Confusing multiplication of Fomal Power Series

Question

I have the following product I want to multiply:

$$\left(\frac{1}{1!}+\frac{z^3}{4!}+\frac{z^6}{7!}+\frac{z^9}{10!}+...\right)\left(\frac{A(0)}{0!}+\frac{A(1)z}{1!}+\frac{A(2)z^2}{2!}+\frac{A(3)z^3}{3!}+...\right)$$

If I multiply the first few terms I get

$$\frac{A(0)}{0!1!}+\frac{A(1)}{1!1!}z+\frac{A(2)}{2!1!}z^2+\left(\frac{A(0)}{0!4!}+\frac{A(3)}{3!1!}\right)z^3+\left(\frac{A(1)}{1!4!}+\frac{A(4)}{4!1!}\right)z^4$$ $$+\left(\frac{A(2)}{2!4!}+\frac{A(5)}{5!1!}\right)z^5+\left(\frac{A(0)}{0!7!}+\frac{A(3)}{3!4!}+\frac{A(6)}{6!1!}\right)z^6+...$$

I'm trying to get this into a double summation but with the periodicity of 3 here, I"m having trouble getting the form down. I'm thinking I can use a floor function in the bound, so something of the form

$$\sum_{n=0}^\infty\sum_{k=0}^{\lfloor\frac{n}{3}\rfloor}\frac{A(k)}{k!(*)!}z^n$$

I'm having an issue nailing down the $(*)$. I think it is the bound of the inner sum plus 1. Is there a way of doing this without the floor function perhaps using modular arithmetic or mod notation instead?

Answer 1

Here is a variation which uses the floor function only as upper limit of the inner sum. We apply the Cauchy product formula of two power series.

We obtain \begin{align*} \left(\sum_{k=0}^\infty\right.&\left.\frac{z^{3k}}{(3k+1)!}\right)\left(\sum_{j=0}^\infty A(j)\frac{z^j}{j!}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{{3k+j=n}\atop{k\geq 0, j\geq 0}}\frac{1}{(3k+1)!}\frac{A(j)}{j!}\right)z^n\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{{n-3k\geq 0}\atop{k\geq 0}}\frac{1}{(3k+1)!}\frac{A(n-3k)}{(n-3k)!}\right)z^n\tag{2}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{3}\right\rfloor}\frac{1}{(3k+1)!}\frac{A(n-3k)}{(n-3k)!}\right)z^n\tag{3}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{3}\right\rfloor}\binom{n}{3k+1}\frac{A(n-3k)}{n-3k}\right)\frac{z^n}{n!}\tag{4}\\ \end{align*}

Comment:

• In (1) we apply the Cauchy product formula of power series. Since the left series provides contributions of the form $z^{3k}$ and the right series contributions of the form $z^j$ we respect this by the condition $3k+j=n$. We also have to state that both, $k$ and $j$ are non-negative.

• In (2) it is more convenient to replace $j$ than to replace $k$. This way we substitute \begin{align*} j=n-3k \end{align*} and since $j\geq 0$ we have to state now $n-3k\geq 0$.

• In (3) we do just a reformulation of the conditions for the index $k$, since we have \begin{align*} 0\leq k\leq \left\lfloor\frac{n}{3}\right\rfloor \end{align*}

• In (4) we use the identity \begin{align*} \binom{n}{3k+1}=\frac{n!}{(3k+1)!(n-3k-1)!} \end{align*} and get the standard representation $$\sum_{n=0}^\infty a_n\frac{z^n}{n!}$$ for exponential generating functions.