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Given the following product: $$\displaystyle P(N)=\prod_{k=1}^N\frac{k^a}{a^k}$$ it can be expressed as: $$P(N)=\Gamma(N+1)^a\frac{1}{a^{\frac{1}{2}(N+1)^2-\frac{1}{2}N-\frac{1}{2}}}$$ I have to find the maximum of $P(N)$ vs. $a$. The derivative of this function vs. $N$ gives: $$P'(N)=-\frac{1}{2}\frac{\Gamma(N+1)^a(-2a\Psi(N+1)+2\ln(a)N+\ln(a))}{a^{\frac{1}{2}N(N+1)}}$$ but I don't see how to solve $P'(N)=0$

Thanks

Martin Sleziak
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Riccardo.Alestra
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  • So for some fixed $a$, you are trying to find the value of $N$ which maximises $f$? Also, is $N$ real or a positive integer (implied by your sum expression)? – Macavity Oct 17 '13 at 16:59

1 Answers1

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From the summation expression I assume $N \in \mathbb{N}$, and from your attempt at $P'(N)$ I assume you are trying to maximise w.r.t. $N$. If this is the case, then there is an alternative to finding turning points of the Gamma involved expression.

If $a>0$ is fixed and we are trying to find $N^*\in \mathbb{N}$ which maximises $P(N)$, we have: $$P(N)=\prod_{k=1}^N\frac{k^a}{a^k} = \frac{(N!)^a}{a^{\frac{N(N+1)}{2}}}$$ $$\text{Let }r(N) = \frac{P(N+1)}{P(N)}= \frac{(N+1)^a}{a^{N+1}}$$

This ratio tells us whether at $N$ the function is increasing ($r>1$) or decreasing ($r < 1$). So we look at when $r > 1 \iff (N+1)^a > a^{N+1} \iff \sqrt[N+1]{N+1} > \sqrt[a]{a}$.

The function $f(x) = \sqrt[x]{x}$ has its maximum at $x=e$ and is decreasing on both sides. So eventually $r(N)< 1$ for all large $N$. Looking at where $a$ falls, we can use this to find the turning points of $r(N)$ and hence the maximum of $P(N)$.

For e.g. if $a > e$, we have $N^*$ could be $1, 2$ or the first integer larger than $a$.

Macavity
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