# proof of multiplication formula for gamma function

Define the function $f$ as

 $f(z)={n^{nz}\prod\limits_{k=0}^{n-1}\Gamma\left(z+{k\over n}\right)\over\Gamma% (nz)}$

By the functional equation of the gamma function,

 $f(z+1)={n^{n}n^{nz}\left(\prod\limits_{m=0}^{n-1}\Gamma\left(z+{m\over n}% \right)\right)\prod\limits_{k=0}^{n-1}\left(z+{k\over n}\right)\over\Gamma(nz)% \prod_{k=0}^{n-1}(nz+k)}=f(z)$

Hence $f$ is a periodic function of $z$. However, for large values of $z$, we can apply the Stirling approximation formula to conclude

 $f(z)={(2\pi)^{n/2}n^{nz}\prod\limits_{k=0}^{n-1}\left[e^{-z-k/n}(z+k/n)^{z+k/n% -1/2}+O(e^{-z}(z+k/n)^{z+k/n-3/2})\right]\over(2\pi)^{1/2}e^{-nz}(nz)^{nz-1/2}% +O(e^{-nz}(nz)^{nz-3/2})}=$
 $(2\pi)^{(n-1)/2}n^{1/2}\,{\prod\limits_{k=0}^{n-1}\left[e^{-k/n}(z+k/n)^{z+k/n% -1/2}+O((z+k/n)^{z+k/n-3/2})\right]\over z^{nz-1/2}+O(z^{nz-3/2})}=$
 $(2\pi)^{(n-1)/2}n^{1/2}\,{z^{1/2}\prod\limits_{k=0}^{n-1}\left[e^{-k/n}\left(1% +{k\over nz}\right)^{z+k/n-1/2}z^{k/n-1/2}+O((z+k/n)^{k/n-3/2})\right]\over 1+% O(z^{-1})}$

Note that

 $\prod\limits_{k=0}^{n-1}e^{-k/n}=e^{-\sum_{k=0}^{n-1}k/n}=e^{(1-n)/2}$
 $z^{1/2}\prod\limits_{k=0}^{n-1}z^{k/n-1/2}=z^{1/2}z^{\sum_{k=0}^{n-1}(k/n-1/2)% }=z^{1/2+(n-1)/2-n/2}=1$

Also,

 $\left(1+{k\over nz}\right)^{z}=e^{k/n}+O(z^{-1})$

Hence, $f(z)=(2\pi)^{(n-1)/2}n^{1/2}+O(z^{-1})$. Now, the only way for a function to be periodic and have a definite limit is for that function to be constant. Therefore, $f(z)=(2\pi)^{(n-1)/2}n^{1/2}$. Writing out the definition of $f$ and rearranging gives the multiplication formula.

Title proof of multiplication formula for gamma function ProofOfMultiplicationFormulaForGammaFunction 2013-03-22 14:44:10 2013-03-22 14:44:10 rspuzio (6075) rspuzio (6075) 9 rspuzio (6075) Proof msc 33B15 msc 30D30