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I have a problem with this calculation. I don't understand the step from the second to the third line in the calculation. I just can't bring out the $(x^{n}-1)$. It is always $x^n$ when I calculate it? Could anyone show me how to calculate this step?

Thanks!

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mannequin
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    Just recall that $c_k = \sum\limits_{i+j \equiv k \bmod n} f_i g_j$. So you actually have to include both the sums over $i+j=k$ and $i+j=k+n$. A quick $\sum_{i+j=k+n}f_ig_j - \sum_{i+j=k+n}f_ig_j$ inside the parentheses and you have your $(x^n-1)$. – Hans Sep 24 '16 at 20:23
  • I kind of don't understand the modulo part of $c_{k}$ in the sum. Or to be more precise I don't understand the notation of the sum with no upper bound and a lower bound that includes a modulo operation. Could you shed some light on this? Thanks for any help! – mannequin Sep 24 '16 at 22:53
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    Basically, the notation with no upper bound is just a handy way to say "sum over everything that verifies that given relation". Just like in the sum $\sum_{i+j=k}f_ig_j$ where we consider every $(i,j)$ so that $i+j=k$, in the sum $\sum\limits_{i+j \equiv k \bmod n} f_i g_j$ we consider every $(i,j)$ that verifies $i+j \equiv k \bmod n$. Now, recall that $i+j \equiv k \bmod n \iff i+j = k + qn \ \mathrm{for} \ q \in \mathbb{Z}$. In this situation, we also have $0 \leq i,j \leq n-1$ and $0 \leq k \leq n-1$, so $i+j=k$ and $i+j=k+n$ are the only cases to consider. – Hans Sep 24 '16 at 23:05
  • OK. The way I wrote it down is $\sum_{k=0}^{n}\sum_{j=0}^{k}(a_{k-j}+a_{k+n+j})b_{j} + (x^{n}-1) \dots$. I wonder if there is any geometric interpretation of the result $f*g \equiv fg \mod x_{n}-1$. I feel like it would be an intuitive way to think of the convolution process. According to Stone-Weierstraß every continuous function defined on a closed interval can be approximated by a polynomial. Then the convolution would be just a natural way to do multiplication on functions. Does my train of thought make any sense? – mannequin Sep 25 '16 at 17:13

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