[Math Talk #1]. Periodic Functions and Fundamental Period

mathsolver (53)in #math • 8 years ago (edited)

Readers who are interested in mathematics already know what periodic functions are. But to be complete, here is the formal definition of periodic function.

Definition 1. (Periodic Function)

A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is said to be periodic if there exists $T>0$ such that

$f(x + T) = f(x)$
for all $x \in \mathbb{R}$ . Any such $T>0$ is called a period of $f$ .

The most famous periodic functions are trigonometric functions, such as $\sin x$
or $\cos x$ ; both are $2\pi$ -periodic functions. Now the question is, does there exist $0<T<2\pi$ such that

$\sin(x + T) = \sin x$
for all $x \in \mathbb{R}$ ? Well if does, then by trigonometric identity,
$\sin(x+T) = \sin x \cos T + \cos x \sin T = \sin x$
so that $\cos T = 1\ \land \ \sin T = 0 \implies T = 2n\pi$ for integer $n$ . Therefore, there is no such $T$ between $0$ and $2\pi$ . This implies that $2\pi$ is the smallest period of $\sin x$ (similar reasoning, $\cos x$ as well).

From this example, we can naturally define the fundamental period as follows.

Definition 2. (Fundamental Period)

For periodic function $f$ , we define

$P := \inf \left\{ T>0: T\text{ is period of }f\right\}$
to be fundamental period of $f$ if $P > 0$ .

The existence of $P$ is clear by completeness of Real numbers; since the set of periods of periodic function is bounded below by $0$ . The real question is then,

Is $P > 0$ always true for all periodic functions?
For the particular case when the functions are trigonometric, the answer was YES; we've just showed that $P = 2\pi$ . However, in general the answer is NO!

Example 1. Periodic Functions with no Fundamental Period

Consider

$f(x)=\begin{cases} 1 & (x \in \mathbb{Q}, \text{rational})\\ 0 & (x \in \mathbb{R} \setminus \mathbb{Q}, \text{irrational})\\ \end{cases}$ Take $r$ to be any positive rational number. Then $f(x + r) = f(x)$ for all $x \in \mathbb{R}$ regardless of the fact that $x$ is rational or not. (Check it!). This means that the whole set of positive rational numbers is period of $f$ ; $P = 0$

From this example, we can deduce some important facts about periodic functions not having fundamental periods. What does it mean for a function to be periodic with no fundamental period? Well, it is stated in the following theorem.

Theorem 1.

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is non-constant, continuous at one point $x=a$ and periodic, then $f$ has fundamental period. Furthermore, the fundamental period itself is also a period of $f$ .

Proof. Assume that $f$ has no fundamental period. Then we can construct a strictly decreasing sequence of periods of $f$ ; namely $\left\{ T_{n} \right\}$ such that $T_{n} \rightarrow 0$ . Define

$g_{n}(a):=\sup_{t \in [a, a+T_{n}]} f(t)-\inf_{t \in [a, a+T_{n}]} f(t)$
The function $g_{n}(a)$ is well defined, since $[a, a + T_{n}]$ is compact and $f$ is continuous. Since $T_{n}$ is a period, supremum and infinium of $f$ at $[a, a + T_{n}]$ is equal to supremum and infimum at whole set $\mathbb{R}$ . Now, taking $n \rightarrow \infty$ , by continuity of $f$ , $g_{n}(a) \rightarrow 0$ . This implies that the global oscillation of $f$ is zero; a constant function. This is contradiction; $f$ should have fundamental period $P$ . Again take a sequence of periods of $f$ ; $\left\{ P_{n} \right\}$ such that $P_{n} \rightarrow P$ . Using continuity,

$f(a+P) = f \left( \lim_{n \rightarrow \infty} (a + P_{n}) \right) = \lim_{n \rightarrow \infty} f(a + P_{n}) = \lim_{n \rightarrow \infty} f(a)=f(a)$
and we are done.

This theorem tells us that except for constant function, any example of periodic functions with no fundamental period should be discontinuous everywhere (as we have seen in Example 1). But we are not done yet. The final question remains.

Classification of Periodic Functions???

The question is directly related to examining the set

$\left\{ T>0: f(x + T) = f(x) \text{ for } \forall x \in \mathbb{R} \right\}$
We only examined the** infimum**, not the whole set itself. This will be discussed in Math talk #2.

Conclusion of Math Talk #1

Periodic functions not always have fundamental period.
If it does not have fundamental period, then it should be discontinuous everywhere or constant.

#mathematics #science #steemstem

8 years ago in #math by mathsolver (53)

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