proving a beta identity

**galactus** · July 5th 2008, 04:31 AM

Hey All:

I was looking at a Beta function identity in a math physics book I have and

it said to show ${\beta}(n,1/2)=2^{2n-1}{\beta}(n,n)$ .

I done that OK. But....how does one prove
${\beta}(n,1/2)=\frac{2^{2n}(n!)^{2}}{n(2n)!}$ ?.

I know the various gamma and beta identities, but got a little stuck on

where this comes from.

I know ${\Gamma}(n+1)=n!, \;\ {\Gamma}(2n+1)=(2n)!$ and so forth.

I tried various ways, but failed to get it to come together.

Where is the world do those factorials come from in that beta identity. I am missing something. Probably obvious. Always is.

**Moo** · July 5th 2008, 04:41 AM

Hello,

What about using the identity $\beta(m,n)=2 \int_0^{\pi/2} (\sin \theta)^{2m-1} (\cos \theta)^{2n-1} \ d\theta$ (I haven't seen a proof of it yet, but perhaps it is possible by substituting $t=\cos^2 \theta$ in $\beta(m,n)=\int_0^1 t^{m-1}(1-t)^{n-1} \ dt$ )

--> $\beta(n,n)=2 \int_0^{\pi/2} \left(\frac{\sin 2 \theta}{2}\right)^{2n-1} \ d\theta$

$\beta(n,1/2)=2^{2n} \int_0^{\pi/2} \left(\frac{\sin 2 \theta}{2}\right)^{2n-1} \ d\theta$

I think you can prove what you want from it (by induction e.g.)

Edit : héhé, I stress the "I think"

**Moo** · July 5th 2008, 05:41 AM

Ok, here is another (and, in my opinion, simpler !) way.

We want to show this :

$\beta(n,1/2)=\frac{2^{2n}(n!)^2}{n(2n)!}=\frac{2^{2n}n!(n-1)!}{(2n)!}$

$=\frac{2^{2n} \Gamma(n+1)\Gamma(n)}{\Gamma(2n+1)}$

But we know that $\beta(x,y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$ (there is a proof of it in wikipedia).

Therefore, we want to prove that $\beta(n,1/2)=2^{2n} \beta(n,n+1)$

We know that $\beta(n,1/2)=2^{2n-1} \beta(n,n)$

------------------------------
So if we can prove that $\beta(n,n)=2 \beta(n,n+1)$ , we're done.

$\beta(n,n+1)=\int_0^1 t^n(1-t)^{n-1} \ dt=\int_0^1 t^{n-1}(1-t)^n \ dt$ (symmetry of beta function).

So $\begin{aligned} 2 \beta(n,n+1)&=\int_0^1 t^n(1-t)^{n-1} + t^{n-1}(1-t)^n \ dt \\ \\<br /> &=\int_0^1 t^{n-1}(1-t)^{n-1} [t+1-t] \ dt \\ \\<br /> &=\beta(n,n) \quad \blacksquare \end{aligned}$

**galactus** · July 5th 2008, 05:52 AM

Thanks Moo, I actually got that far and didn't think of induction. Thanks.

I was looking at the gamma identities.

I knew that $\frac{(n!)^{2}}{(2n)!}=\frac{n{\Gamma}(n){\Gamma}( n+1)}{{\Gamma}(2n+1)}$

I tried plugging in ${\beta}(n,1/2)=\frac{{\Gamma}(n){\Gamma}(1/2)}{{\Gamma}(n+1/2)}$

And knew that ${\Gamma}(n+1/2)=\frac{{\Gamma}(n){\Gamma}(1/2)}{{\beta}(n,1/2)}={\Gamma}(n+1/2)$

I had trouble tying them together with that factorial-laden identity.

**Moo** · July 6th 2008, 06:57 AM

Simplest one

$\beta(n,1/2)=2^{2n-1} \beta(n,n)$

$\beta(n,n)=\frac{\Gamma(n) \Gamma(n)}{\Gamma(2n)}=\frac{(n-1)!(n-1)!}{(2n-1)!}$

$=\frac{(n!)^2}{n^2} \cdot \frac{2n}{(2n)!}=\frac{2(n!)^2}{n(2n)!}$

**galactus** · July 6th 2008, 08:54 AM

Thanks Moo, I actually just got the same result that way. My rpoblem was, for some reason, I failed to see that

$(2n-1)!=\frac{(2n)!}{2n}$ right off the bat. That made all the difference. I knew it was something in front of my nose. Always is.

I am glad to see we jive. Appears to be the best way to go about it. I like all your methods. Good girl.

Regarding your second post, I hadn't thought of it that way. That was a good
observation.

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