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Problem of the Week #4 - April 23rd, 2012

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Chris L T521

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Jan 26, 2012
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Here's the fourth problem of the week. Hopefully we'll see some people contribute solutions this time around...last week was a little disappointing in that regard. :-/

This week's problem was again proposed by yours truly.

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Problem: Let $g$ be the entire function given by

\[g(z) = e^{z\gamma}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)e^{-z/k},\]

where $\gamma=\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right)$ is Euler's constant. Prove the recurrence formula

\[(z+1)g(z+1)=g(z).\]

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Remember to read the POTW submission guidlines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
The week 4 university POTW was answered correctly by Sudharaka. Here's the solution I had:

Proof: We observe that


\[\begin{aligned}(z+1)g(z+1) &= (z+1)e^{(z+1)\gamma}\prod_{k=1}^{\infty}\left(1+ \frac{z+1}{k} \right) e^{-(z+1)/k}\\&=(z+1)e^{z\gamma}e^{\gamma} \prod_{k=1}^{\infty} \left(1+ \frac{z+1}{k} \right)e^{-z/k}\cdot\prod_{k=1}^{\infty} e^{-1/k}\\ &= e^{\gamma} \prod_{k=1}^{\infty} e^{-1/k} \cdot (z+1)e^{z\gamma} \prod_{k=1}^{\infty} \frac{k+1}{k} \left(1+ \frac{z}{k+1} \right) e^{-z/k}\\ &= e^{\gamma} \prod_{k=1}^{\infty} \frac{k+1}{k}e^{-1/k} \cdot e^{z\gamma} \prod_{k=1}^{\infty} \left(1+ \frac{z}{k} \right) e^{-z/k}\\ &= e^{\gamma} \prod_{k=1}^{\infty} \frac{k+1}{k}e^{-1/k} g(z)\end{aligned}\]


All that is left to show is that


\[e^{\gamma}\prod_{k=1}^{\infty}\frac{k+1}{k}e^{-1/k}=1.\]


Note that


\[\begin{aligned}\prod_{k=1}^{\infty}\frac{k+1}{k}e^{-1/k} & = \prod_{k=1}^{\infty}\frac{k+1}{k}\cdot\prod_{k=1}^{\infty}e^{-1/k}\\ &= \prod_{k=1}^{\infty}\frac{k+1}{k}\cdot \exp\left(-\sum_{k=1}^{\infty}\frac{1}{k}\right)\\ &= \lim_{n\to\infty}\left[\prod_{k=1}^n \frac{k+1}{k}\cdot \exp\left(-\sum_{k=1}^n \frac{1}{k}\right)\right]\\ &= \lim_{n\to\infty}\left[(n+1)\exp\left(-\sum_{k=1}^n \frac{1}{k}\right)\right]\\ &= \lim_{n\to\infty}\exp\left(\ln(n+1)-\sum_{k=1}^n\frac{1}{k}\right).\end{aligned}\]


Since $\ln(n+1)\sim\ln n$ as $n\to\infty$, we now see that


\[\lim_{n\to\infty}\exp\left(\ln(n+1)-\sum_{k=1}^n\frac{1}{k}\right)\sim\lim_{n\to\infty}\exp\left(\ln(n)-\sum_{k=1}^n\frac{1}{k}\right)=e^{-\gamma}.\]


Therefore, we have


\[e^{\gamma}\prod_{k=1}^{\infty}\frac{k+1}{k}e^{-1/k}=e^{\gamma}e^{-\gamma}=1\]


and thus $(z+1)g(z+1)=g(z)$. Q.E.D.

I've also included Sudharaka's solution below since it was more tidy than mine.

Consider, \((z+1)g(z+1)\)

\[(z+1)g(z+1) = (z+1) \; \mbox{e}^{\gamma (z+1)} \; \prod_{k=1}^{\infty} \left(1 + \frac{z+1}{k}\right) \; \mbox{e}^{-(z+1)/k}~~~~~~~~(1)\]




The infinite product expression for the Gamma function is,


\[\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{k}\right)^{-1} e^{z/k}\]


\[\Rightarrow \frac{1}{\Gamma(z)} = z \; \mbox{e}^{\gamma z} \; \prod_{k=1}^{\infty} \left(1 + \frac{z}{k}\right) \; \mbox{e}^{-z/k}=zg(z)~~~~~~~~(2)\]


\[\Rightarrow \frac{1}{\Gamma(z+1)} = (z+1) \; \mbox{e}^{\gamma (z+1)} \; \prod_{k=1}^{\infty} \left(1 + \frac{z+1}{k}\right) \; \mbox{e}^{-(z+1)/k}\]


Since, \(\Gamma (z+1)=z\;\Gamma (z)\)


\[\frac{1}{z\;\Gamma(z)} = (z+1) \; \mbox{e}^{\gamma (z+1)} \; \prod_{k=1}^{\infty} \left(1 + \frac{z+1}{k}\right) \; \mbox{e}^{-(z+1)/k}~~~~~~~~(3)\]


By (2) and (3);


\[g(z)= (z+1) \; \mbox{e}^{\gamma (z+1)} \; \prod_{k=1}^{\infty} \left(1 + \frac{z+1}{k}\right) \; \mbox{e}^{-(z+1)/k}\]


Now by (1) we get the recurrence relation,


\[(z+1)g(z+1) = g(z)\]
 
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