# Solving for a variable under $\Gamma(z)$

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.

-
@DJC Why is there a $2 + 2/k$ in there? Did you mean $1 + 2/k$? – user38268 Sep 28 '11 at 3:48
I will delete my other comment. Less explicitly, but helpful (I hope): The Beta-function is related to the Gamma function by $$B(x_1,x_2) = \frac{\Gamma(x_1) \Gamma(x_2)}{\Gamma(x_1 + x_2)}.$$ Setting $x_1 = x_2 = 1 + \frac{1}{k}$ gives $$\frac{1}{x} = \left(2 + \frac{2}{k}\right) B\left(1 + \frac{1}{k}, 1 +\frac{1}{k}\right).$$ – JavaMan Sep 28 '11 at 3:59
Thanks, I wills see what I can work with from there. – user38268 Sep 28 '11 at 4:46
Considering how difficult it is to solve similar equations for Factorials, solving these gamma function equations are certainly not possible analytically. For most purposes using Stirling's approximation to solve for $x$ numerically should suffice. – Ragib Zaman Sep 28 '11 at 7:55
A closed-form solution looks unlikely to me. You'll definitely have to use a numerical method like Newton-Raphson for this. – J. M. Sep 28 '11 at 9:29

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: $$\frac{4^u}{\sqrt{\pi}} \frac{\Gamma\left(u + \frac{1}{2}\right)}{\Gamma\left(u + 1\right)} \sim 1 + \frac{\pi^2}{6} u^2 - 2 \zeta(3) u^3 + \frac{19 \pi^4}{360} u^4 + o(u^4)$$