Let $p_{n,K}(x)$ be the polynomial of degree $n$ that minimizes $\epsilon_{n,K} = ||p_{n,K}(x) - sin(x)||_\infty$ on the interval $[-K, K]$. Question: what is the asymptotic behavior of $\epsilon_{n,K}$ where $n \to \infty$?
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$\begingroup$ Should "the polynomial that minimizes" be "the polynomial of degree $n$ that minimizes"? $\endgroup$– Steven LandsburgCommented Sep 15, 2018 at 18:14
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1 Answer
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The rate of polynomial approximation $$E_{n}(f,\mathcal{K}):=\inf\{\max_{z\in \mathcal{K}}|f(z)-P_{n}(z)|,~\text{deg}~P_{n}\leq n\}$$ to an entire function $f$ on a compact set $\mathcal{K}$ of the complex plane of positive capacity was derived by A. Batyrev in
Batyrev, A. V., On the best approximation of analytic functions by polynomials. Doklady Akad. Nauk SSSR (N.S.) 76, (1951) 173--175.
His theorem states that $$ \operatorname { limsup } _ { n \rightarrow \infty } n \left[ E _ { n } ( f , \mathcal{K} ) \right] ^ { \rho / n } = e \rho \tau ^ { \rho } \sigma. $$ Here $\tau$ is the capacity of $\mathcal{K}$, $\rho$ and $\sigma$ are the order and type of $f$ respectively.
Since the segment $\mathcal{K}=[-K,K]$ has capacity $K/2$, and the sine function is of order 1 and type 1, one gets that $E_{n}(\sin,\mathcal{K})$ decreases like $$\left(\frac{eK}{2n}\right)^{n},$$ as $n$ tends to infinity.
Another reference, possibly easier to get, is
Giroux A., Approximation of entire functions over bounded domains. J. Approx. Theory 28 (1980), no. 1, 45-53.
which extends Batyrev's result to $L^p$ norms, $2\leq p\leq\infty$.
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$\begingroup$ Are you sure that this paper contains a complete proof? Sometimes short papers in Doklady were just research announcements. Also, since older Russian journals can be hard for many people to obtain, do you know if the proof is written down somewhere else (preferably in English)? $\endgroup$ Commented Sep 16, 2018 at 15:42
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1$\begingroup$ @Nate Eldredge I don't have the paper by Batyrev so I can't say, but several subsequent papers cite his paper for the mentioned result. I have added a reference to one of these papers in the answer. In particular, it has a proof of Batyrev's result. $\endgroup$– user111Commented Sep 16, 2018 at 19:36
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$\begingroup$ Thank you for the answer! It is kind of surprising to me that simple Taylor approximation is not too far away from this bound. The error is $K^n/n!$, which is roughly $(eK/n)^n / \sqrt{2\pi n}$. Well, still there is a $2^n$ factor difference. $\endgroup$– hao chenCommented Sep 19, 2018 at 17:12