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I am working on a physics problem that results in the expression

$$ f(x) = \frac{ _2F_1\left( \begin{matrix} \begin{matrix} a & b \end{matrix} \\ \frac{a+b}{2} + x \end{matrix} \hspace{6pt} \frac{1}{2} \right) }{ _2F_1\left( \begin{matrix} \begin{matrix} a - 1 & b \end{matrix} \\ \frac{a+b}{2} + x \end{matrix} \hspace{6pt} \frac{1}{2} \right) } - 1 $$

For physical reasons, I know that $f(x)$ has to be odd as a function of $x$ (but note that $x$ is not the usual argument of the hypergeometric function here). This also seems to be true numerically.

Main question: Is there a simple hypergeometric identity that shows that $f(x)$ is odd in $x$?

Other comments:

  • Obviously, you can show this by actually doing a series expansion in $x$, but this is cumbersome.

  • In my particular case, $a$ happens to be a negative integer, and $b$ happens to be a positive integer. I don't know if this is relevant (i.e., whether $f(x)$ would still be odd if they were not integers or if they both had the same sign). However, I suspect that this will probably be true for any $a$ and $b$, since I don't think I have used the fact that $a$ and $b$ are integers at any point in the derivation.

  • By bringing the $1$ into the fraction and applying a contiguous relation, you can re-cast this into a lot of different forms. Perhaps the simplest is to say that $g(x)$ is odd in $x$, where $g(x)$ is given by

$$ g(x) = \left.\frac{d}{dz}\right|_{z=\frac{1}{2}} \log {_2}F_1\left( \begin{matrix} \begin{matrix} a - 1 & b \end{matrix} \\ \frac{a+b}{2} + x \end{matrix} \hspace{6pt} z \right) $$

  • It seems like the easiest way to prove something like this would be if there was an identity (perhaps a quadratic transformation) that provided a relationship between $$ {_2}F_1\left( \begin{matrix} \begin{matrix} a & b \end{matrix} \\ c \end{matrix} \hspace{6pt} \frac{1}{2} \right) $$ and $$ {_2}F_1\left( \begin{matrix} \begin{matrix} a & b \end{matrix} \\ a + b - c \end{matrix} \hspace{6pt} \frac{1}{2} \right) $$ I have not been able to find anything along these lines. If anyone has a copy of Erdelyi's book with all the identities and is willing to look for one, I'd be very appreciative.
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    $\begingroup$ see dlmf.nist.gov/15.8.E7 $\endgroup$ – Nemo Apr 9 '18 at 20:34
  • $\begingroup$ @Nemo I think that does it. Is there a way for me to accept a comment as an answer? $\endgroup$ – sasquires Apr 11 '18 at 16:55

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