\documentclass[a4paper]{scrartcl} \usepackage{amssymb, amsmath} % needed for math \usepackage[utf8]{inputenc} % this is needed for umlauts \usepackage[english]{babel} % this is needed for umlauts \usepackage[T1]{fontenc} % this is needed for correct output of umlauts in pdf \usepackage[margin=2.5cm]{geometry} %layout \usepackage{hyperref} % links im text \usepackage{braket} % needed for \Set \usepackage{parskip} \usepackage[colorinlistoftodos]{todonotes} \usepackage{pgfplots} \pgfplotsset{compat=1.7,compat/path replacement=1.5.1} \usepackage{tikz} \title{Minimal distance to a cubic function} \author{Martin Thoma} \hypersetup{ pdfauthor = {Martin Thoma}, pdfkeywords = {}, pdftitle = {Minimal Distance} } \def\mdr{\ensuremath{\mathbb{R}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Begin document % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \maketitle \begin{abstract} In this paper I want to discuss how to find all points on a a cubic function with minimal distance to a given point. \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 $d: \mdr^2 \times \mdr^2 \rightarrow \mdr_0^+$ be the euklidean distance of two points: \[d \left ((x_1, y_1), (x_2, y_2) \right) := \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\] Now there is finite set of points $x_1, \dots, x_n$ such that \[\forall \tilde x \in \mathbb{R} \setminus \{x_1, \dots, x_n\}: d(P, (x_1, f(x_1))) = \dots = d(P, (x_n, f(x_n))) < d(P, (\tilde x, f(\tilde x)))\] \section{Minimal distance to a constant function} Let $f(x) = c$ with $c \in \mdr$ be a function. \todo[inline]{add image} Then $(x_P,f(x_P))$ has minimal distance to $P$. Every other point has higher distance. \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 function. \todo[inline]{add image} Now you can drop a perpendicular through $P$ on $f(x)$. The slope $f_\bot$ of the perpendicular is $- \frac{1}{m}$. Then: \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\\ 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. \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 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{Number of points with minimal distance} It is obvious that a quadratic function can have two points with minimal distance. For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} \approx (2.179, 2.179^2)$ has minimal distance to $P$, but also $P_{f,2}\approx (-2.179, 2.179^2)$. Obviously, there cannot be more than three points with minimal distance. But can there be three points? \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=-0.7, % start the diagram at this x-coordinate xmax= 0.7, % end the diagram at this x-coordinate ymin=-0.25, % start the diagram at this y-coordinate ymax= 0.5, % 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=-0.7:0.7, thick,samples=50, orange] {x*x}; \draw (axis cs:0,0.5) circle[radius=0.5]; \draw[red, thick] (axis cs:0,0.5) -- (axis cs:0.101,0.0102); \draw[red, thick] (axis cs:0,0.5) -- (axis cs:-0.101,0.0102); \draw[red, thick] (axis cs:0,0.5) -- (axis cs:0,0); \addlegendentry{$f(x)=x^2$} \end{axis} \end{tikzpicture} \caption{3 points with minimal distance?} \todo[inline]{Is this possible?} \end{figure} \subsection{Calculate points with minimal distance} \todo[inline]{Write this} \section{Minimal distance to a cubic function} \subsection{Number of points with minimal distance} \todo[inline]{Write this} \subsection{Calculate points with minimal distance} \todo[inline]{Write this} \end{document}