# Expanding an expression using Taylor's series

We've been attempting to expand an expression with Taylor's Theorem but can't quite make the math work out.

$$\frac{f\left(x_n\right)}{f'\left(x_n\right)}= \frac{1}{m}\frac{f^{(m)}\left(\xi _n\right)}{f^{(m)}\left(\sigma _n\right)}\left(x_n-r\right)$$

That's $f'$ at $x_n$ above. We're expanding about root $r$ of multiplicity $m \geq 2$.

How do we get from the left side to the right? For a simple root we know that

$$f(x_n) = f(r) + f'(\psi_n)(x_n-r)$$ ($f(r) = 0$, as it is a root) according to Taylor's theorem.

We can't figure out how to do this for a root of greater multiplicity.

If your $f(x)$ has an $n$-fold root at $x=\xi$, then presumably you can perform a factorization like $f(x)=(x-\xi)^n g(x)$, where $g(\xi)$ is nonzero... –  Ｊ. Ｍ. Sep 16 '11 at 1:07
So, what happens if you Taylor-expand $(x-\xi)^n g(x)$ about $x=\xi$? –  Ｊ. Ｍ. Sep 16 '11 at 2:53