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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 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 257 8 "MATH 149" }}{PARA 261 "" 0 "" {TEXT -1 0 "" }{TEXT 258 10 "LABO RATORY" }{TEXT 304 1 " " }{TEXT 305 12 "ASSIGNMENT 6" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 259 15 "NEWTO N'S METHOD" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 276 139 " In past laboratory assignments we have encountered equations for which there are no gene ral formulas for the solution. (For example, " }{TEXT 306 0 "" }{TEXT 277 32 "x = 1 + sin ( 2x ) ). However, " }{TEXT 278 6 "MAPLE " } {TEXT 279 166 "can usually find an approximation of the solution to an y number of decimal places using the \"fsolve\" command. This laborato ry investigates one of the several methods " }{TEXT 303 6 "MAPLE " } {TEXT 302 27 "uses in executing \"fsolve\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 1 " " }{TEXT 264 202 "Newton's Me thod is a technique for approximating the solutions of an equation of \+ the form f ( x ) = 0. It is based on the observation that, in most c ases, if we make a fairly good initial guess, say " }{XPPEDIT 265 0 " x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 0 "" }{TEXT 266 94 " , to the zero of a function f , then the x-intercept of the tangent line to the g raph of " }{TEXT 261 1 " " }{TEXT 280 12 "f at ( " }{XPPEDIT 267 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT 268 7 ", f ( " }{XPPEDIT 269 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT 270 22 " ) ), which we call " } {XPPEDIT 271 0 "x[2]" "6#&%\"xG6#\"\"#" }{TEXT 272 47 " , is a better \+ estimate of the zero than was " }{XPPEDIT 273 0 "x[1]" "6#&%\"xG6#\" \"\"" }{TEXT 274 59 " . We repeat the process, finding the next appro ximation " }{XPPEDIT 18 0 "x[3];" "6#&%\"xG6#\"\"$" }{TEXT 283 60 " , the x-intercept of the tangent to the graph of f at ( " }{XPPEDIT 18 0 "x[2];" "6#&%\"xG6#\"\"#" }{TEXT -1 1 " " }{TEXT 281 5 ", f (" } {XPPEDIT 18 0 "x[2];" "6#&%\"xG6#\"\"#" }{TEXT -1 1 " " }{TEXT 282 73 ") ). Repeating this process n-times, we found in class that the (n \+ + 1)" }{TEXT 263 2 "st" }{TEXT 275 28 " approximation is given by:" } }{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 47 " \+ " }{XPPEDIT 18 0 "x[n+1]" " 6#&%\"xG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "x[n] " "6#&%\"xG6#%\"nG" }{TEXT -1 9 " - f ( " }{XPPEDIT 18 0 "x[n]" "6#& %\"xG6#%\"nG" }{TEXT -1 6 " ) / " }{TEXT 262 2 " " }{TEXT -1 6 "f ' \+ ( " }{XPPEDIT 18 0 "x[n]" "6#&%\"xG6#%\"nG" }{TEXT -1 5 " ) ." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 284 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 285 1 " " }{TEXT 288 8 "Example " }}{PARA 260 " " 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 20 " Find all numbe rs " }{TEXT 287 2 "x " }{TEXT -1 40 " in the interval [ 0 , 1 ] for which " }{XPPEDIT 18 0 "[cos(3*Pi*x/2)]^2" "6#*$7#-%$cosG6#**\"\"$\" \"\"%#PiGF*%\"xGF*\"\"#!\"\"F-" }{TEXT -1 6 " = " }{TEXT 286 1 "x" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 289 72 "First we put the equation in the form f ( x ) = 0, setting f ( x ) = " }{TEXT -1 1 " " }{XPPEDIT 18 0 "[cos(3*Pi*x/2)]^2;" "6#*$7#-%$cos G6#**\"\"$\"\"\"%#PiGF*%\"xGF*\"\"#!\"\"F-" }{TEXT -1 1 " " }{TEXT 290 96 "- x , and we plot the graph of f in order to see the approxi mate locations of the zeros of f." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f:=x->x-(cos(3*Pi*x/2))^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&9$\"\"\"*$)- %$cosG6#,$*&#\"\"$\"\"#F.*&%#PiGF.F-F.F.F.F8F.!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "plot(f,0..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7en7$$\"\"!F)$!\"\"F) 7$$\"3emmm;arz@!#>$!3'[!*)=a=*on*!#=7$$\"3[LL$e9ui2%F/$!34a$*o\"Q.zA*F 27$$\"3nmmm\"z_\"4iF/$!3IYNM0g4Z&)F27$$\"3[mmmT&phN)F/$!3'>85RP*H#p(F2 7$$\"3CLLe*=)H\\5F2$!3IGH)[[z&)p'F27$$\"3gmm\"z/3uC\"F2$!3my=sAzGxcF27 $$\"3%)***\\7LRDX\"F2$!3eu#zFcw'\\XF27$$\"3]mm\"zR'ok;F2$!3#eY,LRXYM$F 27$$\"3w***\\i5`h(=F2$!3[a\"ehQmI9#F27$$\"3WLLL3En$4#F2$!3O.UgZ*y)z%*F /7$$\"3qmm;/RE&G#F2$\"3a:#QbvGxz$!#?7$$\"3\")*****\\K]4]#F2$\"35)>S%\\ %\\'R5F27$$\"3$******\\PAvr#F2$\"3Y<\\!*=+]7.%Q%GK F2$\"3[]n?$[CS?$F27$$\"3?LLL347TLF2$\"3Cc6,Mi)4M$F27$$\"3QLL$3xxlV$F2$ \"3r_(*[(\\DHT$F27$$\"3,LLLLY.KNF2$\"3;VVVxRhWMF27$$\"3mm;HdO2VOF2$\"3 vGQN.U`JMF27$$\"3w***\\7o7Tv$F2$\"3[\"*R6W!fgO$F27$$\"3Im;HK5S_QF2$\"3 ?7h[NM#fE$F27$$\"3'GLLLQ*o]RF2$\"3]w\")RH<%z7$F27$$\"3A++D\"=lj;%F2$\" 3&QjaFa.Hq#F27$$\"31++vV&RC19)QK=F_o7$$\"3&em;zRQb@&F2$!3wAP%QM6>$zF/7$$\"3\\***\\(=>Y2aF2$!3 ORXf>#4eY\"F27$$\"39mm;zXu9cF2$!3pSZ(>e%zA@F27$$\"3l******\\y))GeF2$!3 c?+5'*>!=p#F27$$\"3'*)***\\i_QQgF2$!3**e/H3?N5JF27$$\"34**\\(=-N(RhF2$ !3/@C'Gs^iD$F27$$\"3@***\\7y%3TiF2$!31/*[zrs?O$F27$$\"3;**\\P4kh`jF2$! 3DdE15>LIMF27$$\"35****\\P![hY'F2$!3mCt!GnH[W$F27$$\"3)em;/risc'F2$!3( yKXwl53T$F27$$\"3kKLL$Qx$omF2$!3X%H*y$F27$$\"3!)*****\\P+V)oF2$!3,1LZFv)3,$F27$$\"3?mm\"zpe*zqF2$!37;'> (\\7]XDF27$$\"3%)*****\\#\\'QH(F2$!3o\\!R#R!>x&=F27$$\"3GKLe9S8&\\(F2$ !3%zf5t!pdc5F27$$\"3R***\\i?=bq(F2$!3o'fJ\"yAMN$)F_o7$$\"3\"HLL$3s?6zF 2$\"3IGMMuuYS(*F/7$$\"3a***\\7`Wl7)F2$\"3=h<;JDBe@F27$$\"3#pmmm'*RRL)F 2$\"3vbgue')zOLF27$$\"3Qmm;a<.Y&)F2$\"3\"z(*Qiz\\;a%F27$$\"3=LLe9tOc() F2$\"3u\"oc\\5rup&F27$$\"3u******\\Qk\\*)F2$\"3/OuQN]N$p'F27$$\"3CLL$3 dg6<*F2$\"3\\$R$HQrj@xF27$$\"3ImmmmxGp$*F2$\"3h:8(em?;^)F27$$\"3A++D\" oK0e*F2$\"37MSf#pj[>*F27$$\"3A++v=5s#y*F2$\"3o\"*)e57\\#y'*F27$$\"\"\" F)F]]l-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F(-%+AXESLABELSG6$Q!6\"Fi]l-%% VIEWG6$;F(F]]l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 215 "In \+ order to estimate the location of the three zeros of f in the inte rval [ 0 , 1 ] through Newton's Method, we begin by defining the fun ction g , which will be used to compute the successive approximation s." }}{PARA 260 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "g:= x-> x-f(x)/D(f)(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&9$\"\"\"*&-%\"fG6#F-F .--%\"DG6#F1F2!\"\"F7F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 294 32 "T he value of g for any x is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "g(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&%\"xG\"\"\"*&,&F$F%* $)-%$cosG6#,$**\"\"$F%\"\"#!\"\"%#PiGF%F$F%F%F0F%F1F%,&F%F%**F/F%F*F%- %$sinGF,F%F2F%F%F1F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 308 97 "Now to obtain the first zero of f in [ 0 , \+ 1 ] , we make an initial guess using the graph for " }{TEXT 309 67 "re ference, and obtain each successive approximation by evaluating " } {TEXT 307 1 " " }{TEXT 310 139 "g at the previous estimate. We will \+ want to use decimal representations for all of the approximations, a nd therefore will use \"evalf \"." }{TEXT -1 1 " " }}{PARA 260 "" 0 " " {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 34 "From the graph, it ap pears that " }{XPPEDIT 18 0 "x[1]" "6#&%\"xG6#\"\"\"" }{TEXT -1 38 " = 0.25 is a reasonable first guess." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 260 "> " 0 "" {MPLTEXT 1 0 12 "x[1]:= 0.25;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"\"$\"#D!\"#" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 30 "For every positive integer n, " }{TEXT 311 1 " " }{TEXT -1 2 " " }{XPPEDIT 18 0 "x[n+1]" "6#&%\"xG6#,&%\"nG \"\"\"F(F(" }{TEXT -1 4 " = " }{TEXT 312 1 "g" }{TEXT -1 2 "( " } {XPPEDIT 18 0 "x[n]" "6#&%\"xG6#%\"nG" }{TEXT -1 14 " ). Therefore," } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x[2]:=evalf(g(x[1]));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"#$\"+Og'4E#!#5" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x[3]:=evalf(g(x[2]));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"$$\"+Mb^xA!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x[4]:=evalf(g(x[3]));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"%$\"+98exA!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x[5]:=evalf(g(x[4]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"&$\"+:8exA!#5" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "x[6]:=evalf(g(x[5]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"'$\"+;8exA!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "x[7]:=evalf(g(x[6]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"($\"+:8exA!#5" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 0 "" }{TEXT 292 0 "" }{TEXT -1 115 "Thus, it appears that the first zero of f in the interval [ 0 , 1 ] is 0.227758132 to nine d ecimal places." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 293 0 "" }{TEXT -1 1 " " }{TEXT 296 16 " The \"for / do \"" }{TEXT 297 74 " command allows us to calculate any nu mber of approximations in one step:" }{TEXT -1 0 "" }{TEXT 295 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "for n from 1 to 8 \ndo \n \+ x[n+1]:=evalf(g(x[n])) \nod;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\" xG6#\"\"#$\"+Og'4E#!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\" \"$$\"+Mb^xA!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"%$\"+9 8exA!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"&$\"+:8exA!#5 " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"'$\"+;8exA!#5" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"($\"+:8exA!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\")$\"+;8exA!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"*$\"+:8exA!#5" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }{TEXT 298 25 "Even better, we can tell " }{TEXT 291 5 "MAPLE" }{TEXT 301 35 " to calculate approximations until " } {XPPEDIT 18 0 "x[n];" "6#&%\"xG6#%\"nG" }{TEXT -1 1 " " }{TEXT 299 30 " and the next approximation, " }{XPPEDIT 18 0 "x[n+1];" "6#&%\"xG6#, &%\"nG\"\"\"F(F(" }{TEXT -1 0 "" }{TEXT 300 104 " , are equal in a pre determined number of decimal places. In this example we specify 15 dec imal places; " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 313 133 "NOTE THAT THE NUMBER OF DIGITS IN MAPLE DECIMAL EXPRESS IONS MUST BE RESET TO BE AT LEAST TWO MORE THAN THE DESIRED LEVEL OF A CCURACY." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=17;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'Dig itsG\"#<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 314 46 "Now we can define th e commands to be executed:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "k:=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "while (abs(x[k+1]-x[k]) > 10^(-15) ) do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "k:=k+1:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " x[k+1]:=evalf(g(x[k])):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"$$\"2\"fD@Lb^xA!#<" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6 #\"\"%$\"2$48.:8exA!#<" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"% " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"&$\"2.o(3;8exA!#<" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"'$\"2.o(3;8exA!#<" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 138 "From the graph, i t appears that the second zero in the interval [0 , 1] is approximat ely 0.5, and, in fact, it is exactly 0.5, since " }{XPPEDIT 18 0 "[ cos(3*Pi/4)]^2" "6#*$7#-%$cosG6#*(\"\"$\"\"\"%#PiGF*\"\"%!\"\"\"\"#" } {TEXT -1 2 "= " }{XPPEDIT 18 0 "1/2;" "6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 2 " ." }}{PARA 260 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 316 35 "To approximate the final zero of " }{TEXT 315 1 "f" } {TEXT 317 161 " in [ 0 , 1 ], we first make an initial guess by exa mining the graph of f , and then we apply Newton's Method to obtain s uccessive approximations as before. " }}{PARA 0 "" 0 "" {TEXT -1 6 " \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=10;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'DigitsG\"#5" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 13 "plot(f,0..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7en7$$\"\"!F)$!\"\"F)7$$\"3emmm ;arz@!#>$!3'[!*)=a=*on*!#=7$$\"3[LL$e9ui2%F/$!34a$*o\"Q.zA*F27$$\"3nmm m\"z_\"4iF/$!3IYNM0g4Z&)F27$$\"3[mmmT&phN)F/$!3'>85RP*H#p(F27$$\"3CLLe *=)H\\5F2$!3IGH)[[z&)p'F27$$\"3gmm\"z/3uC\"F2$!3my=sAzGxcF27$$\"3%)*** \\7LRDX\"F2$!3eu#zFcw'\\XF27$$\"3]mm\"zR'ok;F2$!3#eY,LRXYM$F27$$\"3w** *\\i5`h(=F2$!3[a\"ehQmI9#F27$$\"3WLLL3En$4#F2$!3O.UgZ*y)z%*F/7$$\"3qmm ;/RE&G#F2$\"3a:#QbvGxz$!#?7$$\"3\")*****\\K]4]#F2$\"35)>S%\\%\\'R5F27$ $\"3$******\\PAvr#F2$\"3Y<\\!*=+]7.%Q%GKF2$\"3[]n? $[CS?$F27$$\"3?LLL347TLF2$\"3Cc6,Mi)4M$F27$$\"3QLL$3xxlV$F2$\"3r_(*[( \\DHT$F27$$\"3,LLLLY.KNF2$\"3;VVVxRhWMF27$$\"3mm;HdO2VOF2$\"3vGQN.U`JM F27$$\"3w***\\7o7Tv$F2$\"3[\"*R6W!fgO$F27$$\"3Im;HK5S_QF2$\"3?7h[NM#fE $F27$$\"3'GLLLQ*o]RF2$\"3]w\")RH<%z7$F27$$\"3A++D\"=lj;%F2$\"3&QjaFa.H q#F27$$\"31++vV&RC19)Q K=F_o7$$\"3&em;zRQb@&F2$!3wAP%QM6>$zF/7$$\"3\\***\\(=>Y2aF2$!3ORXf>#4e Y\"F27$$\"39mm;zXu9cF2$!3pSZ(>e%zA@F27$$\"3l******\\y))GeF2$!3c?+5'*>! =p#F27$$\"3'*)***\\i_QQgF2$!3**e/H3?N5JF27$$\"34**\\(=-N(RhF2$!3/@C'Gs ^iD$F27$$\"3@***\\7y%3TiF2$!31/*[zrs?O$F27$$\"3;**\\P4kh`jF2$!3DdE15>L IMF27$$\"35****\\P![hY'F2$!3mCt!GnH[W$F27$$\"3)em;/risc'F2$!3(yKXwl53T $F27$$\"3kKLL$Qx$omF2$!3X%H*y$F27 $$\"3!)*****\\P+V)oF2$!3,1LZFv)3,$F27$$\"3?mm\"zpe*zqF2$!37;'>(\\7]XDF 27$$\"3%)*****\\#\\'QH(F2$!3o\\!R#R!>x&=F27$$\"3GKLe9S8&\\(F2$!3%zf5t! pdc5F27$$\"3R***\\i?=bq(F2$!3o'fJ\"yAMN$)F_o7$$\"3\"HLL$3s?6zF2$\"3IGM MuuYS(*F/7$$\"3a***\\7`Wl7)F2$\"3=h<;JDBe@F27$$\"3#pmmm'*RRL)F2$\"3vbg ue')zOLF27$$\"3Qmm;a<.Y&)F2$\"3\"z(*Qiz\\;a%F27$$\"3=LLe9tOc()F2$\"3u \"oc\\5rup&F27$$\"3u******\\Qk\\*)F2$\"3/OuQN]N$p'F27$$\"3CLL$3dg6<*F2 $\"3\\$R$HQrj@xF27$$\"3ImmmmxGp$*F2$\"3h:8(em?;^)F27$$\"3A++D\"oK0e*F2 $\"37MSf#pj[>*F27$$\"3A++v=5s#y*F2$\"3o\"*)e57\\#y'*F27$$\"\"\"F)F]]l- %'COLOURG6&%$RGBG$\"*++++\"!\")F(F(-%+AXESLABELSG6$Q!6\"Fi]l-%%VIEWG6$ ;F(F]]l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 318 48 "A rea sonable first estimate would appear to be " }{XPPEDIT 18 0 "z[1];" "6 #&%\"zG6#\"\"\"" }{TEXT 319 31 " = 0.75. We will again use a " } {TEXT 320 10 "\"while/do\"" }{TEXT 321 64 " command to generate a sol ution accurate to 8 decimal places." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "z[1]:=0.75;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#\"\"\"$\"#v!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z[2]:=evalf(g(z[1]));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#\"\"#$\"+jR.Rx!#5" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 5 "k:=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 " while (abs(z[k+1]-z[k]) > 10^(-8) ) do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "k:=k+1;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "z[k+1]:=evalf(g(z [k])):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6 #\"\"$$\"+oW[Ax!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"$" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#\"\"%$\"+&o=Cs(!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"zG6#\"\"&$\"+\"o=Cs(!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 322 66 "Thus the third zero rounded to eight decimal places \+ is 0.77224187." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "40" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }