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Representing a real periodic valued function with Fourier series

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Hey, the question i have been given reads:

    By a simple change of variables, show that if g(x) is a periodic real valued function with
    period L it can be represented as

    g(x)~ ∑n=-∞ An exp(-2[itex]\pi[/itex]inx/L)

    where the complex constants An are given by

    LAm =[L/2,-L/2] ∫ g(x)exp(-2[itex]\pi[/itex]imx/L) dx

    2. Relevant equations

    N/A


    3. The attempt at a solution

    I used the fact that the generic Fourier series has the form ∑n=-∞ An exp(-2[itex]\pi[/itex]inx/L) and then used the fact that L is the period to rewrite g(x) in the required form.

    Then i used

    Lbm=[L,-L] ∫ g(x)sin(m[itex]\pi[/itex]x/L)dx and

    Lam=[L,-L] ∫ g(x)cos(m[itex]\pi[/itex]x/L)dx

    I added these two to give me

    L(bm+am)=[L,-L] ∫ g(x)(sin(m[itex]\pi[/itex]x/L)+cos(m[itex]\pi[/itex]x/L))dx


    Then i used 0.5(am +bm)=Am to give me

    2LAm=[L,-L] ∫ g(x)(sin(m[itex]\pi[/itex]x/L)+cos(m[itex]\pi[/itex]x/L))dx

    Any help or indication of where i'm going wrong/ right would be a lot of help. Thanks in advance, John :). Also, if you don't understand any of my notation let me know and i'll try and explain it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 13, 2014 #2

    jbunniii

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    I think one of the exponentials should not have the negative sign in the argument. Is there a typo?

    This is true but not quite what you need. The goal is to use the fact that ##\cos(x) + i\sin(x) = e^{ix}##, so try introducing a factor of ##i## in the appropriate place before adding.
     
  4. Mar 13, 2014 #3
    Yes, it's a typo. The exponential in the g(x) term is meant to be positive. Thanks.

    Ahh thanks Am = (am +ibm) not (am+bm)

    So now i use the relations

    Libm=[L,-L] ∫ g(x)isin(mπx/L)dx and

    Lam= [L,-L] ∫g(x)cos(mπx/L)dx

    To give me

    L(am+ibm)=[L,-L] ∫g(x)*e^(im[itex]\pi[/itex]x/L) dx

    and then as Am = (am +ibm)

    this becomes

    L( Am)=[L,-L] ∫g(x)*e^(im[itex]\pi[/itex]x/L) dx

    My initial thoughts were initially to use symmetry to change the limits to get the missing factors but i don't think that would get me the missing -2 factor in the exponential. Any ideas?
     
  5. Mar 13, 2014 #4

    jbunniii

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    Hmm, shouldn't there be factors of 2 in the arguments to the cosine and sine? Like this:

     
  6. Mar 13, 2014 #5
    Reading through my course notes they are written here as they are printed. However what your suggesting does make sense, I might be using the wrong equations for this scenario but i'm unsure. I can't see your suggestions in my notes.
     
  7. Mar 13, 2014 #6

    jbunniii

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    Oh, I see: you are taking your integrals from ##-L## to ##L##, an interval of length ##2L##. So the cosine and sine will have the arguments ##(2\pi m x) / (2L)## which reduces to ##\pi m x / L##. So your formulas are correct, assuming the function has period ##2L##. But if the period is ##L## then you should only be integrating from ##-L/2## to ##L/2## (or ##0## to ##L##). In that case, the arguments will be ##2 \pi m x / L##.
     
  8. Mar 13, 2014 #7
    The period of the function is L which means this all works out nicely. Thanks, i'll try and read the question properly next time. Much a appreciated :)
     
  9. Mar 18, 2014 #8
    Hi again, when reviewing my workings I noticed an error. The answer you helped me work out is:

    L-L f(x)eim[itex]\pi*x/L[/itex]

    However the answer requested in the question is:

    L-L f(x)e-im[itex]\pi*x/L[/itex]

    I've looked through my workings multiple times and have failed to spot my error, my only thoughts are setting m=-n or something along those lines.

    Thanks again, John
     
  10. Mar 18, 2014 #9

    jbunniii

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    It depends on how you define the Fourier series. If the series has a negative exponent, then the formula for the coefficients should have a positive exponent, and vice versa.

    In your original problem statement, you defined the Fourier series with a negative exponent:
    $$f(x) = \sum_{n=-\infty}^{\infty}A_n \exp(-2\pi i n x / L)$$
    So the coefficient formula should have a positive exponent:
    $$A_n = \frac{1}{L} \int_{-L/2}^{L/2} f(x) \exp(2\pi i n x / L) dx$$
    If you change the signs of both exponents, it will still be correct.
     
  11. Mar 19, 2014 #10
    Hello again, thanks for you help. I discovered the error was due to the fact that i assumed n<0 which meant Am=0.5(an+ibn) as opposed the condition where n>0 which gives Am=0.5(an-ibn).

    However when working through this again i got the final answer:

    2LAm=[L/2,-L/2] ∫g(x)e(-2[itex]\pi*mx/L)[/itex] dx

    As opposed to

    LAm=[L/2,-L/2] ∫g(x)e(-2[itex]\pi*mx/L)[/itex] dx

    The factor of two was from the 2Am=0(an-ibn), when working through this previously i simply forgot the two, you can see this mistake if you look at my earlier posts.

    I ran through my workings and found no way to cancel this two out, do you have any suggestions/ ideas?

    This question will end eventually! Haha!
     
  12. Mar 19, 2014 #11

    jbunniii

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    I scrolled up but I couldn't see exactly where you obtained ##2A_m = 0(a_n - i b_n)##. (Is there a typo? Doesn't this mean ##A_m = 0## for all ##m##?) Can you show how you got this equation?
     
  13. Mar 19, 2014 #12
    Yet another typo, it should read. Am=0.5(am+ibm)

    I stated it correctly, initially but then in the 8th line of my second post i write:

    "and then as Am = (am +ibm)" which is incorrect according to my notes. using this however would give me the correct answer.
     
  14. Mar 19, 2014 #13

    jbunniii

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    Can we review how you are defining ##a_n## and ##b_n##? Are these the definitions you are using:
    $$a_n = \frac{1}{L}\int_{-L/2}^{L/2} g(x) \cos(2\pi n x / L) dx$$
    $$b_n = \frac{1}{L}\int_{-L/2}^{L/2} g(x) \sin(2\pi n x / L) dx$$
    If so, then
    $$\begin{align}
    a_n + i b_n &= \frac{1}{L} \int_{-L/2}^{L/2} g(x) (\cos(2\pi n x / L) + i \sin(2\pi n x / L)) dx \\
    &= \frac{1}{L} \int_{-L/2}^{L/2} g(x) \exp(2\pi i n x / L) dx \\
    \end{align}$$
    which is indeed the Fourier coefficient ##A_n## assuming the Fourier series is defined as
    $$g(x) = \sum_{n=-\infty}^{\infty} A_n \exp(-2\pi i n x / L)$$
    If your definitions for ##a_n## and ##b_n## are not as above, then that will change things. Indeed, I believe the formulas for ##a_n## and ##b_n## should be multiplied by 2 if they represent the coefficients of the Fourier series when written as follows:
    $$g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \cos(2\pi n x / L) + b_n \sin(2\pi n x / L)]$$
    Can you please clarify whether that is the case?
     
  15. Mar 19, 2014 #14
    The definitions i am using for an and bn are:

    an = [itex]\frac{1}{L}[/itex] ∫[itex]^{L}[/itex][itex]_{-L}[/itex] f(x) cos([itex]\frac{npix}{L}[/itex]) dx

    and

    bn = [itex]\frac{1}{L}[/itex] ∫[itex]^{L}[/itex][itex]_{-L}[/itex] f(x) sin([itex]\frac{npix}{L}[/itex]) dx

    Since I am integrating over the period L, meaning my limits are L/2 to -L/2 rather than L to -L does this mean i get a factor of 2 which means the 2's cancel?

    I am also using An=0.5(an-ibn), where An is indeed the Fourier coefficient.

    John
     
  16. Mar 19, 2014 #15

    jbunniii

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    It looks like those formulas are correct if the period is ##2L##. I can rewrite them as follows to make it more obvious:
    $$a_n = \frac{2}{2L} \int_{-L}^{L} f(x) \cos\left(\frac{2\pi n x}{2L}\right) dx$$
    $$b_n = \frac{2}{2L} \int_{-L}^{L} f(x) \sin\left(\frac{2\pi n x}{2L}\right) dx$$

    If the period of your function is ##L## then you should replace ##2L## by ##L##, and ##L## by ##L/2##, to get
    $$a_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x) \cos\left(\frac{2\pi n x}{L}\right) dx$$
    $$b_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x) \sin\left(\frac{2\pi n x}{L}\right) dx$$
    Using these definitions we can see that you need to define ##A_n = (a_n + i b_n) / 2## to compensate for the ##2/L##.
     
  17. Mar 19, 2014 #16
    Yup that works perfectly. 2's cancel because it's a period of L, not 2L those were just the definitions given to me in my notes.

    Can't thank you enough for your help, cheers :)
     
  18. Mar 19, 2014 #17

    jbunniii

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    I would suggest scratching out the formulas in your notes and replacing them with the ones for period ##L##, otherwise they will probably continue to confuse you in the future. :biggrin: Glad you got it sorted out. Working with Fourier series, one constantly runs up against annoyances with sign errors, factors of ##2## or ##2\pi##, etc., because not everyone uses the same definitions. Just a fact of life.
     
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