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added some ideas for the case of intervalls [a,b] of R

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Martin Thoma 2013-12-12 23:05:04 +01:00
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63
documents/math-minimal-distance-to-cubic-function/constant-functions.tex
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@ -1,12 +1,13 @@
\chapter{Constant functions}
\section{Defined on $\mdr$}
Let $f(x) = c$ with $c \in \mdr$ be a constant function.
Let $f:\mdr \rightarrow \mdr$, $f(x) := c$ with $c \in \mdr$ be a constant function.
\begin{figure}[htp]
\centering
\begin{tikzpicture}
\begin{axis}[
legend pos=north west,
legend cell align=left,
axis x line=middle,
axis y line=middle,
grid = major,
@ -45,4 +46,62 @@ Then $(x_P,f(x_P))$ has
minimal distance to $P$. Every other point has higher distance.
See Figure~\ref{fig:constant-min-distance}.
\section{Defined on a closed interval of $\mdr$}
This means:
\[S_0(f, P) = \Set{x_P} \text{ with } P = (x_P, y_P)\]
\clearpage
\section{Defined on a closed interval $[a,b] \subseteq \mdr$}
Let $f:[a,b] \rightarrow \mdr$, $f(x) := c$ with $a,b,c \in \mdr$ and
$a \leq b$ be a constant function.
\begin{figure}[htp]
\centering
\begin{tikzpicture}
\begin{axis}[
legend pos=north west,
legend cell align=left,
axis x line=middle,
axis y line=middle,
grid = major,
width=0.8\linewidth,
height=8cm,
grid style={dashed, gray!30},
xmin=-5, % start the diagram at this x-coordinate
xmax= 5, % end the diagram at this x-coordinate
ymin= 0, % start the diagram at this y-coordinate
ymax= 3, % end the diagram at this y-coordinate
axis background/.style={fill=white},
xlabel=$x$,
ylabel=$y$,
tick align=outside,
minor tick num=-3,
enlargelimits=true,
tension=0.08]
\addplot[domain=-5:-2, thick,samples=50, red] {1};
\addplot[domain=-1:3, thick,samples=50, green] {1.5};
\addplot[domain=3:5, thick,samples=50, blue, densely dotted] {3};
\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
\addplot[blue, mark = *, nodes near coords=$P_{h,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(3, 3)};
\addplot[green, mark = x, nodes near coords=$P_{g,\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 1.5)};
\addplot[red, mark = *, nodes near coords=$P_{f,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(-2, 1)};
\draw[thick, dashed] (axis cs:2,1.5) -- (axis cs:2,2);
\draw[thick, dashed] (axis cs:2,2) -- (axis cs:-2,1);
\draw[thick, dashed] (axis cs:2,2) -- (axis cs:3,3);
\addlegendentry{$f(x)=1, D = [-5,-2]$}
\addlegendentry{$g(x)=1.5, D = [-1,3]$}
\addlegendentry{$h(x)=3, D = [3,5]$}
\end{axis}
\end{tikzpicture}
\caption{Three constant functions and their points with minimal distance}
\label{fig:constant-min-distance-closed-intervall}
\end{figure}
The point with minimum distance can be found by:
\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
S_0(f,P) &\text{if } S_0(f,P) \cap [a,b] \neq \emptyset \\
\Set{a} &\text{if } S_0(f,P) \ni x_P < a\\
\Set{b} &\text{if } S_0(f,P) \ni x_P > b
\end{cases}\]

1
documents/math-minimal-distance-to-cubic-function/cubic-functions.tex
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@ -214,5 +214,6 @@ initial guess.
\subsubsection{Quadratic minimization}
\todo[inline]{TODO}
\clearpage
\section{Defined on a closed interval of $\mdr$}

66
documents/math-minimal-distance-to-cubic-function/linear-functions.tex
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@ -8,6 +8,7 @@ $t \in \mdr$ be a linear function.
\begin{tikzpicture}
\begin{axis}[
legend pos=north east,
legend cell align=left,
axis x line=middle,
axis y line=middle,
grid = major,
@ -29,7 +30,7 @@ $t \in \mdr$ be a linear function.
\addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
\addlegendentry{$f(x)=\frac{1}{2}x$}
\addlegendentry{$g(x)=-2x+6$}
\addlegendentry{$f_\bot(x)=-2x+6$}
\end{axis}
\end{tikzpicture}
\caption{The shortest distance of $P$ to $f$ can be calculated by using the perpendicular}
@ -50,9 +51,66 @@ is calculated this way:
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 )
\Leftrightarrow x &= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )\label{eq:solution-of-the-linear-problem}
\end{align}
There is only one point with minimal distance. See Figure~\ref{fig:linear-min-distance}.
There is only one point with minimal distance. I'll call the result
from line~\ref{eq:solution-of-the-linear-problem} \enquote{solution of
the linear problem} and the function that gives this solution
$S_1(f,P)$.
\section{Defined on a closed interval of $\mdr$}
See Figure~\ref{fig:linear-min-distance}
to get intuition about the geometry used.
\clearpage
\section{Defined on a closed interval $[a,b] \subseteq \mdr$}
Let $f:[a,b] \rightarrow \mdr$, $f(x) := m\cdot x + t$ with $a,b,m,t \in \mdr$ and
$a \leq b$, $m \neq 0$ be a linear function.
\begin{figure}[htp]
\centering
\begin{tikzpicture}
\begin{axis}[
legend pos=north east,
legend cell align=left,
axis x line=middle,
axis y line=middle,
grid = major,
width=0.8\linewidth,
height=8cm,
grid style={dashed, gray!30},
xmin= 0, % start the diagram at this x-coordinate
xmax= 5, % end the diagram at this x-coordinate
ymin= 0, % start the diagram at this y-coordinate
ymax= 3, % end the diagram at this y-coordinate
axis background/.style={fill=white},
xlabel=$x$,
ylabel=$y$,
tick align=outside,
minor tick num=-3,
enlargelimits=true,
tension=0.08]
\addplot[domain= 2:3, thick,samples=50, red] {0.5*x};
\addplot[domain=-5:5, thick,samples=50, blue, dashed] {-2*x+6};
\addplot[domain=1:1.5, thick, samples=50, orange] {3*x-3};
\addplot[domain=4:5, thick, samples=50, green] {-x+5};
\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
\draw[thick, dashed] (axis cs:2,2) -- (axis cs:1.5,1.5);
\draw[thick, dashed] (axis cs:2,2) -- (axis cs:4,1);
\addlegendentry{$f(x)=\frac{1}{2}x, D = [2,3]$}
\addlegendentry{$f_\bot(x)=-2x+6, D=[-5,5]$}
\addlegendentry{$h(x)=3x-3, D=[1,1.5]$}
\addlegendentry{$h(x)=-x+5, D=[4,5]$}
\end{axis}
\end{tikzpicture}
\caption{Different situations when you have linear functions which
are defined on a closed intervall}
\label{fig:linear-min-distance-closed-intervall}
\end{figure}
The point with minimum distance can be found by:
\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
S_1(f, P) &\text{if } S_1(f, P) \cap [a,b] \neq \emptyset\\
\Set{a} &\text{if } S_1(f, P) \ni x < a\\
\Set{b} &\text{if } S_1(f, P) \ni x > b
\end{cases}\]

BIN
documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf
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8
documents/math-minimal-distance-to-cubic-function/problem-description.tex
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@ -19,3 +19,11 @@ But minimizing $d_{P,f}$ is the same as minimizing $d_{P,f}^2$:
Then: $f'(x_0) = 0$.
\end{theorem}
Let $S_n$ be the function that returns the set of solutions for a
polynomial of degree $n$ and a point:
\[S_n: \Set{\text{Polynomials of degree } n \text{ defined on } \mdr} \times \mdr^2 \rightarrow \mathcal{P}({\mdr})\]
\[S_n(f, P) := \underset{x\in\mdr}{\arg \min d_{P,f}(x)}\]
If possible, I will explicitly give this function.

31
documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex
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@ -56,9 +56,9 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak
This is an algebraic equation of degree 3.
There can be up to 3 solutions in such an equation. Those solutions
can be found with a closed formula.
\todo[inline]{Where are those closed formulas?}
can be found with a closed formula. But not every solution of the
equation given by Theorem~\ref{thm:required-extremum-property}
has to be a solution to the given problem.
\begin{example}
Let $a = 1, b = 0, c= 1, x_p= 0, y_p = 1$.
@ -204,4 +204,27 @@ t &:= \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\\
\end{cases}
\end{align*}
\section{Defined on a closed interval of $\mdr$}
I call this function $S_2: \Set{\text{Quadratic functions}} \times \mdr^2 \rightarrow \mathcal{P}({\mdr})$.
\clearpage
\section{Defined on a closed interval $[a,b] \subseteq \mdr$}
Now the problem isn't as simple as with constant and linear
functions.
If one of the minima in $S_2(P,f)$ is in $[a,b]$, this will be the
shortest distance as there are no shorter distances.
\todo[inline]{
The following IS WRONG! Can I include it to help the reader understand the
problem?}
If the function (defined on $\mdr$) has only one shortest distance
point $x$ for the given $P$, it's also easy: The point in $[a,b]$ that
is closest to $x$ will have the sortest distance.
\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
S_2(f, P) \cap [a,b] &\text{if } S_2(f, P) \cap [a,b] \neq \emptyset \\
\Set{a} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x < a\\
\Set{b} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x > b\\
todo &\text{if } |S_2(f, P)| = 2 \text{ and } S_2(f, P) \cap [a,b] = \emptyset
\end{cases}\]