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My friend came to see me regarding some calculation in probability where he would like to know if it is possible to solve for a variable analytically under the gamma function. By this I mean say we are given the value of some quantity $x$, such that

$x = \dfrac{\Gamma(1 + \frac{2}{k})}{\left(\Gamma(1 + \frac{1}{k})\right)^2}$

I would like to solve this for some real number $k$. I can use the factorial to manipulate this expression and get

$x = \dfrac{(\frac{2}{k})!}{{\left(\left(\frac{1}{k}\right)!\right)}^2 }$

If (a big if) I can make the substitution $n = \frac{1}{k}$, this would be equal to the central binomial coefficient. However I don't think this is possible as $\frac{1}{k}$ is not an integer in general.

What else can I do? Thanks.

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    @D B Lim then series expansion, as proposed by Sasha, should work well enough. Also computer algebra systems has procedures for numerical roots finding.2011-09-28

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Let $u = \frac{1}{k}$. Using duplication formula for the numerator: $ \Gamma\left( 2 u + 1\right) = \frac{4^u}{\sqrt{\pi}} \Gamma\left(u + 1\right)\Gamma\left(u + \frac{1}{2}\right) $ we rewrite your equation as $ x = \frac{4^u}{\sqrt{\pi}} \frac{\Gamma\left(u + \frac{1}{2}\right)}{\Gamma\left(u + 1\right)} $ This form is helpful if $u$ is large, because $\frac{\Gamma\left(u + \frac{1}{2}\right)}{\Gamma\left(u + 1\right)} \sim \sqrt{u} + o(1)$

Thus if $x$ is large, equation becomes $x = 4^u \sqrt{\frac{u}{\pi}}$, which can be solved exactly $u = \frac{W(4 x^2 \pi \log x )}{4 \log 2}$, where $W$ is the Lambert W function.

If $u$ is small, series expansion works well: $ \frac{4^u}{\sqrt{\pi}} \frac{\Gamma\left(u + \frac{1}{2}\right)}{\Gamma\left(u + 1\right)} \sim 1 + \frac{\pi^2}{6} u^2 - 2 \zeta(3) u^3 + \frac{19 \pi^4}{360} u^4 + o(u^4) $