$\left[x+ 1/2\right] + \left[x-1/2\right] + 2\left[-x\right]$.
I have to prove that this is a periodic function, and find it's period.
Note [.] Represents greatest integer function(GIF for short).
I tried solving it by making a graph and found out that the it was a periodic function with period 1.
But is there any other approach which uses the properties of GIF like when we have find period of, for example, $\sin 4x$, we divide the time period i.e., $2π$ by $4$.
On
When there are many functions , and we have to find the period of the whole function , we find the LCM(Least common factor ) of the periods of the functions.
As all the three terms are GIF functions , they all have periods 1.
Therefore , LCM of periods is also , 1.
On
First, let's compute $f(x+T)$, assuming that $T$ is an integer.
\begin{align} f(x + T) &= \left\lfloor (x+T) + 1/2 \right\rfloor + \left\lfloor (x+T) - 1/2 \right\rfloor + 2\left\lfloor -(x+T) \right\rfloor\\ &= \left\lfloor x + 1/2 + T \right\rfloor + \left\lfloor x - 1/2 + T \right\rfloor + 2\left\lfloor -x - T \right\rfloor \\ &= \left\lfloor x + 1/2\right\rfloor + T + \left\lfloor x - 1/2 \right\rfloor + T + 2(\left\lfloor -x \right\rfloor - T) \\ &= \left\lfloor x + 1/2\right\rfloor + \left\lfloor x - 1/2 \right\rfloor + 2\left\lfloor -x \right\rfloor \\ &= f(x) \end{align}
In particular, $f(x+1) = f(x)$.
It's not too much work, now, to show that, for $0 \lt x \lt 1$,
$f(x) = \begin{cases} -3 & \text{If $0 \lt x \lt \frac 12$} \\ -1 & \text{If $\frac 12 \le x \lt 1$} \\ \end{cases}$
So the period of $f(x)$ is $1$.
Yes .
If we use the properties of GIF :
x=[x]+{x}
and [-x]=-[x]-1 (where x is not an integer)
We will be getting;
[x+ 1/2] + [x-1/2] + 2[-x]
=(x+ 1/2) -{x+ 1/2} + (x-1/2)-{x-1/2} + 2(-x+{x}-1)
= -2 + {x+ 1/2}-{x - 1/2} + 2{x} Proceeding in this manner using the properties of GIF will too result in the period coming out to be 1. Hope that helps!!