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@@ -1,4 +1,4 @@
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-\documentclass[a4paper]{scrartcl}
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+\documentclass[a4paper,oneside,DIV15,BCOR12mm]{scrbook}
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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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@@ -16,7 +16,8 @@
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\usepackage{framed}
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\usepackage{nicefrac}
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\usepackage{siunitx}
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-\usepackage{csquotes}
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+\usepackage{csquotes} % enquote
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+\usepackage{microtype} % better document formatting
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Define theorems %
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@@ -54,586 +55,18 @@
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% Begin document %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{document}
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+\pagenumbering{roman}
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+\setcounter{page}{1}
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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 might also help to get all points
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-on the spline with minimal distance.
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-
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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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-
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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 of a point $P$ and a point $\left (x, f(x) \right )$
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-on the graph of $f$:
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-\[d_{P,f} (x) := \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\]
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-
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-Now there is finite set $M = \Set{x_1, \dots, x_n}$ of minima for given $f$ and $P$:
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-\[M = \Set{x \in \mdr | d_{P,f}(x) = \min_{\overline{x} \in \mdr} d_{P,f}(\overline{x})}\]
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-
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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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-
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-\begin{theorem}[Fermat's theorem about stationary points]\label{thm:required-extremum-property}
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- Let $x_0$ be a local extremum of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.
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-
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- Then: $f'(x_0) = 0$.
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-\end{theorem}
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-\clearpage
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-
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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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-
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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, densely dotted] {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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-
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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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-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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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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-
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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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-
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-Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The
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-slope of $f_\bot$ is $- \frac{1}{m}$ and $t_\bot$ can be calculated:\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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-
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-The point $(x, f(x))$ where the perpendicular $f_\bot$ crosses $f$
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-is calculated this way:
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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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-
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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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-
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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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- 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,dotted] { x*x + 2*x};
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- \addplot[domain=-3:3, thick,samples=50, black,dashed] {-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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-
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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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-
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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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-
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-\todo[inline]{Where are those closed formulas?}
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-
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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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-
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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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-
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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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-
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-
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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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-
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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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-
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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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-
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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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-
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-Because:\footnote{The idea why you subtract $\frac{b}{2a}$ within
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-$f$ is that when you subtract something from $x$ before applying
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-$f$ it takes more time ($x$ needs to be bigger) to get to the same
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-situation. So to move the whole graph by $1$ to the left whe have
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-to add $+1$.}
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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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-
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-
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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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-\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \Big (\underbrace{x_P+\frac{b}{2a}}_{=: z},\;\; \underbrace{y_P+\frac{b^2}{4a}-c}_{=: w} \Big )\]
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-
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-\textbf{Case 1:} 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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-
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-Then compute:
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-\begin{align}
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- d_{P,{f_2}}(x) &= \sqrt{(x-0)^2 + (f_2(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 + \big (w^2 - (1-2 a w)^2 \big)}
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-\end{align}
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-
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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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-
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-\textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
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-\begin{align}
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- d_{P,{f_2}}(x) &= \sqrt{(x-z)^2 + (f(x)-w)^2}\\
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- &= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 + w^2)}\\
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- &= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 + w^2}\\
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- 0 &\stackrel{!}{=} \Big(\big(d_{P, {f_2}}(x)\big)^2\Big)' \\
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- &= 4a^2x^3 + 2(1- 2 aw)x +(- 2z)\\
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- &= 2 \left (2a^2x^2 + (1- 2 aw) \right )x - 2z\\
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- \Leftrightarrow 0 &\stackrel{!}{=} (2a^2x^2 + (1- 2 aw)) x - z\\
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- &= 2 a^2 x^3 + (1- 2 aw) x - z\\
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-\Leftrightarrow 0 &\stackrel{!}{=} x^3 + \underbrace{\frac{(1- 2 aw)}{2 a^2}}_{=: \alpha} x + \underbrace{\frac{-z}{2 a^2}}_{=: \beta}\\
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- &= x^3 + \alpha x + \beta\label{eq:simple-cubic-equation-for-quadratic-distance}
|
|
|
-\end{align}
|
|
|
-
|
|
|
-The solution of Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
|
|
|
-is
|
|
|
-\[t := \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\]
|
|
|
-\[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
|
|
|
-
|
|
|
-When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
|
|
|
-you get:\footnote{Remember: $(a-b)^3 = a^3-3 a^2 b+3 a b^2-b^3$}
|
|
|
-\allowdisplaybreaks
|
|
|
-\begin{align}
|
|
|
- 0 &\stackrel{!}{=} \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right )^3 + \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
|
-&= (\frac{t}{\sqrt[3]{18}})^3
|
|
|
- - 3 (\frac{t}{\sqrt[3]{18}})^2 \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}
|
|
|
- + 3 (\frac{t}{\sqrt[3]{18}})(\frac{\sqrt[3]{\frac{2}{3}} \alpha }{t})^2
|
|
|
- - (\frac{\sqrt[3]{\frac{2}{3}} \alpha }{t})^3
|
|
|
- + \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
|
-&= \frac{t^3}{18}
|
|
|
- - \frac{3t^2}{\sqrt[3]{18^2}} \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}
|
|
|
- + \frac{3t}{\sqrt[3]{18}} \frac{\sqrt[3]{\frac{4}{9}} \alpha^2 }{t^2}
|
|
|
- - \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
|
- + \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
|
-&= \frac{t^3}{18}
|
|
|
- - \frac{\sqrt[3]{18} t \alpha}{\sqrt[3]{18^2}}
|
|
|
- + \frac{\sqrt[3]{12} \alpha^2}{\sqrt[3]{18} t}
|
|
|
- - \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
|
- + \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
|
-&= \frac{t^3}{18}
|
|
|
- - \frac{t \alpha}{\sqrt[3]{18}}
|
|
|
- \color{red}+ \frac{\sqrt[3]{2} \alpha^2}{\sqrt[3]{3} t} \color{black}
|
|
|
- - \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
|
- + \color{red}\alpha \color{black} \left (\frac{t}{\sqrt[3]{18}} \color{red}- \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \color{black}\right )
|
|
|
- + \beta\\
|
|
|
-&= \frac{t^3}{18} \color{blue}- \frac{t \alpha}{\sqrt[3]{18}} \color{black}
|
|
|
- - \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
|
- \color{blue}+ \frac{\alpha t}{\sqrt[3]{18}} \color{black}
|
|
|
- + \beta\\
|
|
|
-&= \frac{t^3}{18} - \frac{\frac{2}{3} \alpha^3 }{t^3} + \beta\\
|
|
|
-&= \frac{t^6 - 12 \alpha^3 + \beta 18 t^3}{18t^3}
|
|
|
-\end{align}
|
|
|
-
|
|
|
-Now only go on calculating with the numerator. Start with resubstituting
|
|
|
-$t$:
|
|
|
-\begin{align}
|
|
|
-0 &= (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta)^2 - 12 \alpha^3 + \beta 18 (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta)\\
|
|
|
-&= (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)})^2 +(9\beta)^2 - 12 \alpha^3 -18\cdot 9\beta^2\\
|
|
|
-&= 3 \cdot (4 \alpha^3 + 27 \beta^2) -81 \beta^2 - 12 \alpha^3\\
|
|
|
-&= (4 \alpha^3 + 27 \beta^2) -27 \beta^2 - 4 \alpha^3\\
|
|
|
-&= 0
|
|
|
-\end{align}
|
|
|
-
|
|
|
-\goodbreak
|
|
|
-So the solution is given by
|
|
|
-\begin{align*}
|
|
|
-x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
|
|
|
-w &:= y_P+\frac{b^2}{4a}-c \;\;\; \text{ and } \;\;\; z := x_P+\frac{b}{2a}\\
|
|
|
-\alpha &:= \frac{(1- 2 aw)}{2 a^2} \;\;\;\text{ and }\;\;\; \beta := \frac{-z}{2 a^2}\\
|
|
|
-t &:= \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\\
|
|
|
-\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 = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} &\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=$y$,
|
|
|
- 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}
|
|
|
-\begin{theorem}
|
|
|
- There cannot be an algebraic solution to the problem of finding
|
|
|
- a closest point $(x, f(x))$ to a given point $P$ when $f$ is
|
|
|
- a polynomial function of degree $3$ or higher.
|
|
|
-\end{theorem}
|
|
|
-
|
|
|
-\begin{proof}
|
|
|
- Suppose you could solve the closest point problem for arbitrary
|
|
|
- cubic functions $f = ax^3 + bx^2 + cx + d$ and arbitrary points $P = (x_P, y_P)$.
|
|
|
-
|
|
|
- Then you could solve the following problem for $x$:
|
|
|
- \begin{align}
|
|
|
- 0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'
|
|
|
- &=-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 For any coefficient $\tilde{a} \in \mdr_{> 0}$ of $x^5$ we can choose $a$ such that we get $\tilde{a}$.
|
|
|
- \item For any coefficient $\tilde{b} \in \mdr \setminus \Set{0}$ of $x^4$ we can choose $b$ such that we get $\tilde{b}$.
|
|
|
- \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 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,
|
|
|
- because if there was one, you could use this formula for finding
|
|
|
- roots of polynomials of degree 5. $\qed$
|
|
|
-\end{proof}
|
|
|
-
|
|
|
-
|
|
|
-\subsection{Another approach}
|
|
|
-\todo[inline]{Currently, this is only an idea. It might be usefull
|
|
|
-to move the cubic function $f$ such that $f$ is point symmetric
|
|
|
-to the origin. But I'm not sure how to make use of this symmetry.}
|
|
|
-Just like we moved the function $f$ and the point to get in a
|
|
|
-nicer situation, we can apply this approach for cubic functions.
|
|
|
-
|
|
|
-\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=$y$,
|
|
|
- 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};
|
|
|
- \addplot[domain=-3:3, thick,samples=50, orange] {x*x*x-x};
|
|
|
- \addplot[domain=-3:3, thick,samples=50, blue, dotted] {x*x*x+2*x};
|
|
|
- \addplot[domain=-3:3, thick,samples=50, lime, dashed] {x*x*x+3*x};
|
|
|
- \addlegendentry{$f_1(x)=x^3$}
|
|
|
- \addlegendentry{$f_2(x)=x^3 + x$}
|
|
|
- \addlegendentry{$f_1(x)=x^3 - x$}
|
|
|
- \addlegendentry{$f_2(x)=x^3 + 2 \cdot x$}
|
|
|
- \addlegendentry{$f_2(x)=x^3 + 3 \cdot x$}
|
|
|
- \end{axis}
|
|
|
-\end{tikzpicture}
|
|
|
- \caption{Cubic functions with $b = d = 0$}
|
|
|
-\end{figure}
|
|
|
-
|
|
|
-First, we move $f_0$ by $\frac{b}{3a}$ to the right, so
|
|
|
-
|
|
|
-\[f_1(x) = ax^3 + \frac{b^2 (c-1)}{3a} x + \frac{2b^3}{27 a^2} - \frac{bc}{3a} + d \;\;\;\text{ and }\;\;\;P_1 = (x_P + \frac{b}{3a}, y_P)\]
|
|
|
-
|
|
|
-because
|
|
|
-
|
|
|
-\begin{align}
|
|
|
- f_1(x) &= a \left (x - \frac{b}{3a} \right )^3 + b \left (x-\frac{b}{3a} \right )^2 + c \left (x-\frac{b}{3a} \right ) + d\\
|
|
|
- &= a \left (x^3 - 3 \frac{b}{3a}x^2 + 3 (\frac{b}{3a})^2 x - \frac{b^3}{27a^3} \right )
|
|
|
- +b \left (x^2 - \frac{2b}{3a} x + \frac{b^2}{9a^2} \right )
|
|
|
- +c x - \frac{bc}{3a} + d\\
|
|
|
- &= ax^3 - bx^2 + \frac{b^2}{3a}x - \frac{b^3}{27 a^2}\\
|
|
|
- & \;\;\;\;\;\;+ bx^2 - \frac{2b^2}{3a}x + \frac{b^3}{9a^2}\\
|
|
|
- & \;\;\;\;\;\;\;\;\;\;\;\; + c x - \frac{bc}{3a} + d\\
|
|
|
- &= ax^3 + \frac{b^2}{3a}\left (1-2+c \right )x + \frac{b^3}{9a^2} \left (1-\frac{1}{3} \right )- \frac{bc}{3a} + d
|
|
|
-\end{align}
|
|
|
-
|
|
|
-\subsection{Number of points with minimal distance}
|
|
|
-As this leads to a polynomial of degree 5 of which we have to find
|
|
|
-roots, 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.}
|
|
|
-
|
|
|
-
|
|
|
-\section{Interpolation and approximation}
|
|
|
-\subsection{Quadratic spline interpolation}
|
|
|
-You could interpolate the cubic function by a quadratic spline.
|
|
|
-
|
|
|
-\subsection{Bisection method}
|
|
|
-
|
|
|
-\subsection{Newtons method}
|
|
|
-One way to find roots of functions is Newtons method. It gives an
|
|
|
-iterative computation procedure that can converge quadratically
|
|
|
-if some conditions are met:
|
|
|
-
|
|
|
-\begin{theorem}[local quadratic convergence of Newton's method]
|
|
|
- Let $D \subseteq \mdr^n$ be open and $f: D \rightarrow \mdr^n \in C^2(\mdr)$.
|
|
|
- Let $x^* \in D$ with $f(x^*) = 0$ and the Jaccobi-Matrix $f'(x^*)$
|
|
|
- should not be invertable when evaluated at the root.
|
|
|
-
|
|
|
- Then there is a sphere
|
|
|
- \[K := K_\rho(x^*) = \Set{x \in \mdr^n | \|x- x^*\|_\infty \leq \rho} \subseteq D\]
|
|
|
- such that $x^*$ is the only root of $f$ in $K$. Furthermore,
|
|
|
- the elements of the sequence
|
|
|
- \[ x_{n+1} = x_n - \frac{f'(x_n)}{f(x_n)}\]
|
|
|
- are for every starting value $x_0 \in K$ again in $K$ and
|
|
|
- \[\lim_{n \rightarrow \infty} x_k = x^*\]
|
|
|
- Also, there is a constant $C > 0$ such that
|
|
|
- \[\|x^* - x_{n+1} \| = C \|x^* - x_n\|^2 \text{ for } n \in \mathbb{N}_0\|\]
|
|
|
-\end{theorem}
|
|
|
-
|
|
|
-The approach is extraordinary simple. You choose a starting value
|
|
|
-$x_0$ and compute
|
|
|
-
|
|
|
-\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
|
|
|
-
|
|
|
-As soon as the values don't change much, you are close to a root.
|
|
|
-The problem of this approach is choosing a starting value that is
|
|
|
-close enough to the root. So we have to have a \enquote{good}
|
|
|
-initial guess.
|
|
|
-
|
|
|
-\subsection{Quadratic minimization}
|
|
|
-\todo[inline]{TODO}
|
|
|
-
|
|
|
-\section{Conclusion}
|
|
|
-\todo[inline]{TODO}
|
|
|
+\input{introduction}
|
|
|
+\tableofcontents
|
|
|
+
|
|
|
+\pagenumbering{arabic}
|
|
|
+\setcounter{page}{1}
|
|
|
+\input{problem-description.tex}
|
|
|
+\input{constant-functions.tex}
|
|
|
+\input{linear-functions.tex}
|
|
|
+\input{quadratic-functions.tex}
|
|
|
+\input{cubic-functions.tex}
|
|
|
|
|
|
\end{document}
|
Martin Thoma