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Proving sums of periodic functions need not be periodic(almost periodic)

  1. Jun 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi and thank you for reading this!

    Let [tex]\left.f(x) = cos(x) + cos\left(\pi x\right)[/tex]
    a) show that the equation f(x)=2 has a unique solution.
    b) conclude from part a that f is not periodic. Does this contradict withe the previous exercise that states if:
    [tex]\left.f_{}1,f_{}2,f_{}3...f_{}n[/tex] are T-perioidc functions, then:
    [tex]\left.a_{}1f_{}1+a_{}2f_{}2+...+a_{}nf_{}[/tex] is also T-periodic?


    2. Relevant equations


    3. The attempt at a solution
    So for part a, I did:
    [tex]\left.\int^{T}_{0}2 dx = 2T[/tex]

    no idea if I approached this problem correctly...

    but for part b) I took the integral of:

    [tex]\left.\int^{T}_{0}cos(x) + cos\left(\pi x\right)+2 dx = -sin(T)+2T[/tex]

    but the only other thing I can say is that f is not periodic, this is relatively obvious seeing how the problem at the end says, the function f is called almost periodic.

    Please let me know what's wrong and what should be the correct way of approaching this question
    Thanks!!!!!!!!!!!!!
     
  2. jcsd
  3. Jun 8, 2009 #2

    Mark44

    Staff: Mentor

    Why on earth did you integrate?
    For part a, you want to find the solution of cos(x) + cos(pi*x) = 2. Both functions in this sum have maximum values of 1 and minimum values of -1. For the two functions to have a combined value of 2, they must both have attained values of 1. Where does that happen?
     
  4. Jun 9, 2009 #3
    hm no idea why i integrated...

    x = 0 will give 2, but how do I "show" it? do I graph it and say "look, there's only one solution?" thanks
     
  5. Jun 9, 2009 #4

    Mark44

    Staff: Mentor

    I gave you a hint:
    List all of the values of x for which cosx = 1, and where cos(pi*x) = 1.
     
  6. Jun 9, 2009 #5
    there is only 1 value( 0) because it's a unique solution right....?
     
  7. Jun 9, 2009 #6

    Mark44

    Staff: Mentor

    That's what you have to show.
     
  8. Jun 9, 2009 #7
    so are you suggesting i create a list of values for each term that gives me 1(how many terms is considered adequate?) and say, ok zero is the only value that can be found on both cosx and cos(pi x) that can give both values 1, and as a result, give me the answer 2?
     
  9. Jun 9, 2009 #8

    Mark44

    Staff: Mentor

    Yes. List all the x values for which cos(x) = 1, and list all the x values where cos(pi*x) = 1. List enough of them to convince yourself that the only solution to cos(x) + cos(pi*x) = 2 is x = 0.
     
  10. Jun 9, 2009 #9
    so then on part b, if i combine the 3 terms so i get cos(x) + cos(pi*x) + 2, how should I approach to proving it not periodic? do I try to prove that:

    [tex]\left.\int^{T}_{0}f(x) = \int^{a+T}_{a}f(x)[/tex]

    doesn't hold?
     
  11. Jun 9, 2009 #10

    Mark44

    Staff: Mentor

    You're just dying to integrate something, aren't you?

    The definition of periodicity has nothing to do with integration. A function f is periodic with period T if f(x) = f(x + T) for all values of x.

    How are cos(x) and cos(pi*x) different from functions f1, f2, f3, ... of the previous problem?
     
  12. Jun 9, 2009 #11
    um what I'm thinking right now is that the f1,f2,f3 can have any arbitrary T while the cosine terms are inherently 2 pi periodic? Other than that I'm not sure how else they're different?
     
  13. Jun 9, 2009 #12
    Question: are the periods for cos(x) and cos(pi x) the same? Think about it. Graph them individually and actually measure the period.

    Question: Clearly at x=0 cos(x) + cos(pi x) = 2. This is because cos(x) and cox(pi x) both have local maxima of 1 at x=0. Are there any other values of x for which cos(x) and cos(pi x) both have a local maximum of 1 again? How can you show that? (HINT: find a general expression for x such that cos(x) has a local maximum, and an expression for x such that cos(pi x) has a local maximum, and show that the only x value satisfying both expressions is x=0).
     
  14. Jun 9, 2009 #13
    well to get the local maximum, for cos(x), the x must be a factor of 2 pi,
    In this case I used k as any integer so I will get a local maximum at 2 * pi *k?

    For the other term, I will get a local maximum simply having x be 2*k?
     
  15. Jun 9, 2009 #14

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You mean that x must be a "multiple" of 2pi, not a factor.

    Yes, cos(x) has it maximum value, 1, for x= 2k pi for any integer k.

    cos(pi x) has its maximum value, 1, for pi x= 2m pi or x= 2m for any integer m.

    In order that the value of this function be 2, cos(x) and cos(pi x) must both be 1. For what integers, k and m, is 2k pi= 2m? If k is not 0, you can solve that equation for pi and that should tell you something.
     
  16. Jun 9, 2009 #15
    if I solve for pi, it means uh.... pi = m/k which means that m must be bigger than k by pi which is represented through the pi in cos(pi x). This also means that when a x value is chosen the x value of cos(pi x) must be smaller than the x value in cos(x) due to the extra pi? And so this means the x can not be the same and so the only way to get the 1 for both cos(x) and cos(pi x) is simply when x = 0? Thanks
     
  17. Jun 9, 2009 #16
    The result is right, but the reasoning is wrong.

    If you have that k pi x = mx, then you invariably get to pi = m/k, where m and k are integers. This should strike you as odd... pi is a fraction? How can you choose m and k to make that true... is there a way? Why or why not?

    Clearly, for x = 0, the equation is true. But can you make it true for every x by an appropriate choice of m and k?
     
  18. Jun 9, 2009 #17
    well seeing how m and k are integers, then there's no way to reach a ratio of pi? Thanks
     
  19. Jun 9, 2009 #18
    "well seeing how m and k are integers, then there's no way to reach a ratio of pi?"

    Correct, except that the punctuation mark at the end should be a period, not a question mark, as you are stating a fact you should be aware of before trying to solve this problem.

    So if
    k pi x = m x

    then the only two options are that (1) x = 0 or (2) pi = m/k, and (2) doesn't work, you're only left with (1). So x=0 is the only place where the equation is true.
     
  20. Jun 9, 2009 #19
    i'm slightly confused how this ties in with a function's periodicity?
     
  21. Jun 9, 2009 #20
    A function is periodic iff f(x) = f(x + period) for every x.

    We now know that at x=0 cos(x) + cox(pi x) = 2, and that nowhere else is it equal to 2 (because nowhere else is the equation k pi x = mx true).

    Ergo...
     
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