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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 257 37 "Newton's Method and the
 Secant Method" }}{PARA 0 "" 0 "" {TEXT 256 26 "Mth 351 Oct 3 2001 Map
le 6" }}{PARA 0 "" 0 "" {TEXT 258 16 "Bent E. Petersen" }}{PARA 0 "" 
0 "" {TEXT 259 29 "Filename: 351f2001_newton.mws" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 99 "The four problems below \+
constitute Assignment 1. Your solutions (carefully written) are due Oc
t 10." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "Newton's method and the \+
secant method are just iterative schemes for the functions GN and GS, \+
respectively, where" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "GN:=(f,x)->x-f(x)/(D(f)(x));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#GNGR6$%\"fG%\"xG6\"6$%)operatorG%&arrowGF),&
9%\"\"\"*&-9$6#F.F/--%\"DG6#F2F3!\"\"F8F)F)F)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 41 "GS:=(f,x,xp)->x-f(x)*(x-xp)/(f(x)-f(xp));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#GSGR6%%\"fG%\"xG%#xpG6\"6$%)operato
rG%&arrowGF*,&9%\"\"\"*&*&-9$6#F/F0,&F/F09&!\"\"F0F0,&F3F0-F46#F7F8F8F
8F*F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 356 "Note with these definitions Maple will attempt symbolic \+
evaluations. Even for very simple functions Maple will rapidly grind t
o a halt. Try to force floating point evaluation to keep things reason
able. For example write the initial points in floating point format, t
hat is 1.0 rather than 1, so Maple will use floating point arithmetic \+
at Digits precision." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "f:=x->x^2-2;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)operatorG%&arrowGF(,&*$)9$\"\"#\"
\"\"F1F0!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Digi
ts:=36: GN(f,1.0); for k from 1 to 5 do GN(f,%); od;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#$\"E+++++++++++++++++:!#N" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"Enmmmmmmmmmmmmmmm;9!#N" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"ER!)4XF'o:#R!)4XF'o:UT\"!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#$\"E\"\\8!*)ybHE1\"**ouBc8UT\"!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#$\"EDID]B'*o,)[]4tBc8UT\"!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"E3
)p4Us)o,)[]4tBc8UT\"!#N" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "
a[0]:=0: a[1]:=1.0: for k from 2 to 14 do a[k]:=GS(f,a[k-1],a[k-2]); o
d;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"#$\"E++++++++++++++
+++?!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"$$\"ELLLLLLLLL
LLLLLLLL8!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"%$\"E++++
+++++++++++++9!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"&$\"
Ej9MYTj9MYTj9MYTj99!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"
\"'$\"EV*GH.`A5L<+([ZQ9@99!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"
aG6#\"\"($\"EyQ%QeCM\"GEY?t0iN@99!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>&%\"aG6#\"\")$\"E;?6+5$41#fG&4tBc8UT\"!#N" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"aG6#\"\"*$\"Ez))pt(p)o,)[]4tBc8UT\"!#N" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#5$\"E3)p4Us)o,)[]4tBc8UT\"!#N" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#6$\"E3)p4Us)o,)[]4tBc8UT
\"!#N" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#7$\"\"\"%*undefin
edG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#8$\"\"\"%*undefined
G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#9$\"\"\"%*undefinedG
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
164 "What happened? The successive iterates are so close that we are t
rying to evaluate 0/0 (within the specified precision). Let's increase
 the precision and try again." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "Digits:=69:a[0]:=0: a[1]:=1.
0: for k from 2 to 14 do a[k]:=GS(f,a[k-1],a[k-2]); od;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"#$\"`o+++++++++++++++++++++++++++++
+++++#!#o" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"$$\"`oLLLLLL
LLLLLLLLLLLLLLLLLLLLLLLLLLLL\"!#o" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
&%\"aG6#\"\"%$\"`o+++++++++++++++++++++++++++++++++S\"!#o" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"&$\"`o:MYTj9MYTj9MYTj9MYTj9MYTj9MY
Tj9MYT\"!#o" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"'$\"`o>'fG
b`$[t)[M8*\\ci2G%*GH.`A5L<+([ZQ9@99!#o" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>&%\"aG6#\"\"($\"`oD!)o)*)*)zl.;%R^B7YZz(Q%QeCM\"GEY?t0iN@99!#o
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\")$\"`o<(zo!HH#Gm&QhKH
>?[d,7,+J41#fG&4tBc8UT\"!#o" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG
6#\"\"*$\"`o:bA]vDT\\tMT`2OEU&y))pt(p)o,)[]4tBc8UT\"!#o" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#5$\"`oGD#fB#f.6keeglp'p&y!)p4Us)o,)[
]4tBc8UT\"!#o" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#6$\"`ov!*
ztzm<t![pPv=np&y!)p4Us)o,)[]4tBc8UT\"!#o" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>&%\"aG6#\"#7$\"`ot!*ztzm<t![pPv=np&y!)p4Us)o,)[]4tBc8UT\"!#o" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"#8$\"`ot!*ztzm<t![pPv=np&y
!)p4Us)o,)[]4tBc8UT\"!#o" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#
\"#9$\"\"\"%*undefinedG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 411 "In this example (and generally) the seca
nt method converges more slowly than Newton's method, but still conver
ges very rapidly indeed. For simple roots Newton's method is quadratic
 and so roughly doubles the number of significant figures at each iter
ation (after a while). The secant method has exponent roughly 1.6 and \+
so multiplies the number of significant figures at each step (after a \+
while) by about 1.6 ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=24:" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Consider now p(x) = \+
" }{XPPEDIT 18 0 "x^4-5*x^3+7*x^2-15*x+36;" "6#,,*$%\"xG\"\"%\"\"\"*&
\"\"&F'*$F%\"\"$F'!\"\"*&\"\"(F'*$F%\"\"#F'F'*&\"#:F'F%F'F,\"#OF'" }
{TEXT -1 80 "  which has a double root at  3. With initial guess  2.0 \+
 Newton's method yields" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "p:=x->x^4-5*x^3+7*x^2-15*x+36;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pGR6#%\"xG6\"6$%)operatorG%&arrowG
F(,,*$)9$\"\"%\"\"\"F1*&\"\"&F1)F/\"\"$F1!\"\"*&\"\"(F1)F/\"\"#F1F1*&
\"#:F1F/F1F6\"#OF1F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 
"GN(p,2.0); for k from 1 to 6 do GN(p,%); od;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"9nmmmmmmmmmmE!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
$\"9#\\sY<S_?dhr%G!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"9!yu^xr+#y
xGEH!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"94,?:y#R\")zaP'H!#B" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"97;9;PY()QI-#)H!#B" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"9rd)e`5&3.r/\"*H!#B" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"9O!)ebMAfZB`&*H!#B" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 196 "This is much slower convergenc
e than in the case of a single root - in fact, the convergence is only
 linear. If we use the modified Newton's method, GN2() we will have qu
adratic convergence again." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "GN2:=(f,x)->x-2*f(x)/(D(f)(x));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$GN2GR6$%\"fG%\"xG6\"6$%)operatorG%&
arrowGF),&9%\"\"\"*&*&\"\"#F/-9$6#F.F/F/--%\"DG6#F4F5!\"\"F:F)F)F)" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "GN2(p,2.0); for k from 1 to
 6 do GN2(p,%); od;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"9LLLLLLLLLLL
L!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"9*41*Rb98Wif@I!#B" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"98))Rd9Co875+I!#B" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"9'>n!zO3C-+++I!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#$\"9M$*\\_;++++++I!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"95-N]J#y.
++++$!#B" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"9!3%zSu************H!#B
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
324 "In like manner we have modified Newton methods GNk() where k  is \+
an integer greater than or equal to 1 and GNk() is defined in the obvi
ous way. We then have quadratic convergence in a neighborhood of a roo
t of multiplicity  k.  Unfortunately, we usually have only linear conv
ergence near roots of multiplicities other than k" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 10 "Problem 1." }{TEXT -1 
96 " Investigate (numerically) the behavior of GS2() near a double roo
t and near a simple root. Here" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "GS2:=(f,x,xp)->x-2*f(x)*(x-x
p)/(f(x)-f(xp));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$GS2GR6%%\"fG%\"
xG%#xpG6\"6$%)operatorG%&arrowGF*,&9%\"\"\"*&*(\"\"#F0-9$6#F/F0,&F/F09
&!\"\"F0F0,&F4F0-F56#F8F9F9F9F*F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "Apply  GS2()  to the polynomial
  p(x)  defined above. Also apply  GS(). Compare the results." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 10 "P
roblem 2." }{TEXT -1 26 " Apply Newton's method to " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 14 "p:=x->x^2-x-1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"pGR6#%\"xG6\"6$%)operatorG%&arrowGF(,(*$)9$\"\"#\"
\"\"F1F/!\"\"F1F2F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "to est
imate " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "(sqrt(5)+1)/2: %=
evalf(%,40);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&*$-%%sqrtG6#\"\"&\"
\"\"#F*\"\"#F+F*$\"I?x6QcOMoe/#[[*)\\())R.=;!#R" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 10 "Problem 3." }{TEXT 
-1 87 "  Show  f(x) = x + cos(x)  has a unique root. Show the root lie
s in the interval from  " }{XPPEDIT 18 0 "-Pi/2;" "6#,$*&%#PiG\"\"\"\"
\"#!\"\"F(" }{TEXT -1 112 "  to  0. Compare the bisection method, Newt
on's method (initial guess 0) and the secant method (initial guesses \+
" }{XPPEDIT 18 0 "-Pi/2;" "6#,$*&%#PiG\"\"\"\"\"#!\"\"F(" }{TEXT -1 
71 "  and 0) for estimating this root. According to Maple the root is \+
about" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Digits:=40: alpha \+
= fsolve(x+cos(x)=0,x); Digits:=10:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#/%&alphaG$!IM,/M(Qn(37`lT1;:K8&3R(!#S" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 69 "For Newton's method and the secant method compute the act
ual error  |" }{XPPEDIT 18 0 "alpha-x[n];" "6#,&%&alphaG\"\"\"&%\"xG6#
%\"nG!\"\"" }{TEXT -1 37 "| at each step and compare it with  |" }
{XPPEDIT 18 0 "x[n+1]-x[n];" "6#,&&%\"xG6#,&%\"nG\"\"\"F)F)F)&F%6#F(!
\"\"" }{TEXT -1 2 "|." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 263 11 "Problem 4. " }{TEXT -1 61 " Apply Newton
's method with initial guess 0.5 to the function" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 18 "h:=x->1-1/(1+x^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"hGR6#%\"xG6\"6$%)operatorG%&arrowGF(,&\"\"\"F-*&F-F
-,&F-F-*$)9$\"\"#F-F-!\"\"F4F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 31 "Why is the convergence so slow?" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0 0" 0 }
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