Question

Write a code in C which uses Newton’s method to solve xe^x − 8 = 0. Run your code. Start from the initial guess x0 = 1. How many iterations does it take before you get the exact answer to five decimal places? Run your code. Start from the initial guess x0 = 100. How many iterations does it take before you get the exact answer to five decimal places?

Now write a code which solves xe^x - 8 = 0 by the bisection method. Which method is better for finding roots, Newton or bisection?

Answer 1

Newton's Method-
#include<stdio.h>
#include<math.h>
double f(double x)
{
return x*exp(x)-8;
}
double der_f (double x)
{
return x*exp(x)+exp(x);
}
int main()
{
int i, maxi=1000;
double h, x0, x1, err=0.00001;
printf("\nEnter the value x0\n");
scanf("%lf", &x0);
for (i=1; i<=maxi; i++)
{
h=f(x0)/der_f(x0);
x1=x0-h;
printf(" At Iteration no. %d, x = %lf\n", i, x1);
if (fabs(h) < err)
{
printf("After %d iterations, root = %lf\n", i, x1);
return 0;
}
x0=x1;
}
printf(" The required solution does not converge or iterations are insufficient\n");
}

Screenshots:

Bisection Method:
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
void bisect (float *x, float a, float b, int *i)
{
*x=(a+b)/2;
++(*i);
printf("Iteration number %d X = %7.5f\n", *i, *x);
}
float function (float x)
{
return (x*exp(x)-8);
}
int main ()
{
int i = 0, maxi=1000;
float x, a=-10, b=10,err=0.00001, x1;
bisect (&x, a, b, &i);
do
{
if (function(a)*function(x) < 0)
b=x;
else
a=x;
bisect (&x1, a, b, &i);
if (fabs(x1-x) < err)
{
printf("After %d iterations, the root is = %6.4f\n", i, x1);
return 0;
}
x=x1;
}
while (i < maxi);
printf("The solution does not converge or iterations are not sufficient");
return 1;
}

Screenshots:

Why is newton's method better for finding roots?

Since the bisection method is slower and also requires some knowledge of the function(we need to provide value of a and b), we conclude that newton's method is better for finding roots.