Newton raphson method in c

  👉 NEWTON RAPHSON METHOD:

Let, x0 be an approximation of a root of the given equation f(x)=0, which may be algebraic or transcendental.

Let x0+h be the exact value or the better approximation of the corresponding root,h being a small quantity.

Then,f(x0+h)=0.

Expanding it by Taylor`s theorem,we get,

f(x0+h)=f(x0)+hf `(x0)+h*h f ``(x0)+..........=0

h=-f(x0) / f `(x0)

x1=x0+h

x2=x1+h

condition for convergence of newton raphson method:

 |1-(f `(x)*f `(x))-[ f(x)*f ``(x)] /f `(x)*f `(x)|   <1

|f(x)*f ``(x)|<|f `(x)*f `(x)|

coding:

1. #include<stdio.h>
2. #include<math.h>
3. float fun(float x)
4. {
5.     return x*x-2*x+3;
6. }
7. float diff1fun(float x)
8. {
9.     return 2*x-2;
10. }
11. void main()
12. {
13.     int itr, maxmitr;
14.     float h, x0, x1, error;
15.     printf("\nEnter x0"); 
16.     scanf("%f", &x0);
17.     printf("\nEnter maximum ittreation\n"); 
18.     scanf("%d",&maxmitr);
19.     printf("\nEnter error"); 
20.     scanf("%f",&error);
21. for (itr=1; itr<=maxmitr; itr++)
22.     {
23.         h=fun(x0)/diff1fun(x0);
24.         x1=x0-h;
25.         printf(" At Iteration no. %3d, x = %9.6f\n", itr, x1);
26.         if (fabs(h) < error)
27.         {
28.             printf("After %3d iterations, root = %8.6f\n", itr, x1);
29.             return 0;
30.         }
31.         x0=x1;
32.     }
33.     printf("iterations are insufficient\n");
34.     return 1;
35. }

output: