next (failed?) try to calculate the quadratic

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

@@ -43,6 +43,7 @@ The situation can be seen in Figure~\ref{fig:constant-min-distance}.
     \label{fig:constant-min-distance}
 \end{figure}
 
+The point $(x, f(x))$ with minimal distance can be calculated directly:
 \begin{align}
     d_{P,f}(x) &= \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\\
                &= \sqrt{(x_P^2 - 2x_P x + x^2) + (y_P^2 - 2 y_P c + c^2)} \\

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

@@ -62,7 +62,7 @@ $b, c, d \in \mdr$ be a function.
 
     Then you could solve the following problem for $x$:
     \begin{align}
-        0  &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'
+        0  &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'\\
            &=-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\\

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

@@ -27,7 +27,7 @@ $t \in \mdr$ be a linear function.
             enlargelimits=true,
             tension=0.08]
           \addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
-          \addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
+          \addplot[domain=-5:5, thick,samples=50, blue, dashed] {-2*x+6};
           \addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
           \addplot[blue, nodes near coords=$f_\bot$,every node near coord/.style={anchor=225}] coordinates {(1.5, 3)};
           \addplot[red, nodes near coords=$f$,every node near coord/.style={anchor=225}] coordinates {(0.9, 0.5)};

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

@@ -40,13 +40,13 @@
 \newtheorem{proof}{Proof:}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
-\title{Minimal distance to a cubic function}
+\title{Minimal distance to polynomial functions of degree 3 or less}
 \author{Martin Thoma}
 
 \hypersetup{ 
   pdfauthor   = {Martin Thoma}, 
-  pdfkeywords = {}, 
-  pdftitle    = {Minimal Distance} 
+  pdfkeywords = {minimal distance, polynomial, function, degree 3, cubic, spline}, 
+  pdftitle    = {Minimal distance to polynomial functions of degree 3 or less} 
 }
 
 \def\mdr{\ensuremath{\mathbb{R}}}

documents/math-minimal-distance-to-cubic-function/quadratic-case-2.2.tex

@@ -0,0 +1,50 @@
+\todo[inline]{calculate...}
+
+\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
+     -\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]
+
+\begin{align}
+    x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}}
+           \underbrace{- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}}\\
+         &\hphantom{{}=}+ \underbrace{3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}}
+           \underbrace{- \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}}}
+\end{align}
+
+Now simplify the summands:
+\begin{align}
+    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} &=
+    \left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3\\
+    &= \frac{\alpha^3(1-3i\sqrt{3} - 3 \cdot 3 + \sqrt{27} i)}{12 t^3}\\
+    &= \frac{-8\alpha^3}{12 t^3}\\
+    &= \frac{-2 \alpha^3}{3 t^3}\\
+    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} &=- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)\\
+    &= \frac{-3\alpha^2(1+2\sqrt{3}i-3)(1-i\sqrt{3})t}{t^2 \sqrt[3]{2^4 \cdot 3^2} \cdot 2 \sqrt[3]{2 \cdot 3^2}}\\
+    &= \frac{-3\alpha^2((1+2\sqrt{3}i - 3)+(- i\sqrt{3}+2\cdot 3 + i\sqrt{3}))}{12 t \sqrt[3]{12}}\\
+    &= \frac{-\alpha^2(4+2\sqrt{3}i)}{4t\sqrt[3]{12}}\\
+    &= \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}\\
+    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} &= 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2\\
+    &= \frac{3(1+i\sqrt{3})\alpha (1-2i\sqrt{3} - 3)t}{4 \sqrt[3]{12 \cdot 18^2}}\\
+    &= \frac{3 \alpha t((1-2i\sqrt{3}-3)+(i\sqrt{3} + 2\cdot 3 - 3i\sqrt{3}))}{4 \sqrt[3]{2^2 \cdot 3 \cdot (2 \cdot 3^2)^2}}\\
+    &= \frac{3 \alpha t(4-4\sqrt{3}i)}{4 \cdot 2 \cdot 3 \sqrt[3]{2 \cdot 3^2}}\\
+    &= \frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\\
+    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}} &= - \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3\\
+    &= -\frac{(1-3i\sqrt{3} - 3 \cdot 3 + i \sqrt{27}) t^3}{8 \cdot 18}\\
+    &=- \frac{t^3 (-8)}{8 \cdot 18}\\
+    &= \frac{t^3}{18}
+\end{align}
+
+Now get back to the original equation:
+\begin{align}
+    0 &\stackrel{!}{=} x^3 + \alpha x + \beta \\
+       &= \left (\frac{-2 \alpha^3}{3 t^3}
+        + \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}
+        + \color{blue}\frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\color{black}
+        + \frac{t^3}{18} \right )\\
+    &\hphantom{{}=} + \color{blue}\alpha\color{black} \left (\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
+     \color{blue}-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}} \color{black} \right ) + \beta\\
+    &= \frac{-2 \alpha^3}{3 t^3}
+    + \frac{\alpha^2(2(1+i\sqrt{3})-(2+\sqrt{3}i))}{2t\sqrt[3]{12}}
+    + \frac{t^3}{18}
+    + \beta\\
+    &= \frac{-24 \alpha^3 + (3\sqrt[3]{18}t^2)(\alpha^2\sqrt{3}i) + 2t^3+36 t^3 \beta}{36t^3}
+\end{align}

documents/math-minimal-distance-to-cubic-function/quadratic-case-2.3.tex

@@ -0,0 +1,90 @@
+\todo[inline]{calculate...} 
+
+\[x = \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t}
+     -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
+
+\begin{align}
+    x^3 &= \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} - \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3\\
+    &= \left (\frac{(2\sqrt[3]{18})(1-i \sqrt{3}) \alpha - (\sqrt[3]{12} \cdot t)(1+i\sqrt{3}) t}{\sqrt[3]{12} \cdot t \cdot 2 \cdot \sqrt[3]{18}} \right)^3\\
+    &= \left (\frac{2\sqrt[3]{18}a \alpha (1-i \sqrt{3}) - \sqrt[3]{12} \cdot t^2(1+i\sqrt{3})}{2t \sqrt[3]{2^3 \cdot 3^3}} \right )^3\\
+    &= 12 \cdot \bigg (\frac{\overbrace{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}^{\text{numerator}}}{12t} \bigg )^3
+\end{align}
+
+Now calculate the numerator$^3$. Remember, that $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$
+and $(1 \pm i \sqrt{3})^3 = -8$.
+\begin{align}
+    \left (\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3}) \right )^3 &= 
+    12 \alpha^3 (1-i\sqrt{3})^3 \\
+    &\hphantom{{}=}- 3 \sqrt[3]{12^2} \alpha^2(1-i\sqrt{3})^2 (t^2(1+i \sqrt{3}))\\
+    &\hphantom{{}=}+ 3 \sqrt[3]{12\hphantom{^2}} \alpha\hphantom{^2} (1-i\sqrt{3}) t^4 (1+i\sqrt{3})^2 - t^6 (1+i\sqrt{3})^3\\
+    &= 12 \alpha^3 \cdot (-8) \\
+    &\hphantom{{}=}- 3 \cdot 2 \sqrt[3]{18} \alpha^2(-2(1+i\sqrt{3}))(t^2(1+i \sqrt{3}))\\
+    &\hphantom{{}=}+ 3 \sqrt[3]{12} \alpha (1-i\sqrt{3}) t^4 (-2(1-i\sqrt{3})) - t^6 (-8)\\
+    &= -96 \alpha^3 + 12 \sqrt[3]{18} \alpha^2 t^2 (1+i \sqrt{3})^2\\
+    &\hphantom{{}=}- 6 \sqrt[3]{12} \alpha (1-i\sqrt{3})^2 t^4 +8 t^6\\
+    &= -96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3})\\
+    &\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6
+\end{align}
+\goodbreak
+Now back to the original equation:
+\begin{align}
+0 &\stackrel{!}{=} x^3 + \alpha x + \beta\\
+  &= \frac{-96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3}) + 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6}{12^2 t^3}\\
+  &\hphantom{{}=}+\alpha \left (\sqrt[3]{12} \cdot \frac{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}{12t} \right ) + \beta
+\end{align}
+
+\todo[inline]{the calculation above seems to be wrong / too long. Next try}
+
+When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
+you get:\footnote{Remember, that $(1+i\sqrt{3})^2 = -2 (1-i \sqrt{3})$ and $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$
+and $(1 \pm i \sqrt{3})^3 = -8$}
+\begin{align}
+ 0 &\stackrel{!}{=} \left( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t}
+     -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3 
+     + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
+     + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    &= \frac{(1-i \sqrt{3})^3 \alpha^3}{12 \cdot t^3}
+       - 3 \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \right )^2 \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}
+       + 3 \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \left (\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^2\\
+    &\hphantom{{}=}
+       + \frac{(1+i\sqrt{3})^3 t^3}{2^3 \cdot 18}
+       + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
+       + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    &= \frac{-8 \alpha^3}{12t^3}
+       - 3 \frac{-2(1+i \sqrt{3}) \alpha^2}{\sqrt[3]{12^2} t^2} \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}
+       + 3 \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \frac{-2(1-i\sqrt{3}) t^2}{4\sqrt[3]{18^2}}\\
+    &\hphantom{{}=}
+       + \frac{-8 t^3}{2^3 \cdot 18}
+       + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
+       + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    &= \frac{-2 \alpha^3}{3t^3}
+       + \frac{6 \alpha^2 t (-2)(1-i \sqrt{3})}{(\sqrt[3]{12^2} t^2)(2\sqrt[3]{18})}
+       + \frac{12 \alpha t^2 (1+i \sqrt{3})}{(\sqrt[3]{12} \cdot t)(4\sqrt[3]{18^2)}}\\
+    &\hphantom{{}=}
+       + \frac{- t^3}{18}
+       + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
+       + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    &= \frac{-2 \alpha^3}{3t^3}
+       + \frac{-6 \alpha^2 (1-i \sqrt{3})}{6 \sqrt[3]{12} t}
+       + \frac{3 \alpha t (1+i \sqrt{3})}{6\sqrt[3]{18}}
+       + \frac{- t^3}{18}\\
+    &\hphantom{{}=}+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
+       + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    &= \frac{-2 \alpha^3}{3t^3}
+       + \frac{-\alpha^2 (1-i \sqrt{3})}{\sqrt[3]{12} t}
+       + \frac{\alpha t (1+i \sqrt{3})}{2\sqrt[3]{18}}
+       + \frac{- t^3}{18}\\
+    &\hphantom{{}=}+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
+       + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    &= \frac{12 \cdot (-2 \alpha^3) +(6 \sqrt[3]{18}t^2)(-\alpha^2 (1-i \sqrt{3}))+ (3 \sqrt[3]{12})(\alpha t (1+i \sqrt{3})) + (2t^3)(- t^3)}{36t^3}\\
+    &\hphantom{{}=}+ \frac{(6 \sqrt[3]{18})((1-i \sqrt{3}) \alpha) - (3 \sqrt[3]{12})((1+i\sqrt{3}) t) + 36t^3 \beta}{36t^3}\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\end{align}
+
+\goodbreak
+Now calculate only the numerator:
+\begin{align}
+    0 &\stackrel{!}{=} -12 \alpha^3 - 6 \sqrt[3]{18} t^2 \alpha^2 (1 - i \sqrt{3})
+        + 3 \sqrt[3]{12} \alpha t (1+i\sqrt{3}) - 2t^6\\
+      &\hphantom{{}=} + 6\sqrt[3]{18} \alpha (1- i \sqrt{3})
+        - 3 \sqrt[3]{12} t (1+i \sqrt{3}) + 36 t^3 \beta
+\end{align}

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

@@ -42,7 +42,7 @@ $b, c \in \mdr$ be a quadratic function.
 
 \subsection{Calculate points with minimal distance}
 In this case, $d_{P,f}^2$ is polynomial of degree 4. 
-We use Theorem~\ref{thm:required-extremum-property}:\nobreak
+We use Theorem~\ref{thm:fermats-theorem}:\nobreak
 \begin{align}
     0     &\overset{!}{=} (d_{P,f}^2)'\\
           &= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\
@@ -57,7 +57,7 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak
 This is an algebraic equation of degree 3.
 There can be up to 3 solutions in such an equation. Those solutions
 can be found with a closed formula. But not every solution of the
-equation given by Theorem~\ref{thm:required-extremum-property}
+equation given by Theorem~\ref{thm:fermats-theorem}
 has to be a solution to the given problem.
 
 \begin{example}
@@ -134,7 +134,7 @@ We start with the graph that was moved so that $f_2 = ax^2$.
 
 \textbf{Case 1:} $P$ is on the symmetry axis, hence $x_P = - \frac{b}{2a}$.
 
-In this case, we have already found the solution. If $y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
+In this case, we have already found the solution. If $w = y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
 then there are two solutions:
 \[x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}\]
 Otherwise, there is only one solution $x_1 = 0$.
@@ -206,62 +206,10 @@ $t$:
 \end{align}
 
 \textbf{Case 2.2:}
-\todo[inline]{calculate...} 
-\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
-     -\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]
+\input{quadratic-case-2.2}
 
-\begin{align}
-    x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}}
-           \underbrace{- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}}\\
-         &\hphantom{{}=}+ \underbrace{3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}}
-           \underbrace{- \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}}}
-\end{align}
-
-Now simplify the summands:
-\begin{align}
-    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} &=
-    \left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3\\
-    &= \frac{\alpha^3(1-3i\sqrt{3} - 3 \cdot 3 + \sqrt{27} i)}{12 t^3}\\
-    &= \frac{-8\alpha^3}{12 t^3}\\
-    &= \frac{-2 \alpha^3}{3 t^3}\\
-    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} &=- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)\\
-    &= \frac{-3\alpha^2(1+2\sqrt{3}i-3)(1-i\sqrt{3})t}{t^2 \sqrt[3]{2^4 \cdot 3^2} \cdot 2 \sqrt[3]{2 \cdot 3^2}}\\
-    &= \frac{-3\alpha^2((1+2\sqrt{3}i - 3)+(- i\sqrt{3}+2\cdot 3 + i\sqrt{3}))}{12 t \sqrt[3]{12}}\\
-    &= \frac{-\alpha^2(4+2\sqrt{3}i)}{4t\sqrt[3]{12}}\\
-    &= \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}\\
-    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} &= 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2\\
-    &= \frac{3(1+i\sqrt{3})\alpha (1-2i\sqrt{3} - 3)t}{4 \sqrt[3]{12 \cdot 18^2}}\\
-    &= \frac{3 \alpha t((1-2i\sqrt{3}-3)+(i\sqrt{3} + 2\cdot 3 - 3i\sqrt{3}))}{4 \sqrt[3]{2^2 \cdot 3 \cdot (2 \cdot 3^2)^2}}\\
-    &= \frac{3 \alpha t(4-4\sqrt{3}i)}{4 \cdot 2 \cdot 3 \sqrt[3]{2 \cdot 3^2}}\\
-    &= \frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\\
-    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}} &= - \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3\\
-    &= -\frac{(1-3i\sqrt{3} - 3 \cdot 3 + i \sqrt{27}) t^3}{8 \cdot 18}\\
-    &=- \frac{t^3 (-8)}{8 \cdot 18}\\
-    &= \frac{t^3}{18}
-\end{align}
-
-Now get back to the original equation:
-\begin{align}
-    0 &\stackrel{!}{=} x^3 + \alpha x + \beta \\
-       &= \left (\frac{-2 \alpha^3}{3 t^3}
-        + \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}
-        + \color{blue}\frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\color{black}
-        + \frac{t^3}{18} \right )\\
-    &\hphantom{{}=} + \color{blue}\alpha\color{black} \left (\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
-     \color{blue}-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}} \color{black} \right ) + \beta\\
-    &= \frac{-2 \alpha^3}{3 t^3}
-    + \frac{\alpha^2(2(1+i\sqrt{3})-(2+\sqrt{3}i))}{2t\sqrt[3]{12}}
-    + \frac{t^3}{18}
-    + \beta\\
-    &= \frac{-24 \alpha^3 + (3\sqrt[3]{18}t^2)(\alpha^2\sqrt{3}i) + 2t^3+36 t^3 \beta}{36t^3}
-\end{align}
-
-
-\textbf{Case 2.3:} 
-\todo[inline]{calculate...} 
-
-\[x = \frac{(1-i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
-     -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
+\textbf{Case 2.3:}
+\input{quadratic-case-2.3}
 
 \goodbreak
 So the solution is given by
@@ -298,7 +246,7 @@ 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. 
 
-\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
+\[\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 \\
               \Set{a} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x < a\\
               \Set{b} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x > b\\