Taylor series of the inverse of $x^4+x$ I would like to expand the inverse function of 
$$g(x) := x^4+x $$
in a taylor series at the point x = 0. 
I calculated the first and second derivate at x = 0 with the rule of 
the derivation of an inverse function. Theoretically, this process 
could be continued for higher derivates.
But I would like to have an easier method to calculate higher derivates
of an inverse function in order to calculate the taylor series.
Any ideas ?
 A: In general you may use series reversion to get the taylor expansion of the inverse of a function given by a Taylor series. See for example at Mathworld or in A&S page 16.
But for functions like $\;f(x):=x+x^n\;$ with $n>1$ integer we may obtain the expansion of the inverse in explicit form as shown by Tom Copeland in his paper "Discriminating Deltas, Depressed Equations, and Generalized Catalan Numbers" (just before chapter $5$).
For $n=4$ and using the generating function of OEIS A002293 we 'guess' directly the solution :
$$f^{-1}(x)= \sum_{n=0}^\infty \frac{(-x)^{3n+1}}{3n+1}\binom{4n}{n}$$
and using Wolfram Alpha or by studying the asymptotic of $\binom{4n}{n}\,x^{3n}$ with Stirling's formula we obtain the equivalence :
$$\sqrt{2\,\pi\, n\, 3/4}\;\binom{4n}{n}x^{3n}\sim \frac{(4n/e)^{4n}\,x^{3n}}{(n/e)^{n}(3n/e)^{3n}}\sim \frac{4^{4n}\,x^{3n}}{3^{3n}}\sim \left(\frac{4^4\,x^{3}}{3^3}\right)^n$$
showing that $\,|x|^3$ shouldn't be larger than $\dfrac{3^3}{4^4}=\dfrac{27}{256}$ as quickly found by Mark McClure.
A: This is an answer to Peter's question in the comments, namely, how did I deduce the radius of convergence to be $\sqrt[3]{27/256}$?  I have no proof as the technique is experimental, relying completely on Mathematica.  But, this is how I did it.
First, we invert the series using Mathematica's InverseSeries command.
invSeries = InverseSeries[Series[x + x^4, {x, 0, 30}]]


Next, we use FindSequenceFunction to generate a closed form expression for the non-zero coefficients.
a[n_] = FullSimplify[FindSequenceFunction[
   DeleteCases[CoefficientList[Normal[invSeries], x], 0], n]]


Clearly, this step has some issues.  It doesn't always work and, even when it appears to work, there's no guarantee that it works for all $n$.  Of course, we can check that the formula works for fairly large $n$, but there's still no proof here.  Furthermore, this expression is clearly not as nice as the one found by Raymond!
Now that we have a candidate for the $a_n$s, though, we can easily use a ratio test to find the radius of convergence.
Limit[a[n]/a[n + 1], n -> Infinity]
(* Out: -27/256 *)

Taking the absolute value and accounting for the fact that only every 3rd term is non-zero, we get $\sqrt[3]{27/256}$.
