To what extent is the taylor polynomial the best polynomial approximation? Given a function $f\in\mathscr C^n([a,b])$ and a point $x_0\in [a,b]$, to what extent is the n-th taylor polynomial $T_n(x,x_0)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ the best polynomial approximation of $f$ in $[a,b]$ ? This may seem to be dumb question, but is there a metric $\rho$ on $ C^n([a,b])$ so that $\rho(T_n(x,x_0),f)=\min\{\rho(p,f)\mid \text{p is a polynomial function} \}$ ? 
Thank you
 A: The answer is that the Taylor polynomial is not a very good approximation on the whole of $[a,b]$ in general. Indeed, the rest of the Taylor series converges to 0 on $[a,b]$ if and only if $f$ is analytic, which of course is not always the case. The intuition is that local information near $x_0$ only has no chance of being sufficient for a good approximation on $[a,b]$.
We know that a continuous function can be approximated uniformly on a segment by polynomials, but it is a bit tricky to find which polynomials exactly. Another natural candidate would be interpolation polynomials, but it turns out that they are no good as well (see http://en.wikipedia.org/wiki/Runge%27s_phenomenon). The answer is Bernstein's polynomials (http://en.wikipedia.org/wiki/Bernstein%27s_polynomial_theorem).
A: $T_n(x, x_0)$ is the only polynomial of degree less than or equal to $n$ such that
$$ T_n(x, x_0) - f(x) \in o((x - x_0)^n) $$
or in terms of limits,
$$ \lim_{x \to x_0} \frac{T_n(x, x_0) - f(x)}{(x - x_0)^n} = 0$$
A: Here is a norm on $\mathscr C^n(a,b)$ for which $T_{n}(\cdot,x_0)$  is the best approximation to $f$: 
$$\|f\|_* = \sum_{k=0}^{n} |f^{(k)}(x_0)|+ \sup_{x\in[a,b]}|f^{(n)}(x)-f^{(n)}(x_0)|$$
This is a reasonable norm, which is equivalent to the more usual norms. For any polynomial $p$ of degree at most $n$ we have 
$$\|f-p\|_* = \sum_{k=0}^{n} |f^{(k)}(x_0)-p^{(k)}(x_0)|+ \sup_{x\in[a,b]}|f^{(n)}(x)-f^{(n)}(x_0)|$$
which is minimized exactly when $p=T_{n}(\cdot,x_0)$. 
