Period of the function $\sin (8\pi \{x\})$

Question

My question is to to find the period of $$\sin(8\pi\{x\}),$$ where $\{\cdot\}$-is the fractional part of function.

I know that the period of $\{\cdot\}$ is 1 and the period of $\sin(8\pi x)$ is $1/4$. But how to find the overall period of the given function?

Answer 1

Let $g(f(x))$ be the composition of $g$ and $f$.

Further, suppose that $g(x)$ is an odd function, so that $g(f(x)) = -g(f(-x))$ .

Then it follows , that the period of $g(f(x))$ must be same as period of $f(x)$ because :

$g(f(x+T)) = g(f(x))$

In your example, let $g(x) =$ sin$(8πx)$ and $f(x) =$ {$x$}

As established above, period of $g(f(x))$ is same as that of $f(x)$ . Hence, period of sin$(8π${$x$}) is same as that of {$x$} , which is 1 .

Answer 2

$\sin(8 \pi \{x\}) = \sin(8 \pi ( x - \lfloor x \rfloor)) = \sin(8 \pi x - 8 \pi \lfloor x \rfloor) = \sin(8 \pi x)$,

hence $\sin(8 \pi \{x\})$ has period $\frac{1}{4}$.