C Program code of Bisection method ( by Shrayashi Majumder)
1. Find the root of the equation(x3 - 4x-9=0)
using bisection method.
Code:
#include<stdio.h>
#include<math.h>
double f(double x)
{
return x*x*x-4*x-9;
}
void main()
{
int i=0;
double
mid,a=0.0,b=1.0,x=0.0,s=1.0,s1=1.0;
while(1)
{
s=f(a)*f(b);
s1=f(-a)*f(-b);
if(s<0.0 || s1<0.0)
break;
else
{
a=b;
b++;
}
}
if(s1<0.0)
{
a=-a;
b=-b;
}
printf("Range %f to %f\n
\n",a,b);
printf("\n a b
f(a) f(b) mid");
while(1)
{
mid=(a+b)/2;
printf("\nStep%d: %f %f %f %f %f",i++,a,b,f(a),f(b),mid);
if(f(mid)==0.0)
break;
if(fabs(a-b)<0.0001)
break;
if(f(a)*f(mid)<0.0)
b=mid;
else
a=mid;
}
printf("\nx=%f",mid);
}
Output:
Range 2.000000 to 3.000000
a b
f(a) f(b) mid
Step0: 2.000000 3.000000 -9.000000 6.000000 2.500000
Step1: 2.500000 3.000000 -3.375000 6.000000 2.750000
Step2: 2.500000 2.750000 -3.375000 0.796875 2.625000
Step3: 2.625000 2.750000 -1.412109 0.796875 2.687500
Step4: 2.687500 2.750000 -0.339111 0.796875 2.718750
Step5: 2.687500 2.718750 -0.339111 0.220917 2.703125
Step6: 2.703125 2.718750 -0.061077 0.220917 2.710938
Step7: 2.703125 2.710938 -0.061077 0.079423 2.707031
Step8: 2.703125 2.707031 -0.061077 0.009049 2.705078
Step9: 2.705078 2.707031 -0.026045 0.009049 2.706055
Step10: 2.706055 2.707031 -0.008506 0.009049 2.706543
Step11: 2.706055 2.706543 -0.008506 0.000270 2.706299
Step12: 2.706299 2.706543 -0.004118 0.000270 2.706421
Step13: 2.706421 2.706543 -0.001924 0.000270 2.706482
Step14: 2.706482 2.706543 -0.000827 0.000270 2.706512
x=2.706512
2. Find the root of
equation using bisection method.
Code:
#include<stdio.h>
#include<math.h>
double f(float c3, float c2 ,
float c1 , float cn , double x)
{
double r;
r=c3*x*x*x+c2*x*x+c1*x+cn;
return r;
}
void main()
{
int i=0;
float c3,c2,c1,cn;
double
mid,omid=100.0,a=0.0,b=1.0,s=1.0,s1=1.0,tol;
printf("Enter
coefficients:");
scanf("%f%f%f%f",&c3,&c2,&c1,&cn);
printf("Enter
Tolerance:");
scanf("%lf",&tol);
while(1)
{
s=f(c3,c2,c1,cn,a)*f(c3,c2,c1,cn,b);
s1=f(c3,c2,c1,cn,-a)*f(c3,c2,c1,cn,-b);
if(s<0.0 || s1<0.0)
break;
else
{
a=b;
b++;
}
}
if(s1<0.0)
{
a=-a;
b=-b;
}
printf("Range %f to
%f",a,b);
printf("\n \ta
b f(a) f(b)
mid");
while(1)
{
mid=(a+b)/2;
printf("\nStep %d: %f %f %f %f
%f",i++,a,b,f(c3,c2,c1,cn,a),f(c3,c2,c1,cn,b),mid);
if(f(c3,c2,c1,cn,mid)==0.0)
break;
if(fabs(a-b)<tol)
break;
if(f(c3,c2,c1,cn,a)*f(c3,c2,c1,cn,mid)<0.0)
b=mid;
else
a=mid;
omid=mid;
}
printf("\nx=%f",mid);
}
Output:
Enter coefficients:1 0 -4 -9
Enter Tolerance:0.0001
Range 2.000000 to 3.000000
a
b f(a) f(b)
mid
Step 0: 2.000000 3.000000
-9.000000 6.000000 2.500000
Step 1: 2.500000 3.000000
-3.375000 6.000000 2.750000
Step 2: 2.500000 2.750000
-3.375000 0.796875 2.625000
Step 3: 2.625000 2.750000
-1.412109 0.796875 2.687500
Step 4: 2.687500 2.750000
-0.339111 0.796875 2.718750
Step 5: 2.687500 2.718750
-0.339111 0.220917 2.703125
Step 6: 2.703125 2.718750
-0.061077 0.220917 2.710938
Step 7: 2.703125 2.710938 -0.061077
0.079423 2.707031
Step 8: 2.703125 2.707031
-0.061077 0.009049 2.705078
Step 9: 2.705078 2.707031
-0.026045 0.009049 2.706055
Step 10: 2.706055 2.707031
-0.008506 0.009049 2.706543
Step 11: 2.706055 2.706543
-0.008506 0.000270 2.706299
Step 12: 2.706299 2.706543
-0.004118 0.000270 2.706421
Step 13: 2.706421 2.706543
-0.001924 0.000270 2.706482
Step 14: 2.706482 2.706543
-0.000827 0.000270 2.706512
x=2.706512
