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I am going step by step through a proof by Jitsuro Nagura which can be found here

Nagura shows:

$\frac{\Gamma'}{\Gamma}(x) - \frac{\Gamma'}{\Gamma}(\frac{x+n-1}{n}) = \int_0^\infty\frac{1}{1-e^{-t}}(e^{-\frac{x+n-1}{n}t} - e^{-xt})dt > 0 (x > 1)$

Then he states:

$\frac{1}{n}\log\Gamma(x) - \log\Gamma(\frac{x+n-1}{n})$ is an increasing function when $x \ge 1$.

I am not clear how this follows. I am not clear where $\frac{1}{n}$ follows from. If someone could explain how the conclusion follows, I would greatly appreciate it.

Thanks,

-Larry

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The logarithmic derivative of a function $f$ is $$ \frac{d}{dx}\log f(x) = \frac{f'(x)}{f(x)} $$ and more generally $$ \frac{d}{dx}\log f(ax+b) = a\frac{f'}{f}\Big(ax+b\Big) $$ from which we can see that $$ \frac{\Gamma'}{\Gamma}\big(x\big)-\frac{\Gamma'}{\Gamma}\left(\frac{x+n-1}{n}\right) = \frac{d}{dx}\left[\log \Gamma(x)-n\log \Gamma\left(\frac{x+n-1}{n}\right)\right] $$ The left side (i.e. the derivative) is positive when $x>1$, hence the expression in the square brackets is increasing. Now divide by $n>1$ to get the result.

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  • $\begingroup$ Thanks very much for your answer. In reviewing your reasoning, I am still not clear where the $n$ in $n\log\Gamma(\frac{x+n-1}{n})$ is coming from. By your reasoning above, shouldn't it be $\frac{d}{dx}[\log\Gamma(x) - \frac{1}{n}\log\Gamma(\frac{x+n-1}{n})]$? $\endgroup$ – Larry Freeman Apr 4 '13 at 14:18
  • $\begingroup$ When we take the derivative we get a factor of $1/n$, so we put in a factor of $n$ ahead so that they will cancel. $\frac{d}{dx} n\log\Gamma((x+n-1)/n) = \frac{n}{n} \Gamma'/\Gamma((x+n-1)/n)$. $\endgroup$ – Zander Apr 4 '13 at 14:33

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