Double Summation indexes problem

Question

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.

Answer 1

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*}

Answer 2

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 $j\geq j_0$.