A question about Riemann sums

Question

Take: $$ (u*v)(k) = \sum_{i=-\infty}^\infty u(i)v(k-i). $$

$k$ is there, it's because you want to define $$ \ldots\ldots, (u*v)(-3), (u*v)(-2), (u*v)(-1), (u*v)(0), (u*v)(1), (u*v)(2), (u*v)(3), \ldots\ldots $$ etc. The number in the parentheses is $k$. Thus, for example, when $k=4$, we have \begin{align} (u*v)(4) = \sum_{i=-\infty}^\infty u(i)v(4-i) \end{align} $$ = \cdots\cdots+u(-3)v(7)+u(-2)v(6)+u(-1)v(5)+u(0)v(4)+u(1)v(3)+u(2)v(2) $$ $$ \phantom{={}} {}+u(3)v(1)+ u(4)v(0) + u(5) v(-1) + u(6)v(-2)+u(7)v(-3)+u(8)v(-4)+\cdots\cdots. $$

What does a notation like $u(3)v(1)$ mean? What should I do with the number 3 and the number 1?

How is this a question about Riemann integrals? – mrf Mar 10 '13 at 14:46 — mrf, Mar 10 '13 at 14:46

azimut · Accepted Answer · 2013-03-10T14:52:41.187

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$u$ and $v$ are two fixed functions. As functions, they can be evaluated. The expression $u(3)v(1)$ means: evaluate the function $u$ at $3$, evaluate the function $v$ at $1$, and then multiply those two results.

Example: if $u(x) = x^2$ and $v(x) = x+5$, then $u(3)v(1) = 3^2\cdot(1+5) = 54$.

edited Mar 10 '13 at 14:52

answered Mar 10 '13 at 14:44

azimut

22,696

How? What whould be the result? – user1095340 Mar 10 '13 at 14:44
The result depends on the definition of the functions $u$ and $v$, of course. – azimut Mar 10 '13 at 14:45
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I've added an example. – azimut Mar 10 '13 at 14:46
Check Michael Hardy's answer. He says "The two functions u and v are both there" I guess I suffer a very rare type of math blindness ;) http://math.stackexchange.com/questions/323000/convolution-of-integer-sequences – user1095340 Mar 10 '13 at 14:49
Lets take $u(x)=x-2$, $v(x)=3x^2+x$ – user1095340 Mar 10 '13 at 17:51
Yeah, feel free to repeat this with any pair of functions you like :) – azimut Mar 10 '13 at 18:05
Do those functions converge? Is it possible to get any value out of it? Or do they just diverge? – user1095340 Mar 10 '13 at 18:07
I don't see the connection, is this a new question? If yes, please open a new question for it (and finish this one by accepting my answer). – azimut Mar 10 '13 at 18:09

A question about Riemann sums

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