{VERSION 2 3 "APPLE_PPC_MAC" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "1. Evaluate the following , if they exist. If they are divergent, state clearly why. " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "sum(2^k * 5^(-k-1), k = 2..i nfinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"%\"#v" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "limit((x^2+3*x-4)/(2*x^2+x-3), x=1) ;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# \"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "limit((x^2+3*x-4) /(2*x^2+x-3), x=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\" \"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "(By l'Hopital's rule)" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "limit((x^2+3*x-4)/(2*x^2+x -3), x=-4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "L'Hopital's rule is NOT valid here, because th e denominator does not tend to 0. It is 0 because the numerator ten ds to 0." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "2. Determine whether the following converge. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "int(((3+x)/(3*x^2))^(1/2), x=2..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We ll, that's divergent. But why? Because the integrand is greater than (1/3x)^(1/2), which cannot be integrated:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "int((1/3*x)^(1/2),x=0..R);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"RG#\"\"$\"\"#F'#\"\"\"F(#F(\"\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "which gets big as R -> °." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "sum(1/(sin(k) + sqrt(k)),k=2..infin ity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$sumG6$*$,&-%$sinG6#%\"kG\" \"\"*$F+#F,\"\"#F,!\"\"/F+;F/%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "Maple is telling us she doesn't know the exact answer. B ut as k -> °, we know that" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "limit(sqrt(k)/(sin(k) + sqrt (k)),k=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "So this series converges if and on ly if " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "sum(1/(sqrt(k)),k =2..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$sumG6$*$%\"kG#!\" \"\"\"#/F';F*%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "Here , even if Maple doesn't say so, we know that the series diverges. It \+ is our friend the p-series with p=1/2." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 245 "Problems 3 and 4 are concern ed with estimating the integral of 1/(4+x^2). No credit will be give n for an accurate estimate of this integral, only for the approximatio n,.\n\n\n3. (10 points) In this problem we use Taylor's polynomial a nd series." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "taylor(1/(4+x ^2),x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++%\"xG#\"\"\"\"\"%\"\"!# !\"\"\"#;\"\"##F&\"#k\"\"%-%\"OG6#F&\"\"'" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 40 "int(1/4-(1/16)*x^2+(1/64)*x^4,x=0..1/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"%jP\"&?2$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+'*[$ \\A\"!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "For what positive val ues of x is the Taylor series convergent for int(1/(4+t^2),t=0..x);" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "Answer: \+ SAME AS FOR int(1/(4+x^2), except possibly for the ends of the interv al. Since this is " }}{PARA 0 "" 0 "" {TEXT -1 93 "the sum of a geome tric series in powers ((x/2)^2)^n, we have convergence for -2 < x < 2 . We" }}{PARA 0 "" 0 "" {TEXT -1 98 "actually do a little better, for it converges when x=2,-2 by the alternating series test. Answer:" }} {PARA 0 "" 0 "" {TEXT -1 11 "-2 ² x ² 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "f := x -> 1/(4+x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG:6#%\"xG6\"6$%)operatorG%&arrowGF(*$,&\"\"%\"\"\"*$9$\"\"#F/! \"\"F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "evalf(student[t rapezoid](f(x),x=0..2,4));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+*eqR \"R!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "evalf(student[sim pson](f(x),x=0..2,2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+mmm;R!#5 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Just for fun:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "int(f(x),x=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%#PiG#\"\"\"\"\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+=3*p #R!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "29 0 0 " 0 }{VIEWOPTS 1 1 0 1 1 1803 }