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MA4202 Statistics and Numerical Methods Unit III- Solution of Equations and Eigen Value Problems_ Class Notes




DEPARTMENT OF MATHEMATICS
MA4202 / STATISTICS AND NUMERICAL METHODS
UNIT-3 - Solution of Equations and Eigen Value Problems
CLASS NOTES



I - SOLUTION OF ALGEBRAIC EQUATIONS AND TRANSCENDENTAL EQUATIONS :
FIXED POINT ITERATION METHOD
NEWTON-RAPHSON METHOD (TANGENT METHOD)
f  xn 
xn1  xn  n  0,1,...
f '  xn 
The criterion of convergence of Newton-Raphson Method is f  x  f ''  x   f '  x 
2



The order of convergence of Newton-Raphson method is 2

Merits of Newton’s method of iteration:
 Newton’s method is successfully used to improve the results obtained by other methods.
 It is applicable to the solution of equations involving algebraic functions as well as
transcendental functions.

(i) Derive Newton-Raphson formula for root of any positive integer N and hence find 15
Let x  N  x  N  0
2


Let f  x   x2  N  0; f ( x)  2x
 f  xn   xn2  N  0; f ( xn )  2xn
f ( xn )
By Newton-Raphson Method, xn 1  xn 
f ( xn )
 xn 2  N  2 xn 2  xn 2  N
 xn   
 2 xn  2 xn
x 2  N 1  xn 2 N  1  N
 n       xn  
2 xn 2  xn xn  2  xn 
1 N
The Newton-Raphson iterative formula for  xn  
N is
2 xn 
1 15 
Hence, the Newton-Raphson iterative formula for 15 is  xn  
2 xn 
since 9  3 & 16  4 , The Root lies between 3 and 4. Take x0  4

St. Josephs Institute of Technology_Chennai-119

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,MA4202 Statistics and Numerical Methods Unit III- Solution of Equations and Eigen Value Problems_ Class Notes
1 15  1  15 
x1   x0     4    3.8750
2 x0  2  4
1 15  1  15 
x2   x1     3.8750    3.8730
2 x1  2  3.8750 
1 15  1  15 
x3   x2     3.8730    3.8730
2 x2  2  3.8730 
Second and Third iterations are same, Stop the process and hence the root is 3.8730.
𝟑 𝟑 𝟑
(ii) Derive Newton’s iterative formula for √𝑵 and hence find the value of √𝟐𝟒 & √𝟒𝟏
Let x  3 N  x  N  x  N  0
3 3


Let f  x   x3  N  0; f ( x)  3x2
 f  xn   xn3  N  0; f ( xn )  3xn 2
f ( xn )
The Newton-Raphson formula is xn 1  xn 
f ( xn )
 xn3  N  3xn3  xn3  N
 xn   2 
 3xn  3xn 2
2x 3  N 1  2x 3 N  1  N 
 n 2   n2  2    2 xn  2 
3xn 3  xn xn  3  xn 
1 N 
 2 xn  2 
3
Therefore, the Newton-Raphson iterative formula for N is
3 xn 

1 24 
 2 xn  2 
3
Hence, the Newton-Raphson iterative formula for 24 is
3 xn 

Since the nearest approximate value of 3
24 is 3
27  3 , x0  3

1 24 
xn1   2 xn  2 
3 xn 

1 24  1  24 
x1   2 x0  2    2  3  2   2.8889
3 x0  3   3 
1 24  1  24 
x2   2 x1  2    2  2.8889     2.8845
3 x1  3   2.8889  
2




1 24  1  24 
x3   2 x2  2    2  2.88451    2.8845
3 x2  3   2.88451 
2



Here 3rd and 4th iterations are same we stop the iteration, The root is 2.8845




St. Josephs Institute of Technology_Chennai-119

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,MA4202 Statistics and Numerical Methods Unit III- Solution of Equations and Eigen Value Problems_ Class Notes
1 N 
 2 xn  2 
3
Newton-Raphson iterative formula for N is
3 xn 

1 41 
 2 xn  2 
3
Hence, the Newton-Raphson iterative formula for 41 is
3 xn 

The nearest approximate value of 3
41 is 3
27  3 . Let x0  3

1 41 
xn1   2 xn  2 
3 xn 

1 41  1  41 
x1   2 x0  2    2  3  2   3.5182
3 x0  3  3 

1 41  1  41 
x2   2 x1  2    2  3.5182     3.4493
3 x1  3  3.51822 

1 41  1  41 
x3   2 x2  2    2  3.4493    3.4479
3 x2  3  3.44932 

1 41  1  41 
x4   2 x3  2    2  3.4479     3.4479
3 x3  3  3.44792 

Hence the root is 3.4479.

(iii) Derive Newton’s iterative formula to find the reciprocal of a given number N and
1
hence obtain the value of
19
1 1 1
We need to find the reciprocal of a given number N (i.e) x   N  N 0
N x x
1 1 1
Let f  x    N  f  xn    N  f '  xn   2
x xn xn
f  xn 
Newton’s iterative formula is xn1  xn 
f '  xn 
 1 
 x N  x2  1 
 xn   n   xn  n    N 
 1  1  xn 
 x2 
 n 
1   1  Nxn 
 xn  xn2   N   xn  xn2  
 xn   xn 
 xn  xn 1  Nxn   xn  xn  Nxn2
 2xn  Nxn 2  xn  2  Nxn  ------ (1)
1
To find Substitute N=19 in (1)
19

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, MA4202 Statistics and Numerical Methods Unit III- Solution of Equations and Eigen Value Problems_ Class Notes
xn1  xn  2 19xn 
1 1
The nearest approximate value of is  0.05 .  x0  0.05
19 20
x1  x0  2  19 x0   0.05 2  19  0.05  0.0525
x2  x1  2  19 x1   0.0525 2  19  0.0525  0.0526
x3  x2  2  19 x2   0.0526 2  19  0.0526  0.0526
Here 3rd and 4th iterations are same, stop the iteration. The root is 0.0526
𝟏 𝟏
(iv) Newton-Raphson formula for and hence find
√𝑵 √𝟏𝟓
1 1 1
Let x   x2   x2   0
N N N
1
Taking f ( x)  x 2  ; f ( x )  2 xn 2
N
1
 f ( xn )  xn 2  ; f ( xn )  2 xn
N
f ( xn )
By Newton’s formula, xn 1  xn 
f ( xn )
 2 1  1
 xn  N  2 xn  xn  N
2 2

 xn   
 2 xn  2 xn
 
1
xn 2 
N  1  xn  1   1  x  1 
2
    n 
2 xn 2  xn Nxn  2  Nxn 
1 1 1 
Hence, the Newton-Raphson iterative formula for is  xn  
N 2 Nxn 
1 1 1 
The Newton-Raphson iterative formula for is  xn  
15 2 15 xn 
1 1 1
The nearest value of is   0.25. x0  0.25
15 16 4
1 1  1 1 
x1   x0    0.25    0.2583
2 15x0  2  15  0.25 
1 1  1 1 
x2   x1    0.2583    0.2582
2 15x1  2  15  0.2583 
1 1  1 1 
x3   x2    0.2582    0.2582
2 15 x2  2  15  0.2582  
The last two successive iterations are equal, the root is 0.2582.
1. Find the least positive root of x 4  x  10  0 correct to 2 decimal places using Newton
Raphson method.


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