How would somebody walk through this equation? I'm looking for $q(n)$. If I'm given an input of 10, for example, how would this play out? The two equals signs is throwing me off. Does the result of the far right equation serve as input to the middle equation? If so, where?
I haven't "mathed" in a very long time. Any help and patience would be BEYOND appreciated.
$$\sum_{n=0}^\infty q(n) x^n = \prod_{k=1}^\infty (1 + x^k) = \prod_{k=1}^\infty \frac{1}{1-x^{2k-1}}$$
This is a link to where I found this equation on Wikipedia => https://en.wikipedia.org/wiki/Partition_(number_theory)
The "two equal signs" are saying that all 3 expressions are the same. As a simpler example, consider the following statement:
$$x^2 + 2x + 1 = (x+1)^2 = (x+1)(x+1)$$
We are saying that all 3 of the above are the same.
Now, as for actually computing the $q(n)$, you have two options. You can expand out $\prod (1+x^k)$ or you can expand out $\prod (1 - x^{2k-1})^{-1}$. Either will give you the right answer.
As an example, notice
$$(1+x)(1+x^2)(1+x^3) = 1 + x + x^2 + 2x^3 + x^4 + x^5 + x^6$$
Also notice $q(0) = 1$, $q(1) = 1$, $q(2)=1$, and $q(3)=2$. So the first few coefficients are exactly the partition numbers!
The reason the $x^4$ coefficient is $1$ instead of $2$ is because we haven't used enough terms! If we multiply by $(1+x^4)$ then we find the correct coefficient. Indeed, if we were to multiply all of the $(1+x^k)$ instead of stopping at some finite point, then the coefficients would be correct for every $n$.
For fun, you should (by hand) compute
$$\prod_{k=1}^7 (1+x^k)$$
that is, you should actually expand out
$$(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)(1+x^7)$$
You'll find that the first 7 coefficients of $x^n$ is exactly $q(n)$! The idea is that multiplying by $(1+x^8)$ cannot possibly impact the $x^7$ term. Do you see why?
So, to actually do this in "the real world" you would get a computer to do this for you. A computer algebra system (like sage) will happily tell you the $nth$ coefficient of $\prod(1+x^k)$ by doing the annoying foiling faster than you or I ever could! There are plenty of tutorials online, but if you want me to say how to do it here, feel free to ask and I'll be happy to.
I hope this helps ^_^