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major cleanup
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Martin Thoma
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1 changed files with 84 additions and 57 deletions
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@ -12,6 +12,26 @@
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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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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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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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\newframedtheorem{theorem}{Theorem}[section]
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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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@ -51,15 +71,35 @@ distance of a point to a polynomial of degree 0, 1 and 2.
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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 $d: \mdr^2 \times \mdr^2 \rightarrow \mdr_0^+$
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be the Euklidean distance of two points:
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\[d \left ((x_1, y_1), (x_2, y_2) \right) := \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\]
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be a point. Let $d_{P,f}: \mdr^2 \rightarrow \mdr_0^+$
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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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\[\forall \tilde x \in \mathbb{R} \setminus \{x_1, \dots, x_n\}: d(P, (x_1, f(x_1))) = \dots = d(P, (x_n, f(x_n))) < d(P, (\tilde x, f(\tilde x)))\]
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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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The task is now to find those $x_1, \dots, x_n$ for given $f, P$.
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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? kann ich den irgendwo zitieren? Ich habe ihn aus \href{https://github.com/MartinThoma/LaTeX-examples/tree/master/documents/Analysis I}{meinem Analysis I Skript} (Satz 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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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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@ -98,7 +138,7 @@ Let $f(x) = c$ with $c \in \mdr$ be a constant function.
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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{3 constant functions and their points with minimal distance}
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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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@ -106,6 +146,9 @@ 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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@ -207,6 +250,39 @@ $b, c \in \mdr$ be a quadratic function.
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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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We use Theorem~\ref{thm:minima-of-polynomials}:\nobreak
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\begin{align}
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0 &\stackrel{!}{=} (d_{P,f}^2)'\\
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&= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\label{eq:minimizingFirstDerivative}\\
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&= -2 x_p + 2x -2y_p (2ax+b) + ((ax^2+bx+c)^2)'\\
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&= -2 x_p + 2x -2y_p \cdot 2ax-2 y_p b + (a^2 x^4+2 a b x^3+2 a c x^2+b^2 x^2+2 b c x+c^2)'\\
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&= -2 x_p + 2x -4y_p ax-2 y_p b + (4a^2 x^3 + 6 ab x^2 + 4acx + 2b^2 x + 2bc)\\
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&= 4a^2 x^3 + 6 ab x^2 + 2(1 -2y_p a+ 2ac + b^2)x +2(bc-by_p-x_p)\\
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0 &\stackrel{!}{=}(d_{P,f}^2)''\\
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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. An example is
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\begin{align*}
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a &= 1 & b &= 0 & c &= 1 & x_p &= 0 & y_p &= 1
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\end{align*}
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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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\subsection{Number of points with minimal distance}
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It is obvious that a quadratic function can have two points with
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minimal distance.
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@ -250,58 +326,9 @@ But can there be three points?
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\end{axis}
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\end{tikzpicture}
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\caption{3 points with minimal distance?}
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\todo[inline]{Is this possible? http://math.stackexchange.com/q/553097/6876}
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\end{figure}
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As the point is already given, you want to minimize the following
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function:
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\begin{align}
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d: &\mdr \rightarrow \mdr^+_0\\
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d(x) &= \sqrt{(x_p,y_p),(x,f(x))}\\
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&= \sqrt{(x_p-x)^2 + (y_p - f(x))^2}\\
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&= \sqrt{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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Minimizing $d$ is the same as minimizing $d^2$:
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\begin{align}
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d(x)^2 &= x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2\\
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(d(x)^2)' &= -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)' \label{eq:minimizing}
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\end{align}
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Now we use thet $f(x) = ax^2 + bx + c$:
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\begin{align}
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0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(2ax+b) + ((ax^2+bx+c)^2)'\\
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&= -2 x_p + 2x -2y_p \cdot 2ax-2 y_p b + (a^2 x^4+2 a b x^3+2 a c x^2+b^2 x^2+2 b c x+c^2)'\\
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&= -2 x_p + 2x -4y_p ax-2 y_p b + (4a^2 x^3 + 6 ab x^2 + 4acx + 2b^2 x + 2bc)\\
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&= 4a^2 x^3 + 6 ab x^2 + 2(1 -2y_p a+ 2ac + b^2)x +2(bc-by_p-x_p)\\
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\end{align}
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\subsubsection{Solutions}
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As the problem stated above is a cubic equation, you can solved it
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analytically. But the solutions are not very nice, so I've entered
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\texttt{$0=4*a^2 *x^3 + 6 *a*b *x^2 + 2*(1 -2*e *a+ 2*a*c + b^2)*x +2*(b*c-b*e-d)$}
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with $d := x_p$ and $e := y_p$.
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to \href{http://www.wolframalpha.com/input/?i=0%3D4*a%5E2+*x%5E3+%2B+6+*a*b+*x%5E2+%2B+2*%281+-2*e+*a%2B+2*a*c+%2B+b%5E2%29*x+%2B2*%28b*c-b*e-d%29}{WolframAlpha} to let it solve. The solutions are:
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\textbf{First solution}
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\begin{align*}
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x = &\frac{1}{6 \sqrt[3]{2} a^2} \sqrt[3]{(108 a^4 d+54 a^3 b+\sqrt{(108 a^4 d+54 a^3 b)^2+4 (12 a^3 c-12 a^3 e-3 a^2 b^2+6 a^2)^3})}\\
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&-\frac{12 a^3 c-12 a^3 e-3 a^2 b^2+6 a^2}
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{3 (2^{\frac{2}{3}}) a^2 \sqrt[3]{108 a^4 d+54 a^3 b+\sqrt{(108 a^4 d+54 a^3 b)^2+4 (12 a^3 c-12 a^3 e-3 a^2 b^2+6 a^2)^3}} }-b/(2 a)
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\end{align*}
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So the minimum for $a=1, b=c=d=0$ is:
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\subsection{Calculate points with minimal distance}
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\todo[inline]{Write this}
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\todo[inline]{write this}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Cubic %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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@ -324,7 +351,7 @@ For $b^2 \geq 3ac$
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\todo[inline]{Write this}
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\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:minimizing}:
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take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
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\begin{align}
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0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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