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@@ -50,13 +50,13 @@
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\begin{document}
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\maketitle
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\begin{abstract}
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-When you have a selfdriving car, you have to plan which path you
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-want to take. A reasonable choice for the representation of this
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-path is a cubic spline. But you also have to be able to calculate
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-how to steer to get or to remain on this path. A way to do this
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+When you want to develop a selfdriving car, you have to plan which path
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+it should take. A reasonable choice for the representation of
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+paths are cubic splines. You also have to be able to calculate
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+how to steer to get or to remain on a path. A way to do this
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is applying the \href{https://en.wikipedia.org/wiki/PID_algorithm}{PID algorithm}.
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-But this algorithm needs to know the current error. So you need to
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-be able to get the minimal distance of a point to a cubic spline.
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+This algorithm needs to know the signed current error. So you need to
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+be able to get the minimal distance of a point to a cubic spline combined with the direction (left or right).
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As you need to get the signed error (and one steering direction might
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be prefered), it is not only necessary to
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get the minimal absolute distance, but also to get all points
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@@ -71,7 +71,7 @@ distance of a point to a polynomial of degree 0, 1 and 2.
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\section{Description of the Problem}
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Let $f: \mdr \rightarrow \mdr$ be a polynomial function and $P \in \mdr^2$
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-be a point. Let $d_{P,f}: \mdr^2 \rightarrow \mdr_0^+$
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+be a point. Let $d_{P,f}: \mdr \rightarrow \mdr_0^+$
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be the Euklidean distance $d_{P,f}$ of a point $P$ and a point $\left (x, f(x) \right )$:
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\[d_{P,f} (x) := \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\]
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@@ -89,7 +89,7 @@ But minimizing $d_{P,f}$ is the same as minimizing $d_{P,f}^2$:
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\todo[inline]{Hat dieser Satz einen Namen? Gibt es ein gutes Buch,
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aus dem ich den zitieren kann? Ich habe ihn aus \href{https://github.com/MartinThoma/LaTeX-examples/tree/master/documents/Analysis I}{meinem Analysis I Skript} (Satz 21.5).}
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\begin{theorem}\label{thm:required-extremum-property}
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- Let $x_0$ be a relative extremum of $f$.
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+ Let $x_0$ be a relative extremum of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.
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Then: $f'(x_0) = 0$.
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\end{theorem}
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@@ -265,10 +265,11 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak
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0 &\overset{!}{=} (d_{P,f}^2)'\\
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&= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\
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&= -2 x_p + 2x -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
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- &= -2 x_p + 2x -2y_p (2ax+b) + ((ax^2+bx+c)^2)'\\
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- &= -2 x_p + 2x -2y_p \cdot 2ax-2 y_p b + (a^2 x^4+2 a b x^3+2 a c x^2+b^2 x^2+2 b c x+c^2)'\\
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- &= -2 x_p + 2x -4y_p ax-2 y_p b + (4a^2 x^3 + 6 ab x^2 + 4acx + 2b^2 x + 2bc)\\
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- &= 4a^2 x^3 + 6 ab x^2 + 2(1 -2y_p a+ 2ac + b^2)x +2(bc-by_p-x_p)
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+\Leftrightarrow 0 &\overset{!}{=} -x_p + x -y_p f'(x) + f(x) \cdot f'(x) \rlap{\hspace*{3em}(divide by 2)}\label{eq:minimizingFirstDerivative}\\
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+ &= -x_p + x -y_p (2ax+b) + (ax^2+bx+c)(2ax+b)\\
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+ &= -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)\\
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+ &= -x_p + x -2y_p ax- y_p b + (2a^2 x^3 + 3 ab x^2 + 2acx + b^2 x + bc)\\
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+ &= 2a^2 x^3 + 3 ab x^2 + (1 -2y_p a+ 2ac + b^2)x +(bc-by_p-x_p)
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\end{align}
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%\begin{align}
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@@ -416,7 +417,8 @@ take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
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\begin{align}
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0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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- &= \underbrace{\left (2 f(x) - 2 y_p \right ) \cdot f'(x)}_{\text{Polynomial of degree 5}} + \underbrace{2x - 2 x_p}_{\text{:-(}}
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+ &= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
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+ &= \underbrace{f'(x) \cdot \left (f(x) - y_p \right )}_{\text{Polynomial of degree 5}} + \underbrace{x - x_p}_{\text{:-(}}
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\end{align}
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\todo[inline]{Although general algebraic equations of degree 5 don't
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