General Periodicity Property

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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:

$\forall x \in \R: \map f {x \bmod L} = \map f x$

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:

\(\displaystyle f \left({x + \left({n + 1}\right)L}\right)\) \(=\) \(\displaystyle f \left({x + \left({L + n L}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle f \left({\left({x + L}\right) + n L}\right)\) Complex Addition is Associative
\(\displaystyle \) \(=\) \(\displaystyle f \left({x + L}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle f \left({x}\right)\) Basis for the Induction


Case 2

If $n \lt 0$, then:

\(\displaystyle f \left({x + n L}\right)\) \(=\) \(\displaystyle f \left({\left({x + n L}\right) - n L}\right)\) by Negative of Negative Number is Positive and Case 1
\(\displaystyle \) \(=\) \(\displaystyle f \left({x + \left({n L - n L}\right)}\right)\) Complex Addition is Associative
\(\displaystyle \) \(=\) \(\displaystyle f \left({x}\right)\)


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.

$\blacksquare$