Solving $x^p-x-1=0$ with Lagrange Inversion Formula

Question

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.

Answer 1

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 $r>4k$

Answer 2

Hint: do a partial fraction decomposition:

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