General Periodicity Property

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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 the period of $f$.


$\forall n \in \Z: \forall x \in X: f \left({x}\right) = f \left({x + n L}\right)$

That is, after every distance $L$, the function $f$ repeats itself.


Let $f: \R \to \R$ be a real function.

Then $f$ is periodic with period $L$ 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.


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 $n \to n + 1$.


\(\displaystyle f \left({x + \left({n + 1}\right)L}\right)\) \(=\) \(\displaystyle f \left({x + \left({L + n L}\right)}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f \left({\left({x + L}\right) + n L}\right)\) $\quad$ 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.