= Newton iteration at a multiple root
{c}
{title2=$e_{k+1}/e_k\to(m-1)/m$}
If a sufficiently smooth <function> has the form $f(x)=(x-\alpha)^m q(x)$ with $m>1$ and $q(\alpha)\ne0$, then the <Newton root-finding iteration> satisfies
$$
x_{k+1}-\alpha=(x_k-\alpha)\left(1-\frac{q(x_k)}{m q(x_k)+(x_k-\alpha)q'(x_k)}\right).
$$
Thus the ratio of successive nonzero errors tends to $(m-1)/m$: the usual iteration has only <linear convergence>. Cancelling the common factor in $f/f'$ produces a continuous extension at the <multiple root>, although the original quotient is undefined there. An implementation should stop upon finding an exact <zero of a function>.
Back to article page
