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I am working through the proof that one can solve quintic equations first by reducing the polynomial to one of the form $x^5-x-t$, and then solving $x^5-x-t=0$ using the Lagrange Inversion Formula on the function $x-x^5$. However, to do this it seems as if I need to be able to compute $n$th derivatives of $$ \left(\frac{x}{x-x^5}\right)^n, $$ which I am unable to get a nice closed form for. Could anyone offer some insight into this? Thanks.

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    $\begingroup$ Lagrange inversion theorem in Wikipedia seems to deal exactly with the example you're willing to cover. $\endgroup$ – mathcounterexamples.net Nov 18 '16 at 22:20
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    $\begingroup$ @mathcounterexamples.net I have looked there. Wikipedia basically says "this can be solved using Lagrange Inversion, and the answer is this." I am looking to actually go through the details. $\endgroup$ – TomGrubb Nov 18 '16 at 22:32
  • $\begingroup$ Oh, wait, did my answer actually solve everything for you? :O $\endgroup$ – Simply Beautiful Art Nov 18 '16 at 22:47
  • $\begingroup$ Well, good for you! (I'm still not done figuring out how to get there) $\endgroup$ – Simply Beautiful Art Nov 18 '16 at 22:57
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Notice that

$$\begin{align}\left(\frac x{x-x^5}\right)^n&=(1-x^4)^{-n}\\&=1+\sum_{k=1}^\infty\frac{\Gamma(n+k)}{k!\Gamma(n)}x^{4k}\end{align}$$

By binomial expansion. Thus,

$$\begin{align}\frac{d^r}{dx^r}\left(\frac x{x-x^5}\right)^n&=\frac{d^r}{dx^r}1+\sum_{k=1}^\infty\frac{\Gamma(n+k)}{k!\Gamma(n)}x^{4k}\\&=\sum_{k=1}^\infty\frac{(4k)!\Gamma(n+k)}{(4k-r)!k!\Gamma(n)}x^{4k-r}\\&=r!\sum_{k=1}^\infty\binom{4k}{r}\binom{n+k}{n}x^{4k-r}\\\end{align}$$

where $\frac1{(4k-r)!}=0$ when $r>4k$

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    $\begingroup$ True. Your idea is better. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 18 '16 at 22:38
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    $\begingroup$ @Martín-BlasPérezPinilla You inspired my brain to think better :D So you can get half credit ;) $\endgroup$ – Simply Beautiful Art Nov 18 '16 at 22:39
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Hint: do a partial fraction decomposition:

$$\frac{x}{x-x^5} = \frac{1}{1-x^4} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-i} + \frac{D}{x+i}.$$

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  • $\begingroup$ Wow, that's a good idea! But how do you tackle it raised to the $n$th power? $\endgroup$ – Simply Beautiful Art Nov 18 '16 at 22:28
  • $\begingroup$ @SimpleArt, multinomial theorem. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 18 '16 at 22:31
  • $\begingroup$ Has to be an easier way IMO $\endgroup$ – Simply Beautiful Art Nov 18 '16 at 22:36

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