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@ -196,7 +196,7 @@ 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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\begin{theorem}[local quadratic convergence of Newton's method\footnotemark]
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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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@ -211,6 +211,8 @@ if some conditions are met:
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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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\footnotetext{Translated from German to English from lecture notes of "Numerische Mathematik für die Fachrichtung Informatik
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und Ingenieurwesen" by Dr. Weiß, KIT}
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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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@ -226,6 +228,8 @@ initial guess.
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\subsubsection{Muller's method}
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Muller's method was first presented by David E. Muller in 1956.
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\todo[inline]{Paper? Might this be worth a try?}
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\subsubsection{Bisection method}
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The idea of the bisection method is the following:
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@ -242,6 +246,8 @@ Now three cases can occur:
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\item[Case 3] $\sgn(b) = \sgn(\frac{a+b}{2})$: Continue searching in $[a, \frac{a+b}{2}]$
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\end{enumerate}
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\todo[inline]{Which intervall can I choose? How would I know that there is exactly one root?}
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\subsubsection{Bairstow's method}
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Cite from Wikipedia:
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The algorithm first appeared in the appendix of the 1920 book "Applied Aerodynamics" by Leonard Bairstow. The algorithm finds the roots in complex conjugate pairs using only real arithmetic.
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