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solved quadratic case

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Martin Thoma 2013-12-05 23:27:17 +01:00
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documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex
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@ -14,6 +14,7 @@
\usepackage{tikz}
\usepackage[framed,amsmath,thmmarks,hyperref]{ntheorem}
\usepackage{framed}
\usepackage{nicefrac}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define theorems %
@ -24,6 +25,7 @@
\theorembodyfont{\normalfont} % nicht mehr kursiv
\def\mdr{\ensuremath{\mathbb{R}}}
\renewcommand{\qed}{\hfill\blacksquare}
\newframedtheorem{theorem}{Theorem}
\newframedtheorem{lemma}[theorem]{Lemma}
@ -271,7 +273,7 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak
&= -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)
&= 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}
%\begin{align}
@ -304,11 +306,24 @@ As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
\subsection{Number of points with minimal distance}
It is obvious that a quadratic function can have two points with
minimal distance.
\subsubsection{Two points with minimal distance}
Quadratic functions can have two points with minimal distance.
For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} = (\sqrt{\frac{9}{2}}, \frac{9}{2})$
has minimal distance to $P$, but also $P_{f,2} = (-\sqrt{\frac{9}{2}}, \frac{9}{2})$.
has minimal distance to $P$, but also $P_{f,2} = (-\sqrt{\frac{9}{2}}, \frac{9}{2})$:
\begin{proof}
\begin{align}
d_{P,f}(x) &= \sqrt{(x-x_P)^2 + (f(x)-y_p)^2}\\
&= \sqrt{x^2 + (x^2-5)^2}\\
&= \sqrt{x^2 + x^4-10x^2+25}\\
&= \sqrt{x^4 -9x^2 + 25}\\
&= \sqrt{x^4 -9x^2 + \frac{81}{4}+\frac{19}{4}}\\
&= \sqrt{\left (x^2 - \frac{9}{2} \right )^2 + \frac{19}{4}}
\end{align}
Obviously, $d_{P,f}$ is minimal for $x = \pm \sqrt{\frac{9}{2}} \qed$
\end{proof}
\begin{figure}[htp]
\centering
@ -344,6 +359,7 @@ has minimal distance to $P$, but also $P_{f,2} = (-\sqrt{\frac{9}{2}}, \frac{9}{
\caption{Two points with minimal distance}
\end{figure}
\subsubsection{Three points with minimal distance}
As discussed before, there cannot be more than 3 points on the graph
of $f$ next to $P$.
@ -390,6 +406,40 @@ $-\frac{b}{2a}$}
\caption{3 points with minimal distance?}
\end{figure}
When move the $f$ and $P$ simultaneously in $x$ direction, you will not change the
results.
First of all, you move $f_0$ by $\frac{b}{2a}$, 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:
\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:
\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]
As $f(x) = ax^2$ is symmetric to the $y$ axis, only points
$P = (0, y_p)$ could possilby have three minima.
Then compute:
\begin{align}
d_{P,f}(x) &= \sqrt{(x-x_P)^2 + (f(x)-y_p)^2}\\
&= \sqrt{x^2 + (ax^2-y_p)^2}\\
&= \sqrt{x^2 + a^2 x^4-2ay_p x^2+y_p^2}\\
&= \sqrt{a^2 x^4 + (1-2ay_p) x^2 + y_p^2}\\
&= \sqrt{\left (a^2 x^2 + \frac{1-2 a y_p}{2} \right )^2 + y_p^2 - (1-2 a y_p)^2}\\
&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a y_p \right )^2 + (y_p^2 - (1-2 a y_p)^2)}\\
\end{align}
For $y_p \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
For all other points, there are exactly two minima.
\clearpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Cubic %