Who has hint how to prove: $\sum_{n=0}^N \sum_{k=1}^N g_k y_n = \sum_{n=1}^N \sum_{k=1}^n g_k y_{n-k}+\sum_{n=1}^N\sum_{k=1}^n g_n y_{N-k+1}$

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So is $g_{k}$ a sequence, and $y_{n}(x)$ a sequence in $n$, given some parameter $x$? –  Daniel Littlewood Nov 20 '12 at 20:42
Now is better. x is not important. I want to prove the summation. –  TheStudent Nov 20 '12 at 20:59
I also presume it should be $\sum_{n=1}^{N} \sum_{k=1}^{n} g_{k}y_{n-k}$? –  Daniel Littlewood Nov 20 '12 at 21:04

It's almost always helpful to expand both sides to see where all the terms come from:

LHS: We can factor out $g_{k}$ from the inner sum, since it is independent of both $n$ and $N$. Thus, we have $g_{1}(y_{0}+y_{1}+\ldots+y_{N})+g_{2}(y_{0}+\ldots+y_{N})+\ldots+g_{N}(y_{0}+\ldots +y_{N})$, which we can easily see is $$y_{0}(g_{1}+g_{2}+\ldots+g_{N})+y_{1}(g_{1}+\ldots+g_{N})+\ldots+y_{N}(g_{1}+\ldots+g_{N})$$

RHS: I think your first sum is incorrect; it should read $\sum_{n=1}^{N} \sum_{k=1}^{n} g_{k}y_{n-k}$. If we expand this first sum, where each set of square brackets contains the next value for $n$ (the first set for $n=1$, the next for $n=2$, etc.), we get the following:$[g_{1}y_{0}]+[g_{1}y_{1}+g_{2}y_{0}]+[g_{1}y_{2}+g_{2}y_{1}+g_{3}y_{0}]+\ldots+[g_{1}y_{N-1}+g_{2}y_{N-2}+\ldots g_{N}y_{0}]$
Which, after rearranging, is equal to $y_{0}(g_{1}+g_{2}+\ldots+g_{N})+y_{1}(g_{1}+g_{2}+\ldots+g_{N-1})+\ldots y_{N-1}g_{1}$
Before expanding the second sum, let's see what we have. Each of the $y_{k}$ Has $N-k$ of the terms we want it to be multiplied by. So $y_{0}$ is being multiplied by all the $g_{k}$ we require, $y_{1}$ by all but $g_{N}$, and so on. Let us move onto the second sum.

Expanding similarly, $[g_{1}y_{N}]+[g_{2}y_{N}+g_{2}y_{N-1}]+\ldots+[g_{N}y_{N}+g_{N}y_{N-1}+\ldots g_{N}y_{1}]$. By rearranging, we have $y_{1}g_{N}+y_{2}(g_{N-1}+g_{N})+\ldots+y_{N}(g_{1}+g_{2}+\ldots+g_{N})$

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My proof was not as elegant as proof of @Daniel Littlewood. It was very "flowery" but I think it is intelligible. –  Elias Nov 20 '12 at 22:40
Thank you! :-) I like your proof too. –  Daniel Littlewood Nov 21 '12 at 20:33

We have \begin{align} \sum_{k=1}^N \sum_{n^\prime=0}^N g_k y_n & = \sum_{\substack{1\leq n\leq N\\ 1\leq k \leq N}} g_k y_n \\ & = \sum_{(k,n)\in G} g_k y_n \\ \end{align}

Here $G=\{ (k,n) \in\mathbb{N}\times\mathbb{N} : 1\leq n\leq N, 1\leq k \leq N\}$. Set \begin{align} G_1 &=\{ (k,n) \in\mathbb{N}\times\mathbb{N} : 1\leq n\leq k, 1\leq k \leq N, \} \\ & \\ G_2 &= \{ (k,n) \in\mathbb{N}\times\mathbb{N} : k+1\leq n\leq N, 1\leq k \leq N\}.\\ \end{align}

Then $G=G_1\cup G_2$, $G_1\cap G_2=\emptyset$ and \begin{align} \sum_{k=1}^N \sum_{n=0}^N g_k y_n & = \sum_{(k,n)\in G} g_k y_n \\ & = \sum_{(k,n)\in G_1} g_k y_n + \sum_{(k,n)\in G_2} g_k y_n \\ & = \sum_{\substack{1\leq n\leq k\\ 1\leq k \leq N}} g_k y_n + \sum_{\substack{k+1\leq n\leq N\\ 1\leq k \leq N}} g_k y_n \\ & = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n\leq k} g_k y_n + \sum_{1\leq k \leq N}\;\;\sum_{k+1\leq n\leq N} g_k y_n \\ & = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n\leq k} g_k y_n + \sum_{1\leq k \leq N}\;\;\sum_{1\leq n-k\leq N-k} g_k y_n \\ \end{align}

Making the "change of variables" $n=n^\prime+k$: \begin{align} \sum_{k=1}^N \sum_{n=0}^N g_k y_n & = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n\leq k} g_k y_n + \sum_{1\leq k \leq N}\;\;\sum_{1\leq n^\prime \leq N-k} g_{k^\prime} y_{n^\prime+k} \\ \end{align}

The change in notation $n=n^\prime$ does not interfere in the value of the sum we now \begin{align} \sum_{k=1}^N \sum_{n^\prime=0}^N g_k y_n & = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n^\prime\leq k} g_k y_{n^\prime} + \sum_{1\leq k \leq N}\;\;\sum_{1\leq n^\prime \leq N-k} g_{k^\prime} y_{n^\prime+k} \\ \end{align}

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