Numerical Techniques

1. Unit – IV: Numerical Techniques
•Zeroes of transcendental and polynomial equations using
•Bisection method,
•Regula-falsi method
•Newton-Raphson method,
• Rate of convergence.
• Interpolation:
• Finite differences,
•Newton’s forward and backward interpolation,
• Lagrange’s and Newton’s divided difference formula for
unequal intervals.
1

2. A root, r, of function f occurs when f(r) = 0.
For example:
f(x) = x2 – 2x – 3
has two roots at r = -1 and r = 3.
f(-1) = 1 + 2 – 3 = 0
f(3) = 9 – 6 – 3 = 0
We can also look at f in its factored form.
f(x) = x2 – 2x – 3 = (x + 1)(x – 3)
2

3. Bisection Method
Initial two values of x taken: x1=a & x2=b such that if y(a) is
+ve then y(b) is –ve. Then new value c=(a+b)/2
a bc
f(a)>0
f(b)<0
f(c)>0
If c is +ve, then replace the value of a by c. If c is –
ve, then replace the value of b by c.
Then continue to find the next value of c (End of
1st iteration) with c=(a+b)/2
Guaranteed to converge to a root if one exists
within these initial two values.
3

4. Regula Falsi
a bc
 
 
 
( ) ( )
( ) ( )
( ) ( )
( ) 0 ( )
0 ( )
( ) ( )
( )
( ) ( )
f a f b
y x f b x b
a b
f a f b
y c f b c b
a b
a b
f b c b
f a f b
f b a b
c b
f a f b

  


   


  


 

f(c)<0
4

5. Newton-Raphson Method
• Only one current guess of x .
• Consider some point x0.
– If we approximate f(x) as a line about x0, then we
can again solve for the root of the line.
0 0 0( ) ( )( ) ( )l x f x x x f x  
 Solving, leads to the following iteration:
0
1 0
0
1
( ) 0
( )
( )
( )
( )
i
i i
i
l x
f x
x x
f x
f x
x x
f x


 

 
 5

6. Unit – V: Numerical Techniques –II
•Solution of system of linear equations,
•Matrix Decomposition methods,
•Jacobi method,
• Gauss- Seidal method.
• Numerical differentiation,
•Numerical integration,
•Trapezoidal rule,
•Simpson’s one third and three-eight rules,
• Solution of ordinary differential equations (first
order,second order and simultaneous)
•by Euler’s, Picard’s and
• fourth-order Runge- Kutta methods.
6

7. Systems of Non-linear Equations
Consider the set of equations:
 
 
 
1 1 2
2 1 2
1 2
, , , 0
, , , 0
, , , 0
n
n
n n
f x x x
f x x x
f x x x



K
K
M
K
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
The system of equations:
N N
N N
N N NN N N
a T a T a T C
a T a T a T C
a T a T a T C
   
   
   
L
L
M M M M M
L
A total of N algebraic equations for the N nodal points and the system can be
expressed as a matrix formulation: [A][T]=[C]
11 12 1 1 1
21 22 2 2 2
1 2
= , ,
N
N
N N NN N N
a a a T C
a a a T C
where A T C
a a a T C
     
     
      
     
     
     
L
L
M M M M M M
L
7

8. 8
Jacobi Iteration
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa







2211
22222121
11212111















0
0
2
0
1
0
nx
x
x
x

)(
1 0
1
0
2121
11
1
1 nn xaxab
a
x  






  

 

1
1 1
1 1 i
j
n
ij
k
jij
k
jiji
ii
k
i xaxab
a
x
)(
1 0
11
0
22
0
11
1
 nnnnnn
nn
n xaxaxab
a
x 
)(
1 0
2
0
323
0
1212
22
1
2 nn xaxaxab
a
x  

9. 9
Gauss-Siedel Method
33
2321313
1
22
3231212
2
11
3132121
1
a
xaxab
x
a
xaxab
x
a
xaxab
x






3333232131
2323222121
1313212111
bxaxaxa
bxaxaxa
bxaxaxa




Now we can start the solution process by choosing
guesses for the x’s. A simple way to obtain initial
guesses is to assume that they are zero. These zeros
can be substituted into x1 equation to calculate a
new x1=b1/a11.

10. 10
Gauss-Seidel (GS) iteration
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa







2211
22222121
11212111















0
0
2
0
1
0
nx
x
x
x







  

 

1
1 1
11 1 i
j
n
ij
k
jij
k
jiji
ii
k
i xaxab
a
x
)(
1 0
1
0
2121
11
1
1 nn xaxab
a
x  
)(
1 1
11
1
22
1
11
1
 nnnnnn
nn
n xaxaxab
a
x 
)(
1 0
2
0
323
1
1212
22
1
2 nn xaxaxab
a
x  
Use the latest
update

11. 11
Difference between the Gauss-Seidel and the Jacobi iterative methods
The Gauss-Seidel method The Jacobi iteration method

12. Example
Solve the following system of equations using (a) the Jacobi methos, (b) the
Gauss Seidel iteration method.
4 2 11
2 0 3
2 4 16
X Y Z
X Y Z
X Y Z
  
   
  
,
* ,
(a) Jacobi method: use initial guess X0=Y0=Z0=1,
stop when maxXk-Xk-1,Yk-Yk-1,Zk-Zk-1  0.1
First iteration:
X1= (11/4) - (1/2)Y0 - (1/4)Z0 = 2
Y1= (3/2) + (1/2)X0 = 2
Z1= 4 - (1/2) X0 - (1/4)Y0 = 13/4
Reorganize into new form:
X =
11
4
-
1
2
Y -
1
4
Z
Y =
3
2
+
1
2
X + 0 * Z
Z = 4 -
1
2
X -
1
4
Y
4 2 1 11
1 2 0 3
2 1 4 16
X
Y
Z
     
      
     
          
12

13. Second iteration: use the iterated values X1=2, Y1=2, Z1=13/4
X2 = (11/4) - (1/2)Y1 - (1/4)Z1 = 15/16
Y2 = (3/2) + (1/2)X1 = 5/2
Z2 = 4 - (1/2) X1 - (1/4)Y1 = 5/2
Final solution [1.014, 2.02, 2.996] Exact solution [1, 2, 3]
5 4 5 4 5 4
Converging Process:
13 15 5 5 7 63 93 133 31 393
[1,1,1], 2,2, , , , , , , , , ,
4 16 2 2 8 32 32 128 16 128
519 517 767
, , . Stop the iteration when
512 256 256
max , , 0.1X X Y Y Z Z
       
              
 
  
   
13

14. 14
Example (cont.)
(b) Gauss-Seidel iteration: Substitute the iterated values into
the iterative process immediately after they are computed.
0 0 0
1 0 0
1 1
1 1 1
Use initial guess X 1
11 1 1 3 1 1 1
, , 4
4 2 4 2 2 2 4
11 1 1
First iteration: X = ( ) ( ) 2
4 2 4
3 1 3 1 5
(2)
2 2 2 2 2
1 1 1 1 5 19
4 4 (2)
2 4 2 4 2 8
5 19
Converging process: [1,1,1], 2, ,
2 8
Y Z
X Y Z Y X Z X Y
Y Z
Y X
Z X Y
  
       
  
    
 
       
 


29 125 783 1033 4095 24541
, , , , , ,
32 64 256 1024 2048 8192
The iterated solution [1.009, 1.9995, 2.996] and it converges faster
    
         
Immediate substitution

15. 15
Numerical Differentiation: [Find dy/dx or f’(x)]
Forward Difference
Backward Difference
h
hxfhxf
xf
2
)()(
)(

Central Difference
h
hxfhxf
xf
2
)()(
)(


h
xfhxf
xf
)()(
)(


3-points Difference Formulas (More Accurate )
h
xfhxfhxf
xf
2
)(3)(4)2(
)(

Forward Difference :
Backward Difference:
h
xfhxfhxf
xf
2
)()(4)2(3
)(


2-Points Difference Formulas

16. 16
Example:
Find estimates for the derivative with the data
set given to the right. We notice the x values are
evenly spaced.
Two-Point Estimates:
i xi yi
1 3 1
2 5 4
3 7 3
2
1
57
43
)5('
2
3
35
14
)3('









f
f
2
1
57
43
)7('
2
3
35
14
)5('









f
f
Forward Backward
2
1
4
2
37
13
)5(' 


f
Central
Three Point Estimates:
2
5
4
10
37
)1(3)4(43
)3(' 


f
2
3
4
6
37
)1(1)4(4)3(3
)7('






f
Forward Backward
  1
4
4
1)4(23
2
1
)5('' 2


f
Second Derivative (h=2)

17. 17
Simpson’s 1/3rd rule
n
n
n
nn
b
a
n
b
a
xaxaxaaxf
dxxfdxxfI





1
110)(
)()(

Trapezoidal rule
 
2
)()(4)(
3
210
ab
hxfxfxf
h
I





















1
1
0
12110
)(2)()(
2
2
)()(
2
)()(
2
)()(
n
i
in
nn
xfxfxf
n
ab
I
xfxf
h
xfxf
h
xfxf
hI 
 )()(3)(3)(
8
3
3210 xfxfxfxf
h
I Simpson’s 3/8th rule 3
ab
h



18. 1818
Example 21.7
2
)()(
2
)()(
2
)()( 121
2
10
1
nn
n
xfxf
h
xfxf
h
xfxf
hI





 

x f(x) x f(x)
0.0 0.2 0.44 2.842
0.12 1.309 0.54 3.507
0.22 1.305 0.64 3.181
0.32 1.743 0.70 2.363
0.36 2.074 0.80 0.232
0.40 2.456
594.112975.01307.00905.0
2
363.2232.0
10.0
2
181.3363.0
06.0
2
309.1305.1
10.0
2
2.0309.1
12.0










I
Data for
f(x)= 0.2+25x-200x2+675x3-900x4+400x5
Using Trapezoidal Rule

19. 19
Example:   x
xexf  Find the approximate value of  2f 
1.9 12.703199
2.0 14.778112
2.1 17.148957
2.2 19.855030
x  xf
with 10.h
Using the 1st Three-point formula:
   

 032310.22855030.19
148957.174778112.143
2.0
1
)2.2()1.2(4)2(3
1.02
1
2




 ffff
        hxfhxfxf
h
xf 243
2
1
0000 
Using the 2nd Three-point formula:
   
 
228790.22
703199.12148957.17
2.0
1
)9.1()1.2(
1.02
1
2




 fff
The exact value of   22.167168:is2f 
      hxfhxf
h
xf  000
2
1

20. 20
by dividing the interval into 20 subintervals.Find 
π
sin
0
(x)dx
20210
20
20
20
.....,,,,,
π
π





k
k
khax
n
ab
h
n
k
 
   
9958861
20
20
40
2
2
19
1
1
10
.
πsin
π
sinsin
π
)()()sin(
π


























k
n
k
k
k
bfxfaf
h
dxx
 
0000062
20
12
4
20
2
20
60
10
1
9
10
.
)πsin(
π)(
sin
π
sinsin
π
)sin(
π









 















k
k
k
k
dxx
By Simpson’s Rule
By Trapizoidal Rule
Example-2

21. 21
Numerical Solutions for Partial Differential
Equations by 4th order Runga-Kutta Method
hkkkkyy ii )22(
6
1
43211 
),(
)
2
1
,
2
1
(
)
2
1
,
2
1
(
),(
34
23
12
1
hkyhxfk
hkyhxfk
hkyhxfk
yxfk
where
ii
ii
ii
ii




Consider
Exact Solution
Initial condition :
Step size :
2
tx
dt
dx

t2
22 ettx 
  10 x
1.0h       
 
 
    
  104829.122
6
109499.0104988.1,1.01.0,
104988.02/.1,05.01.0
2
1
,
2
1
10475.005.1,05.01.0
2
1
,
2
1
1.0011.01,01.0,
432101
34
223
12
2
1





























kkkk
h
xx
fkxhtfhhk
kfkxhtfhhk
fkxhtfhhk
fxtfhhk