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Fourier Series Coefficient Symmetries

  1. Mar 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ## f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) ##
    What can be said about the coefficients ##a_n## and ##b_n## in the following cases?
    a) f(x) = f(-x)
    b) f(x) = - f(-x)
    c) f(x) = f(π/2+x)
    d) f(x) = f(π/2-x)
    e) f(x) = f(2x)
    f) f(x) = f(-x) = f(π/2-x)

    2. Relevant equations
    Sine is odd and cosine is even. Even functions can only contain cosine terms and odd functions only sine terms in their Fourier Series.

    3. The attempt at a solution
    Here is how far I have gotten:
    a) All b=0 because even function
    b) All a=0 because odd function
    c) Funtion has period π/2 (?)
    d) Funtion has period π/2-2x (?)
    e) Function has period x (not neccessarily the shortest period)
    f) All b=0 because even function.

    But here I am stuck, because I have no idea how to relate the coefficients to the period in c)-f), and in c)-e) I don't know if the funtion is odd or even.
    Many thanks for any help! :)
     
  2. jcsd
  3. Mar 28, 2017 #2
    Use the identities for sin and cos:
    ## sin(x+y)=sin(x)cos(y)+sin(y)cos(x) ##
    ## cos(x+y)=cos(x)cos(y)-sin(x)sin(y) ##

    for the different for terms in the sum. What is happening with ##a_n## and ##b_n## for odd and even values of ##n##?
    By putting different values for ##y## you can study the different cases in c), d), e) and f).
     
  4. Mar 29, 2017 #3
    Hello! Thanks for your help!
    Alright, I tried using the formulas in case c):
    ##\sin(\pi/2 +x) = \cos x##
    ##\cos(pi/2 +x) = -\sin x##
    which is as expected. However, I am confused about how this should be applied to the problem. We have a general function ##f(x)## and I can't really see how the above properties reveal anything about the coefficients.
     
  5. Mar 29, 2017 #4
    Yes, but this is only valid for ##n=1##. What do you get for ##cos n(\pi/2+x)## and ##sin n(\pi/2+x)##, for ##n## odd and even? You can then write the Fourier serie for ##f(\pi/2+x)## and simplify the sum. An advice is to split into odd and even ##n##. I hope this helps.
     
  6. Mar 29, 2017 #5
    Alrigth, I feel like the solution is close but I am still not really there. I have found the following:
    ##\sin n(\pi/2 +x) = \pm \cos nx## for odd n
    ##\sin n(pi/2 +x) = \pm \sin nx## for even n
    ##\cos n(\pi/2 +x) = \pm \cos nx## for even n
    ##\cos n(pi/2 +x) = \pm \sin nx## for odd n

    But to me these are still "just" some trigonometric identities which I don't know how to apply to the function.
    ## f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) ##
     
  7. Mar 29, 2017 #6
    '
    You can write the above relations as
    ##\sin n(\pi/2+x)=(-1)^{(n-1)/2}\cos nx ##
    ##\sin n(pi/2 +x) = (-1)^{n/2} \sin nx## for even n
    ##\cos n(\pi/2 +x) = (-1)^{(n-1)/2} \cos nx## for even n
    ##\cos n(pi/2 +x) =-(-1)^{n/2} \sin nx## for odd n

    Furthermore, we can write
    ##f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos n(\pi/2+x) + b_n \sin n(\pi/2+x)) ##,

    Can you now see how to use the above trignometric identies?

    It is also convenient to split the sum into odd and even values of ##n##, i.e. something like ##\sum_{n=1}^{\infty}=\sum_{odd\: n}+\sum_{even\: n}##. It is then convenient to introduce a variable ##k## through ##n=2k+1## (for the sum over odd ##n##) and similarly ##n=2k## (for even n).
     
  8. Mar 29, 2017 #7
    Hmm, I think there migth be an algebraic miss in the exponents in your previous post. shouldn't it be:
    ##\sin n(\pi/2+x)=(-1)^{n/2}\cos nx ## for odd n
    ##\sin n(\pi/2 +x) = (-1)^{(n-1)/2} \sin nx## for even n
    ##\cos n(\pi/2 +x) = (-1)^{n/2} \cos nx## for even n
    ##\cos n(\pi/2 +x) =-(-1)^{(n-1)/2} \sin nx## for odd n?

    Nevertheless,
    ##f(x) = f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n' \cos n(\pi/2+x) + b_n' \sin n(\pi/2+x)) ##,
    Supposed we have odd n, that would imply: ##a_n' \cos n(\pi/2 +x) = a_n'( -(-1)^{(n-1)/2} \sin nx)## and ##b_n' \sin n(\pi/2+x)= b_n'(-1)^{n/2}\cos nx ##. But exactly what is this telling us? Just that:
    ##a_n = b_n'(-1)^{n/2}## and ##b_n = -a_n'(-1)^{(n-1)/2)}##? (I use prime to note that it is the coefficients for ##f(\pi/2+x)##, and unprimed for ##f(x)##)
    Is this helpful at all?
     
    Last edited: Mar 29, 2017
  9. Mar 29, 2017 #8
    Yes, I did a typo while writing my previous post. Thanks for pointing this out. First, first I don't think you should not introduce other coefficients for ##f(\pi/2+x)## since the function ##f## is the same on both sides. By comparing term by term the series for ##f(\pi/2+x)## and ##f(x)## you get as said
    ##-(-1)^{(n-1)/2}a_n=b_n## and ##(-1)^{(n-1)/2}b_n=a_n## if ##n## is odd. But this can only be satisfied if ##a_n=b_n=0##. Now, for even ##n## one has ##(-1)^{n/2}a_n=a_n## and ##(-1)^{n/2}b_n=b_n##. So, we must have ##(-1)^{n/2}=1## or ##a_n=b_n=0##. Therefore, we have only non-zero ##a_n## and ##b_n## for ##n## being a multiple of 4. We have thus
    ##f(x)=a'_0/2+\sum_{n=1}^{\infty}(a'_n\cos(4nx)+b'_n\sin(4nx))##, where
    ##a'_n=a_{4n}## and ##b'_n=b_{4n}##. As is discussed on the page http://mathworld.wolfram.com/FourierSeries.html this correspond to a function with period ##\pi/2## which you already concluded from the beginning. This was maybe a long way to see this. But, I hope you are now able to proceed with the other parts of the exercise.
     
  10. Apr 5, 2017 #9
    Thank you very much for the help! I managed to solve all the problems except for the case when ## f(x) = f(2x)##, because here the trigonometric relations are much more complicated with squares and mixed terms. Do you have any last advice for this case?
     
    Last edited: Apr 5, 2017
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