Proving a function is peridoic

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Show that cos(x) + cos(\alphax) is periodic if \alpha is a rational number.


Ok So I don't think I have ever done a question proving periodicity. But by definition a function f is periodic if:

f(x + p) = f(x)

so then:

f(x+p) = cos(x +p) + cos(\alpha(x+p))
= [cos(x)cos(p)-sin(x)sin(p)] + [cos(\alphax)cos(\alphap)-sin(\alphax)sin(\alphap)


and this is where I am stuck, so from what I've learned I have to solve for p is some fashion, I'm just not sure how,
 
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hi trap101! :smile:

hint: cos(0) + cos(\alpha0) = 2 :wink:
 
trap101 said:
Show that cos(x) + cos(\alphax) is periodic if \alpha is a rational number.

The first thing to do with a rational number is write it as \alpha = r/s for co-prime integers r and s.

Ok So I don't think I have ever done a question proving periodicity. But by definition a function f is periodic if:

f(x + p) = f(x)

so then:

f(x+p) = cos(x +p) + cos(\alpha(x+p))
= [cos(x)cos(p)-sin(x)sin(p)] + [cos(\alphax)cos(\alphap)-sin(\alphax)sin(\alphap)


and this is where I am stuck, so from what I've learned I have to solve for p is some fashion, I'm just not sure how,

You already know that \cos x has period 2\pi and \cos(\alpha x) has period (2\pi/\alpha). So you need to find p such that x + p = x + 2n\pi and x + p = x + 2m\pi/\alpha for integers n and m.
 
You have expressions for f(x) [given], and f(x+p) [derived above].
Now assume that alpha is rational, m/n, and then see if there is any period p which satisfies the condition f(x)=f(x+p).
 
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