{VERSION 6 0 "IBM INTEL NT" "6.0" } {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 "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 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 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1 " -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal " -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 } 3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 256 "" 0 "" {TEXT 265 0 "" }{TEXT -1 0 "" }{TEXT 266 15 "Newton's Method" }}{PARA 0 "" 0 "" {TEXT -1 559 "As discussed in clas s, Newton's method provides an iterative process to find the solutions of equations that cannot be solved analytically. It gets a little cum bersome repeating the process with your calculator but a computer and \+ software packages such as Maple, allow you to repeat these calculation s and arrive at an approximate solution very quickly. This is one topi c in a branch of mathematics called Numerical Analysis. In fact, when you type in a command like \"fsolve\" in Maple, it uses a routine bas ed on the very process we will explore in this lab." }{TEXT 264 0 "" } }{SECT 1 {PARA 3 "" 0 "" {TEXT -1 58 "Newton's Method For Approximatin g The Roots of Polynomials" }}{EXCHG {PARA 0 "" 0 "" {TEXT 268 10 "Def intion:" }{TEXT -1 12 " a number " }{XPPEDIT 18 0 "x[o];" "6#&%\"xG6 #%\"oG" }{TEXT -1 13 " is called a " }{TEXT 267 6 "root " }{TEXT -1 27 "of a function f(x) if f(" }{XPPEDIT 18 0 "x[o];" "6#&%\"xG6#%\" oG" }{TEXT -1 7 ") = 0. " }}{PARA 0 "" 0 "" {TEXT -1 107 "When we are \+ seeking the root of a function, we are trying to find a value where th e function equals zero. " }}{PARA 0 "" 0 "" {TEXT -1 12 "Let f(t) = \+ " }{XPPEDIT 18 0 "t^3+t-1;" "6#,(*$%\"tG\"\"$\"\"\"F%F'F'!\"\"" } {TEXT -1 232 " . From the graph below we see that f(t) = 0 at only on e place: near x = 0.7. We will approximate the exact value of this \+ root using Newton's method. First, define the function and plot it to \+ obtain a first guess at the root. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := t -> t^3 + t - 1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,(*$)9$\"\"$\"\"\"F1F/F 1F1!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(f(t) , t = -1..1);" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$!\"\"\"\"!$!\"$F*7$$!3ommm;p0k&*!# =$!3a1DU38CJG!#<7$$!3wKL$3F37$$!3[++](y$pZiF0$!3GUW=W) R'o=F37$$!33LLL$yaE\"eF0$!3CK:f(Qdwx\"F37$$!3hmmm\">s%HaF0$!3/KodVN+.< F37$$!3Q+++]$*4)*\\F0$!3S!>I/WnYi\"F37$$!39+++]_&\\c%F0$!3%)ok[*yB;b\" F37$$!31+++]1aZTF0$!3#*4;#oW+h[\"F37$$!3umm;/#)[oPF0$!33>ND=qOI9F37$$! 3hLLL$=exJ$F0$!3ow$yd5'Ho8F37$$!3*RLLLtIf$HF0$!3%QkvR))**)=8F37$$!3]++ ]PYx\"\\#F0$!33P52X([YE\"F37$$!3EMLLL7i)4#F0$!3\"o*RA%*[5>7F37$$!3c*** *\\P'psm\"F0$!3OiY&*>;Or6F37$$!3')****\\74_c7F0$!3I_KwUfjF6F37$$!3)3LL L3x%z#)!#>$!3UN6%oKiL3\"F37$$!3KMLL3s$QM%Fer$!3FiugN._V5F37$$!3]^omm;z r)*!#@$!39&pG,=()4+\"F37$$\"3%pJL$ezw5VFer$!3!RG5i9A\"o&*F07$$\"3s*)** *\\PQ#\\\")Fer$!3Ou#ziCk'z\"*F07$$\"3GKLLe\"*[H7F0$!3gI$4wMD>v)F07$$\" 3I*******pvxl\"F0$!3iDk:s\\m'H)F07$$\"3#z****\\_qn2#F0$!3UtYbk)eO$yF07 $$\"3U)***\\i&p@[#F0$!3;t4)o&)**[O(F07$$\"3B)****\\2'HKHF0$!3\\4cCWXd: oF07$$\"3ElmmmZvOLF0$!3QhS;9Kt\"H'F07$$\"3i******\\2goPF0$!3/$f)f?$php &F07$$\"3UKL$eR<*fTF0$!3-2^vGD@?^F07$$\"3m******\\)Hxe%F0$!3'yL&)G[ymW %F07$$\"3ckm;H!o-*\\F0$!3Ijr[^n,nPF07$$\"3y)***\\7k.6aF0$!3IYMT)=\\Y+$ F07$$\"3#emmmT9C#eF0$!3K!)\\wAuv.AF07$$\"33****\\i!*3`iF0$!3;ga!H!H)=I \"F07$$\"3%QLLL$*zym'F0$!3;Rv$R?\"RvOFer7$$\"3wKLL3N1#4(F0$\"3!Q)\\4Mc %=f'Fer7$$\"3Nmm;HYt7vF0$\"3GmF'\\#3,` " 0 "" {MPLTEXT 1 0 9 "t1 := .7:" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Now we define the iteration functi on, " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "newton := t -> t - \+ (t^3 + t - 1)/(3*t^2 + 1):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "and plug in our first guess. The result is our second guess. " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "t2 := newton(t1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t2G$\"+J4\"f#o!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Now keep going. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "t3 := newton(t2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#t3G$\"+I'yK#o!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "t4 : = newton(t3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t4G$\"+Q!yK#o!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "t5 := newton(t4);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t5G$\"+Q!yK#o!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 367 "Notice we have had no change in the first 10 d ecimal places between t4 and t5 so this is our approximation to the so lution. It can be assumed then that the error in this appoximation is less than 10^(-9) or 0.000000001. That's pretty close. As a check, \+ you can plug your approximation into the original equation and this sh ould result in a number close to zero. " }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 6 "f(t5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!\"\"!#5 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "This is pretty close to zero. Let's see what Maple gives us for this solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sol := fsolve(f(t) = 0, t);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%$solG$\"+Q!yK#o!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "This is the same number we found. Maple basically does \+ the same sequence of calculations performed above in a program called \+ \"fsolve\". A sequence of calculations like this is referred to as an " }{TEXT 269 9 "algorithm" }{TEXT -1 29 " and every step is called an " }{TEXT 270 11 "iteration. " }{TEXT -1 141 " If we want a better app roximation using Newton's method, we need to increase the accuracy of \+ our calculuations as shown in the next example." }}}}{SECT 1 {PARA 3 " " 0 "" {TEXT -1 28 "Newton's Method to Evaluate " }{XPPEDIT 18 0 "Pi; " "6#%#PiG" }{TEXT -1 3 " " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Th is little program finds the numerical approximation to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 78 " by solving the equation sin x = 0, b y Newton's method near the solution x = " }{XPPEDIT 18 0 "Pi;" "6#%#Pi G" }{TEXT -1 64 ". This may seem redundant as we can request Maple to do this by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf(Pi,16) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"1$z*e`EfTJ!#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 177 "This is Maple's numerical estimation to \+ 16 digits, and we'll see if we can recreate this calculation using New ton's Method. I will define some of the parameters in the program:" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "\"" }{TEXT 262 1 "N" }{TEXT -1 26 "\" defines the Newton step " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#, &%\"nG\"\"\"F(F(" }{TEXT -1 5 " = N(" }{XPPEDIT 18 0 "x[n];" "6#&%\"xG 6#%\"nG" }{TEXT -1 72 "), by N(x) = x - f(x)/f'(x). With f(x) = sin x and f'(x) = cos(x). " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "\"" } {TEXT 257 9 "tolerance" }{TEXT -1 32 "\" is how close to the solution \+ (" }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 89 ") we want to get. It s hould be noted that we can't really measure the difference between " } {XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 84 " because this would be cheating. R ecall, it is our goal to numerically approximate " }{XPPEDIT 18 0 "Pi; " "6#%#PiG" }{TEXT -1 82 " by the solution of our equation sin(x) = 0. So to compare our approximations to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG " }{TEXT -1 407 " is really using the answer to find out how close to \+ the answer we are. Do you get the contradiction here? Therefore we \+ will measure the difference between consecutive approximations and mak e sure this is less than the tolerance. In this case we set \"toleran ce\" to 10^(-16) = .0000000000000001. Using this tolerance, we can ass ume our resulting approximation is withing 10^(-15) of the exact solut ion. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "In the program we will \+ calculate the difference between consecutive approximations by \"" } {TEXT 258 10 "difference" }{TEXT -1 119 "\" and compare it to \"tolera nce\". But to start we must assign an unacceptable value to \"differe nce\" by difference = 1. " }{TEXT 259 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 152 "In case the iterations fail to converge, we want a safeg uard against an infinite loop. We do this by setting a maximum on the number of iterations by \"" }{TEXT 260 13 "maxiterations" }{TEXT -1 87 "\". If we iterate more than 20 times we will set \"tolerance\" = 0 which exits the loop. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "We wi ll count the iterations with \"" }{TEXT 261 1 "n" }{TEXT -1 73 "\". We initialize the value to one and increase it by one with each step. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "Finally, before starting the i terations, we make an initial guess at our solution. Since we want the solution near " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 28 " we make an intial guess of " }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" } {TEXT -1 7 " = 3. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Now we run the iterations until the difference between " }{XPPEDIT 18 0 "x[n+1]; " "6#&%\"xG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x[ n];" "6#&%\"xG6#%\"nG" }{TEXT -1 164 " is less than \"tolerance\" or \+ \"n\" is greater than \"maxiterations\". In the former case we have a solution in the latter we have avoided a potential runaway program. \+ " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "\"" }{TEXT 263 5 "tempx" } {TEXT -1 14 "\" is actually " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#,&% \"nG\"\"\"F(F(" }{TEXT -1 283 ". We save it as a temporary x value to check the difference between x and tempx, if we are not close enough \+ we then assign x the value of tempx, contiunuing until difference is l ess than tolerance or iterations is greater than 20. The final value \+ of x is then our approximation to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" } {TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 205 "If you type thi s in yourself you should press \"shift\" and \"enter\" at the same tim e to avoid evaluating the current input. Also, the colon's (as opposed to semicolons) prevent the printing of each result. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "N := x -> x - sin(x)/cos(x): # \+ " }{TEXT -1 25 "Defines the Newton step " }{MPLTEXT 1 0 36 "\ntoleran ce := 10^(-16): # " }{TEXT -1 24 "Sets the tolerance level" }{MPLTEXT 1 0 36 "\ndifference := 1: # " }{TEXT -1 36 "Initializes the value for difference" }{MPLTEXT 1 0 36 "\nmaxiteratio ns := 20: # " }{TEXT -1 23 "Maximum number of steps" } {MPLTEXT 1 0 36 "\nn := 1: # " }{TEXT -1 12 " step number " }{MPLTEXT 1 0 36 "\nx := 3: # \+ " }{TEXT -1 13 "initial guess" }{MPLTEXT 1 0 37 " \nwhile difference \+ > tolerance #" }{TEXT -1 51 " keep doing the below while differen ce > tolerance" }{MPLTEXT 1 0 35 "\ndo # " }{TEXT -1 32 " what to do in the \"while\" loop" }{MPLTEXT 1 0 35 " \n tempx := evalf(N(x),16): #" }{TEXT -1 20 " Newton's step fo r " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 1 " " }{MPLTEXT 1 0 45 "\n difference := evalf(abs(tempx - x),16): # " }{TEXT -1 20 " difference between " }{XPPEDIT 18 0 "x[n];" "6#&%\"xG 6#%\"nG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#,&%\" nG\"\"\"F(F(" }{TEXT -1 0 "" }{MPLTEXT 1 0 35 "\n n := n + 1: \+ #" }{TEXT -1 14 " add one to n" }{MPLTEXT 1 0 35 "\n if \+ (n < maxiterations) #" }{TEXT -1 48 " check to see if we've d one too many iterations" }{MPLTEXT 1 0 35 "\n then x := tempx: \+ #" }{TEXT -1 14 " If not, set " }{XPPEDIT 18 0 "x[n+1];" "6#& %\"xG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x[n];" "6# &%\"xG6#%\"nG" }{TEXT -1 1 " " }{MPLTEXT 1 0 35 "\n else differe nce = 0: #" }{TEXT -1 52 " If so, set difference = 0 (which exit s while loop)" }{MPLTEXT 1 0 35 "\n print('Failed'): # " }{TEXT -1 22 " and print a message " }{MPLTEXT 1 0 35 "\n fi: \+ #" }{TEXT -1 25 " ends the \"if\" statement" } {MPLTEXT 1 0 36 " \n print(n): #" }{TEXT -1 23 " print the step number" }{MPLTEXT 1 0 35 "\n print(evalf(x,16)): \+ #" }{TEXT -1 8 " print " }{XPPEDIT 18 0 "x[n];" "6#&%\"xG6#% \"nG" }{TEXT -1 0 "" }{MPLTEXT 1 0 35 "\nod: \+ #" }{TEXT -1 23 " ends the \"while\" loop" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"1yU2VlaUJ!# :" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"1x/I`EfTJ!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\" %" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"1$z*e`EfTJ!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"1$z*e` EfTJ!#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "Notice it only took 5 iterations until the difference between approximations was less than \+ 10^(-15) and this agrees with Maple's approximation to " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 51 " for 16 digits. As a further check, e valuate f(x):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "sin(x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Mki%Q#!#D" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "That's pretty close to zero, but not exactly. Why not? " }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Assignment. " }}{PARA 0 "" 0 "" {TEXT -1 101 "There are no graphs necessary for this problem . However a graph might be useful to answer # 2.b(iii)." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 13 "Problem #1: " }{TEXT -1 176 "Use the intermediate value thereom to prove there is a number tha t is exactly one more than it's cube and use Newton's method to find t his number correct to 9 decimal places. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 273 14 "Problem # 2: " }{TEXT 274 115 "Use Newton's method to find the solution to sin( x) given the following initial guesses. Then answer the questions. " } }{PARA 0 "" 0 "" {TEXT 271 2 "a)" }{TEXT -1 25 " initial guess = 0.2 " }}{PARA 0 "" 0 "" {TEXT -1 64 " (i) What solution to sin(x) = 0 is the algorithm finding. " }}{PARA 0 "" 0 "" {TEXT -1 117 " ( ii) How many iterations does it take before there is no change in the approximations' first 15 decimal places?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 272 1 "b" }{TEXT -1 26 ") initial guess \+ = 1.58079" }}{PARA 0 "" 0 "" {TEXT -1 93 " (i) What solution to si n(x) = 0 is the algorithm finding. Give this answer in terms of " } {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 4 ". " }}{PARA 0 "" 0 "" {TEXT -1 117 " (ii) How many iterations does it take before there is no change in the approximations' first 15 decimal places?" }} {PARA 0 "" 0 "" {TEXT -1 39 " (iii) Why is this number so large? " }}}}}{MARK "2" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }