{VERSION 3 0 "SGI MIPS UNIX" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }{PSTYLE " Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 3 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Warning" 2 7 1 {CSTYLE "" -1 -1 "" 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple O utput" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "with(MS):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "This example tries to find the linear rec urrence relation for the top of the Euler numbers " }{XPPEDIT 18 0 "f (x) = 2/(exp(x)+exp(-x));" "6#/-%\"fG6#%\"xG*&\"\"#\"\"\",&-%$expG6#F' F*-F-6#,$F'!\"\"F*F2" }{XPPEDIT 18 0 " = sum(c[i]*x^i/i!,i = 0 .. in finity);" "6#/%!G-%$sumG6$*&*&&%\"cG6#%\"iG\"\"\")%\"xGF-F.F.-%*factor ialG6#F-!\"\"/F-;\"\"!%)infinityG" }{XPPEDIT 18 0 " = sum(b[i]*x^i/i !,i = 0 .. infinity)/sum(d[j]*x^j/j!,j = 0 .. infinity);" "6#/%!G*&-%$sumG6$*&*&&%\"bG6#%\"iG\"\"\")%\"xGF.F/F/-%*factorialG6#F.!\"\"/F.;\" \"!%)infinityGF/-F'6$*&*&&%\"dG6#%\"jGF/)F1FAF/F/-F36#FAF5/FA;F8F9F5" }{TEXT -1 142 " given the bottom linear recurrence relation, when mult isectioning by 4 at 0. As the function is being multisectioned by 4 \+ at 0, only those " }{XPPEDIT 18 0 "b[i];" "6#&%\"bG6#%\"iG" }{TEXT -1 7 " where " }{XPPEDIT 18 0 "i = mod(0,4);" "6#/%\"iG-%$modG6$\"\"!\" \"%" }{TEXT -1 12 " are needed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "bot := bottom/ms/linalg/fft2(exp(x)+exp(-x),f,x,4);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$botG6&/-%\"fG6#%\"xG,&-F(6#,&F*\"\" \"!\")F/\"%C5-F(6#,&F*F/!\"%F/!#[F(F*7+/-F(6#\"\"!\"#;/-F(6#F/F;/-F(6# \"\"#F;/-F(6#\"\"$F;/-F(6#\"\"%!$G\"/-F(6#\"\"&F;/-F(6#\"\"'F;/-F(6#\" \"(F;/-F(6#\"\")\"&K%=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "B ot := egf/makeproc(bot):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "Now \+ " }{XPPEDIT 18 0 "b[i] = sum(binomial(i,j)*c[i-j]*d[j],j = 0 .. i);" " 6#/&%\"bG6#%\"iG-%$sumG6$*(-%)binomialG6$F'%\"jG\"\"\"&%\"cG6#,&F'F0F/ !\"\"F0&%\"dG6#F/F0/F/;\"\"!F'" }{TEXT -1 30 " from equation \\ref\{eq :eq1\}. " }{XPPEDIT 18 0 "b[i];" "6#&%\"bG6#%\"iG" }{TEXT -1 36 ". A n upper bound for the number of " }{XPPEDIT 18 0 "b[i];" "6#&%\"bG6#% \"iG" }{TEXT -1 11 " needed is " }{XPPEDIT 18 0 "4*2^3*2*2+2 = 130;" " 6#/,&**\"\"%\"\"\"*$\"\"#\"\"$F'\"\"#F'\"\"#F'F'\"\"#F'\"$I\"" }{TEXT -1 28 " by Lemma \\ref\{lem:d pow P\}." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "F := i -> add(binomial(i,j)*euler(i-j)*Bot(j),j=0.. i);" }}{PARA 7 "" 1 "" {TEXT -1 42 "Warning, j in call to add is n ot local" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FGR6#%\"iG6\"6$%)opera torG%&arrowGF(-%$addG6$*(-%)binomialG6$9$%\"jG\"\"\"-%&eulerG6#,&F3F5F 4!\"\"F5-%$BotG6#F4F5/F4;\"\"!F3F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "for i from 4 to 130 by 4 do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " b[i/4] := F(i):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "rec := recurre nce/solve/linalg(b,f,x,4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$recG /-%\"fG6#%\"xG,(-F'6#,&F)\"\"\"!#7F.\"$D'-F'6#,&F)F.!\")F.!$6'-F'6#,&F )F.!\"%F.!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "This could have a lso been discovered by using some of the other built in functions." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "top/ms/naive(2,exp(x)+exp (-x),f,x,4,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&/-%\"fG6#%\"xG,(-F%6 #,&F'\"\"\"!#7F,\"$D'-F%6#,&F'F,!\")F,!$6'-F%6#,&F'F,!\"%F,!#8F%F'7./- F%6#\"\"!\"#;/-F%6#F,F=/-F%6#\"\"#F=/-F%6#\"\"$F=/-F%6#\"\"%!#[/-F%6# \"\"&F=/-F%6#\"\"'F=/-F%6#\"\"(F=/-F%6#\"\")!%3U/-F%6#\"\"*F=/-F%6#\"# 5F=/-F%6#\"#6F=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "top/ms/ linalg/sym(2,exp(x)+exp(-x),f,x,4,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&/-%\"fG6#%\"xG,(-F%6#,&F'\"\"\"!#7F,\"$D'-F%6#,&F'F,!\")F,!$6'-F %6#,&F'F,!\"%F,!#8F%F'7./-F%6#\"\"!\"#;/-F%6#F,F=/-F%6#\"\"#F=/-F%6#\" \"$F=/-F%6#\"\"%!#[/-F%6#\"\"&F=/-F%6#\"\"'F=/-F%6#\"\"(F=/-F%6#\"\")! %3U/-F%6#\"\"*F=/-F%6#\"#5F=/-F%6#\"#6F=" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "This method is automated with the given function below." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "top/ms/linalg/know(Bot, euler, f, x, 4, 2, 16, 2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&/-%\"fG 6#%\"xG,(-F%6#,&F'\"\"\"!#7F,\"$D'-F%6#,&F'F,!\")F,!$6'-F%6#,&F'F,!\"% F,!#8F%F'7./-F%6#\"\"!F=/-F%6#F,F=/-F%6#\"\"#!#;/-F%6#\"\"$F=/-F%6#\" \"%F=/-F%6#\"\"&F=/-F%6#\"\"'\"$W*/-F%6#\"\"(F=/-F%6#\"\")F=/-F%6#\"\" *F=/-F%6#\"#5\"%/>/-F%6#\"#6F=" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Which all give the same result." }}{PARA 0 "" 0 "" {TEXT -1 535 "Now \+ determine the linear recurrence relation multisectioned by 4 at 2. Ta king advantage of the fact that we know what the linear recurrence rel ation most likely is, all that is really needs to done is to determine the initial values, and see if the linear recurrence relation is corr ect. By looking at the recurrence that for the top multisectioned by \+ 4 at 0 that there are only about 12 terms needed. Calculate the first 32 terms for when the function is multisectioned by 4 at 2, and see i f this linear recurrence relation holds.\n" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 78 "initial := [seq(op([f(4*i)=F(4*i),f(4*i+1)=0,f(4*i+ 2)=0,f(4*i+3)=0]),i=0..8)];" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%(init ialG7F/-%\"fG6#\"\"!\"#;/-F(6#\"\"\"F*/-F(6#\"\"#F*/-F(6#\"\"$F*/-F(6# \"\"%!#[/-F(6#\"\"&F*/-F(6#\"\"'F*/-F(6#\"\"(F*/-F(6#\"\")!%3U/-F(6#\" \"*F*/-F(6#\"#5F*/-F(6#\"#6F*/-F(6#\"#7\"&KS*/-F(6#\"#8F*/-F(6#\"#9F*/ -F(6#\"#:F*/-F(6#F+\"(s'=8/-F(6#\"#F*/-F(6#\" #?!))GEs(/-F(6#\"#@F*/-F(6#\"#AF*/-F(6#\"#BF*/-F(6#\"#C\"*_J+d#/-F(6# \"#DF*/-F(6#\"#EF*/-F(6#\"#FF*/-F(6#\"#G\",#*4RoY%/-F(6#\"#HF*/-F(6#\" #IF*/-F(6#\"#JF*/-F(6#\"#K!-o(QW)fy/-F(6#\"#LF*/-F(6#\"#MF*/-F(6#\"#NF *" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "egf/clean(rec, f, x, initial);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&/-%\"fG6#%\"xG,(-F%6#,&F '\"\"\"!#7F,\"$D'-F%6#,&F'F,!\")F,!$6'-F%6#,&F'F,!\"%F,!#8F%F'7./-F%6# \"\"!\"#;/-F%6#F,F=/-F%6#\"\"#F=/-F%6#\"\"$F=/-F%6#\"\"%!#[/-F%6#\"\"& F=/-F%6#\"\"'F=/-F%6#\"\"(F=/-F%6#\"\")!%3U/-F%6#\"\"*F=/-F%6#\"#5F=/- F%6#\"#6F=" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 230 "When cleaning up a ll of the terms, (getting rid of the terms that can be calculated base d on the linear recurrence relation) then fewer than the 32 terms are \+ left. Hence, this linear recurrence relation is most probably correct . " }}{PARA 0 "" 0 "" {TEXT -1 54 "This could have done this with the \+ automated function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "top /ms/know(rec, Bot, euler, f, x, 4, 0, 16);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&/-%\"fG6#%\"xG,(-F%6#,&F'\"\"\"!#7F,\"$D'-F%6#,&F'F,!\" )F,!$6'-F%6#,&F'F,!\"%F,!#8F%F'7./-F%6#\"\"!\"#;/-F%6#F,F=/-F%6#\"\"#F =/-F%6#\"\"$F=/-F%6#\"\"%!#[/-F%6#\"\"&F=/-F%6#\"\"'F=/-F%6#\"\"(F=/-F %6#\"\")!%3U/-F%6#\"\"*F=/-F%6#\"#5F=/-F%6#\"#6F=" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Whic h gives the same result." }}}}{MARK "1 0 4" 128 }{VIEWOPTS 1 1 0 1 1 1803 }