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Proof of sinusoidal periodicity

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that
    f\left(x\right) = \cos(x) + \cos\left(\alpha x \right)
    where alpha is a rational number, is a periodic function.

    EDIT: Also, what is it's period?

    2. Relevant equations
    f\left(x\right + p) = f\left(x\right)
    trig identities

    3. The attempt at a solution
    First, I used the definition of periodicity, then trig identities and term collection to get
    \cos x \cos p - \sin x \sin p - \cos x = \cos \alpha x - \cos \alpha x \cos \alpha p + \sin \alpha x \sin \alpha p
    Since p is a constant (if it exists) and the left side is periodic by definition, the right side and hence the function must be periodic as well, yes?

    EDIT: Now for the period, it would seem to need to be greater than 2 pi due to the naked cosine out in front. So, I would suppose that it would be something like [ltex]2 \pi + \frac{2 \pi}{\alpha}[/ltex], right?

    EDIT2: WTH is going on, the board keeps eating my latex! (much later) and then it starts working again. How bizarre.
    Last edited: Oct 23, 2009
  2. jcsd
  3. Oct 23, 2009 #2
    Is alpha an integer, a fraction, or a real number?
  4. Oct 23, 2009 #3
    Duh, the one thing I forgot to put in the post. It's a rational number.
  5. Oct 23, 2009 #4
    OK, good. I'm not sure where applying the trig identity could lead you, but I don't think it will help.

    By way of a hint, for
    [ltex]f\left(x\right) = \cos(x) + \cos\left(\alpha x \right)[/ltex]
    to be periodic
    [ltex]\cos\left(\alpha x \right)[/ltex]
    will each have an integral number of cycles in some unknown interval.
  6. Oct 23, 2009 #5
    The point behind applying the identity was to separate p from the interior of the function and allow me to rearrange the definition so that one side was in terms only of x and the other in terms of alpha x. Since I know that the basic functions are 2 pi periodic...

    Anyways, thinking about it your way, I have
    [ltex]n = \frac{k}{\alpha}[/ltex]
    where n is an integer (the number of cycles the plain cos has gone through) and k is also an integer referring to the number of cycles that cos alpha x has gone through. Since alpha is a rational number, it can be decomposed into the general form
    where p and q are integers. Since k and n are both integers, k = p since otherwise alpha would not reduce to an integer and hence the function must be periodic. QED. Huh, that was more straightforward than I thought.
  7. Oct 23, 2009 #6
    I didn't quite follow all of that but I think you've get the notion.

    We could change the form of the equation and write things out in a way that's a little easier on the eyes.

    [tex]f(y) = cos \left(2 \pi \frac{y}{M} \right) + cos \left( 2\pi \frac{y}{N} \right)[/tex]

    When y = pM and y = qN each will have cycled an integral number of times. In this form the problem would have been easier to solve, I think. I'm using variables that are all integers.

    To put it into the form of the given equation

    [tex]2 \pi \frac{y}{M} = x [/tex]


    [tex]2 \pi \frac{y}{N} = \alpha x \ .[/tex]

    What is alpha in terms of M and N?
    Last edited: Oct 23, 2009
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