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I have reduced this problem (thanks @Mhenni) to the following (which needs to be proved):

$$\prod_{k=1}^n\frac{\Gamma(3k)\Gamma\left(\frac{k}{2}\right)}{2^k\Gamma\left(\frac{3k}{2}\right)\Gamma(2k)}=\prod_{k=1}^n\frac{2^k(1+k)\Gamma(k)\Gamma\left(\frac{3(1+k)}{2}\right)}{(1+3k)\Gamma(2k)\Gamma\left(\frac{3+k}{2}\right)}.$$

As you see it's quite a mess. Hopefully one can apply some gamma-identities and cancel some stuff out. I have evaluated both products for large numbers and I know that the identity is true, I just need to learn how to manipulate those gammas.

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  • $\begingroup$ You can cancel the $\frac{1}{\Gamma(2k)}$ from both sides, and simplify the integer valued gamma functions and also $\prod_{k=1}^n2^k=2^{\frac{n^2+n}{2}}$, to be honest I think you could easily simplify alot of this... $\endgroup$ – Ethan May 26 '13 at 7:26
  • $\begingroup$ another approach may be to take $\ln$ of both sides and then use the fact that $$ \ln(\Gamma(z+1))= \ln(\Gamma(z)) - \ln(z)$$ turn the product into a sum and solve ? $\endgroup$ – Ahmed Masud May 26 '13 at 7:38
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Put all gamma functions to one side. Then

  • $\Gamma(2k)$ cancels out.

  • Using gamma function duplication formula, one can replace $$\frac{1}{1+k}\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{k+3}{2}\right)=\frac12\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{k}{2}+\frac12\right)=2^{-k}\sqrt{\pi}\,\Gamma(k).$$

  • Similarly, $$\frac{1}{1+3k}\Gamma\left(\frac{3k}{2}\right)\Gamma\left(\frac{3k+3}{2}\right)= \frac12\Gamma\left(\frac{3k}{2}\right)\Gamma\left(\frac{3k}{2}+\frac12\right)= 2^{-3k}\sqrt{\pi}\,\Gamma(3k). $$

  • The identity follows immediately.

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  • $\begingroup$ Ah, the duplication formula... thanks! $\endgroup$ – 77474 May 26 '13 at 7:53

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