# Can $\{1, x^2, x^3, x^4, …\}$ approximate $x$ on $[0,1]$?

Can $$\{1, x^2, x^3, x^4, ...\}$$ approximate $$x$$ on $$[0,1]$$?

Here is an attempt:

Let $$\mathcal{A}$$ be the linear span of our set $$\{x^0, x^2, x^3, x^4, ...\}$$. $$\mathcal{A}$$ is a vector subspace and separates points. It is also a subalgebra since $$\sum_{i \neq 1}^{n} a_i x^i \times \sum_{i \neq 1}^{m} b_i x^i$$ does not generate any $$x^1$$ term. $$[0,1]$$ is compact, and $$x^0 = 1 \in \mathcal{A}$$. By the Stone-Weierstrass theorem, $$\mathcal{A} = C([0,1])$$. Conclude that our initial set can approximate $$x$$.

This is my first time using the Stone-Weierstrass theorem and I think I made a mistake, since the same argument goes through for all sets of the form $$\{1, x^k, x^{k+1}, x^{k+2}, ...\}$$ or say polynomials with even powers. It also seems to contradict the fact that the $$\{1, x, x^2, ... \}$$ form a basis in $$L^2([0,1])$$.

Any help will be greatly appreciated!

• Regarding the even powers case: the basic Weierstrass theorem implies that $\sqrt{t}$ can be uniformly approximated within any $\epsilon > 0$ by some polynomial function of $t$ for $t \in [0,1]$; now substitute $t=x^2$ for $x \in [0,1]$. – Daniel Schepler Feb 6 at 0:49
• Are you trying to approximate $x$, or are you trying to approximate $x$ on $[0,1]$? Two different things, one easy, one impossible. – Gerry Myerson Feb 6 at 1:10
• Sorry I meant on [0,1]! – is it normal Feb 6 at 1:16
• Not relevant for your Stone-Weierstrass question as such, but you might be interested in the Müntz–Szász theorem. I first saw it (and a nifty proof) in Green Rudin. – peter a g Feb 6 at 2:45
• BTW, if I understand your contradiction comment: "basis" here is not "basis" as in algebra (finite sums!) - so there is no contradiction. – peter a g Feb 6 at 3:32

Just for fun, an explicit sequence of polynomials that approximates $$\sqrt{x}$$ on $$[0,1]$$: \begin{align*}f_n(x) &= \frac2{\pi}-\frac1{\pi}\sum_{k=1}^n \frac1{k^2-\frac14}T_k(1-2x)\\ &= \frac{-2}{(2n+1)\pi}\sum_{j=0}^n (-4x)^j\cdot\binom{n+j}{2j}\cdot\frac1{2j-1}\end{align*} These $$f_n$$ are polynomials of degree $$n$$, and $$\max_{x\in [0,1]}\left|f_n(x)-\sqrt{x}\right| = f(0) = \frac2{(2n+1)\pi}$$. By the way, the $$T_k$$ in the first expression for $$p$$ are the Chebyshev polynomials.
When I first came up with these, I wrote out an approximation of $$f_{32}$$ (the first to get within $$0.01$$), with each coefficient to five significant digits. The largest coefficients were just short of $$10^{20}$$.
Of course, $$f_n(x^2)$$ would then be a polynomial of degree $$2n$$ with no linear term approximating $$x$$ to within $$\frac{2}{(2n+1)\pi}$$ on the interval.
This came out of a discussion of various approaches to finding an approximation to $$\sqrt{x}$$, and how to minimize the degree of the resulting polynomials for a given accuracy. Other approaches mentioned included the Taylor polynomial centered at $$1$$ (required degree $$\frac{1}{\pi\cdot\epsilon^2}$$), the iteration $$P_{n+1}(x)=P_n(x)+\frac{x-P_n(x)^2}{2}$$ (required degree $$2^{c/\epsilon}$$ for a constant $$c$$), and the Bernstein polynomials $$g_n(x)=\sum_{k=0}^n\sqrt{\frac kn}\binom nk x^k(1-x)^{n-k}$$ (required degree approximately $$\frac1{13.26\cdot \epsilon^2}$$).
Any of these methods suffice if all you need is the existence of a decent approximation, of course, and all could be converted to the problem here by the substitution $$x\to x^2$$.