Implementation of Newton Rapson method in C.

Here is the code..

//AIM:TO SOLVE GIVEN EQUATION BY NEWTON RAPHSON METHOD

#include<stdio.h>

float fx(float );

float fx1(float );

void main()

{

float a,b,x,i;

int j,n;

printf("the given equation fx = (x*x)-12\n");

for(i=1;fx(i)<0;)

i+=0.2;

a=i;

for(i=1;fx(i)>0;)

i-=0.2;

b=i;

x= (a+b)/2.0;

printf("enter no. of itrations\n");

scanf("%d",&n);

for(j=0;j<n;j++)

{

x=(x-(fx(x)/fx1(x)));

printf("root after %d iteration is %f",(j+1),x);

}

printf("hence the root is %f",x);

}

float fx(float d)

{

return(((d*d)-12));

}

float fx1(float e)

{

return((2*e));

}

For Turbo C users.

//AIM:TO SOLVE GIVEN EQUATION BY NEWTON RAPHSON METHOD

#include<stdio.h>

#include<conio.h>

float fx(float );

float fx1(float );

void main()

{

float a,b,x,i;

int j,n;

clrscr();

printf("the given equation fx = (x*x)-12\n");

for(i=1;fx(i)<0;)

i+=0.2;

a=i;

for(i=1;fx(i)>0;)

i-=0.2;

b=i;

x= (a+b)/2.0;

printf("enter no. of itrations\n");

scanf("%d",&n);

for(j=0;j<n;j++)

{

x=(x-(fx(x)/fx1(x)));

printf("root after %d iteration is %f",(j+1),x);

}

printf("hence the root is %f",x);

getch();

}

float fx(float d)

{

return(((d*d)-12));

}

float fx1(float e)

{

return((2*e));

}