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Topic: fourier transform of distributions (Read 1370 times) 

trusure
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fourier transform of distributions
« on: Mar 16^{th}, 2010, 1:57pm » 
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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 16^{th}, 2010, 1:57pm by trusure » 
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MonicaMath
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Re: fourier transform of distributions
« Reply #1 on: Mar 17^{th}, 2010, 8:40am » 
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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.


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Obob
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Re: fourier transform of distributions
« Reply #2 on: Mar 17^{th}, 2010, 10:50am » 
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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.


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