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{SECT 0 {PARA 260 "" 0 "" {TEXT 258 43 "NEWTON'S METHOD FOR SOLVING EQ
UATIONS  - II" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" 
{TEXT 257 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
256 19 "Calculus I  Project" }}{SECT 0 {PARA 3 "" 0 "" {TEXT 266 10 "O
bjective:" }}{PARA 0 "" 0 "" {TEXT -1 67 " To learn Newton's Method fo
r approximating solutions of equations." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 92 "Using Maple will help you to quickly
 compute the successive approximations of the solutions." }}}{SECT 0 
{PARA 3 "" 0 "" {TEXT 270 0 "" }{TEXT 256 14 "Solved Example" }{TEXT 
-1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
36 "Solve the equation  x = 1 + sin(2x)." }}}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 271 0 "" }{TEXT 
272 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 264 "There is no general technique to solve this equation and
 we must settle for an approximate solution. The method that we will a
pply is attributed to Newton and uses the idea that the tangent line t
o a curve closely approximates the curve near the point of tangency." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "
" }{TEXT 256 19 "The Newton's Method" }}{PARA 0 "" 0 "" {TEXT -1 150 "
Suppose we have a function  f  and we want to solve equation  f(x) = 0
. To use Newton's method we must make an initial approximation of the \+
solution, " }{XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 26 ". T
he next approximation, " }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" }
{TEXT -1 10 ",  is the " }}{PARA 0 "" 0 "" {TEXT -1 62 " x-intercept  \+
of the tangent line to  y =  f(x) at the point (" }{XPPEDIT 18 0 "x[0]
,f(x[0]);" "6$&%\"xG6#\"\"!-%\"fG6#&F$6#F&" }{TEXT -1 35 "). We need t
o find the formula for " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }
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ure,  " }{XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 14 "  is  5
  and  " }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" }{TEXT -1 16 "  is \+
about  4.3)" }}{PARA 12 "" 1 "" {TEXT -1 36 "The slope of the tangent \+
line is    " }{XPPEDIT 18 0 "Delta*y/(Delta*x);" "6#*(%&DeltaG\"\"\"%
\"yGF%*&F$F%%\"xGF%!\"\"" }{TEXT -1 10 "  =  f ' (" }{XPPEDIT 18 0 "x[
0];" "6#&%\"xG6#\"\"!" }{TEXT -1 6 ")  =  " }{XPPEDIT 18 0 "(f(x[0])-0
)/(x[0]-x[1]);" "6#*&,&-%\"fG6#&%\"xG6#\"\"!\"\"\"F+!\"\"F,,&&F)6#F+F,
&F)6#F,F-F-" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "Solving for
  " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 5 " :   " }
{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" }{TEXT -1 3 " = " }{XPPEDIT 
18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 6 " -  f(" }{XPPEDIT 18 0 "x[
0];" "6#&%\"xG6#\"\"!" }{TEXT -1 9 ") / f ' (" }{XPPEDIT 18 0 "x[0];" 
"6#&%\"xG6#\"\"!" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 38 "Now w
e repeat the process  to obtain  " }{XPPEDIT 18 0 "x[2];" "6#&%\"xG6#
\"\"#" }{TEXT -1 7 " from  " }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"
" }{TEXT -1 5 ",    " }{XPPEDIT 18 0 "x[3];" "6#&%\"xG6#\"\"$" }{TEXT 
-1 9 "  from   " }{XPPEDIT 18 0 "x[2];" "6#&%\"xG6#\"\"#" }{TEXT -1 6 
", etc." }}{PARA 0 "" 0 "" {TEXT -1 20 "Write a formula of  " }
{XPPEDIT 18 0 "x[2];" "6#&%\"xG6#\"\"#" }{TEXT -1 15 "  in terms of  \+
" }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" }{TEXT -1 3 ",  " }
{XPPEDIT 18 0 "x[3];" "6#&%\"xG6#\"\"$" }{TEXT -1 16 "  in terms  of  \+
" }{XPPEDIT 18 0 "x[2];" "6#&%\"xG6#\"\"#" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "It follows that   \+
          " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#,&%\"nG\"\"\"F(F(" }
{TEXT -1 5 "  =  " }{XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 
7 "  -  f(" }{XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 9 ") / \+
f ' (" }{XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 1 ")" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 97 "We will \+
apply this formula to solve our equation. We first put the equation in
 the form  f(x) = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 20 "f:=x->1+sin(2*x)-x;\n" }}{PARA 11 "" 1 "" 
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "df:=x->D(f)(x);\n" }}{PARA 
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{MPLTEXT 1 0 9 "x[0]:=2;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6
#\"\"!\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "x[1]:=g(x[0]
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$\"+p_eQ7!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"#$\"+IYT
&Q\"!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "for n from 2 to
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#\"\"$$\"+9ONx8!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"%$
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1 "" {XPPMATH 20 "6#$\"\"!F$" }}}{PARA 0 "" 0 "" {TEXT -1 33 "Yes, 1.3
77336877 is the solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 
1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "w[0]:=0;\n" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"wG6#\"\"!F'" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 41 "for n from 0 to 10 do w[n+1]:=g(w[n])od;\n" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"wG6#\"\"\"$!\"\"\"\"!" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>&%\"wG6#\"\"#$!+evLZS!#5" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>&%\"wG6#\"\"$$!+:NW(>#!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"wG6#\"\"%$\"*i6!eN!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"wG6#\"\"&$!+a;#\\;#!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"wG6#\"\"'$\"*NkKy\"!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"wG6#\"\"($!+!>g5;\"!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"wG6#\"\")$!+3mZjb!#5" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>&%\"wG6#\"\"*$\"+(4%y_^!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>&%\"wG6#\"#5$\"+[Bb')H!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"wG
6#\"#6$\"+eM()>b!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 106 "The values show no signs of stabilizing. Look at the \+
graph of the function to understand what is going on." }}{PARA 0 "" 0 
"" {TEXT -1 20 "The tangent line at " }{XPPEDIT 18 0 "x[0];" "6#&%\"xG
6#\"\"!" }{TEXT -1 246 " = 2 is a good approximation of the function n
ot only  at the point of tangency but also at  the  x-intercept of the
 function. On the other hand the tangent line for  x = 0 bears no rese
mblance to the function at the  x-intercept of the function." }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 3 "" 0 "" {TEXT -1 46 "_________
_____________________________________" }}{PARA 12 "" 1 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 0 "" }
{TEXT 259 0 "" }{TEXT 260 10 "ASSIGNMENT" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{SECT 0 {PARA 259 "" 0 "" {TEXT -1 0 "" }{TEXT 261 0 "" }{TEXT 
262 0 "" }{TEXT 263 0 "" }{TEXT 264 10 "Problem 1." }}{PARA 0 "" 0 "" 
{TEXT -1 56 " Using Newton's Method and Maple, solve the equation    \+
" }{XPPEDIT 18 0 "x^3-4*x^2-1 = 0;" "6#/,(*$%\"xG\"\"$\"\"\"*&\"\"%F(*
$F&\"\"#F(!\"\"F(F-\"\"!" }{TEXT -1 8 "  with  " }{XPPEDIT 18 0 "x[0];
" "6#&%\"xG6#\"\"!" }{TEXT -1 4 " = 5" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 265 10 "Problem 2." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 267 0 "" }{TEXT 268 39 " Go bac
k to the equation in Problem 1. " }}{PARA 0 "" 0 "" {TEXT 269 9 " a)  \+
Let " }{XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 85 " = 2. Wha
t seems to be happening? Sketch the first three iterations on the grap
h  of " }}{PARA 0 "" 0 "" {TEXT -1 11 "       y = " }{XPPEDIT 18 0 "x^
3-4*x^2-1" "6#,(*$%\"xG\"\"$\"\"\"*&\"\"%F'*$F%\"\"#F'!\"\"F'F," }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " b) Let " }{XPPEDIT 18 0 "
x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 19 " = 0. What happens?" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
256 10 "Problem 3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 44 "Solve x cos(x) = sin(x) + 1  with  0 < x < 2" }{XPPEDIT 
18 0 "pi;" "6#%#piG" }{TEXT -1 48 ". Plot the graph to find a good sta
rting point. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{MARK "14 2 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 
2 33 1 1 }