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Remove trailing spaces

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The commands

find . -type f -name '*.md' -exec sed --in-place 's/[[:space:]]\+$//' {} \+

and

find . -type f -name '*.tex' -exec sed --in-place 's/[[:space:]]\+$//' {} \+

were used to do so.
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Martin Thoma 2015-10-14 14:25:34 +02:00
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28
documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex
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@ -1,6 +1,6 @@
\chapter{Quadratic functions}
\section{Defined on $\mdr$}
Let $f:\mdr \rightarrow \mdr, f(x) = a \cdot x^2 + b \cdot x + c$ with $a \in \mdr \setminus \Set{0}$ and
Let $f:\mdr \rightarrow \mdr, f(x) = a \cdot x^2 + b \cdot x + c$ with $a \in \mdr \setminus \Set{0}$ and
$b, c \in \mdr$ be a quadratic function.
\begin{figure}[htp]
@ -25,8 +25,8 @@ $b, c \in \mdr$ be a quadratic function.
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, 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,dotted] { x*x + 2*x};
\addplot[domain=-3:3, thick,samples=50, black,dashed] {-x*x + 6};
@ -35,13 +35,13 @@ $b, c \in \mdr$ be a quadratic function.
\addlegendentry{$f_3(x)=x^2+x$}
\addlegendentry{$f_4(x)=x^2+2x$}
\addlegendentry{$f_5(x)=-x^2+6$}
\end{axis}
\end{axis}
\end{tikzpicture}
\caption{Quadratic functions}
\end{figure}
\subsection{Calculate points with minimal distance}
In this case, $d_{P,f}^2$ is polynomial of degree $n^2 = 4$.
In this case, $d_{P,f}^2$ is polynomial of degree $n^2 = 4$.
We use Theorem~\ref{thm:fermats-theorem}:\nobreak
\begin{align}
0 &\overset{!}{=} (d_{P,f}^2)'\\
@ -80,7 +80,7 @@ Example~\ref{ex:false-positive}.
\subsection{Number of points with minimal distance}
\begin{theorem}
A point $P$ has either one or two points on the graph of a
A point $P$ has either one or two points on the graph of a
quadratic function $f$ that are closest to $P$.
\end{theorem}
@ -93,8 +93,8 @@ minimal distance.
In the following, I will do some transformations with $f = f_0$ and
$P = P_0$. This will make it easier to calculate the minimal distance
points. Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does
not change the minimum distance. Furthermore, we can find the
points. Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does
not change the minimum distance. Furthermore, we can find the
points with minimum distance on the moved situation and calculate
the minimum points in the original situation.
@ -104,7 +104,7 @@ First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
Because:\footnote{The idea why you subtract $\frac{b}{2a}$ within
$f$ is that when you subtract something from $x$ before applying
$f$ it takes more time ($x$ needs to be bigger) to get to the same
situation. In consequence, if we want to move the whole graph by 1
situation. In consequence, if we want to move the whole graph by 1
to the left, we have to add $+1$.}
\begin{align}
f(x-\nicefrac{b}{2a}) &= a (x-\nicefrac{b}{2a})^2 + b (x-\nicefrac{b}{2a}) + c\\
@ -117,7 +117,7 @@ to the left, we have to add $+1$.}
Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \Big (\underbrace{x_P+\frac{b}{2a}}_{=: z},\;\; \underbrace{y_P+\frac{b^2}{4a}-c}_{=: w} \Big )\]
As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
$P = (0, w)$ could possilby have three minima.
Then compute:
@ -132,7 +132,7 @@ Then compute:
This means, the term
\[a^2 x^2 + (\nicefrac{1}{2a}- w)\]
has to get as close to $0$ as possilbe when we want to minimize
has to get as close to $0$ as possilbe when we want to minimize
$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{\frac{\frac{1}{2a} - w}{a}}$.
$\qed$
@ -176,7 +176,7 @@ I will make use of the following identities:
(a-b)^3 &= a^3-3 a^2 b+3 a b^2-b^3
\end{align*}
\textbf{Case 2.1:}
\textbf{Case 2.1:}
\input{quadratic-case-2.1}
\goodbreak
\textbf{Case 2.2:}
@ -211,12 +211,12 @@ 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
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.
is closest to $x$ will have the sortest distance.
\[\underset{x\in[a,b]}{\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 \\