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Fourier transformation and test function -- Question in the derivation

  1. Dec 31, 2015 #1
    How does it work? (The derivative rules of FT)
    We look at $$F[x(t)]=\hat{x}(f)$$
    $$\mathcal{l} \text{ is a distribution, with}\tilde{x}=tx(t)$$
    $$\mathcal{F}[Dl(x)]=\mathcal{F}l'(x)=2\pi il(\mathcal{F}\tilde{x})=2\pi i \mathcal{F}l(\tilde{x})$$
    Till here I fully understand. But next step:
    $$=2\pi i fFl(x)$$
    How do they drag out f and make x~=tx(t) to x?

    **Well, of course this is just a proof of derivative rule of FT, which is not hard
    $$\mathcal{F}x'=2\pi i f \hat{x}$$
  2. jcsd
  3. Jan 5, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Jan 12, 2016 #3
    I solve it on my own! :)
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