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 $\tau$.
[Math] Newton’s method and derivative is 0 at root
numerical methods
Best Answer
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.