Summation over multiple arguments

Question

This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:

$$\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$

(it's the Ising model for 3 lattice sites).

I don't understand how I can calculate this in a brute force way, since the sums for $s_1$ and $s_3$ only apply to one part of the equation to be summed over.

(I know you can simplify it and end up with a much nicer expression in terms of cosh)

Help! Thanks.

Answer 1

Proceeding from right to left:

$$\begin{align*}\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =\sum_{s_1=\pm1} \sum_{s_2=\pm1}e^{-{s_1s_2}}\left[ e^{{s_2}}+e^{-{s_2}} \right]\\ &=\sum_{s_1=\pm1}\left( e^{{s_1}}\left[ e^{{-1}}+e^{1} \right]+e^{-{s_1}}\left[ e^{1}+e^{-{1}} \right]\right)\\ &=\sum_{s_1=\pm1}\left( [e^{{s_1}}+e^{-{s_1}}]\left[ e^{{-1}}+e^{1} \right]\right)\\ &=[e^{{-1}}+e^{1}]\left[ e^{{-1}}+e^{1} \right]+[e^{{1}}+e^{-1}]\left[ e^{{-1}}+e^{1} \right]\\ &=2[e^{{-1}}+e^{1}]^2\\ &=4(\cosh(2)+1) \end{align*}$$

Answer 2

A slightly alternate approach than in the other answer (dividing up the sum differently):

Starting with the inside sum of $$\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$ we can factor out the term that doesn't depend on $s_3$ to get $$\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}= \sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}\sum_{s_3=\pm1} e^{-{s_2s_3}}. $$ Now, let's look at the exponent of the internal sum, since $s_2$ is either $+1$ or $-1$, then $-s_2$ is either $+1$ or $-1$. Multiplying by $s_3$ which is either $+1$ or $-1$ results in one of each of $+1$ and $-1$. Therefore, the sum simplifies to $$ \sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}\sum_{s_3=\pm1} e^{-{s_2s_3}}= \sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})\sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}. $$ By applying the same argument as above, we see that the exponent of this $e$ is one of each of $+1$ and $-1$, so we get $$ (e+e^{-1})\sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}=(e+e^{-1})\sum_{s_1=\pm1} (e+e^{-1})=(e+e^{-1})^2\sum_{s_1=\pm1}1. $$ Since there are only two values for $s_1$, we get that this simplifies to $2(e+e^{-1})^2$, which can then be simplified in terms of $\cosh$.