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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

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up vote 2 down vote accepted

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?

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thanks a lot,could you tell how can USE this fact to solve given problem?for example $[2x]$ ,how could i write this in terms of minimum distance? – dato datuashvili Jun 17 '12 at 13:15
@dato: Look at some examples. As $x$ increases from $1$ to $2$, for instance, $2x$ increases from $2$ to $4$; what does $[2x]$ do? It’s $0$ when $2x$ is $2,3$, or $4$; where is it $1/2$? Where is it increasing, and where is falling? Make a rough graph. As a further hint, how do the periods of $\sin x$ and $\sin 2x$ compare? This situation is very similar. – Brian M. Scott Jun 17 '12 at 13:19
so does it means that in general sin(nx) has period n? or nx has period n? – dato datuashvili Jun 17 '12 at 20:55
@dato: Stop and think: does $\sin x$ have period $1$? No: it has period $2\pi$. The function $\sin 2x$ oscillates twice as fast, so it has period $\pi$. Similarly, $\sin 3x$ has period $\frac{2\pi}3$. I wasn’t saying that $[nx]$ was exactly like $\sin nx$, just that if you understand how the period of $\sin nx$ is related to that of $\sin x$, you should be able to see how the period of $[nx]$ is related to that of $[x]$. – Brian M. Scott Jun 18 '12 at 8:05

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