Can anyone explain me how to calculate this expression? $$\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}$$ Thank you, I tried with doble factorial, but don´t know really how to continue.
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What do you mean by "calculating this expression"? Do you mean you want to minimize this expression? – uriyabsc Dec 21 '20 at 08:30
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Almost as @J.G. answered, for large values of $n$, using Stirlin approximation twice and continuing with Taylor series $$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}=\sqrt{\frac n 2}\left(1-\frac{3}{4 n}-\frac{7}{32 n^2}-\frac{9}{128 n^3}+\frac{59}{2048 n^4}+O\left(\frac{1}{n^5}\right)\right)$$ which shows a relative error lower than $0.1$% as soon as $n>2$, lower than $0.01$% as soon as $n>3$, lower than $0.001$% as soon as $n>5$.
If you wish a very good approximation $$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}\sim\sqrt{\frac n 2}\frac{1-\frac{441823}{287784 n}+\frac{119909}{54816 n^2}-\frac{1473029}{877056 n^3} } {1-\frac{225985}{287784 n}+\frac{697315}{383712 n^2}-\frac{7699031}{18418176 n^3} }$$
J.G.
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Claude Leibovici
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Could you provide a proof of source for that last approximation? How well does it perform? – J.G. Dec 21 '20 at 09:57
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@J.G. The idea is simple : expand as far as you can the series, divide it by $\sqrt{\frac n 2}$ and transform the result as an $[p,p]$ Padé approximant. – Claude Leibovici Dec 21 '20 at 10:28
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1Thanks, that addresses my first question. It belongs in your answer, especially if you include an error bound. – J.G. Dec 21 '20 at 10:29
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As for all $[p,q]$ Padé approximant, it is better the $O(x^{p+q+1}$. For this case, it is precisely $\frac{16}{561 n^{13/2}}$ – Claude Leibovici Dec 21 '20 at 10:39
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It's already in the most helpful form. But its large-$n$ asymptotic behaviour is as $\sqrt{n/2}$, by linearizing $\ln\Gamma(z)$ viz.$$\begin{align}\ln\Gamma(n/2)-\ln\Gamma((n-1)/2)&\approx\frac{\ln\Gamma((n+1)/2)-\ln\Gamma((n-1)/2)}{2}\\&=\frac{\ln((n-1)/2)}{2}\\&=\ln\sqrt{(n-1)/2}.\end{align}$$
J.G.
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If your question is connected to your previous one, you can leave the result as shown.
tommik
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