A polynomial of $n^{th}$ degree can be expressed as a product of its roots:
$$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=a_n(x-R_1)(x-R_2)...(x-R_n)$$
I'm wondering about a general rule for all functions, and in particular, transcendentals.
The image below shows what I'm looking at for $\sin x$. It looks pretty close between $6\pi$ and $7\pi$, and goes wonky to the left and right. (Reminds me of $x\sin x$.)
Is there a product that satisfies the following?
$$\sin x =\lim_{n\rightarrow \infty}\prod_{i=1}^{n} f(n)(x-n\pi)$$
I engineered the scaling factor for my function in the illustration. I'm guessing it depends on $n$, so I put it to the right of the $\Pi$ as a function of $n$. How would we determine the scaling factor?
There is indeed a product formula for $\sin$: $$\sin x=x\prod_{n=1}^\infty \left( 1-\frac{x^2}{n^2\pi^2}\right) $$ For any continuous function on a closed interval, there is a sequence of polynomials converging to that function. You can extend this to all of $\mathbb R$ by choosing larger and larger closed intervals.