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I fear this may be pedestrian but after consulting Abramowitz and Stegun, as well as wikipedia and wolfram, I could not find the answer, though I'm sure it may be simple for someone more gifted than I in analysis :)

I need to know how to simplify, $$ \prod_{i=0}^N\; (1-i\,\epsilon) $$

I also did not see a way of shaping this into either of the trigonometric functions sin or cos...

P.S. $N$ becomes very big, like Avogadro's number big, so I believe I may be able to use either a finite or an infinite product...?

P.P.S. $\epsilon \ll 1$

P.P.P.S $i \neq \sqrt{-1}$ - it is only an index ( a poor choice too )

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I don't think the infinite product converges. – Daniel Littlewood Jan 19 '13 at 20:23
You might want to use \prod and \ll. – Did Jan 19 '13 at 23:12
up vote 2 down vote accepted

If $N\epsilon \ll 1$, using that $\ln (1+x) \approx x$ for small $x$:

$\prod_{n=0}^N (1-n\epsilon) = \exp(\ln(\prod_{n=0}^N (1-n\epsilon))) = \exp(\sum_{n=0}^N \ln(1-n\epsilon)) \approx \exp(\sum_{n=0}^N (-n\epsilon)) = \exp(-\epsilon N(N+1)/2)$

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I was wondering how it would change if instead of $1-n\,\epsilon$, it was $V-n\,\epsilon$ where $V >> 1$. I guess it would be $\exp(\sum\,(\ln(V) - n\,v_0\,(d/dV)[\ln(V)\])$ – nate Jan 19 '13 at 23:24
A related question... of course $1>>n\,\epsilon$, so how can you get rid of it? I know $\ln$ is pretty flat but.... – nate Jan 19 '13 at 23:33
It does work here. If you're bothered by the derivative, you can always say $\ln(V-n\epsilon)=\ln(V*(1-n\epsilon/V))$ and use the fact that $\ln a*b = \ln a + \ln b$. (But it's essentially the same here) – Guest 86 Jan 20 '13 at 22:55
actually I had just forgot about that finite sum $\sum_i^N=N(N+1)/2$... Tell the truth I am trying to get the "4" in the reduced volume van der Waal equation from counting microstates ;) ($ V-4*v_0$) – nate Jan 21 '13 at 2:47

Maple says $$ \prod_{k=1}^N(1 - k a) = \frac{(-1)^{N} a^{N} \Gamma \Bigl(\frac{N a + a - 1}{a}\Bigr)}{\Gamma \Bigl(\frac{a - 1}{a}\Bigr)} $$ But I don't know if you think the Gamma function is "simple" or not.

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