Prove Convolution is Commutative

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The discussion focuses 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) = (-1/2π)∫_{-π}^{π} f(t)g(t-u) dt. A participant suggests interchanging variables to show that (f*g)(u) equals (g*f)(u) but struggles with the integration limits and variable substitution. There is a suggestion to verify the integrand, proposing that it should be f(t)g(u-t) instead of f(t)g(t-u). The conversation highlights the need for clarity in the problem statement and the correct formulation of the convolution integral.
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Homework Statement



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

Define the convolution of f and g by

<br /> (f*g)(u)=(\frac{-1}{2\pi})\int_{-\pi}^{\pi}f(t)g(t-u)dt.<br />

Show that
<br /> (f*g)(u)=(g*f)(u)<br />

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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