The Newton-Raphson method

The idea

Newton-Raphson improves a root estimate using the tangent: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ . It usually converges fast, but fails when $f'(x_n) = 0$ (a flat tangent) and can jump to a different root if the start is poor.

Differentiate $f$ .
Substitute $x_n$ into $f$ and $f'$ .
Apply the formula; repeat with the new value.

Watch out If $f'(x_0) = 0$ , the tangent never meets the $x$ -axis and the method fails.

See it worked

Worked example

Use Newton-Raphson on $f(x) = x^2 - 2$ with $x_0 = 1$ to find $x_1$ and $x_2$ .

$f'(x) = 2x$ .
$x_1 = 1 - \frac{-1}{2} = 1.5$ .
$x_2 = 1.5 - \frac{0.25}{3} = 1.41667$ .

Answer $x_1 = 1.5$ , $x_2 = 1.41667$

Your turn

Question 1Non-calculator3 marks

$f(x) = x^3 - 2x - 5$ . Use one step of Newton-Raphson from $x_0 = 2$ .

Question 2Calculator allowed3 marks

$f(x) = e^x - 4$ . Use one step of Newton-Raphson from $x_0 = 1$ , giving $x_1$ to 4 decimal places.

Question 3Non-calculator3 marks

For $f(x) = x^3 - 3x + 1$ , which of the starting values $x_0 = 0$ , $1$ or $2$ makes Newton-Raphson fail at the first step?

In past papers

16 official questions in 16 papers since 2018. The latest:

All A-level Maths past papers

A-level Maths Pro, Pearson Edexcel 9MA0

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