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From Fourier Series to Fourier Transforms

  1. Aug 30, 2011 #1
    Hello,

    I am trying to formalize the logical steps to prove that the Fourier Series of a function with period[itex]\rightarrow \infty[/itex] leads to the Fourier transform. So let's have the Fourier series:

    [tex]f(x)=\sum_{n=-\infty}^{+\infty}c_n e^{i\cdot \frac{2\pi n}{L}x}[/tex]

    where L is the period of the function f.
    Many texts simply say that when L tends to infinity, cn becomes a continuous function [itex]c(n)[/itex] and the summation becomes an integral.

    [tex]f(x) = \int_{-\infty}^{+\infty}c(k) e^{i\cdot k x}dk[/tex]

    Unfortunately they do not explain why, and they do not mention what is the logical step that allows one to switch from the discrete cn to the continuous c(k), and from the summation to an integral with dk.

    Any hint?
     
  2. jcsd
  3. Aug 30, 2011 #2
    Last edited by a moderator: Apr 26, 2017
  4. Aug 31, 2011 #3
    Uhm...so far, no replies.
    I made the following attempt:

    the Fourier coefficients cn are given by:

    [tex]c_n = \frac{1}{T}\int_{-T/2}^{T/2}f(x)e^{-i\frac{2\pi}{T}nx}dx[/tex]

    Plugging the cn's into the first equation in my first post one gets:

    [tex]f(x) = \frac{1}{T} \sum_{n=-\infty}^{+\infty} \left( \int_{-T/2}^{T/2}f(x)e^{-i\frac{2\pi}{T}nx}dx \right) e^{i\frac{2\pi}{T}nx}[/tex]

    Now we introduce the quantity [itex]k_n=n/T[/itex], and observe that [itex]\Delta k_n=k_{n+1}-k_n = 1/T[/itex].

    Clearly, when [itex]T\rightarrow +\infty[/itex], we have [itex]\Delta k_n \rightarrow 0[/itex], and we replace the discrete quantity kn with a continuous variable u (this step is not rigorous!). As a consequence we must also replace the Fourier coefficients [itex]\{c_n\}[/itex] with a continuous variable c(u) (this step is not rigorous!)

    In doing so we obtain:

    [tex]f(x) = \Delta k_n \sum_{n=-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty}f(x)e^{-i\ 2\pi u x}dx \right) e^{i 2\pi k_n x} = \Delta k_n \sum_{n=-\infty}^{+\infty} c(k_n) e^{i 2\pi k_n x}[/tex]

    The rightmost term is a Riemann sum, so we write:

    [tex]f(x) = \int_{-\infty}^{+\infty} c(u) e^{i 2\pi u x}du[/tex]

    Which is the continuous inverse Fourier transform of c(u)
    I hope someone else will put this into a more rigorous form and spot possible mistakes.
     
    Last edited: Aug 31, 2011
  5. Sep 1, 2011 #4
    Your argument has the right idea. Let [itex]f\in C^\infty_c (\mathbf{R})[/itex] and choose [itex]\epsilon>0[/itex] sufficiently small so that [itex]\mathrm{supp}(f) \subset (-1/\epsilon, 1/\epsilon)[/itex]. Then [itex]f[/itex] has the convergent Fourier series [tex] f(x) = \sum_{n\in \mathbf{Z}} c_n e^{\mathrm{i} \pi \epsilon n x} [/tex]where
    [tex] c_n = \frac{\epsilon}{2} \int e^{-\mathrm{i} \pi \epsilon n x} f(x)\, \mathrm{d}x. [/tex]
    Set [itex]K_\epsilon = \{ k\in \mathbf{R} : k/\pi \epsilon \in \mathbf{Z} \}[/itex]. Then this can be written
    [tex] f(x) = \sum_{k \in K_\epsilon} \frac{\hat{f}(k)e^{\mathrm{i} kx}}{2\pi } (\pi \epsilon) \qquad (*) [/tex]
    where
    [tex] \hat{f}(k) = \int e^{-\mathrm{i} kx} f(x)\, \mathrm{d}x. [/tex]
    Note that [itex]\mathbf{R}[/itex] is the disjoint union of intervals of length [itex](\pi \epsilon)[/itex] centered about the points in [itex]K_\epsilon[/itex], it follows that [itex](*)[/itex] is the Riemann sum for the integral
    [tex] \frac{1}{2\pi} \int e^{\mathrm{i} kx} \hat{f}(k)\, \mathrm{d} k. [/tex]
    Since [itex]f\in C^\infty_c (\mathbf{R})[/itex] it follows that [itex]\hat{f}(k)e^{\mathrm{i} kx}[/itex] is integrable, so the Riemann sums converge to the integral as [itex]\epsilon\rightarrow 0[/itex], so you are done. The result extends to larger function spaces using standard density results.

    Reed & Simon [Methods of Mathematical Physics, Vol II] use this approach to prove the Fourier inversion theorem. It allows you to prove many of the "usual" results for Fourier transforms using the corresponding results from Fourier series.
     
  6. Sep 1, 2011 #5
    Hi Anthony!

    Thanks a lot for providing a rigorous approach to this problem!
    The explanation is now much clearer and easier to follow.
     
  7. Sep 1, 2011 #6
    No problem - glad to be of service.
     
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