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MULLER'S METHOD

Posted By: Kian Evans     Category: C Programming     Views: 13134

Write a program of MULLER'S METHOD.

Code for MULLER'S METHOD in C Programming

#include<stdio.h>
#include<conio.h>
#include<math.h>
#define ESP 0.001
#define F(x) (x)*(x)*(x) + 2*(x)*(x) + 10*(x) - 20

void main()
{
  double x1,x2,x3,x4_1,x4_2,fx1,fx2,fx3,
     h1,h2,h3_1,h3_2,h4,D,d1,d2,a1,a2,a0;
  int i=1;
  clrscr();
  printf("\nEnter the value of x1: ");
  scanf("%lf",&x1);
  printf("\nEnter the value of x2: ");
  scanf("%lf",&x2);
  printf("\nEnter the value of x3: ");
  scanf("%lf",&x3);
  fx1 = F(x1);
  printf("\n\n f(x1) = %lf",fx1);
  getch();
  fx2 = F(x2);
  printf("\n\n f(x2) = %lf",fx2);
  getch();
  fx3 = a0 = F(x3);
  printf("\n\n f(x3) = %lf",fx3);
  getch();
  h1 = x1-x3;
  h2 = x2-x3;

  d1 = fx1-fx3;
  d2 = fx2-fx3;

  D = h1*h2*(h1-h2);

  a1 = (d2*h1*h1 - d1*h2*h2)/D;
  a2 = (d1*h2 - d2*h1)/D;

  h3_1 = -((2*a0)/(a1 + sqrt(fabs(a1*a1 - (4*a2*a0)))));
  h3_2 = -((2*a0)/(a1 - sqrt(fabs(a1*a1 - (4*a2*a0)))));

  if( (a1 + sqrt(fabs(a1*a1 - (4*a2*a0)))) >
         ((a1 - sqrt(fabs(a1*a1 - (4*a2*a0))))) )
  {
   h4 = h3_1;
  }
  else
  {
   h4 = h3_2;
  }
  x4_1 = x3 + h4;
  printf("\n\n\n x4 = %lf \n",x4_1);

  x1=x2;
  x2=x3;
  x3=x4_1;
  printf("\n\nx1 = %lf",x1);
  printf("\n\nx2 = %lf",x2);
  printf("\n\nx3 = %lf",x3);
  getch();

  do
  {
   fx1 = F(x1);
   fx2 = F(x2);
   fx3 = a0 = F(x3);

   h1 = x1-x3;
   h2 = x2-x3;

   d1 = fx1-fx3;
   d2 = fx2-fx3;

   D = h1*h2*(h1-h2);

   a1 = (d2*h1*h1 - d1*h2*h2)/D;
   a2 = (d1*h2 - d2*h1)/D;

   h3_1 = -((2*a0)/(a1 + sqrt(fabs(a1*a1 - (4*a2*a0)))));
   h3_2 = -((2*a0)/(a1 - sqrt(fabs(a1*a1 - (4*a2*a0)))));

   if( (a1 + sqrt(fabs(a1*a1 - (4*a2*a0)))) >
         (a1 - sqrt(fabs(a1*a1 - (4*a2*a0)))) )
   {
    h4 = h3_1;
   }
   else
   {
    h4 = h3_2;
   }
   x4_2 = x3 + h4;
   printf("\n\n\n x4 = %lf \n",x4_2);
   getch();
   if(fabs(x4_1 - x4_2) < ESP)
   {
    printf("\n\nREAL ROOT = %.3lf",x4_2);
    i=0;
   }
   else
   {
     x4_1=x4_2;
     x1=x2;
     x2=x3;
     x3=x4_1;
     printf("\n\nx1 = %lf",x1);
     printf("\n\nx2 = %lf",x2);
     printf("\n\nx3 = %lf",x3);
   }
  }while(i!=0);
getch();
}

/*
____________________________

OUT PUT
____________________________


Enter the value of x1: 0

Enter the value of x2: 1

Enter the value of x3: 2


f(x1) = -20.000000

f(x2) = -7.000000

f(x3) = 16.000000


x4 = 1.354066


x1 = 1.000000

x2 = 2.000000

x3 = 1.354066


x4 = 1.368647


x1 = 2.000000

x2 = 1.354066

x3 = 1.368647


x4 = 1.368808


REAL ROOT = 1.369


*/
  
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Kian Evans
Kian Evans author of MULLER'S METHOD is from London, United Kingdom.
 
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