# Newton's method and derivative is 0 at root?

Newton's method of root-finding uses the derivative at $\tau$ to find the value of $x_i$ on the next iteration. But I have read that a function can still have a root if $f'(\tau) = 0$..is there something special about the root in this case? I can't see how Newton's method can work if the derivative is $0$ at $\tau$.

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It usually works, but very inefficiently (i.e., with very slow convergence). – Harald Hanche-Olsen Oct 11 '12 at 19:46
Is there something special about the root at $\tau$?? Does it have multiplicity > 1? – dukenukem Oct 11 '12 at 19:47
Well, yes, if $f(\tau)=f'(\tau)=0$ (and $f$ is analytic at $\tau$) then by definition, the zero has multiplicity greater than 1. – Harald Hanche-Olsen Oct 11 '12 at 19:50
Try thinking about the graphical interpretation of the situation for say $f(x)=x^2$ (for which you have $0=f(0)=f'(0)$). – anon Oct 11 '12 at 19:51
Usually the multiplicity of a root of a sufficiently differentiable function is defined by how many of it and its derivatives vanish at a point - it isn't based off of anything else like an outside theorem. If you're talking about polynomials, then the compatibility of this definition can be derived by setting $p(x)=(x-p)^eg(x)$, where $g(p)\ne0$, and then iteratively differentiating and evaluating with standard rules. – anon Oct 11 '12 at 19:54

To see how Newton's method typically handles a root of multiplicty $n$, look at the toy example $f(x)=x^n$: Then $$x_{k+1}=x_k-\frac{x_k^n}{nx_k^{n-1}}=\Bigl(1-\frac1n\Bigr)x_k,$$ so that $$x_k=\Bigl(1-\frac1n\Bigr)^k,$$ which converges much more slowly than the typical quadratic convergence of Newton's method when applied to a simple zero.