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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 ?

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    $\begingroup$ Try looking at en.wikipedia.org/wiki/Lagrange_inversion_theorem $\endgroup$ Feb 20, 2014 at 10:05
  • $\begingroup$ A good approach, but the formula still seems to be quite complicated. $\endgroup$
    – Peter
    Feb 20, 2014 at 10:11
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    $\begingroup$ Series reversion is available in CAS and produces (using Lagrange inversion) : $$x - x^4 + 4\,x^7 - 22\,x^{10} + 140\,x^{13} - 969\,x^{16} + 7084\,x^{19} - 53820\,x^22 + 420732\,x^{25}+O(x^{28})$$ These coefficients appear in OEIS A002293 and OEIS A137211 (Generalized $4$-Catalan numbers?). $\endgroup$ Feb 20, 2014 at 10:22
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    $\begingroup$ The radius of convergence is not large. I get $\sqrt[3]{27/256}$. $\endgroup$ Feb 20, 2014 at 10:30
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    $\begingroup$ See also mathoverflow.net/questions/118532/…. $\endgroup$ Oct 15, 2014 at 19:02

2 Answers 2

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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.

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  • $\begingroup$ Great answer! Thank you very much! $\endgroup$
    – Peter
    Feb 20, 2014 at 22:53
  • $\begingroup$ Thanks @Peter! I'll have to look at the details of Tom Copeland's paper myself. Cheers, $\endgroup$ Feb 20, 2014 at 22:57
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    $\begingroup$ BTW, the ratio 27/256 is precisely that for the coefficients of the discriminant curve for the equation. $\endgroup$ Feb 23, 2014 at 17:49
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    $\begingroup$ The limit for the convergence of the inverse series for $x^n + x$ for $n>1$ should be $(\frac{(n-1)^{n-1}}{n^n})^{\frac{1}{n-1}}=(n-1) (\frac{1}{n})^{\frac{n}{n-1}}$. $\endgroup$ Aug 2, 2016 at 4:10
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    $\begingroup$ For an asymptotic approach to the convergence, see p. 775 of "Function Series, Catalan Numbers, and Random Walks on Trees" by Bajunaid, Cohen, Colonna, and Singman $\endgroup$ Dec 25, 2019 at 16:58
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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}]]

enter image description here

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]]

enter image description here

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 $\sqrt[3]{27/256}$.

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  • $\begingroup$ Useful method (in many cases) using Mma. Thanks to share! Cheers, $\endgroup$ Feb 21, 2014 at 8:46
  • $\begingroup$ @MarkMcClure: Also, see my comments on discriminants above. $\endgroup$ Feb 24, 2014 at 1:06

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