# Find the period of the function

1. Sep 15, 2005

### quasar987

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 $\Leftrightarrow b(x) = b(x+L) \Leftrightarrow cos(f(x)) = cos(f(x+L)) \Leftrightarrow f(x+L)=f(x)+2n\pi, \ \ n\in \mathbb{Z}$

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

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".

2. Sep 15, 2005

### AKG

You will end up with something like b being periodic iff

$$f(x) + 2\pi n = f(x + L) \forall x$$

But then f is clearly just the function defined by:

$$f(x) = 2\pi nx/L + C$$

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:

$$f(x) = 2\pi nx/L + C$$

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