mirror of
https://github.com/MartinThoma/LaTeX-examples.git
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476 lines
20 KiB
TeX
476 lines
20 KiB
TeX
\documentclass[a4paper]{scrartcl}
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\usepackage{amssymb, amsmath} % needed for math
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\usepackage{mathtools} % \xRightarrow
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\usepackage[utf8]{inputenc} % this is needed for umlauts
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\usepackage[english]{babel} % this is needed for umlauts
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\usepackage[T1]{fontenc} % this is needed for correct output of umlauts in pdf
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\usepackage[margin=2.5cm]{geometry} %layout
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\usepackage{hyperref} % links im text
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\usepackage{braket} % needed for \Set
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\usepackage{parskip}
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\usepackage[colorinlistoftodos]{todonotes}
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\usepackage{pgfplots}
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\pgfplotsset{compat=1.7,compat/path replacement=1.5.1}
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\usepackage{tikz}
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\usepackage[framed,amsmath,thmmarks,hyperref]{ntheorem}
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\usepackage{framed}
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\usepackage{nicefrac}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Define theorems %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\theoremstyle{break}
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\setlength\theoremindent{0.7cm}
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\theoremheaderfont{\kern-0.7cm\normalfont\bfseries}
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\theorembodyfont{\normalfont} % nicht mehr kursiv
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\def\mdr{\ensuremath{\mathbb{R}}}
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\renewcommand{\qed}{\hfill\blacksquare}
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\newframedtheorem{theorem}{Theorem}
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\newframedtheorem{lemma}[theorem]{Lemma}
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\newtheorem{plaindefinition}{Definition}
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\newenvironment{definition}{\begin{plaindefinition}}{\end{plaindefinition}}
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\newenvironment{definition*}{\begin{plaindefinition*}}{\end{plaindefinition*}}
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\newtheorem{example}{Example}
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\theoremstyle{nonumberplain}
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\newtheorem{proof}{Proof:}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\title{Minimal distance to a cubic function}
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\author{Martin Thoma}
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\hypersetup{
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pdfauthor = {Martin Thoma},
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pdfkeywords = {},
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pdftitle = {Minimal Distance}
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}
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\def\mdr{\ensuremath{\mathbb{R}}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Begin document %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{document}
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\maketitle
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\begin{abstract}
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When you want to develop a selfdriving car, you have to plan which path
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it should take. A reasonable choice for the representation of
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paths are cubic splines. You also have to be able to calculate
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how to steer to get or to remain on a path. A way to do this
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is applying the \href{https://en.wikipedia.org/wiki/PID_algorithm}{PID algorithm}.
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This algorithm needs to know the signed current error. So you need to
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be able to get the minimal distance of a point to a cubic spline combined with the direction (left or right).
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As you need to get the signed error (and one steering direction might
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be prefered), it is not only necessary to
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get the minimal absolute distance, but also to get all points
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on the spline with minimal distance.
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In this paper I want to discuss how to find all points on a cubic
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function with minimal distance to a given point.
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As other representations of paths might be easier to understand and
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to implement, I will also cover the problem of finding the minimal
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distance of a point to a polynomial of degree 0, 1 and 2.
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\end{abstract}
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\section{Description of the Problem}
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Let $f: \mdr \rightarrow \mdr$ be a polynomial function and $P \in \mdr^2$
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be a point. Let $d_{P,f}: \mdr \rightarrow \mdr_0^+$
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be the Euklidean distance $d_{P,f}$ of a point $P$ and a point $\left (x, f(x) \right )$:
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\[d_{P,f} (x) := \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\]
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Now there is \todo{Should I proof this?}{finite set} $x_1, \dots, x_n$ such that
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\[\forall \tilde x \in \mathbb{R} \setminus \{x_1, \dots, x_n\}: d_{P,f}(x_1) = \dots = d_{P,f}(x_n) < d_{P,f}(\tilde x)\]
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Essentially, you want to find the minima $x_1, \dots, x_n$ for given
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$f$ and $P$.
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But minimizing $d_{P,f}$ is the same as minimizing $d_{P,f}^2$:
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\begin{align}
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d_{P,f}(x)^2 &= \sqrt{(x_P - x)^2 + (y_P - f(x))^2}^2\\
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&= x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2
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\end{align}
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\todo[inline]{Hat dieser Satz einen Namen? Gibt es ein gutes Buch,
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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).}
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\begin{theorem}\label{thm:required-extremum-property}
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Let $x_0$ be a relative extremum of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.
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Then: $f'(x_0) = 0$.
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\end{theorem}
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%bzw. 22.3
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%\begin{theorem}[Minima of polynomial functions]\label{thm:minima-of-polynomials}
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% Let $n \in \mathbb{N}, n \geq 2$, $f$ polynomial function of
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% degree $n$, $x_0 \in \mathbb{R}$, \\
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% $f'(x_0) = f''(x_0) = \dots = f^{(n-1)} (x_0) = 0$
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% and $f^{(n)} > 0$.
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%
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% Then $x_0$ is a local minimum of $f$.
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%\end{theorem}
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Constant functions %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Minimal distance to a constant function}
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Let $f(x) = c$ with $c \in \mdr$ be a constant function.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=north west,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-5, % start the diagram at this x-coordinate
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xmax= 5, % end the diagram at this x-coordinate
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ymin= 0, % start the diagram at this y-coordinate
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ymax= 3, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-5:5, thick,samples=50, red] {1};
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\addplot[domain=-5:5, thick,samples=50, green] {2};
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\addplot[domain=-5:5, thick,samples=50, blue] {3};
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\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
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\addplot[blue, mark = *, nodes near coords=$P_{h,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 3)};
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\addplot[green, mark = x, nodes near coords=$P_{g,\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 2)};
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\addplot[red, mark = *, nodes near coords=$P_{f,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 1)};
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\draw[thick, dashed] (axis cs:2,0) -- (axis cs:2,3);
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\addlegendentry{$f(x)=1$}
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\addlegendentry{$g(x)=2$}
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\addlegendentry{$h(x)=3$}
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\end{axis}
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\end{tikzpicture}
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\caption{Three constant functions and their points with minimal distance}
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\label{fig:constant-min-distance}
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\end{figure}
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Then $(x_P,f(x_P))$ has
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minimal distance to $P$. Every other point has higher distance.
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See Figure~\ref{fig:constant-min-distance}.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Linear functions %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Minimal distance to a linear function}
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Let $f(x) = m \cdot x + t$ with $m \in \mdr \setminus \Set{0}$ and
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$t \in \mdr$ be a linear function.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=north east,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin= 0, % start the diagram at this x-coordinate
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xmax= 5, % end the diagram at this x-coordinate
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ymin= 0, % start the diagram at this y-coordinate
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ymax= 3, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
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\addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
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\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
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\addlegendentry{$f(x)=\frac{1}{2}x$}
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\addlegendentry{$g(x)=-2x+6$}
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\end{axis}
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\end{tikzpicture}
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\caption{The shortest distance of $P$ to $f$ can be calculated by using the perpendicular}
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\label{fig:linear-min-distance}
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\end{figure}
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Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The slope of $f_\bot$
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is $- \frac{1}{m}$. Now you can calculate $f_\bot$:\nobreak
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\begin{align}
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f_\bot(x) &= - \frac{1}{m} \cdot x + t_\bot\\
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\Rightarrow y_P &= - \frac{1}{m} \cdot x_P + t_\bot\\
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\Leftrightarrow t_\bot &= y_P + \frac{1}{m} \cdot x_P
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\end{align}
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Now find the point $(x, f(x))$ where the perpendicular crosses the function:
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\begin{align}
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f(x) &= f_\bot(x)\\
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\Leftrightarrow m \cdot x + t &= - \frac{1}{m} \cdot x + \left(y_P + \frac{1}{m} \cdot x_P \right)\\
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\Leftrightarrow \left (m + \frac{1}{m} \right ) \cdot x &= y_P + \frac{1}{m} \cdot x_P - t\\
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\Leftrightarrow x &= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )
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\end{align}
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There is only one point with minimal distance. See Figure~\ref{fig:linear-min-distance}.
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Quadratic functions %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Minimal distance to a quadratic function}
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Let $f(x) = a \cdot x^2 + b \cdot x + c$ with $a \in \mdr \setminus \Set{0}$ and
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$b, c \in \mdr$ be a quadratic function.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=north west,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-0.25, % start the diagram at this y-coordinate
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ymax= 9, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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%xticklabels={-2,-1.6,...,7},
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%yticklabels={-8,-7,...,8},
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-3:3, thick,samples=50, red] {0.5*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x};
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\addplot[domain=-3:3, thick,samples=50, blue] {x*x + x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x + 2*x};
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\addplot[domain=-3:3, thick,samples=50, black] {-x*x + 6};
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\addlegendentry{$f_1(x)=\frac{1}{2}x^2$}
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\addlegendentry{$f_2(x)=x^2$}
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\addlegendentry{$f_3(x)=x^2+x$}
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\addlegendentry{$f_4(x)=x^2+2x$}
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\addlegendentry{$f_5(x)=-x^2+6$}
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\end{axis}
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\end{tikzpicture}
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\caption{Quadratic functions}
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\end{figure}
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\subsection{Calculate points with minimal distance}
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In this case, $d_{P,f}^2$ is polynomial of degree 4.
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We use Theorem~\ref{thm:required-extremum-property}:\nobreak
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\begin{align}
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0 &\overset{!}{=} (d_{P,f}^2)'\\
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&= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\
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&= -2 x_p + 2x -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
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\Leftrightarrow 0 &\overset{!}{=} -x_p + x -y_p f'(x) + f(x) \cdot f'(x) \rlap{\hspace*{3em}(divide by 2)}\label{eq:minimizingFirstDerivative}\\
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&= -x_p + x -y_p (2ax+b) + (ax^2+bx+c)(2ax+b)\\
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&= -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)\\
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&= -x_p + x -2y_p ax- y_p b + (2a^2 x^3 + 3 ab x^2 + 2acx + b^2 x + bc)\\
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&= 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}
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\end{align}
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%\begin{align}
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% 0 &\overset{!}{=}(d_{P,f}^2)''\\
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% &= 2 - 2y_p f''(x) + \left ( 2 f(x) \cdot f'(x) \right )' \rlap{\hspace*{3em}(Eq. \ref{eq:minimizingFirstDerivative})}\\
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% &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x) \cdot f'(x) + f(x) \cdot f''(x) \right ) \rlap{\hspace*{3em}(product rule)}\\
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% &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x)^2 + f(x) \cdot f''(x) \right )\\
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% &= 12a^2 x^2 + 12abx + 2(1 -2y_p a+ 2ac + b^2)
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%\end{align}
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This is an algebraic equation of degree 3.
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There can be up to 3 solutions in such an equation. Those solutions
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can be found with a closed formula.
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\todo[inline]{Where are those closed formulas?}
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\begin{example}
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Let $a = 1, b = 0, c= 1, x_p= 0, y_p = 1$.
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So $f(x) = x^2 + 1$ and $P(0, 1)$.
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\begin{align}
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0 &\stackrel{!}{=} 4 x^3 - 2x\\
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&=2x(2x^2 - 1)\\
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\Rightarrow x_1 &= 0 \;\;\; x_{2,3} = \pm \frac{1}{\sqrt{2}}
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\end{align}
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As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
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\end{example}
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\subsection{Number of points with minimal distance}
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\begin{theorem}
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A point $P$ has either one or two points on the graph of a
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quadratic function $f$ that are closest to $P$.
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\end{theorem}
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In the following, I will do some transformations with $f = f_0$ and
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$P = P_0$ .
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Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does
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not change the minimum distance. Furthermore, we can find the
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points with minimum distance on the moved situation and calculate
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the minimum points in the original situation.
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First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
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\[f_1(x) = ax^2 - \frac{b^2}{4a} + c \;\;\;\text{ and }\;\;\; P_1 = \left (x_p+\frac{b}{2a},\;\; y_p \right )\]
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Because:
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\begin{align}
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f(x-\nicefrac{b}{2a}) &= a (x-\nicefrac{b}{2a})^2 + b (x-\nicefrac{b}{2a}) + c\\
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&= a (x^2 - \nicefrac{b}{a} x + \nicefrac{b^2}{4a^2}) + bx - \nicefrac{b^2}{2a} + c\\
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&= ax^2 - bx + \nicefrac{b^2}{4a} + bx - \nicefrac{b^2}{2a} + c\\
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&= ax^2 -\nicefrac{b^2}{4a} + c
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\end{align}
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Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
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\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]
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As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
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$P = (0, w)$ could possilby have three minima.
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Then compute:
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\begin{align}
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d_{P,{f_2}}(x) &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
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&= \sqrt{x^2 + (ax^2-w)^2}\\
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&= \sqrt{x^2 + a^2 x^4-2aw x^2+w^2}\\
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&= \sqrt{a^2 x^4 + (1-2aw) x^2 + w^2}\\
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&= \sqrt{\left (a^2 x^2 + \frac{1-2 a w}{2} \right )^2 + w^2 - (1-2 a w)^2}\\
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&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + (w^2 - (1-2 a w)^2)}\\
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\end{align}
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The term
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\[a^2 x^2 + (\nicefrac{1}{2}-a w)\]
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should get as close to $0$ as possilbe when we want to minimize
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$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
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For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.
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So the solution is given by
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\begin{align*}
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x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
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\underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
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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} \\
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x_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\
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x_1 = x_S &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c \leq \frac{1}{2a} \\
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x_1 = todo &\text{if } x_P \neq x_S
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\end{cases}
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\end{align*}
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Cubic %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Minimal distance to a cubic function}
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Let $f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$ be a cubic function
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with $a \in \mdr \setminus \Set{0}$ and
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$b, c, d \in \mdr$ be a function.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
|
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legend pos=south east,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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|
width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-3, % start the diagram at this y-coordinate
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|
ymax= 3, % end the diagram at this y-coordinate
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|
axis background/.style={fill=white},
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|
xlabel=$x$,
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|
ylabel=$y$,
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|
%xticklabels={-2,-1.6,...,7},
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|
%yticklabels={-8,-7,...,8},
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|
tick align=outside,
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|
minor tick num=-3,
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|
enlargelimits=true,
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|
tension=0.08]
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|
\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x*x};
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\addplot[domain=-3:3, thick,samples=50, blue] {x*x*x+2*x*x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x+x};
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|
\addlegendentry{$f_1(x)=x^3$}
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|
\addlegendentry{$f_2(x)=x^3 + x^2$}
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\addlegendentry{$f_2(x)=x^3 + 2 \cdot x^2$}
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|
\addlegendentry{$f_1(x)=x^3 + x$}
|
|
\end{axis}
|
|
\end{tikzpicture}
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|
\caption{Cubic functions}
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|
\end{figure}
|
|
|
|
%
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|
%\subsection{Special points}
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|
%\todo[inline]{Write this}
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|
%
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|
%\subsection{Voronoi}
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|
%
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|
%For $b^2 \geq 3ac$
|
|
%
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|
%\todo[inline]{Write this}
|
|
|
|
\subsection{Calculate points with minimal distance}
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|
When you want to calculate points with minimal distance, you can
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|
take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
|
|
|
|
\begin{align}
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|
0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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|
&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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|
&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
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|
&= \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}
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|
0 &\stackrel{!}{=} f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
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|
&= \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}
|