# Newton-Raphson Method – Algorithm, Implementation in C With Solved Examples

Numerical Methods & Algorithms / Friday, October 12th, 2018

To solve non-linear function of the real variable x we have already learned Bisection method and Iteration method, in this article we are going to learn Newton-Raphson method to solve the same.

# Newton-Raphson Method or Method of Tangent

Let x0 be an approximate root of the equation f(x) = 0. Suppose x1 =x0 + h be the exact root of the equation, where h is the correction of the root. Then f(x1) =0.

Using Taylor’s series, f(x1) = f(x0 + h) is expanded in the following form
$f\left( {{x}_{0}} \right)+hf’\left( {{x}_{0}} \right)+\frac{{{h}^{2}}}{2!}f”\left( {{x}_{0}} \right)+….=0$
Neglecting the second and higher order derivatives the above equation reduces to
$f\left( {{x}_{0}} \right)+hf’\left( {{x}_{0}} \right)=0\Rightarrow h=-\frac{f\left( {{x}_{0}} \right)}{f’\left( {{x}_{0}} \right)}$
$Hence,{{x}_{1}}={{x}_{0}}+h={{x}_{0}}-\frac{f\left( {{x}_{0}} \right)}{f’\left( {{x}_{0}} \right)}$
To compute the value of h, the second and higher powers of h are neglected so the value of
$h=-\frac{f\left( {{x}_{0}} \right)}{f’\left( {{x}_{0}} \right)}$
is not exact, it is an approximate value. So, x1 obtained is not a root of the equation, but it is a better approximation of x than x0.

In general,
${{x}_{n+1}}={{x}_{n}}-\frac{f\left( {{x}_{n}} \right)}{f’\left( {{x}_{n}} \right)}…………(1)$

This expression generates a sequence of approximate values x1, x2, ….,xn,… each successive tern of which is closer to the exact value of the root x than its predecessor. The method will terminate when |xn+1 – xn| becomes very small. In Newton-Raphson method the arc of the curve y = f(x) is replaced by a tangent to the curve, hence this method is sometimes called the method of tangents.

## Implementation of Newton-Raphson method in C

Output
Enter the initial guess x0 : 1.1
One root is 1.53209 obtained at 7 th iteration

 Example 01

Find a real root of the equation x2 + 2x = 2 correct to three significant figures by Newton-Raphson method.

Solution:

Let f(x) = x2 + 2x -2
Therefore f’(x) = 2x + 2
$\text{So the iteration formula}~{{x}_{n+1}}={{x}_{n}}-\frac{f\left( {{x}_{n}} \right)}{f’\left( {{x}_{n}} \right)},~\text{gives}$
${{x}_{n+1}}={{x}_{n}}-\frac{{{x}_{n}}^{2}+2{{x}_{n}}-2}{2{{x}_{n}}+2}=\frac{{{x}_{n}}^{2}+2}{2{{x}_{n}}+2}\left( n=0,1,2,… \right)$
$\text{Now }f\left( 0 \right)\text{ }=\text{ }-\text{2 }<0,\text{ }f\left( \text{1} \right)\text{ }=\text{ 1}>0$
So a real root lies between 0 and 1. But for quick convergence we take x0 = 0.5. Then we have
${{x}_{1}}=\frac{{{x}_{0}}^{2}+2}{2{{x}_{0}}+2}=0.75$
and similarly we get
${{x}_{2}}=0.7321,{{x}_{3}}=0.7320,{{x}_{4}}=0.7320$
Thus the required real root is 0.732, correct to three significant figures.

 Example 02

Use Newton-Raphson method to find a positive root of the equation ex – 3x = 0, correct to four decimal places.

Solution:

Let f(x) = ex – 3x
Therefore f’(x) = ex – 3
So the iteration formula for Newton-Raphson method
${{x}_{n+1}}={{x}_{n}}-\frac{f\left( {{x}_{n}} \right)}{f’\left( {{x}_{n}} \right)},~\text{gives}$
${{x}_{n+1}}={{x}_{n}}-\frac{{{e}^{{{x}_{n}}}}-3{{x}_{n}}}{{{e}^{{{x}_{n}}}}-3}=\frac{\left( {{x}_{n}}-1 \right){{e}^{{{x}_{n}}}}}{{{e}^{{{x}_{n}}}}-3}\left( n=0,1,2,… \right)$
$\text{Now }f\left( 1 \right)\text{ }=\text{ }-0.2817\text{ }<0,\text{ }f\left( 2 \right)\text{ }=\text{ 1}\text{.3890}>0$
So a real root lies between 1 and 2. Taking x0 = 2, we have
${{x}_{1}}=\frac{\left( 2-1 \right){{e}^{2}}}{{{e}^{2}}-3}=1.68352$
${{x}_{2}}=\frac{\left( 1.68352-1 \right){{e}^{1.68352}}}{{{e}^{1.68352}}-3}=1.54348$
and similarly we get
${{x}_{3}}=1.51349,{{x}_{4}}=1.51214,{{x}_{5}}=1.51213$
Hence a positive real root of the given equation correct to four decimal places is 1.5121

 Example 03

Using Newton-Raphson method, find the value of
$\sqrt[4]{12}$

Solution:
$\text{Let}~x=\sqrt[4]{12}\Rightarrow {{x}^{4}}=12$
$\text{So let}~~f(x)={{x}^{4}}-12\Rightarrow f'(x)=4{{x}^{3}}$
So the iteration formula for Newton-Raphson method
${{x}_{n+1}}={{x}_{n}}-\frac{f\left( {{x}_{n}} \right)}{f’\left( {{x}_{n}} \right)},~\text{gives}$
${{x}_{n+1}}={{x}_{n}}-\frac{{{x}_{n}}^{4}-12}{4{{x}_{n}}^{3}}=\frac{3{{x}_{n}}^{4}+12}{4{{x}_{n}}^{3}}\left( n=0,1,2,… \right)$
$\text{Now }f\left( 0 \right)\text{ }=\text{ }-12\text{ }<0,\text{ }f\left( 1 \right)\text{ }=\text{ }-\text{110, }f\left( 2 \right)\text{ }=4>0$
So a real root lies between 1 and 2. Taking x0 = 1.5, we have
${{x}_{1}}=\frac{3{{x}_{0}}^{4}+12}{4{{x}_{0}}^{3}}=2.0138$
and similarly we get
${{x}_{2}}=1.8777,{{x}_{3}}=1.8614,{{x}_{4}}=1.8612$
$\text{Hence }\sqrt[4]{12}=1.861,~\text{correct to four significant figures}.$

<< Previous  Next>>