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find . -type f -name '*.md' -exec sed --in-place 's/[[:space:]]\+$//' {} \+
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226 lines
9.8 KiB
TeX
226 lines
9.8 KiB
TeX
\chapter{Quadratic functions}
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\section{Defined on $\mdr$}
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Let $f:\mdr \rightarrow \mdr, f(x) = a \cdot x^2 + b \cdot x + c$ with $a \in \mdr \setminus \Set{0}$ and
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$b, c \in \mdr$ be a quadratic 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=north west,
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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=-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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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,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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\addlegendentry{$f_4(x)=x^2+2x$}
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\addlegendentry{$f_5(x)=-x^2+6$}
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\end{axis}
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\end{tikzpicture}
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\caption{Quadratic functions}
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\end{figure}
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\subsection{Calculate points with minimal distance}
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In this case, $d_{P,f}^2$ is polynomial of degree $n^2 = 4$.
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We use Theorem~\ref{thm:fermats-theorem}:\nobreak
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\begin{align}
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0 &\overset{!}{=} (d_{P,f}^2)'\\
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&= 2x -2 x_p -2y_p f'(x) + \left (f(x)^2 \right )'\\
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&= 2x -2 x_p -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
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\Leftrightarrow 0 &\overset{!}{=} x -x_p -y_p f'(x) + f(x) \cdot f'(x) \rlap{\hspace*{3em}(divide by 2)}\label{eq:minimizingFirstDerivative}\\
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&= x -x_p -y_p (2ax+b) + (ax^2+bx+c)(2ax+b)\\
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&= 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)\\
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&= x -x_p -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)\label{eq:quadratic-derivative-eq-0}
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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. But not every solution of the
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equation given by Theorem~\ref{thm:fermats-theorem}
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has to be a solution to the given problem as you can see in
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Example~\ref{ex:false-positive}.
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\goodbreak
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\begin{example}\label{ex:false-positive}
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Let $a = 1, b = 0, c=-1, x_p= 0, y_p = 1$.
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So $f(x) = x^2 - 1$ and $P(0, 1)$.
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\begin{align}
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\xRightarrow{\text{Equation}~\ref{eq:quadratic-derivative-eq-0}} 0 &\stackrel{!}{=} 2x^3 - 3x\\
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&= x(2x^2-3)\\
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\Rightarrow x_{1,2} &= \pm \sqrt{\frac{3}{2}} \text{ and } x_3 = 0\\
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d_{P,f}(x_3) &= \sqrt{0^2 + (-1-1)^2} = 2\\
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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}\\
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&= \sqrt{\nicefrac{3}{2}+\nicefrac{1}{4}} \\
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&= \sqrt{\nicefrac{7}{4}}\\
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\end{align}
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This means $x_3$ is not a point of minimal distance, although
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$(d_{P,f}(x_3))' = 0$.
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\end{example}
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\subsection{Number of points with minimal distance}
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\begin{theorem}
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A point $P$ has either one or two points on the graph of a
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quadratic function $f$ that are closest to $P$.
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\end{theorem}
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\begin{proof}
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The number of closests points of $f$ cannot be bigger than 3, because
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Equation~\ref{eq:quadratic-derivative-eq-0} is a polynomial function
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of degree 3. Such a function can have at most 3 roots. As $f$ has
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at least one point on its graph, there is at least one point with
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minimal distance.
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In the following, I will do some transformations with $f = f_0$ and
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$P = P_0$. This will make it easier to calculate the minimal distance
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points. Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does
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not change the minimum distance. Furthermore, we can find the
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points with minimum distance on the moved situation and calculate
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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:\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. In consequence, if we want to move the whole graph by 1
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to the left, we have 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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&= ax^2 - bx + \nicefrac{b^2}{4a} + bx - \nicefrac{b^2}{2a} + c\\
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&= ax^2 -\nicefrac{b^2}{4a} + c
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\end{align}
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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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\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \Big (\underbrace{x_P+\frac{b}{2a}}_{=: z},\;\; \underbrace{y_P+\frac{b^2}{4a}-c}_{=: w} \Big )\]
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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-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 x^2 + \frac{1-2 a w}{2a} \right )^2 + w^2 - \left (\frac{1-2 a w}{2a} \right )^2}\\
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&= \sqrt{\left (a x^2 + \nicefrac{1}{2a}- w \right )^2 + \left (w^2 - \left (\frac{1-2 a w}{2a} \right )^2 \right)}
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\end{align}
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This means, the term
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\[a^2 x^2 + (\nicefrac{1}{2a}- w)\]
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has to get as close to $0$ as possilbe when we want to minimize
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$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
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For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{\frac{\frac{1}{2a} - w}{a}}$.
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$\qed$
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\end{proof}
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\subsection{Solution formula}
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We start with the graph that was moved so that $f_2 = ax^2$.
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\textbf{Case 1:} $P$ is on the symmetry axis, hence $x_P = - \frac{b}{2a}$.
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In this case, we have already found the solution. If $w = y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
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then there are two solutions:
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\[x_{1,2} = \pm \sqrt{\frac{\frac{1}{2a} - w}{a}}\]
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Otherwise, there is only one solution $x_1 = 0$.
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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_2(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{!}{=} \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^3 + (1- 2 aw)x \right ) - 2z\\
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\Leftrightarrow 0 &\stackrel{!}{=} 2a^2x^3 + (1- 2 aw) x - z\\
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\stackrel{a \neq 0}{\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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Let $t$ be defined as
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\[t := \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\]
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\subsubsection{Analyzing $t$}
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\input{analyzing-t.tex}
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\subsubsection{Solutions of $x^3 + \alpha x + \beta$}
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I will make use of the following identities:
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\begin{align*}
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(1-i \sqrt{3})^2 &= -2 (1+i \sqrt{3})\\
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(1+i \sqrt{3})^2 &= -2 (1-i \sqrt{3})\\
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(1 \pm i \sqrt{3})^3 &= -8\\
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(a-b)^3 &= a^3-3 a^2 b+3 a b^2-b^3
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\end{align*}
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\textbf{Case 2.1:}
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\input{quadratic-case-2.1}
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\goodbreak
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\textbf{Case 2.2:}
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\input{quadratic-case-2.2}
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\textbf{Case 2.3:}
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\input{quadratic-case-2.3}
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\goodbreak
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So the solution is given by
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\todo[inline]{NO! Currently, there are erros in the solution.
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Check $f(x) = x^2$ and $P=(-2,4)$. Solution should be $x_1 = -2$, but it isn't!}
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\begin{align*}
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x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
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w &:= y_P+\frac{b^2}{4a}-c \;\;\; \text{ and } \;\;\; z := x_P+\frac{b}{2a}\\
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\alpha &:= \frac{1- 2 aw}{2 a^2} \;\;\;\text{ and }\;\;\; \beta := \frac{-z}{2 a^2}\\
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t &:= \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\\
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\underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
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x_1 = +\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S \text{ and } &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c > \frac{1}{2a} \\
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x_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\
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x_1 = x_S &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c \leq \frac{1}{2a} \\
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x_1 = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} &\text{if } x_P \neq x_S
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\end{cases}
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\end{align*}
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\clearpage
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\section{Defined on a closed interval $[a,b] \subseteq \mdr$}
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Now the problem isn't as simple as with constant and linear
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functions.
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If one of the minima in $S_2(P,f)$ is in $[a,b]$, this will be the
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shortest distance as there are no shorter distances.
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\todo[inline]{
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The following IS WRONG! Can I include it to help the reader understand the
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problem?}
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If the function (defined on $\mdr$) has only one shortest distance
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point $x$ for the given $P$, it's also easy: The point in $[a,b]$ that
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is closest to $x$ will have the sortest distance.
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\[\underset{x\in[a,b]}{\arg \min d_{P,f}(x)} = \begin{cases}
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S_2(f, P) \cap [a,b] &\text{if } S_2(f, P) \cap [a,b] \neq \emptyset \\
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\Set{a} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x < a\\
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\Set{b} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x > b\\
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todo &\text{if } |S_2(f, P)| = 2 \text{ and } S_2(f, P) \cap [a,b] = \emptyset
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\end{cases}\]
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