byronsit / sqpep_solver

for sQPEP_Solver

sQPEP_Solver

The corresponding paper is currently under review. Detailed information and the full paper will be shared once the review process is complete.

Proof the Lemma

Given the description of the theorem, our objective is to demonstrate that for a polynomial function $f(x_1, x_2, \ldots, x_m) = C \prod_{i=1}^m x_i^{p_i}$, the product of its $n^{th}$ order partial derivatives with respect to all variables and $vec(S^{\otimes n})$ is equal to $n!$ times the function itself. Here, $C$ is a constant, $p_i$ are non-negative integers, and $\sum_{i=1}^m p_i = n$.

To establish this, it is essential to consider all possible $n^{th}$ order partial derivatives. For each specific set of $(k_1, k_2, \ldots, k_m)$ that satisfies $\sum_{i=1}^m k_i = n$, the $n^{th}$ order mixed partial derivative of $f$ (provided that all $k_i! \leq! p_i$) is given by:

$$ \frac{\partial^n f}{\partial x_1^{k_1} \partial x_2^{k_2} \ldots \partial x_m^{k_m}}! =! C \cdot \prod_{i=1}^m \frac{p_i!}{(p_i - k_i)!} \cdot x_i^{p_i - k_i}, $$

This is due to the fact that for each variable $x_i$, if we take the $k_i^{th}$ partial derivative, the remaining power is $p_i - k_i$, accompanied by a coefficient $\frac{p_i!}{(p_i - k_i)!}$. Should $k_i! >! p_i$, the result is zero, in accordance with the definition of partial derivatives.

Next, we turn our attention to $vec(S^{\otimes n})$. This vector comprises all elements of $S^{\otimes n}$, where $S$ is the set formed by the variables of $f$. Consequently, each element within $vec(S^{\otimes n})$ corresponds to the coefficient of a specific $n^{th}$ order partial derivative, represented by the multinomial coefficient $\binom{n}{k_1, k_2, \ldots, k_m}$. This coefficient is the number of ways to distribute $k_i$ derivatives to each $x_i$ from the $n$ derivatives. Here, the multinomial coefficient $\binom{n}{k_1, k_2, \ldots, k_m}$ is defined as $\frac{n!}{k_1! k_2! \ldots k_m!}$.

Now, we compute the $\mathbf{D}^{(n)} \cdot vec(S^{\otimes n})$:

$$ \mathbf{D}^{(n)} \cdot vec(S^{\otimes n}) = \sum_{\sum_{i=1}^m k_i = n} \left( C \cdot \prod_{i=1}^m \frac{p_i!}{(p_i - k_i)!} \cdot x_i^{p_i - k_i} \right) \cdot \left( \frac{n!}{k_1! k_2! \ldots k_m!} \right). $$

Because we have a complete permutation of $n!$, it will be multiplied by each $\frac{p_i!}{(p_i - k_i)!}$. However, to compute the dot product, we must also consider the corresponding $x_i^{k_i}$ terms in $vec(S^{\otimes n})$. It is noted that the product of these terms yields the powers of the original polynomial $f$. This is due to each term $x_i^{p_i - k_i} \cdot x_i^{k_i}$ ultimately resulting in $x_i^{p_i}$. Consequently, we obtain:

$$ \mathbf{D}^{(n)} \cdot vec(S^{\otimes n}) = C \cdot n! \cdot \prod_{i=1}^m x_i^{p_i} = n! \cdot f. eee $$