General Periodicity Property
Contents
Theorem
Let $f: X \to X$ be a periodic function, where $X$ is either the set of real numbers $\R$ or the set of complex numbers $\C$.
Let $L$ be a periodic element of $f$.
Then:
- $\forall n \in \Z: \forall x \in X: f \left({x}\right) = f \left({x + n L}\right)$
That is, after every distance $L$, the function $f$ repeats itself.
Corollary
Let $f: \R \to \R$ be a real function.
Then $L$ is a periodic element of $f$ if and only if:
- $f \left({x \bmod L}\right) = f \left({x}\right)$
for all $x \in \R$, where $x \bmod L$ is the modulo operation.
Proof
Let $X = \mathbb C$.
There are two cases to consider: either $n$ is not negative, or it is negative.
Since the Natural Numbers are Non-Negative Integers, the case where $n \ge 0$ will be proved using induction.
Case 1
Basis for the Induction
The case for which $n = 0$ is trivial, because:
- $x + 0 \cdot L = x$
Induction Hypothesis
For some $n \in \Z_{\ge 0}$, suppose that:
- $f \left({x}\right) = f \left({x + n L}\right)$
Induction Step
For the induction step, let $n \to n + 1$.
Then:
\(\displaystyle f \left({x + \left({n + 1}\right)L}\right)\) | \(=\) | \(\displaystyle f \left({x + \left({L + n L}\right)}\right)\) | $\quad$ | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle f \left({\left({x + L}\right) + n L}\right)\) | $\quad$ Complex Addition is Associative | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle f \left({x + L}\right)\) | $\quad$ Induction Hypothesis | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle f \left({x}\right)\) | $\quad$ Basis for the Induction | $\quad$ |
Case 2
If $n \lt 0$, then:
\(\displaystyle f \left({x + n L}\right)\) | \(=\) | \(\displaystyle f \left({\left({x + n L}\right) - n L}\right)\) | $\quad$ by Negative of Negative Number is Positive and Case 1 | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle f \left({x + \left({n L - n L}\right)}\right)\) | $\quad$ Complex Addition is Associative | $\quad$ | |||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle f \left({x}\right)\) | $\quad$ | $\quad$ |
Combining the results above, it is seen that for all $n \in \Z$:
- $f \left({x}\right) = f \left({x + n L}\right)$
The proof for when $X = \R$ is nearly identical to the above proof.
$\blacksquare$