Newton-Raphson Method

       §       

PROGRAMMING IN C: ASSIGNMENT

PRATYUSH DAS

BEE 3^RD YEAR

Electrical Engineering Department, Jadavpur University

-------Contents-------

         §         Gauss-elimination method

         §         Gauss-Seidel method

         §         Bisection method

         §         False position method

         §         Secant method

         §         Newton-Raphson method  ...............................................................................

......................................................................... 

/* Newton Raphson Method*/

#include<stdio.h>

#include<math.h>

#define f(x) (x*x*x-4*x-9) // any equation can be taken here

#define f1(x) (3*x*x-4)

void main()

{

      int itr,N;

      float h,x0,x1,err;

         printf("------------NEWTON-RAPHSON METHODOLOGY------------\n\n\n");

      printf("enter x0,allowed error,maximum iterations\n");

      scanf("%f %f %d",&x0,&err,&N);

         printf("\n\n");

      for(itr=1;itr<=N;itr++)

      {

      h=f(x0)/f1(x0);

      x1=x0-h;

      printf("iteration no. %3d  gives ,x= %9.6f\n",itr,x1);

      if(fabs(h)<err)

      {

      printf("after %3d iterations,""root is  \n    x=%8.6f\n",itr,x1);

      getch();

      exit(1);

      }

      x0=x1;

      }

      printf("iterations not sufficient, solution does not converge\n");

      getch();

}

ª       Results: (Newton-Raphson method)

 ------------NEWTON-RAPHSON METHODOLOGY-------

enter x0,allowed error,maximum iterations

2    .005    66

iteration no.   1  gives ,x=  3.125000

iteration no.   2  gives ,x=  2.768530

iteration no.   3  gives ,x=  2.708196

iteration no.   4  gives ,x=  2.706529

after   4 iterations,root is

    x=2.706529