Newton-Raphson Method

This method is generally used to improve the result obtained by one of the previous methods. Let x_o be an approximate root of f(x) = 0 and let x₁ = x_o + h be the correct root so that f(x₁) = 0. Expanding f(x_o + h) by Taylor’s series, we obtain

f(x_o) + hf’(x_o) + h²f’’(x_o)/2! +-----= 0

Neglecting the second and higher order derivatives, we have

f(x_o) + hf’(x_o) = 0

Which gives

h = - f(x_o)/f’(x_o)

A better approximation than x_o is therefore given be x₁, where

x₁ = x_o – f(x_o)/f’(x_o)

Successive approximations are given by x₂, x₃,----,x_n+1, where

x_n+1 = x_n – f(x_n)/f’(x_n)

Which is the Newton-Raphson formula.

Example: - Find a real root of the equation x³ -5x + 3 = 0.

Solⁿ: -

Let, f(x) = x³ -5x + 3 = 0

f’(x) = 3x² - 5

Choosing x_o = 1

Step-1:

f(x_o) = -1

f(x_o) = -2

So, x₁ =1 – ½ = 0.5

Step-2:

f(x₁) = 0.625

f’(x₁) = -4.25

x₂ = 0.5 + 0.625/4.25 = 0.647

Step-3:

f(x₂) = 0.0356

f’(x₂) = -4.139

X3 = 0.647+ 0.0356/4.1393 = 0.6556

Step-4:

f(x₃) = 0.0037

f’(x₃) = -3.7105

x₄ = 0.6556 + 0.0037/3.7105 = 0.6565

The successive approximation 0.6565