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documents/math-minimal-distance-to-cubic-function/cubic-functions.tex
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@ -1,8 +1,8 @@
\chapter{Cubic functions}
\section{Defined on $\mdr$}
Let $f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$ be a cubic function
with $a \in \mdr \setminus \Set{0}$ and
$b, c, d \in \mdr$ be a function.
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]
\centering
@ -50,8 +50,8 @@ $b, c, d \in \mdr$ be a function.
%\todo[inline]{Write this}
\subsection{Calculate points with minimal distance}
\begin{theorem}
There cannot be an algebraic solution to the problem of finding
\begin{theorem}\label{thm:no-finite-solution}
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}
@ -71,29 +71,29 @@ $b, c, d \in \mdr$ be a function.
General algebraic equations of degree 5 don't have a solution formula.\footnote{TODO: Quelle}
Although here seems to be more structure, the resulting algebraic
equation can be almost any polynomial of degree 5:\footnote{Thanks to Peter Košinár on \href{http://math.stackexchange.com/a/584814/6876}{math.stackexchange.com} for this one}
equation can be almost any polynomial of degree 5:\footnote{Thanks to Peter Košinár on \href{http://math.stackexchange.com/a/584814/6876}{math.stackexchange.com} for the idea.}
\begin{align}
0 &\stackrel{!}{=} f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
&= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{\tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{=: \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{\tilde{d}} x^2 \\
& &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{=: \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{=: \tilde{f}}\\
&= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{= \tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{= \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{= \tilde{d}} x^2 \\
& &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{= \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{= \tilde{f}}\\
0 &\stackrel{!}{=} \tilde{a}x^5 + \tilde{b}x^4 + \tilde{c}x^3 + \tilde{d}x^2 + \tilde{e}x + \tilde{f}
\end{align}
\begin{enumerate}
\item For any coefficient $\tilde{a} \in \mdr_{> 0}$ of $x^5$ we can choose $a$ such that we get $\tilde{a}$.
\item For any coefficient $\tilde{b} \in \mdr \setminus \Set{0}$ of $x^4$ we can choose $b$ such that we get $\tilde{b}$.
\item With $c$, we can get any value of $\tilde{c} \in \mdr$.
\item With $d$, we can get any value of $\tilde{d} \in \mdr$.
\item With $y_p$, we can get any value of $\tilde{e} \in \mdr$.
\item With $x_p$, we can get any value of $\tilde{f} \in \mdr$.
\item For any coefficient $\tilde{a} \in \mdr_{> 0}$ of $x^5$ we can choose $a := \frac{1}{3} \sqrt{\tilde{a}}$ such that we get $\tilde{a}$.
\item For any coefficient $\tilde{b} \in \mdr \setminus \Set{0}$ of $x^4$ we can choose $b := \frac{1}{5a} \cdot \tilde{b}$ such that we get $\tilde{b}$.
\item With $c := -2b^2 + \frac{1}{4a} \tilde{c}$, we can get any value of $\tilde{c} \in \mdr$.
\item With $d := -bc + a y_p + \frac{1}{a} \tilde{d}$, we can get any value of $\tilde{d} \in \mdr$.
\item With $y_p := \frac{1}{2b}(2bd + c^2)\cdot \tilde{e}$, we can get any value of $\tilde{e} \in \mdr$.
\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
degree 5. The second one is necessary, to get a high range of
$\tilde{e}$.
This means, that there is no 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. $\qed$
@ -101,9 +101,6 @@ $b, c, d \in \mdr$ be a function.
\subsection{Another approach}
\todo[inline]{Currently, this is only an idea. It might be usefull
to move the cubic function $f$ such that $f$ is point symmetric
to the origin. But I'm not sure how to make use of this symmetry.}
Just like we moved the function $f$ and the point to get in a
nicer situation, we can apply this approach for cubic functions.
@ -144,7 +141,7 @@ nicer situation, we can apply this approach for cubic functions.
\caption{Cubic functions with $b = d = 0$}
\end{figure}
First, we move $f_0$ by $\frac{b}{3a}$ to the right, so
First, we move $f_0$ by $\frac{b}{3a}$ in $x$ direction, so
\[f_1(x) = ax^3 + \frac{b^2 (c-1)}{3a} x + \frac{2b^3}{27 a^2} - \frac{bc}{3a} + d \;\;\;\text{ and }\;\;\;P_1 = (x_P + \frac{b}{3a}, y_P)\]
@ -161,6 +158,19 @@ because
&= ax^3 + \frac{b^2}{3a}\left (1-2+c \right )x + \frac{b^3}{9a^2} \left (1-\frac{1}{3} \right )- \frac{bc}{3a} + d
\end{align}
The we move it in $y$ direction by $- (\frac{2b^3}{27 a^2} - \frac{bc}{3a} + d)$:
\[f_2(x) = ax^3 + \frac{b^2 (c-1)}{3a} x \;\;\;\text{ and }\;\;\;P_2 = (x_P + \frac{b}{3a}, y_P - (\frac{2b^3}{27 a^2} - \frac{bc}{3a} + d))\]
Multiply everything by $\sgn(a)$:
\[f_3(x) = \underbrace{|a|}_{=: \alpha}x^3 + \underbrace{\frac{b^2 (c-1)}{3|a|}}_{=: \beta} x \;\;\;\text{ and }\;\;\;P_2 = (x_P + \frac{b}{3a}, \sgn(a) (y_P - \frac{2b^3}{27 a^2} + \frac{bc}{3a} - d))\]
Now the problem seems to be much simpler. The function $\alpha x^3 + \beta x$
with $\alpha > 0$ is centrally symmetric to $(0, 0)$.
\todo[inline]{Und weiter?}
\subsection{Number of points with minimal distance}
As this leads to a polynomial of degree 5 of which we have to find
roots, there cannot be more than 5 solutions.
@ -211,9 +221,6 @@ As soon as the values don't change much, you are close to a root.
The problem of this approach is choosing a starting value that is
close enough to the root. So we have to have a \enquote{good}
initial guess.
\subsubsection{Quadratic minimization}
\todo[inline]{TODO}
\clearpage
\subsubsection{Muller's method}