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many improvements (theorem-proof-structure for constant function; corrected errors)

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Martin Thoma 2013-12-21 19:10:35 +01:00
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documents/math-minimal-distance-to-cubic-function/linear-functions.tex
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@ -1,7 +1,13 @@
\chapter{Linear function}
\section{Defined on $\mdr$}
Let $f(x) = m \cdot x + t$ with $m \in \mdr \setminus \Set{0}$ and
$t \in \mdr$ be a linear function.
\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 the points $(x, f(x))$ with minimal distance are 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
@ -39,30 +45,32 @@ $t \in \mdr$ be a linear function.
\label{fig:linear-min-distance}
\end{figure}
Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The
slope of $f_\bot$ is $- \frac{1}{m}$ and $t_\bot$ can be calculated:\nobreak
\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
\end{align}
\begin{proof}
Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The
slope of $f_\bot$ is $- \frac{1}{m}$ and $t_\bot$ can be calculated:\nobreak
\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
\end{align}
The point $(x, f(x))$ where the perpendicular $f_\bot$ crosses $f$
is calculated this way:
\begin{align}
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 )\label{eq:solution-of-the-linear-problem}
\end{align}
The point $(x, f(x))$ where the perpendicular $f_\bot$ crosses $f$
is calculated this way:
\begin{align}
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 )\label{eq:solution-of-the-linear-problem}
\end{align}
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
$S_1(f,P)$.
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
$S_1(f,P)$.
See Figure~\ref{fig:linear-min-distance}
to get intuition about the geometry used.
See Figure~\ref{fig:linear-min-distance}
to get intuition about the geometry used. $\qed$
\end{proof}
\clearpage
\section{Defined on a closed interval $[a,b] \subseteq \mdr$}