# Logarithm of Absolute Value of 2 times Sine of pi x is Replicative Function

## Theorem

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

$\forall x \in \R: f \left({x}\right) = \log \, \left \vert{2 \sin \pi x}\right\vert$

Then $f$ is a replicative function.

## Proof

We have that:

 $\displaystyle \sum_{k \mathop = 0}^{n - 1} f \left({x + \frac k n}\right)$ $=$ $\displaystyle \sum_{k \mathop = 0}^{n - 1} \log \, \left \vert{2 \sin \pi \left({x + \frac k n}\right)}\right\vert$ $\displaystyle$ $=$ $\displaystyle \log \prod_{k \mathop = 0}^{n - 1} \left \vert{2 \sin \pi \left({x + \frac k n}\right)}\right\vert$ Sum of Logarithms

Thus to demonstrate that $f$ is replicative, it is sufficient to demonstrate that:

$\displaystyle \prod_{k \mathop = 0}^{n - 1} \left({2 \sin \pi \left({x + \frac k n}\right)}\right) = 2 \sin \pi n x$

We have:

 $\displaystyle 2 \sin \theta$ $=$ $\displaystyle \frac {e^{i \theta} - e^{-i \theta} } i$ $\displaystyle$ $=$ $\displaystyle \frac {e^{i \theta} e^{-i \theta} - e^{-i \theta} e^{-i \theta} } {i e^{-i \theta} }$ $\displaystyle$ $=$ $\displaystyle \left({1 - e^{-2 i \theta} }\right) \left({-i e^{i \theta} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({1 - e^{-2 i \theta} }\right) \left({e^{i \theta} e^{-i \pi / 2} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({1 - e^{-2 i \theta} }\right) \left({e^{i \theta - i \pi / 2} }\right)$ $\displaystyle \leadsto \ \$ $\displaystyle \prod_{k \mathop = 0}^{n - 1} \left({2 \sin \pi \left({x + \frac k n}\right)}\right)$ $=$ $\displaystyle \prod_{k \mathop = 0}^{n - 1} \left({1 - e^{-2 \pi \left({x + i k / n}\right)} }\right) \left({e^{\pi \left({x - \left({1 / 2}\right) + \left({k / n}\right)}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({1 - e^{-2 \pi n x} }\right) \left({e^{\pi \left({n x - 1 / 2}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle 2 \sin n x$

$\blacksquare$