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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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    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).2011-09-16

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