what is the fundamental period of $\sin{x}$ and $\sin[x]$

Question

The question was which of the following is a periodic function, in which two of the possible answers which I thought of were $\sin{\{x\}}$ and $\sin[x]$ where $\{.\}$ and $[.]$ are fractional part and greatest integer functions respectively. The answer is the former one. But my question is why is it not the latter one. I mean $\sin()$ is a periodic function. At some point it will repeat itself the the latter case too.

Please also give the periods of both the functions(or whichever is periodic) please. Thanks in advance.

Answer 1

The period of $\sin(x)$ is $2\pi$.

For $f(x)=\sin\lfloor x\rfloor$, note that $f(0)=0$ only for $x\in[0,1)$. For other values, $f(x) = \sin(n), n\in \Bbb N$. But $\sin(x) = 0 \iff x = k\pi, k \in \Bbb Z$. Thus $f(x)$ is never zero again, or the function is not periodic.

Answer 2

As we increase x, $Sin[x]$ will be $sin(1)$ for sometime then $sin(2)$ then $sin(3)$ and so on..

Now for sin[x] to be periodic , $sin( a) =sin( b)$ at least once where a and b are 2 different integers. For this , $a=n*(irrational number) + (-1)^n*b$ where n is any integer. This equality can never hold. A contradiction.

Irrational no. Refers to ${\pi}$.

Answer 3

If I understand your question correctly, you're asking whether $\sin[x]$ is periodic, where $[x]$ denotes the fractional part of $x$, and your argument is that since $\sin x$ is periodic and $[x]$ is periodic, the composition $\sin[x]$ should be periodic when the periods overlap.

However, $\sin x$ has period $2\pi$ and $[x]$ has period $1$. If they have a common period $T$, we would need $T = 2\pi n = 1 m$ for some integers $n,m$. But that would mean $\pi = \frac{m}{2n}$, i.e., that $\pi$ is rational, which we know not to be true.

So two periodic functions don't necessarily ever have their periods overlap -- in particular, if the ratio between the periods is irrational, they won't.