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added section about newtons method

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Martin Thoma 2013-12-12 00:39:37 +01:00
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@ -16,6 +16,7 @@
\usepackage{framed}
\usepackage{nicefrac}
\usepackage{siunitx}
\usepackage{csquotes}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Define theorems %
@ -519,6 +520,9 @@ $b, c, d \in \mdr$ be a function.
\subsection{Another approach}
\todo[inline]{Currently, this is only an idea. It might be usefull
to move the cubic function $f$ such that $f$ is point symmetric
to the origin. But I'm not sure how to make use of this symmetry.}
Just like we moved the function $f$ and the point to get in a
nicer situation, we can apply this approach for cubic functions.
@ -576,11 +580,9 @@ because
&= 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
\end{align}
\todo[inline]{Which way to move might be clever?}
\subsection{Number of points with minimal distance}
As there is an algebraic equation of degree 5, there cannot be more
than 5 solutions.
As this leads to a polynomial of degree 5 of which we have to find
roots, there cannot be more than 5 solutions.
\todo[inline]{Can there be 3, 4 or even 5 solutions? Examples!
After looking at function graphs of cubic functions, I'm pretty
@ -590,9 +592,38 @@ 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.}
\section{Bisection method}
\section{Newtons method}
\todo[inline]{When does Newtons method converge? How fast?
How to choose starting point?}
One way to find roots of functions is Newtons method. It gives an
iterative computation procedure that can converge quadratically
if some conditions are met:
\begin{theorem}[local quadratic convergence of Newton's method]
Let $D \subseteq \mdr^n$ be open and $f: D \rightarrow \mdr^n \in C^2(\mdr)$.
Let $x^* \in D$ with $f(x^*) = 0$ and the Jaccobi-Matrix $f'(x^*)$
should not be invertable when evaluated at the root.
Then there is a sphere
\[K := K_\rho(x^*) = \Set{x \in \mdr^n | \|x- x^*\|_\infty \leq \rho} \subseteq D\]
such that $x^*$ is the only root of $f$ in $K$. Furthermore,
the elements of the sequence
\[ x_{n+1} = x_n - \frac{f'(x_n)}{f(x_n)}\]
are for every starting value $x_0 \in K$ again in $K$ and
\[\lim_{n \rightarrow \infty} x_k = x^*\]
Also, there is a constant $C > 0$ such that
\[\|x^* - x_{n+1} \| = C \|x^* - x_n\|^2 \text{ for } n \in \mathbb{N}_0\|\]
\end{theorem}
The approach is extraordinary simple. You choose a starting value
$x_0$ and compute
\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
As soon as the values don't change much, you are close to a root.
The problem of this approach is choosing a starting value that is
close enough to the root. So we have to have a \enquote{good}
initial guess.
\section{Quadratic minimization}
\todo[inline]{TODO}