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I should solve $\pi^{n/2} / \Gamma(n/2 + 1) = 1$. Therefore, I need to know other forms of $\Gamma(n/2)$ or $(n/2)!$. I have already checked the Mathematica and MathWorld, very well. But unfortunately, I had not any progress so far.

The best way that I have done is to find the most similar statistical distribution, which is a mixture of Levy distribution and uniform distribution.

Then I tried to fit the best curve to recognize the parameters. Then I changed the parameters with their rationalized one -I mean one integer divided by another integer-, and replaced them in mixture distribution and then I fully simplified the final expression completely.At the end, I verified it to measure the residuals and I repeated the above procedure again and again to find the better and better result.

However, it is a approximately solution, but I need a exact one.

David G. Stork
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Dr. Saeed Forouzesh
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  • Hello, welcome to MSE. Please use $ signs to format your mathematics. – uniquesolution Nov 10 '18 at 18:33
  • Is $n$ an integer? If yes, you just have to consider when $n$ is even and when $n$ is odd – Jakobian Nov 10 '18 at 18:58
  • No. It could be belongs to Integer or Real numbers and omit the complex numbers. Actually, I know the exact solution, which is 12.76405293503267913265321996..., but I don't know how to reach that. – Dr. Saeed Forouzesh Nov 10 '18 at 19:32
  • "$12.76405293503267913265321996...$" is not an exact solution. It is a close approximation, but until you specify what all "..." stands for, it is not exact. – Paul Sinclair Nov 11 '18 at 03:56

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You could get a nice approximation of the solution.

Let $m=\frac n2$ which means that you want to solve the equation $$m! =\pi ^m$$

If you look at this question of mine, you will see a magnificent approximation by @robjohn.

Applied to you case $(a=\pi,k=0)$ $$m=-\frac{\log (2 \pi^2 )}{2 W\left(-\frac{\log (2 \pi^2 )}{2 e \pi}\right)}-\frac{1}{2}$$ that is to say $$n=-\frac{\log (2 \pi^2 )}{ W\left(-\frac{\log (2 \pi^2 )}{2 e \pi}\right)}-1$$ which is $\approx 12.748$ (relative error of $0.12$%).

Claude Leibovici
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