minor changes

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

@@ -3,7 +3,7 @@ When you want to develop a selfdriving car, you have to plan which path
 it should take. A reasonable choice for the representation of
 paths are cubic splines. You also have to be able to calculate
 how to steer to get or to remain on a path. A way to do this
-is applying the \href{https://en.wikipedia.org/wiki/PID_algorithm}{PID algorithm}.
+is by applying the \href{https://en.wikipedia.org/wiki/PID_algorithm}{PID algorithm}.
 This algorithm needs to know the signed current error. So you need to 
 be able to get the minimal distance of a point (the position of the car)
 to a cubic spline (the prefered path)
@@ -13,7 +13,7 @@ be prefered), it is not only necessary to
 get the minimal absolute distance, but might also help to get all points
 on the spline with minimal distance.
 
-In this paper I want to discuss how to find all points on a cubic 
+In this paper, I want to discuss how to find all points on a cubic 
 function with minimal distance to a given point.
 As other representations of paths might be easier to understand and
 to implement, I will also cover the problem of finding the minimal
@@ -21,7 +21,7 @@ distance of a point to a polynomial of degree 0, 1 and 2.
 
 While I analyzed this problem, I've got interested in variations
 of the underlying PID-related problem. So I will try to give
-robust and easy-to-implement algorithms to calculated the distance
+robust and easy-to-implement algorithms to calculate the distance
 of a point to a (piecewise or global) defined polynomial function
 of degree $\leq 3$.
 

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

@@ -46,30 +46,18 @@
 \end{figure}
 
 \begin{proof}
-    Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The 
-    slope of $f_\bot$ is $- \frac{1}{m}$ and $t_\bot$ can be calculated:\nobreak
+    With Theorem~\ref{thm:fermats-theorem} you get:
     \begin{align}
-                     f_\bot(x) &= - \frac{1}{m} \cdot x + t_\bot\\
-        \Rightarrow        y_P &= - \frac{1}{m} \cdot x_P + t_\bot\\
-        \Leftrightarrow t_\bot &= y_P + \frac{1}{m} \cdot x_P
+        0 &\stackrel{!}{=} (d_{P,f}(x)^2)'\\
+        &= 2(x-x_P) + 2 (f(x) - y_P)f'(x)\\
+        \Leftrightarrow 0 &\stackrel{!}{=} x - x_P + (f(x) - y_P) f'(x)\\
+        &= x- x_P + (mx+t - y_P)\cdot m\\
+        &= x (m+1) + m(t-y_P) - x_P\\
+        \Leftrightarrow x &\stackrel{!}{=} \frac{x_p - m(t-y_p)}{m^2+1}\\
+        &= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )\label{eq:solution-linear-r}
     \end{align}
-
-    The point $(x, f(x))$ where the perpendicular $f_\bot$ crosses $f$
-    is calculated this way:
-    \begin{align}
-        f(x) &= f_\bot(x)\\
-        \Leftrightarrow m \cdot x + t &= - \frac{1}{m} \cdot x + \left(y_P + \frac{1}{m} \cdot x_P \right)\\
-        \Leftrightarrow \left (m + \frac{1}{m} \right ) \cdot x &= y_P + \frac{1}{m} \cdot x_P - t\\
-        \Leftrightarrow x &= \frac{m}{m^2+1} \left ( y_P + \frac{1}{m} \cdot x_P - t \right )\label{eq:solution-of-the-linear-problem}
-    \end{align}
-
-    There is only one point with minimal distance. I'll call the result
-    from line~\ref{eq:solution-of-the-linear-problem} \enquote{solution of 
-    the linear problem} and the function that gives this solution 
-    $S_1(f,P)$.
-
-    See Figure~\ref{fig:linear-min-distance}
-    to get intuition about the geometry used. $\qed$
+    It is obvious that a minium has to exist, the $x$ from Equation~\ref{eq:solution-linear-r}
+    has to be this minimum.
 \end{proof}
 \clearpage
 

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

@@ -1,7 +1,7 @@
 \chapter{Description of the Problem}
 Let $f: D \rightarrow \mdr$ with $D \subseteq \mdr$ be a polynomial function and $P \in \mdr^2$
 be a point. Let $d_{P,f}: \mdr \rightarrow \mdr_0^+$
-be the Euklidean distance of a point $P$ to a point $\left (x, f(x) \right )$
+be the Euklidean distance of $P$ to a point $\left (x, f(x) \right )$
 on the graph of $f$:
 \[d_{P,f} (x) := \sqrt{(x - x_P)^2 + (f(x) - y_P)^2}\]
 

documents/math-minimal-distance-to-cubic-function/quadratic-case-2.1.tex

@@ -1,5 +1,5 @@
 $4 \alpha^3 + 27 \beta^2 \geq 0$:
-The solution of Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
+One solution of Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
 is
 \[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
 

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

@@ -1,26 +1,28 @@
-\todo[inline]{calculate...} 
-
+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} \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
+    &= \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 the numerator$^3$:
+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 \cdot 2 \sqrt[3]{18} \alpha^2(-2(1+i\sqrt{3}))(t^2(1+i \sqrt{3}))\\
+    &\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 + 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 + 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}

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

@@ -45,12 +45,12 @@ In this case, $d_{P,f}^2$ is polynomial of degree 4.
 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 )'\\
-          &= -2 x_p + 2x -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
-\Leftrightarrow 0 &\overset{!}{=} -x_p + x -y_p f'(x) + f(x) \cdot f'(x) \rlap{\hspace*{3em}(divide by 2)}\label{eq:minimizingFirstDerivative}\\
-          &= -x_p + x -y_p (2ax+b) + (ax^2+bx+c)(2ax+b)\\
-          &= -x_p + x -y_p \cdot 2ax- y_p b + (2 a^2 x^3+2 a b x^2+2 a c x+ab x^2+b^2 x+bc)\\
-          &= -x_p + x -2y_p ax- y_p b + (2a^2 x^3 + 3 ab x^2 + 2acx + b^2 x + bc)\\
+          &= 2x -2 x_p -2y_p f'(x) + \left (f(x)^2 \right )'\\
+          &= 2x -2 x_p -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
+\Leftrightarrow 0 &\overset{!}{=} x -x_p -y_p f'(x) + f(x) \cdot f'(x) \rlap{\hspace*{3em}(divide by 2)}\label{eq:minimizingFirstDerivative}\\
+          &= x -x_p -y_p (2ax+b) + (ax^2+bx+c)(2ax+b)\\
+          &= x -x_p -y_p \cdot 2ax- y_p b + (2 a^2 x^3+2 a b x^2+2 a c x+ab x^2+b^2 x+bc)\\
+          &= x -x_p -2y_p ax- y_p b + (2a^2 x^3 + 3 ab x^2 + 2acx + b^2 x + bc)\\
           &= 2a^2 x^3 + 3 ab x^2 + (1 -2y_p a+ 2ac + b^2)x +(bc-by_p-x_p)\label{eq:quadratic-derivative-eq-0}
 \end{align}
 
@@ -68,12 +68,11 @@ has to be a solution to the given problem.
         &= x(2x^2-3)\\
         \Rightarrow x_{1,2} &= \pm \sqrt{\frac{3}{2}} \text{ and } x_3 = 0\\
         d_{P,f}(x_3) &= \sqrt{0^2 + (-1-1)^2} = 2\\
-        d_{P,f} \left (+ \sqrt{\frac{3}{2}} \right ) &= \sqrt{\left (\sqrt{\frac{3}{2} - 0} \right )^2 + \left (\frac{1}{2}-1 \right )^2}\\
+        d_{P,f} \left (\pm \sqrt{\frac{3}{2}} \right ) &= \sqrt{\left (\sqrt{\frac{3}{2} - 0} \right )^2 + \left (\frac{1}{2}-1 \right )^2}\\
             &= \sqrt{\nicefrac{3}{2}+\nicefrac{1}{4}} \\
             &= \sqrt{\nicefrac{7}{4}}\\
-            &= d_{P,f} \left (- \sqrt{\frac{3}{2}} \right )
     \end{align}
-    This means $x_3$ is not a point of minimal distance, but with
+    This means $x_3$ is not a point of minimal distance, although
     $(d_{P,f}(x_3))' = 0$.
 \end{example}
 
@@ -125,8 +124,8 @@ Then compute:
     &= \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 x^2 + \frac{1-2 a w}{2a} \right )^2 + w^2 - (\frac{1-2 a w}{2a})^2}\\
-    &= \sqrt{\left (a x^2 + \nicefrac{1}{2a}- w \right )^2 + \big (w^2 - (\frac{1-2 a w}{2a})^2 \big)}
+    &= \sqrt{\left (a x^2 + \frac{1-2 a w}{2a} \right )^2 + w^2 - \left (\frac{1-2 a w}{2a} \right )^2}\\
+    &= \sqrt{\left (a x^2 + \nicefrac{1}{2a}- w \right )^2 + \left (w^2 - \left (\frac{1-2 a w}{2a} \right )^2 \right)}
 \end{align}
 
 The term