Cauchy product of multivariate formal power series

Question

Cauchy product of two univariate formal power series is pretty straight forward. If

$$A=\sum_{i=0}^\infty a_i x^i \, ,$$ and $$B=\sum_{j=0}^\infty b_j x^j \, ,$$ then $$A \times B = \sum_{k=0}^\infty \left(\sum_{l=0}^k a_l b_{k-l} \right)x^k \, ,$$

where $\times$ is the Cauchy product.

How can we calculate the Cauchy product of two multivariate infinite power series of the form:

$$\Phi=\sum_{i_1=0}^\infty \ldots \sum_{i_n=0}^\infty \phi_{i_1,\ldots,i_n}\prod_{j=1}^n x_j^{i_j} \, .$$

P.S.

I will try to answer my question here, maybe it will help to understand the form of solution I'm looking for. Consider $\boldsymbol{\phi}_{p_1,\ldots,p_n}$ and $\boldsymbol{\psi}_{q_1,\ldots,q_n}$ to be n dimensional arrays (consider NumPy's ndarrays for example). Also consider the operators

1. $\boldsymbol{\phi}^F$ to be a flipped version of $\boldsymbol{\phi}$ in all directions from 1 to n.
2. $\langle \boldsymbol{\phi}, \boldsymbol{\psi} \rangle_\mathrm{F}$ is the Frobenius inner product of two nD arrays of the same shape
3. $\left\{ \boldsymbol{\phi} \right\}_{k_1,\ldots,k_n}$ is a slice of the nD array from ${0,\ldots,0}$ to ${k_1,\ldots,k_n}$

then

$$\Omega=\Phi \times \Psi \equiv \boldsymbol{\omega}$$

I think the solution be of the form:

$$\omega_{k_1,\ldots,k_n}=\langle \left\{ \boldsymbol{\phi'} \right\}_{k_1,\ldots,k_n}, \left\{ \left\{ \boldsymbol{\psi'} \right\}_{k_1,\ldots,k_n}\right\}^F \rangle_\mathrm{F}\, , \, \, 0\leq k_j \leq o_j=p_j+q_j-1 $$

Answer 1

Let $$\Phi=\sum_{i_1=0}^\infty \ldots \sum_{i_n=0}^\infty \phi_{i_1,\ldots,i_n}\prod_{j=1}^n x^{i_j}, \\ \Psi=\sum_{i_1=0}^\infty \ldots \sum_{i_n=0}^\infty \psi_{i_1,\ldots,i_n}\prod_{j=1}^n x^{i_j} .$$ Then $$ \Phi \cdot \Psi= \sum_{i_1,i_2,\ldots,i_n=0}^\infty \left( \sum_{\{a_s+b_s=i_s \}}\phi_{a_1,\ldots,a_n} \psi_{b_1,\ldots,b_n}\right) \prod_{j=1}^n x^{i_j} $$

Answer 2

Assuming $\Phi_1$ has $m$ multivariate terms while $\Phi_2$ has $n$ multivariate terms, let

$$A_p = \left\{(i_1,\ldots, i_m,j_1,\ldots, j_n) \in \mathbb{N}^{m+n} \mid \sum_{k=1}^m i_k + \sum_{k=1}^n j_k = p\right\}$$

This would be the set of all $m+n$-tuples that add up to $p$ where each component is at least zero (basically, the set of all integral solutions to the associated Diophantine equation).

Then, the product:

$$\Phi_1\times \Phi_2 = \sum_{k=0}^\infty \sum_{(i_1,\ldots, i_m,j_1,\ldots, j_n)\in A_k}\phi_{i_1,\ldots, i_m}\phi_{j_1,\ldots, j_n}x^k$$

Edit: Actually, there is no reason to do this the way I was trying to do it. I can use your original notation and it should work fine.

$$\sum_{i_1=0}^\infty\cdots \sum_{i_n=0}^\infty \sum_{j_1=0}^\infty \cdots \sum_{j_n=0}^\infty \phi_{i_1,\ldots, i_n}\psi_{j_1,\ldots, j_n}\prod_{k=1}^n x^{i_k+j_k}$$