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many changes; added proof
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Martin Thoma
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@ -64,7 +64,7 @@ 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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get the minimal absolute distance, but might also help to get all points
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on the spline with minimal distance.
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In this paper I want to discuss how to find all points on a cubic
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@ -77,37 +77,24 @@ 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 $d_{P,f}: \mdr \rightarrow \mdr_0^+$
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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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be the Euklidean distance of a point $P$ and a point $\left (x, f(x) \right )$
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on the graph of $f$:
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\[d_{P,f} (x) := \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\]
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Now there is \todo{Should I proof this?}{finite set} $x_1, \dots, x_n$ such that
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\[\forall \tilde x \in \mathbb{R} \setminus \{x_1, \dots, x_n\}: d_{P,f}(x_1) = \dots = d_{P,f}(x_n) < d_{P,f}(\tilde x)\]
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Now there is finite set $M = \Set{x_1, \dots, x_n}$ of minima for given $f$ and $P$:
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\[M = \Set{x \in \mdr | d_{P,f}(x) = \min_{\overline{x} \in \mdr} d_{P,f}(\overline{x})}\]
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Essentially, you want to find the minima $x_1, \dots, x_n$ for given
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$f$ and $P$.
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But minimizing $d_{P,f}$ is the same as minimizing $d_{P,f}^2$:
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\begin{align}
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d_{P,f}(x)^2 &= \sqrt{(x_P - x)^2 + (y_P - f(x))^2}^2\\
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&= x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2
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\end{align}
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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 a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.
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\begin{theorem}[Fermat's theorem about stationary points]\label{thm:required-extremum-property}
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Let $x_0$ be a local 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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%bzw. 22.3
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%\begin{theorem}[Minima of polynomial functions]\label{thm:minima-of-polynomials}
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% Let $n \in \mathbb{N}, n \geq 2$, $f$ polynomial function of
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% degree $n$, $x_0 \in \mathbb{R}$, \\
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% $f'(x_0) = f''(x_0) = \dots = f^{(n-1)} (x_0) = 0$
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% and $f^{(n)} > 0$.
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%
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% Then $x_0$ is a local minimum of $f$.
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%\end{theorem}
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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@ -140,7 +127,7 @@ Let $f(x) = c$ with $c \in \mdr$ be a constant function.
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tension=0.08]
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\addplot[domain=-5:5, thick,samples=50, red] {1};
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\addplot[domain=-5:5, thick,samples=50, green] {2};
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\addplot[domain=-5:5, thick,samples=50, blue] {3};
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\addplot[domain=-5:5, thick,samples=50, blue, densely dotted] {3};
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\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
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\addplot[blue, mark = *, nodes near coords=$P_{h,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 3)};
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\addplot[green, mark = x, nodes near coords=$P_{g,\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 2)};
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@ -199,15 +186,16 @@ $t \in \mdr$ be a linear function.
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\label{fig:linear-min-distance}
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\end{figure}
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Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The slope of $f_\bot$
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is $- \frac{1}{m}$. Now you can calculate $f_\bot$:\nobreak
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Now you can drop a perpendicular $f_\bot$ through $P$ on $f(x)$. The
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slope of $f_\bot$ is $- \frac{1}{m}$ and $t_\bot$ can be calculated:\nobreak
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\begin{align}
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f_\bot(x) &= - \frac{1}{m} \cdot x + t_\bot\\
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\Rightarrow y_P &= - \frac{1}{m} \cdot x_P + t_\bot\\
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\Leftrightarrow t_\bot &= y_P + \frac{1}{m} \cdot x_P
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\end{align}
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Now find the point $(x, f(x))$ where the perpendicular crosses the function:
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The point $(x, f(x))$ where the perpendicular $f_\bot$ crosses $f$
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is calculated this way:
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\begin{align}
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f(x) &= f_\bot(x)\\
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\Leftrightarrow m \cdot x + t &= - \frac{1}{m} \cdot x + \left(y_P + \frac{1}{m} \cdot x_P \right)\\
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@ -235,24 +223,22 @@ $b, c \in \mdr$ be a quadratic function.
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-0.25, % start the diagram at this y-coordinate
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ymax= 9, % end the diagram at this y-coordinate
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ymin=-0.25, % start the diagram at this y-coordinate
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ymax= 9, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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%xticklabels={-2,-1.6,...,7},
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%yticklabels={-8,-7,...,8},
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-3:3, thick,samples=50, red] {0.5*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x};
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\addplot[domain=-3:3, thick,samples=50, blue] {x*x + x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x + 2*x};
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\addplot[domain=-3:3, thick,samples=50, black] {-x*x + 6};
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\addplot[domain=-3:3, thick,samples=50, red] {0.5*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] { x*x};
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\addplot[domain=-3:3, thick,samples=50, blue] { x*x + x};
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\addplot[domain=-3:3, thick,samples=50, orange,dotted] { x*x + 2*x};
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\addplot[domain=-3:3, thick,samples=50, black,dashed] {-x*x + 6};
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\addlegendentry{$f_1(x)=\frac{1}{2}x^2$}
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\addlegendentry{$f_2(x)=x^2$}
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\addlegendentry{$f_3(x)=x^2+x$}
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@ -277,15 +263,6 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak
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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)\label{eq:quadratic-derivative-eq-0}
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\end{align}
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%\begin{align}
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% 0 &\overset{!}{=}(d_{P,f}^2)''\\
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% &= 2 - 2y_p f''(x) + \left ( 2 f(x) \cdot f'(x) \right )' \rlap{\hspace*{3em}(Eq. \ref{eq:minimizingFirstDerivative})}\\
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% &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x) \cdot f'(x) + f(x) \cdot f''(x) \right ) \rlap{\hspace*{3em}(product rule)}\\
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% &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x)^2 + f(x) \cdot f''(x) \right )\\
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% &= 12a^2 x^2 + 12abx + 2(1 -2y_p a+ 2ac + b^2)
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%\end{align}
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This is an algebraic equation of degree 3.
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There can be up to 3 solutions in such an equation. Those solutions
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can be found with a closed formula.
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@ -323,7 +300,11 @@ the minimum points in the original situation.
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First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
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\[f_1(x) = ax^2 - \frac{b^2}{4a} + c \;\;\;\text{ and }\;\;\; P_1 = \left (x_p+\frac{b}{2a},\;\; y_p \right )\]
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Because:
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Because:\footnote{The idea why you subtract $\frac{b}{2a}$ within
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$f$ is that when you subtract something from $x$ before applying
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$f$ it takes more time ($x$ needs to be bigger) to get to the same
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situation. So to move the whole graph by $1$ to the left whe have
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to add $+1$.}
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\begin{align}
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f(x-\nicefrac{b}{2a}) &= a (x-\nicefrac{b}{2a})^2 + b (x-\nicefrac{b}{2a}) + c\\
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&= a (x^2 - \nicefrac{b}{a} x + \nicefrac{b^2}{4a^2}) + bx - \nicefrac{b^2}{2a} + c\\
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@ -333,19 +314,19 @@ Because:
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Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
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\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]
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\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \Big (\underbrace{x_P+\frac{b}{2a}}_{=: z},\;\; \underbrace{y_P+\frac{b^2}{4a}-c}_{=: w} \Big )\]
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\textbf{Case 1:} As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
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$P = (0, w)$ could possilby have three minima.
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Then compute:
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\begin{align}
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d_{P,{f_2}}(x) &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
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d_{P,{f_2}}(x) &= \sqrt{(x-0)^2 + (f_2(x)-w)^2}\\
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&= \sqrt{x^2 + (ax^2-w)^2}\\
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&= \sqrt{x^2 + a^2 x^4-2aw x^2+w^2}\\
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&= \sqrt{a^2 x^4 + (1-2aw) x^2 + w^2}\\
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&= \sqrt{\left (a^2 x^2 + \frac{1-2 a w}{2} \right )^2 + w^2 - (1-2 a w)^2}\\
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&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + (w^2 - (1-2 a w)^2)}\\
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&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + \big (w^2 - (1-2 a w)^2 \big)}
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\end{align}
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The term
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@ -357,37 +338,24 @@ For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \
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\textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
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\begin{align}
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d_{P,{f_2}}(x) &= \sqrt{(x-z)^2 + (f(x)-w)^2}\\
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&= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 - w^2)}\\
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&= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 - w^2}\\
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0 &\stackrel{!}{=} d_{P, {f_2}}'(x) \\
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&= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 + w^2)}\\
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&= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 + w^2}\\
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0 &\stackrel{!}{=} \Big(\big(d_{P, {f_2}}(x)\big)^2\Big)' \\
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&= 4a^2x^3 + 2(1- 2 aw)x +(- 2z)\\
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&= 2 \left (2a^2x^2 + (1- 2 aw) \right )x - 2z\\
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\Leftrightarrow z &\stackrel{!}{=} (2a^2x^2 + (1- 2 aw)) x\\
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\Leftrightarrow 0 &\stackrel{!}{=} 2 a^2 x^3 + (1- 2 aw)x - z\label{line:solution-equation}
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\Leftrightarrow 0 &\stackrel{!}{=} (2a^2x^2 + (1- 2 aw)) x - z\\
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&= 2 a^2 x^3 + (1- 2 aw) x - z\\
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\Leftrightarrow 0 &\stackrel{!}{=} x^3 + \underbrace{\frac{(1- 2 aw)}{2 a^2}}_{=: \alpha} x + \underbrace{\frac{-z}{2 a^2}}_{=: \beta}\\
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&= x^3 + \alpha x + \beta\label{eq:simple-cubic-equation-for-quadratic-distance}
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\end{align}
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The solution for this equation was computated with Wolfram|Alpha.
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I will only verify that the solution is correct. As there is only
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one solution in this case, we only have to check this one.
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The solution of Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
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is
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\[t := \sqrt[3]{\sqrt{3 \cdot (4a^3 + 27 b^2)} -9b}\]
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\[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} a }{t}\]
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\begin{align}
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t &:= \sqrt[3]{108a^4z + \sqrt{\num{11664}a^8 z^2 + 864 a^6 (1-2aw)^3}}\\
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x &= \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t}\\
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\xRightarrow{x \text{ in line }\ref{line:solution-equation}} 0 &\stackrel{!}{=} 2a^2 \left ( \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3 + (1-2aw) (\frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t}) - z\\
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&= 2a^2 \underbrace{\left ( \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3}_{=: \alpha} + \frac{t \cdot (1-2aw)}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2} (1-2aw)^2}{t} - z
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\end{align}
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\todo[inline]{verify this solution}
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Now compute $\alpha$. We know that $(a-b)^3 = a^3 - 3a^2 b +3ab^2 - b^3$:
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\begin{align}
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\alpha &= \left (\frac{t}{6 \sqrt[3]{2} a^2} \right )^3 - 3 \left (\frac{t}{6 \sqrt[3]{2} a^2} \right )^2 \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right ) + 3 \left (\frac{t}{6 \sqrt[3]{2} a^2} \right ) \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right )^2 - \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3\\
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&= \frac{t^3}{216 \cdot 2 \cdot a^6} - \frac{3t^2}{36 \sqrt[3]{4} a^4} \cdot \frac{\sqrt[3]{2}(1-2aw)}{t}
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+ \frac{3t}{6 \sqrt[3]{2} a^2} \cdot \frac{\sqrt[3]{4} (1-2aw)^2}{t^2} - \frac{2 (1-2aw)^3}{t^3}\\
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&= \frac{t^3}{432a^6} - \frac{t (1-2aw)}{12\sqrt[3]{2} a^4} + \frac{\sqrt[3]{2} (1-2aw)^2}{2ta^2} - \frac{2 (1-2aw)^3}{t^3}\\
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&= \frac{t^3}{432a^6} + \frac{t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2}{12\sqrt[3]{2} a^4} - \frac{2 (1-2aw)^3}{t^3}\\
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&= \frac{t^3- 2 (1-2aw)^3 \cdot 432a^6}{432a^6 t^3} + \frac{t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2}{12\sqrt[3]{2} a^4}\\
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&= \frac{\sqrt[3]{2}(t^3- 2 (1-2aw)^3 \cdot 432a^6) + 36 a^2 t^3 (t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2)}{432 \sqrt[3]{2} a^6 t^3}\\
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&= \frac{\sqrt[3]{2} t^3 - \sqrt[3]{2} (1-2aw)^3 \cdot 864a^6 + 36 a^2 t^3 (2awt^1-t^2 + 6 \sqrt[3]{4} a^2 (1-2aw)^2)}{6 \sqrt[3]{2} a^2 t \cdot 2a^2 \cdot 36 a^2 t^2}
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\end{align}
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\goodbreak
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So the solution is given by
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@ -421,15 +389,13 @@ $b, c, d \in \mdr$ be a function.
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-3, % start the diagram at this y-coordinate
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ymax= 3, % end the diagram at this y-coordinate
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-3, % start the diagram at this y-coordinate
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ymax= 3, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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%xticklabels={-2,-1.6,...,7},
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%yticklabels={-8,-7,...,8},
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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@ -458,6 +424,53 @@ $b, c, d \in \mdr$ be a function.
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%\todo[inline]{Write this}
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\subsection{Calculate points with minimal distance}
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\begin{theorem}
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There cannot be an algebraic solution to the problem of finding
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a closest point $(x, f(x))$ to a given point $P$ when $f$ is
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a polynomial function of degree $3$ or higher.
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\end{theorem}
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\begin{proof}
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Let $g : \mdr \rightarrow \mdr$ be a polynomial of degree 5
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\[g(x) = \tilde{a} x^5 + \tilde{b} x^4 + \tilde{c} x^3 + \tilde{d} x^2 + \tilde{e} x + \tilde{f}\]
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with $\tilde{a} \in \mdr_{> 0},\; \tilde{b} \in \mdr \setminus \Set{0}$ and $\tilde{c}, \tilde{d}, \tilde{e}, \tilde{f} \in \mdr$.
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Then, according to the Abel-Ruffini theorem, the equation
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\[g(x) = 0\]
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cannot be solved algebraicly.
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%But lets define $a := \frac{\sqrt{\tilde{a}}}{3}$, $b := \frac{\tilde{b}}{5a}$,
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%$c : = \frac{\frac{\tilde{c}}{2} - b^2}{2a}$, $d := \frac{\tilde{d}}{3} - bc + $
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So you can find $a, b, c, d, x_p, y_p$ such that
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\begin{align}
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g(x) &= \underbrace{3 a^2}_{= \tilde{a}} x^5 + \underbrace{5ab}_{= \tilde{b}}x^4 + \underbrace{2(2ac + b^2 )}_{= \tilde{c}}x^3 &+& \underbrace{3(ad+bc-ay_p)}_{= \tilde{d}} x^2 \\
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& &+& \underbrace{(2 b d+c^2+1-2 b y_p)}_{= \tilde{e}}x+\underbrace{c d-c y_p-x_p}_{= \tilde{f}}\\
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&= f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
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&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p
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\end{align}
|
||||
|
||||
And
|
||||
\begin{align}
|
||||
g(x) &\stackrel{!}{=}0\\
|
||||
\Leftrightarrow 0 &\stackrel{!}{=} 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
|
||||
&= -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
|
||||
&= \left ((x-x_p)^2 \right )' + \left ( (f(x) - y_p)^2 \right )'\\
|
||||
&= \left ((x-x_p)^2 + (f(x) - y_p)^2 \right )'\\
|
||||
&= (d_{P,f}(x)^2)'
|
||||
\end{align}
|
||||
|
||||
So the problem of finding a closest point $(x, f(x))$ on a
|
||||
cubic function $f$ to $P$ is essentially the same as finding
|
||||
a root of a polynomial function of degree 5. As this cannot
|
||||
be solved algebraicly, the problem of finding such a point
|
||||
can also not be solved algebraicly.$\qed$
|
||||
\end{proof}
|
||||
|
||||
\todo[inline]{Start with theorem that this problem is not solvable
|
||||
with analytics only. Use a general 5th degree function and show
|
||||
that it can be mapped to a $f$ and $P$ instance.}
|
||||
|
||||
When you want to calculate points with minimal distance, you can
|
||||
take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
|
||||
|
||||
|
|
@ -510,15 +523,13 @@ nicer situation, we can apply this approach for cubic functions.
|
|||
width=0.8\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=-3, % start the diagram at this y-coordinate
|
||||
ymax= 3, % end the diagram at this y-coordinate
|
||||
xmin=-3, % start the diagram at this x-coordinate
|
||||
xmax= 3, % end the diagram at this x-coordinate
|
||||
ymin=-3, % start the diagram at this y-coordinate
|
||||
ymax= 3, % 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,
|
||||
|
|
@ -569,6 +580,14 @@ chose the cubic function $f$ and $P$.
|
|||
I'm also pretty sure that there is no polynomial (no matter what degree)
|
||||
that has more than 3 solutions.}
|
||||
|
||||
\todo[inline]{If there is no closed form solution, I want to
|
||||
describe a numerical solution. I guess Newtons method might be good.}
|
||||
\section{Newtons method}
|
||||
\todo[inline]{When does Newtons method converge? How fast?
|
||||
How to choose starting point?}
|
||||
|
||||
\section{Quadratic minimization}
|
||||
\todo[inline]{TODO}
|
||||
|
||||
\section{Conclusion}
|
||||
\todo[inline]{TODO}
|
||||
|
||||
\end{document}
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue