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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".
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".