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added argument why there is no closed form solution

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Martin Thoma 2013-11-30 18:34:24 +01:00
parent ba59e009cf
commit fb0c7e39cb
2 changed files with 29 additions and 6 deletions
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documents/math-minimal-distance-to-cubic-function/Makefile
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@ -5,4 +5,4 @@ make:
make clean
clean:
rm -rf $(TARGET) *.class *.log *.aux *.out
rm -rf $(TARGET) *.class *.log *.aux *.out *.thm

33
documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex
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@ -23,6 +23,8 @@
\theoremheaderfont{\kern-0.7cm\normalfont\bfseries}
\theorembodyfont{\normalfont} % nicht mehr kursiv
\def\mdr{\ensuremath{\mathbb{R}}}
\newframedtheorem{theorem}{Theorem}
\newframedtheorem{lemma}[theorem]{Lemma}
\newtheorem{plaindefinition}{Definition}
@ -418,14 +420,35 @@ take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
&= \underbrace{f'(x) \cdot \left (f(x) - y_p \right )}_{\text{Polynomial of degree 5}} + \underbrace{x - x_p}_{\text{:-(}}
&= \underbrace{f'(x) \cdot \left (f(x) - y_p \right )}_{\text{Polynomial of degree 5}} + x - x_p
\end{align}
\todo[inline]{Although general algebraic equations of degree 5 don't
have a solution formula, the special structure might give the
possibilty to find a closed form solution. But I don't know how.
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}
This is a first-order nonlinear ordinary differential equation - does this help?}
\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}}\\
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 With $a$, we can get any value of $\tilde{a} \in \mdr \setminus \Set{0}$.
\item With $b$, we can get any value of $\tilde{b} \in \mdr \setminus \Set{0}$.
\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 $\tilde{f} \in \mdr$.
\end{enumerate}
The first restriction only 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
finding the closest points on a cubic function to a given point.
\subsection{Number of points with minimal distance}
As there is an algebraic equation of degree 5, there cannot be more