Math factoid of the day

Here's an odd tidbit that I came up with: a way to write a multinomial expansion with all coefficients equal to 1. It turns out this factorization was not very useful for my particular problem, but I thought maybe it could be useful to someone, so I'm posting it here.

First, a quick bit of background math. An ordinary binomial is just a sum of two variables raised to some power, $(a+b)^N$. The binomial theorem is a way of calculating the coefficients of the series expansion of a binomial,
\[ (a+b)^N = \sum_{i=0}^N \frac{N!}{i!(N-i)!} a^i b^{N-i}, \]
where $\frac{N!}{i!(N-i)!}$ is called the $i^{\text{th}}$ binomial coefficient. There are well-known ways of writing similar sums for longer multinomials, and obtaining the cofficients of the sum that way.

I had an unusual problem involving a sum that was kind of like a multinomial sum, but without the coefficients. That is, in the two-variable (I guess it would be called 'unit-coefficient binomial?') case, I had a sum of the form
\[ \sum_{i=0}^N a^i b^{N-i}, \]
which is on first glance actually a simpler sum, but has the great disadvantage that it can't be easily condensed into a binomial $(a+b)^N$. Since I was ultimately interested in taking derivatives of combinations of these types of sums, I thought it would be nice to have a way of expressing the sum more compactly.

There probably is a more efficient way of doing this, but the method I came up is to rewrite the sum as a product of relatively simple matrices. It seems to work for arbitrarily long multinomials raised to any power. It's probably easiest to explain by example. For instance, consider a cubed unit-coefficient binomial (that is, the series expansion of $(a+b)^3$ with all coefficients equal to one),
\[ a^{3} + a^{2} b + a b^{2} + b^3 = \sum_{i=0}^3 a^i b^{3-i}. \]
For some reason, my mind tends to try and turn any math problem into an exercise in linear algebra. So naturally, I noticed that this sum may be rewritten as an inner product of vectors, with elements given by powers of $a$ and $b$:
\[ \sum_{i=0}^3 a^i b^{3-i} = \begin{bmatrix} a^3 & a^2 & a & 1 \end{bmatrix} \begin{bmatrix} 1 \\ b \\ b^2 \\ b ^ 3 \end{bmatrix}. \]
This works for higher powers than three; the only difference is that the vectors are longer. This immediately suggests that a unit-coefficient trinomial (here, $(a+b+c)^2$ with all coefficients equal to one),
\[ a^{2} + a b + b^{2} + a c + b c + c^2 = \sum_{i=0}^2 \sum_{j=0}^i a^j b^{i-j} c^{2-i}, \]
may be rewritten by inserting a triangular Toeplitz matrix between the two vectors:
\[ \sum_{i=0}^2 \sum_{j=0}^i a^j b^{i-j} c^{2-i} = \begin{bmatrix} a^2 & a & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ b & 1 & 0 \\ b^2 & b & 1 \end{bmatrix} \begin{bmatrix} 1 \\ c \\ c^2 \end{bmatrix}. \]
In this case, the unit-coefficient trinomial is squared (that is, the upper limit on the sum is $2$), so it is a $3 \times 3$ matrix; in general, the size of the Toeplitz matrix would be $(N+1) \times (N+1)$, and each vector would also of course be $N+1$ dimensional.

This factorization holds generally for unit-coefficient multinomial expansions. For example, the cube of the unit-coefficient quadrinomial $\left( a + b + c + d \right)^3$ factors into a pair of $4 \times 4$ matrices:
\[ \sum_{i=0}^3 \sum_{j=0}^i \sum_{\ell =0}^{3-i} a^j b^{i-j} c^{\ell} d^{3-i-\ell} = \begin{bmatrix} a^3 & a^2 & a & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ b & 1 & 0 & 0 \\ b^2 & b & 1 & 0\\ b^3 & b^2 & b & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ c & 1 & 0 & 0 \\ c^2 & c & 1 & 0\\ c^3 & c^2 & c & 1 \end{bmatrix} \begin{bmatrix} 1 \\ d \\ d^2 \\ d^3 \end{bmatrix}. \]
For larger multinomials, each extra term in the unit-coefficient multinomial results in an extra Toeplitz matrix in the above product, with the dimensions of the matrices and vectors set by upper limit on the sum (as before). No idea if this factorization strategy is useful (turns out in my case it wasn't), but I thought it was sort of neat regardless...