Second-order partial derivative of $S = \sum_{k=0}^{2n}\left(\sum_{i=\max(0,k-n)}^{\min(n,k)}a_{i}a_{k-i}\right)$

Question

Given the following finite sum:

$S = \sum_{k=0}^{2n}\left(\sum_{i=\max(0,k-n)}^{\min(n,k)}a_{i}a_{k-i}\right)$

From this summation, I want to calculate explicitly each element $(i,j)$, i.e. the partial derivatives, in the Hessian (and with that find an analytical expression), $H_{i,j}$.

The Hessian is calculated in the following way: \begin{equation} \mathbf{H}=\left[\begin{array}{cccc} \frac{\partial^2 S}{\partial a_0^2} & \frac{\partial^2 S}{\partial a_0 \partial a_1} & \cdots & \frac{\partial^2 S}{\partial a_0 \partial a_n} \\ \frac{\partial^2 S}{\partial a_1 \partial a_0} & \frac{\partial^2 S}{\partial a_1^2} & \cdots & \frac{\partial^2 S}{\partial a_1 \partial a_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial^2 S}{\partial a_n \partial a_0} & \frac{\partial^2 S}{\partial a_n \partial a_1} & \cdots & \frac{\partial^2 S}{\partial a_n^2} \end{array}\right] . \end{equation}

Example for $n=2$, the result is (without using an analytical expression): \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \end{bmatrix}

I want a similar result (i.e. an analytical expression) like in this topic to give an idea: Differentiation of a double summation.

Answer 1

Note that with respect to your previous post $S$ admits due to symmetry the significant simpler representation \begin{align*} \color{blue}{S}&=\sum_{k=0}^{2n}\sum_{i=\max(0,k-n)}^{\min(n,k)}a_{i}a_{k-i}\\ &=\left(\sum_{k=0}^na_k\right)^2\\ &\,\,\color{blue}{=\sum_{k=0}^na_k^2+2\sum_{0\leq k<l\leq n}a_ka_l}\tag{1} \end{align*}

From (1) it is easily seen that \begin{align*} \color{blue}{\mathbf{H}}&= \begin{bmatrix} \frac{\partial^2 }{\partial a_r \partial a_s}S \end{bmatrix}_{0\leq r,s\leq n}\\ &= \begin{bmatrix} \frac{\partial^2}{\partial a_r \partial a_s} \left(\sum_{k=0}^na_k^2+2\sum_{0\leq k<l\leq n}a_ka_l\right) \end{bmatrix}_{0\leq r,s\leq n}\tag{1}\\ &\,\,\color{blue}{= \begin{bmatrix} 2 \end{bmatrix}_{0\leq r,s\leq n}}\\ \end{align*}

Answer 2

Take the first derivative

$$\frac{\partial S}{\partial a_m}=2\left(\sum_{k=0}^na_k\right)\sum_{k=0}^n\frac{\partial a_k}{\partial a_m}=2\left(\sum_{k=0}^na_k\right)\sum_{k=0}^n\delta_{km}=2\sum_{k=0}^na_k$$

where $\delta_{km}$ is Kronecker delta.

$$\frac{\partial^2 S}{\partial a_m\partial a_p}=2\sum_{k=0}^n \frac{\partial a_k}{\partial a_p}=2\sum_{k=0}^n\delta_{kp}=2$$