\begin{frame} \frametitle{Newton's Method} \begin{block}{Newton's Method} Let $f(x)$ be a function, and $x_1$ and approximation of a root $r$. \pause\medskip We compute a sequence $x_2,x_3,x_4,\ldots$ of approximations by \begin{talign} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \end{talign} \pause The hope is that $x_2,x_3,\ldots$ get closer and closer to the root $r$. \\\pause \onslide<18->{\alert{However, this does not always work.}} \end{block} \pause \only<-17>{ \begin{exampleblock}{} Let $x_1 = 2$. Find the 3rd approximation to the root of $x^2-1$. \pause \begin{talign} f'(x) &= 2x\\[-.75ex] \mpause[1]{x_2 &=} \mpause{ x_1 - \frac{f(x_1)}{f'(x_1)} } \mpause{= 2 - \frac{f(2)}{f'(2)} } \mpause{= 2 - \frac{3}{4} } \mpause{= \frac{5}{4} = 1.25 } \\[-.5ex] \mpause{x_3 &=} \mpause{ x_2 - \frac{f(x_2)}{f'(x_2)} } \mpause{= \frac{5}{4} - \frac{f(\frac{5}{4})}{f'(\frac{5}{4})} } \mpause{= \frac{5}{4} - \frac{\left(\frac{5}{4}\right)^2-1}{\frac{10}{4}} } \mpause{= \frac{41}{40} = 1.025 }\\[-3.75ex] \end{talign} \pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause The sequence $x_1,x_2,x_3,\ldots$ gets closer and closer to the root $1$. \end{exampleblock} } \only<18-25>{ \pause[18] \begin{exampleblock}{} Let $x_1 = 1$. Find the 2nd approximation to the root of $\sqrt[3]{x}$. \pause \begin{minipage}{.49\textwidth} \begin{talign} f'(x) &= \frac{1}{3\sqrt[3]{x^2}} \\ \mpause[1]{x_2 &=} \mpause{1 - \frac{f(1)}{f'(1)} } \mpause{= 1 - \frac{1}{\left(\frac{1}{3}\right)} } \mpause{= -2 } \end{talign} \pause\pause\pause\pause\pause \end{minipage}~% \begin{minipage}{.49\textwidth} \smallskip \begin{center} \scalebox{.8}{ \begin{tikzpicture}[default,baseline=1cm] \def\diabordery{.25cm} \def\diaborderx{.25cm} \diagram{-3}{2}{-1.4}{1.3}{1} \diagramannotatez \begin{scope}[ultra thick] \draw[cgreen,ultra thick] plot[smooth,domain=0:2,samples=200] function{x**(1/3.)}; \draw[cgreen,ultra thick] plot[smooth,domain=-3:0,samples=200] function{-(-x)**(1/3.)}; \node[include=cgreen] (r) at (0,0) {}; \node[anchor=south east] at (r) {$r$}; \end{scope} \draw[gray] (1,-.2) -- node[at start,below,black] {$x_1$} node[include=cred,at end] {} (1,1); \tangent{4.2cm}{1cm}{pow(\x,1/3)}{1} \draw[gray] (-2,-.2) -- node[at end,above,black] {$x_2$} (-2,.2); \end{tikzpicture} } \end{center} \end{minipage} \pause\vspace{.2ex} Note that $x_2 = -2$ is further away from the root $0$ than $x_1 = 1$. \end{exampleblock} } \only<26>{ \vspace{1.2cm} \begin{exampleblock}{} For more complicated examples see \begin{itemize} \item Chapter 4.8, Examples 1,2 and 3 \end{itemize} \end{exampleblock} } \vspace{10cm} \end{frame}