Iterative method for solving non-linear equations

 Source:

 #include<stdio.h>

 #include<stdlib.h>

 #include<conio.h>

 #include<math.h>

 #define f(x) ((log10(x)+7)/2)

 #define f1(x) (1/(x*log (10)*2))

 float error;

 void Iterative_Driver(float);

 void Iterative_Method(float);                

 void main()

 {

      long float x;

      int d;

      clrscr();

      printf("\nEnter the approximate root: - ");

      scanf("%lf",&x);

      printf("\nEnter the degree of precision required: - ");

      scanf("%d",&d);

      error=0.5*pow(10,-d);

      Iterative_Driver(x);

      getch();

 }

/*function to check for conditions and then calls iterative method*/

 void Iterative_Driver(float approot)

 {

       if(f(approot)==0.0f)

       {

         printf("\nThe no. entered is the root\n");

         return;

       }

       if(fabs(f1(approot))>=1.0)

       {

         printf("\nCondition of convergence not satisfied.Try again!\n");

         return;

       }

       else

        Iterative_Method(approot);

 }

/*function to demonstrate the repeated substitution method*/

void Iterative_Method(float root)

{

  float Xi,Xi_1,Ri,Ri_1;

  int count=0,flag=1;

  printf("\n\nIteration(i)        Xi                  Relative Error      Order of Convergence\n");

  printf("========================================\n\n\n");

  do

  {

       if(count==0)

        Xi=root;

       else

        Xi=f(Xi_1);

       printf("%-20d",count+1);

       printf("%-20lf",Xi);

       if(count)

       {

           printf("%-20lf",fabs(Xi-Xi_1));

           if(fabs(Xi-Xi_1)<error)

            flag=0;

       }

       else

           printf("%-20s","-----");

       if(count>1)

           printf("%-18lf",log(fabs(Xi-Xi_1))/log(Ri_1));

       else

           printf("%-18s","-----");

       if(count>=1)

           Ri_1=fabs(Xi-Xi_1);

       printf("\n\n");

       Xi_1=Xi;

       count++;

       if(!flag)

          break;

  }while(1);

  printf("\n\nThe root is : %.4f",Xi);

}

Output