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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 O utput" -1 12 1 {CSTYLE "" -1 -1 "Times" 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 "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 } {PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 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 "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT 256 24 "Calculus Single Variable " }{TEXT 259 1 " " }{TEXT 258 37 " \nBrian E. Blank and Steven G. Kran tz" }{TEXT 257 45 "\n\nSection 9.1\nInfinite Series - Introduction\n" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 15 "1. Partial Sums" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Given a sequence " }{XPPEDIT 18 0 "\{a[n], n = 1 .. infinity\};" "6#<$&%\"aG6#%\"nG/F';\" \"\"%)infinityG" }{TEXT -1 53 ", we can form an associated sequence of partial sums " }{XPPEDIT 18 0 "\{S[N], N = 1 .. infinity\};" "6#<$&% \"SG6#%\"NG/F';\"\"\"%)infinityG" }{TEXT -1 7 " with " }{XPPEDIT 18 0 "S[N];" "6#&%\"SG6#%\"NG" }{TEXT -1 11 " defined by" }}{PARA 0 "" 0 "" {TEXT -1 31 " " }{XPPEDIT 18 0 "S[N] \+ = Sum(a[n],n = 1 .. N);" "6#/&%\"SG6#%\"NG-%$SumG6$&%\"aG6#%\"nG/F.;\" \"\"F'" }{TEXT -1 1 " " }{TEXT 288 1 "." }}{PARA 3 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "We perform the same construction eve n if the sequence " }{XPPEDIT 18 0 "\{a[n]\};" "6#<#&%\"aG6#%\"nG" } {TEXT -1 32 " does not begin with the index " }{XPPEDIT 18 0 "n = 1; " "6#/%\"nG\"\"\"" }{TEXT -1 63 ". For example, the first three part ial sums of the sequence " }{XPPEDIT 18 0 "\{1/(n-Pi), n = 4 .. infin ity\};" "6#<$*&\"\"\"F%,&%\"nGF%%#PiG!\"\"F)/F';\"\"%%)infinityG" } {TEXT -1 10 " are " }{XPPEDIT 18 0 "S[1] = 1/(4-Pi);" "6#/&%\"SG6 #\"\"\"*&F'F',&\"\"%F'%#PiG!\"\"F," }{TEXT -1 4 ", " }{XPPEDIT 18 0 "S[2] = 1/(4-Pi)+1/(5-Pi);" "6#/&%\"SG6#\"\"#,&*&\"\"\"F*,&\"\"%F*%#Pi G!\"\"F.F**&F*F*,&\"\"&F*F-F.F.F*" }{TEXT -1 8 ", and " }{XPPEDIT 18 0 "S[3] = 1/(4-Pi)+1/(5-Pi)+1/(6-Pi);" "6#/&%\"SG6#\"\"$,(*&\"\"\"F *,&\"\"%F*%#PiG!\"\"F.F**&F*F*,&\"\"&F*F-F.F.F**&F*F*,&\"\"'F*F-F.F.F* " }{TEXT 289 3 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "A " }{TEXT 405 18 "geometric sequence" }{TEXT -1 3 " " }{TEXT 398 3 "..." }{TEXT -1 2 " " }{XPPEDIT 18 0 "a[n-1];" "6#&%\"aG 6#,&%\"nG\"\"\"F(!\"\"" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "a[n];" "6#&% \"aG6#%\"nG" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "a[n+1];" "6#&%\"aG6#,&% \"nG\"\"\"F(F(" }{TEXT -1 2 ", " }{TEXT 399 4 " ..." }{TEXT -1 43 " \+ is one in which there is a single value " }{XPPEDIT 18 0 "r;" "6#%\"rG " }{TEXT -1 31 " equal to the all the ratios " }{TEXT 400 3 "..." } {TEXT -1 4 " , " }{XPPEDIT 18 0 "a[n]/a[n-1];" "6#*&&%\"aG6#%\"nG\"\" \"&F%6#,&F'F(F(!\"\"F," }{TEXT -1 4 ", " }{XPPEDIT 18 0 "a[n+1]/a[n] ;" "6#*&&%\"aG6#,&%\"nG\"\"\"F)F)F)&F%6#F(!\"\"" }{TEXT -1 3 ", " } {TEXT 401 3 "..." }{TEXT -1 21 " of successive terms" }{TEXT 402 1 ". " }{TEXT -1 6 " If " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 24 " i s the first term and " }{XPPEDIT 18 0 "r;" "6#%\"rG" }{TEXT -1 91 " is the ratio, then the geometric sequence is\n\n \+ " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 4 ", \+ " }{XPPEDIT 18 0 "a*r;" "6#*&%\"aG\"\"\"%\"rGF%" }{TEXT -1 4 ", " } {XPPEDIT 18 0 "a*r^2;" "6#*&%\"aG\"\"\"*$%\"rG\"\"#F%" }{TEXT -1 3 ", \+ " }{XPPEDIT 18 0 "a*r^3;" "6#*&%\"aG\"\"\"*$%\"rG\"\"$F%" }{TEXT -1 4 ", " }{XPPEDIT 18 0 "a*r^4;" "6#*&%\"aG\"\"\"*$%\"rG\"\"%F%" } {TEXT -1 4 ", " }{TEXT 404 3 "..." }{TEXT -1 2 " " }{TEXT 403 1 ". " }{TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "A " }{TEXT 407 13 "geomet ric sum" }{TEXT -1 6 " or " }{TEXT 406 16 "geometric series" }{TEXT -1 69 " is the sum of consecutive terms of a geometric sequence. Thus , if " }{XPPEDIT 18 0 "M;" "6#%\"MG" }{TEXT -1 7 " and " }{XPPEDIT 18 0 "N;" "6#%\"NG" }{TEXT -1 21 " are integers with " }{XPPEDIT 18 0 "M < N;" "6#2%\"MG%\"NG" }{TEXT -1 6 ", then" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }}{PARA 0 "" 0 "" {TEXT -1 30 " \+ " }{XPPEDIT 18 0 "Sum(a*r^n,n = M .. N);" "6#-%$SumG6$*& %\"aG\"\"\")%\"rG%\"nGF(/F+;%\"MG%\"NG" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 22 "is a geometric series." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 291 30 "Example 1.1: Geometric Series\n" }{TEXT -1 7 " Let " } {XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 7 " and " }{XPPEDIT 18 0 "r; " "6#%\"rG" }{TEXT -1 22 " be constants with " }{XPPEDIT 18 0 "r <> 1;" "6#0%\"rG\"\"\"" }{TEXT 290 1 "." }{TEXT -1 17 " Calculate the \+ " }{XPPEDIT 18 0 "N^th;" "6#)%\"NG%#thG" }{TEXT -1 19 " partial sum \+ of " }{XPPEDIT 18 0 "\{a*r^n, n = 0 .. infinity\};" "6#<$*&%\"aG\"\" \")%\"rG%\"nGF&/F);\"\"!%)infinityG" }{TEXT -1 1 "," }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 292 9 "Solution:" }{TEXT -1 21 " \n\nThe sequence is " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 4 ", " }{XPPEDIT 18 0 "a*r;" "6#*&%\"aG\"\"\"%\"rGF%" }{TEXT -1 4 ", " }{XPPEDIT 18 0 "a*r^2;" "6#*&%\"aG\"\"\"*$%\"rG\"\"#F%" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "a*r^3;" "6#*&%\"aG\"\"\"*$%\"rG\"\"$F%" } {TEXT -1 4 ", " }{XPPEDIT 18 0 "a*r^4;" "6#*&%\"aG\"\"\"*$%\"rG\"\"% F%" }{TEXT -1 4 ", " }{TEXT 385 3 "..." }{TEXT -1 2 " " }{TEXT 384 1 "." }}{PARA 0 "" 0 "" {TEXT -1 20 "The first term is " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 21 ", the second term is " }{XPPEDIT 18 0 "a*r;" "6#*&%\"aG\"\"\"%\"rGF%" }{TEXT -1 21 ", the third term is \+ " }{XPPEDIT 18 0 "a*r^2;" "6#*&%\"aG\"\"\"*$%\"rG\"\"#F%" }{TEXT -1 34 ", and, in general, the power of " }{XPPEDIT 18 0 "r;" "6#%\"rG" }{TEXT -1 54 " is one less than the index of the term. Thus, the " }{XPPEDIT 18 0 "N^th;" "6#)%\"NG%#thG" }{TEXT -1 10 " term is " } {XPPEDIT 18 0 "a*r^(N-1);" "6#*&%\"aG\"\"\")%\"rG,&%\"NGF%F%!\"\"F%" } {TEXT 395 1 "." }{TEXT -1 5 " \n\n" }}{PARA 0 "" 0 "" {TEXT -1 45 "A dding the first N terms and factoring the " }{XPPEDIT 18 0 "a;" "6#% \"aG" }{TEXT -1 26 ", we have\n\n " }{XPPEDIT 18 0 "S[N] = a*(1+r+r^2+r^3+` ... `+r^(N-1));" "6#/&%\"SG6#%\"NG*&%\"aG\"\"\",.F *F*%\"rGF**$F,\"\"#F**$F,\"\"$F*%&~...~GF*)F,,&F'F*F*!\"\"F*F*" } {TEXT 293 1 "." }{TEXT -1 28 " \n\nNotice that\n " } {XPPEDIT 18 0 "r*S[N] = a*(r+r^2+r^3+` ... `+r^(N-1)+r^N);" "6#/*&%\"r G\"\"\"&%\"SG6#%\"NGF&*&%\"aGF&,.F%F&*$F%\"\"#F&*$F%\"\"$F&%&~...~GF&) F%,&F*F&F&!\"\"F&)F%F*F&F&" }{TEXT 294 1 "." }{TEXT -1 21 " \n\nIt fo llows that " }{XPPEDIT 18 0 "S[N]-r*S[N] = a*(1-r^N);" "6#/,&&%\"SG6# %\"NG\"\"\"*&%\"rGF)&F&6#F(F)!\"\"*&%\"aGF),&F)F))F+F(F.F)" }{TEXT -1 10 " , or " }{XPPEDIT 18 0 "S[N] = a*(1-r^N)/(1-r)" "6#/&%\"SG6#% \"NG*(%\"aG\"\"\",&F*F*)%\"rGF'!\"\"F*,&F*F*F-F.F." }{TEXT -1 1 " " } {TEXT 295 1 "." }{TEXT -1 63 "\n\nIn other words,\n \+ " }{XPPEDIT 18 0 "Sum(a*r^n,n = 0 .. N-1) = \+ a*(1-r^N)/(1-r);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\"nGF)/F,;\"\"!,&% \"NGF)F)!\"\"*(F(F),&F)F))F+F1F2F),&F)F)F+F2F2" }{TEXT -1 1 " " } {TEXT 296 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 139 " \+ \+ " }{TEXT 396 3 " " }{XPPEDIT 397 1 "Omega;" "6#%&OmegaG " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }} {PARA 0 "" 0 "" {TEXT -1 91 "It is convenient to have a generalization of the formula found in Example 1.1. First, let " }{XPPEDIT 18 0 "L \+ = N-1;" "6#/%\"LG,&%\"NG\"\"\"F'!\"\"" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "k = n;" "6#/%\"kG%\"nG" }{TEXT -1 1 " " }{TEXT 388 1 ":" }{TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 45 " \+ " }{XPPEDIT 18 0 "Sum(a*r^k,k = 0 .. L) = a*(1-r^(L+1) )/(1-r);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\"kGF)/F,;\"\"!%\"LG*(F(F), &F)F))F+,&F0F)F)F)!\"\"F),&F)F)F+F5F5" }{TEXT -1 1 " " }{TEXT 386 1 ". " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Next, for any integer " }{XPPEDIT 18 0 "M;" "6#%\"MG" } {TEXT -1 28 ", multiply both sides by " }{XPPEDIT 18 0 "r^M;" "6#)% \"rG%\"MG" }{TEXT -1 1 " " }{TEXT 387 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 " \+ " }{XPPEDIT 18 0 "Sum(a*r^(k+M),k = 0 .. L) = a*r^M*(1-r^( L+1))/(1-r);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG,&%\"kGF)%\"MGF)F)/F-;\" \"!%\"LG**F(F))F+F.F),&F)F))F+,&F2F)F)F)!\"\"F),&F)F)F+F8F8" }{TEXT -1 1 " " }{TEXT 389 1 "." }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "Finally, let " }{XPPEDIT 18 0 "N = \+ L+M;" "6#/%\"NG,&%\"LG\"\"\"%\"MGF'" }{TEXT -1 54 " and change the in dex of summation for the series to " }{XPPEDIT 18 0 "n;" "6#%\"nG" } {TEXT -1 7 " where " }{XPPEDIT 18 0 "n = k+M;" "6#/%\"nG,&%\"kG\"\"\"% \"MGF'" }{TEXT -1 31 " . Since the initial value of " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 5 " is " }{XPPEDIT 18 0 "M;" "6#%\"MG" } {TEXT -1 24 " and the final value is " }{XPPEDIT 18 0 "N;" "6#%\"NG" } {TEXT -1 24 ", we have, for integers " }{XPPEDIT 18 0 "M;" "6#%\"MG" } {TEXT -1 5 " and " }{XPPEDIT 18 0 "N;" "6#%\"NG" }{TEXT -1 7 " with \+ " }{XPPEDIT 18 0 "M < N;" "6#2%\"MG%\"NG" }{TEXT -1 1 " " }{TEXT 391 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 410 36 "General Formula For \+ a Geometric Sum:" }{TEXT -1 4 " " }{XPPEDIT 18 0 "Sum(a*r^n,n = M . . N) = a*r^M*(1-r^(N-M+1))/(1-r);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\" nGF)/F,;%\"MG%\"NG**F(F))F+F/F),&F)F))F+,(F0F)F/!\"\"F)F)F6F),&F)F)F+F 6F6" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "a*(r^M-r^(N+1))/(1-r);" "6#*( %\"aG\"\"\",&)%\"rG%\"MGF%)F(,&%\"NGF%F%F%!\"\"F%,&F%F%F(F-F-" }{TEXT 390 2 " ." }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 393 12 "Example 1.2:" }{TEXT -1 13 " Calculate " }{XPPEDIT 18 0 "5/9+5/27+5/81+5/243+5/729;" "6#,,*&\"\"&\"\"\"\"\"*!\"\"F&*&F%F& \"#FF(F&*&F%F&\"#\")F(F&*&F%F&\"$V#F(F&*&F%F&\"$H(F(F&" }{TEXT -1 1 " \+ " }{TEXT 392 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 394 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 37 "We can add this up in the normal way:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "5/9+5/2 7+5/81+5/243+5/729;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"$0'\"$H(" }} }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "Or, we c an use the formula derived before this example with " }{XPPEDIT 18 0 "a = 5;" "6#/%\"aG\"\"&" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "r = 1/3;" " 6#/%\"rG*&\"\"\"F&\"\"$!\"\"" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "M = 2; " "6#/%\"MG\"\"#" }{TEXT -1 9 ", and " }{XPPEDIT 18 0 "N = 6;" "6#/ %\"NG\"\"'" }{TEXT -1 1 " " }{TEXT 408 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "formula := Sum(a*r^n, n = M .. N) = a*(r^M-r^(N+1))/( 1-r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(formulaG/-%$SumG6$*&%\"aG \"\"\")%\"rG%\"nGF+/F.;%\"MG%\"NG*(F*F+,&)F-F1F+)F-,&F2F+F+F+!\"\"F+,& F+F+F-F8F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "subs( \{a=5, \+ r=1/3, M=2, N=6\} , formula);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$S umG6$,$*&\"\"&\"\"\")#F*\"\"$%\"nGF*F*/F.;\"\"#\"\"'#\"$0'\"$H(" }}} {PARA 0 "" 0 "" {TEXT -1 138 " \+ \+ " }{TEXT 409 3 " " }{XPPEDIT 257 1 "Omeg a;" "6#%&OmegaG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 49 "2. Partial Sums of Collapsing or \+ Telescoping Sums" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "If " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 23 " is a \+ function and " }{XPPEDIT 18 0 "a[n] = f(n)-f(n+1);" "6#/&%\"aG6#%\" nG,&-%\"fG6#F'\"\"\"-F*6#,&F'F,F,F,!\"\"" }{TEXT -1 22 ", then we sa y that " }{XPPEDIT 18 0 "Sum(a[n],n = 1 .. N);" "6#-%$SumG6$&%\"aG6#% \"nG/F);\"\"\"%\"NG" }{TEXT -1 8 " is a " }{TEXT 298 10 "collapsing " }{TEXT -1 5 ", or " }{TEXT 297 11 "telescoping" }{TEXT -1 85 ", seri es.\nNotice that the cancellation that occurs when consecutive terms a re added: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 " " }{XPPEDIT 18 0 "a[n]+a[n+1];" "6#,&&%\"aG6#%\"nG\"\"\"&F%6#,&F'F(F(F(F(" }{TEXT -1 7 " = " }{TEXT 370 1 "(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "f(n)-f (n+1);" "6#,&-%\"fG6#%\"nG\"\"\"-F%6#,&F'F(F(F(!\"\"" }{TEXT -1 1 " " }{TEXT 369 1 ")" }{TEXT -1 7 " + " }{TEXT 372 1 "(" }{TEXT -1 1 " \+ " }{XPPEDIT 18 0 "f(n+1)-f(n+2);" "6#,&-%\"fG6#,&%\"nG\"\"\"F)F)F)-F%6 #,&F(F)\"\"#F)!\"\"" }{TEXT -1 1 " " }{TEXT 373 1 ")" }{TEXT -1 9 " \+ = " }{XPPEDIT 18 0 "f(n)-f(n+2);" "6#,&-%\"fG6#%\"nG\"\"\"-F%6#,&F 'F(\"\"#F(!\"\"" }{TEXT -1 1 " " }{TEXT 371 1 "." }}{PARA 0 "" 0 "" {TEXT -1 10 " \nSo,\n\n " }{XPPEDIT 18 0 "Sum(a[n],n = 1 .. N);" "6# -%$SumG6$&%\"aG6#%\"nG/F);\"\"\"%\"NG" }{TEXT -1 5 " = " }{TEXT 299 1 "(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "f(1)-f(2);" "6#,&-%\"fG6#\"\"\"F '-F%6#\"\"#!\"\"" }{TEXT -1 1 " " }{TEXT 300 1 ")" }{TEXT -1 3 " + " } {TEXT 301 1 "(" }{XPPEDIT 18 0 "f(2)-f(3);" "6#,&-%\"fG6#\"\"#\"\"\"-F %6#\"\"$!\"\"" }{TEXT 302 1 ")" }{TEXT -1 3 " + " }{TEXT 303 1 "(" } {XPPEDIT 18 0 "f(3)-f(4);" "6#,&-%\"fG6#\"\"$\"\"\"-F%6#\"\"%!\"\"" } {TEXT 304 1 ")" }{TEXT -1 3 " + " }{TEXT 305 3 "..." }{TEXT -1 3 " + \+ " }{TEXT 306 1 "(" }{XPPEDIT 18 0 "f(N-1)-f(N);" "6#,&-%\"fG6#,&%\"NG \"\"\"F)!\"\"F)-F%6#F(F*" }{TEXT 307 1 ")" }{TEXT -1 3 " +" }{TEXT 308 1 "(" }{XPPEDIT 18 0 "f(N)-f(N+1);" "6#,&-%\"fG6#%\"NG\"\"\"-F%6#, &F'F(F(F(!\"\"" }{TEXT 309 1 ")" }{TEXT -1 19 " \n\nsimplifies to " }{XPPEDIT 18 0 "f(1)-f(N+1);" "6#,&-%\"fG6#\"\"\"F'-F%6#,&%\"NGF'F'F'! \"\"" }{TEXT -1 11 " . \n\nThus," }}{PARA 0 "" 0 "" {TEXT -1 29 " \+ " }{XPPEDIT 18 0 "Sum(f(n)-f(n+1),n = 1 .. N) = f(1)-f(N+1);" "6#/-%$SumG6$,&-%\"fG6#%\"nG\"\"\"-F)6#,&F+F,F,F,!\" \"/F+;F,%\"NG,&-F)6#F,F,-F)6#,&F3F,F,F,F0" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 310 12 "Example 2.1:" } {TEXT -1 17 " Calculate the " }{XPPEDIT 18 0 "342^th;" "6#)\"$U$%#th G" }{TEXT -1 20 " partial sum of " }{XPPEDIT 18 0 "\{1/n/(n+1), n \+ = 1 .. infinity\};" "6#<$*(\"\"\"F%%\"nG!\"\",&F&F%F%F%F'/F&;F%%)infin ityG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 312 9 "Solution:" }{TEXT -1 18 " Notice that " }{XPPEDIT 18 0 "1/n/(n+1) = 1/n-1/(n+1);" "6#/*(\"\"\"F%%\"nG!\"\",&F&F%F%F%F',& *&F%F%F&F'F%*&F%F%,&F&F%F%F%F'F'" }{TEXT -1 41 ". Using the collapsin g sum formula with " }{XPPEDIT 18 0 "f(n) = 1/n;" "6#/-%\"fG6#%\"nG*& \"\"\"F)F'!\"\"" }{TEXT -1 15 " , we have\n\n " }{XPPEDIT 18 0 "S[34 2];" "6#&%\"SG6#\"$U$" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "Sum(1/n/(n+1 ),n = 1 .. 342);" "6#-%$SumG6$*(\"\"\"F'%\"nG!\"\",&F(F'F'F'F)/F(;F'\" $U$" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "Sum(f(n)-f(n+1),n = 1 .. 342); " "6#-%$SumG6$,&-%\"fG6#%\"nG\"\"\"-F(6#,&F*F+F+F+!\"\"/F*;F+\"$U$" } {TEXT -1 4 " = " }{XPPEDIT 18 0 "f(1)-f(343);" "6#,&-%\"fG6#\"\"\"F'- F%6#\"$V$!\"\"" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "1-1/343;" "6#,&\"\" \"F$*&F$F$\"$V$!\"\"F'" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "342/343;" " 6#*&\"$U$\"\"\"\"$V$!\"\"" }{TEXT 311 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "Here is a verification with " } {TEXT 313 5 "Maple" }{TEXT -1 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "Sum(1/n/(n+1), n = 1 .. 342) = sum(1/n/(n+1), n = 1 .. 342);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*&\"\"\"F(*&%\"nGF(,&F*F(F(F (F(!\"\"/F*;F(\"$U$#F/\"$V$" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "(Sometimes " } {TEXT 314 5 "Maple" }{TEXT -1 59 " does calculations like this by brut e force. In this case, " }{TEXT 315 5 "Maple" }{TEXT -1 34 " knows the general simplification:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Sum(1/n/(n+1), n = 1 .. N) = sum(1/ n/(n+1), n = 1 .. N);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*& \"\"\"F(*&%\"nGF(,&F*F(F(F(F(!\"\"/F*;F(%\"NG,&*&F(F(,&F/F(F(F(F,F,F(F (" }}}{PARA 0 "" 0 "" {TEXT -1 139 " \+ \+ " }{TEXT 411 3 " " }{XPPEDIT 257 1 "Omega;" "6#%&OmegaG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 18 "3. Infinite Series" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Given a sequence " }{XPPEDIT 18 0 " \{a[n], n = 1 .. infinity\};" "6#<$&%\"aG6#%\"nG/F';\"\"\"%)infinityG " }{TEXT -1 46 " with its associated sequence of partial sums " } {XPPEDIT 18 0 "\{S[N], N = 1 .. infinity\};" "6#<$&%\"SG6#%\"NG/F';\" \"\"%)infinityG" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 31 "we def ine the infinite series " }{XPPEDIT 18 0 "Sum(a[n],n = 1 .. infinity) ;" "6#-%$SumG6$&%\"aG6#%\"nG/F);\"\"\"%)infinityG" }{TEXT -1 64 " by \+ \n\n " } {XPPEDIT 18 0 "Sum(a[n],n = 1 .. infinity) = limit(S[N],N = infinity); " "6#/-%$SumG6$&%\"aG6#%\"nG/F*;\"\"\"%)infinityG-%&limitG6$&%\"SG6#% \"NG/F5F." }{TEXT -1 11 " \n" }}{PARA 0 "" 0 "" {TEXT -1 25 " provided the limit exists" }{TEXT 318 1 "." }{TEXT -1 44 " In this cas e we say the infinite series is " }{TEXT 316 10 "convergent" }{TEXT 321 1 "." }{TEXT -1 24 " Otherwise we say it is " }{TEXT 317 9 "diverg ent" }{TEXT 319 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 320 12 "Example 3.1:" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 41 "Estimate the partial sums of t he series " }{XPPEDIT 18 0 "Sum(n/(2*n+1),n = 1 .. infinity);" "6#-%$ SumG6$*&%\"nG\"\"\",&*&\"\"#F(F'F(F(F(F(!\"\"/F';F(%)infinityG" } {TEXT -1 1 " " }{TEXT 418 1 "." }{TEXT -1 32 " Show that the series \+ diverges" }{TEXT 422 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 1 " " }{TEXT 419 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{XPPEDIT 18 0 "N^th ;" "6#)%\"NG%#thG" }{TEXT -1 17 " partial sum is " }{XPPEDIT 18 0 "S[ N];" "6#&%\"SG6#%\"NG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "1/3;" "6#*& \"\"\"F$\"\"$!\"\"" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "2/5;" "6#*&\" \"#\"\"\"\"\"&!\"\"" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "3/7;" "6#*&\" \"$\"\"\"\"\"(!\"\"" }{TEXT -1 6 " + " }{TEXT 420 4 " ..." }{TEXT -1 6 " + " }{XPPEDIT 18 0 "(N-1)/(2*N-1);" "6#*&,&%\"NG\"\"\"F&!\" \"F&,&*&\"\"#F&F%F&F&F&F'F'" }{TEXT -1 4 " + " }{XPPEDIT 18 0 "N/(2*N +1);" "6#*&%\"NG\"\"\",&*&\"\"#F%F$F%F%F%F%!\"\"" }{TEXT -1 1 " " } {TEXT 421 1 "." }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "To get so me sense of these sums we calculate several" }{TEXT 442 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "seq(Sum(n/(2*n+1), n = 1..N) = eval f(add(n/(2*n+1), n = 1..N)), N = 1 .. 10); " }}{PARA 12 "" 1 "" {XPPMATH 20 "6,/-%$SumG6$*&%\"nG\"\"\",&*&\"\"#F)F(F)F)F)F)!\"\"/F(;F) F)$\"+LLLLL!#5/-F%6$F'/F(;F)F,$\"+LLLLtF2/-F%6$F'/F(;F)\"\"$$\"+iZ!>; \"!\"*/-F%6$F'/F(;F)\"\"%$\"+1#\\jg\"FB/-F%6$F'/F(;F)\"\"&$\"+hY*31#FB /-F%6$F'/F(;F)\"\"'$\"+AJVADFB/-F%6$F'/F(;F)\"\"($\"+*y*4*)HFB/-F%6$F' /F(;F)\"\")$\"+C!)ofMFB/-F%6$F'/F(;F)\"\"*$\"+NAPLRFB/-F%6$F'/F(;F)\"# 5$\"+6Fc4WFB" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "It appears that " } {XPPEDIT 18 0 "N/3 <= S[N];" "6#1*&%\"NG\"\"\"\"\"$!\"\"&%\"SG6#F%" } {TEXT -1 1 " " }{TEXT 435 1 "." }{TEXT -1 45 " If this were true we \+ could conclude that " }{XPPEDIT 18 0 "limit(S[N],N = infinity) = infi nity;" "6#/-%&limitG6$&%\"SG6#%\"NG/F*%)infinityGF," }{TEXT -1 48 ", \+ which would show that the series is divergent" }{TEXT 440 1 "." } {TEXT -1 39 " Upon closer observation we see that " }{TEXT 441 5 "ev ery" }{TEXT -1 43 " term of the series is no smaller than 1/3" } {TEXT 439 1 "." }{TEXT -1 36 " We prove this as follows. Adding " } {XPPEDIT 18 0 "2*n;" "6#*&\"\"#\"\"\"%\"nGF%" }{TEXT -1 34 " to each \+ side of the inequality " }{XPPEDIT 18 0 "1 <= n;" "6#1\"\"\"%\"nG" } {TEXT -1 15 " results in " }{XPPEDIT 18 0 "2*n+1 <= 3*n;" "6#1,&*& \"\"#\"\"\"%\"nGF'F'F'F'*&\"\"$F'F(F'" }{TEXT 438 2 " ." }{TEXT -1 48 " Dividing each side of the last inequality by " }{XPPEDIT 18 0 "3*( 2*n+1);" "6#*&\"\"$\"\"\",&*&\"\"#F%%\"nGF%F%F%F%F%" }{TEXT -1 36 " g ives us the required inequality " }{XPPEDIT 18 0 "1/3 <= n/(2*n+1);" "6#1*&\"\"\"F%\"\"$!\"\"*&%\"nGF%,&*&\"\"#F%F)F%F%F%F%F'" }{TEXT -1 1 " " }{TEXT 437 1 "." }{TEXT -1 67 " Thus,\n\n \+ " }{XPPEDIT 18 0 "N/3;" "6#*&%\"NG\" \"\"\"\"$!\"\"" }{TEXT -1 6 " = " }{XPPEDIT 18 0 "Sum(1/3,n = 1 .. \+ N);" "6#-%$SumG6$*&\"\"\"F'\"\"$!\"\"/%\"nG;F'%\"NG" }{TEXT -1 2 " " }{XPPEDIT 18 0 "`` <= ``;" "6#1%!GF$" }{TEXT -1 1 " " }{XPPEDIT 18 0 " Sum(n/(2*n+1),n = 1 .. N);" "6#-%$SumG6$*&%\"nG\"\"\",&*&\"\"#F(F'F(F( F(F(!\"\"/F';F(%\"NG" }{TEXT -1 7 " = " }{XPPEDIT 18 0 "S[N];" "6# &%\"SG6#%\"NG" }{TEXT -1 16 " \n\nfor every " }{XPPEDIT 18 0 "N;" " 6#%\"NG" }{TEXT -1 69 " , which, as we have seen, implies the divergen ce of the given series" }{TEXT 436 1 "." }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 " \+ \+ " }{TEXT 412 3 " \+ " }{XPPEDIT 257 1 "Omega;" "6#%&OmegaG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 35 "4. Collapsing or Telescoping Series" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 374 9 "Theorem: " }{TEXT -1 9 " \n\nIf " }{XPPEDIT 18 0 "limit(f(x),x = infinity) = L;" "6#/-%&li mitG6$-%\"fG6#%\"xG/F*%)infinityG%\"LG" }{TEXT -1 12 " , then " } {XPPEDIT 18 0 "Sum(``*(f(n)-f(n+1))*``,n = 1 .. infinity) = ` `*f(1)- L;" "6#/-%$SumG6$*(%!G\"\"\",&-%\"fG6#%\"nGF)-F,6#,&F.F)F)F)!\"\"F)F(F )/F.;F)%)infinityG,&*&%#~~GF)-F,6#F)F)F)%\"LGF2" }{TEXT -1 1 " " } {TEXT 375 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 377 6 "Proof:" }}{PARA 0 "" 0 " " {TEXT -1 13 "The series " }{XPPEDIT 18 0 "Sum(``*(f(n)-f(n+1))*``, n = 1 .. infinity);" "6#-%$SumG6$*(%!G\"\"\",&-%\"fG6#%\"nGF(-F+6#,&F- F(F(F(!\"\"F(F'F(/F-;F(%)infinityG" }{TEXT -1 38 " telescopes. We h ave seen that its " }{XPPEDIT 18 0 "N^th;" "6#)%\"NG%#thG" }{TEXT -1 13 " partial sum " }{XPPEDIT 18 0 "S[N];" "6#&%\"SG6#%\"NG" }{TEXT -1 14 " is given by " }{XPPEDIT 18 0 "S[N] = f(1)-f(N+1);" "6#/&%\"SG6#% \"NG,&-%\"fG6#\"\"\"F,-F*6#,&F'F,F,F,!\"\"" }{TEXT -1 12 " . Therefore " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 " \+ " }{XPPEDIT 18 0 "Sum(``*(f(n)-f(n+1))*``,n = 1 .. infinity); " "6#-%$SumG6$*(%!G\"\"\",&-%\"fG6#%\"nGF(-F+6#,&F-F(F(F(!\"\"F(F'F(/F -;F(%)infinityG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "limit(S[N],N = in finity);" "6#-%&limitG6$&%\"SG6#%\"NG/F)%)infinityG" }{TEXT -1 7 " = " }{XPPEDIT 18 0 "limit((f(1)-f(N+1))*``,N = infinity);" "6#-%&limi tG6$*&,&-%\"fG6#\"\"\"F+-F)6#,&%\"NGF+F+F+!\"\"F+%!GF+/F/%)infinityG" }{TEXT -1 7 " = " }{XPPEDIT 18 0 "f(1)-L;" "6#,&-%\"fG6#\"\"\"F'% \"LG!\"\"" }{TEXT -1 1 " " }{TEXT 376 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 378 12 "Example 4.1:" }{TEXT -1 23 " Show t hat the series " }{XPPEDIT 18 0 "Sum(1/n/(n+1),n = 1 .. infinity);" "6 #-%$SumG6$*(\"\"\"F'%\"nG!\"\",&F(F'F'F'F)/F(;F'%)infinityG" }{TEXT -1 51 " is convergent. What is the value of the series?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 383 9 "Solution:" } {TEXT -1 19 " As we have seen, " }{XPPEDIT 18 0 "Sum(1/n/(n+1),n = 1 \+ .. infinity)" "6#-%$SumG6$*(\"\"\"F'%\"nG!\"\",&F(F'F'F'F)/F(;F'%)infi nityG" }{TEXT -1 26 " is the collapsing series " }{XPPEDIT 18 0 "Sum(f (n)-f(n+1)*``,n = 1 .. infinity);" "6#-%$SumG6$,&-%\"fG6#%\"nG\"\"\"*& -F(6#,&F*F+F+F+F+%!GF+!\"\"/F*;F+%)infinityG" }{TEXT -1 6 " with " } {XPPEDIT 18 0 "f(n) = 1/n;" "6#/-%\"fG6#%\"nG*&\"\"\"F)F'!\"\"" } {TEXT 379 2 " ." }{TEXT -1 7 " \nThe " }{XPPEDIT 18 0 "N^th;" "6#)%\" NG%#thG" }{TEXT -1 14 " partial sum " }{XPPEDIT 18 0 "S[N];" "6#&%\"S G6#%\"NG" }{TEXT -1 58 " is given by\n\n \+ " }{XPPEDIT 18 0 "S[N] = f(1)-f(N+1);" "6#/&%\"SG6#%\"NG ,&-%\"fG6#\"\"\"F,-F*6#,&F'F,F,F,!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "1-1/(N+1);" "6#,&\"\"\"F$*&F$F$,&%\"NGF$F$F$!\"\"F(" }{TEXT 380 1 "." }{TEXT -1 56 " \n\nClearly\n \+ " }{XPPEDIT 18 0 "limit(S[N],N = infinity) = 1;" "6#/-%&limit G6$&%\"SG6#%\"NG/F*%)infinityG\"\"\"" }{TEXT 381 1 "." }{TEXT -1 12 " \+ \n\nThus, " }{XPPEDIT 18 0 "Sum(1/n/(n+1),n = 1 .. infinity)" "6#-% $SumG6$*(\"\"\"F'%\"nG!\"\",&F(F'F'F'F)/F(;F'%)infinityG" }{TEXT -1 15 " converges and " }{XPPEDIT 18 0 "Sum(1/n/(n+1),n = 1 .. infinity) \+ = 1;" "6#/-%$SumG6$*(\"\"\"F(%\"nG!\"\",&F)F(F(F(F*/F);F(%)infinityGF( " }{TEXT 382 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 " \+ \+ " }{TEXT 423 3 " " }{XPPEDIT 256 1 "Omega;" "6#%&OmegaG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 424 12 "Example 4.2:" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 42 "Calculate the partial sums of the serie s " }{XPPEDIT 18 0 "Sum(n/(n+1)!,n = 1 .. infinity);" "6#-%$SumG6$*&% \"nG\"\"\"-%*factorialG6#,&F'F(F(F(!\"\"/F';F(%)infinityG" }{TEXT -1 1 " " }{TEXT 426 1 "." }{TEXT -1 33 " Evaluate the sum of the series " }{TEXT 430 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 427 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{XPPEDIT 18 0 "N^th;" "6#)%\"NG %#thG" }{TEXT -1 17 " partial sum is " }{XPPEDIT 18 0 "S[N];" "6#&%\" SG6#%\"NG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "1/2!;" "6#*&\"\"\"F$-%* factorialG6#\"\"#!\"\"" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "2/3!;" "6# *&\"\"#\"\"\"-%*factorialG6#\"\"$!\"\"" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "3/4!;" "6#*&\"\"$\"\"\"-%*factorialG6#\"\"%!\"\"" }{TEXT -1 6 " \+ + " }{TEXT 428 4 " ..." }{TEXT -1 6 " + " }{XPPEDIT 18 0 "(N-1)/ N!;" "6#*&,&%\"NG\"\"\"F&!\"\"F&-%*factorialG6#F%F'" }{TEXT -1 4 " + \+ " }{XPPEDIT 18 0 "N/(N+1)!;" "6#*&%\"NG\"\"\"-%*factorialG6#,&F$F%F%F% !\"\"" }{TEXT -1 1 " " }{TEXT 429 1 "." }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "Notice that the " } {XPPEDIT 18 0 "n^th;" "6#)%\"nG%#thG" }{TEXT -1 30 " term of the seri es satisfies" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " " }{XPPEDIT 18 0 "n/(n+1)! = (``*(n+1)-1) /(n+1)!;" "6#/*&%\"nG\"\"\"-%*factorialG6#,&F%F&F&F&!\"\"*&,&*&%!GF&,& F%F&F&F&F&F&F&F+F&-F(6#,&F%F&F&F&F+" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "1/n!-1/(n+1)!;" "6#,&*&\"\"\"F%-%*factorialG6#%\"nG!\"\"F%*&F%F%-F' 6#,&F)F%F%F%F*F*" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "f(n)-f(n+1);" "6 #,&-%\"fG6#%\"nG\"\"\"-F%6#,&F'F(F(F(!\"\"" }{TEXT -1 1 " " }{TEXT 431 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "where" }{TEXT 433 2 " " }{XPPEDIT 18 0 "f(n) = 1/n!;" "6#/-%\"fG6#%\"nG*&\"\"\"F)-%*factorialG6#F'!\"\" " }{TEXT 432 3 " . " }{TEXT -1 18 " It follows that " }{XPPEDIT 18 0 "S[N] = f(1)-f(N+1);" "6#/&%\"SG6#%\"NG,&-%\"fG6#\"\"\"F,-F*6#,&F'F,F, F,!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "1/1!-1/(N+1)!;" "6#,&*&\"\" \"F%-%*factorialG6#F%!\"\"F%*&F%F%-F'6#,&%\"NGF%F%F%F)F)" }{TEXT -1 8 " and \n\n" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }{XPPEDIT 18 0 " Sum(n/(n+1)!,n = 1 .. infinity) = limit(S[N],N = infinity);" "6#/-%$Su mG6$*&%\"nG\"\"\"-%*factorialG6#,&F(F)F)F)!\"\"/F(;F)%)infinityG-%&lim itG6$&%\"SG6#%\"NG/F8F1" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "limit((1- 1/(N+1)!)*``,N = infinity);" "6#-%&limitG6$*&,&\"\"\"F(*&F(F(-%*factor ialG6#,&%\"NGF(F(F(!\"\"F/F(%!GF(/F.%)infinityG" }{TEXT -1 6 " = 1 " }{TEXT 434 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 138 " \+ \+ " }{TEXT 425 3 " " }{XPPEDIT 256 1 "Omega;" "6#%&OmegaG " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 19 "5. Geometric Series" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 5 "Let " }{XPPEDIT 18 0 "a;" "6#%\"aG" } {TEXT -1 7 " and " }{XPPEDIT 18 0 "r;" "6#%\"rG" }{TEXT -1 22 " be \+ constants with " }{XPPEDIT 18 0 "r <> 1;" "6#0%\"rG\"\"\"" }{TEXT 322 1 "." }{TEXT -1 67 " The series,\n \n \+ " }{XPPEDIT 18 0 "a+a*r+a*r^2+a*r^3;" "6#,*%\"a G\"\"\"*&F$F%%\"rGF%F%*&F$F%*$F'\"\"#F%F%*&F$F%*$F'\"\"$F%F%" }{TEXT -1 4 " + " }{TEXT 324 3 "..." }{TEXT -1 15 " , \n\nthat is, " } {XPPEDIT 18 0 "Sum(a*r^n,n = 0 .. infinity);" "6#-%$SumG6$*&%\"aG\"\" \")%\"rG%\"nGF(/F+;\"\"!%)infinityG" }{TEXT -1 10 ", is an " }{TEXT 326 27 "(infinite) geometric series" }{TEXT 325 1 "." }{TEXT -1 25 " \+ We have seen that its " }{XPPEDIT 18 0 "N^th;" "6#)%\"NG%#thG" } {TEXT -1 15 " partial sum " }{XPPEDIT 18 0 "S[N];" "6#&%\"SG6#%\"NG " }{TEXT -1 14 " is given by\n" }}{PARA 0 "" 0 "" {TEXT -1 43 " \+ " }{XPPEDIT 18 0 "S[N] = a*(1-r^N )/(1-r)" "6#/&%\"SG6#%\"NG*(%\"aG\"\"\",&F*F*)%\"rGF'!\"\"F*,&F*F*F-F. F." }{TEXT -1 1 " " }{TEXT 323 1 "." }{TEXT -1 15 "\n\nNotice that " }{XPPEDIT 18 0 "limit(S[N],N = infinity);" "6#-%&limitG6$&%\"SG6#%\"NG /F)%)infinityG" }{TEXT -1 28 " exists if and only if " }{XPPEDIT 18 0 "abs(r) < 1;" "6#2-%$absG6#%\"rG\"\"\"" }{TEXT 327 1 "." }{TEXT -1 15 " In this case " }{XPPEDIT 18 0 "limit(S[N],N = infinity) = a/( 1-r);" "6#/-%&limitG6$&%\"SG6#%\"NG/F*%)infinityG*&%\"aG\"\"\",&F/F/% \"rG!\"\"F2" }{TEXT -1 48 " . In summary, \n\n\n \+ " }{XPPEDIT 18 0 "Sum(a*r^n,n = 0 .. infinity);" "6#-%$SumG6$* &%\"aG\"\"\")%\"rG%\"nGF(/F+;\"\"!%)infinityG" }{TEXT -1 5 " " } {TEXT 329 8 "diverges" }{TEXT -1 9 " if " }{XPPEDIT 18 0 "1 <= ab s(r);" "6#1\"\"\"-%$absG6#%\"rG" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "and \n\n \+ " }{XPPEDIT 18 0 "Sum(a*r^n,n = 0 .. infinity) = a/ (1-r);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\"nGF)/F,;\"\"!%)infinityG*&F (F),&F)F)F+!\"\"F3" }{TEXT -1 16 " if " }{XPPEDIT 18 0 "ab s(r) < 1;" "6#2-%$absG6#%\"rG\"\"\"" }{TEXT 328 1 "." }}{PARA 0 "" 0 " " {TEXT -1 4 "\n\n\n\n" }}{PARA 0 "" 0 "" {TEXT 340 17 "A Generalizati on:" }}{PARA 0 "" 0 "" {TEXT -1 26 "By factoring we see that " } {XPPEDIT 18 0 "Sum(a*r^n,n = M .. infinity) = r^M*Sum(a*r^n,n = 0 .. i nfinity);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\"nGF)/F,;%\"MG%)infinityG *&)F+F/F)-F%6$*&F(F))F+F,F)/F,;\"\"!F0F)" }{TEXT -1 1 " " }{TEXT 339 1 "." }{TEXT -1 68 " Therefore, we have\n\n\n \+ " }{XPPEDIT 18 0 "Sum(a*r^n,n = M .. infinity) = \+ a*r^M/(1-r);" "6#/-%$SumG6$*&%\"aG\"\"\")%\"rG%\"nGF)/F,;%\"MG%)infini tyG*(F(F))F+F/F),&F)F)F+!\"\"F4" }{TEXT -1 1 " " }{TEXT 338 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 54 "6. Expressing a Repeating Decimal as a Rational Number" }}{PARA 0 "" 0 "" {TEXT 330 22 "\nExercise 20, \+ page 628" }}{PARA 0 "" 0 "" {TEXT -1 9 "Express " }{TEXT 331 16 "0.98 398398398..." }{TEXT -1 23 " as a rational number " }{XPPEDIT 18 0 "m /n;" "6#*&%\"mG\"\"\"%\"nG!\"\"" }{TEXT -1 6 " with " }{XPPEDIT 18 0 " m;" "6#%\"mG" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "n;" "6#%\"nG" } {TEXT -1 11 " integers." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 353 9 "Solution:" }{TEXT -1 8 " We have" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 " 0.98 398398398 ,,, = " }{XPPEDIT 18 0 "983;" "6#\"$$)*" }{TEXT -1 1 " \+ " }{TEXT 354 1 "(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "1/(10^3);" "6#*&\" \"\"F$*$\"#5\"\"$!\"\"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "1/(10^6);" " 6#*&\"\"\"F$*$\"#5\"\"'!\"\"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "1/(10^ 9);" "6#*&\"\"\"F$*$\"#5\"\"*!\"\"" }{TEXT -1 3 " + " }{TEXT 352 3 ".. ." }{TEXT -1 2 " " }{TEXT 355 1 ")" }{TEXT 356 1 "." }{TEXT -1 2 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 " ." }}{PARA 0 "" 0 "" {TEXT -1 21 "We apply the formula " }{XPPEDIT 18 0 " Sum(a*r^n,n = M .. infinity) = a*r^M/(1-r);" "6#/-%$SumG6$*&%\"aG\"\" \")%\"rG%\"nGF)/F,;%\"MG%)infinityG*(F(F))F+F/F),&F)F)F+!\"\"F4" } {TEXT -1 7 " with " }{XPPEDIT 18 0 "a = 1;" "6#/%\"aG\"\"\"" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "M = 1;" "6#/%\"MG\"\"\"" }{TEXT -1 9 ", an d " }{XPPEDIT 18 0 "r = 1/(10^3);" "6#/%\"rG*&\"\"\"F&*$\"#5\"\"$!\" \"" }{TEXT 357 1 "." }{TEXT -1 12 " \nWe find " }{XPPEDIT 18 0 "1/(1 0^3);" "6#*&\"\"\"F$*$\"#5\"\"$!\"\"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "1/(10^6);" "6#*&\"\"\"F$*$\"#5\"\"'!\"\"" }{TEXT -1 4 " + " } {XPPEDIT 18 0 "1/(10^9);" "6#*&\"\"\"F$*$\"#5\"\"*!\"\"" }{TEXT -1 3 " + " }{TEXT 359 3 "..." }{TEXT -1 6 " = " }{XPPEDIT 18 0 "``*1/(10^ 3)*``/(1-1/(10^3));" "6#*,%!G\"\"\"F%F%*$\"#5\"\"$!\"\"F$F%,&F%F%*&F%F %*$F'F(F)F)F)" }{TEXT -1 1 " " }{TEXT 358 1 " " }{TEXT -1 1 "=" } {TEXT 361 2 " " }{XPPEDIT 18 0 "1/999;" "6#*&\"\"\"F$\"$***!\"\"" } {TEXT 360 2 " ." }{TEXT -1 45 " Thus, \n 0.98398398398 ,, , = " }{XPPEDIT 18 0 "983;" "6#\"$$)*" }{TEXT -1 1 " " }{TEXT 362 1 "(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "1/999;" "6#*&\"\"\"F$\"$***!\"\" " }{TEXT -1 1 " " }{TEXT 363 1 ")" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "983/999;" "6#*&\"$$)*\"\"\"\"$***!\"\"" }{TEXT -1 1 " " }{TEXT 364 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "It would be only a little bit of extra work if the \+ repetition did not begin right after the decimal." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 350 8 "Example:" }}{PARA 0 "" 0 "" {TEXT -1 9 "Express " }{TEXT 351 16 "0.97398398398..." }{TEXT -1 23 " as a rational number " }{XPPEDIT 18 0 "m/n;" "6#*&%\"mG\"\"\"%\" nG!\"\"" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "m;" "6#%\"mG" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 11 " integers." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 334 10 "\nSoluti on:" }{TEXT -1 8 " We have" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " 0.97398398398 ,,, = " }{XPPEDIT 18 0 "97/100; " "6#*&\"#(*\"\"\"\"$+\"!\"\"" }{TEXT -1 5 " + " }{XPPEDIT 18 0 "398 /(10^5);" "6#*&\"$)R\"\"\"*$\"#5\"\"&!\"\"" }{TEXT -1 3 " + " } {XPPEDIT 18 0 "398/(10^8);" "6#*&\"$)R\"\"\"*$\"#5\"\")!\"\"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "398/(10^11);" "6#*&\"$)R\"\"\"*$\"#5\"#6!\" \"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "398/(10^14);" "6#*&\"$)R\"\"\"*$ \"#5\"#9!\"\"" }{TEXT -1 3 " + " }{TEXT 332 3 "..." }{TEXT -1 5 " = \+ " }{XPPEDIT 18 0 "97/100;" "6#*&\"#(*\"\"\"\"$+\"!\"\"" }{TEXT -1 5 " \+ + " }{XPPEDIT 18 0 "398/(10^5);" "6#*&\"$)R\"\"\"*$\"#5\"\"&!\"\"" } {TEXT -1 1 " " }{TEXT 335 1 "(" }{TEXT -1 5 "1 + " }{XPPEDIT 18 0 "1/ (10^3);" "6#*&\"\"\"F$*$\"#5\"\"$!\"\"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "1/(10^6);" "6#*&\"\"\"F$*$\"#5\"\"'!\"\"" }{TEXT -1 3 " + " } {XPPEDIT 18 0 "1/(10^9);" "6#*&\"\"\"F$*$\"#5\"\"*!\"\"" }{TEXT -1 3 " + " }{TEXT 333 3 "..." }{TEXT -1 2 " " }{TEXT 336 1 ")" }{TEXT 337 1 "." }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 " ." }}{PARA 0 "" 0 "" {TEXT -1 21 "We apply the formu la " }{XPPEDIT 18 0 "Sum(a*r^n,n = M .. infinity) = a*r^M/(1-r);" "6#/ -%$SumG6$*&%\"aG\"\"\")%\"rG%\"nGF)/F,;%\"MG%)infinityG*(F(F))F+F/F),& F)F)F+!\"\"F4" }{TEXT -1 7 " with " }{XPPEDIT 18 0 "a = 1;" "6#/%\"aG \"\"\"" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "M = 0;" "6#/%\"MG\"\"!" } {TEXT -1 9 ", and " }{XPPEDIT 18 0 "r = 1/(10^3);" "6#/%\"rG*&\"\" \"F&*$\"#5\"\"$!\"\"" }{TEXT 341 1 "." }{TEXT -1 16 " \nWe find 1 + \+ " }{XPPEDIT 18 0 "1/(10^3);" "6#*&\"\"\"F$*$\"#5\"\"$!\"\"" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "1/(10^6);" "6#*&\"\"\"F$*$\"#5\"\"'!\"\"" } {TEXT -1 4 " + " }{XPPEDIT 18 0 "1/(10^9);" "6#*&\"\"\"F$*$\"#5\"\"*! \"\"" }{TEXT -1 3 " + " }{TEXT 343 3 "..." }{TEXT -1 6 " = " } {XPPEDIT 18 0 "1*``/(1-1/(10^3));" "6#*(\"\"\"F$%!GF$,&F$F$*&F$F$*$\"# 5\"\"$!\"\"F+F+" }{TEXT -1 1 " " }{TEXT 342 1 " " }{TEXT -1 1 "=" } {TEXT 345 2 " " }{XPPEDIT 18 0 "1000/999;" "6#*&\"%+5\"\"\"\"$***!\" \"" }{TEXT 344 2 " ." }{TEXT -1 45 " Thus, \n 0.973983983 98 ,,, = " }{XPPEDIT 18 0 "97/100;" "6#*&\"#(*\"\"\"\"$+\"!\"\"" } {TEXT -1 5 " + " }{XPPEDIT 18 0 "398/(10^5);" "6#*&\"$)R\"\"\"*$\"#5 \"\"&!\"\"" }{TEXT -1 1 " " }{TEXT 346 1 "(" }{TEXT -1 1 " " } {XPPEDIT 18 0 "1000/999;" "6#*&\"%+5\"\"\"\"$***!\"\"" }{TEXT -1 1 " \+ " }{TEXT 347 1 ")" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "97301/99900;" " 6#*&\"&,t*\"\"\"\"&+***!\"\"" }{TEXT -1 1 " " }{TEXT 348 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "As a chec k, we ask " }{TEXT 349 5 "Maple" }{TEXT -1 59 " for a 15 decimal place numerical evaluation of our answer:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(97301/99900, 15);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"0%)R)R)R)R(*!#:" }}}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 39 "7. The Multiplier Effect (in Economics)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 21 "Exercise 43, page 629" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 315 "Suppose that every dollar that we spend gives rise (through wages , profits, etc.) to 90 cents for someone else to spend. That 90 cents \+ will generate a further 81 cents for spending, and so on. How much spe nding will result from the purchase of a $16000 automobile? This pheno menon is known as the multiplier effect." }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 59 "The spending, including the init ial purchase of the car, is" }}{PARA 0 "" 0 "" {TEXT -1 22 "\n \+ 16000 + " }{TEXT 276 1 "(" }{TEXT -1 6 "16000 " }{XPPEDIT 18 0 "9 /10" "6#*&\"\"*\"\"\"\"#5!\"\"" }{TEXT -1 1 " " }{TEXT 277 1 ")" } {TEXT -1 5 " + " }{TEXT 278 1 "(" }{TEXT -1 6 "16000 " }{XPPEDIT 18 0 "9/10" "6#*&\"\"*\"\"\"\"#5!\"\"" }{TEXT -1 1 " " }{TEXT 279 2 ")(" }{TEXT -1 1 " " }{XPPEDIT 18 0 "9/10;" "6#*&\"\"*\"\"\"\"#5!\"\"" } {TEXT -1 1 " " }{TEXT 280 1 ")" }{TEXT -1 5 " + " }{TEXT 286 1 "(" } {TEXT 281 1 "(" }{TEXT -1 6 "16000 " }{XPPEDIT 18 0 "9/10" "6#*&\"\"* \"\"\"\"#5!\"\"" }{TEXT -1 1 " " }{TEXT 282 2 ")(" }{TEXT -1 1 " " } {XPPEDIT 18 0 "9/10;" "6#*&\"\"*\"\"\"\"#5!\"\"" }{TEXT -1 1 " " } {TEXT 283 1 ")" }{TEXT -1 1 " " }{TEXT 285 2 ")(" }{TEXT -1 1 " " } {XPPEDIT 18 0 "9/10;" "6#*&\"\"*\"\"\"\"#5!\"\"" }{TEXT -1 1 " " } {TEXT 284 1 ")" }{TEXT -1 4 " + " }{TEXT 287 4 " ..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "or\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "Sum(16000*(9/10)^k, k=0..infinity) = sum(16 000*(9/10)^k, k=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$S umG6$,$*&\"&+g\"\"\"\")#\"\"*\"#5%\"kGF*F*/F/;\"\"!%)infinityG\"'++;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 23 " 8. Two Trains and A Fly" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 464 "An eastbound and a westbound train head toward eac h other on the same track. The speed of each train is 120 miles per ho ur. When the trains are 120 miles apart, a fly departs the front of th e eastbound train and flies to the front of the westbound train, where upon the fly reverses direction and flies back to the eastbound train. If the speed of the fly is 90 miles per hour and the fly repeats this process until the two trains collide, how far does the fly fly?" } {TEXT -1 3 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Let " }{XPPEDIT 18 0 "d[n ];" "6#&%\"dG6#%\"nG" }{TEXT -1 68 " denote the distance between the t wo trains when the fly begins its " }{XPPEDIT 18 0 "n^th;" "6#)%\"nG%# thG" }{TEXT -1 7 " trip" }{TEXT 267 1 "." }{TEXT -1 70 " This is the distance the fly and the oncoming train cover during the " }{XPPEDIT 18 0 "n^th;" "6#)%\"nG%#thG" }{TEXT -1 7 " trip" }{TEXT 268 1 "." } {TEXT -1 16 " We are given " }{XPPEDIT 18 0 "d[1] = 120;" "6#/&%\"dG 6#\"\"\"\"$?\"" }{TEXT -1 6 " miles" }{TEXT 269 1 "." }{TEXT -1 8 " L et " }{XPPEDIT 18 0 "f[n];" "6#&%\"fG6#%\"nG" }{TEXT -1 46 " denote the distance the fly travels in its " }{XPPEDIT 18 0 "n^th;" "6#)%\" nG%#thG" }{TEXT -1 30 " trip between the two trains" }{TEXT 270 1 ". " }{TEXT -1 103 " During each trip the fly and the oncoming train appr oach each other at 90 + 60, or 150, miles per hour" }{TEXT 271 1 "." } {TEXT -1 139 " Since the speed of the fly, namely 90, is 3/5 of the sp eed at which the fly and the oncoming train near each other, namely 15 0, we have " }{XPPEDIT 18 0 "f[n] = 3/5*d[n];" "6#/&%\"fG6#%\"nG*(\" \"$\"\"\"\"\"&!\"\"&%\"dG6#F'F*" }{TEXT 265 1 "." }{TEXT -1 18 " Als o, the time " }{XPPEDIT 18 0 "t[n];" "6#&%\"tG6#%\"nG" }{TEXT -1 9 " o f the " }{XPPEDIT 18 0 "n^th;" "6#)%\"nG%#thG" }{TEXT -1 22 " trip \+ is given by " }{XPPEDIT 18 0 "t[n] = d[n]/150;" "6#/&%\"tG6#%\"nG*&& %\"dG6#F'\"\"\"\"$]\"!\"\"" }{TEXT -1 6 " hours" }{TEXT 264 1 "." } {TEXT -1 15 " During the " }{XPPEDIT 18 0 "n^th;" "6#)%\"nG%#thG" } {TEXT -1 78 " trip the two trains approach each other at 60 + 60, o r 120, miles per hour" }{TEXT 272 1 "." }{TEXT -1 60 " The distance b y which they close on each other during the " }{XPPEDIT 18 0 "n^th;" " 6#)%\"nG%#thG" }{TEXT -1 23 " trip is therefore " }{XPPEDIT 18 0 " 120*d[n]/150;" "6#*(\"$?\"\"\"\"&%\"dG6#%\"nGF%\"$]\"!\"\"" }{TEXT -1 7 " , or " }{XPPEDIT 18 0 "4*d[n]/5;" "6#*(\"\"%\"\"\"&%\"dG6#%\"nGF% \"\"&!\"\"" }{TEXT -1 9 ", miles" }{TEXT 273 1 "." }{TEXT -1 18 " T hus, we have " }{XPPEDIT 18 0 "d[n+1] = d[n]-4*d[n]/5;" "6#/&%\"dG6# ,&%\"nG\"\"\"F)F),&&F%6#F(F)*(\"\"%F)&F%6#F(F)\"\"&!\"\"F2" }{TEXT -1 11 " , or " }{XPPEDIT 18 0 "d[n+1] = d[n]/5;" "6#/&%\"dG6#,&%\"nG \"\"\"F)F)*&&F%6#F(F)\"\"&!\"\"" }{TEXT -1 1 " " }{TEXT 266 1 "." } {TEXT -1 21 " It follows that " }{XPPEDIT 18 0 "d[2] = 120/5;" "6# /&%\"dG6#\"\"#*&\"$?\"\"\"\"\"\"&!\"\"" }{TEXT -1 5 ", " }{XPPEDIT 18 0 "d[3] = d[2]/5;" "6#/&%\"dG6#\"\"$*&&F%6#\"\"#\"\"\"\"\"&!\"\"" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "120/(5^2);" "6#*&\"$?\"\"\"\"*$\"\"& \"\"#!\"\"" }{TEXT -1 27 " , and, in general, " }{XPPEDIT 18 0 "d[n] = 120/(5^(n-1));" "6#/&%\"dG6#%\"nG*&\"$?\"\"\"\")\"\"&,&F'F*F*! \"\"F." }{TEXT -1 1 " " }{TEXT 261 1 "." }{TEXT -1 9 " Thus, " } {XPPEDIT 18 0 "f[n] = 3/5*d[n];" "6#/&%\"fG6#%\"nG*(\"\"$\"\"\"\"\"&! \"\"&%\"dG6#F'F*" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "360/(5^n);" "6#* &\"$g$\"\"\")\"\"&%\"nG!\"\"" }{TEXT -1 2 " " }{TEXT 262 1 "." } {TEXT -1 31 " The distance the fly flies is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 " \+ " }{XPPEDIT 18 0 "Sum(f[n],n = 1 .. infinity);" "6#-%$SumG6$ &%\"fG6#%\"nG/F);\"\"\"%)infinityG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " Sum(360/(5^n),n = 1 .. infinity);" "6#-%$SumG6$*&\"$g$\"\"\")\"\"&%\"n G!\"\"/F+;F(%)infinityG" }{TEXT -1 5 " = 90" }{TEXT 263 1 "." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 " " 0 "" {TEXT 274 626 "A simpler way to solve this problem is to notice that, since the two trains approach each other at 120 miles per hour \+ and since they are 120 miles apart when the fly sets off, they will co llide one hour later. At a constant speed of 90 miles per hour, the fl y flies 90 miles in this time. When the Hungarian-American methematici an John von Neumann, famous for his computational abilities among othe r things, was asked this problem, he responded with the correct answer almost instantaneously. The poser of the problem, deprived of his amu sement, said \"You must know the trick.\" Replied von Neumann, \"I jus t summed the series.\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 22 "9. The Harmonic Series" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "The so-called " }{TEXT 365 15 "harmonic series" }{TEXT -1 2 " " }{XPPEDIT 18 0 "Sum(1/n,n = \+ 1 .. infinity);" "6#-%$SumG6$*&\"\"\"F'%\"nG!\"\"/F(;F'%)infinityG" } {TEXT -1 57 " is known to diverge, as we will soon see for ourselves " }{TEXT 443 1 "." }{TEXT -1 7 " \nThe " }{XPPEDIT 18 0 "N^th;" "6#)% \"NG%#thG" }{TEXT -1 58 " partial sum of the harmonic series is usuall y denoted by " }{XPPEDIT 18 0 "H[N];" "6#&%\"HG6#%\"NG" }{TEXT -1 1 " \+ " }{TEXT 366 1 ":" }{TEXT -1 3 " \n " }}{PARA 0 "" 0 "" {TEXT -1 46 " \+ " }{XPPEDIT 18 0 "H[N] = \+ Sum(1/n,n = 1 .. N);" "6#/&%\"HG6#%\"NG-%$SumG6$*&\"\"\"F,%\"nG!\"\"/F -;F,F'" }{TEXT -1 1 " " }{TEXT 444 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "Figure 1 below shows how " } {XPPEDIT 18 0 "H[N];" "6#&%\"HG6#%\"NG" }{TEXT -1 81 " can be interp reted geometrically as the area of rectangles each having base 1." }} {PARA 13 "" 1 "" {GLPLOT2D 380 380 380 {PLOTDATA 2 "60-%'CURVESG6%7Y7$ $\"\"\"\"\"!F(7$$\"3kmmT&)G\\a5!#<$\"3!4f8,OJK[*!#=7$$\"3FLL$3x&)*36F. $\"3\"=73)[#[s,*F17$$\"3!****\\ilyM;\"F.$\"3Kc`'p'[\"\\f)F17$$\"3bmmmT :(z@\"F.$\"3ml3tdCP5#)F17$$\"34+]7y%*z78F.$\"3<8'4&[&4th(F17$$\"3SLLe9 ui29F.$\"3qNfWIE:/rF17$$\"3?+](oMrU^\"F.$\"3#3kF:COQg'F17$$\"3ymm;z_\" 4i\"F.$\"3sAub+RNphF17$$\"3&ommm6m#G(z[F17$$\"3smm\"z/3 uC#F.$\"3])R=YDq&\\WF17$$\"3o***\\7LRDX#F.$\"3!>?nrd1u2%F17$$\"3%om;zR 'okEF.$\"3W)>EC#oy_PF17$$\"3I++D1J:wGF.$\"3cj7\\mj'oZ$F17$$\"3oLLL3En$ 4$F.$\"3;:(*[#4/CB$F17$$\"3#pmmT!RE&G$F.$\"3cSgn&\\&*Q/$F17$$\"3D+++D. &4]$F.$\"31]^@#Hnj&GF17$$\"3;+++vB_vspa#F17$$\"3Inm\"z*ev:TF.$\"3$\\Et+M(oHCF17$$\"3(RLL$347 TVF.$\"30'[`\"=Db.BF17$$\"3nLLLLY.KXF.$\"353RmqW^1AF17$$\"3k***\\7o7Tv %F.$\"3f')=uR>W.@F17$$\"3kLLL$Q*o]\\F.$\"3MUt822#*>?F17$$\"3m++D\"=lj; &F.$\"3?BW%p\"ofN>F17$$\"3S++vV&RHN%fh=F17$$\"3CML$e9Ege& F.$\"3SW5*4Q\"=!z\"F17$$\"3]LLeR\"3Gy&F.$\"3te8WlOEHh,%3r)3;F17$$\"3:++v=>Y2kF .$\"3ij80k.og:F17$$\"3Znm;zXu9mF.$\"3*p.:1\"G%o0/8F17$$\"3Y+++v.I%)yF.$\"3ANuLcLMo7F17$$\"3?mm\"zpe*z!)F.$ \"3FE1q)3IwB\"F17$$\"3;,++D\\'QH)F.$\"3aWo!p\\5d?\"F17$$\"31KLe9S8&\\) F.$\"39J5!oYWr<\"F17$$\"3h,+D1#=bq)F.$\"3%G.Wfp'p[6F17$$\"3!QLL$3s?6*) F.$\"3Rkd77C=A6F17$$\"3a***\\7`Wl7*F.$\"3[<$)*!#>7$$\"3zmmmw(Gp.\"F_[l$\"3`dmp.R'Qk*Fb[l7$$\"3-+] 7oK0e5F_[l$\"3c)>ym)*>8X*Fb[l7$$\"37+](=5s#y5F_[l$\"3!R**[&Rp4u#*Fb[l7 $$\"#6F*$\"3;\"4444444*Fb[l-%'COLOURG6&%$RGBG$\")!\\DP\"!\")F[]l$\")vi obF]]l-%*THICKNESSG6#\"\"#-%%TEXTG6(7$$\"#_!\"\"$\"#X!\"#Q(y~=~1/x6\"% +ALIGNBELOWG%*ALIGNLEFTGFg\\l-%%FONTG6%%&TIMESG%%BOLDG\"#9-%)POLYGONSG 6$7&F'7$$Fc]lF*F(7$F]_l$F*F*7$F(F__l-%&COLORG6&Fj\\l$\"#'*F]^l$\"#(*F] ^l$\"$***!\"$-Fi^l6$7&7$F]_l$\"+++++]!#57$$\"\"$F*F_`l7$Fc`lF__lF^_lFa _l-Fi^l6$7&7$Fc`l$\"+LLLLLFa`l7$$\"\"%F*Fj`l7$F]alF__lFe`lFa_l-Fi^l6$7 &7$F]al$\"+++++DFa`l7$$\"\"&F*Fdal7$FgalF__lF_alFa_l-Fi^l6$7&7$Fgal$\" +++++?Fa`l7$$\"\"'F*F^bl7$FablF__lFialFa_l-Fi^l6$7&7$Fabl$\"+nmmm;Fa`l 7$$\"\"(F*Fhbl7$F[clF__lFcblFa_l-Fi^l6$7&7$F[cl$\"+H9dG9Fa`l7$$\"\")F* Fbcl7$FeclF__lF]clFa_l-Fi^l6$7&7$Fecl$\"++++]7Fa`l7$$\"\"*F*F\\dl7$F_d lF__lFgclFa_l-Fi^l6$7&7$F_dl$\"+66666Fa`l7$$\"#5F*Ffdl7$FidlF__lFadlFa _l-Fi^l6$7&7$Fidl$\"+++++5Fa`l7$Fc\\lF`el7$Fc\\lF__lF[elFa_l-%+AXESLAB ELSG6$Q\"xF_^lQ!F_^l-%%VIEWG6$;F__lFc\\l;F__lF(" 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" "Curv e 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Cur ve 11" "Curve 12" }}}{PARA 0 "" 0 "" {TEXT -1 157 " \+ \n \+ \+ " }{TEXT 445 8 "Figure 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 264 " The first rectan gle has height 1/1. Its base is the interval [1,2]. It therefore has a rea 1. The second rectangle has height 1/2. Its base is the interval [ 2,3], which has length 1. The area of the second rectangle is therefor e 1/2. In general, the height of the " }{XPPEDIT 18 0 "n^th;" "6#)%\"n G%#thG" }{TEXT -1 97 " rectangle is 1/n and the base is the interval [n , n+1], which has length 1. The area of the " }{XPPEDIT 18 0 "n^t h;" "6#)%\"nG%#thG" }{TEXT -1 29 " rectangle is therefore 1/n. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "We see th at" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 19 " " }{XPPEDIT 18 0 "ln(N +1);" "6#-%#lnG6#,&%\"NG\"\"\"F(F(" }{TEXT -1 10 " = " } {XPPEDIT 18 0 "Int(1/x,x = 1 .. N+1);" "6#-%$IntG6$*&\"\"\"F'%\"xG!\" \"/F(;F',&%\"NGF'F'F'" }{TEXT -1 7 " < " }{XPPEDIT 18 0 "Sum(1/n,n = 1 .. N);" "6#-%$SumG6$*&\"\"\"F'%\"nG!\"\"/F(;F'%\"NG" }{TEXT -1 6 " = " }{XPPEDIT 18 0 "H[N];" "6#&%\"HG6#%\"NG" }{TEXT -1 1 " " } {TEXT 446 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "It follows that " }{XPPEDIT 18 0 "limit(H[N],N = infinit y) = infinity;" "6#/-%&limitG6$&%\"HG6#%\"NG/F*%)infinityGF," }{TEXT -1 41 ", which shows that the harmonic series " }{XPPEDIT 18 0 "Sum( 1/n,n = 1 .. infinity);" "6#-%$SumG6$*&\"\"\"F'%\"nG!\"\"/F(;F'%)infin ityG" }{TEXT -1 10 " diverges." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 413 12 "Example 9.1:" }{TEXT -1 1 " " }{TEXT 415 23 "(Exercise 4 0, page 629)" }{TEXT -1 24 " \nUse the inequality " }{XPPEDIT 18 0 "x/2 < ln(1+x);" "6#2*&%\"xG\"\"\"\"\"#!\"\"-%#lnG6#,&F&F&F%F&" } {TEXT -1 13 " for 0 < " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 23 " < 1 to show that " }{XPPEDIT 18 0 "Sum(ln(1+1/n),n = 1 .. in finity);" "6#-%$SumG6$-%#lnG6#,&\"\"\"F**&F*F*%\"nG!\"\"F*/F,;F*%)infi nityG" }{TEXT -1 16 " is divergent." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 414 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "First of all, let us get a sen se of this series by calculating several of the partial sums. Here are the first 25:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "seq(Sum(ln(1 +1/n),n = 1..infinity) = add(evalf(ln(1+1/n)), n = 1..N), N=1..25);" } }{PARA 12 "" 1 "" {XPPMATH 20 "6;/-%$SumG6$-%#lnG6#,&\"\"\"F+*&F+F+%\" nG!\"\"F+/F-;F+%)infinityG$\"+1=ZJp!#5/F$$\"+*G7')4\"!\"*/F$$\"+hVH'Q \"F8/F$$\"+7zV4;F8/F$$\"+p%fF8/F$$\"+U:Wz?F8/F$$ \"+yXA(>#F8/F$$\"+%4&e-BF8/F$$\"+u_*yR#F8/F$$\"+^m!\\[#F8/F$$\"+e$\\\\ c#F8/F$$\"+It0REF8/F$$\"+,-03FF8/F$$\"+A()esFF8/F$$\"+WL@LGF8/F$$\"+d< P!*GF8/F$$\"+z*QW%HF8/F$$\"+tAt&*HF8/F$$\"+PC_WIF8/F$$\"+`C/\"4$F8/F$$ \"+:U\\NJF8/F$$\"+IQ0yJF8/F$$\"+De()=KF8/F$$\"+Ql4eKF8" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 266 "The partial sums in crease because each new term added is positive. However, the rate at w hich the partial sums increase is decreasing. The numerical evidence f or convergence or divergence is not especially clear. The analytic app roach we now take will be convincing." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 35 "The following graph shows that " }{XPPEDIT 18 0 "x/2 < ln(1+x);" "6#2*&%\"xG\"\"\"\"\"#!\"\"-%#lnG6#,&F &F&F%F&" }{TEXT -1 13 " for 0 < " }{XPPEDIT 18 0 "x;" "6#%\"xG" } {TEXT -1 6 " < 1 " }{TEXT 416 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "plot([ln(1+x),x/2], x = 0 .. 1, color = [NAVY,SIENNA], thickness =2);" }}{PARA 13 "" 1 "" {GLPLOT2D 375 375 375 {PLOTDATA 2 "6'-%'CURVE SG6$7S7$$\"\"!F)F(7$$\"3emmm;arz@!#>$\"3o+;Hz#*Hc@F-7$$\"3[LL$e9ui2%F- $\"3Cvs9f\\Q&*RF-7$$\"3nmmm\"z_\"4iF-$\"3V!)[]e.,CgF-7$$\"3[mmmT&phN)F -$\"3Nzg%G7[`-)F-7$$\"3CLLe*=)H\\5!#=$\"3ifDyp?=y**F-7$$\"3gmm\"z/3uC \"FB$\"3UnYqLh_v6FB7$$\"3%)***\\7LRDX\"FB$\"3/'RE2)QEc8FB7$$\"3]mm\"zR 'ok;FB$\"3Z-#))4G4)R:FB7$$\"3w***\\i5`h(=FB$\"3(>7m*eNZ>FB7$$\"3qmm;/RE&G#FB$\"3aT,pXR:e?FB7$$\"3\")***** \\K]4]#FB$\"3p)zVUu&>KAFB7$$\"3$******\\PAvr#FB$\"3S9'e5kcRS#FB7$$\"3) ******\\nHi#HFB$\"3aY)*R@YtmDFB7$$\"3jmm\"z*ev:JFB$\"3s7O*za\"H7FFB7$$ \"3?LLL347TLFB$\"3i*z`2qfE)GFB7$$\"3,LLLLY.KNFB$\"32ef'G\"RB%*f80%FB7$$\"3&em;zRQb@&FB$\"3'G**Rsv?t>%FB7$$\" 3\\***\\(=>Y2aFB$\"3&)HU_#RoEK%FB7$$\"39mm;zXu9cFB$\"3c<90-aIcWFB7$$\" 3l******\\y))GeFB$\"3_A[(eA:Df%FB7$$\"3'*)***\\i_QQgFB$\"3Fqhc\\$)*Rs% FB7$$\"3@***\\7y%3TiFB$\"3qUn$QO!f\\[FB7$$\"35****\\P![hY'FB$\"3%*3MPj a@()\\FB7$$\"3kKLL$Qx$omFB$\"36%e&)\\h#G4^FB7$$\"3!)*****\\P+V)oFB$\"3 3Voz_7*zB&FB7$$\"3?mm\"zpe*zqFB$\"3wCp3sn?``FB7$$\"3%)*****\\#\\'QH(FB $\"3QTP,qrmxaFB7$$\"3GKLe9S8&\\(FB$\"31c2hHpP$f&FB7$$\"3R***\\i?=bq(FB $\"3fNqi5E\"Hr&FB7$$\"3\"HLL$3s?6zFB$\"30')Rd\\_TGeFB7$$\"3a***\\7`Wl7 )FB$\"3g#)HL'>Bz%fFB7$$\"3#pmmm'*RRL)FB$\"3cEz8@*)ohgFB7$$\"3Qmm;a<.Y& )FB$\"3I:km:vqwhFB7$$\"3=LLe9tOc()FB$\"3)=9e)=>[*G'FB7$$\"3u******\\Qk \\*)FB$\"3[e_iT/+#R'FB7$$\"3CLL$3dg6<*FB$\"3-i%QA.?#3lFB7$$\"3ImmmmxGp $*FB$\"3N)4(HQh.6mFB7$$\"3A++D\"oK0e*FB$\"3l@k8\"\\2&>nFB7$$\"3A++v=5s #y*FB$\"3V^^#)yyBAoFB7$$\"\"\"F)$\"3'GX*f0=ZJpFB-%'COLOURG6&%$RGBG$\") !\\DP\"!\")Fjz$\")viobF\\[l-F$6$7SF'7$F+$\"3ILLL3x&)*3\"F-7$F1$\"3umm \"H2P\"Q?F-7$F6$\"3MLL$eRwX5$F-7$F;$\"3CLL$3x%3yTF-7$F@$\"3=mm\"z%4\\Y _F-7$FF$\"3)HL$eR-/PiF-7$FK$\"3A***\\il'pisF-7$FP$\"3`KLe*)>VB$)F-7$FU $\"3!))**\\7`l2Q*F-7$FZ$\"3smm;/j$o/\"FB7$Fin$\"3NLL3_>jU6FB7$F^o$\"3! *****\\i^Z]7FB7$Fco$\"3'*****\\(=h(e8FB7$Fho$\"3)*****\\P[6j9FB7$F]p$ \"3KL$e*[z(yb\"FB7$Fbp$\"3gmm;a/cq;FB7$Fgp$\"3]mmm;t,mFB7$Ffq$\"35+]i!f#=$3#FB7$F[r$\"3/+](=x pe=#FB7$F`r$\"3smm\"H28IH#FB7$Fer$\"3km;zpSS\"R#FB7$Fjr$\"3GLL3_?`(\\# FB7$F_s$\"3#HLe*)>pxg#FB7$Fds$\"3u**\\Pf4t.FFB7$Fis$\"32LLe*Gst!GFB7$F ^t$\"3#)*****\\#RW9HFB7$Fct$\"3[***\\7j#>>IFB7$Fht$\"3h**\\i!RU07$FB7$ F]u$\"3b***\\(=S2LKFB7$Fbu$\"3Kmmm\"p)=MLFB7$Fgu$\"3!*****\\(=]@W$FB7$ F\\v$\"35L$e*[$z*RNFB7$Fav$\"3#*****\\iC$pk$FB7$Ffv$\"39m;H2qcZPFB7$F[ w$\"3q**\\7.\"fF&QFB7$F`w$\"3Ymm;/OgbRFB7$Few$\"3y**\\ilAFjSFB7$Fjw$\" 3YLLL$)*pp;%FB7$F_x$\"3?LL3xe,tUFB7$Fdx$\"3em;HdO=yVFB7$Fix$\"3))***** \\#>#[Z%FB7$F^y$\"3immT&G!e&e%FB7$Fcy$\"3;LLL$)Qk%o%FB7$Fhy$\"37+]iSjE !z%FB7$F]z$\"35+]P40O\"*[FB7$Fbz$\"3++++++++]FB-Fgz6&FizF][l$\")%yg>%F \\[lFjz-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$Q\"x6\"Q!F^el-%%VIEWG6$;F(F bz%(DEFAULTG" 1 2 0 1 10 2 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 32 "We can derive the inequality " }{XPPEDIT 18 0 "x/2 < ln(1+x);" "6#2*&%\"xG\"\"\"\"\"#!\"\"-%#lnG6#,&F&F&F%F&" }{TEXT -1 13 " for 0 < " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 26 " < 1 \+ by noticing that " }{XPPEDIT 18 0 "1+t < 2;" "6#2,&\"\"\"F%%\"tGF%\" \"#" }{TEXT -1 12 " for 0 < " }{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 21 " < 1 and therefore\n\n " }{XPPEDIT 18 0 "1/2 < 1/(1+t);" "6#2*& \"\"\"F%\"\"#!\"\"*&F%F%,&F%F%%\"tGF%F'" }{TEXT -1 11 " for 0 < " } {XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 4 " < 1" }{TEXT 417 1 "." } {TEXT -1 69 " We integrate each side of the last inequality over the interval " }{XPPEDIT 18 0 "[0, x];" "6#7$\"\"!%\"xG" }{TEXT -1 8 " \+ , 0 < " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 16 " < 1, to obtain " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "int(1/2, t = 0 ..x) < int(1/ (1+t), t = 0 .. x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#2,$*&\"\"#!\"\" %\"xG\"\"\"F)-%#lnG6#,&F)F)F(F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 15 "It follows that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Sum(ln(1+1/n), n = 1 .. N) > Sum(1/2/n, n = 1 .. N); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#2-%$SumG6$,$*&\"\"\"F)*&\"\"#F)%\" nGF)!\"\"F)/F,;F)%\"NG-F%6$-%#lnG6#,&F)F)*&F)F)F,F-F)F." }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Sum(ln(1+ 1/n), n = 1 .. N) > (1/2)*Sum(1/n, n = 1 .. N);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#2,$*&#\"\"\"\"\"#F'-%$SumG6$*&F'F'%\"nG!\"\"/F-;F'%\"NG F'F'-F*6$-%#lnG6#,&F'F'F,F'F/" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 8 "Thus, " }{XPPEDIT 18 0 "H[N]/2 < Sum(ln( 1+1/n),n = 1 .. N);" "6#2*&&%\"HG6#%\"NG\"\"\"\"\"#!\"\"-%$SumG6$-%#ln G6#,&F)F)*&F)F)%\"nGF+F)/F4;F)F(" }{TEXT -1 26 " . Since the partial s ums " }{XPPEDIT 18 0 "H[N];" "6#&%\"HG6#%\"NG" }{TEXT -1 127 " of the \+ harmonic series tend to infinity, we conclude the same for the partial sums of the given series. It therefore diverges." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 148 " \+ \+ " }{TEXT 478 1 " " }{XPPEDIT 479 0 "Omega;" "6#%&OmegaG" }{TEXT 480 2 " " }{TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 447 12 "E xample 9.2:" }{TEXT -1 1 " " }{TEXT 449 23 "(Exercise 49, page 629)" } {TEXT -1 25 " \nHomer Woodman has a " }{TEXT 450 5 "large" }{TEXT -1 81 " supply of wooden dominoes. They are 1 inch long and 3/16 inch thick.\nHe stacks " }{XPPEDIT 18 0 "N;" "6#%\"NG" }{TEXT -1 31 " of \+ them so that, for each 2 " }{XPPEDIT 18 0 "`` <= ``;" "6#1%!GF$" } {XPPEDIT 18 0 "n;" "6#%\"nG" }{XPPEDIT 18 0 "`` <= ``;" "6#1%!GF$" } {TEXT -1 1 " " }{XPPEDIT 18 0 "N;" "6#%\"NG" }{TEXT -1 8 ", the " } {XPPEDIT 18 0 "n^th;" "6#)%\"nG%#thG" }{TEXT -1 60 " domino, countin g from the bottom of the stack, protrudes " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 " \+ " }{XPPEDIT 18 0 "1/2/(N-n+1);" "6#*(\"\"\"F$\"\"#!\"\",(%\"NGF$% \"nGF&F$F$F&" }{TEXT -1 8 " inch" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 20 "over the end of the " }{XPPEDIT 18 0 "( n-1)^st;" "6#),&%\"nG\"\"\"F&!\"\"%#stG" }{TEXT -1 30 " domino. See \+ Figure 2. \n\n\n\n " }}{PARA 13 "" 1 "" {GLPLOT2D 380 380 380 {PLOTDATA 2 "6>-%)POLYGONSG6#7&7$$\"\"!F)F(7$$\"\"\"F)F(7$F+$\"++++v=! #57$F(F.-F$6#7&7$$\"+LLL$3#!#6F.7$$\"+LL$3-\"!\"*F.7$F:$\"++++]PF07$F6 F>-F$6#7&7$$\"+xjCdUF8F>7$$\"+kCdU5F7$FH$\"++++DcF07$FEFK-F$6#7&7$$ \"+]O(*HlF8FK7$$\"+O(*Hl5FF0Ffq7$$\"+#f 9w>\"Fi#F0F`s7$$\"+Ez>i7F'\\\\B%F0Fcv7$$\"+i\\\\B9F'\\\\t%F0F`w7$$\"+i\\\\t9FF " 0 "" {MPLTEXT 1 0 47 "eqn : = 1 + sum(1/2/(N-k) , k = 1 .. N-1) = 120;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqnG/,&\"\"\"F'-%$sumG6$,$*&F'F'*&\"\"#F',&%\"NGF'% \"kG!\"\"F'F2F'/F1;F',&F0F'F'F2F'\"$?\"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "\n \nThis equation is equivalent to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "eqn2 := sum(1/(N-k) , k = 1 .. N-1) = 238;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%eqn2G/-%$sumG6$*&\"\"\"F*,&%\"NGF*%\"kG!\"\"F./F-;F* ,&F,F*F*F.\"$Q#" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "or, setting " }{XPPEDIT 18 0 "n = N-k;" "6#/%\"nG,&%\"NG \"\"\"%\"kG!\"\"" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "eqn3 := sum(1/n , n = 1 .. 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Other Exercises from the Text" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 472 22 "Exercise 51 (page 630)" }{TEXT -1 29 "\n\011\nPlot the partial sums of " }{XPPEDIT 18 0 "sum(exp(-n)/n, n = 1 .. infinity);" "6#-%$sumG6$*&-%$expG6#,$%\"nG!\"\"\"\"\"F+F,/F+; F-%)infinityG" }{TEXT 473 1 "." }{TEXT -1 125 " From your plot does it appear that the given series converges? If so, then approximately wha t number does it converge to?\n\011\n" }}{PARA 0 "" 0 "" {TEXT 474 9 " Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "S[1] := evalf(exp(-1)):\nfor N from 2 to 20 do\nS[N] \+ := evalf(S[N-1] + exp(-N)/N): \nend do:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot([seq([N,S[N]], N = 1 .. 20)], style = POINT);" } }{PARA 13 "" 1 "" {GLPLOT2D 538 226 226 {PLOTDATA 2 "6&-%'CURVESG6$767 $$\"\"\"\"\"!$\"3!******>T%zyO!#=7$$\"\"#F*$\"3%)*****z#3ZbVF-7$$\"\"$ F*$\"3'******HsF9_%F-7$$\"\"%F*$\"3=+++?o@nXF-7$$\"\"&F*$\"3w*****Rr#p !e%F-7$$\"\"'F*$\"3C+++oR#[e%F-7$$\"\"(F*$\"3w*****plEhe%F-7$$\"\")F*$ \"35+++&)fa'e%F-7$$\"\"*F*$\"3E+++2Jo'e%F-7$$\"#5F*$\"3A+++2&Gne%F-7$$ \"#6F*$\"3E+++!pVne%F-7$$\"#7F*$\"3y******4)[ne%F-7$$\"#8F*$\"3!****** *[0v'e%F-7$$\"#9F*$\"3=+++V6v'e%F-7$$\"#:F*$\"3')*****pM^ne%F-7$$\"#;F *$\"35+++<9v'e%F-7$$\"#F*$\"3A+++_9v'e%F-7$$\"#?F*$\"3u*****HX^ne%F--%'COLOURG6&%$ RGBG$FX!\"\"$F*F*F]r-%&STYLEG6#%&POINTG-%+AXESLABELSG6$Q!6\"Fer-%%VIEW G6$%(DEFAULTGFjr" 1 5 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "S[19], S[2 0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+_9v'e%!#5$\"+`9v'e%F%" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "The graph ical evidence suggests that the series converges to a number not too m uch bigger than " }{XPPEDIT 18 0 ".4586751453;" "6#-%&FloatG6$\"+`9v' e%!#5" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "In fact, the series can be summed exactly:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sum(exp(-n)/n,n=1..infinity);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#,&\"\"\"F(*&F(F(-%$expG6#F(! \"\"F-F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf( -ln(1-1/ exp(1)) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+a9v'e%!#5" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 151 " \+ \+ " } {TEXT 483 1 " " }{XPPEDIT 257 0 "Omega;" "6#%&OmegaG" }{TEXT 484 2 " \+ " }{TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 477 24 "Exercise 52. (page 630)" }{TEXT -1 1 "\011" }}{PARA 0 " " 0 "" {TEXT -1 26 "Plot the partial sums of " }{XPPEDIT 18 0 "sum(sq rt(n)/n!,n = 1 .. infinity);" "6#-%$sumG6$*&-%%sqrtG6#%\"nG\"\"\"-%*fa ctorialG6#F*!\"\"/F*;F+%)infinityG" }{TEXT 475 1 "." }{TEXT -1 126 " \+ From your plot does it appear that the given series converges? If so, \+ then approximately what number does it converge to?\n\011\n" }}{PARA 0 "" 0 "" {TEXT 476 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "S[1] := evalf(sqrt(1)/1!):\n for N from 2 to 20 do\nS[N] := evalf(S[N-1] + sqrt(N)/N!): \nend do:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot([seq([N,S[N]], N = 1 .. 20)], style = POINT);" }}{PARA 13 "" 1 "" {GLPLOT2D 538 226 226 {PLOTDATA 2 "6&-%'CURVESG6$767$$\"\"\"\"\"!F(7$$\"\"#F*$\"3-+++\"y1rq \"!#<7$$\"\"$F*$\"3))*****f\">y&*>F07$$\"\"%F*$\"3++++\\_6z?F07$$\"\"& F*$\"3?+++\\\"\\x4#F07$$\"\"'F*$\"3/+++=7:,@F07$$\"\"(F*$\"34+++phn,@F 07$$\"\")F*$\"3++++=ju,@F07$$\"\"*F*$\"37+++&ea<5#F07$$\"#5F*$\"3!)*** **fXb<5#F07$$\"#6F*$\"3,+++Rbv,@F07$$\"#7F*$\"39+++Ybv,@F07$$\"#8F*$\" 3y*****pab<5#F07$$\"#9F*F`o7$$\"#:F*F`o7$$\"#;F*F`o7$$\"#F*F`o7$$\"#?F*F`o-%'COLOURG6&%$RGBG$FV!\"\"$F*F*F]q-%&STYL EG6#%&POINTG-%+AXESLABELSG6$Q!6\"Feq-%%VIEWG6$%(DEFAULTGFjq" 1 5 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 13 "S[19], S[20];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+Zbv,@!\"*F#" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 96 "The graphical evidence suggests that the \+ series converges to a number not too much bigger than " }{XPPEDIT 18 0 "2.101755547;" "6#-%&FloatG6$\"+Zbv,@!\"*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 502 5 "Maple" }{TEXT -1 91 " cannot sum this series exactly, but its numerical evaluation i s in line with our evidence:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "ev alf(sum(sqrt(n)/n!,n = 1 .. infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+[bv,@!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 155 " \+ \+ " }{TEXT 485 1 " " }{XPPEDIT 257 0 "Ome ga;" "6#%&OmegaG" }{TEXT 486 2 " " }{TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 2 "\011\n" }}{PARA 0 "" 0 "" {TEXT -1 2 "\011\n" }}{PARA 0 "" 0 "" {TEXT 489 24 "Exercise 53. (page 630)" }{TEXT -1 1 "\011" } }{PARA 0 "" 0 "" {TEXT -1 26 "Plot the partial sums of " }{XPPEDIT 18 0 "sum((1.1/n)^n,n = 1 .. infinity);" "6#-%$sumG6$)*&-%&FloatG6$\"# 6!\"\"\"\"\"%\"nGF,F./F.;F-%)infinityG" }{TEXT 487 1 "." }{TEXT -1 126 " From your plot does it appear that the given series converges? \+ If so, then approximately what number does it converge to?\n\011\n" }} {PARA 0 "" 0 "" {TEXT 488 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "S[1] := 1.1:\nfor N from 2 to 20 do\nS[N] := evalf(S[N-1] + (1.1/N)^N): \nend do:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot([seq([N,S[N]], N = 1 .. 20)], \+ style = POINT);" }}{PARA 13 "" 1 "" {GLPLOT2D 538 226 226 {PLOTDATA 2 "6&-%'CURVESG6$767$$\"\"\"\"\"!$\"33+++++++6!#<7$$\"\"#F*$\"33+++++]-9 F-7$$\"\"$F*$\"3%******fH'z^9F-7$$\"\"%F*$\"33+++Pa^d9F-7$$\"\"&F*$\"3 '********zI!e9F-7$$\"\"'F*$\"3)******4xo!e9F-7$$\"\"(F*$\"3!******p8r! e9F-7$$\"\")F*$\"31+++l72e9F-7$$\"\"*F*$\"37+++r72e9F-7$$\"#5F*FT7$$\" #6F*FT7$$\"#7F*FT7$$\"#8F*FT7$$\"#9F*FT7$$\"#:F*FT7$$\"#;F*FT7$$\"#F*FT7$$\"#?F*FT-%'COLOURG6&%$RGBG$FX!\"\"$F*F*Fgp- %&STYLEG6#%&POINTG-%+AXESLABELSG6$Q!6\"F_q-%%VIEWG6$%(DEFAULTGFdq" 1 5 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "S[19], S[20];" }}{PARA 11 " " 1 "" {XPPMATH 20 "6$$\"+r72e9!\"*F#" }}}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 96 "The graphical evidence suggests that \+ the series converges to a number not too much bigger than " } {XPPEDIT 18 0 "1.458071271;" "6#-%&FloatG6$\"+r72e9!\"*" }{TEXT -1 2 " . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "Map le cannot sum this series exactly, but its numerical evaluation is in \+ line with our evidence:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "evalf(s um((1.1/n)^n,n = 1 .. infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$ \"+r72e9!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 " \+ \+ " }{TEXT 490 1 " " }{XPPEDIT 256 0 "Omega ;" "6#%&OmegaG" }{TEXT 491 2 " " }{TEXT -1 4 " " }}{PARA 0 "" 0 " " {TEXT -1 2 "\011\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 494 24 "Exercise 54. (page 630)" }{TEXT -1 1 "\011" }}{PARA 0 " " 0 "" {TEXT -1 26 "Plot the partial sums of " }{XPPEDIT 18 0 "sum(si n(n)/(n^3),n = 1 .. infinity);" "6#-%$sumG6$*&-%$sinG6#%\"nG\"\"\"*$F* \"\"$!\"\"/F*;F+%)infinityG" }{TEXT 492 1 "." }{TEXT -1 126 " From yo ur plot does it appear that the given series converges? If so, then ap proximately what number does it converge to?\n\011\n" }}{PARA 0 "" 0 " " {TEXT 493 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "S[1] := sin(1.0):\nfor N from 2 to \+ 50 do\nS[N] := evalf(S[N-1] + sin(N)/N^3): \nend do:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot([seq([N,S[N]], N = 1 .. 50)], style \+ = POINT);" }}{PARA 13 "" 1 "" {GLPLOT2D 538 226 226 {PLOTDATA 2 "6&-%' CURVESG6$7T7$$\"\"\"\"\"!$\"3W+++[)4ZT)!#=7$$\"\"#F*$\"3o*****>jJ8b*F- 7$$\"\"$F*$\"3U+++-$)f.'*F-7$$\"\"%F*$\"3[*****>\"zM&[*F-7$$\"\"&F*$\" 3!*******pRj3%*F-7$$\"\"'F*$\"33+++n!)p&R*F-7$$\"\"(F*$\"3q*****\\>_[T *F-7$$\"\")F*$\"3s*****zfvTV*F-7$$\"\"*F*$\"3x*****4!)G)R%*F-7$$\"#5F* $\"3=+++!f)QM%*F-7$$\"#6F*$\"3^*****f^voU*F-7$$\"#7F*$\"3@+++\\.xB%*F- 7$$\"#8F*$\"3$)*****p!GoD%*F-7$$\"#9F*$\"3l*****>*GHH%*F-7$$\"#:F*$\"3 e*****4n>7V*F-7$$\"#;F*$\"3@+++#y;0V*F-7$$\"#F*$\"3R+++\\2\\F%*F-7$$\"#?F*$\"3(**** **4$>jG%*F-7$$\"#@F*$\"3,+++\\``H%*F-7$$\"#AF*$\"3`+++Oq_H%*F-7$$\"#BF *$\"3Q+++J:$)G%*F-7$$\"#CF*$\"3:+++ak(G%*F-7$$\"#SF*$\"3_*****4!e$)G%*F-7$$\"#TF*$\"3R+ ++'y7)G%*F-7$$\"#UF*$\"3(*******y!*oG%*F-7$$\"#VF*$\"3L+++iWeG%*F-7$$ \"#WF*$\"3=+++SleG%*F-7$$\"#XF*$\"3F+++=*z'G%*F-7$$\"#YF*$\"3$)*****\\ cs(G%*F-7$$\"#ZF*$\"3%)*****pY%yG%*F-7$$\"#[F*$\"3u********\\rG%*F-7$$ \"#\\F*$\"36+++KRjG%*F-7$$\"#]F*$\"3x*****>%HhG%*F--%'COLOURG6&%$RGBG$ FX!\"\"$F*F*Fc[l-%&STYLEG6#%&POINTG-%+AXESLABELSG6$Q!6\"F[\\l-%%VIEWG6 $%(DEFAULTGF`\\l" 1 5 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "S[49], S[5 0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+KRjG%*!#5$\"+UHhG%*F%" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "The graph ical evidence suggests that the series converges to about 0" } {XPPEDIT 18 0 ".94286;" "6#-%&FloatG6$\"&'G%*!\"&" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 503 5 "Maple" } {TEXT -1 74 " cannot sum this series exactly and does not return a num erical evaluation" }{TEXT 495 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "evalf(sum(sin(n)/n^3,n = 1 .. infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$sumG6$*&-%$sinG6#%\"nG\"\"\"F*!\"$/F*;F+%)infinityG " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 " \+ \+ \+ " }{TEXT 496 1 " " }{XPPEDIT 256 0 "Omega;" "6#%&OmegaG" } {TEXT 497 2 " " }{TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 500 24 "Exercise 55. \+ (page 630)" }{TEXT -1 1 "\011" }}{PARA 0 "" 0 "" {TEXT -1 22 "The geom etric series " }{XPPEDIT 18 0 "sum((1/(1+10^(-50)))^n,n = 1 .. infini ty);" "6#-%$sumG6$)*&\"\"\"F(,&F(F()\"#5,$\"#]!\"\"F(F.%\"nG/F/;F(%)in finityG" }{TEXT 498 2 " " }{TEXT -1 26 "converges, but its sum " } {XPPEDIT 18 0 "S;" "6#%\"SG" }{TEXT -1 16 " is very large" }{TEXT 501 1 "." }{TEXT -1 52 " By about how much does the millionth partial sum " }{XPPEDIT 18 0 "S[1000000];" "6#&%\"SG6#\"(+++\"" }{TEXT -1 29 " differ from the full sum " }{XPPEDIT 18 0 "S;" "6#%\"SG" } {TEXT -1 20 "? How large must " }{XPPEDIT 18 0 "N;" "6#%\"NG" } {TEXT -1 15 " be so that " }{XPPEDIT 18 0 "S[N];" "6#&%\"SG6#%\"NG " }{TEXT -1 22 " is within 0.1 of " }{XPPEDIT 18 0 "S;" "6#%\"SG" }{TEXT -1 6 " ?\n \011\n" }}{PARA 0 "" 0 "" {TEXT 499 9 "Solution:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Let " } {XPPEDIT 18 0 "r = 1/(1+10^(-50));" "6#/%\"rG*&\"\"\"F&,&F&F&)\"#5,$\" #]!\"\"F&F," }{TEXT -1 1 " " }{TEXT 504 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "The difference between the full sum and the partial sum " }{XPPEDIT 18 0 "S[10^6];" "6#&%\"SG6#*$\"# 5\"\"'" }{TEXT -1 7 " is " }{XPPEDIT 18 0 "sum(r^n,n = 10^6+1 .. in finity);" "6#-%$sumG6$)%\"rG%\"nG/F(;,&*$\"#5\"\"'\"\"\"F/F/%)infinity G" }{TEXT -1 12 " , or " }{XPPEDIT 18 0 "r^(10^6+1)/(1-r);" "6#* &)%\"rG,&*$\"#5\"\"'\"\"\"F*F*F*,&F*F*F%!\"\"F," }{TEXT -1 1 " " } {TEXT 505 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "We will write this difference as " }{XPPEDIT 18 0 "r = 1 /(1+a^(-b));" "6#/%\"rG*&\"\"\"F&,&F&F&)%\"aG,$%\"bG!\"\"F&F," }{TEXT -1 14 " and fill in " }{XPPEDIT 18 0 "a = 10;" "6#/%\"aG\"#5" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "b = 50;" "6#/%\"bG\"#]" }{TEXT -1 7 " l ater" }{TEXT 507 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "r := 1/(1+a^(-b));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG*&\"\"\"F&,&F&F&)%\"aG,$%\"bG!\"\"F&F," }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "The difference \+ " }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 22 " is then g iven by " }{XPPEDIT 18 0 "epsilon = r^c/(1-r);" "6#/%(epsilonG*&)%\" rG%\"cG\"\"\",&F)F)F'!\"\"F+" }{TEXT -1 10 " where " }{XPPEDIT 18 0 "c = 10^6+1;" "6#/%\"cG,&*$)\"#5\"\"'\"\"\"F*F*F*" }{TEXT -1 25 " w ill be filled in later" }{TEXT 506 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "epsilon := r^c/(1-r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(epsilonG*&)*&\"\"\"F(,&F(F()%\"aG,$%\"bG!\"\"F(F.%\"cGF(,&F(F (F'F.F." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "After some algebraic simplification we find " }{XPPEDIT 18 0 " epsilon = a^(b*c)/((a^b+1)^(c-1));" "6#/%(epsilonG*&)%\"aG*&%\"bG\"\" \"%\"cGF*F*),&)F'F)F*F*F*,&F+F*F*!\"\"F0" }{TEXT -1 35 " . Let us tes t this equation with " }{TEXT 508 5 "Maple" }{TEXT 509 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "testeq( epsilon = a^(b*c)/(a^b+1) ^(c-1) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Set " }{XPPEDIT 18 0 "E = ln(epsilon);" "6#/%\"EG-%#lnG6#%(epsilonG" }{TEXT -1 55 " in or der to facilitate working with very large numbers" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "E := ln( a^(b*c)/(a^b+1)^(c-1) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG-%#lnG6#*&)%\"aG*&%\"bG\"\"\"%\"cGF-F-),&)F* F,F-F-F-,&F.F-F-!\"\"F3" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 35 "Simplifying the logarithm we find " }{XPPEDIT 18 0 "E = b*c*ln(a)-(c-1)*ln(a^b+1);" "6#/%\"EG,&*(%\"bG\"\"\"%\"cGF(-%#l nG6#%\"aGF(F(*&,&F)F(F(!\"\"F(-F+6#,&)F-F'F(F(F(F(F0" }{TEXT -1 1 " " }{TEXT 510 1 ":" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "e := b*c*ln(a)-(c-1)*ln(a^b+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%\"eG,&*(%\"bG\"\"\"%\"cGF(-%#lnG6#%\"aGF(F(*&,&F)F(F(!\"\"F(-F+6#,& )F-F'F(F(F(F(F0" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "Next we numerically evaluate the logarithm of the error" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "log_difference := evalf( sub s(\{a=10, b=50, c = 10^6+1\} , e), 100);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%/log_differenceG$\"ipXDg)*z6SinQ[I'y$[]Rm;!)['Q9Uu]0!Q5#=Uts&* *3?%G-(\\YDH^6!#\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 22 "Thus, the difference " }{XPPEDIT 18 0 "exp(log_differe nce);" "6#-%$expG6#%/log_differenceG" }{TEXT -1 22 " between the full \+ sum " }{XPPEDIT 18 0 "S;" "6#%\"SG" }{TEXT -1 32 " and the millionth p artial sum " }{XPPEDIT 18 0 "S[1000000];" "6#&%\"SG6#\"(+++\"" } {TEXT -1 11 " is about" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "exp(log_difference);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+.&*******\"#S" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "After adding a million terms of the series we a re still about " }{XPPEDIT 18 0 "10^50;" "6#*$)\"#5\"#]\"\"\"" }{TEXT -1 24 " away from the full sum." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 28 "Next we consider how large " }{XPPEDIT 18 0 "N;" "6#%\"NG" }{TEXT -1 20 " must be so that " }{XPPEDIT 18 0 "S[N];" "6#&%\"SG6#%\"NG" }{TEXT -1 22 " is within 0.1 of " } {XPPEDIT 18 0 "S;" "6#%\"SG" }{TEXT -1 16 " . \nFirst set " } {XPPEDIT 18 0 "delta;" "6#%&deltaG" }{TEXT -1 42 " to be the natural \+ logarithm of the error" }{TEXT 511 1 ":" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "delta := ln(0.1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%&deltaG$!+$4&e-B!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "The natural logarithm of the difference between the \+ full sum and the " }{XPPEDIT 18 0 "N^th;" "6#)%\"NG%#thG" }{TEXT -1 16 " partial sum is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Delta := subs(\{a=10, b=50, c = N+1\}, e) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&DeltaG,&*(\"#]\"\"\",&%\"NGF(F (F(F(-%#lnG6#\"#5F(F(*&F*F(-F,6#\"T,++++++++++++++++++++++++\"F(!\"\" " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "We se t " }{XPPEDIT 18 0 "Delta = delta;" "6#/%&DeltaG%&deltaG" }{TEXT -1 16 " and solve for " }{XPPEDIT 18 0 "N;" "6#%\"NG" }{TEXT 513 1 "." } {TEXT -1 137 " In order to facilitate the numerical root-finding algor ithm, we first run a loop to determine a reasonable interval in which \+ to search. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "for j from 50 to 60 do\nevalf(subs(N=10^j, Delta - de lta),100), `N`=j;\nend do;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"SWJW2 b+Q5#=Uts&**3?%G-FuR=V;\"!#Z/%\"NG\"#]" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"R9Vu]0!Q5#=Uts&**3?%G-FuR=V2\"!#Y/%\"NG\"#^" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"PJW2b+Q5#=Uts&**3?%G-FuR=V'*y\"ylsU+\"*z:xHd-;oD))!#W/ %\"NG\"#`" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!Pc#\\%*>'*y\"ylsU+\"*z: xHd-;oD))*!#V/%\"NG\"#a" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!PE\\%*>'* y\"ylsU+\"*z:xHd-;oD))**!#U/%\"NG\"#b" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!P$\\%*>'*y\"ylsU+\"*z:xHd-;oD))***!#T/%\"NG\"#c" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!P\\%*>'*y\"ylsU+\"*z:xHd-;oD))****!#S/%\"NG\"#d" } }{PARA 11 "" 1 "" {XPPMATH 20 "6$$!PX*>'*y\"ylsU+\"*z:xHd-;oD))*****!# R/%\"NG\"#e" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!P%*>'*y\"ylsU+\"*z:xH d-;oD))******!#Q/%\"NG\"#f" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$!P*>'*y \"ylsU+\"*z:xHd-;oD))*******!#P/%\"NG\"#g" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "We should search in the interval " }{XPPEDIT 18 0 "[10^52, 10^53];" "6#7$*$)\"#5\"#_\"\"\"*$)F&\"#`F(" }{TEXT -1 33 " . Let us narrow it d own futher\n\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "for j from 1 to 10 do\nevalf(subs(N=j*10^52, Delta - delta),100), `N`=j*10^52;\n end do;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$\"PJW2b+Q5#=Uts&**3?%G-FuR =V'*y\"ylsU+\"*z:xHd-;oD)!#X/%\"NG\"V+++++++++ +++++++++++++++++#" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$!Qpb#\\%*>'*y\" ylsU+\"*z:xHd-;oD=!#X/%\"NG\"V++++++++++++++++++++++++++$" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$!Qpb#\\%*>'*y\"ylsU+\"*z:xHd-;oDG!#X/%\"NG\"V+ +++++++++++++++++++++++++%" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$!Qpb#\\ %*>'*y\"ylsU+\"*z:xHd-;oDQ!#X/%\"NG\"V++++++++++++++++++++++++++&" }} {PARA 12 "" 1 "" {XPPMATH 20 "6$$!Qpb#\\%*>'*y\"ylsU+\"*z:xHd-;oD[!#X/ %\"NG\"V++++++++++++++++++++++++++'" }}{PARA 12 "" 1 "" {XPPMATH 20 "6 $$!Qpb#\\%*>'*y\"ylsU+\"*z:xHd-;oDe!#X/%\"NG\"V+++++++++++++++++++++++ +++(" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$!Qpb#\\%*>'*y\"ylsU+\"*z:xHd- ;oDo!#X/%\"NG\"V++++++++++++++++++++++++++)" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$!PdD\\%*>'*y\"ylsU+\"*z:xHd-;oDy!#W/%\"NG\"V++++++++++ ++++++++++++++++*" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$$!PdD\\%*>'*y\"yl sU+\"*z:xHd-;oD))!#W/%\"NG\"W++++++++++++++++++++++++++5" }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "We should search i n the interval " }{XPPEDIT 18 0 "[10^52, 2*10^52];" "6#7$*$\"#5\"#_*& \"\"#\"\"\"*$)F%F&F)F)" }{TEXT -1 32 " . Let us narrow it down futher \n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "Digits := 70;\nfsolve( Delta = delta, N, 10^52 .. 2*1 0^52);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'DigitsG\"#q" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#$\"ao'y$[]Rm;!)['Q9Vu]0!Q5#=Uts&**3?%G-FuR=V<\"! #<" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 " T his figure represents the number of terms needed for an error no great er than 0" }{TEXT 514 1 "." }{TEXT -1 2 "1 " }{TEXT 515 1 "." }}{PARA 0 "" 0 "" {TEXT -1 155 " \+ \+ " }{TEXT 517 1 " " }{XPPEDIT 256 0 "Omega;" "6#%&OmegaG" }{TEXT 518 2 " " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 21 "Code f or The Figures " }}{PARA 3 "" 0 "" {TEXT 367 20 " Code for Figure 1: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 405 "for j from 1 to 10 do\nR[j] := plottools[rectangle]([j,1/j],[ j+1,0], color=COLOR(RGB,.96,.97,.999)):\nr[j] := plottools[rectangle]( [j,1/(j+1)],[j+1,0]):\nend do:\nhyperbola := plot(1/x,x=1..11, thickne ss=2, color = NAVY):\ncurveLabel := plots[textplot]([5.2,0.45,`y = 1/x `],align=\{BELOW,LEFT\}, color=NAVY, font = [TIMES, BOLD, 14]):\nplots [display](hyperbola, curveLabel , seq(R[j],j=1..10), view=[0..11,0..1] );" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT 368 18 "Code for Figure 2:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 376 "N := 25: h := 3/16:\nL[1] := 0: B[1] := 0: R[1] := 1: T[1] := h:\nfor n from 1 to N do\ndomino[n] := plottools[rectangle]([L[n],B[n]],[R[n],T[n]]) :\nif n < N then\nL[n+1] := L[n]+1/(2*(N-n)): \nB[n+1] := B[n]+h: \nR [n+1] := R[n]+1/(2*(N-n)): \nT[n+1] := T[n]+h:\nfi:\nod: \nplots[dis play](seq(domino[n],n=1..N),scaling=constrained,tickmarks=[3,0],color= COLOR(RGB,.96,.96,.96)):\n\n" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 32 "Copyright and Author Information " }}{PARA 0 "" 0 "" {TEXT -1 100 "\nWorksheet Title: BlankKrantz-09_1R 8.mws A Maple Release 8 worksheet.\n\nAuthor: Brian E. Blank " } }{PARA 0 "" 0 "" {TEXT -1 30 "Date Created: 6 November 2007" }}{PARA 0 "" 0 "" {TEXT -1 482 "Date Last Revised: 11 November 2007\n\nThis do cument may not be distributed by any medium,\nincluding print, disk, a nd electronic transfer, without\nprior written permission of the autho r.\n\nFor more information, please contact the author:\n \n De partment of Mathematics, \n Washington University in St. Louis\n \+ St. Louis, MO 63130\n \n Telephone: (314) 935-6763\n \+ e-mail: brian@math.wustl.edu\n\nCopyright: \251 2007 Bria n E. Blank, All Rights Reserved.\n" }}{PARA 3 "" 0 "" {TEXT -1 0 "" } }}}{MARK "10" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }