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                                                                                                  

False Position Method/Regula Falsi

       §       /* False position method*/

#include<stdio.h>

#include<math.h>

#define f(x) (x*x*x-4*x*x-9*x-12) // any equation can be used

void main()

{

       void false(float*,float,float,float,float,int*);

    int itr=0,N;

       float x,x0,x1,x2,x3,err;

       printf("----------METHOD OF FALSE POSITION FOR SOLVING NONLINEAR EQUATIONS-----------\n\n\n");

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

       scanf("%f%f%f%d",&x0,&x1,&err,&N);

       false(&x2,x1,x0,f(x0),f(x1),&itr);

       do

       {

         if(f(x0)*f(x2)<0)

      x1=x2;

      else

      x0=x2;

         false(&x3,x0,x1,f(x0),f(x1),&itr);

         if(fabs(x3-x2)<err)

      {

         printf("after %d th iterations root=%6.4f\n",itr,x3);

      getch();

      return 0;

      }

      x2=x3;

        }while(itr<N);

      printf("solution has more error for given number of iteration");

      return 1;

      getch();

}

// now function false//

void false(float *x,float x0,float x1,float fx0,float fx1,int *itr)

{

     *x=x0-((x1-x0)/(fx1-fx0))*fx0;

     ++(*itr);

     printf("iteration no.%3d x=%7.5f\n",*itr,*x);

}

ª       Results: (False-position Method)

----------METHOD OF FALSE POSITION FOR SOLVING NONLINEAR EQUATIONS-----------

enter starting values,max error,max iteration

2 9 .00005 99

iteration no.  1 x=8.24000

iteration no.  2 x=2.98913

iteration no.  3 x=3.99729

iteration no.  4 x=4.81304

iteration no.  5 x=5.33790

iteration no.  6 x=5.62234

iteration no.  7 x=5.76140

iteration no.  8 x=5.82585

iteration no.  9 x=5.85496

iteration no. 10 x=5.86796

iteration no. 11 x=5.87373

iteration no. 12 x=5.87629

iteration no. 13 x=5.87743

iteration no. 14 x=5.87793

iteration no. 15 x=5.87815

iteration no. 16 x=5.87825

iteration no. 17 x=5.87829

after 17 th iterations root=5.87828970