Solution of equation in one variable, Lecture 1. Numerical analysis

Date:14/02/2011
Monday
Numerical Analysis
3 credits
course conducted by Mr. Babul Hasan(BH)

Numerical Analysis

Definition: Numerical Analysis involves the study of methods of computing numeric data. It produces a sequences of approximations to many problems. So it is the questions of rate of accuracy.

We seek results in numerical form. To provide efficient numerical methods for obtaining numerical results to the mathematical problems we proceed as follows:

Step 1: We first start with an initial approximation

Step 2: Compute

Step 3: After some iterations we get the desired results. Since the data used are only approximate, being correct to some desired decimal places, the computation result always have errors.

By using it we will solve –

1.    1.   Non Linear equations

2.    2.   Interpolation/ Extrapolation

3.   3.    Numerical Differentiation/ Integration

4.    4.   System  of  linear equations

5.   5.    Differential equations

1.      Non Linear Equation

To find the root of f(x) =0------------------------------------------------------- (1)

If f(x) is a polynomial of degree 2 or 3 then we can solve it easily by exact methods, But f(x) is a polynomial of higher degree or it contains transcendental functions (e.g. 1+cos x, cos x+ tan x, e^x+cos x etc) recourse must be taken to find the root of f(x)=0 by approximation methods.

The methods are :

i.       i) Bisection method

ii.    ii)  Iteration method

iii.    iii) False position method

iv.   iv) Newton-Raphson’s method

v.    v)   Secant method and any others

i) Bisection Method:

If f(x) is differentiable and continuous on (a,b) and has opposite sign at a & b then there must be at least one point c for which f(c)=0

Suppose f(a) is positive & f(b) is negetive. Let the initial approximation be x_o

X_o=(a+b)/2

Then there may be three cases:

Case one:  f(x_o)=0 , Then stop and x_o  is the desired  root. [x_o=c]

Case two:  f(x_o)>0 ,  Then replace ‘a’ by x_o and find the new approximation  x₁=( x_o+b)/2

Case three:  f(x_o)<0,  Then  replace ‘b’ by x_o and find new approximation    x₁=( x_o+a)/2

Then repeat the process until the desired root is obtained .

Problem: Find the real root of  x³-x-1=0 by Bisection method. Correct to 2 decimal places.

Solution:  f(1)= -ve

                 f(2)= +ve

so root lies between 1 and 2.

Let, x_o= (1+2)/2,   then   x_o=1.5

f(x_o)= f(1.5) = (1.5)³-1.5-1

            =0.875

A root lies between 1 and 1.5

Let  x₁ = (1.5+1)/2 = 1.25,  then  x₁=1.25

f(x₁)= f(1.25) =-0.29

 So a root  lies between 1.25 and 1.5

Let,  x₂=(1.25+1.5)/2 =  1.375,  then  x₂=1.375

f(x₂)= f(1.375) = 0.22