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Thread: Gamma Function logarithmic derivative

  1. #1
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    Gamma Function logarithmic derivative

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

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

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

    Now take $\displaystyle \log $'s of both sides.

    $\displaystyle -\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.

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

    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 $\displaystyle \frac{\Gamma'}{\Gamma}(z) = O(log|z|)$ as $\displaystyle z \to \infty$
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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 $\displaystyle \frac{\Gamma'}{\Gamma}(z) = O(log|z|)$ as $\displaystyle z \to \infty$
    $\displaystyle \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) $

    $\displaystyle = 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 $\displaystyle x\in\mathbb{R} $ and $\displaystyle x>0 $, $\displaystyle \log(x) = \sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n}+O(1) $. (Ask if you'd like to see why.)

    Thus $\displaystyle \frac{\Gamma'(z)}{\Gamma(z)} = O(\log(z)) $ as $\displaystyle z\to \infty $. (I omitted the absolute value since $\displaystyle 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 $\displaystyle 1 $ to approximate $\displaystyle \int_{1}^x \frac{1}{t}dt $, where $\displaystyle x\in\mathbb{N} $.
    We know that $\displaystyle \int_{1}^x \frac{1}{t}dt = \log(x) $, but since $\displaystyle \frac{1}{t} $ is monotonically decreasing, RRAM $\displaystyle <\log(x)< $ LRAM.

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

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