LaTeX-examples/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex

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\usepackage[framed,amsmath,thmmarks,hyperref]{ntheorem}
\usepackage{framed}
\usepackage{nicefrac}

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\def\mdr{\ensuremath{\mathbb{R}}}
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\title{Minimal distance to a cubic function}
\author{Martin Thoma}

\hypersetup{ 
  pdfauthor   = {Martin Thoma}, 
  pdfkeywords = {}, 
  pdftitle    = {Minimal Distance} 
}

\def\mdr{\ensuremath{\mathbb{R}}}

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% Begin document                                                    %
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\begin{document}
\maketitle
\begin{abstract}
When you want to develop a selfdriving car, you have to plan which path 
it should take. A reasonable choice for the representation of
paths are cubic splines. You also have to be able to calculate
how to steer to get or to remain on a path. A way to do this
is applying the \href{https://en.wikipedia.org/wiki/PID_algorithm}{PID algorithm}.
This algorithm needs to know the signed current error. So you need to 
be able to get the minimal distance of a point to a cubic spline combined with the direction (left or right).
As you need to get the signed error (and one steering direction might
be prefered), it is not only necessary to
get the minimal absolute distance, but also to get all points
on the spline with minimal distance.

In this paper I want to discuss how to find all points on a cubic 
function with minimal distance to a given point.
As other representations of paths might be easier to understand and
to implement, I will also cover the problem of finding the minimal
distance of a point to a polynomial of degree 0, 1 and 2.
\end{abstract}

\section{Description of the Problem}
Let $f: \mdr \rightarrow \mdr$ be a polynomial function and $P \in \mdr^2$
be a point. Let $d_{P,f}: \mdr \rightarrow \mdr_0^+$
be the Euklidean distance $d_{P,f}$ of a point $P$ and a point $\left (x, f(x) \right )$:
\[d_{P,f} (x) := \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\]

Now there is \todo{Should I proof this?}{finite set} $x_1, \dots, x_n$ such that 
\[\forall \tilde x \in \mathbb{R} \setminus \{x_1, \dots, x_n\}: d_{P,f}(x_1) = \dots = d_{P,f}(x_n) < d_{P,f}(\tilde x)\]

Essentially, you want to find the minima $x_1, \dots, x_n$ for given 
$f$ and $P$.
But minimizing $d_{P,f}$ is the same as minimizing $d_{P,f}^2$:
\begin{align}
    d_{P,f}(x)^2    &= \sqrt{(x_P - x)^2 + (y_P - f(x))^2}^2\\
                &= x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2
\end{align}

\todo[inline]{Hat dieser Satz einen Namen? Gibt es ein gutes Buch,
aus dem ich den zitieren kann? Ich habe ihn aus \href{https://github.com/MartinThoma/LaTeX-examples/tree/master/documents/Analysis I}{meinem Analysis I Skript} (Satz 21.5).}
\begin{theorem}\label{thm:required-extremum-property}
    Let $x_0$ be a relative extremum of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.

    Then: $f'(x_0) = 0$.
\end{theorem}

%bzw. 22.3
%\begin{theorem}[Minima of polynomial functions]\label{thm:minima-of-polynomials}
%    Let $n \in \mathbb{N}, n \geq 2$, $f$ polynomial function of 
%    degree $n$, $x_0 \in \mathbb{R}$,  \\
%    $f'(x_0) = f''(x_0) = \dots = f^{(n-1)} (x_0) = 0$
%    and $f^{(n)} > 0$.
%
%    Then $x_0$ is a local minimum of $f$.
%\end{theorem}
\clearpage

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% Constant functions                                                %
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\section{Minimal distance to a constant function}
Let $f(x) = c$ with $c \in \mdr$ be a constant function. 

\begin{figure}[htp]
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            legend pos=north west,
            axis x line=middle,
            axis y line=middle,
            grid = major,
            width=0.8\linewidth,
            height=8cm,
            grid style={dashed, gray!30},
            xmin=-5, % start the diagram at this x-coordinate
            xmax= 5, % end   the diagram at this x-coordinate
            ymin= 0, % start the diagram at this y-coordinate
            ymax= 3, % end   the diagram at this y-coordinate
            axis background/.style={fill=white},
            xlabel=$x$,
            ylabel=$y$,
            tick align=outside,
            minor tick num=-3,
            enlargelimits=true,
            tension=0.08]
          \addplot[domain=-5:5, thick,samples=50, red] {1};
          \addplot[domain=-5:5, thick,samples=50, green] {2};
          \addplot[domain=-5:5, thick,samples=50, blue] {3};
          \addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
          \addplot[blue, mark = *, nodes near coords=$P_{h,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 3)};
          \addplot[green, mark = x, nodes near coords=$P_{g,\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 2)};
          \addplot[red, mark = *, nodes near coords=$P_{f,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 1)};
          \draw[thick, dashed] (axis cs:2,0) -- (axis cs:2,3);
          \addlegendentry{$f(x)=1$}
          \addlegendentry{$g(x)=2$}
          \addlegendentry{$h(x)=3$}
        \end{axis} 
    \end{tikzpicture}
    \caption{Three constant functions and their points with minimal distance}
    \label{fig:constant-min-distance}
\end{figure}

Then $(x_P,f(x_P))$ has
minimal distance to $P$. Every other point has higher distance.
See Figure~\ref{fig:constant-min-distance}.

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% Linear functions                                                  %
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\section{Minimal distance to a linear function}
Let $f(x) = m \cdot x + t$ with $m \in \mdr \setminus \Set{0}$ and 
$t \in \mdr$ be a linear function.

\begin{figure}[htp]
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            legend pos=north east,
            axis x line=middle,
            axis y line=middle,
            grid = major,
            width=0.8\linewidth,
            height=8cm,
            grid style={dashed, gray!30},
            xmin= 0, % start the diagram at this x-coordinate
            xmax= 5, % end   the diagram at this x-coordinate
            ymin= 0, % start the diagram at this y-coordinate
            ymax= 3, % end   the diagram at this y-coordinate
            axis background/.style={fill=white},
            xlabel=$x$,
            ylabel=$y$,
            tick align=outside,
            minor tick num=-3,
            enlargelimits=true,
            tension=0.08]
          \addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
          \addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
          \addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
          \addlegendentry{$f(x)=\frac{1}{2}x$}
          \addlegendentry{$g(x)=-2x+6$}
        \end{axis} 
    \end{tikzpicture}
    \caption{The shortest distance of $P$ to $f$ can be calculated by using the perpendicular}
    \label{fig:linear-min-distance}
\end{figure}

Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The slope of $f_\bot$
is $- \frac{1}{m}$. Now you can calculate $f_\bot$:\nobreak
\begin{align}
                 f_\bot(x) &= - \frac{1}{m} \cdot x + t_\bot\\
    \Rightarrow        y_P &= - \frac{1}{m} \cdot x_P + t_\bot\\
    \Leftrightarrow t_\bot &= y_P + \frac{1}{m} \cdot x_P
\end{align}

Now find the point $(x, f(x))$ where the perpendicular crosses the function:
\begin{align}
    f(x) &= f_\bot(x)\\
    \Leftrightarrow m \cdot x + t &= - \frac{1}{m} \cdot x + \left(y_P + \frac{1}{m} \cdot x_P \right)\\
    \Leftrightarrow \left (m + \frac{1}{m} \right ) \cdot x &= y_P + \frac{1}{m} \cdot x_P - t\\
    \Leftrightarrow x &= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )
\end{align}

There is only one point with minimal distance. See Figure~\ref{fig:linear-min-distance}.
\clearpage
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% Quadratic functions                                               %
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\section{Minimal distance to a quadratic function}
Let $f(x) = a \cdot x^2 + b \cdot x + c$ with $a \in \mdr \setminus \Set{0}$ and 
$b, c \in \mdr$ be a quadratic function.

\begin{figure}[htp]
    \centering
\begin{tikzpicture}
    \begin{axis}[
        legend pos=north west,
        axis x line=middle,
        axis y line=middle,
        grid = major,
        width=0.8\linewidth,
        height=8cm,
        grid style={dashed, gray!30},
        xmin=-3,     % start the diagram at this x-coordinate
        xmax= 3,    % end   the diagram at this x-coordinate
        ymin=-0.25,     % start the diagram at this y-coordinate
        ymax= 9,   % end   the diagram at this y-coordinate
        axis background/.style={fill=white},
        xlabel=$x$,
        ylabel=$y$,
        %xticklabels={-2,-1.6,...,7},
        %yticklabels={-8,-7,...,8},
        tick align=outside,
        minor tick num=-3,
        enlargelimits=true,
        tension=0.08]
      \addplot[domain=-3:3, thick,samples=50, red] {0.5*x*x}; 
      \addplot[domain=-3:3, thick,samples=50, green] {x*x}; 
      \addplot[domain=-3:3, thick,samples=50, blue] {x*x + x};
      \addplot[domain=-3:3, thick,samples=50, orange] {x*x + 2*x};
      \addplot[domain=-3:3, thick,samples=50, black] {-x*x + 6};
      \addlegendentry{$f_1(x)=\frac{1}{2}x^2$}
      \addlegendentry{$f_2(x)=x^2$}
      \addlegendentry{$f_3(x)=x^2+x$}
      \addlegendentry{$f_4(x)=x^2+2x$}
      \addlegendentry{$f_5(x)=-x^2+6$}
    \end{axis} 
\end{tikzpicture}
    \caption{Quadratic functions}
\end{figure}

\subsection{Calculate points with minimal distance}
In this case, $d_{P,f}^2$ is polynomial of degree 4. 
We use Theorem~\ref{thm:required-extremum-property}:\nobreak
\begin{align}
    0     &\overset{!}{=} (d_{P,f}^2)'\\
          &= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\
          &= -2 x_p + 2x -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
\Leftrightarrow 0 &\overset{!}{=} -x_p + x -y_p f'(x) + f(x) \cdot f'(x) \rlap{\hspace*{3em}(divide by 2)}\label{eq:minimizingFirstDerivative}\\
          &= -x_p + x -y_p (2ax+b) + (ax^2+bx+c)(2ax+b)\\
          &= -x_p + x -y_p \cdot 2ax- y_p b + (2 a^2 x^3+2 a b x^2+2 a c x+ab x^2+b^2 x+bc)\\
          &= -x_p + x -2y_p ax- y_p b + (2a^2 x^3 + 3 ab x^2 + 2acx + b^2 x + bc)\\
          &= 2a^2 x^3 + 3 ab x^2 + (1 -2y_p a+ 2ac + b^2)x +(bc-by_p-x_p)\label{eq:quadratic-derivative-eq-0}
\end{align}

%\begin{align}
%    0     &\overset{!}{=}(d_{P,f}^2)''\\
%          &= 2 - 2y_p f''(x) + \left ( 2 f(x) \cdot f'(x) \right )' \rlap{\hspace*{3em}(Eq. \ref{eq:minimizingFirstDerivative})}\\
%          &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x) \cdot f'(x) + f(x) \cdot f''(x) \right ) \rlap{\hspace*{3em}(product rule)}\\
%          &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x)^2 + f(x) \cdot f''(x) \right )\\
%          &= 12a^2 x^2 + 12abx + 2(1 -2y_p a+ 2ac + b^2)
%\end{align}


This is an algebraic equation of degree 3.
There can be up to 3 solutions in such an equation. Those solutions
can be found with a closed formula.

\todo[inline]{Where are those closed formulas?}

\begin{example}
    Let $a = 1,  b = 0,  c= 1, x_p= 0, y_p = 1$.
    So $f(x) = x^2 + 1$ and $P(0, 1)$.

\begin{align}
    0 &\stackrel{!}{=} 4 x^3 - 2x\\
      &=2x(2x^2 - 1)\\
    \Rightarrow x_1 &= 0 \;\;\; x_{2,3} = \pm \frac{1}{\sqrt{2}}
\end{align}

As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
\end{example}


\subsection{Number of points with minimal distance}
\begin{theorem}
    A point $P$ has either one or two points on the graph of a 
    quadratic function $f$ that are closest to $P$.
\end{theorem}

In the following, I will do some transformations with $f = f_0$ and
$P = P_0$ .

Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does 
not change the minimum distance. Furthermore, we can find the 
points with minimum distance on the moved situation and calculate
the minimum points in the original situation.

First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
\[f_1(x) = ax^2 - \frac{b^2}{4a} + c \;\;\;\text{ and }\;\;\; P_1 = \left (x_p+\frac{b}{2a},\;\; y_p \right )\]

Because:
\begin{align}
    f(x-\nicefrac{b}{2a}) &= a (x-\nicefrac{b}{2a})^2 + b (x-\nicefrac{b}{2a}) + c\\
    &= a (x^2 - \nicefrac{b}{a} x + \nicefrac{b^2}{4a^2}) + bx - \nicefrac{b^2}{2a} + c\\
    &= ax^2 - bx + \nicefrac{b^2}{4a} + bx - \nicefrac{b^2}{2a} + c\\
    &= ax^2 -\nicefrac{b^2}{4a} + c
\end{align}


Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]

As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points 
$P = (0, w)$ could possilby have three minima.

Then compute:
\begin{align}
  d_{P,{f_2}}(x)  &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
    &= \sqrt{x^2 + (ax^2-w)^2}\\
    &= \sqrt{x^2 + a^2 x^4-2aw x^2+w^2}\\
    &= \sqrt{a^2 x^4 + (1-2aw) x^2 + w^2}\\
    &= \sqrt{\left (a^2 x^2 + \frac{1-2 a w}{2} \right )^2 + w^2 - (1-2 a w)^2}\\
    &= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + (w^2 - (1-2 a w)^2)}\\
\end{align}

The term 
\[a^2 x^2 + (\nicefrac{1}{2}-a w)\]
should get as close to $0$ as possilbe when we want to minimize 
$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.

So the solution is given by

\begin{align*}
x_S &:= - \frac{b}{2a}\\
\underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
     x_1 = +\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S \text{ and }   &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c >  \frac{1}{2a} \\
     x_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\
     x_1 = x_S   &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c \leq  \frac{1}{2a} \\
     x_1 = todo   &\text{if } x_P \neq x_S
    \end{cases}
\end{align*}

\clearpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Cubic                                                             %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Minimal distance to a cubic function}
Let $f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$ be a cubic function
with $a \in \mdr \setminus \Set{0}$ and 
$b, c, d \in \mdr$ be a function.

\begin{figure}[htp]
    \centering
\begin{tikzpicture}
    \begin{axis}[
        legend pos=south east,
        axis x line=middle,
        axis y line=middle,
        grid = major,
        width=0.8\linewidth,
        height=8cm,
        grid style={dashed, gray!30},
        xmin=-3,     % start the diagram at this x-coordinate
        xmax= 3,    % end   the diagram at this x-coordinate
        ymin=-3,     % start the diagram at this y-coordinate
        ymax= 3,   % end   the diagram at this y-coordinate
        axis background/.style={fill=white},
        xlabel=$x$,
        ylabel=$y$,
        %xticklabels={-2,-1.6,...,7},
        %yticklabels={-8,-7,...,8},
        tick align=outside,
        minor tick num=-3,
        enlargelimits=true,
        tension=0.08]
      \addplot[domain=-3:3, thick,samples=50, red] {x*x*x}; 
      \addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x*x};
      \addplot[domain=-3:3, thick,samples=50, blue] {x*x*x+2*x*x};
      \addplot[domain=-3:3, thick,samples=50, orange] {x*x*x+x}; 
      \addlegendentry{$f_1(x)=x^3$}
      \addlegendentry{$f_2(x)=x^3 + x^2$}
      \addlegendentry{$f_2(x)=x^3 + 2 \cdot x^2$}
      \addlegendentry{$f_1(x)=x^3 + x$}
    \end{axis} 
\end{tikzpicture}
    \caption{Cubic functions}
\end{figure}

%
%\subsection{Special points}
%\todo[inline]{Write this}
%
%\subsection{Voronoi}
%
%For $b^2 \geq 3ac$
%
%\todo[inline]{Write this}

\subsection{Calculate points with minimal distance}
When you want to calculate points with minimal distance, you can 
take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:

\begin{align}
    0  &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
       &= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
       &= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
       &= \underbrace{f'(x) \cdot \left (f(x) - y_p \right )}_{\text{Polynomial of degree 5}} + x - x_p
\end{align}

General algebraic equations of degree 5 don't have a solution formula.\footnote{TODO: Quelle}
Although here seems to be more structure, the resulting algebraic
equation can be almost any polynomial of degree 5:\footnote{Thanks to Peter Košinár on \href{http://math.stackexchange.com/a/584814/6876}{math.stackexchange.com} for this one}

\begin{align}
    0  &\stackrel{!}{=} f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
    &= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{\tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{=: \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{\tilde{d}} x^2 \\
    & &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{=: \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{=: \tilde{f}}\\
    0 &\stackrel{!}{=} \tilde{a}x^5 + \tilde{b}x^4 + \tilde{c}x^3 + \tilde{d}x^2 + \tilde{e}x + \tilde{f}
\end{align}

\begin{enumerate}
    \item With $a$, we can get any value of $\tilde{a} \in \mdr \setminus \Set{0}$.
    \item With $b$, we can get any value of $\tilde{b} \in \mdr \setminus \Set{0}$.
    \item With $c$, we can get any value of $\tilde{c} \in \mdr$.
    \item With $d$, we can get any value of $\tilde{d} \in \mdr$.
    \item With $y_p$, we can get any value of $\tilde{e} \in \mdr$.
    \item With $x_p$, we can get any value of $\tilde{f} \in \mdr$.
\end{enumerate}

The first restriction only guaratees that we have a polynomial of 
degree 5. The second one is necessary, to get a high range of
$\tilde{e}$.

This means, that there is no solution formula for the problem of 
finding the closest points on a cubic function to a given point.

\subsection{Number of points with minimal distance}
As there is an algebraic equation of degree 5, there cannot be more
than 5 solutions.
\todo[inline]{Can there be 3, 4 or even 5 solutions? Examples!

After looking at function graphs of cubic functions, I'm pretty 
sure that there cannot be 4 or 5 solutions, no matter how you 
chose the cubic function $f$ and $P$.

I'm also pretty sure that there is no polynomial (no matter what degree)
that has more than 3 solutions.}

\todo[inline]{If there is no closed form solution, I want to 
describe a numerical solution. I guess Newtons method might be good.}
\end{document}