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219 lines
8.8 KiB
TeX
219 lines
8.8 KiB
TeX
\chapter{Cubic functions}
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\section{Defined on $\mdr$}
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Let $f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$ be a cubic function
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with $a \in \mdr \setminus \Set{0}$ and
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$b, c, d \in \mdr$ be a function.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=south east,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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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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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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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] {x*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x*x};
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\addplot[domain=-3:3, thick,samples=50, blue] {x*x*x+2*x*x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x+x};
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\addlegendentry{$f_1(x)=x^3$}
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\addlegendentry{$f_2(x)=x^3 + x^2$}
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\addlegendentry{$f_2(x)=x^3 + 2 \cdot x^2$}
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\addlegendentry{$f_1(x)=x^3 + x$}
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\end{axis}
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\end{tikzpicture}
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\caption{Cubic functions}
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\end{figure}
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%
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%\section{Special points}
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%\todo[inline]{Write this}
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%
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%\section{Voronoi}
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%
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%For $b^2 \geq 3ac$
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%
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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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Suppose you could solve the closest point problem for arbitrary
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cubic functions $f = ax^3 + bx^2 + cx + d$ and arbitrary points $P = (x_P, y_P)$.
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Then you could solve the following problem for $x$:
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\begin{align}
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0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'\\
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&=-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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&= 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}} + x - x_p
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\end{align}
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General algebraic equations of degree 5 don't have a solution formula.\footnote{TODO: Quelle}
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Although here seems to be more structure, the resulting algebraic
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equation can be almost any polynomial of degree 5:\footnote{Thanks to Peter Košinár on \href{http://math.stackexchange.com/a/584814/6876}{math.stackexchange.com} for this one}
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\begin{align}
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0 &\stackrel{!}{=} f'(x) \cdot \left (f(x) - y_p \right ) + (x - x_p)\\
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&= \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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0 &\stackrel{!}{=} \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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\end{align}
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\begin{enumerate}
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\item For any coefficient $\tilde{a} \in \mdr_{> 0}$ of $x^5$ we can choose $a$ such that we get $\tilde{a}$.
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\item For any coefficient $\tilde{b} \in \mdr \setminus \Set{0}$ of $x^4$ we can choose $b$ such that we get $\tilde{b}$.
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\item With $c$, we can get any value of $\tilde{c} \in \mdr$.
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\item With $d$, we can get any value of $\tilde{d} \in \mdr$.
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\item With $y_p$, we can get any value of $\tilde{e} \in \mdr$.
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\item With $x_p$, we can get any value of $\tilde{f} \in \mdr$.
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\end{enumerate}
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The first restriction guaratees that we have a polynomial of
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degree 5. The second one is necessary, to get a high range of
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$\tilde{e}$.
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This means, that there is no solution formula for the problem of
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finding the closest points on a cubic function to a given point,
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because if there was one, you could use this formula for finding
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roots of polynomials of degree 5. $\qed$
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\end{proof}
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\subsection{Another approach}
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\todo[inline]{Currently, this is only an idea. It might be usefull
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to move the cubic function $f$ such that $f$ is point symmetric
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to the origin. But I'm not sure how to make use of this symmetry.}
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Just like we moved the function $f$ and the point to get in a
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nicer situation, we can apply this approach for cubic functions.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=south east,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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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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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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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] {x*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x-x};
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\addplot[domain=-3:3, thick,samples=50, blue, dotted] {x*x*x+2*x};
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\addplot[domain=-3:3, thick,samples=50, lime, dashed] {x*x*x+3*x};
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\addlegendentry{$f_1(x)=x^3$}
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\addlegendentry{$f_2(x)=x^3 + x$}
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\addlegendentry{$f_1(x)=x^3 - x$}
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\addlegendentry{$f_2(x)=x^3 + 2 \cdot x$}
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\addlegendentry{$f_2(x)=x^3 + 3 \cdot x$}
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\end{axis}
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\end{tikzpicture}
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\caption{Cubic functions with $b = d = 0$}
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\end{figure}
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First, we move $f_0$ by $\frac{b}{3a}$ to the right, so
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\[f_1(x) = ax^3 + \frac{b^2 (c-1)}{3a} x + \frac{2b^3}{27 a^2} - \frac{bc}{3a} + d \;\;\;\text{ and }\;\;\;P_1 = (x_P + \frac{b}{3a}, y_P)\]
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because
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\begin{align}
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f_1(x) &= a \left (x - \frac{b}{3a} \right )^3 + b \left (x-\frac{b}{3a} \right )^2 + c \left (x-\frac{b}{3a} \right ) + d\\
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&= a \left (x^3 - 3 \frac{b}{3a}x^2 + 3 (\frac{b}{3a})^2 x - \frac{b^3}{27a^3} \right )
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+b \left (x^2 - \frac{2b}{3a} x + \frac{b^2}{9a^2} \right )
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+c x - \frac{bc}{3a} + d\\
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&= ax^3 - bx^2 + \frac{b^2}{3a}x - \frac{b^3}{27 a^2}\\
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& \;\;\;\;\;\;+ bx^2 - \frac{2b^2}{3a}x + \frac{b^3}{9a^2}\\
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& \;\;\;\;\;\;\;\;\;\;\;\; + c x - \frac{bc}{3a} + d\\
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&= ax^3 + \frac{b^2}{3a}\left (1-2+c \right )x + \frac{b^3}{9a^2} \left (1-\frac{1}{3} \right )- \frac{bc}{3a} + d
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\end{align}
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\subsection{Number of points with minimal distance}
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As this leads to a polynomial of degree 5 of which we have to find
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roots, there cannot be more than 5 solutions.
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\todo[inline]{Can there be 3, 4 or even 5 solutions? Examples!
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After looking at function graphs of cubic functions, I'm pretty
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sure that there cannot be 4 or 5 solutions, no matter how you
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chose the cubic function $f$ and $P$.
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I'm also pretty sure that there is no polynomial (no matter what degree)
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that has more than 3 solutions.}
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\subsection{Interpolation and approximation}
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\subsubsection{Quadratic spline interpolation}
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You could interpolate the cubic function by a quadratic spline.
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\subsubsection{Bisection method}
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\todo[inline]{TODO}
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\subsubsection{Newtons method}
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One way to find roots of functions is Newtons method. It gives an
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iterative computation procedure that can converge quadratically
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if some conditions are met:
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\begin{theorem}[local quadratic convergence of Newton's method]
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Let $D \subseteq \mdr^n$ be open and $f: D \rightarrow \mdr^n \in C^2(\mdr)$.
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Let $x^* \in D$ with $f(x^*) = 0$ and the Jaccobi-Matrix $f'(x^*)$
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should not be invertable when evaluated at the root.
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Then there is a sphere
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\[K := K_\rho(x^*) = \Set{x \in \mdr^n | \|x- x^*\|_\infty \leq \rho} \subseteq D\]
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such that $x^*$ is the only root of $f$ in $K$. Furthermore,
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the elements of the sequence
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\[ x_{n+1} = x_n - \frac{f'(x_n)}{f(x_n)}\]
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are for every starting value $x_0 \in K$ again in $K$ and
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\[\lim_{n \rightarrow \infty} x_k = x^*\]
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Also, there is a constant $C > 0$ such that
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\[\|x^* - x_{n+1} \| = C \|x^* - x_n\|^2 \text{ for } n \in \mathbb{N}_0\|\]
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\end{theorem}
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The approach is extraordinary simple. You choose a starting value
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$x_0$ and compute
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\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
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As soon as the values don't change much, you are close to a root.
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The problem of this approach is choosing a starting value that is
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close enough to the root. So we have to have a \enquote{good}
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initial guess.
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\subsubsection{Quadratic minimization}
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\todo[inline]{TODO}
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\clearpage
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\section{Defined on a closed interval $[a,b] \subseteq \mdr$}
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