General Periodicity Property

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

Let $L$ be a periodic element of $f$.


$\forall n \in \Z: \forall x \in X: \map f x = \map f {x + n L}$

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


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.


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:

$\map f x = \map f {x + n L}$

Induction Step

For the induction step, let $n \to n + 1$.


\(\ds \map f {x + \paren {n + 1} L}\) \(=\) \(\ds \map f {x + \paren {L + n L} }\)
\(\ds \) \(=\) \(\ds \map f {\paren {x + L} + n L}\) Real Addition is Associative
\(\ds \) \(=\) \(\ds \map f {x + L}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \map f x\) Definition of Periodic Element

Case 2

If $n < 0$, then:

\(\ds \map f {x + n L}\) \(=\) \(\ds \map f {\paren {x + n L} - n L}\) Negative of Negative Number is Positive and Case 1
\(\ds \) \(=\) \(\ds \map f {x + \paren {n L - n L} }\) Real Addition is Associative
\(\ds \) \(=\) \(\ds \map f x\)

Combining the results above, it is seen that for all $n \in \Z$:

$\map f x = \map f {x + n L}$