5 =ftaylor4 x ( ) 1.25 0.43 -4.01 -18.49 -53.03 -121.25 =ftaylor5 x ( ) 1.25 0.43 -4.01 -18.49 -53.03 -121.25 =f x ( ) 1.25 0.43 -4.01 -18.49 -53.03 -121.25 = h 1 2 3 4 5 = Table 2.2 Errors associated with the different orders of approximation zeroth order first order second order third order fourth order fifth order x 0 1 2 3 4 5 = err0 x ( ) 0 -0.82 -5.26 -19.74 -54.28 -122.5 = err1 x ( ) 0 -0.57 -4.76 -18.99 -53.28 -121.25 = err2 x ( ) 0 -0.32 -3.76 -16.74 -49.28 -115 = err3 x ( ) 0 -0.15 -2.4 -12.15 -38.4 -93.75 = err4 x ( ) 0 0 0 0 0 0 = err5 x ( ) 0 0 0 0 0 0 = h 1 2 3 4 5 = PROBLEMS 2.1. Using the Taylor series expansion for cos x, which is given as f(x) = cos x = 1- x 2 / 2 + x 4 / 24 , determine the one-term, two-term and three-term approximations of f(x0 + h ) , where x0 = 0 Chapter 2: Introduction to Numerical Methods 49
- 60. rad and h= 0. 1, 0.2 ....1.0 rad , and compare these with the exact solution. Using Mathcad, generate plots of the various Taylor series approximations and associated errors as functions of the independent variable x. 2.2 Develop a Taylor series expansion of the following function: f(x) = x 5 - 6 x 4 + 3x 2 + 9 . Use x =3 as the base and h as the increment. Using Mathcad, evaluate the series for h= 0.1, 0.2....1.0, adding terms incrementally as in Problem 2.1. Compare the various Taylor series approximations obtained with true values in a table. Generate plots of the approximations and associated errors as functions of x . 2.3 Given the following function: f(x) = x 3 - 3 x 2 + 5 x + 10 , determine f ( x0 + h ) with the help of a Taylor series expansion, where x 0 = 2 and h = 0.4. Compare the true value of f ( 2.4 ) with estimates obtained by resorting to (a) one term only (b) two terms (c) three terms and (d) four terms of the series. 2.4 Given the following function f x ( ) 3 x 3 ⋅ 6 x 2 ⋅ − 15 x ⋅ + 25 + = use a Taylor series expansion to determine the zeroth, first, second and third order approximations of f(x0+h) where x 0 = 2 and h = 0.5. Compare these with the exact solution. 2.5 By developing a Taylor series expansion for f(x) = e x about x = 0 , determine the fourth-order approximation of e 2.5 and compare it with the exact solution. 2.6. By developing a Taylor series expansion for f(x) = ln(2-x) about x = 0, determine the fourth-order approximation of ln (0.5) and compare it with the exact solution 50 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 61. 2.7. By developinga Taylor series expansion for f(x) = x3 e - 5x about x = 1, determine the third-order approximation of f(1.2) and compare it with the exact solution. 2.8. By developing a Taylor series expansion for f(x) = e cos x about x =0 , determine the fourth- order approximation of f (2π) and compare it with the exact solution. . 2.9. By developing a Taylor series expansion for f(x) = (x - 2) 1/2 about x = 2, determine the third-order approximation of f (2.2), that is, (0.2) 1/2 , and compare it with the exact solution. 2.10. Given the function f(x) = x2 - 5 x 0.5 + 6 , use a Taylor series expansion to determine the first, second , third and fourth order approximations of f (2.5 ) by resorting to x 0 = 2 and h = 0.5 . Compare these with the exact solution. 2.11. Given the function f(x) = 6 x3 - 9 x2 +25 x + 40 , use a Taylor series expansion to determine the zeroth, first, second and third order approximations of f (x0 + h ) where x 0 = 3 and h = 1 . Compare these with the exact solution. 2.12. Given the function f(x) = 4 x4 - 7 x3 + 5 x2 - 6 x + 90 use a Taylor series expansion to determine the zeroth, first, second , third and fourth order Chapter 2: Introduction to Numerical Methods 51
- 62. approximations of f(x0 + h ) ) where x 0 = 3 and h = 0.5 . Compare these with the exact solution. Calculate errors abnd generate calculations to three decimal places. 2.13. Given the function f(x) = 8 x3 - 10 x2 + 25 x + 45 , use a Taylor series expansion to determine the zeroth, first, second and third order approximations of f (x0 + h ) where x 0 = 2 and h = 1 . Compare these with the exact solution. 2.14. Given the function f(x) = 1 + x + x2 / 2! + x3 / 3! +x4 / 4! use a Taylor series expansion to determine the zeroth, first, second , third and fourth order approximations of f (x0 + h ) where x 0 = 0 and h = 0.5 . Compare these with the exact solution. Generate answers correct to four decimal places. 2.15. Given the function f(x) = x + x3 / 3 + 2 x5 / 15 use a Taylor series expansion to determine the zeroth, first, second , third and fourth order approximations of f (x0 + h ) where x 0 = 0 and h = 0.8 . Compare these with the exact solution by computing percentage errors. Generate answers correct to four decimal places. 2.16. Given the function f(x) = sin (x) use a Taylor series expansion to determine the fifth order approximation of f (x0 + h ) where x 0 = 0 and h = 0.2 radians . Compare your answer with the true value. Generate answers correct to four decimal places. 2.17. Given the function f(x) = 3 x2 - 6 x 0.5 + 9 , use a Taylor series expansion to determine the zeroth, first, second , third and fourth order approximations of f (x0 + h ) whereo x 0 = 3 and h = 1 . Compare these with the exact solution by computing percentage errors. Generate answers correct to four decimal places. 52 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 63. C H AP T E R 3 ROOTS OF EQUATIONS 3.1 INTRODUCTION In many problems occurring in science and engineering, it is often necessary to find roots or zeros of equations that are nonlinear. Nonlinear equations have no closed-form solutions except in some very special cases, and thus, computer methods are indispensable in their solution. Some examples of equations whose roots may need to be found are: 1 4 x ⋅ + 16 x 2 ⋅ − 3 x 3 ⋅ + 3 x 4 ⋅ + 0 = <-- a polynomial, a characteristic equation, for instance f x ( ) α − 0 = <-- α is a number and f(x) is a function of x tan α ( ) tanh 2 x ⋅ ( ) = <-- a transcendental equation 3.2 METHODS AVAILABLE There are several methods available for finding roots of equations. Some of these are: (1) Direct Search, which is not a very efficient technique, (2) Bisection (3) False Position (4) Newton-Raphson (5) Secant Method (6) Bairstow's Method, which is applicable only to polynomials and (6) Successive Iteration or Fixed Point Iteration method . In this chapter, however, only the Bisection, False Position, Newton-Raphson, Secant and Successive Iteration methods will be addressed along with the functions used in Mathcad to find roots. 3.3 BISECTION METHOD This method can be resorted to when there is only one root occurring in a given range of x. The method involves investigating a given range to seek a root and then bisecting the region successively until a root is found. Other names for this technique are Interval halving, Binary Chopping and Bolzano's Method. Procedure for Finding Roots 1. Choose starting and end points xstart and xend 2. Compute: f(xstart) and f(xend) 3. Compute: f(xstart) times f(xend) Chapter 3: Roots of Equations 53
- 64. 4. If theabove product is negative, then the root lies between xstart and xend. If this product is positive, reselect xstart and xend. 5. If f(xtart)* f(xend) < 0, compute the mid-point of the xstart-xend range. Call it "xmid1" and repeat above steps, i.e., compute f xstart ( ) f xmid1 ( ) ⋅ and f xmid1 ( ) f xend ( ) ⋅ 6. If f(xstart)*f(xmid1) < 0, the root lies between xstart and xmid1. If (xmid1)*f(xend) < 0, then the root lies between xmid1 and xend. 7. Repeat the above procedure until convergence at a root value occurs. Figure 3.1. Bisection Method Computation of Error and Convergence Criterion A convergence criterion has to be followed in order to determine if a root has indeed been found. This is expressed in terms of the error ε or the percentage relative error ε rel which are defined as 54 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 65. ε xmidi 1 +xmidi − = (3.1) (3.2) εrel xmidi 1 + xmidi − xmidi 1 + 100 ⋅ = percent where xmid i+1 and xmid i are the midpoints in the current and previous iterations. While, in general, the relative error should not be greater than 5 %, an error of 0.01 % is the largest that is tolerable for some classes of problems, that require immense precision. The true error , ε True , is an indicator of the real accuracy of a solution and can be evaluated only if the true solution, x True , is known. It is defined as εTrue xTrue xmidi − xTrue 100 ⋅ = (3.3) Calculation of the true error clearly requires knowledge of the true solution, which, in general, will not be known to us. Therefore, the quantity ε rel may have to be mostly used to determine the error associated with a solution process. Advantages and Disadvantages of Bisection While Bisection is a simple, robust technique for finding one root in a given interval, when the root is known to exist and it works even for non-analytic functions, its convergence process is generally slow, making it a somewhat inefficient procedure. Sometimes, a singularity may be identified as if it were a root, since the method does not distinguish between roots and singularities, at which the function would go to infinity. Therefore, as the method proceeds, a check must be made to see if the absolute value of [f(xend)-f(xstart)], in fact, converges to zero. If this quantity diverges, the method is chasing a singularity rather than a root . When there are multiple roots, Bisection is not a desirable technique to use, since the function may not change signs at points on either side of the roots. Therefore, a graph of the function must first be drawn before proceeding to do the calculations. Example 3.1 Obtain a root of f(x) in the range of 4 < x < 20 f(x)= ( 750.5 / x ) [ 1 - exp( -0.15245x) ] - 40 Chapter 3: Roots of Equations 55
- 66. Let us generatea table for x between 4 and 20 and draw a graph using Mathcad to explore where the root may lie. f x ( ) 750.5 x 1 e 0.15245 − x ⋅ − ( ) ⋅ 40 − := x 4 8 , 20 .. := 5 10 15 20 10 − 0 10 20 30 40 50 Graph of f(x) versus x f x ( ) x x 0 0 1 2 3 4 4 8 12 16 20 = f x ( ) 0 0 1 2 3 4 45.6584 26.1051 12.5031 2.8146 -4.2539 = The above table and graph suggest that the root lies between xstart=16 and xend= 20. In an attempt to minimize the number of iterations needed, we will obtain a root value that is correct only to two decimal places, in this case. Iteration # 1: xstart 16 := xend 20 := xmid xstart xend + 2 := xmid 18 = f xstart ( ) f xmid ( ) ⋅ 2.7773 − = f xmid ( ) f xend ( ) ⋅ 4.1976 = The root must lie beteen above xstart and xmid. Thus the new xstart is 16 and the new xend is 18 Iteration # 2: xstart 16 := xend 18 := xmid xstart xend + 2 := xmid 17 = f xstart ( ) f xmid ( ) ⋅ 2.366 = f xmid ( ) f xend ( ) ⋅ 0.8295 − = The root must lie beteen above xmid and xend. Thus the new xstart is 17 and the new xend is 18 Iteration # 3: xstart 17 := xend 18 := xmid xstart xend + 2 := xmid 17.5 = f xstart ( ) f xmid ( ) ⋅ 0.0761 − = f xmid ( ) f xend ( ) ⋅ 0.0893 = 56 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 67. The root mustlie beteen above xstart and xmid. Thus the new xstart is 17 and the new xend is 17.5 Iteration # 4: xstart 17. := xend 17.5 := xmid xstart xend + 2 := xmid 17.25 = f xstart ( ) f xmid ( ) ⋅ 0.3115 = f xmid ( ) f xend ( ) ⋅ 0.0335 − = The root must lie beteen above xmid and xend. Thus the new xstart is 17.25 and the new xend is 17.5 Iteration # 5: xstart 17.25 := xend 17.5 := xmid xstart xend + 2 := xmid 17.375 = f xstart ( ) f xmid ( ) ⋅ 0.0515 = f xmid ( ) f xend ( ) ⋅ 0.0126 − = The root must lie beteen above xmid and xend. Thus the new xstart is 17.38 and the new xend is 17.5 Iteration # 6: xstart 17.38 := xend 17.5 := xmid xstart xend + 2 := xmid 17.44 = f xstart ( ) f xmid ( ) ⋅ 2.5064 10 3 − × = f xmid ( ) f xend ( ) ⋅ 1.7497 − 10 3 − × = The root must lie beteen above xmid and xend. Thus the new xstart is 17.44 and the new xend is 17.5 Iteration # 7: xstart 17.44 := xend 17.5 := xmid xstart xend + 2 := xmid 17.47 = f xstart ( ) f xmid ( ) ⋅ 6.893 − 10 4 − × = f xmid ( ) f xend ( ) ⋅ 3.2291 10 3 − × = The root must lie beteen above xstart and xmid. Thus the new xstart is 17.44 and the new xend is 17.47 Iteration # 8: xstart 17.44 := xend 17.47 := xmid xstart xend + 2 := xmid 17.455 = f xstart ( ) f xmid ( ) ⋅ 1.5821 − 10 4 − × = f xmid ( ) f xend ( ) ⋅ 2.9199 10 4 − × = Chapter 3: Roots of Equations 57
- 68. The root mustlie beteen above xstart and xmid. Thus the new xstart is 17.44 and the new xend is 17.455 Iteration # 9: xstart 17.44 := xend 17.455 := xmid xstart xend + 2 := xmid 17.447 = f xstart ( ) f xmid ( ) ⋅ 1.0756 10 4 − × = f xmid ( ) f xend ( ) ⋅ 4.5564 − 10 5 − × = a check to see if this is small f xstart ( ) 0.0193 = f xmid ( ) 5.5657 10 3 − × = <-- Since f(xmid) is a very small quantitity, it is reasonable to assume that the xmid obtained in iteration # 9 is, in fact, the required root. The % error in this case is, then error 17.447 17.455 − 17.447 100 ⋅ := error 0.0459 − = (percent) Thus, the solution which has been obtained after 9 iterations is : x = 17.447 3.4 THE REGULA FALSI OR THE FALSE POSITION METHOD The False Position method is similar to the Bisection method in that the size of the interval containing the root is reduced with every step of the iteration process until the root is found. The main difference is that while the interval size in the Bisection method is reduced by bisecting it in each step of the iteration , this reduction of interval size is achieved by a linear interpolation fitting the two end points. While the Bisection method is reliable, it is slower than the False Position method in achieving convergence. The equation of a straight line connecting two points (xstart,ystart) and (xend, yend) is y yend yend ystart − xend xstart − x xend − ( ) ⋅ + = (3.4) 58 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 69. f(xstart) f(xend) f(x) x xstart xnew Root xend (xstart, f(xstart)) (xend, f(xend)) Figure3.2. Regula Falsi method The procedure for finding roots then is as follows. ( See Figure 3.2 ) 1. Choose the starting and ending points xstart and xend as in the Bisection method. 2. Compute f(xstart) and f(xend). Make sure that f(xstart) times f(xend) is still a negative product. If it is not, then there is no root between xstart and xend. 3. The new end point xnew is located by setting y = 0 in Equation 3.4 and solving for x. xnew xend xend xstart − yend ystart − yend ⋅ − = (3.5) 4. If f(xstart)*f(xnew) < 0, the root lies between xstart and xnew. However, if f(xnew)*f(xend) < 0, the root lies between xnew and xend. 5. Repeat the above procedure until convergence takes place. Chapter 3: Roots of Equations 59
- 70. A disadvantage ofthis method is that sometimes a root is approached from only one side and, thus, one end of the interval does not change at all in successive iterations. This is called stagnation of an end point. This is not desirable since it slows down the convergence process especially when the initial interval is very large or when the function is highly nonlinear. Examples 3.2 and 3.3 show the occurrence of stagnation. Example 3.2 Obtain the required root for the function of Example 3.1 by the False Position method correct to four decimal places. The starting and ending points were 16 and 20 respectively. f x ( ) 750.5 x 1 e 0.15245 − x ⋅ − ( ) ⋅ 40 − := Iteration # 1: xstart 16 := xend 20 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.8146 = yend 4.2539 − = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 17.5927 = f xstart ( ) f xnew ( ) ⋅ 0.73 − = f xnew ( ) f xend ( ) ⋅ 1.1033 = The root must lie beteen above xstart and xnew. Thus the new xstart is 16 and the new xend is 17.5927 Iteration # 2: xstart 16 := xend 17.5927 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.8146 = yend 0.2593 − = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 17.4584 = f xstart ( ) f xnew ( ) ⋅ 0.0404 − = f xnew ( ) f xend ( ) ⋅ 3.7171 10 3 − × = The root must lie between above xstart and xnew. Thus the new xstart is 16 and the new xend is 17.4584 Iteration # 3: xstart 16 := xend 17.4584 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.8146 = yend 0.0144 − = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 17.451 = f xstart ( ) f xnew ( ) ⋅ 2.2312 − 10 3 − × = f xnew ( ) f xend ( ) ⋅ 1.143 10 5 − × = The root must lie beteen above xstart and xnew. Thus the new xstart is 16 and the new xend is 17.451 60 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 71. Iteration # 4:xstart 16 := xend 17.451 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.8146 = yend 8.5304 − 10 4 − × = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 17.4506 = f xstart ( ) f xnew ( ) ⋅ 1.3196 − 10 4 − × = f xnew ( ) f xend ( ) ⋅ 3.9995 10 8 − × = The root must lie beteen above xstart and xnew. Thus the new xstart is 16 and the new xend is 17.4506 Iteration # 5: xstart 16 := xend 17.4506 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.8146 = yend 1.1956 − 10 4 − × = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 17.4505 = f xstart ( ) f xnew ( ) ⋅ 1.8496 − 10 5 − × = f xnew ( ) f xend ( ) ⋅ 7.8568 10 10 − × = The root must lie beteen above xstart and xnew. Thus the new xstart is 16 and the new xend is 17.4505 Iteration # 6: xstart 16 := xend 17.4505 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.8146 = yend 6.3811 10 5 − × = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 17.4505 = f xstart ( ) f xnew ( ) ⋅ 9.8712 10 6 − × = f xnew ( ) f xend ( ) ⋅ 2.238 10 10 − × = f xnew ( ) 3.5072 10 6 − × = The above process indicates that in this case, the root is being approached only from the side of the original starting point and therefore stagnation has occurred. Since the xnew computed in the sixth iteration is the same as the one in the fifth, and f(xnew) is a very small quantitity, it is reasonable to say that the xnew (=17.4505) computed above is, in fact, the required root. Example 3.3 Obtain a root of f(x) by the False Position method in the range of 0 < x < 10 correct to three decimal places. f x ( ) x 3 9 x ⋅ + 200 − := Iteration # 1: xstart 0 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := Chapter 3: Roots of Equations 61
- 72. ystart 200 − = yend890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 1.8349 = f xstart ( ) f xnew ( ) ⋅ 3.5462 10 4 × = f xnew ( ) f xend ( ) ⋅ 1.578 − 10 5 × = The root must lie beteen above xnew and xend. Thus the new xstart is 1.835 and the new xend is 10 Iteration # 2: xstart 1.835 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 177.3061 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 3.1914 = f xstart ( ) f xnew ( ) ⋅ 2.4605 10 4 × = f xnew ( ) f xend ( ) ⋅ 1.2351 − 10 5 × = The root must lie between above xnew and xend. Thus the new xstart is 3.191 and the new xend is 10 Iteration # 3: xstart 3.191 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 138.7887 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 4.1096 = f xstart ( ) f xnew ( ) ⋅ 1.2992 10 4 × = f xnew ( ) f xend ( ) ⋅ 8.3312 − 10 4 × = The root must lie beteen above xnew and xend. Thus the new xstart is 4.11 and the new xend is 10 Iteration # 4: xstart 4.11 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 93.5835 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 4.6704 = f xstart ( ) f xnew ( ) ⋅ 5.2493 10 3 × = f xnew ( ) f xend ( ) ⋅ 4.9922 − 10 4 × = The root must lie beteen above xnew and xstart. Thus the new xstart is 4.67 and the new xend is 10 Iteration # 5: xstart 4.67 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := 62 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 73. ystart 56.1224 − = yend890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 4.9862 = f xstart ( ) f xnew ( ) ⋅ 1.7487 10 3 × = f xnew ( ) f xend ( ) ⋅ 2.7732 − 10 4 × = The root must lie beteen above xnew and xend. Thus the new xstart is 4.986 and the new xend is 10 Iteration # 6: xstart 4.986 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 31.1731 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.1557 = f xstart ( ) f xnew ( ) ⋅ 516.0963 = f xnew ( ) f xend ( ) ⋅ 1.4735 − 10 4 × = The root must lie beteen above xnew and xend. Thus the new xstart is 5.156 and the new xend is 10 Iteration # 7: xstart 5.156 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 16.5272 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.2443 = f xstart ( ) f xnew ( ) ⋅ 141.6022 = f xnew ( ) f xend ( ) ⋅ 7.6254 − 10 3 × = The root must lie beteen above xnew and xend. Thus the new xstart is 5.244 and the new xend is 10 Iteration # 8: xstart 5.244 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 8.5964 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.2895 = f xstart ( ) f xnew ( ) ⋅ 37.8307 = f xnew ( ) f xend ( ) ⋅ 3.9167 − 10 3 × = The root must lie beteen above xnew and xend. Thus the new xstart is 5.290 and the new xend is 10 Iteration # 9: xstart 5.290 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := Chapter 3: Roots of Equations 63
- 74. ystart 4.3541 − = yend890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3129 = f xstart ( ) f xnew ( ) ⋅ 9.6414 = f xnew ( ) f xend ( ) ⋅ 1.9707 − 10 3 × = The root must lie beteen above xnew and xend. Thus the new xstart is 5.313 and the new xend is 10 Iteration # 10: xstart 5.313 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 2.2078 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3246 = f xstart ( ) f xnew ( ) ⋅ 2.4707 = f xnew ( ) f xend ( ) ⋅ 995.9944 − = The root must lie beteen above xnew and xend. Thus the new xstart is 5.325 and the new xend is 10 Iteration # 11: xstart 5.325 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 1.0813 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3307 = f xstart ( ) f xnew ( ) ⋅ 0.5916 = f xnew ( ) f xend ( ) ⋅ 486.9595 − = The root must lie beteen above xnew and xend. Thus the new xstart is 5.331 and the new xend is 10 Iteration # 12: xstart 5.331 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 0.5163 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3337 = f xstart ( ) f xnew ( ) ⋅ 0.1348 = f xnew ( ) f xend ( ) ⋅ 232.3231 − = The root must lie beteen above xnew and xend. Thus the new xstart is 5.334 and the new xend is 10 Iteration # 13: xstart 5.334 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := 64 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 75. ystart 0.2334 − = yend890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3352 = f xstart ( ) f xnew ( ) ⋅ 0.0275 = f xnew ( ) f xend ( ) ⋅ 104.9753 − = The root must lie beteen above xnew and xend. Thus the new xstart is 5.335 and the new xend is 10 Iteration # 14: xstart 5.335 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 0.139 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3357 = f xstart ( ) f xnew ( ) ⋅ 9.7667 10 3 − × = f xnew ( ) f xend ( ) ⋅ 62.5217 − = The root must lie beteen above xnew and xend. Thus the new xstart is 5.336 and the new xend is 10 Iteration # 15: xstart 5.336 := xend 10 := ystart f xstart ( ) := yend f xend ( ) := ystart 0.0446 − = yend 890 = xnew xend xend xstart − yend ystart − yend ⋅ − := xnew 5.3362 = f xstart ( ) f xnew ( ) ⋅ 1.0062 10 3 − × = f xnew ( ) f xend ( ) ⋅ 20.0659 − = f xnew ( ) 0.0225 − = The above process indicates that in this case, stagnation occurs as well, but the root is being approached only from the side of the original ending point. Since the xnew computed in the fifteenth iteration is about the same as the one in the fourteenth, it is reasonable to say that the xnew ( = 5.336 ) computed above is, in fact, the required root. 3.5 NEWTON-RAPHSON METHOD This is the most widely used iterative method for locating roots. In this method, an initial approximation of the root must be assumed, and calculations are started with a "good initial guess" . If this initial guess is not a good one, then divergence may occur. By starting with an approximation of the root value , x i , and constructing a tangent to the function curve at x i , an improved guess x i+1 can be determined as shown in Figure 3.3. From Figure 3.3, the slope of the function f(x) at x i can be seen to be Chapter 3: Roots of Equations 65
- 76. fprime xi ( ) fxi ( ) − xi 1 + xi − = (3.6) which gives the new improved value of the guess as xi 1 + xi f xi ( ) fprime xi ( ) − = g xi ( ) = (3.7) where fprime( x i ) is the slope of the function at x i Figure 3.3. Newton-Raphson method Procedure for Finding Roots 1. Make a good initial guess. Call it X Old. 2. Improve the guess using XNew XOld f XOld ( ) fprime XOld ( ) − = g XOld ( ) = (3.8) 66 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 77. 3. Keep improvingthe guess using Equation (3.8). 4. Solution is done when a new improved value X New is almost equal to the previous value X Old Advantages and Disadvantages of the Method While the Newton-Raphson method is faster than the Bisection method, is applicable to the complex domain as well and can be extended to simultaneous nonlinear equations, it may not converge in some situations. The solution may oscillate about a local maximum or minimum, and if an initial estimate is chosen such that the derivative becomes zero at some point in the iteration process, then a division by zero takes place and convergence will never occur. Although convergence will occur quite rapidly if the initial estimate is sufficiently close to the root, it is possible for it to be be slow when it is far from the root. Also, if the roots are complex, they will never be generated with real initial guesses. A worthwhile feature of the Newton-Raphson method is that the numerical process will correct itself automatically for minor errors. Thus, any errors that are made in computing the next guess will simply generate a different point for drawing the tangent line and will not have any effect on the final answer. Convergence Criterion for the Newton-Raphson Method It can be mathematically shown that in order for the Newton-Raphson method to converge to a real root, the absolute value of the derivative of the g(xi) of Equation (3.7) must always be less than 1 , that is, g xi ( ) 1 < This condition must be satisfied if convergence is to be attained. However, in some cases, it may not hold for the initial guess. Example 3.4 Using the Newton- Raphson method, solve : f(x)= x3 - 4.2 x -8.5 =0 for a real root between 2 and 3. Let us first draw a graph of the function in the given range f x ( ) x 3 4.2 x ⋅ − 8.5 − := x 2 2.1 , 3 .. := Chapter 3: Roots of Equations 67
- 78. 2 2.2 2.42.6 2.8 10 − 5 − 0 5 10 f x ( ) x The derivative of the given function is fprime x ( ) 3 x 2 ⋅ 4.2 − := , and thus, fdblprime x ( ) 6 x ⋅ := , g x ( ) x f x ( ) fprime x ( ) − := , gprime x ( ) 1 fprime x ( ) 2 f x ( ) fdblprime x ( ) ⋅ − fprime x ( ) 2 − := where fdblprime(x) is the second derivative of f(x) and gprime(x) is the derivative of g(x). Start with an initial estimate of 2.5 xold 2.5 := xnew g xold ( ) := xnew 2.732 = gprime xnew ( ) 0.0206 = xold xnew := xnew g xold ( ) := xnew 2.7091 = gprime xnew ( ) 2.1887 10 4 − × = xold xnew := xnew g xold ( ) := xnew 2.7088 = gprime xnew ( ) 2.3959 10 8 − × = xold xnew := xnew g xold ( ) := xnew 2.7088 = gprime xnew ( ) 0 = Clearly, the root is the converged value 2.7088, and convergence has been established with the absolute value of the slope of g(x) being less than 1 in all the iterations. Example 3.5 Using the Newton- Raphson method, solve : f(x)= x3 +7x2+19x+13 = 0 for all roots, real as well as complex. Let us first draw a graph of the function 68 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 79. f x ( )x 3 7 x 2 ⋅ + 19 x ⋅ + 13 + := which suggests that there is a real root between 0 and -2.5 x 5 − 4.999 − , 0 .. := 4 − 2 − 0 40 − 30 − 20 − 10 − 0 10 20 f x ( ) x The derivative of the given function is fprime x ( ) 3 x 2 ⋅ 14 x ⋅ + 19 + := with fdblprime x ( ) 6 x ⋅ := , g x ( ) x f x ( ) fprime x ( ) − := , gprime x ( ) 1 fprime x ( ) 2 f x ( ) fdblprime x ( ) ⋅ − fprime x ( ) 2 − := where fdblprime(x) is the second derivative of f(x) and gprime(x) is the first derivative of g(x). Let us start with an initial estimate of -2.5 in an attempt to find the real root. xold 2.5 − := xnew g xold ( ) := xnew 0.1818 − = gprime xnew ( ) 0.0389 = xold xnew := xnew g xold ( ) := xnew 0.7721 − = gprime xnew ( ) 0.095 = xold xnew := xnew g xold ( ) := xnew 0.9768 − = gprime xnew ( ) 0.0164 = xold xnew := xnew g xold ( ) := xnew 0.9997 − = gprime xnew ( ) 1.9933 10 4 − × = Chapter 3: Roots of Equations 69
- 80. xold xnew := xnewg xold ( ) := xnew 1 − = gprime xnew ( ) 2.6521 10 8 − × = xold xnew := xnew g xold ( ) := xnew 1 − = gprime xnew ( ) 0 = From the above calculations, it is clear that the real root is the converged value of (-1). Let us now determine the complex roots by starting with an initial estimate of (-4+i) xold 4 − i + := xnew g xold ( ) := xnew 2.9024 − 1.122i + = xold xnew := xnew g xold ( ) := xnew 2.2579 − 2.5753i + = xold xnew := xnew g xold ( ) := xnew 2.636 − 1.9589i + = xold xnew := xnew g xold ( ) := xnew 3.0346 − 1.9196i + = xold xnew := xnew g xold ( ) := xnew 2.9957 − 2.001i + = xold xnew := xnew g xold ( ) := xnew 3 − 2i + = xold xnew := xnew g xold ( ) := xnew 3 − 2i + = <-- Converged Clearly, if we had started with an initial estimate of (-4-i ), we would have obtained the complex conjugate of the above complex root, as demonstrated below. However, there is no need to do this every time, since a complex root will always have a complex conjugate associated with it. xold 4 − i − := xnew 3 − 2i + = xnew g xold ( ) := xold xnew := xnew g xold ( ) := xnew 2.2579 − 2.5753i − = 70 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 81. xold xnew := xnewg xold ( ) := xnew 2.636 − 1.9589i − = xold xnew := xnew g xold ( ) := xnew 3.0346 − 1.9196i − = xold xnew := xnew g xold ( ) := xnew 2.9957 − 2.001i − = xold xnew := xnew g xold ( ) := xnew 3 − 2i − = xold xnew := xnew g xold ( ) := xnew 3 − 2i − = <-- Converged 3.6 USE OF MATHCAD'S root AND polyroots FUNCTIONS Use of root The function root (f(z),z) returns the value of "z" at which the function f(z) goes to zero. Here, both f(z) and z are to be scalar quantities. The procedure to be used will be clear from the following example. Let us find the root of the function in Example 3.4 which is f x ( ) x 3 4.2 x ⋅ − 8.5 − := A guess value for the root must be first defined : x 2.5 := Type the following to define " a" as the root of the given function a root f x ( ) x , ( ) := Chapter 3: Roots of Equations 71
- 82. Type " a=" to see the root computed. a 2.7088 = Use of polyroots The function polyroots is a function in Mathcad which will return all roots of a polynomial at the same time, be they real or complex. It does not require a guess value. polyroots (v) returns all roots of an nth degree polynomial whose coefficients are given in the vector "v" with (n+1) components. By choosing "matrix" from the "insert" menu, type in a vector "v" as shown , beginning with the constant term, making sure that you insert all coefficients even if they are zero. Then, polyroots(v) returns all roots at once. The following steps will find the root of the polynomial of Example 3.4. f x ( ) x 3 4.2 x ⋅ − 8.5 − := <-- Given polynomial This is a vector of the coefficients of the polynomial whose roots are to be found. Include all coefficients, even zeros. Begin with the constant term. <-- v 8.5 − 4.2 − 0 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := polyroots v ( ) 1.3544 − 1.1417i − 1.3544 − 1.1417i + 2.7088 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = <-- Returns all roots at the same time. 3.7 SECANT METHOD The Secant method is very similar to the Newton-Raphson method. The one difference is that while the derivative fprime(x) is evaluated analytically in the Newton-Raphson method, it is determined numerically in the Secant method . From Figure 3.4, it is clear that an approximation for the slope of the given function f(x) at x i can be written as fprime xi ( ) = f xi 1 − ( ) f xi ( ) − ( ) − xi xi 1 − − (3.9) 72 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 83. Figure 3.4. Secantmethod By the Newton-Raphson method, a new improved guess of the root is generated using Equation 3.7, which is xi 1 + xi f xi ( ) fprime xi ( ) − = (3.10) Combining Equations 3.9 and 3.10 yields the equation used in the Secant method, which is xi 1 + xi f xi ( ) xi 1 − xi − ( ) ⋅ f xi 1 − ( ) f xi ( ) − − = (3.11) Note that the format of Equation 3.11 does utilize the derivative but computes it numerically rather than analytically. However, two initial estimates x i-1 and xi of the solution will be required to start the iteration process as can be seen from Equation 3.11. Procedure for Finding Roots 1. Make initial estimates x 0 and x 1. 2. With i= 1, improve the guess x 2 using Equation 3.11, which is Chapter 3: Roots of Equations 73
- 84. xi 1 + xi fxi ( ) xi 1 − xi − ( ) ⋅ f xi 1 − ( ) f xi ( ) − − = 3. Continue obtaining improved estimates x3, x4 ...... using the above equation. 4. Solution is done when a new improved value x i+1 is almost equal to the previous value x i Advantages and Disadvantages of the Secant Method The Secant method has a definite advantage when the derivative of the given function is difficult or time-consuming to evaluate analytically. Among the disadvantages are convergence to an unintended root at times, and divergence from the root if the initial guesses are bad. Also, if f(x) is far from linear near the root, the iterations may start to yield points far away from the actual root. Example 3.6 Obtain a real root of f(x)= 1.15 x3 + 1.2 x2 - 3.5 x - 3.12 between 0 and 10 by the Secant Method. We first draw a graph of the given function to see approximately where a real root lies in the range given. f x ( ) 1.15 x 3 ⋅ 1.2 x 2 ⋅ + 3.5 x ⋅ − 3.12 − := x 0 0.01 , 3 .. := 0 1 2 3 10 − 0 10 20 30 Graph of functon for x= 0 to 3 f x ( ) x From the plot, it appears like the root lies between 1 and 3. We will, therefore, choose x0 to be 1 and x1 to be 2 74 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 85. We will resortto Mathcad's vector notation ( subscript form ) in the use of the Secant Method to determine the root of the given function f(x). To type a subscript, use the left bracket " [ "and put an integer in the placeholder. Vector elements in Mathcad are ordinarily numbered starting with the first element numbered as the zeroth element. A look at the Secant Method suggests that this feature relates well with its recursive formula. Put in "N" as the expected number of iterations. Then i goes from 1 to N . To find the required root, begin with the initial estimates x0 and x1 and use the recursive formula of the Secant Method to obtain improved estimates. Continue until convergence to a solution takes place. N 6 := i 1 N .. := x0 1 := x1 2 := xi 1 + xi f xi ( ) xi 1 − xi − ( ) ⋅ f xi 1 − ( ) f xi ( ) − − := x2 1.523926 = x3 1.662739 = x4 1.6996 = x5 1.6953 = x6 1.6954 = x7 1.6954 = xi 1 − 0 0 1 2 3 4 5 1 2 1.5239 1.6627 1.6996 1.6953 = xi 0 0 1 2 3 4 5 2 1.5239 1.6627 1.6996 1.6953 1.6954 = xi 1 + 0 0 1 2 3 4 5 1.5239 1.6627 1.6996 1.6953 1.6954 1.6954 = <---- Convergence has occurred here Let us solve this problem by Newton Raphson Method, using subscripts and compare answers f x ( ) 1.15 x 3 ⋅ 1.2 x 2 ⋅ + 3.5 x ⋅ − 3.12 − := fprime x ( ) 3.45 x 2 ⋅ 2.4 x ⋅ + 3.50 − := N 4 := i 1 N .. := x1 2 := <--starting estimate xi 1 + xi f xi ( ) fprime xi ( ) − := x2 1.743046 = x3 1.696894 = x4 1.6954 = x5 1.6954 = Chapter 3: Roots of Equations 75
- 86. xi 0 0 1 2 3 2 1.743 1.6969 1.6954 = xi 1 + 0 0 1 2 3 1.743 1.6969 1.6954 1.6954 = <----Convergencehas occurred here. Let us verify the above root by resorting to the polyroots function f x ( ) 1.15 x 3 ⋅ 1.2 x 2 ⋅ + 3.5 x ⋅ − 3.12 − := v 3.12 − 3.5 − 1.2 1.15 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := polyroots v ( ) 1.8941 − 0.8448 − 1.6954 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = which indicates that there is one root between x = 0 and x = 10. Example 3.7 Obtain all roots, real as well as complex, of the function f(x) of Example 3.5 by the Secant method. f x ( ) x 3 7 x 2 ⋅ + 19 x ⋅ + 13 + := To determine the real root, we will consider initial estimates of 1 and 2 as follows. N 9 := i 1 N .. := x0 1 := x1 2 := xi 1 + xi f xi ( ) xi 1 − xi − ( ) ⋅ f xi 1 − ( ) f xi ( ) − − := x2 0.148936 = x3 0.267833 − = x4 0.7285 − = x5 0.9224 − = x6 0.9904 − = x7 0.9996 − = x8 0.999998 − = x9 1 − = x10 1 − = 76 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 87. xi 1 − 0 0 1 2 3 4 5 6 7 8 1 2 0.1489 -0.2678 -0.7285 -0.9224 -0.9904 -0.9996 -1 = xi 0 0 1 2 3 4 5 6 7 8 2 0.1489 -0.2678 -0.7285 -0.9224 -0.9904 -0.9996 -1 -1 =xi 1 + 0 0 1 2 3 4 5 6 7 8 0.1489 -0.2678 -0.7285 -0.9224 -0.9904 -0.9996 -1 -1 -1 = <--- Convergence has occurred here To find the complex roots, let us use initial estimates of (-4+i) and (-4+2i). x0 4 − i + := x1 4 − 2i + := N 7 := j 1 N .. := xj 1 + xj f xj ( ) xj 1 − xj − ( ) ⋅ f xj 1 − ( ) f xj ( ) − − := x2 3.211618 − 1.456432i + = x3 2.929182 − 1.735276i + = x4 2.9136 − 2.0766i + = x5 3.0167 − 1.9899i + = x6 3.0009 − 2.0008i + = x7 3 − 2i + = x8 3 − 2i + = xj 1 − 4 − i + 4 − 2i + 3.2116 − 1.4564i + 2.9292 − 1.7353i + 2.9136 − 2.0766i + 3.0167 − 1.9899i + 3.0009 − 2.0008i + ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = xj 4 − 2i + 3.2116 − 1.4564i + 2.9292 − 1.7353i + 2.9136 − 2.0766i + 3.0167 − 1.9899i + 3.0009 − 2.0008i + 3 − 2i + ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = xj 1 + 3.2116 − 1.4564i + 2.9292 − 1.7353i + 2.9136 − 2.0766i + 3.0167 − 1.9899i + 3.0009 − 2.0008i + 3 − 2i + 3 − 2i + ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Chapter 3: Roots of Equations 77
- 88. Convergence has occurredwith a complex root determined as ( - 3 + 2i ). Clearly, the other complex root will be the complex conjugate (-3 - 2i) . Resorting to the polyroots function will confirm the roots generated above. f x ( ) x 3 7 x 2 ⋅ + 19 x ⋅ + 13 + := v 13 19 7 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := polyroots v ( ) 3 − 2i + 3 − 2i − 1 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 3.8 METHOD OF SUCCESSIVE SUBSTITUTION This method encompasses all iteration schemes for finding roots of nonlinear functions. Applications of this technique can be found in the use of the Newton-Raphson method and the Secant method. Another name for this method is Fixed Point Iteration. In this method, the equation f(x) = 0 is written in the form x = g(x). Thus, if x1 is an assumed initial approximation to the root, then a successive approximation can be generated using x i+ 1 = g ( x i ) (3.12) Iteration must be pursued until convergence to a root is indicated Advantages and Disadvantages of the Iteration Method While the method provides a simple computational technique for finding roots of equations, offers flexibility in the selection of the function g(x), and is easy to program, it is still possible for the solution to diverge with improper choices of the function. To insure convergence of the iteration, the function selected must be such that the absolute value of its slope is always less than unity. Otherwise, divergence will occur. Example 3.8 The function y(x)= x2 - 3 x + e x - 4 has two roots, one positive and the other negative. Determine the negative root by the method of successive substitution. Rewrite the given equation as follows y x ( ) x 2 3 x ⋅ − e x + 4 − := 78 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 89. g x ( ) x 2 e x +4 − 3 := x g x ( ) := gprime x ( ) 1 3 2 x ⋅ e x + ( ) ⋅ := Let the initial guess be 0. Using the above recursive relationship, we find revised values of x until convergence occurs as shown below. x 0 := g x ( ) 1 − = Iteration # 1: Iteration # 2: x g x ( ) := x 1 − = g x ( ) 0.8774 − = gprime x ( ) 0.544 = Iteration # 3: x g x ( ) := x 0.8774 − = g x ( ) 0.9381 − = gprime x ( ) 0.4463 = x g x ( ) := x 0.9381 − = g x ( ) 0.9095 − = gprime x ( ) 0.495 = x g x ( ) := x 0.9095 − = g x ( ) 0.9233 − = gprime x ( ) 0.4721 = x g x ( ) := x 0.9233 − = g x ( ) 0.9167 − = gprime x ( ) 0.4832 = x g x ( ) := x 0.9167 − = g x ( ) 0.9199 − = gprime x ( ) 0.4779 = x g x ( ) := x 0.9199 − = g x ( ) 0.9184 − = gprime x ( ) 0.4804 = x g x ( ) := x 0.9184 − = g x ( ) 0.9191 − = gprime x ( ) 0.4792 = x g x ( ) := x 0.9191 − = g x ( ) 0.9188 − = gprime x ( ) 0.4798 = Iteration # 11: x g x ( ) := x 0.9188 − = g x ( ) 0.9189 − = gprime x ( ) 0.4795 = Note that convergence has taken place at x = -0.919, and the absolute value of the slope of the function selected is less than unity throughout the iteration process. i 0 1 , 9 .. := x0 0 := xi 1 + g xi ( ) := Chapter 3: Roots of Equations 79
- 90. x 0 0 1 2 3 4 5 6 7 8 9 10 0 -1 -0.8774 -0.9381 -0.9095 -0.9233 -0.9167 -0.9199 -0.9184 -0.9191 -0.9188 = <--- Convergence occurshere ---> y x10 ( ) 5.0333 − 10 4 − × = 3.9 MULTIPLE ROOTS AND DIFFICULTIES IN COMPUTATION When the given function makes contact with the x- axis and has a slope of zero, multiple roots can occur as indicated in Figure 3.5. These roots are difficult to compute by the methods discussed in this chapter for the following reasons. 1. In the case of the Bisection method, the function does not change sign at the root. 2. In the case of Newton-Raphson and Secant methods, the derivative at the multiple root is zero. Figure 3.5. Case of multiple roots Because the Newton-Raphson and Secant methods in their orginal forms are methods that resort to linear convergence, they cannot be employed to generate multiple roots. The following modification to the original Newton-Raphson equation has been suggested [ 4 ]. 80 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 91. xi 1 + xim f xi ( ) fprime xi ( ) ⋅ − = (3.13) where m is the multiplicity of the root ( for example, m = 2 for a double root ). However, this may not be a very practical route since it assumes prior knowledge about the multiplicity of a root. Another approach, also suggested [ 4 ] involves defining a new function u(x) u x ( ) f x ( ) fprime x ( ) = (3.14) which has the same roots as the original given function f(x), and then resorting to the following alternative form of the Newton-Raphson formula xi 1 + xi u xi ( ) uprime xi ( ) − = (3.15) where uprime(x) represents the derivative of the new function u(x). Substitution of Equation (3.14) into the Newton-Raphson equation, which is now Equation (3.15), yields the desired formula xi 1 + xi f xi ( ) fprime xi ( ) ⋅ fprime xi ( )2 f xi ( ) fdoubleprime xi ( ) ⋅ − − = (3.16) where fprime and fdoubleprime represent first and second derivatives respectively. The modified formula for the Secant Method that will be suitable for the generation of multiple roots can be obtained by simply substituting u(x) for f(x) in the Secant method formula giving xi 1 + xi u xi ( ) xi 1 − xi − ( ) ⋅ u xi 1 − ( ) u xi ( ) − − = (3.17) Example 3.9 The function f(x)= x3 - 4 x 2 + 5 x - 2 = (x-1) (x-1) (x-2) Chapter 3: Roots of Equations 81
- 92. has a doubleroot at 1 and a single root at 2, which we will verify using Equation (3.16). f x ( ) x 3 4 x 2 ⋅ − 5 x ⋅ + 2 − := The first and second derivatives of the given function are fprime x ( ) 3 x 2 ⋅ 8 x ⋅ − 5 + := fdoubleprime x ( ) 6 x ⋅ 8 − := In order to determine the multiple root by the modified form of the Newton-Raphson formula , let us employ Equation (3.16) and start with an initial estimate of 0.25. xi 1 + xi f xi ( ) fprime xi ( ) ⋅ fprime xi ( )2 f xi ( ) fdoubleprime xi ( ) ⋅ − − = xold 0.25 := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1.0841 = xold xnew := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1.0042 = xold xnew := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1 = xold xnew := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1 = It is clear that the multiple root is the converged value of 1. To determine the single root, the above approach can still be employed, but with a different starting estimate, as shown below. xold 2.50 := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1.8182 = 82 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 93. xold xnew := xnewxold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1.9101 = xold xnew := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1.9809 = xold xnew := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 1.9992 = xold xnew := xnew xold f xold ( ) fprime xold ( ) ⋅ fprime xold ( ) 2 f xold ( ) fdoubleprime xold ( ) ⋅ − − := xnew 2 = Thus, the single root is the converged value of 2. The following steps will now demonstrate the application of the modified Secant Method formula, which is Equation 3.17, to the determination of the multiple root at 1. f x ( ) x 3 4 x 2 ⋅ − 5 x ⋅ + 2 − := fprime x ( ) 3 x 2 ⋅ 8 x ⋅ − 5 + := u x ( ) f x ( ) fprime x ( ) := i 0 5 .. := xi 1 + xi u xi ( ) xi 1 − xi − ( ) ⋅ u xi 1 − ( ) u xi ( ) − − := i 1 − x0 0.25 := x1 0.26 := x2 x1 u x1 ( ) x0 x1 − ( ) ⋅ u x0 ( ) u x1 ( ) − − := x2 1.0835 = x3 x2 u x2 ( ) x1 x2 − ( ) ⋅ u x1 ( ) u x2 ( ) − − := x3 0.9802 = x4 x3 u x3 ( ) x2 x3 − ( ) ⋅ u x2 ( ) u x3 ( ) − − := x4 0.9991 = x5 x4 u x4 ( ) x3 x4 − ( ) ⋅ u x3 ( ) u x4 ( ) − − := x5 1 = <--- Convergence at the double root x =1 Application of the polyroots and root functions confirms the correctness of the above analysis. Chapter 3: Roots of Equations 83
- 94. v 2 − 5 4 − 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := polyroots v () 0.9995 1.0005 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Guess x 0.5 := a root f x ( ) x , ( ) := generates a 1 = Guess x 3 := a root f x ( ) x , ( ) := generates a 2 = 3.10 SOLUTION OF SYSTEMS OF NONLINEAR EQUATIONS The methods discussed so far deal with finding the roots of functions of only a single variable. In certain situations, it may be necessary to solve nonlinear equations in two or more variables. In these cases, iteration can be resorted to as presented in the following example. Example 3.10 Solve: 1.1 x2 +1.2 xy = 10.5 ; y + 3.2 x y2 = 58.2 Let us put the given equations in the form x 10.5 1.1 x 2 ⋅ − 1.2 y ⋅ = y 58.2 3.2 x ⋅ y 2 ⋅ − = and make the following initial guesses: x 1 := y 3 := Then, x 10.5 1.1 x 2 ⋅ − 1.2 y ⋅ := and y 58.2 3.2 x ⋅ y 2 ⋅ − := yield x 2.6111 = y 17 − = A second iteration gives 84 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 95. x 10.5 1.1 x 2 ⋅ − 1.2y ⋅ := x 0.1471 − = y 58.2 3.2 x ⋅ y 2 ⋅ − := y 194.214 = Notice that the approach resorted to is diverging and will not yield a solution. Thus, the computations must be repeated with equations set up in a different format as shown below. Iteration # (1) x 1 := y 3 := x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 2.5045 = y 58.2 y − 3.2 x ⋅ := y 2.6244 = (2) x 2.5045 = y 2.6244 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.5411 = y 58.2 y − 3.2 x ⋅ := y 3.357 = (3) x 1.5411 = y 3.357 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.9753 = y 58.2 y − 3.2 x ⋅ := y 2.9456 = (4) x 1.9753 = y 2.9456 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.7883 = y 58.2 y − 3.2 x ⋅ := y 3.1073 = (5) x 1.7883 = y 3.1073 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8664 = y 58.2 y − 3.2 x ⋅ := y 3.0372 = (6) x 1.8664 = y 3.0372 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8335 = y 58.2 y − 3.2 x ⋅ := y 3.0663 = (7) x 1.8335 = y 3.0663 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8473 = y 58.2 y − 3.2 x ⋅ := y 3.054 = (8) x 1.8473 = y 3.054 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8415 = y 58.2 y − 3.2 x ⋅ := y 3.0591 = (9) x 1.8415 = y 3.0591 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8439 = y 58.2 y − 3.2 x ⋅ := y 3.057 = (10) x 1.8439 = y 3.057 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8429 = y 58.2 y − 3.2 x ⋅ := y 3.0579 = Chapter 3: Roots of Equations 85
- 96. (11) x 1.8429 =y 3.0579 = x 10.5 1.2 x ⋅ y ⋅ − 1.1 := x 1.8433 = y 58.2 y − 3.2 x ⋅ := y 3.0575 = Notice that after ten iterations convergence has taken place yielding the solution as : x =1.843 and y = 3.058. This example illustrates a very serious disadvantage associated with the successive substitution method which is the dependence of convergence on the format in which the equations are put and utilized in the iteration process. Also , even in situations where a converged solution can be attained, initial estimates that are fairly close to the true solution must be resorted to, because, otherwise, divergence may occur and a solution may never be obtained. The results of the iterative process are summarized in the following table. ---------------------------------------------------------------------------------------------------------------------------- Iteration # x y ------------------------------------------------------------------------------------------------------------------------------ 0 init=1.0 3.00 1 2.505 2.624 2 1.541 3.357 3 1.975 2.946 4 1.788 3.107 5 1.866 3.037 6 1.833 3.066 7 1.847 3.054 8 1.841 3.059 9 1.844 3.057 10 1.843 3.058 11 1.843 3.058 ---------------------------------------------------------------------------------------------------------------------------------- 3.11 SOLVING SYSTEMS OF EQUATIONS USING MATHCAD'S Given AND Find FUNCTIONS Mathcad allows you to solve a system of 50 simultaneous equations in 50 unknowns. The procedure is as follows: 1. Provide initial guesses for all unknowns. These initial guesses give Mathcad a place to start searching for the solution. 2. Type the word Given. This tells Mathcad that what follows is a system of equations. Given can be typed in any combination of upper and lower case letters, and in any font. However, it should not be typed in a text region. 3. Then type the equations and inequalities in any order below the word Given . Make sure you separate the left and right sides of an equation by an " = ". Press [Ctrl]= to type " = ". You can of course separate the left and right sides of an inequality with the appropriate symbol. 4. a:= Find(x,y.....) returns the solution to the system of equations. 86 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 97. Example 3.11. Using theGiven and Find functions, solve 1.5 x 2 ⋅ 2 y 2 ⋅ + 6.5 = . x y + 2.75 = Provide initial guesses: x 1 := y 1 := Given 1.5 x 2 ⋅ 2 y 2 ⋅ + 6.5 = x y + 2.75 = a Find x y , ( ) := a 1.5 1.25 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <--- Solution Alternatively, do the following sequence to obtain xval and yval as the solution Given 1.5 x 2 ⋅ 2 y 2 ⋅ + 6.5 = x y + 2.75 = xval yval ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find x y , ( ) := xval 1.5 = yval 1.25 = <--Solution 3.12. APPLICATIONS IN ROOT-FINDING The following are examples of situations where roots of nonlinear equations need to be evaluated. 3.12.1 Maximum Design Load for a Column A steel pipe with an outside diameter of 10.70 inches and a wall thickness of 0.375 inches is to serve as a pin-ended column that is 22 feet long . It supports an eccentric load at an eccentricity of e = 1.5 inches from the column center-line. The maximum load that the column can carry with a factor of safety of 2.75 is to be determined. The modulus of elasticity E for steel is 30 x 10 6 psi and its yield strength, σyld , is 40,000 psi . The equation for the stress level σ max resulting from the application of the maximum load P max is [ 3,18 ] Chapter 3: Roots of Equations 87
- 98. (3.18) σmax Pmax A 1 e c ⋅ K 2 sec Pmax A E ⋅ L 2K ⋅ ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ = where A is the column cross-sectional area, e is the eccentricity at which the load is applied, as shown in Figure 3.6 , c is the distance of the outermost fiber in the cross-section from the neutral axis , K is the radius of gyration , L is the column length and E is the modulus of elasticity. This relationship between σ max and P max is not linear and there is no proportionality between the applied load and the stress that is produced. Thus, it must be solved iteratively using trial and error. How this can be accomplished with Mathcad is demonstrated below. Figure 3.6. Eccentrically Loaded Column Here, σyld 40000. := (psi) E 30 10 6 ⋅ := (psi) FS 2.75 := do 10.70 := t 0.375 := di do 2 t ⋅ − := di 9.95 = <--- all dimensions in inches 88 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 99. e 1.50 := L264. := c do 2 := c 5.35 = I π 64 do 4 di 4 − ( ) ⋅ := ( in 4 ) A π 4 do 2 di 2 − ( ) ⋅ := (in 2 ) A 12.1639 = I 162.3057 = K I A := K 3.6528 = (in ) With a factor of safety of 2.75, the maximum stress level that the column can have is σmax σyld FS := σmax 1.4545 10 4 × = psi The corresponding maximum allowable load may be determined from Equation (3.18) using the root function as follows. g Pmax ( ) Pmax E A ⋅ L 2 K ⋅ ⋅ := f Pmax ( ) σmax Pmax A 1 e c ⋅ K 2 sec g Pmax ( ) ( ) ⋅ + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ − := Pmax 125000. := (lbs) <--Guess value for the root a root f Pmax ( ) Pmax , ( ) := a 1.0221 10 5 × = (lbs) <-- Maximum allowable load column can carry 3.12.2. Natural Frequencies of Vibration of a Uniform Beam For a uniform beam that is clamped at one end and free at the other, a shown in Figure 3.7 , the natural frequencies of vibration [20] are given by the equation 1 cos βprime ( ) cosh βprime ( ) ⋅ + 0 = (3.19) Chapter 3: Roots of Equations 89
- 100. where (3.20) βprime β L ⋅ = andβ ρ ω 2 ⋅ EI ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 4 = (3.21) in which ρ is the mass per unit length of the beam, EI is the flexural rigidity or bending stiffness about the neutral axis , and ω refers to the natural frequencies . For a beam with a length of 20 feet, an EI value of 16 x 10 6 lb-in 2 , and a weight per unit length of 0.03 lb/ in , the natural frequencies may be determined as shown below. Figure 3.7. A Fixed-Free Beam ρwt 0.03 := lb/in --> ρ ρwt 12 ⋅ 32.2 := --> ρ 0.0112 = (lb- ft -2 sec 2) EI 16 10 6 ⋅ := lb-in 2 --> EI 16 10 6 ⋅ 144 := --> EI 1.1111 10 5 × = (lb- ft 2 ) L 20 := ft βprime 1 := <-- Guess value for root f βprime ( ) 1 cos βprime ( ) cosh βprime ( ) ⋅ + := a root f βprime ( ) βprime , ( ) := a 1.8751 = 90 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 101. β a L := β 0.0938 = ω EIβ 4 ⋅ ρ := ω 27.7106 = (rad/sec) <-- First or fundamental natural frequency βprime 5 := <-- Guess value for next root f βprime ( ) 1 cos βprime ( ) cosh βprime ( ) ⋅ + := a root f βprime ( ) βprime , ( ) := a 4.6941 = β a L := β 0.2347 = ω EI β 4 ⋅ ρ := ω 173.6595 = (rad/sec) <-- Second natural frequency Similarly by making appropriate guesses, higher natural frequencies can be determined using the root function. Some of these are listed below. Third natural frequency: 486.3 rad/sec Fourth natural frequency: 952.8 rad/sec Fifth natural frequency: 1575.1 rad/sec Sixth natural frequency: 2353 rad/sec Seventh natural frequency: 3286 rad/sec. 3.12.3. Solving the Characteristic Equation in Control Systems Engineering A characteristic equation is an algebraic equation that is formulated from the differential equation or equations of a control system [14] . Its solution, which often requires evaluation of the roots of a polynomial of degree higher than two, is crucial in determining system stability and assessing system transient response in terms of its time constant, natural frequencies , damping qualities etc. An application involving the use of Mathcad' s polyroots function in determining the roots of a characteristic polynomial of a control system is presented below. Chapter 3: Roots of Equations 91
- 102. For the unity- feedback system shown in Figure 3.8, the characteristic polynomial is s 3 12 s 2 ⋅ + 20 s ⋅ + K + 0 = where K is a variable system parameter that can be selected based on the performance requirements that may be prescribed. For varying values of this parameter, the roots of the characteristic polynomial can be evaluated as shown below, and the system transient response and stability studied , following which, an appropriate range for K may be recommended. Figure 3.8. Feedback Control System Block Diagram K 1 := <-- Negative real roots indicate a stable system v 1 20 12 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := r polyroots v ( ) := r 10.0125 − 1.9359 − 0.0516 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = K 10 := v 10 20 12 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := <-- Negative real parts of root indicate a stable system r polyroots v ( ) := r 10.1216 − 0.9392 − 0.3255i − 0.9392 − 0.3255i + ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 92 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 103. <-- Negative real parts ofroot indicate a stable system K 100 := v 100 20 12 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := r polyroots v ( ) := r 11.0084 − 0.4958 − 2.9729i + 0.4958 − 2.9729i − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = <-- Positive real parts of root indicate an unstable system K 400 := v 400 20 12 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := r polyroots v ( ) := r 12.8628 − 0.4314 5.5598i + 0.4314 5.5598i − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 3.12.4. Horizontal Tension in a Uniform Cable Flexible cables are often used in suspension bridges, transmission lines, telephone lines , mooring lines and many other applications. A cable hanging under its own weight takes the shape of a catenary in which the deflection y along the span is given by the following relationship [ 8 ] y H w cosh w x ⋅ H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ = (3.22) where x is the horizontal distance along the span a shown in Figure 3.9 , H is the horizontal component of tension in the cable and w is the weight per unit length of the cable. Figure 3.9. Catenary cable Chapter 3: Roots of Equations 93
- 104. The constant His obtained by using the boundary condition that y = h at x = L/2, which gives h H w cosh w L ⋅ 2 H ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ + (3.23) Thus, the root of the following equation must be evaluated by using a method such as the Secant method h g H ( ) = (3.24) where g H ( ) H w cosh w L ⋅ 2 H ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ = (3.25) For a cable with a span of L= 22 meters, a weight per unit length of w =5.5 N/m and a maximum sag of h= 6.5 meters, h 6.5 := w 5.5 := L 22 := The intersection of the plot of g(H) with h= 6.5 in Figure 3.10 suggests that f(H) has a root at about H = 50 Newtons. g H ( ) H w cosh w L ⋅ 2 H ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ := H 10 11 , 200 .. := f H ( ) h H w cosh w L ⋅ 2 H ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ − := 94 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 105. 0 50 100150 200 0 100 200 300 400 h g H ( ) H Figure 3.10. Plot of g(H) The use of the Secant method in obtaining the required root is shown below. N 7 := i 1 N .. := H0 25 := H1 35 := <--initial required guesses Hi 1 + Hi f Hi ( ) Hi 1 − Hi − ( ) ⋅ f Hi 1 − ( ) f Hi ( ) − − := <--recursive formula of Secant method H2 41.1846 = H3 49.0109 = H4 53.8261 = H5 55.9053 = H6 56.2878 = <-- improved estimates H7 56.31 = H8 56.3102 = Hi 1 − 0 0 1 2 3 4 5 6 25 35 41.1846 49.0109 53.8261 55.9053 56.2878 = Hi 0 0 1 2 3 4 5 6 35 41.1846 49.0109 53.8261 55.9053 56.2878 56.31 = Hi 1 + 0 0 1 2 3 4 5 6 41.1846 49.0109 53.8261 55.9053 56.2878 56.31 56.3102 = <----Convergence has occurred here Mathcad's root function can also be used to determine the root of f(H) . This is done below. Guess: H 35 := a root f H ( ) H , ( ) := a 56.3102 = Thus, by solving f(H)=0, the horizontal tension in the cable has been found as H= 56.31 N. Chapter 3: Roots of Equations 95
- 106. PROBLEMS 3.1. Obtain aroot of the following polynomial lying in the range 1 to 3 by the Bisection method f x ( ) x 3 3.45 x 2 ⋅ − 3.35 x ⋅ + 0.934 − = Using Mathcad, draw a graph of the polynomial for the given range of x. Label it and give it a title . Verify your answer by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3.2. Estimate a root of the polynomial x 3 3.5 x 2 ⋅ − 3.28 x ⋅ + 0.924 − between x = 2 and x = 3 using the Bisection method. Generate an answer that is good to three decimal places. 3.3 Estimate a root of the polynomial x 3 6.1 x 2 ⋅ − 11.26 x ⋅ + 6.336 − between x = 1.5 and x = 2 using the Bisection method. 3.4. Obtain all roots of the following polynomial lying in the range -5.5 to 4.0 by the Newton-Raphson method f x ( ) 1.1 x 3 ⋅ 1.15x 2 + 17.1 x ⋅ − 15.2 + = Using Mathcad, draw a graph of the polynomial for the given range of x. Label it and give it a title . Verify your answers by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3.5. Obtain all roots of the following polynomial lying in the range -2.5 to 1.5 by the Newton-Raphson method f x ( ) x 3 4.25 x 2 ⋅ + 4.05 − = Using Mathcad, draw a graph of the polynomial for the given range of x. Label it and give it a title . Verify your answers by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3.6. Estimate a root of the polynomial 96 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 107. x 3 2 x 2 ⋅ − 5x ⋅ − 6 + between x = 3 and x = 4 using the Newton-Raphson method. 3.7. Using the Newton-Raphson method, estimate a real root of the polynomial f x ( ) e x − ( ) x 3 − = between x = 0 and x =1. Obtain an answer that is correct to at least three decimal places. 3.8. Using the Newton-Raphson method, estimate a real root of the polynomial f x ( ) x 3 5 cos x ( ) ⋅ − 1 − = between x = 1 and x =2. Obtain an answer that is correct to at least three decimal places. 3.9. Using the Newton-Raphson method, estimate a real root of the polynomial f x ( ) cos x ( ) x − = between x = 0 and x =4. Obtain an answer that is correct to three decimal places. 3.10. Using the Newton-Raphson method, estimate a real root of the polynomial f x ( ) e x − x 2 − = between x = 0 and x =1. Obtain an answer that is correct to four decimal places. 3.11. Obtain all roots of the following polynomial lying in the range 0 to 1.0 by the Secant method f x ( ) x 3 1.72 x 2 ⋅ − 0.85 x ⋅ + 0.110 − = Using Mathcad, draw a graph of the polynomial for the given range of x. Label it and give it a title . Verify your answers by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3.12. Obtain all roots of the following polynomial lying in the range 0.0 to 1.0 by the Chapter 3: Roots of Equations 97
- 108. Secant method f x () x 4 3.05 x 3 ⋅ − 2.75 x 2 ⋅ + 0.43 x ⋅ + 0.45 − = Using Mathcad, draw a graph of the polynomial for the given range of x.Label it and give it a title . Verify your answers by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3. 13. Using the Secant method, estimate a real root of the polynomial x 3 x 2 + 17 x ⋅ − 15 + between x = 1.5 and x =4. Start with initial estimates of 1.5 and 4.0. 3.14. Using the Secant method, estimate a real root of the polynomial x 4 2 x 3 ⋅ − 15 x 2 ⋅ − 4 x ⋅ − 20 + between x = 2 and x =6. Start with initial estimates of 4.0 and 4.5. 3.15. Using the Secant method, estimate a real root of the polynomial f x ( ) x 2 sin x ( ) − 1 − = between x = 1 and x =2. 3.16. Using the Secant method, estimate a real root of the polynomial f x ( ) 5 x 4 ⋅ 2 x 3 ⋅ − 25 x 2 ⋅ − 6 x ⋅ − 45 + = between x = 1 and x =2. Generate an answer that is correct to four decimal places. 3.17. Using the Secant method, estimate a real root of the polynomial f x ( ) 2 cos x ( ) ⋅ e x − = between x = 0 and x =1. Generate an answer that is correct to three decimal places. 3.18. Using the Secant method, estimate a real root of the polynomial 98 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 109. f x ( )e x − x 2 − = between x = 0 and x =1. Generate an answer that is correct to three decimal places. 3.19. Using the Secant method, estimate a real root of the polynomial f x ( ) x 4 5 x 3 ⋅ − 40 x 2 ⋅ − 12 x ⋅ − 24 + = between x = 0 and x =2. Generate an answer that is correct to threer decimal places. 3.20. Obtain a root of the following polynomial lying in the range -10 to 10 by the method of Successive Substitution f x ( ) x 3 0.25 x 2 ⋅ + 0.75 x ⋅ − 0.350 − := Using Mathcad, draw a graph of the polynomial for the given range of x. Label it and give it a title. Notice what this plot suggests in terms of the interval where the roots may lie. Then, draw a second graph for a narrower range (-0.60 to 1.40). Start with an initial estimate of 0.9. Verify your answer by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3.21. Obtain a root of the following polynomial lying in the range --5 to 25 by the method of Successive Substitution f x ( ) x 3 17.45 x 2 ⋅ − 82.89 x ⋅ − 84.50 − = Using Mathcad, draw a graph of the polynomial for the given range of x. Label it and give it a title . Start with a reasonable initial estimate. Verify your answer by using the root function. Also obtain all the roots of the polynomial with the polyroots function. 3.22. Obtain the solution of the following system of nonlinear equations by iteration e x 1.1 y ⋅ − 0 = and 1.1 x ⋅ y ⋅ e 1.05 x ⋅ − 0 = Start with an initial estimate of y= 2.2. Verify your answers with the Given and Find functions. 3.23. Using the Given and Find functions of Mathcad, obtain the solution to the following system of equations. 3.25x- 1.01y + 2.1z = -3.72 ; 2.12x + 3.15 y - 2.2z = -18.25 ; -x + 2.22 y + 4.15 z = -6 Chapter 3: Roots of Equations 99
- 110. 3.24. The dynamicresponse of an underdamped second-order system, which is a common mathematical model in control system analysis, to a unit step function is given by [ 14 ] c t ( ) 1 1 1 ζ 2 − e ζ − ωn ⋅ t ⋅ ⋅ cos ωn 1 ζ 2 − ⋅ t ⋅ ϕ − ⎛ ⎝ ⎞ ⎠ ⎡ ⎣ ⎤ ⎦ ⋅ − = where ζ is the system damping ratio, ω n is the undamped natural frequency, and φ is ϕ atan ζ 1 ζ 2 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = A precise analytical relationship between the damping ratio and the rise time which is the time required for the response to go from 10 % of the final value to 90 % ( that is, from 0.1 to 0.9 in this case ) cannot be determined . However, using Mathcad's root function, the normalized rise time which is ω n times the rise time, can be found for a specified damping ratio ζ by solving for the values of ω n t that yield c(t) = 0.9 and c(t)= 0.1 . Subtracting these two values yields the normalized rise time for the value of ζ considered . Using a range of damping ratios from 0.1, 0.2, 0.3... .. 0.9 , construct a plot depicting normalized rise time as a function of the damping ratio . 3.25. The frequency equation for the free undamped longitudinal vibration of a slender bar of length L and mass m with one end fixed and with the other end carrying a mass M is given by [ 21 ] β tan β ( ) ⋅ α = where β ω L c0 ⋅ = c0 E ρ = Figure P 3.25. 100 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 111. α m M = in which Eis Young's modulus for the material of the bar, ρ is the mass per unit volume and ω is the natural frequency . For a steel bar with mass ratio α = 0.70, E = 30 x 10 6 psi , L = 98 in., and ρ = 7.35x 10 -4 lb-sec 2 / in 4 , determine the first three natural frequencies. 3.26. For a uniform beam of length L with one end fixed and the other simply supported, the natural frequencies of transverse vibration are given by the relationship [ 21 ] tan k L ⋅ ( ) tanh k L ⋅ ( ) = where k 2 ω ρ A ⋅ E I ⋅ ⋅ = Figure P 3.26 in which ω is the natural frequency , ρ is the mass density of the beam material, A is the area of cross section of the beam , E is Young's modulus, and I is the moment of inertia of the beam cross section. Determine the first three natural frequencies of a 36 in. long steel beam whose area of cross section A is 12.50 in 2 , moment of inertia I is 12.75 in 4, and ρ = 7.35x 10 -4 lb-sec 2 / in 4 . E for steel is 30 x 10 6 psi. Chapter 3: Roots of Equations 101
- 112. NOTES: 102 AN INTRODUCTIONTO NUMERICAL METHODS USING MATHCAD
- 113. C H AP T E R 4 MATRICES AND LINEAR ALGEBRA 4.1 BASIC MATRIX OPERATIONS Basic matrix operations are addition, subtraction and multiplication. Determination of the inverse of a matrix is linked with the concepts of transpose, minor, cofactor and adjoint. Addition and Subtraction: Two matrices with the same number of rows and columns can be added by adding the corresponding terms. Similarly, subtraction is done by subtracting the corresponding terms Multiplication: If the product of two matrices A and B is C , the element Cij of C is obtained by multiplying the elements of the ith row of A by the elements of the j th column of B in accordance with Cij k aik bkj ⋅ ( ) ∑ = (4.1) Transpose: The transpose of a matrix is one in which the rows and the columns are interchanged Minor: The minor M ij of the element a ij is the determinant obtained by deleting the i th row and the j th column from the given determinant. If A 2 5 1 3 7 6 4 9 8 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := , then, M12 5 1 9 8 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := , that is, M12 31 = Cofactor: The cofactor C ij of the element a ij is given by: Cij 1 − ( ) i j + Mij ⋅ = (4.2) Chapter 4: Matrices and Linear Algebra 103
- 114. For the abovecase, C12 1 − ( ) 3 31 ( ) ⋅ := , that is, C12 31 − = Adjoint Matrix: The adjoint of a square matrix is the transpose of the matrix of its cofactors. Inverse of a Matrix: The inverse A -1 of a matrix A must satisfy the relationship A A -1 = I , (4.3) where I is the Identity or the Unit matrix which is a matrix with "1" along the diagonal and zero elsewhere. The inverse A -1 is calculated as follows A -1 = [adjoint (A)] / Det (A) , (4.4) where Det (A) is the determinant of the given matrix, which must be non-zero for the inverse to exist. Example 4.1 Determine the inverse of the following matrix A A 1 1 1 2 3 0 3 2 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := The determinant of this matrix is A 1 − = The minors of A are M11 3 0 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M12 1 1 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M13 1 1 3 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M21 2 0 3 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M22 1 1 3 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M23 1 1 2 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := 104 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 115. M31 2 3 3 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M32 1 1 3 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := M33 1 1 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := Thatis, M11 12 = M12 2 = M13 3 − = M21 8 = M22 1 = M23 2 − = M31 5 − = M32 1 − = M33 1 = Multiplying the minors by 1 − ( ) i j + will yield the cofactor matrix C as : C 12 8 − 5 − 2 − 1 1 3 − 2 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := The adjoint matrix is the transpose of the cofactor matrix, and the inverse of the matrix A will now be the adjoint matrix divided by the determinant of A. InverseA adj A ( ) A = or, InverseA 12 2 − 3 − 8 − 1 2 5 − 1 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 1 − := InverseA 12 − 2 3 8 1 − 2 − 5 1 − 1 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The correctness of the result can now be verified as follows. Multiplying A by its inverse obtained above gives 1 1 1 2 3 0 3 2 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ 12 − 2 3 8 1 − 2 − 5 1 − 1 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⋅ 1 0 0 0 1 0 0 0 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 4.2 USE OF MATHCAD IN PERFORMING MATRIX OPERATIONS To put in a matrix in your document, choose "Matrix" from the "Insert" menu, then enter Chapter 4: Matrices and Linear Algebra 105
- 116. the number ofrows and columns, click on "Insert" and start putting in the numbers in the placeholders, as shown in Figure 4.1. How to perform matrix operations will be clear from the following examples. Figure 4.1. Matrix operations in Mathcad 106 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 117. Given the twomatrices A and B: A 1 1 7 0 3 8 2 2 9 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := B 3 1 1 2 2 5 1 3 9 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Addition: A B + 4 2 8 2 5 13 3 5 18 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Subtraction: A B − 2 − 0 6 2 − 1 3 1 1 − 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Multiplication: <--Use the asterisk ( * ) to multiply two matrices A B ⋅ 5 8 38 12 18 75 19 28 112 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = <-- Order in which multiplication is done is important. Note that [A] [B] is not equal to [B][A] B A ⋅ 12 24 69 14 30 87 19 33 93 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Scalar subtraction: A 5 − 4 − 4 − 2 5 − 2 − 3 3 − 3 − 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Chapter 4: Matrices and Linear Algebra 107
- 118. Scalar multiplication: (Use the asterisk "*" ) A 5 ⋅ 5 5 35 0 15 40 10 10 45 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Matrix inverse: Use : ^ -1 A 1 − 0.733 − 0.333 − 0.867 1.067 − 0.333 0.533 0.4 0 0.2 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Determinant of A : A 15 − = Transpose of A: Use [Ctrl]1 A T 1 0 2 1 3 2 7 8 9 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Matrix subscripts: Use " [ " to put in subscripts. Note that, in this case, subscripts will go from 0 to 2. because, in Mathcad, vector and matrix elements are ordinarily numbered starting with row zero and column zero. To make " one " the " ORIGIN " , go to the Tools menu, choose Worksheet Options , as shown in Figure 4.2 , and change the array origin from 0 to 1. 4.3 SOLUTION OF LINEAR ALGEBRAIC EQUATIONS BY USING THE INVERSE For a system of linear algebraic equations that can be written in matrix form as [A]{X} = {B} (4.5) where the matrix [A] and the column vector {B} contain known constants, the inverse of the matrix [A] can be utilized in the determination of the solution vector {X} as follows {X} = [A] -1 {B} (4.6) 108 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 119. Figure 4.2. Changingthe array origin in Mathcad Chapter 4: Matrices and Linear Algebra 109
- 120. Example 4.2 Solve: 2x+ 4 y + z = -11 -x + 3 y -2 z = -16 2x - 3y + 5 z = 21 Putting the given equations in the form [A] {X}= {B}, where A 2 1 − 2 4 3 3 − 1 2 − 5 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := B 11 − 16 − 21 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := The inverse of [A] can be easily obtained as A 1 − 0.474 0.053 0.158 − 1.211 − 0.421 0.737 0.579 − 0.158 0.526 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = and [A] multiplied by [A]-1 will yield the identity matrix as shown A A 1 − ⋅ 1 0 0 0 1 0 0 0 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The solution can then be generated by performing the matrix operation X A 1 − B ⋅ := yielding X 2 4 − 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = 4.4 SOLUTION OF LINEAR ALGEBRAIC EQUATIONS BY CRAMER'S RULE As an alternative to the above procedure, the solution {X} to the matrix equation (Equation 4.5) 110 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 121. [A] {X} ={B} can be computed using Cramer's rule as shown below Letting A A11 A21 A31 A12 A22 A32 A13 A23 A33 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = B B1 B2 B3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = the solution {X} can be computed using X1 B1 B2 B3 A12 A22 A32 A13 A23 A33 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ A11 A21 A31 A12 A22 A32 A13 A23 A33 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = X2 A11 A21 A31 B1 B2 B3 A13 A23 A33 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ A11 A21 A31 A12 A22 A32 A13 A23 A33 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = (4.7) X3 A11 A21 A31 A12 A22 A32 B1 B2 B3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ A11 A21 A31 A12 A22 A32 A13 A23 A33 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = For Example 4.2, A 2 1 − 2 4 3 3 − 1 2 − 5 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := B 11 − 16 − 21 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := D A := D 19 = Chapter 4: Matrices and Linear Algebra 111
- 122. , x B1 B2 B3 A1 2 , A2 2 , A32 , A1 3 , A2 3 , A3 3 , ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ D := y A1 1 , A2 1 , A3 1 , B1 B2 B3 A1 3 , A2 3 , A3 3 , ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ D := z A1 1 , A2 1 , A3 1 , A1 2 , A2 2 , A3 2 , B1 B2 B3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ D := where the matrix elements have been specified in the format required by Mathcad. These calculations result in x 2 = y 4 − = z 1 = 4.5 SOLUTION OF LINEAR ALGEBRAIC EQUATIONS USING THE FUNCTION lsolve The lsolve function in Mathcad can be used to solve a system of linear algebraic equations . Given: [A] {X}= {B}, the function lsolve (A,B) will return the solution vector {X}, provided the matrix [A] is not singular or nearly singular. For Example 4.2, A 2 1 − 2 4 3 3 − 1 2 − 5 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = B 11 − 16 − 21 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = X lsolve A B , ( ) := gives X 2 4 − 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Example 4.3 Application of Matrices to Electrical Circuit Analysis. Formulate a set of linear equations that represent the relationship between voltage, current and resistances for the circuit shown in Figure 4.3. Solve the simultaneous equations generated using (1) the inverse (2) Cramer's Rule and (3) the Mathcad function lsolve 112 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 123. Figure 4.3. Electricalcircuit example The governing equations are: i1 + i2 = i3 7 i1 + 3 i3 = 6 7 i1 - 5 i2 = 6 These must be put in the form: [A] {X} = {B} and solved for {X} Using the inverse A 1 7 7 1 0 5 − 1 − 3 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := B 0 6 6 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := <--These are the currents i1, i2 and i3 X A 1 − B ⋅ := X 0.6761 0.2535 − 0.4225 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Using Cramer's Rule L 0 6 6 1 0 5 − 1 − 3 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := M 1 7 7 0 6 6 1 − 3 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := N 1 7 7 1 0 5 − 0 6 6 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := i1 L A := i2 M A := i3 N A := i1 0.676 = i2 0.254 − = i3 0.4225 = Chapter 4: Matrices and Linear Algebra 113
- 124. Using Mathcad Functionlsolve X lsolve A B , ( ) := X 0.676 0.254 − 0.423 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = <--These are the currents i1, i2 and i3 4.6 THE EIGENVALUE PROBLEM Eigenvalue problems occur in a variety of physical situations. Some examples of eigenvalue problems are given below. 1. Determination of natural frequencies and mode shapes of oscillating systems. 2. Computation of principal stresses and principal directions 3. Computation of principal moments of inertia and principal axes. 4. Buckling of structures. 5. Oscillations of electrical networks. The eigenvalue problem can be mathematically stated as [A] {X} = λ {X} (4.8) where [A] is a known square matrix, {X} is a column vector of unknowns, and λ is an unknown scalar quantity. The solution of the eigenvalue problem involves the determination of vectors {X} and associated constants λ satisfying equation (4.8). These vectors {X} are termed eigenvectors. With each eigenvector is associated a constant λ which is called an eigenvalue. Equation (4.8) can be written in the form [A- λ I ] {X} = {0} (4.9) A trivial solution to the above is {X} = 0 which is not of interest to us. The condition for the existence of a non-trivial solution is det ( A- λ I ) = 0 (4.10) 114 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 125. Expansion of theabove determinant produces a polynomial in λ. For instance, if [A] is a 3x 3 matrix, then Equation (4.10) would give a cubic in λ. This polynomial is called the characteristic polynomial and its roots are called eigenvalues. Associated with each eigenvalue is a solution vector {X} called the eigenvector. It can be shown that the adjoint of matrix (A- λiI) contains columns, each of which is the ith eigenvector {X} i associated with the eigenvalue λ i multiplied by an arbitrary constant. Thus, to determine an eigenvector corresponding to an eigenvalue λi , compute the adjoint of (A- λiI) and choose any column of it as the eigenvector. Example 4.4 Compute the eigenvalues and eigenvectors of 17 45 6 − 16 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ A 17 45 6 − 16 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := The [A- λI] matrix can be written as A λ I ⋅ − 17 λ − 45 6 − 16 − λ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = A non-trivial solution is obtained by setting the determinant of [ A - λ I ] equal to zero 17 λ − 45 6 − 16 − λ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 0 = which gives λ 2 - λ - 2 = 0, or, λ 1 = 2, λ 2 = -1. The adjoint of the matrix (A- λ I ), in this case, is 16 − λ − 45 − 6 17 λ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ For λ 1 = 2, the corresponding eigenvector is the first or second column of the above adjoint matrix with λ = λ 1 which gives Chapter 4: Matrices and Linear Algebra 115
- 126. {X} 1 = 18 − 45 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ or 2 5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ or 0.4 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ as the firsteigenvector. Similarly, the second eigenvector is obtained by putting λ = λ 2 in any one column of the adjoint matrix which gives {X} 2 = 15 − 45 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ or, 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ or 0.333 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.7 SOLVING THE EIGENVALUE PROBLEM WITH MATHCAD The Mathcad function eigenvals(A) returns a vector containing the eigenvalues of the matrix A Let A 1 1 1 1 2 0 1 2 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Then, eigenvals A ( ) 3.879 0.468 1.653 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The Mathcad function eigenvecs(A) returns a matrix containing normalized eigenvectors of the square matrix A. The nth column of this matrix is an eigenvector corresponding to the nth eigenvalue returned by eigenvals(A). eigenvecs A ( ) 0.433 − 0.755 − 0.493 − 0.923 − 0.127 0.364 0.535 − 0.746 − 0.397 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The Mathcad function eigenvec(A,z) returns a matrix containing the normalized eigenvector corresponding to the eigenvalue z of the square matrix A. For example, the eigenvector associated with the eigenvalue 0.468 is 116 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 127. eigenvec A 0.468 , () 0.923 0.127 − 0.364 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = For Example 4.4, the matrix given was A 17 45 6 − 16 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := The eigenvalues and eigenvectors obtained with Mathcad are eigenvals A ( ) 2 1 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = eigenvecs A ( ) 0.371 0.928 0.316 0.949 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = or 0.4 1 0.333 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.8 APPLICATION OF THE EIGENVALUE PROBLEM TO VIBRATION ENGINEERING Figure 4.4. Mathematical model of two-story building Consider the vibration problem of a two -story building which is mathematically modeled as the two degree of freedom system shown in Figure 4.4, in which the building is represented by lumped masses m and 2m and springs k and 2k . After drawing suitable free body diagrams of the two masses m and 2m, the governing Chapter 4: Matrices and Linear Algebra 117
- 128. differential equations ofmotion can be seen to be 2 m d2x1 /dt2 + 3 k x1 - k x2 = 0 m d2x2 /dt2 +k ( x2 - x1 ) = 0 (4.11) Expressing the above equations in matrix form as .. (4.12) [M] {x} + [K ] {x} = {0} where [M] is the mass or inertia matrix , and [K] is the stiffness matrix, and assuming that {x} = {A}sin ωt (4.13) we obtain a set of linear homogeneous algebraic equations that can be written in matrix form as (4.14) ([K]- ω2 [M ] ) {A} = 0 For a non-trivial solution, the determinant | K- ω2 M | must be zero. The problem can also be posed slightly differently in the familiar form |A-λI| = 0 (4.15) where the λ 's ( λ = ω2) are the eigenvalues of the matrix [A] , which, in this case, is [A] = [M] -1 [K] (4.16) To find the natural frequencies and mode shapes in this case, we need to determine the eigenvalues and eigenvectors of the matrix [A]. The square roots of the eigenvalues will then give the natural frequencies and the corresponding eigenvectors will be the mode shapes. Letting m = 1, and k =1 in this problem for convenience m 1 := k 1 := Then, the mass and stiffness matrices M 2 m ⋅ 0 0 m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := K 3 k ⋅ k − k − k ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := 118 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 129. give M 2 0 0 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = K 3 1 − 1 − 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = The [A]matrix, which is called the dynamic matrix in the language of vibrations is , A M 1 − K ⋅ := which gives A 1.5 1 − 0.5 − 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = The eigenvalues of [A] are obtained from setting the determinant of [A-λI] equal to zero, as follows A λ I ⋅ − 0 = This yields 1.5 λ − ( ) 1 λ − ( ) ⋅ 0.5 − 0 = , or, λ1 = 0.5 , and λ2 = 2 The adjoint of the matrix [A-λI] is 1 λ − ( ) 1 0.5 1.5 λ − ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ In Vibration Engineering, eigenvectors are termed as natural modes or mode shapes, and these are simply the configurations that the system takes when vibrating at a natural frequency. Because the system considered here is a two-degree-of-freedom system, there will be two natural frequencies and correspondingly two modes. While the natural frequencies are simply the square roots of the eigenvalues, the associated eigenvectors or modeshapes can be determined by putting λ = λ1(for the first mode ) and λ = λ2 (for the second mode ) in either column of the adjoint of [A- λ I] This procedure generates the following mode shapes FirstMode 0.5 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := and SecondMode 1 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := The determination of eigenvalues and eigenvectors using Mathcad is done below. Chapter 4: Matrices and Linear Algebra 119
- 130. lambda eigenvals A () := lambda 2 0.5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = By default, Mathcad arrays begin at element zero. To change this to begin at element 1, choose "Worksheet Options" from the " Tools" menu, click on the "Built-In Variables" tab and set "ORIGIN" to 1. i 1 2 .. := ωi lambdai := ωi 1.414 0.707 = In Vibration Engineering, the lowest natural frequency is always considered as the first or fundamental natural frequency. In this case, the natural frequencies are ω2 0.707 = <--First natural frequency in rads/sec ω1 1.414 = <--Second natural frequency in rads/sec The eigenvectors or mode shapes can be determined, using the Mathcad function eigenvecs as follows X eigenvecs A ( ) := This gives the following matrix of eigenvectors which is referred to as the modal matrix in Vibrations language X 0.707 0.707 − 0.447 0.894 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <- Modal Matrix A close inspection of the analysis will show that it is not possible to obtain unique values for the eigenvector components. Therefore, it is traditional to select a value of unity for one of the components of an eigenvector and compute the other one accordingly . This is called normalization and should be clear from the calculation shown below. The first natural mode, corresponding to the first natural frequency is 120 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 131. FirstMode X1 2 , 0.8944 X2 2 , 0.8944 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ :=FirstMode 0.5 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = while the second natural mode, corresponding to the second natural frequency is SecondMode X1 1 , 0.7071 − X2 1 , 0.7071 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := SecondMode 1 − 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = These mode shapes are sketched in Figure 4.5. Figure 4.5. Modes shapes of the two-story building Example 4.5. Investigate the free vibration of a three-story building mathematically modeled as the three-degree-of-freedom system shown in Figure 4.6. Chapter 4: Matrices and Linear Algebra 121
- 132. Figure 4.6. Mathematicalmodel of three-story building The equations of motion of the mathematical model resorted to can be shown to be m1 2 t x1 d d 2 ⋅ k1 x1 ⋅ + k2 x2 x1 − ( ) ⋅ − 0 = m2 2 t x2 d d 2 ⋅ k2 x2 x1 − ( ) ⋅ + k3 x3 x2 − ( ) ⋅ − 0 = (4.17) m3 2 t x3 d d 2 ⋅ k3 x3 x2 − ( ) ⋅ + 0 = These can be put in the following matrix format [M] d2 {x} /dt2 + {K] { x } = { 0} (4.18) where 122 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 133. [M] = m1 0 0 0 m2 0 0 0 m3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ (4.19) [K] = k1k2 + ( ) k2 − 0 k2 − k2 k3 + ( ) k3 − 0 k3 − k3 ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ Letting m1 = 2x 10 6 kg , m2 = 14x 10 5 kg , m3 = 7x10 5 kg; and k1 = 12x10 8 N/ m, k2= 8x 10 8 N/m , k3 = 4x 10 8 N/ m , the [A] matrix can be computed, using Mathcad , as follows m1 2 10 6 ⋅ := m2 14 10 5 ⋅ := m3 7 10 5 ⋅ := k1 12 10 8 ⋅ := k2 8 10 8 ⋅ := k3 4 10 8 ⋅ := M m1 0 0 0 m2 0 0 0 m3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := K k1 k2 + ( ) k2 − 0 k2 − k2 k3 + ( ) k3 − 0 k3 − k3 ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := M 2 10 6 ⋅ 0 0 0 1.4 10 6 ⋅ 0 0 0 0.7 10 6 ⋅ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := K 2 10 9 × 8 − 10 8 × 0 8 − 10 8 × 1.2 10 9 × 4 − 10 8 × 0 4 − 10 8 × 4 10 8 × ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = A M 1 − K ⋅ := A 1 10 3 × 571.429 − 0 400 − 857.143 571.429 − 0 285.714 − 571.429 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The eigenvalues and the eigenvectors are lambda eigenvals A ( ) := lambda 1.495 10 3 × 761.106 172.146 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Chapter 4: Matrices and Linear Algebra 123
- 134. Note that Mathcadhas found the highest eigenvalue first. The natural frequencies in Hz and the corresponding eigenvectors are computed below. i 1 2 , 3 .. := fi lambdai 2 π ⋅ := f1 6.154 = f2 4.391 = f3 2.088 = (Hz) X eigenvecs A ( ) := X 0.566 − 0.701 0.434 − 0.467 − 0.279 − 0.839 0.267 0.552 0.79 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The normalized eigenvectors or mode shapes can be determined as shown. Xnorm X1 1 , X1 1 , X2 1 , X1 1 , X3 1 , X1 1 , X1 2 , X1 2 , X2 2 , X1 2 , X3 2 , X1 2 , X1 3 , X1 3 , X2 3 , X1 3 , X3 3 , X1 3 , ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Xnorm 1 1.238 − 0.766 1 0.597 1.799 − 1 2.07 2.962 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Mode No: 3 2 1 Figure 4.7 shows the natural modes of the building as it vibrates at its natural frequencies. Figure 4.7. Mode shapes of three-story building 124 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 135. 4.9 APPLICATION OFTHE EIGENVALUE PROBLEM TO STRESS ANALYSIS- DETERMINATION OF PRINCIPAL STRESSES AND PRINCIPAL DIRECTIONS If [σ ] is the state of stress at a point in a material given by the stress components σxx , σyy, σzz, σxy , σyz and σzx , the principal stresses are the eigenvalues λ of the stress matrix and the principal directions, which are the x, y and z components of the unit vectors along the normals to the principal planes , are the eigenvectors {η} of the stress matrix. Thus , the eigenvalue problem in this case is stated as [σ] - λ [ I ]) { η } = {0} (4.20) where the eigenvalues λ and the eigenvectors η are to be determined. Example 4.6 The two-dimensional state of stress in a mechanical element, which is shown in Figure 4.8, is given by the components: σ xx = 80 MPa, σyy = 0 , and σ xy = 50 MPa. Determine the principal stresses and their associated directions. Figure 4.8. Two-dimensional state of stress Because the stress matrix is symmetrical, σ yx must be equal to σ xy. Thus, the matrix of stresses is Chapter 4: Matrices and Linear Algebra 125
- 136. σ 80 50 50 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := (MPa) The principalstresses are the eigenvalues of the given stress matrix while the principal directions are the associated eigenvectors. These are. prinstress eigenvals σ ( ) := prinstress 104.031 24.031 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = (MPa) prindir eigenvecs σ ( ) := prindir 0.901 0.433 0.433 − 0.901 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = The principal direction associated with the maximum principal stress ( 104.031 MPa ) can be found in the form of a unit vector along the normal to the principal plane. Its x and y components , prindir1, are computed below. Figure 4.9 shows the principal stresses on the element and the associated principal directions. prindir1 eigenvec σ 104.031 , ( ) := prindir1 0.901 0.433 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Figure 4.9. Principal stresses and principal directions for two-dimensional stress example Example 4.7 The three-dimensional state of stress in a mechanical element, which is shown in Figure 4.10, is given by: σxx= 1000 psi, σyy = -1000 psi, σxy = 500 psi , σ yz = -200 psi, σ zx =100 psi , σ zz = 250 psi Determine the principal stresses and their associated directions. 126 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 137. Figure 4.10. Three-dimensionalstate of stress The stress matrix can be written as (psi) σ 1000 500 100 500 1000 − 200 − 100 200 − 250 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := The principal stresses and principal directions can be determined using Mathcad as shown below psi prinstress eigenvals σ ( ) := prinstress 1.152 − 10 3 × 1.121 10 3 × 280.707 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = prindir eigenvecs σ ( ) := prindir 0.23 − 0.961 0.154 0.973 0.224 0.06 0.024 − 0.163 − 0.986 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = If only the principal direction associated with one of the principal stresses , say 280.707 psi, is required, it can be found as follows. Chapter 4: Matrices and Linear Algebra 127
- 138. prindir2 eigenvec σ280.707 , ( ) := These are the the x and y components of the unit vector along the normal to the principal plane associated with the principal stress, 280.707 psi prindir2 0.024 − 0.163 − 0.986 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = <----- The principal stresses and their directions are shown in Figure 4.11. Note the orthogonality of the principal stress directions with respect to one another. acos 0.9730 ( ) 180 π ⋅ 13.344 = deg acos 0.224 ( ) 180 π ⋅ 77.056 = deg acos 0.06 ( ) 180 π ⋅ 86.56 = deg Figure 4.11. Principal stresses and principal directions for three-dimensional state of stress. 4.10 REPEATED ROOTS IN THE DETERMINANTAL EQUATION When equal roots are encountered in the determinantal equation, the associated eigenvectors are not unique and thus a linear combination of these eigenvectors may also be 128 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 139. considered as eigenvectors.The following examples will illustrate this situation. Example 4.8 For the given matrix [A] A 0 1 − 1 1 − 0 1 1 1 0 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := the eigenvalues and eigenvectors are found using Mathcad as <-- double root at λ = 1 lambda eigenvals A ( ) := lambda 1 2 − 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = X eigenvecs A ( ) := X 0.816 0.408 − 0.408 0.577 − 0.577 − 0.577 0.226 − 0.793 0.566 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = It is clear that the first and third eigenvectors corresponding to the repeated root λ = 1 are arbitrary, since they are both different , while the one corresponding to λ = - 2 is 1 − 1 − 1 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Example 4.9 The following problem is, clearly, a case of repeated roots with a double root at λ = 1 Chapter 4: Matrices and Linear Algebra 129
- 140. A 1 0 0 1 − 1 0 0 1 1 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := The eigenvalues andeigenvectors are lambda eigenvals A ( ) := lambda 1 1 1 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = X eigenvecs A ( ) := X 1 0 0 1 0 0 0.218 − 0.436 − 0.873 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Note that the first and second eigenvectors corresponding to the repeated (double) root at λ = 1 are, in fact , arbitrary since they have two components that are zero, while the one corresponding to λ = -1 is the third one which is 1 − 2 − 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Example 4.10 For the following matrix [A] A 14 − 0 1 1 2 0 0 0 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := the eigenvalues and eigenvectors are 130 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 141. eigenvals A ( ) 2 14 − 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ =eigenvecs A ( ) 0 0 1 0.998 0 0.062 − 0 7.105 10 15 − × 1 − ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = This is, again, a case of repeated roots in which the first and third eigenvectors corresponding to the repeated roots are arbitrary. Thus, there is no unique eigenvector correponding to the repeated root . 4.11 SOLUTION OF NONLINEAR SIMULTANEOUS EQUATIONS There are no direct methods available for solving nonlinear simultaneous equations like you have in the case of linear algebraic equations. In general, iterative procedures must be resorted to, requiring initial guesses as a starting point. The following procedure utilizes an adaptation of the Taylor series approach in obtaining a solution to a set of nonlinear equations. Consider the following two nonlinear equations: f(x,y)= 0 ; g(x,y) = 0 (4.21) which have the exact solution: x = x r and y = y r . If x = x i and y = y i are approximate solutions, a Taylor series approximation about ( x i , y i ) gives f( x r , y r ) = f(x i , y i ) + h ( partial deriv of f (x,y) with respect to x at (x i ,y i) ) + j ( partial deriv of f(x,y) with respect to y at (x i , y i) ) + ... = 0 g (x r , y r ) = g( xi , y i ) + h ( partial deriv of g (x,y ) with respect to x at (xi , yi ) ) + j ( partial deriv of g (x, y ) with respect to y at ( xi , yi ) ) + ....... = 0 (4.22) where x r = x i + h , and y r = y i + j . (4.23) If only linear terms are retained in the Taylor series, two linear equations are generated in the two unknowns h and j, where h and j are approximations to ( x r - x i ) and ( yr - y i ) . Defining a, b, c and d as the partial derivatives in Equations (4.22) and (4.23), p as f(x i , y i ) and q as g( xi , y i ) , these equations become Chapter 4: Matrices and Linear Algebra 131
- 142. ah + bj= -p and ch + d j = -q (4.24) When these two simultaneous equations are solved for h and j , we generate new improved values for x and y, say x i+1 and y i+1 , where , x i + 1 = x i + h and y i+1 = yi + j . These new improved values replace the old values x i and y i in the next step in an effort to obtain still better solutions. This procedure is repeated until convergence occurs. The following example will clarify the procedure. Example 4.11 Determine the solution to: f(x,y) = x2 + 3 y 2 - 54 = 0 , g(x, y ) = -4 x 2 + 3x y - 2 y + 15 = 0 Iteration Number 1 f x y , ( ) x 2 3 y 2 ⋅ + 54 − := g x y , ( ) 4 − x 2 ⋅ 3 x ⋅ y ⋅ + 2 y ⋅ − 15 + := partfx x y , ( ) 2 x ⋅ := partfy x y , ( ) 6 y ⋅ := partgx x y , ( ) 8 − x ⋅ 3 y ⋅ + := partgy x y , ( ) 3 x ⋅ 2 − := where the above are partial derivatives of f(x,y) and g(x,y) with respect to x and y. xold 1.5 := yold 2.0 := <-- Assumed starting values of solution p f xold yold , ( ) := q g xold yold , ( ) := a partfx xold yold , ( ) := b partfy xold yold , ( ) := c partgx xold yold , ( ) := d partgy xold yold , ( ) := a 3 = b 12 = c 6 − = d 2.5 = p 39.75 − = q 11 = Now solve the equations obtained from the linear Taylor series approximations, namely a h + b j = -p and c h + d j = -q in the form: [A] {H} = {P} using lsolve as follows A a c b d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := P p − q − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := H lsolve A P , ( ) := H 2.91 2.585 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h H1 := j H2 := h 2.91 = j 2.585 = xnew xold h + := ynew yold j + := xnew 4.41 = ynew 4.585 = 132 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 143. Iteration Number 2 xoldxnew := yold ynew := p f xold yold , ( ) := q g xold yold , ( ) := a partfx xold yold , ( ) := b partfy xold yold , ( ) := c partgx xold yold , ( ) := d partgy xold yold , ( ) := a 8.821 = b 27.509 = c 21.528 − = d 11.231 = p 28.516 = q 11.312 − = H lsolve A P , ( ) := A a c b d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := P p − q − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := H 2.91 2.585 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = h H1 := j H2 := h 2.91 = j 2.585 = xnew xold h + := ynew yold j + := xnew 7.321 = ynew 7.17 = Following the same procedure as above, new improved values for x and y can be generated in iterations 3, 4 and 5 as Iteration Number 3: h 0.1166 − := j 0.0728 − := --> xnew 3.3804 := ynew 3.7684 := Iteration Number 4: h 2.1741 − 10 3 − ⋅ := j 6.5547 − 10 4 − ⋅ := --> xnew 3.3782 := ynew 3.7677 := Iteration Number 5: h 2.1741 − 10 3 − ⋅ := j 6.5547 − 10 4 − ⋅ := --> xnew 3.3782 := ynew 3.7677 := f xnew ynew , ( ) 1.075 − 10 3 − × = a check to see if a solution has indeed been achieved <--- g xnew ynew , ( ) 2.085 − 10 4 − × = Thus, it is clear that convergence has occurred in the fourth iteration because the values of the functions f ( x, y ) and g (x, y ) are zero and the answers (xnew and ynew) obtained in the fifth iteration are exactly the same as in the fourth. The solution is : x= 3.3782 and y = 3.7677. These answers may now be verified by using the given and find commands. x 1 := y 1 := <--- Initial guesses required by Mathcad Given x 2 3 y 2 ⋅ + 54.0 − 0 = 4 − x 2 ⋅ 3 x ⋅ y ⋅ + 2 y ⋅ − 15.0 + 0 = <---- Press "Ctrl=" to type "=" xval yval ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find x y , ( ) := xval 3.378 = yval 3.768 = <--- Solution Chapter 4: Matrices and Linear Algebra 133
- 144. PROBLEMS 4.1. Given thematrix 2 − 1 0 1 − 1 3 5 − 4 3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ find its inverse or else show that it is singular. Do calculations on a calculator. Also check your answers with Mathcad 4.2. Given the matrix 0 1 1 0 12 2 − 1 − 0 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find its inverse or else show that it is singular. Do calculations on a calculator. Also check your answers with Mathcad. 4.3. Solve the following by Cramer's rule: (a) x + y = 4 ; 2x - y = 2 (b) 5x - 3 y = 37 ; -2x + 7 y = - 38 4.4. Solve the following system of equations 8x-y-z =4 ; x +2 y -3 z = 0 ; 2 x -y +4 z =5 Do this problem on a calculator. Check your answers with (1) Mathcad, using the inverse and (2) Mathcad's lsolve function. 4.5. Solve the following system of equations : x + y + z = 6 ; 2 x - y + 3z = 9 ; x + 2 y - z = 2 Do this problem on a calculator. Check your answers with (1) Mathcad, using the inverse and (2) Mathcad's lsolve function. 134 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 145. 4.6. Solve thefollowing system of equations : x + y + 3 z = 96 ; 2 x - y + 7z = 6 ; x + 2 y - 5z = 2 Do this problem on a calculator. Check your answers with (1) Mathcad, using the inverse and (2) Mathcad's lsolve function. 4.7. Solve the following by Cramer's rule: x + y - 3z = 0 ; y - 4z = 0 ; x- y - z =5 Do this problem on a calculator. Check your answers with (1) Mathcad, using Cramer's rule and (2) Mathcad's lsolve function. 4.8. Find the eigenvalues of the matrix 1 2 3 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Corresponding to each eigenvalue, find an eigenvector. Do this problem on a calculator. Check your answers with Mathcad, using the eigenvals, eigenvecs and eigenvec functions 4.9. Find the eigenvalues of the following matrix. Find an eigenvector corresponding to each eigenvalue. 5 − 1 0 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Do this problem on a calculator. Check your answers with Mathcad, using the eigenvals, eigenvecs and eigenvec functions . 4.10. Determine the eigenvalues and eigenvectors of the following matrices. (a) 3 4 4 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ (b) 5 0 0 5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ (c) 6 0 0 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ (d) 4 5 2 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ (d) 10 0 0 10 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ (e) 6 8 2 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Chapter 4: Matrices and Linear Algebra 135
- 146. 4.11. The differentialequations of motion of the spring-mass system shown are: m 2 t x1 d d 2 ⋅ 2 k ⋅ x1 ⋅ k x2 ⋅ − ( ) + 0 = and 2 m ⋅ 2 t x2 d d 2 ⋅ 2 k ⋅ x2 ⋅ k x1 ⋅ − ( ) + 0 = Figure P 4.11 Obtain the natural frequencies ( eigenvalues ) and mode shapes ( eigenvectors) of the system for m= 1 lb--sec 2 / ft and k = 1 lb/ ft 4.12 The two-dimensional state of stress in psi at a point on a machine element is prescribed as: σ 12 4 4 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Obtain the eigenvalues (principal stresses) and their associated directions (eigenvectors). 4.13. For the two pendulums shown which are coupled by means of a spring "k" , the governing differential equations are: m L 2 ⋅ 2 t θ1 d d 2 ⋅ m − g ⋅ L ⋅ θ1 ⋅ k a 2 ⋅ θ1 θ2 − ( ) ⋅ − = and m L 2 ⋅ 2 t θ2 d d 2 ⋅ m − g ⋅ L ⋅ θ2 ⋅ k a 2 ⋅ θ2 θ1 − ( ) ⋅ − = 136 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 147. Determine the naturalfrequencies (eigenvalues) and mode shapes (eigenvectors) of the system for m = 1 lb- sec 2 / ft , L = 1 ft, g= 32.2 ft/ sec 2, k = 1 lb/ ft and a = 0.5 ft Figure P 4.13 4.14. The spring-mass system shown has the following mass and stiffness characteristics: M= 2 m ⋅ 0 0 0 m 0 0 0 2 m ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ K= 4 k ⋅ k − 0 k − 2 k ⋅ k − 0 k − 4 k ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Figure P4.14 Use Mathcad to determine its natural frequencies (eigenvalues) and mode shapes (eigenvectors) if m=1 kg and k= 1N/m Chapter 4: Matrices and Linear Algebra 137
- 148. 4.15. Determine theroots of the nonlinear simultaneous equations f(x,y) = x 2 + y 2 -4.5 = 0 and g(x,y) = x 3 - 1.5 y = 0 using the Taylor series approach. Check your answer using the Given and Find functions of Mathcad 4. 16. Determine the roots of the nonlinear simultaneous equations f(x,y) = x 3 + y 3 - 5.5 = 0 and g(x,y) = x 4 - 3 y2 = 0 using the Taylor series approach. Check your answer using the Given and Find functions of Mathcad . 4.17. A two-degree-of-freedom mathematical model of an automobile suspension system is shown . With the system modeled as a rigid bar of mass " m " and mass moment of inertia " I " connected to springs , the governing differential equations are: W g 2 t x t ( ) d d 2 ⋅ 4000 x t ( ) ⋅ + 2000 θ t ( ) ⋅ + 0 = R eference C .G . x K 1 K 2 L 1 L 2 and I 2 t θ t ( ) d d 2 ⋅ 2000 x t ( ) ⋅ + 65000. θ t ( ) ⋅ + 0 = Figure P 4.17 where "x" represents the linear up and down ( bounce) motion and "θ" represents the angular motion (pitch) of the system . Determine the natural frequencies (eigenvalues ) and mode shapes ( eigenvectors ) if W = 3000 lbs , I = 300 lb-ft-sec 2 , and g = acceleration due to gravity = 32.2 ft / sec 2 138 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 149. 4.18. The governingdifferential equations of motion of the double pendulum shown in terms of angular displacements θ1(t) and θ2(t) are L 2 t θ1 d d 2 ⋅ 2 g ⋅ θ1 ⋅ + g θ2 ⋅ − 0 = and L 2 t θ2 d d 2 ⋅ 2 g ⋅ θ2 ⋅ + 2 g ⋅ θ1 ⋅ − 0 = 1 L L m m x 1 x 2 2 , where "g" is the acceleration due to gravity = 32.2 ft/ sec 2 and L is the length of each pendulum. Using matrix methods , determine the natural frequencies(eigenvalues) and mode-shapes (eigenvectors) of the system for L = 1 ft. Figure P 4.18 4.19. An automobile is math-modeled as the two-degree-of-freedom system shown above in Figure P 4.17. Its up and down linear motion ( bounce ) is represented by the coordinate x(t) while its angular motion ( pitch) is represented by the coordinate θ(τ). The governing differential equations of motion are: m 2 t x d d 2 ⋅ K1 K2 + ( ) x ⋅ K1 L1 ⋅ K2 L2 ⋅ − ( ) θ ⋅ − ⎡ ⎣ ⎤ ⎦ + 0 = JO 2 t θ d d 2 ⋅ K1 L1 ⋅ K2 L2 ⋅ − ( ) x ⋅ K1 L1 2 ⋅ K2 L2 2 ⋅ + ⎛ ⎝ ⎞ ⎠ θ ⋅ + ⎡ ⎣ ⎤ ⎦ + 0 = Determine the natural frequencies (eigenvalues) and mode-shapes (eigenvectors) of an automobile with the following data: m = 100 lb-sec2/ ft , JO = 1600 lb-ft-sec2 , L1 = 5.5 ft , L2 = 4.5 ft, K1= 2600 lbs / ft , K 2 = 2400 lbs/ ft. Chapter 4: Matrices and Linear Algebra 139
- 150. 4.20. A rigidrod of negligible mass and length 2L is pivoted at the middle point and is constrained to move in the vertical plane by springs and masses shown. The governing differential equations of motion of the system are given as: m 2 t x d d 2 ⋅ 2 k ⋅ x ⋅ + k L ⋅ θ ⋅ + 0 = , 4 m ⋅ L 2 ⋅ 2 t θ d d 2 ⋅ k L ⋅ x ⋅ + 2 k ⋅ L 2 ⋅ θ ⋅ + 0 = K K L L (t) 2 m 2 m m K x (t) Figure P 4.20 Determine the natural frequencies (eigenvalues) and mode-shapes (eigenvectors) of the system for m = 1 kg, L = 1 meter, and k = 1 N/m . 4.21. The governing differential equations of motion of the double pendulum shown in Figure P 4.18 in terms of coordinates of coordinates x1(t) and x2 (t) are L 2 t x1 d d 2 ⋅ 3 g ⋅ x1 ⋅ + g x2 ⋅ − 0 = , L 2 t x2 d d 2 ⋅ g x2 ⋅ + g x1 ⋅ − 0 = where "g" is the acceleration due to gravity = 32.2 ft/ sec 2 and L is the length of each pendulum. Determine the natural frequencies (eigenvalues) and mode-shapes (eigenvectors) of the system for L = 1 ft. 4. 22. Determine the roots of the nonlinear simultaneous equations f(x,y) = x 2 + 2 x y - 64 = 0 and g(x,y) = 3 x 2 + 6 y2 - 264 = 0 using the Taylor series approach. Obtain answers that are correct to two decimal places. Use starting values xold= 3 and yold = 5 . Check your answer using the Given and Find functions of Mathcad . 140 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 151. C H AP T E R 5 NUMERICAL INTERPOLATION In any field where measured or statistical data is involved, a need to interpolate between given data points is bound to exist. Because measured or available data is , typically, not provided in the form of an analytically determined function , the process known as interpolation must be resorted to in order to obtain function values at points other thn the given data points. This process involves the generation of a curve that must pass through the given data points and its use in determining the function value at any intermediate point on this curve. As will be seen in the sections following, a polynomial fitting n data points will be of order (n-1), that is one less than the number of data points given. Thus, four data points will generate a cubic while three data points will give a quadratic and so on. 5.1 LINEAR INTERPOLATION Given two data points ( x k , y k ) and (x k+1 , y k+1 ), as shown in the figure below, the connection between the coordinates of a point ( x, y) lying between the data points can be expressed by the linear relationship y yk xk xk 1 + − ( ) ⋅ y k 1 + ( ) x xk − ( ) ⋅ − xk xk 1 + − = (5.1) 5.2 THE METHOD OF UNDETERMINED COEFFICIENTS Given the data points ( x k , y k ), k= 0...n, it is required to find an interpolating polynomial P n ( x ) such that the following constraint equations are satisfied Pn x0 ( ) y0 = Chapter 5: Numerical Interpolation 141
- 152. Pn x1 ( )y1 = (5.2) Pn xn ( ) yn = Assuming the interpolating polynomial P n ( x ) is of the form Pn x ( ) a0 a1 x ⋅ + a2 x 2 ⋅ + ............ + an x n ⋅ + = , (5.3) Equations (5.2) yield the following matrix equation 1 1 1 x0 x1 xn xo ( )2 x1 ( )2 xn ( )2 x0 ( )n x1 ( )n xn ( )n ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ a0 a1 an ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⋅ y0 y1 yn ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = (5.4) Once the equations represented by the matrix equation (5.4) are solved for the n unknowns a0, a1,..... , a n , the interpolating polynomial P n ( x ) satisfying the constraint equations is completely determined. In general, the higher the order of the interpolating polynomial, the more accurate would be the results of the interpolation process. EXAMPLE 5.1 Determine the polynomial f(x) that passes thru (0,2), (1,2.25) and (2,5.82), and compute f(0.5) and f(1.5). The given data is put in as follows. Go to the Insert menu, select Matrix, put in the appropriate number of rows and columns and then fill in the placeholders. 142 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 153. X 0 1 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Y 2 2.25 5.82 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := The orderof the polynomial fitting 3 data points is 2 and the form, thus, is f(x) = a + b x + c x 2 . Therefore, the matrix equation that needs to be solved for the coefficients a, b and c is 1 1 1 0 1 2 0 2 1 2 2 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ a b c ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ ⋅ 2 2.25 5.82 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = Using Cramer's rule, the solution to the above equation can be written as : a = m1/ D , b= m2/D and c = m3/D, where m1 2 2.25 5.82 0 1 2 0 1 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := m2 1 1 1 2 2.25 5.82 0 1 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := m3 1 1 1 0 1 2 2 2.25 5.82 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := D 1 1 1 0 1 2 0 1 4 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := a m1 D := b m2 D := c m3 D := a 2 = b 1.41 − = c 1.66 = Thus, the interpolating polynomial generated is f x ( ) a b x ⋅ + c x 2 ⋅ + := with a 2 = b 1.41 − = c 1.66 = It is plotted in Figure 5.1 below, which gives Chapter 5: Numerical Interpolation 143
- 154. f 0.5 ( )1.71 = f 1.5 ( ) 3.62 = x 0 0.1 , 2 .. := 0 0.5 1 1.5 2 1 2 3 4 5 6 Given data points Interpolation Given data points Interpolation Interpolated function Yi f x ( ) Xi x , Figure 5.1. Interpolated function using undetermined coefficient method 5.3 THE GREGORY-NEWTON INTERPOLATING POLYNOMIAL The form of the nth order Gregory-Newton interpolating polynomial is f x ( ) a1 a2 x x1 − ( ) ⋅ + a3 x x1 − ( ) ⋅ x x2 − ( ) ⋅ + a4 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ + ....... an x x1 − ( ) ⋅ x x2 − ( ) ⋅ .... x xn 1 − − ( ) + an + 1 x x1 − ( ) ⋅ x x2 − ( ) ⋅ .... x xn − ( ) + + ... = (5.5) where the a' s are unknown coefficients that can be determined as was done in the method of undetermined coefficients. That is , f xk ( ) yk = k 1 ..n 1 + , := .. (5.6) where the xk and yk are the given data points that the interpolating polynomial must pass through. 144 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 155. To obtain asecond order polynomial, put x= x1 , x= x 2 and x = x 3 in Equation (5.5) yielding a1 y1 = a2 y2 y1 − x2 x1 − = (5.7) a3 y3 a1 − a2 x3 x1 − ( ) ⋅ − x3 x1 − ( ) x3 x2 − ( ) ⋅ = The interpolating polynomial is then determined as f x ( ) a1 a2 x x1 − ( ) ⋅ + a3 x x1 − ( ) ⋅ x x2 − ( ) ⋅ + = (5.8) which is a second order polynomial passing through the three data points ( x1, y1 ), (x2, y2 ) and (x3, y3 ). For obtaining a third order polynomial, a fourth data point will be needed , and a fourth order polynomial will require the use of a fourth as well as a fifth data point . In general, determining an nth order polynomial , then, will require (n+1 ) data points to be prescribed. Thus, the coefficients a4 , a5 and a6 can be determined using a4 y4 a1 − a2 x4 x1 − ( ) ⋅ − a3 x4 x1 − ( ) ⋅ x4 x2 − ( ) ⋅ − x4 x1 − ( ) x4 x2 − ( ) ⋅ x4 x3 − ( ) ⋅ = (5.9) a5 y5 a1 − a2 x5 x1 − ( ) ⋅ − a3 x5 x1 − ( ) ⋅ x5 x2 − ( ) ⋅ − a4 x5 x1 − ( ) ⋅ x5 x2 − ( ) ⋅ x5 x3 − ( ) ⋅ − x5 x1 − ( ) x5 x2 − ( ) ⋅ x5 x3 − ( ) ⋅ x5 x4 − ( ) ⋅ = (5.10) Chapter 5: Numerical Interpolation 145
- 156. a6 y6 a1 − a2x6 x1 − ( ) ⋅ − a3 x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ − a4 x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ − 1 − ( ) a5 ⋅ x6 x1 − ( ) x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ x6 x4 − ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ⋅ + ... x6 x1 − ( ) x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ x6 x4 − ( ) ⋅ x6 x5 − ( ) ⋅ = (5.11) Example 5.2 - Given three data points: X: 100 200 400 Y: 975 1575 2054 Determine the Gregory -Newton Polynomial f(x) and evaluate f(251) The form of this polynomial will be : y = f(x) = a1 + a2(x-X1 + a3(x-X1)(x-X2) with a1 = Y1 a2 = (Y2-Y1) / (X2-X1) a3 = [ Y3- a1 -a2(X3-X1)] /[ (X3-X1)(X3-X2)] The data is put in in vector form as follows. X 100 200 400 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Y 975 1575 2054 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := and the coefficients a1, a2 and a3 are computed using a1 Y1 := a2 Y2 Y1 − X2 X1 − := a3 Y3 a1 − a2 X3 X1 − ( ) ⋅ − X3 X1 − ( ) X3 X2 − ( ) ⋅ := 146 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 157. a1 975 = a26 = a3 0.012 − = Thus, the interpolating polynomial generated is f x ( ) a1 a2 x X1 − ( ) ⋅ + a3 x X1 − ( ) ⋅ x X2 − ( ) ⋅ + := which is plotted in Figure 5.2 below, giving f 251 ( ) 1.788 10 3 × = x 100 105 , 400 .. := 100 150 200 250 300 350 400 500 1000 1500 2000 2500 Given data points Interpolation Given data points Interpolation Gregory-Newton Interpolating Polynomial x- values Function values Yi f x ( ) Xi x , Figure 5.2. Interpolated function obtained using Gregory-Newton polynomial method Example 5.3 Given: x: 0 100 200 300 400 500 y: 104.2 97.3 48.1 23.7 8.6 5.2 Determine a fifth order Gregory-Newton interpolating polynomial f(x) and evaluate f(150 ), f(350) and f (450). Chapter 5: Numerical Interpolation 147
- 158. Because six datapoints are given, the interpolating polynomial in this case will be of fifth order, and will have the following form f(x) = a1 + a2(x-x1) + a3(x-x1)(x-x2) + a4(x-x1)(x-x2)(x-x3) +a5(x-x1)(x-x2)(x-x3)(x-x4) + a6x-x1)(x-x2)(x-x3)(x-x4)(x-x5) where a1, a2 .... are computed as shown below. x1 0 := x2 100 := x3 200 := x4 300 := x5 400 := x6 500 := y1 104.2 := y2 97.3 := y3 48.1 := y4 23.7 := y5 8.6 := y6 5.2 := a1 y1 := a2 y2 y1 − x2 x1 − := a3 y3 a1 − a2 x3 x1 − ( ) ⋅ − x3 x1 − ( ) x3 x2 − ( ) ⋅ := a4 y4 a1 − a2 x4 x1 − ( ) ⋅ − a3 x4 x1 − ( ) ⋅ x4 x2 − ( ) ⋅ − x4 x1 − ( ) x4 x2 − ( ) ⋅ x4 x3 − ( ) ⋅ := a5 y5 a1 − a2 x5 x1 − ( ) ⋅ − a3 x5 x1 − ( ) ⋅ x5 x2 − ( ) ⋅ − a4 x5 x1 − ( ) ⋅ x5 x2 − ( ) ⋅ x5 x3 − ( ) ⋅ − x5 x1 − ( ) x5 x2 − ( ) ⋅ x5 x3 − ( ) ⋅ x5 x4 − ( ) ⋅ := a6 y6 a1 − a2 x6 x1 − ( ) ⋅ − a3 x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ − a4 x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ − 1 − ( ) a5 ⋅ x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ x6 x4 − ( ) ⋅ + ... x6 x1 − ( ) x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ x6 x4 − ( ) ⋅ x6 x5 − ( ) ⋅ := The above equations generate the following computed coefficients a1 104.2 = a2 0.069 − = a3 2.115 − 10 3 − × = a4 1.118 10 5 − × = a5 3.442 − 10 8 − × = a6 8.375 10 11 − × = which are inserted into the interpolating polynomial f x ( ) a1 a2 x x1 − ( ) ⋅ + a3 x x1 − ( ) ⋅ x x2 − ( ) ⋅ + a4 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ + a5 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ + ... a6 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ x x5 − ( ) ⋅ + ... := 148 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 159. yielding f 150 ( )70.68 = f 350 ( ) 16.435 = f 450 ( ) 1.84 = The interpolated curve f(x) is shown below in Figure 5.3 along with the given data points X 0 100 200 300 400 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Y 104.2 97.3 48.1 23.7 8.6 5.2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := x 0 5 , 500 .. := 0 100 200 300 400 500 0 50 100 150 Given data points Interpolation Given data points Interpolation Interpolated function( Gregory-Newton) x-values Function values Yi f x ( ) Xi x , Figure 5.3. Fifth order Gregory-Newton interpolating polynomial 5.4 INTERPOLATION USING FINITE DIFFERENCES A finite difference table can furnish a way of developing an interpolating polynomial so long as the known values of the independent variable are equally spaced. For an increment Δx in the independent variable x, the first finite difference of the dependent variable y = f(x) is Chapter 5: Numerical Interpolation 149
- 160. Δf = f(x+Δx)-f(x) (5.12) The second finite difference is Δ 2 f = Δ[Δf] =[f(x+2Δx)-f(x+Δx)]-[f(x+Δx)- f(x)] (5.13) Similarly, the (n-1)th finite difference will be Δn f = Δ[Δ n-1 f ] (5.14) The finite differences as derived above are arranged in a table as shown in Table 5.1 which can be used to generate an interpolating polynomial, as will be demonstrated later. TABLE 5.1 Table of finite differences -------------------------------------------------------------------------------------------------------------------------------- x f x ( ) Δf Δ 2 f Δ 3 f Δ n f ------------------------------------------------------------------------------------------------------------------------------- x f x ( ) Δf x ( ) x Δx + f x Δx + ( ) Δ 2 f x ( ) Δf x Δx + ( ) Δ 3 f x ( ) x 2 Δx ⋅ + f x 2 Δx ⋅ + ( ) Δ 2 f x Δx + ( ) Δf x 2 Δx ⋅ + ( ) x 3 Δx ⋅ + f x 3 Δx ⋅ + ( ) ------------------------------------------------------------------------------------------------------------------------------------ 150 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 161. An alternate representationof the above is shown in Table 5.2, in which the independent variable goes from x0 to x3 and successive finite differences up to the fourth are generated TABLE 5.2 Alternate form of finite-difference table ------------------------------------------------------------------------------------------------------------------------------------ xi yi Δyi Δ 2 yi Δ 3 yi Δ 4 yi ---------------------------------------------------------------------------------------------------------------------------------- x0 y0 Δy0 y1 y0 − = x1 y1 Δ 2 y0 Δy1 Δy0 − = Δy1 y2 y1 − = Δ 3 y0 Δ 2 y1 Δ 2 y0 − = x2 y2 Δ 2 y1 Δy2 Δy1 − = Δ 4 y0 Δ 3 y1 Δ 3 y0 − = Δy2 y3 y2 − = Δ 3 y1 Δ 2 y2 Δ 2 y1 − = x3 y3 Δ 2 y2 Δy3 Δy2 − = ------------------------------------------------------------------------------------------------------------------------------------ 5.5 NEWTON'S METHOD UTILIZING FINITE DIFFERENCES Let ( x0, y0 ), (x1, y1 ) , ( x2, y2 )and ( x3, y3 ) be four given data points for which a third degree Newton interpolating polynomial utilizing finite differences has to be determined . Assume the interpolating polynomial to be f x ( ) C0 C1 x x0 − ( ) ⋅ + C2 x x0 − ( ) ⋅ x x1 − ( ) ⋅ + C3 x x0 − ( ) ⋅ x x1 − ( ) ⋅ x x2 − ( ) ⋅ + = (5.15) Chapter 5: Numerical Interpolation 151
- 162. where the C'sare undetermined coefficients that must be computed using the following constraint equations that the polynomial must satisfy . y0 f x0 ( ) = y1 f x1 ( ) = y2 f x2 ( ) = y3 f x3 ( ) = (5.16) which yield y0 C0 = y1 C0 C1 x1 x0 − ( ) ⋅ + = (5.17) y2 C0 C1 x2 x0 − ( ) ⋅ + C2 x2 x0 − ( ) ⋅ x2 x1 − ( ) ⋅ + = y3 C0 C1 x3 x0 − ( ) ⋅ + C2 x3 x0 − ( ) ⋅ x3 x1 − ( ) ⋅ + C3 x3 x0 − ( ) ⋅ x3 x1 − ( ) ⋅ x3 x2 − ( ) ⋅ + = Using the finite differences x1 x0 − Δx = x2 x0 − 2Δx = x3 x0 − 2Δx = (5.18) the undetermined coefficients are obtained as C0 y0 = C1 Δy0 Δx = (5.19) C2 Δ 2 y0 2! Δx 2 ⋅ = C3 Δ 3 y0 3! Δx 3 ⋅ = Defining x x0 − n Δx ⋅ = x x1 − n 1 − ( ) Δx ⋅ = (5.20) x x2 − n 2 − ( ) Δx ⋅ = 152 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 163. where n x x0 − Δx = (5.21) theinterpolating polynomial of equation (5.15) can now be written as f x ( ) f x0 ( ) n Δf x0 ( ) ⋅ + n n 1 − ( ) ⋅ 2! Δ 2 ⋅ f x0 ( ) + n n 1 − ( ) ⋅ n 2 − ( ) ⋅ 3! Δ 3 ⋅ f x0 ( ) + ............... + = (5.22) From the above process, it can be seen that the form of an mth order interpolating polynomial will be f x ( ) f x0 ( ) n Δf x0 ( ) ⋅ + n n 1 − ( ) ⋅ 2! Δ 2 ⋅ f x0 ( ) + n n 1 − ( ) ⋅ n 2 − ( ) ⋅ 3! Δ 3 ⋅ f x0 ( ) + ...... + n n 1 − ( ) ⋅ . n m − 1 + ( ) m! Δ m ⋅ f x0 ( ) + ... = (5.23) Example 5.4 The following data is given on the radiation flux R in gram-calories per square centimeter per day for a certain month in a given year as a function of latitude L (deg N.) L 0 20 40 60 80 R 895 858 722 489 235 Generate the finite difference table and obtain an interpolating polynomial using Newton's method. Estimate the flux R at a latitude of 45 degrees. The following finite difference table can be generated using the given data. Chapter 5: Numerical Interpolation 153
- 164. ________________________________________________ Lat R deltaR del 2 R del 3 R del 4 R ____________________________________________________ 0 895 -37 20 858 -99 -136 2 40 722 -97 74 -233 76 60 489 -21 -254 80 235 ____________________________________________________ The interpolating poly can now be written as: R(L) = R(L0) + n del R(L0) + (1/2) n(n-1) del 2 R(L0) + (1/6) n(n-1)(n-2)del 3 R ( L0) + (1/24) n(n-1)(n-2)(n-3) del 4 R(L0) where R(L0)= 895; del R (L0)=-37; del 2 R(L0)= -99; del 3 R(L0)=2; del 4 R(L0)=74; n=( L- L0)/ del L = ( L- 0)/ 20 = L/20 Calculations are done below and the interpolating polynomial plotted in Figure 5.4 RL0 895 := L0 0 := delRL0 37 − := del2RL0 99 − := del3RL0 2 := del4RL0 74 := delL 20 := n L ( ) L L0 − delL := R L ( ) RL0 n L ( ) delRL0 ⋅ + n L ( ) n L ( ) 1 − ( ) ⋅ del2RL0 ⋅ 2 + n L ( ) n L ( ) 1 − ( ) ⋅ n L ( ) 2 − ( ) ⋅ del3RL0 ⋅ 6 + n L ( ) n L ( ) 1 − ( ) ⋅ n L ( ) 2 − ( ) ⋅ n L ( ) 3 − ( ) ⋅ del4RL0 ⋅ 24 + ... := 154 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 165. These are thegiven data points, in vector form, needed for the plot below LAT 0 20 40 60 80 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := FLUX 895 858 722 489 235 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := <--- L 0 1 , 80 .. := 0 10 20 30 40 50 60 70 80 200 400 600 800 1000 Given data points Interpolation Given data points Interpolation Interpolated function by Newton's meth Latitude- degrees North Radiation flux FLUX R L ( ) LAT L , Figure 5.4. The Newton interpolating polynomial The interpolation yields R 45 ( ) 671.14 = 5.6 THE LAGRANGE INTERPOLATING POLYNOMIAL In many practical problems, data measurements are recorded at unequal spacings or intervals. In such situations, the Lagrangian interpolation method offers a viable means of deducing an interpolating polynomial connecting the dependent variable with the independent variable at intervals that are not necessarily constant. By this method, given a data sample (x1, y1); (x2, y2); .......... ( x n , y n) Chapter 5: Numerical Interpolation 155
- 166. an interpolating polynomialcan be generated using f x ( ) 1 n i wi x ( ) . yi ⋅ ( ) ∑ = = (5.24) where wi x ( ) 1 n j x xj − ( ) ∏ = 1 n j xi xj − ( ) ∏ = = j i ≠ (5.25) For example, if three data points are prescribed, n is 3 and the interpolating polynomial will be f x ( ) 1 3 i wi x ( ) .yi ( ) ∑ = = with w1 x ( ) x x2 − ( ) x x3 − ( ) ⋅ x1 x2 − ( ) x1 x3 − ( ) ⋅ = w2 x ( ) x x1 − ( ) x x3 − ( ) ⋅ x2 x1 − ( ) x2 x3 − ( ) ⋅ = (5.26) w3 x ( ) x x1 − ( ) x x2 − ( ) ⋅ x3 x1 − ( ) x3 x2 − ( ) ⋅ = Example 5.5 Given the following data x: 96 207 375 450 y: 932 850 767 1235 Determine the Lagrange Interpolating Polynomial f(x) that passes through the above points . Determine f(102) , f(322) and f(415) 156 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 167. The given datais x1 96 := x2 207 := x3 375 := x4 450 := y1 932 := y2 850 := y3 767 := y4 1235 := The Lagrange interpolating polynomial, in this case, will be a cubic of the following form f x ( ) w1 x ( ) y1 ⋅ w2 x ( ) y2 ⋅ + w3 x ( ) y3 ⋅ + w4 x ( ) y4 ⋅ + = where w1 x ( ) x x2 − ( ) x x3 − ( ) ⋅ x x4 − ( ) ⋅ x1 x2 − ( ) x1 x3 − ( ) ⋅ x1 x4 − ( ) ⋅ := w2 x ( ) x x1 − ( ) x x3 − ( ) ⋅ x x4 − ( ) ⋅ x2 x1 − ( ) x2 x3 − ( ) ⋅ x2 x4 − ( ) ⋅ := w3 x ( ) x x1 − ( ) x x2 − ( ) ⋅ x x4 − ( ) ⋅ x3 x1 − ( ) x3 x2 − ( ) ⋅ x3 x4 − ( ) ⋅ := w4 x ( ) x x1 − ( ) x x2 − ( ) ⋅ x x3 − ( ) ⋅ x4 x1 − ( ) x4 x2 − ( ) ⋅ x4 x3 − ( ) ⋅ := Note that the equations for w1(x), w2(x) etc have to be written in Mathcad as shown above and not in the form of Equation 5.25. Then, the interpolating polynomial is generated using f x ( ) w1 x ( ) y1 ⋅ w2 x ( ) y2 ⋅ + w3 x ( ) y3 ⋅ + w4 x ( ) y4 ⋅ + := and is shown in Figure 5.5. It yields f 102 ( ) 940.053 = f 322 ( ) 683.419 = f 415 ( ) 955.728 = X 96 207 375 450 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := Y 932 850 767 1235 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := <--- Given data points in vector form, used in the plot below Chapter 5: Numerical Interpolation 157
- 168. x 96 97 ,450 .. := 100 150 200 250 300 350 400 450 600 800 1000 1200 1400 Given data points Interpolation Given data points Interpolation Lagrange Interpolating Polynomial X- Values Function values Y f x ( ) X x , Figure 5.5. The Lagrange interpolating polynomial 5.7 INTERPOLATION USING LINEAR, QUADRATIC AND CUBIC SPLINES. For n data points that are provided, the highest- order interpolating polynomial that can be generated will be of order (n-1) . Although, in general, the accuracy of the interpolation process increases with the order of the polynomial , there are situations when the accuracy can, in fact, decrease with the polynomial order. This can happen when the measured data reflects abrupt changes in the dependent variable values for steady changes in the independent variable . In these cases, accuracy can be improved by resorting to lower order polynomials, commonly referred to as splines . Splines normally used are linear, quadratic and cubic. Linear Splines: A linear spline results from connecting adjacent data points with straight lines. Given n data points (x1, y1), (x2, y2), ( x3, y3 ),..... (x n, y n ) , the interpolation function for x between x1 and x2 is f x ( ) y1 x x1 − x2 x1 − y2 y1 − ( ) ⋅ + = (5.27) Similarly, for x between x2 and x3 the interpolation function will be 158 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 169. f x ( )y2 x x2 − x3 x2 − y3 y2 − ( ) ⋅ + = (5.28) It follows ,then, that for x between x n-1 and x n f x ( ) yn 1 − x xn 1 − − xn xn 1 − − yn yn 1 − − ( ) ⋅ + = (5.29) The resulting linear spline, however, must satisfy the following constraint equations f xi ( ) fi 1 + xi ( ) = for i 1 = 2..... , n 1 − , (5.30) Quadratic Splines: Quadratic splines are of the form : ax 2 + bx + c The conditions to be satisfied are (1) The spline function must pass through the given data points. (2) The spline is to be continuous at the interior data points . (3) The second derivative of the spline between the first two data points is to be zero. Cubic Splines: Cubic splines have the form: ax 3 + b x 2 + cx + d and must satisfy the following conditions (1) The spline function must pass through the given data points (2) The first derivatives of the spline must be continuous. (3) The second derivatives of the spline must also be continuous. (4) The second derivative must be zero at the first and second data points. 5.8 INTERPOLATION WITH MATHCAD Interpolation in Mathcad allows you to either connect the data points with straight lines ( linear interpolation ) or with sections of a cubic polynomial ( cubic spline interpolation ). These interpolation functions return a curve passing through the points you specify. Linear interpolation: This is done using the linterp function as shown below. linterp(vx,vy,x): Uses the data vectors vx and vy to return a linearly interpolated value of y corresponding to x . The quantities vx and vy must be specified as vectors and vx must contain real values in ascending order. The linterp function is intended for interpolation and not for extrapolation. Chapter 5: Numerical Interpolation 159
- 170. Cubic spline interpolation:Cubic spline interpolation enables you to pass a curve through a set of points such that the first and second derivatives of the curve are continuous at each point. Mathcad does this by taking three adjacent points and constructing a cubic polynomial that passes through these points. These cubics are then connected together to make up the interpolated curve. In order to fit a cubic spline curve through a set of points, the procedure shown below must be followed: 1. Create the vectors vx and vy containing the x and y coordinates of the points of interest through which you want to go through. The elements of vx must be put in in ascending order. 2. Generate the vector vs:= cspline( vx,vy). This vector vs is a vector of intermediate results to be used with interp. 3. To evaluate the spline at a point x1, do the following: interp(vs,vx,vy,x1) . Note: Steps 2 and 3 can be combined by doing : interp(cspline(vx,vy),vx,vy,x1) Mathcad has two other cubic spline functions as given below: lspline(vx,vy) : This generates a spline curve that approaches a straight line at the endpoints. pspline(vx,vy) : This generates a spline curve that approaches a parabola at the endpoints. The function interp(vs,vx,vy,x) returns the interpolated y value corresponding to x. The vector vs is a vector of intermediate results obtained by utilizing the option of using lspline, pspline or cspline on the vectors of given data , namely, vx and vy . For Example 5.1 , the vx and vy vectors are vx 0 1 2 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := vy 2 2.25 5.82 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := As shown below, the linterp function is used to do a linear interpolation. The resulting spline function is drawn in Figure 5.6 and the values of the interpolated function at x= 0.5 and 1.5 are: 160 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 171. x 0 0.1 ,2 .. := 0 0.5 1 1.5 2 2 4 6 vy linterp vx vy , x , ( ) vx x , Figure 5.6. Interpolated function obtained with linterp linterp vx vy , 0.5 , ( ) 2.125 = linterp vx vy , 1.5 , ( ) 4.035 = The interp function along with lspline, pspline and cspline can be utilized as follows to generate cubic splines vs1 lspline vx vy , ( ) := vs2 pspline vx vy , ( ) := vs3 cspline vx vy , ( ) := f1 x ( ) interp vs1 vx , vy , x , ( ) := f2 x ( ) interp vs2 vx , vy , x , ( ) := f3 x ( ) interp vs3 vx , vy , x , ( ) := flin x ( ) linterp vx vy , x , ( ) := Values of the interpolated cubic splines at x= 0.5 and 1.5 are f1 0.5 ( ) 1.814 = f2 0.5 ( ) 1.71 = f3 0.5 ( ) 1.502 = f1 1.5 ( ) 3.724 = f2 1.5 ( ) 3.62 = f3 1.5 ( ) 3.828 = flin 0.5 ( ) 2.125 = flin 1.5 ( ) 4.035 = Figure 5.7 shows a comparison of the various splines generated using Mathcad Chapter 5: Numerical Interpolation 161
- 172. x 0 0.1 ,2 .. := 0 0.5 1 1.5 2 1 2 3 4 5 6 Given data points Using lspline Using pspline Using cspline Using linterp Given data points Using lspline Using pspline Using cspline Using linterp Comparison of Mathcad splines x- Values function values vy f1 x ( ) f2 x ( ) f3 x ( ) flin x ( ) vx x , Figure 5.7. Comparison of spline functions obtained with Mathcad For the problem of Example 5.2, interpolations using the Mathcad linterp and interp functions yield the following. The results are sketched in Figure 5.8. vx 100 200 400 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := vy 975 1575 2054 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := vs cspline vx vy , ( ) := fcspline x ( ) interp vs vx , vy , x , ( ) := flin x ( ) linterp vx vy , x , ( ) := fcspline 251 ( ) 1.737 10 3 × = flin 251 ( ) 1.697 10 3 × = 162 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 173. x 100 105 ,400 .. := 100 150 200 250 300 350 400 800 1000 1200 1400 1600 1800 2000 2200 Given data points Mathcad's cubic spline Mathcad's linear interpolation with linterp Given data points Mathcad's cubic spline Mathcad's linear interpolation with linterp Interpolation with Mathcad x- values function values vy fcspline x ( ) flin x ( ) vx x , Figure 5.8. Interpolation with Mathcad For the problem of Example 5.3 , interpolation using the Mathcad spline functions is shown below vx 0 100 200 300 400 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := <---vx and vy vectors vy 104.2 97.3 48.1 23.7 8.6 5.2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := flin x ( ) linterp vx vy , x , ( ) := Linear interpolation values at x= 150, x= 350 and x= 450 are Chapter 5: Numerical Interpolation 163
- 174. linterp vx vy ,150 , ( ) 72.7 = linterp vx vy , 350 , ( ) 16.15 = linterp vx vy , 450 , ( ) 6.9 = The linearly interpolated function is shown in Figure 5.9. x 0 5 , 500 .. := 0 100 200 300 400 500 0 50 100 150 Given data points Mathcad's linear interpolation Given data points Mathcad's linear interpolation x- values Interpolated values vy linterp vx vy , x , ( ) vx x , Figure 5.9. Linear interpolation for Example 5.3 The Gregory Newton polynomial for this problem has been derived earlier and is the function f(x) shown below . x1 0 := x2 100 := x3 200 := x4 300 := x5 400 := x6 500 := <--data points y1 104.2 := y2 97.3 := y3 48.1 := y4 23.7 := y5 8.6 := y6 5.2 := a1 104.2 := a2 0.069 − := a3 2.115 − 10 3 − ⋅ := <--Coefficients obtained in Example 5.3 a4 1.118 10 5 − ⋅ := a5 3.442 − 10 8 − ⋅ := a6 8.375 10 11 − ⋅ := f x ( ) a1 a2 x x1 − ( ) ⋅ + a3 x x1 − ( ) ⋅ x x2 − ( ) ⋅ + a4 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ + a5 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ + ... a6 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ x x5 − ( ) ⋅ + ... := vs1 lspline vx vy , ( ) := vs2 pspline vx vy , ( ) := vs3 cspline vx vy , ( ) := 164 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 175. <-- Mathcad splines flin x () linterp vx vy , x , ( ) := flspline x ( ) interp vs1 vx , vy , x , ( ) := fpspline x ( ) interp vs2 vx , vy , x , ( ) := fcspline x ( ) interp vs3 vx , vy , x , ( ) := Figure 5.10 shows a comparison of the Mathcad spline functions with the Gregory-Newton interpolation f(x) vx 0 100 200 300 400 500 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vy 104.2 97.3 48.1 23.7 8.6 5.2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := <--- Given data in vector form x 0 5 , 500 .. := 0 100 200 300 400 500 0 50 100 150 Given data points Linterp interp/lspline interp/pspline interp/cspline Gregory-Newton interpolation Given data points Linterp interp/lspline interp/pspline interp/cspline Gregory-Newton interpolation Interpolation with Mathcad Splines and Gregory-Newton interpolation x values Interpolated values vy flin x ( ) flspline x ( ) fpspline x ( ) fcspline x ( ) f x ( ) vx x , Figure 5.10. Comparison of Mathcad splines with the Gregory-Newton interpolating polynomial The Mathcad interpolated values at x = 150, 350 and 450 are shown below. flin 150 ( ) 72.7 = flin 350 ( ) 16.15 = flin 450 ( ) 6.9 = flspline 150 ( ) 73.947 = flspline 350 ( ) 15.3 = flspline 450 ( ) 5.721 = fpspline 150 ( ) 73.207 = fpspline 350 ( ) 15.425 = fpspline 450 ( ) 5.069 = Chapter 5: Numerical Interpolation 165
- 176. fcspline 150 ( )72.343 = fcspline 350 ( ) 15.48 = fcspline 450 ( ) 4.645 = In the following paragraphs, Mathcad splines for the problem of Example 5.5 are generated and compared with the Lagrange interpolating polynomial obtained earlier. The data given is x1 96 := x2 207 := x3 375 := x4 450 := y1 932 := y2 850 := y3 767 := y4 1235 := w1 x ( ) x x2 − ( ) x x3 − ( ) ⋅ x x4 − ( ) ⋅ x1 x2 − ( ) x1 x3 − ( ) ⋅ x1 x4 − ( ) ⋅ := w2 x ( ) x x1 − ( ) x x3 − ( ) ⋅ x x4 − ( ) ⋅ x2 x1 − ( ) x2 x3 − ( ) ⋅ x2 x4 − ( ) ⋅ := w3 x ( ) x x1 − ( ) x x2 − ( ) ⋅ x x4 − ( ) ⋅ x3 x1 − ( ) x3 x2 − ( ) ⋅ x3 x4 − ( ) ⋅ := w4 x ( ) x x1 − ( ) x x2 − ( ) ⋅ x x3 − ( ) ⋅ x4 x1 − ( ) x4 x2 − ( ) ⋅ x4 x3 − ( ) ⋅ := f x ( ) w1 x ( ) y1 ⋅ w2 x ( ) y2 ⋅ + w3 x ( ) y3 ⋅ + w4 x ( ) y4 ⋅ + := <-- Lagrange interpolating polynomial f 102 ( ) 940.053 = f 322 ( ) 683.419 = f 415 ( ) 955.728 = The Mathcad vx and vy vectors are vx x1 x2 x3 x4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := vy y1 y2 y3 y4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := vs1 lspline vx vy , ( ) := vs2 pspline vx vy , ( ) := vs3 cspline vx vy , ( ) := flspline x ( ) interp vs1 vx , vy , x , ( ) := fpspline x ( ) interp vs2 vx , vy , x , ( ) := <-- Mathcad splines fcspline x ( ) interp vs3 vx , vy , x , ( ) := flin x ( ) linterp vx vy , x , ( ) := A comparison of the Mathcad spline functions generated with linterp, lspline, pspline and cspline functions is shown in Figure 5.11. 166 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 177. x 96 97 ,450 .. := 100 150 200 250 300 350 400 450 600 700 800 900 1000 1100 1200 1300 Given data points interp/lspline interp/pspline interp/cspline linterp Lagrangian interpolation Given data points interp/lspline interp/pspline interp/cspline linterp Lagrangian interpolation Lagrange Poly/ Mathcad Interpolations compared x- Values y- Values vy flspline x ( ) fpspline x ( ) fcspline x ( ) flin x ( ) f x ( ) vx x , Figure 5.11. Comparison of Mathcad splines with Lagrangian interpolation The various interpolations resorted to give the following interpolated function values for x = 102, 322 and 415. x=102: f 102 ( ) 940.053 = flin 102 ( ) 927.568 = flspline 102 ( ) 930.335 = fpspline 102 ( ) 932.981 = fcspline 102 ( ) 940.053 = x=322: f 322 ( ) 683.419 = flin 322 ( ) 793.185 = flspline 322 ( ) 669.542 = fpspline 322 ( ) 684.098 = fcspline 322 ( ) 683.419 = x=415: f 415 ( ) 955.728 = flin 415 ( ) 1.017 10 3 × = flspline 415 ( ) 985.191 = Chapter 5: Numerical Interpolation 167
- 178. fpspline 415 ( )962.582 = fcspline 415 ( ) 955.728 = 5.9. APPLICATIONS IN NUMERICAL INTERPOLATION In the examples that follow, data from various practical areas such as design of machinery, noise control and vibration analysis, is utilized in interpolation procedure s to generate function values at points other than the given data points. 5.9.1. Stress-Strain Data for Titanium The stress versus strain data given in the table below is obtained from a tensile test of annealed A-40 titanium [ 18 ]. A cubic spline can be developed using Mathcad as follows and the stresses interpolated for any values of strain in the range provided. The cubic spline interpolation generated together with the given data points is presented in Figure 5.12. Notice that the spline generated passes through all the given data points as is expected of an interpolating polynomial. Strain (in/in) Stress (kpsi) <--Given stress versus strain data σ 0 5 10 15 20 25 30 34.9 39.9 44.9 49.9 60.9 73.3 78.2 86.2 94.5 98.7 118.2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := ε 0 0.00030 0.00060 0.00090 0.00120 0.00175 0.00220 0.00285 0.00349 0.00469 0.00698 0.01610 0.04807 0.07676 0.14364 0.26677 0.34470 0.72202 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vε ε := Vσ σ := Vs cspline Vε Vσ , ( ) := 168 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 179. fspline ε ( )interp Vs Vε , Vσ , ε , ( ) := <-- Generates a cubic spline function for given data fspline 0.004 ( ) 42.699 = <-- interpolated values of stress at strains of 0.004 and 0.65 fspline 0.65 ( ) 114.503 = i 1 18 .. := ε 0 0.00010 , 0.722 .. := 1 .10 4 1 .10 3 0.01 0.1 1 1 10 100 1 .10 3 fspline ε ( ) Vσi ε Vεi , Figure 5.12. Mathcad cubic spline for given stress-strain data 5.9.2. Notch Sensitivity of Aluminum The fatigue stress concentration factor K f is used with the nominal stress σ 0 to compute the maximum resulting stress σ max at a discontinuity in a machine part via the following relationship [18] . σmax Kf σ0 ⋅ = (5.31) The fatigue stress concentration factor K f , which is, in fact , just a reduced value of the geometrical stress concentration factor K t, is a function of the sensitivity of the material of which the part is made to notches ( known as notch sensitivity q ). These quantities are connected by the following relationship . Chapter 5: Numerical Interpolation 169
- 180. Kf 1 qKt 1 − ( ) ⋅ + = (5.32) Notch sensitivity charts which are graphs of notch sensitivity versus notch radius are available for metals and alloys subjected to reversed bending or reversed axial loads. The data given below provides notch sensitivity (q) versus notch radius (r) information for an aluminum alloy and can be used to generate an interpolation from which notch sensitivity values at points other than the given data points can be readily obtained. r ( in) 0.015 0.04 0.06 0.10 0.12 0.14 0.16 q 0.25 0.55 0.65 0.75 0.775 0.80 0.825 Since seven data points are given , the interpolating polynomial that can be generated will be of order 6 , having the form f r ( ) a b r ⋅ + c r 2 ⋅ + d r 3 ⋅ + e r 4 ⋅ + f r 5 ⋅ + g r 6 ⋅ + = Resorting to the method of undetermined coefficients results in the following matrix equation M C ⋅ A = where M 1 1 1 1 1 1 1 0.015 0.04 0.06 0.10 0.12 0.14 0.16 0.015 2 0.04 2 0.06 2 0.10 2 0.12 2 0.14 2 0.16 2 0.015 3 0.04 3 0.06 3 0.10 3 0.12 3 0.14 3 0.16 3 0.015 4 0.04 4 0.06 4 0.10 4 0.12 4 0.14 4 0.16 4 0.015 5 0.04 5 0.06 5 0.10 5 0.12 5 0.14 5 0.16 5 0.015 6 0.04 6 0.06 6 0.10 6 0.12 6 0.14 6 0.16 6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := C a b c d e f g ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = A 0.25 0.55 0.65 0.75 0.775 0.80 0.825 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := The coefficients in the column vector {C} can be obtained using the Mathcad function lsolve C lsolve M A , ( ) := 170 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 181. C 0.187 − 40.462 922.431 − 1.27 10 4 × 1.01 − 10 5 × 4.22810 5 × 7.169 − 10 5 × ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = a C1 := b C2 := c C3 := d C4 := e C5 := f C6 := g C7 := Thus, the interpolating polynomial f(x) is completely determined with a 0.187 − = b 40.462 = c 922.431 − = d 1.27 10 4 × = e 1.01 − 10 5 × = f 4.228 10 5 × = g 7.169 − 10 5 × = f r ( ) a b r ⋅ + c r 2 ⋅ + d r 3 ⋅ + e r 4 ⋅ + f r 5 ⋅ + g r 6 ⋅ + := <---- interpolating polynomial obtained by the method of undetermined coefficients Mathcad cubic spline interpolation can also be employed as shown below and the resulting plot compared with the results of the undetermined coefficients method and with the given data points. This is done in in Figure 5.13. Vr 0.015 0.04 0.06 0.10 0.12 0.14 0.16 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vq 0.25 0.55 0.65 0.75 0.775 0.80 0.825 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vs cspline Vr Vq , ( ) := fspline r ( ) interp Vs Vr , Vq , r , ( ) := Chapter 5: Numerical Interpolation 171
- 182. fspline 0.13 ( )0.787 = fspline 0.15 ( ) 0.813 = f 0.15 ( ) 0.814 = fspline 0.15 ( ) 0.813 = r 0.015 0.016 , 0.16 .. := i 1 7 .. := 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.2 0.4 0.6 0.8 Given data points Interpolation by undet coeff method Mathcad cubic spline interpolation Given data points Interpolation by undet coeff method Mathcad cubic spline interpolation Mathcad interpolation/ notch sensitivity Notch radius, r, inches Notch sensitivity, q Vqi f r ( ) fspline r ( ) Vri r , Figure 5.13. Interpolation of notch sensitivity data 5.9.3 Speech Interference Level In the field of Noise and Vibration Control, it often becomes necessary to determine the effect of steady background noise on speech communication in a given setting. The preferred speech interference level (PSIL) was established in an effort to study this effect under the constraint that speech sounds would not be allowed to be reflected back to the listener [ 2, 10 ]. For effective communication at a given voice level , the maximum distance that there can be between the speaker and the listener is a function of the preferred speech interference level existing at the location. The following data which is provided for communication at the level of a normal male voice can be utilized in an interpolation scheme to determine the maximum PSIL permitted for a given distance between the speaker and the listener. Dist (ft) 0 2 4 6 8 10 12 PSIL(dB) 83 62.5 56 52.5 50 47.5 46.5 172 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 183. Because there are7 data points prescribed, the interpolating polynomial to be generated will be of order 6 and if a Gregory-Newton interpolating polynomial is to be determined , it will have the form f x ( ) a1 a2 x x1 − ( ) ⋅ + a3 x x1 − ( ) ⋅ x x2 − ( ) ⋅ + a4 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ + a5 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ + ... a6 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ x x5 − ( ) ⋅ + ... a7 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ x x5 − ( ) ⋅ x x6 − ( ) ⋅ + ... = where the a' s are unknown coefficients to be determined. The given data is x1 0 := x2 2 := x3 4 := x4 6 := x5 8 := x6 10 := x7 12 := y1 83 := y2 62.5 := y3 56. := y4 52.5 := y5 50 := y6 47.5 := y7 46.5 := and the unknown coefficients can be determined using a1 y1 := a2 y2 y1 − x2 x1 − := a3 y3 a1 − a2 x3 x1 − ( ) ⋅ − x3 x1 − ( ) x3 x2 − ( ) ⋅ := a4 y4 a1 − a2 x4 x1 − ( ) ⋅ − a3 x4 x1 − ( ) ⋅ x4 x2 − ( ) ⋅ − x4 x1 − ( ) x4 x2 − ( ) ⋅ x4 x3 − ( ) ⋅ := a5 y5 a1 − a2 x5 x1 − ( ) ⋅ − a3 x5 x1 − ( ) ⋅ x5 x2 − ( ) ⋅ − a4 x5 x1 − ( ) ⋅ x5 x2 − ( ) ⋅ x5 x3 − ( ) ⋅ − x5 x1 − ( ) x5 x2 − ( ) ⋅ x5 x3 − ( ) ⋅ x5 x4 − ( ) ⋅ := Term4 a4 x6 x1 − ( ) x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ⋅ := a6 y6 a1 − a2 x6 x1 − ( ) ⋅ − a3 x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ − Term4 − 1 − ( ) a5 ⋅ x6 x1 − ( ) ⋅ x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ x6 x4 − ( ) ⋅ + ... x6 x1 − ( ) x6 x2 − ( ) ⋅ x6 x3 − ( ) ⋅ x6 x4 − ( ) ⋅ x6 x5 − ( ) ⋅ := Chapter 5: Numerical Interpolation 173
- 184. Termno4 a4 x7x1 − ( ) ⋅ x7 x2 − ( ) x7 x3 − ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ⋅ := a7 y6 a1 − a2 x7 x1 − ( ) ⋅ − a3 x7 x1 − ( ) ⋅ x7 x2 − ( ) ⋅ − Termno4 − 1 − ( ) a5 ⋅ x7 x1 − ( ) x7 x2 − ( ) ⋅ x7 x3 − ( ) ⋅ x7 x4 − ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ⋅ + ... a6 x7 x1 − ( ) x7 x2 − ( ) ⋅ x7 x3 − ( ) ⋅ x7 x4 − ( ) ⋅ x7 x5 − ( ) ⋅ 1 − ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ⋅ + ... x7 x1 − ( ) x7 x2 − ( ) ⋅ x7 x3 − ( ) ⋅ x7 x4 − ( ) ⋅ x7 x5 − ( ) ⋅ x7 x6 − ( ) ⋅ := which give a1 83 = a2 10.25 − = a3 1.75 = a4 0.229 − = a5 0.023 = a6 2.083 − 10 3 − × = a7 2.279 10 4 − × = The interpolating polynomial obtained by the Gregory-Newton method can now be obtained using f x ( ) a1 a2 x x1 − ( ) ⋅ + a3 x x1 − ( ) ⋅ x x2 − ( ) ⋅ + a4 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ + a5 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ + ... a6 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ x x5 − ( ) ⋅ + ... a7 x x1 − ( ) ⋅ x x2 − ( ) ⋅ x x3 − ( ) ⋅ x x4 − ( ) ⋅ x x5 − ( ) ⋅ x x6 − ( ) ⋅ + ... := Mathcad cubic spline interpolation can also be employed as shown below and the resulting plot compared with the results of the Gregory-Newton method and with the given data points. This is done in in Figure 5.14. Vx x1 x2 x3 x4 x5 x6 x7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy y1 y2 y3 y4 y5 y6 y7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vx 0 2 4 6 8 10 12 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy 83 62.5 56 52.5 50 47.5 46.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vs cspline Vx Vy , ( ) := fspline x ( ) interp Vs Vx , Vy , x , ( ) := x 0 1 , 12 .. := i 1 2 , 7 .. := 174 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 185. 0 2 46 8 10 12 40 50 60 70 80 90 Given data points Gregory-Newton interpolation Mathcad's interp/cspline Given data points Gregory-Newton interpolation Mathcad's interp/cspline Interpolation- speech communication Distance from speaker to listener (ft) Preferred speech interference level ( dB Vyi f x ( ) fspline x ( ) Vxi x , Figure 5.14. Interpolation of speech communication data 5.9.4. Load-Deflection Data for Elastomeric Mounts Elastomeric mounts are employed when small electrical and mechanical devices have to be isolated from high forcing frequencies [ 9, 10 ]. They are especially useful in the protection of delicate electronic instruments . Mount performance characteristics in the form of load versus static deflection data are provided by manufacturers of vibration control products to enable a designer to select an appropriate isolator for a given application. The following data is available for an elastomeric mount. Deflection (cm): 0 0 0.02 0.04 0.06 0.08 0.10 Load (N): 5 11 17.5 23.5 28 30.5 There are six data points prescribed. Thus, if a Lagrangian interpolating polynomial is to be generated, it will be of order 5, having the form L D ( ) 1 6 i wi D ( ) . Li ⋅ ( ) ∑ = = where Chapter 5: Numerical Interpolation 175
- 186. wi D ( ) 1 6 j DDj − ( ) ∏ = 1 6 j Di Dj − ( ) ∏ = = j i ≠ Here, D1 0 := D2 0.02 := D3 0.04 := D4 0.06 := D5 0.08 := D6 0.10 := L1 5 := L2 11. := L3 17.5 := L4 23.5 := L5 28 := L6 30.5 := w1 D ( ) D D2 − ( ) D D3 − ( ) ⋅ D D4 − ( ) ⋅ D D5 − ( ) ⋅ D D6 − ( ) ⋅ D1 D2 − ( ) D1 D3 − ( ) ⋅ D1 D4 − ( ) ⋅ D1 D5 − ( ) ⋅ D1 D6 − ( ) ⋅ := w2 D ( ) D D1 − ( ) D D3 − ( ) ⋅ D D4 − ( ) ⋅ D D5 − ( ) ⋅ D D6 − ( ) ⋅ D2 D1 − ( ) D2 D3 − ( ) ⋅ D2 D4 − ( ) ⋅ D2 D5 − ( ) ⋅ D2 D6 − ( ) ⋅ := w3 D ( ) D D1 − ( ) D D2 − ( ) ⋅ D D4 − ( ) ⋅ D D5 − ( ) ⋅ D D6 − ( ) ⋅ D3 D1 − ( ) D3 D2 − ( ) ⋅ D3 D4 − ( ) ⋅ D3 D5 − ( ) ⋅ D3 D6 − ( ) ⋅ := w4 D ( ) D D1 − ( ) D D2 − ( ) ⋅ D D3 − ( ) ⋅ D D5 − ( ) ⋅ D D6 − ( ) ⋅ D4 D1 − ( ) D4 D2 − ( ) ⋅ D4 D3 − ( ) ⋅ D4 D5 − ( ) ⋅ D4 D6 − ( ) ⋅ := w5 D ( ) D D1 − ( ) D D2 − ( ) ⋅ D D3 − ( ) ⋅ D D4 − ( ) ⋅ D D6 − ( ) ⋅ D5 D1 − ( ) D5 D2 − ( ) ⋅ D5 D3 − ( ) ⋅ D5 D4 − ( ) ⋅ D5 D6 − ( ) ⋅ := w6 D ( ) D D1 − ( ) D D2 − ( ) ⋅ D D3 − ( ) ⋅ D D4 − ( ) ⋅ D D5 − ( ) ⋅ D6 D1 − ( ) D6 D2 − ( ) ⋅ D6 D3 − ( ) ⋅ D6 D4 − ( ) ⋅ D6 D5 − ( ) ⋅ := The interpolating polynomial obtained by Lagrangian interpolation may now be obtained using L D ( ) L1 w1 D ( ) ⋅ L2 w2 D ( ) ⋅ + L3 w3 D ( ) ⋅ + L4 w4 D ( ) ⋅ + L5 w5 D ( ) ⋅ + L6 w6 D ( ) ⋅ + := giving L 0.07 ( ) 25.986 = 176 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 187. Mathcad cubic splineinterpolation can also be employed as shown below and the resulting plot compared with the results of the Lagrangian interpolation and with the given data points. This is done in in Figure 5.15 vD D1 D2 D3 D4 D5 D6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vL L1 L2 L3 L4 L5 L6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vs cspline vD vL , ( ) := fspline D ( ) interp vs vD , vL , D , ( ) := D 0 0.01 , 0.1 .. := i 1 2 , 6 .. := 0 0.02 0.04 0.06 0.08 0.1 0 10 20 30 40 Given data points Lagrangian interpolation Mathcad cubic spline interpolation Given data points Lagrangian interpolation Mathcad cubic spline interpolation Interpolation-Load/deflect data Static deflection (cm) Load (N) vLi L D ( ) fspline D ( ) vDi D , Figure 5.15. Interpolation of load-deflection data for elastomeric mount PROBLEMS In problems 5.1 through 5.15, draw graphs of the interpolating polynomial generated . Check your results against functions obtained with Mathcad splines . Present these comparisons as Mathcad plots with proper labels and titles. 5.1. Using the method of undetermined coefficients, derive an interpolating polynomial f(x) of the form "a+ bx " , for the following data. Determine the interpolated value of f (0.52). Chapter 5: Numerical Interpolation 177
- 188. x 0.45 0.550.65 0.75 y 0.075 0.136 0.227 0.372 5.2. For the same data as in problem 1, use the method of undetermined coefficients to derive an interpolating polynomial f(x) of form " a+bx + cx2 " . Calculate f (0.63). 5.3. Using the method of undetermined coefficients, derive an interpolating polynomial f(x) for the data given below. x 0 20 40 y 892 853 725 Estimate f (37). 5.4. The following table gives the values of cos(x) for selected angles (x) in degrees x 10 12 14 16 f(x)= cos(x) 0.0.98481 0.978150 .9703 0.96126 Using the method of undetermined coefficients, derive a cubic interpolating polynomial for the given interval . Estimate the value of cos x for x = 12.75 deg and x = 13.5 deg. Compare these results with the true values. 5.5. Given the following data set : x 0.50 0.55 0.60 0.65 y 0.5823 0.6124 0.6435 0.7869 derive an interpolating polynomial f(x) using the Gregory-Newton method. Estimate f(0.535) and f(0.625). Compare the accuracy of the estimated values if the true values are 0.6022 and 0.6978 respectively. 5.6. The deflection, δ, of a structure under loading is measured at a point for four different values of the applied force,F. The measued data is given below. F (kN) 0.5 1.5 2.0 2.5 δ (cm) 3.12 5.25 7.32 10.63 Using the Gregory-Newton method, compute the deflection for F = 1.75 kN. 178 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 189. 5.7. The amplitudeof vibration of an automobile in the vertical direction, after passing over a road bump is found to be as follows. Time, t (seconds) 0 0.65 1.30 1.95 Amplitude, A (mm) 5.25 2.50 0.80 0.40 Using the Gregory-Newton method, compute the amplitude for t = 1.80 seconds. 5.8. The Charpy test [ 18 ] provides material toughness data under dynamic conditions. In this test, which is helpful in comparing several materials and in determining low-temperature brittleness and impact strength, the specimen is struck by a pendulum released from a fixed height and the energy absorbed by the specimen, termed the impact value , computed from the height of the swing after fracture. The following table gives the impact value, V, as a function of temperature, T, for a certain material. Using the Gregory-Newton method, derive an interpolating polynomial and estimate V for T = -25 deg. F and - 75 deg. F T (deg F) -200 -150 -100 -50 0 V (ft-lbs) 0 1.5 3 10 35 5.9. The data in the following table are the distances traveled by a motorcycle, which was stationary at time t = 0, for selected times. time, t (seconds) 0 2 4 6 8 10 Distance, D (ft) 0 10 50 150 330 610 Generate a finite-difference table and obtain an interpolating polynomial using Newton's method. Estimate the distance D traveled at t = 7 seconds. 5.10. For the given data, generate both a finite-difference table and an interpolating polynomial. Angle, θ, (degrees) Cosine (θ) 0 1.00000 10 0.98481 20 0.93969 30 0.86603 40 0.76604 50 0.64279 60 0.50000 Evaluate the error of the estimated value of cos (25 degrees). The true value is given to be 0.90631. Chapter 5: Numerical Interpolation 179
- 190. 5.11. The followingdata gives the notch sensitivity , q , of a steel as a functionof the notch radius , r [ 18 ]. Compute a finite-difference table and derive an interpolating polynomial by Newton's method. Estimate q for r = 0.11 inches. q 0.45 0.65 0.75 0.78 0.785 r ( in) 0.01 0.05 0.09 0.13 0.17 5.12. The deflection, δ, of a structure under loading is measured at a point for four different values of the applied force, F. The measured data is as follows. F (kn) 0.50 1.50 2.0 2.50 δ (cm) 3.0 5.2 7.3 10.5 Using Lagrangian interpolation , compute the deflection, δ, for F = 2.2 kN. 5.13. When a vibration problem is solved with nonlinearities included, the natural frequencies of vibration become dependent on the amplitudes of vibration [ 9,16,20,21 ]. The following table relates the nonlinear natural frequency / linear natural frequency ratio, r, of a rotating blade vibrating in the plane of rotation to its vibration amplitude , A. A (in) 10 8.5 5 0 r (%) 99.9325 99.9575 99.9921 100 Derive a Lagrange interpolating polynomial and compute r for A = 4.25 in . 5.14. The behavior of a mooring line employed to control the excursions of a floating ocean structure resembles that of a nonlinear spring with tension-displacement characteristics which depend upon its length, weight, elastic properties, anchor holding capacities and water depth [ 22 ]. In the table below, the horizontal component of mooring line tension , H, is given as a function of the horizontal distance, X, between the ends of the line. X (ft) 440 470 485 492.5 495 498 H (Kips)0 5 10 42 150 440 Using Lagrangian interpolation, generate an interpolating polynomial and compute H for X= 487 ft. 180 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 191. 5.15. The followingtable gives the pressure (P) versus temperature (T) relationship in the liquid-vapor region for water [ 17 ] T(deg Kelvin) 403.15 453.15 503.15 553.15 633.15 P (MPa) 0.2701 1.0021 2.795 6.412 18.651 Using Lagrangian interpolation, generate an interpolating polynomial and estimate P for T= 525 deg. K. 5.16. While the value of the endurance limit, S end , of a material is based on its tensile strength, S ult , it is also dependent on the condition of its surface[19 ]. The following data relates to a machined , unnotched specimen subjected to reversed bending. S ult ( kpsi): 60 80 100 140 180 220 S end (kpsi): 22 30 38 50 60 64 Develop linear and cubic splines using linterp and interp for the above data. Also , graph the functions generated with proper labels and a title. Chapter 5: Numerical Interpolation 181
- 192. NOTES: 182 AN INTRODUCTIONTO NUMERICAL METHODS USING MATHCAD
- 193. C H AP T E R 6 CURVE-FITTING 6.1 THE NEED TO FIT A FUNCTION TO MEASURED DATA Regression analysis plays an important role in analysis and interpretation of experimental as well as statistical data and in their correlation with mathematical models. It involves the determination of a function f(x) that would " best fit " a bunch of experimentally measured values. ( x1, y1), (x2, y2) ....( xn,yn ). This function can be a linear function, a polynomial, a nonlinear function, an exponential function , or a linear combination of known functions. Some examples are shown in Figure 6.1. In curve- fitting a bunch of data points, typically the number of data points would be much larger than the number of undetermined coefficients in a given problem. Thus, there will be discrepancies between the function f(x) determined and the data points given and it is very rare for a curve-fit to go through all the given data points exactly. However, these differences are minimized by an adaptation of the Method of Least Squares. Figure 6.1. Examples of functions employed in curve-fitting Chapter 6: Curve-Fitting 183
- 194. Handling of WeightedData When dealing with engineering, scientific and statistical data, situations may be encountered when certain data may be considered to be more accurate than other data and consequently must be assigned more weight in the process of generating a curve-fit. One way to include this weighting is to make multiple inclusions of the associated data point in the regression analysis. For example, if the following data is given and the point (2, 20) is to be assigned a "weighting factor" of 3, this data point must simply be considered thrice in coming up with a curve-fit as shown below. x: 1 2 3 4 y: 10 20 30 40 <--Given data x 1 2 2 2 3 4 y: 10 20 20 20 30 40 <--- Information to be used for curve-fitting 6.2 THE METHOD OF LEAST SQUARES With a well-chosen approximating function, a " least squares " fit will yield a good representation of experimental data. Suppose you want to measure the distance between two points in a field, and let us say that you do this " n " times. You will come up with " n " measurements which are likely to be somewhat different from one another. Let these be d1, d2, d3,.....dn . If the true distance is "D", then the sum of the squares of the deviations from the true distance "D" is S d1 D − ( )2 d2 D − ( )2 + ................ + dn D − ( )2 + = (6.1) This sum " S " will be a maximum or a minimum when D S d d 0 = which yields 1 n i di ( ) ∑ = n D ⋅ − 0 = (6.2) or, D 1 n i di ∑ = n = (6.3) 184 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 195. For S tobe a minimum, 2 D S d d 2 0 > , which it is since it comes out to be 2n. Thus, S will be a mimimum when D i di ∑ n = , that is, if " n " measurements are taken, the true distance will be the arithmetic mean of the n measurements if the sum of the squares of the deviations is to be a minimum. 6.3 STRAIGHT LINE REGRESSION Let us assume that a plot of the given data suggests that we should fit it with a linear function. Let this function be f x ( ) C1 C2 x ⋅ + = (6.4) where C1 and C2 are coefficients to be determined. If the given data is (x1,y1), ( x2, y2).................(xn, yn), then the quantity "S", which is the sum of the squares of the deviations of the measured values (y) from the function values f(x) is S y1 C1 C2 x1 ⋅ + ( ) − ⎡ ⎣ ⎤ ⎦ 2 y2 C1 C2 x2 ⋅ + ( ) − ⎡ ⎣ ⎤ ⎦ 2 + .......... + yn C1 C2 xn ⋅ + ( ) − ⎡ ⎣ ⎤ ⎦ 2 + = (6.5) For S to be a minimum, the partial derivatives of S with respect to C1 and C2 must be zeros, which leads to n C1 ⋅ 1 n i xi ∑ = C2 ⋅ + 1 n i yi ∑ = = (6.6) 1 n i xi ∑ = C1 ⋅ 1 n i xi ( )2 ∑ = C2 ⋅ + 1 n i xi yi ⋅ ( ) ∑ = = Chapter 6: Curve-Fitting 185
- 196. In matrix form,this is n 1 n i xi ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C1 C2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 n i yi ∑ = 1 n i xi yi ⋅ ( ) ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = (6.7) Using Cramer's rule or the inverse of the matrix on the right hand side of the above equation or the Mathcad function lsolve , the undetermined coefficients C1 and C2 can now be found. Example 6.1 Given the following data: xi yi 0.10 65.85 0.15 65.20 0.20 55.51 0.25 50.43 0.30 45.97 0.35 33.25 0.40 34.33 0.45 29.76 0.50 23.89 0.55 23.76 0.60 18.99 Fit a straight line y= f(x)= C1 + C2 x to the above data. In this case, C1 and C2 are determined using n 1 n i xi ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C1 C2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 n i yi ∑ = 1 n i xi yi ⋅ ( ) ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = 186 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 197. Put in thegiven data as : j 1 2 , 10 .. := x1 0.10 := xj 1 + xj 0.05 + := n 11 := with y 65.85 65.20 55.51 50.43 45.97 33.25 34.33 29.76 23.89 23.76 18.99 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Defining: d n 1 n i xi ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := m 1 n i yi ∑ = 1 n i xi yi ⋅ ( ) ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := q n 1 n i xi ∑ = 1 n i yi ∑ = 1 n i xi yi ⋅ ( ) ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := C1 m d := C1 75.497 = C2 q d := C2 99.618 − = the fitted straight line function is then f x ( ) C1 C2 x ⋅ + := which is compared with the given data in Figure 6.2. At x= 0.20, 0.42 and 0.52, the fitted function yields . f 0.20 ( ) 55.574 = f 0.42 ( ) 33.658 = f 0.52 ( ) 23.696 = Chapter 6: Curve-Fitting 187
- 198. i 1 11 .. := 0.10.2 0.3 0.4 0.5 0.6 0 50 100 Given data Fitted function Given data Fitted function Comparison- Fitted Function with Data x-values y and f(x) values yi f x ( ) xi x , Figure 6.2. Curve-fit with a linear function 6.4 CURVE- FITTING WITH A QUADRATIC FUNCTION For some types of data, it may be appropriate to resort to a quadratic function of the form: f(x)= C1 + C2 x2 (6.8) In this case, then , S will be S y1 C1 C2 x1 ( )2 ⋅ + ⎡ ⎣ ⎤ ⎦ − ⎡ ⎣ ⎤ ⎦ 2 y2 C1 C2 x2 ( )2 ⋅ + ⎡ ⎣ ⎤ ⎦ − ⎡ ⎣ ⎤ ⎦ 2 + .................. + yn C1 C2 xn ( )2 ⋅ + ⎡ ⎣ ⎤ ⎦ − ⎡ ⎣ ⎤ ⎦ 2 + = (6.9) Taking the partial derivatives of S with respect to C1 and C2 yields (6.10) n 1 n i xi ( )2 ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ( )4 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C1 C2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 n i yi ∑ = 1 n i xi ( )2 yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = 188 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 199. which can besolved for the C's using algebraic or matrix methods or by accessing the Mathcad function lsolve. The above can be expressed as [M]{C} = {A}, (6.11) where (6.12) M n 1 n i xi ( )2 ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ( )4 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = , A 1 n i yi ∑ = 1 n i xi ( )2 yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = Example 6.2 A data set is given below . x y 0.5 0.51 1.0 2.35 1.5 7.54 2.0 13.23 2.5 17.65 3.0 24.21 3.5 28.94 4.0 37.63 4.5 58.32 5.0 63.21 Determine the quadratic function that will best fit the data given Put in the given data : i 1 2 , 10 .. := x1 0.5 := xi 1 + xi 0.5 + := n 10 := y1 0.51 := y2 2.35 := y3 7.54 := y4 13.23 := y5 17.65 := y6 24.21 := y7 28.94 := y8 37.63 := y9 58.32 := y10 63.21 := In this problem, the [M} and {A} matrices are Chapter 6: Curve-Fitting 189
- 200. M n 1 n i xi ( )2 ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ()4 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := A 1 n i yi ∑ = 1 n i xi ( )2 yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := Use lsolve to compute the C's: C lsolve M A , ( ) := C 0.779 2.554 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = C1 0.779 = C2 2.554 = The quadratic function fitted to the given data is f(x), f x ( ) C1 C2 x 2 ⋅ + := which is now plotted and compared with the given data in Figure 6.3 x 0.5 0.51 , 5.0 .. := i 1 2 , 10 .. := x1 0.5 := xi 1 + xi 0.5 + := xi 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 = yi 0.51 2.35 7.54 13.23 17.65 24.21 28.94 37.63 58.32 63.21 = f xi ( ) 1.417 3.333 6.525 10.994 16.74 23.763 32.063 41.639 52.493 64.623 = <--- Given data compared with generated function 190 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 201. 0 1 23 4 5 0 50 100 Curve-fitting with quadratic function Given Data Points Curve-fitting with quadratic function Given Data Points Comparison-Fitted Function with Data x-Values Data (y) and f(x) Values f xi ( ) yi xi Figure 6.3. Curve-fit with quadratic function 6.5 CURVE-FIT WITH A POWER FUNCTION For some data types, it may be appropriate to do a curve-fit with a function of the form f(x)= A x B (6.13) where A and B are undetermined coefficients to be evaluated on a least squares basis. Taking the logarithm of the above, we obtain ln (f) = ln (A) +B ln (x) (6.14) or, F = C + B X (6.15) where F= ln f , C = ln (A) and X = ln (x) (6.16) Thus, now the problem has been reduced to one of linear regression and can be handled as done earlier for a linear fit. Chapter 6: Curve-Fitting 191
- 202. Example 6.3 For thedata set given below, find a power function that will serve as a least squares fit. x y 0.18 4.5921 0.32 5.2345 0.65 6.2321 1.12 6.5467 1.65 7.2976 2.34 7.7654 2.52 7.7237 2.85 8.2543 3.12 7.9765 3.61 8.4328 3.92 9.6322 4.51 8.8764 4.72 9.0276 5.25 9.2568 6.75 8.6432 7.85 10.2137 Using f(x)= y(x) = A x B , F=Y= C+B X, where F=Y = ln (y) , C = ln (A) and X= ln (x), we have the following matrix equation to solve for C and B n 1 n i Xi ∑ = 1 n i Xi ∑ = 1 n i Xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 n i Yi ( ) ∑ = 1 n i Yi ( ) Xi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = (n=16) (6.17) Put in the given data as shown below and calculate the Y's and X's using Yi ln yi ( ) := Xi ln xi ( ) := n 16 := i 1 2 , 16 .. := 192 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 203. x 0.18 0.32 0.65 1.12 1.65 2.34 2.52 2.85 3.12 3.61 3.92 4.51 4.72 5.25 6.75 7.85 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := y 4.5921 5.2345 6.2321 6.5467 7.2976 7.7654 7.7237 8.2543 7.9765 8.4328 9.6322 8.8764 9.0276 9.2568 8.6432 10.2137 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Yi lnyi ( ) := Xi ln xi ( ) := Equation 6.17 is: [D] { K} = {P}, where P 1 n i Yi ( ) ∑ = 1 n i Yi ( ) Xi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := D n 1 n i Xi ∑ = 1 n i Xi ∑ = 1 n i Xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := and {K} is the vector of unknown coefficients C and B The constants C and B can be obtained using lsolve as shown below, and then the coefficients A and B of the power function can be calculated K lsolve D P , ( ) := K 1.883 0.2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <------ This means: C= 1.883 and B = 0.2 C K1 := B K2 := C 1.883 = B 0.2 = Chapter 6: Curve-Fitting 193
- 204. A e C := A6.571 = B 0.2 = The fit generated is plotted and compared with the given data in Figure 6.4 f x ( ) A x B ⋅ := 1 2 3 4 5 6 7 8 0 5 10 Least Squares/Power Function Fit Given Data Points Least Squares/Power Function Fit Given Data Points Least Squares-Power Function Fit x-Values Function Values f xi ( ) yi xi Figure 6.4. Curve-fit with a power function 6.6. CURVE-FITTING WITH AN EXPONENTIAL FUNCTION Function of Form A e Bx Exponential functions appear frequently in the analysis of business, science as well as engineering problems and are often employed to curve-fit experimental and statistical data. The function, in this case, has the form . (6.18) f x ( ) A e B x ⋅ ⋅ = where A and B are undetermined coefficients to be evaluated on a least squares basis. Taking the logarithm of the above gives ln f ( ) ln A ( ) B x ⋅ + = (6.19) or, F C B x ⋅ + = (6.20) where 194 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 205. F ln F () = and C ln A ( ) = (6.21) Thus, now the problem is again reduced to one of linear regression and can be handled as done earlier for a linear fit. Example 6.4 For the data set given below, find an exponential function that will serve as a least squares fit. x y 0 16.45 1 4.47 2 1.06 3 0.45 4 0.15 5 0.03 Using f(x)= y(x) = A e Bx , F=Y= C+B x , where F=Y = ln (y) , and C = ln (A) we have the following matrix equation to solve for C and B n 1 n i xi ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 n i Yi ( ) ∑ = 1 n i Yi ( ) xi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = (n=6) (6.22) which can be solved for C and B. Put in the given data and calculate the natural logs of the y-values i 1 2 , 6 .. := i 1 2 , 6 .. := n 6 := y 16.45 4.47 1.06 0.45 0.15 0.03 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := x 0 1 2 3 4 5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Yi ln yi ( ) := Chapter 6: Curve-Fitting 195
- 206. The matrix inthe left hand side of Equation (6.22) and the column vector in the right hand side are: P 1 n i Yi ( ) ∑ = 1 n i Yi ( ) xi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := D n 1 n i xi ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := The unknown constants C and B can be computed using lsolve and the coefficient A can then be calculated as shown . K lsolve D P , ( ) := K 2.733 1.216 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <------That is, C= 2.733 and B = -1.216 C K1 := B K2 := C 2.733 = B 1.216 − = A e C := A 15.384 = f x ( ) A e B x ⋅ ⋅ := The curve-fit and its natural log are generated below and compared with the given data in Figures 6.5 (a) and 6.5(b) i 1 6 .. := A 15.384 = B 1.216 − = fi f xi ( ) := logfunci ln fi ( ) := 0 2 4 4 2 0 2 4 Data points Curve-fit Data points Curve-fit Natural Log of Curve- Fit Natural Log of Function Values Yi logfunci xi f x ( ) A e B x ⋅ ⋅ := 0 2 4 0 10 Analysis-Generated Fit Given Data Points Analysis-Generated Fit Given Data Points f z ( ) yi z xi , Figure 6.5 (a) Curve-fit with an exponential function of form A e Bx Figure 6.5 (b) Natural log of Curve-Fit 196 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 207. Function of FormA x e Bx There are times when exponential functions of the following form appear in the analysis of practical problems f(x)= y(x) = A x e Bx (6.23) where A and B are undetermined coefficients to be evaluated on a least squares basis. Taking the logarithm of the above, we obtain ln (y ) = ln (f) = ln(A) + ln (x) + Bx or, ln ( y/x) = ln ( A) + Bx Z = C + B x (6.24) where, Z= ln (y/x) and C = ln (A) (6.25) Thus, now the problem is reduced again to one of linear regression and can be handled as done earlier for a linear fit. Example 6.5 For the stress versus strain data given below, find an exponential function of form S = A e exp ( B e ) that will serve as a least squares fit. Strain (e) 300 800 1300 1800 2300 2800 (micro-in/in) Stress (S) 1656 3521 4079 4354 4300 3675 (psi) Using f(e)= y(e) = A e exp(Be), Z= C + B e , where Z = ln(S/e) and C= ln (A), we have the following matrix equation to solve for C and B Here, the x-coordinate is "e" and the y - coordinate is "S" n 1 n i ei ∑ = 1 n i ei ∑ = 1 n i ei ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 n i Zi ∑ = 1 n i Zi ei ⋅ ( ) ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = (6.26) Chapter 6: Curve-Fitting 197
- 208. Put in thegiven data and calculate the Z's : i 1 2 , 6 .. := e1 300 := e2 800 := e3 1300 := e4 1800 := e5 2300 := e6 2800 := S1 1656 := S2 3521 := S3 4079 := S4 4354 := S5 4300 := S6 3675 := Here, n 6 := Zi ln Si ei ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := Letting P 1 n i Zi ( ) ∑ = 1 n i Zi ( ) ei ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := D n 1 n i ei ∑ = 1 n i ei ∑ = 1 n i ei ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := the constants C and B of Equation (6.26) can be computed as shown below K lsolve D P , ( ) := K 1.906 5.721 − 10 4 − × ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <------That is, C= 1.906 and B = -5.7215x 10(-4) C K1 := B K2 := C 1.906 = B 5.721 − 10 4 − × = A exp C ( ) := A 6.725 = The exponential fit, then, is f e ( ) A e ⋅ exp B e ⋅ ( ) ⋅ := which is plotted and compared with the given data in Figure 6.6 e 300 301 , 2800 .. := i 1 2 , 6 .. := 198 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 209. e1 300 := e2800 := e3 1300 := e4 1800 := e5 2300 := e6 2800 := <--Given Data Points S1 1656 := S2 3521 := S3 4079 := S4 4354 := S5 4300 := S6 3675 := 500 1000 1500 2000 2500 1000 2000 3000 4000 5000 Analysis-Generated Fit Given Data Points Analysis-Generated Fit Given Data Points Curve-Fit with Exponential Function Strain ( e) Values Stress ( S) Values f z ( ) Si z ei , Figure 6.6. Curve -fit with exponential function A x e Bx 6.7 CURVE-FITTING WITH A LINEAR COMBINATION OF KNOWN FUNCTIONS In fitting a function to given data points, a linear combination of known functions may have to be used at times. Given a set of data points: (x1, y1), (x2,y2) ......... ( xn, yn), It is required to find a function: f(x) = C1 f1(x) + C2 f2 (x) + ......... C m fm(x) (6.27) where the f i (x) are prescribed functions and the C's are undetermined coefficients to be evaluated. The total number of data points given is "n" while the total number of prescribed functions to be utilized is "m" . In general, then, the function value at x i will be Chapter 6: Curve-Fitting 199
- 210. f xi ( ) 1 m j Cjfj xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = = (6.28) In this case, the sum of the squares of the deviations from the true values y i will be: s y1 f x1 ( ) − ( )2 y2 f x2 ( ) − ( )2 + ...................... + yn f xn ( ) − ( )2 + = (6.29) For S to be a minimum , the partial derivatives of S with respect to C1, C2 ....... C m must be zero and the second partials must be positive. These requirements lead to the following matrix equation in the unknowns, C 1, C2 ...... C m [M] {C} = {L} (6.30) where C C1 C2 Cm ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := , L 1 n i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ yi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i f2 xi ( ) ⎡ ⎣ ⎤ ⎦ yi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i fm xi ( ) ⎡ ⎣ ⎤ ⎦ yi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := , and 200 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 211. M 1 n i f1 xi ( ) ⎡ ⎣⎤ ⎦ 2 ∑ = 1 n i f2 xi ( ) ⎡ ⎣ ⎤ ⎦ f1 xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i fm xi ( ) ⎡ ⎣ ⎤ ⎦ f1 xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ f2 xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i f2 xi ( ) ⎡ ⎣ ⎤ ⎦ 2 ∑ = 1 n i fm xi ( ) ⎡ ⎣ ⎤ ⎦ f2 xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ fm xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i f2 xi ( ) ⎡ ⎣ ⎤ ⎦ fm xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i fm xi ( ) ⎡ ⎣ ⎤ ⎦ 2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := The above equations can now be solved for the C 's. Example 6.6. Fit the data points: x: 1.0 2.0 3.0 4.0 5.0 y: 5.75 10.75 12.65 29.95 49.35 with a function of the form y x ( ) C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + = , where f1 x ( ) 1 = , and f2 x ( ) x 3 = In this case, n= 5 and m = 2. Equation (6.30), then, gives the following matrix equation in C1 and C2 1 5 i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ 2 ∑ = 1 5 i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ f2 xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 5 i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ f2 xi ( ) ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 5 i f2 xi ( ) ⎡ ⎣ ⎤ ⎦ 2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ C1 C2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 5 i f1 xi ( ) ⎡ ⎣ ⎤ ⎦ yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 5 i f2 xi ( ) ⎡ ⎣ ⎤ ⎦ yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = The given data is Chapter 6: Curve-Fitting 201
- 212. x1 1 := x22.0 := x3 3.0 := x4 4.0 := x5 5.0 := y1 5.75 := y2 10.75 := y3 12.65 := y4 29.95 := y5 49.35 := The subscipts 1 and 2 are literal subscripts , and do not denote vector components. These are done by including a period in the variable name . Whatever follows the period then becomes the subscript. For example , "vel init" is created by typing a period after "vel " and then typing "init" i 1 2 , 5 .. := f1 x ( ) 1 := f2 x ( ) x 3 := <---- Letting M 1 5 i f1 xi ( )2 ∑ = 1 5 i f2 xi ( ) f1 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 5 i f1 xi ( ) ( ) f2 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 5 i f2 xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := L 1 5 i f1 xi ( ) ( ) yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 5 i f2 xi ( ) ( ) yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := the constants C 1 and C2 can be computed as shown C lsolve M L , ( ) := C1 C1 := C2 C2 := C 5.931 0.35 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = C1 5.931 = C2 0.35 = The fit yielded by this analysis is: y= f(x) = 5.931 + 0.350 x 3 which is plotted in Figure 6.7 and compared with the given data. x1 1 := x2 2.0 := x3 3.0 := x4 4.0 := x5 5.0 := <--- Given data points y1 5.75 := y2 10.75 := y3 12.65 := y4 29.95 := y5 49.35 := 202 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 213. yanal x ( )C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + := <--- Curve-fit generated by analysis yanal 1 ( ) 6.281 = yanal 2 ( ) 8.733 = <---y-values generated by curve-fit yanal 3 ( ) 15.386 = z 1 1.1 , 5 .. := i 1 5 .. := 1 2 3 4 5 0 20 40 60 Results of Analysis Given data points Results of Analysis Given data points Analysis - Generated Fit/ Given Data X- Values Y- Values yanal z ( ) yi z xi , Figure 6.7. Curve-fit with a linear combination of known functions 6.8 CURVE-FITTING WITH POLYNOMIALS If an mth order polynomial seems to be a good fit to the data points given, the procedure used in fitting a linear combination of known functions can still be resorted to. In this case, f 1 ( x) = 1, f 2 (x) = x , f 3 ( x ) = x 2, f 4 ( x ) = x 3 , ............. , f m+1 (x ) = x m (6.31) f(x) = C1 f 1 (x ) + C 2 f 2 ( x ) +....................... C m+1 f m+1 ( x ) (6.32) in which the C 's must be solved for using [M] {C} = {L} (6.33) Chapter 6: Curve-Fitting 203
- 214. where C C1 C2 Cm 1 + ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := M n 1 n i xi ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ()m ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ( )m ∑ = 1 n i xi ( )m 1 + ∑ = 1 n i xi ( )m 2 + ∑ = 1 n i xi ( )2 m ⋅ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := L 1 n i yi ∑ = 1 n i xi yi ⋅ ( ) ∑ = 1 n i xi ( )m yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := , where the ( x i, y i ) are the "n" given data points. Example 6.7. Given the data points x: 0 1 2 3 4 5 y: 0 8.47 17.48 19.57 14.69 11.23 Obtain a curve-fit of the form y x ( ) f x ( ) = C1 C2 x ⋅ + C3 x 2 ⋅ + = Here, n = 6 and m = 2 and the matrix equation to be solved is [M]{C} = {L} 204 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 215. where [M] isa known 3x3 matrix and {L} is a known column vector as shown below. The The column vector {C} , which involves the undetermined coefficients, is C1 C2 C3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ The given data is n 6 := x1 0 := x2 1 := x3 2 := x4 3 := x5 4 := x6 5 := y1 0 := y2 8.47 := y3 17.48 := y4 19.57 := y5 14.69 := y6 11.23 := The matrices M and L are L 1 n i yi ∑ = 1 n i xi yi ⋅ ( ) ∑ = 1 n i xi ( )2 yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := M n 1 n i xi ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ( )3 ∑ = 1 n i xi ( )2 ∑ = 1 n i xi ( )3 ∑ = 1 n i xi ( )4 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := The unknown constants are determined below using lsolve C lsolve M L , ( ) := C 0.444 − 12.484 2.057 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = C1 C1 := C2 C2 := C3 C3 := C1 0.444 − = C2 12.484 = C3 2.057 − = The polynomial fitting the given data , then, is Chapter 6: Curve-Fitting 205
- 216. f x ( )C1 C2 x ⋅ + C3 x 2 ⋅ + := which yields f 1 ( ) 9.982 = f 2 ( ) 16.294 = f 4 ( ) 16.574 = f 3.5 ( ) 18.047 = f 4 ( ) 16.574 = f 5 ( ) 10.542 = Figure 6.8 shows a comparison of the curve-fit generated with the given data points i 1 6 .. := x1 0 := x2 1 := x3 2 := x4 3 := y1 0 := y2 8.47 := y3 17.48 := y4 19.57 := <---- Given data x5 4 := x6 5 := y5 14.69 := y6 11.23 := z 0 0.01 , 5 .. := 0 1 2 3 4 5 0 5 10 15 20 Sec order Poly Given data points Sec order Poly Given data points Curve Fitted with Sec order Poly X- Values Y-Values f z ( ) yi z xi , Figure 6.8. Curve-fit with polynomial 206 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 217. 6.9 USE OFMATHCAD'S REGRESSION FUNCTIONS FOR CURVE-FITTING 6.9.1 Linear Regression with Mathcad In Mathcad, linear regression is done on data represented by vectors vx and vy by using the " slope" and the " intercept " functions . The function slope(vx,vy) returns a scalar, which is the slope of the least-squares fitted straight line . The function intercept(vx,vy) returns a scalar which is the y- axis intercept of this line. Alternatively, the function line(vx,vy) can be used, which returns a vector containing the y-intercept as well as the slope of the regression line. Use of these functions is demonstrated below. Example 6.8. Determine a linear function that would fit the data of Example 6.1 using Mathcad's linear regression For this example, the given data is i 1 2 , 11 .. := x1 0.10 := xi 1 + xi 0.05 + := y1 65.85 := y2 65.20 := y3 55.51 := y4 50.43 := y5 45.97 := y6 33.25 := y7 34.33 := y8 29.76 := y9 23.89 := y10 23.76 := y11 18.99 := n 11 := The function generated in the example was f(x) with C1 and C2 shown below C1 75.497 := C2 99.618 − := f x ( ) C1 C2 x ⋅ + := The data points are entered in as the vx and vy vectors in Mathcad vx 0.1 0.15 0.2 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vy 65.85 65.20 55.51 50.43 45.97 33.25 34.33 29.76 23.89 23.76 18.99 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Chapter 6: Curve-Fitting 207
- 218. The " slope" and the " intercept " functions are used as follows to generate the slope m and the intercept c m slope vx vy , ( ) := m 99.618 − = c intercept vx vy , ( ) := c 75.497 = Alternatively, use line(vx,vy) to generate the above quantities. c m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ line vx vy , ( ) := m 99.618 − = c 75.497 = The straight line fit generated by Mathcad is then ylinmathcad x ( ) m x ⋅ c + := ylinmathcad 0.55 ( ) 20.707 = Figure 6.9 shows a comparison of the data with the curve-fits generated 0.1 0.2 0.3 0.4 0.5 0.6 0 20 40 60 80 Given data- Fitted Function - f(x) Mathcad Linear Regression Given data- Fitted Function - f(x) Mathcad Linear Regression Comparison- Fitted Function with Data X- Values Y and f(x) values yi f x ( ) ylinmathcad x ( ) xi x , Figure 6.9. Curve-fit with Mathcad's linear regression functions 208 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 219. 6.9.2 Nonlinear Regressionwith Mathcad In Mathcad, the functions "regress" and "loess" are used along with "interp" to fit a polynomial through given data points. The " regress " function fits a single polynomial of any desired order to fit all given data points. In practice, however, there is no need to fit a polynomial of order higher than 4 ( n = 4 ). This function will not work very well if your data does not fit into a single polynomial. Unlike "regress", the " loess " function generates different second-order polynomials for different regions of the curve . It does this by examining data in small regions. The argument "span" controls the size of this region. With larger "span " , "loess" becomes equivalent to n = 2. A good value to use is n = 0.75 . While smaller values of "span " will make the fitted curve track the data fluctuations in a more precise manner, a larger value of " span " , in general , will generate a smoother fit. The " interp" function is the same as the one we resorted to when doing interpolation. It returns the interpolated y- values corresponding to the x values. In the examples that follow, the vector " vs1" relates to "regress" while " vs2" is related to the use of the " loess " function. The formats of these functions are given below. regress(vx,vy,n): vx and vy are are vectors containing the given data points and n is the polynomial order required. loess(vx, vy, span): vx and vy are are vectors containing the given data points and span controls the size of the regions for which second-order polynomials are to be generated. interp(vs,vx,vy,x): This returns the interpolated y value for x. Here, vs is the vector that is generated through the use of loess or regress and vx and vy are as defined above Example 6.9 Do a curve-fit on the data of Example 6.2, using Mathcad's nonlinear regression. The given data is . i 1 2 , 10 .. := x1 0.5 := xi 1 + xi 0.5 + := y1 0.51 := y2 2.35 := y3 7.54 := y4 13.23 := y5 17.65 := y6 24.21 := y7 28.94 := y8 37.63 := y9 58.32 := y10 63.21 := Chapter 6: Curve-Fitting 209
- 220. The function generatedin the example was a quadratic function of the form f(x)= C1+ C2 x 2 with C1 0.779 := C2 2.554 := f x ( ) C1 C2 x 2 ⋅ + := The data points given are put in as the vx and vy vectors shown below i 1 10 .. := vxi xi := vyi yi := Define : vs1 regress vx vy , 2 , ( ) := vs2 loess vx vy , 1. , ( ) := f1 x ( ) interp vs1 vx , vy , x , ( ) := f2 x ( ) interp vs2 vx , vy , x , ( ) := A comparison of curve-fits generated for the given data is presented in Figure 6.10 and in Table 6.1 x1 0.5 := xi 1 + xi 0.5 + := z 0.5 0.6 , 5 .. := z 0.5 0.51 , 5.0 .. := i 1 10 .. := y1 0.51 := y2 2.35 := y3 7.54 := y4 13.23 := y5 17.65 := <-- Given data y6 24.21 := y7 28.94 := y8 37.63 := y9 58.32 := y10 63.21 := 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 50 100 Given Data Points Curve-Fit with Quadratic Curve-Fit with Mathcad regress function Curve-Fit with Mathcad loess function Given Data Points Curve-Fit with Quadratic Curve-Fit with Mathcad regress function Curve-Fit with Mathcad loess function Comparison-Fitted Function with Data X-Values Data (Y) and f(x) Values yi f z ( ) f1 z ( ) f2 z ( ) xi z , Figure 6.10. Curve-fitting with Mathcad's nonlinear regression functions 210 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 221. TABLE 6.1 Comparison ofgiven data with curve-fits generated Given data Quadratic Fit Mathcad -regress Mathcad - loess xi 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 = yi 0.51 2.35 7.54 13.23 17.65 24.21 28.94 37.63 58.32 63.21 = f xi ( ) 1.417 3.333 6.525 10.995 16.741 23.765 32.066 41.643 52.498 64.629 = f1 xi ( ) 7.709 7.375 8.732 11.782 16.523 22.956 31.081 40.898 52.407 65.608 = f2 xi ( ) 7.654 7.437 8.881 11.946 16.818 23.167 31.243 40.985 52.4 65.52 = 6.9.3 Use of the Function linfit The Mathcad function "linfit" evaluates the undetermined coefficients C1, C2 ..... in a regression function that is a linear combination of known functions f1(x), f2(x)..... .It has the form C1 f1 (x) + C2 f2(x) + ....... . The function linfit(vx,vy,F) returns a vector that contains the coefficients C 1, C2 ..... in the linear combination of known functions that would best fit the given data based on the least squares principle. Here, "F" is a vector representing the functions that are linearly combined to generate the best fit to the given data and vx and vy are vectors representing the given data. Use of linfit requires that there be at least as many data points as there are terms in F. The use of " linfit" is illustrated in the following steps, as it is applied to the solution of Example 6.6 . Chapter 6: Curve-Fitting 211
- 222. The given datapoints are: x1 1.0 := x2 2.0 := x3 3.0 := x4 4.0 := x5 5.0 := y1 5.75 := y2 10.75 := y3 12.65 := y4 29.95 := y5 49.35 := The curve-fit of Example 6.6 was obtained as a linear combination of the form: yanal(x)=C1 f1(x) + C2 f2(x) with C1 = 5.931, C2 = 0.35, f1(x) =1 and f2(x)= x3 C1 5.931 := C2 0.35 := f1 x ( ) 1 := f2 x ( ) x 3 := yanal x ( ) C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + := The vx and vy vectors, in this case, are vx x := vy y := Put in "F" as a vector of functions used in the linear combination resorted to F x ( ) 1 x 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := <--- vx 1 2 3 4 5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = vy 5.75 10.75 12.65 29.95 49.35 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = The vector "S" returned contains the coefficients C 1, C2 ........ S linfit vx vy , F , ( ) := S 5.9309 0.3502 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <---- C1 S1 := C1 5.9309 = C2 S2 := C2 0.3502 = The function " g(x) " is the linear combination of the known functions resorted to. Notice that here both F(x) and S are vectors . 212 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 223. g x ( )F x ( ) S ⋅ := which means: g(x)= C1 f1(x) + C2 f2(x) x 1.0 1.01 , 5 .. := The Mathcad ( linfit ) results and the linear combination analysis results are presented in Figure 6.11 and compared with the given data points. j 1 2 , 5 .. := 1 1.5 2 2.5 3 3.5 4 4.5 5 0 20 40 Given Data Points Analysis Results (Linear Combination ) Regression with linfit Given Data Points Analysis Results (Linear Combination ) Regression with linfit Comparison of results x- values y-values vy j yanal x ( ) g x ( ) vxj x , Figure 6.11. Comparison of linfit results with given data 6.9.4 Use of the Function genfit The function genfit(vx,vy,vg,F) returns a vector that contains the parameters that make a nonlinear function f of x and n parameters u1, u2 ... un best fit the data represented in vectors vx and vy. Use of this function requires that there be at least as many data points as parameters. Here, vg is an n -element vector of guess values for the parameters u1, u2 ... un , and F is a column vector containing the function f and its partial derivatives with respect to the n parameters u1, u2 ... un. While it is often faster and less sensitive to poor guess values, this process may fail to converge in situations where the derivative vectors are done incorrectly. The method also permits a solution that employs numerical approximations for the parameter derivatives. To change methods, right-click on the genfit function and select the desired method from the menu. The use of this Mathcad function is best illustrated by an example as shown below. Chapter 6: Curve-Fitting 213
- 224. Example 6.11 Given thefollowing data:x: 0.3, 0.4, 1, 1.4, 2, 4 y: 9.4 11.2, 5, 3 , 6, 0 Fit the data with a function of the form: exp ( u1 + u2 x + u3 x2 ) Assemble the column vector F as follows. This is a column vector with: 1st row: Function fitting data 2nd row: Partial derivative of function with resp to u1 3rd row: Partial derivative of function with respect to u2 4th row: Partial derivative of function with respect to u3 <--- F x u , ( ) e u1 u2 x ⋅ + u3 x 2 ⋅ + e u1 u2 x ⋅ + u3 x 2 ⋅ + x e u1 u2 x ⋅ + u3 x 2 ⋅ + ⋅ x 2 e u1 u2 x ⋅ + u3 x 2 ⋅ + ⋅ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := * Put in the given data points as shown vx 0.3 0.4 1 1.4 2 4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := * vy 9.4 11.2 5 3 6 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := * Fill in the vector of guess values for u1, u2, u3 ..... vg 1 0 1 − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := * The function genfit now computes the parameters u1, u2, u3 ..... P genfit vx vy , vg , F , ( ) := * 214 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 225. That is, u1= 2.5654, u2 = -0.7881 , u3 = 0.0364, as calculated by genfit. P 2.5654 0.7881 − 0.0364 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = * <----- u1 P1 := * u2 P2 := * u3 P3 := * The function that best fits the given data is now This form also can be used since it represents the zeroth column of the vector F(x,u ) g x ( ) F x P , ( )1 := * <-- g x ( ) e u1 u2 x ⋅ + u3 x 2 ⋅ + := * or It is plotted in Figure 6.12 and is compared with the given data points i 0 6 .. := * x 0.3 0.31 , 4 .. := * 0 0.5 1 1.5 2 2.5 3 3.5 4 0 5 10 15 Given data points Curve-fit with genfit Given data points Curve-fit with genfit Comparison of curve-fit with given data vyi g x ( ) vxi x , Figure 6.12. Curve-fitting with genfit 6.9.5 Use of the Mathcad functions logfit, lnfit, pwrfit and expfit Mathcad's logfit The function logfit(vx,vy,vg) returns a vector containing the coefficients a, b and c for a Chapter 6: Curve-Fitting 215
- 226. logarithmic curve ofthe form a ln(x + b ) + c that best fits the real data values represented by vectors vx and vy. The use of this function requires that there be at least three data points The vector vg is a three-element vector of real guess values for the parameters a, b and c. If a logarithmic fit is desired that is different from the above form, then use genfit or linfit. As an example, for the data represented by vectors vx and vy below and with the use of the guess vector vg as given vx 0 1 2 3 4 5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vy 4 5 5.5 5.8 6.2 6.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vg 2 1 5 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := the curve-fit returned by Mathcad's logfit is a b c ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ logfit vx vy , vg , ( ) := a 1.388 = b 1.0437 = c 3.9536 = which is the function f(x )= 1.388 ln (x + 1.044 ) + 3.954, the plot of which is drawn below. f x ( ) a ln x b + ( ) ⋅ c + := x 0 0.01 , 5. .. := 0 1 2 3 4 5 4 6 8 Given data points Mathcad's logfit Given data points Mathcad's logfit vy f x ( ) vx x , 216 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 227. Mathcad's lnfit The functionlnfit(vx,vy) returns a vector containing the coefficients a and b for a logarithmic curve of the form a ln(x ) + b that best fits the real data values represented by vectors vx and vy. The use of this function requires that there be at least two data points. If a logarithmic fit is desired that is different from the above form, then use genfit or linfit. For the data represented by the following vx and vy vectors, vx 1 2 3 4 5 6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vy 5 6.5 7.25 7.80 8.25 8.63 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := the curve-fit returned by Mathcad's lnfit is a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ lnfit vx vy , ( ) := a 2.0048 = b 5.04 = which is the function f(x ) = 2.005 ln (x) + 5.04, the plot of which is shown below. f x ( ) a ln x ( ) ⋅ b + := x 1 1.01 , 6. .. := 1 2 3 4 5 6 4 6 8 10 Given data points Mathcad's lnfit Given data points Mathcad's lnfit vy f x ( ) vx x , Chapter 6: Curve-Fitting 217
- 228. Mathcad's pwrfit The functionpwrfit(vx,vy,vg) returns a vector containing the coefficients a, b and c for a power curve of the form a x b + c that best fits the real data values represented by vectors vx and vy. The use of this function requires that there be at least three data points and the vx values must all be greater than or equal to zero. Negative x-values are not appropriate in this setting because raising them to an arbitrary power can produce complex results that will not correspond to the real y-values. If you need to fit a power function to data in the left-half plane, shift it so that all x-values are positive, then adjust the fitted function accordingly to obtain correct results. The vector vg is a three-element vector of real guess values for the parameters a, b and c . If a logarithmic fit is desired that is different from the above form, then use genfit. For the data points represented by the vectors vx and vy as given below, and with the use of the vector vg of guess values shown for the coefficients a, b and c.. vx 0 1 2 3 4 5 6 7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vy 5 9 50 192 517 1123 2121 3635 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vg 3 4 10 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := the curve-fit returned by Mathcad's pwrfit is a b c ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ pwrfit vx vy , vg , ( ) := giving a 3.9977 = b 3.5003 = c 4.9612 = which is the function f(x ) = 3.998 x 3.5 + 4.961 , the plot of which is sketched below. f x ( ) a x b ⋅ c + := x 0 0.01 , 7. .. := 218 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 229. 0 1 23 4 5 6 7 0 2000 4000 Given data points Mathcad's pwrfit Given data points Mathcad's pwrfit vy f x ( ) vx x , Mathcad's expfit The function expfit(vx,vy, [vg]) returns a vector containing the coefficients a, b and c for an exponential curve of the form A exp (b x ) + C that best fits the real data values represented by vectors vx and vy. The use of this function requires that there be at least three data points and the vx values must all be greater than or equal to zero. If a fit has be done on data that has negative x-values, then the data must be shifted to the positive axis. . Otherwise, a curve-fit with erroneous values may be generated. The vector vg is a three-element vector of real guess values for the parameters A, b and C, but it is optional, and does not have to be used. However, if it is not used, then expfit generates a guess that fits a line to the logs of vy. If an exponential fit is desired that is different from the above form, then use genfit . For the data points represented by the vectors vx and vy as given below, and with the use of the vector vg of guess values shown for the coefficients a, b and c ,. vg 3 2 10 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := vx 0 1 2 3 4 5 6 7 8 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vy 21 24.892 31.31 41.89 59.334 88.095 135.513 213.693 342.589 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := the curve-fit returned by Mathcad's expfit is defined by Chapter 6: Curve-Fitting 219
- 230. A b C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ expfit vx vy ,vg , ( ) := giving A 6 = b 0.5 = C 14.9999 = Or, if vg is not used, resort to A b C ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ expfit vx vy , ( ) := again, giving A 6 = b 0.5 = C 14.9999 = which leads to the curve-fit f(x ) = 6 exp (0.5 x ) + 15 , the plot of which is sketched below. f x ( ) A e b x ⋅ ⋅ C + := x 0 0.01 , 8 .. := 0 1 2 3 4 5 6 7 8 0 200 400 Given data points Mathcad's expfit Given data points Mathcad's expfit vy f x ( ) vx x , 6.9.6 More Examples with Mathcad Example 6.12 Using Mathcad's linear regression, generate a curve-fit in the form of a power function for the data of Example 6.3 and compare with the given data and the analytical results of Example 6.3. 220 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 231. The function generatedby analysis in Example 6.3 was: f x ( ) A x B ⋅ = with A 6.571 := B 0.2 := f x ( ) A x B ⋅ := The data of Example 6.3 is shown below n 16 := i 1 2 , 16 .. := y 4.5921 5.2345 6.2321 6.5467 7.2976 7.7654 7.7237 8.2543 7.9765 8.4328 9.6322 8.8764 9.0276 9.2568 8.6432 10.2137 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := <-- Given data x 0.18 0.32 0.65 1.12 1.65 2.34 2.52 2.85 3.12 3.61 3.92 4.51 4.72 5.25 6.75 7.85 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := In Mathcad, linear regression with the slope and intercept functions can be performed on this problem or the Mathcad function pwrfit can be resorted to and curve-fits obtained as shown in the following steps. As was done in Section 6.5, we define: i 1 2 , 16 .. := Chapter 6: Curve-Fitting 221
- 232. Yi ln yi () := Xi ln xi ( ) := vXi Xi := vYi Yi := <-- Define vectors vX and vY The slope and the intercept of the new linear relationship Y= CSTAR+BSTAR*X can now be computed as BSTAR slope vX vY , ( ) := CSTAR intercept vX vY , ( ) := BSTAR 0.1996 = CSTAR 1.8827 = ASTAR e CSTAR := A 6.571 = Using Mathcad's pwrfit function on the given data points with the guess vector vg 6 0.1 0.1 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := yields a b c ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ pwrfit x y , vg , ( ) := That is, a 9.4201 = b 0.142 = c 2.7907 − = powerfit x ( ) a x b ⋅ c + := The various curve-fits generated are now fmcad x ( ) ASTAR x BSTAR ⋅ := <--Fit generated with Mathcad's linear regression (slope and intercept functions) powerfit x ( ) a x b ⋅ c + := <--Fit generated with Mathcad's pwrfit function f x ( ) A x B ⋅ := <--- Fit generated with power function (Example 6.3) 222 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 233. and these arecompared and presented along with the given data in Figure 6.13. i 1 2 , 16 .. := 1 2 3 4 5 6 7 8 4 5 6 7 8 9 10 11 Given Data Points Power Function Fit Mathcad's Linear Regression Mathcad's pwrfit function Given Data Points Power Function Fit Mathcad's Linear Regression Mathcad's pwrfit function Curve-Fitting with Power Function x- values y- values yi f z ( ) fmcad z ( ) powerfit z ( ) xi z , Figure 6.13. Curve-fitting with a power function Chapter 6: Curve-Fitting 223
- 234. Example 6.13 Using Mathcad'slinear regression, generate a curve-fit in the form of the exponential function A exp ( Bx ) for the data of Example 6.4 . Compare this with the given data and with the analytical results of Example 6.4 The data given is i 1 2 , 6 .. := x 0 1 2 3 4 5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := y 16.45 4.47 1.06 0.45 0.15 0.03 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := The function generated in Example 6.4 was : n 6 := f x ( ) A e B x ⋅ ⋅ = with A 15.384 := B 1.216 − := f x ( ) A e B x ⋅ ⋅ := <-- Analysis-generated fit As was done in Section 6.6 , define fi f xi ( ) := logfunci ln fi ( ) := Yi ln yi ( ) := <-- log of the y-coordinates of given data In Mathcad, linear regression with the slope and intercept functions or with the line function can be performed on this problem and the curve-fit obtained as shown in the following steps i 1 6 .. := vx x := vY Y := BSTAR slope vx vY , ( ) := CSTAR intercept vx vY , ( ) := 224 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 235. CSTAR BSTAR ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ line vx vY , () := BSTAR 1.2164 − = CSTAR 2.7334 = ASTAR e CSTAR := ASTAR 15.3844 = Then, Mathcad -generated fit using slope and intercept or the line function fmcad x ( ) ASTAR e BSTAR x ⋅ ⋅ := <--- Mathcad's expfit function can also be used to generate an exponential fit of the form a exp(bx) + c as shown below a b c ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ expfit x y , ( ) := a 16.4172 = b 1.3222 − = c 0.0391 = exponfit x ( ) a e b x ⋅ ⋅ c + := <-- Mathcad -generated fit using the expfit function Figure 6.14 shows a comparison of the Mathcad fits with the results of Example 6.4 and the given data . z 0 0.1 , 5 .. := i 1 6 .. := f x ( ) A e B x ⋅ ⋅ := 0 1 2 3 4 5 0 10 20 Given Data Points Analysis-Generated Fit Mathcad's Linear Regression Mathcad's expfit function Given Data Points Analysis-Generated Fit Mathcad's Linear Regression Mathcad's expfit function Curve-Fitting with Exponential Function x- values y- values yi f z ( ) fmcad z ( ) exponfit z ( ) 0.05 − xi z , Figure 6.14. Curve-fit with an exponential function ( AeBx) using Mathcad's linear regression Chapter 6: Curve-Fitting 225
- 236. Example 6.14 Using Mathcad'slinear regression, generate a fit in the form of the exponential function "Ax exp(Bx)" for the problem of Example 6.5 . Compare this with the results of Example 6.5 and with the given data . The data given is i 1 2 , 6 .. := e1 300 := e2 800 := e3 1300 := e4 1800 := e5 2300 := e6 2800 := S1 1656 := S2 3521 := S3 4079 := S4 4354 := S5 4300 := S6 3675 := The function generated in Example 6.5 was f e ( ) A e ⋅ exp B e ⋅ ( ) ⋅ = with A 6.7248 := B 5.7205 − 10 4 − ⋅ := f e ( ) A e ⋅ exp B e ⋅ ( ) ⋅ := n 5 := Zi ln Si ei ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := Si ei exp Zi ( ) ⋅ := e 300 301 , 3000 .. := In Mathcad, linear regression can be performed on this problem and the curve-fit obtained as shown in the following steps, : e1 300 := e2 800 := e3 1300 := e4 1800 := e5 2300 := e6 2800 := S1 1656 := S2 3521 := S3 4079 := S4 4354 := S5 4300. := S6 3675 := i 1 2 , 6 .. := As was done in Example 6.5, define vei ei := vZi Zi := Zi ln Si ei ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := 226 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 237. BSTAR slope vevZ , ( ) := CSTAR intercept ve vZ , ( ) := BSTAR 5.7205 − 10 4 − × = CSTAR 1.9058 = ASTAR exp CSTAR ( ) := ASTAR 6.7248 = fmcad e ( ) ASTAR e ⋅ exp BSTAR e ⋅ ( ) ⋅ := <-- Mathcad - generated fit f e ( ) A e ⋅ exp B e ⋅ ( ) ⋅ := <-- Analysis-generated fit Figure 6.15 shows a comparison of the Mathcad-generated fit using linear regression with the analysis- generated fit of Example 6.5 and with the given data . z 300 301 , 2800 .. := 500 1000 1500 2000 2500 3000 1000 2000 3000 4000 5000 Given Data Points Analysis -generated fit Mathcad's Linear Regression Given Data Points Analysis -generated fit Mathcad's Linear Regression Comparison of Curve-Fit Results Strain (e ) Values Stress (S) Values Si f z ( ) fmcad z ( ) ei z , Figure 6.15. Curve-fit with an exponential function (AxeBx) using Mathcad's linear regression Chapter 6: Curve-Fitting 227
- 238. Example 6.15 Using Mathcad,generate a nonlinear fit for the data of Example 6.6. Compare these results with the fit generated in Example 6.6 by using a linear combination of functions and the given data. x 1 1 := x 2 2.0 := x 3 3.0 := x 4 4.0 := x 5 5.0 := <--- Given data y 1 5.75 := y 2 10.75 := y 3 12.65 := y 4 29.95 := y 5 49.35 := The fit generated in Example 6.6 was yanal x ( ) C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + = with f1 x ( ) 1 := f2 x ( ) x 3 := C1 5.931 := C2 0.35 := Using Mathcad' s interp and loess functions , curve-fitting with a polynomial can be done on this problem as shown in the following steps. i 1 2 , 5 .. := x 1 1.1 , 5.5 .. := x 1 1 := x 2 2.0 := x 3 3.0 := x 4 4.0 := x 5 5.0 := y 1 5.75 := y 2 10.75 := y 3 12.65 := y 4 29.95 := y 5 49.35 := vx i x i := vy i y i := vs loess vx vy , 1 , ( ) := floess x ( ) interp vs vx , vy , x , ( ) := yanal x ( ) C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + := floess i floess x i ( ) := yanal i yanal x i ( ) := Figure 6.16 compares the Mathcad fit with the given data and the linear combination method yanal x ( ) C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + := floess x ( ) interp vs vx , vy , x , ( ) := z 1 1.01 , 5.0 .. := i 1 5 .. := 1 1.5 2 2.5 3 3.5 4 4.5 5 0 20 40 60 Given Data Points Linear Combination Mathcad loess function Given Data Points Linear Combination Mathcad loess function Comparison- Mathcad loess / anal method X- values Y- Values yi yanal z ( ) floess z ( ) xi z , Figure 6.16. Comparison of Mathcad fit with linear combination method 228 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 239. Example 6.16 Using Mathcad,generate a nonlinear fit for the data of Example 6.7. Compare these results with the fit of form y = f(x) = C1 + C2 x + C3 x 2 generated in Example 6.7 and with the given data . x 1 0 := x 2 1 := x 3 2 := x 4 3 := x 5 4 := x 6 5 := <--Given data y 1 0 := y 2 8.47 := y 3 17.48 := y 4 19.57 := y 5 14.69 := y 6 11.23 := n 6 := The fit generated in Example 6.7 was f x ( ) C1 C2 x ⋅ + C3 x 2 ⋅ + = with C1 0.444 − := C2 12.484 := C3 2.057 − := Using Mathcad's interp and loess functions , curve-fitting the given data with a polynomial can be done as shown below. i 1 2 , 6 .. := vx i x i := vy i y i := vs loess vx vy , 1 , ( ) := floess x ( ) interp vs vx , vy , x , ( ) := floess i floess x i ( ) := f x ( ) C1 C2 x ⋅ + C3 x 2 ⋅ + := f i f x i ( ) := The Mathcad results generated are compared with the data given and with the polynomial method in Figure 6.17. i 1 6 .. := z 0 0.01 , 5 .. := f x ( ) C1 C2 x ⋅ + C3 x 2 ⋅ + := floess x ( ) interp vs vx , vy , x , ( ) := 0 1 2 3 4 5 0 10 20 Given Data Points Polynomial Method Mathcad loess Function Given Data Points Polynomial Method Mathcad loess Function Comparison- Polynomial Method/ Mathcad X- Values Y- Values yi f z ( ) floess z ( ) xi z , Figure 6.17. Comparison of Mathcad fit with polynomial method Chapter 6: Curve-Fitting 229
- 240. Example 6.17 Fit thedata points : x: 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 y: 0.59 0.83 0.90 0.95 1.02 1.43 1.75 3.15 with a function of the form : y = C1 f 1 (x ) + C2 f 2 ( x) +C 3 f 3(x) + C4 f4 (x) where f 1 (x ) = 1 , f 2 ( x ) = x2 , f 3 ( x) = sin (2x) , f 4 ( x) = exp( 0.95x ) In this case, we have the following matrix equation [M] {C} = {L} , where {C} is the column vector of unknown coefficients C1, C2, C3 and C4 , and [M] and {L} are known matrices as generated below. The given data is N 8 := x1 0.1 := x2 0.2 := x3 0.3 := x4 0.4 := x5 0.5 := x6 0.6 := x7 0.7 := x8 0.8 := y1 0.59 := y2 0.83 := y3 0.90 := y4 0.95 := y5 1.02 := y6 1.43 := y7 1.75 := y8 3.15 := i 1 2 , 8 .. := f1 x ( ) 1 := f2 x ( ) x 2 := <--- The subscipts 1 and 2 are literal subscripts , and do not denote vector components. These are done by including a period in the variable name . Whatever follows the period then becomes the subscript. For example , "vel init" is created by typing a period after "vel" and then typing "init" f3 x ( ) sin 2 x ⋅ ( ) := f4 x ( ) exp 0.95 x ⋅ ( ) := In this case, the M and L matrices are 230 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 241. N 8 := M 1 N i f1 xi ()2 ∑ = 1 N i f2 xi ( ) f1 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f3 xi ( ) f1 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f4 xi ( ) f1 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f1 xi ( ) ( ) f2 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f2 xi ( )2 ∑ = 1 N i f3 xi ( ) f2 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f4 xi ( ) f2 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f1 xi ( ) ( ) f3 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f2 xi ( ) ( ) f3 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f3 xi ( )2 ∑ = 1 N i f4 xi ( ) f3 xi ( ) ( ) ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f1 xi ( ) ( ) f4 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f2 xi ( ) ( ) f4 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f3 xi ( ) ( ) f4 xi ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f4 xi ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := L 1 N i f1 xi ( ) ( ) yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f2 xi ( ) ( ) yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f3 xi ( ) ( ) yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 N i f4 xi ( ) ( ) yi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := The unknown coefficients in {C} can be computed using the mathcad function lsolve as follows C lsolve M L , ( ) := yielding C 45.612 − 49.735 − 16.7 − 45.46 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = Chapter 6: Curve-Fitting 231
- 242. The constants C1 through C4 are the components of the column vector {C} generated above. Thus, C1 C1 := C2 C2 := C3 C3 := C4 C4 := C1 45.612 − = C2 49.735 − = C3 16.7 − = C4 45.46 = The fit yielded by the analysis is, then, yanal x ( ) C1 f1 x ( ) ⋅ C2 f2 x ( ) ⋅ + C3 f3 x ( ) ⋅ + C4 f4 x ( ) ⋅ + := and has the form -45.612 -49.735 x 2 -16.7 sin (2x)+ 45.46 exp(0.95x)I The use of " linfit" as applied to this problem is illustrated in the following steps. The vx and vy vectors, in this are vx x := vy y := Put in "F" as a vector of functions used in the linear combination resorted to F x ( ) 1 x 2 sin 2 x ⋅ ( ) exp 0.95 x ⋅ ( ) ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := <--- vx 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = vy 0.59 0.83 0.9 0.95 1.02 1.43 1.75 3.15 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = S linfit vx vy , F , ( ) := The vector "S" returned contains the coefficients C 1, C2 ........ <---- S 45.612 − 49.735 − 16.7 − 45.46 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = 232 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 243. C1 S1 := C2S2 := C3 S3 := C4 S4 := C1 45.612 − = C2 49.735 − = C3 16.7 − = C4 45.46 = The function " g(x) " is then the linear combination of the known functions resorted to. Notice that both F(x) and S are vectors here. g x ( ) F x ( ) S ⋅ := which means: g(x)= C1 f1(x) + C2 f2(x) + C3 f3(x) + C4 f4 (x) The Mathcad ( linfit ) results and the linear combination analysis results are both presented in Figure 6.18 and compared with the given data points. j 1 2 , 8 .. := x 0.1 0.101 , 0.80 .. := 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 1 2 3 4 Given Data Points Analysis Results (Linear Combination ) Regression with linfit Given Data Points Analysis Results (Linear Combination ) Regression with linfit Comparison of results x- values y-values vy j yanal x ( ) g x ( ) vxj x , Figure 6.18. Curve-fit with a linear combination of functions Chapter 6: Curve-Fitting 233
- 244. Example 6.18 Given thefollowing data:x: 0,1,2,3,4,5,6,7 y: 3,6.2,9.75,20.13,24.25,33.21,37.54,41.95 Using Mathcad's genfit, fit the data with a function of the form: f(x )= u1sin(2x)+u2cos(4x)+u3x, where the u's are constants to be determined using the least squares principle. Put in the vector of the desired function and its partial derivatives, the given data points and the guess vector as follows A column vector with: 1st row: Function to fit data 2nd row: Partial derivative with respect to u1 3rd row: Partial derivative with respect to u2 4th row: Partial derivative with respect to u3 F x u , ( ) u1 sin 2 x ⋅ ( ) ⋅ u2 cos 4 x ⋅ ( ) ⋅ + u3 x ⋅ + sin 2 x ⋅ ( ) cos 4 x ⋅ ( ) x ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := * <--- vx 0 1 2 3 4 5 6 7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := * vy 3 6.2 9.75 20.13 24.25 33.21 37.54 41.95 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := * <----- Given data points vg 1 2 3 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := * <--- vector of guess values for u1, u2, u3..... The function genfit now returns a vector of the u- coefficients that will best fit the given data. P genfit vx vy , vg , F , ( ) := * 234 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 245. That is ,u1 = 1.935, u2 = 3.175, u3= 6.251, as calculated by genfit. P 1.935 3.175 6.251 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ = * u1 P1 := u2 P2 := u3 P3 := The function that best fits the given data is now f x ( ) F x P , ( )1 := * or, use, f x ( ) u1 sin 2 x ⋅ ( ) ⋅ u2 cos 4 x ⋅ ( ) ⋅ + u3 x ⋅ + := It is plotted in Figure 6.19 and is compared with the given data points i 1 8 .. := * x 0 0.1 , 7 .. := * 0 1 2 3 4 5 6 7 0 10 20 30 40 50 Given data points Curve-fit obtained Given data points Curve-fit obtained Comparison of curve-fit with given data vyi f x ( ) vxi x , Figure 6.19. Curve-fitting with Mathcad's genfit Chapter 6: Curve-Fitting 235
- 246. 6.10 APPLICATIONS INCURVE-FITTING In the world of engineering, science, and business, it is very common to see a function in the form of data prescribed at many points, and in addition, there may also be considerable scatter in the available data . The methods discussed in the earlier sections of this chapter will now be used to generate appropriate curve-fits to actual data from several practical applications. 6.10.1 Fatigue Failure Curve for Loading in the Finite Life Range The fatigue failure curve of a material provides useful information pertinent to the design of machinery for finite life [ 19 ]. It is essentially a relationship between the completely reversed applied stress, S, and fatigue life, L , which is measured in terms of the number of stress reversals to failure. The following data pertains to a steel with an endurance limit of 40 kpsi and an ultimate strength of 90 kpsi. It is required to curve-fit the data with a suitable function. L ( million cycles) Stress (kpsi) vL 0.02 0.03 0.04 0.05 0.075 0.1 0.2 0.4 0.6 0.8 1.0 1.1 1.2 1.3 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := vS 80 70 65 61 57 55 50 45 42 40.5 40.1 40 39.9 39.85 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := From the plot of the given data shown in Figure 6.20, it appears that a power function would furnish a reasonable curve-fit. This has the form S(L)= A*LB , 236 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 247. where A andB are undetermined coefficients to be evaluated on a least squares basis. i 1 14 .. := 0 0.2 0.4 0.6 0.8 1 1.2 40 60 80 Given Data Given Data Given Data Points Given L- values (million cycles ) Given S- values (kpsi) vSi vLi Figure 6.20. Fatigue life data Using S(L) = A L B , Sstar= C+ B . Lstar , where Lstar= ln (L) , C = ln (A) and Sstar= ln (S), the matrix equation [D]{X} = {P} is obtained , where {X} is the column vector containing the unknowns C and B. n 14 := Sstar ln vS ( ) := Lstar ln vL ( ) := D X ⋅ P = D n 1 n i Lstar i ∑ = 1 n i Lstar i ∑ = 1 n i Lstar i ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := P 1 n i Sstari ∑ = 1 n i Sstari Lstar i ⋅ ( ) ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := The unknown constants C and B can be computed using lsolve as shown below K lsolve D P , ( ) := K 3.684 0.155 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <------Thus, C= 3.684 and B = -0.155 C K1 := B K2 := C 3.684 = B 0.155 − = A e C := A 39.808 = Chapter 6: Curve-Fitting 237
- 248. The curve-fit generatedby the linear regression analysis is now S L ( ) A L B ⋅ := with A 39.808 = Mathcad's linear regression can also be performed on this problem and the curve-fit obtained as shown in the following steps. vLstar ln vL ( ) := vSstar ln vS ( ) := C1 B1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ line vLstar vSstar , ( ) := B1 0.155 − = C1 3.684 = A1 e C1 := A1 39.808 = fmcad L ( ) A1 L B1 ⋅ := <--- Mathcad -generated fit Figure 6.21 shows a comparison of the curve-fits generated with the given data points . L 0.02 0.03 , 1.3 .. := i 1 14 .. := 0 0.2 0.4 0.6 0.8 1 1.2 20 40 60 80 Given data points Analysis-generated fit Mathcad's linear regression Given data points Analysis-generated fit Mathcad's linear regression Curve-Fitting with Power Function L- values (million cycles) S- values (kpsi) vSi S L ( ) fmcad L ( ) vLi L , Figure 6.21. Fatigue failure curve for loading in the finite life range 238 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 249. 6.10.2. Temperature Responseof an Object Placed in a Hot Stream of Air The following data pertains to the temperatures of a solid steel sphere suspended in a hot stream of air measured at several instants of time [ 17 ]. It is required to curve-fit the data with a suitable function. t= Time (secs) T=Temp (deg C) t=Time(Secs) T=Temp (deg C) 0.0 14 500 51 100 28 600 54 200 37.5 700 55 300 44 800 57.8 400 48 900 59.1 The given data is put in as follows and plotted in Figure 6.22. n 10 := t1 0 := t2 100 := t3 200 := t4 300 := t5 400 := t6 500 := t7 600 := T1 14 := T2 28 := T3 37.5 := T4 44 := T5 48 := T6 51 := T7 54 := t8 700 := t9 800 := t10 900 := T8 55 := T9 57.8 := T10 59.1 := i 1 10 .. := 0 200 400 600 800 0 20 40 60 Given data points Given data points Ti ti Figure 6.22. Given temperature versus time data A glance at the plot of Figure 6.22 suggests that a function of the following form would provide a reasonable fit to the given data f t ( ) A B e t − τ ⋅ + = where f(t) is a linear combination of functions in which A and B are constants to be determined on a least squares basis. From the data plot, it appears that the temperature profile attains steady state in approximately 1200 seconds, which , when equated to four time constants, will yield a time constant, τ, of 300 seconds. Thus, the coefficient of t in the exponential term can be selected as (-1/300). Chapter 6: Curve-Fitting 239
- 250. In this case,the matrix equation to be solved is [M] {C} = {A} in which {M] and {A} contain the functions f1 t ( ) 1 := f2 t ( ) e 1 − 300 t ⋅ := and M 1 n i f1 ti ( )2 ∑ = 1 n i f1 ti ( ) f2 ti ( ) ⋅ ( ) ∑ = 1 n i f1 ti ( ) f2 ti ( ) ⋅ ( ) ∑ = 1 n i f2 ti ( )2 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := , C C1 C2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = , A 1 n i f1 ti ( ) Ti ⋅ ( ) ∑ = 1 n i f2 ti ( ) Ti ⋅ ( ) ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := In the above, note that n = 10 and m =2, and the unknown constants are C1 and C2 which are determined below using lsolve. C lsolve M A , ( ) := A C1 := B C2 := C 60.477 45.967 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = The function fitting the given data , then, is f t ( ) A B e 1 − 300 t ⋅ ⋅ + := which is shown in Figure 6.23 . f 0 ( ) 14.511 = f 200 ( ) 36.877 = f 800 ( ) 57.283 = f 900 ( ) 58.189 = Mathcad's function linfit can also be used to evaluate the undetermined coefficients A and B in the regression function employed which is a linear combination of the functions selected for the curve-fit. This can be done as shown in the following steps. vt t := vT T := F t ( ) 1 e 1 − 300 t ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := S linfit vt vT , F , ( ) := 240 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 251. S 60.477 45.967 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = <-- Thisvector returns the coefficients A and B which are referred to here as Astar and Bstar Astar S1 := Bstar S2 := Astar 60.477 = Bstar 45.967 − = The function g(t) which is a scalar product of the vectors F(t) and S will then yield the linear combination employed g t ( ) F t ( ) S ⋅ := That is, g t ( ) Astar f1 t ( ) ⋅ Bstar f2 t ( ) ⋅ + := The linfit results and those of the linear combination analysis are presented in Figure 6.23 and compared with the given data . f t ( ) A B e 1 − 300 t ⋅ ⋅ + := g t ( ) F t ( ) S ⋅ := i 1 10 .. := t 0.5 1.0 , 900 .. := 0 100 200 300 400 500 600 700 800 900 0 20 40 60 Given data points Analysis fit linfit results Given data points Analysis fit linfit results Comparison-curve-fits with data time , seconds temperature, deg. C vTi f t ( ) g t ( ) vti t , Figure 6.23. Temperature response of an object placed in a hot stream of air Chapter 6: Curve-Fitting 241
- 252. 6.10.3. The Effectof Operating Temperature on the Strength of a Mechanical Element When a mechanical element is subjected to reversed stresses at temperatures below room temperature, there is a strong possibility of the occurrence of brittle fracture . On the other hand, when the operating temperatures are higher than room temperature, the yield strength of the material drops off very rapidly with increase in temperature, and yielding can take place . The temperature-corrected value of tensile strength, then, to be used in design calculations is obtained by multiplying the tensile strength at room temperature by a factor K d which is a function of the temperature of the operating environment. The following data was collected from tests done on carbon and alloy steels and shows the effect of operating temperature on tensile strength [ 18 ]. An appropriate curve-fit to the given data is sought. Temp., deg F. K d Temp.,deg F. Kd 0 0.975 600 0.963 100 1.008 700 0.927 200 1.020 800 0.872 300 1.024 900 0.797 400 1.018 1000 0.698 500 0.995 1100 0.567 The given data points, which are plotted in Figure 6.24 below, are put in as follows. n 12 := T1 0 := T2 100 := T3 200 := T4 300 := T5 400 := T6 500 := T7 600 := T8 700 := T9 800 := T10 900 := T11 1000 := T12 1100 := Kd1 0.975 := Kd2 1.008 := Kd3 1.020 := Kd4 1.024 := Kd5 1.018 := Kd6 0.995 := Kd7 0.963 := Kd8 0.927 := Kd9 0.872 := Kd10 0.797 := Kd11 0.698 := Kd12 0.567 := 242 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 253. 0 200 400600 800 1000 0.5 1 Given data points Given data points Given data points Temperature, deg. F Factor Kd Kdi Ti i 1 12 .. := Figure 6.24. Given Kd versus temperature data We will fit the given data with a second-order polynomial of the following form. Kd(T) = f(T) = C1 + C2 T + C3T 2 Here, n = 12 and m = 2 and the matrix equation to be solved is: [M]{C} = {A}, where [M] is a known 3x3 matrix and {A} is a known column vector as shown below. The column vector {C} , which involves the undetermined coefficients, is C C1 C2 C3 ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ = The [M] and {A} matrices are Chapter 6: Curve-Fitting 243
- 254. M n 1 n i Ti ∑ = 1 n i Ti ( )2 ∑ = 1 n i Ti ∑ = 1 n i Ti ( )2 ∑ = 1 n i Ti ()3 ∑ = 1 n i Ti ( )2 ∑ = 1 n i Ti ( )3 ∑ = 1 n i Ti ( )4 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := A 1 n i Kdi ∑ = 1 n i Ti Kdi ⋅ ( ) ∑ = 1 n i Ti ( )2 Kdi ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := Using the Mathcad function lsolve gives C lsolve M A , ( ) := C1 C1 := C2 C2 := C3 C3 := C1 0.968 = C2 4.199 10 4 − × = C3 6.966 − 10 7 − × = f T ( ) C1 C2 T ⋅ + C3 T 2 ⋅ + := f 200 ( ) 1.024 = f 700 ( ) 0.921 = f 900 ( ) 0.782 = The regress and loess functions in Mathcad can also be employed as shown below to generate curve-fits to the given data . i 1 12 .. := vTi Ti := vKdi Kdi := vs1 regress vT vKd , 2 , ( ) := vs2 loess vT vKd , 2 , ( ) := freg T ( ) interp vs1 vT , vKd , T , ( ) := floess T ( ) interp vs2 vT , vKd , T , ( ) := Figure 6.25 shows a comparison of the various curve-fits generated with the given data points. f T ( ) C1 C2 T ⋅ + C3 T 2 ⋅ + := T 0 1 , 1100 .. := i 1 12 .. := 244 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 255. 0 200 400600 800 1000 0.4 0.6 0.8 1 1.2 Given data points Analysis-generated function Mathcad regress Mathcad loess Given data points Analysis-generated function Mathcad regress Mathcad loess Curve-fits generated Temperature, deg F Multiplying factor Kd vKdi f T ( ) freg T ( ) floess T ( ) vTi T , Figure 6.25. Effect of operating temperature on the strength of a mechanical element 6.10.4. Drop-Testing of Packaged Articles An interesting parameter that often warrants investigation is the height from which a package can be dropped before it suffers any accountable damage, which can be assessed from the maximum acceleration imparted to the package at the end of the drop [ 20 ]. This is a question that arises when packages must be properly cushioned before shipment to another location. The maximum acceleration to gravity ratio, a , has been found to be a function of the ratio, h , of twice the distance dropped to the static deflection of the package. The following a versus h data is provided , using which a reasonable curve-fit is to be generated . h a h a ________________________ 1 1.4 9 3.25 2 1.8 10 3.3 3 2 11 3.55 4 2.25 12 3.65 5 2.65 13 3.75 6 2.8 14 3.85 7 2.9 15 4.2 8 3.01 16 4.3 _______________________________ Chapter 6: Curve-Fitting 245
- 256. These data pointsare plotted in Figure 6.26. n 16 := i 1 2 , 16 .. := h1 1 := hi 1 + hi 1 + := a1 1.4 := a2 1.8 := a3 2. := a4 2.25 := a5 2.65 := a6 2.8 := a7 2.9 := a8 3.01 := a9 3.25 := a10 3.3 := a11 3.55 := a12 3.65 := a13 3.75 := a14 3.85 := a15 4.2 := a16 4.3 := We will generate a curve-fit of the form: a = f(h) = C1 + C2 h + C3 h 2 + C 4 h 3 Here, n = 16 and m = 3 and the matrix equation to be solved is: [M]{C} = {A}, 0 2 4 6 8 10 12 14 16 1 2 3 4 5 Given data points Given data points Given data points (2h/ static deflection ) ratio (max acceleration/ gravity) ratio ai hi Figure 6.26. Given acceleration ratio versus height ratio data 246 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 257. where [M] isa known 4x4 matrix and {A} is a known column vector as shown below. The column vector {C} , which involves the undetermined coefficients, is C C1 C2 C3 C4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = M n 1 n i hi ∑ = 1 n i hi ( )2 ∑ = 1 n i hi ( )3 ∑ = 1 n i hi ∑ = 1 n i hi ( )2 ∑ = 1 n i hi ( )3 ∑ = 1 n i hi ( )4 ∑ = 1 n i hi ( )2 ∑ = 1 n i hi ( )3 ∑ = 1 n i hi ( )4 ∑ = 1 n i hi ( )5 ∑ = 1 n i hi ( )3 ∑ = 1 n i hi ( )4 ∑ = 1 n i hi ( )5 ∑ = 1 n i hi ( )6 ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := A 1 n i ai ∑ = 1 n i hi ai ⋅ ( ) ∑ = 1 n i hi ( )2 ai ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = 1 n i hi ( )3 ai ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ := C lsolve M A , ( ) := C 0.997 0.433 0.029 − 9.472 10 4 − × ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = C1 C1 := C2 C2 := C3 C3 := C4 C4 := C1 0.997 = C2 0.433 = C3 0.029 − = C4 9.472 10 4 − × = f h ( ) C1 C2 h ⋅ + C3 h 2 ⋅ + C4 h 3 ⋅ + := f 15 ( ) 4.114 = f 9 ( ) 3.219 = The Mathcad regress and loess functions also can be utilized as shown below to generate curve-fits to the given data. Chapter 6: Curve-Fitting 247
- 258. i 1 2 ,16 .. := vhi hi := vai ai := vs1 regress vh va , 3 , ( ) := vs2 loess vh va , 2 , ( ) := freg h ( ) interp vs1 vh , va , h , ( ) := floess h ( ) interp vs2 vh , va , h , ( ) := h 1 1.1 , 16 .. := i 1 2 , 16 .. := Figure 6.27 shows a comparison of the various curve-fits obtained with the given data points. 0 2 4 6 8 10 12 14 16 1 2 3 4 5 Given data points Analysis-generated fit Mathcad 's regress function Mathcad's loess function Given data points Analysis-generated fit Mathcad 's regress function Mathcad's loess function Curve-fits generated (2 h / static deflection ) ratio (max acceleration/gravity ) ratio vai f h ( ) freg h ( ) floess h ( ) vhi h , Figure 6.27. Acceleration ratio versus height ratio curve-fit PROBLEMS In all problems, also obtain results with the Mathcad functions, slope, intercept, regress, loess, interp, expfit, and pwrfit as appropriate and compare the results of your regression analysis with the given data and results generated with Mathcad. Show this comparison by means of Mathcad plots with proper labels, titles and traces. 6.1 Given the following data. x: 1.3 3.2 4.6 6.1 y: 52 46 38 26 fit a straight line y= f(x)= C1 + C2 x to the above data. Determine y (5.1) 248 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 259. 6.2. Given: x: 01 2 3 4 y: 11.25 12.34 18.56 28.69 44.73 determine a least squares quadratic fit. Determine y ( 3.75 ) 6.3. Generate a least squares fit of the form f(x) = AxB to the following data x: 1 2 4 5 y: 2 10 60 110 Apply a weighting factor of 3 to the last data point and determine f(3.5) 6.4. The thermal conductivity, K, of a material varies with temperature, T, as follows T ( deg Kelvin) 200 600 1000 1400 K (W/(cm-deg Kelvin) 1.0 0.40 0.30 0.25 Determine a least squares fit of the form K = ATB. Determine K for T = 1200 deg Kelvin. Obtain an answer that is good to two decimal places. 6.5. For the data set given below, determine a power function that will serve as a least squares fit. x: 1 2 3 4 5 6 7 8 y: 0.43 0.71 0.94 1.07 1.19 1.27 1.32 1.47 Compute y( 4.5) 6.6. For the data set given below, find an exponential function of form : y= f(x) = A exp(Bx) that will serve as a least squares fit. x: 0 1 2 2.75 3.5 y: 2.5 6.75 15.42 28.86 52.03 Determine y( 3.0) Chapter 6: Curve-Fitting 249
- 260. 6.7. For thedata set given below, find an exponential function of form : y= f(x) = A exp(Bx) that will serve as a least squares fit. x: 0 1 2 3 4 y: 2.01 1.37 0.83 0.55 0.33 Determine y(1.5) 6.8 For the experimental data given below Strain, x, ( μ-in/in) 40 50 70 90 Displacement, y, (inches) 0.015 0.028 0.045 0.072 Determine a least squares fit of the form f(x)= Ax e (Bx). Determine f(45) and f(60) 6.9. For the strain ( e) / displacement (d) data shown below: e: 43 57 68 84 99 115 (micro-in/in) d: 0.009 0.017 0.029 0.043 0.067 0.097 (inches) determine a least squares fit using an approximation of the form: e= A x exp(-Bx) Determine d(72) 6.10. The temperature of a a hot surface varies sinusoidally with time. The data which is in terms of temperatures, T (deg C), measured at several time intervals , τ , where τ is the number of hours from midnight, is given below. τ (hours) 2 4 8 T (deg C) 80 120 105 Obtain a curve fit to the given data of the form T = A sin ( πτ/12) + B cos ( πτ/12) Determine T (6 hours). 250 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 261. 6.11. Fit thedata points given below with a function of the form y(x) = C1 f1(x) + C2 f2(x) where f1(x) = x and f2(x) = x3. Compute y(3.75). x: 2.25 3.30 4.40 y: 9.75 15.60 28.27 6.12. The following data pertains to the temperatures of an object suspended in a hot stream of air measured at several instants of time t, Time (seconds) 0.0 400 1200 T, Temperature (deg C) 20 50 61.5 Curve-fit the data with a function of the form f t ( ) A B e 0.0033 − t ⋅ ( ) ⋅ + = and estimate the temperature attained by the object at time t = 600 seconds 6. 13. Fit a linear combination of known functions f(x) = A + B x + C sin (πx) + D sin (3πx ) to x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 ( ) = y 0.5541 2.1456 2.8021 2.5645 3.2458 2.9765 2.8521 2.4379 1.7872 0.1315 ( ) = Determine y ( 0.85) 6.14. Given the 4 data points x: 0 2.5 4.75 6.26 y: 0 12.45 22.67 36.89 obtain a curve-fit of the form: y = f(x) = C1 + C2 x + C3 x 2 + C 4 x 3 . Apply a weighting factor of 2 to the last two data points. Determine y ( 3.5). 6.15. Data defining the stress (S) versus strain (e) curve for an aluminum alloy is given below. Chapter 6: Curve-Fitting 251
- 262. Strain,e, ( %)Stress,S, (Kpsi) 10 63 20 63.6 40 62 60 60 80 58 100 56 120 52 140 48 150 47 Using an approximating polynomial of the form S C1 C2 e ⋅ + C3 e 2 ⋅ + = obtain a least squares fit to the given data. Determine S(105%) 6. 16. The following data relating the distance traveled by an object to time was obtained from a test in an experimental test track . The distance traveled, S, is in meters and the time , t, is in seconds. Time, t (secs) 0.1 0.5 1.5 3.0 Distance, S (m) 0.26 1.56 7.41 30.90 Generate a curve-fit of the form S(t) = A + B t + C t 2. Compute S for t= 2 seconds. 6.17. The pressure drag , D P, on an object can be reduced in comparison with the total drag , DT, by streamlining it , that is, making its length, L, in the direction of flow larger with respect to its maximum thickness or diameter, D . [ 17 ] . Obtain an appropriate curve-fit to the following data which is provided to demonstrate the effect of streamlining on the pressure drag of a body of symmetrical airfoil cross section. L D 1 2 3 4 5 6 7 8 9 10 11 ( ) = DP DT 1.25 0.65 0.42 0.25 0.20 0.10 0.08 0.06 0.04 0.02 0.01 ( ) = Determine D P / DT for L / D = 5.5 6.18. The following data , in which the time, t, is given in seconds provides the 252 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 263. response c (t ) of a control system to a step input. t 0 0.75 1.50 2 3 4 5 6 ( ) := c t ( ) 0 1.30 1.40 1.10 0.85 1.02 1.01 1.009 ( ) := Obtain a least-squares fit of the form. c(t)= A+ B e -t sin (2.828 t ) + C e -t cos ( 2.828 t ) and determine c at t = 3.5 seconds. 6.19. The Brinell hardness number, Bhn , for steel [ 18 ] is given as a function of the tempering temperature, T, in the following table T, deg F. 400 600 800 1000 1200 Bhn 540 500 445 375 310 Obtain an appropriate curve-fit to the given data and determine the Brinell hardness number for T = 850 deg F. 6.20. The average strain rate used in obtaining stress/ strain data for a material specimen is about 0.001 in / (in-sec ). However, when the strain rate, SR, is increased, as it happens, under conditions of impact loading, the strength of the material also increases. [ 18 ] The following data shows the influence of the increase in strain rate on the yield strength, SYLD, of a steel specimen SR 10 6 − 10 5 − 10 4 − 10 2 − 1 10 2 10 3 ( ) = in/(in-sec) SYLD 30 28 30 37 45 65 78 ( ) = kpsi Obtain an appropriate least-squares fit to the given data and determine SYLD corresponding to an SR of 50. 6.21. The endurance limit of a steel specimen is, typically, a function of the condition of its surface [ 19 ]. For an unnotched, ground specimen in reversed bending, the endurance limit, S end , is related to its tensile strength, S ult , as given in the following table of data. Sult 60 80 100 120 140 160 180 200 220 ( ) = kpsi Send 25 35 45 55 65 72 80 85 90 ( ) = kpsi Chapter 6: Curve-Fitting 253
- 264. Obtain a reasonablecurve-fit to the given data. Determine S end corresponding to S ult = 135 kpsi . 6.22. Expected voice levels required for speech communication with various back- ground noise levels and separations between the speaker and the listener [ 10 ] are given in the following table of data . Obtain appropriate curve-fits to the given data in the form of separate curves for (a) Peak shouting (b) Shouting (c) Very loud voice (d) Raised voice and (e) Normal voice. Determine the voice levels required in the following situations. (1) Speaker listener distance = 2.5 m , background noise level = 68 dBA (2) Speaker listener distance = 4.5 m , background noise level = 86 dBA (3) Speaker listener distance = 8.5 m , background noise level = 45 dBA (4) Speaker listener distance = 6.5 m , background noise level = 65 dBA BACKGROUND NOISE LEVELS (dBA) Speaker /listener distance ( m ) Normal Voice Raised Voice Very Loud Voice Shouting Peak Shouting 0.25 0.5 1.0 2 3 4 5 6 7 8 9 9.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 90 80 72 63 57 54.5 52 50 49.5 49 48 47.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 62 60 57 56.5 55 54 53.5 52.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 95 88 78 72 68 66 64 62.5 61.5 60 58 57.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 100 92 84 77 74 72 68 67.5 66 65 64.5 64 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 118 112 103 96 92.5 90 88 86.5 86 85 84.5 84 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 254 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 265. C H AP T E R 7 NUMERICAL DIFFERENTIATION 7.1 INTRODUCTION TO NUMERICAL DIFFERENTIATION AND THE USE OF THE MATHCAD DERIVATIVE OPERATORS In the process of generating a solution to a science or an engineering problem, the need to come up with a numerical estimate of a derivative is often encountered. We will discuss two methods of computing derivatives as follows 1. Method of finite differences 2. Interpolating polynomial method. These methods will be discussed in detail in the paragraphs following. To compute the derivative of a function using the Mathcad derivative operator, first define the point x at which the derivative is to be computed . Then click on the derivative operator in the calculus palette of Figure 7.1 and fill in the placeholders as required. Press the equals sign to see the derivative. 7.2 METHOD OF FINITE DIFFERENCES In traditional calculus , a derivative of a function f(x) is defined as x f x ( ) d d 0 ∆x f x ∆x + ( ) f x ( ) − ∆x lim → = (7.1) In numerical methods, taking the above limit as Δx goes to zero is not possible , and, although, Δx can be very small, it certainly cannot be zero. The process by which this increment Δx is made small and utilized in calculations paves the way for the method of finite differences. If the dependent variable x is incremented by Δx, as shown in Figure 7.2 , the difference between the function value at (x+Δx ) and that at x is called a forward difference . In this case, the derivative of the function f(x) at x is x f x ( ) d d f x ∆x + ( ) f x ( ) − ∆x = (7.2) However, if the dependent variable is incremented by - Δx , the difference between the function values is a backward difference, and the derivative , in this case is x f x ( ) d d f x ( ) f x ∆x − ( ) − ∆x = (7.3) Chapter 7: Numerical Differentiation 255
- 266. Figure 7.1. Useof the Mathcad derivative operators 256 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 267. Figure 7.2. Finitedifferences If a two-step difference is employed, the method of computing derivatives is called the two-step method. An estimate of the derivative, in this case, is x f x ( ) d d f x ∆x + ( ) f x ∆x − ( ) − 2 ∆x ⋅ = (7.4) For obtaining an estimate of the second derivative of f(x), we will consider the following Taylor series expansions. f x ∆x + ( ) f x ( ) ∆x ( ) x f x ( ) d d ⋅ + ∆x 2 2! 2 x f x ( ) ................................................................. + ( d d 2 ⋅ + = (7.5) f x ∆x − ( ) f x ( ) ∆x ( ) x f x ( ) d d ⋅ − ∆x 2 2! 2 x f x ( ) ............................................................. + ( d d 2 ⋅ + = (7.6) Adding Equations (7.5) and (7.6) with terms beyond the second derivative terms ignored will give an estimate of the second derivative as 2 x f x ( ) ( d d 2 1 ∆x 2 f x ∆x + ( ) 2 f x ( ) ⋅ − f x ∆x − ( ) + ( ) ⋅ = (7.7) By following a similar procedure, the third and fourth derivatives can also be estimated as Chapter 7: Numerical Differentiation 257
- 268. 3 x f x ( ) d d 31 ∆x 3 f x 2 ∆x ⋅ + ( ) 2 f x ∆x + ( ) ⋅ − 2 f x ∆x − ( ) ⋅ + f x 2 ∆x ⋅ − ( ) − ( ) ⋅ = (7.8) 4 x f x ( ) d d 4 1 ∆x 4 f x 2 ∆x ⋅ + ( ) 4 f x ∆x + ( ) ⋅ − 6 f x ( ) ⋅ + 4 f x ∆x − ( ) ⋅ − f x 2 ∆x ⋅ − ( ) + ( ) ⋅ = (7.9) Example 7.1. Using the data given, compute the first derivative at x= 4 using the forward, backward as well as the two-step finite difference approximations. Also compute the second derivative at x= 4 . x 2 3 4 5 6 f(x) 0.703 1.212 1.456 1.723 1.853 The first derivative at x=4 using Equation (7.2) , which is the forward difference approximation is fprimefwd 1.723 1.456 − 5 4 − := fprimefwd 0.267 = The first derivative at x=4 using Equation (7.3) , which is the backward difference approximation, is fprimebckwd 1.456 1.212 − 4 3 − := fprimebckwd 0.244 = The first derivative at x=4 using Equation (7.4) , which is the two-step difference approximation, is fprimetwostep 1.723 1.212 − 5 3 − := fprimetwostep 0.2555 = The second, third and fourth derivatives at x =4 are given by Equations (7.7), (7.8) and (7.9), and these are ∆x 1 := fdblprime 1 ∆x 2 1.723 2 1.456 ⋅ − 1.212 + ( ) ⋅ := fdblprime 0.023 = ftripleprime 1 ∆x 3 1.853 2 1.723 ⋅ − 2 1.212 ⋅ + 0.703 − ( ) ⋅ := ftripleprime 0.128 = 258 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 269. ffourprime 1 ∆x 4 1.853 4 1.723 ⋅ −6 1.456 ⋅ + 4 1.212 ⋅ − 0.703 + ( ) ⋅ := ffourprime 0.448 − = 7.3. INTERPOLATING POLYNOMIAL METHOD In the interpolating polynomial method, an interpolating polynomial passing through given data points is determined by any of the methods discussed in Chapter 5. Then, the polynomial generated can be differentiated as is done in traditional calculus. If the polynomial derived , for instance, is f x ( ) a0 a1 x ⋅ + a2 x 2 ⋅ + a3 x 3 ⋅ + ............ + an x n ⋅ + = (7.10) then, its derivative can be obtained as x f x ( ) d d a1 2 a2 ⋅ x ⋅ + 3 a3 ⋅ x 2 ⋅ + .............. + n an ⋅ x n 1 − ⋅ + = (7.11) Example 7.2 Given the following data : x 2 3 4 5 6 y 0.80354 1.23591 1.62312 1.89765 1.96789 Use the Method of Undetermined Coefficients to fit an interpolating polynomial to the given data and estimate: yprime(4.5), yprime(2.7), ydblprime(4.5) and ydblprime(2.7). Since 5 data points are given, we can generate a 4 th order polynomial. Let y(x) = f(x)= a + bx + cx2 + dx3 + e x4 Substituting the given values of (x,y) in the assumed polynomial above yields the following matrix equation: [M]{A}= {V} where Chapter 7: Numerical Differentiation 259
- 270. M 1 1 1 1 1 2 3 4 5 6 2 2 3 2 4 2 5 2 6 2 2 3 3 3 4 3 5 3 6 3 2 4 3 4 4 4 5 4 6 4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := V 0.80354 1.23591 1.62312 1.89765 1.96789 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Using theMathcad function lsolve, the unknown coefficients in the column vector {A} can be found A lsolve M V , ( ) := A 0.04705 − 0.40726 7.43375 10 3 − × 2.79917 10 3 − × 1.00375 − 10 3 − × ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = The coefficients a, b, c, and d of the assumed polynomial are a A0 := a 0.04705 − = c A2 := c 7.43375 10 3 − × = d A3 := d 2.79917 10 3 − × = b A1 := b 0.40726 = e A4 := e 1.00375 − 10 3 − × = The polynomial derived is, then, y x ( ) a b x ⋅ + c x 2 ⋅ + d x 3 ⋅ + e x 4 ⋅ + := where a, b, c,d and e are as given above and with first and second derivatives as follows First derivative: yprime x ( ) b 2 c ⋅ x ⋅ + 3 d ⋅ x 2 ⋅ + 4 e ⋅ x 3 ⋅ + := yprime 4.5 ( ) 0.27835 = yprime 2.7 ( ) 0.42959 = Second derivative: ydblprime x ( ) 2 c ⋅ 6 d ⋅ x ⋅ + 12 e ⋅ x 2 ⋅ + := ydblprime 4.5 ( ) 0.15347 − = ydblprime 2.7 ( ) 0.02759 − = 260 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 271. Use of theMathcad derivative operators yields: x 4.5 := x y x ( ) d d 0.27835 = 2 x y x ( ) ( ) d d 2 0.15347 − = x 2.7 := x y x ( ) d d 0.42959 = 2 x y x ( ) ( ) d d 2 0.02759 − = Notice that the Mathcad derivative operators have yielded the same results as those obtained through the manual process of differentiation. These are presented in Figures 7.3 and 7.4. x 2.0 2.1 , 6.0 .. := 2 3 4 5 6 0.2 − 0 0.2 0.4 0.6 First deriv/ inter poly First deriv/Mathcad A comparison of results X-Values First derivative values yprime x ( ) x y x ( ) d d x Figure 7.3 Graph of first derivative 2 3 4 5 6 0.4 − 0.3 − 0.2 − 0.1 − 0 0.1 Second deriv/Inter poly Second deriv/ Mathcad A comparison of results X-Values Second derivative values ydblprime x ( ) 2 x y x ( ) ( ) d d 2 x Figure 7.4. Graph of second derivative Chapter 7: Numerical Differentiation 261
- 272. The interpolating polynomialcan also be generated by using the Mathcad function interp with lspline, pspline or cspline, and then numerically differentiated by using the Mathcad derivative operator, as shown below. j 0 4 .. := Vyj Vj := <-- Generate the Vx and Vy vectors for given data Vy 0.80354 1.23591 1.62312 1.89765 1.96789 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vx 2 3 4 5 6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vs cspline Vx Vy , ( ) := fspline x ( ) interp Vs Vx , Vy , x , ( ) := <--- interpolating polynomial generated x 4.5 := x fspline x ( ) d d 0.2786 = 2 x fspline x ( ) ( ) d d 2 0.15547 − = x 2.7 := x fspline x ( ) d d 0.43082 = 2 x fspline x ( ) ( ) d d 2 0.02671 − = The Mathcad results appear to be in reasonable agreement with those of the Undetermined Coefficients method. 7.4. APPLICATIONS IN NUMERICAL DIFFERENTIATION The following examples illustrate situations where numerical differentiation can be applied and used to advantage in generating realistic solutions to practical problems. 7.4.1. Determination of Velocities and Accelerations from Given Displacement Data The following position versus time data for a train has been experimentally obtained. Time t (secs) 0 10 20 30 35 40 Position s ( m) 0 75 175 360 480 600 262 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 273. Determine the velocityand acceleration of the train as functions of time. Estimate the velocity and acceleration of the train at t = 32 seconds We first use the method of undetermined coefficients to fit an interpolating polynomial to the given data. Since six data points are given, a fifth order polynomial of the following form will be generated. s(t) = f(t)= a + bt + ct2 + dt3 + e t4 + f t 5 Substituting the given values of (t,s) in the assumed polynomial above yields the following matrix equation: [M]{A} = {V} where V 0 75 175 360 480 600 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := M 1 1 1 1 1 1 0 10 20 30 35 40 0 2 10 2 20 2 30 2 35 2 40 2 0 3 10 3 20 3 30 3 35 3 40 3 0 4 10 4 20 4 30 4 35 4 40 4 0 5 10 5 20 5 30 5 35 5 40 5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Using the Mathcad function lsolve, the unknown coefficients in the column vector {A} can be found A lsolve M V , ( ) := A 0 9.41429 0.36131 − 0.01667 7.7381 10 5 − × 4.52381 − 10 6 − × ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Chapter 7: Numerical Differentiation 263
- 274. The coefficients a,b, c, d. e and f of the assumed polynomial are a A0 := a 0 = c A2 := c 0.36131 − = d A3 := d 0.01667 = b A1 := b 9.41429 = e A4 := e 7.7381 10 5 − × = f A5 := f 4.52381 − 10 6 − × = The polynomial derived is, then, s t ( ) a b t ⋅ + c t 2 ⋅ + d t 3 ⋅ + e t 4 ⋅ + f t 5 ⋅ + := with first and second derivatives as follows First derivative: velocity t ( ) b 2 c ⋅ t ⋅ + 3 d ⋅ t 2 ⋅ + 4 e ⋅ t 3 ⋅ + 5 f ⋅ t 4 ⋅ + := velocity 32 ( ) 23.91516 = (m/sec) Second derivative: acceler t ( ) 2 c ⋅ 6 d ⋅ t ⋅ + 12 e ⋅ t 2 ⋅ + 20 f ⋅ t 3 ⋅ + := acceler 32 ( ) 0.46351 = (m/sec2) Use of the Mathcad derivative operators yields: t 32 := t s t ( ) d d 23.91516 = m/sec 2 t s t ( ) ( ) d d 2 0.46351 = (m/sec2) As is expected, the results of the Mathcad derivative operators are the same as those obtained through the process of manual differentiation of the interpolating polynomial . They are presented in Figure 7.5. 264 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 275. t 0 1 ,40 .. := 0 10 20 30 40 0 200 400 600 Position versus time Time , secs Position, meters s t ( ) t 0 10 20 30 40 5 10 15 20 25 Interpolating polynomial method Mathcad's derivative operator Velocity versus time Time, secs Velocity, m/sec velocity t ( ) t s t ( ) d d t 0 10 20 30 40 1.5 − 1 − 0.5 − 0 0.5 1 Interpolating polynomial method Mathcad's derivative operator Acceleration versus time Time, sec Acceleration, m/sec2 acceler t ( ) 2 t s t ( ) d d 2 t Figure 7.5. Position, velocity and acceleration versus time Chapter 7: Numerical Differentiation 265
- 276. 7.4.2. Determination ofShock Absorber Parameters , and Damper and Spring Restoring Forces from Given Vehicle Displacement Data The following displacement (x) versus time (t) data is provided for a 200 kg motorcycle with a shock absorber . t,sec 0 0.5 1.0 1.5 2.0 2.5 x,mm 0 0.23 0 -0.075 0.025 0 Generate an interpolating polynomial for the displacement data given and obtain velocity and acceleration profiles using Mathcad. Determine (a) the stiffness and the damping constant of the shock absorber and (b) the damper and spring restoring forces of the shock absorber as functions of time . The given data is plotted in Figure 7.6 below. i 1 2 , 6 .. := t1 0 := ti 1 + ti 0.5 + := x1 0 := x2 0.23 := x3 0 := x4 0.075 − := x5 0.025 := x6 0 := Vxi xi := Vti ti := 0 0.5 1 1.5 2 2.5 0.1 0 0.1 0.2 Given data Given data Given data Time, sec Displacement, mm Vxi Vti Figure 7.6. Given data 266 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 277. As done inthe preceding problem, we first use the method of undetermined coefficients to fit an interpolating polynomial to the given data. Since six data points are prescribed , a fifth order polynomial of the following form can be generated. x(t) = f(t)= a + bt + ct2 + dt3 + e t4 + f t 5 Substituting the given values of (t,x) in the assumed polynomial above yields the following matrix equation: [M]{A}= {V} where M 1 1 1 1 1 1 0 0.5 1 1.5 2 2.5 0 2 0.5 2 1 2 1.5 2 2 2 2.5 2 0 3 0.5 3 1 3 1.5 3 2 3 2.5 3 0 4 0.5 4 1 4 1.5 4 2 4 2.5 4 0 5 0.5 5 1 5 1.5 5 2 5 2.5 5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := V 0 0.23 0 0.075 − 0.025 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := A lsolve M V , ( ) := A 0 1.7375 3.69917 − 2.65167 0.76333 − 0.07333 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = The coefficients a, b, c, d. e and f of the assumed polynomial are a A1 := a 0 = c A3 := c 3.69917 − = d A4 := d 2.65167 = b A2 := b 1.737 = e A5 := e 0.76333 − = f A6 := f 0.073 = Chapter 7: Numerical Differentiation 267
- 278. The interpolated polynomialwhich is x t ( ) a b t ⋅ + c t 2 ⋅ + d t 3 ⋅ + e t 4 ⋅ + f t 5 ⋅ + := is plotted in Figure 7.7 below. As follows, Mathcad's cspline function can also be used with interp to generate a cubic spline interpolation which is compared with the undetermined coefficients method and with the given data . i 1 2 , 6 .. := Vs cspline Vt Vx , ( ) := Vt 0 0.5 1 1.5 2 2.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vx 0 0.23 0 0.075 − 0.025 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = fspline t ( ) interp Vs Vt , Vx , t , ( ) := t 0 0.01 , 2.5 .. := 0 0.5 1 1.5 2 2.5 0 0.1 0.2 Given data points Undetermined coefficients method Mathcad's cubic spline interpolation Given data points Undetermined coefficients method Mathcad's cubic spline interpolation Motorcycle displacement profile Time, seconds Displacement, mm Vxi x t ( ) fspline t ( ) Vti t , Figure 7.7. Interpolating polynomial generated 268 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 279. Using Mathcad's derivativeoperators, the velocity and the acceleration of the vehicle are also plotted as functions of time in Figure 7.8. 0 0.5 1 1.5 2 2.5 8 6 4 2 0 2 Displacement, mm Velocity, mm/sec Acceleration, mm/sec2 Displacement, mm Velocity, mm/sec Acceleration, mm/sec2 x t ( ) t x t ( ) d d 2 t x t ( ) d d 2 t Figure 7.8. Displacement, velocity and acceleration profiles The system settling time is the time taken for the free vibration of the motorcycle to completely die down. This , in this case, is 2.5 seconds. Since the settling time is approximately equal to four time constants for a 2 % error, the system time constant τ can be computed. τ 2.5 4 := τ 0.625 = seconds The displacement profile also indicates a natural period, τ d , of one oscillation to be approximately two seconds τd 2 := seconds yielding the natural frequency of the system as ωd 2 π τd ⋅ := ωd 3.142 = rad/sec Chapter 7: Numerical Differentiation 269
- 280. As shown below, the damping, c , and stiffness, k , furnished by the shock absorber can be computed using the pertinent formulas available in textbooks in Vibrations and Controls [ 6, 9, 14, 16, 20, 21]. Damping factor x undamped natural frequency = 1/ system time constant : ζωn 1 τ := ζωn 1.6 = Since ζωn and ωd are known , the damping factor ζ can be calculated. ωd ωn 1 ζ 2 − ⋅ = ωd 2 π τd ⋅ := ωd 3.142 = rad/sec ζ ζωn 2 ωd 2 ζωn 2 + := ζ 0.454 = The undamped natural frequency of the system along with the stiffness and damping coefficients of the shock absorber can now be computed using ωn ζωn ζ := ωn 3.526 = rad/sec The stiffness and damping coefficients of the shock absorber can now be determined. m 200 := kg k m ωn 2 ⋅ := k 2.486 10 3 × = N/m c 2 m ⋅ ωn ⋅ ζ ⋅ := c 640 = N-sec/ m The shock absorber spring and damper restoring forces which can now be calculated as functions of time using 270 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 281. Fspring t ( )k x t ( ) 1000 ⋅ := Fdamper t ( ) c t x t ( ) 1000 d d ⋅ := are plotted in Figure 7.10. t 0 0.1 , 2.5 .. := 0 0.5 1 1.5 2 2.5 0.5 0 0.5 1 1.5 Shock absorber spring/damp rest forces Time, seconds Force, Newtons Fdamper t ( ) Fspring t ( ) t Figure 7.9. Shock absorber spring and damper restoring forces PROBLEMS 7.1. Given the following function θ 4 5 , 6 , 7 , 8 , ( ) = Cos θ ( ) 0.99756 0.99619 , 0.99452 , 0.99252 , 0.99027 , ( ) = determine the first derivative of cos(θ ) at θ = 7 degrees, using forward, backward and two-step finite difference approximations. Compare the results with the true value. Chapter 7: Numerical Differentiation 271
- 282. Also obtain ananswer using cspline/interp and the Mathcad derivative operator. Obtain the Mathcad interpolated function and its derivative in the given range and graph them with proper labels and titles. 7.2. Given the following x 1.0 1.2 1.4 1.6 1.8 2.0 f(x) 1.1 0.9283 0.8972 0.9015 0.9422 1.1 compute first and second derivatives at x= 1.4, using finite differences . Also obtain answers using cspline/interp and the Mathcad derivative operators. Obtain the Mathcad interpolated function and its first and second derivatives in the given range and graph them with proper labels and titles. 7.3. Given f(x) versus x as follows x 1 2 3 4 5 f(x) 5.25 155.2 458.1 1682 6489 fit an interpolating polynomial using the method of undetermined coefficients and estimate the first derivative at x=3.5 and x=4.5. Also obtain answers using cspline/interp and the Mathcad derivative operator. Graph the polynomials obtained in the given range . Show a comparison of the first derivative obtained by the undetermined coefficient method with that generated using Mathcad. 7.4. Fit an interpolating polynomial to the data of Problem 7.2. Estimate first and second derivatives at x= 1.4. Also obtain answers using cspline/interp and the Mathcad derivative operators. Graph the polynomials obtained in the given range . Show a comparison of the first derivative and second derivative obtained by the undetermined coefficient method with those generated using Mathcad. 7.5. The horizontal and vertical positions of a weather balloon with respect to a station are furnished below at various instants of time. 272 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 283. t (seconds) horiz distance x (ft) verticaldistance, y (ft) 0 8 16 24 32 40 48 54 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 6.4 25.6 57.6 102.4 160 230.4 291.6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 1 2 3 4 5 6 7 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Using the method of undetermined coefficients, fit interpolating polynomials to the given data and generate x versus time and y versus time curves. Estimate horizontal and vertical velocities as well as horizontal and vertical accelerations at t = 4.5 seconds. Also obtain answers using cspline/interp and the Mathcad derivative operators. Show a comparison of all results generated with the help of appropriate plots. 7.6. The deflection curve of a cantilevered beam is given by the following data, in which x is the distance in feet from the fixed end and y is the deflection in feet at x . x 0 1 2 3 4 5 6 7 8 ( ) = y 0 0.015 0.05 0.1 0.2 0.3 0.4 0.5 0.6 ( ) = Using Mathcad's cspline/interp functions, generate an interpolating polynomial for the given data , and obtain the beam slope at x = 5.5 ft and x= 6.5 ft. Also plot the deflection versus x and slope versus x curves 7.7. In a slider-crank mechanism , the crank angle θ and the position x of the slider are prescribed below at various instants of time. The position x is in feet while the angle θ is in radians θ 0 0.393 0.785 1.178 1.571 1.963 2.356 2.749 3.142 ( ) := x 0.667 0.65 0.604 0.539 0.471 0.412 0.368 0.342 0.333 ( ) := Chapter 7: Numerical Differentiation 273
- 284. Figure P 7.7 Usingcspline/interp, generate an interpolating polynomial for the given data. If the rotational speed of the crank is ω 2 = 1000 RPM, counterclockwise, and it is maintained constant , estimate the velocity and acceleration of the slider at θ = π/2 and 3π/4 radians. Plot the position x, velocity dx/dt and acceleration d2x/ dt 2 as functions of the crank angle θ. 7,8 The data in the following table are the distances (D) traveled by a motorcycle for selected times (t) t(seconds): 0 2 4 6 8 10 D (ft): 0 10 50 150 330 610 (a) Determine the velocity and acceleration of the motorcycle at t = 4 , 6 and 8 seconds, using finite differences. (b) Using only three relevant data points, determine the velocity and acceleration at t = 6 seconds by the Interpolating Polynomial method. Compare these results with those of Part (a). . 7.9. The deflection curve of a cantilevered beam is given by the following data, in which x is the distance in feet from the fixed end and y is the deflection at x. x: 1 4 8 y: 0.015 0.20 0.60 Using the Interpolating Polynomial method, compute the first and second derivatives of y(x) at x= 3 ft and x = 6 ft. 7.10. The displacements of an instrument subjected to a random vibration test are given below . Time, t (seconds): 0.05 0.10 0.15 0.20 0.25 0.30 Displacement, y (in): 0.144 0.172 0.213 0.296 0.070 0.085 274 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 285. Using finite differences,determine the velocity and acceleration of the instrument at t= 0.15, 0.20 and 0.25 seconds. Chapter 7: Numerical Differentiation 275
- 286. NOTES: 276 AN INTRODUCTIONTO NUMERICAL METHODS USING MATHCAD
- 287. C H AP T E R 8 NUMERICAL INTEGRATION 8.1 INTRODUCTION TO NUMERICAL INTEGRATION AND THE USE OF THE MATHCAD INTEGRAL OPERATORS In traditional calculus, an integral of a function f(x) between limits a and b is the shaded area shown in Figure 8.1 and is defined as I a b x f x ( ) ⌠ ⎮ ⌡ d = (8.1) Figure 8.1. Integral of a function This analytical solution is possible only when f(x) is explicitly known. When the evaluation of this integral becomes either impossible or extremely difficult , it is advantageous to resort to numerical techniques. In many science and engineering problems, data is often presented as a given bunch of numbers for given values of the independent variable. In this case too, the evaluation of the integral, is done by a numerical process. There are several methods available for evaluating an integral numerically. However, only the following will be discussed here. 1. The interpolating polynomial method. Chapter 8: Numerical Integration 277
- 288. 2. Trapezoidal rule. 3.Simpson's rules. 4. Romberg integration. To compute the definite integral of a function usiong the Mathcad integral operator, click on the integral operator in the calculus palette shown in Figure 8.2 . Then, start filling in the placeholders as required and press the equals sign to see the result. Mathcad resorts to Romberg integration to approximate the integral of an expression over an interval of real numbers. Figure 8.2. Use of the Mathcad Integral Operator 278 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 289. 8.2 THE INTERPOLATINGPOLYNOMIAL METHOD In this method, an interpolating polynomial passing through the data points is generated as was done in the chapter on numerical differentiation and then the polynomial is integrated anaytically over the prescribed limits of integration. If the polynomial generated is of the form y x ( ) a bx + c x 2 ⋅ + d x 3 ⋅ + .......... + = , (8.2) its integral is I a x ⋅ b x 2 2 ⋅ + c x 3 3 ⋅ + d x 4 4 ⋅ + ............ + = (8.3) which can be evaluated analytically over any prescribed integration limits. Example 8.1. Given the data of Example 7.2 of Chapter 7, which is x 2 3 4 5 6 y(x) 0.80354 1.23591 1.62312 1.89765 1.96789 estimate the integral of y(x) over the range x = 2 to x = 6. The interpolating polynomial generated in Example 7.2 for the data given was y x ( ) a b x ⋅ + c x 2 ⋅ + d x 3 ⋅ + e x 4 ⋅ + = with a 0.04705 − := b 0.407 := c 7.43375 10 3 − ⋅ := d 2.79917 10 3 − ⋅ := e 1.00375 − 10 3 − ⋅ := y x ( ) a b x ⋅ + c x 2 ⋅ + d x 3 ⋅ + e x 4 ⋅ + := Then, the required integral is 2 6 x y x ( ) ⌠ ⎮ ⌡ d 6.1803 = Chapter 8: Numerical Integration 279
- 290. 8.3 TRAPEZOIDAL RULE Figure8.3. The trapezoidal rule The trapezoidal rule estimates the integral of a function on the basis of linear interpolation between the given data points . If the integral of a given function f(x) is to be evaluated between limits a and b , the interval from a to b is divided into n equal subintervals each of width H, where H b a − n = (8.4) The desired integral which is, in fact, the sum of the areas in the subintervals under the curve f(x) is then ( See Figure 8.3. ) I 1 2 H ⋅ 1 n i f xi 1 + ( ) f xi ( ) + ( ) ∑ = ⋅ = (8.5) where the (x i , y i ) are the x and y- coordinates of the given data points at the beginning and end of each subinterval. An alternative form of the above integral is I 1 2 h ⋅ 1 m 1 − ⋅ i f xi 1 + ( ) f xi ( ) + ( ∑ = ⋅ = (8.6) 280 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 291. where m ,now, is the number of data points given on the curve and h is, accordingly, h b a − m 1 − = (8.7) For example, if five data points are prescribed as shown in Figure 8.4, there will be four segments and the trapezoidal rule will estimate the integral of the function from the first to the last data point as I 1 2 h ⋅ y1 2 y2 ⋅ + 2 y3 ⋅ + 2 y4 ⋅ + y5 + ( ) ⋅ = (8.8) Figure 8.4. Trapezoidal rule with five data points If the intervals are not of equal width, the following equation should be used in place of Equation (8.6). I 1 2 1 m 1 − i xi 1 + xi − ( ) f xi 1 + ( ) f xi ( ) + ( ) ⋅ ⎡ ⎣ ⎤ ⎦ ∑ = ⋅ = (8.9) Chapter 8: Numerical Integration 281
- 292. Example 8.2. Given: θ (degrees):0 15 30 45 60` 75 90 Sin(2θ): 0 0.5 0.86603 1 0.86603 0.5 0 Use the trapezoidal rule to determine the area under the curve from θ = 0 to 90 degrees. Compare the result with the true value of the integral. The subinterval width, in this case , is h 15 π 180 ⋅ := h 0.2618 = The y-coordinates of the 7 given data points are y1 0 := y2 0.5 := y3 0.86603 := y4 1.0 := y5 0.86603 := y6 0.5 := y7 0.0 := The value of the desired integral can be estimated by the trapezoidal rule as Int 0.5 h ⋅ y1 2 y2 ⋅ + 2 y3 ⋅ + 2 y4 ⋅ + 2 y5 ⋅ + 2 y6 ⋅ + y7 + ( ) ⋅ := Int 0.97705 = The true value and the error between the true and estimated values of the integral are Trueval 1.0 := %err Trueval Int − Trueval 100 ⋅ := %err 2.2949 = Using Mathcad, the following estimate of the integral is obtained: x 0 0.2618 , 1.5708 .. := y x ( ) sin 2 x ⋅ ( ) := y x ( ) 0 0.5 0.866 1 0.866 0.5 -6 -7.3464·10 = 0 1.5708 θ sin 2 θ ⋅ ( ) ⌠ ⎮ ⌡ d 1 = 282 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 293. 8.4 SIMPSON'S RULES Simpson'srules are two in number. They are the Simpson's one-third rule and the Simpson's three-eighth rule. The first is based on the area under a parabola passing through three equally spaced points, while the second is based on the area under a third-order interpolating polynomial passing through 4 points . 8.4.1 Simpson's One-Third Rule It can be easily shown that the area under a parabola passing through the equally spaced points ( x1, y1 ), ( x2, y2) and ( x3, y3 ), shown in Figure 8.5 is A h 3 y1 4 y2 ⋅ + y3 + ( ) ⋅ = (8.10) Figure 8.5. Basis of Simpson's one-third rule In general, if the area under a curve is divided into n segments, there are (n+1 ) points on the curve x1 y1 , ( ) x2 y2 , ( )................ xn 1 + yn 1 + , ( ) If each segment is of width "h" and there are "n" segments, "n" being an even number, the area under the parabola passing through the first three points will be Chapter 8: Numerical Integration 283
- 294. A1 h 3 y1 4 y2 ⋅ +y3 + ( ) ⋅ = (8.11) Similarly, the areas under the parabolas passing through the third, fourth and fifth points, and through the last three points on the curve will be A2 h 3 y3 4 y4 ⋅ + y5 + ( ) ⋅ = ........................................ (8.12) Alast h 3 yn 1 − 4 yn ⋅ + yn 1 + + ( ) ⋅ = The integral according to Simpson's one-third rule is the sum of the above areas and can be written as I h 3 y1 ⋅ 4 ieven yi ∑ ⋅ + 2 ioddexceptlast yi ∑ ⋅ + yn 1 + + = (8.13) For example, if there are seven data points, this integral can be estimated as I h 3 y1 4 y2 ⋅ + 2 y3 ⋅ + 4 y4 ⋅ + 2 y5 ⋅ + 4 y6 ⋅ + y7 + ( ) ⋅ = (8.14) Figure 8.6 shows a comparison between the function approximations obtained with the trapezoidal rule and Simpson's rule, and the consequent integrals which are the respective areas under these functions. Figure 8.6. Comparison between the trapezoidal and Simpson's one-third rules 284 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 295. Example 8.3. For thegiven data, compute the area under the curve by Simpson's one-third rule and compare the estimated value with the true value Theta: 0 15 30 45 60 75 90 Sin(2θ) 0 0.5 0.86603 1 0.86603 0.5 0. The interval width in this case is h 15 π 180. ⋅ := h 0.2618 = The y-coordinates of the given data points are y1 0 := y2 0.5 := y3 0.86603 := y4 1.0 := y5 0.86603 := y6 0.5 := y7 0.0 := An estimate of the desired integral by Simpson's one-third rule is then Int1 h y1 4 y2 ⋅ + 2 y3 ⋅ + 4 y4 ⋅ + 2 y5 ⋅ + 4 y6 ⋅ + y7 + 3 ⋅ := Int1 1.0004 = The true value of the integral is Trueval 0 π 2 θ sin 2 θ ⋅ ( ) ⌠ ⎮ ⎮ ⌡ d ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Trueval 1.0 := The percentage error between the true and estimated values is Chapter 8: Numerical Integration 285
- 296. %err1 Trueval Int1 − Trueval 100 ⋅ := %err10.0433 − = 8.4.2 Simpson's three-eighth rule Simpson's 3/8 th rule is based on the area under a third order interpolating polynomial passing through four points . If these points are (x1, y1 ) , (x2, y2 ), (x3, y3 ) and (x4, y4 ) as shown in Figure 8.7, then this area can be shown to be A 3 8 h ⋅ y1 3 y2 ⋅ + 3 y3 ⋅ + y4 + ( ) ⋅ = (8.15) Figure 8.7. Basis of Simpson's three-eighth rule If the area under a curve is divided into n segments of equal width h giving (n+1) points on the curve, Equation (8.15)) will give the area under the third-order interpolating polynomial passing through the first four points . Similarly, the area under the polynomial passing through the fourth, fifth, sixth and seventh points will be 286 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 297. A2 3 8 y4 3 y5 ⋅ +3 y6 ⋅ + y7 + ( ) ⋅ h ⋅ = (8.16) and that under the polynomial passing through the last four points will be Alast 3 8 yn 2 − 3 yn 1 − ⋅ + 3 yn ⋅ + yn 1 + + ( ) ⋅ h ⋅ = (8.17) The integral according to Simpson's 3/8 th rule is then the sum of the above areas and can be written as I 3 8 h ⋅ y1 3 y2 ⋅ + 3 y3 ⋅ + 2 y4 ⋅ + 3 y5 ⋅ + 3 y6 ⋅ + 2 y7 ⋅ + 3 y8 ⋅ + 3 y9 ⋅ + 2 y10 ⋅ + .....yn 1 + + ( ) ⋅ = (8.18) 8.4.3 Limitations of Simpson's Rules While the one-third rule requires the number of segments under the curve to be even, the mininum number being two, the three-eighth rule requires the number of segments to be a number divisible by three , the minimum number being three. That is, for computation by the one-third rule, the number of segments can be 2, 4, 6 and so on while the number of segments required by the 3/8 th rule is 3, 6, 9, 12 etc. Example 8.4. For the data of Example 8.3, compute the area under the curve by Simpson's three-eighth rule and compare the estimated value with the true value Theta: 0 15 30 45 60 75 90 Sin(2θ): 0 0.5 0.86603 1 0.86603 0.5 0. The interval width is , as before, h 15 π 180. ⋅ := h 0.2618 = and the y-coordinates of the given data points are y1 0 := y2 0.5 := y3 0.86603 := y4 1.0 := y5 0.86603 := Chapter 8: Numerical Integration 287
- 298. y6 0.5 := y70.0 := The integral , according to Simpson's three-eighth rule, can be estimated as Int2 3 8 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ h ⋅ y1 3 y2 ⋅ + 3 y3 ⋅ + 2 y4 ⋅ + 3 y5 ⋅ + 3 y6 ⋅ + y7 + ( ) ⋅ := Int2 1.001 = The true value, as computed in Example 8.3 , is Trueval 1.0 := and the percentage error is %err2 Trueval Int2 − Trueval 100 ⋅ := %err2 0.1008 − = 8.5 ROMBERG INTEGRATION Romberg integration is a very powerful and efficient method of computing an integral. It uses a combination of the trapezoidal rule with an extrapolation technique known as Richardson extrapolation, and provides much better accuracy than the trapezoidal rule. (a) (b) Figure 8.8. Basis of Romberg integration 288 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 299. The procedure forcomputing the integral of f(x) over a prescribed range of x= a to x= b by the Romberg method is as follows. 1. For a range of integration from x =a to x= b, a shown in Figure 8.8 (a) ,obtain the first trapezoidal estimate I 0,1 of the integral of the given function f(x) by the trapezoidal rule as (8.19) I0 1 , 1 2 h ⋅ f a ( ) ( f b ( ) + [ ] ⋅ = where h b a − = (8.20) 2. Divide the given range into two segments as shown in Figure 8.8 (b) and obtain the second trapezoidal estimate I 1,1 as I1 1 , h 2 y1 2 y2 ⋅ + y3 + ( ) ⋅ = (8.21) where h b a − 2 = y1 f a ( ) = <--Value of function at the beginning of the range of integration (8.22) y2 f a h + ( ) = <--Value of function at the middle of the range of integration y3 f a 2 h ⋅ + ( ) = <--Value of function at the end of the range of integration 3 Now divide the given range further into four segments and obtain the third trapezoidal estimate I 2,1 with h = one-half of the previous h = (b-a)/4 . 4. Obtain a fourth trapezoidal estimate I 3,1with eight segments and h = (b-a)/8 5. Obtain a fifth trapezoidal estimate I 4,1 with sixteen segments and h = (b-a)/16 6. Using the Richardson extrapolation formula which is Chapter 8: Numerical Integration 289
- 300. Ii j , Iij 1 − , 1 4 j 1 − 1 − Ii j 1 − , Ii 1 − j 1 − , − ( ) ⋅ + = (8.23) compute the other unknown components of the matrix [I] , namely , I1 2 , I2 2 , I2 3 , I3 2 , I3 3 , I3 4 , I4 2 , I4 3 , I4 4 , I4 5 , ............ and fill in the Romberg table which is shown in Table 8.1. Keep filling in the table until the integral values converge , i.e, I3 3 , I3 2 , = or I4 3 , I4 2 , = or I4 4 , I4 3 , = whichever occurs earlier. TABLE 8.1 The Romberg table R I0 1 , I1 1 , I2 1 , I3 1 , I4 1 , I5 1 , I1 2 , I2 2 , I3 2 , I4 2 , I5 2 , I2 3 , I3 3 , I4 3 , I5 3 , I3 4 , I4 4 , I5 4 , I4 5 , I5 5 , I5 6 , ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := The procedure of this method should be clear from the examples that follow. 290 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 301. Example 8.5 Use Rombergintegration to find the integral of exp (- x 3 ) between the limits of a= 0.1 and b= 1.75 Use an "h" equal to (b-a) to begin with. The given function and the integration limits are f x ( ) e x 3 − := a 0.1 := b 1.75 := Computation of the first trapezoidal estimate : f a ( ) 0.999 = f b ( ) 4.70385 10 3 − × = h b a − := h 1.65 = I0 1 , 0.5 h ⋅ f a ( ) f b ( ) + ( ) ⋅ := I0 1 , 0.82806 = <--- First trapezoidal estimate Computation of the second trapezoidal estimate: In this case, the interval size must be halved, giving y1, y2 and y3 as the function values at the beginning, middle and end of the integration range. h b a − 2 := h 0.825 = y1 f a ( ) := y1 0.999 = y2 f a h + ( ) := y2 0.45319 = y3 f a 2 h ⋅ + ( ) := y3 4.70385 10 3 − × = Second trapezoidal estimate I1 1 , 0.5 h ⋅ y1 2 y2 ⋅ + y3 + ( ) ⋅ := I1 1 , 0.78791 = <-- Computation of the third trapezoidal estimate: The "h" must be halved again giving h=(b-a)/4 h h 2 := h 0.4125 = In this case, the computation of the number of subintervals to be used in the integration process (n), the coordinates of the resulting data points and the third trapezoidal estimate is done as shown below n b a − h := n 4 = i 1 n 1 + ( ) .. := xi a i 1 − ( ) h ⋅ + := yi f xi ( ) := Chapter 8: Numerical Integration 291
- 302. xi 0.1 0.5125 0.925 1.3375 1.75 = yi 0.999 0.87406 0.45319 0.09139 -3 4.70385·10 = I2 1 ,0.5 h ⋅ y1 2 2 4 j yj ∑ = ⋅ + y5 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I2 1 , 0.7922 = <--Third trapezoidal estimate Computations of the fourth and fifth trapezoidal estimates: These are very similar to that of the third estimate and are presented below Fourth estimate: h h 2 := h 0.20625 = n b a − h := n 8 = i 1 2 , n 1 + ( ) .. := xi a i 1 − ( ) h ⋅ + := yi f xi ( ) := xi 0.1 0.30625 0.5125 0.71875 0.925 1.13125 1.3375 1.54375 1.75 = yi 0.999 0.97169 0.87406 0.68983 0.45319 0.23511 0.09139 0.02525 -3 4.70385·10 = I3 1 , 0.5 h ⋅ y1 2 2 8 j yj ∑ = ⋅ + y9 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I3 1 , 0.79249 = <--Fourth trapezoidal estimate Fifth estimate: h h 2 := h 0.10313 = n b a − h := n 16 = i 1 2 , n 1 + ( ) .. := xi a i 1 − ( ) h ⋅ + := yi f xi ( ) := 292 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 303. xi 0.1 0.20313 0.30625 0.40937 0.5125 0.61563 0.71875 0.82188 0.925 1.02812 1.13125 1.23438 1.3375 1.44063 1.54375 1.64688 = yi 0.999 0.99165 0.97169 0.93369 0.87406 0.7919 0.68983 0.57398 0.45319 0.3373 0.23511 0.15247 0.09139 0.05029 0.02525 0.01149 = y16 0.01149 = y17 4.7038510 3 − × = I4 1 , 0.5 h ⋅ y1 2 2 16 j yj ∑ = ⋅ + y17 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I4 1 , 0.79253 = <--Fifth trapezoidal estimate Using the Richardson extrapolation formula for improved estimates, which is j 2 3 , 4 .. := i 1 2 , 4 .. := Ii j , Ii j 1 − , 1 4 j 1 − 1 − Ii j 1 − , Ii 1 − j 1 − , − ( ) ⋅ + := or, Ii j , 4 j 1 − Ii j 1 − , ⋅ Ii 1 − j 1 − , − 4 j 1 − 1 − := , improved estimates of the integral can be computed as shown below. I2 2 , 0.79363 = I2 3 , 0.7949 = I3 2 , 0.79258 = I3 3 , 0.79251 = I3 4 , 0.79247 = I4 2 , 0.79254 = I4 3 , 0.79254 = I4 4 , 0.79254 = Chapter 8: Numerical Integration 293
- 304. However, the Rombergprocedure requires the following quantities to be zeros. I1 3 , 0 := I1 4 , 0 := I2 4 , 0 := Shown to the left is the Romberg Table The zeros shown in the first column of matrix, however, are to be ignored. <-- I 0 0 0 0 0 0.82806 0.78791 0.7922 0.79249 0.79253 0 0.77452 0.79363 0.79258 0.79254 0 0 0.7949 0.79251 0.79254 0 0 0 0.79247 0.79254 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = The Romberg table of integrals is cleaned up and shown below: TABLE 8.2 The Romberg table for the integration of exp( - x 3 ) ____________________________________________________________________________ 0.82806 0.78791 0.77452 0.7922 0.79363 0.79490 0.79249 0.79258 0.79251 0.79247 0.79253 0.79254 0.79254 0.79254 ______________________________________________________________________________ Notice that convergence has taken place at an integral value of 0.79254 as shown in the tabular presentation. Use of the Mathcad integral operator gives 0.1 1.75 x f x ( ) ⌠ ⎮ ⌡ d 0.79254 = where the function f(x) was already defined earlier. 294 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 305. Example 8.6 Use Rombergintegration to find the integral of ln (1+ x 3) between the limits of a=1.0 and b=4.5. Use an "h" value equal to (1/2) (b-a) to begin with . Computation of first trapezoidal estimate: a 1.0 := b 4.5 := f x ( ) ln 1 x 3 + ( ) := h b a − 2 := h 1.75 = y1 f a ( ) := y1 0.69315 = y2 f a h + ( ) := y2 3.08177 = y3 f a 2 h ⋅ + ( ) := y3 4.52315 = I1 1 , 0.5 h ⋅ y1 2 y2 ⋅ + y3 + ( ) ⋅ := I1 1 , 9.95735 = <--First trapezoidal estimate Computation of second trapezoidal estimate: h h 2 := h 0.875 = n b a − h := n 4 = i 1 2 , n 1 + ( ) .. := xi a i 1 − ( ) h ⋅ + := yi f xi ( ) := xi 1 1.875 2.75 3.625 4.5 = yi 0.69315 2.02707 3.08177 3.88434 4.52315 = I2 1 , 0.5 h ⋅ y1 2 2 4 j yj ∑ = ⋅ + y5 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I2 1 , 10.15116 = <---- Second trapezoidal estimate Computation of third trapezoidal estimate: h h 2 := h 0.4375 = n b a − h := n 8 = i 1 2 , n 1 + ( ) .. := xi a i 1 − ( ) h ⋅ + := yi f xi ( ) := Chapter 8: Numerical Integration 295
- 306. xi 1 1.4375 1.875 2.3125 2.75 3.1875 3.625 4.0625 4.5 = yi 0.69315 1.37888 2.02707 2.59275 3.08177 3.50812 3.88434 4.2202 4.52315 = I3 1 , 0.5h ⋅ y1 2 2 8 j yj ∑ = ⋅ + y9 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I3 1 , 10.19431 = <--- Third trapezoidal estimate Computation of fourth trapezoidal estimate: h h 2 := h 0.21875 = n b a − h := n 16 = i 1 2 , n 1 + ( ) .. := xi a i 1 − ( ) h ⋅ + := yi f xi ( ) := xi 1 1.21875 1.4375 1.65625 1.875 2.09375 2.3125 2.53125 2.75 2.96875 3.1875 3.40625 3.625 3.84375 4.0625 4.28125 = yi 0.69315 1.03328 1.37888 1.7126 2.02707 2.32028 2.59275 2.84597 3.08177 3.30193 3.50812 3.70182 3.88434 4.0568 4.2202 4.3754 = y16 4.3754 = y17 4.52315 = 296 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 307. I4 1 , 0.5h ⋅ y1 2 2 16 j yj ∑ = ⋅ + y17 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I4 1 , 10.20455 = <--- Fourth trapezoidal estimate Use of the Richardson extrapolation formula for improved estimates: j 2 3 , 4 .. := i 1 2 , 4 .. := Ii j , Ii j 1 − , 1 4 j 1 − 1 − Ii j 1 − , Ii 1 − j 1 − , − ( ) ⋅ + := <---- Richardson extrapolation formula for improved estimates or Ii j , 4 j 1 − Ii j 1 − , ⋅ Ii 1 − j 1 − , − 4 j 1 − 1 − := I2 2 , 10.21576 = I3 2 , 10.20869 = I3 3 , 10.20822 = I4 2 , 10.20796 = I4 3 , 10.20791 = I4 4 , 10.20791 = The Romberg procedure requires the following quantities to be zeros. I0 1 , 0 := I1 2 , 0 := I1 3 , 0 := I1 4 , 0 := I2 3 , 0 := I2 4 , 0 := I3 4 , 0 := Thus, the estimates computed can be assembled in the tabular form (Romberg Table ) shown below. The zeros shown in the matrix are to be ignored. I 0 0 0 0 0 0 9.957348 10.151155 10.194307 10.204548 0 0 10.215758 10.20869 10.207962 0 0 0 10.208219 10.207914 0 0 0 0 10.207909 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Chapter 8: Numerical Integration 297
- 308. The Romberg tableof integrals is cleaned up and shown below : TABLE 8.3 The Romberg Table for the integral of ln (1+x3) _____________________________________________________________________________ 9.957348 10.151155 10.215758 10.194307 10.20869 10.208219 10.204548 10.207962 10.207914 10.207909 ______________________________________________________________________________ The tabular presentation of integral values shows that convergence has taken place at an integral value of 10.2079 Use of the Mathcad integral operator yields Shown on the left is the integral obtained with Mathcad where the function f(x) was already defined earlier. 1.0 4.5 x f x ( ) ⌠ ⎮ ⌡ d 10.20791 = <---- Example 8.7. Use Romberg integration to find the integral of the given function f(x) between the limits of a= 0 and b= 8. Begin with h= (b-a) to compute the first trapezoidal estimate I 0,1 . Compute four trapezoidal estimates, then complete the Romberg table. x f(x) 0 3.2 0.50 2.6 1.0 1.7 1.50 1.0 2.0 0.62 2.50 0.45 3.0 0.40 3.50 0.27 4.0 0.20 298 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 309. Computation of thefirst trapezoidal estimate: a 0 := b 4 := y1 3.2 := y2 0.20 := h b a − := h 4 = I0 1 , 0.5 h ⋅ y1 y2 + ( ) ⋅ := I0 1 , 6.8 = <-- FIRST TRAPEZOIDAL ESTIMATE Computation of the second trapezoidal estimate h h 2 := h 2 = n b a − h := n 2 = x1 0 := y1 3.2 := x2 2 := y2 0.62 := x3 4 := y3 0.20 := I1 1 , 0.5 h ⋅ y1 2 y2 ⋅ + y3 + ( ) ⋅ := I1 1 , 4.64 = <-- SEC TRAPEZOIDAL ESTIMATE Computation of the third trapezoidal estimate: h h 2 := h 1 = n b a − h := n 4 = x1 0 := y1 3.2 := x2 1. := y2 1.7 := x3 2.0 := y3 0.62 := x4 3.0 := y4 0.40 := x5 4.0 := y5 0.20 := I2 1 , 0.5 h ⋅ y1 2 2 4 j yj ∑ = ⋅ + y5 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I2 1 , 4.42 = <-- THIRD TRAPEZOIDAL ESTIMATE Computation of the fourth trapezoidal estimate: n b a − h := n 4 = h h 2 := h 0.5 = x1 0 := y1 3.2 := x2 0.50 := y2 2.60 := x3 1. := y3 1.70 := x4 1.50 := y4 1.0 := x5 2.0 := y5 0.62 := x6 2.50 := y6 0.45 := x7 3. := y7 0.40 := x8 3.50 := y8 0.27 := x9 4. := y9 0.20 := I3 1 , 0.5 h ⋅ y1 2 2 8 j yj ∑ = ⋅ + y9 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ := I3 1 , 4.37 = <-- FOURTH TRAPEZOIDAL ESTIMATE Chapter 8: Numerical Integration 299
- 310. Use of theRichardson formula to obtain improved estimates: j 2 4 .. := i 1 2 , 3 .. := Ii j , 4 j 1 − Ii j 1 − , ⋅ Ii 1 − j 1 − , − 4 j 1 − 1 − := I1 2 , 3.92 = I2 2 , 4.34667 = I2 3 , 4.37511 = I3 1 , 4.37 = I3 2 , 4.35333 = I3 3 , 4.35378 = I3 4 , 4.35344 = I1 3 , 0 := I1 4 , 0 := I2 4 , 0 := The Romberg procedure, in this case, requires these quantities that are shown on the left to be zeros. <--- I0 2 , 0 := I0 3 , 0 := I0 4 , 0 := I4 1 , 0 := I4 2 , 0 := I4 3 , 0 := I4 3 , 0 := I4 4 , 0 := Shown to the left is the Romberg Table The zeros shown in the matrix, however, are to be ignored. I 0 0 0 0 0 6.8 4.64 4.42 4.37 0 0 3.92 4.34667 4.35333 0 0 0 4.37511 4.35378 0 0 0 0 4.35344 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <-- The Romberg table of integrals is cleaned up and shown below : TABLE 8.4 The Romberg Table for the integral of given function of Example 8.7 _____________________________________________________________________________ 6.8 4.64 3.92 4.42 4.34667 4.37511 4.37 4.35333 4.35378 4.35344 ______________________________________________________________________________ The tabular presentation of integral values shows that convergence has taken place at an integral value of approximately 4.353. 300 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 311. 8.6 APPLICATIONS INNUMERICAL INTEGRATION Applications of numerical integration can be encountered in situations dealing with location of centroids, determination of moments of inertia of cross sectional shapes and computation of periods of oscillations in nonlinear vibration problems. Some examples are dealt with below. 8.6.1 Centroid of a Rod Bent into the Shape of a Parabola The length of an element dL of a rod bent into the shape of the parabola shown in Figure 8.8 can be expressed in terms of the differentials dx and dy as [ 8 ] dL dx 2 dy 2 + = or 1 y x d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + dy ⋅ (8.24) Since x 2 y 2 ⋅ = , y x d d 4 y × = and dL 1 16 y 2 ⋅ + dy ⋅ = y x ( ) x 2 := x 0 0.001 , 0.5 .. := 0 0.1 0.2 0.3 0.4 0 0.2 0.4 bent shape of rod bent shape of rod Rod of parabolic shape: x = 2 y .y x , meters y, meters y x ( ) x Figure 8.8. Rod bent into a parabolic shape The length of the rod , then, can be written as L L 1 ⌠ ⎮ ⎮ ⌡ d = or L 0 0.5 y 1 16 y 2 ⋅ + ⌠ ⎮ ⌡ d := L 0.739 = meters Chapter 8: Numerical Integration 301
- 312. The location ofthe centroid ( x CG , y CG ) can then be calculated using xCG L x ⌠ ⎮ ⎮ ⌡ d L = yCG L y ⌠ ⎮ ⎮ ⌡ d L = or, xCG 1 L 0 0.5 y 2 y 2 ⋅ 1 16 y 2 ⋅ + ⋅ ⌠ ⎮ ⌡ d ⋅ := yCG 1 L 0 0.5 y y 1 16 y 2 ⋅ + ⋅ ⌠ ⎮ ⌡ d ⋅ := giving xCG 0.205 = m yCG 0.287 = m 8.6.2. Moment of Inertia of a Beam of Semi-Elliptic Cross Section For the semi-elliptic beam cross section shown in Figure 8.9 , its moments of inertia about the x and y axes can be calculated using [ 8 ] Ixx A y 2 ⌠ ⎮ ⎮ ⌡ d = Iyy A x 2 ⌠ ⎮ ⎮ ⌡ d = (8.25) where dA 2 x ⋅ dy ⋅ = with x y ( ) 1 y 2 b 2 − a ⋅ = and a 4 := in b 2 := in dA y dx ⋅ = with y x ( ) 1 x 2 a 2 − b ⋅ := x 4 − 3.99 − , 4. .. := 4 2 0 2 4 0 2 Semi-elliptic cross section y x ( ) x The moments of inertia , I xx and I yy, can now be computed as follows,using Equations (8.25) Ixx 2 a b ⋅ 0 b y y 2 b 2 y 2 − ⋅ ⌠ ⎮ ⌡ d ⋅ := Figure 8.9. Beam cross section Iyy b a a − a x x 2 a 2 x 2 − ⋅ ⌠ ⎮ ⌡ d ⋅ := giving Ixx 12.566 = in 4 and Iyy 50.266 = in 4 302 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 313. 8.6.3. Launch ofa Projectile A projectile is launched with an initial vertical velocity vy init from a height y 0 above the ground. Neglecting air resistance, the relationship between its vertical velocity v y at any time t and the height y above the ground that it attains on the way up is given by [ 8 ] vy vyinit 2 2 G ⋅ y0 y − ( ) ⋅ + = (8.26) where G is the acceleration due to gravity ( 32.2 ft/ sec 2 or 9.81 m/ sec 2 ) . Equating the above quantity to dy/dt , and then integrating with respect to time from 0 to t seconds, the following expression can be derived for the time t taken by the projectile to attain the height y above the ground t y0 y y 1 vyinit 2 2 G ⋅ y0 y − ( ) ⋅ + ⌠ ⎮ ⎮ ⎮ ⌡ d = (8.27) The time taken to reach the maximum height , y max , is then the above integral with y max as the upper limit of integration , where ymax vyinit 2 2 G ⋅ y0 + = (8.28) For an initial velocity of 55 ft/ sec at an angle of 65 degrees from a height of 3.5 ft above ground, the time taken to attain the maximum height is calculated below using Equations (8.27) and (8.28). vinit 55 := ft/sec θ 65 deg ⋅ := sin θ ( ) 0.906 = G 32.2 := ft/ sec 2 vyinit vinit sin θ ( ) ⋅ := vyinit 49.847 = ft/sec y0 3.5 := ft ymax vyinit 2 2 G ⋅ y0 + := ymax 42.083 = ft tmaxheight y0 ymax y 1 vyinit 2 2 G ⋅ y0 y − ( ) ⋅ + ⌠ ⎮ ⎮ ⎮ ⌡ d := tmaxheight 1.548 = sec Chapter 8: Numerical Integration 303
- 314. 8.6.4. Large Oscillationsof a Simple Pendulum A simple pendulum, which is a mass attached to one end of a light rigid rod, is shown in Figure 8.10. Figure 8.10. Vibration of a simple pendulum The large-amplitude motion of this pendulum is governed by the following nonlinear differential equation [ 6,9,16,20 ]. 2 t θ t ( ) d d 2 G L sin θ t ( ) ( ) ⋅ + 0 = (8.29) where L is the length of the rod and G is the acceleration due to gravity . However, if the amplitudes of vibration are considered small, the above differential equation assumes its linear form 2 t θ t ( ) d d 2 G L θ t ( ) ⋅ + 0 = (8.30) and the motion, then, has a natural period τlinear 2 π ⋅ L G ⋅ := seconds (8.31) For the nonlinear problem , it can be shown that the period of one oscillation, which is now a function of the amplitude of vibration , is given by the integral 304 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 315. τnonlinear A ( )2 L G ⋅ 0 A π 180. ⋅ θ 1 sin A π 360 ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⌠ ⎮ ⎮ ⎮ ⎮ ⎮ ⌡ d ⋅ := seconds (8.32) where A is the maximum vibration amplitude in degrees that the pendulum attains. The corresponding linear and nonlinear natural frequencies in Hertz can be computed using flinear 1 τlinear := fnonlinear A ( ) 1 τnonlinear A ( ) := (8.33) In a nonlinear problem, the percentage variation between the linear and nonlinear natural frequencies is typically expressed in the following form Percentdiff A ( ) fnonlinear A ( ) flinear − flinear 100 ⋅ := (8.34) This variation, which is plotted in Figure 8.11 below as a function of the maximum amplitude of vibration of the pendulum, confirms that there is a sharp increase in the variation from the linear natural frequency with increase in the amplitude of vibration. A 0.1 0.2 , 30 .. := 5 10 15 20 25 30 0 0.5 1 1.5 2 Variation of nat freq with amplitude Amplitude A, degrees % variation from linear nat frequency Percentdiff A ( ) A Figure 8.11. Variation of nonlinear natural frequency wiith amplitude Chapter 8: Numerical Integration 305
- 316. PROBLEMS 8.1. Use thetrapezoidal rule to estimate the integral of x (2/3) over the range a=0 to b=4.0 for interval widths of h= 1.0, 0.5 and 0.25. Also obtain an answer using the Mathcad integral operator. 8.2. Using Simpson's rules, with h= 0.25, 0.5 and 1.5, obtain estimates of the following integral: I 0 3 x x 4 3 ⌠ ⎮ ⎮ ⎮ ⌡ d = Also obtain an answer using the Mathcad integral operator 8.3. Using Simpson's rules, with h = 4, 8 and 10 subdivisions of the given range of x, obtain estimates of the following integral: 1 4 x 1 6 5 e 2 x ⋅ ⋅ + ⌠ ⎮ ⎮ ⎮ ⌡ d Also obtain an answer using the Mathcad integral operator 8.4. Use Romberg integration to find the integral of x3 exp ( 2x ) between the limits of a= 0 and b= 2.0. Use an "h" equal to (b-a) to begin with to compute the first trapezoidal estimate. Also verify your answer using the Mathcad integral operator. 8.5. Use Romberg integration to find the integral of (1/x) between the limits of a= 1.0 and b= 2.0. Use an "h" equal to (b-a) to begin with. The completion of three rows of the Romberg table should be sufficient. 8.6. Use Romberg integration to find the integral of ln(x) between the limits of a= 1.0 and b= 2.0. Use an "h" equal to (b-a) to begin with. Compute 3 trapezpoidal estimates before completing the Romberg table. 8.7. Use Romberg integration to find the integral of 3 / (1 + x 2 ) between the limits of a= 0 and b= 4.0. Use an "h" equal to (b-a) to begin with. Compute 3 trapezpoidal estimates before completing the Romberg table. 306 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 317. 8.8. Use Rombergintegration to find the integral of ( x e 2x ) between the limits of a= 0 and b= 0.5. Use an "h" equal to (b-a) to begin with. Compute 3 trapezpoidal estimates before completing the Romberg table. 8.9. Use Romberg integration to find the integral of x sin (x) between the limits of a= 1.0 and b= 2.0. Use an "h" equal to (b-a) to begin with. Compute an answer that is good to at least 3 decimal places. 8.10. Use Romberg integration to find the integral of the following function f(x) between the limits of a= 1.8 and b= 3.4. Begin with h= (b-a)/2 = 0.8 , that is , use the values marked by asterisks to compute the first trapezoidal estimate I 1, 1 . Also verify your result by resorting to Mathcad cubic spline interpolation and the Mathcad integral operator. ______________________________________________________________________ x f(x) x f(x) x f(x) x f(x) ---------------------------------------------------------------------------------------------------------------------------- 1.6 4.751 1.8 6.152* 2.0 7.412 2.2 9.038 2.4 11.151 2.6 13.475* 2.8 16.521 3.0 20.08 3.2 24.721 3.4 30.114* 3.6 36.617 3.8 44.701 ______________________________________________________________________ 8.11. Use Romberg integration to find the integral of the following function f(x) between the limits of a= 0.3 and b= 1.1. Begin with h= (b-a) to compute the first trapezoidal estimate I 0,1 . Compute four trapezoidal estimates, then complete the Romberg table. x f(x) 0.3 55 0.4 42 0.5 37 0.6 35 0.7 39 0.8 48 0.9 53 1.0 51 1.1 39 8.12. Use Romberg integration to find the integral of the function f(x) given in the following table between the limits of a= 0 and b= 4. Begin with h= (b-a) to compute the first trapezoidal estimate I 0,1 Compute four trapezoidal estimates, then complete the Romberg table. Obtain an answer that is good to at least two decimal places. Chapter 8: Numerical Integration 307
- 318. x f(x) 0 3 0.502.4 1.0 1.5 1.50 0.92 2.0 0.60 2.50 0.41 3.0 0.30 3.50 0.23 4.0 0.18 8.13. Use Romberg integration to find the integral of the following function f(x) between the limits of a= 0 and b= 0.80. Begin with h= (b-a) to compute the first trapezoidal estimate I 0,1 . Compute four trapezoidal estimates, then complete the Romberg table. x f(x) 0 0 0.1 0.12 0.2 0.30 0.3 0.55 0.4 0.89 0.5 1.36 0.6 1.99 0.7 2.84 0.8 3.96 8.14. Use Romberg integration to determine the integral of the following function f(x) between the limits of a= 1.0 and b= 2.0. Use an "h" equal to (b-a) to begin with. Compute three trapezoidal estimates and then complete the Romberg table. Obtain an answer that is good to at least three decimal places. x f(x) 1 0 1.25 0.2231 1.50 0.4055 1.75 0.5596 2.00 0.6931 8.15. The time, T, required to raise or lower the water surface in a tank with inflow as well as outflow through an orifice at the bottom is given by the following integral 308 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 319. T H1 H2 h A Q k h 0.5 ⋅ − ⌠ ⎮ ⎮ ⎮ ⌡ d = whereH1 and H2 are the initial and final heights of the water surface above the base of the tank , A is the cross sectional area of the tank , Q is the inflow rate, and k is a constant that depends on the coefficient of discharge and the area of the orifice. Determine how long it will take for the water level to fall 10 ft in a tank 50 ft long by 30 ft wide with vertical sides, when water enters the tank at the rate of 6 cu. ft / sec and the initial level of the water surface in the tank is 16 ft above the bottom. Take k to be 3. 8.16. The motion of bodies moving with moderate speed in fluids such as water or air is resisted by a damping force that is proportional to the square of the speed. In order to estimate the resonant amplitude of the motion, the equivalent linear viscous damping, C eq , is needed which is found by equating the energy dissipated by the viscous damping in one cycle to that of the nonviscous damping [ 20 ]. The equivalent viscous damping, in this case , can be shown to be Ceq 2 a ⋅ ω ⋅ X ⋅ π 0 π θ sin θ ( ) 3 ⌠ ⎮ ⌡ d ⋅ = where a is a proportionality constant that depends on the body geometry and the properties of the fluid medium , ω is the frequency of excitation and X is the amplitude of the resulting vibration. Determine C eq if a = 0.85 N-sec 2 / m , ω = 550 rads/ sec and X = 16 mm. 8.17. The moment of inertia of a sphere of radius R about a diametral axis is given by the integral [ 8 ] Idia 1 2 ρ ⋅ π ⋅ R − R x R ( 2 x 2 − ⎡ ⎣ ⎤ ⎦ 2 ⌠ ⎮ ⎮ ⌡ d ⋅ = where ρ is the mass density of the material of the sphere. Estimate I dia for a 1 m diameter sphere made of steel with ρ = 7.8 Mg / m 3 8.18. The shear stress distribution in a beam cross section subjected to a shear force V is given by [ 18 ] Chapter 8: Numerical Integration 309
- 320. τ V Q ⋅ I b ⋅ = whereV is the shear force acting on the cross section, I is the moment of inertia of the entire cross section about the neutral axis , b is the cross sectional width at the level of interest in the cross section, and Q is the first moment of the part of the area of the cross section above the level of interest about the neutral axis. For a circular cross section of radius R , the shear stress at a height y1 above the neutral axis is given by the integral τ V I 2 ⋅ R 2 y1 2 − ⋅ 2 ⋅ y1 R y R 2 y 2 − y ⋅ ⌠ ⎮ ⎮ ⌡ d ⋅ = Determine the shear stress distribution in the case of a beam with a circular cross section of radius 1 inch and subjected to a shear force of magnitude 500 lbs 8.19. The time taken by an object to slide down under the action of gravity from a point O (0,0) on a curved ramp described by the function y = y(x) to a point B is given by the integral T 0 xB x 1 x y x ( ) d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + 2 G ⋅ y x ( ) ⋅ ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d = where G is the acceleration due to gravity ( 32.2 ft / sec 2 ). Estimate the time required for a sack to slide from the top O ( 0,0) of a ramp described by y(x) = A x 2 + B x ( with A= 0.02551, B= 0.01 ) to its bottom at x = 28 ft. 310 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 321. C H AP T E R 9 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS 9.1 INTRODUCTION A differential equation is an equation that expresses a relationship between a dependent variable y , an independent variable x and derivatives of y . The order of the differential equation is the order of the highest derivative in the equation. The following equation has only the first derivative of y in it and is, therefore, a first order differential equation. x y d d f x y , ( ) = (9.1) while one that has both first and second derivatives or just the second derivative is a second order differential equation. Its general form is given below. 2 x y d d 2 f x y , x y d d , ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = (9.2) Applications of differential equations can be seen in heat transfer, electrical circuit analysis, vibration analysis , control systems analysis and a host of other areas in science as well as engineering. The two broad classes of problems involving differential equations are (1) initial value problems and (2) boundary value problems. In an initial value problem, conditions are specified at only one value of the independent variable. An example of an initial value problem is given below. m 2 t y d d 2 ⋅ c t y d d ⋅ + k y ⋅ + f t ( ) = y 0 ( ) dispinit = t y 0 ( ) d d velinit = (9.3) Although the independent variable in the above equation represents time, it can, just as well, represent a spatial coordinate also such as x . In a boundary value problem, conditions are prescribed at two values of the independent variable, as in the following. A 2 x y d d 2 ⋅ B x y d d ⋅ + C y ⋅ + g x ( ) = y 0 ( ) y0 = y L ( ) yL = (9.4) There are many methods of solving differential equations numerically. However, our discussion is going to be limited to the Taylor series method, the Euler and Modified Euler methods, and the Runge-Kutta methods. Chapter 9: Numerical Solution of Ordinary Differential Equations 311
- 322. 9.2 TAYLOR SERIESMETHOD Given y= y(x) and the value of the independent variable at x= x0 as y(x0), a Taylor series will give the value of y at any value of the dependent variable x as y x ( ) y x0 ( ) x x0 − ( ) x y 0 ( ) d d ⋅ + x x0 − ( )2 2! 2 x y 0 ( ) d d 2 ⋅ + ................... + = (9.5) If the problem to be solved is a first- order differential equation of the form x y d d f x ( ) = (9.6) with initial condition y y0 = at x x0 = , (9.7) the Taylor series expansion of Equation (9.5) can, clearly, be made use of in generating a solution to the given differential equation. The following example should clarify this issue. Example 9.1 Given the differential equation: dy/dx = 7 x4 - 3 x2 +9 , with initial condition y = 4 at x = 1 , estimate values of the function using a Taylor series expansion for x = 1 to x = 1.3 . Take Δx to be 0.05. Compare the accuracy of the solution by comparing the computed values with the true values computed from the true solution, which is: 312 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 323. y = (7/5)x5- x3 +9x- (27/5) The initial condition prescribed is x0 1 := y0 4.0 := The first, second, third, fourth and fifth derivatives of f(x) are yprime x ( ) 7 x 4 ⋅ 3 x 2 ⋅ − 9 + := <--First derivative (given) ydblprime x ( ) 28 x 3 ⋅ 6 x ⋅ − := <--Second derivative ytrprime x ( ) 84 x 2 ⋅ 6 − := <--Third derivative (9.8) y4prime x ( ) 168 x ⋅ := <--Fourth derivative y5prime x ( ) 168. := <--Fifth derivative y6prime x ( ) 0 := <--Sixth derivative The Taylor series of Equation (9.5) expanded to seven terms up to the sixth derivative term is y x ( ) y0 x x0 − ( ) yprime x0 ( ) ⋅ + 1 2! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 2 ⋅ ydblprime x0 ( ) ⋅ + 1 3! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 3 ⋅ ytrprime x0 ( ) ⋅ + 1 4! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 4 ⋅ y4prime x0 ( ) ⋅ 1 5! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 5 ⋅ y5prime x0 ( ) ⋅ + 1 6! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 6 ⋅ y6prime x ( ⋅ + + . := The true solution to the differential equation is given as (9.9) ytrue x ( ) 7 5 x 5 ⋅ x 3 − 9 x ⋅ + 27 5 − := (9.10) Using Equations (9.8) and (9.9), the Taylor series solution to the given differential equation can be generated and compared with the true solution of Equation (9.10) as shown below. Chapter 9: Numerical Solution of Ordinary Differential Equations 313
- 324. x 1.0 1.05 ,1.3 .. := <--Range of interest h 0.05 := <--Increment for x n 1.3 1.0 − h := n 6 = <--Number of segments or subintervals in range of interest %error x ( ) ytrue x ( ) y x ( ) − ytrue x ( ) 100 ⋅ := <--Percent error between true and computed values , function of x. i 1 2 , n 1 + .. := xi x0 i 1 − ( ) h ⋅ + := <--x-values at which solution is to be computed %error i %error xi ( ) := <--Percent error between true and computed values at x= x i yi y xi ( ) := <--Computed Taylor solution at x= x i ytruei ytrue xi ( ) := <--True solution at x= x i i 1 2 3 4 5 6 7 = xi 1 1.05 1.1 1.15 1.2 1.25 1.3 = yi 4 4.67917 5.42371 6.24503 7.15565 8.16934 9.3011 = ytruei 4 4.67917 5.42371 6.24503 7.15565 8.16934 9.3011 = %error i 0 -14 1.898·10 -14 -1.638·10 -14 -1.422·10 -14 -2.482·10 0 0 = In what follows, several Taylor series solutions with varying number of terms are developed and compared in Figure 9.1 with the true solution. Notice that because the sixth derivative is zero, the six-term solution which is the same as the one with seven terms produces no error at all. 314 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 325. y6 x ( )y0 x x0 − ( ) yprime x0 ( ) ⋅ + 1 2! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 2 ⋅ ydblprime x0 ( ) ⋅ + 1 3! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 3 ⋅ ytrprime x0 ( ) ⋅ + ... 1 4! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 4 ⋅ y4prime x0 ( ) ⋅ 1 5! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 5 ⋅ y5prime x0 ( ) ⋅ + + ... := <--Six-term solution y2 x ( ) y0 x x0 − ( ) yprime x0 ( ) ⋅ + := <--Two-term solution y1 x ( ) y0 := <--One-term solution ytrue x ( ) 7 5 x 5 ⋅ x 3 − 9 x ⋅ + 27 5 − := <--True solution x 1.0 1.05 , 1.5 .. := 1 1.05 1.1 1.15 1.2 1.25 0 5 10 15 20 Six term/ seven term solution Two-term solution One-term solution True solution Six term/ seven term solution Two-term solution One-term solution True solution Comparison of computed y with true y X- values Y- values y6 x ( ) y2 x ( ) y1 x ( ) ytrue x ( ) x Figure 9.1. Comparison of Taylor series solutions with true solution Chapter 9: Numerical Solution of Ordinary Differential Equations 315
- 326. Percentage errors forthe various approximations are generated below and compared in Figure 9.2. Notice that the error between the six-term/ seven -term approximation and the true solution is zero. In general, the error is sizeable in the range of interest if anything short of a five-term approximation is employed. %error6 x ( ) ytrue x ( ) y6 x ( ) − ytrue x ( ) 100 ⋅ := %error2 x ( ) ytrue x ( ) y2 x ( ) − ytrue x ( ) 100 ⋅ := %error1 x ( ) ytrue x ( ) y1 x ( ) − ytrue x ( ) 100 ⋅ := x 1.0 1.01 , 3. .. := 1 1.05 1.1 1.15 1.2 1.25 20 0 20 40 60 80 100 Six/Seven terms Two terms One term Six/Seven terms Two terms One term Comparison of Errors X- Values Percentage Error %error6 x ( ) %error2 x ( ) %error1 x ( ) x Figure 9.2. Comparison of errors generated with varying number of terms in Taylor series 316 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 327. 9.3 EULER'S METHOD Euler'smethod is a simple, programmable, one-step method which uses information from the beginning of an interval to estimate the function value at the end of that interval . It is based on the forward difference equation for the first derivative only , without requiring the inclusion of higher order derivatives . Using the forward difference representation , the derivative of a function f(x) is x f x ( ) d d f x Δx + ( ) f x ( ) − Δx = (9.11) If the slope of a curve y(x) at xn is known, say y' n , the function value y n+1 at an adjacent point x n+1 is given by y n+1 = y n + h y' n with h = x n+1 - xn (9.12) so long as h, which is the difference between x n+1 and xn is small. ( See Figure 9.3) Figure 9.3. Basis of Euler's method If the problem to be solved is the following first-order initial value problem Chapter 9: Numerical Solution of Ordinary Differential Equations 317
- 328. x y x ( ) d d fx y , ( ) = (9.13) with initial condition y 0 ( ) y0 = (9.14) a numerical solution can be generated using Equation (9.12) which gives yn 1 + yn h f xn yn , ( ) ⋅ + = (9.15) Now, given the initial condition (9.14), Equation 9.15 can be used to do the computations one step at a time until the solution at a desired x value is generated, which is why, this method is also referred to as Euler's one-step method . It is to be noted that the smaller the h, the more accurate will be the results, and , clearly, the method is not going to be very accurate for large values of h. Example 9.2 will illustrate the procedure of Euler's method. Because the relationship of Equation 9.12 is based on a straight line segment between ( xn , y n ) and ( x n+1 , y n+1 ), the error due to the use of Euler's method is (y predicted - y true) as shown in Figure 9.4. An approximate estimate of this error is given by the second order term of the Taylor series expansion which is ε x x0 − ( )2 2! 2 x y d d 2 ⋅ = (9.16) Figure 9.4. Error of Euler's method 318 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 329. Example 9.2 Solve: dy/dx= y' =f(x,y)=-4x3+18x2-22x +9 by Euler's method. The prescribed initial conditions are: x=0, y=2, and the range for the solution is to be x=0 to 4.0 with a step size of h= 0.5. The function f(x,y) which is the derivative of y(x) is given as f x y , ( ) 4 − x 3 ⋅ 18 x 2 ⋅ + 22 x ⋅ − 9 + := The initial condition for this initial-value problem is prescribed as x0 0 := y0 2 := By integrating the given equation, its true solution can be easily seen to be ytrue x ( ) x 4 − 6 x 3 ⋅ + 11 x 2 ⋅ − 9 x ⋅ + 2 + := The procedure of Euler's method is utilized as shown below to generate solutions at x = 0.5 and 1.0. f x0 y0 , ( ) 9 = h 0.5 := x1 0.5 := y1 y0 h f x0 y0 , ( ) ⋅ + := y1 6.5 = f x1 y1 , ( ) 2 = x2 1.0 := y2 y1 h f x1 y1 , ( ) ⋅ + := y2 7.5 = f x2 y2 , ( ) 1 = The percentage errors between the true values at x = 0.5 and 1.0 and those computed by Euler's method are: ytrue x1 ( ) 4.438 = error ytrue x1 ( ) y1 − ytrue x1 ( ) 100 ⋅ := error 46.479 − = ytrue x2 ( ) 5 = error ytrue x2 ( ) y2 − ytrue x2 ( ) 100 ⋅ := error 50 − = Similarly, solutions as well as errors at x = 1.5, 2.0,.....4 can also be generated easily Chapter 9: Numerical Solution of Ordinary Differential Equations 319
- 330. Alternatively, Euler solutionsas well as the associated percentage errors at x = 0.5, 1.0, 1.5,........4.0. can be generated by resorting to the following procedure . x0 0 := y0 2 := <-- Prescribed initial condition xlast 4 := <--End-point of prescribed range of solution h 0.5 := <-- Step size f x y , ( ) 4 − x 3 ⋅ 18 x 2 ⋅ + 22 x ⋅ − 9 + := <--- Prescribed derivative of y(x) at (x,y) f x0 y0 , ( ) 9 = <-- Derivative value at starting point Number of points beyond the starting point at which solution is desired N xlast x0 − h := N 8 = <-- i 0 1 , N .. := xi 1 + xi h + := <--Formulas to be used in Euler's method yi 1 + yi h f xi yi , ( ) ⋅ + := ytruei ytrue xi ( ) := <---True solution value at x i error i ytruei yi − ytruei 100 ⋅ := <-- Percentage Error between true value and value computed by Euler's method The computed and true solutions and the error due to Euler's method for the range of interest are tabulated and shown in Table 9.1 below. 320 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 331. TABLE 9.1 Comparison betweentrue and computed solutions and error due to Euler's method x-value Computed solution True solution Percentage error xi 0 0.5 1 1.5 2 2.5 3 3.5 4 = yi 2 6.5 7.5 8 9.5 12 14 12.5 3 = ytruei 2 4.438 5 5.938 8 10.438 11 5.938 -10 = error i 0 -46.479 -50 -34.737 -18.75 -14.97 -27.273 -110.526 130 = Mathcad's cubic spline interpolation can now be resorted to as shown below to generate a plot of the computed solution . This solution is compared with the true solution in Figure 9.5. The errors due to Euler's method for the range of interest are presented in Figure 9.6, which shows that the Euler error for this problem can be as large as 98 % , thus making the method extremely inaccurate. Vxi xi := Vyi yi := ytrue x ( ) x 4 − 6 x 3 ⋅ + 11 x 2 ⋅ − 9 x ⋅ + 2 + := Vx 0 0.5 1 1.5 2 2.5 3 3.5 4 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy 2 6.5 7.5 8 9.5 12 14 12.5 3 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <-- Vx and Vy vectors for Euler solution generated Vs cspline Vx Vy , ( ) := fspline x ( ) interp Vs Vx , Vy , x , ( ) := <--Cubic spline solution generated with interp/ cspline ytrue x ( ) x 4 − 6 x 3 ⋅ + 11 x 2 ⋅ − 9 x ⋅ + 2 + := <-- True solution Chapter 9: Numerical Solution of Ordinary Differential Equations 321
- 332. %error x ( ) ytruex ( ) fspline x ( ) − ytrue x ( ) 100 ⋅ := <--Percentage error between computed and true solutions, function of x x 0 0.01 , 4 .. := 0 0.5 1 1.5 2 2.5 3 3.5 4 0 5 10 15 true solution Euler solution true solution Euler solution Comparison- Euler with True solution X-VALUES Y-VALUES ytrue x ( ) fspline x ( ) x Figure 9.5. Comparison of Euler solution with true solution 0 0.5 1 1.5 2 2.5 3 3.5 100 50 0 ERROR VERSUS x X-VALUES PERCENTAGE ERROR %error x ( ) x Figure 9.6. Error due to Euler's method 322 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 333. 9.4 MODIFIED EULER'SMETHOD Euler's method which was discussed in the previous section does not require the use of higher-order derivatives but apparently yields very inaccurate answers unless the step size used is very very small. The modified Euler's method , which is an attempt at improving the accuracy of the Euler solution, is quite similar and is still a one-step method but uses an average slope over the segment or step rather than that at the beginning of the segment . Thus, in this "modified" method, Equation (9.15) is replaced by yn 1 + yn h f xn 1 + ( ) f xn ( ) + 2 ⋅ + = (9.17) Another way to look at this method is to note that it employs the trapezoidal rule as shown in the sketch below to compute the integral xn xn 1 + x f x ( ) ⌠ ⎮ ⌡ d The procedural steps to be followed in this method are given below. 1. Determine slope dy/dx at the beginning of the interval , x0 2. Estimate y at x1 using Euler's formula (Equation 9.15) yn 1 + yn h f xn yn , ( ) ⋅ + = 3. Evaluate slope dy/dx at the end of the interval x1, i.e., evaluate f(x1, y1 ). 4. Determine average of the slopes at the beginning and end of the interval, i.e, f x0 y0 , ( ) f x1 y1 , ( ) + 2 Chapter 9: Numerical Solution of Ordinary Differential Equations 323
- 334. 5. Compute arevised estimate of y 1 using y1 y0 h f x0 y0 , ( ) f x1 y1 , ( ) + 2 ⋅ + = 6. Keep improving the estimate " y 1 " until convergence to a solution takes place . Example 9.3 Using the modified Euler method, solve: dy/dx= f(x,y) = -3x -y; y(0) = -1.5; h= 0.1; Range of x: 0 to 0.5 The function, f(x,y) which is the derivative of y(x) is given as f x y , ( ) 3 − x ⋅ y − := h 0.1 := The initial condition prescribed is x0 0. := y0 1.5 − := The procedure of the modified Euler method is utilized as shown in the following steps to generate numerical solutions at x= 0.1 and 0.2. Solutions for x= 0.3, 0.4 and 0.5 are obtained in a similar manner. The quantity y1c refers to the revised estimate of the solution in each iteration . x1 0.1 := f x0 y0 , ( ) 1.5 = y1 y0 h f x0 y0 , ( ) ⋅ + := y1 1.35 − = f x1 y1 , ( ) 1.05 = f x1 y1 , ( ) f x0 y0 , ( ) + 2 1.275 = y1c y0 f x1 y1 , ( ) f x0 y0 , ( ) + 2 h ⋅ + := y1c 1.373 − = y1 y1c := f x1 y1 , ( ) 1.073 = f x1 y1 , ( ) f x0 y0 , ( ) + 2 1.286 = y1c y0 f x1 y1 , ( ) f x0 y0 , ( ) + 2 h ⋅ + := y1c 1.371 − = y1 y1c := f x1 y1 , ( ) 1.071 = f x1 y1 , ( ) f x0 y0 , ( ) + 2 1.286 = 324 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 335. y1c y0 f x1y1 , ( ) f x0 y0 , ( ) + 2 h ⋅ + := y1c 1.371 − = y1 y1c := f x1 y1 , ( ) 1.071 = f x1 y1 , ( ) f x0 y0 , ( ) + 2 1.286 = y1c y0 f x1 y1 , ( ) f x0 y0 , ( ) + 2 h ⋅ + := y1c 1.371 − = y1 y1c := x2 0.2 := f x1 y1 , ( ) 1.071 = y2 y1 h f x1 y1 , ( ) ⋅ + := y2 1.264 − = f x2 y2 , ( ) 0.664 = f x2 y2 , ( ) f x1 y1 , ( ) + 2 0.868 = y2c y1 f x2 y2 , ( ) f x1 y1 , ( ) + 2 h ⋅ + := y2c 1.285 − = y2 y2c := f x2 y2 , ( ) 0.685 = f x2 y2 , ( ) f x1 y1 , ( ) + 2 0.878 = y2c y1 f x2 y2 , ( ) f x1 y1 , ( ) + 2 h ⋅ + := y2c 1.284 − = y2 y2c := f x2 y2 , ( ) 0.684 = f x2 y2 , ( ) f x1 y1 , ( ) + 2 0.878 = y2c y1 f x2 y2 , ( ) f x1 y1 , ( ) + 2 h ⋅ + := y2c 1.284 − = y2 y2c := f x2 y2 , ( ) 0.684 = f x2 y2 , ( ) f x1 y1 , ( ) + 2 0.878 = y2c y1 f x2 y2 , ( ) f x1 y1 , ( ) + 2 h ⋅ + := y2c 1.284 − = y2 y2c := At x3, x4, and x5 , the solutions can be similarly generated as : Chapter 9: Numerical Solution of Ordinary Differential Equations 325
- 336. x3 0.3 := y31.2328 − := x4 0.4 := y4 1.2154 − := x5 0.5 := y5 1.2283 − := The results obtained by this method are presented in Table 9.2 in a tabular format for x- values ranging from 0 to 0.5 with a step size of 0.1. Converged values of the solution are pointed out in the " corrected yn " column . TABLE 9.2 Calculations by Modified Euler method corrected avg slope f(xn,yn) yn xn yn at beginning and end 1.5 -1.5 0 -1.3725 1.275 1.05 -1.35 0.1 -1.3714 1.2862 1.0725 -1.3725 -1.3714 1.2857 1.0714 -1.3714 converged -1.3714 1.2857 1.0714 -1.3714 -1.2846 0.8679 0.6643 -1.2643 0.2 -1.2836 0.878 0.6846 -1.2846 -1.2837 0.8775 0.6836 -1.2836 converged -1.2837 0.8776 0.6837 -1.2837 -1.2337 0.4995 0.3153 -1.2153 0.3 -1.2338 0.5087 0.3337 -1.2337 converged -1.2338 0.5082 0.3328 -1.2328 -1.2162 0.1662 -0.0004 -1.1996 0.4 -1.2154 0.1745 0.0162 -1.2162 converged -1.2154 0.1741 0.0154 -1.2154 -1.229 -0.1353 -0.2861 -1.2139 0.5 -1.2282 -0.1278 -0.271 -1.229 -1.2283 -0.1282 -0.2718 -1.2282 converged -1.2283 -0.1282 -0.2717 -1.2283 The x values and the computed "y" values are assembled in the form of vectors Vx and Vy in order to generate a cubic spline interpolation as follows which is presented as a plot of the solution in Figure 9.7. 326 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 337. <--Vx and Vyvectors for modified Euler solution generated Vx x0 x1 x2 x3 x4 x5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vx 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy y0 y1 y2 y3 y4 y5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy 1.5 − 1.371 − 1.284 − 1.233 − 1.215 − 1.228 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vs cspline Vx Vy , ( ) := fspline x ( ) interp Vs Vx , Vy , x , ( ) := <-- cubic spline interpolation for the generated solution A solution by the Euler method is generated below in order to obtain a comparison between the Modified Euler solution and the not-so-accurate Euler solution x0 0 := y0 1.5 − := <-- Prescribed initial condition xlast 0.5 := <--End-point of prescribed range of solution h 0.1 := <-- Step size f x y , ( ) 3 − x ⋅ y − := <--- Prescribed derivative of y(x) at (x,y) f x0 y0 , ( ) 1.5 = <-- Derivative value at starting point Number of points beyond the starting point at which solution is desired N xlast x0 − h := N 5 = <-- i 0 1 , N .. := xi 1 + xi h + := <--Formulas of Euler's method yi 1 + yi h f xi yi , ( ) ⋅ + := Chapter 9: Numerical Solution of Ordinary Differential Equations 327
- 338. x-value Computed solution xi 0 0.1 0.2 0.3 0.4 0.5 =yi -1.5 -1.35 -1.245 -1.181 -1.152 -1.157 = <--- Euler solution Mathcad's cubic spline interpolation is now used below to generate a plot of the computed Euler solution, which is compared with the Modified Euler solution in Figure 9.7. Vxeuleri xi := Vyeuleri yi := Vxeuler 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vyeuler 1.5 − 1.35 − 1.245 − 1.181 − 1.152 − 1.157 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <-- Vx and Vy vectors for Euler solution generated Vseuler cspline Vxeuler Vyeuler , ( ) := Cubic spline solution using the Euler method generated with interp / cspline fsplineeuler x ( ) interp Vseuler Vxeuler , Vyeuler , x , ( ) := <--- x 0 0.01 , 0.5 .. := 0 0.1 0.2 0.3 0.4 1.6 1.4 1.2 1 Euler solution Modified Euler solution Euler solution Modified Euler solution Euler/ Mod Euler solutions compared X- Values Y- Values fsplineeuler x ( ) fspline x ( ) x Figure 9.7. Comparison of the modified Euler solution with the Euler solution . 328 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 339. 9.5 RUNGE- KUTTAMETHODS Runge-Kutta methods are among the most popular techniques employed in the numerical solution of ordinary differential equations. These methods use the y-value at the beginning of an interval along with some representative slope over the interval to determine the y-value at the end of the interval. Special cases of the method are Euler's method , which uses the slope at the beginning of the interval and the modified Euler method which uses the average slope over the segment . There are several orders of Runge-Kutta methods from the first order to the eighth depending on the number of iterations involved in the solution -generating process. However, only the fourth-order Runge-Kutta method will be disussed in detail in this chapter. All of these still resort only to the use of the first derivative and do not require the involvement of higher order derivatives. While the second-order Runge-Kutta method is essentially the same as the modified Euler method, and offers the same order of accuracy, the main difference is that it resorts only to two iterations. The third-order Runge-Kutta method resorts to Simpson's one-third rule , which is a higher-order integration scheme than the trapezoidal rule , and uses slopes at the beginning, middle and end of the segment in order to estimate the y-value at the end of the segment . As its name clearly suggests , the fourth-order method does require four iterations to determine the y-value at the segment end. Because higher-order integration schemes are employed, the Runge-Kutta fourth-order method gives very accurate results. Orders of the method that are higher than fourth are not common . Advantages of Runge-Kutta methods include ease of programming, good stability characteristics, flexibility of step size, and self-starting capability. 9.5.1 Fourth-order Runge-Kutta Method The fourth-order Runge-Kutta method comes in two popular versions . The first version is based on Simpson's one-third rule while the second is based on Simpson's three-eighth rule. The sequence of calculations to be used in these versions is given below. Procedure for version 1- based on Simpson's one-third rule: Figure 9.8. Basis of the fourth-order Runge-Kutta method utilizing Simpson's one- third rule Chapter 9: Numerical Solution of Ordinary Differential Equations 329
- 340. Given : dy/dx= f(x,y) = Slope , y= y0 at x= x0 ( Initial condition ), h = interval width. Determine: y1 , which is y ( x 1 ). ( See Figure 9.8) 1. Compute h times slope K1 at ( x0, y0): K1= h. f(x0,y0) 2. Compute y at midpoint of interval x0 to x1 using the Euler formula yMP y0 1 2 K1 ⋅ + = (9.18) 3. Compute K2 at x = x MP, y= yMP K2 h f xMP yMP , ( ) ⋅ = 4. Obtain a revised estimate of y MP with y0, K2 and (1/2)h yMP y0 1 2 K2 ⋅ + = (9.19) 5. Compute revised estimate of K at mid-interval, using above value of yMP K3 h f xMP yMP , ( ) ⋅ = (9.20) 6 Compute y1 using: . y1 y0 K3 + = (9.21) 7. Compute K at end of interval: K4 h f x1 y1 , ( ) ⋅ = (9.22) 8. Compute revised estimate of y1 using: y1 y0 1 6 K1 2 K2 ⋅ + 2 K3 ⋅ + K4 + ( ) ⋅ + = (9.23) per Simpson's 1/3 rd rule. Notice the four estimates of slope being used 9. Repeat above steps for computing y2, y3 etc. For example, to compute y2, repeat the procedure with (x1,y1) replacing (x0,y0) and (x2,y2) replacing (x1,y1). Procedure for version 2- based on Simpson's three-eighth rule : The procedure for version 2 is very similar to version 1, except that the range x0 to x1 is 330 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 341. subdivided into threeregions , the starting and ending points of which are x0, x13 , x23 and x1. The slopes at these intermediate points within the range x0 to x1 are computed and revised until y1 is determined as shown in the following summary of calculations. Summary of calculations for versions 1 and 2: VERSION 1- based on Simpson's one-third rule: K1 h f x0 y0 , ( ) ⋅ = yMP y0 1 2 K1 ⋅ + = K2 h f xMP yMP , ( ) ⋅ = yMP y0 1 2 K2 ⋅ + = Revised yMP (using h times the new slope) (9.24) <----- K3 h f xMP yMP , ( ) ⋅ = y1 y0 K3 + = K4 h f x1 y1 , ( ) ⋅ = y1 y0 1 6 K1 2 K2 ⋅ + 2 K3 ⋅ + K4 + ( ) ⋅ + = VERSION 2- based on Simpson's three-eighth rule: x13 x0 1 3 h ⋅ + = x23 x0 2 3 h ⋅ + = K1 h f x0 y0 , ( ) ⋅ = y13 y0 1 3 K1 ⋅ + = K2 h f x13 y13 , ( ) ⋅ = y23 y0 1 3 K1 ⋅ + 1 3 K2 ⋅ + = Note that y0+(1/3)K1 is, in fact, y13 (9.25) K3 h f x23 y23 , ( ) ⋅ = y1adj y0 K1 + K2 − K3 + = <--- " K1 - K2 + K3 " represents h multiplied by a representative average slope over segment x0 -> x1 K4 h f x1 y1adj , ( ) ⋅ = y1 y0 1 8 K1 3 K2 ⋅ + 3 K3 ⋅ + K4 + ( ) ⋅ + = Chapter 9: Numerical Solution of Ordinary Differential Equations 331
- 342. Example 9.4 Solve: dy/dx= f(x,y) = -2x 2 - y ; Initial condition : y(0) = -1.75 ; h = 0.1; Range of x : 0 to 0.5 Version 1 - based on Simpson's one-third rule: The given information is. f x y , ( ) 2 − x 2 ⋅ y − := h 0.1 := x0 0 := y0 1.75 − := Calculations for x= x1 and x= x2 are shown below x1 0.1 := f x0 y0 , ( ) 1.75 = k1 h f x0 y0 , ( ) ⋅ := k1 0.175 = ymp y0 0.5 k1 ⋅ + := ymp 1.6625 − = xmp x0 0.5 h ⋅ + := xmp 0.05 = k2 h f xmp ymp , ( ) ⋅ := k2 0.166 = ymp y0 0.5 k2 ⋅ + := ymp 1.6671 − = k3 h f xmp ymp , ( ) ⋅ := k3 0.166 = y1 y0 k3 + := y1 1.584 − = k4 h f x1 y1 , ( ) ⋅ := k4 0.156 = This formula is based on <-- Simpson's one-third rule y1 y0 1 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 2 k2 ⋅ + 2 k3 ⋅ + k4 + ( ) ⋅ + := y1 1.584 − = x2 0.2 := f x1 y1 , ( ) 1.564 = k1 h f x1 y1 , ( ) ⋅ := k1 0.156 = ymp y1 0.5 k1 ⋅ + := ymp 1.50591 − = xmp x1 0.5 h ⋅ + := xmp 0.15 = k2 h f xmp ymp , ( ) ⋅ := k2 0.146 = ymp y1 0.5 k2 ⋅ + := ymp 1.5111 − = k3 h f xmp ymp , ( ) ⋅ := k3 0.147 = y2 y1 k3 + := y2 1.438 − = k4 h f x2 y2 , ( ) ⋅ := k4 0.136 = y2 y1 1 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 2 k2 ⋅ + 2 k3 ⋅ + k4 + ( ) ⋅ + := y2 1.438 − = Calculations for x= 0.3, 0.4 and 0.5 are similar to the above and lead to 332 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 343. x3 0.3 := y31.31316 − := x4 0.4 := y4 1.21178 − := x5 0.5 := y5 1.13531 − := In order to generate a cubic spline interpolation for the Runge-Kutta solution, the x and y values are put into vectors Vx and Vy. The resulting y is plotted as a function of x in Figure 9.9. <-- Vx and Vy vectors for the Runge-Kutta results generated Vx x0 x1 x2 x3 x4 x5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy y0 y1 y2 y3 y4 y5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vx 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy 1.75 − 1.584 − 1.438 − 1.313 − 1.212 − 1.135 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vs cspline Vx Vy , ( ) := cubic spline interpolation for the solution generated fspline1 x ( ) interp Vs Vx , Vy , x , ( ) := <----- Version 2 - based on Simpson's three-eighth rule: f x y , ( ) 2 − x 2 ⋅ y − := h 0.1 := x0 0 := y0 1.75 − := f x0 y0 , ( ) 1.75 = x1 0.1 := k1 h f x0 y0 , ( ) ⋅ := k1 0.175 = y13 y0 0.33333 k1 ⋅ + := y13 1.69167 − = x13 x0 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ h ⋅ + := x13 0.033 = k2 h f x13 y13 , ( ) ⋅ := k2 0.169 = x23 x0 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ h ⋅ + := x23 0.067 = y23 y0 k1 3 + k2 3 + := y23 1.635 − = k3 h f x23 y23 , ( ) ⋅ := k3 0.163 = x1 x0 h + := y1adj y0 k1 + k2 − k3 + := y1adj 1.581 − = k4 h f x1 y1adj , ( ) ⋅ := k4 0.156 = This formula is based on <-- Simpson's 3/8 th rule y1 y0 1 8 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 3 k2 ⋅ + 3 k3 ⋅ + k4 + ( ) ⋅ + := y1 1.584 − = x2 0.2 := Chapter 9: Numerical Solution of Ordinary Differential Equations 333
- 344. k1 h fx1 y1 , ( ) ⋅ := k1 0.156 = y13 y1 0.33333 k1 ⋅ + := y13 1.53212 − = x13 x1 1 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ h ⋅ + := x13 0.133 = k2 h f x13 y13 , ( ) ⋅ := k2 0.15 = x23 x1 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ h ⋅ + := x23 0.167 = y23 y1 k1 3 + k2 3 + := y23 1.482 − = k3 h f x23 y23 , ( ) ⋅ := k3 0.143 = x2 x1 h + := y2adj y1 k1 + k2 − k3 + := y2adj 1.435 − = k4 h f x2 y2adj , ( ) ⋅ := k4 0.135 = This formula is based on <--- Simpson's 3/8 th rule y2 y1 1 8 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 3 k2 ⋅ + 3 k3 ⋅ + k4 + ( ) ⋅ + := y2 1.438 − = Calculations for x = 0.3 , 0.4 and 0.5 are again very similar to the above and lead to: x3 0.3 := y3 1.31361 − := x4 0.4 := y4 1.21238 − := x5 0.5 := y5 1.13605 − := Generation of the Vx and Vy vectors and fspline(x) as in version 1, leads to the graph in Figure 9.9 which compares the two solutions of the fourth-order Runge-Kutta method. These plots definitely indicate that the two solutions are very very close to each other. Vs cspline Vx Vy , ( ) := Vx x0 x1 x2 x3 x4 x5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy y0 y1 y2 y3 y4 y5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vx 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy 1.75 − 1.584 − 1.438 − 1.314 − 1.212 − 1.136 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = fspline2 x ( ) interp Vs Vx , Vy , x , ( ) := fspline2 0.5 ( ) 1.136 − = x 0 0.01 , 0.5 .. := 0 0.1 0.2 0.3 0.4 2 1.5 1 Version 1-one-third rule Version 2-three-eighth rule Version 1-one-third rule Version 2-three-eighth rule Fourth Order R/K Method X-VALUES Y-VALUES fspline1 x ( ) fspline2 x ( ) x Figure 9.9 Solution by fourth-order Runge-Kutta method 334 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 345. 9.5.2 Mathcad Solutionsto a First-Order Differential Equation The Mathcad function " rkfixed" is a general-purpose Runge-Kutta solver which can handle both nth order differential equations with initial conditions as well as systems of differential equations. The arguments to the function rkfixed are rkfixed( y,x1, x2, npoints, D) where y is a vector of n initial values, n being the order of the differential equation or the size of the system of equations being solved. In this case, where we have a first-order differential equation, the vector reduces to one point. which is the initial value of y . The quantities x1 and x2 are the endpoints of the interval over which the solution is to be generated, and npoints is the number of points beyond the initial point at which the solution is to be approximated. This controls the number of rows ( 1+ npoints ) in the matrix that rkfixed returns. Here, D(x,y) is an n -element vector function containing the first derivatives of the unknown functions. and, for a first-order differential equation, this will be just one function , namely the given derivative. Another Mathcad function that can be used is "Rkadapt". While "rkfixed" evaluates the solution at equally spaced intervals , " Rkadapt " examines how fast the solution is changing and adjusts the step size accordingly. Although "Rkadapt " may use non-uniform step sizes in generating the solution, it will still return a solution at equally spaced points. The arguments used by " Rkadapt " are the same as those of " rkfixed " . To illustrate the use of these Mathcad functions, the Mathcad solution to the problem of Example 9.1 presented in the section on the Taylor series method is outlined below. Define the initial y value: Use the left bracket " [ " to make the subscript: y0 4.0 := Put in the first derivative that is given: yprime = 7x4 -3 x 2 + 9 D x y , ( ) 7 x 4 ⋅ 3 x 2 ⋅ − 9 + := Evaluate the solution at 50 points between 1.0 and 1.3 Z rkfixed y 1.0 , 1.3 , 50 , D , ( ) := Chapter 9: Numerical Solution of Ordinary Differential Equations 335
- 346. i 0 rowsZ ( ) 1 − .. := Plot the solution. Use Ctrl + 6 to get the superscript. ( See Figure 9.10). To use Rkadapt , the procedure, which is presented below , is essentially the same as with the use of rkfixed. Define the initial y value: Use the left bracket " [ " to make the subscript: y0 4.0 := Put in the first derivative that is given: yprime = 7x4 - 3 x 2 + 9 D x y , ( ) 7 x 4 ⋅ 3 x 2 ⋅ − 9 + := Evaluate the solution at 50 points between 1.0 and 1.3 Y Rkadapt y 1.0 , 1.3 , 50 , D , ( ) := i 0 rows Y ( ) 1 − .. := Plot the solution. Use Ctrl + 6 to get the superscript. (See Figure 9.10 ). The solutions to the differential equation obtained with rkfixed and Rkadapt are presented in Figure 9.10 and compared with the Taylor series solution, which is redone here in order to obtain its plot. x0 1 := y0 4.0 := yprime x ( ) 7 x 4 ⋅ 3 x 2 ⋅ − 9 + := ydblprime x ( ) 28 x 3 ⋅ 6 x ⋅ − := ytrprime x ( ) 84 x 2 ⋅ 6 − := y4prime x ( ) 168 x ⋅ := y5prime x ( ) 168. := y6prime x ( ) 0 := x 1.0 1.05 , 1.3 .. := 336 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 347. y x ( )y0 x x0 − ( ) yprime x0 ( ) ⋅ + 1 2! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 2 ⋅ ydblprime x0 ( ) ⋅ + 1 3! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 3 ⋅ ytrprime x0 ( ) ⋅ + ... 1 4! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 4 ⋅ y4prime x0 ( ) ⋅ 1 5! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 5 ⋅ y5prime x0 ( ) ⋅ + + ... 1 6! ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ x x0 − ( ) 6 ⋅ y6prime x0 ( ) ⋅ + ... := <-- Taylor series solution xi Y 0 〈 〉 ( )i := yi y xi ( ) := 1 1.05 1.1 1.15 1.2 1.25 4 6 8 10 Rkadapt solution rkfixed solution Taylor series solution Rkadapt solution rkfixed solution Taylor series solution Comparison of solutions X-Values Y-Values Y 1 〈 〉 ( )i Z 1 〈 〉 ( )i yi Y 0 〈 〉 ( )i Figure 9.10. Comparison of Mathcad solutions with the Taylor series solution As can be seen from the solution plots, the Mathcad solutions agree well with the Taylor series solution. Example 9.5 Using the Mathcad functions, solve the problem of Example 9.3 presented in the section on the modified Euler method, namely, dy/dx= f(x,y) = -3x -y; y(0) = -1.5 ; h= 0.1; Range of x: 0 to 0.5 The Mathcad statements necessary for the computation of the solution with rkfixed as well as Rkadapt are Chapter 9: Numerical Solution of Ordinary Differential Equations 337
- 348. y0 1.5 − := Dx y , ( ) 3 − x ⋅ y − := Z rkfixed y 0. , 0.5 , 50 , D , ( ) := i 0 rows Z ( ) 1 − .. := Y Rkadapt y 0. , 0.5 , 50 , D , ( ) := The modified Euler solution is partially redone here in the following steps in order to generate a solution plot. Vx 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy 1.5 − 1.3714 − 1.2837 − 1.2328 − 1.2154 − 1.2283 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vx and Vy vectors for the modified Euler solution <--- Vs cspline Vx Vy , ( ) := fspline x ( ) interp Vs Vx , Vy , x , ( ) := <---- modified Euler solution x 0 0.01 , 0.5 .. := xi Y 0 〈 〉 ( )i := fsplinei fspline xi ( ) := Figure 9.11 shows a comparison of the Mathcad solutions with the modified Euler solution. The agreement between the results is, clearly, very good. 0 0.1 0.2 0.3 0.4 1.5 1.4 1.3 rkfixed solution Rkadapt solution modified Euler solution rkfixed solution Rkadapt solution modified Euler solution Comparison of solutions X-VALUES Y-VALUES Z 1 〈 〉 ( )i Y 1 〈 〉 ( )i fsplinei Z 0 〈 〉 ( )i Figure 9.11. Comparison of Mathcad solutions with modified Euler method 338 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 349. Example 9.6 Using theMathcad functions , solve the problem of Example 9.4 presented in the sub-section on the Runge-Kutta fourth order method, which is repeated below. Solve: dy/dx = f(x,y) = - 2x 2 - y ; Initial condition : y(0) = -1.75 ; h = 0.1; Range of x: 0 to 0.5 The Mathcad statements needed to solve the problem are y0 1.75 − := D x y , ( ) 2 − x 2 ⋅ y − := Z rkfixed y 0. , 0.5 , 50 , D , ( ) := Y Rkadapt y 0. , 0.5 , 50 , D , ( ) := i 0 rows Z ( ) 1 − .. := The Runge-Kutta solutions (Versions 1 and 2 ) are repeated here in the following steps in order to generate solution plots in Figure 9.12 for comparison with the solutions generated with the Mathcad functions rkfixed and Rkadapt. Vx and Vy vectors for the Runge-Kutta solution (Version 1 - one-third rule) Vx1 0.0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy1 1.75 − 1.58412 − 1.43786 − 1.31316 − 1.21178 − 1.13531 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := <------ Vs1 cspline Vx1 Vy1 , ( ) := fspline1 x ( ) interp Vs1 Vx1 , Vy1 , x , ( ) := <-- R/K solution (Version 1) fspline1i fspline1 Z 0 〈 〉 ( )i ⎡ ⎣ ⎤ ⎦ := Vx2 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy2 1.75 − 1.58426 − 1.43815 − 1.31361 − 1.21238 − 1.13605 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := <---- Vx and Vy vectors for the Runge-Kutta solution (Version 2- three-eighth rule ) Chapter 9: Numerical Solution of Ordinary Differential Equations 339
- 350. Vs2 cspline Vx2Vy2 , ( ) := fspline2 x ( ) interp Vs2 Vx2 , Vy2 , x , ( ) := <--R/K solution- version 2 x 0 0.01 , 0.5 .. := fspline2i fspline2 Z 0 〈 〉 ( )i ⎡ ⎣ ⎤ ⎦ := 0 0.1 0.2 0.3 0.4 0.5 1.8 1.6 1.4 1.2 1 rkfixed solution Rkadapt solution version 1- one-third rule version 2- three-eighth rule rkfixed solution Rkadapt solution version 1- one-third rule version 2- three-eighth rule Comparison of solutions x- values y-values Z 1 〈 〉 ( )i Y 1 〈 〉 ( )i fspline1i fspline2i Z 0 〈 〉 ( )i Figure 9.12 Comparison of Mathcad solutions with Runge-Kutta fourth order methods The functions " rkfixed " and " Rkadapt " generate the solution to the given differential equation for a number of uniformly spaced x values in the integration range bounded by x1 and x2 . If you are interested in seeing the solution only at a value of x, say x2 , it is preferable to resort to the function " rkadapt " . Although " rkfixed " and " Rkadapt " will give you the solution at x2 , they also go through a lot of unnecessary work returning intermediate values of the solution at x- values in which you may not have any interest at all. The use of the function " rkadapt " is illustrated below: rkadapt ( y, x1, x2, acc, D, kmax, save) where y is a vector of initial values, x1 and x2 are the endpoints of the interval of interest over which the solution to the differential equation is to be generated. The initial values in " y " are the values, then, at x1. The parameter "acc" controls the accuracy of the solution. 340 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 351. Small values of" acc " yield accurate solutions. A value of " acc " around 0.001 is recommended. The parameter " kmax " is the maximum number of intermediate points at which the solution will be computed. This value places an upper limit on the number of rows of the matrix returned by " rkadapt" . The parameter " save " is the smallest allowable spacing between the x- values at which the solutions are to be computed . This command computes y values for the x- values given in the first column. Y rkadapt y 0 , 0.5 , 0.001 , D , 50 , 0.001 , ( ) := <---- x y This is the solution returned by "rkadapt" for the interval x= 0 to 0.5, with acc= 0.001. The solution is generated at kmax = 50 points in the interval where the maximum allowable spacing between the x-values at which the solution is to be computed is save = 0.001. The last row is the solution generated by "Rkadapt" at the endpoint x= 0.5 Y 0 1 0 1 2 3 4 0 -1.75 -3 5·10 -1.741272 0.03 -1.698298 0.155 -1.501116 0.5 -1.135307 = <---- 9.6 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS Sometimes, in engineering and the physical sciences, the need to solve a system of first-order ordinary differential equations is encountered. Such a system is dy1/dx = f ( x, y1, y2, ....... yn) dy2/dx = g ( x, y1, y2, ....... yn) .................................................. dyn/dx = z ( x, y1, y2, ....... yn) (9.26) with n initial conditions specified at the starting value of the independent variable x . This system of equations can be easily solved using any of the methods discussed in sections 9.2- 9.5. However, here, the application of the the fourth order Runge-Kutta procedure to obtain a solution to a problem involving a pair of first-order differential equations will be illustrated. Notice that in a problem involving a pair of differential equations , where x and y are the dependent variables and t is the independent variable , x and y calculations have to be alternately performed as illustrated in the following example. Chapter 9: Numerical Solution of Ordinary Differential Equations 341
- 352. Example 9.7. Using theRunge-Kutta fourth-order method, obtain a solution to dx/dt = f(x,y,t) = xy2 + t; dy/dt= g(x,y,t) = ty + x2 for t = 0 to t= 0.5 seconds. The initial conditions are given as: x(0)= -1; y (0) = 1 Use a time increment of h = 0.1 second. Note that both x and y are functions of the independent variable " t". Thus, in this case, dx/dt = f(x,y,t ) = xy2 + t , and dy/dt = g(x,y,t) = ty + x2 Version 1 of the Runge-Kutta Method which utilizes Simpson's one-third rule, will then resort to the following sequential equations to compute x and y at t1, where t1 = t0 + h., and h is the increment or step size k1 = h f(x0, y0, t0) ; m1= h g(x0,y0, t0) ; tmp = t0 + 0.5 h ; xmp= x0+ 0.5 k1; ymp= y0+0.5 m1 k2= h f(xmp, ymp, tmp) ; m2 = h g( xmp, ymp, tmp ) xmp* = x0 + 0.5 k2 ; ymp* = y0+ 0.5 m2 k3 = h f ( xmp*, ymp*, tmp ) ; m3 = h g ( xmp*, ymp*, tmp ) x1 = x0 + k3 ; y1 = y0 + m3 ; t1 = t0 + h k4 = h f ( x1, y1, t1 ) ; m4 = h g ( x1, y1, t1 ) x1= x0 + (1/6) [ k1 +2 k2 + 2 k3 + k4 ] ; y1 = y0 + (1/6 ) [ m1 + 2 m2 + 2 m3 + m4 ] In order to compute the solution (x1,y1) at t1, we start with known values x0, y0 and t0 as shown above. Similarly, to compute the solution (x2, y2) at t2, the above procedure needs to be repeated with ( x1, y1 , t1) replacing (x0, y0,t0) and (x2,y2,t2) replacing (x1,y1,t1). Define the functions f(x,y,t) and g(x,y,t), the step size and the initial conditions:. 342 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 353. f x y ,t , ( ) x y 2 ⋅ t + := g x y , t , ( ) t y ⋅ x 2 + := <--Define functions f(x,y,t) and g(x,y,t) h 0.1 := <--Time increment or step size t0 0 := x0 1 − := y0 1 := <-- Given initial conditions Calculations for generating solution at t1: t1 t0 h + := t1 0.1 = k1 h f x0 y0 , t0 , ( ) ⋅ := k1 0.1 − = m1 h g x0 y0 , t0 , ( ) ⋅ := m1 0.1 = tmp t0 0.5 h ⋅ + := tmp 0.05 = xmp x0 0.5 k1 ⋅ + := xmp 1.05 − = ymp y0 0.5 m1 ⋅ + := ymp 1.05 = k2 h f xmp ymp , tmp , ( ) ⋅ := k2 0.110763 − = m2 h g xmp ymp , tmp , ( ) ⋅ := m2 0.1155 = xmpstar x0 0.5 k2 ⋅ + := xmpstar 1.055381 − = ympstar y0 0.5 m2 ⋅ + := ympstar 1.05775 = k3 h f xmpstar ympstar , tmp , ( ) ⋅ := k3 0.11308 − = m3 h g xmpstar ympstar , tmp , ( ) ⋅ := m3 0.116672 = x1 x0 k3 + := x1 1.11308 − = y1 y0 m3 + := y1 1.116672 = t1 t0 h + := t1 0.1 = k4 h f x1 y1 , t1 , ( ) ⋅ := k4 0.128796 − = m4 h g x1 y1 , t1 , ( ) ⋅ := m4 0.135061 = Chapter 9: Numerical Solution of Ordinary Differential Equations 343
- 354. x1 x0 1 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 2k2 ⋅ + 2 k3 ⋅ + k4 + ( ) ⋅ + := x1 1.112747 − = y1 y0 1 6 m1 2 m2 ⋅ + 2 m3 ⋅ + m4 + ( ) ⋅ + := y1 1.116567 = Calculations for generating solution at t2: t2 t1 h + := t2 0.2 = k1 h f x1 y1 , t1 , ( ) ⋅ := k1 0.128729 − = m1 h g x1 y1 , t1 , ( ) ⋅ := m1 0.134986 = tmp t1 0.5 h ⋅ + := tmp 0.15 = xmp x1 0.5 k1 ⋅ + := xmp 1.177111 − = ymp y1 0.5 m1 ⋅ + := ymp 1.184061 = k2 h f xmp ymp , tmp , ( ) ⋅ := k2 0.150031 − = m2 h g xmp ymp , tmp , ( ) ⋅ := m2 0.15632 = xmpstar x1 0.5 k2 ⋅ + := xmpstar 1.187762 − = ympstar y1 0.5 m2 ⋅ + := ympstar 1.194727 = k3 h f xmpstar ympstar , tmp , ( ) ⋅ := k3 0.154538 − = m3 h g xmpstar ympstar , tmp , ( ) ⋅ := m3 0.158999 = x2 x1 k3 + := x2 1.267285 − = y2 y1 m3 + := y2 1.275566 = k4 h f x2 y2 , t2 , ( ) ⋅ := k4 0.186196 − = m4 h g x2 y2 , t2 , ( ) ⋅ := m4 0.186112 = 344 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 355. x2 x1 1 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 2k2 ⋅ + 2 k3 ⋅ + k4 + ( ) ⋅ + := x2 1.266757 − = y2 y1 1 6 m1 2 m2 ⋅ + 2 m3 ⋅ + m4 + ( ) ⋅ + := y2 1.27519 = Caculations for t= 0.3, 0.4 and 0.5 seconds can be performed in a similar way , giving the following results. t3 0.3 := x3 1.506 − := y3 1.5 := t4 0.4 := x4 1.945 − := y4 1.848 := t5 0.5 := x5 3.025 − := y5 2.522 := Plots of the results of the Runge-Kutta process can be obtained by an interpolation procedure involving the generation of the Vt, Vx and Vy vectors and the spline functions xspline(t) and yspline(t) as shown below. Vt t0 t1 t2 t3 t4 t5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vx x0 x1 x2 x3 x4 x5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy y0 y1 y2 y3 y4 y5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vt 0 0.1 0.2 0.3 0.4 0.5 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vx 1 − 1.112747 − 1.266757 − 1.506 − 1.945 − 3.025 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy 1 1.116567 1.27519 1.5 1.848 2.522 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vsx cspline Vt Vx , ( ) := Vsy cspline Vt Vy , ( ) := Chapter 9: Numerical Solution of Ordinary Differential Equations 345
- 356. xspline t ( )interp Vsx Vt , Vx , t , ( ) := yspline t ( ) interp Vsy Vt , Vy , t , ( ) := The Mathcad solution process using " rkfixed " and " Rkadapt " is done below. t 0 0.1 , 0.5 .. := f x y , t , ( ) x y 2 ⋅ t + := g x y , t , ( ) t y ⋅ x 2 + := Define initial conditions: x 1 − 1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := <-- Define initial conditions x(0) = x0 = -1 and y(0) = y1 = 1. Define derivatives: Note that the vector function "D" has 2 rows in this case. If the problem had involved 3 first order eqs, there would be 3 rows. < First derivative < Second derivative <--- D t x , ( ) x0 x1 ( )2 ⋅ t + t x1 ⋅ x0 ( )2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ := Evaluate solution at 50 points between 0 and 0.5. The following commands will generate the solution matrix which will contain 3 columns- the "t" values, the corresponding "x" values and the "y " values. Z rkfixed x 0 , 0.5 , 50 , D , ( ) := Y Rkadapt x 0 , 0.5 , 50 , D , ( ) := i 0 rows Z ( ) 1 − .. := xsplinei xspline Z 0 〈 〉 ( )i ⎡ ⎣ ⎤ ⎦ := ysplinei yspline Z 0 〈 〉 ( )i ⎡ ⎣ ⎤ ⎦ := The Mathcad solution generated through the above procedure is compared with the analytical Runge-Kutta results in Figures 9.13 and 9.14. 346 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 357. 0 0.1 0.20.3 0.4 4 3 2 1 rkfixed solution Rkadapt solution R/K 4th order- rkfixed solution Rkadapt solution R/K 4th order- System of first order differential equations Time, t, seconds x- values Z 1 〈 〉 ( )i Y 1 〈 〉 ( )i xsplinei Z 0 〈 〉 ( )i Figure 9.13. Solution to differential equation pair- x (t) 0 0.1 0.2 0.3 0.4 1 1.5 2 2.5 rkfixed solution Rkadapt solution R/K 4th order rkfixed solution Rkadapt solution R/K 4th order System of first-order differential equations Time , t , seconds y- Values Z 2 〈 〉 ( )i Y 2 〈 〉 ( )i ysplinei Z 0 〈 〉 ( )i Figure 9.14. Solution to differential equation pair - y (t) Chapter 9: Numerical Solution of Ordinary Differential Equations 347
- 358. t x(t) y(t)t x(t) y(t) ( obtained with rkfixed ) (obtained with Rkadapt ) Z 0 1 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 -1 1 0.01 -1.010102 1.010151 0.02 -1.020419 1.020611 0.03 -1.030966 1.031388 0.04 -1.041759 1.042491 0.05 -1.052815 1.05393 0.06 -1.064154 1.065717 0.07 -1.075794 1.077862 0.08 -1.087758 1.090377 0.09 -1.100067 1.103275 0.1 -1.112747 1.11657 0.11 -1.125822 1.130277 0.12 -1.139322 1.144412 0.13 -1.153276 1.158991 0.14 -1.167716 1.174032 0.15 -1.182678 1.189556 = Y 0 1 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 -1 1 0.01 -1.010102 1.010151 0.02 -1.020419 1.020611 0.03 -1.030966 1.031388 0.04 -1.041759 1.042491 0.05 -1.052815 1.05393 0.06 -1.064154 1.065717 0.07 -1.075794 1.077862 0.08 -1.087758 1.090377 0.09 -1.100067 1.103275 0.1 -1.112747 1.11657 0.11 -1.125822 1.130277 0.12 -1.139322 1.144412 0.13 -1.153276 1.158991 0.14 -1.167716 1.174032 0.15 -1.182678 1.189556 = Note that the first column of the above matrices is the time " t " column, the second is the solution "x" while the third is the solution "y". Using " rkadapt " , the solution can be found at any point in time as follows. This command generates the solution to the differential equation at the endpoint t = 0.5 secs with acc = 0.001, kmax = 50 ( maximum number of intermediate points at which the solution will be approximated) and save = 0.00001 which is the smallest spacing between the t- values at which the solutions are to be computed. Value rkadapt x 0 , 0.5 , 0.001 , D , 50 , 0.00001 , ( ) := <-- 348 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 359. t x y Note:The first column contains the t-values, the second column contains the x-values, and the third contains the y- values Value 0 5 10 3 − × 0.03 0.155 0.374387 0.5 1 − 1.005025 − 1.030966 − 1.190367 − 1.800212 − 3.026613 − 1 1.005038 1.031388 1.197505 1.741152 2.524711 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <--- It is noted that the Runge-Kutta- Version 1 solution agrees reasonably well with the Mathcad solutions. 9.7 SOLUTION OF HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS In many areas of engineering and physics, the solution of higher-order differential equations is often required. Examples can be encountered in vibration engineering, control systems engineering , electrical circuit analysis etc. The general form of a second-order differential equation is d2y/dx2 = f ( x, y, dy/dx ) (9.27) A more special form often encountered in vibrations, control systems, and electrical circuit analysis is a d2y/dx2 + b dy/dx + c y = q(x) (9.28) with initial conditions y(0) and dy/dx(0) specified. Although any of the methods discussed in sections 9.2-9.5 can certainly be used to obtain solutions to Equation 9.28, it must first be converted to a system of first-order differential equations as follows. dy/dx = z = f ( x,y, z) dz/dx = [q(x) - b z - c y] / a = g (x,y, z ) (9.29) Chapter 9: Numerical Solution of Ordinary Differential Equations 349
- 360. with the initialconditions now becoming y(0) and z(0) . Thus, if a solution to the second-order differential equation, which is now represented by the two first-order equations of Equations 9.29 , is to be generated by Version 1 of the fourth-order Runge-Kutta method, the calculations shown in the following procedure must be resorted to. k1 = h f (y0, z0, x0) ; m1= h g (y0,z0, x0) ; xmp = x0 + 0.5 h ; ymp= y0+ 0.5 k1; zmp= z0+0.5 m1 k2= h f (ymp, zmp, xmp) ; m2 = h g ( ymp, zmp, xmp ) ymp* = y0 + 0.5 k2 ; zmp* = z0+ 0.5 m2 k3 = h f ( ymp*, zmp*, xmp ) ; m3 = h g ( ymp*, zmp*, xmp ) y1 = y0 + k3 ; z1 = z0 + m3 ; x1 = x0 + h k4 = h f ( y1, z1, x1 ) ; m4 = h g ( y1, z1, x1 ) y1= y0 + (1/6) [ k1 +2 k2 + 2 k3 + k4 ] ; z1 = z0 + (1/6 ) [ m1 + 2 m2 + 2 m3 + m4 ] (9.30) where h is the increment or step size. Note that in order to generate the solution (y1, z1), we started with known values ( x0, y0, z0 ). Similarly, to obtain ( y2, z2) , the above procedure must be repeated with (x1,y1,z1) replacing ( x0, y0,z0) and (x2, y2,z2) replacing (x1,y1,z1). Example 9.8 Using the Runge-Kutta fourth-order method, obtain a solution to m d2y/dt2 +c dy/dt + k y = F(t) ; m= 1kg, c= 3 N-sec/m, k = 9 N/m , F(t)= 9 N. for t = 0 to t= 3 seconds. The initial conditions are given as: y(0)=0; dy/dt (0) = 0 . Use a time increment of 0.5 seconds. 350 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 361. The above differentialequation represents the equation of motion of a linear spring-mass-damper system subjected to a step forcing function as shown in the above figure. As mentioned earlier , application of the fourth order Runge- Kutta Method to the solution of a second-order differential equation is quite similar to that of a first order differential equation , except that the following two first-order equations must now be considered. dy/dt = z ; and m dz/dt + c z +k y = F(t) Note that, now, both y and z are functions of the independent variable " t". Thus, in this case, dy/dt = f( y,z, t ) = z , and dz/dt = g(y,z,t) =[F(t) - ( k y + c z)] /m Version 1 of the R-K procedure will then resort to the following equations to compute y and z at t1: k1 = h f(y0, z0, t0) ; m1= h g(y0,z0, t0) ; tmp = t0 + 0.5 h ; ymp= y0+ 0.5 k1; zmp= z0+0.5 m1 k2= h f(ymp, zmp, tmp) ; m2 = h g( ymp, zmp, tmp ) ymp* = y0 + 0.5 k2 ; zmp* = z0+ 0.5 m2 k3 = h f ( ymp*, zmp*, tmp ) ; m3 = h g ( ymp*, zmp*, tmp ) y1 = y0 + k3 ; z1 = z0 + m3 ; t1 = t0 + h k4 = h f ( y1, z1, t1 ) ; m4 = h g ( y1, z1, t1 ) y1= y0 + (1/6) [ k1 +2 k2 + 2 k3 + k4 ] ; z1 = z0 + (1/6 ) [ m1 + 2 m2 + 2 m3 + m4 ] Chapter 9: Numerical Solution of Ordinary Differential Equations 351
- 362. F t ( )9 := m 1 := c 3 := k 9 := <--Given data f y z , t , ( ) z := g y z , t , ( ) F t ( ) c z ⋅ − k y ⋅ − m := <--Define functions f(y,z,t) and g(y,z,t) h 0.5 := <--Time increment or step size t0 0 := y0 0 := z0 0 := <-- Given initial conditions Calculations for generating solution at t1: t1 t0 h + := t1 0.5 = k1 h f y0 z0 , t0 , ( ) ⋅ := m1 h g y0 z0 , t0 , ( ) ⋅ := tmp t0 0.5 h ⋅ + := k1 0 = m1 4.5 = ymp y0 0.5 k1 ⋅ + := zmp z0 0.5 m1 ⋅ + := k2 h f ymp zmp , tmp , ( ) ⋅ := m2 h g ymp zmp , tmp , ( ) ⋅ := k2 1.125 = m2 1.125 = ympstar y0 0.5 k2 ⋅ + := zmpstar z0 0.5 m2 ⋅ + := k3 h f ympstar zmpstar , tmp , ( ) ⋅ := m3 h g ympstar zmpstar , tmp , ( ) ⋅ := k3 0.28125 = m3 1.125 = y1 y0 k3 + := z1 z0 m3 + := t1 t0 h + := k4 h f y1 z1 , t1 , ( ) ⋅ := m4 h g y1 z1 , t1 , ( ) ⋅ := k4 0.5625 = m4 1.546875 = y1 y0 1 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 2 k2 ⋅ + 2 k3 ⋅ + k4 + ( ) ⋅ + := y1 0.5625 = 352 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 363. z1 z0 1 6 m1 2m2 ⋅ + 2 m3 ⋅ + m4 + ( ) ⋅ + := z1 1.757813 = Calculations for generating solution at t2: t2 t1 h + := t2 1 = k1 h f y1 z1 , t1 , ( ) ⋅ := m1 h g y1 z1 , t1 , ( ) ⋅ := tmp t1 0.5 h ⋅ + := k1 0.878906 = m1 0.667969 − = ymp y1 0.5 k1 ⋅ + := zmp z1 0.5 m1 ⋅ + := k2 h f ymp zmp , tmp , ( ) ⋅ := m2 h g ymp zmp , tmp , ( ) ⋅ := k2 0.711914 = m2 2.144531 − = ympstar y1 0.5 k2 ⋅ + := zmpstar z1 0.5 m2 ⋅ + := k3 h f ympstar zmpstar , tmp , ( ) ⋅ := m3 h g ympstar zmpstar , tmp , ( ) ⋅ := k3 0.342773 = m3 0.661377 − = y2 y1 k3 + := z2 z1 m3 + := t2 t1 h + := k4 h f y2 z2 , t2 , ( ) ⋅ := m4 h g y2 z2 , t2 , ( ) ⋅ := k4 0.548218 = m4 1.218384 − = y2 y1 1 6 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k1 2 k2 ⋅ + 2 k3 ⋅ + k4 + ( ) ⋅ + := y2 1.151917 = z2 z1 1 6 m1 2 m2 ⋅ + 2 m3 ⋅ + m4 + ( ) ⋅ + := z2 0.508118 = Caculations for t = 1.5,2.0,2.5 and 3.0 can be performed in a similar way , giving the following results. t3 1.5 := y3 1.166 := z3 0.342 − := t4 2 := y4 1.006 := z4 0.24 − := Chapter 9: Numerical Solution of Ordinary Differential Equations 353
- 364. t5 2.5 := y50.938 := z5 2.889 10 3 − ⋅ := t6 3 := y6 0.94 := z6 0.019 − := Plots of the results of the Runge-Kutta process can be obtained by an interpolation procedure involving the generation of the Vt, Vy and Vz vectors and the spline functions yspline(t) and zspline(t) as shown below. Vt t0 t1 t2 t3 t4 t5 t6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vy y0 y1 y2 y3 y4 y5 y6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vz z0 z1 z2 z3 z4 z5 z6 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ := Vt 0 0.5 1 1.5 2 2.5 3 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vy 0 0.5625 1.151917 1.166 1.006 0.938 0.94 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vz 0 1.757813 0.508118 0.342 − 0.24 − 2.889 10 3 − × 0.019 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Vsy cspline Vt Vy , ( ) := Vsz cspline Vt Vz , ( ) := yspline t ( ) interp Vsy Vt , Vy , t , ( ) := zspline t ( ) interp Vsz Vt , Vz , t , ( ) := The Mathcad solution process using " rkfixed " and " Rkadapt " is done below. t 0 0.1 , 3 .. := F t ( ) 9 := m 1 := c 3 := k 9 := f y z , t , ( ) z := g y z , t , ( ) F t ( ) c z ⋅ − k y ⋅ − m := Define initial conditions: 354 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 365. Define initial conditionsy(0) = y0 = 0 and z(0) = y1 = 0. y0 is initial displacement and y 1 is initial velocity y 0 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := <-- Define first and second derivatives: Note that the vector function "D" has 2 rows in this case. If the problem had involved 3 first order eqs, there would be 3 rows. < First derivative < Second derivative D t y , ( ) y1 F t ( ) c y1 ⋅ − k y0 ⋅ − m ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Evaluate solution at 100 points between 0 and 3: The following commands will generate the solution matrix which will contain 3 columns- the "t" values, the corresponding "y" values and the "dy/dt " values. Z rkfixed y 0 , 3.0 , 100 , D , ( ) := Y Rkadapt y 0 , 3.0 , 100 , D , ( ) := i 0 rows Z ( ) 1 − .. := ysplinei yspline Z 0 〈 〉 ( )i ⎡ ⎣ ⎤ ⎦ := zsplinei zspline Z 0 〈 〉 ( )i ⎡ ⎣ ⎤ ⎦ := The displacement and velocity profiles generated are compared in Figures 9.15 and 9.16. _________________________________________________________________________________ N O T E S: Chapter 9: Numerical Solution of Ordinary Differential Equations 355
- 366. 0 0.5 11.5 2 2.5 3 0 0.5 1 1.5 rkfixed solution Rkadapt solution R/K 4th order- rkfixed solution Rkadapt solution R/K 4th order- Sec-order differential equation solution TIME DISPLACEMENT Z 1 〈 〉 ( )i Y 1 〈 〉 ( )i ysplinei Z 0 〈 〉 ( )i Figure 9.15. Solution to second- order differential equation - displacement as a function of time 0 0.5 1 1.5 2 2.5 3 0 1 2 rkfixed solution Rkadapt solution R/K 4th order rkfixed solution Rkadapt solution R/K 4th order Second-order differential equation solution Time values Velocity ( dy/dt) Values Z 2 〈 〉 ( )i Y 2 〈 〉 ( )i zsplinei Z 0 〈 〉 ( )i Figure 9.16. Solution to second-order differential equation- velocity as a function of time 356 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 367. t y(t) ydot(t) (obtained with rkfixed ) Z 0 1 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 0 0.03 -3 3.9285·10 0.257858 0.06 0.015229 0.491527 0.09 0.033181 0.701279 0.12 0.057071 0.887549 0.15 0.086205 1.050917 0.18 0.119904 1.192089 0.21 0.157516 1.311883 0.24 0.198412 1.41121 0.27 0.241994 1.491065 0.3 0.287692 1.552504 0.33 0.334971 1.596638 0.36 0.383329 1.624613 0.39 0.432298 1.637605 0.42 0.481447 1.636801 0.45 0.53038 1.623397 = t y(t) ydot(t) ( obtained with Rkadapt ) Y 0 1 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 0 0.03 -3 3.928548·10 0.257858 0.06 0.01523 0.491526 0.09 0.033181 0.701278 0.12 0.057071 0.887549 0.15 0.086205 1.050917 0.18 0.119904 1.192088 0.21 0.157516 1.311882 0.24 0.198413 1.411209 0.27 0.241994 1.491064 0.3 0.287692 1.552503 0.33 0.334971 1.596637 0.36 0.383329 1.624612 0.39 0.432298 1.637603 0.42 0.481447 1.6368 0.45 0.53038 1.623396 = Note that the first column of the above matrices is the time " t " column, the second is the displacement " y " and the last is velocity " dy/dt " Chapter 9: Numerical Solution of Ordinary Differential Equations 357
- 368. Using " rkadapt" , the solution can be found at any point in time as follows. This command generates the solution to the differential equation at the endpoint t = 0.1 secs with acc = 0.001, kmax = 100 ( maximum number of intermediate points at which the solution will be approximated) and save = 0.00001 which is the smallest spacing between the t- values at which the solutions are to be computed. Value rkadapt y 0 , 0.1 , 0.001 , D , 100 , 0.00001 , ( ) := Time Displacement Velocity First column is the "time" column. Second column is "Displacement" . Third is " Velocity ". Value 0 1.638135 10 5 − × 8.192772 10 5 − × 4.096596 10 4 − × 2.048319 10 3 − × 0.010242 0.051208 0.1 0 1.207549 10 9 − × 3.020221 10 8 − × 7.54885 10 7 − × 1.884157 10 5 − × 4.671742 10 4 − × 0.011197 0.040519 0 1.474285 10 4 − × 7.372589 10 4 − × 3.684671 10 3 − × 0.018378 0.090759 0.42554 0.765952 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <-- It is noted that the Runge-Kutta- Version 1 solution agrees reasonably well with the Mathcad solutions. 9.8 BOUNDARY-VALUE PROBLEMS AND THE SHOOTING METHOD As mentioned in the introduction, a boundary value problem is one in which conditions are prescribed at two different values of the independent variable. An example of a boundary-value problem is 2 x y d d 2 C y ⋅ + F x ( ) = , y 0 ( ) 0 = , y L ( ) 2 = (9.31) Because the two conditions given in Equations 9.31 deal with different values of the independent variable x , this problem cannot be solved by the methods discussed so far. Other methods, therefore, have to be resorted to in order to obtain a solution. One such method is the Shooting Method in which a boundary-value problem is converted to a pseudo- initial-value problem involving a trial and error approach that can make use of any of the methods discussed in this chapter. Since only one initial condition is prescribed at x= 0 , the second initial condition, which is not given, must be assumed at x = 0, in order to start the 358 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 369. solution process. Uponcompletion of the solution process, a check must be made to determine if the condition prescribed at the point x = L is satisfied or not. If not, a second guess must be made and the procedure redone. If the second guess still does not satisfy the prescribed end condition, a third guess has to be estimated. This can be accomplished by noting the difference between the given and the computed conditions at x = L and resorting to a linear interpolation process that utilizes the results of the first two trials. This iteration process must be repeated until the prescribed end condition is , in fact, satisfied. In this method, it is important for the analyst to be skillful at estimating the unprescribed initial condition properly at the starting point in order to maintain efficiency and avoid divergence. Because the method is analogous to aiming properly and shooting at a given object, it has been appropriately labelled as the shooting method. When higher-order differential equations must be solved involving several boundary conditions, this method has proven to be quite uncouth and laborious and, thus, inefficient. Although any of the methods discussed in this chapter can be employed in the solution of a boundary-value problem by the shooting method. the following example will illustrate its application with Mathcad's Rkadapt and rkadapt functions. Example 9.9 Solve: d2y/ dx2 = M/EI = (30 x - 1.5 x^2)/ EI , where EI = 121,500 kips-ft 2 The boundary conditions prescribed are: x = 0, y = 0 and x= 20, y =0 This differential equation represents the relationship between the bending moment M and the deflection "y" at a distance "x" along the span from the left hand end of a 20 foot long simply supported beam loaded with a uniformly distributed load of 3 kips/ ft. as shown in Figure 9.17. The slope and deflection at any point " x " can be obtained by solving the above differential equation. Consider the two equations: dy/dx = z ; and (9.32) dz/dx = M/EI = (30 x - 1.5 x^2 )/ EI Note that, now, both y and z are functions of the dependent variable " x". Thus, in this case, dy/dx = f ( x, y, z ) = z , (9.33) and dz/dx = g( x, y, z ) =[30 x - 1.5 x ^2 ] / EI As said earlier, the shooting method is a trial- and- error method used in the solution of boundary-value problems. Although two boundary conditions are prescribed here, there is only one initial condition ( x= 0, y = 0 ) provided to begin the solution process. The initial Chapter 9: Numerical Solution of Ordinary Differential Equations 359
- 370. slope ( derivative) at x= 0 is not known. Thus, the second initial condition, namely, the value of dy/dx at x = o has to be assumed so that the solution process can be initiated. With these initial values of y and dy/dx , the solution " y" at x= 20 ft must then be computed by any appropriate method such as Modified Euler or Runga- Kutta. If this solution does not agree with the true value of y at x= 20 ft , which, in this case is zero , the initial condition dy/dx at x= 0 used must be revised and a new solution y at x= 20 computed. This process has to be repeated until the solution obtained at x= 20 ft agrees with the actual condition prescribed at that point on the beam. In the following paragraphs, an iteration process utilizing Mathcad's Rkadapt and rkadapt functions is presented by which a solution to this boundary value problem is arrived at : EI 121500 := <-- Given EI value in kips-ft 2 Figure 9.17. A simply supported beam with a uniformly distributed load Define initial conditions: Define initial conditions y(0) = y0 = 0 and z(0) = y1 = -1.000 (assumed). y0 is y at x = 0, and y 1 is slope (dy/dx) at x = 0. <--- y 0 y1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := Define first and second derivatives: (Note that the vector function "D" has 2 rows in this case. If the problem had involved 3 first order equations, there would be 3 rows.) < First derivative < Second derivative D x y , ( ) y1 30 x ⋅ 1.5 x 2 ⋅ − EI ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := 360 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 371. Using " rkadapt" : Because y1 which is the slope at x = 0 is not known, a reasonable starting value has to be assumed. This command generates the solution to the differential equation at the endpoint x= 20 Ft with acc = 0.001, kmax = 1000 ( max number of intermediate points at which the solution will be approximated) and save = 0.00001 which is the smallest spacing between the x- values at which the solutions are to be computed. y1 1.0 − := Value1 rkadapt y 0 , 20 , 0.001 , D , 1000 , 0.00001 , ( ) := <-- First column is the " x" column Second column is "y " Third is " dy/dx " Value1 0 0.2 1.2 6.2 20 0 0.2 − 1.199931 − 6.191712 − 19.835391 − 1 − 0.999995 − 0.999829 − 0.996235 − 0.983539 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <-- The above computation indicates that the deflection at x = 20 Ft generated with the assumed dy/dx value at x= 0 is non-zero. Therefore, try a new value of y1: y1 0.5 − := Value2 rkadapt y 0 , 20 , 0.001 , D , 1000 , 0.00001 , ( ) := Value2 0 0.2 1.2 6.2 20 0 0.1 − 0.599931 − 3.091712 − 9.835391 − 0.5 − 0.499995 − 0.499829 − 0.496235 − 0.483539 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = The above analysis has yielded a deflection of 9.835 ft at the end x= 20 Ft. With the assumed value of y1, the deflection at the end is , clearly, still non-zero. Therefore, a third value of y1 must be tried which can be arrived at by a process of linear interpolation utilizing the results of the first two trials. The formula arrived at that must be used for estimating the next guess is the following. Chapter 9: Numerical Solution of Ordinary Differential Equations 361
- 372. New value ofy1 to be tried = [second value of the first derivative y1 assumed times the first deflection value at the end computed - first value of the first derivative y1 assumed times the second deflection value at the end computed ] / [ first deflection value at the end computed - the second deflection value at the end computed ] . That is, y1 Value20 2 , Value14 1 , ⋅ Value10 2 , Value24 1 , ⋅ − Value14 1 , Value24 1 , − := y1 8.23045 − 10 3 − × = Then, the solution with the new guess is Value3 rkadapt y 0 , 20 , 0.001 , D , 1000 , 0.00001 , ( ) := which yields Note the negligible deflection computed at the end with the current assumed value of the slope (derivative) at x= 0 <---- Value3 0 0.2 1.2 6.2 20 0 1.645763 − 10 3 − × 9.807565 − 10 3 − × 0.042741 − 1.922074 − 10 15 − × 8.230453 − 10 3 − × 8.225547 − 10 3 − × 8.059786 − 10 3 − × 4.465547 − 10 3 − × 8.230453 10 3 − × ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Now , using Rkadapt , the solution can be evaluated at 100 points between 0 and 20 with the above initial slope value put in which led to the computation of the true displacement (y= 0) at the end x= 20 Ft. The following commands will then generate the solution matrix [ Z ] which will contain 3 columns- the "x" values, the corresponding "y" values and the "dy/dx " values. Z Rkadapt y 0 , 20 , 100 , D , ( ) := i 0 rows Z ( ) 1 − .. := The results of the above Mathcad computation are plotted in Figures 9.18 and 9.19. 362 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 373. 0 10 20 0.01 0 0.01 BeamSlope ( dy/dx ) x- Values dy/ dx values Z 2 〈 〉 ( )i Z 0 〈 〉 ( )i 0 10 20 0.06 0.04 0.02 0 Beam Displacement (y) x- Values (Feet) Y- values ( Feet ) Z 1 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.18. Beam displacement along span Figure 9.19. Beam slope along span 9.9 APPLICATIONS IN NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS 9.9.1 Response of an Electric R-L Circuit to a Unit-Step Voltage Input Consider the electric circuit shown in Figure 9.20 which is subjected to a unit step input voltage . Figure 9.20. R-L circuit with a unit-step voltage input Summing voltages around the loop, the governing differential equation for the circuit can be seen to be R I t ( ) ⋅ L t I t ( ) d d ⋅ + u t ( ) = (9.34) where R is the resistance of the circuit, L its inductance and I(t) the current in it at any time t . With no initial energy in the network, the following initial condition is applicable I 0 ( ) 0 = (9.35) Chapter 9: Numerical Solution of Ordinary Differential Equations 363
- 374. The Mathcad statementsthat are needed to generate the solution using Rkadapt are R 1 := L 1 := I0 0 := D t I , ( ) 1 R I ⋅ − L := Z Rkadapt I 0. , 8. , 50 , D , ( ) := i 0 rows Z ( ) 1 − .. := 0 1 2 3 4 5 6 0 0.5 1 Current in a an electric circuit Time, seconds Current, amperes Z 1 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.21. Response of R-L circuit to unit step input voltage The Mathcad solution generated is shown in Figure 9.21. Notice how the current attains its steady-state value in a time t equal to 4 τ, where τ is the system time constant, which, in this case, is L/ R or 1 second. 9.9.2. Deflection Curve of a Cantilevered Beam with a Uniformly Distributed Load Consider a uniform beam fixed at one end at free at the other as shown with a uniformly distributed load of w per unit length. The bending moment at a distance x from the left hand end of the beam is given by M x ( ) w L ⋅ x ⋅ w L 2 2 ⋅ − w x 2 2 ⋅ − = (9.36) Figure 9.22. Cantilevered beam with uniformly distributed load 364 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 375. The displacement atx is then given by 2 x y x ( ) d d 2 M x ( ) EI = (9.37) where E is Young's modulus of elasticity for the material of the beam, and I is the moment of inertia of the beam cross section about its neutral axis. In order to determine the deflection curve of the beam, Equation (9.37) must be integrated and the appropriate boundary conditions applied , which are y 0 ( ) 0 = , x y 0 ( ) d d 0 = For a beam of length L= 165 inches, w = 12 lb/in, EI = 75 x 10 6 lb- in 2 , Equation (9.37) can be solved with Mathcad's Rkadapt function as follows. L 165 := EI 75 10 6 ⋅ := w 12 := y0 0 := y1 0 := Define initial conditions to start the solution process: y(0)= 0 , dy/dx (0) = 0 ( y 0 is the displacement at x=0 and y 1 is the slope at x= 0 ) y y0 y1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := <--- <--Bending moment expression at x M x ( ) w L ⋅ x ⋅ w L 2 2 ⋅ − w x 2 2 ⋅ − := D x y , ( ) y1 M x ( ) EI ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := <--Define first and second derivatives Z Rkadapt y 0 , L , 100 , D , ( ) := <------ Evaluates solution at 100 points between 0 and L with the given initial conditions i 0 rows Z ( ) 1 − .. := The results of the Mathcad computation are shown in Figures 9.23 and 9.24, with the maximum deflection and slope occurring at the free end as anticipated. Chapter 9: Numerical Solution of Ordinary Differential Equations 365
- 376. 0 100 0.2 0.1 0 Slope alongbeam span Distance from left end Slope Z 2 〈 〉 ( )i Z 0 〈 〉 ( )i 0 50 100 150 20 10 0 Deflection along beam span Distance from left end, inches Deflection , inches Z 1 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.23. Beam deflection curve Figure 9.24. Beam slope along span 9.9.3. Temperature Response of a Solid Steel Ball Placed in a Hot Stream Of Air When a body which is initially at a uniform temperature T 0 is suddenly placed in a hot fluid, the rate at which the temperature T(t) of the body changes with time is given by the differential equation [ 17 ] ρ c ⋅ Lc ⋅ t T t ( ) d d ⋅ h Tinf T t ( ) − ( ) ⋅ = (9.38) where ρ is the mass density of the body, c is its specific heat at constant pressure, T inf is the temperature of the surrounding fluid, h is the average convection heat transfer coefficient, and L c is the characteristic length. For a solid steel sphere of diameter 2 cm initially at 10 deg C., and placed in a stream of hot air at temperature, T inf = 75 deg C. , with h =25 W/ m 2 . K, we have d 0.02 := m ρ 7854 := kg/ m 3 c 434 := J/ kg . K h 25 := W/ m 2 . K Tinf 75 := deg. K T0 10 := deg. K V π d 3 6 ⋅ := A π d 2 ⋅ := Lc V A := Lc 3.333333 10 3 − × = m and the Mathcad statements needed to generate the temperature response of the steel ball to the hot stream of air are 366 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 377. T0 10 := Dt T , ( ) h ρ c ⋅ Lc ⋅ Tinf T − ( ) ⋅ := Z Rkadapt T 0. , 2000 , 100 , D , ( ) := i 0 rows Z ( ) 1 − .. := 0 500 1000 1500 2000 0 50 100 Temp response of steel ball Time, seconds Temperature, deg. C. Z 1 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.25. Temperature response of steel ball The Mathcad solution generated, which is plotted in Figure 9.25 shows how the steel ball attains its steady-state temperature of 75 deg C. in a time t equal to 4τ, where τ is the system time constant which, in this case, is ρ c ⋅ Lc h ⋅ = 454.5 seconds. 9.9.4. Nonlinear Vibration of a Simple Pendulum The natural frequency of large oscillation of a simple pendulum, which is a mass attached to one end of a light rigid rod, has been generated in Section 6 of Chapter 8 as a function of the maximum amplitude of the pendulum. The nonlinear behavior of the pendulum is governed by the differential equation [ 6, 9, 16, 20 ] 2 t θ t ( ) d d 2 G L sin θ t ( ) ( ) ⋅ + 0 = (9.39) where L is the length of the pendulum and G is the acceleration due to gravity . Chapter 9: Numerical Solution of Ordinary Differential Equations 367
- 378. For a pendulumof of length L= 12 inches, with a maximum initial amplitude of A= 60 degrees and zero initial velocity, the initial conditions are defined by θ 0 ( ) 0 = t θ 0 ( ) d d 0 = (9.40) Equation (9.39) can then be solved with Mathcad's Rkadapt function as follows. L 12 := in G 386.4 := in/ sec2 A 60 π 180 ⋅ := θ0 A := θ1 0 := Define initial conditions to start the solution process: θ(0)= A , dθ/dt (0) = 0 ( θ 0 is the angular displacement at t=0 and θ 1 is the angular velocity at t= 0 ) <---- θ θ0 θ1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := D t θ , ( ) θ1 G − L sin θ0 ( ) ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := <---- Define first and second derivatives Z Rkadapt θ 0 , 5 , 100 , D , ( ) := <---- Evaluates solution at 100 points between t=0 and t=5 seconds with the given initial conditions Results are generated below with Mathcad for several initial amplitudes of the pendulum ranging from 10 degrees to 120 degrees and these are plotted in Figure 9.26. Amplitude = 10 degrees: A 10 π 180 ⋅ := θ0 A := θ1 0 := θ θ0 θ1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := D1 t θ , ( ) θ1 G − L sin θ0 ( ) ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Z1 Rkadapt θ 0 , 5 , 100 , D1 , ( ) := Amplitude = 50 degrees: A 50 π 180 ⋅ := θ0 A := θ1 0 := θ θ0 θ1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := D2 t θ , ( ) θ1 G − L sin θ0 ( ) ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Z2 Rkadapt θ 0 , 5 , 100 , D2 , ( ) := 368 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 379. Amplitude = 90degrees: A 90 π 180 ⋅ := θ0 A := θ1 0 := θ θ0 θ1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := D3 t θ , ( ) θ1 G − L sin θ0 ( ) ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Z3 Rkadapt θ 0 , 5 , 100 , D3 , ( ) := Amplitude = 120 degrees: A 120 π 180 ⋅ := θ0 A := θ1 0 := θ θ0 θ1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := D4 t θ , ( ) θ1 G − L sin θ0 ( ) ⋅ ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ := Z4 Rkadapt θ 0 , 5 , 100 , D4 , ( ) := i 0 rows Z ( ) 1 − .. := 0 1 2 3 4 5 100 0 100 200 A= 10 degrees A= 50 degrees A= 90 degrees A= 120 degrees A= 10 degrees A= 50 degrees A= 90 degrees A= 120 degrees Nonlinear pendulum response Time, seconds Angular position , degrees Z1 1 〈 〉 ( )i 180 π ⋅ Z2 1 〈 〉 ( )i 180 π ⋅ Z3 1 〈 〉 ( )i 180 π ⋅ Z4 1 〈 〉 ( )i 180 π ⋅ Z 0 〈 〉 ( )i Figure 9.26. Dynamic response of nonlinear pendulum The graphs plotted in Figure 9.26 clearly indicate that there is an increase in the magnitude of the period of vibration of the nonlinear pendulum with increase in the oscillation amplitude . Also, as can be seen from the response plots , because there is no damping present in the system, the free vibration of the pendulum never decays out. Chapter 9: Numerical Solution of Ordinary Differential Equations 369
- 380. 9.9.5. Transient Vibrationof a Spring-Mass-Damper System Excited by a Pulse Function The differential equation of the spring-mass-damper system of Figure 9.27 excited by the pulse function F(t) shown is M 2 t x d d 2 ⋅ c t x d d ⋅ + k x ⋅ + F t ( ) = (9.41) Figure 9.27. Spring-mass-damper system excited by pulse function For a system with m= 3 kg, c =12 N-sec/ m, k= 120 N/m, and initially at rest, the solution to Equation (9.41) can be obtained using Mathcad as follows M 3 := c 12 := k 120 := <--Specify the parameters of the given system xinit 0 := vinit 0 := <-- Specify initial displacement and velocity Express initial conditions as a vector y , the first component being the initial displacement and the second being the initial velocity. <---- y xinit vinit ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := These steps create the forcing function F(t) using linear interpolation and the " if " function. if(cond,tval,fval) returns tval if cond is true but returns fval if cond is false . <----- vt 0 0.5 1.0 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := vf 0 2. 0.5 0 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ := G t ( ) linterp vt vf , t , ( ) := F t ( ) if t 2.0 > 0 , G t ( ) , ( ) := t 0 0.001 , 2 .. := 370 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 381. 0 0.5 11.5 2 0 1 2 Plot of exciting pulse Time, seconds Force, Newtons F t ( ) t Set up the derivative vector function. In this case, it has two rows . If the problem had involved three first-order equations, there would be three rows <-- D t y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − ( ) ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 4.0 := N 500 := <--- Specify the time period in seconds and the number of points beyond the initial point at which the solution is to be approximated . This command will generate the solution at N points between 0 and T seconds using the Rkadapt function. The solution matrix will contain three columns- the "t " values, the corresponding "y" and the "v" values . Z Rkadapt y 0 , T , N , D , ( ) := <---- The results can be plotted as shown below. The first column ( the zeroth one according to Mathcad ) is a vector of time values, the second ( first one according to Mathcad ) is a column of displacements while the last is a column of velocities. i 0 N .. := ti Zi 0 , := xi Zi 1 , := <---- Note that the t's are in the zeroth column of "Z" , while the displacements "x" and the velocities " v" are in the first and second columns respectively vi Zi 2 , := The displacement and velocity time histories plotted in Figures 9.28 and 9.29 indicate that, because of the presence of damping in the system, the response, which is in fact transient in nature, dies out soon after the forcing function is withdrawn at t = 2 seconds. Chapter 9: Numerical Solution of Ordinary Differential Equations 371
- 382. 0 0.5 11.5 2 2.5 3 3.5 4 0 0.01 0.02 Disp Time Hist Obt with Rkadapt Time, seconds Displacement , m xi ti Figure 9.28. Displacement response of spring-mass-damper system 0 0.5 1 1.5 2 2.5 3 3.5 4 0.05 0 0.05 Velocity Time Hist Obt with Rkadapt Time, seconds Velocity , m/ sec vi ti Figure 9.29. Velocity response of spring-mass-damper system 372 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 383. 9.9.6. Nonlinear Vibrationof a Damped System with a Hardening Spring. Consider the following nonlinear equation known as Duffing's equation [ 6, 16, 20, 21 ], which is the differential equation of motion of a spring-mass-damper system with a hardening spring M 2 t x d d 2 ⋅ c t x d d ⋅ + k x ⋅ + μ x 3 ⋅ + F t ( ) = (9.42) where the last term on the right hand side of the equation is due to the effect of the hardening spring in which the spring stiffness increases as the load on it increases. The term F(t) on the right hand side of the equation is the exciting force that the system is subjected to. We will utilize Mathcad to obtain solutions to Equation (9.42) for m = 1.2 kg, c= 0.6 N-sec/m , k= 1.2 N/ m and F(t) = (a) a step function and then (b) a ramp function for values of μ ranging from 0.01 to 1.0. The results will then be compared with those for μ =0 (a) Step input with μ =0, 0.01, 0.1, 1.0: M 1.2 := k 1.2 := c 0.6 := xinit 0 := vinit 0 := <-- Specify initial displacement and velocity Express initial conditions as a vector y , the first component being the initial displacement and the second being the initial velocity. y xinit vinit ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := <---- F t ( ) Φ t ( ) := This creates a unit step function (Φ(t)= unit step or Heaviside function ) <---- 0 0.2 0.4 0.6 0.8 0.998 0.999 1 1.001 Unit step input Time, seconds F(t), Newtons F t ( ) t t 0 0.1 , 1.0 .. := Chapter 9: Numerical Solution of Ordinary Differential Equations 373
- 384. μ = 0.01: Dt y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 0.01 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := <-- Set up the derivative vector function. T 20.0 := N 500 := <--- Specify the time period in seconds and the number of points beyond the initial point at which the solution is to be approximated . Z1 Rkadapt y 0 , T , N , D , ( ) := <--- Generate the solution at N points between 0 and T seconds. The solution matrix will contain three columns- the "t " values, the corresponding "y" and the "v" values as shown below μ = 0.1: D t y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 0.1 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z2 Rkadapt y 0 , T , N , D , ( ) := μ = 1.0: D t y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 1.0 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z3 Rkadapt y 0 , T , N , D , ( ) := 374 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 385. μ = 0.0: Dt y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − ( ) ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z4 Rkadapt y 0 , T , N , D , ( ) := The results can be plotted as shown below. The first column ( the zeroth one according to Mathcad ) is a vector of time values, the second ( first one according to Mathcad ) is a column of displacements while the last is a column of velocities. i 0 N .. := ti Z1i 0 , := xlowmui Z1i 1 , := xmidmui Z2i 1 , := Note that the t's are in the zeroth column of the "Z1 " , "Z2", "Z3" and "Z4 " matrices , while the displacements "x" and the velocities " v" are in the first and second columns respectively xhighmui Z3i 1 , := <----- xzeromui Z4i 1 , := 0 5 10 15 20 0 0.5 1 1.5 mu=0.01 mu=0.1 mu=1.0 mu= 0 mu=0.01 mu=0.1 mu=1.0 mu= 0 Step Response Time, seconds Displacement, m xlowmui xmidmui xhighmui xzeromui ti Figure 9.30. Response of system with hardening spring to unit step input Chapter 9: Numerical Solution of Ordinary Differential Equations 375
- 386. (b) Ramp inputwith μ = 0, 0.01, 0.1, 1.0: M 1.2 := k 1.2 := c 0.6 := μ 1.0 := xinit 0 := vinit 0 := <-- Specify initial displacement and velocity Express initial conditions as a vector y , the first component being the initial displacement and the second being the initial velocity. y xinit vinit ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := <--- F t ( ) Φ t ( ) t ⋅ := <--This creates the unit ramp function t 0 0.1 , 1.0 .. := 0 0.2 0.4 0.6 0.8 0 0.5 1 The unit ramp input Time, seconds F(t), Newtons F t ( ) t mu = 0.01: D t y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 0.01 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z1 Rkadapt y 0 , T , N , D , ( ) := 376 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 387. mu= 0.1: D ty , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 0.1 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z2 Rkadapt y 0 , T , N , D , ( ) := mu= 1.0: D t y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 1.0 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z3 Rkadapt y 0 , T , N , D , ( ) := mu= 0.0: D t y , ( ) y1 1 M F t ( ) c y1 ⋅ − k y0 ⋅ − 0.0 y0 ( )3 ⋅ − ⎡ ⎣ ⎤ ⎦ ⋅ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ := T 20.0 := N 500 := Z4 Rkadapt y 0 , T , N , D , ( ) := The results are generated below and are plotted in Figure 9.31. i 0 N .. := ti Z1i 0 , := xlowmui Z1i 1 , := xmidmui Z2i 1 , := xhighmui Z3i 1 , := xzeromui Z4i 1 , := Chapter 9: Numerical Solution of Ordinary Differential Equations 377
- 388. 0 5 1015 20 0 5 10 15 20 mu= 0.01 mu= 0.1 mu= 1.0 mu = 0 mu= 0.01 mu= 0.1 mu= 1.0 mu = 0 Reponse to ramp input Time, seconds Displacement , m xlowmui xmidmui xhighmui xzeromui ti Figure 9.31. Response of system with hardening spring to a unit ramp input 9.9.7. Temperature Distribution in the Wall of a Pipe Carrying a Hot Fluid The differential equation relating temperature in the wall of a metal pipe to radial distance is [ 17 ] r 2 r T r ( ) d d 2 ⋅ t T r ( ) d d + 0 = (9.43) where T(r) is the temperature in the pipe wall at a radial distance r from the pipe centerline. The inner radius of the pipe is r i and the outer radius is r o . The temperature of the fluid inside the pipe is T f while the temperature of the outer wall of the pipe is T o. Although two boundary conditions are prescribed here, namely the temperatures at the inner and outer walls of the pipe, there is only one initial condition available, that is , T(r i )= T f to begin the solution process. The initial temperature gradient, namely, dT/dr , is not known. Thus, in order to resort to the Shooting Method, this initial temperature gradient must be assumed so that the process of generating a solution may be initiated. The solution T at ro must then be determined with these initial values of T and dT/dr ( at r i ). If the solution obtained does not match up with the given condition at r = r o , the assumed initial condition must be revised and a new solution T(r o ) determined. This process must be repeated until the solution obtained at r o agrees with the prescribed condition . As discussed in Section 9.7, Equation (9.43) must first be converted to the following two first- order differential equations before a solution can be obtained. r T d d 0 = , r z d d 1 − r z ⋅ = (9.44) 378 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 389. In the followingparagraphs, an iteration process utilizing Mathcad's Rkadapt and rkadapt functions is presented by which a solution to this boundary value problem is arrived at for r i = 1 cm, r o = 2 cm, T f = 650 deg C. and T o = 10 deg C. ri 1 := ro 2 := Tf 650 := TO 10 := <-- Given data Define initial conditions: initgrad 100 − := <--- Define initial conditions T 0 = T(ri) = Tf = 650 deg C. and dT/dr (r i ) = z(ri) = initgrad. T 650 initgrad ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := Define first and second derivatives: Note that the vector function "D" has 2 rows in this case. If the problem had involved 3 first order equations, there would be 3 rows. < First derivative < Second derivative D r T , ( ) initgrad 1 − r initgrad ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := The use of the Mathcad function " rkadapt " is demonstrated below. Because initgrad which is the temperature gradient at r = r i is not known, a reasonable starting value has been assumed. The following command generates the solution to the differential equation at the endpoint r= r o with acc = 0.001, kmax = 1000 ( max number of intermediate points at which the solution will be approximated) and save = 0.00001 which is the smallest spacing between the r- values at which the solutions are to be computed. . Value1 rkadapt T ri , ro , 0.001 , D , 1000 , 0.00001 , ( ) := First column is the " r" column Second column is "T " Third is " dT/dr " Value1 1 1.01 1.06 1.31 2 650 649 644 619 550 100 − 99.005 − 94.173 − 72.997 − 30.685 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = <-- Chapter 9: Numerical Solution of Ordinary Differential Equations 379
- 390. The above computationindicates that, with the assumed dT/dr value at r= r i , a temperature of 550 deg. C. is generated at r = ro , which is not the prescribed temperature. Therefore, try a new value of initgrad: initgrad 200 − := D r T , ( ) initgrad 1 − r initgrad ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := T 650 initgrad ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := Value2 rkadapt T ri , ro , 0.001 , D , 1000 , 0.00001 , ( ) := Value2 1 1.01 1.06 1.31 2 650 648 638 588 450 200 − 198.01 − 188.346 − 145.995 − 61.369 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = The above analysis has yielded a temperature of 450 deg. C . at the outer wall of the pipe. With the second assumed value of initgrad, the temperature of the outer pipe wall is , clearly, still not what it should be . Therefore, a third value of initgrad must be tried which can be arrived at by a process of linear interpolation utilizing the results of the first two trials. The formula generated that must be used for estimating the next guess is given below. Tdiff1 firstTouter Touterprescribed − = Tdiff2 secondTouter Touterprescribed − = guess secguess Tdiff1 ⋅ firstguess Tdiff2 ⋅ − firstTouter secondTouter − = That is, initgrad Value20 2 , Value14 1 , TO − ( ) ⋅ Value10 2 , Value24 1 , TO − ( ) ⋅ − Value14 1 , Value24 1 , − := initgrad 640 − = 380 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 391. Then, the solutionwith the new initgrad is initgrad 640 − = T 650 initgrad ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := D r T , ( ) initgrad 1 − r initgrad ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Value3 rkadapt T ri , ro , 0.001 , D , 1000 , 0.00001 , ( ) := which yields Note that the temperature generated at the outer wall now matches the prescribed temperature (10 deg. C. ). <---- Value3 1 1.01 1.06 1.31 2 650 643.6 611.6 451.6 10 640 − 633.632 − 602.708 − 467.183 − 196.382 − ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = Now , using Rkadapt , the solution can be evaluated at 100 points between r= 1 and r = 2 cm with the above initial dT/dr value put in which led to the computation of the true temperature at the outer wall . The following commands will then generate the solution matrix [ Z ] which will contain 3 columns- the "r" values, the corresponding "T" values and the "dT/dr " values. initgrad 640 − = Z Rkadapt T ri , ro , 100 , D , ( ) := D r T , ( ) initgrad 1 − r initgrad ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := i 0 rows Z ( ) 1 − .. := The results of the above Mathcad computation are plotted in Figures 9.32 and 9.33. Chapter 9: Numerical Solution of Ordinary Differential Equations 381
- 392. 1 1.2 1.41.6 1.8 0 200 400 600 Temperature in pipe wall r, cm Temperature, deg. C. Z 1 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.32. Temperature in pipe outer wall 1 1.2 1.4 1.6 1.8 800 600 400 200 0 Temp gradient in pipe wall r, cm dT/dr, (deg.C./ cm) Z 2 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.33. Temperature gradient in pipe outer wall 9.9.8 Response of an R-L Circuit with a Nonlinear Resistor Consider a series R-L circuit having a nonlinear resistor R with the voltage drop across it is given by VR R I t ( ) I t ( ) 3 − ( ) ⋅ = (9.45) 382 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 393. Summing voltages aroundthe loop, the governing differential equation for the circuit is R I t ( ) I t ( ) 3 − ( ) ⋅ L t I t ( ) d d ⋅ + 0 = (9.46) where R is the resistance of the circuit, L its inductance and I(t) the current in it at any time t . If R= 2.25 Ω, L= 1.10 H and the initial current in the circuit is I(0) = 0.75 amperes, the following Mathcad statements will be needed to generate the solution using Rkadapt and compare it with that for the corresponding linear problem. R 2.25 := L 1.10 := I0 0.75 := D1 t I , ( ) 0 R I I 3 − ( ) ⋅ − L := Z Rkadapt I 0. , 3 , 100 , D1 , ( ) := <-- Nonlinear resistor D2 t I , ( ) 0 R I ⋅ − L := Y Rkadapt I 0. , 3 , 100 , D2 , ( ) := <--Linear resistor i 0 rows Z ( ) 1 − .. := 0 0.5 1 1.5 2 2.5 3 0 0.5 Nonlinear resistor Linear resistor Nonlinear resistor Linear resistor Current in R-L circuit Time, seconds Current, amperes Z 1 〈 〉 ( )i Y 1 〈 〉 ( )i Z 0 〈 〉 ( )i Figure 9.34. Response of R-L circuit with a nonlinear resistor The Mathcad solutions generated are shown in Figure 9.34. Notice how the current decays down to a zero value in a time t approximately equal to 4 τ, where τ is the system time constant, which, in this case, is L / R or 0.5 seconds . Chapter 9: Numerical Solution of Ordinary Differential Equations 383
- 394. 9.9.9. The Effectof Damping on the Step Response of a Second-Order Control System The general differential equation governing the behavior of a second-order control system subjected to a unit step input is [ 14 ] 2 t y t ( ) d d 2 2 ζ ⋅ ωn ⋅ t y t ( ) d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ωn 2 y t ( ) ⋅ + ωn 2 = (9.47) where y(t) is the system response to the unit step input, ζ is the system damping ratio, and ω n is the natural frequency of the undamped system. We want to investigate the dynamic response of the system to different levels of damping obtained by varying ζ. If ωn = 5 rad/sec and ζ is varied from 0.1 to1.4, the following Mathcad statements will be needed to generate the rseponses . ωn 5 := y 0 0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ := ζ1 0.1 := D1 t y , ( ) y1 ωn 2 2 ζ1 ⋅ ωn ⋅ y1 ⋅ − ωn 2 y0 ⋅ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Z1 Rkadapt y 0 , 10 , 1000 , D1 , ( ) := ζ2 0.2 := D2 t y , ( ) y1 ωn 2 2 ζ2 ⋅ ωn ⋅ y1 ⋅ − ωn 2 y0 ⋅ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Z2 Rkadapt y 0 , 10 , 1000 , D2 , ( ) := ζ3 0.6 := D3 t y , ( ) y1 ωn 2 2 ζ3 ⋅ ωn ⋅ y1 ⋅ − ωn 2 y0 ⋅ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Z3 Rkadapt y 0 , 10 , 1000 , D3 , ( ) := ζ4 0.8 := D4 t y , ( ) y1 ωn 2 2 ζ4 ⋅ ωn ⋅ y1 ⋅ − ωn 2 y0 ⋅ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Z4 Rkadapt y 0 , 10 , 1000 , D4 , ( ) := 384 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 395. ζ5 1.0 := D5t y , ( ) y1 ωn 2 2 ζ5 ⋅ ωn ⋅ y1 ⋅ − ωn 2 y0 ⋅ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Z5 Rkadapt y 0 , 10 , 1000 , D5 , ( ) := ζ6 1.4 := D6 t y , ( ) y1 ωn 2 2 ζ6 ⋅ ωn ⋅ y1 ⋅ − ωn 2 y0 ⋅ − ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ := Z6 Rkadapt y 0 , 10 , 1000 , D6 , ( ) := i 0 rows Z1 ( ) 1 − .. := 0 2 4 6 8 10 0 0.5 1 1.5 Damp ratio=0.1 Damp ratio= 0.2 Damp ratio= 0.6 Damp ratio=0.8 Damp ratio=1.0 Damp ratio= 1.4 Damp ratio=0.1 Damp ratio= 0.2 Damp ratio= 0.6 Damp ratio=0.8 Damp ratio=1.0 Damp ratio= 1.4 Second-order system response Time, seconds Response Z1 1 〈 〉 ( )i Z2 1 〈 〉 ( )i Z3 1 〈 〉 ( )i Z4 1 〈 〉 ( )i Z5 1 〈 〉 ( )i Z6 1 〈 〉 ( )i Z1 0 〈 〉 ( )i Figure 9.35. Response of second-order system to unit step input with various levels of damping The response of the second-order system to a unit step input , which is plotted in Figure 9.35 for varying levels of damping, indicates that as ζ varies from 0.1 to 1.0, the peak overshoots as well as the time to achieve steady-state diminish, with the transient oscillations dying down faster as the system damping is increased. An overdamped system (ζ>1.0) , however, is the most sluggish in attaining steady-state. Chapter 9: Numerical Solution of Ordinary Differential Equations 385
- 396. PROBLEMS In all problems,also obtain numerical solutions using the Mathcad functions rkfixed and Rkadapt and graph them with proper labels , traces and titles 9.1 Given the differential equation dy/dx = ( 1 - 2 x) e -2x with initial condition: y = 3 at x =0 , estimate values of the function using a Taylor series expansion for x = 0 to x = 0.5 . Take Δx to be 0.10. Construct a Mathcad plot to determine the accuracy of the Taylor series solution by comparing the computed values with the true values computed from the true solution, which is y =3 + x e-2x . Also plot the percentage error as a function of x . 9.2 Given the differential equation dy/dx = e x + x e x+ 2 x with boundary condition y = 0 at x = 0 , estimate values of the function using a Taylor series expansion for x = 0 to x = 1.0 . Take Δx to be 0.20. Construct a Mathcad plot to determine the accuracy of the Taylor series solution by comparing the computed values with the true values computed from the true solution, which is y = x e x + x 2 . Also plot the percentage error as a function of x . 9.3. Given the differential equation dy/dx = x + y +xy , with boundary condition y = 1 at x = 0 , estimate values of the function using a 5-term Taylor series expansion for the range x = 0.1 to x = 0.5 . Take Δx to be 0.10. 9.4. Solve Problem 9.1 using Euler's method. The step size is to be h= 0.1 and the range for the solution is to be x= 0 to x= 1. Compare the Euler solution with the true solution y =3 + x e-2x by constructing a Mathcad plot. Also plot the percentage error as a function of x . 9.5. Using Euler's method, solve dy/dx = y' = f(x,y) = 3 x 2 / 2y The initial condition given is y=3 at x = 2 , and the step size, h is to be 0.1. The range for the solution is to be : x= 2 to x = 3. By constructing a Mathcad plot, compare the Euler solution with the true solution y 1 x 3 + = . Also plot the percentage error as a function of x. 386 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 397. 9.6. Given thefollowing information: dy/dt = -y 2 + 2, y(t=0) = 10 , h = 0.1. (a) Determine y(0.1), using the modified Euler method. The answer should be correct to 3 decimal places. (b) Continue the computation process until t= 0.5 is reached. (c) Use Mathcad to generate a plot of y(t) in the range : t= 0 to 0.5. Compare with results of "rkfixed" and "Rkadapt" . 9.7. Given the differential equation dy/dt = y2 + t2 , with initial condition y = 0 at t=1, determine (a) y(1.1) and y(1.2) by Euler's Method using an interval size h = 0.1. (b) y(1.1) and y(1.2) by Modified Euler's Method, using an interval size h= 0.1 again 9.8. Given the differential equation dy/dt = y2 + t3 , with initial condition y = 1 at t=2, determine (a) y(2.1) and y(2.2) by Euler's Method using an interval size h = 0.1 (b) y(2.1) by Modified Euler's Method, using an interval size h= 0.1 9.9. Given the differential equation dy/dt = y + t2 , with initial condition y = 2 at t=1, determine (a) y(1.1) and y(1.2) by Euler's Method using an interval size h = 0.1 (b) y(1.1) by Modified Euler's Method, using an interval size h= 0.1 9.10. Given the differential equation dy/dx = y + x5 , with initial condition y = 3 at x =2 determine (a) y(2.1) and y(2.2) by Euler's Method using an interval size h = 0.1 (b) y(2.1) by Modified Euler's Method, using an interval size h= 0.1. Obtain answers that are good to at least three decimal places 9.11. Given the differential equation dy/dx = 4x/y - xy3 , with initial condition y = 3 at x =0, determine (a) y(0.1) and y(0.2) by Euler's Method using an interval size h = 0.1 Chapter 9: Numerical Solution of Ordinary Differential Equations 387
- 398. (b) y(0.1) byModified Euler's Method, using an interval size h= 0.1.Obtain answers that are good to at least three decimal places. 9.12. Given the differential equation dy/dx = y + x 2 , with initial condition y = 0.515 at x = 1, determine y at x = 1.2 by Modified Euler's Method, using an interval size of h = 0.2 . 9.13. Given the differential equation dy/dt + y = 1 , with initial condition y = 0 at t =0, determine (a) y(0.10) and y(0.20) by Euler's Method using an interval size h =0.10 (b) y(0.10) by Modified Euler's Method, using an interval size h= 0.10 . You will require about 4- 5 iterations to converge to a solution. Obtain answers that are good to at least three decimal places. 9.14. Given the differential equation dy/dx = y+2x5, with initial condition y = 3.525 at x =2, determine (a) y(2.1) and y(2.2) by Euler's Method using an interval size h = 0.1. (b) y(2.1) by Modified Euler's Method, using an interval size h= 0.1.Obtain answers that are good to at least three decimal places. 9.15. Given the differential equation dy/dx = -xy2 , with initial condition y = 2 at x =1, determine (a) y(1.1) and y(1.2) by Euler's Method using an interval size h = 0.1.(b) y(1.1) by Modified Euler's Method, using an interval size h= 0.1.Obtain answers that are good to at least three decimal places. 9.16. Given the differential equation dy/dx = -xy2 , with initial condition y = 0.3333 at x = 2, determine (a) y(2.2) and y(2.4) by Euler's Method using an interval size h = 0.2. (b) y(2.2) by Modified Euler's Method, using an interval size h= 0.2. Obtain answers that are good to four decimal places. 9.17. Given the initial-value problem 388 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 399. dy/dx = f(x,y)= x 3 - 2 y 2 ; Initial condition : y(x=0) = 1; h = step size =0.3 , obtain the solution at x = 0.3, using Version 1 of the Runge-Kutta method, which is based on Simpson's one-third rule. 9.18.. Given the initial-value problem dy/dx = 4 x 3 - 9 x 2 + 4 x - 1 , y= 5 at x = 1 , h= 0.2 , obtain a solution by the Runge-Kutta fourth-order method for the range x =1 to 2. Compare the solution generated solution generated with the true solution which is y = x 4 - 3 x 3 + 2 x 2 - x + 6. Using Mathcad, plot the results of the Runge-Kutta solution and compare with the results of rkfixed and Rkadapt . 9.19. Using the Runge-Kutta fourth order method (Version2- that employs Simpson's three-eighth rule),.obtain a solution at x= 0.1 and x= 0.2 to the differential equation dy/dx = 4x/y - xy with initial condition y(0)= 3. Use a step size of h = 0.1 9.20. Given dy/dx = ( 3 x 3 + 6 x y 2 - x) / ( 2 y ) , y = 0.707 at x = 0 , h= 0.1 , obtain a solution by the fourth-order Runge-Kutta method for the range x=0 to 0.5. Compare the solution generated with the true solution, which is ln (x 2 + 2 y 2 ) - 3 x 2 . Using Mathcad, plot the results of the Runge-Kutta solution and compare with the results of rkfixed and Rkadapt . 9.21. The governing differential equation of a series R-L circuit with a nonlinear resistor is L dy/dt = - R { y - y 3 } where R and L are the resistance and inductance in the circuit, and y(t) is the current. The prescribed initial condition is: at t = o, y = 0.75 (ampere). Using Version 1 (that employs Simpson's 1/3rd rule) as well as Version 2 ( that employs Simpson's 3/8th rule ) of the Runge-Kutta fourth order method, obtain solutions at t= 0.5 seconds, that is, determine y( 0.5) and dy/dt (0.5). Use a step size of h = 0.5 seconds. Take R = 2.25 Ohms, and L = 1.10 Henries. Obtain answers that are good to at least two decimal places. 9.22. Given the initial value problem Chapter 9: Numerical Solution of Ordinary Differential Equations 389
- 400. c dy/dx +k y = F(x) with c= 0.4 , k= 1, F(x)=1 and initial condition: y(0)=0. Using Version 1 ( that employs Simpson's 1/3 rd rule ) as well as Version 2 ( that employs Simpson's 3/8 th rule ) of Runga-Kutta's fourth order method, obtain solutions at t= 0.5 sec, i.e, find y(0.5) and dy/dx ( 0.5 ). Use a step size of h= 0.5 9.23. Using the Runge-Kutta fourth-order method, obtain a solution to dx/dt = f(x,y,t) = xy3 + t2; dy/dt= g(x,y,t) = ty + x3 for t = 0 to t= 1 second. The initial conditions are given as: x(0)=0; y (0) = 1. Use a time increment of 0.2 second. 9.24. Given the initial value problem d2y/dt2 = (y / (e t + 1) , y(0)=1, and dy/dt (0) =0 , obtain a solution at t= 0.1 , i.e., find y(0.1) and dy/dt (0.1), using Version 1 of Runge-Kutta's fourth order method . Use a step size of 0.1 9.25. Given the initial value problem: d2y/ dt2 + 2 dy/dt + 4 y =0, y(0)=2, and dy/dt (0) =0 obtain a solution at t= 0.1 , i.e., find y(0.1) and dy/dt (0.1), using Version 1 of Runge-Kutta's fourth order method . Use a step size of 0.1 . 9.26. Given the initial value problem m d2y/dt2 +c dy/dt + k y3 = F(t) with m= 1, c= 0.5 , k= 9, and F(t)=9 t and initial condns: y(0)=0; dy/dt (0) = 1 ,obtain a solution at t= 0.5 seconds, i.e, find y(0.5) and dy/dt ( 0.5 ) using Runga-Kutta's fourth order method. Use a step size of h= 0.5 9.27. Given the initial value problem d2 y/ dt2 + 3 dy/dt + 4 y = 3 and initial conditions y(t=0) = 0 , dy/dt(t=0) =1, obtain a solution at t = 0.5 seconds , i.e, determine y ( 0.5 ) and dy/dt ( 0.5 ) using Runge-Kutta's fourth order method. Use a step size of h = 0.5. Obtain answers that are good to at least three decimal places. 390 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 401. 9.28. Given theinitial value problem d2 y/ dt2 + 4 dy/dt + 6 y = 6 and initial conditions y(t=0) = 0 , dy/dt(t=0) =2, obtain a solution at t = 0.5 seconds , i.e, determine y ( 0.5 ) and dy/dt ( 0.5 ) using Runge-Kutta's fourth order method. Use a step size of h = 0.5. Obtain answers that are good to at least three decimal places. 9.29. Given the initial value problem d2 y/ dt2 + 2 dy/dt + 3 y = 3 and initial conditions y(t=0) = 0 , dy/dt(t=0) =1, obtain a solution at t = 0.5 seconds , i.e, determine y ( 0.5 ) and dy/dt ( 0.5 ) using Runge-Kutta's fourth order method. Use a step size of h = 0.5. Obtain answers that are good to at least three decimal places. 9.30. Given the initial value problem d2y/ dt2 + 4 dy/dt +6 y =0, y(0)=0, and dy/dt (0) =2 , obtain a solution at t= 0.1 , i.e., find y(0.1) and dy/dt (0.1), using Version 1 of Runge-Kutta's fourth order method). Use a step size of 0.1 9.31. The differential equation for a series R-C circuit with an applied voltage v(t) is R i t ( ) ⋅ t vC 0 ( ) + 1 C i t ( ) ⋅ ⌠ ⎮ ⎮ ⎮ ⌡ d + v t ( ) = where v C (0) is the initial voltage across the capacitor and i(t) is the resulting current in the circuit. Figure P 9.31 Obtain a numerical solution to this differential equation for R = 1 Ω , C = 1 F, and v(t) = u(t), a unit step function . Assume that there is no initial energy in the network . Chapter 9: Numerical Solution of Ordinary Differential Equations 391
- 402. 9.32. The differentialequation governing the vertical fall of an object in air from a height is M t v t ( ) d d ⋅ C v t ( ) 2 ⋅ + M G ⋅ = where M is the mass of the object, C is a proportionality constant that defines the aerodynamic drag on the object which is proportional to the square of the velocity, v(t) is the vertical velocity ( positive downward ) the object acquires at time t measured from the start of the fall and G is the acceleration due to gravity . Obtain and plot the numerical solution of this differential equation over 30 seconds for M = 100 kg, and C = 0.30 N- sec 2 / m 2 . Assume that the initial vertical velocity of the object at t= 0 is zero . 9.33. In a tank with inflow as well as outflow, the relationship between the liquid level or head h (t) and the rate of inflow q in (t) is given by the following differential equation A t h t ( ) d d ⋅ h t ( ) R + qin t ( ) = where A is the cross sectional area of the tank and R is the hydraulic resistance which is the change in head required to cause a unit change in the rate of outflow . Figure P 9.33 If the hydraulic resistance is given by R C h t ( ) ⋅ = ,where C is a constant, obtain a numerical solution to the differential equation for A = 0.04 m 2 , q in = 5 x 10 - 4 m 3 / sec and C = 3500 . The initial level in the tank is h = 0 at t= 0. Plot h versus time and determine how long it will take for the level in the tank to rise to 2.5 m . Also determine the steady state value of h and the time taken to achieve it. 9.34. The large-angle deflection of a cantilevered beam of length L loaded with a vertical 392 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 403. load P atthe free end is given by 2 x y x ( ) d d 2 P − L x − ( ) ⋅ E I ⋅ 1 x y x ( ) d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 3 2 ⋅ = where y(x) is the beam deflection at a distance x from the left hand end which is the cantilevered end, L is its length, I is the moment of inertia of the cross- section about the neutral axis , and E is the modulus of elasticity. Using Mathcad, obtain numerical solutions to this differential equation and generate deflection curves for a beam with a load P = 10 kips, length L = 15 feet , and moment of inertia I = 200 in 4 in the following cases. Figure P 9.34 a) The beam is made of steel, with modulus of elasticity E = 30 x 10 6 psi. (b) The beam is made of wood, with modulus of elasticity E = 1.8 x 10 6 psi. Compare your results in the two cases with those for small-angle bending (linear beam theory ) generated by ignoring the effect of the nonlinearity in the right hand side of the equation. 9.35. The differential equation governing the behavior of a control system is 2 t y t ( ) d d 2 6 t y t ( ) d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ + 9 y t ( ) ⋅ + u t ( ) = Using Mathcad, obtain and plot the numerical solution to the differential equation for zero initial conditions. What is the steady-state value of the response and how long does it take to achieve it ? 9.36. A spring-mass-damper system is subjected to a force pulse F(t) which is one half of a sine wave. The governing differential equation is, thus, Chapter 9: Numerical Solution of Ordinary Differential Equations 393
- 404. M 2 t x t ( ) d d 2 ⋅C t x t ( ) d d ⋅ + K x t ( ) ⋅ + F t ( ) = where M is the mass , C is the damping coefficient, K is the spring constant and x(t) is the displacement of the mass at time t. The forcing function F(t) is F t ( ) F0 sin ωf t ⋅ ( ) ⋅ = 0 t ≤ 1 ≤ F t ( ) 0 := t 1 ≥ 0 0.2 0.4 0.6 0.8 10 0 10 Time, sec Force F(t) 10 sin π t ⋅ ( ) ⋅ t Figure P 9.36 Using Mathcad, obtain and plot the numerical solution ( displacement and velocity ) to the above differential equation for M= 1 kg, C = 1.25 N-sec / m, K = 10 N / m , F0 = 10 N, ω f = π, and zero initial conditions. How long does it take for the transient vibration to die out? 9.37. The system of Problem 9.36 is now excited by the trapezoidal pulse F(t) shown below. Using Mathcad, obtain and plot the numerical solution (displacement and velocity ). How long does it take for the transients to die out ? 394 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 405. Figure P 9.37 9.38.The Van der Pol oscillator is a spring-mass system with a damping mechanism that depends on position as well as velocity . The motion of this system is a sustained oscillation that is independent of initial conditions and is governed by the following differential equation 2 t x d d 2 μ t x d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 x 2 − ( ) ⋅ − x + F t ( ) = Using Mathcad, obtain numerical solutions to this equation for μ = 0.01, 0.1 , 1 , when F(t) is a unit step function, and the initial conditions are zero. 9.39. Verify, using Mathcad, that the motion of the Van der Pol oscillator of Problem 9.38 subjected to a unit step forcing function is a sustained oscillation that is independent of initial conditions. Consider four sets of initial conditions as follows : (1) t= 0, x= 0 , dx/dt =0 (2) t = 0 , x=0 , dx/dt = 10 (3) t=0 , x=2, dx/dt = 0 (4) t=0 , x=2, dx/dt =10. Obtain numerical solutions for μ = 1 only. 9.40. The differential equation for a series R-L-C circuit with an applied voltage v(t) is R i t ( ) ⋅ L t i t ( ) d d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ + t vC 0 ( ) + 1 C i t ( ) ⋅ ⌠ ⎮ ⎮ ⎮ ⌡ d + v t ( ) = where v C (0) is the initial voltage across the capacitor and i(t) is the resulting current in the circuit. Chapter 9: Numerical Solution of Ordinary Differential Equations 395
- 406. Figure P 9.40 UsingMathcad, obtain a numerical solution to this differential equation for R = 1 Ω , L = 1 H , C = 0.04 F, and v(t) = u(t), a unit step function . Assume that there is initially no current in the circuit. 9.41. A simply supported steel beam of length L is loaded with a distributed load that increases from zero at the left hand end to a maximum value of w 0 at the right hand end. Its deflection curve is given by the following differential equation 2 x y x ( ) d d 2 M x ( ) E I ⋅ = where EI is the bending stiffness of the beam and the bending moment M(x) at x is M x ( ) w0 x ⋅ 6 L ⋅ L 2 x 2 − ( ) ⋅ = Figure P 9.41 Using the Shooting Method, obtain a numerical solution to this boundary value problem in the form of deflection and slope curves for w 0 = 2.5 kips/ ft, L = 15 ft , EI= 2.5x 10 9 lbs- in 2 9.42. A cylindrical metal pipe of inner radius r i and outer radius r o transports cold air at T f 396 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 407. degrees C ina building which has a room temperature of T R degrees C. The differential equation that relates the temperature in the pipe wall to radial distance is known to be r 2 r T r ( ) d d 2 ⋅ r T r ( ) d d + 0 = By employing the Shooting Method, obtain the numerical solution to the differential equation in the form of temperature and temperature gradient profiles in the pipe wall for T f = 2 deg C , T R = 30 deg C , r i = 2 cm and r o = 4 cm. Chapter 9: Numerical Solution of Ordinary Differential Equations 397
- 408. NOTES: 398 AN INTRODUCTIONTO NUMERICAL METHODS USING MATHCAD
- 409. BIBLIOGRAPHY 1. Ayyub, BilalM. and McCuen, Richard H., Numerical Methods for Engineers, Prentice- Hall , Inc., 1996. 2. Bell, Lewis H. and Bell, Douglas H., Industrial Noise Control Fundamentals and Applications, Second Edition, Marcell Dekker, Inc., 1994. 3. Cernica, John J., Strength of Materials, Holt, Rinehart and Winston, Inc., 1966. 4. Chapra, Stephen C. and Canale, Raymond P., Numerical Methods for Engineers, Fourth Edition, McGraw Hill Book Company, 2002. 5. Deutschman, Aaron D., Michels, Walter J. and Wilson, Charles E., Machine Design, Macmillan Publishing Company, Inc., 1975. 6. Dimargonas, Andrew D., Vibration for Engineers, Second Edition, Prentice-Hall, Inc., 1996. 7. Gerald, Curtis F. and Wheatley, Patrick O., Applied Numerical Analysis, Fifth Edition, Addison-Wesley Publishing Company, 1994. 8. Hibbeler, Russel C., Engineering Mechanics, Tenth Edition, Pearson Prentice-Hall, 2004. 9. Inman, Daniel J., Engineering Vibration, Second Edition, Prentice- Hall Inc., 2001. 10. Irwin, J. D., and Graf, E. R. , Industrial Noise and Vibration Control, Prentice- Hall , Inc., 1979 . 11. James, M. L. , Smith, G.M. and Wolford , J. C., Applied Numerical Methods for Digital Computation, Fourth Edition, Harper Collins College Publishers, 1993. 12. Juvinall, Robert C. and Marshek, Kurt M, Fundamentals of Machine Component Design, Third Edition, John Wiley and Sons, 2000. 13. Nakamura, Shoichiro, Numerical Analysis and Graphic Visualization with MATLAB, Prentice-Hall, Inc., 1996 . 14. Nise, Norman S. , Control Systems Engineering, Fourth Edition , John Wiley and Sons, Inc., 2004 15. Norton, Robert L., Machine Design- An Integrated Approach, Third Edition, Pearson Prentice-Hall, 2006 399
- 410. 16. Rao, SingiresuS., Mechanical Vibrations, Fourth Edition, Pearson Prentice Hall, 2004 17. Schmidt, Frank W., Henderson, Robert E. and Wolgemuth, Carl H., Introduction to Thermal Sciences, John Wiley and Sons, Inc., 1984. 18. Shigley, Joseph E., and Mischke, Charles R., Mechanical Engineering Design, Fifth Edition, McGraw-Hill Book Company, 1989. 19. Spotts, M. F., Shoup, T. E. and Hornberger, L. E. , Design of Machine Elements, Eighth Edition, Pearson Prentice- Hall, 2004 20. Thomson, William T. and Dahleh, Marie Dillon, Theory of Vibration with Applications, Fifth Edition, Prentice-Hall, Inc., 1998. 21. Volterra, Enrico, and Zachmanoglou, E. C., Dynamics of Vibrations, Charles E. Merrill Books, Inc., Columbus, Ohio, 1965. 22. Wilson, James F. , Ed., Dynamics of Offshore Structures, John Wiley and Sons, 1984. 400 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 411. INDEX Accuracy, 43 Analytical methods,43 Approximation, 44 Animation, 7 Applications, curve-fitting, 236-248 matrices, 117-128 numerical differentiation, 262-271 numerical integration, 301-305 numerical interpolation , 168-177 numerical solution of ordinary differential Equations , 363- 385 roots of equations, 87-95 Backward differences, 255 Bisection method, 53 advantages and disadvantages , 55 convergence criterion, 54 finding roots, 53 Boundary conditions, differential equations, 311,358,365,378 Boundary value problems, differential equations, 311 shooting method, 358,378 Characteristic equation, 53,91 Cofactor, 103 Column vector, 108,114 Convergence, Bisection method, 54 Iteration method, 78,84 Modified Euler’s method, 324 Newton-Raphson method , 67 Romberg integration, 290 Secant method, 74 Successive substitution method, 78,86 Cramer’s rule, 110 cspline function. 160 Cubic spline, 159-160 Mathcad’s lspline function, 160 Mathcad’s pspline function, 160 Mathcad’s cspline function, 160 in interpolation, 159 Curve-fitting (or regression analysis), 183 applications, 236-248 least squares method, 184 linear regression , 191,195,197 regression with Mathcad, 207 expfit function, 219 genfit function, 213 intercept function , 207 interp function , 209 line fuinction, 208 linear regression, 207 linfit function , 211 lnfit function , 217 loess function , 209 logfit function , 215 nonlinear regression, 209 pwrfit function. 218 regress function , 209 slope function , 207 straight line regression, 185 weighted data, 184 with exponential function, 194 with linear combination of functions, 199 with polynomials, 203 with power function, 191 with quadratic function , 188 Degrees of freedom, 117,121 Dependent variable, differential equations, 311 Determinant, 104 Differential equations, ordinary, ( see numerical solution of ordinary differential equations ), 311 Differentiation, (see Numerical differentiation ), 255 Eigenvalue problem, 114,116, 117 in stress analysis, 125 Index 401
- 412. in vibration engineering,117 solving with Mathcad , 116 eigenvals function, 116 eigenvec function, 116 eigenvecs function , 116 Error, Euler’s method, 318 relative error, 44 Taylor series, 44 Taylor series solution, differential equation, 312 true error, 44 Euler’s method, 317 expfit function , 219 Exponential function,curve-fitting, 194 False position or Regula falsi method, 58 find function , 86 Finite differences, differentiation, 255 interpolation , 149 First-order differential equation, 311 Euler’s method, 317 4th-order Runge-Kutta method, Simpson’s 1/3 rule, 329 Simpson’s 3/8 rule, 330 Mathcad solution, 335, 346, 354 Modified Euler’s method, 323 Runge-Kutta methods, 329 Taylor series method, 312 Forward differences, 255 given function , 85 Gregory-Newton method, 144 Higher order differential equations, 349 Independent variable, differential equations, 311-312 Initial conditions, differential equations, 311 Initial-value problems, 311 Integration, ( see Numerical integration), 277 intercept function, 207 interp function , 160 Interpolating polynomial, 141- 177, 259,279 Interpolation, ( see Numerical interpolation ), 141 Inverse of matrix, 104,108 Inversion, matrix, 104,108 Iteration method , 78,84 Lagrangian interpolation, 155 Least squares method, 184 Linear algebraic equations, 108-114 Linear combination of functions, 199 Linear interpolation, 141 Linear interpolation with Mathcad, 159 Linear spline, 158 line function , 207 linfit function ,211 linterp function , 159 lnfit function, 217 loess function, 209 logfit function , 215 lsolve function , 112 lspline function, 160 Mathcad, advantages, 7 building math expressions, 21 computation , 10 defining functions, 19 defining range variables, 23 editing expressions, 21 entering math, 16 entering text, 16 exact answers , 3 feedback, 4 formatting math, 35 formatting results and graphs, 36 formatting text, 35 functions, 19 graphing of functions, 5 graphics, 5 402 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 413. graphs, 5, 28 livemath, 3 Mathcad screen, 1 matrices, 24, 103 palettes, 2,13 plotting of data, 5, 28 regions, 15 styles, 17 templates, 17 toolbars, 13 tutorial, 7 using units, 39 variables, 18 vectors, 24 window, 13 Mathcad functions, cspline , 160 eigenvals, 116 eigenvec, 116 eigenvecs, 116 expfit, 219 find , 86 genfit, 213 given , 86 intercept, 207 interp, 209 linfit, 211 linterp, 159 lnfit, 217 loess, 209 logfit, 215 lsolve, 112 lspline , 160 polyroots, 71, 72 pspline, 160 pwrfit, 218 regress, 209 rkadapt, 340,348,360 Rkadapt, 335, 346, 354 rkfixed , 335,346, 354 root, 71 slope, 207 Matrices, 103 addition, 103 adjoint, 104 cofactor, 103 Cramer’s rule, 110 determinant, 103 eigenvalues, 114-131 identity matrix, 104,114 inverse, 104 minor, 103 multiplication, 103 nonlinear simultaneous equations, 131 operations, 103 stress analysis application, 125 transpose, 103 unit matrix , 104 use of eigenvals, 116 use of eigenvec, 116 use of eigenvecs, 116 use of lsolve, 112 vibration engineering application, 117-124 Modified Euler’s method, 323 Mode shapes, 118-121,124 Multiple roots, 80 Natural frequencies, 118-119,124 Natural modes, 120, 124 Newton-Raphson method, 65 Advantages, 67 Disadvantages, 67 Newton interpolating polynomial, 151 Newton interpolation, 151 Normal modes, see natural modes, 120,124 Normalization , 120 Numerical methods, 43 Numerical differentiation, 255 applications, 262- 271 calculus definition , 255 finite differences, 255 first derivative, 255, 257 interpolating polynomial method, 259 Mathcad derivative operators, 255,256 second derivative, 257-259 Numerical integration , 277 applications, 301- 305 Index 403
- 414. interpolating polynomial method, 279 Mathcad’sintegral operators, 277, 278 Romberg integration, 288 Simpson’s 1/3 rule, 283 Simpson’s 3/8 rule, 286 trapezoidal rule , 280 Numerical interpolation, 141 applications, 168-177 cubic spline interpolation, 158-160 finite differences, 149 Gregory-Newton method, 144 interpolation with Mathcad, 159 Lagrangian interpolation, 155 linear interpolation, 141 linear spline, 158 Newton’s method , 151 quadratic spline, 158 undetermined coefficients, 141 use of cspline , 160 use of lspsline, 160 use of pspline , 160 use of interp , 160 use of linterp , 159 Numerical solution of ordinary differential equations, 311 applications , 363-385 boundary value problems, 311 Euler’s method, 317 first -order equations, 311,335 higher- order equations, 349 initial value problems, 317 Mathcad solution to first order equation, 335 Mathcad solution to second order equation, 346 modified Euler’s method, 323 Runge-Kutta fourth order method, 329 Simpson’s 1/3 rule, 330 Simpson’s 3/8 rule, 331 Runge-Kutta methods, 329 second- order equations, 349- 363 systems of ordinary differential equations, 349 shooting method, 358,378 Taylor series method, 312 use of rkfixed & Rkadapt, 335,346, 354 use of rkadapt , 340,348,360 One-step method, 317,323 Order, differential equation, 311 Ordinary differential equations ( See numerical solution of ordinary differential equations , 311 Orthogonality, 128 Polynomial, 53,141,144,149,155,203 polyroots function , 72 Power function, curve-fitting, 191 Principal directions, 125 Principal stresses, 125 pspline function , 160 pwrfit function, 218 Quadratic function, curve-fitting, 188 Quadratic spline, interpolation, 158 Regula falsi or false position method, 58 regress function , 209 Regression analysis ( see curve-fitting), 183 Relative error, 44 rkadapt function , 340,348,360 Rkadapt function , 336,346, 354 rkfixed function , 335,346, 354 Romberg integration , 288 root , 71 Root-finding, 53 applications, 87-95 bisection method, 53 False position or Regula falsi method, 58 iteration method, 78,84 multiple roots , 80 404 AN INTRODUCTION TO NUMERICAL METHODS USING MATHCAD
- 415. Newton-Raphson method, 65 Regulafalsi or false position method, 58 secant method, 72 successive substitution, 78 using Mathcad, 71,86 using root, 71 using polyroots , 72 Runge-Kutta methods, 329 first order differential equation, 329 first order method, 329 fourth order method, 329,341, 350 second-order differential equation, 349-363 second order method, 329 Simpson’s 1/3 rule, 330 Simpson’s 3/8 rule, 331 Secant method, 72 Second order differential equation, 311, 349- 363 Shooting method, boundary value problems, 31, 358, 378 Simpson’s one-third rule, differential equation, 329 numerical integration, 283 Simpson’s three-eighth rule, differential equation, 330 numerical integration, 286 slope function , 207 Successive substitution method, 78 Systems of ordinary differential equations, 349 Taylor series, basis of approximation in numerical analysis, 44 differential equation solution, 312 Trapezoidal rule, integration, 280 True error, 44 Undetermined coefficients, curve-fitting, 185-206 interpolation, 141 Weighting factors, curve-fitting, 184 Index 405