Approximating $x^2$ by odd powers of $x$ on $\mathbb{R}^+$ By some Stone-Weierstrass argument, odd functions on $[-R,R]$ are approximated by power series in odd powers of $x$. This means e.g. that we can approximate $x^2$ on $[0,R]$ by an odd power series.
Suppose that for fixed $k$, we want to find the coefficients $\alpha_i$ that minimise the $L^\infty$ distance from $\sum_{i=1}^{k} \alpha_i x^{2i-1}$ to  $x^2$ on $[0,R]$, these will be functions of $R$. For instance, I believe the lowest two approximations are:
$$f_1(x) = (2\sqrt{2} - 2)R x$$
$$f_2(x) = \frac{1}{\sqrt{3} R} x^3 + \frac{R}{\sqrt{3}} x$$
To find these, I essentially waded through the messy algebra. I wonder whether there are some observations that would make it easy to solve for the best coefficients, or whether all hope is lost and they become non-algebraic for the higher approximations - I have been trying but keep getting lost in the algebra. 
 A: The solution to the $\,L^\infty\,$ distance problem
involves the Chebyshev polynomials
of the first kind $\,T_n(x).\,$
The Wikipedia article states

Chebyshev polynomials are important in approximation theory because the roots of $T_n(x)$, which are also called Chebyshev nodes, are used as nodes in polynomial interpolation. The resulting interpolation polynomial minimizes the problem of Runge's phenomenon and provides an approximation that is close to the polynomial of best approximation to a continuous function under the maximum norm. 

Thus, the answer to your problem is to expand the function in
a infinite series of Chebyshev polynomials with odd indices.
By truncating this infinite series we will get the best polynomial
apporximation under the maximum norm of a given degree.
To determine the coefficients in the Chebyshev series we can use
the orthogonality of Chebyshev polynomials. That is, if
$$ \langle f(x), g(x)\rangle := \frac4\pi \int_0^1 f(x)g(x)/\sqrt{1-x^2} dx \tag1 $$ then
$\, \langle T_n(x), T_m(x)\rangle = 0\,$ if $\,n\ne m\,$ while
$\, \langle T_n(x), T_n(x)\rangle = 1\,$ if $\,n\gt 0\,$ and
$\, \langle T_0(x), T_0(x)\rangle = 2.\,$ Using these facts we find
$$ x^2 = \frac{8R}{\pi} \left(\frac{T_1(x)}{3} + 
 \frac{T_3(x)}{15} - \frac{T_5(x)}{105} + \dots\right).
\tag2 $$
The general series using OEIS sequence A061550 is
$$ x^2 = \frac{8R}{\pi} \left(\sum_{n=0}^\infty 
  \frac{(-1)^{n+1} T_{2n+1}(x)}{(2n-1)(2n+1)(2n+3)} \right). \tag3 $$
