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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.

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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

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