How to use Taylor Series to find infinite power of $\sin x$

Question

I am told that the Taylor Series for $\sin(x)$ is...

$$p(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!}$$

I know that the product for a finite polynomial has the form...

$$ p(x) = c\prod_{i = 1}^n(x - a_i) $$

How can one use the Taylor Series to calculate $c$?

I know that $\sin(\pi z)$ regardless of what $z$ is will equate to 0 for all $z\in\Bbb{Z}$, so I can rewrite $p(x)$ as...

$$ p(x) = c\prod_{i = 1}^n(x - \pi z) $$

But I still cannot figure out how $c$ is determined. Can someone please explain?

Answer 1

You have $p(x) = c(x-a_1)\cdots(x-a_n) = cx^n\pm \cdots\pm ca_1\cdots a_n$. So, $c$ is the coefficient in front of the highest power of $x$ in the Taylor series truncation. In your case above $c = -\frac 1{7!}$.

Answer 2

Just wanted to add that for the full expansion of $\sin(x)$, the expression is not possible. You have to truncate it like in the answer by @amsmath.

It seems like $c$ can't have a finite value. Let's substitute $x=\frac{\pi}{2}$ into your equation. We get $p(\frac{\pi}{2})=1$. This implies:

$$1=c \prod\limits_{z=-\infty}^{\infty} (\frac{\pi}{2}-z\pi)$$ $$=> \frac{1}{c} = \dots\frac{5\pi}{2} \frac{3\pi}{2} \frac{\pi}{2} \frac{-\pi}{2} \frac{-3\pi}{2}\dots$$

It is quite clear that the RHS blows up. So we must have $c=0$, meaning the expression is not possible.