Why is sequence $\sum_{k = 0}^{n}\frac{(-1)^{k}}{(2k + 1)!}x^{2k + 1}$ not a sequence of polynomials converging to $f(x) = \sin x$?

Question

It is a fact that $\sin x = \sum_{k = 0}^{\infty}\frac{(-1)^{k}}{(2k + 1)!}x^{2k + 1}$. It would appear that $\sum_{k = 0}^{n}\frac{(-1)^{k}}{(2k + 1)!}x^{2k + 1}$ is a sequence of polynomials converging uniformly on $\mathbb{R}$ to $\sin x$, since $\lim_{n \rightarrow \infty}\sum_{k = 0}^{n}\frac{(-1)^{k}}{(2k + 1)!}x^{2k + 1} = \sin x$. However, there is no sequence of polynomials converging uniformly on $\mathbb{R}$ to $\sin x$ (proof). Can anyone resolve this misunderstanding? Thanks!

Answer 1

The problem is the word uniformly, which does not hold over all of $\mathbb{R}$.

What happens is that you need more and more terms of the Taylor polynomial to properly approximate $\sin(x)$ (to the same $\epsilon$) the further away from $x=0$ you go, and therefore convergence $N(\epsilon)$ depends on $x$ - not uniform.

Answer 2

If a sequence of polynomial functions $ \left(P_{n}=\sum\limits_{k=0}^{n}{\left(x\mapsto a_{k}x^{k}\right)}\right)_{n} $ converges uniformly, on $ \mathbb{R} $, to $ P $, then $ P $ is a polynomial.

Since $ \sin{x} $ is not a polynomial. $ \left(\sum\limits_{k=0}^{n}{\left(x\mapsto\frac{\left(-1\right)^{k}}{2k+1}x^{2k+1}\right)}\right)_{n} $ does not converge uniformly, on $ \mathbb{R} $, to $ \sin{x} \cdot $