I'm trying to get the coefficient of $x^6$ of this polynomial product:
$$x^2(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4).$$
I know with infinite series, you can use the closed form solution of the series to calculate the coefficient, but I haven't been able to figure out a way to calculate the coefficient for a finite series.
I hope there is a better way than doing the algebra.
On
Computing the coefficient directly doesn’t seem all that bad to me since you’re only interested in one term of the product. Its coefficient can be found without multiplying the whole mess out. Your function is $x^2$ times some polynomial in $x$, so you just need to count the number of ways that you can choose monomials from the other three terms in the product such that the sum of their exponents is $4$. The seven ways to do that are pretty easy to find by inspection.
On
What is being asked is to find the coefficient of $x^4$ of $$(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4),$$ which, when multiplied by $x^2$ leads to the coefficient of $x^6$ overall. This can be obtained by expansion: \begin{align} P &= (1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4) \\ &= (1+x+x^2+x^3+x^4+x^5)(1+x+2 x^2 + x^3 + 2 x^4 + x^5 + x^6) \\ &= 1 + 2 x + 4 x^2 + 5 x^3 + 7 x^4 + \mathcal{O}(x^5) \end{align} Now, $x^2 \, P = x^2 + 2 x^3 + 4 x^4 + 5 x^5 + 7 x^6 + \mathcal{O}(x^7)$.
There's a closed form for a finite series, too, e.g., $$1+x+\cdots+x^5={1-x^6\over1-x}$$ So your product is $$x^2{(1-x^6)^2(1-x^3)\over(1-x)^3(1+x)}$$ which is $$x^2(1-x^3-2x^6+2x^9+x^{12}-x^{15})(1+3x+6x^2+10x^3+\cdots)(1-x+x^2-x^3+\cdots)$$ Since you're only after the coefficient of $x^6$, you can simplify this a little to $$x^2(1-x^3)(1+3x+6x^2+10x^3+15x^4)(1-x+x^2-x^3+\cdots)$$ but this doesn't look any easier than what you call "doing the algebra".