Polynomial approximation of circle or ellipse Trying again, with a somewhat simpler sounding question, since my previous one (Generalizations of equi-oscillation criterion) got zero response:
Let $F:[0,1] \to R^2$ be a parametric polynomial curve of degree $m$. I want to adjust the coefficients of $F$ to make it lie as closely as possible to the unit circle, $C$. More specifically, I want to have $F(0) = (1,0)$, and $F(1) = (0,1)$, and I want the maximum distance 
$$E_1 = \max \{dist(F(t),C): t \in [0,1]\}$$
to be minimized. 
It would be OK to minimize the following error $E_2$ instead, if it's easier:
$$ E_2 = \max \{ \big| F_x^2(t) + F_y^2(t) - 1 \big|: t \in [0,1]\}$$
What if I replace the unit circle by an ellipse (for example).
Two specific questions:
(1) Can we prove that the best approximation is equi-oscillatory, in some sense ?
(2) How can the best approximation be computed ?
This site http://spencermortensen.com/articles/bezier-circle/ has a good result for the case of a circle with ($m=3$).
You could also think of this as a question about Bezier curves of degree $m$. Clearly, it would be a good idea to set the first control point equal to $(1,0)$, and the last one equal to $(0,1)$. Then, we just need to adjust the other control points until we get an optimal fit.
A simple algebraic version of the problem: Find two polynomials $x(t)$ and $y(t)$ such that $x(0)=1$, $x(1)=0$, $y(0)=0$, $y(1)=1$, and $x(t)^2 + y(t)^2 - 1$ is small for all $t \in [0,1]$.
 A: A proposal: Present your circular arc in the form
$$\eqalign{
x(t)&={1\over\sqrt{2}}(\cos t-\sin t) ={1\over\sqrt{2}}\left(1-t-{t^2\over2}+{t^3\over6}+{t^4\over 24}-{t^5\over120}-\ldots\right) \cr
y(t)&={1\over\sqrt{2}}(\cos t+\sin t) ={1\over\sqrt{2}}\left(1+t-{t^2\over2}-{t^3\over6}+{t^4\over 24}+{t^5\over120}-\ldots\right) \cr}\qquad\left(-{\pi\over4}\leq t\leq{\pi\over4}\right)$$
and use as many terms as are needed for the required precision. The following figure shows an overlay of the exact circle and the above polynomial approximation up to ${t^5\over120}$:
A: The complete answer may be hard, but contrary to the OP, I see no difficulty in stating the question and generalizing the concept of equi-oscillation to the two-dimensional case. Let $V_n$ denote the affine space of all pairs $v=(x_n,y_n)$ of polynomials each of degree $\leq n$ with $v(0)=(1,0)$ and $v(1)=(0,1)$ . Note that $V_n$ has dimension $2(n-1)$. For $v=(x_n,y_n)\in V$ put
$$
|| v ||=\sup_{t\in[0,1]} \bigg| x_n(t)^2+y_n(t)^2-1\bigg|
$$
Then define $\mu_n={\sf inf}(|| v |||, v \in V_n)={\sf min}(|| v |||, v \in V_n)$. Say that a $v\in V_n$ is optimal if $||v||=\mu_n$.
 Definition  Let $\varepsilon>0$. We say that a pair $v=(x_n,y_n)\in V_n$ is $\varepsilon$-oscillating if there is an increasing sequence of $1+{\sf dim}(V_n)$ elements $t_1<t_2< \ldots<t_{1+{\sf dim}(V_n)}$ in $[0,1]$ such that the sequence $w_i=x_n(t_i)^2+y_n(t_i)^2-1$ is alternating (i.e. $w_{i+1}=-w_i$ for all $i$) and $|w_i|=\varepsilon$ for all $i$.
Note that  $1+{\sf dim}(V_n)=2n-1$ above.
 Conjecture 1  A $v\in V_n$ is optimal if and only if it is $\mu_n$-oscillating.
 Conjecture 2  There is a unique optimal solution for each $n$.
Those conjectures are true when $n=2$. Indeed, in this case a generic $v\in V_2$ can
be written $$v(a,b,t)=(1-t+at(1-t),t+bt(1-t))$$. Let us also put
$$
F(a,b,t)=1-t+at(1-t))^2+(t+bt(1-t))^2-1 \tag{1}
$$
Let $\theta$ be the largest real root of $T=X^4+4X^3-8X^2+8X-4$, so that $\theta$ is approximately
$0,85 \ldots$. I claim then that $v(\theta,\theta,.)$ is the unique optimal solution. 
Let $G(t)=F(\theta,\theta,t)$ and 
$$\mu=G(1/2)=\frac{\theta^2}{8}+\frac{\theta}{2}-\frac{1}{2} \approx 0,015 \tag{3} $$
We then have the identities
$$
\mu-G(t)=\big(t-\frac{1}{2}\big)^2 \bigg(\frac{\theta^2}{2}+2\theta-2+2\theta^2t(1-t) \bigg) \tag{4}
$$
$$
\mu+G(t)=2\theta^2 \bigg(t(1-t)-\frac{\theta-1}{2\theta^2} \bigg)^2 \tag{5}
$$
Note that the polynomial $Q(t)=t(1-t)-\frac{\theta-1}{2\theta^2}$ has two roots in $[0,1]$,
$\alpha$ and $1-\alpha$ where $\alpha \approx 0,12 \ldots$. This shows that $G$ is $\mu$-oscillating.
Let $(a,b)$ be any pair such that $||v(a,b,.)|| \leq \mu $. Then
$$
F(a,b,\alpha) \geq -\mu, \ F(a,b,\frac{1}{2}) \leq \mu, \ F(a,b,1-\alpha) \geq -\mu \tag{6}
$$
Those three inequalities alone will suffice to force $(a,b)=(\theta,\theta)$, showing 
optimality and unicity. Indeed, consider in the plane points $A,B,C,S$ with the following 
coordinates :
$$
A(-\frac{1}{\alpha},-\frac{1}{1-\alpha}), \ B(-2,-2), \ C(-\frac{1}{1-\alpha},-\frac{1}{\alpha}), \
S(\theta,\theta)
$$
Also, consider the disk $D_A$ with center $A$ and radius $\frac{\sqrt{\mu}}{\alpha(1-\alpha)}$,
the disk $D_B$ with center $B$ and radius $4\sqrt{1+\mu})$, and the disk $D_C$ with 
center $C$ and radius $\frac{\sqrt{\mu}}{\alpha(1-\alpha)}$.
Then a pair $(a,b)$ satisfies (6) iff the point with such coordinates is inside $D_C$
  but outside $D_A$ and $D_B$. The figure below shows that $S$ is the only such point, qed. 

A: Maybe you can get your answer here:
https://www.jimherold.com/computer-science/best-fit-bezier-curve
http://www.google.com.br/url?sa=t&rct=j&q=Optimal+fitting+bezier+curves&source=web&cd=1&cad=rja&ved=0CCsQFjAA&url=https%3A%2F%2Fece.uwaterloo.ca%2F~dwharder%2FMaplesque%2FBezier%2FBestFittingBezierCurves.ppt&ei=tnXxUK_UDZOe8gTko4H4Cw&usg=AFQjCNFfkMNXviJSzFCBIqbTL3EWZcgTpA
http://www.google.com.br/url?sa=t&rct=j&q=Optimal+fitting+bezier+curves&source=web&cd=9&cad=rja&ved=0CGkQFjAI&url=http%3A%2F%2Fwww.dtic.mil%2Fcgi-bin%2FGetTRDoc%3FAD%3DADA350611&ei=tnXxUK_UDZOe8gTko4H4Cw&usg=AFQjCNGW_AP5QddfGugn_5WzoIUDIMJ7ng
A: As a start, there is the well-known rational exact fit of
$(2t/(1+t^2), (1-t^2)/(1+t^2)$.
(Because $(2t)^2 + (1-t^2)^2
= 4t^2 + 1 - 2t^2 + t^4
= 1 + 2t^2 + t^4
= (1+t^2)^2$).
If you write
$1/(1+t^2) = 1-t^2+t^4 ...\pm t^{2k} ...$,
you can get an initial polynomial approximation.
