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Half-range sine expansion

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I have to find the half-range sine expansion of a function f(x) = 1 for 0<x<2. My question is: This function is not piecewise smooth, so why does the book ask me to do this?
  2. jcsd
  3. Sep 20, 2008 #2


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    Hi Niles! :smile:

    It looks piecewise smooth to me …

    what definition of "piecewise smooth" are you using? :smile:
  4. Sep 21, 2008 #3
    That f and f' must be piecewise continuous, i.e. that there are a finite number of discontinuities, where the limits exists.

    f is "piecewise" continuous, so that is OK. But f' is defined on the interval 0 < x < 2 with no discontinuities?
    Last edited: Sep 21, 2008
  5. Sep 21, 2008 #4


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    f' = 0 on the interval 0 < x < 2 … that's continuous, isn't it? :wink:
  6. Sep 21, 2008 #5
    Yes, but if there are no discontinuities for either f or f', then f is simply just smooth? Does this count as f being piecewise smooth?
  7. Sep 21, 2008 #6
    That's right. In this case, there are two pieces, and in each piece, f and f' are both continuous.
  8. Sep 21, 2008 #7
    According to my book, if a function f is defined on an interval 0 < x < p, and if f is piecewise smooth, then it has a sine series expansion.

    In this case f is defined on 0 < x < 2. but it is only smooth - not piecewise smooth - on this interval?
  9. Sep 21, 2008 #8


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    Everything that's smooth is a fortiori (all the more so) piecewise smooth! :biggrin:
  10. Sep 21, 2008 #9
    So "smooth > piecewise smooth"?

    I just thought that there had to be some discontinuities.
  11. Sep 21, 2008 #10


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    Generally speaking "piecewise smooth" does include "smooth". But here, since you are talking about a "half-range sine expansion", you are assuming that the function is odd (as is sin(nx) for all x). That is, the full range is -2 to 2 and f(x)= -1 for -2< x< 0, 1 for 0< x< 2 and is periodic over all real numbers with period 4. That is surely "piecewise" smooth.
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