got case 2.2 correct

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

@@ -40,8 +40,23 @@ Now get back to the original equation:
     &\hphantom{{}=} + \alpha \left (\color{red}\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \color{black}
      \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}
+    &= \frac{-12 \alpha^3 + t^6+18 t^3 \beta}{18t^3}
+\end{align}
+
+Now continue with only the numerator
+\begin{align}
+    0 &\stackrel{!}{=}
+    - 12 \alpha^3 
+    + (\sqrt{3(4 \alpha^3 + 27 \beta^2)}-9\beta)^2
+    + 18 (\sqrt{3(4 \alpha^3 + 27 \beta^2)} - 9 \beta) \beta\\
+    &= 
+    \color{red}- 12 \alpha^3 \color{black}+
+    \left (
+        3(\color{red}4 \alpha^3\color{black} + \color{blue}27 \beta^2 \color{black})
+        \color{orange}- 2 \cdot \sqrt{3(4 \alpha^3 + 27 \beta^2)} \cdot 9\beta\color{black}
+        + \color{blue}81 \beta^2\color{black}
+    \right )\\
+    &\hphantom{{}=}+ 18 \beta (\color{orange}\sqrt{3(4 \alpha^3 + 27 \beta^2)}\color{black} \color{blue}- 9 \beta\color{black}) 
 \end{align}

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

@@ -2,90 +2,5 @@ One solution is
 \[x = \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t}
      -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
 
-We will verify it in multiple steps. First, get $x^3$:
-\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} t \cdot 2 \sqrt[3]{18}} \right)^3\\
-    &= \left (\frac{2\sqrt[3]{18}\alpha (1-i \sqrt{3}) - \sqrt[3]{12} t^2(1+i\sqrt{3})}{2t \cdot 6} \right )^3\\
-    &= \bigg (\frac{\overbrace{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2 (1+i\sqrt{3})}^{\text{numerator}}}{\sqrt[3]{12^2} t} \bigg )^3
-\end{align}
-
-Now calculate numerator$^3$:
-\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 \sqrt[3]{12^2} \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 + 6 \sqrt[3]{12^2} \alpha^2 t^2 (1+i \sqrt{3})^2\\
-    &\hphantom{{}=}- 6 \sqrt[3]{12} \alpha t^4 (1-i\sqrt{3})^2 +8 t^6\\
-    &= -96 \alpha^3 - 12 \sqrt[3]{12^2} \alpha^2 t^2 (1-i \sqrt{3})\\
-    &\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +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}
+The verification of this case is pretty much the same as for
+case 2.2.

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

@@ -171,7 +171,7 @@ I will make use of the following identities:
 
 \textbf{Case 2.1:} 
 \input{quadratic-case-2.1}
-
+\goodbreak
 \textbf{Case 2.2:}
 \input{quadratic-case-2.2}