added argument why there is no closed form solution

documents/math-minimal-distance-to-cubic-function/Makefile

@@ -5,4 +5,4 @@ make:
 	make clean
 
 clean:
-	rm -rf  $(TARGET) *.class *.log *.aux *.out
+	rm -rf  $(TARGET) *.class *.log *.aux *.out *.thm

documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex

@@ -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}
+
+\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}
 
-This is a first-order nonlinear ordinary differential equation - does this help?}
+\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