Newton on x³−3x+1

Text

Apply Newton’s method to $f(x)=x^3-3x+1=0$ with $x_0=0{,}5$: perform four iterations and compare with bisection.

Solution

We have $f(x)=x^3-3x+1$ and $f'(x)=3x^2-3$, hence $x_{n+1} = x_n - \frac{x_n^3-3x_n+1}{3x_n^2-3}.$ Starting from $x_0=0{,}5$:

$n$	$x_n$
$0$	$0{,}5$
$1$	$0{,}333333$
$2$	$0{,}347222$
$3$	$0{,}347296$
$4$	$0{,}347296$

Newton already stabilises at the third iteration on the value $\boxed{x\approx 0{,}347296}$. Bisection on $[0,1]$, which halves the error at each step, requires about $20$ iterations to reach the same precision (six decimal places): Newton is drastically faster because its convergence is quadratic.

Links

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