General Periodicity Property
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Theorem
Let $f: \R \to \R$ be a periodic real function.
Let $L$ be a periodic element of $f$.
Then:
- $\forall n \in \Z: \forall x \in X: \map f x = \map f {x + n L}$
That is, after every distance $L$, the function $f$ repeats itself.
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
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 $n \to n + 1$.
Then:
\(\ds \map f {x + \paren {n + 1} L}\) | \(=\) | \(\ds \map f {x + \paren {L + n L} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f {\paren {x + L} + n L}\) | 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 $n \in \Z$:
- $\map f x = \map f {x + n L}$
$\blacksquare$