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minor changes

Martin Thoma %!s(int64=11) %!d(string=hai) anos
pai
achega
03d2d98754

+ 3 - 3
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
 it should take. A reasonable choice for the representation of
 paths are cubic splines. You also have to be able to calculate
 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
 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 
 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)
 be able to get the minimal distance of a point (the position of the car)
 to a cubic spline (the prefered path)
 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
 get the minimal absolute distance, but might also help to get all points
 on the spline with minimal distance.
 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.
 function with minimal distance to a given point.
 As other representations of paths might be easier to understand and
 As other representations of paths might be easier to understand and
 to implement, I will also cover the problem of finding the minimal
 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
 While I analyzed this problem, I've got interested in variations
 of the underlying PID-related problem. So I will try to give
 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 a point to a (piecewise or global) defined polynomial function
 of degree $\leq 3$.
 of degree $\leq 3$.
 
 

+ 10 - 22
documents/math-minimal-distance-to-cubic-function/linear-functions.tex

@@ -46,30 +46,18 @@
 \end{figure}
 \end{figure}
 
 
 \begin{proof}
 \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}
     \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}
     \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}
 \end{proof}
 \clearpage
 \clearpage
 
 

BIN=BIN
documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf


+ 1 - 1
documents/math-minimal-distance-to-cubic-function/problem-description.tex

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

+ 1 - 1
documents/math-minimal-distance-to-cubic-function/quadratic-case-2.1.tex

@@ -1,5 +1,5 @@
 $4 \alpha^3 + 27 \beta^2 \geq 0$:
 $4 \alpha^3 + 27 \beta^2 \geq 0$:
-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
 is
 \[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
 \[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
 
 

+ 11 - 9
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}
 \[x = \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t}
      -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
      -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
 
 
+We will verify it in multiple steps. First, get $x^3$:
 \begin{align}
 \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\\
     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}
 \end{align}
 
 
-Now calculate the numerator$^3$:
+Now calculate numerator$^3$:
 \begin{align}
 \begin{align}
     \left (\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3}) \right )^3 &= 
     \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 \\
     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]{12^2} \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\\
     &\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) \\
     &= 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)\\
     &\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})\\
     &= -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
     &\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6
 \end{align}
 \end{align}

+ 10 - 11
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
 We use Theorem~\ref{thm:fermats-theorem}:\nobreak
 \begin{align}
 \begin{align}
     0     &\overset{!}{=} (d_{P,f}^2)'\\
     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}
           &= 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}
 \end{align}
 
 
@@ -68,12 +68,11 @@ has to be a solution to the given problem.
         &= x(2x^2-3)\\
         &= x(2x^2-3)\\
         \Rightarrow x_{1,2} &= \pm \sqrt{\frac{3}{2}} \text{ and } x_3 = 0\\
         \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}(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{3}{2}+\nicefrac{1}{4}} \\
             &= \sqrt{\nicefrac{7}{4}}\\
             &= \sqrt{\nicefrac{7}{4}}\\
-            &= d_{P,f} \left (- \sqrt{\frac{3}{2}} \right )
     \end{align}
     \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
     $(d_{P,f}(x_3))' = 0$.
     $(d_{P,f}(x_3))' = 0$.
 \end{example}
 \end{example}
 
 
@@ -125,8 +124,8 @@ Then compute:
     &= \sqrt{x^2 + (ax^2-w)^2}\\
     &= \sqrt{x^2 + (ax^2-w)^2}\\
     &= \sqrt{x^2 + a^2 x^4-2aw x^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{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}
 \end{align}
 
 
 The term 
 The term