moved argument to the correct location; consistency for titles; added proof to theorem (only two solutions for quadratic problem)

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

@@ -116,3 +116,11 @@ The point with minimum distance can be found by:
   \Set{a} &\text{if } S_0(f,P) \ni x_P < a\\
   \Set{b} &\text{if } S_0(f,P) \ni x_P > b
     \end{cases}\]
+
+Because:
+\begin{align}
+    \underset{x\in[a,b]}{\arg \min d_{P,f}(x)} &= \underset{x\in[a,b]}{\arg \min d_{P,f}(x)^2}\\
+   &=\underset{x\in[a,b]}{\arg \min} x^2 - 2x_P x + (x_P^2 + y_P^2 - 2 y_P c + c^2)\\
+   &=\underset{x\in[a,b]}{\arg \min} x^2 - 2 x_P x + x_P^2\\
+   &=\underset{x\in[a,b]}{\arg \min} (x-x_P)^2
+\end{align}

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

@@ -216,4 +216,4 @@ initial guess.
 \todo[inline]{TODO}
 \clearpage
 
-\section{Defined on a closed interval of $\mdr$}
+\section{Defined on a closed interval $[a,b] \subseteq \mdr$}

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

@@ -29,6 +29,8 @@ $t \in \mdr$ be a linear function.
           \addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
           \addplot[domain=-5:5, thick,samples=50, blue] {-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)};
           \addlegendentry{$f(x)=\frac{1}{2}x$}
           \addlegendentry{$f_\bot(x)=-2x+6$}
         \end{axis} 
@@ -115,10 +117,4 @@ The point with minimum distance can be found by:
    \Set{b} &\text{if } S_1(f, P) \ni x > b
     \end{cases}\]
 
-Because:
-\begin{align}
-    \underset{x\in[a,b]}{\arg \min d_{P,f}(x)} &= \underset{x\in[a,b]}{\arg \min d_{P,f}(x)^2}\\
-   &=\underset{x\in[a,b]}{\arg \min} x^2 - 2x_P x + (x_P^2 + y_P^2 - 2 y_P c + c^2)\\
-   &=\underset{x\in[a,b]}{\arg \min} x^2 - 2 x_P x + x_P^2\\
-   &=\underset{x\in[a,b]}{\arg \min} (x-x_P)^2\\
-\end{align}
+\todo[inline]{argument? proof?}

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

@@ -80,6 +80,7 @@ As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
     quadratic function $f$ that are closest to $P$.
 \end{theorem}
 
+\begin{proof}
 In the following, I will do some transformations with $f = f_0$ and
 $P = P_0$ .
 
@@ -107,7 +108,7 @@ to add $+1$.}
 Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
 \[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \Big (\underbrace{x_P+\frac{b}{2a}}_{=: z},\;\; \underbrace{y_P+\frac{b^2}{4a}-c}_{=: w} \Big )\]
 
-\textbf{Case 1:} As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points 
+As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points 
 $P = (0, w)$ could possilby have three minima.
 
 Then compute:
@@ -125,6 +126,18 @@ The term
 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 $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.
+$\qed$
+\end{proof}
+
+\subsection{Solution formula}
+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 $y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
+then there are two solutions:
+\[x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}\]
+Otherwise, there is only one solution $x_1 = 0$.
 
 \textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
 \begin{align}
@@ -140,10 +153,12 @@ For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \
     &= x^3 + \alpha x + \beta\label{eq:simple-cubic-equation-for-quadratic-distance}
 \end{align}
 
+Let $t$ be defined as
+\[t := \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\]
+
+\textbf{Case 2.1:} $4 \alpha^3 + 27 \beta^2 \geq 0$:
 The solution of Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
 is
-\todo[inline]{Can $4 \alpha^3 + 27 \beta^2$ be negative for $\alpha=\frac{1-2aw}{2a^2}$ and $\beta = \frac{-z}{2a^2}$?}
-\[t := \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\]
 \[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
 
 When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
@@ -190,6 +205,16 @@ $t$:
 &= 0
 \end{align}
 
+\textbf{Case 2.2:} TODO
+
+\[x = \frac{(1+i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
+     -\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]
+
+\textbf{Case 2.3:} TODO
+
+\[x = \frac{(1-i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
+     -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
+
 \goodbreak
 So the solution is given by
 \todo[inline]{NO! Currently, there are erros in the solution.