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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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34
documents/math-minimal-distance-to-cubic-function/cubic-functions.tex
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@ -1,7 +1,7 @@
\chapter{Cubic functions}
\section{Defined on $\mdr$}
Let $f:\mdr \rightarrow \mdr, f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$
be a cubic function with $a \in \mdr \setminus \Set{0}$ and
Let $f:\mdr \rightarrow \mdr, f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$
be a cubic function with $a \in \mdr \setminus \Set{0}$ and
$b, c, d \in \mdr$.
\begin{figure}[htp]
@ -26,15 +26,15 @@ $b, c, d \in \mdr$.
minor tick num=-3,
enlargelimits=true,
tension=0.08]
\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x*x};
\addplot[domain=-3:3, thick,samples=50, blue] {x*x*x+2*x*x};
\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x+x};
\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x+x};
\addlegendentry{$f_1(x)=x^3$}
\addlegendentry{$f_2(x)=x^3 + x^2$}
\addlegendentry{$f_2(x)=x^3 + 2 \cdot x^2$}
\addlegendentry{$f_1(x)=x^3 + x$}
\end{axis}
\end{axis}
\end{tikzpicture}
\caption{Cubic functions}
\end{figure}
@ -51,7 +51,7 @@ $b, c, d \in \mdr$.
\subsection{Calculate points with minimal distance}
\begin{theorem}\label{thm:no-finite-solution}
There cannot be a finite, closed form solution to the problem of finding
There cannot be a finite, closed form solution to the problem of finding
a closest point $(x, f(x))$ to a given point $P$ when $f$ is
a polynomial function of degree $3$ or higher.
\end{theorem}
@ -89,11 +89,11 @@ $b, c, d \in \mdr$.
\item With $x_p := cd - c y_P+\tilde{f}$, we can get any value of $\tilde{f} \in \mdr$.
\end{enumerate}
The first restriction guaratees that we have a polynomial of
The first restriction guaratees that we have a polynomial of
degree 5. The second one is necessary, to get a high range of
$\tilde{e}$.
This means that there is no finite solution formula for the problem of
This means that there is no finite solution formula for the problem of
finding the closest points on a cubic function to a given point,
because if there was one, you could use this formula for finding
roots of polynomials of degree 5. $\qed$
@ -101,7 +101,7 @@ $b, c, d \in \mdr$.
\subsection{Another approach}
Just like we moved the function $f$ and the point to get in a
Just like we moved the function $f$ and the point to get in a
nicer situation, we can apply this approach for cubic functions.
\begin{figure}[htp]
@ -126,9 +126,9 @@ nicer situation, we can apply this approach for cubic functions.
minor tick num=-3,
enlargelimits=true,
tension=0.08]
\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x};
\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x-x};
\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x-x};
\addplot[domain=-3:3, thick,samples=50, blue, dotted] {x*x*x+2*x};
\addplot[domain=-3:3, thick,samples=50, lime, dashed] {x*x*x+3*x};
\addlegendentry{$f_1(x)=x^3$}
@ -136,7 +136,7 @@ nicer situation, we can apply this approach for cubic functions.
\addlegendentry{$f_1(x)=x^3 - x$}
\addlegendentry{$f_2(x)=x^3 + 2 \cdot x$}
\addlegendentry{$f_2(x)=x^3 + 3 \cdot x$}
\end{axis}
\end{axis}
\end{tikzpicture}
\caption{Cubic functions with $b = d = 0$}
\end{figure}
@ -176,8 +176,8 @@ As this leads to a polynomial of degree 5 of which we have to find
roots, there cannot be more than 5 solutions.
\todo[inline]{Can there be 3, 4 or even 5 solutions? Examples!
After looking at function graphs of cubic functions, I'm pretty
sure that there cannot be 4 or 5 solutions, no matter how you
After looking at function graphs of cubic functions, I'm pretty
sure that there cannot be 4 or 5 solutions, no matter how you
chose the cubic function $f$ and $P$.
I'm also pretty sure that there is no polynomial (no matter what degree)
@ -193,7 +193,7 @@ You could interpolate the cubic function by a quadratic spline.
\subsubsection{Newtons method}
One way to find roots of functions is Newtons method. It gives an
iterative computation procedure that can converge quadratically
iterative computation procedure that can converge quadratically
if some conditions are met:
\begin{theorem}[local quadratic convergence of Newton's method\footnotemark]
@ -201,7 +201,7 @@ if some conditions are met:
Let $x^* \in D$ with $f(x^*) = 0$ and the Jaccobi-Matrix $f'(x^*)$
should not be invertable when evaluated at the root.
Then there is a sphere
Then there is a sphere
\[K := K_\rho(x^*) = \Set{x \in \mdr^n | \|x- x^*\|_\infty \leq \rho} \subseteq D\]
such that $x^*$ is the only root of $f$ in $K$. Furthermore,
the elements of the sequence
@ -233,7 +233,7 @@ Muller's method was first presented by David E. Muller in 1956.
\subsubsection{Bisection method}
The idea of the bisection method is the following:
Suppose you know a finite intervall $[a,b]$ in which you have
Suppose you know a finite intervall $[a,b]$ in which you have
exactly one root $r \in (a,b)$ with $f(r) = 0$.
Then you can half that interval: