Results 1 to 6 of 6

Math Help - proving a beta identity

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    proving a beta identity

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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"
    Last edited by Moo; July 5th 2008 at 05:02 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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)!}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    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.
    Last edited by galactus; July 6th 2008 at 09:04 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving an identity.
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 12th 2011, 05:22 PM
  2. Proving identity
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 1st 2010, 11:03 AM
  3. Proving Identity
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 8th 2009, 09:30 PM
  4. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 01:30 PM
  5. Proving Beta Distribution Variance
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 14th 2008, 01:43 PM

Search Tags


/mathhelpforum @mathhelpforum