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trusure
Newbie  Gender: Posts: 24 fourier transform  of distributions   « on: Mar 16th, 2010, 1:57pm » Quote Modify

the Fourier transform of a derivative is:
(d/dx(f))^ = i s (f^)(s)

Now I have the function
f(t)={ sin(t)  , if -pi/2 <=t<=pi/2
0        , else
}

how to prove that
(d/dx(f))^ = i s (f^)(s)

in distribution sense; I mean applying the notation
<f^, phi> = <f, (phi)^>
(or any other rules like : <f ', phi> = - <f, (phi)'>    )

for any test function phi

thanks

 « Last Edit: Mar 16th, 2010, 1:57pm by trusure » IP Logged
MonicaMath
Newbie  Gender: Posts: 43 Re: fourier transform  of distributions   « Reply #1 on: Mar 17th, 2010, 8:40am » Quote Modify

Did you find the FT for your function? I will give it to you

f^(s)=sqrt(2/pi) i s cos(s.pi/2)/(s^2 - 1)

use this now. IP Logged
Obob
Senior Riddler     Gender: Posts: 489 Re: fourier transform  of distributions   « Reply #2 on: Mar 17th, 2010, 10:50am » Quote Modify

Rather than work with this particular function, you can do it for any (tempered) distribution f as follows (writing D for d/ds and p = phi for the test function to save space):

<i s f^, p> = <f^, i s p> = <f, (i s p)^> = - <f, D(p^)> = <Df, p^> = <(Df)^, p>,

which says that (Df)^ = i s f^.  The crucial step here is <f, (i s p)^> = - <f, D(p^)>, which follows from (i s p)^ = - D(p^).  This is just a statement about test functions; you can prove it by writing out the definition of the Fourier transform and differentiating under the integral. IP Logged

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