How is this Statistics Summation Simplified?

[olechka722]

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!

 

[statdad]

Remember that

$$\sum_{n=0}^m \frac{m!}{n!(m-n)!} 1^{(m-n)} b^n = (1 + b)^m$$

In your sum
$$\exp\left((M-2N)a\right) = \exp\left(Ma\right) \cdot \left(\exp(-2a)\right)^N$$

so your sum is

$$\sum_{N=0}^M {\frac{M!}{N!(M-N)!} \exp\left((M-2N)a\right)} <br /> = e^{Ma} \sum_{N=0}^M {\frac{M!}{N!(M-N)!} \left(e^{-2a}\right)^N <br /> = e^{Ma} \left(1 + e^{-2a}\right)^M$$

For the second one - look at the definition of the hyperbolic function as exponentials, and rearrange terms.

 

[olechka722]

Thanks so much! That really clears things up.