mirror of
https://github.com/MartinThoma/LaTeX-examples.git
synced 2025-04-25 06:18:05 +02:00
258 lines
11 KiB
TeX
258 lines
11 KiB
TeX
\chapter{Quadratic functions}
|
|
\section{Defined on $\mdr$}
|
|
Let $f(x) = a \cdot x^2 + b \cdot x + c$ with $a \in \mdr \setminus \Set{0}$ and
|
|
$b, c \in \mdr$ be a quadratic function.
|
|
|
|
\begin{figure}[htp]
|
|
\centering
|
|
\begin{tikzpicture}
|
|
\begin{axis}[
|
|
legend pos=north west,
|
|
axis x line=middle,
|
|
axis y line=middle,
|
|
grid = major,
|
|
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=-0.25, % start the diagram at this y-coordinate
|
|
ymax= 9, % end the diagram at this y-coordinate
|
|
axis background/.style={fill=white},
|
|
xlabel=$x$,
|
|
ylabel=$y$,
|
|
tick align=outside,
|
|
minor tick num=-3,
|
|
enlargelimits=true,
|
|
tension=0.08]
|
|
\addplot[domain=-3:3, thick,samples=50, red] {0.5*x*x};
|
|
\addplot[domain=-3:3, thick,samples=50, green] { x*x};
|
|
\addplot[domain=-3:3, thick,samples=50, blue] { x*x + x};
|
|
\addplot[domain=-3:3, thick,samples=50, orange,dotted] { x*x + 2*x};
|
|
\addplot[domain=-3:3, thick,samples=50, black,dashed] {-x*x + 6};
|
|
\addlegendentry{$f_1(x)=\frac{1}{2}x^2$}
|
|
\addlegendentry{$f_2(x)=x^2$}
|
|
\addlegendentry{$f_3(x)=x^2+x$}
|
|
\addlegendentry{$f_4(x)=x^2+2x$}
|
|
\addlegendentry{$f_5(x)=-x^2+6$}
|
|
\end{axis}
|
|
\end{tikzpicture}
|
|
\caption{Quadratic functions}
|
|
\end{figure}
|
|
|
|
\subsection{Calculate points with minimal distance}
|
|
In this case, $d_{P,f}^2$ is polynomial of degree 4.
|
|
We use Theorem~\ref{thm:required-extremum-property}:\nobreak
|
|
\begin{align}
|
|
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)\\
|
|
&= 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}
|
|
|
|
This is an algebraic equation of degree 3.
|
|
There can be up to 3 solutions in such an equation. Those solutions
|
|
can be found with a closed formula. But not every solution of the
|
|
equation given by Theorem~\ref{thm:required-extremum-property}
|
|
has to be a solution to the given problem.
|
|
|
|
\begin{example}
|
|
Let $a = 1, b = 0, c= 1, x_p= 0, y_p = 1$.
|
|
So $f(x) = x^2 + 1$ and $P(0, 1)$.
|
|
|
|
\begin{align}
|
|
0 &\stackrel{!}{=} 4 x^3 - 2x\\
|
|
&=2x(2x^2 - 1)\\
|
|
\Rightarrow x_1 &= 0 \;\;\; x_{2,3} = \pm \frac{1}{\sqrt{2}}
|
|
\end{align}
|
|
|
|
As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
|
|
\end{example}
|
|
|
|
|
|
\subsection{Number of points with minimal distance}
|
|
\begin{theorem}
|
|
A point $P$ has either one or two points on the graph of a
|
|
quadratic function $f$ that are closest to $P$.
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
In the following, I will do some transformations with $f = f_0$ and
|
|
$P = P_0$ .
|
|
|
|
Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does
|
|
not change the minimum distance. Furthermore, we can find the
|
|
points with minimum distance on the moved situation and calculate
|
|
the minimum points in the original situation.
|
|
|
|
First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
|
|
\[f_1(x) = ax^2 - \frac{b^2}{4a} + c \;\;\;\text{ and }\;\;\; P_1 = \left (x_p+\frac{b}{2a},\;\; y_p \right )\]
|
|
|
|
Because:\footnote{The idea why you subtract $\frac{b}{2a}$ within
|
|
$f$ is that when you subtract something from $x$ before applying
|
|
$f$ it takes more time ($x$ needs to be bigger) to get to the same
|
|
situation. So to move the whole graph by $1$ to the left whe have
|
|
to add $+1$.}
|
|
\begin{align}
|
|
f(x-\nicefrac{b}{2a}) &= a (x-\nicefrac{b}{2a})^2 + b (x-\nicefrac{b}{2a}) + c\\
|
|
&= a (x^2 - \nicefrac{b}{a} x + \nicefrac{b^2}{4a^2}) + bx - \nicefrac{b^2}{2a} + c\\
|
|
&= ax^2 - bx + \nicefrac{b^2}{4a} + bx - \nicefrac{b^2}{2a} + c\\
|
|
&= ax^2 -\nicefrac{b^2}{4a} + c
|
|
\end{align}
|
|
|
|
|
|
Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
|
|
\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \Big (\underbrace{x_P+\frac{b}{2a}}_{=: z},\;\; \underbrace{y_P+\frac{b^2}{4a}-c}_{=: w} \Big )\]
|
|
|
|
As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
|
|
$P = (0, w)$ could possilby have three minima.
|
|
|
|
Then compute:
|
|
\begin{align}
|
|
d_{P,{f_2}}(x) &= \sqrt{(x-0)^2 + (f_2(x)-w)^2}\\
|
|
&= \sqrt{x^2 + (ax^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{\left (a^2 x^2 + \frac{1-2 a w}{2} \right )^2 + w^2 - (1-2 a w)^2}\\
|
|
&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + \big (w^2 - (1-2 a w)^2 \big)}
|
|
\end{align}
|
|
|
|
The term
|
|
\[a^2 x^2 + (\nicefrac{1}{2}-a w)\]
|
|
should get as close to $0$ as possilbe when we want to minimize
|
|
$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
|
|
For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.
|
|
$\qed$
|
|
\end{proof}
|
|
|
|
\subsection{Solution formula}
|
|
We start with the graph that was moved so that $f_2 = ax^2$.
|
|
|
|
\textbf{Case 1:} $P$ is on the symmetry axis, hence $x_P = - \frac{b}{2a}$.
|
|
|
|
In this case, we have already found the solution. If $y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
|
|
then there are two solutions:
|
|
\[x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}\]
|
|
Otherwise, there is only one solution $x_1 = 0$.
|
|
|
|
\textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
|
|
\begin{align}
|
|
d_{P,{f_2}}(x) &= \sqrt{(x-z)^2 + (f(x)-w)^2}\\
|
|
&= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 + w^2)}\\
|
|
&= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 + w^2}\\
|
|
0 &\stackrel{!}{=} \Big(\big(d_{P, {f_2}}(x)\big)^2\Big)' \\
|
|
&= 4a^2x^3 + 2(1- 2 aw)x +(- 2z)\\
|
|
&= 2 \left (2a^2x^2 + (1- 2 aw) \right )x - 2z\\
|
|
\Leftrightarrow 0 &\stackrel{!}{=} (2a^2x^2 + (1- 2 aw)) x - z\\
|
|
&= 2 a^2 x^3 + (1- 2 aw) x - z\\
|
|
\Leftrightarrow 0 &\stackrel{!}{=} x^3 + \underbrace{\frac{1- 2 aw}{2 a^2}}_{=: \alpha} x + \underbrace{\frac{-z}{2 a^2}}_{=: \beta}\\
|
|
&= x^3 + \alpha x + \beta\label{eq:simple-cubic-equation-for-quadratic-distance}
|
|
\end{align}
|
|
|
|
Let $t$ be defined as
|
|
\[t := \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\]
|
|
|
|
\textbf{Case 2.1:} $4 \alpha^3 + 27 \beta^2 \geq 0$:
|
|
The solution of Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
|
|
is
|
|
\[x = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}\]
|
|
|
|
When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
|
|
you get:\footnote{Remember: $(a-b)^3 = a^3-3 a^2 b+3 a b^2-b^3$}
|
|
\allowdisplaybreaks
|
|
\begin{align}
|
|
0 &\stackrel{!}{=} \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right )^3 + \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
&= (\frac{t}{\sqrt[3]{18}})^3
|
|
- 3 (\frac{t}{\sqrt[3]{18}})^2 \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}
|
|
+ 3 (\frac{t}{\sqrt[3]{18}})(\frac{\sqrt[3]{\frac{2}{3}} \alpha }{t})^2
|
|
- (\frac{\sqrt[3]{\frac{2}{3}} \alpha }{t})^3
|
|
+ \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
&= \frac{t^3}{18}
|
|
- \frac{3t^2}{\sqrt[3]{18^2}} \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t}
|
|
+ \frac{3t}{\sqrt[3]{18}} \frac{\sqrt[3]{\frac{4}{9}} \alpha^2 }{t^2}
|
|
- \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
+ \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
&= \frac{t^3}{18}
|
|
- \frac{\sqrt[3]{18} t \alpha}{\sqrt[3]{18^2}}
|
|
+ \frac{\sqrt[3]{12} \alpha^2}{\sqrt[3]{18} t}
|
|
- \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
+ \alpha \left (\frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \right ) + \beta\\
|
|
&= \frac{t^3}{18}
|
|
- \frac{t \alpha}{\sqrt[3]{18}}
|
|
\color{red}+ \frac{\sqrt[3]{2} \alpha^2}{\sqrt[3]{3} t} \color{black}
|
|
- \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
+ \color{red}\alpha \color{black} \left (\frac{t}{\sqrt[3]{18}} \color{red}- \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} \color{black}\right )
|
|
+ \beta\\
|
|
&= \frac{t^3}{18} \color{blue}- \frac{t \alpha}{\sqrt[3]{18}} \color{black}
|
|
- \frac{\frac{2}{3} \alpha^3 }{t^3}
|
|
\color{blue}+ \frac{\alpha t}{\sqrt[3]{18}} \color{black}
|
|
+ \beta\\
|
|
&= \frac{t^3}{18} - \frac{\frac{2}{3} \alpha^3 }{t^3} + \beta\\
|
|
&= \frac{t^6 - 12 \alpha^3 + \beta 18 t^3}{18t^3}
|
|
\end{align}
|
|
|
|
Now only go on calculating with the numerator. Start with resubstituting
|
|
$t$:
|
|
\begin{align}
|
|
0 &= (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta)^2 - 12 \alpha^3 + \beta 18 (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta)\\
|
|
&= (\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)})^2 +(9\beta)^2 - 12 \alpha^3 -18\cdot 9\beta^2\\
|
|
&= 3 \cdot (4 \alpha^3 + 27 \beta^2) -81 \beta^2 - 12 \alpha^3\\
|
|
&= (4 \alpha^3 + 27 \beta^2) -27 \beta^2 - 4 \alpha^3\\
|
|
&= 0
|
|
\end{align}
|
|
|
|
\textbf{Case 2.2:} TODO
|
|
|
|
\[x = \frac{(1+i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
|
|
-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]
|
|
|
|
\textbf{Case 2.3:} TODO
|
|
|
|
\[x = \frac{(1-i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
|
|
-\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
|
|
|
|
\goodbreak
|
|
So the solution is given by
|
|
\todo[inline]{NO! Currently, there are erros in the solution.
|
|
Check $f(x) = x^2$ and $P=(-2,4)$. Solution should be $x_1 = -2$, but it isn't!}
|
|
\begin{align*}
|
|
x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
|
|
w &:= y_P+\frac{b^2}{4a}-c \;\;\; \text{ and } \;\;\; z := x_P+\frac{b}{2a}\\
|
|
\alpha &:= \frac{1- 2 aw}{2 a^2} \;\;\;\text{ and }\;\;\; \beta := \frac{-z}{2 a^2}\\
|
|
t &:= \sqrt[3]{\sqrt{3 \cdot (4 \alpha^3 + 27 \beta^2)} -9\beta}\\
|
|
\underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
|
|
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} \\
|
|
x_2 = -\sqrt{a (y_p + \frac{b^2}{4a} - c) - \frac{1}{2}} + x_S\\
|
|
x_1 = x_S &\text{if } x_P = x_S \text{ and } y_p + \frac{b^2}{4a} - c \leq \frac{1}{2a} \\
|
|
x_1 = \frac{t}{\sqrt[3]{18}} - \frac{\sqrt[3]{\frac{2}{3}} \alpha }{t} &\text{if } x_P \neq x_S
|
|
\end{cases}
|
|
\end{align*}
|
|
|
|
I call this function $S_2: \Set{\text{Quadratic functions}} \times \mdr^2 \rightarrow \mathcal{P}({\mdr})$.
|
|
\clearpage
|
|
|
|
\section{Defined on a closed interval $[a,b] \subseteq \mdr$}
|
|
Now the problem isn't as simple as with constant and linear
|
|
functions.
|
|
|
|
If one of the minima in $S_2(P,f)$ is in $[a,b]$, this will be the
|
|
shortest distance as there are no shorter distances.
|
|
|
|
\todo[inline]{
|
|
The following IS WRONG! Can I include it to help the reader understand the
|
|
problem?}
|
|
|
|
If the function (defined on $\mdr$) has only one shortest distance
|
|
point $x$ for the given $P$, it's also easy: The point in $[a,b]$ that
|
|
is closest to $x$ will have the sortest distance.
|
|
|
|
\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
|
|
S_2(f, P) \cap [a,b] &\text{if } S_2(f, P) \cap [a,b] \neq \emptyset \\
|
|
\Set{a} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x < a\\
|
|
\Set{b} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x > b\\
|
|
todo &\text{if } |S_2(f, P)| = 2 \text{ and } S_2(f, P) \cap [a,b] = \emptyset
|
|
\end{cases}\]
|