{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }1 0 0 0 6 6 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 4 4 0 0 0 0 0 0 -1 0 } {PSTYLE "Bullet Item" 0 15 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 3 3 0 0 0 0 0 0 15 2 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 1 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "" 18 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 4 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 4 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE " " 4 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 4 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 4 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 4 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 33 "Solving f(a) = b: Newton's Meth od" }}{PARA 257 "" 0 "" {TEXT -1 9 "Jim Herod" }}{PARA 258 "" 0 "" {TEXT -1 22 "School of Mathematics " }}{PARA 259 "" 0 "" {TEXT -1 12 " Georgia Tech" }}{PARA 260 "" 0 "" {TEXT -1 22 "Atlanta, Ga 30332 0160 " }}{PARA 261 "" 0 "" {TEXT -1 3 "USA" }}{PARA 262 "" 0 "" {TEXT -1 21 "herod@math.gatech.edu" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 12 "Background: " } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 28 "Equations for tangent lines." }}{PARA 0 "" 0 "" {TEXT -1 141 "Suppose that f is a function with derivative, f '(a), at the poin t a in its domain. Then the line tangent to the graph of f at a has eq uation" }}{PARA 0 "" 0 "" {TEXT -1 62 " \+ y = f '(a) (x - a) + f(a)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 95 "If f '(a) 0, this line crosses the x-axi s where y = 0 and, at this point, x = a - f(a)/f '(a)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 27 "Explanation of the Problem:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 562 "We try to find a number c so that f(c) = 0. The beginning of Newton's method is to make a guess for wha t the value of c might be. Perhaps the guess will be based on the appe arance of the graph, or on the context in which the problem arose. The idea of Newton's method is the construction of a procedure for findin g an improved guess for what c should be. The choice for an improved g uess is the point at which the the tangent line intersects the x-axis. We will see that this guess does not always work. However the followi ng picture makes the method believable." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f:=x->exp(x)-2;\nxo:=3;\nm:= D(f)(xo);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "J:=plot([x,f( x),x=0..3.5],color=BLACK):\nK:=plot([[xo,0],[xo,f(xo)]],color=RED):\nL :=plot([x,m*(x-xo)+f(xo),x=1.5..3.5],color=BLUE):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 17 "display(\{J,K,L\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 222 "The continuation of the \+ procedure is to take this intersection of the tangent line as a second guess, and repeat the process. The goal is to repeat the process unti l a zero of f is approximated within the accuracy desired." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 27 "Illustration of \+ the Method:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 259 " We define a function f for which we will approximate a ze ro and graph three iterates. First, we define the function f for which we want to find a zero and we define from this f a function that comp utes the point where the tangent line crosses the x-axis." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "f:=x->exp(x)-2;\ng:=evalf@unapply(a -f(a)/D(f)(a),a);" }}}{PARA 0 "" 0 "" {TEXT -1 176 "We generate a sequ ence of iterates which we call x[p], The number x[0] is the first gues s for a zero. The number x[p] is the p-th approximation, as computed b y Newton's method." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "x[0]:=3 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x[1]:=g(x[0]);x[2]:=g( x[1]);" }}}{PARA 0 "" 0 "" {TEXT -1 175 "One could continue this proce ss to compute as many interates as desired , We draw a graph to illust rate what has been done. Different iterates are drawn with different c olors." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "p:=plot(f(x),x=-0. .3,color=BLACK):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 167 "q1:=x0 ->plot([[x0,0],[x0,f(x0)],[g(x0),0]],color=RED);\nq2:=x0->plot([[x0,0] ,[x0,f(x0)],[g(x0),0]],color=BLUE);\nq3:=x0->plot([[x0,0],[x0,f(x0)],[ g(x0),0]],color=GREEN);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 " display(\{p,q1(x[0]),q2(x[1]),q3(x[2])\});" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 259 26 "Construction of a routine:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 " We could make a procedure which will continue the Newton iteration process until one of four th ings has occurred:" }}{PARA 15 "" 0 "" {TEXT -1 153 "A pre-set number \+ of itereations has been performed, even though the distance between su ccessive iterates did not acheived the specified error tollerance." }} {PARA 15 "" 0 "" {TEXT -1 61 "The difference between successive iterat es fails to decrease." }}{PARA 15 "" 0 "" {TEXT -1 96 "The tangent lin e for one iterate has slope so small that this tangent line is nearly \+ horizontal." }}{PARA 15 "" 0 "" {TEXT -1 51 "A zero has been obtained \+ within specified accuracy." }}{PARA 0 "" 0 "" {TEXT -1 49 "This last r equirement achieves the desired goal. " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 24 "Exer cise for the reader:" }}{PARA 0 "" 0 "" {TEXT -1 26 "Find a number x s uch that " }{XPPEDIT 18 0 "exp(2*x)=3*x+3/2*(1-ln(3/2)" "6#/-%$expG6#* &\"\"#\"\"\"%\"xGF),&*&\"\"$F)F*F)F)*(F-F)F(!\"\",&F)F)-%#lnG6#*&F-F)F (F/F/F)F)" }{TEXT -1 2 ".." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 20 "Rate of convergence:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 127 "Suppose f is a function \+ which has two derivatives on an interval containing the zero c of f. \+ Suppose also that f '(c) 0. If " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\" \"\"" }{TEXT -1 79 " is sufficiently close to c, then repeated applica tion of the Newton iterations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 35 " " } {XPPEDIT 18 0 "z[n+1]=z[n]" "6#/&%\"zG6#,&%\"nG\"\"\"F)F)&F%6#F(" } {TEXT -1 5 " - f(" }{XPPEDIT 18 0 "z[n]" "6#&%\"zG6#%\"nG" }{TEXT -1 6 ")/f '(" }{XPPEDIT 18 0 "z[n]" "6#&%\"zG6#%\"nG" }{TEXT -1 1 ")" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "yields a \+ sequence of approximations " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 15 ", ... for which" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " | " }{XPPEDIT 18 0 "z[n+1]" "6#&%\"zG6#,&%\"nG \"\"\"F(F(" }{TEXT -1 14 " - c | " }{XPPEDIT 18 0 "(1/2)^n" "6# )*&\"\"\"F%\"\"#!\"\"%\"nG" }{TEXT -1 5 " | " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 7 " - c |." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 21 "Pat hological Example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "Look at the graph of f(x) = " }{XPPEDIT 18 0 "x*exp(-x^2 )" "6#*&%\"xG\"\"\"-%$expG6#,$*$F$\"\"#!\"\"F%" }{TEXT -1 80 ". It is seen from the graph that zero is the only number c such that f(c) = 0 . " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "with(plots):\nplot(x*e xp(-x^2),x=-2..2);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 " " {TEXT -1 185 " According to the statement for convergence rates \+ given above, there is an interval about zero for which the Newton sche me converges to a root. We illustrate this in the following. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f:=x->x*exp(-x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "g :=evalf@unapply(x-f(x)/D(f)(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x[0]:=1/3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x[1]:=g(x[0]);x[2]:=g(x[1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "p:=plot(f(x),x=-1/2..1/2,color=BLACK):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 170 "q1:=x0->plot([[x0,0],[x0,f(x0)],[g (x0),0]],color=BLACK):\nq2:=x0->plot([[x0,0],[x0,f(x0)],[g(x0),0]],col or=BLACK):\nq3:=x0->plot([[x0,0],[x0,f(x0)],[g(x0),0]],color=BLACK):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "display(\{p,q1(x[0]),q2(x [1]),q3(x[2])\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 " For larger initial guess, we guess that the iterates might increase without bound." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x[0]:=.52;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x[1]:=g (x[0]);x[2]:=g(x[1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "p: =plot(f(x),x=-3..3,color=BLACK):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 168 "q1:=x0->plot([[x0,0],[x0,f(x0)],[g(x0),0]],color=BLA CK):\nq2:=x0->plot([[x0,0],[x0,f(x0)],[g(x0),0]],color=RED):\nq3:=x0-> plot([[x0,0],[x0,f(x0)],[g(x0),0]],color=GREEN):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plots[display](\{p,q1(x[0]),q2(x[1]),q3(x[2]) \});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 301 " We see from this \+ function and from these two examples of starting values that the quest ion of convergence with a specified initial guess has several posibili ties. The posibilities will divide the positive numbers into 4 subinte rvals. We will see that these intervals can be delineated as follows " }}{PARA 15 "" 0 "" {TEXT -1 55 "starting with initial value x in the interval 0 < x < " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 53 " produces an iteration scheme that converges to zero." }}{PARA 15 "" 0 "" {TEXT -1 46 "starting with initial value x in the interval \+ " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 7 " < x < " } {XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 59 " produces an iter ation scheme that increases without bound." }}{PARA 15 "" 0 "" {TEXT -1 47 "starting with intitial value x in the interval " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 7 " < x < " }{XPPEDIT 18 0 "z[3] " "6#&%\"zG6#\"\"$" }{TEXT -1 60 " produces an interation scheme that \+ decreases without bound." }}{PARA 15 "" 0 "" {TEXT -1 38 "starting wit h initial value such that " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6#\"\"$" } {TEXT -1 63 " < x produces an iteration scheme that increases without \+ bound." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 29 "We find each of these points:" }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 15 "" 0 "" {TEXT -1 10 "The poi nt " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 26 " has the p roperty that if " }{XPPEDIT 18 0 "x[0]" "6#&%\"xG6#\"\"!" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 6 " then " } {XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 103 ". That is, starting at this initial value leads to a two-cycle for the iterates, Initial points i n (0, " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 19 ") conve rge to zero." }}{PARA 15 "" 0 "" {TEXT -1 10 "The point " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 26 " has the property that if \+ " }{XPPEDIT 18 0 "x[0]" "6#&%\"xG6#\"\"!" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 11 " then f ' (" }{XPPEDIT 18 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT -1 44 ") = 0. Starting at this initi al value makes " }{XPPEDIT 18 0 "x[3]" "6#&%\"xG6#\"\"$" }{TEXT -1 31 " undefined. Initial points in (" }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\" \"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 51 ") leads to a sequence that increases without bound." }}{PARA 15 "" 0 "" {TEXT -1 10 "The point " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6# \"\"$" }{TEXT -1 26 " has the property that if " }{XPPEDIT 18 0 "x[0] " "6#&%\"xG6#\"\"!" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6 #\"\"$" }{TEXT -1 10 " then f '(" }{XPPEDIT 18 0 "x[0]" "6#&%\"xG6#\" \"!" }{TEXT -1 28 ") = 0. Initial points in (" }{XPPEDIT 18 0 "z[2] " "6#&%\"zG6#\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6# \"\"$" }{TEXT -1 36 ") eventually decrease without bound." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 " Note that becau se of the change in character for the iterates on both sides of " } {XPPEDIT 18 0 "z[p]" "6#&%\"zG6#%\"pG" }{TEXT -1 158 ", it is not suff icient to approximate any of these bifurcation points. They must be fo und exactly; that is, the bifurcation points must be found analyticall y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 " \+ We first find " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 27 ". With g as defined above, " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\" \"" }{TEXT -1 29 " is a point w with g(g(w))=w." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 29 "g:=unapply(a-f(a)/D(f)(a),a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "g(g(w))-w;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve(%=0,w);\nz1:=1/2;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "We find there are three real numbe rs for which g(g(w)) = w. We choose the positive one." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x[0]:=1/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x[1]:=g(x[0]);x[2]:=g(x[1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "p:=plot(f(x),x=-3..3,color=BLACK):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 170 "q1:=x0->plot([[x0,0],[x0,f(x0)],[g (x0),0]],color=BLACK);\nq2:=x0->plot([[x0,0],[x0,f(x0)],[g(x0),0]],col or=BLACK);\nq3:=x0->plot([[x0,0],[x0,f(x0)],[g(x0),0]],color=BLACK);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plots[display](\{p,q1(x[0 ]),q2(x[1]),q3(x[2])\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 18 " We next find " }{XPPEDIT 18 0 "z[2]" "6#&%\"z G6#\"\"#" }{TEXT -1 36 ". We first find the where f''(x) =0." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(f(x),x)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%,x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "gg:=eva l@unapply(x-f(x)/D(f)(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ss:=solve(g(x)= -sqrt(2)/2,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "z2:=simplify(ss[1]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 18 " We next find " }{XPPEDIT 18 0 "z [3]" "6#&%\"zG6#\"\"$" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "z3:=sqrt(2)/2; gg(z3);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "While the three points " } {XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6#\"\"$" }{TEXT -1 61 " are determined analytically, we give \+ decimal approximations." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "m ap(evalf,[0, z1, z2, z3]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 176 "Finally, we illustrate the dynamics by choosin g random numbers, albeit rational, in each of the four intervals and o bserve the dynamics. We choose a number in the interval (0, " } {XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 23 ") and compute it erates." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "with(stats[random ]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "z:=uniform[0,z1](1); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "z, g(z),g(g(z)),g(g(g(z )));" }}}{PARA 0 "" 0 "" {TEXT -1 37 "We choose a number in the interv al ( " }{XPPEDIT 18 0 "z[1]" "6#&%\"zG6#\"\"\"" }{TEXT -1 2 ", " } {XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 27 ") and compute the iterates." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z:=uniform[z1, z2](1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "z, g(z),g(g(z)), g(g(g(z))),g(g(g(g(z))));g(g(g(g(g(z)))));" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 37 "We choose a number in the interval ( " }{XPPEDIT 18 0 "z[2]" "6#&%\"zG6#\"\"#" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG6#\"\"$" }{TEXT -1 27 ") and \+ compute the iterates." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z:= uniform[z2,z3](1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "z, g( z),g(g(z)),g(g(g(z))),g(g(g(g(z))));" }}}{PARA 0 "" 0 "" {TEXT -1 37 " We choose a number in the interval ( " }{XPPEDIT 18 0 "z[3]" "6#&%\"zG 6#\"\"$" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "z[3]+1" "6#,&&%\"zG6#\"\"$\" \"\"F(F(" }{TEXT -1 27 ") and compute the iterates." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "z:=uniform[z3,1+z3](1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "z, g(z),g(g(z)),g(g(g(z))),g(g(g(g(z))));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 19 "Reference Material:" }}{PARA 0 "" 0 "" {TEXT -1 72 "(1) A statement of Newton's method is on page x xx of Stanley Grossman's " }{TEXT 263 8 "Calculus" }{TEXT -1 17 " (Fif th Edition)." }}{PARA 0 "" 0 "" {TEXT -1 75 "(2) A suggestion of a pro of for the the convergence rate is on page 200 of " }{TEXT 264 29 "The Elements of Real Analysis" }{TEXT -1 62 " (Second Edition), Robert G. Bartle, John WIley & Sons (1976)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }