Solution of equation in one variable, Lecture 2, Numerical analysis

Date: 28.02.2011
Problem: By using bisecting method to find a real root of  f(x)= e^x-3x²correct two decimal points.


FIXED POINT  ITERATION  METHOD/ ITERATION METHOD

This method is used to find the numerical solution to  polynomial  and other  equations such as

                                    f(x)=0......................................(1)

we start an initial approximation x_o to the solution of   f(x)=0 and apply it to a procedure which gives a new approximation is normally a better one. Use this approximation to find a new one. This procedure is known as iteration method.

METHOD IN DETAILS:
Rewrite the given equation   f(x)=0 in the form

                              x=Φ(x).....................(2)

 let x_{o be the initial approximation to the desired root. Substitute  it on the right hand side of  (1).
we get the next approximation}
x₁=Φ(x_o)
x₂=Φ(x₁)
x₃=Φ(x₂)
x₄=Φ(x₃)
.
.
.
x_n=Φ(x_n-1)    


_{stop when} │x_n-x_n-1│< tolerance

Condition : This iteration method or fixed point iteration method is applicable when

│ϕ’(x)│< 1

Example: Use iteration method find a real root of  f(x)= x³+x²-1 correct to 3 decimal places.

Solution:  Let us have the eqution f(x)= x³+x²-1.    if   f(x)= 0   then      
                                                                    x³+x²-1 =0
                                                                   
We get if x=0 then f(x)= -ve and  x=1 then f(x)= +ve  then by bisection method, there must be solution of  f(x) in between 0 and 1 that is on the interval (0,1).
Let the solution x_o=0.75 ( changeable, u can take different value between 0 and 1 )

x₁=Φ(x_o) =Φ(0.75)=0.7559

x₂=Φ(x₁)=Φ(0.7559)=0.7546

x₃=Φ(x₂)= Φ(0.7546)=0.7549

...........................................................
..........................................................


we get │x₃-x₂│=0.7549 - 0.7546
                              =0.0003

the result 0.0003 says that 0.7549 or 0.755 is the desired result.

Therefore the desired root is 0.755

Example 2: Use iteration method find a real root of  f(x)= 2x-cos x -3  correct to 3 decimal places.

Let us have the equation  f(x)= 2x-cos x -3=0
                                            
                                                   

To find the solution by iteration method we need to show that the given function satisfy the condition of iteration method. That is  



 │ϕ’(x)= -(sin x)/2│<1 . For any value of x this condition is satisfied.


Now let the solution be x_o= 1.5 , Because for the value of x in radian unit the value of  f(x) becomes negative when x=1 and positive when x=2.

x₁=Φ(x_o) =Φ(1.5)=1.535368601

x₂=Φ(x₁)=Φ(1.535368601)=1.517710158

x₃=Φ(x₂)= Φ(1.517710158)=1.526530619

...........................................................

we get │x₃-x₂│=1.526530619 - 1.517710158
                              =0.008820461
the result 0.008820461 says that 1.526530619 or simply 1.527 is the desired result.

Therefore the desired root is1.527

(To be continued)