Approximating $1/x$ by a polynomial on $[0,1]$

Question

For every $\varepsilon > 0$, is there a polynomial of $x^4$ without constant term, i.e., $p(x^4) = a_1 x^4 + a_2 x^8 + \cdots +a_n x^{4n}$, such that $$\|p(x^4)x^2 - x\| < \varepsilon $$ for every $x \in [0,1]$?

Answer 1

Of course there is. Let $P$ approximate on $[0, 1]$ with error no greater than $\varepsilon$ the function $$f(x) = \min\{\varepsilon^{-5}, x^{-5/4}\} ,$$ and define $p(x) = x P(x)$. If $x \geqslant \varepsilon^4$, then $f(x^4) = x^{-5}$ and hence $$|p(x^4) x^2 - x| = x^6 |P(x^4) - f(x^4)| \leqslant x^6 \varepsilon \leqslant \varepsilon .$$ On the other hand, if $x < \varepsilon^4$, we simply have $f(x^4) = \varepsilon^{-5}$, and hence $$\begin{aligned}|p(x^4)x^2 - x| & \leqslant x^6 |P(x^4) - f(x^4)| + x^6 |f(x^4)| + |x| \\ & \leqslant \varepsilon^{24} \varepsilon + \varepsilon^{24} \varepsilon^{-5} + \varepsilon^4 \leqslant \varepsilon ,\end{aligned}$$ provided that $\varepsilon$ is small enough.