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Is a function really equal to its Fourier series?

  1. Oct 2, 2015 #1
    Suppose all Dirichlet conditions are met and we have a function that has jump discontinuities.

    Dirichlet's theorem says that the series converges to the midpoint of the values at the jump discontinuity.

    What bothers me then is: Dirichlet's theorem is basically telling us the series isn't the same as the function, precisely because it converges to the midpoint and doesn't itself have that jump discontinuity!

    So, are we being easy with the equality sign? Whats going on? Is a function really equal to its Fourier series representation?
     
  2. jcsd
  3. Oct 2, 2015 #2

    Geofleur

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    I think we might say, in such a case, that the function is equal to its Fourier series representation everywhere except on a set "of measure zero"; then the equal sign is really an equivalence relation. Hopefully a more mathematically rigorous person can confirm.
     
  4. Oct 2, 2015 #3

    FactChecker

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    The infinite series does have a jump discontinuity. It is only the value at the point of the jump that may be different.
     
  5. Oct 2, 2015 #4
    ....................º---------------

    ................... ..

    ---------------º


    So my series gives a graph like the one I've drawn above? Namely, my series has a jump discontinuity but its value at the discontinuity is given by the dot I've drawn as the midpoint?
     
  6. Oct 2, 2015 #5
    ... For some reason I cant draw correctly here.

    Edit: I think it looks okay now.
     
  7. Oct 2, 2015 #6

    FactChecker

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  8. Oct 2, 2015 #7
    But still it isn't exactly the same as the function. So can we conclude that its an "easy-type" of equality?
     
  9. Oct 2, 2015 #8
    By the way, summing continuous functions to produce a jump discontinuity seems non-intuitive to me. So I was wondering how can this be proven? Which theorem/concept should I look out for?

    Thanks.
     
  10. Oct 2, 2015 #9

    FactChecker

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    The definition of "equal" is not "easy". The theorem does not say the series equals the function at the points of discontinuity. The way the continuous series converges to a discontinuous function is that the partial sums get steeper and steeper at the discontinuity. The higher frequencies have steep slopes and can get add up to approach a step.

    This should get you started: https://en.wikipedia.org/wiki/Dirichlet_conditions
     
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