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PROBLEM SET 4\\[.1in]
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Math 505-Fall 2002\\
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\begin{enumerate}
\item 
 Let the scalar function $f(x)$ have a zero of multiplicity $p$
at $x = x^*$, i.e. $f^{(k)}(x^*) = 0$, $k=0,1,\ldots,p-1$;
$f^{(p)} \ne 0$.
  \begin{enumerate}
  \item Show that Newton's method converges linearly (for sufficiently close 
     $x^0$) with asymptotic convergence factor $(p-1)/p$.
  \item If the Newton iterates are converging to this root of multiplicity
     $p$ show how to estimate $p$ from the computed quantities.
  \item If the integer $p$ is known show that quadratic convergence can be
     restored by using the modified Newton's method:
     \[ x^{n+1} = x^n-p \frac{f(x^n)}{f^\prime(x^n)} \ . \]
  \end{enumerate}
\item
 To compute the square root of $a > 0$ Newton's method is applied to
  $x^2 - a = 0$. This yields the sequence:
     \[ x_{n+1} = \frac{1}{2} \left( x_n + \frac{a}{x_n}\right) \ ,
       \ n = 0,1, \ldots \ .  \]
  We consider three choices for the initial iterate $x_0$: \\
(i) $x_0 = \alpha $, (ii) $x_0 = \beta + a$, (iii) $x_0 = \gamma + 
     a \delta$. \\
   \begin{enumerate}
   \item Find the constants $\alpha,  \beta, \gamma, \delta$ so that
         $x_0$ in each case best approximates $\sqrt{a}$ in the 
         interval $1/16 \le a \le 1$ in the Chebyshev sense, i.e.
         so that 
\[ \max_{\frac{1}{16} \le a \le 1} \left| x_0 - \sqrt{a} \right|
= \mbox{minimal} \ . \]
   \item In each case estimate how many iterations it would take to 
         produce an approximation to $\sqrt{a}$ to full single
         precision in IEEE arithmetic. That is, so that
         \[\frac{x_n-\sqrt{a}}{\sqrt{a}} \le 2^{-24} \ , \ 
         \mbox{for all } a \in \left[ \frac{1}{16}, 1\right] \ . \]
   \end{enumerate}
\item 
     \begin{enumerate}
     \item With given points ${\bf x}^{n}$, $n = 0,1,2$ in the plane
(${\bf x} \in {\bf R}^2$) define ${\bf x}^{n+1}$ as the intersection of the
planes $z = L_1( x, y)$ and $z = L_2(x,y)$ 
with $z = 0$ where
$L_1(x,y)$ is the plane through the three points 
$({\bf x}^i,z^i_1)$, $i= n, n-1, n-2$ and $z^i_1 = f_1({\bf x}^i)$.
Similarly for $L_2(x,y)$ in terms of $z^i_2=f_2({\bf x}^i)$. 
Find the formula for the components of ${\bf x}^{n+1}$.
     \item When will this iteration procedure fail to converge to a root
           of the system $\left(f_1({\bf x}^i),f_2({\bf x}^i)\right)$?
           Is there a two-dimensional analog of the secant method
           that will always converge?
     \item Use this method to compute three iterates in atempting to
           solve:
  \[ (x+i y) = e^{(x+i y)} \ . \]
  Take ${\bf x}_0 = (0.4,1.4)$, ${\bf x}_1 = (0.2,1.4)$,
${\bf x}_2 = (0.3,1.1)$. Will this method always converge for this function?
     \end{enumerate}
\end{enumerate}
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