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Calculus by Apostol Exercise 2.8 number 30

  1. Mar 13, 2015 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations


    3. The attempt at a solution
    I have no idea on how to start proving this, but I know the theorem is stating that the integral of a translated periodic function is the same with the integral of the periodic function without translation, is this concept correct? What I am thinking is to use the translation property of integrals to manipulate the equation but so far no success. Any hints or suggestions?
     
  2. jcsd
  3. Mar 13, 2015 #2

    BvU

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    That's the theorem. It'scorrect, but the exercise you are dealing with is asking you to prove that theorem. So you can't use it. You'll have to do something with the exact mening of 'preriodic' and 'integral'.

    Hint: ##\int_e^g = \int_e^f +\int_f^g##
     
  4. Mar 13, 2015 #3
    I still can't think on how to start with that hint...
     
  5. Mar 13, 2015 #4

    Ray Vickson

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    Nevertheless, PF rules forbid us from helping until you have done some of your own work on the problem.
     
  6. Mar 13, 2015 #5
    Let f be a periodic function with period p. ∫f(x)dx 0 to p is equal to ∫f(x+p)dx 0 to p since f is periodic. ∫f(x+p)dx 0 to p is equal to ∫f(x)dx np to np+p from the translation property. We can find a unique interval say, Np to Np+p where it contains a. Let Np=a, then ∫f(x)dx Np to Np+p is equal to ∫f(x)dx a to a+p. Is this rigorous enough?
     
  7. Mar 13, 2015 #6

    Ray Vickson

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    I don't understand the argument. Anyway, understanding and "intuition" are more important than "rigor" at this stage; after you see the reasons, they can easily be translated into rigorous statements. I suggest you draw a plot of y = f(x) over two periods, from 0 to 2p, and take the case where f >= 0, so the integral is the area under the graph. Now take 0 < a < p and look (on your graph) at the two areas (from 0 to p and from a to a+p). Can you now see what is happening?
     
  8. Mar 13, 2015 #7
    Should I draw random function? Why did you use a limit of integration from 0 to 2p? The given is 0 to p.
     
  9. Mar 13, 2015 #8

    Ray Vickson

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    You should draw whatever kind of periodic function you like; the argument will not depend on the details.

    Why go from x = 0 to x = 2p? The reason is that in the question you want to also go from x =a to x = a+p, where a > 0, so you want to look at parts of the graph y = f(x) for x > p. However, if 0 < a < p you do not need to go past x = 2p.

    Just sit down and start to work things through. That way you will understand completely and you will be able to answer your own questions.
     
  10. Mar 14, 2015 #9

    HallsofIvy

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    A simple change of variable should help you.
     
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