Expression of coefficients of a product of Dirichlet polynomials

Question

Suppose we have two Dirichlet polynomials:

$$ f_1(s) = \sum_{n=1}^{m} \frac{a_n}{n^s} \\ f_2(s) = \sum_{n=1}^{m} \frac{b_n}{n^s} $$

Their product will also be a Dirichlet polynomial:

$$ f_1(s)f_2(s) = \sum_{n=1}^{m}\sum_{k=1}^{m} \frac{a_n b_k}{(nk)^s} = \sum_{n=1}^{m^2} \frac{c_n}{n^s}. $$

Is there a general way to express $c_n$ through $a_n$ and $b_n$? It appears that a variant of Dirichlet convolution can be used, that is,

$$ c_n = \sum_{d:\: d|n,\: d \leq m, \: n/d \leq m} a_d b_{\frac{n}{d}} $$

if $n \in \{lk : l=\overline{1,m}, k=\overline{1,m}\}$, and $c_n = 0$ otherwise, but such description looks too unwieldy and complex. Is there a simpler way?

Answer 1

Note: There is no simpler way compared to what you have already indicated. But, in fact it's not more complicated than the presumably more familiar Cauchy product of polynomials. Let's make a comparison head-to-head:

We consider two Dirichletpolynomials \begin{align*} f_1(s)=\sum_{n=1}^m\frac{a_n}{n^s} \qquad \text{ and } \qquad f_2(s)=\sum_{n=1}^m\frac{b_n}{n^s}\qquad\qquad (m\geq 1) \end{align*} Multiplication gives \begin{align*} f_1(s)f_2(s)&=\left(\sum_{k=1}^m\frac{a_{k}}{k^s}\right)\left(\sum_{l=1}^m\frac{b_{l}}{l^s}\right)\\ &=\sum_{n=1}^{m^2}\left(\sum_{{1\leq k,l\leq m}\atop{k\cdot l=n}}a_{k}b_{l}\right)\frac{1}{n^s}\\ &=\sum_{n=1}^{m^2}\left(\sum_{{k=1}\atop{k|n}}^{m}a_kb_{\frac{n}{k}}\right)\frac{1}{n^s} \end{align*}

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Now we consider two polynomials of degree $m\geq 1$ \begin{align*} h_1(z)=\sum_{n=1}^mc_mz^n \qquad \text{ and } \qquad h_2(z)=\sum_{n=1}^md_mz^m \end{align*} and multiplication gives \begin{align*} h_1(z)h_2(z)&=\left(\sum_{k=1}^mc_kz^{k}\right)\left(\sum_{l=1}^md_lz^{l}\right)\\ &=\sum_{n=1}^{2m}\left(\sum_{{1\leq k,l\leq m}\atop{k+ l=n}}c_kd_l\right)z^n\\ &=\sum_{n=1}^{2m}\left(\sum_{k=1}^{m}c_kd_{n-k}\right)z^n \end{align*}