<s>[INST] Inverse function of a polynomial What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance!
 [/INST] This is an experimental way of working out the inverse.
We can treat the polynomial like an expansion
\begin{equation}
f(x) = -1 + x + 0x^2 + 2x^3 + 0x^4 + x^5 + 0x^6 + 0x^7 + \cdots
\end{equation}
then we can perform a Series Reversion on this to give the inverse series (as an infinite expansion) \begin{equation}
f^{-1}(x) = (1+x) -2(1+x)^3 +11(1+x)^5-80(1+x)^7+665(1+x)^9-\cdots
\end{equation}
at a glance this doesn't seem to helpful, but if we search the coefficients in the OEIS we seem to get a hit! It would appear (conjecture) that the general coefficient is then \begin{equation}
a(n) = \binom{5n+1}{n}\frac{(-1)^n}{2n+1}
\end{equation}
and we could write that \begin{equation}
f^{-1}(x) = \sum_{n=0}^\infty \binom{5n+1}{n}\frac{(-1)^n}{2n+1}(x+1)^{2n+1}
\end{equation}
if we evaluate that in Mathematica, it gives \begin{equation}
f^{-1}(x)=(1+x)\;_4F_3\left(\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5}\bigg|\frac{3}{4},\frac{5}{4},\frac{6}{4}\bigg|-\frac{5^5}{4^4}(1+x)^2 \right)
\end{equation}
a generalised hypergeometric function. If we plot the composition of these two i.e $f(f^{-1}(x))$ or $f^{-1}(f(x))$ the plot seems to indicate \begin{equation}
f(f^{-1}(x))=f^{-1}(f(x))=x
\end{equation}
of course this is not a proof, and evaluating the inverse function may become numerically unstable if $x$ becomes too large.
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