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I have the following optimization problem: $$\max_{x}~\frac{A^{x}}{\Gamma(1+x)\Gamma(1-x)}$$ such that $A > 0$ and $x \in (0,1)$. How can we obtain the optimal $x$?

My attempt: I wrote $\Gamma(1+x)\Gamma(1-x) = \frac{\pi x}{\sin(\pi x)}$ and equate the first derivative of $\frac{A^{x}}{\Gamma(1+x)\Gamma(1-x)}$ with respect to $x$ to $0$. I got the following: $$\frac{A^x \log(A)\sin(\pi x)}{(\pi x)}+ \frac{A^x \cos(\pi x)}{x}-\frac{A^x \sin(\pi x)}{(\pi x^2)} = 0. \tag1$$ I could not find the optimal $x$ from ($1$).

1 Answers1

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You have done a mistake; it should be $\max_x A^x \frac{sin(\pi x)}{\pi x}$ what you wanted to differentiale and equate to 0.

kryomaxim
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  • Thanks. I have edited the question. Still I cannot simplify (1) to obtain the solution. –  Jan 03 '17 at 20:20
  • You can try to find a solution by setting x as a solution of $cos(\pi x) = 0$. Then solve for the integer $n$. – kryomaxim Jan 03 '17 at 20:26
  • This will give the solution $x = 1/2$ as $x \in (0,1)$. –  Jan 03 '17 at 20:30
  • Such an equation I would solve numerically or by using a function plot tool (I think there will be free online plotters) for the general case. – kryomaxim Jan 03 '17 at 20:32
  • Yes. I did that already. I am more interested in an analytical solution if there is any. –  Jan 03 '17 at 20:36