Let $\mathbb{R}_k[x_1, \dots x_n]$ denote the ring of polynomials of degree $k=0, 1 \dots d$.
A polynomial is said homogeneous if the non-zero terms all have the same degree. Any homogeneous polynomial can be written as
$$p(x) = \sum_{i_1, \dots , i_k =1}^n \textbf{W}_{i_1, \dots, i_k}\, x_{i_1} \dots x_{i_k}$$
where $\textbf{W}_{i_1, \dots , i_k}$ is its coefficient tensor. I'm having some issues to fully understand these tensor generalization of polynomial coefficient.
if I consider $p \in \mathbb{R}[x_1, \dots , x_n]$ I usually write it as
$$a_nx^{n} + a_{n-1}x^{n-1} + \dots + a_1x + a_0\, \, \, , \, a_i \in \mathbb{R}$$.
So how should I generalize the coefficients belonging to a higher order tensor? I think I'm getting confused between the rank of the tensor and the degree of polynomial...
I'm trying to do an example, let's say that $k=2$ and $n=3$ so we have
\begin{align} \sum_{i_1, i_2=1}^3 W_{i_1i_2}x_{i_1}x_{i_2}&= \sum_{i_1=1}^3\left(W_{i_11}x_{i_1}x_{1} + W_{i_12}x_{i_1}x_{2}+W_{i_13}x_{i_1}x_{3}\right)\\ &=W_{11}x_{1}x_{1} + W_{12}x_{1}x_{2}+W_{13}x_{1}x_{3}+W_{21}x_{2}x_{1} + W_{22}x_{2}x_{2}+W_{23}x_{2}x_{3} \end{align}
So in this case $\textbf{W} \in \mathbb{R}^{2 \times 3}$ so it is a 2-rank tensor while $n$ is the degree of polynomial.. but how does this thing work more in detail? ^^
Should I assume now that the degree of each monomial is 3?
Why the $\textbf{W}$ shape results in being tensor rank $\times$ polynomial degree?
Many thanks,
James