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I wanted to show the relation between convolution and polynomial multiplication like it is showed here. I saw in this thread a similar approach to mine but with same degrees for both polynomials. So what I try to show is that $$p*q \equiv pq \mod x^{n}-1 $$ but I don't really know how I can continue my calculation to get to this form (but I don't really get this form like in the threads I linked).

$ p(x)q(x) = \\ \sum_{k=0}^{m+n}(\sum_{j=0}^k a_{k-j}b_{j})x^k = \\ \sum_{k=0}^{m}(\sum_{j=0}^k a_{k-j}b_{j})x^k + \sum_{k=m}^{m+n}(\sum_{j=0}^k a_{k-j}b_{j})x^k = \\ \sum_{k=0}^{m}(\sum_{j=0}^k a_{k-j}b_{j})x^k + \sum_{k=0}^{n}(\sum_{j=m}^{k+m} a_{k-j+m}b_{j-m})x^{k+m} = \\ \sum_{k=0}^{m}(\sum_{j=0}^k a_{k-j}b_{j})x^k + x^m(\sum_{k=0}^{n-1}(\sum_{j=m}^{k+m} a_{k-j+m}b_{j-m})x^{k}+x^{n}\sum_{j=m}^{n+m}a_{n-j+m}b_{j-m}) $

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mannequin
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1 Answers1

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Pad the coefficients with zeroes. You should be aware that you need to do this to even define the convolution due to the $a_{k-j}$ terms in your sum. That is, if $\deg f = N$ and $ \deg g = N+M$, then write $f(x) = \sum_{n=0}^Na_n x^n + \sum_{n=N+1}^{N+M} 0x^n$ and use the known result.

Calvin Khor
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  • So if I have two polynomials one with degree m and the other with degree n, then I should pad the coefficients of the smaller degree polynomial with zeros until it matches the degree of the other polynomial, right? – mannequin Sep 24 '16 at 14:19
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    @mannequin that's right. – Calvin Khor Sep 24 '16 at 14:20
  • Ok thanks, I have a question about the known result, because I don't see how the person gets $(x^n)-1$ i the last line. I get $x^n$ instead. – mannequin Sep 24 '16 at 19:31
  • @mannequin the left term is not exactly $f*g$, which is why you don't just take out the right term (which gives you $x^n$). You need to add and subtract $∑_{i+j = k+n} f_i g_j $. – Calvin Khor Sep 24 '16 at 21:58