Double Summation indexes problem I have the following sum:
\begin{equation}
\sum_{j=0}^{a}
    \sum_{k=0}^{n-2j} c_{jk}\,\,
    x^{\,j+k}
\end{equation}
Where $a=\lfloor n/2\rfloor$. I want to convert the previous sum to other like:
\begin{equation}
\sum_{l=0}^n a_lx^l
\end{equation}
but I have been struggling in the intent to writte the coefs. $a_l$ in term of the coefs $c_{ij}$. Any clue how to solve this problem will be highly appreciated.
 A: Define two sets of pairs of natural numbers to serve as indices for summation:
\begin{gather*}
K = \{ (j, k) \colon j \geqslant 0, \ 0 \leqslant k \leqslant n - 2j \}, \\
L = \{ (j, l) \colon 0 \leqslant j \leqslant \min\{l, n - l\} \}.
\end{gather*}
Verify that these two functions are well defined, and are inverse to each other:
\begin{gather*}
\lambda \colon K \to L, \ (j, k) \mapsto (j, j + k), \\
\kappa \colon L \to K, \ (j, l) \mapsto (j, l - j), \\
\kappa \circ \lambda = 1_K, \\
\lambda \circ \kappa = 1_L.
\end{gather*}
Write the function to be summed as:
$$
f \colon K \to \mathbb{C}, \ (j, k) \mapsto c_{jk}x^{j+k}.
$$
Then:
\begin{align*}
\sum_{j=0}^{\left\lfloor n/2\right\rfloor}\sum_{k=0}^{n-2j} c_{jk}x^{j+k}
& = \sum_{(j,k)\in K}f(j, k) \\
& = \sum_{(j,l)\in L}f(\kappa(j, l)) \\
& = \sum_{(j,l)\in L}f(j,l-j) \\
& = \sum_{(j,l)\in L}c_{j,l-j}x^l \\
& = \sum_{l=0}^n\left(\sum_{j=0}^{\min\{l,n-l\}}c_{j,l-j}\right)x^l.
\end{align*}
A: I defined: $\bar{c}_{jk}=c_{jk}\,\,\theta\left(n-2j-k\right)$, where $\theta(x)$ is the step function ($\theta(x)=1$ if $x\geq0$ otherwise $\theta(x)=0$). So:
\begin{equation}
\begin{array}{c}
\displaystyle
    \sum_{j=0}^{a}
    \sum_{k=0}^{n-2j} c_{jk}\,\,
    x^{\,(j+k)}=
    \sum_{j=0}^{a}
    \sum_{k=0}^{n} \bar{c}_{jk}\,
    x^{\,(j+k)} \underset{(1)}{=}
    \sum_{j=0}^{a}
    \sum_{l=0}^{a+n} \bar{c}_{j,l-j}\theta(l-j)\,x^{\,l} \underset{(2)}{=}
    \\ \\ \displaystyle
    \sum_{l=0}^{n}\sum_{j=0}^{a} \bar{c}_{j,l-j}\theta(l-j)\,
    x^{\,l}\underset{(3)}{=}
    \sum_{l=0}^{n}\sum_{j=0}^{j_0} c_{j,l-j}
    x^{\,l}=\sum_{l=0}^{n}a_lx^{
l}\,.
\end{array}
\end{equation}
Where in step (1) I defined  $l:=j+k$. Its clear that $0\leq l \leq a+n$. In (2) If $l\geq n$ then $\theta(n-l-j)=0$. In (3) I defined $j_0:=min\{a,n-l,l
\}=min\{n-l,l
\}$ and I used that $\theta(n-l-j)=0$ if $j\geq j_0$.
