# 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

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Interesting question! One things to remark is that the Taylor polynomial is definitely not the best approximation in the sup norm on $\mathcal C[a,b]$. This was studied by Tchebychev, who showed that for each $n$, there is a best such approximation of degree $n$. It is usually significantly more accurate than the Taylor polynomial. See for example Chapters 43-45 (especially 44) of Körner's "Fourier analysis". – Andrés Caicedo Feb 9 '13 at 17:20

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 $p=T_{n}(\cdot,x_0)$.

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+1 This is very interesting, I had never seen such a norm. Can you provide a reference where I can find that norm defined? – Adrián Barquero Feb 11 '13 at 17:06
@AdriánBarquero Sure. Here it is in amsrefs format: \bib\{298965}{misc}{ title={To what extent is the taylor polynomial the best polynomial approximation?}, author={5PM (http://math.stackexchange.com/users/53153/5pm)}, note={URL: http://math.stackexchange.com/q/298965 (version: 2013-02-11)}, eprint={http://math.stackexchange.com/q/298965}, organization={Mathematics} } // but seriously, I don't recall seeing it anywhere. – user53153 Feb 11 '13 at 17:22
I see, so you came up with it? That's really clever. Thanks =) – Adrián Barquero Feb 11 '13 at 18:03
I may be misunderstanding something here but if I set f = p in your metric the first term is zero but the second term is not zero. But isn't the norm of 0 supposed to be zero? – Michael Smith Feb 12 '13 at 22:38
@MichaelSmith If $f$ is a polynomial of degree at most $n$, then $f^{(n)}$ is a constant function. – user53153 Feb 12 '13 at 22:41

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).

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The Runge phenomenon does not mean that "interpolation polynomials ... are no good". It shows the limitations of interpolation with equidistant nodes. Interpolation with Chebyshev nodes yields uniform approximation, and the Wikipedia article you cited mentions this. – user53153 Feb 9 '13 at 19:20
you're right. my point was just that this is a non-trivial question, and that naive answer may be false. – Glougloubarbaki Feb 10 '13 at 3:48

$T_n(x, x_0)$ 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,

$$\lim_{x \to x_0} \frac{T_n(x, x_0) - f(x)}{(x - x_0)^n} = 0$$

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