removed unnecessary stuff; began developing solution formula

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

@@ -306,110 +306,20 @@ As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
 
 
 \subsection{Number of points with minimal distance}
-\subsubsection{Two points with minimal distance}
-Quadratic functions can have two points with minimal distance. 
-
-For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} = (\sqrt{\frac{9}{2}}, \frac{9}{2})$
-has minimal distance to $P$, but also $P_{f,2} = (-\sqrt{\frac{9}{2}}, \frac{9}{2})$:
-
-\begin{proof}
-    \begin{align}
-      d_{P,f}(x)  &= \sqrt{(x-x_P)^2 + (f(x)-y_p)^2}\\
-        &= \sqrt{x^2 + (x^2-5)^2}\\
-        &= \sqrt{x^2 + x^4-10x^2+25}\\
-        &= \sqrt{x^4 -9x^2 + 25}\\
-        &= \sqrt{x^4 -9x^2 + \frac{81}{4}+\frac{19}{4}}\\
-        &= \sqrt{\left (x^2 - \frac{9}{2} \right )^2 + \frac{19}{4}}
-    \end{align}
-
-    Obviously, $d_{P,f}$ is minimal for $x = \pm \sqrt{\frac{9}{2}} \qed$
-\end{proof}
-
-\begin{figure}[htp]
-    \centering
-    \begin{tikzpicture}
-        \begin{axis}[
-            %legend pos=north west,
-            axis x line=middle,
-            axis y line=middle,
-            grid = major,
-            width=0.6\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= 0,     % start the diagram at this y-coordinate
-            ymax= 5,   % 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, orange] {x*x};
-          \draw (axis cs:0,5) circle[radius=2.17];
-          \draw[red, thick] (axis cs:0,5) -- (axis cs:2.121,4.5);
-          \draw[red, thick] (axis cs:0,5) -- (axis cs:-2.121,4.5);
-          \addlegendentry{$f(x)=x^2$}
-        \end{axis} 
-    \end{tikzpicture}
-    \caption{Two points with minimal distance}
-\end{figure}
-
-\subsubsection{Three points with minimal distance}
-As discussed before, there cannot be more than 3 points on the graph
-of $f$ next to $P$.
-
-\todo[inline]{But can there be three points? O.b.d.A: $a > 0$.
-As $c$ is moves the curve only up and down, we can o.b.d.A assume
-that $c=0$. 
+\begin{theorem}
+    A point $P$ has either one or two points on the graph of a 
+    quadratic function $f$ that are closest to $P$.
+\end{theorem}
 
-$x=-\frac{b}{2a}$ is the minimum of $f$. If there are 3 points, this will
-be one of them. The other two ones are symmetric by an axis through
-$-\frac{b}{2a}$}
+In the following, I will do some transformations with $f = f_0$ and
+$P = P_0$ .
 
-\begin{figure}[htp]
-    \centering
-    \begin{tikzpicture}
-        \begin{axis}[
-            legend pos=north west,
-            axis x line=middle,
-            axis y line=middle,
-            grid = major,
-            width=0.8\linewidth,
-            height=8cm,
-            grid style={dashed, gray!30},
-            xmin=-0.7,     % start the diagram at this x-coordinate
-            xmax= 0.7,    % end   the diagram at this x-coordinate
-            ymin=-0.25,     % start the diagram at this y-coordinate
-            ymax= 0.5,   % 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=-0.7:0.7, thick,samples=50, orange] {x*x};
-          \draw (axis cs:0,0.5) circle[radius=0.5];
-          \draw[red, thick] (axis cs:0,0.5) -- (axis cs:0.101,0.0102);
-          \draw[red, thick] (axis cs:0,0.5) -- (axis cs:-0.101,0.0102);
-          \draw[red, thick] (axis cs:0,0.5) -- (axis cs:0,0);
-          \addlegendentry{$f(x)=x^2$}
-        \end{axis} 
-    \end{tikzpicture}
-    \caption{3 points with minimal distance?}
-\end{figure}
+Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does 
+not change the minimum distance. Furthermore, we can find the 
+points with minimum distance on the moved situation and calculate
+the minimum points in the original situation.
 
-When move the $f$ and $P$ simultaneously in $x$ direction, you will not change the
-results. 
-
-First of all, you move $f_0$ by $\frac{b}{2a}$, so
+First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
 \[f_1(x) = ax^2 - \frac{b^2}{4a} + c \;\;\;\text{ and }\;\;\; P_1 = \left (x_p+\frac{b}{2a},\;\; y_p \right )\]
 
 Because:
@@ -425,20 +335,28 @@ 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 $f(x) = ax^2$ is symmetric to the $y$ axis, only points 
-$P = (0, y_p)$ could possilby have three minima.
+$P = (0, w)$ could possilby have three minima.
 
 Then compute:
 \begin{align}
-  d_{P,f}(x)  &= \sqrt{(x-x_P)^2 + (f(x)-y_p)^2}\\
-    &= \sqrt{x^2 + (ax^2-y_p)^2}\\
-    &= \sqrt{x^2 + a^2 x^4-2ay_p x^2+y_p^2}\\
-    &= \sqrt{a^2 x^4 + (1-2ay_p) x^2 + y_p^2}\\
-    &= \sqrt{\left (a^2 x^2 + \frac{1-2 a y_p}{2} \right )^2 + y_p^2 - (1-2 a y_p)^2}\\
-    &= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a y_p \right )^2 + (y_p^2 - (1-2 a y_p)^2)}\\
+  d_{P,f}(x)  &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
+    &= \sqrt{x^2 + (ax^2-w)^2}\\
+    &= \sqrt{x^2 + a^2 x^4-2aw x^2+w^2}\\
+    &= \sqrt{a^2 x^4 + (1-2aw) x^2 + w^2}\\
+    &= \sqrt{\left (a^2 x^2 + \frac{1-2 a w}{2} \right )^2 + w^2 - (1-2 a w)^2}\\
+    &= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + (w^2 - (1-2 a w)^2)}\\
 \end{align}
 
-For $y_p \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
-For all other points, there are exactly two minima.
+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.
+
+So the solution is given by
+
+\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
+     x_1 = todo \text{ and } x_2 = todo   &\text{if } x_P = - \frac{b}{2a} \text{ and } y_p + \frac{b^2}{4a} - c >  \frac{1}{2a} \\
+     x = todo   &\text{if } x_P = - \frac{b}{2a} \text{ and } y_p + \frac{b^2}{4a} - c \leq  \frac{1}{2a} \\
+     x = todo   &\text{if } x_P \neq - \frac{b}{2a}
+    \end{cases}\]
 
 \clearpage
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