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Continuity and periodic functions

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data
    We have a piecewise continous function and T-periodic function f and we have that:
    [tex]
    F(a) = \int_a^{a + T} {f(x)dx}
    [/tex]

    I have to show that F is diferentiable at a if f is continous at a.

    My attempt so far:

    I have showed that F is continous for all a. If we look at one piece where f is continous for a, we have that if F'(a) = f(a):

    [tex]
    \frac{1}{h}\int_a^{a + h} {f(x)dx = \frac{{\int_x^{a + h} {f(x)dx - } \int_x^a {f(x)dx} }}{h}}
    [/tex]

    When h -> 0, then the above goes to f(a) - this is just the Fundemental Theorem of Calculus.

    Now I have that:

    [tex]
    \frac{{F(a + h) - F(a)}}{h} = \frac{{\int_{a + h}^{a + h + T} {f(x)dx - } \int_a^{a + T} {f(x)dx} }}{h} = \frac{{\int_{a + T}^{a + h + T} {f(x)dx - } \int_a^{a + h} {f(x)dx} }}{h}
    [/tex]

    But where to go from here?
     
  2. jcsd
  3. Sep 8, 2008 #2

    tiny-tim

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    Hi Niles! :smile:

    Hint: f is T-periodic. :wink:

    (hmm … doesn't this work even if f is not continuous? :confused:)
     
  4. Sep 8, 2008 #3
    Then it goes to zero as h -> 0. The limit exists, and hence F is differnetiable at all points a, where f is diff. - correct? :smile:

    Also, does this mean that F'(a) = 0?
     
  5. Sep 8, 2008 #4

    tiny-tim

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    Actually, it's 0 anyway, isn't it?

    That was what was puzzling me. :confused:
    Yup! :biggrin:

    though the actual value was obvious from the definition, wasn't it? :wink:
     
  6. Sep 8, 2008 #5
    Great - I have two final (easy!) questions.

    1) I have to conclude that F is piece-wise constant.

    2) I have to show that F is constant

    My answers:

    1) Ok, it's kinda obvious that F must be constant. But I can't find the argument for F being piecewise constant, because I have shown that F is continous for all a in a previous question. Just because f is P.C., then it doesn't mean that F is it also.

    2) If F is constant (and periodic and piecewise continous), then it must be constant for all a, since it is bounded. This question was easy - #1 is annoying me :grumpy:
     
  7. Sep 8, 2008 #6

    tiny-tim

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    Isn't that obvious from the previous formula for F(a + h) - F(a)? :smile:

    Or am I missing something? :confused:
     
  8. Sep 8, 2008 #7
    Question nr. 2 was the easy one. Question nr. 1 was what I couldn't do :smile:
     
  9. Sep 8, 2008 #8

    tiny-tim

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    Doesn't piecewise constant just mean a multi-step step-function?

    So any constant function is piecewise constant? :confused:
     
  10. Sep 8, 2008 #9
    Hmm, I guess you are right.

    I have one final question, then I'll let you off the hook (I made a joke :rofl:).

    In my first post, in the last expression, I changed the limits of integration - here it is:

    [tex]

    \frac{{F(a + h) - F(a)}}{h} = \frac{{\int_{a + h}^{a + h + T} {f(x)dx - } \int_a^{a + T} {f(x)dx} }}{h} = \frac{{\int_{a + T}^{a + h + T} {f(x)dx - } \int_a^{a + h} {f(x)dx} }}{h}

    [/tex]

    Is the last step even necessary?
     
  11. Sep 8, 2008 #10

    tiny-tim

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    Hi Niles! :smile:

    Yes, I think it is …

    it's only the periodicity of f that makes it zero …

    and the periodicity doesn't obviously apply until the last step. :wink:
     
  12. Sep 8, 2008 #11
    Hmm, I can't quite see that. Do you mean that:

    [tex]
    \int_{a + T}^{a + h + T} {f(x) = \int_a^{a + h} {f(x)} } ?
    [/tex]
     
    Last edited: Sep 8, 2008
  13. Sep 8, 2008 #12

    tiny-tim

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    Yes, because the substitution y = x + T makes it

    [tex]\int_a^{a + h} f(y)\,dy\ =\ \int_a^{a + h} f(x)\,dx[/tex] :smile:
     
  14. Sep 9, 2008 #13
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