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First of all: I am aware of the thread Weierstrass approximation does not hold on the entire Real Line.

My question is just that if we have a function like $sin(x)$ that can be approximated by its Taylor series, then the converge of the partial sums(which are polynomials basically) is uniform. Therefore, since $sin$ is no polynomial, I do not understand where I am wrong.

Weierstrass approximation does not hold on the entire Real Line

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    $\begingroup$ Every non-constant polynomial is unbounded, hence the approximation of $\sin x$ by any sequence of polynomials cannot be uniform. $\endgroup$ – Daniel Fischer Nov 21 '13 at 22:04
  • $\begingroup$ but is it not true that the taylor series for the sine function converges uniformly? $\endgroup$ – user66906 Nov 21 '13 at 22:06
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    $\begingroup$ I don't think the partial sums converge uniformly. Given any polynomial $f:\mathbb{R}\rightarrow\mathbb{R}$ of degree 1 or greater, $\lim_{x\rightarrow\infty}f(x)=\pm\infty$. So if $\{f_{n}\}\rightarrow \sin(x)$ we certainly cannot find a $\delta$ so that $|f_{n}(x)-\sin(x)|<\varepsilon$ for all $x\in\mathbb{R}$. $\endgroup$ – Eric Nov 21 '13 at 22:08
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    $\begingroup$ No. It converges uniformly on every bounded subset of $\mathbb{R}$ (or more generally, of $\mathbb{C}$), but not on the entire line. $\endgroup$ – Daniel Fischer Nov 21 '13 at 22:08
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The question has been answered in comments: a power series $\sum_{n=0}^\infty a_n x^n$ converges uniformly on $\mathbb R$ if and only if there is $N$ such that $a_n=0$ for all $n>N$.

In this context it is natural to mention a theorem due to Carleman (1927): For every continuous function $f:\mathbb R\to\mathbb C$ and for every continuous positive function $\epsilon: \mathbb R\to (0,\infty)$ there exists an entire function $g:\mathbb C\to\mathbb C$ such that $$|f(x)-g(x)|<\epsilon (x) \quad \text{for all } \ x\in\mathbb R$$

As a special case (taking $\epsilon(x)\equiv \epsilon$) we find that every continuous function is a uniform limit of entire functions. But the theorem also allows for $\epsilon$ to decay to zero as a function of $x$, as fast as you want.

Of course, Carleman's theorem does not say anything new about $\sin x$, since it is itself entire.

For more on approximation on unbounded subsets of $\mathbb C$, see Lectures on complex approximation by D. Gaier.

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