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In the solution of a problem in physics I came across the expression $$g(x)=\frac{f(x)}{1-f(x)}, \quad f(x)={}_2F_1\left(\frac12,1;\frac d2;-x^2\right)$$ where $d$ is a integer. I was wondering if this can be expressed as a hypergeometric series or not. In literature, people argue that this is clearly impossible, never give a proof though.

If we expand the denominator, we get a series $\sum_{k=1}^\infty f(x)^k$ and therefore can express the term $f(x)^k$ as a ${}_{p}F_{q}$ hypergeometric function with coefficients $p$ and $q$ increasing with $k$. This is probably not the way to go, then.

I know hypergeometric functions have a quite rich algebra, but I couldn't find any helpful identity in my textbooks (Gradshteyn and Ryzhik, Abramowitz and Stegun, Rainville or Erdélyi). Does someone know a technique or a more solid reference that could answer the question and possibly the expression of $g$ if there is such an hypergeometric expansion ?

Tom-Tom
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  • For $d=3$, Wolfram Alpha gave $f(x) = \frac{\arctan(x)}{x}$ and I found this list of identities to confirm the result. However, other $d$ values gave quite different results. – Anant Apr 23 '14 at 16:51
  • @Anant. Thank you for searching. The function $g$ with $d=3$ is exactly the result I am fighting with, but I already knew $f$. Is there a hypergeometric expansion for $g(x)=\arctan x/(x-\arctan x)$ ? – Tom-Tom Apr 23 '14 at 21:53
  • Quite late to the show but ... for x=0 the denominator is zero and the numerator is one; for all hypergeometric functions. Thus, g(0)=inf wouldn't qualify. But allowing a little more flexibility in the formulation raises a whole realm of interesting possibilities. – rrogers Feb 18 '21 at 15:18

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