# How to find expression for this tensor?

First consider 1D case. Suppose I would like to compute \begin{alignat}{2} m_p = \sum_{x=1}^{N} x^p f(x), \qquad \forall\; p = 1, 2, \dots, P \end{alignat} I know it can be expressed as a matrix product as follows \begin{alignat}{2} \underbrace{ \begin{bmatrix} m_1 \\ m_2 \\ \vdots \\ m_P \end{bmatrix} }_{:=\, \boldsymbol{m}} = \underbrace{ \begin{bmatrix} 1^1 & 2^1 & \dots & N^1 \\ 1^2 & 2^2 & \dots & N^2 \\ \vdots & \vdots & \ddots & \vdots \\ 1^P & 2^P & \dots & N^P \end{bmatrix} }_{ :=\, X} \underbrace{ \begin{bmatrix} f(1) \\ f(2) \\ \vdots \\ f(N) \end{bmatrix} }_{:=\, \boldsymbol{f}} \end{alignat}

Now generalizing this to 2D case, I know that \begin{alignat}{2} m_{pq} = \sum_{x=1}^{N} \sum_{y=1}^{M} x^p y^q f(x, y), \qquad \forall\; p = 1, 2, \dots, P;\; q = 1, 2, \dots, Q \end{alignat} can be expressed as \begin{alignat}{2} &\underbrace{ \begin{bmatrix} m_{11} & m_{12} & \dots & m_{1Q} \\ m_{21} & m_{22} & \dots & m_{2Q} \\ \vdots & \vdots & \ddots & \vdots \\ m_{P1} & m_{P2} & \dots & m_{PQ} \end{bmatrix} }_{ :=\, M} =\\ &=\underbrace{ \begin{bmatrix} 1^1 & 2^1 & \dots & N^1 \\ 1^2 & 2^2 & \dots & N^2 \\ \vdots & \vdots & \ddots & \vdots \\ 1^P & 2^P & \dots & N^P \end{bmatrix} }_{ :=\, X} \underbrace{ \begin{bmatrix} f(1,1) & f(1,2) & \dots & f(1,M) \\ f(2,1) & f(2,2) & \dots & f(2,M) \\ \vdots & \vdots & \ddots & \vdots \\ f(N,1) & f(N,2) & \dots & f(N,M) \end{bmatrix} }_{ :=\, F} \underbrace{ \begin{bmatrix} 1^1 & 2^1 & \dots & M^1 \\ 1^2 & 2^2 & \dots & M^2 \\ \vdots & \vdots & \ddots & \vdots \\ 1^Q & 2^Q & \dots & M^Q \end{bmatrix}^{\top} }_{ :=\, Y^{\top}} \end{alignat} Now my question is how does this generalize to 3D case? That is how can I express in terms of vectors/matrices/tensors the tensor \begin{alignat}{2} \begin{pmatrix} m_{pqr} \end{pmatrix}_{p=1, q=1, r=1}^{p=P, q=Q, r=R} \end{alignat} whose elements are given as \begin{alignat}{2} m_{pqr} = \sum_{x=1}^{N} \sum_{y=1}^{M} \sum_{z=1}^{K} x^p y^q z^r f(x, y, z), \qquad \forall\; p = 1, 2, \dots, P;\; q = 1, 2, \dots, Q;\; r = 1, 2, \dots, R. \end{alignat}

1D: $m_p = X_{pi} f_i$ with $X_{pi} = i^p$, $f_i = f(i)$
2D: $m_{pq} = X_{pi} Y_{qj} f_{ij}$ with $X_{pi}$ as above, $Y_{qj} = j^q$, $f_{ij} = f(i,j)$.
3D: $m_{pqr} = X_{pi} Y_{qj} Z_{rk} f_{ijk}$ with $Z_{rk} = k^r$, $f_{ijk} = f(i,j,k)$.
For the 3D case, while the $X$, $Y$, and $Z$ objects could be written as matrices, the $m_{pqr}$ and $f_{ijk}$ are not easily expressed that way.