2.18 Periodic Functions

In physics and engineering we see periodic phenomena such as vibrations, the motion of the tide, planetary, and alternating current (AC).

We say a function $f$ is periodic with period $T$ if $x+T$ lies in the domain of $f$ whenever $x$ lies in the domain of $f$ and if for every $x$ in the domain of $f$
\[
f(x+T)=f(x)
\]
  •  The condition “$x+T$ lies in the domain of $f$ whenever $x$ lies in the domain of $f$” does not say that $f$ needs to be defined for all $x$. For example,
  •  The period of a periodic function is not unqiue. In fact, if $f$ is periodic with $T$, it is also periodic with periods $2T$, $3T,\dots$ , or $-T$ and $-2T,\dots$ because
    \[
    f(x+2T)=f((x+T)+T)=f(x+T)=f(x)
    \] and
    \[
    f(x-T)=f(x-T+T)=f(x),
    \] and so on. In general:
If $f$ is periodic with period $T$ then
\[
f(x+nT)=f(x)
\] for every integer $n$.

 

The smallest positive period of a periodic function (if exists) is called the (fundamental) period of the function.

 

  • A constant function $f(x)\equiv3$ is periodic, and every number $T$ is its period
    \[
    f(x+T)=f(x)=3.
    \] Therefore, there is no smallest period, and hence $f$ does not have a fundamental period.
Example84
Determine the fundamental period of $f(x)=2x-\lfloor2x\rfloor$.
Solution
Let $T$ be the period of $f$. Thus
\begin{align*}
f(x+T) & =f(x)\\
2(x+T)-\lfloor2(x+T)\rfloor & =2x-\lfloor2x\rfloor\\
\cancel{2x}+2T-\lfloor2(x+T)\rfloor & =\cancel{2x}-\lfloor2x\rfloor
\end{align*}
Therefore
\[
2T=\lfloor2(x+T)\rfloor-\lfloor2x\rfloor
\] Because both $\lfloor2(x+T)\rfloor$ and $\lfloor2x\rfloor$ are integers, $2T$ must be an integer too. To determine the fundamental period,we choose $2T$ to be the smallest positive integer, 1:
\[
2T=1\Rightarrow T=\frac{1}{2}.
\] That is, the fundamental period of $f$ is $1/2$. The graph of this
function is shown below.
Example85
Show that $f(x)=x^{2}-\lfloor x^{2}\rfloor$ is not periodic.
Solution
Assume $f$ is periodic with period $T$. Then
\[
f(x+T)=f(x)
\] \[
\Rightarrow(x+T)^{2}-\lfloor(x+T)^{2}\rfloor=x^{2}-\lfloor x^{2}\rfloor
\] Expanding $(x+T)^{2}$, we get
\[
\cancel{x^{2}}+2xT+T^{2}-\lfloor(x+T)^{2}\rfloor=\cancel{x^{2}}-\lfloor x^{2}\rfloor
\] or
\[
2xT+T^{2}=\lfloor(x+T)^{2}\rfloor-\lfloor x^{2}\rfloor
\] The right-hand side is always an integer, so the left-hand side for every real number $x$ must be an integer too, and that is impossible. [For any number $T$, $2xT+T^2$ is a linear function of $x$ that can take any real value not only integers]. The graph of this function is shown below.