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Change of integration order?

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data

    On this site,
    http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node31.html#eq:completeness_with_function
    they changed integration order with e^(-ikt) integrated wrt t to t is now a constant and it's now integrated wrt k. How can one do this?

    As stated from this sentence "By interchanging integration order and letting one has"
     
  2. jcsd
  3. Apr 13, 2007 #2
    [tex]f(x) \propto \int e^{ikx} \int e^{-ikt} f(t) dt dk = \int \int e^{ikx} e^{-ikt} f(t) dt dk = \int \int e^{ikx} e^{-ikt} f(t) dk dt [/tex]

    There's nothing wrong with changing the order of integration here, think back of integrals as Riemann sums, changing the order in which you sum terms doesn't matter.

    (You were going to integrate both exp(ikx) and exp(-ikt) over k anyway, so why not do that first and then integrate over t on which only the latter term depends).
     
    Last edited: Apr 13, 2007
  4. Apr 13, 2007 #3
    Right, but why is the f (hat) function in terms of t?

    f is in terms of x and k only.
     
  5. Apr 13, 2007 #4

    HallsofIvy

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    k and t are "dummy" variables. They can call them whatever they wish!
     
  6. Apr 13, 2007 #5
    So only x counts. This brings about the question, what is the point of the fourier transform.
     
  7. Apr 13, 2007 #6
    The point?

    Um, well, from my relatively limited experience in partial differential equations, in Fourier space differentiation and antidifferentiation break down into multiplication and division. Which is always good I suppose.
     
  8. Apr 13, 2007 #7

    HallsofIvy

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    I'm not sure what you mean by that. Certainly in the integral you show, f is a function of x only. However, you have not said what it is you want to do with the Fourier Transform (the function inside the integral). Surely "the point of the Fourier transform" depends on that!
     
  9. Apr 13, 2007 #8
    Take equation (234) from here and there, replace [tex]\hat{f}(k)[/tex] by equation (233).
    Then you almost have equation (235), except that you have to change
    the order of integration.

    The function [tex]\hat{f}(k)[/tex] is indeed a function of k, but it is not the function f, whose variable is x. Maybe you are a little confused about the meaning of [tex]\hat{f}(k)[/tex].

    Think of the Fourier expansion as sort of linear combination of basis-vectors.

    Let me try to explain:
    Take the vectors in [tex]R^3[/tex].

    Every vector can be represented as a linear combination of the basis-vectors [tex](1,0,0)[/tex], [tex](0,1,0)[/tex] and [tex](0,0,1)[/tex].
    For example:
    Take the vector [tex]v=(9,8,4)[/tex]. This vector can be written as a linear combination of the basisvectors vectors
    [tex]u_1=(1,0,0)[/tex]
    [tex]u_2=(0,1,0)[/tex]
    [tex]u_3=(0,0,1)[/tex]
    in the following way:

    [tex](9,8,4) = 9 \cdot (1,0,0) + 8 \cdot (0,1,0) + 4 \cdot (0,0,1)[/tex]
    [tex](9,8,4) = 9 \cdot u_1 + 8 \cdot u_2 + 4 \cdot u_3[/tex]
    Let's call the numbers 9,8,4 in front of the basis-vectors coefficients.
    [tex](9,8,4) = c_1 \cdot u_1 + c_2 \cdot u_2 + c_3 \cdot u_3[/tex]
    where [tex]c_1 = 9[/tex], [tex]c_2=8[/tex] and [tex]c_3=4[/tex]

    What if I choose different basis vectors, for example:
    [tex]b_1=(3,0,0)[/tex]
    [tex]b_2=(0,2,0)[/tex]
    [tex]b_3=(0,0,2)[/tex]

    Then our vector [tex]v=(9,8,4)[/tex] can be written as:
    [tex](9,8,4) = 3 \cdot b_1 + 4 \cdot b_2 + 2 \cdot b_3[/tex]
    Thus, the coefficients are 3,4,2 for our new basis-vectors
    [tex]b_1[/tex],[tex]b_2[/tex] and [tex]b_3[/tex].

    In general, you can write a vector [tex]v[/tex] as a linear combination of
    basis-vectors [tex]b_k[/tex], where in front of the basis-vectors you have the coefficients [tex]c_k[/tex].

    [tex]v = \sum_{k=1}^{3} c_k b_k[/tex]

    In our last example we had
    [tex]b_1=(3,0,0)[/tex]
    [tex]b_2=(0,2,0)[/tex]
    [tex]b_3=(0,0,2)[/tex]

    together with the coefficients:
    [tex]c_1=3[/tex]
    [tex]c_2=4[/tex]
    [tex]c_3=2[/tex]

    Just check the formula

    [tex]v = \sum_{k=1}^{3} c_k b_k[/tex]

    by plugging in the above values:

    [tex]v = c_1 \cdot b_1 + c_2 \cdot b_2 + c_3 \cdot b3[/tex]
    [tex]v = 3 \cdot b_1 + 4 \cdot b_2 + 2 \cdot b_3[/tex]
    [tex]=3 \cdot (3,0,0) + 4 \cdot (0,2,0) + 2 \cdot (0,0,2)[/tex]

    ----------

    So, every vector can be represented by some basis vectors [tex]b_k[/tex]
    as

    [tex]v = \sum_{k=1}^{3} c_k b_k[/tex]


    More general, if we have a vector with n entries instead of 3, we write

    [tex]v = \sum_{k=1}^{n} c_k b_k = c_1 b_1 + c_2 b_2 + ... + c_n b_n [/tex]

    Now, let's make a step from the discrete to the continuous case.
    Say we have a function [tex]f(x)[/tex] and we also want to ask, whether it's possible to represent [tex]f(x)[/tex] as a linear combination of ''basis-vectors''.

    Question: Is it possible to write

    [tex]f(x) = \sum_{k=1}^n c_k b_k[/tex]

    Let us be more specific and ask: Can I write [tex]f(x)[/tex] as a sum
    of [tex]b_k = e^{ikx}[/tex]? So my new basis-vectors are [tex]b_k=e^{ikx}[/tex].
    The question then becomes:

    Question: Is it possible to write

    [tex]f(x) = \sum_{k=1}^n c_k e^{ikx}[/tex]

    Indeed, it is possible, with a correction for the values of k.
    Instead of going from k=1 to n, we use infinitely many basis-vectors, and
    write [tex]k=- \infty[/tex] to [tex]k=+ \infty[/tex].

    Corrected version:

    [tex]f(x) = \sum_{k=-\infty}^{+\infty} c_k e^{ikx}[/tex]

    The only question is, how do the coefficients
    [tex]c_k[/tex] look like?

    Have a look at Wikipedia
    or here on page 2 of the pdf. It shows how the coefficients can be calculated.

    To finally get to your Fourier integral, we replace the sum by an integral

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

    Now, on your ohio-website, [tex]c(k)[/tex] is [tex]\hat{f}(k)[/tex], see equation (234) on the ohio-website.

    Thus, [tex]\hat{f}(k)[/tex] plays the role of the coefficients.

    ----------

    Note 1 (on how to get from the Fourier series to the Fourier integral):
    The quote is taken from here.

    Also see here
    for the transition from the Fourier series to the Fourier integral.

    Note 2: Summary:
    In summary, consider the Fourier integral as a linear combination of basis vectors [tex]e^{ikx}[/tex] with the
    coefficients [tex]c(k)[/tex] (or [tex]\hat{f}(k)[/tex]).

    Note 3 (on applications of the Fourier transform:
    Have a look at What is a Fourier Transform and what is it used for?.

    Applications of Fourier Transform
    in Communications, Astronomy, Geology and Optics


    Note 4 (functions as basis-vectors?)
    You might ask "Why can I consider the functions [tex]e^{ikx}[/tex] as
    basis-vectors?"

    The functions [tex]e^{ikx}[/tex] fulfill some properties similar to those of
    basis vectors (1,0,0),(0,1,0),(0,0,1)from [tex]R^3[/tex].
    The functions are orthonormal, that is they are
    orthogonal to each other and they are normalized to 1 (delta-function?).
    See

    Fourier Analysis on page 3.

    here on page 8.

    Orthonormal functions: Definition on Wolfram mathworld

    http://mathworld.wolfram.com/OrthonormalBasis.html

    Note 5: Some examples of coefficients
    Examples of Fourier transforms can be found here

    Note 6: Java Applets
    Approximation of a function by a Fourier transform
    This applet shows you how you can approximate a function by a Fourier transform. The more
    coefficients you use, the better the approximation becomes.

    Applet: Rectangular pulse approximation by Fourier Transform
    This applet shows how you approximate a rectangular shaped pulse. If you don't use enough
    "basis-vectors" (bandwidth is limited), then the pulse will not look rectangular anymore. This has applications in
    electronics where you want to transmit a signal but the bandwidth is limited.
     
    Last edited: Apr 13, 2007
  10. Apr 13, 2007 #9
    I see. From all the Fourier transforms I have seen they use x as the integration variable in f (hat) which is confusing. f(hat) is not related to x in f(x) the original function but merely computes the coefficients of the basis vectors e^(ikx) for each k from -infinity to +infinity.
     
  11. Apr 13, 2007 #10
    One thing I find fascinating is that there is a dirac delta function lurking inside every fourier transform.
     
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