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Question:

I wish to find the inverse of the following function:
$f(x)=\frac{1}{2}\left(\arctan(x) + \ln\left(\sqrt{\frac{1+x}{1-x}}\right)\right)$
This is the equation for a radial null geodesic in a particular version of of the $AdS_2$-Schwarzschild metric. I want to invert the above function. I can plot as a parametric function $(f(x),x)$ (which I cannot post because of my non-existent reputation). It is monotonically increasing and bounded on $[0,\infty)$. I am sure that an inverse function exists but cannot be written in terms of elementary functions. What I want to obtain is at least the exact coefficients of the power series for the inverse function (an analytic expression for all coefficients would be nice but I am not that hopeful).

Attempt:

I tried to compute the coefficients of the power series of the inverse function using the Lagrange inversion theorem. I was able to find the first $100$ non-zero coefficients. However, the resulting power series was only convergent on $[0,1]$. I want a power series that converges over a larger interval. This should be possible since the inverse of $f(x)$ as plotted parametrically is bounded and without singularities on $[0,\infty)$. Is there a way to find such a power series?

mrmrob003
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    The convergence of a power series usually has to do with the nearness of poles (undefined points) of the function. That means that you can't really get a better converging power series around the same point, though around a different point (farther from the poles) you might have a larger interval. – weux082690 Jul 02 '15 at 16:08
  • In general, the power series of a function is unique, so there isn't another choice of power series. – Michael Burr Jul 02 '15 at 16:26
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    Per @MichaelBurr, expanding a smooth function around a given point gives a unique power series, whose radius of convergence extends to the nearest singularity in the complex plane. Better convergence is often available in a continued fraction expansion. – hardmath Jul 02 '15 at 16:44
  • $f(x)=\dfrac{\arctan x+\text{arctanh }x}2=\displaystyle\sum_{n=0}^\infty\frac{x^{4n+1}}{4n+1}$ – Lucian Jul 02 '15 at 17:24
  • @Lucian, thank you for simplifying my expression. I am aware of its power series expansion. Do you, however, know the power series for the inverse of $f(x)$? – mrmrob003 Jul 03 '15 at 06:18
  • Using the power series for $f(x)$ one can obtain the power series for the inverse function. This is called series reversion (I learned today) and there is an article about it on MathWorld: http://mathworld.wolfram.com/SeriesReversion.html – weux082690 Jul 06 '15 at 19:26
  • This is the wolfram language function “inversebetaregularized”. A series expansion is here – Тyma Gaidash Jun 01 '23 at 12:56

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