Proof of sinusoidal periodicity

[truth is life]

Homework Statement

Prove that
[ltex]
f\left(x\right) = \cos(x) + \cos\left(\alpha x \right)
[/ltex]
where alpha is a rational number, is a periodic function.

EDIT: Also, what is it's period?

Homework Equations

[ltex]
f\left(x\right + p) = f\left(x\right)
[/ltex]
trig identities

The Attempt at a Solution

First, I used the definition of periodicity, then trig identities and term collection to get
[ltex]
\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
[/ltex]
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.

 

[Phrak]

Is alpha an integer, a fraction, or a real number?

 

[truth is life]

Is alpha an integer, a fraction, or a real number?

Duh, the one thing I forgot to put in the post. It's a rational number.

 

[Phrak]

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
\cos(x)[/ltex]<br /> and<br /> [ltex]\cos\left(\alpha x \right)[/ltex]<br /> will each have an integral number of cycles in some unknown interval.

 

[truth is life]

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
\cos(x)[/ltex]<br /> and<br /> [ltex]\cos\left(\alpha x \right)[/ltex]<br /> will each have an integral number of cycles in some unknown interval.

<br /> <br /> 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...<br /> <br /> Anyways, thinking about it your way, I have<br /> [ltex]n = \frac{k}{\alpha}[/ltex]<br /> 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<br /> [ltex]\frac{p}{q}[/ltex]<br /> 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.

 

[Phrak]

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.

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

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

2 \pi \frac{y}{M} = x

and

2 \pi \frac{y}{N} = \alpha x \ .

What is alpha in terms of M and N?