Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

Thank you for any help you can give.

share|cite|improve this question
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
I'm afraid I'm not sure what that means. Can you explain a bit more? – Xorlev Sep 16 '11 at 2:49
So, what happens if you Taylor-expand $(x-\xi)^n g(x)$ about $x=\xi$? – J. M. Sep 16 '11 at 2:53
f(x) = f(r) + f^m(psi)*(x_n-r)^m/m!, theoretically. However, we do not understand what the similar expansion is for f'(x). At least not one that will not result in things canceling and us ending up with f^m(psi)/f^m(sigma). – Xorlev Sep 16 '11 at 4:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.