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Take:

$u(x)$ and $v(x)$ to be integer polynomials, and then interpret them as sequences in the obvious way: i.e. you put the $i$th term to be the coefficient of $x^i$. Then you'll find that $u\ast v$ is the sequence representing the coefficients of $u(x)v(x)$!

Example: $u(x)=x-2$, $v(x)=3x^2+x$. The sequences are: $$ \dots u_{-2}=u_{-1}=0;\ \ u_0=-2;\ \ u_1=1; \ \ u_2=u_3=\dots=0 $$ $$ \dots v_{-2}=v_{-1}=v_0=0;\ \ v_1=1;\ \ v_2=3;\ \ v_3=v_4=\dots=0 $$

The convoluted sequence is:

$$ \dots (v\ast u)_{-2}=(v\ast u)_{-1}=(v\ast u)_0=0;\ \ (v\ast u)_1=-2;\ \ (v\ast u)_2=-5;\ \ (v\ast u)_3=3;\ \ (v\ast u)_4=(v\ast u)_5=\dots=0 $$

And that turns out to be exactly the sequence for $u(x)v(x)=3x^3-5x^2-2x$.

Why is $ (v*u)_1=-2$? We have $v_1=1$ and $u_1=1$, and $1*1=1$.

Zev Chonoles
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user1095340
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1 Answers1

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Remember that $\ast$ denotes convolution, not multiplication. Therefore $$\begin{align*}(u\ast v)_1&=\cdots+(u_1v_{1-1})+(u_0v_{1-0})+(u_{-1}v_{1-(-1)})+\cdots\\\\ &=\cdots+(1\cdot 0)+(-2\cdot 1)+(0\cdot 3)+\cdots\\\\ &=-2\end{align*}$$

Zev Chonoles
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  • Where can I find this rule in the first place? $$\begin{align}(u\ast v)1&=\cdots+(u_1v{1-1})+(u_0v_{1-0})+(u_{-1}v_{1-(-1)})+\cdots\\ \end{align}$$ – user1095340 Mar 31 '13 at 23:24
  • That follows directly the definition of $\ast$ (look at the Wikipedia page I linked to), $$(u\ast v)n=\cdots+(u_1v{n-1})+(u_0v_{n-0})+(u_{-1}v_{n-(-1)})+\cdots$$ Didn't your teacher / book mention what they intended the symbol "$\ast$" to mean anywhere? – Zev Chonoles Mar 31 '13 at 23:34
  • yes, the wikipedia page does explain that "" means convolution, but I can't find the $(uv)_n$ trick there.. How is it called? – user1095340 Mar 31 '13 at 23:45
  • Let "$f$" represent "$u$", let "$g$" represent "$v$", and use the notation $a_n$ for a sequence instead of $a[n]$ as they do in the Wikipedia article. – Zev Chonoles Mar 31 '13 at 23:48
  • Please quote the article. – user1095340 Mar 31 '13 at 23:58
  • And is this (coefficient method) really the best way to solve the equation? – user1095340 Apr 01 '13 at 00:03
  • The exact section I linked to states that $$(f * g)[n]\ \stackrel{\mathrm{def}}{=}\ \sum_{m=-\infty}^\infty f[m], g[n - m]$$ They are simply using different letters and different notation for sequences, as I just explained. – Zev Chonoles Apr 01 '13 at 00:12
  • Also, this is ultimately exactly the same computation you would do any other way of multiplying two polynomials (the extra 0's on either side are irrelevant). Whatever you find easier is fine. – Zev Chonoles Apr 01 '13 at 00:14
  • Always have done it this way: $$ (x-2)(3x^2+x)=3x^3+x^2-2*3x^2-2x= 3x^3+x^2-6x^2-2x= 3x^3-5x^2-2x $$. Method above confused me. – user1095340 Apr 01 '13 at 00:39
  • What's the benefit of the method in the post above? – user1095340 Apr 01 '13 at 15:40