Submitted By—Mr. Arun Kumar Singh (MCA 2nd SEM) (Bisection method orBolzano method) Introduction – This method is based on the repeated application of intermediate value property or theorem. If we know that a root of f(x) =0 lies in the interval I0= (a0, b0), we bisect I0 at the point M1=(a0+b0)/2.denote by I1 the interval (a0,m1) if f(a0)*f(m1)<0 or the interval (m1,b0) if f(m1)*f(b0)<0. Therefore the interval I1 also contain the root .we bisect the interval I1 and get subinterval I2 at whose end point f(x) takes the value of f(x) takes the value of opposite sign and therefore contain the root .continuing this procedure ,we obtain a sequence of nested sets of sub interval . Simplification— Let the function be continuous between a and b .For definiteness, let f (a) be (-) and f (b) be (+) .then the first approximation to the root is – X1=1/2(a+b). If f(x1)=0,then x1 is a root of f(x)=0 otherwise ,the root lies between a and x1 or x1 and b according as f(x1) is (+)ve or (-)ve .then we bisect the interval as before and continue the process until the root is found to desired accuracy . In the adjoining figure, f(x1) is (+) ve so that the root lies between a and x1. Then second approximation to the root is x2=1/2(a+x1). If f(x2) is (-)ve; the root is lies between x1 and x2 .then the third approximation to the root is x3=1/2(x1+x2) and so on. Program— Bisection method through the ‘C’ #include<stdio.h> #include<conio.h> #include<math.h> float funx (float x) { float s; s=((x*x*x)-(5*x)+1); return(s); } //main function */ void main() { float x1, x2, x3,eps,max; int i; clrscr(); printf("enter the initial interval\n"); scanf("%f%f",&x1,&x2); printf("the value of possible error epsilon"); scanf("%f",&eps); max=(log(x2-x1)-log(eps))/log(2); for(i=0;i<max;i++) { x3=(x1+x2)/2; if(funx(x1)*funx(x3)<0) x2=x3; else x1=x3; printf("function value is =%f\n",funx(x3)); printf("root is =%f\n",x3); } getch(); } Output of the program – Enter the initial interval 0 1 the value of possible error epsilon.01 function value is =-1.375000 root is =0.500000 function value is =-0.234375 root is =0.250000 function value is =0.376953 root is =0.125000 function value is =0.069092 root is =0.187500 function value is =-0.083282 root is =0.218750 function value is =-0.007244 root is =0.203125 function value is =0.030888 root is =0.195312 Question— Perform five iteration of the bisection method to obtain the smallest positive root of the equation. F(x) = x3-5x+1=0 Solution— Since f(0)>0 and f(1)<0,the smallest positive root lies in the interval (0,1). Takinga0=0 and b0=1, we get m1`=(a0+b0)/2 m1=(0+1)/2 m1=0.5 f(m1)=-1.375 now we check the condition f(a0)*f(m1)<0 thus ,the root lies in the interval (0,0.5).Tkinga1=0,b0=0.5 We get m2=(a1+b1)/2 m2=(0+0.5)/2 m2=0.25 f(m2)=f(0.25)=-0.234375 and f(a1)*f(m2)<0. Thus the root lies in the interval (0, 0.25). the sequence of interval is given in follows— Kak-1bk-1mkf(mk)*f(ak-1) __ 1010.5< 0 200.50.25< 0 300.250.125> 0 40.1250.250.1875> 0 50.18750.250.21875< 0 Hence the root lies in (0.1875, 0.21875).the approximation root is taken as the midpoint of this interval, that is 0.203125.

—Mr. Arun Kumar SinghSubmitted By(MCA 2nd SEM) (

Bisection method orBolzano)methodIntroduction –This method is based on the repeated application of intermediate value property or theorem.

If we know that a root of f(x) =0 lies in the interval I0= (a0, b0), we bisect I0 at the point

M1=(a0+b0)/2.denote by I1 the interval (a0,m1) if

or the interval (m1,b0) iff(a0)*f(m1)<0f(m1)*f(b0)<0.Therefore the interval I1 also contain the root .we bisect the interval I1 and get subinterval I2 at whose end point f(x) takes the value of f(x) takes the value of opposite sign and therefore contain the root .continuing this procedure ,we obtain a sequence of nested sets of sub interval .

Simplification—

Let the function be continuous between a and b .For definiteness, let f (a) be (-) and f (b) be (+) .then the first approximation to the root is –

X1=1/2(a+b).

If f(x1)=0,then x1 is a root of f(x)=0 otherwise ,the root lies between a and x1 or x1 and b according as f(x1) is (+)ve or (-)ve .then we bisect the interval as before and continue the process until the root is found to desired accuracy .

In the adjoining figure, f(x1) is (+) ve so that the root lies between a and x1. Then second approximation to the root is x2=1/2(a+x1).

If f(x2) is (-)ve; the root is lies between x1 and x2 .then the third approximation to the root is x3=1/2(x1+x2) and so on.

Program—Bisection method through the ‘C’

#include<stdio.h>

#include<conio.h>

#include<math.h>

float funx (float x)

{

float s;

s=((x*x*x)-(5*x)+1);

return(s);

}

//main function */

void main()

{

float x1, x2, x3,eps,max;

int i;

clrscr();

printf("enter the initial interval\n");

scanf("%f%f",&x1,&x2);

printf("the value of possible error epsilon");

scanf("%f",&eps);

max=(log(x2-x1)-log(eps))/log(2);

for(i=0;i<max;i++)

{

x3=(x1+x2)/2;

if(funx(x1)*funx(x3)<0)

x2=x3;

else

x1=x3;

printf("function value is =%f\n",funx(x3));

printf("root is =%f\n",x3);

}

getch();

}

Output of the program–Enter the initial interval

0

1

the value of possible error epsilon.01

function value is =-1.375000

root is =0.500000

function value is =-0.234375

root is =0.250000

function value is =0.376953

root is =0.125000

function value is =0.069092

root is =0.187500

function value is =-0.083282

root is =0.218750

function value is =-0.007244

root is =0.203125

function value is =0.030888

root is =0.195312

Question—Perform five iteration of the bisection method to obtain the smallest positive root of the equation.

F(x) = x3-5x+1=0

Solution—Since f(0)>0 and f(1)<0,the smallest positive root lies in the interval (0,1).

Taking a0=0 and b0=1, we get

m1`=(a0+b0)/2

m1=(0+1)/2

m1=0.5

f(m1)=-1.375

now we check the condition f(a0)*f(m1)<0

thus ,the root lies in the interval (0,0.5).Tking a1=0,b0=0.5

We get

m2=(a1+b1)/2

m2=(0+0.5)/2

m2=0.25

f(m2)=f(0.25)=-0.234375 and f(a1)*f(m2)<0.

Thus the root lies in the interval (0, 0.25). the sequence of interval is given in follows—

K ak-1 bk-1 mk f(mk)*f(ak-1)

__

1 0 1 0.5 < 0

2 0 0.5 0.25 < 0

3 0 0.25 0.125 > 0

4 0.125 0.25 0.1875 > 0

5 0.1875 0.25 0.21875 < 0

Hence the root lies in (0.1875, 0.21875).the approximation root is taken as the midpoint of this interval, that is 0.203125.