{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "Hyperlink" -1 17 "" 0 1 0 128 128 1 2 0 1 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Bullet Item " -1 15 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 } 1 1 0 0 3 3 1 0 1 0 2 2 15 2 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT 256 44 "lab14.mws --- Numerical Evaluation of Series" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "restart;\nwith( plots ):" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 12 "Lab Overview" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 646 "Th e typical textbook problem with series asks you to determine if a spec ific series converges or diverges. Rarely are you asked to find (even approximately) the sum of a convergent series. This lab is designed \+ to fill this gap. You will approximate the sum of convergent series a nd see some of the different ways a series can fail to converge. For \+ the Lab Questions you will be asked to analyze 4 series. For extra cr edit, you are asked to explain how rearranging the terms in the altern ating harmonic series can change the sum of the series. If you pay at tention to the presentation in lab on Tuesday, you should see how thi s can be done." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 215 "Continuing the practice from recent weeks, this lab assi gnment includes a separate and explicit integration problem. In fact, \+ this week there are two integrals for a total of 4 points. You are enc ouraged to use the " }{TEXT 258 11 "Integration" }{TEXT -1 9 " maplet \+ [" }{URLLINK 17 "Maplet Viewer" 4 "http://www.math.sc.edu/~meade/math1 42-S03/maplets/Calculus1Maplets/Integration.maplet" "" }{TEXT -1 2 "][ " }{URLLINK 17 "MapleNet" 4 "http://maple.math.sc.edu/maplenet/meade/C alculus1Maplets/Integration.html" "" }{TEXT -1 28 "] to help with this problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 257 77 "Deadline for submitting a lab solution is midn ight, Thursday, April 24, 2003." }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Exam ple 1: " }{XPPEDIT 18 0 "Sum( k/(k+1), k=1..infinity )" "6#-%$SumG6$*& %\"kG\"\"\",&F'F(F(F(!\"\"/F';F(%)infinityG" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "The terms in this series are " }{XPPEDIT 18 0 "a[k] = k/(k+1)" "6#/&%\"aG6#%\"kG*&F'\"\"\",&F'F)F)F)!\"\"" } {TEXT -1 48 ". If this is entered as a Maple function, using" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "a := k -> k/(k+1);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "then the " }{XPPEDIT 18 0 "n" "6#% \"nG" }{TEXT -1 35 "th partial sum can be defined to be" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "s := n -> add( a(k), k=1..n );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "For example, the first 10 terms in the series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "terms := [seq([k,a(k)], k=1..10)]:\nconvert( terms, matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "and the corresponding partial sums for th is series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "partial_su ms := [seq([k,s(k)], k=1..10)]:\nconvert( partial_sums, matrix );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "To see these values as floating po int numbers, use " }{HYPERLNK 17 "evalf" 2 "evalf" "" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "convert( evalf( terms ), \+ matrix )," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "convert( evalf( partia l_sums ), matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "The relati onship between the sequences of terms and partial sums should be viewe d graphically:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "plot( [ \+ terms, partial_sums ],\n style=point, color=[green,red],\n l egend=[\"a[k] - term\", \"s[k] - partial sum\"] );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "Notice how the terms (in green) level off around 1 and the partial sums increase (by about 1). Because the terms do n ot converge to zero, this series diverges by the " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 13 "th Term Test." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "L := Li mit( a(k), k=infinity ):\nL = value( L );" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 48 "S := Sum( a(k), k=1..infinity ):\nS = value( S );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Example 2: " }{XPPEDIT 18 0 "Sum(1/k,k = 1 .. infini ty);" "6#-%$SumG6$*&\"\"\"F'%\"kG!\"\"/F(;F'%)infinityG" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "The terms in the harmonic ser ies are " }{XPPEDIT 18 0 "a[k] = 1/k;" "6#/&%\"aG6#%\"kG*&\"\"\"F)F'! \"\"" }{TEXT -1 48 ". If this is entered as a Maple function, using" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "a := k -> 1/k;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "then the " }{XPPEDIT 18 0 "n" "6#% \"nG" }{TEXT -1 35 "th partial sum can be defined to be" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "s := n -> add( a(k), k=1..n );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "For example, the first 10 terms in the series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "terms := [seq([k,a(k)], k=1..10)]:\nconvert( terms, matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "and the corresponding partial sums for th is series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "partial_su ms := [seq([k,s(k)], k=1..10)]:\nconvert( partial_sums, matrix );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "To see these values as floating po int numbers, use " }{HYPERLNK 17 "evalf" 2 "evalf" "" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "convert( evalf( terms ), \+ matrix )," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "convert( evalf( partia l_sums ), matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "The relati onship between the sequences of terms and partial sums should be viewe d graphically:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "plot( [ \+ terms, partial_sums ],\n style=point, color=[green,red],\n l egend=[\"a[k] - term\", \"s[k] - partial sum\"] );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "It is clear that the terms of this series converg e to zero:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "L := Limit( a (k), k=infinity ):\nL = value( L );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "While the partials sums are increasing, it appears as though th ey " }{TEXT 261 5 "might" }{TEXT -1 185 " be leveling off. If so, the n the series converges; if not, then the series diverges. Maybe we ca n answer this quesiton by looking at additional terms in the sequence \+ of partial sums:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "partia l_sum2 := [ seq([k,s(k)], k = [ 10*($1..10), 200*($1..4), 1000*($1..10 )]) ]:\nconvert( evalf(partial_sum2), matrix );" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 41 "plot( evalf(partial_sum2), style=point );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 238 "The partial sums with up to 10,00 0 terms do not provide an answer to the question. The partial sums ar e still increasing but VERY slowly. Until we determine what happens w ith these partial sums we cannot decide if this series converges." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 193 "Of cours e, you have previously learned that the harmonic series. The typical \+ explanation is that it is always possible to group the terms so that e ach group sums to at least 1/2. For example," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "for i from 1 to 10 do\n lo := 1+2^(i-1);\n hi := 2^i;\n print( Sum( a(k), k=lo..hi ) = evalf(add( a(k), k=lo..hi ) ) )\nend do:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "In general, for a ny integer " }{XPPEDIT 18 0 "i" "6#%\"iG" }{TEXT -1 17 ", the sum of t he " }{XPPEDIT 18 0 "2^(i-1)" "6#)\"\"#,&%\"iG\"\"\"F'!\"\"" }{TEXT -1 22 " terms beginning with " }{XPPEDIT 18 0 "a[2^(i-1)+1] = 1/(2^(i- 1)+1)" "6#/&%\"aG6#,&)\"\"#,&%\"iG\"\"\"F,!\"\"F,F,F,*&F,F,,&)F),&F+F, F,F-F,F,F,F-" }{TEXT -1 18 " must be at least " }{XPPEDIT 18 0 "1/2" " 6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 2 ": " }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "Sum( 1/k, k=2^(i-1)+1..2^i )" "6#-%$SumG6$*&\" \"\"F'%\"kG!\"\"/F(;,&)\"\"#,&%\"iGF'F'F)F'F'F')F.F0" }{TEXT -1 3 " > \+ " }{XPPEDIT 18 0 "1/2" "6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 2 " " }} {PARA 0 "" 0 "" {TEXT -1 69 "Because of this it is always possible to \+ increase the partial sum by " }{XPPEDIT 18 0 "1/2" "6#*&\"\"\"F$\"\"#! \"\"" }{TEXT -1 63 ". This would not be possible if the harmonic seri es converged." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "S := Sum( \+ a(k), k=1..infinity ):\nS = value( S );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Example 3: \+ " }{XPPEDIT 18 0 "Sum(3*n/(n^3+1),n = 1 .. infinity);" "6#-%$SumG6$*( \"\"$\"\"\"%\"nGF(,&*$F)F'F(F(F(!\"\"/F);F(%)infinityG" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT 259 4 "Note" }{TEXT -1 26 ": This is #5d fro m Exam 3." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "The terms in this ser ies are " }{XPPEDIT 18 0 "a[n] = 3*n/(n^3+1);" "6#/&%\"aG6#%\"nG*(\"\" $\"\"\"F'F*,&*$F'F)F*F*F*!\"\"" }{TEXT -1 48 ". If this is entered as a Maple function, using" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "a := n -> 3*n/(n^3+1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "then th e " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 35 "th partial sum can be d efined to be" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "s := n -> a dd( a(k), k=1..n );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "For exampl e, the first 10 terms in the series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "terms := [seq([k,a(k)], k=1..10)]:\nconvert( terms, m atrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "and the corresponding partial sums for this series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "partial_sums := [seq([k,s(k)], k=1..10)]:\nconvert( p artial_sums, matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "To see \+ these values as floating point numbers, use " }{HYPERLNK 17 "evalf" 2 "evalf" "" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "convert( evalf( terms ), matrix ),\nconvert( evalf( partial_sums ) , matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "The relationship b etween the sequences of terms and partial sums should be viewed graphi cally:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "plot( [ terms, p artial_sums ],\n style=point, color=[green,red],\n legend=[ \"a[k] - term\", \"s[k] - partial sum\"] );" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 46 "Notice how the terms (in green) converge to 0." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "L := Limit( a(n), n=infinity ):\nL = value( L );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 392 "The part ial sums increase and appear to level off. Based on our experience wi th the harmonic series, we know to be somewhat skeptical of \"apparent leveling off\". Instead of looking at additional partial sums - whic h will never provide a conclusive decision concerning the convergence \+ or divergence of a series - let's try to use some of the convergence t ests for series with positive terms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 13 "Integral Tes t" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 193 "The plot of the first 10 ter ms of the series suggests that these terms are monotonically decreasin g to zero. The fact that the terms form a decreasing sequence can be \+ confirmed by noting that " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "Diff( a(x),x ) = simplify(diff( a(x), x ));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "which is negative for all " }{XPPEDIT 18 0 "x" "6#% \"xG" }{TEXT -1 46 " > 1. Thus, the Integral Test can be applied." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "The improper integral" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "q1 := Int( a(x), x=1..infinity ):\nq1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "is difficult to evaluate by hand. (Note \+ that " }{XPPEDIT 18 0 "x^3+1" "6#,&*$%\"xG\"\"$\"\"\"F'F'" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(x+1)*(x^2-x+1)" "6#*&,&%\"xG\"\"\"F&F&F&,(*$F %\"\"#F&F%!\"\"F&F&F&" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(x+1)*((x-1/2 )^2+3/4)" "6#*&,&%\"xG\"\"\"F&F&F&,&*$,&F%F&*&F&F&\"\"#!\"\"F,F+F&*&\" \"$F&\"\"%F,F&F&" }{TEXT -1 215 " so an antiderivative can be found us ing partial fractions with two terms. The details of this calculation will be omitted -- until you do it for Extra Credit.) Maple can eval uate this integral without difficulty:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "q1 = value( q1 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Because this improper integral is finite, the series converges \+ by the Integral Test." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 21 "Limit Comparison Test" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "Another method to determine the c onvergence of this series is the Limit Comparison Test. Observe that \+ the highest degree term in the numerator is " }{XPPEDIT 18 0 "3*n" "6# *&\"\"$\"\"\"%\"nGF%" }{TEXT -1 51 " and the highest degree term in th e denominator is " }{XPPEDIT 18 0 "n^3" "6#*$%\"nG\"\"$" }{TEXT -1 23 ". This suggests using " }{XPPEDIT 18 0 "b[n] = 3*n/n^3" "6#/&%\"bG6# %\"nG*(\"\"$\"\"\"F'F**$F'F)!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "3 /n^2" "6#*&\"\"$\"\"\"*$%\"nG\"\"#!\"\"" }{TEXT -1 30 " in the Limit C omparison Test." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "b := n - > 3/n^2:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "To confirm that this \+ is a valid choice of a comparison series, note that" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "LCT_Limit := Limit( a(n)/b(n), n=infinity ):\nLCT_Limit = value(LCT_Limit);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "Now, because the series " }{XPPEDIT 18 0 "Sum( b[n], n=1..infin ity )" "6#-%$SumG6$&%\"bG6#%\"nG/F);\"\"\"%)infinityG" }{TEXT -1 17 " \+ is a convergent " }{XPPEDIT 18 0 "p" "6#%\"pG" }{TEXT -1 13 "-series w ith " }{XPPEDIT 18 0 "p=2" "6#/%\"pG\"\"#" }{TEXT -1 32 " > 1, both se ries must converge." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} }{SECT 1 {PARA 4 "" 0 "" {TEXT -1 35 "Approximating the Sum of the Ser ies" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "Now that we know the serie s converges, the question is: What is the sum of this series? Maple w ill tell us that" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "S := Su m( a(n), n=1..infinity ):\nS = value( S );\n`` = evalf( S );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "The exact formula requires some explanation:" }}{PARA 15 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "gamma" "6#%&gammaG" }{TEXT -1 4 " is " }{HYPERLNK 17 "Euler's constant" 2 "gamma" "" }{TEXT -1 21 " \+ and is approximately" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "gam ma = evalf( gamma );" }}}{EXCHG {PARA 15 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "I = sqrt(`-1`)" "6#/%\"IG-%%sqrtG6#%#-1G" }{TEXT -1 7 " , the " }{HYPERLNK 17 "imaginary unit" 2 "I" "" }{TEXT -1 5 ", and" } }{PARA 15 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Psi" "6#%$PsiG" } {TEXT -1 8 " is the " }{HYPERLNK 17 "digamma" 2 "digamma" "" }{TEXT -1 142 " function (a special function defined by a complicated formula ; you don't need - and probably don't care - to know more about this f unction!)." }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 299 "The floating point approximation is not difficult to und erstand. Because the terms are all positive, each parial sum will be \+ less than -- and converging to -- the sum of the series. To conclude, we will determine the number of terms needed to approximate the sum o f this series to 2 decimal places" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 188 "for N from 0 do\n k := 2^N:\n app := evalf( s(k) ) :\n err := evalf( S - app ):\n print( nprintf( \"S[%5d] = %8.5f, Err or = %8.5f\", k, app, err ) );\n if err<0.005 then break end if;\nend do:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 341 "That is, somewhere betwe en 512 and 1024 terms are needed to approximate this integral to 2 dec imal places. To narrow this range, repeat the above loop starting wit h the partial sum with 512 terms. (Repeating this process several tim es should lead you to the conclusion that 600 terms are needed to appr oximate the sum to 2 decimal places.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Example 4: " }{XPPEDIT 18 0 "Sum((-1)^n*(1/2)^(2*n+1)/(2*n+1)!,n = 1 .. infinity );" "6#-%$SumG6$*(),$\"\"\"!\"\"%\"nGF))*&F)F)\"\"#F*,&*&F.F)F+F)F)F)F )F)-%*factorialG6#,&*&F.F)F+F)F)F)F)F*/F+;F)%)infinityG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 260 4 "Note" }{TEXT -1 51 ": This is the co rrected version of #5e from Exam 3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "The terms in this alternating ser ies are " }{XPPEDIT 18 0 "u[n] = (-1)^n*a[n];" "6#/&%\"uG6#%\"nG*&),$ \"\"\"!\"\"F'F+&%\"aG6#F'F+" }{TEXT -1 7 " where " }{XPPEDIT 18 0 "a[n ] = (1/2)^(2*n+1)/(2*n+1)!;" "6#/&%\"aG6#%\"nG*&)*&\"\"\"F+\"\"#!\"\", &*&F,F+F'F+F+F+F+F+-%*factorialG6#,&*&F,F+F'F+F+F+F+F-" }{TEXT -1 48 " . If this is entered as a Maple function, using" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "u := n -> (-1)^n*a(n);\na := n -> (1/2)^(2*n+ 1)/(2*n+1)!;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "then the nth part ial sum can be defined to be" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "s := n -> add( u(k), k=0..n );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "For example, the first 11 terms in the series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "pos_terms := [seq([k,a(k)], k=0..10 )]:\nterms := [seq([k,u(k)], k=0..10)]:\nconvert( terms, matrix );" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "and the corresponding partial sum s for this series are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "pa rtial_sums := [seq([k,s(k)], k=1..10)]:\nconvert( partial_sums, matrix );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "To see these values as flo ating point numbers, use " }{HYPERLNK 17 "evalf" 2 "evalf" "" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "convert( evalf( \+ terms ), matrix ),\nconvert( evalf( partial_sums ), matrix );" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "The relationship between the seque nces of terms and partial sums should be viewed graphically:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "plot( [ terms, partial_sums ],\n style=point, color=[green,red],\n legend=[\"u[k] - ter m\", \"s[k] - partial sum\"] );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 311 "Notice how the terms (in green) immediately decay to a level wher e they become lost in the horizontal axis and the partial sums appear \+ to be essentially constant with a value slightly smaller than -0.02. \+ While these are strong indications that this series converges, the Alt ernating Series Test should be used." }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 23 "Alternating Series Test" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "The Alternating Series Test applies only when the sequence of p ositive terms, " }{XPPEDIT 18 0 "a[n]" "6#&%\"aG6#%\"nG" }{TEXT -1 103 ", must be decreasing and converge to zero. The limit is easy to \+ determine: positive integer powers of " }{XPPEDIT 18 0 "1/2" "6#*&\"\" \"F$\"\"#!\"\"" }{TEXT -1 56 " converge to zero and the factorial grow s without bound." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "L_pos : = Limit( a(n), n=infinity ):\nL_pos = value(L_pos);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "To see that the terms are decreasing for all po sitive integers (" }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 20 " >= 1), \+ observe that" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "a[n+ 1] = (1/2)^(2*(n+1)+1)/(2*(n+1)+1)!" "6#/&%\"aG6#,&%\"nG\"\"\"F)F)*&)* &F)F)\"\"#!\"\",&*&F-F),&F(F)F)F)F)F)F)F)F)-%*factorialG6#,&*&F-F),&F( F)F)F)F)F)F)F)F." }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 4 " = \+ " }{XPPEDIT 18 0 "(1/2)^(2*n+3)/(2*n+3)!;" "6#*&)*&\"\"\"F&\"\"#!\"\", &*&F'F&%\"nGF&F&\"\"$F&F&-%*factorialG6#,&*&F'F&F+F&F&F,F&F(" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 30 " \+ = " }{XPPEDIT 18 0 "(1/2)^(2*n+1)*(1/2)^2/((2*n+1)!*(2*n+2)*(2*n+3)); " "6#*()*&\"\"\"F&\"\"#!\"\",&*&F'F&%\"nGF&F&F&F&F&*$*&F&F&F'F(F'F&*(- %*factorialG6#,&*&F'F&F+F&F&F&F&F&,&*&F'F&F+F&F&F'F&F&,&*&F'F&F+F&F&\" \"$F&F&F(" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 18 " \+ = " }{XPPEDIT 18 0 "(1/2)^2/((2*n+2)*(2*n+3));" "6#*&*&\"\"\"F% \"\"#!\"\"F&*&,&*&F&F%%\"nGF%F%F&F%F%,&*&F&F%F+F%F%\"\"$F%F%F'" } {TEXT -1 1 " " }{XPPEDIT 18 0 "a[n]" "6#&%\"aG6#%\"nG" }{TEXT -1 1 " \+ " }}{PARA 256 "" 0 "" {TEXT -1 4 " < " }{XPPEDIT 18 0 "a[n]" "6#&%\"a G6#%\"nG" }{TEXT -1 19 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "Now, the Alternating Series Tes t tells us that this series converges." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 10 "Ratio Test " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "From the analysis that the seq uence of absolute values, a[n], is decreasing, we compute" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "rho = Limit(abs(u[n+1])/abs(u[n]),n = infinity);" "6#/%$rhoG-%&Limi tG6$*&-%$absG6#&%\"uG6#,&%\"nG\"\"\"F1F1F1-F*6#&F-6#F0!\"\"/F0%)infini tyG" }{TEXT -1 2 " " }}{PARA 256 "" 0 "" {TEXT -1 4 " = " }{XPPEDIT 18 0 "Limit(a[n+1]/a[n],n = infinity);" "6#-%&LimitG6$*&&%\"aG6#,&%\"n G\"\"\"F,F,F,&F(6#F+!\"\"/F+%)infinityG" }{TEXT -1 2 " " }}{PARA 256 "" 0 "" {TEXT -1 26 " = " }{XPPEDIT 18 0 "Limit ( (1/2)^2/(2*n+2)/(2*n+3), n=infinity )" "6#-%&LimitG6$*(*&\"\"\"F(\" \"#!\"\"F),&*&F)F(%\"nGF(F(F)F(F*,&*&F)F(F-F(F(\"\"$F(F*/F-%)infinityG " }{TEXT -1 2 " " }}{PARA 256 "" 0 "" {TEXT -1 27 " = 0. \+ " }}{PARA 0 "" 0 "" {TEXT -1 8 "Because " }{XPPEDIT 18 0 "rho " "6#%$rhoG" }{TEXT -1 71 " = 0 < 1, the Ratio Test tells us that the \+ series converges absolutely." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 35 "Approximating the Sum \+ of the Series" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "As with Example \+ 3, knowing that the series converges we want to know an accurate appro ximation to the sum of the series. Maple will tell us that" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "S := Sum( u(n), n=0..infinit y ):\nS = value( S );\n`` = evalf( S );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "This exact v alue should not be a complete surprise. Note that this series has onl y odd terms and the general form is that of " }{XPPEDIT 18 0 "sin(x)" "6#-%$sinG6#%\"xG" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "x=1/2" "6#/%\" xG*&\"\"\"F&\"\"#!\"\"" }{TEXT -1 3 ". " }}{PARA 257 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 408 "The floating point approxima tion is not difficult to understand. The method introduced in Example 3 can be used here with one modification. Because the series contains both positive and negative terms, the partial sums could be larger or smaller than the exact sum. An absolute value will take care of this . To determine the number of terms needed to approximate the sum of t his series to 8 decimal places" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 231 "Digits := 20:\nfor N from 0 do\n k := 2^N:\n app := evalf( \+ s(k) ):\n err := evalf( abs(S - app) ):\n print( nprintf( \"S[%5d] = %14.11f, Error = %14.11f\", k, app, err ) );\n if err<0.000000005 th en break end if;\nend do:\nDigits := 10:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "Wow! This series converges so fast that the partial sum with 4 terms provides an estimate that is accurate to 8 decimal place s." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 118 "In fact, because this series is an alternating se ries, the Alternating Series Test tells us that the partial sum with \+ " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 79 " terms differs from the e xact sum by no more than the next term in the series: " }{XPPEDIT 18 0 "a[n+1]" "6#&%\"aG6#,&%\"nG\"\"\"F(F(" }{TEXT -1 41 ". This suggest s considering the equation" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eqn := a(n+1)=0.000000005;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "The solution to this equation is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve( eqn, n );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Recalling that " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 40 " must \+ be an integer, round up to obtain " }{XPPEDIT 18 0 "n=4" "6#/%\"nG\"\" %" }{TEXT -1 40 ", in agreement with the previous search." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 204 "Note how the addi tional information provided by the Alternating Series Test provides a \+ much more efficient method for finding the number of terms needed to a pproximate a series with a prescribed accuracy." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Lab Questions" }}{PARA 0 "" 0 "" {TEXT -1 88 "For each of the following s eries, explain why the series converges. For #1, #3, and #4:" }} {PARA 0 "" 0 "" {TEXT -1 47 " i) approximate the series to 3 decimal \+ places" }}{PARA 0 "" 0 "" {TEXT -1 70 " ii) determine the number of te rms needed to obtain the estimate in i)" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 11 "1. [2 pts] " }{XPPEDIT 18 0 "Sum( 1/n^4 , n=1..infinity ) " "6#-%$SumG6$*&\"\"\"F'*$%\"nG\"\"%!\"\"/F);F'%)infinityG" }{TEXT -1 1 " " }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 11 "2. [2 pts] " }{XPPEDIT 18 0 "Sum( 1/(n*(ln(n))^2), n=2..infinity )" "6#-%$SumG6$*&\"\"\"F'*&% \"nGF'*$-%#lnG6#F)\"\"#F'!\"\"/F);F.%)infinityG" }{TEXT -1 1 " " }}} {SECT 1 {PARA 4 "" 0 "" {TEXT -1 11 "3. [2 pts] " }{XPPEDIT 18 0 "Sum( (exp(n)-1)/(5^n+1), n=0..infinity )" "6#-%$SumG6$*&,&-%$expG6#%\"nG\" \"\"F,!\"\"F,,&)\"\"&F+F,F,F,F-/F+;\"\"!%)infinityG" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 11 "4. [2 pts] " }{XPPEDIT 18 0 "Sum( (n^5+3*n^4+2) /(n^8+2*n^5+1), n=0..infinity )" "6#-%$SumG6$*&,(*$%\"nG\"\"&\"\"\"*& \"\"$F+*$F)\"\"%F+F+\"\"#F+F+,(*$F)\"\")F+*&F0F+*$F)F*F+F+F+F+!\"\"/F) ;\"\"!%)infinityG" }{TEXT -1 1 " " }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 23 "5. [Extra Credit: 2pts]" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Explain how the terms in the alternating harmonic series can be rearr anged so the sequence of partial sums converges to " }{XPPEDIT 18 0 "1 /2" "6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 51 ". Give the first 25 terms i n the rearranged series." }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 31 "6. \+ [Integration Problem: 2 pts]" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "Sh ow all steps in the evaluation of the indefinite integral " }{XPPEDIT 18 0 "Int(1/(x*ln(x)^2),x);" "6#-%$IntG6$*&\"\"\"F'*&%\"xGF'*$-%#lnG6# F)\"\"#F'!\"\"F)" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "C-1/ln(x);" "6#,&% \"CG\"\"\"*&F%F%-%#lnG6#%\"xG!\"\"F+" }{TEXT -1 123 " . You may use a ny technology to help solve this problem. Your answer, however, must \+ explain all steps in the evaluation." }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 31 "7. [Integration Problem: 2 pts]" }}{EXCHG {PARA 0 "" 0 " " {TEXT -1 60 "Show all steps in the evaluation of the indefinite inte gral " }{XPPEDIT 18 0 "Int(3*x/(x^3+1),x = 1 .. infinity) = 1/3*3^(1/2 )*Pi+ln(2);" "6#/-%$IntG6$*(\"\"$\"\"\"%\"xGF),&*$F*F(F)F)F)!\"\"/F*;F )%)infinityG,&**F)F)F(F-)F(*&F)F)\"\"#F-F)%#PiGF)F)-%#lnG6#F5F)" } {TEXT -1 122 ". You may use any technology to help solve this problem . Your answer, however, must explain all steps in the evaluation." }} }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }