Statistics summation question:

  • Thread starter olechka722
  • Start date
  • #1
This equation comes out of deriving the canonical partition function for some system. However, the question is more math based. I am having trouble understanding the simplification that was performed in the text:

∑ from N=0 to M of: (M!exp((M-2N)a))/(N!(M-N)!) supposedly becomes

exp(Ma)(1+exp(-2a))^M.... I tried to look at the first few terms and see how I can simplify this, and no dice... Anyone have any ideas? a is just a constant.

Also, the next step is that the above becomes (2cosh(a))^M Which is great, except, huh? I'm more of a scientist than a math person, so I apologize if I am missing something elementary.

Thanks!
 

Answers and Replies

  • #2
statdad
Homework Helper
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Remember that

[tex]
\sum_{n=0}^m \frac{m!}{n!(m-n)!} 1^{(m-n)} b^n = (1 + b)^m
[/tex]

In your sum
[tex]
\exp\left((M-2N)a\right) = \exp\left(Ma\right) \cdot \left(\exp(-2a)\right)^N
[/tex]

so your sum is

[tex]
\sum_{N=0}^M {\frac{M!}{N!(M-N)!} \exp\left((M-2N)a\right)}
= e^{Ma} \sum_{N=0}^M {\frac{M!}{N!(M-N)!} \left(e^{-2a}\right)^N
= e^{Ma} \left(1 + e^{-2a}\right)^M
[/tex]

For the second one - look at the definition of the hyperbolic function as exponentials, and rearrange terms.
 
  • #3
Thanks so much!! That really clears things up.
 

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