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Generalised Fourier Series

  1. Aug 2, 2016 #1
    1. The problem statement, all variables and given/known data

    By applying the Gram–Schmidt procedure to the list of monomials 1, x, x2, ..., show that the first three elements of an orthonormal basis for the space L2 (−∞, ∞) with weight function ##w(x) = \frac{1}{\sqrt{\pi}} e^{-x^2} ##

    are ##e_0(x)=1## , ##e_1(x)= 2x## ,##e_2(x)= \frac{1}{\sqrt{2}} (2x^2−1)##

    Explain why any well-behaved function f : R → C may be expressed as a series of the form

    ## f(x)= \frac{1}{\sqrt{2 \pi}} \sum {a_n e_n(x) e^{- \frac{1}{2} x^2}} ## where the sum runs from 0 to infinity


    2. Relevant equations

    3.The attempted solution

    I have done the first part to do with the Gram–Schmidt procedure, using the usual formula, but cannot understand how to do the second part of the question

    My initial thought was to try to isolate one of the co-efficients, ##a_m## and show that this is the same as the 'normal' formula for ##a_m## when using a weight function of one- but this didn't seem to work out.

    What i'm confused about it where the ##\frac{1}{\sqrt{2 \pi}}## has come from, since the basis functions are already normalised, and why the square root of the weight function is included in the summation. I get the feeling this is what the question is wanting me to show is okay, so to speak, but I can't quite see it.

    Any help would be greatly appreciated :)

    (note: i did initially type out all my working, but my computer couldn't cope with it, and kept freezing- sorry)
     
    Last edited: Aug 2, 2016
  2. jcsd
  3. Aug 2, 2016 #2

    Krylov

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    Sorry, but I am confused about what your weight function is? Could you clarify that?
    EDIT: I suppose it is ##w(x) = \tfrac{1}{\sqrt{\pi}} e^{-x^2}##, correct?
     
    Last edited: Aug 2, 2016
  4. Aug 2, 2016 #3
    How careless of me. Yes, sorry. I have fixed it in the original post as well- thank you :)
     
  5. Aug 2, 2016 #4

    Krylov

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    In any case, the hint I would give is this: Let ##H## be ##L^2(\mathbb{R})## with the usual inner product ##\langle, \rangle## and let ##H_w## be ##L^2(\mathbb{R})## with the weighted inner product ##\langle, \rangle_w##. Note that ##H \subset H_w##. You have obtained an orthormal basis of ##H_w##, namely ##\{e_n\}##.

    How do you obtain from this an orthonormal basis for ##H##? Once this is done, think about what the Fourier series of ##f \in H## with respect to that latter basis is.
     
  6. Aug 2, 2016 #5

    Ray Vickson

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    I think that the last formula (in the problem statement) should be that
    $$f(x) = \sum_{n=0}^{\infty} a_n e_n(x)$$
    For example,
    $$x^2 = \frac{1}{2} e_0(x) + \frac{1}{\sqrt{2}} e_2(x) $$,
    so ##a_0 = 1/2, a_1 = 0## and ##a_2 = 1/\sqrt{2}## for ##f(x) = x^2##. If you included the factor ##\exp(-x^2/2)## in the expansion (as you did in your formula) then your ##a_n## could not be constant: they would need to contain the compensating factor ##\exp(+x^2/2)##.

    Basically, the claim is (or should be) that ##f = \sum_n \langle \: e_n,f \, \rangle \: e_n##. The weight function ##w(x)## is used for defining the inner product ##\langle f,g \rangle##.
     
    Last edited: Aug 2, 2016
  7. Aug 2, 2016 #6
    It's not- that's what is confusing me? :)
     
  8. Aug 2, 2016 #7

    Ray Vickson

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    OK: I am saying that is SHOULD BE. So, you were being asked to show something that is wrong. Don't just take my work for it; look it up in your textbook, or do an on-line search under "orthonormal expansions" or something along those lines. That might give you the ammunition you need if you have to prove that the problem statement is incorrect.
     
  9. Aug 2, 2016 #8
    So would you say that the basis would ##e_n(x) \sqrt{w(x)}## be the new basis for ##L^2(\mathbb{R})## ?
     
  10. Aug 2, 2016 #9

    Ray Vickson

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    No. The basis is ##\{ e_n(x) \}##. The claim is that for a wide range of functions ##f(x)## you can have ##f(x) = \sum_n c_n e_n(x)##. The issue is how to compute the coefficients ##c_n##, and you do that using the inner product ##c_n = \langle f , e_n \rangle##.
     
  11. Aug 2, 2016 #10

    Krylov

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    Come on, the question from the OP is correct and it is not difficult to prove, see post #4.
    Yes, perhaps up to some fixed numerical constant. There are two spaces in play here: One with an ordinary inner product (##H##), the other with a weighted inner product (##H_w##). It is important to keep these spaces apart.
     
  12. Aug 2, 2016 #11
    What about that divided by its modulus such that it's an orthonormal basis?
     
  13. Aug 2, 2016 #12

    Krylov

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    Yes, in the notation of post #4 you have:
    $$
    \langle e_m \sqrt{w}, e_n \sqrt{w} \rangle = \int{e_m(x)\sqrt{w} e^{\ast}_n(x) \sqrt{w}\,dx} = \int{e_m(x)e^{\ast}_n(x)w(x)\,dx} = \langle e_m, e_n \rangle_w = \delta_{mn}
    $$
    so in fact you can choose that normalization constant equal to one. :wink:
    So, you have found an orthonormal basis for ##H##. Now expand any "well-behaved" (i.e. living in ##H##) function ##f## in terms of that basis.
     
  14. Aug 2, 2016 #13
    Is this a rigorous enough proof? (I'm not doubting, just checking)
     
  15. Aug 2, 2016 #14

    Krylov

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    You could remark that ##\{\sqrt{w}e_n\}## is also spanning, because...? (This is almost trivial, so I would not substract points if you would omit it, but by making this remark you do show that you know the definition of "orthonormal basis" completely.)
     
  16. Aug 2, 2016 #15

    Ray Vickson

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    The weight function appears in the expression ##\langle f, g \rangle = \int_R w(x) f(x) g(x) \, dx## The basis functions are the ##e_n(x)##, not ##\sqrt{w(x)} e_n(x)##. That is clear from the first part of what the question asks, where it says "show that the first three elements of an orthonormal basis for the space
    ##L2 (−\infty,\infty)## with weight function ##w(x)=1/\sqrt{\pi}\, e^{-x^2/2}## are ##e_0(x) = 1##, ##e_1(x) = 2x## and ##e_2(x) = 1/\sqrt{2} \; (x^2 - 1)##." It is stating very clearly that the basis functions are ##e_0, e_1, e_2, \ldots##, not ##\sqrt{w} e_0, \sqrt{w} e_1, \sqrt{w} e_2, \ldots ##.

    The (typogaphical?) error that the person setting the problem has made came after that, or maybe the OP copied it incorrectly. Choosing the inner product as ##\langle f,g \rangle = \int w f g \, dx## means that if we use ##a_n = \langle f, e_n \rangle## for ##n = 0,1,2,\ldots## we will be minimizing
    $$\text{squared error} = E \equiv \int_R w(x) \left[ f(x) - \sum_{n=0}^{\infty} c_n e_n(x) \right]^2 \ dx$$
    Completeness would imply that ##S = 0##, but that is a whole other issue that the OP was not being asked to address.
     
  17. Aug 2, 2016 #16

    Krylov

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    • I am not contesting that the basis functions for the space given in the OP's problem are the ##e_n##.
    • I am merely stating that there is ANOTHER space (namely: ##L_2(\mathbb{R})## with the usual inner product) for which ##\{\sqrt{w} e_n\}## is an orthonormal basis).
    • Then, using this second statement, I (provide hints to) solve the exercise.
    Please read my posts before claiming that there is something wrong with them or with the OP's statement of the problem. There it nothing wrong with either.
     
  18. Aug 2, 2016 #17

    Ray Vickson

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    I agree completely that in the ordinary Hilbert space ##L_2(R)## with ##\langle f,g \rangle = \int_R f g \, dx## that the functions ##w\, e_n## do, indeed, form an orthonormal basis. However, I maintain that this is not related to the OP's problem, where it is stated that "his" ##L_2## uses the weight function ##w(x)##. (Although it does not specificaly say so in the problem I presume that in the course notes that is explained as defining an inner product. In all sources I have ever seen, that is how it is applied.) So, we are really talking about two different ##L_2##s here.

    No matter what ##L_2## we are using, there clearly IS something wrong with the second part of the OP's statement; it just simply contradicts the first part. That is the reason I gave the very simple example that ##x^2 = (1/2) e_0(x) + (1/\sqrt{2}) e_2(x)##. This last equation is absolutely correct, and follows from elementary algebra. If we regard ##x^2## as a "well-behaved" function, then it is clear that the expansion of that ##f## at least does not have the form given in the last part of the problem statement; it has the form you would get by leaving out the ##w(x)## factor in the terms of the expansion.

    I did not invent any of this; it is widely documented in standard sources, such as
    http://mathworld.wolfram.com/GeneralizedFourierSeries.html
    which gives exactly the same expansion formula that I did.
     
  19. Aug 2, 2016 #18

    Krylov

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    Yes, one of these sources includes my post #4 where I defined the inner product ##\langle, \rangle_w## in ##H_w##.
    This is what I have been saying all along. I called one ##H_w## and the other ##H##, again see post #4.
    The function ##x \mapsto x^2## is not in ##H##, so in that sense it is not well-behaved. (It simply blows up.)
    The expansion identity in the OP is correct for ##f \in H##.
     
  20. Aug 2, 2016 #19

    Ray Vickson

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    Of course the function ##x^2## is not in the "ordinary" ##L_2## space ##H## (with ##||f||^2 = \int_R |f|^2 \, dx##), but it IS in the "new" ##L_2## that uses ##w## in the definition of inner product. In fact, the very "basis" functions ##e_0, e_2, e_2, \ldots## given in the first part of the problem are not even in the ##H##, so for the ordinary Hilbert space ##H## the OP would be asked to use basis functions that are not actually members of the vector space itself! I think it makes more sense (and is, I maintain, in the spirit of the problem) to use the "modified" ##L_2## with weight ##w## is the inner product, and to expand in terms of simple functions that are actually members of the space itself. The remaining issue is whether the second part of the problem has an error, typographical or otherwise. In ##H## it does not, while in ##L_2## it does.

    I have said all I intend to say on this subject.
     
  21. Aug 3, 2016 #20

    Krylov

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    Assuming that @Physgeek64 is still interested in a solution to the exercise as he originally posted it, let me resume what we have now (using the notation of post #4):
    • In the weighted space ##H_w## we have the orthonormal basis ##\{e_n\}##.

    • In the "ordinary" space ##H## we have the orthonormal system ##\{\sqrt{w}e_m\}##.

    • It is almost trivial to check that this latter system is in fact a orthonormal basis for ##H##. Namely, it remains to check that it is spanning, For this, just verify that if ##f \in H## is such that ##\langle f, \sqrt{w}e_n\rangle = 0## for all ##n##, then it follows that ##f = 0## identically. (To do this, write the inner product ##\langle f, \sqrt{w}e_n\rangle## in ##H## as an inner product in ##H_w## instead and use that ##\{e_n\}## is an orthonormal basis for ##H_w##.)

    • Now you may conclude that every ##f \in H## (which, one could argue, makes it "well-behaved", for example for certain QM problems) can be written as the series that appears in your exercise. (Maybe up to a mutiplicative numerical constant that you can incorporate into the coefficients ##a_n##.)
    Please write this into a tidy proof. Done.
     
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