MODULE 5A
NUMERICAL METHODS – 1
Finding root of a Polynomial Equation
f(x)= 
+ 
+
+ ……+ 
x + 
= 0 or a root of a transcendental Equation Which is combination of Polynomial Exponential function Trigonometric function etc. e.g 
= x , 
2 = 0 . 
- x
= 0 are all transcendental Equation.
Bisection method 
This method is based on the repeated application of the intermediate value theorem to obtain an approximation to the root of the Equation f(x) = 0. Suppose f(a) f(b) 
0 then a root lies between a and b . The mean value 
= 
is taken as first approximate value of the root since f(a) and f(b) are of opposite sign . Let us consider the case f(a) is negative and f(b) is positive.
If f(
) = 0 then obviously x = 
is a root of the Equation f(x) = 0 otherwise the root lies between 
And b if f(
) is negative and the root lies between a and 
is positive.
Find real root of the Equation 
– 2x – 5 = 0 correct to three decimal places.
And here f(x )= 
f(1) = 1-2-5 = -6
f(2) = 8-4 -5 = -1
f(3) = 27 – 6 – 5 = 1.6
Therefore at least one root lies between 2 and 3 and we take 
= 
= 2.5
F(2.5) = 
– 2(2.5) – 5 = 2.5 (6.25) – 10 = 15.625 – 10 = 5.625 = +ve
Hence root lies between 2 and 2.5.
The interval is now (2, 2.5) hence 
= 
= 2.25
Now f(2.25) = 
– 2(2.25) – 5
= 1.890625 = +ve
Root lies in 
then 
= 
= 2.125
→ Newton- Raphson method → 
= 
- 
Where n = 0,1 , 2 -----
e.g using Newton Raphson method obtain the real root of the Equation x
+ 
= 0
Ans → Let f(x) = x
+ 
(x) = x
Let us take initial value of the root of f(x) = 0 as 
= 
= 0 as 
= 
f(
) = -1 
(
) = -
By N-R-M
= 
- 
= 
- 
= 
- 
- 
n = 0,1 ,2 , --------
= 3.1416 - 
= 2.8233
Similarly, 
= 2.7986 & 
= 2.7984
Repeating the process, we get 
= 2.798
Regula falsi method 
x = 
Find the approximate value of the root of the Equation 
x – 1 = 0 near x = 1
Using this method of falsi position two times
Ans 
Given f(x) = 
f(1)= 1 + 1 -1 = 1
f(0.5) = (
+ (0.5) – 1 = - 0.375
The root lies between 0.5 and 0.1
Let 
= 0.5 
= 
= 
= 
= 0.6363
Now f(0.6363) = -0.1061
f(1) = 1
Root lies between 0.6363 and 1
= 0.6363 
= 1
= 
= 
= 0.6712
f(0.6712) = - 0.0264 f(1) = 1
= 
= 
= 0.6797
Model question
(1) Solve the following Equation by N-R-M
(2) Solve by bisection method up to three iteration
- 5x + 3 = 0
(3) Solve the equation by Regula falsi method
= 0
Finite difference 
(i) Forward difference
= 
- 
= 
- 
(r = 0, 1 , 2 ……)
(ii) Backward difference
= 
=
(r = 1.2.3 ……)
(iii) Central difference 
_ 
(r = 0,1,2 …..)
= 
= 
= 
= 
Relation between operators 
(1) Shift operator (E) 
E f(x) = f(x+h)
E – 1
E = 
= 1 - 
- 
= 
= 
(2) Average 
(3) Relation between 
= 1 + 
= 
(4) Relation between E and D 
E = 
1 + 
= 
Newton Forward differences 
Y(
+ rh) = 
+ 
+ ….
Newton Backward difference 
Y(x) = y (
) = 
+ 
+ 
+ …….
The population of a city in the decimal census is given in the following table
Year X | 1891 | 1901 | 1911 | 1921 | 1931 |
Population Y (in thousand) | 46 | 66 | 81 | 93 | 101 |
Approximate the population of the city in the years 1895 and 1925
Ans 
Difference Table
X | Y |  |  |  |  |
1891 | 46 | 20 |
|
|
|
1901 | 66 | 15 | -5 | -2 |
|
1911 | 81 | 12 | -3 | -1 | -3 |
1921 | 93 | 8 | -4 |
|
|
1931 | 101 |
|
|
|
|
Compute the population in 1895
Using Newton forward differences formula
Y(x) = 
+ 
+ 
+ …
Let 
= 1891 h = 10 X = 
= 1895
r = 
y(1895) = 46 + 0.4(20)+ 
(-5)+ 
(2) + 
= 54.85 thousand (Approx)
Compute the population in 1925 using backward difference formula
Y(x) = 
+ r
+ ……..
Let 
= 1931 r = 
Y(1925) = 101 + (0.6) 8 + 
+ 
= 96.84 thousand only.
Interpolation with unequal intervals
Newton divided difference formula 
f(
) = 
f(
) = 
(x) = f(
) + (x - 
)f(
) + (x -
)(x - 
)f(
) + ……
e.g using divided difference interpolation
Find f(x)
x | 0 | 1 | 2 | 4 | 5 | 6 |
Y = f(x) | 1 | 14 | 15 | 5 | 6 | 19 |
The divided difference table
x | 4 | d.d of order1 | d.d of order2 | d.d of order3 | d.d of order4 | d.d of order5 |
0 | 1 |
|
|
|
|
|
1 | 14 | 13 | -6 | 1 |
|
|
2 | 15 | 1 | -2 | 1 | 0 |
|
3 | 5 | -5 | 2 | 1 | 0 | 0 |
4 | 6 | 1 | 6 |
|
|
|
5 |
| 13 |
|
|
|
|
6 | 19 |
|
|
|
|
|
Using formula
Y = f(x)= 
+ (x - 
)
+ (x - 
) (x- 
)f
+ (x - 
)(x - 
)(x - 
) f
+…….
= 1 + (x - 0)(13) + (x - 0)(x-1)(-6)+ (x - 0)(x-1)(x -2)(1) + (x -0)(x-1)(x - 2)(x - 4)(0) + (x -0)(x - 1)(x -1) (x - 2)(x - 4)(x -5)0+ 0
= 1 + 13x + (
- x)(-6)+ (
+ 2x)(1) + 0
= 1 + 21x - 9
f(x) = 
Using Lagranye’s interpolation Solve the following data
x | 300 | 304 | 305 | 307 |
 | 2.477 | 2.482 | 2.484 | 2.487 |
Calculate the approximate value of 
Ans. Using formula
y = 
= 
+ 
+ 
= 
(2.477) + 
(2.482) + 
(2.484) + 
(2.4871)
= 
(2.477) +
(2.4829) + 
(2.4843) + 
(2.4871)
= 1.2739 + 4.9658 – 4.4717 + 0.7106
= 2.4786
Newton forward difference formula
Y = 
+ P
+ 
+ 
+ ….
= 
= 
( 
= 
Where h = 
= 
Backward difference 
= 
( 
= 
(1) Find 
from the following table
x | 3 | 5 | 11 | 27 | 34 |
f(x) | -13 | 23 | 899 | 17315 | 35606 |
x | 1 | 1.2 | 1.4 | 1.6 | 1.8 | 2 | 2.2 |
4 | 2.71 | 3.32 | 4.05 | 4.95 | 6.04 | 7.3 | 9.05 |
(2)
Find 
and 
at x = 1.2 , x = 1.6 and x = 2.2
Numerical integration 
f(x) by interoperating Polynomial 
(x) and obtain
Which Will be taken as a approximate value of 
. This is also called quadrature.
Trapezoidal rule 
= h 
where h = 
Simpson’s 
rule 
= 
Simpson’s 
rule 
= 
e.g Solve 
by (i) Trapezoidal rule , (ii) Simpson’s 
rule ,(iii) Simpson’s 
rule
Ans 
Let h = 1 or Let n = 6
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
f(x) =  | 1 | 0.500 | 0.200 | 0.100 | 0.05 | 0.038 | 0.027 |
|  |  |  |  |  |  |  |
By Trapezoidal rule
I = 
= h 
= 1
= 1.41
By Simpson’s rule
I = 
= 
= 1.366
Simpson’s 
rule 
I = 
= 
= 1.357
TEXTBOOKS/REFERENCES:
- P. KANDASAMY, K. THILAGAVATHY, K. GUNAVATHI, NUMERICAL METHODS, S. CHAND COMPANY, 2ND EDITION, REPRINT 2012.
- S.S. SASTRY, INTRODUCTORY METHODS OF NUMERICAL ANALYSIS, PHI, 4TH EDITION, 2005.
- ERWIN KREYSZIG, ADVANCED ENGINEERING MATHEMATICS, 9TH EDITION, JOHN WILEY SONS, 2006.
- B.S. GREWAL, HIGHER ENGINEERING MATHEMATICS, KHANNA PUBLISHERS, 35TH EDITION, 2010.
