Factorize product of a power series and polynomials

Question

I have a product of an infinite power series ($e^x$ power series) with a polynomial, and I want to multiply each other to obtain the coefficients in a single infinite power series, i.e. I want to obtain the numbers $c_k$ that satisfy:

$$ \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) \left( \sum_{m=0}^{a} a_m x^m \right) = \sum_{k=0}^{\infty} c_k x^k $$

Supposing that the $a_m$ is infinite with $a_m = 0$ if $m > k$, and using the Cauchy product I get:

$$ \sum_{n=0}^{\infty} \sum_{m=0}^{n-k} \frac{x^n}{n!} a_{n-m}x^{n-m} = \sum_{n=0}^{\infty} \sum_{m=0}^{n-k} \frac{1}{n!} a_{n-m}x^{2n-m} $$

but I don't know if I can go any further. Is there any way to factorize this expression? Do I need an analytic expression for the sum $\sum\limits_{m=0}^{n-k} \frac{1}{n!} a_{n-m}x^{2n-m}$ ?

Answer 1

We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain with $k\geq 0$:

\begin{align*} \color{blue}{c_k}&=[x^k]\left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) \left( \sum_{m=0}^{a} a_m x^m \right) \\ &=\sum_{n=0}^k\frac{1}{n!}[x^{n-k}]\sum_{m=0}^a a_mx^m\tag{1}\\ &=\sum_{n=0}^k\frac{1}{(k-n)!}[x^k]\sum_{m=0}^a a_mx^m\tag{2}\\ &\,\,\color{blue}{=\sum_{n=0}^k\frac{1}{(k-n)!}a_k[[a> k]]}\tag{3}\\ \end{align*}

Comment:

• In (1) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We also set the upper limit of the outer sum to $k$ since indices greater than $k$ do not contribute to the coefficient of $x^k$.

• In (2) we change the order of summation of the outer sum $n\to k-n$ for easier selection of the coefficient in the next step.

• In (3) we select the coefficient of $x^k$. We also use the Iverson brackets to indicate that $c_k=0$ if $a>k$.