In numerical analysis, Newton’s method is named after Isaac Newton and Joseph Raphson. This method is to find successively better approximations to the roots (or zeroes) of a real-valued function.

The method starts with a function f defined over the real numbers x, the function’s derivative f’, and an initial guess $x_{0}$ for a root of the function f. If the function satisfies the assumptions made in the derivation of the formula and the initial guess is close, then a better approximation $x_{1}$.

In general solving an equation f(x) = 0 is not easy, though we can do it in simple cases like find roots of quadratics. If the function is complicated we can approximate the solution using an iterative procedure also known as a numerical method. One simple method is called Newtonâ€™s Method.

The formula for Newton’s method is given as,

\[\large x_{1}=x_{0}-\frac{f(x_{0})}{{f}'{(x_{0})}}\]

Where,

f($x_{0}$) is a function at $x_{0}$, f'(x) is the first derivative of the function at $x_{0}$, $x_{0}$ is the initial value.

Solved Examples

**QuestionÂ 1: **Estimate the positive root of the equation x^{2Â }– 2 = 0 by using Newton’s method. Begin with x_{0}Â = 2 and compute x_{1Â }1

**S****olution:**

Given measures are,

f(x) = x^{2Â }– 2 = 0, x_{0}Â = 2

Newton’s method formula is: x^{1}Â = x_{0Â }–Â $\frac{f(x-{0})}{f'(x_{0})}$

To calculate this we have to find out the first derivative f'(x)

f'(x) = 2x

So, at x_{0},

f(x_{0}) = 2^{2Â }– 2 = 4 – 2 = 2

f'(x_{0}) = 2 $\times$ 2 = 4

The formula becomes

x_{1}Â = 2 –Â $\frac{2}{4}$ = $\frac{6}{4}$ = $\frac{3}{2}$