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I'm trying to find a gamma rep for $ 15 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot ... $


Steps so far:

It's a simple sequence of $ n \cdot (n-2) \cdot (n-4) \cdot (n-6) \cdot (n-8)... $ and so on.

Allegedly I am supposed to find an equivalent expression using factorials and powers of 2

$$= n\cdot(n-2(1)) \cdot (n-2(2)) \cdot (n-2(3)) \cdot (n-2(4)) \cdot ... $$ I know that $$\Gamma(n) = (n-1)!$$ and $$ n\Gamma(n) = n \cdot (n-1)! $$

I've fiddled and monkeyed with this for a while now. How in the world the above sequence translates into a gamma/factorial equivalent I have no idea.


Update: I'm not even sure how powers of two are involved.

It seems like a simpler expression can be constructed, but I believe the point of the exercise is to transform it into gamma functions, which is where I start drawing a blank.

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  • $\begingroup$ Notice that it is $15!$ or $16!$ divided by the product of several even numbers $\endgroup$ – Henry Apr 6 '16 at 22:35
  • $\begingroup$ 15!! (long live the semi-factorial!) $\endgroup$ – user2469 Apr 6 '16 at 23:58
  • $\begingroup$ I see that but how do you prove it? Working with what Henry said: $$ \frac {n!}{(n-1) \cdot (n-3) \cdot (n-5)...} $$ where $ n = 15$. Why can't I figure this out? It seems so easy. $\endgroup$ – Zainy Apr 7 '16 at 3:33
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$$ \begin{align} (2n-1)(2n-3)(2n-5)\cdots1 &=\frac{2n(2n-1)(2n-2)\cdots1}{2n(2n-2)(2n-4)\cdots2}\\ &=\frac{(2n)!}{2^nn!}\\ &=\frac{\Gamma(2n+1)}{2^n\Gamma(n+1)}\\ &=\frac{\Gamma(2n)}{2^{n-1}\Gamma(n)}\\ \end{align} $$ This is also known as $(2n-1)!!$ which is the Double Factorial.

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I think I've worked an answer:

$$ S = 15 \cdot 13 \cdot 11 \cdot ...$$

$$ \rightarrow S = \frac {15!}{(n-1)\cdot(n-3)\cdot(n-5)\cdot(n-7)\cdot...}$$

Looking at Denominator: $$ (n-1)\cdot(n-3)\cdot(n-5)\cdot(n-7)\cdot(n-9)\cdot(n-11)\cdot(n-13) $$ $$ =(n-2z+1) $$

This was the error on my part, not noticing the $z$ term

$$\rightarrow S=\lim_{n\to15}\frac{n!}{(n-2z+1)} $$

Still working on how to transform into Gamma functions...

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