Cube root of 10 with Newton

Text

Compute $\sqrt[3]{10}$ with four iterations of Newton’s method, taking $f(x)=x^3-10$ and $x_0=2$.

Solution

With $f(x)=x^3-10$ and $f'(x)=3x^2$, the iteration is $x_{n+1} = x_n - \frac{x_n^3-10}{3x_n^2} = \frac{2x_n}{3}+\frac{10}{3x_n^2}.$ Starting from $x_0=2$:

$n$	$x_n$
$0$	$2$
$1$	$2{,}166667$
$2$	$2{,}154504$
$3$	$2{,}154435$
$4$	$2{,}154435$

The value stabilises at $\boxed{\sqrt[3]{10}\approx 2{,}154435}$, correct to six decimal places already from the third iteration: once again quadratic convergence (roughly) doubles the number of correct digits at each step.

Links

Topics: Continuity
Concepts: Quadratic convergence · Newton-Raphson method
Methods: Newton-Raphson
Skills: Calculating · Using formulae
Exercise types: Computing expressions