Return to Revisions

Of course there is. Let $p$ approximate on $[0, 1]$ with error no greater than $\varepsilon$ the function $$f(x) = \min\{1/\varepsilon, x^{-1/4}\} .$$ If $x \geqslant \varepsilon^4$, then $f(x) = x^{-1/4}$ and hence $$|p(x^4) x^2 - x| = x^2 |p(x^4) - f(x^4)| \leqslant x^2 \varepsilon \leqslant \varepsilon .$$ On the other hand, if $x < \varepsilon^4$, we simply have $f(x) = 1 / \varepsilon$, and hence $$\begin{aligned}|p(x^4)x^2 - x| & \leqslant x^2 |p(x^4) - f(x^4)| + |(1 / \varepsilon) x^2| + |x| \\ & \le \varepsilon^8 \varepsilon + (1/\varepsilon) \varepsilon^8 + \varepsilon^4 \leqslant \varepsilon ,\end{aligned}$$ provided that $\varepsilon$ is small enough.