03 Chapter MATLAB finite precision arithmatic

Chapter 3
F I N I T E P R E C I S I O N A R I T H M E T I C :
3 . 1 N U M E R I C A L V E R S U S S Y M B O L I C C A L C U L A T I O N
3 . 2 F I N D I N G R O O T S O F N O N L I N E A R E Q U A T I O N S - ( 1 ) F I X E D P O I N T , B I S E C T I O N M E T H O D S ( 2 ) N E W T O N M E T H O D S
4/5/2016 DR. MOHAMMED DANISH/ UNIKL-MICET 1

Numerical versus symbolic calculations
For numeric calculations
>>r=0.78;
>>area=pi*r^2
area=
1.7672
For symbolic calculations
>>syms r
>>syms area
Or
>> syms r area

Symbolic Mathematics
In MATLAB we can use symbols to perform mathematical computations as well as
numbers.
These features are contained in the Symbolic Math Toolbox. They are based upon
Maple 8 a software package published by Waterloo Maple, Inc.
We will learn how to define expressions symbolically and then manipulate them
with MATLAB functions.
Symbolic manipulations are often very useful in engineering problem solving and
compliments solutions to problems with numbers.

Example of symbolic calculation
Before jumping into the details, look at some simple examples,
>> a = sym (‘x - 2’); % a and b are sym variables, they
are
>> b = sym (‘2*x + 3’); % strings of characters
>> y = a*b % y is another string of characters
y =
(x-2)*(2*x+3)
>> expand(y)
ans =
2*x^2 - x - 6

Another Example
Consider another example,
>> S = sym(‘(x^2 - 3*x – 10)/(x + 2)’);
>> simplify(S)
ans =
x – 5
Note that the symbolic expression is defined by the sym function, the argument of
the sym function is a string of characters defining the expression symbolically
enclosed in single quotes.

One More Example
Suppose you want to solve the equation D = D0*exp(-Q/RT) for Q. This equation
describes the rate of diffusion.
>>X = sym(‘D=D0*exp(-Q/RT)’);
>>solve (X,’Q’) % Q is in single quotes to tell MATLAB
ans = % that it is a symbol
-log(D/D0)*RT
Note in MATLAB log represents the natural logarithm, normally written as ln in
mathematics, log10 represents the logarithm to the base 10 and log2 represents
the logarithm to the base 2.

Defining Symbolic Expressions and
Variables
There are two methods,
◦ Either create the complex symbolic expression all at once using the sym command as we did in the
examples,
◦ Or, use the syms command to list all the symbolic variables individually and then compose the
expression using algebraic operators like, * or + and -, etc.
>> syms a x y;
>>S = x^2 -2*y^2 + 3*a;
◦ These two MATLAB statements define the variables a, x, and y as symbolic
variables, and then creates the symbolic expression S.
◦ Note that in the first case using the sym function, only S would appear in the
Workspace Window, while in the second case, S, a, x, and y all appear in the
Workspace Window.

Plotting Symbolic Expressions
MATLAB provides an easy way to plot symbolic expressions of a single variable, it is
called ezplot. Suppose S is a symbolic expression of x, then
>>ezplot(S, [xmin, xmax])
will plot S between the limits of xmin and xmax.
>> ezplot(S) always uses the range [-2 *pi, 2*pi]
You can use the commands xlabel, ylabel, and title to add x and y labels, etc., in the
usual way. Also use grid on, hold on and hold off, subplots, etc.

Finding Roots of Equations
In this section we are examining equations with one
independent variable.
These equations may be linear or non-linear
Non-linear equations may be polynomials or generally non-
linear equations
A root of the equation is simply a value of the independent
variable that satisfies the equation

Classification of Equations
Linear: independent variable appears to the first power only, either alone or
multiplied by a constant
Nonlinear:
◦ Polynomial: independent variable appears raised to powers of positive integers
only
◦ General non-linear: all other equations

Finding Roots of Equations
As with our method for solving simultaneous non-linear equations, we often set
the equation to be equal to zero when the equation is satisfied
Example:
If we say that
then when f(y) =0, the equation is satisfied

Bisection methods for solving equation
Consider an initial interval of ylower = -10 to yupper = 10
Since the signs are opposite, we know that the method will converge to a root of the
equation
The value of the function at the midpoint of the interval is:

The method can be better understood by looking at a graph of the function:

Now we eliminate half of the interval, keeping the half where the sign of f(midpoint) is opposite the
sign of f(endpoint)
In this case, since f(ymid) = -6 and f(yupper) = 64, we keep the upper half of the interval, since the
function crosses zero in this interval

New interval: ylower = 0, yupper = 10, ymid = 5
Function values:
Since f(ylower) and f(ymid) have opposite signs, the lower half of the interval is kept

At each step, the difference between the high and low values of y is compared to 2*
(allowable error)
If the difference is greater, than the procedure continues
Suppose we set the allowable error at 0.0005. As long as the width of the interval
is greater than 0.001, we will continue to halve the interval
When the width is less than 0.001, then the midpoint of the range becomes our
answer

First iteration
If signs are same (+ product),
eliminate lower half of interval.
Is interval width narrow
enough to stop?
Evaluate function at
lower and mid values.
Initial
Guesses

Second iteration
New Interval (if statements
based on product at the end
of previous row)
lower and mid values.
Is interval width narrow
enough to stop? If signs are different (- product),
eliminate upper half of interval.

Continue until interval width < 2*error (16 iterations)
Answer:
y = 0.857

MATLAB code for bisection methods
Draw a flow chart of the algorithm used to find a root of an equation using the bisection method
Write the MATLAB code to determine a root of
within the interval x = 0 to 10

Input lower and upper
limits low and high
Define tolerance tol
while high-low > 2*tol
mid = (high+low)/2
lower limit and
midpoint:
fl = f(low), fm = f(mid)
fl*fm > 0?
Keep upper half of
range:
low = mid
Keep lower half
of range:
high = mid
Display root (mid)
YESNO

Defining function as equation
Consider defining the function
as a MATLAB function “fun1”
This will allow our bisection program to be used on other functions without editing the program
– only the MATLAB function needs to be modified

MATLAB code generation
function y = fun1(x)
y = exp(x) - 15*x -10;
Check values at x = 0 and x = 10:
>> fun1(0)
ans =
-9
>> fun1(10)
ans =
2.1866e+004
Different signs, so a root exists within this range

Set tolerance to 0.00001;
answer will be accurate to 5
decimal places

Find root
>> bisect
Enter the lower limit 0
Enter the upper limit 10
Root found: 4.3135

What if no root exist?
Try interval of 0 to 3:
>> bisect
Root found: 3.0000
This value is not a root – we might want to add a check to see if the converged value is a root

Modified code
Add “solution tolerance” (usually looser than convergence tolerance):
Add check at end of program:

Check revised code
>> bisect
No root found

Numerical tools
Of course, Excel and MATLAB have built-in tools for finding roots of equations
However, the examples we have considered illustrate an important concept
about non-linear solutions:
Remember that there may be many roots to a non-linear equation.
Even when specifying an interval to be searched, keep in mind
that there may be multiple solutions (or no solution) within the
interval.

Perform symbolic computations
Derivative of single –variable expressions
Partial derivatives
Second and higher order derivatives
Mixed derivatives

Expression with one variable
Differentiate the symbolic expression, use the diff command. Example is: >>syms x >>f=
sin(x)^2; >>diff(f) ans=2*cos(x)*sinf(x).

c =
0.7000
fc =
-0.0648
c =
0.8000
fc =
0.1033
c =
0.7500
fc =
0.0183
c =
0.7250
fc =
-0.0235
a=.5; b=.9;
u=a-cos(a);
v= b-cos(b);
for i=1:5
c=(a+b)/2
fc=c-cos(c)
if u*fc<0
b=c ; v=fc;
else
a=c; u=fc;
end
end

Newton methods
Newton’s Method (also know as the Newton-Rapshon Method) is another widely-used
algorithm for finding roots of equations
In this method, the slope (derivative) of the function is calculated at the initial guess
value and projected to the x-axis
The corresponding x-value becomes the new guess value
The steps are repeated until the answer is obtained to a specified tolerance

Newton-Raphson methods
Find a root of this equation:
The first derivative is:
Initial guess value: x = 10

First approximation method
For x=10
This is new value of x for next approximation.

Next approximation

Continuous iterations
Methods quickly converses to:

If different initial values were suppose different roots were found

MATLAB code for Newton-Raphson
methods
Draw a flow chart of Newton’s Method
Write the MATLAB code to apply Newton’s Method to the previous example:

Define converge
tolerance tol
while abs(f(x)) > tol
Input initial guess x
Calculate f(x)
Calculate slope fpr(x)
x = x – f(x)/fpr(x)
Calculate f(x)
Output root x

MATLAB code
MATLAB functions defining the function and its derivative:

fzero function in MATLAB
The MATLAB function fzero finds the root of a function, starting with a guess
value.
Example: function fun1 defines this equation:

fzero command in MATLAB
The function fzero has arguments of the name of the function to be evaluated and a guess value:
>> fzero('fun1',10)
ans =
5.6577
Or the name of function to be evaluated and a range of values to be considered:
>> fzero('fun1',[4 10])
ans =
5.6577

Slides End here
Dr. Mohammed Danish
Sr. Lecturer, Malaysian Institute of Chemical and bioengineering
Technology (MICET)-UniKL, Alor Gajah-78000, Melaka, Malaysia

03 Chapter MATLAB finite precision arithmatic

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03 Chapter MATLAB finite precision arithmatic