Subtract Half is Replicative Function

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Theorem

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

$\forall x \in \R: f \left({x}\right) = x - \dfrac 1 2$


Then $f$ is a replicative function.


Proof

\(\displaystyle \sum_{k \mathop = 0}^{n - 1} f \left({x + \frac k n}\right)\) \(=\) \(\displaystyle \sum_{k \mathop = 0}^{n - 1} \left({x - \frac 1 2 + \frac k n}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle n x - \frac n 2 + \frac 1 n \sum_{k \mathop = 0}^{n - 1} k\)
\(\displaystyle \) \(=\) \(\displaystyle n x - \frac n 2 + \frac 1 n \frac {n \left({n - 1}\right)} 2\) Closed Form for Triangular Numbers
\(\displaystyle \) \(=\) \(\displaystyle n x - \frac n 2 + \frac n 2 - \frac 1 2\)
\(\displaystyle \) \(=\) \(\displaystyle n x - \frac 1 2\)
\(\displaystyle \) \(=\) \(\displaystyle f \left({n x}\right)\)

Hence the result by definition of replicative function.

$\blacksquare$


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