Results 1 to 4 of 4

Math Help - fourier transform/commutation relation

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    133

    fourier transform/commutation relation

    Hello all,

    I have been trying to show that:

    (\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)

    I have done the following so far:

    (\digamma D_{c}f)(\gamma)= \int_{ -\infty}^{\infty}(D_{c}f)(x)e^{-2\pi ix\gamma}dx=\int_{-\infty}^{\infty}\frac{1}{\sqrt{c}}f(\frac{x}{c})e^  {-2\pi ix\gamma}dx

    Now if we put y=\tfrac{x}{c} we get:

    \int_{-\infty}^{\infty}\frac{1}{\sqrt{c}}f(y)e^{-2\pi iyc\gamma}cdy=\sqrt{c}\int_{-\infty}^{\infty}f(y)e^{-2\pi iyc\gamma}dy

    How do I continue so that I end up with what I intented to show?

    Thanks a million
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by surjective View Post
    I have been trying to show that:

    (\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)

    How do I continue so that I end up with what I intented to show?
    What exactly did you intend to show? Rewrite (D_{\frac{1}{c}}(\mathcal{F}f))(\gamma) as an integral, and compare with what you got.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    133

    fourier tranform/commutation relation

    Hey,

    I wanted to show (intended) that:

    (\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)


    The right side is what I want to arrive at, i.e. I would like to continue from where I left off and arrive at the right-hand-side instead of "going backwards". I seem to be stuck !!?????
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by surjective View Post
    Hey,

    I wanted to show (intended) that:

    (\digamma D_{c}f)(\gamma)= (D_{\frac{1}{c}}\digamma f) (\gamma)


    The right side is what I want to arrive at
    I know that; all I wanted you to notice is that you have arrived to the right side. It is just written in a different way. Just like you wrote the left side as an integral, you can do the same for the right side. You'll end up exactly with what your change of variable lead you to. Just one additional "hint" (not that you did it): D_{\frac{1}{c}}(g)(\gamma)=\sqrt{c}g(\gamma c), and take g(\gamma)=\mathcal{F}f(\gamma)=\int\cdots.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Commutation Relation
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 27th 2010, 12:06 PM
  2. Commutation relation of operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 2nd 2010, 01:48 AM
  3. relation between DFT and continuous Fourier Transform
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: October 21st 2009, 09:51 AM
  4. Commutation relation - angular momentum operator
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: April 26th 2009, 11:57 PM
  5. Replies: 0
    Last Post: April 23rd 2009, 05:44 AM

Search Tags


/mathhelpforum @mathhelpforum