Prove Convolution is Commutative

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SUMMARY

The discussion centers on proving the commutativity of convolution for two continuous, periodic functions \( f \) and \( g \) defined on the interval \([-π, π]\). The convolution is defined as \((f*g)(u) = \left(\frac{-1}{2\pi}\right) \int_{-\pi}^{\pi} f(t) g(t-u) dt\). Participants suggest interchanging variables to demonstrate that \((f*g)(u) = (g*f)(u)\), with a focus on the correct formulation of the integrand, which should be \( f(t)g(u-t) \) instead of \( f(t)g(t-u) \).

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  • Understanding of convolution operations in functional analysis
  • Knowledge of periodic functions and their properties
  • Familiarity with integral calculus, specifically definite integrals
  • Ability to manipulate variables within integrals
NEXT STEPS
  • Study the properties of convolution in functional analysis
  • Learn about periodic functions and their applications in signal processing
  • Explore variable substitution techniques in integral calculus
  • Investigate the implications of commutativity in convolution operations
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Students and educators in mathematics, particularly those focusing on functional analysis, signal processing, and integral calculus. This discussion is beneficial for anyone seeking to understand the properties of convolution and its applications.

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Homework Statement



Let f,g be two continuous, periodic functions bounded by
[tex] [-\pi,\pi] [/tex]

Define the convolution of f and g by

[tex] (f*g)(u)=(\frac{-1}{2\pi})\int_{-\pi}^{\pi}f(t)g(t-u)dt.[/tex]

Show that
[tex] (f*g)(u)=(g*f)(u)[/tex]

The Attempt at a Solution



I think the way I'm supposed to do this is by interchanging variables, but I'm stuck. If I let k=t-u and try to switch the variables around, I end up with (-1/2pi) times the integral of g(k)f(k+u)dk. Am I doing this wrong? Is there a better way to solve this?
 
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can you check the integral of f(t)g(t-u) or f(t)g(u-t)
 
Are you sure you stated the problem correctly? Shouldn't the integrand in the convolution be f(t)g(u-t)? That might help.
 

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