Find the period of the function $y=[2x]-3*[4x]$ suppose that we have function
  $y=[2x]-3*[4x]$
here   $[*]$ denotes  as a minimum distance till   integer.
we are required to find period of this function,first of all  i am confused in terms of what does mean  minimum distance till  integer?could  you explain me it?thanks  guys  ,i know generally  term of period,namely  $f(x+T)=f(x)$,but here could not find  way how i can continue  to solve this task.
thanks a lot of
 A: It’s just the distance to the nearest integer.
Let $x$ be a real number. If $x$ is an integer, the smallest distance from $x$ to any integer is the distance from $x$ to $x$, which is $0$. Otherwise, $x$ lies between two consecutive integers, say $n<x<n+1$. Clearly $x$ is closer to $n$ and $n+1$ than it is to any other integer, so it’s just a matter of deciding which is closer. The distance from $x$ to $n$ is $x-n$; the distance from $x$ to $n+1$ is $n+1-x$. If $$x-n<n+1-x\;,$$ then $x$ is closer to $n$, and the minimum distance from $x$ to any integer is $x-n$. If $$n+1-x<x-n\;,$$ then $x$ is closer to $n+1$, and the minimum distance from $x$ to any integer is $n+1-x$. If $$n+1-x=x-n\;,$$ then both distances are $1/2$: $x$ is equidistant from $n$ and $n+1$, and its distance from the nearest integers is $1/2$.
As an example, as $x$ goes from $1$ to $2$, its minimum distance from an integer starts at $0$, increases to $1/2$ when $x=3/2$, and then decreases back to $0$. This cycle repeats as $x$ increases from $2$ to $3$. Thus, the nearest integer function itself has a period of $1$.
Added: This is very similar to finding the period of $\sin 2x+3\sin 4x$, for instance. Notice that the factor of $3$ in the second term doesn’t affect the period at all: it just changes the $y$-value. Thus, you’re really concerned with the periods of $[2x]$ and $[4x]$ and how they interact.
As $x$ increases from $1$ to $2$, say, $2x$ increases from $2$ to $4$. $[2x]$ starts out at $0$ when $2x=2$, rises to $1/2$ when $2x=5/2$ and falls back to $0$ at $2x=3$, then rises again to $1/2$ when $2x=7/2$ and finally falls back to $0$ when $2x=4$. In other words, the function goes through two full cycles. It finished the first cycle at $2x=3$, i.e., when $x=3/2$. The cycle started at $x=1$, so what is the period of $[2x]$?
You can analyze $[4x]$ with the same kind of reasoning. In both cases it helps to make a rough graph of the function: it’s all straight line segments, so no artistic ability is needed!
Finally, ask yourself whether the periods mesh. Does one of $[2x]$ and $[4x]$ go through a complete cycle or several complete cycles while the other is going through just one?
