f ** g=int_0^(t) alpha e^(-beta(t−tau)) y(tau)d tau. Does anyone know how one would take the derivative of this expression? In general, are there rules for taking derivative of f ** g, for some given f,g?

drogaid1d8

drogaid1d8

Answered question

2022-11-16

Let f ( t ) = α e β t , where α , β are constants
Let g ( t ) = y ( t )
Then the resulting convolution f g is:
f g = 0 t α e β ( t τ ) y ( τ ) d τ
Does anyone know how one would take the derivative of this expression?
In general, are there rules for taking derivative of f g, for some given f , g?

Answer & Explanation

cismadmec

cismadmec

Beginner2022-11-17Added 22 answers

It suffices that one of the two functions is in L 1 and the other is in L C 1 and its derivative is in L . More generally:
If f L 1 ( R d ) and g C k ( R d ), such that D α g L ( R d ) for all | α | k, then f g C k ( R d ) and: D α ( f g ) = f D α g
Where α = ( α 1 , , α d ) N d , | α | = α 1 + + α d , and:
D α g := | α | g x 1 α 1 x d α d
This can be proved by induction on k. To prove the case k=1, we differentiate under the integral sign.
Yaretzi Mcconnell

Yaretzi Mcconnell

Beginner2022-11-18Added 3 answers

If f and g are C 1 and f , g are both integrable, then:
( f g ) = f g = f g .
Basically, whenever f g and f g are well-defined, one has the equality above.

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