Euler Formula for Sine Function/Real Numbers/Proof 3

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

for all $x \in \R$.


We have that $\sin x$ has a power series representation:

$\sin x = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots$

The roots of sine are the numbers $k \pi$, where $k$ is any integer.

From the Polynomial Factor Theorem, the following might be true:

$\displaystyle \sin x = A x \prod \left({1 - \frac x {k \pi} }\right)$

where the product is taken over all $n \in \Z \setminus \left\{{0}\right\}$, and $A$ is some constant.

The intuition is as follows.

\(\displaystyle \sin x\) \(=\) \(\displaystyle \ldots \left({1 - \frac x {2 \pi} }\right) \left({1 - \frac x \pi}\right) A x \left({1 + \frac x \pi}\right) \left({1 + \frac x {2 \pi} }\right)\cdots\)
\(\displaystyle \) \(=\) \(\displaystyle A x \left({1 - \frac {x^2} {\pi^2} }\right) \left({1 - \frac {x^2} {2^2 \pi^2} }\right) \left({1 - \frac {x^2} {3^2 \pi^2} }\right) \cdots\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\sin x} x\) \(=\) \(\displaystyle A \left({1 - \frac {x^2} {\pi^2} }\right) \left({1 - \frac {x^2} {2^2 \pi^2} }\right) \left({1 - \frac {x^2} {3^2 \pi^2} }\right) \cdots\) for $x \ne 0$.

That $\dfrac {\sin x} x \to 1$ as $x \to 0$ is a well known limit.

Letting $x$ tend to $0$ in the above equation implies that $A = 1$.

We now formalize the above claims.