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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 $t \in [0,1]$.

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I get a 404 from your URL, but I guess you mean this one: math.stackexchange.com/q/147311 –  Rahul Jan 6 '13 at 3:56
    
Correct. I'll see if I can fix it. Thanks. –  bubba Jan 6 '13 at 4:25
    
I think an absolute value would be in order in the definition of $E_2$ : you want $\max \{ \big| F_x^2(t) + F_y^2(t) - 1 \big|: t \in [0,1]\}$ not $\max \{ F_x^2(t) + F_y^2(t) - 1: t \in [0,1]\}$. –  Ewan Delanoy Jan 8 '13 at 8:36
    
@Ewan -- you're right. I'll correct the question. My apologies. –  bubba Jan 8 '13 at 13:23

5 Answers 5

up vote 6 down vote accepted
+100

There is a general procedure for a circle that minimizes $\max |p^2+q^2-1|$. In what follows it is convenient to make two slight modifications to your problem:

  1. The parameterization will be defined on an interval $[-\alpha, \alpha] \subset [-1,1]$ rather than $[0,1]$.
  2. The approximation is for another arc of length $\tfrac{\pi}{2}$.

The first modification is only a parameter substitution and the second a rotation so solving this problem will clearly also solve the original problem.

Let $n > 0$ be some degree. Then we want to find polynomials $p$ and $q$ of degree at most $n$ such that $\max |p^2+q^2-1|$ is minimized over the interval $[-1,1]$. Note that $p^2+q^2-1$ will have degree at most $2n$. It is well known that the Chebyshev polynomial of the first kind $T_{2n}$ solves the minimax problem on $[-1,1]$ among all polynomials of degree at most $2n$ with leading coefficient $2^{2n-1}$. Therefore the best solution would be of the form

$$ p^2+q^2 = 1 + \lambda T_{2n} $$

for some parameter $\lambda \in [0,1)$ that should be as small as possible. Such a solution can indeed be found based on the explicit factorization

$$ \begin{eqnarray} 1 + \frac{T_{2n}(x)}{\cosh(2n\,t)} &=& \frac{2^{2n-1}}{\cosh(2n\, t)} \prod_{k=1}^{2n}\left(x - \cos\left(\tfrac{(2k-1) \pi}{2n} - t\,i\right)\right) \\ &=& \frac{2^{2n-1}}{\cosh(2n\, t)} \prod_{k=1}^{2n}\left(x - \cos\left(\tfrac{(2k-1) \pi}{2n}\right)\cosh(t) - \sin\left(\tfrac{(2k-1) \pi}{2n}\right) \sinh(t)\,i\right) \end{eqnarray}$$

for all $t \in \mathbb{R}$. (See the edit below for a derivation of this factorization.) Since the LHS is a real polynomial, the roots in the expression above have conjugate symmetry. Now take $t > 0$ and define the complex polynomial $r(x)$ of degree $n$ by including only the roots in the upper half plane:

$$ r(x) = \frac{2^n}{\sqrt{2 \cosh(2n\,t)}} \prod_{k=1}^{n}\left(x - \cos\left(\tfrac{(2k-1) \pi}{2n} - t\,i\right)\right).$$

Then $r(-x) = (-1)^n \,\overline{r}(x)$ where $\overline{r}$ means $r$ with conjugated coefficients. Write $r = p + q\,i$ for real polynomials $p$ and $q$ then

$$p^2+q^2 = (p + q\,i)(p - q\, i) = r \,\overline{r} = 1 + \frac{T_{2n}}{\cosh(2n\, t)}.$$ So in fact we found a one-parameter family of good approximations to the circle on the interval $[-1,1]$. (Larger $t$ give better approximations.) The question is what this parameter signifies and which value of $t$ actually solves our problem as stated.

Since all roots of $r$ are in the upper half plane its restriction to $\mathbb{R}$ is a complex curve with strictly increasing argument, so it wraps the real line around the origin in counter clockwise direction. One can show that $p$ and $q$ have degrees $n$ and $n-1$ respectively, both have only real roots and the roots of $q$ separate those of $p$. The parameter $t$ controls how "fast" the curve wraps around the origin, where greater values of $t$ give "slower" curves that stay closer to the unit circle on $[-1,1]$.

The only restrictions that prevent us from taking aribrarily large values of $t$ are that the curve must begin and end on the unit circle and cover an angle of exactly $\tfrac{\pi}{2}$ over some interval contained in $[-1,1]$. The largest zeroes of $T_{2n}$ are $\pm \cos(\tfrac{\pi}{4n})$ so if $\alpha = \cos(\tfrac{\pi}{4n})$ and we restrict $r$ to the interval $[-\alpha, \alpha]$ then at least it begins and ends on the unit circle. Since $r(-\alpha) = (-1)^n \, \overline{r}(\alpha)$ the end point is obtained by reflecting the starting point in the real axis (for even $n$) or imaginary axis (for odd $n$). Therefore if we take $t$ maximal such that $r(\alpha)^2$ is purely imaginary, so $p(\alpha) = \pm q(\alpha)$, then the curve spans an angle of precisely $\tfrac{\pi}{2}$.

To summarize the results of this fairly long story:

  1. To get a degree $n$ approximation consider the polynomial $r(x)$ defined above that depends on an auxiliary real parameter $t>0$.
  2. Let $\alpha = \cos(\tfrac{\pi}{4n})$. Determine the maximal value $t>0$ for which $r(\alpha)^2 \in i\,\mathbb{R}$.
  3. Then $|r(-\alpha)| = |r(\alpha)| = 1$ and $\left|\,|r(x)|^2 - 1\,\right| \leq \tfrac{1}{\cosh(2n \, t)}$ for $x \in [-\alpha, \alpha]$. Moreover $r$ is the best approximation of a quarter circle on the interval $[-\alpha, \alpha]$.

The first few approximations found this way are:

$$ \begin{array}{c|c|c|c} n & \alpha & t & \max \, \left| \, |r|^2-1 \, \right| \\ \hline 1 & 0.70710678 & 0.65847895 & 0.5 \\ 2 & 0.92387953 & 1.2149195 & 0.015505028 \\ 3 & 0.96592583 & 1.5982608 & 0.0001368784 \\ 4 & 0.98078528 & 1.8786122 & 5.9437783 \times 10^{-7} \\ 5 & 0.98768834 & 2.098419 & 1.5406786 \times 10^{-9} \\ 6 & 0.99144486 & 2.2789419 & 2.6561189 \times 10^{-12} \\ 7 & 0.99371221 & 2.4320125 & 3.2665949 \times 10^{-15} \\ 8 & 0.99518473 & 2.5648448 & 3.0106701 \times 10^{-18} \\ 9 & 0.9961947 & 2.6821493 & 2.1570627 \times 10^{-21} \\ 10 & 0.99691733 & 2.7871679 & 1.2359399 \times 10^{-24} \end{array} $$

Edit: Here's how the factorization can be found. The key property is that $$T_{2n}\left(\frac{x+x^{-1}}{2}\right) = \frac{x^{2n}+x^{-2n}}{2}$$ for all $x \neq 0$. So $$1 + \frac{T_{2n}\left(\frac{x+x^{-1}}{2}\right)}{\cosh(2n\,t)} = \frac{2 \cosh(2n\,t) + x^{2n}+x^{-2n}}{2 \cosh(2n\,t)}$$ and this vanishes when $x^{2n} = -e^{\pm 2n \, t}$. The roots are therefore $$ x= \exp\left(\frac{(2k-1)\pi\,i}{2n} \pm t\right)$$ for $k \in \{1, \dotsc, 2n\}$ and for such a root we have $$\frac{x+x^{-1}}{2} = \cos\left(\frac{(2k-1)\pi}{2n}\mp t\, i\right).$$ Since $\cos$ is even this gives $2n$ roots of the polynomial $$1+\frac{T_{2n}}{\cosh(2n \, t)}$$ of degree $2n$. After fixing the proper leading coefficient the factorization follows.

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Amazing. Thanks very much. The factorization of $1 + \lambda T_{2n}$ looks like magic -- where did that come from? Actually, any polynomial that's non-negative can be written as the sum of two squares, using complex conjugate factors, as you did. Your breakthrough, it seems to me, is that you were somehow able to see what these factors are, explicitly. More comments later, after I've had a chance to read more carefully. Thanks again. –  bubba Jan 14 '13 at 2:13
    
Another interesting thing: for representing a circular quadrant, polynomials of degree 7 are just as good as your computer's sine and cosine functions. –  bubba Jan 14 '13 at 2:26
    
@bubba Finding an explicit factorization was a pleasant surprise. It is actually not even that hard. I'll try to edit it in some time today. –  WimC Jan 14 '13 at 6:17
    
I still don't see why the factorisation is correct. But, time's running out on the bounty, so I'll take your word for it. Please explain it if you have time. Also, if you have code for calculating the approximations, I'd like to see it. Any language is OK. If you send me your e-mail address, I'll explain why this problem is important, if you're interested. Thanks very much, again. Impressive work. –  bubba Jan 14 '13 at 13:08
    
@bubba I added a section about the factorization method. You can contact me on the e-mail address in my profile. I'd like to hear more about this problem, sure. I do have some code to share to approximate the parameter $t$ (which I used to produce the table.) I did not generate the actual polynomial coefficients but that could be done if needed. –  WimC Jan 14 '13 at 19:52

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 ${t^5\over120}$:enter image description here

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Thanks, Christian. Best attempt so far. But at $t = \pm \pi/4$ your degree 5 approximation gives an error of $E_1 = 0.0002536$. The reference I cited gets a smaller maximum error of $E_1 = 0.00019$, using a curve of degree 3. The basic problem is that your approximation is very good need the middle of the curve but much worse near the ends. –  bubba Jan 13 '13 at 8:05
    
@bubba: $E1=0.00019$ seems to be the optimum for degree ≤3. My approach is not optimal, but has simple rational coefficients, and the error can be made as small as desired by choosing the degree appropriately. – –  Christian Blatter Jan 14 '13 at 9:52

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.

enter image description here

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Thanks for formalising the problem statement. Generalising the concept of equi-oscillation to the case of circle approximations is straightforward, as you say. It's a bit less clear in the case of ellipses and other curves. If $f(x,y) = 0$ is a curve, we could look at equi-oscillations of the function $t \mapsto f(x(t),y(t))$. These wouldn't really be equi-distance oscillations, but would still be of interest. As you say, this is all generalization of Chebyshev's theorems. I'm surprised that generalisations have not been published already. Or have they ?? –  bubba Jan 13 '13 at 12:40
    
@bubba : please read my update. I agree with you that this has probably been studied already. If you don’t obtain further information here, you should certainly ask on MathOverflow. –  Ewan Delanoy Jan 13 '13 at 14:55
    
I accepted Wim's answer, but I appreciate your contribution, too. Thanks very much. –  bubba Jan 14 '13 at 13:11

Maybe you can get your answer here:

http://jimherold.com/2012/04/20/least-squares-bezier-fit/

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

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The last reference in the list is the more promising one. –  Ewan Delanoy Jan 12 '13 at 18:01
    
The first and third links didn't work for me, but I looked at the second one. Then I tried a small experiment with least-squares fitting of a cubic. So, in effect, I was doing fitting in the $L_2$ norm, rather than the $L_\infty$ norm (in some sense). The results were pretty good. Almost as good as the ones in the reference I cited, but not quite. Better than the Taylor series approaches that others have suggested, anyway. Are there any known theoretical results telling us that the $L_2$ fit will be almost as good as the $L_\infty$ fit ?? –  bubba Jan 13 '13 at 9:20
    
Got to the other two references. All three deal with least-squares fitting. It may well be true that lsq fitting is almost as good as minimax fitting, but it's not obvious to me. To do lsq fitting, you have to assign parameter values to the data points. How you do this makes a big difference to the accuracy of the fit. The third paper considers this issue; the other two gloss over it. But, thanks, anyway. –  bubba Jan 14 '13 at 13:19

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.

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Good when $t$ is close to zero, but terrible when $t$ is near 1 (as one would expect from a Maclaurin series). In particular, this doesn't satisfy my requirement that $F(1) =(0,1)$. –  bubba Jan 6 '13 at 5:57
    
A Taylor series expansion around $t=0.5$ would be a lot better, but still not very good. –  bubba Jan 6 '13 at 6:58

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