Arraryeldergox2

2022-06-26

Is there a rational function $f\left(x\right)\in \mathbb{Q}\left(x\right)$ such that $\sqrt{2}\le f\left(x\right)\le \sqrt[3]{2x}$ for all $x\ge \sqrt{2}$?
My thoughts : it is easy to find such an f if we relax the conditions to $f\left(x\right)\in \mathbb{R}\left(x\right)$ (take f constant equal to $\sqrt{2}$), however no easy perturbation of this solution seems to solve the original problem. Clearly f must have zero degree in x and can be written in the form $x+\left({x}^{2}-2\right)g\left(x\right)$ where g is another rational function. Then I am stuck.

massetereqe

Expert

The simplest function I can find seems $f\left(x\right)=\frac{2x+2}{x+2}$. It is easily verified this satisfies the double inequality for $x\ge \sqrt{2}$.
This was obtained by looking at linear approximants, and then setting the conditions $f\left(\sqrt{2}\right)=\sqrt{2},f\left(x\right)\ge \sqrt{2}$ and finally the RHS inequality.
Even among linear functions, there are of course many choices, many of which can be got from the form
$f\left(x\right)=\frac{ax+2b}{bx+a}$, with
$\left(a,b\right)\in \left\{\left(3,2\right),\left(5,2\right),\left(5,3\right),\left(7,3\right),\left(7,4\right),...\right\}$
... which leads to the thought $\frac{\left(2n±1\right)x+2n}{nx+2n±1}$ could work for $1 (not checked).

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