Replicative Function of x minus Floor of x is Replicative

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Theorem

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

Let $f$ be a replicative function.


Let $g: \R \to \R$ be the real function defined as:

$\map g x = \map f {x - \floor x}$


Then $g$ is also a replicative function.


Proof

Lemma

Let $x \in \R$.

Suppose $x - \floor x < \dfrac 1 n$.

Then:

$\floor {x + \dfrac k n} = \dfrac {\floor {n x} } n$

for any $0 \le k \le n - 1$.

$\Box$


First observe:

$\ds \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n} = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n - \floor {x + \frac k n} }$
\(\ds \map g {n x}\) \(=\) \(\ds \map f {n x - \floor {n x} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \map f {\frac {n x - \floor {n x} } n + \frac k n}\) Definition of Replicative Function
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \map f {x - \frac {\floor {n x} } n + \frac k n}\)

We need to show that the two final sums are equal.


Suppose:

$\dfrac j n \le x - \floor x < \dfrac {j + 1} n$

for some integer $j$.

By Real Number minus Floor:

$0 \le x - \floor x < 1$

This gives:

$0 \le j \le n - 1$


Let $y = x - \dfrac j n$.

Then:

$\floor x \le x - \dfrac j n = y$

By Number not less than Integer iff Floor not less than Integer:

$\floor x \le \floor y$

On the other hand:

$y \le x$

which gives

$\floor y \le \floor x$

So $\floor y = \floor x$.


We now have:

$\dfrac j n \le x - \floor x = y + \dfrac j n - \floor y < \dfrac {j + 1} n$

which gives:

$0 \le y - \floor y < \dfrac 1 n$

which satisfies the condition for the lemma.


Hence:

\(\ds \map f {y + \frac k n - \floor {y + \frac k n} }\) \(=\) \(\ds \map f {y - \frac {\floor {n y} } n + \frac k n}\) by the lemma
\(\ds \) \(=\) \(\ds \map f {x - \dfrac j n - \frac {\floor {n x - j} } n + \frac k n}\)
\(\ds \) \(=\) \(\ds \map f {x - \dfrac j n - \frac {\floor {n x} } n + \dfrac j n + \frac k n}\) Floor of Number plus Integer
\(\ds \) \(=\) \(\ds \map f {x - \frac {\floor {n x} } n + \frac k n}\)


We also have:

\(\ds \sum_{k \mathop = 0}^{n - 1} \map f {y + \frac k n - \floor {y + \frac k n} }\) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \map f {x - \frac j n + \frac k n - \floor {x - \frac j n + \frac k n} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{j - 1} \map f {x + \frac {k - j} n - \floor {x + \frac {k - j} n} } + \sum_{k \mathop = j}^{n - 1} \map f {x + \frac {k - j} n - \floor {x + \frac {k - j} n} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{j - 1} \map f {x + \frac {k - j} n + 1 - \floor {x + \frac {k - j} n} - 1} + \sum_{k \mathop = 0}^{n - j - 1} \map f {x + \frac k n - \floor {x + \frac k n} }\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{j - 1} \map f {x + \frac {k - j} n + 1 - \floor {x + \frac {k - j} n + 1} } + \sum_{k \mathop = 0}^{n - j - 1} \map f {x + \frac k n - \floor {x + \frac k n} }\) Floor of Number plus Integer
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{j - 1} \map f {x + \frac {n - j + k} n + 1 - \floor {x + \frac {n - j + k} n} } + \sum_{k \mathop = 0}^{n - j - 1} \map f {x + \frac k n - \floor {x + \frac k n} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = n - j}^{n - 1} \map f {x + \frac k n + 1 - \floor {x + \frac k n} } + \sum_{k \mathop = 0}^{n - j - 1} \map f {x + \frac k n - \floor {x + \frac k n} }\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n - \floor {x + \frac k n} }\)

Hence the result.

$\blacksquare$


Sources

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