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Math Help - Beta function?

  1. #1
    Super Member fardeen_gen's Avatar
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    Beta function?

    Prove that:
    1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" alt="n\:\int_{0}^{1} x^m \1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" />
    Is this correct?

    L.H.S
    = n.B(m, n -1) , where B(x, y) is Beta function
    = n. m!(n-1)!/((m + n - 1 + 1)!)
    = n(n-1)!m!/((m+n)!)
    = R.H.S
    Last edited by fardeen_gen; February 10th 2009 at 01:08 PM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by fardeen_gen View Post
    Prove that:
    1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" alt="n\:\int_{0}^{1} x^m \1-x)^{n-1}dx = \frac{n!\:m!}{(m+n)!}" />
    Is this correct?

    L.H.S
    = n.B(m, n -1) , where B(x, y) is Beta function
    = n. m!(n-1)!/((m + n - 1 + 1)!)
    = n(n-1)!m!/((m+n)!)
    = R.H.S
    that's not a proof! you're using what you're supposed to prove! do it by induction: fix an m and do the induction over n. nothing to prove for n = 1. so suppose the claim is true for n \geq 1.

    let I(m,n)=\int_0^1 x^m (1-x)^{n-1} \ dx. since x^m(1-x)^{n-1}=x^{m+1}(1-x)^{n-1} + x^m(1-x)^n, we have: I(m,n)=I(m+1,n)+I(m,n+1). \ \ \ \ \ \ (1)

    on the other hand by integration by parts: I(m+1,n)=\frac{m+1}{n}I(m,n+1), which with (1) gives us: I(m,n+1)=\frac{n}{m+n+1}I(m,n)=\frac{1}{m+n+1} \cdot \frac{n!m!}{(m+n)!}=\frac{1}{n+1} \cdot \frac{(n+1)!m!}{(m+n+1)!}. \ \Box
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