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 $x \in [0,1]$?

  • $\begingroup$ 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
  • 4
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.