In this program, you will implement the Newton-Raphson method for finding the root of the nonlinear equations in the C Programming language. The Newton-Raphson method is also known as Newton Method. you can find the root of the equation by taking the initial guess value.
#include <stdio.h>
#include <math.h>
#include<stdlib.h>
//equation f(x)
float f(float x) {
// change the equation for solving another problem
return pow(x, 2) - 6;
}
//derivative of equation i.e f'(x)
float derivative(float x) {
// write the derivative of your equation
return 2 * x;
}
int main() {
float x;
int n, i;
printf("Enter initial guess value: ");
scanf("%f", &x);
printf("Enter number of iterations: ");
scanf("%d", &n);
for (i = 1; i <= n; i++)
{
if (derivative(x) == 0) {
// f'(x) should not be 0
printf("Division by Zero is not allowed.");
exit(0);
}
//Newton Raphson formula
x = x - f(x) / derivative(x);
// fabs() method will return positive value
printf("\nIteration %d and value %f", i, fabs(x));
}
printf("\n approximate root: %f \n", fabs(x));
return 0;
}
Enter initial guess value: 1 Enter number of iterations: 5 Iteration 1 and value 3.500000 Iteration 2 and value 2.607143 Iteration 3 and value 2.454256 Iteration 4 and value 2.449494 Iteration 5 and value 2.449490 approximate root: 2.449490
x and n. derivative(x) is zero or not. if 0 then the Newton-Raphson method will get failed. x = x - f(x) / derivative(x) this is a formula of the Newton-Raphson method, it will calculate the root of an equation. For example, let's find the root of x2 -6=0. if your initial guess will be 1 then f(x)=-2 and derivative(x)=2.f(2) and derivative(2), the function will return value and you will again substitute value in the newton formula. you will do this up to n time.If your guess value is closer to the root the Newton-Raphson method will take less time otherwise it will calculate but take more time.
