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Math Help - Gamma Function logarithmic derivative

  1. #1
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    Gamma Function logarithmic derivative

    Show that -\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})

    Hint: use Weierstrass product formula \frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Show that -\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})

    Hint: use Weierstrass product formula \frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}
    \frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}

    Now take  \log 's of both sides.

     -\log\Gamma(s) = \log s +\gamma s+\sum_{n=1}^\infty \left(\log\left(1+\frac{s}{n}\right)-\frac{s}{n}\right)

    Now differentiate both sides.

    -\frac{\Gamma'(s)}{\Gamma(s)}=\frac{1}{s} + \gamma + \sum_{n=1}^\infty(\frac{1}{s+n}-\frac{1}{n})

    This process is called logarithmic differentiation.
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    Oh nice it was easy, somehow I havent heard that term before. Here yet another problem:

    Show that, for real z > 0, we have \frac{\Gamma'}{\Gamma}(z) = O(log|z|) as z \to \infty
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Oh nice it was easy, somehow I havent heard that term before. Here yet another problem:

    Show that, for real z > 0, we have \frac{\Gamma'}{\Gamma}(z) = O(log|z|) as z \to \infty
     \frac{\Gamma'(z)}{\Gamma(z)} = O(1) + \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{z+n}\right) = O\left(\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{\lfloor z\rfloor+n}\right)\right)

     = O\left(\sum_{n=1}^{\lfloor z \rfloor} \frac{1}{n}\right) since the previous sum is telescoping.

    Now it's a well known fact that for  x\in\mathbb{R} and  x>0 ,  \log(x) = \sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n}+O(1) . (Ask if you'd like to see why.)

    Thus  \frac{\Gamma'(z)}{\Gamma(z)} = O(\log(z)) as  z\to \infty . (I omitted the absolute value since  z > 0 .)
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    Ok thanks, could you say something about the log x thing? A reference is enough or just give the general idea, I dont need details.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Ok thanks, could you say something about the log x thing? A reference is enough or just give the general idea, I dont need details.
    Perform LRAM and RRAM with step size  1 to approximate  \int_{1}^x \frac{1}{t}dt , where  x\in\mathbb{N} .
    We know that  \int_{1}^x \frac{1}{t}dt = \log(x) , but since  \frac{1}{t} is monotonically decreasing, RRAM  <\log(x)< LRAM.

    Now RRAM  = \sum_{n=2}^x \frac{1}{n}
    LRAM  = \sum_{n=1}^{x-1} \frac{1}{n}

    Another way to prove this is by applying the Euler Maclaurin formula to  \sum_{n\leq x} \frac{1}{n} , if you are familiar.
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