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I need to compute the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2n)$ in order to use the Stirling's formula to calculate, but I am stuck. Thanks!

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I am a little bit confused now... should the last term of $w(n/2+1/2)$ be $(n+1-2n)$ or $(n+1-2k)$?

J.doe
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  • Do you mean, you want to estimate this product using Stirling's Formula? – Travis Willse Feb 18 '16 at 23:11
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    When $n$ is odd the product has to be zero since $0$ is a factor so you only need to consider $n=2m$ even. – Winther Feb 18 '16 at 23:12
  • Oops yes I know that n should be even, but what's next. Is there any formula for it? Thanks – J.doe Feb 18 '16 at 23:16
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    Try to take $n=2m$ and multiply by $2^m m! = 2m\cdot (2m-2) \cdots 2$. Do a similar thing for the negative terms. You should get a pretty compact answer for the product. – Winther Feb 18 '16 at 23:17
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    multiplied by that, I'll get $(2m+1)!$ and after that I'll divide the whole thing by that and I'll get a formula. Is that what you mean? – J.doe Feb 18 '16 at 23:20
  • Yes. Call the product $S$ then $S(2^m m!) = (2m+1)! (-1)(-3)\cdots (1-2m)$. Now you can do the same thing with the remaining terms and solve for $S$ in the end. – Winther Feb 18 '16 at 23:22
  • Smart! Thanks a lot! – J.doe Feb 18 '16 at 23:23
  • What's $2^mm!$$2^mm!$? – J.doe Feb 18 '16 at 23:31
  • I get $(n+1)!n!/(16^nn!)^2$. Is this correct? – J.doe Feb 18 '16 at 23:35
  • @J.doe just to make sure, is the final term in your multiplication supposed to be $(n+1-2n) = (1-n)$ or $(n+1-2k)$ for some constant $k$? – Brevan Ellefsen Feb 18 '16 at 23:39
  • Oops it should be $(n+1-2k)$. How do I do that then? – J.doe Feb 18 '16 at 23:42
  • hmm... if that's the case then WA gives the result $$(-2)^k (n+1) \left(\frac{1-n}{2}\right)_k = \frac{(-2)^k (n+1) \Gamma(\frac{j-n+1}{2})}{\Gamma(\frac{1-n}{2})}$$ where $(a)_k$ is the Pochammer symbol, AKA rising factorial – Brevan Ellefsen Feb 18 '16 at 23:45
  • What does that greek letter mean? – J.doe Feb 18 '16 at 23:46
  • That Greek letter is the Gamma function. It is the extension of the factorial function to values other than integers; it's one of the most important non-elementary functions in all of mathematics, showing up everywhere. – Brevan Ellefsen Feb 18 '16 at 23:47
  • it is also shifted by a factor of one, i.e. $\Gamma(n) = (n-1)!$ for all whole numbers. Just use that definition to adjust Stirling's Approximation for the gamma function and you should have your answer... now to prove the statement above will take me a bit. I would guess the formula above simplies a bit – Brevan Ellefsen Feb 18 '16 at 23:48
  • So for $\gamma((1-n)/2)= ((-n-1)/2)!$? – J.doe Feb 18 '16 at 23:52
  • Note that when you adjust Stirling's Approximation you get that $\Gamma(z) \sim \sqrt{\frac{2\pi}{z}} \left(\frac{z}{e}\right)^z$ – Brevan Ellefsen Feb 18 '16 at 23:54
  • What are you using the letter $\gamma$ to represent?? Generally $\gamma(n)$ is the falling factorial. Also, where did you get this problem from? Is there a specific way you are supposed to solve this? – Brevan Ellefsen Feb 18 '16 at 23:55
  • I've attached a photo in the question. I tried to solve $w(n/2+1/2)$ and I am stuck. – J.doe Feb 19 '16 at 00:01
  • I presume this is a tie-in to this question posted earlier?? Note that duplicating questions is frowned upon, and may lead to the closure of one of the questions. Again, is this for a class or for personal study? Context is important for knowing how to solve the problem, and this site is not for doing your homework. – Brevan Ellefsen Feb 19 '16 at 00:05
  • I was just trying to clarify my question so I attached it. Anyways thank you. :) – J.doe Feb 19 '16 at 00:07
  • You are welcome! I actually recommend leaving the picture there, it adds context. Now, I'm not sure if that post earlier was yours or not, but I would link to it anyway and just explain that this goes deeper into one aspect of that problem; this will make the question meet site guidelines. I'm more than anything just trying to make your question not get closed (I'm also working on an answer atm) – Brevan Ellefsen Feb 19 '16 at 00:11
  • You are also working on an answer on this problem? Great I get to double check my answer then! Thanks! – J.doe Feb 19 '16 at 00:22

3 Answers3

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Well, the comment thread here is long, and may have answers. I didn't check.

For $n=2m$ even, and $k$ not so large that factors wander into negative territory, then you have:

$$\begin{align} &\phantom{{}={}}(n+1)(n-1)\cdots(n+1-2k)\\ &=(2m+1)(2m-1)\cdots(2m+1-2k)\\ &=\frac{(2m+1)(2m)(2m-1)(2m-2)\cdots(2m+1-2k)(2m-2k)(2m-2k-1)(2m-2k-2)\cdots3\cdot2\cdot1}{\left[(2m)(2m-2)\cdots(2m-2k)\right]\left[(2m-2k-1)(2m-2k-2)\cdots3\cdot2\cdot1\right]}\\ &=\frac{(2m+1)!}{2^{k+1}\left[(m)(m-1)\cdots(m-k)\right]\left[(2m-2k-1)(2m-2k-2)\cdots3\cdot2\cdot1\right]}\\ &=\frac{(2m+1)!}{2^{k+1}\left[(m)(m-1)\cdots(m-k)\right](2m-2k-1)!}\\ &=\frac{(2m+1)!(m-k-1)!}{2^{k+1}m!(2m-2k-1)!}\\ &=\frac{(n+1)!\left(\frac{n}{2}-k-1\right)!}{2^{k+1}\left(\frac{n}{2}\right)!\left(n-2k-1\right)!}\\ \end{align}$$

2'5 9'2
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    Note in the comments that the OP said the product is $(n+1)(n-1)\cdots(n+1-2k)$ for some constant $k$, not $(n+1)(n-1)\cdots(n+1-2n)$ as he originally wrote – Brevan Ellefsen Feb 19 '16 at 00:28
  • Thank you but I actually got it wrong in the question. I am supposed to find the product of $(n+1)(n-1)(n-3)...(n+1-2k)$. Thanks for the answer tho! – J.doe Feb 19 '16 at 00:28
  • OK, well I edited a bit then. And now it's only a partial answer for even $n$. – 2'5 9'2 Feb 19 '16 at 00:38
  • How am I supposed to apply the Stirling's formula to $(n/2-k-1)!$? – J.doe Feb 19 '16 at 00:52
  • @J.doe You just do. If you have some version of Sterling's formula for $n!$, then put "$n/2-k-1$" everywhere that formula has "$n$". – 2'5 9'2 Feb 19 '16 at 01:01
  • Oh okay! Thanks! – J.doe Feb 19 '16 at 01:03
  • I am sorry that you might be correct before, it should be n instead of k. And I get $(n+1)!n!/((\sqrt2^nn/2)!)^2$. Is this the correct answer? – J.doe Feb 19 '16 at 01:47
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While @alex.gordan beat me to a full proof that's simpler than mine functionally, I figured I would share the way I approached this problem!

First note that $(a)_k = a(a+1)(a+2)\cdots(a+k-1)$. Thus, $$\begin{align}\left(\frac{1-n}{2}\right)_k & = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+1\right)\cdots\left(\frac{1-n}{2}+k\right) \\ & = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+\frac{2}{2}\right)\left(\frac{1-n}{2}+\frac{4}{2}\right)\cdots\left(\frac{1-n}{2}+\frac{2k-2}{2}\right)\\ & = \left(\frac{1-n}{2}\right)\left(\frac{3-n}{2}\right)\left(\frac{5-n}{2}\right)\cdots\left(\frac{2k-1-n}{2}\right) \\& = \frac{(1-n)(3-n)(5-n)\cdots(2k-1-n)}{2^k}\end{align}$$
Let's now adjust this to fit your product. First let's remove the denominator
$$2^k\left(\frac{1-n}{2}\right)_k =(1-n)(3-n)(5-n)\cdots(2k-1-n)$$
Now let's pull out $k$ negative signs to flip each term $$2^k(-1)^k\left(\frac{1-n}{2}\right)_k =(n-1)(n-3)(n-5)\cdots(n+1-2k)$$
Now we simplify the powers and multiply by $(n+1)$ $$(-2)^k(n+1)\left(\frac{1-n}{2}\right)_k =(n+1)(n-1)(n-3)(n-5)\cdots(n+1-2k)$$
We now use the definition $(a)_k = \frac{\Gamma(a+k)}{\Gamma(a)}$, let $n = 2m$, and transform this into $$\frac{(-2)^k (n+1) \Gamma(\frac{2k+1-n}{2})}{\Gamma(\frac{1-n}{2})}$$
$$=(-2)^k (n+1)\frac{\Gamma(k-m+1/2)}{\Gamma(1/2-m)}$$ Now using the fact that $\Gamma(z) \sim \sqrt{\frac{2\pi}{z}} \left(\frac{z}{e}\right)^z$ (known as Stirling's approximation)
$$\sim (-2)^k (n+1)\frac{\sqrt{\frac{2\pi}{k-m+1/2}} \left(\frac{k-m+1/2}{e}\right)^{k-m+1/2}}{\sqrt{\frac{2\pi}{1/2-m}} \left(\frac{1/2-m}{e}\right)^{1/2-m}}$$ $$= \frac{(-2)^k (n+1)(k-m+1/2)}{e^k}\sqrt{\frac{1/2-m}{k-m+1/2}} \left(\frac{(k-m+1/2)}{(1/2-m)}\right)^{1/2-m}$$

Brevan Ellefsen
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Here is a closed form formula for the finite product

$$ p=2^n(1-n)\frac{ \Gamma( \frac{(3+n) }{2}) } {\Gamma( \frac{(3-n) }{2}) }. $$

Mhenni Benghorbal
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