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Math Help - Gamma Function

  1. #1
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    Gamma Function

    How is \Gamma(50\frac{1}{2}) evaluated iteratively?
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Cairo View Post
    How is \Gamma(50\frac{1}{2}) evaluated iteratively?
    Please explain the notation 50\frac{1}{2}... is it a [real] number?...

    Kind regards

    \chi \sigma
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    Sorry.

    Yes it is real - as in \Gamma(n+\frac{1}{2}).
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Cairo View Post
    How is \Gamma(50\frac{1}{2}) evaluated iteratively?
    I'm not quite sure how 'iteratively' you want it. But, you can use the formula \displaystyle \Gamma\left(z+\frac{1}{2}\right)=\frac{\Gamma(2z)}  {\Gamma(z)}2^{1-2z}\sqrt{\pi}. If you'd like to know a proof of this, I can type one up.
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Cairo View Post
    Sorry.

    Yes it is real - as in \Gamma(n+\frac{1}{2}).
    From the property ...

    \displaystyle \Gamma(1+x) = x\ \Gamma(x) (1)

    ... You derive that...

    \displaystyle \Gamma(2+x) = x\ (1+x)\ \Gamma(x)

    \displaystyle \Gamma(3+x) = x\ (1+x)\ (2+x)\ \Gamma(x)

    ...

    \displaystyle \Gamma(n+x) = x\ (1+x)\ ... \ (n-1+x)\ \Gamma(x) (2)

    Now if You set in (2) x=\frac{1}{2} and take into account that is \Gamma(\frac{1}{2})= \sqrt{\pi} You obtain...

    \displaystyle \Gamma(n+\frac{1}{2}) = \frac{1\ 3\ 5\ ...(2n-1)}{2^{n}}\ \sqrt{\pi}

    Kind regards

    \chi \sigma
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    Quote Originally Posted by Drexel28 View Post
    I'm not quite sure how 'iteratively' you want it. But, you can use the formula \displaystyle \Gamma\left(z+\frac{1}{2}\right)=\frac{\Gamma(2z)}  {\Gamma(z)}2^{1-2z}\sqrt{\pi}. If you'd like to know a proof of this, I can type one up.
    Thanks for this Drexel.

    I managed to find something similar on Wikipaedia, but it is the proof I am after.

    Can you supply one?
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    Thanks chisigma.

    But is that the same expression that Drexel gets?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Cairo View Post
    Thanks for this Drexel.

    I managed to find something similar on Wikipaedia, but it is the proof I am after.

    Can you supply one?
    Certainly. But, my original proof didn't work out. So I found an alternate...but it's fairly technical. You might find \chi\sigma's suggestion more useful. Also, I did find an alternate proof on page 22 of this book (a very good book by the wayP may be found for a more general result. Regardless...


    For z>0 define \displaystyle f(z)=\frac{\Gamma(2z)}{\Gamma\left(z+\frac{1}{2}\r  ight)}2^{1-2z}\sqrt{\pi}. Note firstly then that \displaystyle f(1)=\frac{\Gamma(2)}{\Gamma(\frac{1}{2})}2^{-1}\sqrt{\pi}=1. Moreover,


    \displaystyle \begin{aligned}f(z+1) &=\frac{\Gamma(2z+2)}{\Gamma\left(\left(z+\frac{1}  {2}\right)+1\right)}2^{1-2z-2}\sqrt{\pi}\\ &= \frac{1}{4}\frac{\left(2z+1\right)\left(2z\right)}  {z+\frac{1}{2}}\frac{\Gamma(2z)}{\Gamma\left(z+\fr  ac{1}{2}\right)}2^{1-2z}\sqrt{\pi}\\ &=z\frac{\Gamma(2z)}{\Gamma\left(z+\frac{1}{2}\rig  ht)}2^{1-2z}\sqrt{\pi}\\ &=zf(z)\end{aligned}


    Finally, note that \log(f(z))=\log\left(\Gamma(2z)\right)-\log\left(\Gamma\left(z+\frac{1}{2}\right)\right)+  \left(1-2z\right)\log(2)+\frac{1}{2}\log(\pi). So that \frac{d^2}{dz^2}\log(f(z))=4\psi'(2z)-\psi\left(z+\frac{1}{2}\right) where \psi is the digamma function. But, a quick check (using digamma's series expression) shows then that 4\psi'(2z)-\psi\left(z+\frac{1}{2}\right)>0 and thus \log(f(z)) is convex, and so f(z) is log convex. The result then follows from the Bohr-Mollerup theorem.
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