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128 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 3" -1 5 1 {CSTYLE " " -1 -1 "Times" 1 12 128 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Err or" -1 8 1 {CSTYLE "" -1 -1 "Courier" 1 10 255 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 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 Output" -1 12 1 {CSTYLE "" -1 -1 "Ti mes" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 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 "Bullet Item" -1 15 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 3 3 1 0 1 0 2 2 15 2 }{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 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 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 }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 34 "The secant method for root-findin g" }}{PARA 0 "" 0 "" {TEXT -1 37 "by Peter Stone, Nanaimo, B.C., Canad a" }}{PARA 0 "" 0 "" {TEXT -1 19 "Version: 24.3.2007" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 28 "load root-finding procedures" }}{PARA 0 "" 0 "" {TEXT -1 17 "The Maple m-file " }{TEXT 275 7 "roots.m" }{TEXT -1 38 " \+ contains the code for the procedures " }{TEXT 0 6 "secant" }{TEXT -1 5 " and " }{TEXT 0 11 "secant_step" }{TEXT -1 25 " used in this worksh eet. " }}{PARA 0 "" 0 "" {TEXT -1 121 "It can be read into a Maple ses sion by a command similar to the one that follows, where the file path gives its location." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "read \"K:\\\\Maple/procdrs/roots.m\";" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 44 "The iterative formula for the sec ant method " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 41 "Suppose that we have two approximations " } {XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" }{TEXT -1 12 " for a root " }{TEXT 269 1 "r" }{TEXT -1 4 " of " }{XPPEDIT 18 0 "f(x) = 0;" "6#/-%\"fG6#% \"xG\"\"!" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 5 "Let " } {XPPEDIT 18 0 "y[0] = f(x[0]);" "6#/&%\"yG6#\"\"!-%\"fG6#&%\"xG6#F'" } {TEXT -1 6 " and " }{XPPEDIT 18 0 "y[1] = f(x[1]);" "6#/&%\"yG6#\"\" \"-%\"fG6#&%\"xG6#F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 45 "W e could regard the line joining the points (" }{XPPEDIT 18 0 "x[0],y[0 ];" "6$&%\"xG6#\"\"!&%\"yG6#F&" }{TEXT -1 7 ") and (" }{XPPEDIT 18 0 " x[1],y[1];" "6$&%\"xG6#\"\"\"&%\"yG6#F&" }{TEXT -1 51 ") as being a li near approximation for the function " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6 #%\"xG" }{TEXT -1 6 " near " }{XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x[1];" "6#&%\"xG6#\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 27 "This line has the equation " } {XPPEDIT 18 0 "y-y[1] = ``((y[0]-y[1])/(x[0]-x[1]))*(x-x[1]);" "6#/,&% \"yG\"\"\"&F%6#F&!\"\"*&-%!G6#*&,&&F%6#\"\"!F&&F%6#F&F)F&,&&%\"xG6#F2F &&F76#F&F)F)F&,&F7F&&F76#F&F)F&" }{TEXT -1 15 ". It meets the " } {TEXT 286 1 "x" }{TEXT -1 19 " axis at the point " }{XPPEDIT 18 0 "x[2 ] = x[1]-``((x[0]-x[1])/(y[0]-y[1]))*y[1];" "6#/&%\"xG6#\"\"#,&&F%6#\" \"\"F+*&-%!G6#*&,&&F%6#\"\"!F+&F%6#F+!\"\"F+,&&%\"yG6#F4F+&F:6#F+F7F7F +&F:6#F+F+F7" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 22 "We can ta ke the value " }{XPPEDIT 18 0 "x[2];" "6#&%\"xG6#\"\"#" }{TEXT -1 40 " to be a new approximation for the root " }{TEXT 268 1 "r" }{TEXT -1 2 ". 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" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 266 25 "gen eral iterative formula" }{TEXT -1 26 " for the secant method is " }} {PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "x[n+1] = x[n]-``((x[n -1]-x[n])/(f(x[n-1])-f(x[n])))*f(x[n]);" "6#/&%\"xG6#,&%\"nG\"\"\"F)F) ,&&F%6#F(F)*&-%!G6#*&,&&F%6#,&F(F)F)!\"\"F)&F%6#F(F6F),&-%\"fG6#&F%6#, &F(F)F)F6F)-F;6#&F%6#F(F6F6F)-F;6#&F%6#F(F)F6" }{TEXT -1 2 ", " }} {PARA 257 "" 0 "" {TEXT -1 1 " " }{TEXT 287 22 "______________________ " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 37 "which will give a seq uence of values " }}{PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 " x[0],x[1],x[2],` . . . `,x[n],x[n+1],` . . . `;" "6)&%\"xG6#\"\"!&F$6# \"\"\"&F$6#\"\"#%(~.~.~.~G&F$6#%\"nG&F$6#,&F0F)F)F)F-" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "conve rging to the root " }{TEXT 271 1 "r" }{TEXT -1 17 " of the equation " }{XPPEDIT 18 0 "f(x) = 0" "6#/-%\"fG6#%\"xG\"\"!" }{TEXT -1 16 ", prov ided that " }{XPPEDIT 18 0 "x[0];" "6#&%\"xG6#\"\"!" }{TEXT -1 33 " is sufficiently close the root " }{TEXT 270 1 "r" }{TEXT -1 24 ", and t hat the function " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 44 " is sufficiently well-behaved near the root." }}{PARA 0 "" 0 "" {TEXT -1 90 "This method can be regarded as being a modification of Ne wton's method in which the ratio " }{XPPEDIT 18 0 "(f(x[n-1])-f(x[n])) /(x[n-1]-x[n]);" "6#*&,&-%\"fG6#&%\"xG6#,&%\"nG\"\"\"F-!\"\"F--F&6#&F) 6#F,F.F-,&&F)6#,&F,F-F-F.F-&F)6#F,F.F." }{TEXT -1 53 " is a numerical approximation for the derivative of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG 6#%\"xG" }{TEXT -1 4 " at " }{XPPEDIT 18 0 "``(x[n],y[n]);" "6#-%!G6$& %\"xG6#%\"nG&%\"yG6#F)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 21 "Introductory example " }}{PARA 0 "" 0 "" {TEXT -1 24 "The following procedure " }{TEXT 0 16 "nextsecantapprox" }{TEXT -1 55 " can be used to perform one step of the secant method. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "nextsecantapprox := proc(a,b)\n local fa,fb;\n fa := f(a);\n \+ fb := f(b);\n evalf(b - fb*(a - b)/(fa - fb));\nend proc:" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "For exam ple, if we take the two starting approximations 1 and 2 for the calcul ation of the positive root of " }{XPPEDIT 18 0 "f(x)=x^2-2" "6#/-%\"f G6#%\"xG,&*$F'\"\"#\"\"\"F*!\"\"" }{TEXT -1 49 ", the next approximati on by the secant method is " }{TEXT 272 1 "c" }{TEXT -1 11 " from . . \+ ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "f := x -> x^2-2;\na := 1;\nb := 2;\nc := nextsecantap prox(a,b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)op eratorG%&arrowGF(,&*$)9$\"\"#\"\"\"F1F0!\"\"F(F(F(" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"aG\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"b G\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\"+LLLL8!\"*" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Rename " } {TEXT 273 1 "a" }{TEXT -1 5 " and " }{TEXT 274 1 "b" }{TEXT -1 24 " an d repeat the process." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "a := b;\nb := c;\nc := nextsecantapprox(a ,b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG\"\"#" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"bG$\"+LLLL8!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"cG$\"+++++9!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 86 "Carry on in this manner always retaining the two mos t recently computed approximations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 131 "a := b;\nb := c;\nc := next secantapprox(a,b);\na := b;\nb := c;\nc := nextsecantapprox(a,b);\na : = b;\nb := c;\nc := nextsecantapprox(a,b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\"+LLLL8!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"bG$\"+++++9!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\"+YTj99 !\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\"+++++9!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"+YTj99!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\"+Q9@99!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"aG$\"+YTj99!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"+Q9@99 !\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\"+iN@99!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "The last value \+ gives the decimal value of " }{XPPEDIT 18 0 "sqrt(2)" "6#-%%sqrtG6#\" \"#" }{TEXT -1 22 " correct to 10 digits." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "evalf(sqrt(2)); " } }{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+iN@99!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 259 "With each iteration abov e there are two function evaluations, but with each iteration there is a duplication of an evaluation which was performed in the previous it eration. The procedure in the next section only requires one function \+ evaluation per iteration." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 65 "A procedure for graphing successive stages of the secant \+ method: " }{TEXT 0 11 "secant_step" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 14 "The procedure " }{TEXT 0 11 "secant_step " }{TEXT -1 71 " enables the progress of the secant method to be obse rved graphically." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 18 "secant_step: usage" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 278 18 "Calling Sequence:\n" }}{PARA 0 "" 0 "" {TEXT 279 2 " " }{TEXT -1 53 " secant_step( eqn, \+ rng ) or secantstep( eqn, rng ) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 256 "" 0 "" {TEXT -1 11 "Parameters:" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT 23 11 " eqn - " }{TEXT -1 66 " an equation or expression involving a single variable, say x," }} {PARA 0 "" 0 "" {TEXT -1 26 " " }{TEXT 266 2 "OR" }{TEXT -1 34 " a function of the form x -> f(x)," }}{PARA 0 "" 0 "" {TEXT -1 96 " where f(x) evaluates t o a real or complex floating point number." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{TEXT 23 8 "rng - " }{TEXT 281 89 "a range x=a..b (or \+ simply a..b when the1st argument is a procedure) where a and b are two " }}{PARA 0 "" 0 "" {TEXT 282 181 " distinct initi al approximations for the root, and x is the variable appearing in the 1st argument.\n a and b may or may not bracket th e root." }}{PARA 0 "" 0 "" {TEXT -1 8 " " }}{PARA 256 "" 0 "" {TEXT -1 12 "Description:" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 14 "The procedure " }{TEXT 0 11 "secant_step" }{TEXT -1 93 " performs a single step of the secant method and returns the se quence consisting of the pair " }{XPPEDIT 18 0 "b,c;" "6$%\"bG%\"cG" } {TEXT -1 8 ", where " }{TEXT 288 1 "c" }{TEXT -1 25 " is the new appro ximation" }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "c = b-f(b )*``((a-b)/(f(a)-f(b)))" "6#/%\"cG,&%\"bG\"\"\"*&-%\"fG6#F&F'-%!G6#*&, &%\"aGF'F&!\"\"F',&-F*6#F1F'-F*6#F&F2F2F'F2" }{TEXT -1 3 ". " }} {PARA 0 "" 0 "" {TEXT -1 43 "A picture is drawn to illustrate the step . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 280 8 "Opt ions:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 " draw=true/false" }}{PARA 0 "" 0 "" {TEXT -1 79 "This option determines whether to draw the picture. The default is \"draw=true\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "max_ratio=n or \+ maxratio=n" }}{PARA 0 "" 0 "" {TEXT -1 17 "If the new value " }{TEXT 289 1 "c" }{TEXT -1 13 " is close to " }{TEXT 290 1 "a" }{TEXT -1 5 " \+ (or " }{TEXT 291 1 "b" }{TEXT -1 53 "), the picture will exclude the o ther outlying point " }{TEXT 292 1 "b" }{TEXT -1 5 " (or " }{TEXT 293 1 "a" }{TEXT -1 26 ") if the distance between " }{TEXT 294 1 "a" } {TEXT -1 5 " and " }{TEXT 295 1 "b" }{TEXT -1 43 " is \"max_ratio\" ti mes the distance between " }{TEXT 296 1 "a" }{TEXT -1 5 " and " } {TEXT 297 1 "c" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 31 "The d efault is \"max_ratio=10\". " }}{PARA 0 "" 0 "" {TEXT -1 73 "Increasin g max_ratio means that an outlier is more likely to be included." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "color=c o r colour=c" }}{PARA 0 "" 0 "" {TEXT -1 228 "If c is a list of up to 4 \+ colours, these colours will be applied in respective order to the curv e, the secant line, the ordinate of the initial approximation, and the 3 points shown. A single colour is applied to the curve only." }} {PARA 0 "" 0 "" {TEXT -1 46 "The default is \"colour=[red,green,blue,n avy]\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "thickness=t" }}{PARA 0 "" 0 "" {TEXT -1 152 "If t is is a list of \+ 1 or 2 positive integers, then they will be applied in respective orde r to specify the thickness of the curve and the secant line. " }} {PARA 0 "" 0 "" {TEXT -1 103 "A single thickness is applied to both th e curve and the tangent line. The default is \"thickness=[1,2]\"." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 266 16 "How to activate:" }{TEXT 256 1 "\n" } {TEXT -1 154 "To make the procedure active open the subsection, place \+ the cursor anywhere after the prompt [ > and press [Enter].\nYou can \+ then close up the subsection." }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 27 " secant_step: implementation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "newton_step" {MPLTEXT 1 0 7114 "# to allow for differe nt spellings\nsecantstep := proc() secant_step(args[1..nargs]) end:\n \nsecant_step := proc(ff,rng)\n local rg,a,b,fa,fb,faa,fbb,x,w2,h2,x L,xR,yT,yB,xrange,\n yrange,m,graphs,pts,ords,f,fn,lmr,sf,proctyp e,vars,\n Options,i,clr,colr,thk,thik,drawpic,lft,rgt,d,h,c,cc,\n r,s,e,fe,mxr;\n\n if nargs<2 then\n error \"at least 2 ar guments are required; the basic syntax is: 'secant_step(f(x),x=a..b)'. \"\n end if; \n\n if type(ff,procedure) then\n if nops([op (1,eval(ff))])<>1 then\n error \"the 1st argument, %1, is inva lid .. it should be a procedure with a single argument\",ff;\n en d if;\n proctype := true;\n if type(rng,realcons..realcons) \+ then\n rg := rng\n else\n error \"the 2nd argumen t, %1, is invalid .. when the 1st argument is a procedure, the 2nd arg ument should be a range of real constants\",rng;\n end if;\n el if type(ff,algebraic) or type(ff,equation) then\n if type(ff,equa tion) then\n lmr := lhs(ff)-rhs(ff);\n sf := traperror (simplify(lmr));\n if sf<>lasterror then\n f := sf; \n else\n f := lmr;\n end if;\n else\n f := ff;\n end if;\n vars := indets(f,name) minus i ndets(f,realcons);\n if nops(vars)<>1 then \n if not has( indets(f),\{Int,Sum\}) then\n error \"the 1st argument, %1, is invalid .. it should be an expression or an equation which depends only on a single variable\",ff;\n end if;\n end if;\n \+ if type(rng,name=realcons..realcons) then\n proctype := fal se;\n x := op(1,rng);\n if not member(x,vars) then\n \+ error \"the 1st argument, %1, is invalid .. it should be an \+ expression or an equation which depends only on the variable %2\",ff,x ;\n end if;\n rg := op(2,rng);\n else\n e rror \"the 2nd argument, %1, is invalid .. it should have the form 'x= a..b', to provide two real number starting values\",rng;\n end if ;\n else\n error \"the 1st argument, %1, is invalid .. it shoul d be an algebraic expression in a single variable, an equation in a si ngle variable, or a procedure with a single real argument\",ff;\n en d if;\n\n # Get the options.\n # Set the default value to start wi th.\n drawpic := false;\n mxr := 10;\n clr := [COLOR(RGB,1,0,0), COLOR(RGB,0,1,0),COLOR(RGB,0,0,1), COLOR(RGB,.137,.137,.557)];\n thk := [1,2];\n if nargs>2 then\n Options:=[args[3..nargs]];\n \+ if not type(Options,list(equation)) then\n error \"each opti onal argument must be an equation\"\n end if;\n if hasoption (Options,'draw','drawpic','Options') then\n if drawpic<>true t hen drawpic := false end if;\n end if;\n if hasoption(Option s,'maxratio','mxr','Options') then\n if not type(mxr,posint) t hen\n error \"\\\"maxratio\\\" must be a positive integer\" ;\n end if;\n elif hasoption(Options,'max_ratio','mxr','O ptions') then\n if not type(mxr,posint) then\n erro r \"\\\"max_ratio\\\" must be a positive integer\";\n end if; \n end if;\n if hasoption(Options,'color','colr','Options') \+ or\n hasoption(Options,'colour','colr','Options') then\n \+ if type(colr,list) then\n for i from 1 to min(nops(colr) ,4) do\n clr[i] := `plot/color`(colr[i]);\n e nd do;\n else\n clr[1] := `plot/color`(colr);\n \+ end if;\n end if;\n if hasoption(Options,'thickness','t hik','Options') then\n if type(thik,list) then\n fo r i from 1 to min(nops(thik),2) do\n thk[i] := thik[i]; \n end do;\n else\n thk := [thik,thik]; \n end if;\n end if;\n if nops(Options)>0 then\n \+ error \"%1 is not a valid option for %2 .. the recognised options are \\\"draw\\\", \\\"max_ratio\\\", or (\\\"maxratio\\\"),\\\"colour \\\", or (\\\"color\\\") and \\\"thickness\\\"\",op(1,Options),procnam e;\n end if;\n end if;\n\n if proctype then\n fn := ff; \n else\n # Evaluate any real constants in f\n fn := unapp ly(evalf(f),x);\n end if;\n\n a := evalf(op(1,rg));\n b := evalf (op(2,rg));\n if a=b then\n error \"distinct starting values ar e required\"\n end if;\n\n fa := traperror(evalf(fn(a)));\n if f a=lasterror or not type(fa,numeric) then\n error \"evaluation fai led at %1\",evalf(a);\n end if;\n fb := traperror(evalf(fn(b)));\n if fb=lasterror or not type(fb,numeric) then\n error \"evaluat ion failed at %1\",evalf(b);\n end if;\n\n d := fb - fa;\n if d= 0 then error \"zero denominator obtained in secant formula\" end if;\n h := fb*(b - a)/d;\n c := b - h; \n\n if drawpic then\n \+ # recalulate for the picture\n Digits := max(Digits,15);\n \+ faa := traperror(evalf(fn(a)));\n if faa<>lasterror and type(faa ,numeric) then\n fa := faa;\n end if;\n fbb := trape rror(evalf(fn(b)));\n if fbb<>lasterror and type(fbb,numeric) the n\n fb := fbb;\n end if;\n h := fb*(b-a)/(fb-fa);\n \+ m := (fb-fa)/(b-a);\n cc := b - h;\n r := min(abs(a-cc) ,abs(b-cc));\n s := abs(a-b);\n if s rgt then\n xrange := lft..rgt;\n yT := max(0,fa, fb);\n yB := min(0,fa,fb);\n h2 := (yT-yB)/2;\n \+ yrange := yB-h2..yT+h2;\n graphs := plot(['fn'( x),m*(x-a)+fa],x=xrange,yrange,\n color=[op(1,clr),op(2,c lr)],thickness=thk);\n ords := CURVES([[a,0],[a,fa]],[[b,0] ,[b,fb]],LINESTYLE(2),op(3,clr));\n pts := POINTS([a,0],[a, fa],[b,0],[b,fb],[cc,0],SYMBOL(CIRCLE),op(4,clr)),\n POINTS([ a,0],[a,fa],[b,0],[b,fb],[cc,0],SYMBOL(CROSS),op(4,clr)),\n P OINTS([a,0],[a,fa],[b,0],[b,fb],[cc,0],SYMBOL(DIAMOND),op(4,clr));\n \+ print(PLOT(pts,ords,op(graphs)));\n else\n \+ WARNING(\"the range for the plot is empty\");\n end if;\n \+ else\n if abs(a-cc)rgt then\n xrange := lft. .rgt;\n yT := max(0,fe);\n yB := min(0,fe);\n \+ h2 := (yT-yB)/2;\n yrange := yB-h2..yT+h2;\n \+ graphs := plot(['fn'(x),m*(x-e)+fe],x=xrange,yrange,\n \+ color=[op(1,clr),op(2,clr)],thickness=thk,\n title=` The picture shows just the latest two points`);\n ords := C URVES([[e,0],[e,fe]],LINESTYLE(2),op(3,clr));\n pts := POIN TS([e,0],[e,fe],[cc,0],SYMBOL(CIRCLE),op(4,clr)),\n POINTS([e ,0],[e,fe],[cc,0],SYMBOL(CROSS),op(4,clr)),\n POINTS([e,0],[e ,fe],[cc,0],SYMBOL(DIAMOND),op(4,clr));\n print(PLOT(pts,or ds,op(graphs)));\n else\n WARNING(\"the range for t he plot is empty\");\n end if;\n end if;\n end if;\n \+ b,c;\nend proc:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Examples are given in the next section." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT 0 11 "secant_step" }{TEXT -1 10 ": e xamples" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT 266 4 "Note" }{TEXT -1 53 ": The examples in this section r equire the procedure " }{TEXT 0 11 "secant_step" }{TEXT -1 2 ". " }} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 1" }}{PARA 0 "" 0 "" {TEXT -1 60 "We can graph a single the secant method step for computin g " }{XPPEDIT 18 0 "sqrt(2);" "6#-%%sqrtG6#\"\"#" }{TEXT -1 14 " as a root of " }{XPPEDIT 18 0 "x^2-2;" "6#,&*$%\"xG\"\"#\"\"\"F&!\"\"" } {TEXT -1 6 " . . ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 50 "f := x->x^2-2:\nsecant_step(f(x),x=1..2,draw =true);" }}{PARA 13 "" 1 "" {GLPLOT2D 360 270 270 {PLOTDATA 2 "6*-%'PO INTSG6)7$$\"\"\"\"\"!F)7$F'$!\"\"F)7$$\"\"#F)F)7$F.F.7$$\"0LLLLLLL\"!# 9F)-%'SYMBOLG6#%'CIRCLEG-%&COLORG6&%$RGBG$\"$P\"!\"$F=$\"$d&F?-F$6)F&F *F-F0F1-F66#%&CROSSGF9-F$6)F&F*F-F0F1-F66#%(DIAMONDGF9-%'CURVESG6&7$F& F*7$F-F0-%*LINESTYLEG6#F/-F:6&F+5F4$!/IvQ$)>'***F47$$\"0++]Z/N/\"F4$!/(\\l5%)4 6*F47$$\"0++]$fC&3\"F4$!/)fcg7CA)F47$$\"0L$ez6:B6F4$!/d+yUJ&Q(F47$$\"0 nm;=C#o6F4$!/Cp8E__jF47$$\"0nmm#pS17F4$!/o!HFBeW&F47$$\"0+]i`A3D\"F4$! /74G)HWN%F47$$\"0nmm(y8!H\"F4$!/-!>fUaN$F47$$\"0+]i.tKL\"F4$!/r33,$QA# 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1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve 6" }}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"aG%\"bG6$$\"+YT j99!\"*$\"+Q9@99F*" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 2" }}{PARA 0 "" 0 "" {TEXT 276 8 "Question" }{TEXT -1 3 ": \+ " }}{PARA 0 "" 0 "" {TEXT -1 33 "(a) Plot a graph of the function " } {XPPEDIT 18 0 "f(x)=sin(19*x)/5+x^2-2" "6#/-%\"fG6#%\"xG,(*&-%$sinG6#* &\"#>\"\"\"F'F/F/\"\"&!\"\"F/*$F'\"\"#F/F3F1" }{TEXT -1 15 " to check \+ that " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 28 " has a ze ro in the interval " }{XPPEDIT 18 0 "1<=x" "6#1\"\"\"%\"xG" }{XPPEDIT 18 0 "``<=2" "6#1%!G\"\"#" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 65 "(b) Illustrate the first 5 steps of the solution of the equatio n " }{XPPEDIT 18 0 "sin(19*x)/5+x^2-2=0" "6#/,(*&-%$sinG6#*&\"#>\"\"\" %\"xGF+F+\"\"&!\"\"F+*$F,\"\"#F+F0F.\"\"!" }{TEXT -1 27 " by the secan t method with " }{XPPEDIT 18 0 "1 <= x;" "6#1\"\"\"%\"xG" }{XPPEDIT 18 0 "`` <= 2;" "6#1%!G\"\"#" }{TEXT -1 53 " as the starting interval. You may use the procedure " }{TEXT 0 11 "secant_step" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 39 "(c) Find the solution of the equation " }{XPPEDIT 18 0 "sin(19*x)/5+x^2-2=0" "6#/,(*&-%$sinG6#*&\"#>\"\"\" %\"xGF+F+\"\"&!\"\"F+*$F,\"\"#F+F0F.\"\"!" }{TEXT -1 28 " which lies i n the interval " }{XPPEDIT 18 0 "1<=x" "6#1\"\"\"%\"xG" }{XPPEDIT 18 0 "``<=2" "6#1%!G\"\"#" }{TEXT -1 50 " by the secant method correct to about 10 digits. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 277 8 "Solution" }{TEXT -1 4 ": " }}{PARA 0 "" 0 "" {TEXT -1 36 "(a) The following graph shows that " }{XPPEDIT 18 0 "f(x)=sin(19* x)/5+x^2-2" "6#/-%\"fG6#%\"xG,(*&-%$sinG6#*&\"#>\"\"\"F'F/F/\"\"&!\"\" F/*$F'\"\"#F/F3F1" }{TEXT -1 29 " has a zero in the interval " } {XPPEDIT 18 0 "1<=x" "6#1\"\"\"%\"xG" }{XPPEDIT 18 0 "``<=2" "6#1%!G\" \"#" }{TEXT -1 2 ". 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" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 31 "We can check this result using " }{TEXT 0 6 "fsolve" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "fsolve(sin(19*x)/5+x^2-2 = 0,x=1..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+mDqj8!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 61 " A procedure implementing the secant method for root finding: " }{TEXT 0 6 "secant" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 13 "secant: u sage" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 18 "Calling Sequence:\n" }}{PARA 0 "" 0 "" {TEXT -1 21 " s ecant( eqn, rng ) " }{TEXT 262 1 "\n" }{TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 11 "Parameters:" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }} {PARA 0 "" 0 "" {TEXT 23 10 " eqn - " }{TEXT -1 63 " an equation o r expression involving a single variable, say x," }}{PARA 0 "" 0 "" {TEXT -1 22 " " }{TEXT 266 2 "OR" }{TEXT -1 73 " \+ a function of the form x -> f(x), where f(x) evaluates to a real numbe r," }}{PARA 0 "" 0 "" {TEXT -1 22 " " }{TEXT 266 2 "OR" }{TEXT -1 71 " a numerical procedure which evaluates to a real \+ floating point number." }}{PARA 0 "" 0 "" {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{TEXT 23 8 "rng - " }{TEXT 263 89 "a range \+ x=a..b (or simply a..b when the1st argument is a procedure) where a an d b are two" }}{PARA 0 "" 0 "" {TEXT 264 181 " dis tinct initial approximations for the root, and x is the variable appea ring in the 1st argument.\n a and b may or may not bracket the root." }{TEXT -1 0 "" }{TEXT 265 2 " " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 12 "Description:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "The proce dure " }{TEXT 0 6 "secant" }{TEXT -1 30 " attempts to locate a root of " }{XPPEDIT 18 0 "f(x) = 0;" "6#/-%\"fG6#%\"xG\"\"!" }{TEXT -1 84 " b y the secant method when given two distinct initial approximations fo r the root.\n" }}{PARA 0 "" 0 "" {TEXT 261 8 "Options:" }{TEXT -1 1 " \n" }}{PARA 0 "" 0 "" {TEXT -1 148 "maxiterations=n or maxiter=n\nThis option can be used to override the default value of Digits*5 for the \+ maximum number of iterations to be performed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "precision=fixed/variable " }}{PARA 0 "" 0 "" {TEXT -1 309 "If the computed value of the functio n exhibits a loss of significant digits as the successive approximatio ns converge the root then the working precision is increased to compen sate for this. This feature can be turned off via the option \"precisi on=fixed\". The default for this option is \"precision=variable\". " } }{PARA 0 "" 0 "" {TEXT -1 51 "The abreviated form \"prcsn=fixed\" may \+ also be used." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "info=true/false/0/1/2/3\n\"info=0\" is the same as \"inf o=false\" and \"info=1\" is the same as \"info=true\"." }}{PARA 0 "" 0 "" {TEXT -1 123 "This option allows the progress of the computation \+ to be monitored by printing the result of each secant step as it occur s." }}{PARA 0 "" 0 "" {TEXT -1 81 "With the option \"info= 2\" the exp ressions for function being used are also given." }}{PARA 0 "" 0 "" {TEXT -1 224 "The option \"info= 3\" provides additional information r egarding the value of the function and the correction term at each ste p, together with information regarding any change in the working preci sion used in the computation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 16 "How to a ctivate:" }{TEXT 256 1 "\n" }{TEXT -1 155 "To make the procedure activ e, open the subsection, place the cursor anywhere after the prompt [ > and press [Enter].\nYou can then close up the subsection." }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 21 "secant: implemetation" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7804 "secant : = proc(ff,rng)\n local Options,a,b,c,d,h,fa,fb,fc,eps,saveDigits,i,m axit,prntflg,\n x,f,fn,rg,t1,t2,sf,lmr,proctype,vars,workingDigit s,extraDigits,\n adjustDigits,eps2,prsn,triedzero,small,f0;\n\n \+ if nargs<2 then\n error \"at least 2 arguments are required; the basic syntax is: 'secant(f(x),x=a..b)'.\"\n end if; \n\n if ty pe(ff,procedure) then\n if nops([op(1,eval(ff))])<>1 then\n \+ error \"the 1st argument, %1, is invalid .. it should be a procedur e with a single argument\",ff;\n end if;\n proctype := true; \n if type(rng,realcons..realcons) then\n rg := rng\n \+ else\n error \"the 2nd argument, %1, is invalid .. when the \+ 1st argument is a procedure, the 2nd argument should be a range of rea l constants\",rng;\n end if;\n elif type(ff,algebraic) or type( ff,equation) then\n if type(ff,equation) then\n lmr := lh s(ff)-rhs(ff);\n sf := traperror(simplify(lmr));\n if \+ sf<>lasterror then\n f := sf;\n else\n f := lmr;\n end if;\n else\n f := ff;\n end i f;\n vars := indets(f,name) minus indets(f,realcons);\n if n ops(vars)<>1 then \n if not has(indets(f),\{Int,Sum\}) then\n \+ error \"the 1st argument, %1, is invalid .. it should be an expression or an equation which depends only on a single variable\",f f;\n end if;\n end if;\n if type(rng,name=realcons.. realcons) then\n proctype := false;\n x := op(1,rng); \n if not member(x,vars) then\n error \"the 1st arg ument, %1, is invalid .. it should be an expression or an equation whi ch depends only on the variable %2\",ff,x;\n end if;\n \+ rg := op(2,rng);\n else\n error \"the 2nd argument, %1, \+ is invalid .. it should have the form 'x=a..b', to provide two real nu mber starting values\",rng;\n end if;\n else\n error \"the 1st argument, %1, is invalid .. it should be an algebraic expression \+ in a single variable, an equation in a single variable, or a procedure with a single real argument\",ff;\n end if;\n \n # Get the opti ons \"maxiterations\" and \"info\".\n # Set the default values to st art with.\n maxit := Digits*5;\n prntflg := 0;\n prsn := 1;\n \+ if nargs>2 then\n Options:=[args[3..nargs]];\n if not type(O ptions,list(equation)) then\n error \"each optional argument m ust be an equation\"\n end if;\n if hasoption(Options,'maxit erations','maxit','Options') then\n if not type(maxit,posint) \+ then\n error \"\\\"maxiterations\\\" must be a positive int eger\"\n end if;\n elif hasoption(Options,'maxiter','maxi t','Options') then\n if not type(maxit,posint) then\n \+ error \"\\\"maxiter\\\" must be a positive integer\"\n end \+ if;\n end if;\n if hasoption(Options,'precision','prsn','Opt ions') then\n if not member(prsn,\{'fixed','variable'\}) then \n error \"\\\"precision\\\" must be 'fixed' or 'variable' \"\n end if;\n if prsn='fixed' then prsn := 0 else prs n := 1 end if;\n elif hasoption(Options,'prcsn','prsn','Options') then\n if not member(prsn,\{'fixed','variable'\}) then\n \+ error \"\\\"prcsn\\\" must be 'fixed' or 'variable'\"\n \+ end if;\n if prsn='fixed' then prsn := 0 else prsn := 1 end i f;\n end if;\n if hasoption(Options,'info','prntflg','Option s') then\n if not member(prntflg,\{true,false,0,1,2,3\}) then \n error \"\\\"info\\\" must be false <-> 0, true <-> 1,2 o r 3\"\n end if;\n if prntflg=false then prntflg := 0\n elif prntflg=true then prntflg := 1 end if; \n end if;\n if nops(Options)>0 then\n if type(op(1,Options),'name'=' range') \n or type(op(1,Options),'range' ) then\n error \"%1 is not a valid argument for %2 .. only \+ one range is allowed as an argument\",op(1,Options),procname;\n \+ else\n error \"%1 is not a valid option for %2 .. the rec ognised options are \\\"maxiterations\\\",(or \\\"maxiter\\\"),\\\"pre cision\\\",(or \\\"prcsn\\\") and \\\"info\\\"\",op(1,Options),procnam e;\n end if;\n end if;\n end if;\n\n # Increase preci sion for the computation\n saveDigits := Digits;\n extraDigits := \+ min(iquo(iquo(Digits,5)+1,2)+3,8);\n workingDigits := Digits + extra Digits;\n Digits := workingDigits;\n\n if proctype then\n fn \+ := ff;\n else\n # Evaluate any real constants in f\n fn := unapply(evalf(f),x);\n if prntflg>1 then\n print(`Attemp ting to calculate a zero of`);\n print(f); \n print(`b y the secant method`);\n print(``);\n end if;\n end if; \n if prntflg>2 then\n print(`** working precision is `||Digits ||` digits **`);\n end if;\n\n a := evalf(min(op(rg)));\n b := e valf(max(op(rg)));\n if a=b then\n error \"distinct starting va lues are required\"\n end if;\n\n fa := traperror(evalf(fn(a)));\n if fa=lasterror or not type(fa,numeric) then\n error \"evaluat ion failed at %1\",evalf[saveDigits](a);\n end if;\n fb := traperr or(evalf(fn(b)));\n if fb=lasterror or not type(fb,numeric) then\n \+ error \"evaluation failed at %1\",evalf[saveDigits](b);\n end if ;\n if prntflg>2 then\n print(`first value`=evalf[workingDigits ](fa));\n print(`second value`=evalf[workingDigits](fb));\n end if;\n\n eps := Float(1,-saveDigits-min(iquo(Digits,10),2));\n eps 2 := Float(1,-iquo(saveDigits,2));\n small := min(abs(a),abs(b))*Flo at(1,-trunc(saveDigits*.75)-1);\n triedzero := false;\n\n for i fr om 1 to maxit do\n d := fb - fa;\n if d=0 then error \"zero \+ denominator obtained in secant formula\" end if;\n h := fb*(b-a)/ d;\n if prntflg>2 then\n print(`correction`=evalf[working Digits](-h));\n print(``);\n end if;\n c := b-h;\n \+ if prntflg>0 then \n print(`approximation `||i||` -> `, evalf[workingDigits](c))\n end if;\n if prntflg>2 then print (``) end if;\n if abs(h)<=eps*abs(c) then\n if max(abs(a) ,abs(b))<=eps*abs(d) then\n WARNING(\"a large change in val ues suggests that the result may be unreliable\")\n end if;\n \+ Digits := saveDigits;\n return evalf(c);\n end if ;\n \n a := b;\n fa := fb;\n b := c;\n fb := t raperror(evalf(fn(b)));\n if fb=lasterror or not type(fb,numeric) then\n error \"evaluation failed at %1\",evalf(b,saveDigits); \n end if;\n if prntflg>2 then\n print(`value`=fb)\n end if;\n if prsn=1 and fb<>0 then\n adjustDigits : = extraDigits-length(SFloatMantissa(fb));\n if adjustDigits>0 \+ and (abs(h)<=eps2*abs(c) or abs(fb)2 then\n print(`** increasing working precision \+ to `||Digits||` digits **`); \n end if;\n \+ if not proctype then\n fn := unapply(evalf(f),x);\n \+ end if;\n fb := traperror(evalf(fn(b)));\n \+ if fb=lasterror or not type(fb,numeric) then\n error \"failed to evaluate function at %1\",evalf[saveDigits](b);\n \+ end if;\n if prntflg>2 then\n print(`valu e`=evalf[workingDigits](fb))\n end if;\n end if;\n \+ end if;\n if i>6 and not triedzero \n and abs(c)0 then\n print( `The values appear to be converging to 0`);\n print(``); \n end if;\n return 0.0\n end if;\n \+ triedzero := true;\n end if;\n end do:\n Digits := saveDi gits;\n print(`last iteration gives`,evalf(c));\n error \"reached \+ max, %1, iterations without convergence\",maxit;\nend proc:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Examp les are given in the next section." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 " " {TEXT 0 6 "secant" }{TEXT -1 10 ": examples" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Exampl e 1 " }{TEXT 304 28 ".. choice of starting values" }}{PARA 0 "" 0 "" {TEXT -1 75 "Using the secant method is like using the bisection metho d in that we need " }{TEXT 266 19 "two starting values" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Thes e two starting values should be entered as a Maple range for the secon d argument of the procedure " }{TEXT 0 6 "secant" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 170 "It is no t necessary for the two starting values to bracket the desired root ho wever. They can usually simply be two values which are reasonably clos e to the desired root." }}{PARA 0 "" 0 "" {TEXT -1 28 "( Incorporating the option \"" }{TEXT 275 9 "info=true" }{TEXT -1 9 "\" causes " } {TEXT 0 6 "secant" }{TEXT -1 45 " to show all the successive approxima tions.) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "secant(x^2-2,x=-2..-1,info=true);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6$%7approximation~1~~->~~~G$!/LLLLLL8!#8" }}{PARA 11 " " 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$!/9dG9dG9!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$!/$[M5$z89!#8" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$!/\\ZQ9@99!# 8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~->~~~G$!/*)oiN @99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~6~~->~~~G$!/ JPiN@99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~7~~->~~~ G$!/JPiN@99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+iN@99!\"*" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "secant(x^2-2,x=-1.4..-1.3,info=true);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~1~~->~~~G$!/[\"[\"[\"[T\"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$!/VRn#)=99!#8" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$!/r*pb8UT\"! #8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$!/JPiN @99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~->~~~G$!/ JPiN@99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+iN@99!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 2 " }{TEXT 303 93 ".. the secant method can be used when a derivative is not available for \+ the function involved" }}{PARA 0 "" 0 "" {TEXT -1 209 "The secant meth od generally converges about as fast as Newton's method, but also has \+ the advantage that it can be used in situations where the function who se zero is sought does not have a symbolic derivative." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "The \"black box func tion\" " }{TEXT 275 2 "Wn" }{TEXT -1 60 " defined in the following sub section has real number values " }{XPPEDIT 18 0 "Wn(x);" "6#-%#WnG6#% \"xG" }{TEXT -1 6 " for " }{XPPEDIT 18 0 "-exp(-1) <= x;" "6#1,$-%$ex pG6#,$\"\"\"!\"\"F*%\"xG" }{TEXT -1 17 ", approximately. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 40 "code for \"black box\" numerical function " }{TEXT 0 2 "Wn" } {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1659 "Wn := proc(x::realcons)\n local xx,eps,saveD igits,doW,val,p,q,maxit;\n\n if x=0 then return 0. end if;\n saveD igits := Digits;\n Digits := Digits+min(iquo(Digits,3),5);\n xx := evalf(x);\n if xx<-.3678794411714423215955237701615 or xx945 then\n s := ln(x);\n s := s-ln(s)+(.2 507906353+.1169032816e-6*x)/\n (1.+(.8463520229e-6+.7588315 325e-22*x)*x);\n elif x>45 then\n s := ln(x);\n s := s-ln(s)+.3;\n elif x>2.567437424 then\n s := (.398584 0902+(.3564649982+.1739375228e-2*x)*x)/\n (1.+.1364948 306*x) \n elif x>-.367879441171442321595523770161 then \n \+ s := 0.1020602722*x+1.162467682-1.961004179/\n (x+1.908142 587-0.1691574945/\n (x+0.7869827766-0.01043395288/(0.468054 5525+x)));\n end if;\n\n # solve x=s*exp(s) for s by Halley' s method \n for i to maxit do\n p := exp(s);\n t \+ := s*p-x;\n h := t/(p*(s+1)-1/2*(s+2)*t/(s+1));\n s := s-h;\n if abs(h)<=eps*abs(s) then break end if;\n end do ;\n s;\n end proc;\n\n p := ilog10(saveDigits);\n q := Floa t(saveDigits,-p);\n maxit := trunc((p+(.02331061386+.1111111111*q))* 2.095903274)+2;\n eps := Float(3,-saveDigits-1);\n if Digits<=trun c(evalhf(Digits)) and \n ilog10(xx)trunc(evalhf(DBL_MIN_10_EXP)) then\n \+ val := evalhf(doW(xx,eps,maxit))\n else\n val := doW(xx,eps,m axit)\n end if;\n evalf[saveDigits](val);\nend proc:" }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "xx := 1.3;\nWn(xx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#xxG$\" #8!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+k%=8n'!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "Once the function has been assigned or \+ loaded, its graph can be plotted, but the expression " }{TEXT 275 5 "W n(x)" }{TEXT -1 31 " must be enclosed in quotes as " }{TEXT 0 7 "'Wn(x )'" }{TEXT -1 4 " or " }{TEXT 0 7 "'Wn'(x)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 10 "The graph " }{XPPEDIT 18 0 "y = Wn(x);" "6#/%\"yG -%#WnG6#%\"xG" }{TEXT -1 43 " is plotted below along with the graph of " }{XPPEDIT 18 0 "y = x^2;" "6#/%\"yG*$%\"xG\"\"#" }{TEXT -1 1 "." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "plot(['Wn'(x),x^2],x=-0.2..2,y=-0.2..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 360 270 270 {PLOTDATA 2 "6&-%'CURVESG6$7S7$$!\"#!\"\"$!+=5r \"f#!#57$$!+3EY?:F-$!+;T&[#=F-7$$!+)o>K5\"F-$!+Rw7]7F-7$$!*'Q')RjF-$!+ h!\\\\y'!#67$$!*,Fkh\"F-$!+\"p2Kk\"F=7$$\"*\"F-$\"+::&Q2\"F-7$$\"+v+Ji;F-$\"+1FXR9F- 7$$\"+Lo`F@F-$\"+T7`!y\"F-7$$\"+Q(zgg#F-$\"+vsF5@F-7$$\"+*e!eFIF-$\"+z X7&Q#F-7$$\"+:24-NF-$\"+i\\.zEF-7$$\"+D#\\&yRF-$\"+V(y$fHF-7$$\"+&G0xV %F-$\"+)>rp@$F-7$$\"+vHma[F-$\"+]&)>TMF-7$$\"+)*fY]`F-$\"+zt*op$F-7$$ \"+$>w/x&F-$\"+y5*\\!RF-7$$\"+)*y/fiF-$\"+q`/QTF-7$$\"+Vm^\"p'F-$\"+-? 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\"3JCa!))oEG#GFcdl7$$\"3hL$3#43SEDKz$3HJFcdl7$$\"3lLLeD`l<=Fcdl$\"3Xa9i#)3(QI$Fcdl7$$\"35n mm3LCh=Fcdl$\"3QUk0amAkMFcdl7$$\"3F+]()*=Fcdl$\"3xnXaw[QROFcdl7$$ \"3W+]7C')>_>Fcdl$\"3([aN!o%z5\"QFcdl7$Ffz$\"\"%Fhz-F\\[l6&F^[lFa[lF_[ lFa[l-%+AXESLABELSG6$Q\"x6\"Q\"yFc[m-%%VIEWG6$;F(Ffz;F($\"\"\"Fhz" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "T he equation " }{XPPEDIT 18 0 "Wn(x) = x^2;" "6#/-%#WnG6#%\"xG*$F'\"\"# " }{TEXT -1 21 " has a solution near " }{TEXT 299 1 "x" }{TEXT -1 8 " \+ = 0.65." }}{PARA 0 "" 0 "" {TEXT -1 81 "We calculate this solution to \+ an accuracy of about 10 digits using the procedure " }{TEXT 0 6 "secan t" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "secant('Wn(x)'=x^2,x=0.64..0.65,info=true);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~1~~->~~~G$\"/::uFsHl !#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$\"/M- ?U=Hl!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$ \"/#HSS'=Hl!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~- >~~~G$\"/?>/k=Hl!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation ~5~~->~~~G$\"/?>/k=Hl!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+/k=Hl! #5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "Ne wton's method cannot be used here." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "newton('Wn'(x)=x^2,x=0.65,in fo=true);" }}{PARA 8 "" 1 "" {TEXT -1 56 "Error, (in newton) failed to evaluate derivative at .65\n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 87 "The bisection method does work, but a sta rting range which brackets the root is needed." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "bisect('Wn(x )'=x^2,x=0.65..0.66);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+/k=Hl!#5 " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "The \+ the \"black box\" function W is actually the " }{TEXT 0 8 "LambertW" } {TEXT -1 31 " function, which Maple \"knows\"." }}{PARA 0 "" 0 "" {TEXT -1 110 "Since Maple also knows the derivative of this function, \+ we can use Newton's method if we replace the function " }{TEXT 275 5 " Wn(x)" }{TEXT -1 4 " by " }{TEXT 275 11 "LambertW(x)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "newton(LambertW(x)=x^2,x=0.65);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+/k=Hl!#5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "Since the LambertW function is the inverse of the functio n " }{XPPEDIT 18 0 "y = x*exp(x);" "6#/%\"yG*&%\"xG\"\"\"-%$expG6#F&F' " }{TEXT -1 105 ", another way to solve the original equation is to fi nd the intersection point of the two inverse graphs." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "newton(exp (x)*x=sqrt(x),x=0.65);\nsqrt(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 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45.000000 0 0 "Curve \+ 1" "Curve 2" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Examp le 3 " }{TEXT 302 70 ".. a problem with the terminating the iterations for the secant method" }}{PARA 0 "" 0 "" {TEXT 285 8 "Question" } {TEXT -1 4 ": " }}{PARA 0 "" 0 "" {TEXT -1 37 "(a) Plot the graphs o f the functions " }{XPPEDIT 18 0 "f(x)=8*exp(-x)" "6#/-%\"fG6#%\"xG*& \"\")\"\"\"-%$expG6#,$F'!\"\"F*" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g (x) = sin(2*x)+3;" "6#/-%\"gG6#%\"xG,&-%$sinG6#*&\"\"#\"\"\"F'F.F.\"\" $F." }{TEXT -1 64 " in the same picture to show their single point of \+ intersection." }}{PARA 0 "" 0 "" {TEXT -1 45 "(b) Find the single solu tion of the equation " }{XPPEDIT 18 0 "8*exp(-x) = sin(2*x)+3" "6#/*& \"\")\"\"\"-%$expG6#,$%\"xG!\"\"F&,&-%$sinG6#*&\"\"#F&F+F&F&\"\"$F&" } {TEXT -1 47 " correct to 10 digits by the secant method. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 284 8 "Solution" } {TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "(a) " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot([8* exp(-x),sin(2*x)+3],x=-1..4,y=0..10);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6$7U7$$!\"\"\"\"!$\"33Oswiaiu@!#; 7$$!3kLL$e9r]X*!#=$\"3/8#=lt%Hf?F-7$$!3Fnmm\"HU,\"*)F1$\"3oeK&pa!3]>F- 7$$!3d**\\P4E+O%)F1$\"3q+E2Esxf=F-7$$!3)HL$3FH'='zF1$\"3/wpkCclt#eF1$\"3C.Mg$\\l>V\"F-7$$ !3SLL3_!4Nv%F1$\"3gbUkXG'oG\"F-7$$!3YmmTg(fHw$F1$\"3m5H#oh-b;\"F-7$$!3 ^++vVLIPFF1$\"37[pYX\"))=0\"F-7$$!3#em;/,oln\"F1$\"3G#e\\:eX-Y*!#<7$$! 3Y&***\\(oWB>'!#>$\"3tgbi&=Z5^)F[o7$$\"3C%ommTIOo%F_o$\"3;nc/C([Rj(F[o 7$$\"3eML$3_>jU\"F1$\"3.xb4x]eOpF[o7$$\"3E,++D;v/DF1$\"34g=R7lWFiF[o7$ $\"3y+++v=h(e$F1$\"37D1M9#H$)e&F[o7$$\"3V+++v$[6j%F1$\"3b*Q`-(GcM]F[o7 $$\"3EMLe*[z(ybF1$\"3'\\OJ_))z$zXF[o7$$\"3knmmTXg0nF1$\"3%zuk9&\\P\"4% F[o7$$\"3OommmJ17zPi))=PF[o7$$\"3U++D1Mcq()F1$\"3b(Q95_X!GLF [o7$$\"3'fmmm\"pW`(*F1$\"3#F[o7$$\"3um;zpSS\"R\"F[o$\"3'R!fOZm!)*)>F[o7$$\"3GLL3_?` (\\\"F[o$\"3_&\\326_%*y\"F[o7$$\"3fL$e*)>pxg\"F[o$\"3U(f%[q@n-;F[o7$$ \"33+]Pf4t.tc-c9F[o7$$\"3uLLe*Gst!=F[o$\"3A].'Q!zn78F[o7$$ \"30+++DRW9>F[o$\"3)*QL%yI!Rz6F[o7$$\"3:++DJE>>?F[o$\"3!34L;)35i5F[o7$ $\"3F+]i!RU07#F[o$\"3)R%oh%pcF1 7$$\"3[m;H2qcZFF[o$\"3QfS:FpoE^F17$$\"3O+]7.\"fF&GF[o$\"3FQC-#R&z9YF17 $$\"3Ymm;/OgbHF[o$\"3U3D<8)yP;%F17$$\"3w**\\ilAFjIF[o$\"3w0Xme8wQPF17$ $\"3yLLL$)*pp;$F[o$\"3cgnS,][qLF17$$\"3)RL$3xe,tKF[o$\"3(=N\\E;e8.$F17 $$\"3Cn;HdO=yLF[o$\"3'R\\?AA[(GFF17$$\"3a+++D>#[Z$F[o$\"3g*)RTFyQxCF17 $$\"3SnmT&G!e&e$F[o$\"3*3eN#phk)F1$ \"3yB=#>))yB+#F[o7$F?$\"3C\"3Q%eFB+?F[o7$$!3amTN'4`_p(F1$\"3#e;cE&Q]+? 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" }{TEXT -1 7 ".8 and " }{XPPEDIT 18 0 "x=1" "6#/%\"xG\"\"\"" }{TEXT -1 23 " for the secant method." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "secant(8*exp(-x)=sin(2*x)+3, x=0.8..1,info=true);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximatio n~1~~->~~~G$\"/!4f\"*zrb'!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7appr oximation~2~~->~~~G$\"/+4K[#H6(!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$ %7approximation~3~~->~~~G$\"/#oq%ytvp!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$\"/ivyhSqp!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~->~~~G$\"/J`9bZqp!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~6~~->~~~G$\"/K?\"[v/(p!#9" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~7~~->~~~G$\"/K?\"[v/(p !#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+\"[v/(p!#5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 4 "Note" }{TEXT -1 89 " : It is important to take starting values which are reasonably close t o the desired root." }}{PARA 0 "" 0 "" {TEXT -1 83 "The result of the \+ following calculation is clearly incorrect. What has gone wrong? " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "secant(8*exp(-x)=sin(2*x)+3, x=0.9..1.8,info=true);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximat ion~1~~->~~~G$!.KXagXj$!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approx imation~2~~->~~~G$\"/lbKa\"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$% 7approximation~3~~->~~~G$\"/)pa-f#*H\"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$!/OD1zi;H!#7" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~->~~~G$\".qa-f#*H\"!#7" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~6~~->~~~G$\"/*oa-f#*H \"!#8" }}{PARA 7 "" 1 "" {TEXT -1 77 "Warning, a large change in value s suggests that the result may be unreliable\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+D!f#*H\"!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "The function used by " }{TEXT 0 6 "secant" }{TEXT -1 4 " is " } {XPPEDIT 18 0 "f(x) = 8*exp(-x)-sin(2*x)-3;" "6#/-%\"fG6#%\"xG,(*&\"\" )\"\"\"-%$expG6#,$F'!\"\"F+F+-%$sinG6#*&\"\"#F+F'F+F0\"\"$F0" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "When " }{XPPEDIT 18 0 "a =-29.16627906314" "6#/%\"aG,$-%&FloatG6$\" .9j!zi;H!#6!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "b=1.29925902548 " "6#/%\"bG-%&FloatG6$\"-[D!f#*H\"!#6" }{TEXT -1 31 ", the difference \+ in the values " }{XPPEDIT 18 0 "f(a)" "6#-%\"fG6#%\"aG" }{TEXT -1 5 " \+ and " }{XPPEDIT 18 0 "f(b)" "6#-%\"fG6#%\"bG" }{TEXT -1 4 " is " } {TEXT 266 10 "very large" }{TEXT -1 55 ". This causes the change, or c orrection, to be made in " }{TEXT 298 1 "b" }{TEXT -1 9 ", namely " } {XPPEDIT 18 0 "h= ``((f(b)-f(a))/(b-a))*f(b)" "6#/%\"hG*&-%!G6#*&,&-% \"fG6#%\"bG\"\"\"-F,6#%\"aG!\"\"F/,&F.F/F2F3F3F/-F,6#F.F/" }{TEXT -1 40 ", to give the next secant approximation " }{XPPEDIT 18 0 "c=b-h" " 6#/%\"cG,&%\"bG\"\"\"%\"hG!\"\"" }{TEXT -1 22 " for the root, to be \+ " }{TEXT 266 8 "so small" }{TEXT -1 101 " that the error criterion for the termination of the secant method is met under false circumstances . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 166 "f := x -> 8*exp(-x)-sin(2*x)-3;\nDigits := 13:\na := -29.16627906314;\nb := 1.29925902548;\nfa := f(a);\nfb := f(b);\nm := (fb-fa)/(b-a);\nh := fb/m;\nc := b-h;\nDigits := 10:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,(*&\"\") \"\"\"-%$expG6#,$9$!\"\"F/F/-%$sinG6#,$*&\"\"#F/F4F/F/F5\"\"$F5F(F(F( " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$!.9j!zi;H!#6" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%\"bG$\"-[D!f#*H\"!#6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#faG$\".%3@$=Sr$\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fbG$!.%*[@+\\L\"!#7" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG $!.`8P)3>7\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hG$\".Lrs)*\\4 \"!#C" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\".za-f#*H\"!#7" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 4 " }{TEXT 301 44 ".. an example where the convergence is slow " }}{PARA 0 "" 0 "" {TEXT -1 125 "The secant method is subject to slow convergence in the \+ case of a multiple root similar to that exhibited by Newton's method. " }}{PARA 0 "" 0 "" {TEXT -1 15 "The polynomial " }{XPPEDIT 18 0 "x^4- 4*x^2+4=(x^2-2)^2" "6#/,(*$%\"xG\"\"%\"\"\"*&F'F(*$F&\"\"#F(!\"\"F'F(* $,&*$F&F+F(F+F,F+" }{TEXT -1 14 " has the root " }{XPPEDIT 18 0 "x = s qrt(2)" "6#/%\"xG-%%sqrtG6#\"\"#" }{TEXT -1 36 ", which is a root of m ultiplicity 2." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 26 "plot(x^4-4*x^2+4,x=-1..2);" }}{PARA 13 "" 1 " " {GLPLOT2D 292 205 205 {PLOTDATA 2 "6%-%'CURVESG6$7_o7$$!\"\"\"\"!$\" \"\"F*7$$!1*****\\P&3Y$*!#;$\"1.XOax,p7!#:7$$!1++Dcx6x()F0$\"1F/ie.(>^ \"F37$$!1,+]iTDP\")F0$\"10/1$F37$$!1++D \"oS:P%F0$\"1J27**e5sKF37$$!1+++v@)*=PF0$\"1*Rg*R F37$$\"19+++DFOB!#=$\"1:INn\"y***RF37$$\"1,++DJL(4$Fbp$\"1bANJN;'*RF37 $$\"1,+++!R5'fFbp$\"1txDK!*z&)RF37$$\"1++v=(4AH*Fbp$\"1yhG!\\Ob'RF37$$ \"1++vV!QBE\"F0$\"1j#3*HS^ORF37$$\"1******\\\"o?&=F0$\"1!>D+MqR'QF37$$ \"1,+vVb4*\\#F0$\"12Pue93aPF37$$\"1,+DJ'=_6$F0$\"1F%)H;WB@OF37$$\"1,+] P%y!ePF0$\"1I@!\\??]X$F37$$\"1,+v=WU[VF0$\"1<0`bCSzKF37$$\"1++]7B>&)\\ F0$\"1))eW-snnIF37$$\"1++v$>:mk&F0$\"1Umy(4!HEGF37$$\"1++DcdQAiF0$\"1p $[ni&=,EF37$$\"1,+]PPBWoF0$\"18tBj-pXBF37$$\"1******\\Nm'[(F0$\"1\\*=` Zc@2#F37$$\"1****\\(yb^6)F0$\"1%)>5))yY*z\"F37$$\"1++vVVDB()F0$\"1c')G !yS_`\"F37$$\"1++]7TW)R*F0$\"1o%*y+@+Z7F37$$\"1+++:K^+5F3$\"1X/?tmZz** F07$$\"1++]7,Hl5F3$\"1dP%[$f'\\[(F07$$\"1+]P4w)R7\"F3$\"1\"3'fR&flU&F0 7$$\"1++]x%f\")=\"F3$\"1zqZY*)pgMF07$$\"1+]P/-a[7F3$\"1F07$$ \"1+](=Yb;J\"F3$\"1AQ8FlP:yFbp7$$\"1+v=7)3DM\"F3$\"1M/R;kM2RFbp7$$\"1+ +]i@Ot8F3$\"1DjSyFy'H\"Fbp7$$\"1](=7$$\"1]i:gI\"=U\"F3$\"1'>fy)*\\]k%F\\z7$$\"1 +]PfL'zV\"F3$\"1Z42L5_)e%Fbq7$$\"1+vouE2p9F3$\"1\\=()))z\">]#Fbp7$$\"1 +++!*>=+:F3$\"1y(ez'*HtF'Fbp7$$\"1+]7ed*>`\"F3$\"1AoUhj;/7F07$$\"1++DE &4Qc\"F3$\"1wwT)e/Z)>F07$$\"1+]P%>5pi\"F3$\"1$3po?yR=%F07$$\"1+vou;!fl \"F3$\"1x#ovo$z0bF07$$\"1+++bJ*[o\"F3$\"11HzQR%p.(F07$$\"1+]7j17=F37$$\"1+]P/)fT(=F3$\"1s&R>a!e(G#F37$$\"1+++b!)[/>F3$\"1FoTZAPZEF37$$ \"1+]i0j\"[$>F3$\"1)[=UaT)RIF37$$\"1](=#HA6^>F3$\"1tBgApmkKF37$$\"1+D \"G:3u'>F3$\"1_)yM=*\\*\\$F37$$\"1]iSwSq$)>F3$\"1/qa:=aWPF37$$\"\"#F*$ \"\"%F*-%'COLOURG6&%$RGBG$\"#5F)F*F*-%+AXESLABELSG6$Q\"x6\"%!G-%%VIEWG 6$;F(Fb`l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 44 "The secant method, with the starting values " } {XPPEDIT 18 0 "x=1.4" "6#/%\"xG-%&FloatG6$\"#9!\"\"" }{TEXT -1 5 " and " }{TEXT 300 1 "x" }{TEXT -1 45 " = 1.5, converges rather slowly to t his root." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "secant(x^4-4*x^2+4,x=1.4..1.5,info=true);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~1~~->~~~G$\"/.?UFP(R\"!#8" } }{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$\"/!e-QaNR \"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$\"/ 1>\\u*\\S\"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~- >~~~G$\"/X/T&pyS\"!#8" 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in att empt to circumvent problems that can arise in such cases. 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{XPPMATH 20 "6#/ %+correctionG$\"/+++gF[Q!#?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%8approximation~27~~->~~~G$\"/$\\fx8UT \"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%&valueG$\"\"$!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%+correctionG$\"/+++G[a6!#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%8approxima tion~28~~->~~~G$\"/@VI\\@99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%&v alueG$\"$[\"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%+correctionG$!/X. ^'o$y6!#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%8approximation~29~~->~~~G$\"/M1_P@99!#8" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/%&valueG$\"\"%!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%+correctionG$!/*))))QYKF$!#@" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%8approximation~30~~->~~~G$\" /4L>P@99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%&valueG$\"\"$!#8" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%+correctionG$!/++++v>)*!#@" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%8app roximation~31~~->~~~G$\"/M8@O@99!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/%&valueG$\"\"#!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%+correctionG$ !/++++&R'>!#?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%!G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%8approximation~32~~->~~~G$\"/%QZU8UT\"!#8" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%&valueG$\"\"#!#8" }}{PARA 8 "" 1 "" {TEXT -1 63 "Error, (in secant) zero denominator obtained in secant formula \n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "Si nce in the last two steps the two values of " }{XPPEDIT 18 0 "x^4-4*x^ 2+4" "6#,(*$%\"xG\"\"%\"\"\"*&F&F'*$F%\"\"#F'!\"\"F&F'" }{TEXT -1 163 " are the same (mainly as a consequence of having only one signifcant digit) the iterations terminate with a zero dominator occurring in th e secant method formula." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Example 5 " }{TEXT 311 93 ".. the secant method is suitab le when the expression for the function involved is complicated" }} {PARA 0 "" 0 "" {TEXT 305 8 "Question" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "If " }{XPPEDIT 18 0 " f(x)=arcsinh(1-sin(x)+x^2)/(sqrt(1+x^2)*ln(1+x*arctan(x)^2+exp(x^2))) " "6#/-%\"fG6#%\"xG*&-%(arcsinhG6#,(\"\"\"F--%$sinG6#F'!\"\"*$F'\"\"#F -F-*&-%%sqrtG6#,&F-F-*$F'F3F-F--%#lnG6#,(F-F-*&F'F-*$-%'arctanG6#F'F3F -F--%$expG6#*$F'F3F-F-F1" }{TEXT -1 44 ", find all the zeros of the 5t h derivative f" }{XPPEDIT 18 0 "``@@5" "6#-%#@@G6$%!G\"\"&" }{TEXT -1 2 " (" }{TEXT 306 1 "x" }{TEXT -1 5 ") of " }{XPPEDIT 18 0 "f(x)" "6#- %\"fG6#%\"xG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 307 8 "Solution" }{TEXT -1 2 ": " }}{PARA 0 "" 0 "" {TEXT -1 5 "Let " }{XPPEDIT 18 0 "g(x)=``" "6#/-%\"gG6#%\"xG%!G " }{TEXT -1 1 "f" }{XPPEDIT 18 0 "``@@5" "6#-%#@@G6$%!G\"\"&" }{TEXT -1 2 " (" }{TEXT 309 1 "x" }{TEXT -1 22 "). The expression for " } {XPPEDIT 18 0 "g(x)" "6#-%\"gG6#%\"xG" }{TEXT -1 24 " is rather compli cated. " }}{PARA 0 "" 0 "" {TEXT -1 67 "The secant method is a suitabl e method to use to find the zeros of " }{XPPEDIT 18 0 "g(x)" "6#-%\"gG 6#%\"xG" }{TEXT -1 94 " because it converges about as rapidly as Newto n's method but does not require the derivative " }{XPPEDIT 18 0 "g*`'` (x)=``" "6#/*&%\"gG\"\"\"-%\"'G6#%\"xGF&%!G" }{TEXT -1 1 "f" } {XPPEDIT 18 0 "`@@`(``,6);" "6#-%#@@G6$%!G\"\"'" }{TEXT -1 2 " (" } {TEXT 308 1 "x" }{TEXT -1 5 ") of " }{XPPEDIT 18 0 "g(x)" "6#-%\"gG6#% \"xG" }{TEXT -1 43 ", which will be even more complicated than " } {XPPEDIT 18 0 "g(x)" "6#-%\"gG6#%\"xG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 104 "f := x -> arcsinh(1-sin(x)+x^2)/(sqrt(1+x^2)* ln(1+x*arctan(x)^2+exp(x^2))):\n'f(x)'=f(x);\ng := (D@@5)(f):" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"fG6#%\"xG*(-%(arcsinhG6#,(\"\"\"F --%$sinGF&!\"\"*$)F'\"\"#F-F-F-,&F-F-F1F-#F0F3-%#lnG6#,(F-F-*&F'F-)-%' arctanGF&F3F-F--%$expG6#F1F-F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 21 "The expression for g(" }{TEXT 310 1 "x" }{TEXT -1 2 ") " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "'g(x)'=g(x);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/-%\"gG6#%\"xG,h \\l*0\"$S#\"\"\"-%(arcsinhG6#,&F+F+*$)F'\"\"#F+F+F+,&F+F+*$)F'\"\"%F+F +#!\"$F2-%#lnG6#,(F+F+*&F'F+)-%'arctanGF&F2F+F+-%$expG6#F0F+F8F'F+,(*$ F>F+F+**F2F+F'F+F?F+F/!\"\"F+*(F2F+F'F+FAF+F+F2F " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(g(x),x= -3..3);" }}{PARA 13 "" 1 "" {GLPLOT2D 524 335 335 {PLOTDATA 2 "6%-%'CU RVESG6$7[t7$$!\"$\"\"!$\"3'pMS-%eF!3&!#=7$$!3!******\\2<#pG!#<$\"3;&[# )>pa(erF-7$$!3#)***\\7bBav#F1$\"3[o1i)4)Gh(*F-7$$!36++]K3XFEF1$\"3AqQ9 UKn$R\"F17$$!3%)****\\F)H')\\#F1$\"3S.%*HR)37*>F17$$!3#****\\i3@/P#F1$ \"3IXn]M.Z\"z#F17$$!3;++Dr^b^AF1$\"39=*ov1&*=n$F17$$!3$****\\7Sw%G@F1$ \"3OWI'HS+IY%F17$$!3*****\\7;)=,?F1$\"35Txh;'Q?F%F17$$!3!****\\([\"[x$ >F1$\"3#oL*y>x^iJF17$$!3/++DO\"3V(=F1$\"3+#fNsO@5k(F-7$$!3)***\\i&G_!4 =F1$!3=h>^U%o5y$F17$$!3#******\\V'zV jo\"F1$!39@#)fO)ej9#Feo7$$!3******\\d;%)G;F1$!3'f6o(>F\\INFeo7$$!3!*** ***\\!)H%*\\\"F1$!376-gKk')zyFeo7$$!3'********=eWV\"F1$!3/$QPho>(35!#: 7$$!3/+++vl[p8F1$!3vM*)R@-6H6Fjp7$$!3'*****\\FN$QN\"F1$!3N9syms.I6Fjp7 $$!36+++![!=Q8F1$!3hVr4@WK;6Fjp7$$!30++]Ku_A8F1$!3!*o\\x0q<'3\"Fjp7$$! 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_xFecnF_xFagnF_xF_x**F_amF_xF]^mF_xFayF_xF^jnF_xF_x*(FeamF_xFecnF_xFfe mF_xFcx**F[amF_xFj\\nF_xFayF_xFe`nF_xF_x**Fe`mF_xFjilF_xFfjmF_xFh`lF_x F_x**F]amF_xFajlF_xFailF_xFgjlF_xFcx>8dp,8**F_`mF_xF\\gmF_xFbclF_xF^jn F_xFcx*(F]amF_xF\\[nF_xF`]nF_xFcx**F^hmF_xFhinF_xFeelF_xFijmF_xF_x*(\" $?#F_xFeflF_xFifmF_xF_x*(FeinF_xF_^lF_xFh]lF_xFcx**F`imF_xFggnF_xFayF_ xF^jnF_xFcx*(FeamF_xFeflF_xFgdmF_xF_x*(Fe`mF_xFeyF_xFg\\nF_xF_x*0F_`mF _xFjglF_xF[clF_xFayF_xFiclF_xFfalF_xFiwF_xFcx*(F[amF_xF][mF_xFh]lF_xFc x*,F]amF_xF\\[nF_xF[^lF_xF[xF_xFh[nF_xF_x>8ep,FFh_mF_xFhgmF_xFi]nF_xFg anF_xFddnF_xF]hnF_xFajnF_x*(F[amF_xF][mF_xFi_nF_xFcx*.F[`mF_xFbflF_xFb [lFcxF^[lFcxFf[mF_xFi[mF_xFcx**Fe`mF_xFbdlF_xFc]nF_xF[jnF_xF_x**FeamF_ xF^\\nF_xF`^mF_xFjgnF_xF_x*(F[amF_xF_^lF_xF`[mF_xFcx*,F[amF_xFb`nF_xFj jlF_xF]blF_xFb[nF_xFcx*.Fe^nF_xF_dlF_xF[flF_xFjjlF_xFb_lF_xFf^mF_xF_x* *F]`mF_xFjilF_xFadnF_xFh`lF_xF_x**Fi`mF_xF^\\nF_xFhdlF_xFd\\nF_xF_x**F [amF_xFg\\mF_xFhimF_xF[\\nF_xFcxFi[oF_xFbwFbwFbw" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "co degen[cost](g(x));\ncodegen[cost](gn);\nsavings=%%-%;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&\"%vA\"\"\"%*additionsGF&F&*&\"$%eF&%*division sGF&F&*&\"%!>#F&%*functionsGF&F&*&\"%,\\F&%0multiplicationsGF&F&" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,.*&\"$$>\"\"\"%(storageGF&F&*&F%F&%,a ssignmentsGF&F&*&\"#5F&%*functionsGF&F&*&\"$%oF&%0multiplicationsGF&F& *&\"$%>F&%*additionsGF&F&*&\"#UF&%*divisionsGF&F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%(savingsG,.*&\"%\"3#\"\"\"%*additionsGF(F(*&\"$U&F(%* divisionsGF(F(*&\"%!=#F(%*functionsGF(F(*&\"%F(%(storageGF(!\"\"*&F4F(%,assignmentsGF(F6" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "All 9 zeros can be calc ulated rapidly by the secant method using the procedure " }{TEXT 275 2 "gn" }{TEXT -1 40 " with suitable starting approximations. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 310 "x1 := secant('gn'(x),x=-1.9..-1.8);\nx2 := secant('gn'(x),x=-1.2. .-1.1);\nx3 := secant('gn'(x),x=-0.8..-0.7);\nx4 := secant('gn'(x),x=- 0.4..-0.3);\nx5 := secant('gn'(x),x=0..0.1);\nx6 := secant('gn'(x),x=0 .4..0.45);\nx7 := secant('gn'(x),x=0.9..1.1);\nx8 := secant('gn'(x),x= 1.8..1.9);\nx9 := secant('gn'(x),x=2.9..3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x1G$!+\\&y.'=!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%#x2G$!+7vR*=\"!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x3G$!+#oJ( yv!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x4G$!+>#*4EN!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x5G$\"+D\"\\VO\"!#6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x6G$\"+%Gil3%!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#x7G$\"+F:.[**!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x8G$\"+]5[^= !\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#x9G$\"+p=%o)G!\"*" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 119 "Newton's method is noticibly slower than the secant method because the express ion for the derivative is so complicated." }}{PARA 0 "" 0 "" {TEXT -1 251 "In addition, there are accuracy problems. Repeating a calculation produces slightly different intermediate results. This may be a resul t of the Maple feature whereby previously callculated numerical values are remembered in case they are needed again." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "st := time() :\nevalf(newton(g(x),x=0,info=true));\ntime()-st;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~1~~->~~~G$\"/d\"GR=>U\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$\"/S/RUVk8!#:" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$\"/lvC\"\\VO \"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$\"/ vcC\"\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~- >~~~G$\"/-dC\"\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+D\"\\VO \"!#6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"%$4(!\"$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "st := tim e():\nevalf(newton(g(x),x=0,info=true));\ntime()-st;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~1~~->~~~G$\"/r\"GR=>U\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$\"/3/RUVk8!#:" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$\"/\"eZ7\\VO \"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$\"/ ncC\"\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~- >~~~G$\"/)pX7\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+D\"\\VO \"!#6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"%pq!\"$" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "st := time ():\nnewton(g(x),x=0,info=true);\ntime()-st;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~1~~->~~~G$\"/t\"GR=>U\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$\"/U/RUVk8!#:" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$\"/%eZ7\\VO \"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$\"/ 'pX7\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~-> ~~~G$\"/%pX7\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+D\"\\VO\" !#6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"%9s!\"$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "Maple can find a derivati ve for the procedure " }{TEXT 275 2 "gn" }{TEXT -1 48 ", but this seem s to slow things down even more. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "st := time():\nnewton('gn'(x ),x=0,info=true);\ntime()-st;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7app roximation~1~~->~~~G$\"/a\"GR=>U\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~2~~->~~~G$\"/F/RUVk8!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~3~~->~~~G$\"/svC\"\\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~4~~->~~~G$\"/qcC\"\\VO\"!#: " }}{PARA 11 "" 1 "" {XPPMATH 20 "6$%7approximation~5~~->~~~G$\"/jcC\" \\VO\"!#:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+D\"\\VO\"!#6" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"&uV\"!\"$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Tasks" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q1 " }}{PARA 0 "" 0 "" {TEXT -1 72 "Use the secant method to find ( corre ct to 15 digits ) all the roots of " }{XPPEDIT 18 0 "f(x) = x*sin(x)-1 ;" "6#/-%\"fG6#%\"xG,&*&F'\"\"\"-%$sinG6#F'F*F*F*!\"\"" }{TEXT -1 27 " which lie in the interval " }{XPPEDIT 18 0 "[0,10]" "6#7$\"\"!\"#5" } {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 56 " Show how you have obtai ned your initial approximations." }}{PARA 0 "" 0 "" {TEXT -1 41 "_____ ____________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 41 "_______________________________ __________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 64 "Find a numerical approximation for the so lution of the equation " }{XPPEDIT 18 0 "3/(1+x^2) = Sn(x);" "6#/*&\" \"$\"\"\",&F&F&*$%\"xG\"\"#F&!\"\"-%#SnG6#F)" }{TEXT -1 8 ", where " } {TEXT 275 2 "Sn" }{TEXT -1 94 " is the numerical \"black box\" functi on defined by the procedure in the following subsection. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 40 "code for \"black box\" numerical function " }{TEXT 0 2 "Sn" } {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1470 "Sn := proc(x::realcons)\n local xx,eps,saveD igits,doS,val,p,q,maxit;\n \n if x=0 then return 0. end if;\n\n \+ doS := proc(x,eps,maxit)\n local t,p,s,u,v,h,i; \n # set up \+ a starting approximation\n if x<2.6 and x>-2.6 then \n s \+ := .9741807232*x-12.95848365/(x+32.34978388/\n (x+.702777931 1/(x-20.37123456/(x+35.08632448/(x-5.546018485/x)))));\n elif x>0 then \n s := x-1.570796327+(.2346004180+.4696458221*x)/\n \+ (1.+(-.6231393515+.4753495862*x)*x);\n else\n s := x+1.570796327+(-.2346004180+.4696458221*x)/\n (1.+(.6 231393515+.4753495862*x)*x);\n end if;\n # solve the equatio n x=s+arctan(s) for s by Halley's method \n for i to maxit do\n \+ t := s+arctan(s)-x;\n p := 1+s^2;\n u := 1+1/p; \n v := -2*s/p^2;\n h := t/(u-1/2*v*t/u);\n s \+ := s-h;\n if abs(h)<=eps*abs(s) then break end if;\n end \+ do;\n s;\n end proc;\n\n p := ilog10(Digits);\n q := Float( Digits,-p);\n maxit := trunc((p+(.02331061386+.1111111111*q))*2.0959 03274)+2;\n saveDigits := Digits;\n Digits := Digits+min(iquo(Digi ts,3),5);\n xx := evalf(x);\n eps := Float(3,-saveDigits-1);\n i f Digits<=trunc(evalhf(Digits)) and \n ilog10(xx)trunc(evalhf(DBL_MIN_10_EX P)) then\n val := evalhf(doS(xx,eps,maxit))\n else\n val : = doS(xx,eps,maxit)\n end if;\n evalf[saveDigits](val);\nend proc: " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 266 5 "Notes" }{TEXT -1 2 ": " }}{PARA 15 "" 0 "" {TEXT -1 11 "When using " }{TEXT 275 5 "Sn(x)" }{TEXT -1 30 " as (part of) an argu ment for " }{TEXT 0 4 "plot" }{TEXT -1 35 ", it must be enclosed in qu otes as " }{TEXT 0 7 "'Sn(x)'" }{TEXT -1 3 " or" }{TEXT 0 8 " 'Sn'(x) " }{TEXT -1 2 ". " }}{PARA 15 "" 0 "" {TEXT -1 13 "The function " } {TEXT 275 5 "Sn(x)" }{TEXT -1 47 " provides a numerical inverse for th e function " }{XPPEDIT 18 0 "g(x)=x+arctan(x)" "6#/-%\"gG6#%\"xG,&F'\" \"\"-%'arctanG6#F'F)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 41 " _________________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 41 "____________ _____________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q3" }}{PARA 0 "" 0 "" {TEXT -1 34 "Use the secant method to find the " }{TEXT 266 37 "least \+ and greatest positive solutions" }{TEXT -1 17 " of the equation " } {XPPEDIT 18 0 "sin(x^3)=x/6*(7-2*x)" "6#/-%$sinG6#*$%\"xG\"\"$*(F(\"\" \"\"\"'!\"\",&\"\"(F+*&\"\"#F+F(F+F-F+" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 103 " Your solutions should be correct to 10 digits. Sho w how you have obtained your initial approximations." }}{PARA 0 "" 0 " " {TEXT -1 41 "_________________________________________" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 41 "__ _______________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q4" }} {PARA 0 "" 0 "" {TEXT -1 45 "This question is concerned with the funct ion " }{XPPEDIT 18 0 "f(x) = 1-exp(-8*(x-1)^2)-exp(-9*(x-2)^2)-exp(-10 *(x-3)^2);" "6#/-%\"fG6#%\"xG,*\"\"\"F)-%$expG6#,$*&\"\")F)*$,&F'F)F)! \"\"\"\"#F)F2F2-F+6#,$*&\"\"*F)*$,&F'F)F3F2F3F)F2F2-F+6#,$*&\"#5F)*$,& F'F)\"\"$F2F3F)F2F2" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 22 " (a) Plot the graph of " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" } {TEXT -1 5 " for " }{XPPEDIT 18 0 "0 <= x;" "6#1\"\"!%\"xG" }{XPPEDIT 18 0 "`` <= 4;" "6#1%!G\"\"%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 26 "(b) Find all the zeros of " }{XPPEDIT 18 0 "f(x)" "6#-%\" fG6#%\"xG" }{TEXT -1 23 " correct to 10 digits. " }}{PARA 0 "" 0 "" {TEXT -1 41 "_________________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 41 "_________________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q5" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} {PARA 0 "" 0 "" {TEXT -1 100 "Find numerical approximations for all th e zeros of the function described in the subsections below. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 48 "Mathema tical expression for a soliton wave s(x) " }}{PARA 0 "" 0 "" {TEXT -1 3 "** " }{XPPEDIT 18 0 "s(x) = sec(arctan(1/2*tan(arctan(1/5*tan(arcta n(exp(-3*x-5/3))-arctan(exp(2*x+5/2))))-arctan(3*tan(arctan(exp(2*x+5/ 2))-arctan(exp(x+5))))))+arctan(exp(2*x+5/2))-arctan(19*tan(arctan(3*t an(arctan(exp(2*x+5/2))-arctan(exp(x+5))))+arctan(7/2*tan(arctan(exp(x +5))-arctan(exp(9/5*x+25/9))))))-arctan(exp(x+5)))^2*(1/2*sec(arctan(1 /5*tan(arctan(exp(-3*x-5/3))-arctan(exp(2*x+5/2))))-arctan(3*tan(arcta n(exp(2*x+5/2))-arctan(exp(x+5)))))^2*(1/5*sec(arctan(exp(-3*x-5/3))-a rctan(exp(2*x+5/2)))^2*(-3*exp(-3*x-5/3)/(1+exp(-3*x-5/3)^2)-2*exp(2*x +5/2)/(1+exp(2*x+5/2)^2))/(1+1/25*tan(arctan(exp(-3*x-5/3))-arctan(exp (2*x+5/2)))^2)-3*sec(arctan(exp(2*x+5/2))-arctan(exp(x+5)))^2*(2*exp(2 *x+5/2)/(1+exp(2*x+5/2)^2)-exp(x+5)/(1+exp(x+5)^2))/(1+9*tan(arctan(ex p(2*x+5/2))-arctan(exp(x+5)))^2))/(1+1/4*tan(arctan(1/5*tan(arctan(exp (-3*x-5/3))-arctan(exp(2*x+5/2))))-arctan(3*tan(arctan(exp(2*x+5/2))-a rctan(exp(x+5)))))^2)+2*exp(2*x+5/2)/(1+exp(2*x+5/2)^2)-19*sec(arctan( 3*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5))))+arctan(7/2*tan(arctan(ex p(x+5))-arctan(exp(9/5*x+25/9)))))^2*(3*sec(arctan(exp(2*x+5/2))-arcta n(exp(x+5)))^2*(2*exp(2*x+5/2)/(1+exp(2*x+5/2)^2)-exp(x+5)/(1+exp(x+5) ^2))/(1+9*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5)))^2)+7/2*sec(arctan (exp(x+5))-arctan(exp(9/5*x+25/9)))^2*(exp(x+5)/(1+exp(x+5)^2)-9/5*exp (9/5*x+25/9)/(1+exp(9/5*x+25/9)^2))/(1+49/4*tan(arctan(exp(x+5))-arcta n(exp(9/5*x+25/9)))^2))/(1+361*tan(arctan(3*tan(arctan(exp(2*x+5/2))-a rctan(exp(x+5))))+arctan(7/2*tan(arctan(exp(x+5))-arctan(exp(9/5*x+25/ 9)))))^2)-exp(x+5)/(1+exp(x+5)^2))/(1+1/16*tan(arctan(1/2*tan(arctan(1 /5*tan(arctan(exp(-3*x-5/3))-arctan(exp(2*x+5/2))))-arctan(3*tan(arcta n(exp(2*x+5/2))-arctan(exp(x+5))))))+arctan(exp(2*x+5/2))-arctan(19*ta n(arctan(3*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5))))+arctan(7/2*tan( arctan(exp(x+5))-arctan(exp(9/5*x+25/9))))))-arctan(exp(x+5)))^2)+12*s ec(arctan(exp(2*x+5/2))-arctan(exp(x+5)))^2*(2*exp(2*x+5/2)/(1+exp(2*x +5/2)^2)-exp(x+5)/(1+exp(x+5)^2))/(1+9*tan(arctan(exp(2*x+5/2))-arctan (exp(x+5)))^2);" "6#/-%\"sG6#%\"xG,&*(-%$secG6#,*-%'arctanG6#*(\"\"\"F 2\"\"#!\"\"-%$tanG6#,&-F/6#*(F2F2\"\"&F4-F66#,&-F/6#-%$expG6#,&*&\"\"$ F2F'F2F4*&FF2-F66#,&-F /6#*&FGF2-F66#,&-F/6#-FC6#,&*&F3F2F'F2F2*&F " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 48 "Maple func tion s for a soliton wave - version 1 " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2047 "s := x -> sec(arctan(1 /2*tan(arctan(1/5*tan(arctan(exp(-3*x-5/3))-arctan(exp(2*x+5/2))))-arc tan(3*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5))))))+arctan(exp(2*x+5/2 ))-arctan (19*tan(arctan(3*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5)))) +arctan(7/\n2*tan(arctan(exp(x+5))-arctan(exp(9/5*x+25/9))))))-arctan( exp(x+5)))^2*(1/2*\nsec(arctan(1/5*tan(arctan(exp(-3*x-5/3))-arctan(ex p(2*x+5/2))))-arctan(3*tan(\narctan(exp(2*x+5/2))-arctan(exp(x+5)))))^ 2*(1/5*sec(arctan(exp(-3*x-5/3))-\narctan(exp(2*x+5/2)))^2*(-3*exp(-3* x-5/3)/(1+exp(-3*x-5/3)^2)-2*exp(2*x+5/2)/\n(1+exp(2*x+5/2)^2))/(1+1/2 5*tan(arctan(exp(-3*x-5/3))-arctan(exp(2*x+5/2)))^2\n)-3*sec(arctan(ex p(2*x+5/2))-arctan(exp(x+5)))^2*(2*exp(2*x+5/2)/(1+exp(2*x +5/2)^2)-ex p(x+5)/(1+exp(x+5)^2))/(1+9*tan(arctan(exp(2*x+5/2))-arctan (exp(x+5)) )^2))/(1+1/4*tan(arctan(1/5*tan(arctan(exp(-3*x-5/3)) -arctan(exp(2*x+ 5/2))))-arctan(3*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5)))))^2)+2*exp (2*x+5/2)/(1+exp(2*x+5/2)^2)- 19*sec(arctan(3*tan(arctan(exp(2*x+5/2)) -arctan(exp(x+5))))+arctan(7/2*tan (arctan(exp(x+5))-arctan(exp(9/5*x+ 25/9)))))^2*(3*sec(arctan(exp\n(2*x+5/2))-arctan(exp(x+5)))^2*(2*exp(2 *x+5/2)/(1+exp(2*x+5/2)^2)-exp(x+5)/(1\n+exp(x+5)^2))/(1+9*tan(arctan( exp(2*x+5/2))-arctan(exp(x+5)))^2)+7/2*sec(\narctan(exp(x+5))-arctan(e xp(9/5*x+25/9)))^2*(exp(x+5)/(1+exp(x+5)^2)-9/5*exp(\n9/5*x+25/9)/(1+e xp(9/5*x+25/9)^2))/(1+49/4*tan(arctan(exp(x+5))-arctan(exp(9 /5*x+25/9 )))^2))/(1+361*tan(arctan(3*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5 )) ))+arctan(7/2*tan(arctan(exp(x+5))-arctan(exp(9/5*x+25/9)))))^2)-exp(x +5)/(1+exp(x+5)^2))/(1+1/16*tan(arctan(1/2*tan(arctan(1/5*tan(arctan ( exp(-3*x-5/3))-arctan(exp(2*x+5/2))))-arctan(3*tan(arctan(exp(2*x+5/2) )-arctan(exp(x+5))))))+arctan(exp(2*x+5/2))-arctan(19*tan(arctan(3*tan (arctan (exp(2*x+5/2))-arctan(exp(x+5))))+arctan(7/2*tan(arctan(exp(x+ 5))-arctan (exp(9/5*x+25/9))))))-arctan(exp(x+5)))^2)+12*sec(arctan(ex p(2*x+5/2)) -arctan(exp(x+5)))^2*(2*exp(2*x+5/2)/(1+exp(2*x+5/2)^2)-ex p(x+5)/(1+exp(x+5)^2)) /(1+9*tan(arctan(exp(2*x+5/2))-arctan(exp(x+5)) )^2):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 60 "Maple func tion (procedure) sn for a soliton wave - version 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "This version was construc ted \"by hand\". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1043 "sn := proc(x)\n local t1,t2,t3,t4,at1,at2 ,at3,at4,st1,st2,st3,st4,tn12,\n tn23,tn34,s12,s23,s34,q1,q2,q3,q4,a 1,a2,a3,sa1,sa2,sa3,\n u1,u2,u3,u4,u5,u6;\n\nt1 := exp(-3*x-5/3);\nt 2 := exp(2*x+5/2);\nt3 := exp(x+5);\nt4 := exp(9/5*x+25/9);\n\nat1 := \+ arctan(t1);\nat2 := arctan(t2);\nat3 := arctan(t3);\nat4 := arctan(t4) ;\n\ntn12 := tan(at1-at2);\ntn23 := tan(at2-at3);\ntn34 := tan(at3-at4 );\n\ns12 := sec(at1-at2)^2;\ns23 := sec(at2-at3)^2;\ns34 := sec(at3-a t4)^2;\n\nst1 := t1^2;\nst2 := t2^2;\nst3 := t3^2;\nst4 := t4^2;\n\nq1 := t1/(1+st1);\nq2 := t2/(1+st2);\nq3 := t3/(1+st3);\nq4 := t4/(1+st4 );\n\na1 := arctan(3*tn23)+arctan(7/2*tn34);\na2 := arctan(tn12/5)-arc tan(3*tn23);\na3 := arctan(tan(a2)/2)+at2-arctan(19*tan(a1))-at3;\n\ns a1 := sec(a1)^2;\nsa2 := sec(a2)^2;\nsa3 := sec(a3)^2;\n\nu1 := 1+9*tn 23^2;\nu2 := 1+tn12^2/25;\nu3 := 1+49/4*tn34^2;\nu4 := 1+tan(a3)^2/16; \nu5 := 1+tan(a2)^2/4;\nu6 := 1+361*tan(a1)^2;\n\nsa3*(1/2*sa2*(1/5*s1 2*(-3*q1-2*q2)/u2-3*s23*(2*q2-q3)/u1)/u5+2*q2\n-19*sa1*(3*s23*(2*q2-q3 )/u1+7/2*s34*(q3-9/5*q4)/u3)/u6-q3)/u4+12*s23*(2*q2-q3)/u1;\nend proc: " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 61 "Maple func tion (procedure) sn2 for a soliton wave - version 3" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1047 "sn2 := p roc(x)\nlocal t1,t2,t3,t4,t5,t6,t7,t8,t9,t10,t11,t12,t13,t14,t15,t17, \n t18,t19,t20,t21,t22,t26,t27,t28,t29,t30,t31,t32,t33,t34;\n t30 \+ := exp(2*x+5/2);\n t34 := t30/(1+t30^2);\n t27 := arctan(t30);\n \+ t31 := exp(x+5);\n t28 := arctan(t31);\n t22 := t27-t28;\n t33 \+ := t31/(1+t31^2);\n t10 := 2*t34-t33;\n t19 := sec(t22);\n t20 : = tan(t22);\n t32 := t19^2*t10/(1+9*t20^2);\n t29 := exp(-3*x-5/3) ;\n t26 := exp(9/5*x+25/9);\n t21 := arctan(t29)-t27;\n t18 := t an(t21);\n t17 := t28-arctan(t26);\n t15 := sec(t21);\n t14 := s ec(t17);\n t13 := tan(t17);\n t12 := arctan(3*t20);\n t9 := -arc tan(1/5*t18)+t12;\n t8 := t12+arctan(7/2*t13);\n t7 := tan(t9);\n \+ t6 := sec(t9);\n t5 := sec(t8);\n t4 := tan(t8);\n t3 := -arct an(1/2*t7)-arctan(19*t4)+t22;\n t2 := tan(t3);\n t1 := sec(t3);\n \+ t11 := t1^2*(1/2*t6^2*(1/5*t15^2*(-3*t29/(1+t29^2)-2*t34)/\n ( 1+1/25*t18^2)-3*t32)/(1+1/4*t7^2)-\n 19*t5^2*(3*t32+7/2*t14^2*(t 33-9/5*t26/(1+t26^2))/\n (1+49/4*t13^2))/(1+361*t4^2)+t10)/(1+1/ 16*t2^2)+12*t32\nend proc:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 27 "The code for the procedure " }{TEXT 275 3 "sn2" }{TEXT -1 71 " can be generated automatically as follows, \+ provided that the function " }{TEXT 275 1 "s" }{TEXT -1 48 " from the \+ earlier subsection has been activated." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "sn2 := codegen[makeproc] ([codegen[optimize](s(x),'tryhard')],x):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 34 "Output code for the proced ure sn2 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "eval(sn2);" }}{PARA 2 "" 1 "" {TEXT -1 1157 " proc(x )\nlocal t1, t2, t3, t4, t5, t6, t7, t8, t9, t10, t11, t12, t13, t14, \+ t15, t17, t18, t19,\nt20, t21, t22, t26, t27, t28, t29, t30, t31, t32, t33, t34;\n t30 := exp(2*x + 5/2);\n t34 := t30/(1 + t30^2);\n \+ t27 := arctan(t30);\n t31 := exp(x + 5);\n t28 := arctan(t31) ;\n t22 := t27 - t28;\n t33 := t31/(1 + t31^2);\n t10 := 2*t3 4 - t33;\n t19 := sec(t22);\n t20 := tan(t22);\n t32 := t19^2 *t10/(1 + 9*t20^2);\n t29 := exp(-3*x - 5/3);\n t26 := exp(9/5*x + 25/9);\n t21 := arctan(t29) - t27;\n t18 := tan(t21);\n t1 7 := t28 - arctan(t26);\n t15 := sec(t21);\n t14 := sec(t17);\n \+ t13 := tan(t17);\n t12 := arctan(3*t20);\n t9 := -arctan(1/5* t18) + t12;\n t8 := t12 + arctan(7/2*t13);\n t7 := tan(t9);\n \+ t6 := sec(t9);\n t5 := sec(t8);\n t4 := tan(t8);\n t3 := -ar ctan(1/2*t7) - arctan(19*t4) + t22;\n t2 := tan(t3);\n t1 := sec (t3);\n t11 := t1^2*(1/2*t6^2*\n (1/5*t15^2*(-3*t29/(1 + t29 ^2) - 2*t34)/(1 + 1/25*t18^2) - 3*t32)/(1 + 1/4*t7^2)\n - 19*t 5^2*(3*t32 + 7/2*t14^2*(t33 - 9/5*t26/(1 + t26^2))/(1 + 49/4*t13^2))/( \n 1 + 361*t4^2) + t10)/(1 + 1/16*t2^2) + 12*t32\nend proc" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 283 7 "Remarks" }}{PARA 15 "" 0 "" {TEXT -1 115 "Versions 2 and 3 evaluate m ore efficiently than version 1 because it avoids duplicate evaluation \+ of subexpressions." }}{PARA 15 "" 0 "" {TEXT -1 35 "When using version s 2 and 3 with a " }{TEXT 0 4 "plot" }{TEXT -1 56 " command, or with a root-finding procedure in the forms " }{TEXT 275 5 "sn(x)" }{TEXT -1 5 " and " }{TEXT 275 6 "sn2(x)" }{TEXT -1 28 ", quotes must be used as in " }{TEXT 267 7 "'sn'(x)" }{TEXT -1 4 " or " }{TEXT 267 7 "'sn(x)' " }{TEXT -1 14 ", for example." }}{PARA 15 "" 0 "" {TEXT -1 410 "Maple can find the derivative of all three versions so Newton's method coul d be used. However Newton's method is not as efficient as the secant m ethod firstly because of the time required to obtain the derivative, a nd secondly because of the time required to evaluate both the function and derivative for each iteration. In contrast only one function eval uation per iteration is required by the secant method. " }}{PARA 15 " " 0 "" {TEXT -1 79 "The formula for this function was given with Maple 4 as an example for Maple's " }{TEXT 0 4 "plot" }{TEXT -1 12 " proced ure. " }}{PARA 0 "" 0 "" {TEXT -1 41 "________________________________ _________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 41 "_______________________________________ __" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q6" }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 38 " be defined by the piecewi se formula. " }}{PARA 257 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "f(x) \+ = PIECEWISE([-1, x <= -3],[1/2-3/2*tanh((2*x+5)/(24+4*x^2+20*x)), -3 " 0 " " {MPLTEXT 1 0 168 "f := x -> piecewise(x<=-3,-1,x<-2,1/2-3/2*tanh((2* x+5)/(4*(6+x^2+5*x))),\n x<=2,2,x<3,3/2*tanh((2*x-5)/(4*(6+x^2 -5*x)))+1/2,-1):\n'f(x)'=f(x);\nplot(f(x),x=-5..5);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#/-%\"fG6#%\"xG-%*PIECEWISEG6'7$!\"\"1F'!\"$7$,&#\"\" \"\"\"#F2*&#\"\"$F3F2-%%tanhG6#*&,&*&F3F2F'F2F2\"\"&F2F2,(\"#CF2*&\"\" %F2)F'F3F2F2*&\"#?F2F'F2F2F,F2F,2F'!\"#7$F31F'F37$,&*&#F6F3F2-F86#*&,& *&F3F2F'F2F2F=F,F2,(F?F2*&FAF2FBF2F2*&FDF2F'F2F,F,F2F2F1F22F'F67$F,%*o therwiseG" }}{PARA 13 "" 1 "" {GLPLOT2D 658 264 264 {PLOTDATA 2 "6%-%' CURVESG6$7jp7$$!\"&\"\"!$!\"\"F*7$$!3YLLLe%G?y%!#f_ PF0F+7$$!3K++vo1YZNF0F+7$$!3;LL3-OJNLF0F+7$$!3?mm\"zC!eHKF0F+7$$!3p*** \\P*o%Q7$F0F+7$$!37m\"H#=qYpIF0F+7$$!3+L$3F9(3:IF0F+7$$!3m;z%\\?(*y)HF 0F+7$$!3))*\\(=nsqgHF0$!3]]45s(\\)****!#=7$$!3[\"H2$)H7r%HF0$!3wwZ.mT, '***FZ7$$!35$3F%Ht^LHF0$!3eaH:*))fA(**FZ7$$!3ruoagB#*>HF0$!35-^pos**** )*FZ7$$!3Kmmm\"RFj!HF0$!3[8Q^\"[9RjiFF0$!3?aM9bmx.VFZ7$$ !33LL$e4OZr#F0$!3h`\"R!H1pTAFZ7$$!3a**\\i!\\93m#F0$!3=;^shgR:;!#>7$$!3 UmmT&)G*og#F0$\"3uY)Gm_RZp\"FZ7$$!3IL$3-GrHb#F0$\"3ej!>2)\\%*)R$FZ7$$! 3u*****\\n\\!*\\#F0$\"3(fZg'=)4&G]FZ7$$!3x****\\im!\\W#F0$\"3ywr&oJ4im 'FZ7$$!3y******\\Ow!R#F0$\"390rJ\"Q8AQ)FZ7$$!3#)****\\P1iOBF0$\"3qe;'3 \">\\D5F07$$!3%)*****\\ixCG#F0$\"3\"zBB/G?cB\"F07$$!3>++++3IIAF0$\"3eX VA&>kAY\"F07$$!35+++vR7y@F0$\"3uXk_'))*Q+F07$$!3#******\\KqP2#F0$\"3))\\zux*\\T*>F07$$!3am;a8s+z>F0$\"\" #F*7$$!39LL3-TC%)=F0F]u7$$!3[mmm\"4z)e;F0F]u7$$!3Mmmmm`'zY\"F0F]u7$$!3 #****\\(=t)eC\"F0F]u7$$!3!ommmh5$\\5F0F]u7$$!3S$***\\(=[jL)FZF]u7$$!3) f***\\iXg#G'FZF]u7$$!3ndmmT&Q(RTFZF]u7$$!3%\\mmTg=><#FZF]u7$$!3vDMLLe* e$\\!#?F]u7$$\"3!=nm\"zRQb@FZF]u7$$\"3_,+](=>Y2%FZF]u7$$\"3summ\"zXu9' FZF]u7$$\"3#4+++]y))G)FZF]u7$$\"3H++]i_QQ5F0F]u7$$\"3b++D\"y%3T7F0F]u7 $$\"3+++]P![hY\"F0F]u7$$\"3iKLL$Qx$o;F0F]u7$$\"3Y+++v.I%)=F0F]u7$$\"3! p;Hd&\\@L>F0F]u7$$\"3ML$ek`H@)>F0F]u7$$\"3M;H#o#oe1?F0F]u7$$\"3y*\\(=< T/J?F0$\"3;NnL\"\\*****>F07$$\"3A$3_vS,b0#F0$\"3U,d)*=qP**>F07$$\"3?mm \"zpe*z?F0$\"3eZ5&)oD4!*>F07$$\"3&**\\(oa_VL@F0$\"3GaiC[b3z=F07$$\"3oL $e9\"=\"p=#F0$\"3K&3B8F)*3m\"F07$$\"3Vn\"H#o$)QSAF0$\"3nA(pvYJoT\"F07$ $\"3;,++D\\'QH#F0$\"3?c(zjLm#*=\"F07$$\"3*Q$eR(>#=WBF0$\"3]`Ok!Q!4\")* *FZ7$$\"3/n;zp%*\\%R#F0$\"3IkQz*)fnf#)FZ7$$\"3A+v=Un\"[W#F0$\"35$o\\&* ys*omFZ7$$\"3%HL$e9S8&\\#F0$\"3#**RY*z())f9&FZ7$$\"3!*****\\i+tZDF0$\" 3yUZFkPQfNFZ7$$\"3%om;/6E.g#F0$\"3B'>ELqi)3>FZ7$$\"3yLLLe@#Hl#F0$\"3#o zbu)y7E7Fbq7$$\"3s++D1#=bq#F0$!3ke/]U)*)y'=FZ7$$\"3yL3xc/%pv#F0$!3Oe;9 ..5\\SFZ7$$\"3#om\"H2FO3GF0$!3Y\"\\?m=vSR'FZ7$$\"3')*\\7y&\\yfGF0$!3cu 2w(fXfb)FZ7$$\"3\"HLL$3s?6HF0$!36&*e\\]:*\\\")*FZ7$$\"3jCc,\"zlY#HF0$! 3MZzAX*>E$**FZ7$$\"3N;zptV7QHF0$!3'4;jwW*=%)**FZ7$$\"313-QcHe^HF0$!3e9 FFS!R$)***FZ7$$\"3!)*\\i!R:/lHF0$!3YqY^d!p*****FZ7$$\"3A$3FWqe>*HF0F+7 $$\"3Am;zpe()=IF0F+7$$\"3mK3_+-rsIF0F+7$$\"3a***\\7`Wl7$F0F+7$$\"3enmm m*RRL$F0F+7$$\"3%zmmTvJga$F0F+7$$\"3]MLe9tOcPF0F+7$$\"31,++]Qk\\RF0F+7 $$\"3![LL3dg6<%F0F+7$$\"3%ymmmw(GpVF0F+7$$\"3C++D\"oK0e%F0F+7$$\"35,+v =5s#y%F0F+7$$\"\"&F*F+-%'COLOURG6&%$RGBG$\"#5F,$F*F*Fbcl-%+AXESLABELSG 6$Q\"x6\"Q!Fgcl-%%VIEWG6$;F(Fjbl%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 41 "_________________________________________" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 41 "_________________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 2 "Q7" }}{PARA 0 "" 0 "" {TEXT -1 45 "This question is concer ned with the function " }{XPPEDIT 18 0 "f(x)= (4*sqrt(13+4*x^2-12*x)-8 +tanh(2*x-3))/(8*x-12)" "6#/-%\"fG6#%\"xG*&,(*&\"\"%\"\"\"-%%sqrtG6#,( \"#8F,*&F+F,*$F'\"\"#F,F,*&\"#7F,F'F,!\"\"F,F,\"\")F7-%%tanhG6#,&*&F4F ,F'F,F,\"\"$F7F,F,,&*&F8F,F'F,F,F6F7F7" }{TEXT -1 3 ". " }}{PARA 0 " " 0 "" {TEXT -1 22 "(a) Plot the graph of " }{XPPEDIT 18 0 "f(x)" "6#- %\"fG6#%\"xG" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "-8 <= x;" "6#1,$\"\" )!\"\"%\"xG" }{XPPEDIT 18 0 "`` <= 8;" "6#1%!G\"\")" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 28 "(b) Find the single zero of " } {XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%\"xG" }{TEXT -1 51 " correct to 10 di gits by the secant method method. " }}{PARA 0 "" 0 "" {TEXT -1 31 " \+ You may use the procedure " }{TEXT 0 6 "secant" }{TEXT -1 10 " for t his." }}{PARA 0 "" 0 "" {TEXT -1 75 "(c) Investigate the convergence o f the secant method to the single zero of " }{XPPEDIT 18 0 "f(x)" "6#- %\"fG6#%\"xG" }{TEXT -1 28 " with the starting interval " }{XPPEDIT 18 0 "[1,2]" "6#7$\"\"\"\"\"#" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 31 " You may use the procedure " }{TEXT 0 11 "secant_step " }{TEXT -1 10 " for this." }}{PARA 0 "" 0 "" {TEXT -1 57 "(d) What ha ppens if the starting interval is taken to be " }{XPPEDIT 18 0 "[4,5] " "6#7$\"\"%\"\"&" }{TEXT -1 2 "? " }}{PARA 0 "" 0 "" {TEXT -1 41 "___ ______________________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 41 "__ _______________________________________" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}} }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 0 "" 0 "" {TEXT -1 24 "Code for drawing pictu re" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 755 "x := 'x': fn := x -> x^2-2:\ny0 := fn(1.0): y1 := fn (2.0): \np1 := plot(fn(x),x=0.5..2.5):\np2 := plot([[[1,0],[1,y0]],[[2 ,0],[2,y1]]],linestyle=2,color=black):\np3 := plot([[[1,y0],[evalf(4/3 ),0],[2,y1]]$3],style=point,\n color=navy,symbol=[circle,diamond ,cross]):\np4 := plot([[1,y0],[2,y1]],color=blue):\np5 := plot([[0,0], [4,0]],color=black):\nt1 := plots[textplot]([2.48,3.4,`y = f(x)`],colo r=red,font=[HELVETICA,10]):\nt2 := plots[textplot]([[1.15,-1.15,`(x , y )`],[2.15,2,`(x ,y )`],\n [.98,0.3,`x`],[1.98,-0.2,`x`],[1.28,0 .3,`x`]],font=[HELVETICA,10]):\nt3 := plots[textplot]([[1.16,-1.25,`0 \+ 0`],[2.16,1.9,`1 1`],\n [1.015,0.2,`0`],[2.015,-0.3,`1`],[1.31 5,0.2,`2`]],font=[HELVETICA,8]):\nplots[display]([p1,p2,p3,p4,p5,t1,t2 ,t3],axes=none);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}} {MARK "4 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }