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Give two polynomials with finite degree - I wonder if a multiplication of both polynomials is the same as the continous convolution of both polynomials? Can anybody shed some light on this?

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2 Answers 2

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To elaborate on qbert's answer a bit, suppose you had two polynomials $f$ and $g$ given by

$$ f(x) = \sum_{i=0}^m a_i x^i, \quad g(x) = \sum_{j=0}^n b_j x^j.$$

Then their product is

$$ f(x)g(x) = \sum_{i=0}^m \sum_{j=0}^n a_i b_j x^i x^j.$$

If we wish to write this in standard form (i.e. as a sum of single powers of $x$), then we need to consider powers of degree $i+j = k$. Rewriting slightly,

$$ f(x)g(x) = \sum_{i=0}^m \sum_{j=0}^n a_{k-j} b_j x^k.$$

$k$ ranges from $0$ to $m+n$, but now our sums are coupled since $j$ depends on $k$. This dependence is exactly that $j$ ranges from $0$ to $k$. The easy way to see this is just by writing out a few cases, but logically it follows because $a_0 x^0 b_k x^k$ results in a power of $k$. Thus

$$ f(x)g(x) = \sum_{k=0}^{m+n}\left( \sum_{j=0}^k a_{k-j} b_j\right) x^k.$$

The term inside the parentheses is the discrete convolution of the coefficients.

The fact that convolution shows up when doing products of polynomials is pretty closely tied to group theory and is actually very important for the theory of locally compact abelian groups. It provides a direct avenue of generalization from discrete groups to continuous groups. The discrete convolution is a very important aspect of $\ell^1$, which makes it clear that a lot of the setting for LCA groups is in $L^1$.

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    $\begingroup$ the formula is not accurate, if k>m>n then the sum start from j=max(0,k-m) $\endgroup$ Nov 18, 2018 at 6:24
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It is procedurally the same as discrete convolution, which I think is what you mean.

The Cauchy products is generalizing multiplication of polynomials to multiplication of power series.

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  • $\begingroup$ The discrete convolution just takes place with the coefficients when multiplying two polynomials not the polynomial as a whole if that makes sense. Why I am asking this question is - I recently tried to understand convolution in a more motivated way. Then I noticed that when multiplying polynomials the coefficients do a discrete convolution. Every function on a closed interval can be described by a polynomial. My idea was that convolution is a really natural way of multiplying functions and the motivation comes from multiplying polynomials. $\endgroup$
    – mannequin
    Sep 22, 2016 at 22:07

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