2.3: Secant Method
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- 96042
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Secant Method is second best to Newton’s Method, and is used when a faster convergence than Bisection is desired, but it is too difficult or impossible to take an analytical derivative of the function \(f(x)\). We write in place of \(f^{\prime}\left(x_{n}\right)\),
\[f^{\prime}\left(x_{n}\right) \approx \frac{f\left(x_{n}\right)-f\left(x_{n-1}\right)}{x_{n}-x_{n-1}} \nonumber \]
Starting the Secant Method requires a guess for both \(x_{0}\) and \(x_{1}\).
2.3.1. Estimate \(\sqrt{2}=1.41421356\) using Newton’s Method
The \(\sqrt{2}\) is the zero of the function \(f(x)=x^{2}-2\). To implement Newton’s Method, we use \(f^{\prime}(x)=2 x\). Therefore, Newton’s Method is the iteration
\[x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}} \nonumber \]
We take as our initial guess \(x_{0}=1\). Then
\[\begin{aligned} &x_{1}=1-\frac{-1}{2}=\frac{3}{2}=1.5 \\ &x_{2}=\frac{3}{2}-\frac{\frac{9}{4}-2}{3}=\frac{17}{12}=1.416667, \\ &x_{3}=\frac{17}{12}-\frac{\frac{17^{2}}{12^{2}}-2}{\frac{17}{6}}=\frac{577}{408}=1.41426 . \end{aligned} \nonumber \]
2.3.2. Example of fractals using Newton’s Method
Consider the complex roots of the equation \(f(z)=0\), where
\[f(z)=z^{3}-1 \nonumber \]
These roots are the three cubic roots of unity. With
\[e^{i 2 \pi n}=1, \quad n=0,1,2, \ldots \nonumber \]
the three unique cubic roots of unity are given by
\[1, \quad e^{i 2 \pi / 3}, \quad e^{i 4 \pi / 3} \nonumber \]
With
\[e^{i \theta}=\cos \theta+i \sin \theta, \nonumber \]
and \(\cos (2 \pi / 3)=-1 / 2, \sin (2 \pi / 3)=\sqrt{3} / 2\), the three cubic roots of unity are
\[r_{1}=1, \quad r_{2}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i, \quad r_{3}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i \nonumber \]
The interesting idea here is to determine which initial values of \(z_{0}\) in the complex plane converge to which of the three cubic roots of unity.
Newton’s method is
\[z_{n+1}=z_{n}-\frac{z_{n}^{3}-1}{3 z_{n}^{2}} \nonumber \]
If the iteration converges to \(r_{1}\), we color \(z_{0}\) red; \(r_{2}\), blue; \(r_{3}\), green. The result will be shown in lecture.