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Find the period of the function

  • Thread starter quasar987
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  • #1
quasar987
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I have doubts concerning this problem.

Consider f:R-->R a continuous non-periodic function, and the function b(x) = cos(f(x)). Is b(x) periodic, if so, with what period?

I got...

b(x) is periodic of period L [itex]\Leftrightarrow b(x) = b(x+L) \Leftrightarrow cos(f(x)) = cos(f(x+L)) \Leftrightarrow f(x+L)=f(x)+2n\pi, \ \ n\in \mathbb{Z}[/itex]

So it depends on f wheter b is periodic of not. For exemple, if f(x) = x, then for a given n, [itex] f(x+L) = f(x)+2n\pi \Leftrightarrow x+L=x+2n\pi \Rightarrow 0<L=2n\pi[/itex] and hence the period of b(x) is [itex]2\pi[/itex].


I have faith in what I have done; it's just that it never happened to me in 21 years of life that the answer to a yes/no question in a textbook is "it depends".
 

Answers and Replies

  • #2
AKG
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You will end up with something like b being periodic iff

[tex]f(x) + 2\pi n = f(x + L) \forall x[/tex]

But then f is clearly just the function defined by:

[tex]f(x) = 2\pi nx/L + C[/tex]

for some constant C. Such a function is non-periodic (if n is not zero) and continuous. So yes, it does depend on f. If f is one of the family of functions defined by:

[tex]f(x) = 2\pi nx/L + C[/tex]

for [itex]n \in \mathbb{Z} - \{0\}[/itex], [itex]C \in \mathbb{R}[/itex], [itex]L \in \mathbb{R} - \{0\}[/itex] then the answer is yes (and with period L), and it is no otherwise. I did this in a rush, hopefully it's right.
 

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