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I came across the rather nice identity

\begin{align} &&\frac{(-n)^{n-1} \Gamma (n+1)}{(1-n)_{n-1}}&&\tag{1}&\\ \\ &=&\prod _{k=1}^{n-1} \frac{(k+1) n^2}{n^2-k n}&&\tag{2}&\\ \\ &=&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n} \cdots\frac{n^3- 2n^2}{3 n}\cdot\frac{n^3- n^2}{2 n}\cdot n^2&&\tag{3}&\\ \\ &=&\underbrace{n\cdot n\cdot n\cdots n\cdot n\cdot n}_{n\text{ times}}&&\tag{4}&\\ \\ &=&n^{n}&&\tag{5}&\\ \end{align}

where ${(1-n)_{n-1}}$ in $(1) $ is the Pochhammer symbol, where $(x)_{n}\equiv\dfrac{\Gamma(x+n)}{\Gamma(x)}$.

Interestingly, Mathematica doesn't simplify the expression to steps $(4)$ & $(5).$ How is this identity proven, and for what $n$ is this not the case?

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    $\begingroup$ Exclude $0$ from the natural numbers as $(-0)^{0-1} = \frac{1}{0}$ $\endgroup$ – Axoren Nov 19 '14 at 15:33
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    $\begingroup$ Well, actually, if you include $0$, you end up with an interesting observation. Because when $n = 0$, $(-n)^{n-1}$ AND $n^n$ are both undefined. $\endgroup$ – Axoren Nov 19 '14 at 15:36
  • $\begingroup$ @mvw now updated. $\endgroup$ – martin Nov 20 '14 at 0:43
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Steps (1) to (2)

\begin{align} f(n) &= \frac{(-n)^{n-1} \Gamma(n+1)}{(1-n)_{n-1}} \\ &= \frac{(-n)^{n-1} n!} {\underbrace{(1-n)(1-n+1)(1-n+2)\cdots -2 \cdot -1}_{n-1 \tiny \mbox{ factors}}} \\ &= \frac{(-n)^{n-1} n!} {(1-n)(2-n)(3-n)\cdots -2 \cdot -1} \\ &= \frac{n^{n-1} n!} {(n-1)(n-2)(n-3)\cdots 2 \cdot 1} \\ \end{align}

here we could continue directly with \begin{align} f(n) = \frac{n^{n-1} n!}{(n-1)!} = n^{n-1} n = n^n \end{align}

or complicate things with multiplying nominator and denominator with $n^{n-1}$ \begin{align} f(n) &= \frac{n^{2n-2}n!}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\ &= \frac{(n^2)^{n-1}n!}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\ &= \frac{n^2\cdot 2 n^2 \cdot 3n^2 \cdots (n-2) n^2 \cdot (n-1) n^2 \cdot n}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\ &= \frac{2 n^2 \cdot 3n^2 \cdots (n-2) n^2 \cdot (n-1) n^2 \cdot n^3}{(n^2-n)(n^2-2n)(n^2-3n)\cdots 2n \cdot n} \\ &= \prod_{k=1}^{n-1} \frac{(k+1) n^2}{n^2-kn} \end{align}

Steps $(2)$ to $(5)$

For even $n=2m+2$ and $m\ge 0$ one gets $2m+1$ factors: \begin{align} f(n) =&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n} \cdots \frac{n^3-3n^2}{4n} \cdot \frac{n^3- 2n^2}{3 n}\cdot\frac{n^3- n^2}{2 n}\cdot n^2 \\ =&\underbrace{\frac{2 n^2}{n^2- n}\cdot \underbrace{\frac{3 n^2}{n^2-2 n}\cdot \underbrace{\frac{4 n^2}{n^2-3 n} \cdots \frac{n^2- 3n}{4}} \cdot \frac{n^2- 2n}{3}}\cdot\frac{n^2- n}{2}}_{\tiny \mbox{left factor} \times \mbox{right factor}}\cdot n^2 \\ =&(n^2)^m n^2 \\ =&n^{2m+2} \\ =&n^n \end{align} Note: The underbraces mean that $m$ pairs consisting of a left and a right side factor each, from the outside to the inside, multiply, resulting into $n^2$ each.

For odd $n=2m+3$ and $m\ge 0$ one gets $2m+1+1$ factors: \begin{align} f(n) &=(n^2)^m \cdot \frac{(m+2)n^2}{n^2-(m+1)n} \cdot n^2 \\ &=(n^2)^m \frac{m+2}{(2m+3)-(m+1)} n^3 \\ &=n^{2m+3} \\ &=n^n \end{align} So this seems to hold for $n\ge 2$.

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  • $\begingroup$ thank you - thislooks good :) willl have a read & digest. $\endgroup$ – martin Nov 19 '14 at 20:53
  • $\begingroup$ thanks for answering the question so completely :) $\endgroup$ – martin Nov 20 '14 at 12:24
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    $\begingroup$ a litle bit is missing, not done yet $\endgroup$ – mvw Nov 20 '14 at 12:28
  • $\begingroup$ yes, that is a bit clearer - thank you $\endgroup$ – martin Nov 20 '14 at 12:40
  • $\begingroup$ Ok, done. Thanks for the interesting problem. $\endgroup$ – mvw Nov 20 '14 at 12:48

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