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Fourier Transform and Limit of Euler's Formula

  1. Aug 12, 2012 #1
    Hey Physics Forums,

    Grading an assignment, the current topic is continuous Fourier Transforms. They're trying to prove the convenient property:
    [tex] \mathcal{F} \left[ \frac{d^n}{dx^n} f(x) \right] = (i \omega)^n \mathcal{F} \left[ f(x) \right] [/tex]
    So there's a simple way to get it:
    Let [itex]f(x)[/itex] be represented by it's Inverse Fourier Transform [itex]f(x) = \mathcal{F}^{-1} \left[ g( \omega ) \right] = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega )e^{i \omega x} d \omega [/itex]
    Then,
    [tex] \frac{d^n}{d x^n} f(x) = \frac{d^n}{d x^n} \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega )e^{i \omega x} d \omega = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega ) \frac{d^n}{d x^n}e^{i \omega x} d \omega = (i \omega)^n \frac{1}{2 \pi} \int_{-\infty}^{\infty} g( \omega ) e^{i \omega x} d \omega = (i \omega)^n \mathcal{F} \left[ f(x) \right] [/tex]

    However, I have some people with a different method. They directly transform the entire expression, and apply integration by parts recursively until they obtain:
    [tex] \left. (i \omega)^0 \frac{d^{n-1}}{dx^{n-1}}[f(x)]e^{-i \omega x} + (i \omega)^1 \frac{d^{n-2}}{dx^{n-2}}[f(x)]e^{-i \omega x} + \ldots \hspace{10px} \right|^{\infty}_{-\infty} + (i \omega)^n \int f(x) e^{-i \omega x}dx [/tex]

    The claim is that when the delimiter is applied, [itex]\lim_{x \to \pm \infty} e^{-i \omega x} = 0[/itex]. I'm not sure if this is true. [itex]e^{-i \omega x} = \cos(\omega x) - i \sin(\omega x)[/itex] which clearly oscillates. Can anyone else confirm this?
     
  2. jcsd
  3. Aug 12, 2012 #2

    Mute

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    Homework Helper

    It is not the [itex]\exp(i\omega x)[/itex] term which vanishes at the end points, but the derivatives of f(x). Strictly speaking, only functions which decay to zero as the argument grows large have fourier transforms. Since they decay to zero, their derivatives must decay to zero.

    For generalized functions/distributions like the delta function, the identity the students are asked to prove is basically just defined to match the case of legitimately fourier-transformable functions. (Alternatively, the distribution definition includes integration against a smooth test function which decays to zero at the end points).
     
  4. Aug 12, 2012 #3
    Right, forgot Dirichlet conditions. Thanks for the catch!
     
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