Convolution - prove commutative

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SUMMARY

The commutative property of convolution can be proven using the definitions of convolution for two functions, f and g. The convolution is defined as f*g(x) = ∫₀^∞ f(x-t)g(t)dt and g*f(x) = ∫₀^∞ g(x-u)f(u)du. By applying a change of variables, specifically substituting t with (x-u), one can demonstrate that f*g(x) equals g*f(x), confirming the commutative property.

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Mathematicians, signal processing engineers, and students studying advanced calculus or linear systems will benefit from this discussion on the commutative property of convolution.

benndamann33
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anyone know how to prove that it is commutative...

as if f *g = g*f
 
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Try a change of variables.
 
You might want to start with the definition: the convolution of f and g is
[itex]f*g(x)= \int_0^\infty f(x-t)g(t)dt[/itex] and, of course, [itex]g*f(x)= \int_0^\infty g(x-u)f(u)du[/itex] (I have intentionally used a different variable of integration here). Hmm, in one you have f(x-t) and in the other f(u). Does that substitution benndamann33 mentioned leap to mind?
 

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