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Repeat the process for angles closer \+ to the top, and make your observation." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 43 "Let us introduce the concept of proce dures:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 251 "a procedure takes arguments (which it cannnot modify), and comput es something which gets returned (very much like a FUNCTION subprogram in Fortran). If one wants to return more than one object, then one ha s to place it on the list of global variables." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 130 "Our objective is to put \+ the computation of the list of datapoints into a procedure. 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)Fgfl$\"3')*44`^b5v#Fgfl7$F`[o$\"33+x&pwi\\o#Fgfl7$$\"3O++++++]\")Fgfl $\"3$)*RzP(o%yg#Fgfl7$F`cm$\"3#**f!zQ>,=DFgfl7$$\"3+++++++]#)Fgfl$\"3$ *****Rh$yNT#Fgfl7$Fh[o$\"3'**RW%o+^#H#Fgfl7$$\"3k************\\$)Fgfl$ \"3!**p?-\"\\n_@Fgfl7$Fecm$\"3/+.NgR&>*>Fgfl7$$\"3G************\\%)Fgf l$\"3***f%>.wT3=Fgfl7$F`\\o$\"3!**fh'HPe+;Fgfl7$$\"3s++++++]&)Fgfl$\"3 7+dUY4wn8Fgfl7$Fjcm$\"30+(=vo^/6\"Fgfl7$$\"3O++++++]')Fgfl$\"3J+r_D%\\ uI)F[gl7$Fh\\o$\"3y*pr$)\\tdK&F[gl7$$\"3+++++++]()Fgfl$\"32+'ztrer@#F[ gl7$F_dm$!3O+/wv,Kk%*Ff[n7$$\"3k************\\))Fgfl$!3-+4BfsR'3%F[gl7 $F`]o$!3k*H4\\9Aq7(F[gl7$$\"3G************\\*)Fgfl$!3.+,$poFgfl7$Fidm$!3-+-1u,j! 4#Fgfl7$$\"3+++++++]#*Fgfl$!3<+;Ni/iQAFgfl7$F`^o$!35+e(eZ\")pO#Fgfl7$$ \"3k************\\$*Fgfl$!39+!4M`cyZ#Fgfl7$F^em$!3%)*\\W\"*)*HLd#Fgfl7 $$\"3G************\\%*Fgfl$!39+chI#\\`l#Fgfl7$Fh^o$!3&**p)[rdoDFFgfl7$ $\"3s++++++]&*Fgfl$!3%)*p%o?=#fy#Fgfl7$Fcem$!3)**H.wQ_u$GFgfl7$$\"3O++ ++++]'*Fgfl$!3++/GHh\\\")GFgfl7$F`_o$!3)***p=D06>HFgfl7$$\"3+++++++](* Fgfl$!3))*f:)R'37&HFgfl7$Fhem$!3++$R[&edyHFgfl7$$\"3k************\\)*F gfl$!3\")*H%Gwc)=+$Fgfl7$Fh_o$!3')*z1R<9<-$Fgfl7$$\"3G************\\** Fgfl$!38+(HD)GbQIFgfl7$Fiel$!3')*>iF:?G0$Fgfl-F^fl6&F`flF(FafmF(FdfmFh fm-%+AXESLABELSG6%Q\"t6\"Q!Ffbr-%%FONTG6#%(DEFAULTG-%%VIEWG6$;F(FielF[ cr" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 " For this choice of initial condition the spaceing of " }{TEXT 19 6 "dt =0.2" }{TEXT -1 87 " leads quickly to a big deviation, while the small er spacings seem to work for a while." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 256 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "Repeat the above for " } {TEXT 19 21 "phi0=3.12, 3.13, 3.14" }{TEXT -1 29 ", and make your obse rvations." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Let us now understan d how the results depend on " }{TEXT 19 2 "dt" }{TEXT -1 131 ". Let us simply look at the final point in order to have a fixed time point, a nd we will also generate a more accurate calculation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "dt4:=dt3/2: n4:=trunc(10/dt4): Res4 :=Verlet(3.11,dt4,n4):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "R es1[n1+1],Res2[n2+1],Res3[n3+1],Res4[n4+1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&7$$\"$+\"!\"\"$\"/\"R*GFb\\^!#87$$\"/++++++5!#7$!/l>UYu PIF)7$F+$!/Aw_,#G0$F)7$F+$!/:)\\Qhx0$F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Res_dsolve:=Y(10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%+Res_dsolveG$!3k(>AtY*QgI!#<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "dt:='dt': # unassigned, so that we can use it to label the hor izontal plot axis!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 199 "P_ac c:=plot([[0.1,Res2[n2+1][2]],[0.05,Res3[n3+1][2]],[0.025,Res4[n4+1][2] ]],dt=0..0.1,style=point,symbol=cross,symbolsize=20): P_exact:=plot(Re s_dsolve,dt=0..0.1,color=blue): display(P_acc,P_exact);" }}{PARA 13 " " 1 "" {GLPLOT2D 789 341 341 {PLOTDATA 2 "6&-%'CURVESG6&7%7$$\"3/+++++ ++5!#=$!3)**\\'>UYuPI!#<7$$\"3G+++++++]!#>$!3')*>iF:?G0$F-7$$\"39+++++ ++DF1$!3=+:)\\Qhx0$F--%'COLOURG6&%$RGBG$\"#5!\"\"$\"\"!FAF@-%&STYLEG6# %&POINTG-%'SYMBOLG6$%&CROSSG\"#?-F$6$7S7$F@$!3k(>AtY*QgIF-7$$\"3[mmm;a rz@!#?FO7$$\"3mLL$e9ui2%FTFO7$$\"3-nmm\"z_\"4iFTFO7$$\"3[mmmT&phN)FTFO 7$$\"3KLLe*=)H\\5F1FO7$$\"3omm\"z/3uC\"F1FO7$$\"3#****\\7LRDX\"F1FO7$$ \"3emm\"zR'ok;F1FO7$$\"3'****\\i5`h(=F1FO7$$\"3gLLL3En$4#F1FO7$$\"3wmm ;/RE&G#F1FO7$$\"3A+++D.&4]#F1FO7$$\"3!)*****\\PAvr#F1FO7$$\"3)******\\ nHi#HF1FO7$$\"3jmm\"z*ev:JF1FO7$$\"31LLL347TLF1FO7$$\"3cLLLLY.KNF1FO7$ $\"3!****\\7o7Tv$F1FO7$$\"3sKLL$Q*o]RF1FO7$$\"33++D\"=lj;%F1FO7$$\"33+ +vV&RY2aF1FO7$$\"3qmm;zXu9cF1FO7$$ \"3^******\\y))GeF1FO7$$\"3!)****\\i_QQgF1FO7$$\"3j***\\7y%3TiF1FO7$$ \"3o****\\P![hY'F1FO7$$\"33KLL$Qx$omF1FO7$$\"3k+++v.I%)oF1FO7$$\"3Amm \"zpe*zqF1FO7$$\"37+++D\\'QH(F1FO7$$\"3GKLe9S8&\\(F1FO7$$\"3]++D1#=bq( F1FO7$$\"3>LLL3s?6zF1FO7$$\"3)*)**\\7`Wl7)F1FO7$$\"3[nmmm*RRL)F1FO7$$ \"3Smm;a<.Y&)F1FO7$$\"3-MLe9tOc()F1FO7$$\"3u******\\Qk\\*)F1FO7$$\"3!Q LL3dg6<*F1FO7$$\"3-mmmmxGp$*F1FO7$$\"3!3+]7oK0e*F1FO7$$\"3'****\\(=5s# y*F1FO7$F(FO-F:6&F " 0 "" {MPLTEXT 1 0 15 "A:= 'A': B:='B':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "eq1:=A + B* dt2^2 = Res2[n2+1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/,&% \"AG\"\"\"*&$\"/++++++5!#:F(%\"BGF(F($!/l>UYuPI!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "eq2:=A + B*dt3^2 = Res3[n3+1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,&%\"AG\"\"\"*&$\"/++++++D!#;F(%\"BG F(F($!/Aw_,#G0$!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:= solve(\{eq1,eq2\},\{A,B\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG <$/%\"BG$\"/nUv!o+,#!#8/%\"AG$!/3&HKXy0$F*" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 12 "assign(sol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "A,Res_dsolve; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!/3&HKXy0$! #8$!3k(>AtY*QgI!#<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "The extrapo lation of our two results (1 and 2) does not agree all that well with \+ " }{TEXT 19 6 "dsolve" }{TEXT -1 60 "'s answer. What does it predict f or the answer with spacing " }{TEXT 19 5 "0.025" }{TEXT -1 1 "?" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "A + B * dt4^2,Res4[n4+1][2]; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!/PSI!*ecI!#8$!/:)\\Qhx0$F%" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "Therefore, we can realize that the function is not a perfect second-order parabola in the range " } {TEXT 19 10 "dt=0..0.1 " }{TEXT -1 51 "even without having access to a n accurate solution." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "P_e xtra:=plot(A + B * dt^2, dt=0..0.1, color=magenta):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "display(P_extra,P_acc,P_exact);" }}{PARA 13 "" 1 "" {GLPLOT2D 491 491 491 {PLOTDATA 2 "6'-%'CURVESG6$7S7$$\"\"! F)$!3!)*z]HKXy0$!#<7$$\"3[mmm;arz@!#?$!37urzsd$y0$F,7$$\"3mLL$e9ui2%F0 $!3eCxjB>\"y0$F,7$$\"3-nmm\"z_\"4iF0$!3%RoLw#ywdIF,7$$\"3[mmmT&phN)F0$ !37#)G!)o\\qdIF,7$$\"3KLLe*=)H\\5!#>$!3a!p!44SidIF,7$$\"3omm\"z/3uC\"F E$!33TO'4bKv0$F,7$$\"3#****\\7LRDX\"FE$!3q\"\\:YA@u0$F,7$$\"3emm\"zR'o k;FE$!33$QWnH)GdIF,7$$\"3'****\\i5`h(=FE$!3W!\\U*)yPr0$F,7$$\"3gLLL3En $4#FE$!3!o+Zm@kp0$F,7$$\"3wmm;/RE&G#FE$!3=GauybzcIF,7$$\"3A+++D.&4]#FE $!3c`J6v!)ecIF,7$$\"3!)*****\\PAvr#FE$!3]/!y@!4OcIF,7$$\"3)******\\nHi #HFE$!3Sbn\"y8Ch0$F,7$$\"3jmm\"z*ev:JFE$!3GBd'>'R*e0$F,7$$\"31LLL347TL FE$!3I^Q3m9gbIF,7$$\"3cLLLLY.KNFE$!3)=,E!4xLbIF,7$$\"3!****\\7o7Tv$FE$ !3Q3HChC,bIF,7$$\"3sKLL$Q*o]RFE$!3K<!3Z0$F,7$$\"33++D\"=lj;%FE$!3en yTEhNaIF,7$$\"33++vV&RY2aFE$!35;*oVvn>0 $F,7$$\"3qmm;zXu9cFE$!3UVgk6&3:0$F,7$$\"3^******\\y))GeFE$!3!)Q5,Hf,^I F,7$$\"3!)****\\i_QQgFE$!3)yE@L>;00$F,7$$\"3j***\\7y%3TiFE$!3Qg1!)ye,] IF,7$$\"3o****\\P![hY'FE$!3!y8hJ,T%\\IF,7$$\"33KLL$Qx$omFE$!3!Q\"ef,r! *[IF,7$$\"3k+++v.I%)oFE$!3=/YR)))=$[IF,7$$\"3Amm\"zpe*zqFE$!3))o\"pGpp x/$F,7$$\"37+++D\\'QH(FE$!3Q()>@n;:ZIF,7$$\"3GKLe9S8&\\(FE$!39N+rcLbYI F,7$$\"3]++D1#=bq(FE$!3=$ow:a5f/$F,7$$\"3>LLL3s?6zFE$!3MQwyp[EXIF,7$$ \"3)*)**\\7`Wl7)FE$!3UVgW(oqX/$F,7$$\"3[nmmm*RRL)FE$!3!>)p]%[%)Q/$F,7$ $\"3Smm;a<.Y&)FE$!3)e&oNf[;VIF,7$$\"3-MLe9tOc()FE$!3Ye8JLLVUIF,7$$\"3u ******\\Qk\\*)FE$!3\\+'flXX@Pk3'QIF,7$$\"3/+++++++5!#=$!3%G`'>UYuPIF,-%'COLOUR G6&%$RGBG$\"*++++\"!\")F(F^[l-F$6&7%7$Fez$!3)**\\'>UYuPIF,7$$\"3G+++++ ++]FE$!3')*>iF:?G0$F,7$$\"39+++++++DFE$!3=+:)\\Qhx0$F,-F[[l6&F][l$\"#5 !\"\"F(F(-%&STYLEG6#%&POINTG-%'SYMBOLG6$%&CROSSG\"#?-F$6$7S7$F($!3k(>A tY*QgIF,7$F.Fc]l7$F4Fc]l7$F9Fc]l7$F>Fc]l7$FCFc]l7$FIFc]l7$FNFc]l7$FSFc ]l7$FXFc]l7$FgnFc]l7$F\\oFc]l7$FaoFc]l7$FfoFc]l7$F[pFc]l7$F`pFc]l7$Fep Fc]l7$FjpFc]l7$F_qFc]l7$FdqFc]l7$FiqFc]l7$F^rFc]l7$FcrFc]l7$FhrFc]l7$F ]sFc]l7$FbsFc]l7$FgsFc]l7$F\\tFc]l7$FatFc]l7$FftFc]l7$F[uFc]l7$F`uFc]l 7$FeuFc]l7$FjuFc]l7$F_vFc]l7$FdvFc]l7$FivFc]l7$F^wFc]l7$FcwFc]l7$FhwFc ]l7$F]xFc]l7$FbxFc]l7$FgxFc]l7$F\\yFc]l7$FayFc]l7$FfyFc]l7$F[zFc]l7$F` zFc]l7$FezFc]l-F[[l6&F][lF(F(F^[l-%+AXESLABELSG6%Q#dt6\"Q!F[al-%%FONTG 6#%(DEFAULTG-%%VIEWG6$;F($\"\"\"Fe\\lF`al" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "This means that the simple parabol ic extrapolation in the range " }{TEXT 19 14 "dt=0.05 .. 0.1" }{TEXT -1 68 " has not worked very well! It means that the error as a functio n of " }{TEXT 19 2 "dt" }{TEXT -1 79 " has a large contribution from a higher order. We need to calculate at smaller " }{TEXT 19 2 "dt" } {TEXT -1 88 ", and then do the extrapolation. It also means that the e xtrapolation has to be checked!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 123 "Another alternative to explore is whethe r the initial condition for the recursion isn't responsible for the de viation from " }{TEXT 19 7 "O(dt^2)" }{TEXT -1 62 " behaviour: the way in which the pendulum responds for larger " }{TEXT 19 2 "dt" }{TEXT -1 157 " suggests that energy is not conserved. In some cases the ener gy is too large from the beginning, in others it picks up extra energy after some oscillations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 81 "Let us check for the presence of a linear term. Fi rst we unassign the parameters " }{TEXT 19 1 "A" }{TEXT -1 5 " and " } {TEXT 19 1 "B" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "A:='A': B:='B': C_lin:='C_lin':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "eq1:=A + B*dt2^2 +C_lin*dt2 = Res2[n2+1][2];" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/,(%\"AG\"\"\"*&$\"/++++++5!#:F (%\"BGF(F(*&$F+!#9F(%&C_linGF(F($!/l>UYuPI!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "eq2:=A + B*dt3^2 +C_lin*dt3 = Res3[n3+1][2];" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,(%\"AG\"\"\"*&$\"/++++++D!#;F (%\"BGF(F(*&$\"/++++++]!#:F(%&C_linGF(F($!/Aw_,#G0$!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "eq3:=A + B*dt4^2 +C_lin*dt4 = Res4[ n4+1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/,(%\"AG\"\"\"*&$ \"/+++++]i!# " 0 "" {MPLTEXT 1 0 38 "sol:=solve(\{eq1,eq2,eq 3\},\{A,B,C_lin\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<%/%\"AG $!/^#=gr41$!#8/%&C_linG$\"/++Bm$)y$*!#:/%\"BG$\"/+c+B\"[Q\"F*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "The linear coefficient is not larg e." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(sol);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "A,Res_dsolve; " }}{PARA 11 " " 1 "" {XPPMATH 20 "6$$!/^#=gr41$!#8$!3k(>AtY*QgI!#<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "Our newly extrapolated value does agree better \+ with the 'exact' answer." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "P_extra1:=plot(A + B * dt^2 + C_lin * dt, dt=0..0.1, color=magenta): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "display(P_extra1,P_acc, P_exact);" }}{PARA 13 "" 1 "" {GLPLOT2D 810 312 312 {PLOTDATA 2 "6'-%' CURVESG6$7S7$$\"\"!F)$!3')*4D=gr41$!#<7$$\"3[mmm;arz@!#?$!39([8/fg21$F ,7$$\"3mLL$e9ui2%F0$!3]fjn%Gm01$F,7$$\"3-nmm\"z_\"4iF0$!3Y6&fg'eLgIF,7 $$\"3[mmmT&phN)F0$!3kM%G]>\"4gIF,7$$\"3KLLe*=)H\\5!#>$!3Chdm5]$)fIF,7$ $\"3omm\"z/3uC\"FE$!3)*4io(>'efIF,7$$\"3#****\\7LRDX\"FE$!3'3:a96<$fIF ,7$$\"3emm\"zR'ok;FE$!39xK7jl-fIF,7$$\"3'****\\i5`h(=FE$!3[KoSTXseIF,7 $$\"3gLLL3En$4#FE$!3'QpnG&4SeIF,7$$\"3wmm;/RE&G#FE$!3#)36Y\"30\"eIF,7$ $\"3A+++D.&4]#FE$!3'*o7bN)fx0$F,7$$\"3!)*****\\PAvr#FE$!3y:R43-SdIF,7$ $\"3)******\\nHi#HFE$!33\"3f*[8/dIF,7$$\"3jmm\"z*ev:JFE$!3K)==&=]qcIF, 7$$\"31LLL347TLFE$!3!4(f#39#HcIF,7$$\"3cLLLLY.KNFE$!3AS(*pt8$f0$F,7$$ \"3!****\\7o7Tv$FE$!3Q!GmZ,*\\bIF,7$$\"3sKLL$Q*o]RFE$!3%pL`r!\\5bIF,7$ $\"33++D\"=lj;%FE$!3=)Hte>gY0$F,7$$\"33++vV&RY2aFE$!3SuCB^2&=0$F,7$$\"3qmm;zXu9cFE$!31ESWa*R80$F,7$$\"3^***** *\\y))GeFE$!3_D%oSv*z]IF,7$$\"3!)****\\i_QQgFE$!3uLh&H)*e-0$F,7$$\"3j* **\\7y%3TiFE$!3gj()p)=C(\\IF,7$$\"3o****\\P![hY'FE$!37Hyveq6\\IF,7$$\" 33KLL$Qx$omFE$!3a:o=f&f&[IF,7$$\"3k+++v.I%)oFE$!3A6\"3h!=&z/$F,7$$\"3A mm\"zpe*zqFE$!3?m5pR**QZIF,7$$\"37+++D\\'QH(FE$!3#R\")o\"RNwYIF,7$$\"3 GKLe9S8&\\(FE$!3Cz6LtD;YIF,7$$\"3]++D1#=bq(FE$!3i7gl(RAb/$F,7$$\"3>LLL 3s?6zFE$!3OdPYdY)[/$F,7$$\"3)*)**\\7`Wl7)FE$!3Epa]ZW?WIF,7$$\"3[nmmm*R RL)FE$!3_?uU$=PN/$F,7$$\"3Smm;a<.Y&)FE$!3/))e)))[UG/$F,7$$\"3-MLe9tOc( )FE$!3ol=]U79UIF,7$$\"3u******\\Qk\\*)FE$!3dni:ng[TIF,7$$\"3!QLL3dg6<* FE$!3r`R6QCsSIF,7$$\"3-mmmmxGp$*FE$!3'4(zYDz-SIF,7$$\"3!3+]7oK0e*FE$!3 HiCeaaFRIF,7$$\"3'****\\(=5s#y*FE$!3P?X4oOaQIF,7$$\"3/+++++++5!#=$!3)* *\\'>UYuPIF,-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F^[l-F$6&7%Fdz7$$\"3G+++ ++++]FE$!3')*>iF:?G0$F,7$$\"39+++++++DFE$!3=+:)\\Qhx0$F,-F[[l6&F][l$\" #5!\"\"F(F(-%&STYLEG6#%&POINTG-%'SYMBOLG6$%&CROSSG\"#?-F$6$7S7$F($!3k( >AtY*QgIF,7$F.F`]l7$F4F`]l7$F9F`]l7$F>F`]l7$FCF`]l7$FIF`]l7$FNF`]l7$FS F`]l7$FXF`]l7$FgnF`]l7$F\\oF`]l7$FaoF`]l7$FfoF`]l7$F[pF`]l7$F`pF`]l7$F epF`]l7$FjpF`]l7$F_qF`]l7$FdqF`]l7$FiqF`]l7$F^rF`]l7$FcrF`]l7$FhrF`]l7 $F]sF`]l7$FbsF`]l7$FgsF`]l7$F\\tF`]l7$FatF`]l7$FftF`]l7$F[uF`]l7$F`uF` ]l7$FeuF`]l7$FjuF`]l7$F_vF`]l7$FdvF`]l7$FivF`]l7$F^wF`]l7$FcwF`]l7$Fhw F`]l7$F]xF`]l7$FbxF`]l7$FgxF`]l7$F\\yF`]l7$FayF`]l7$FfyF`]l7$F[zF`]l7$ F`zF`]l7$FezF`]l-F[[l6&F][lF(F(F^[l-%+AXESLABELSG6%Q#dt6\"Q!Fh`l-%%FON TG6#%(DEFAULTG-%%VIEWG6$;F($\"\"\"Fb\\lF]al" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT 257 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 39 "Repeat the calculation with time steps " }{TEXT 19 8 "dt2 ..dt4" }{TEXT -1 218 " which are reduced by a factor of two and observ e whether that leads to further improvement. Then observe whether it w orks for a wide range of initial amplitude values of the pendulum, and for longer integration times." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 255 "We remove the dependence of the error at order dt by correcting the initial condition. The assumption that the first two points of the sequence of angles phi[i] was equal is consis tent with the physical initial condition of zero angular velocity at l evel " }{TEXT 19 5 "O(dt)" }{TEXT -1 62 " only, because it is based on the simple differencing formula " }{TEXT 19 36 "phi'[0] = (phi[1]-phi [0])/dt + O(dt)" }{TEXT -1 110 ". In order to be consistent and incorp orate the initial condition of arbitrary initial angular speed at orde r " }{TEXT 19 7 "O(dt^2)" }{TEXT -1 26 " we use in the first step " } {TEXT 19 45 "phi'[0] = (phi[1] - phi[-1])/(2*dt) + O(dt^2)" }{TEXT -1 130 ". We combine this information with the second derivative replacem ent by a finite difference to have a special formula to evaluate " } {TEXT 19 6 "phi[1]" }{TEXT -1 75 ", and from then on the recursion wor ks as before.We redefine the procedure " }{TEXT 19 6 "Verlet" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "Verlet:=proc(phi 0,phi0dot,dt,N) local i,y; global g,l;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "y[0]:=phi0: y[1]:=dt*phi0dot+evalf(y[0]-0.5*dt^2*g/l*sin(y[0]) ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "for i from 1 to N-1 do:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "y[i+1]:=2*y[i]-y[i-1]+evalf(dt^2*(- g/l*sin(y[i]))); od:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "[seq([i*dt, y[i]],i=0..N)]; end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "dt1 :=0.2: n1:=trunc(10/dt1): Res1:=Verlet(phi0,0,dt1,n1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "dt2:=dt1/2: n2:=trunc(10/dt2): Res2 :=Verlet(phi0,0,dt2,n2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "dt3:=dt2/2: n3:=trunc(10/dt3): Res3:=Verlet(phi0,0,dt3,n3):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Res1[n1+1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"$+\"!\"\"$!/uQ^KU6J!#8" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 11 "Res2[n2+1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #7$$\"/++++++5!#7$!/w%z/Sq0$!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Res3[n3+1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"/++++++5!# 7$!/\")>.\"f-1$!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "PR1:= plot(Res1,color=red,style=point,symbol=diamond,symbolsize=20):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "PR2:=plot(Res2,color=blue,st yle=point,symbol=diamond,symbolsize=20):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "PR3:=plot(Res3,color=green,style=point,symbol=diamond ,symbolsize=20):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Pdsolve :=plot(Y(t),t=0..10,color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "display(Pdsolve,PR1,PR2,PR3);" }}{PARA 13 "" 1 "" {GLPLOT2D 865 429 429 {PLOTDATA 2 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%)Fgfl$\"3#**f2Wg/qo\"Fgfl7$Fc\\o$\"3-+$p>.IUY\"Fgfl7$$\"3s++++++]&)Fg fl$\"3%***3$*zsf;7Fgfl7$F\\dm$\"3s**)>58k^X*F[gl7$$\"3O++++++]')Fgfl$ \"3V+8)\\jb;a'F[gl7$F[]o$\"3'**zW\"=G-wMF[gl7$$\"3+++++++]()Fgfl$\"3=+ 6\"3$))G_KFj`o7$Fadm$!3\"**zs2M%pLGF[gl7$$\"3k************\\))Fgfl$!3% **HnH\\>F#fF[gl7$Fc]o$!3-+=/&y#=s))F[gl7$$\"3G************\\*)Fgfl$!3/ +JNg;yi6Fgfl7$Ffdm$!34+'=DG(R:9Fgfl7$$\"3s++++++]!*Fgfl$!36+P\\eTJV;Fg fl7$F[^o$!37+^6TnHY=Fgfl7$$\"3O++++++]\"*Fgfl$!3A+w&R4A_-#Fgfl7$F[em$! 33+h&fh%o\"=#Fgfl7$$\"3+++++++]#*Fgfl$!3******Qn'owJ#Fgfl7$Fc^o$!3*)* \\6$)G2`V#Fgfl7$$\"3k************\\$*Fgfl$!3!**\\eJg?n`#Fgfl7$F`em$!3 \")*4&=(\\.\"f-1$Fgfl-F^fl6 &F`flF(FcfmF(FffmFjfm-%+AXESLABELSG6%Q\"t6\"Q!F]cr-%%FONTG6#%(DEFAULTG -%%VIEWG6$;F(FielFbcr" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "We notice a substantial improvement in th e solution for " }{TEXT 19 6 "dt=0.2" }{TEXT -1 149 ", although it doe s give us a deviation after the first swing to the opposite side. Let \+ us carry out the accuracy analysis for the improved algorithm." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "A:='A': B:='B':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "eq1:=A + B*dt2^2 = Res2[n2+1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/,&%\"AG\"\"\"*&$\"/++++++5!# :F(%\"BGF(F($!/w%z/Sq0$!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "eq2:=A + B*dt3^2 = Res3[n3+1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/,&%\"AG\"\"\"*&$\"/++++++D!#;F(%\"BGF(F($!/\")>.\"f-1$!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=solve(\{eq1,eq2\},\{ A,B\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<$/%\"BG$\"/LLnO2#H% !#9/%\"AG$!/\\h@@LhI!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 " assign(sol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "A,Res_dsolv e; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!/\\h@@LhI!#8$!3k(>AtY*QgI!#< " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "P_extra:=plot(A + B * d t^2, dt=0..0.1, color=magenta):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "The extrapolation of our two results (1 and 2) does agree better \+ than before with dsolve's answer. What does it predict for the answer \+ with spacing " }{TEXT 19 5 "0.025" }{TEXT -1 32 "? First we compute a \+ new result " }{TEXT 19 4 "Res4" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 28 "Res4:=Verlet(3.11,0,dt4,n4):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "A + B * dt4^2,Res4[n4+1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!/2,nQ1hI!#8$!/-G9#\\51$F%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "Therefore, we can realize that the function is no t a perfect second-order parabola in the range " }{TEXT 19 10 "dt=0..0 .1 " }{TEXT -1 51 "even without having access to an accurate solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "dt5:=dt4/2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$dt5G$\"/+++++]7!#:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 18 "n5:=trunc(10/dt5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#n5G\"$+)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Res5:=Verlet(3.11,0,dt5,n5):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 144 "P_acc:=plot([[dt2,Res2[n2+1][2]],[dt3,Res3[n3+1][2]] ,[dt4,Res4[n4+1][2]],[dt5,Res5[n5+1][2]]],dt=0..0.1,style=point,symbol =cross,symbolsize=20):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "d isplay(P_extra,P_acc,P_exact,title=\"Comparison at t=10 with the Runge -Kutta-Fehlberg result\");" }}{PARA 13 "" 1 "" {GLPLOT2D 960 488 488 {PLOTDATA 2 "6(-%'CURVESG6$7S7$$\"\"!F)$!35+\\h@@LhI!#<7$$\"3[mmm;arz@ !#?$!3%)GBQ#3I81$F,7$$\"3mLL$e9ui2%F0$!3u51!**)\\KhIF,7$$\"3-nmm\"z_\" 4iF0$!3'3/NTd:81$F,7$$\"3[mmmT&phN)F0$!3?Bh'>:-81$F,7$$\"3KLLe*=)H\\5! #>$!3!)R#QZ'[GhIF,7$$\"3omm\"z/3uC\"FE$!3AHnzN`EhIF,7$$\"3#****\\7LRDX \"FE$!3#fP=WcT71$F,7$$\"3emm\"zR'ok;FE$!3)GO$\\!=871$F,7$$\"3'****\\i5 `h(=FE$!3%pKZF/\"=hIF,7$$\"3gLLL3En$4#FE$!3c4$o+)R9hIF,7$$\"3wmm;/RE&G #FE$!3sgV-rz5hIF,7$$\"3A+++D.&4]#FE$!3#QoGImj51$F,7$$\"3!)*****\\PAvr# FE$!3OG42b^,hIF,7$$\"3)******\\nHi#HFE$!3)4un!*fk41$F,7$$\"3jmm\"z*ev: JFE$!3CY5!*\\a\"41$F,7$$\"31LLL347TLFE$!3_FZh$*H&31$F,7$$\"3cLLLLY.KNF E$!3_)>%*Qn'zgIF,7$$\"3!****\\7o7Tv$FE$!3QYY8CssgIF,7$$\"3sKLL$Q*o]RFE $!35;([q@i11$F,7$$\"33++D\"=lj;%FE$!3KxdwxqegIF,7$$\"33++vV&RY2aFE$!3-Kjh\"4x+1$F,7$$\"3qmm;zXu9cFE$!3_1>3I !z*fIF,7$$\"3^******\\y))GeFE$!3e-q^\\Q()fIF,7$$\"3!)****\\i_QQgFE$!37 -\"p:9n(fIF,7$$\"3j***\\7y%3TiFE$!3;kT-5.mfIF,7$$\"3o****\\P![hY'FE$!3 a\"Rq'fv`fIF,7$$\"33KLL$Qx$omFE$!3Iq(*=aNUfIF,7$$\"3k+++v.I%)oFE$!3U[v u`zHfIF,7$$\"3Amm\"zpe*zqFE$!37,'4#*eIF,7$$\"3]++D1#=bq(FE$!3oC-T-Py eIF,7$$\"3>LLL3s?6zFE$!3uJj!H$ekeIF,7$$\"3)*)**\\7`Wl7)FE$!3!*[P+0w\\e IF,7$$\"3[nmmm*RRL)FE$!3U'=L43^$eIF,7$$\"3Smm;a<.Y&)FE$!3Yb*e-U(>eIF,7 $$\"3-MLe9tOc()FE$!31c6S=7/eIF,7$$\"3u******\\Qk\\*)FE$!3#[UBpL%*y0$F, 7$$\"3!QLL3dg6<*FE$!3%yxg/0Ax0$F,7$$\"3-mmmmxGp$*FE$!3I49%oQkv0$F,7$$ \"3!3+]7oK0e*FE$!3kT!\\Gd#RdIF,7$$\"3'****\\(=5s#y*FE$!3u$oJr`Cs0$F,7$ $\"3/+++++++5!#=$!3+nv%z/Sq0$F,-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F^[l- F$6&7&7$Fez$!3))*fZz/Sq0$F,7$$\"3G+++++++]FE$!37+\")>.\"f-1$F,7$$\"39+ ++++++DFE$!3%**>!G9#\\51$F,7$$\"33++++++]7FE$!3-+&zu#eChIF,-F[[l6&F][l $\"#5!\"\"F(F(-%&STYLEG6#%&POINTG-%'SYMBOLG6$%&CROSSG\"#?-F$6$7S7$F($! 3k(>AtY*QgIF,7$F.Fh]l7$F4Fh]l7$F9Fh]l7$F>Fh]l7$FCFh]l7$FIFh]l7$FNFh]l7 $FSFh]l7$FXFh]l7$FgnFh]l7$F\\oFh]l7$FaoFh]l7$FfoFh]l7$F[pFh]l7$F`pFh]l 7$FepFh]l7$FjpFh]l7$F_qFh]l7$FdqFh]l7$FiqFh]l7$F^rFh]l7$FcrFh]l7$FhrFh ]l7$F]sFh]l7$FbsFh]l7$FgsFh]l7$F\\tFh]l7$FatFh]l7$FftFh]l7$F[uFh]l7$F` uFh]l7$FeuFh]l7$FjuFh]l7$F_vFh]l7$FdvFh]l7$FivFh]l7$F^wFh]l7$FcwFh]l7$ FhwFh]l7$F]xFh]l7$FbxFh]l7$FgxFh]l7$F\\yFh]l7$FayFh]l7$FfyFh]l7$F[zFh] l7$F`zFh]l7$FezFh]l-F[[l6&F][lF(F(F^[l-%&TITLEG6#QXComparison~at~t=10~ with~the~Runge-Kutta-Fehlberg~result6\"-%+AXESLABELSG6%Q#dtF`alQ!F`al- %%FONTG6#%(DEFAULTG-%%VIEWG6$;F($\"\"\"Fj\\lFial" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "The parabolic fit is quite consi stent, so why do we deviate so much from Maple's answer??? Try another method in " }{TEXT 19 15 "dsolve[numeric]" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "sol:=dsolve(\{DE,T(0)=phi0,D(T)(0)= 0\},numeric,method=lsode,output=listprocedure):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Y:=eval(T(t),sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%\"YGj,6#%\"tG6'%$resG%%dataG%)solnprocG%)outpointG%%T(t)G6#%inCopyr ight~(c)~2000~by~Waterloo~Maple~Inc.~All~rights~reserved.G6\"C)>%8_Env DSNumericSaveDigitsG%'DigitsG>F4\"#9@%/%-_EnvInFsolveG%%trueG>8'-&%&ev alfG6#F36#9$>F<-F?FA>8%`6$%$GetG%$SetGb6#%+thismoduleG6#%%DataGF06$%$G etG%$SetGF0F0F0F06\"6#%%DataG>8&-_FFFQ6#Q/soln_procedureF0@$4-%%typeG6 $F<.%(numericGC$@.-%'memberG6$FB7+Q&startF0.%&startGQ'methodF0.%'metho dGQ%leftF0Q&rightF0Q)leftdataF0Q*rightdataF0Q+enginedataF0C$@%-F`o6$FB 7$FdoFgo>8$-FW6#-%(convertG6$FB.%'stringG>Fdp-FWFA@&-Fin6$Fdp.%&arrayG O-%%evalG6$Fdp\"\"\"0FdpQ)procnameF0OFdp-F`o6$FB7$Q%lastF0Q(initialF0C $F\\q@$-Fin6$Fdp.%%listGO&Fdp6#\"\"#-F`o6$FB7$.%%lastG.%(initialGC$Fcp Far3-Fin6$F<%\"=G-F`o6$-%$lhsG6#F<7$F_rF_sC$@%-Fin6$-%$rhsGFjsFdr>Fdp- FW6#/F_r7$\"\"!-%#opG6#F`t>Fdp-FW6#/F_r7$FirF`tFar/F8(-%(pointtoG6#&-FY6#Q0soln_proceduresF0FhrO-.FivFAZ%C$> Fdp-FWFjsFgrF0YF0F0F0F06#FgtF0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Res_lsode:=Y(10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*Res_ls odeG$!3i+BrM[%>1$!#<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "P_l sode:=plot(Res_lsode,dt=0..0.1,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "display(P_extra,P_acc,P_lsode,title=\"Comparison a t t=10 with the LSODE result\");" }}{PARA 13 "" 1 "" {GLPLOT2D 909 424 424 {PLOTDATA 2 "6(-%'CURVESG6$7S7$$\"\"!F)$!35+\\h@@LhI!#<7$$\"3[ mmm;arz@!#?$!3%)GBQ#3I81$F,7$$\"3mLL$e9ui2%F0$!3u51!**)\\KhIF,7$$\"3-n mm\"z_\"4iF0$!3'3/NTd:81$F,7$$\"3[mmmT&phN)F0$!3?Bh'>:-81$F,7$$\"3KLLe *=)H\\5!#>$!3!)R#QZ'[GhIF,7$$\"3omm\"z/3uC\"FE$!3AHnzN`EhIF,7$$\"3#*** *\\7LRDX\"FE$!3#fP=WcT71$F,7$$\"3emm\"zR'ok;FE$!3)GO$\\!=871$F,7$$\"3' ****\\i5`h(=FE$!3%pKZF/\"=hIF,7$$\"3gLLL3En$4#FE$!3c4$o+)R9hIF,7$$\"3w mm;/RE&G#FE$!3sgV-rz5hIF,7$$\"3A+++D.&4]#FE$!3#QoGImj51$F,7$$\"3!)**** *\\PAvr#FE$!3OG42b^,hIF,7$$\"3)******\\nHi#HFE$!3)4un!*fk41$F,7$$\"3jm m\"z*ev:JFE$!3CY5!*\\a\"41$F,7$$\"31LLL347TLFE$!3_FZh$*H&31$F,7$$\"3cL LLLY.KNFE$!3_)>%*Qn'zgIF,7$$\"3!****\\7o7Tv$FE$!3QYY8CssgIF,7$$\"3sKLL $Q*o]RFE$!35;([q@i11$F,7$$\"33++D\"=lj;%FE$!3KxdwxqegIF,7$$\"33++vV&R< P%FE$!35,'=f\"=^gIF,7$$\"3gLL$e9Ege%FE$!3+yCJG%H/1$F,7$$\"3ILLeR\"3Gy% FE$!3%R]u))H].1$F,7$$\"3smm;/T1&*\\FE$!35R_n?7EgIF,7$$\"3Smm\"zRQb@&FE $!3mX(p&)fk,1$F,7$$\"3!****\\(=>Y2aFE$!3-Kjh\"4x+1$F,7$$\"3qmm;zXu9cFE $!3_1>3I!z*fIF,7$$\"3^******\\y))GeFE$!3e-q^\\Q()fIF,7$$\"3!)****\\i_Q QgFE$!37-\"p:9n(fIF,7$$\"3j***\\7y%3TiFE$!3;kT-5.mfIF,7$$\"3o****\\P![ hY'FE$!3a\"Rq'fv`fIF,7$$\"33KLL$Qx$omFE$!3Iq(*=aNUfIF,7$$\"3k+++v.I%)o FE$!3U[vu`zHfIF,7$$\"3Amm\"zpe*zqFE$!37,'4#*eIF,7$$\"3]++D1#=bq(FE$! 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What is the moral of th e story? Numerical work is delicate and requires attention. Black-box \+ programs can fail, or at least have only limited accuracy. Of course, \+ we shouldn't assume that our results are perfect either, in fact, all \+ of Maple's methods are of higher order than ours, and do extrapolate t o zero " }{TEXT 19 2 "dt" }{TEXT -1 161 ". The mystery is why they see m to converge to different answers, i.e., answers that deviate in the \+ fourth digit, which is far away from the precision set by the " } {TEXT 19 6 "Digits" }{TEXT -1 22 " environment variable." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Let us write now our own extrapolation procedure. The idea is a s follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Given a time-step " }{TEXT 19 2 "dt" }{TEXT -1 80 " provided as an argument (so that we can change it) generate solutions at level " }{TEXT 19 2 "dt" }{TEXT -1 6 ", and " }{TEXT 19 4 "dt/2" }{TEXT -1 57 " and use these results to extrapolate the answer to zero." }}{PARA 0 "" 0 "" {TEXT -1 210 "We rely on the Verlet procedure to accomplish th e basic integration of the pendulum equation. We also supply the lengt h of the integration interval by specifying the number of steps desire d for the integration:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "E xtra:=proc(phi0,phi0dot,dt,N) local dt2,N2,res,res2,eq1,eq2,A,B,sol;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "res:=Verlet(phi0,phi0dot,dt,N): d t2:=0.5*dt: N2:=2*N: res2:=Verlet(phi0,phi0dot,dt2,N2):" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 30 "eq1:=A + B*dt^2 = res[N+1][2];" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 33 "eq2:=A + B*dt2^2 = res2[N2+1][2];" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 27 "print(res[N+1],res2[N2+1]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "sol:=solve(\{eq1,eq2\},\{A,B\}); assign(sol); A; end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Extra(3.11,0,0.05,100);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$7$$\"$+&!\"#$!/>8Bvaa9!#87$$\"%+]!\"$$!/sdO8FW9F)" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#$!/!f5%f%3W\"!#8" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 12 "Y(100*0.05);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!3*[s:-_M3W\"!#<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "This i s close, but can we do better?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Extra(3.11,0,0.025,200);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$7$ $\"%+]!\"$$!/sdO8FW9!#87$$\"&++&!\"%$!/NKCVpT9F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!/*Q-KN3W\"!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Extra(3.11,0,0.01,500);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$7$$ \"$+&!\"#$!/#GC*[QT9!#87$$\"%+]!\"$$!/1@Z@(4W\"F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!/9ZlX$3W\"!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "The idea behind the extrapolation scheme would be to try it at severa l levels of " }{TEXT 19 2 "dt" }{TEXT -1 57 ", and then decide to acce pt the answer from the smallest " }{TEXT 19 2 "dt" }{TEXT -1 112 "-bas ed result provided the extrapolated answer agreed to within a given to lerance with the result obtained with " }{TEXT 19 4 "2*dt" }{TEXT -1 29 ". We have to find a value of " }{TEXT 19 2 "dt" }{TEXT -1 84 " sma ll enough so that the error terms of higher order than second become n egligible." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 136 "Make explorations of solutions for cases with initial angular velocity, and particularl y for cases where the pendulum goes over the top." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 8 "Project:" }}{PARA 0 "" 0 "" {TEXT -1 141 "Incorporate friction into the Verlet algorithm such t hat the second-order accuracy remains preserved. Compare your extrapol ated results with " }{TEXT 19 6 "dsolve" }{TEXT -1 131 "'s answers and document them. Include a case where the pendulum starts near the top, as well as a case where it goes over the top. " }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "112" 0 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }