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I'm trying to derive the functional equation $2^{2z-1}\Gamma(z)\,\Gamma(z+\frac{1}{2})=\sqrt{\pi}\,\Gamma(2z)$ using Gauss's formula: $$\Gamma(z)=\lim_{n\to\infty}\frac{n!\,\,n^z}{z(z+1)\cdots(z+n)}\,,$$ but it's tricky business.

I've started by just looking at the product $$\frac{n!\,\,n^z}{z(z+1)\cdots(z+n)}\cdot\frac{n!\,\,n^{z+1/2}}{(z+1/2)(z+3/2)\cdots(z+(2n+1)/2)}$$ and the numerator very nicely becomes $(n!\, n^{2z})(n!\,n^{1/2})$, so we have the beginning stages of $\Gamma(2z)$ and $\Gamma(1/2)=\sqrt{\pi}$. But the denominator is killing me! Any advice?

Also, where in the world does this $2^{2z-1}$ come from?

Thanks!

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1  
Just for the record, the formula in the title is known as the duplication formula. –  AD. May 17 '12 at 10:31

2 Answers 2

up vote 4 down vote accepted

With Pochhammer symbol notation, we write

$$\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=\lim_{n\to\infty}\frac{n!\,n^z}{(z)_{n+1}}\frac{n!\,n^{z+1/2}}{(z+1/2)_{n+1}}. \tag{$\circ$}$$

Notice that

$$(z)_{n+1}(z+1/2)_{n+1}=z(z+1)\cdots(z+n)~\times~(z+1/2)(z+3/2)\cdots\left(z+\frac{2n+1}{2}\right) $$

$$=\frac{2z+0}{2}\frac{2z+2}{2}\cdots\frac{2z+2n}{2}~\times~\frac{2z+1}{2}\frac{2z+3}{2}\cdots\frac{2z+(2n+1)}{2}=\frac{(2z)_{2(n+1)}}{2^{2(n+1)}}$$

Therefore for $(\circ)$ we rewrite

$$2^{2(n+1)}\frac{n!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{1}. \tag{$\bullet$}$$

We have an extra power of $2$, the Pochhammer's $\color{Red}2(n+1)$ is not compatible with the numerator's simple $(\color{Red}1\cdot n)!$ and $(\color{Red}1\cdot n)^{2z}$, and we are missing a $(1/2)_{n+1}$ down below. Now we notice that

$$\left(\frac{1}{2}\right)_{n+1}=\frac{1+0}{2}\cdots\frac{1+2n}{2}=\frac{1}{2^{n+1}}\frac{(2n+1)!}{(2\cdot1)\cdots(2\cdot n)}=\frac{(2n+1)!}{2^{2n+1}n!}.$$

We choose to further rewrite $(\bullet)$ as

$$2^{2(n+1)}\frac{n!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{\color{Purple}{(1/2)_{n+1}}}\color{DarkBlue}{\frac{(2n+1)!}{2^{2n+1}n!}}=\color{LimeGreen}2\frac{(2n+1)!\,n^{2z}}{(2z)_{2(n+1)}}\frac{n!\,n^{1/2}}{(1/2)_{n+1}}. \tag{$\triangle$}$$

There is but one thing left to do to make $(\triangle)$ into the form of $\Gamma(2z)\Gamma(1/2)$ (granted, the first $\Gamma$ will have a dummy variable $2n+1$ while the second simply has $n$). The $n^{2z}$ in the left numerator needs to be $(2n+1)^{2z}$, or something asymptotically $\sim$ to it, like - I don't know - $(\color{Orange}2n)^{2z}$?

Now can you tell why the $\color{Orange}2^{\color{Orange}{2z}\color{LimeGreen}{-1}}$ needs to be there for it to work out? ;-)

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Wow, this is fantastic. A+ for effort! –  Bey May 3 '12 at 4:16

Recall that

$$B(x,y)= \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=2\int_0^{\frac {\pi}2}\sin ^{2y-1}\theta\cos^{2x-1}\theta d\theta$$

We now choose $x=y=z$

$$\eqalign{ & B(z,z) = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}} = 2\int_0^{\frac{\pi }{2}} {{{(\sin \theta \cos \theta )}^{2z - 1}}} d\theta \cr & B(z,z) = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}} = 2\int_0^{\frac{\pi }{2}} {{{\left( {\frac{{\sin 2\theta }}{2}} \right)}^{2z - 1}}} d\theta \cr & B(z,z) = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}} = \frac{1}{{{2^{2z - 1}}}}\int_0^{\frac{\pi }{2}} {{{\left( {\sin 2\theta } \right)}^{2z - 1}}} d\left( {2\theta } \right) \cr & B(z,z) = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}} = \frac{1}{{{2^{2z - 1}}}}\int_0^\pi {{{\sin }^{2z - 1}}\phi } d\phi = \frac{2}{{{2^{2z - 1}}}}\int_0^{\pi /2} {{{\sin }^{2z - 1}}\phi } d\phi = \frac{2}{{{2^{2z - 1}}}}B\left( {z,\frac{1}{2}} \right) \cr} $$

But $$B\left( {z,\frac{1}{2}} \right) = \frac{{\Gamma (z)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {z + \frac{1}{2}} \right)}}$$

so we have

$$\frac{2}{{{2^{2z - 1}}}}\frac{{\Gamma (z)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( {z + \frac{1}{2}} \right)}} = \frac{{\Gamma {{(z)}^2}}}{{\Gamma (2z)}}$$

$$\sqrt \pi \Gamma (2z) = {2^{2z - 1}}\Gamma (z)\Gamma \left( {z + \frac{1}{2}} \right)$$

As desired. This works (given the integral definition works) for $\Re(z) >0$.

I can't think of a solution in terms of Gauss' identity, but it seems quite laborious to use it. Also, you'll have to consider a different variable for each of the limits to make it rigorous, so maybe it is not a good idea using it after all.

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Don't we need Re(z)>0 for this to work? –  Bey May 3 '12 at 2:54
1  
@Bey I'll clarify that. Thanks. –  Pedro Tamaroff May 3 '12 at 2:59
    
Thanks for this. The reason I was attempting to do this by Gauss's formula is that the functional equation should be true for for all z =/= 0, -1, ... –  Bey May 3 '12 at 3:04
3  
Downvoter: Did you -1 because of the remark about Gauss' identity, or the fact that this answer does not use it like requested? You should at least explain the vote. I feel "this answer is not useful" is not a fair description, it's just (it seems) not the answer the OP was looking for. –  anon May 3 '12 at 3:20
1  
@anon, yes, I downvoted because I feel that proving the identity via the integral representation of the Beta function is at best tangentially related to the question. –  Antonio Vargas May 3 '12 at 6:12

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