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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 $x \in [0,1]$?

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  • $\begingroup$ Can we start off by approximating $\sqrt{x}$ with a polynomial (Weierstrass), perturb the polynomial so it has zero constant term, and then do some kind of bootstrap or iteration? $\endgroup$ – Yemon Choi Jun 10 at 4:44
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    $\begingroup$ Muentz‘ theorem $\endgroup$ – user131781 Jun 10 at 8:01
  • $\begingroup$ @user131781 Indeed, this occurred to me after my original comment, I'm getting even more forgetful I guess ... $\endgroup$ – Yemon Choi Jun 10 at 15:40
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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.

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  • $\begingroup$ How are you justifying the first step? The function $f$ you are approximating depends on $\epsilon$. $\endgroup$ – Yuval Peres Jun 10 at 13:18
  • $\begingroup$ @YuvalPeres I don't see a problem (perhaps I am being slow) - one fixes epsilon, defines $f=f_\epsilon$, and then chooses some $P$ that approximates $f_\epsilon$ to with $\epsilon$. What goes wrong? $\endgroup$ – Yemon Choi Jun 10 at 19:53
  • $\begingroup$ @YuvalPeres: Neither do I see a problem here: $f$ and $P$ do depend on a (fixed) $\varepsilon$. (But still it is far simpler to apply the Müntz–Szász theorem, as user131781 pointed out in their comment.) $\endgroup$ – Mateusz Kwaśnicki Jun 10 at 20:59
  • $\begingroup$ Indeed there is no problem. $\endgroup$ – Yuval Peres Jun 10 at 21:08

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