wu :: forums
« wu :: forums - fourier transform  of distributions »

Welcome, Guest. Please Login or Register.
Jan 16th, 2022, 12:00pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   hard
(Moderators: SMQ, Icarus, Eigenray, ThudnBlunder, towr, Grimbal, william wu)
   fourier transform  of distributions
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: fourier transform  of distributions  (Read 1370 times)
trusure
Newbie
*





   


Gender: female
Posts: 24
fourier transform  of distributions  
« on: Mar 16th, 2010, 1:57pm »
Quote Quote Modify 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: female
Posts: 43
Re: fourier transform  of distributions  
« Reply #1 on: Mar 17th, 2010, 8:40am »
Quote Quote Modify 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: male
Posts: 489
Re: fourier transform  of distributions  
« Reply #2 on: Mar 17th, 2010, 10:50am »
Quote Quote Modify 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
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board