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Let $\delta, \varepsilon \in (0,1)$. I am interested in a sequence $\{f_n\}$ of polynomial approximations of the square root function $x \to x^{1/2}$ on $[\delta,1]$, of the form $$ f_n(x) = \sum_{i=0}^n \alpha_i x^i $$ which satisfies \begin{align*} \forall x \in [\delta,1],\ \left|f_n(x) - x^{1/2}\right|\ & \leq \varepsilon \\ n & = O\left(\frac{1}{\delta} \log\frac{1}{\varepsilon}\right),\\ \sum_{i=0}^\infty |\alpha_i| & \leq B, \\ \end{align*} where $B$ is a universal constant. (Obviously, I also want the sequence to satisfy $f_n(x) = f_{n-1}(x) + \alpha_n x^n$.)

Does such a sequence of approximations exist?

We can also relax the requirement that $B$ is a constant, and require $\sum_{i=0}^n |\alpha_i|$ to have a bound which is polynomial in $n$.

More generally, I am interested in approximations satisfying the same properties for the function $x \to x^{\alpha}$ where $\alpha \in (0,1)$.

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  • $\begingroup$ I made some of the notation easier to read (at least for me). I think the rest would be clearer if you omitted the sentences with "let $\delta, \varepsilon \in (0,1)$" and "$n=O(\frac{1}{\delta}\log\frac{1}{\varepsilon})$", and just said $|f_n(x)-x^{1/2}| \le e^{-\delta \,k\,n}$ for some universal constant $k$. $\endgroup$
    – user44143
    Commented Jul 26, 2021 at 17:11

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I have a strong impression that something like that has been asked before (perhaps, by somebody else) but it is easier to answer again than to find that old thread.

What you ask for is patently impossible. Indeed, assume that you have a polynomial $f_n(z)$ that approximates $\sqrt z$ on the interval $[\frac 13,1]$ with precision $e^{-cn}$. Consider the domain $\Omega=\{z: \frac 13\le |z|\le 1, 0\le \arg z\le \frac{3\pi}2\}$.

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The function $g(z)=f_n(z)-\sqrt{z}$ is then analytic in $\Omega$, continuous up to the boundary, and bounded by $B_n+1$. By the standard two constant lemma, we have $$ |g(-2/3)|\le [\max_{\partial\Omega\setminus[\frac 13,1]}|g|]^{1-\gamma}[\max_{[\frac 13,1]}|g|]^{\gamma}\le (B_n+1)^{1-\gamma}e^{-\gamma cn} $$ with some constant $\gamma\in(0,1)$ (the harmonic measure of $[\frac 13,1]$ with respect to the domain $\Omega$ and the point $-2/3$). If $B_n$ is subexponential in $n$, the RHS tends to $0$ as $n\to\infty$, so we get $f_n(-2/3)$ close to $i\sqrt{\frac 23}$.

Considering the domain symmetric to $\Omega$ with respect to the real axis, we conclude that $f_n(-2/3)$ must be also very close to $-i\sqrt{\frac 23}$. But those two numbers are rather far apart.

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  • $\begingroup$ Can you provide a statement or reference for the two-constant lemma? $\endgroup$
    – user44143
    Commented Jul 29, 2021 at 7:35
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    $\begingroup$ @MattF. The statement is just as I've written. Usually it is stated on the unit circle for the value at $0$: If $F$ is an analytic function in the disk continuous up to the boundary and $L$ is an arc on the unit circumference $\mathbb T$, then $|F(0)|\le [\max_L |F|]^{m(L)}[\max_{\mathbb T\setminus L}|F|]^{1-m(L)}$ where $m(L)=|L|/(2\pi)$. The case of the general simply connected domain with an arbitrary point is obtained by applying a conformal mapping. It is just about subharmonicity of $\log|F|$, nothing fancy. See encyclopediaofmath.org/wiki/Two-constants_theorem $\endgroup$
    – fedja
    Commented Jul 29, 2021 at 23:56
  • $\begingroup$ Thanks! It is a hundred years old but new to me. $\endgroup$
    – user44143
    Commented Jul 30, 2021 at 12:52
  • $\begingroup$ What about if you restrict to the positive real numbers, ie $\forall x \in [\delta,1]$? Can you restrict the output of $f$ to be real? Or if not, why not? $\endgroup$
    – Andrew Hardy
    Commented Jul 21, 2022 at 18:46

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