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

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Theorem

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

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

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:

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

$\blacksquare$



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