# Polynomial approximation for square root function with fast convergence and bounded coefficients

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 $$\alpha \in (0,1)$$.

• 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 $k$. Jul 26, 2021 at 17:11

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 $$\Omega=\{z: \frac 13\le |z|\le 1, 0\le \arg z\le \frac{3\pi}2\}$$.
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.
• @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 Jul 29, 2021 at 23:56
• 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? Jul 21, 2022 at 18:46