I have a set of points $(x_i,y_i),\ i=1,\ldots,n$ and want to find a quadratic function $f(x) = ax^2 + bx + c$, at the end I need to find $a,b,c$, which has shortest distances to the set of points.

So, the condition is to find $a, b, c$ which do minimize the function
where $d_i^2=(\hat x_i -x_i)^2 + (\hat y_i -y_i)^2$ is the distance from point $(x_i,y_i)$ to the parabola $f(x)$.
and point $(\hat x_i, \hat y_i)$ is the point on the parabola $f(x)$, which has the shortest distance to the point $(x_i,y_i)$


I could constitute formulas for $a,b,c$ for OLS(ordinary least squares) dependent only from $x_i, y_i$ because I had to find a minimum on function $P(a,b,c)$, so it was enough to take a derivative on $a$, $b$, and $c$ and so solve system of three equation with three unknown.

For TLS(total least squares) regression, I think the same is orthogonal regression see Deming regression, the function $P(a,b,c)$ does also depend on unknown $\hat x_i$, which is the $x$ coordinate of the point on the parabola nearest to the point$(x_i,y_i)$.
In the internet I could unfortunately only find formulas for TLS regression for the line $f(x)=ax+b$, but not for parabola.
At the moment I stuck to get the formulas for $a,b,c$ dependent only on $(x_i,y_i)$

What are the formulas (like $a=\sum(x^2_i)+ \sum(y^2_i)$ etc.) and how can I find them.

To understand clearer what I want, here are the formulas for $a$, $b$ and $c$ using OLS regression(they can definitely be simplified, but I had too few time, all $\sum$ are $\sum_{i=1}^n$):

$$ \bbox[5px,border:2px solid darkblue] { \mathbf b = \frac{n \sum (x^2y) - \sum x^2 \sum y + (\sum x^2)^2(M-L) - n(M-L)\sum x^4}{n\sum x^3} / (1 - \frac{E(\sum x^2)^2 - \sum x^2 \sum x}{n\sum X^3} + \frac{E\sum x^4}{\sum x^3}) }$$
$$ \bbox[5px,border:2px solid darkblue] { \mathbf a = b*E + M - L }$$
$$ \bbox[5px,border:2px solid darkblue] { \mathbf c = \frac{\sum y - a \sum x^2 - b \sum x}{n} }$$

$E = \frac{(\sum x^2)^2 -n \sum x^2}{n\sum x^3 - \sum x^2 \sum x}$

$M = \frac{n\sum (xy)}{n\sum x^3 - \sum x^2 \sum x}$

$L = \frac{\sum x \sum y}{n\sum x^3 - \sum x^2 \sum x}$

  • $\begingroup$ Are both x and y subject to uncertainty or just y? $\endgroup$ – Paul Apr 22 at 16:42
  • $\begingroup$ @Paul Since a, b, c are unknown, the points on parabola, which have shortest distance are also uncertain, both x and y, or have I misunderstood the question? $\endgroup$ – Rekshino Apr 22 at 16:46
  • $\begingroup$ The $x_i$ values might be known (e.g fixed time intervals) and the $y_i$ measurements taken at those x values but subject to error. Then normal least squares is appropriate. Are you saying that your $x_i$ values are also subject to error? $\endgroup$ – Paul Apr 22 at 16:55
  • $\begingroup$ You ask the questions.. :) I would say, yes, the $x$ is also subject to error, so the OLS gives not well enough fitting of the curve to the set of points. $\endgroup$ – Rekshino Apr 22 at 17:05
  • $\begingroup$ Do you need the actual derivations or a general solver? $\endgroup$ – Royi Apr 30 at 17:21

$\color{brown}{\textbf{Calculations of the distance.}}$

The square of the distance from the point $\;(X,Y)\;$ of the given set $(X_i,Y_i)$ to the parabola $\,(a,s,v) = a(x-s)^2+v = \pm z^2+v\;$ is $$d^2 = \min\limits_{x\in\mathbb R} \delta(x),$$ where $\;x\;$ is the abscissa of the arbitrary point on the parabola, $$z = \sqrt{|a|}\,(x-s),\quad Z =\sqrt{|a|}\,(X-s).\tag1$$ $$\delta(x) = f(z) = (z-Z)^2+(z^2\pm(v-Y))^2,\tag2$$ $$\dfrac12f'(z) = z-Z+2z(z^2\pm(v-Y)),\tag3$$ If to denote abscissa of the optimal point on the parabola as $\;\hat x,\;$ and $\;\hat z= \sqrt{|a|}\,(\hat x-s),\;$ then $$\begin{cases} 2\hat z^3+(1\pm2(v-Y))\hat z=Z\\[4pt] d^2= (\hat z-Z)^2+(\hat z^2\pm(v-Y))^2. \end{cases}\tag4$$ From $(4)$ should $$d^2\in\big[0,(Z^2\pm(v-Y)^2)^2\big],\tag5$$ $$2d^2 = 2(\hat z- Z)^2 +Z\hat z+2(v-Y)^2\big(2\hat z^2\pm(v-Y))^2-(1\pm2(v-Y))\big)\hat z^2,$$ $$2d^2 = (1\mp2(v-Y)+4(v-Y)^2)\hat z^2 -3 Z\hat z+2Z^2\pm2(v-Y)^4,$$ $$d^2 = p\left(\hat z - \dfrac{3Z}{2p}\right)^2+\dfrac{8p-9}{8p}\,Z^2\pm(v-Y)^4,\tag{6.1}$$ where $$p = 1\mp2(v-Y)+4(v-Y)^2\tag{6.2}$$ Formulas $(5)-(6)$ can simplify the calculations.

Let $$r=\sqrt{\dfrac{|2\pm 4(v-Y)|}{3}}.\tag7$$

If $\;\mathbf{1\pm2(v-Y) \ge 0},\;$ then $$4\hat z^3+3r^2\hat z = 2Z,$$ $$\dfrac{2Z}{r^3} = 4\left(\dfrac{\hat z}r\right)^3 + 3\dfrac{\hat z}r =\sinh\left(3\operatorname{arcsinh}\dfrac{\hat z}r\right),$$ $$\hat z = r\sinh\left(\dfrac13\operatorname{arcsinh}\dfrac{2Z}{r^3}\right).\tag{8.1}$$

If $\;\mathbf{1\pm2(v-Y) \le 0},\;$ then $$4\hat z^2-3r^2\hat z = 2Z,$$ $$\dfrac{2Z}{r^3} = 4\left(\dfrac{\hat z}r\right)^3 - 3\dfrac{\hat z}r =-\sin\left(3\arcsin\dfrac{\hat z}r\right) =\cosh\left(3\operatorname{arccosh}\dfrac{\hat z}r\right),$$ $$\hat z = \begin{cases} -r\sin\left(\dfrac13\arcsin\dfrac{2Z}{r^3}\right),\quad\text{if}\quad 2|Z|\le r^3\\[4pt] r\cosh\left(\dfrac13\operatorname{arccosh}\dfrac{2Z}{r^3}\right),\quad\text{if}\quad 2|Z|\ge r^3 \end{cases}\tag{8.2}$$ On the other hand, the cubic equations $$\hat z^3\pm\dfrac34\,r^2\hat z -\dfrac Z2 = 0$$ have the discriminants $$D = \dfrac{Z^2}{16}\pm\dfrac{r^6}{64},\tag9$$ wherein

  • if $\;D>0,\;$ then the single solution can be obtained by the Cardano's formula;
  • if $\;D\le 0\;\Rightarrow (a<0)\wedge(2|Z|\le r^3),\;$ then there are three real roots, which satisfy the equation $$\dfrac{2Z}{r^3} = 4\left(\dfrac{\hat z}r\right)^3 - 3\dfrac{\hat z}r =\cos\left(3\arccos\dfrac{\hat z}r\right).$$ Therefore, the most accurate solution of cubic is $$\hat z =r\begin{cases} \sqrt[\large3]{-\dfrac Z4+\sqrt D}+\sqrt[\large3]{-\dfrac Z4-\sqrt D},\;\text{if}\; D>0\\[4pt] r\cos\left(\dfrac{2k\pi}3+\dfrac13\arccos\dfrac{2Z}{r^3}\right), \;\text{otherwize}. \end{cases}\tag{10}$$ Parameter $\;k=-1,0,+1\;$ should minimize the distance square $(4.1)$ or $(6.1)$ under the condiion $(5).$

$\color{brown}{\textbf{How does it work?}}$

If $$\binom{X_i}{Y_i}=\left\{\dbinom11,\dbinom22, \dbinom34, \dbinom48, \dbinom5{13}\right\},$$ $$y=\dfrac12x^2+v,$$ The task a=1/2 standing

then for $\;Z_i=\sqrt2\,X_i\;$ formulas $(8.1)$ allow to get quite realistic plot $\;\hat z_i(v) = \sqrt2 \hat X_i\;$ The task a=1/2, hat z(v)

If $$\binom{X_i}{Y_i}=\left\{\dbinom1{15},\dbinom2{14}, \dbinom3{12}, \dbinom48, \dbinom5{3}\right\},$$ $$y=-\dfrac12x^2+v,$$ The task a=-1/2 standing

then for $\;Z_i=\sqrt2\,X_i\;$ formulas $(10.2)$ over the complex numbers allow to get quite understandable plots $\;\hat z_i(v) = \sqrt2 \hat X_i\;$ with $\;k=0, -1, 1\;$

The task a=-1/2, k=0, hat z(v) The task a=-1/2, k=-1, hat z(v) The task a=-1/2, k=1, hat z(v)

To get the parabola parameters, it suffices to use the method of the volumes. Every elementary volume should provide the next conditions:

  • the sign of $\;a\;$ should be hold (the case $\;a=0\;$ can be used for training);
  • the signs of $\;1\pm 2(v-Y)\;$ should be hold at the all given points;
  • the signs of $r^3-2|Z|$ should be hold at the all given points, where $\;1\pm2(v-Y)<0.\;$

Under these conditions, the constant expression of the sums of the squares of the distances should be defined in the each elementary volume of $\;(a,v,s).\;$

Obtained formulas allow to minimize the sums of the distances squares in the every elementary volume (gradient descent recommended) and then to provide the least of them.

  • 1
    $\begingroup$ Yuri, what is $a$, $s$ and $v$ in the distance formula? $\endgroup$ – Rekshino Apr 30 at 8:05
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    $\begingroup$ @Rekshino Parabola via vertex coordinates $\;(s,v).\;$ Looks more usable and can be easily transformed to the common polynomial form. Thanks! $\endgroup$ – Yuri Negometyanov Apr 30 at 8:53
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    $\begingroup$ Is $Z$ the $\hat X$ in my notation? $\endgroup$ – Rekshino Apr 30 at 9:18
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    $\begingroup$ Can you explain the second part of (4), is it also another form of distance ;) ? Usually $d^2= (x-x_0)^2 + (y-y_0)^2$. $\endgroup$ – Rekshino Apr 30 at 16:27
  • 1
    $\begingroup$ @Rekshino Sadly, explicit solutions does not exist. $\endgroup$ – Yuri Negometyanov May 3 at 14:58

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