Bisection Method

       §       /*Bisection Method*/

#include<stdio.h>

#include<math.h>

//#include<conio.h>

#define f(x) (x*x*x-4*x*x-9*x-12)

void main()

{

       void bisect(float *x,float a,float b,int*itr);

       int N,itr=0;

       float a,b,err,x,x1;

       printf("\n -----BISECTION METHOD FOR SOLVING NONLINEAR EQUATIONS-----\n\n

       equation is to be given after define\n\n\n");

       printf("enter the starting values,max error, maximum iteration\n");

       scanf("%f%f%f%d",&a,&b,&err,&N);

       bisect(&x,a,b,&itr);

       if(f(a)*f(b)>0)

       {

printf("\nThe initial values do not enclose any roots,choose different        values\n");

              getch();

              exit(2);

       }

       while(itr<N)

       {

              if(f(a)*f(x)<0)

                     b=x;

              else

                     a=x;

              printf("x1=%f & x2=%f\n",a,b);

              bisect(&x1,a,b,&itr);

             

              if(fabs(x1-x)<err)

              {

              printf("after %dth iterations,root=%6.9f\n",itr,x1);

              getch();

              exit(1);

        }

              else

                     x=x1;

       }

       printf("\nerror and iteration conditions do no match\n");

       getch();

}

              //function bisection//

              void bisect(float *x,float a,float b,int*itr)

              {

                     *x=(a+b)/2;

                     ++(*itr);

                     printf("\niteration no %4d gives\t x=%7.5f\n",*itr,*x);

              }

ª      Results: (Bisection Method)

 -----BISECTION METHOD FOR SOLVING NONLINEAR EQUATIONS-----

 equation is to be given after define

enter the starting values,max error,maximum iteration

2 9 .005 99

iteration no    1 gives  x=5.50000

x1=5.500000 & x2=9.000000

iteration no    2 gives  x=7.25000

x1=5.500000 & x2=7.250000

iteration no    3 gives  x=6.37500

x1=5.500000 & x2=6.375000

iteration no    4 gives  x=5.93750

x1=5.500000 & x2=5.937500

iteration no    5 gives  x=5.71875

x1=5.718750 & x2=5.937500

iteration no    6 gives  x=5.82813

x1=5.828125 & x2=5.937500

iteration no    7 gives  x=5.88281

x1=5.828125 & x2=5.882813

iteration no    8 gives  x=5.85547

x1=5.855469 & x2=5.882813

iteration no    9 gives  x=5.86914

x1=5.869141 & x2=5.882813

iteration no   10 gives  x=5.87598

x1=5.875977 & x2=5.882813

iteration no   11 gives  x=5.87939

after 11th iterations,root=5.879394531

ª      Results: (Bisection Method alternative)

 -----BISECTION METHOD FOR SOLVING NONLINEAR EQUATIONS-----

 equation is to be given after define

enter the starting values,max error, maximum iteration

2 3 .005 98

The initial values do not enclose any roots, choose different values