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Homework Help: Periodic solutions

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi guys

    Say I have an equation of the form

    f(x) = cos(x)+cos(0.2x),

    and I wish to find the solutions x. When I plot this graph, I see multiple solutions, but there is no apparent period for the solutions. Are all the solutions equally valid, or can some be discarded?
  2. jcsd
  3. Mar 12, 2010 #2
    By "solutions", do you mean "zeroes", i.e., solutions of the equation [tex]f(x) = 0[/tex]?

    If this is the case, there is no sense in which any solution is less "equally valid" than any other. A solution of this equation is a solution of the equation.

    However, if it seems to you that the solutions of this particular equation are not periodic, you need to look more closely -- at the equation, not at a graph. The function [tex]f(x) = \cos x + \cos (0.2x)[/tex] is periodic indeed.
  4. Mar 12, 2010 #3
    Hmm, lets say that a solution x0 to f(x) is to be used in another function g(x)=sin(x). Since the zeroes of f(x) are not periodic with 2π, it matters which zero I choose. Is there a way to determine which one?
  5. Mar 12, 2010 #4
    One can't answer this meaningfully without more context.
  6. Mar 12, 2010 #5


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    Science Advisor
    Homework Helper

    Hi Niles! :smile:

    Use one of the standard trigonometric identities to get f(x) as a product of two sines.

    That should give you a period, and slso a pattern within that period. :wink:
  7. Mar 12, 2010 #6


    Staff: Mentor

    Your terminology is very imprecise. There is no such thing as a solution to a function. There is the idea of a solution to an equation, so maybe you mean solutions to the equation f(x) = 0. IOW the x-intercepts of this function.

    If you look at the graph of the function f(x) = cos(x) + cos(.2x), you should be able to see that it is periodic. In fact, its period is exactly the same as the period of cos(.2x). Once you figure out what the period is, it's straightforward to prove that f is periodic with that period, by showing that f(x + P) = f(x) for all real x.
  8. Mar 12, 2010 #7
    You mean two cosines, right? But I understand the thing with the period; my problem is that within that period, there are solutions that are not 2π-periodic with eachother. So when I have a harmonic function g(x), which takes the zeroes of f(x) as arguments, then I am not sure which one to pick.

    Yeah, I agree. It is very imprecise, but that is exactly what I mean. Thanks for claryfing that.
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