added some ideas

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

@@ -15,6 +15,7 @@
 \usepackage[framed,amsmath,thmmarks,hyperref]{ntheorem}
 \usepackage{framed}
 \usepackage{nicefrac}
+\usepackage{siunitx}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % Define theorems                                                   %
@@ -334,7 +335,7 @@ Because:
 Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
 \[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]
 
-As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points 
+\textbf{Case 1:} As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points 
 $P = (0, w)$ could possilby have three minima.
 
 Then compute:
@@ -353,8 +354,43 @@ should get as close to $0$ as possilbe when we want to minimize
 $d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
 For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.
 
-So the solution is given by
+\textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
+\begin{align}
+  d_{P,{f_2}}(x)  &= \sqrt{(x-z)^2 + (f(x)-w)^2}\\
+    &= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 - w^2)}\\
+    &= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 - w^2}\\
+  0 &\stackrel{!}{=} d_{P, {f_2}}'(x) \\
+    &= 4a^2x^3 + 2(1- 2 aw)x +(- 2z)\\
+    &= 2 \left (2a^2x^2 + (1- 2 aw) \right )x - 2z\\
+    \Leftrightarrow z &\stackrel{!}{=} (2a^2x^2 + (1- 2 aw)) x\\
+    \Leftrightarrow 0 &\stackrel{!}{=} 2 a^2 x^3 + (1- 2 aw)x - z\label{line:solution-equation}
+\end{align}
 
+The solution for this equation was computated with Wolfram|Alpha.
+I will only verify that the solution is correct. As there is only
+one solution in this case, we only have to check this one.
+
+\begin{align}
+    t &:= \sqrt[3]{108a^4z + \sqrt{\num{11664}a^8 z^2 + 864 a^6  (1-2aw)^3}}\\
+    x &= \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t}\\
+    \xRightarrow{x \text{ in line }\ref{line:solution-equation}} 0 &\stackrel{!}{=} 2a^2 \left ( \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3 + (1-2aw) (\frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t}) - z\\
+    &= 2a^2 \underbrace{\left ( \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3}_{=: \alpha} +  \frac{t \cdot (1-2aw)}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2} (1-2aw)^2}{t} - z
+\end{align}
+
+Now compute $\alpha$. We know that $(a-b)^3 = a^3 - 3a^2 b +3ab^2 - b^3$:
+\begin{align}
+    \alpha &= \left (\frac{t}{6 \sqrt[3]{2} a^2} \right )^3 - 3 \left (\frac{t}{6 \sqrt[3]{2} a^2} \right )^2 \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right ) + 3 \left (\frac{t}{6 \sqrt[3]{2} a^2} \right ) \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right )^2 - \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3\\
+    &= \frac{t^3}{216 \cdot 2 \cdot a^6} - \frac{3t^2}{36 \sqrt[3]{4} a^4} \cdot \frac{\sqrt[3]{2}(1-2aw)}{t}
+       + \frac{3t}{6 \sqrt[3]{2} a^2} \cdot \frac{\sqrt[3]{4} (1-2aw)^2}{t^2} - \frac{2 (1-2aw)^3}{t^3}\\
+    &= \frac{t^3}{432a^6} - \frac{t (1-2aw)}{12\sqrt[3]{2} a^4} + \frac{\sqrt[3]{2} (1-2aw)^2}{2ta^2} - \frac{2 (1-2aw)^3}{t^3}\\
+    &= \frac{t^3}{432a^6} + \frac{t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2}{12\sqrt[3]{2} a^4} - \frac{2 (1-2aw)^3}{t^3}\\
+    &= \frac{t^3- 2 (1-2aw)^3 \cdot 432a^6}{432a^6 t^3} + \frac{t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2}{12\sqrt[3]{2} a^4}\\
+    &= \frac{\sqrt[3]{2}(t^3- 2 (1-2aw)^3 \cdot 432a^6) + 36 a^2 t^3 (t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2)}{432 \sqrt[3]{2} a^6 t^3}\\
+    &= \frac{\sqrt[3]{2} t^3 - \sqrt[3]{2} (1-2aw)^3 \cdot 864a^6 + 36 a^2 t^3 (2awt^1-t^2 + 6 \sqrt[3]{4} a^2 (1-2aw)^2)}{6 \sqrt[3]{2} a^2 t \cdot 2a^2 \cdot 36 a^2 t^2}
+\end{align}
+
+\goodbreak
+So the solution is given by
 \begin{align*}
 x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
 \underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
@@ -459,6 +495,68 @@ $\tilde{e}$.
 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{Another approach}
+Just like we moved the function $f$ and the point to get in a 
+nicer situation, we can apply this approach for cubic functions.
+
+\begin{figure}[htp]
+    \centering
+\begin{tikzpicture}
+    \begin{axis}[
+        legend pos=south east,
+        axis x line=middle,
+        axis y line=middle,
+        grid = major,
+        width=0.8\linewidth,
+        height=8cm,
+        grid style={dashed, gray!30},
+        xmin=-3,     % start the diagram at this x-coordinate
+        xmax= 3,    % end   the diagram at this x-coordinate
+        ymin=-3,     % start the diagram at this y-coordinate
+        ymax= 3,   % end   the diagram at this y-coordinate
+        axis background/.style={fill=white},
+        xlabel=$x$,
+        ylabel=$y$,
+        %xticklabels={-2,-1.6,...,7},
+        %yticklabels={-8,-7,...,8},
+        tick align=outside,
+        minor tick num=-3,
+        enlargelimits=true,
+        tension=0.08]
+      \addplot[domain=-3:3, thick,samples=50, red] {x*x*x}; 
+      \addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x};
+      \addplot[domain=-3:3, thick,samples=50, orange] {x*x*x-x}; 
+      \addplot[domain=-3:3, thick,samples=50, blue, dotted] {x*x*x+2*x};
+      \addplot[domain=-3:3, thick,samples=50, lime, dashed] {x*x*x+3*x};
+      \addlegendentry{$f_1(x)=x^3$}
+      \addlegendentry{$f_2(x)=x^3 + x$}
+      \addlegendentry{$f_1(x)=x^3 - x$}
+      \addlegendentry{$f_2(x)=x^3 + 2 \cdot x$}
+      \addlegendentry{$f_2(x)=x^3 + 3 \cdot x$}
+    \end{axis} 
+\end{tikzpicture}
+    \caption{Cubic functions with $b = d = 0$}
+\end{figure}
+
+First, we move $f_0$ by $\frac{b}{3a}$ to the right, so
+
+\[f_1(x) = ax^3 + \frac{b^2 (c-1)}{3a} x + \frac{2b^3}{27 a^2} - \frac{bc}{3a} + d \;\;\;\text{ and }\;\;\;P_1 = (x_P + \frac{b}{3a}, y_P)\]
+
+because
+
+\begin{align}
+    f_1(x) &= a \left (x - \frac{b}{3a} \right )^3 + b \left (x-\frac{b}{3a} \right )^2 + c \left (x-\frac{b}{3a} \right ) + d\\
+           &= a \left (x^3 - 3 \frac{b}{3a}x^2 + 3 (\frac{b}{3a})^2 x - \frac{b^3}{27a^3} \right )
+             +b \left (x^2 - \frac{2b}{3a} x + \frac{b^2}{9a^2} \right )
+             +c x - \frac{bc}{3a} + d\\
+            &= ax^3 - bx^2 + \frac{b^2}{3a}x - \frac{b^3}{27 a^2}\\
+            & \;\;\;\;\;\;+ bx^2 - \frac{2b^2}{3a}x + \frac{b^3}{9a^2}\\
+            & \;\;\;\;\;\;\;\;\;\;\;\; + c x - \frac{bc}{3a} + d\\
+            &= 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}
+
+\todo[inline]{Which way to move might be clever?}
+
 \subsection{Number of points with minimal distance}
 As there is an algebraic equation of degree 5, there cannot be more
 than 5 solutions.