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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!

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    $\begingroup$ 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 $x \in [0,1]$. $\endgroup$ – Daniel Schepler Feb 6 at 0:49
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    $\begingroup$ Are you trying to approximate $x$, or are you trying to approximate $x$ on $[0,1]$? Two different things, one easy, one impossible. $\endgroup$ – Gerry Myerson Feb 6 at 1:10
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    $\begingroup$ Sorry I meant on [0,1]! $\endgroup$ – is it normal Feb 6 at 1:16
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    $\begingroup$ 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. $\endgroup$ – peter a g Feb 6 at 2:45
  • $\begingroup$ BTW, if I understand your contradiction comment: "basis" here is not "basis" as in algebra (finite sums!) - so there is no contradiction. $\endgroup$ – peter a g Feb 6 at 3:32
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Yes, your argument is good.

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 $x\to x^2$.

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