TITLE: Why Taylor series for $\sin$ about 0 does not converges uniformly to $\sin$ on $\mathbb{R}$?
QUESTION [0 upvotes]: I need to prove that the Taylor series for $f(x)=\sin(x)$ about 0,$$\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}x^k$$ does not converges uniformly to $f$ on $\mathbb{R}$.

REPLY [3 votes]: Abstract: Taylor series is a series of polynomial.
Assume it converges uniformly to sin, then by definition of uniform converges, there is a polynomial with infinitely many zeros. Or alternatively, since every non-constant polynomial are unbounded on $\mathbb{R}$, but sin is definitely bounded on $\mathbb{R}$. So there is no series of polynomial converges uniformly to sin.There is a contradiction.
Therefore, the Taylor series of sin does not converges uniformly to 
sin on $\mathbb{R}$.
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