mirror of
https://github.com/MartinThoma/LaTeX-examples.git
synced 2025-04-25 14:28:05 +02:00
154 lines
6.5 KiB
TeX
154 lines
6.5 KiB
TeX
\chapter{Linear function}
|
|
\section{Defined on $\mdr$}
|
|
\begin{theorem}[Solution formula for linear functions on $\mdr$]
|
|
Let $f: \mdr \rightarrow \mdr $ be a linear function
|
|
$f(x) := m \cdot x + t$ with $m \in \mdr \setminus \Set{0}$ and
|
|
$t \in \mdr$ be a linear function.
|
|
|
|
Then there is only one point $(x, f(x))$ on the graph of $f$ with
|
|
minimal distance to $P = (x_P, y_P)$. This point is given by
|
|
\[x = \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )\]
|
|
\end{theorem}
|
|
|
|
\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=-5:5, thick,samples=50, red] {0.5*x};
|
|
\addplot[domain=-5:5, thick,samples=50, blue, dashed] {-2*x+6};
|
|
\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
|
|
\newcommand{\R}{0.9}
|
|
\addplot [domain=0:2*pi,samples=50, dotted]({\R*cos(deg(x))+2},{\R*sin(deg(x))+2});
|
|
\addplot[blue, nodes near coords=$f_\bot$,every node near coord/.style={anchor=225}] coordinates {(1.5, 3)};
|
|
\addplot[red, nodes near coords=$f$,every node near coord/.style={anchor=225}] coordinates {(0.9, 0.5)};
|
|
\addlegendentry{$f(x)=\frac{1}{2}x$}
|
|
\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}
|
|
\label{fig:linear-min-distance}
|
|
\end{figure}
|
|
|
|
\begin{proof}
|
|
With Theorem~\ref{thm:fermats-theorem} you get:
|
|
\begin{align}
|
|
0 &\stackrel{!}{=} (d_{P,f}(x)^2)'\\
|
|
&= 2(x-x_P) + 2 (f(x) - y_P)f'(x)\\
|
|
\Leftrightarrow 0 &\stackrel{!}{=} x - x_P + (f(x) - y_P) f'(x)\\
|
|
&= x- x_P + (mx+t - y_P)\cdot m\\
|
|
&= x (m+1) + m(t-y_P) - x_P\\
|
|
\Leftrightarrow x &\stackrel{!}{=} \frac{x_p - m(t-y_p)}{m^2+1}\\
|
|
&= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )\label{eq:solution-linear-r}
|
|
\end{align}
|
|
It is obvious that a minium has to exist, the $x$ from Equation~\ref{eq:solution-linear-r}
|
|
has to be this minimum. $\qed$
|
|
\end{proof}
|
|
\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:
|
|
\[\underset{x\in[a,b]}{\arg \min d_{P,f}(x)} = \begin{cases}
|
|
S_1(f, P) &\text{if } S_1(f, P) \cap [a,b] \neq \emptyset\\
|
|
\Set{a} &\text{if } S_1(f, P) \ni x < a\\
|
|
\Set{b} &\text{if } S_1(f, P) \ni x > b
|
|
\end{cases}\]
|
|
|
|
If $S_1(f, P) \cap [a,b] \neq \emptyset$, then $\underset{x\in[a,b]}{\arg \min d_{P,f}(x)} = S_1(f,P) \cap [a,b]$,
|
|
because $S_1(f,P)$ gives all global minima of $f$. Those are also
|
|
minima for the intervall $[a,b]$. There are not more minima, because
|
|
$S_1$ gives all minima of $P$ to $f$.
|
|
|
|
If $S_1(f, P) \cap [a,b] = \emptyset$, then it is not that simple.
|
|
But we can calculate the distance function:
|
|
|
|
\begin{align}
|
|
d_{P,f}(x) &= \sqrt{(x-x_P)^2 + (f(x) - y_P)^2}\\
|
|
&= \sqrt{(x^2 - 2x x_P + x_P^2) + (mx + (t-y_P))^2}\\
|
|
&= \sqrt{(x^2 - 2x x_P + x_P^2) + m^2 x^2 + 2mx(t-y_P) + (t-y_P)^2}\\
|
|
&= \sqrt{x^2(1+m^2) + x(-2 x_P + 2m(t-y_P)) + (x_P^2 + (t-y_P)^2)}
|
|
\end{align}
|
|
|
|
This function (defined on $\mdr$) is symmetry to the axis
|
|
\begin{align}
|
|
x_S &= - \frac{-2 x_P + 2m(t-y_P)}{2(1+m^2)}\\
|
|
&= \frac{x_P - m(t-y_P)}{1+m^2}\\
|
|
&= \frac{m}{m^2+1} (y_P + \frac{1}{m} x_P - t)
|
|
\end{align}
|
|
|
|
$f$ is on $(-\infty, x_S]$ strictly monotonically decreasing and
|
|
on $[x_S, + \infty)$ strictly monotonically increasing.
|
|
|
|
Thus we can conclude:
|
|
\[\forall x,y \in \mdr: x \leq y < x_S \Rightarrow d_{P,f}(x_S) < d_{P,f}(y) \leq d_{P,f}(x)\]
|
|
\[\forall x,y \in \mdr: x_S < y \leq x \Rightarrow d_{P,f}(x_S) < d_{P,f}(y) \leq d_{P,f}(x)\]
|
|
|
|
When $S_1(f, P) \cap [a,b] = \emptyset$, then you can have two cases:
|
|
\begin{itemize}
|
|
\item $a \leq b < x_S$: $b$ has the shortest distance in $[a,b]$
|
|
on the graph of $f$ to $P$.
|
|
\item $x_S < a \leq b$: $a$ has the shortest distance in $[a,b]$
|
|
on the graph of $f$ to $P$.
|
|
\end{itemize}
|