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

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 $$x \in [0,1]$$?

• 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? – Yemon Choi Jun 10 at 4:44
• Muentz‘ theorem – user131781 Jun 10 at 8:01
• @user131781 Indeed, this occurred to me after my original comment, I'm getting even more forgetful I guess ... – 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.
• How are you justifying the first step? The function $f$ you are approximating depends on $\epsilon$. – Yuval Peres Jun 10 at 13:18
• @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? – Yemon Choi Jun 10 at 19:53
• @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.) – Mateusz Kwaśnicki Jun 10 at 20:59