Compute
$$\text{const} \left( \frac{1}{(1-z_1z_2)(1-z_1^2z_2)(1-z_1)(1-z_2)z_1^{3t}z_2^{2t}} \right) $$
The answer should be a polynomial in $t$. This is exercise 2.38 from Beck's Computing the Continuous Discretely, and the correct answer is
$$\frac{7}{4} t^2 + \frac{5}{2}t + \frac{7+(-1)^t}{8} $$
I'm honestly at a loss here. Can someone please explain to me how to compute the constant term here? What about a situation where we have a greater number of variables and products of functions?
This is closely related to an earlier post of mine at Finding Constant Term in Product of Series, which still doesn't have an answer. I would greatly appreciate help on either or both of these problems, and would greatly appreciate a general procedure for doing this. Thank you so much.
I don’t know of any real short cuts for attacking this. I think you just need to bash it out one variable at a time.
Starting from gt6989b’s answer, we’re trying to find $[x^{3t}y^{2t}] F(x,y)$ with $g(x,y)$ your generating function (replacing $z_1$ and $z_2$ with $x$ and $y$ to reduce visual clutter) and $F(x,y)=x^{3t}y^{2t}g(x,y)$. Starting with $y$ because it looks like it’ll produce fewer terms, compute the partial fraction expansion with respect to $y$, treating $x$ as a constant: $$\frac1{(1-xy)(1-x^2y)(1-x)(1-y)} = {1 \over (1-x)^3(1+x)(1-y)} - {x \over (1-x)^3(1-xy)} + {x^3 \over (1-x)^3(1+x)(1-x^2y)}$$ and extract the required coefficient of $y$: $$\begin{align} [x^{3t}y^{2t}] F(x,y) &= [x^{3t}]{1 \over (1-x)^3(1+x)} - [x^{3t}]{x^{2t+1} \over (1-x)^3} + [x^{3t}]{x^{4t+3} \over (1-x)^3(1+x)} \\ &= [x^{3t}]{1 \over (1-x)^3(1+x)} - [x^{t-1}]{1 \over (1-x)^3} + [x^{-(t+3)}]{1 \over (1-x)^3(1+x)}. \end{align}$$
Having eliminated $y$, we can now proceed with $x$ in the usual manner. Since $t\gt0$ we can discard the last term. The second term can be evaluated immediately: it’s equal to $-\binom{-3}{t-1} = -\binom{2+t-1}2 = -\frac12t(t+1)$. For the first term, compute the partial fraction expansion and extract coefficients. The resulting sum simplifies to $\frac74t^2+\frac52t+\frac{7+(-1)^t}8$ as required.