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2 2 2 0 0 0 1 }{PSTYLE "_pstyle15 " -1 214 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 }0 0 0 -1 -1 -1 1 0 1 0 2 2 -1 1 }{CSTYLE "_cstyle12" -1 213 "Times" 0 1 0 0 0 0 0 0 0 2 2 2 0 0 0 1 }} {SECT 0 {EXCHG {PARA 204 "" 0 "" {TEXT 205 12 "Wei-Chi Yang" }}{PARA 204 "" 0 "" {TEXT 205 23 "Professor of Math/Stats" }}{PARA 204 "" 0 "" {TEXT 205 18 "Radford University" }}{PARA 204 "" 0 "" {TEXT 205 22 "w ww.radford.edu/~wyang" }}{PARA 205 "" 0 "" {TEXT 205 10 "Objective:" } {TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 206 134 "Proving convergence \+ of a sequence; when epsilon is given, how to find appropriate positive integer N. Consider the following sequence " }{TEXT 207 11 "a(n)=f(n) . " }{TEXT 206 13 "Assume that " }{TEXT 207 29 "f(n) convergest to a \+ limit L." }{TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 206 67 "Method 1. \+ With CAS, we can find the least positve integer n so that" }{TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 206 44 "abs(f(n)-L) < epsilon when epsl ion is given." }{TEXT 206 0 "" }}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 30 "f:=x->(x^2+3*x+1 )/(2*x^2+x+4);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 " 6#>I\"fG6\"f*6#I\"xGF%F%6$I)operatorGF%I&arrowGF%F%*&,(*$9$\"\"#\"\"\" F/\"\"$F1F1F1,(F.F0F/F1\"\"%F1!\"\"F%F%F%" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 39 "g:=proc(x,eps) abs(f(x)-(1/2))- eps end;" }{MPLTEXT 1 208 0 "" }}{PARA 207 "> " 0 "" {MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#>I\"gG6\"f*6$I\"xGF%I$epsGF%F%F% F%,&-I$absGI*protectedGF-6#,&-I\"fGF%6#9$\"\"\"#!\"\"\"\"#F4F49%F6F%F% F%" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 9 "g(x ,eps);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#,&-I$a bsGI*protectedGF&6#,&*&,(*$I\"xG6\"\"\"#\"\"\"F,\"\"$F/F/F/,(F+F.F,F/ \"\"%F/!\"\"F/#F3F.F/F/I$epsGF-F3" }{TEXT 20 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 14 "evalf(f(100));" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+\\f&Q7&!#5" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 18 "evalf(f(125)-0.5);" }{MPLTEXT 1 208 0 "" }} {PARA 208 "" 1 "" {XPPMATH 20 "6#$\")8-F**!#5" }{TEXT 20 0 "" }}} {EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 18 "evalf(f(124)-0.5);" } {MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"*nZ1+\"!#5" }{TEXT 20 0 "" }}}{EXCHG {PARA 205 "" 0 "" {TEXT 206 162 "Remark: If w e choose eps = 0.01, then the smallest integer that satisfies g(x) <0 \+ is 125. Let's confirm this by solving the inequalities and plotting f unction g." }{TEXT 206 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 32 "fsolve(g(x,10^(-2)),x,100..200);" }{MPLTEXT 1 208 0 "" }} {PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+'=43C\"!\"(" }{TEXT 20 0 "" }}} {EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 27 "plot(g(x,0.01),x=100..20 0);" }{MPLTEXT 1 208 0 "" }}{PARA 210 "" 1 "" {GLPLOT2D 400 400 400 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$\"$+\"\"\"!$\"3]ui[1\\f&Q#!#?7$$\"3nmm ;arz@5!#:$\"3e8CMn*4Q7#F-7$$\"3QL$e9ui2/\"F1$\"3<#zTV0r[!>F-7$$\"3omm \"z_\"4i5F1$\"3-BqsO\"3zm\"F-7$$\"3ummT&phN3\"F1$\"3&[Lguw7(Q9F-7$$\"3 RL$e*=)H\\5\"F1$\"3<*>Quut$>7F-7$$\"3tm;z/3uC6F1$\"3Q#[s[d8F-7$$\"3aLLe9Ege9F1$!3UjC:*[\\O[\"F-7$$\"3]L$eR\"3Gy 9F1$!3C$HFBs/jf\"F-7$$\"3qmmT5k]*\\\"F1$!3`PgWnZ_9Y2a\"F1$!3%3,ALJi\\$>F-7$$\"3cmm\"zX u9c\"F1$!3'*GRi\"z*QT?F-7$$\"35+++&y))Ge\"F1$!3!3rBk))H%[@F-7$$\"3.++D E&QQg\"F1$!3gQUB2zR]AF-7$$\"34+]7y%3Ti\"F1$!3cP^(zsnlM#F-7$$\"3'**** \\P![hY;F1$!3UoyFh(*e]CF-7$$\"3MLLLQx$om\"F1$!3o?K![hv;a#F-7$$\"33++]P +V)o\"F1$!3'[Y)RBS`OEF-7$$\"3im;zpe*zq\"F1$!3j$y*z+*H/s#F-7$$\"3E++]# \\'QH_`Qw'GwHF-7$$\"3;LL$3s?6z\"F1$!3;$fdyjIl0$F-7$$\"3#***\\7`W l7=F1$!3zTqXYtfQJF-7$$\"3zmmm'*RRL=F1$!3b0$4dF1$!3_71ow;n5NF-7$$\"3vmmmw(Gp$>F1$!334DwI_tw NF-7$$\"3G+]7oK0e>F1$!3)G)G\"o72dk$F-7$$\"39+](=5s#y>F1$!3N#4;h!*[.r$F -7$$\"$+#F*$!3]b%>)o^OyPF--%&COLORG6&%$RGBG$\"#5!\"\"$F*Fa[lFb[l-%+AXE SLABELSG6$Q\"x6\"Q!Fg[l-%%VIEWG6$;$F`[l\"\"\"$\"#?F^\\l;$!2/dD.PW;!R!# >$\"2'*Q#*z5u)3DFd\\l" 1 2 2 0 10 1 2 6 1 4 2 1.0 45.0 45.0 1 0 "Curve 1" }}{TEXT 210 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 13 " g(x,10^(-4));" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6 #,&-I$absGI*protectedGF&6#,&*&,(*$I\"xG6\"\"\"#\"\"\"F,\"\"$F/F/F/,(F+ F.F,F/\"\"%F/!\"\"F/#F3F.F/F/#F3\"&++\"F/" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 32 "fsolve(g(x,10^(-3)),x,200..500) ;" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#-I'fsolveG6 $I*protectedGF&I(_syslibG6\"6%,&-I$absGF&6#,&*&,(*$I\"xGF(\"\"#\"\"\"F 2\"\"$F4F4F4,(F1F3F2F4\"\"%F4!\"\"F4#F8F3F4F4#F8\"%+5F4F2;\"$+#\"$+&" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 19 "evalf( f(1000)-1/2);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6 #$\"+mI()[7!#7" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 19 "evalf(f(1500)-1/2);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 " " {XPPMATH 20 "6#$\"+*fF$G$)!#8" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 19 "evalf(f(1300)-1/2);" }{MPLTEXT 1 208 0 "" } }{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+.!>(3'*!#8" }{TEXT 20 0 "" }}} {EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 19 "evalf(f(1100)-1/2);" } {MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+JCVN6!#7" } {TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 19 "evalf(f (1200)-1/2);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6# $\"+(H%)3/\"!#7" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 19 "evalf(f(1250)-1/2);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+$4!z#***!#8" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 34 "fsolve(g(x,10^(-3)),x,1200..130 0);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+5\")4 \\7!\"'" }{TEXT 20 0 "" }}}{EXCHG {PARA 205 "" 0 "" {TEXT 206 100 "Rem ark: Therefore, if we pick eps = 10^(-3), then the smallest integer th at is required is n = 1250." }{TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 205 9 "Method 2." }{TEXT 206 48 " If abs(f(n)-L)< epsilon, this is equivalent to " }{TEXT 205 41 "b oth f(n)L-epsilon. " }{TEXT 206 44 "We intend to \+ express 'n' in terms of epsilon" }{TEXT 206 0 "" }}{PARA 205 "" 0 "" {TEXT 206 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 28 "solve( (f(x)-1/2)=epsilon,x);" }{MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6$,$*&,(I(epsilonG6\"\"\"#!\"&\"\"\"*$,(*$F&F(!$C\"F&!#_ \"#DF*#F*F(!\"\"F*F&F2#F2\"\"),$*&,(F&F(F)F*F+F*F*F&F2F3" }{TEXT 211 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 95 "expression1:=pro c(epsilon) -1/8*(2*epsilon-5-(-124*epsilon^2-52*epsilon+25)^(1/2))/eps ilon end;" }{MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6#>I, expression1G6\"f*6#I(epsilonGF%F%F%F%,$*&,(9$\"\"#!\"&\"\"\"*$,(*$F,F- !$C\"F,!#_\"#DF/#F/F-!\"\"F/F,F7#F7\"\")F%F%F%" }{TEXT 211 0 "" }}} {EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 21 "expression1(epsilon);" } {MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6#,$*&,(I(epsilon G6\"\"\"#!\"&\"\"\"*$,(*$F&F(!$C\"F&!#_\"#DF*#F*F(!\"\"F*F&F2#F2\"\")" }{TEXT 211 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 95 "expr ession2:=proc(epslion) -1/8*(2*epsilon-5+(-124*epsilon^2-52*epsilon+25 )^(1/2))/epsilon end;" }{MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6#>I,expression2G6\"f*6#I(epslionGF%F%F%F%,$*&,(I(epsilon GF%\"\"#!\"&\"\"\"*$,(*$F,F-!$C\"F,!#_\"#DF/#F/F-F/F/F,!\"\"#F7\"\")F% F%F%" }{TEXT 211 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 21 "expression2(epslion);" }{MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6#,$*&,(I(epsilonG6\"\"\"#!\"&\"\"\"*$,(*$F&F(!$C\"F&!#_ \"#DF*#F*F(F*F*F&!\"\"#F2\"\")" }{TEXT 211 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 28 "evalf(expression1(10^(-2)));" }{MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6#$\"+'=43C\"!\"(" }{TEXT 211 0 "" }}}{EXCHG {PARA 209 "> " 0 "" {MPLTEXT 1 209 28 "evalf(expres sion2(10^(-2)));" }{MPLTEXT 1 209 0 "" }}{PARA 211 "" 1 "" {XPPMATH 20 "6#,$*&,(I(epsilonG6\"$\"\"#\"\"!$!\"&F*\"\"\"*$,(*$F&F)$!$C\"F*F&$ !#_F*$\"#DF*F-#F-F)F-F-F&!\"\"$!++++]7!#5" }{TEXT 211 0 "" }}}{EXCHG {PARA 212 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 203 "Notice that we expressed 'n' in terms of epsilon in eit her expression1 or expression 2. In this case, expression1(10^(-2)) gi ves us what we expected-the least positive to be 125 when epsilon is 1 0^(-2). " }{TEXT 200 0 "" }}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 157 "E xercise 2. We want to prove the following sequence coverges to 0 by us ing epsilon and N idea. Find the least positive integer n so that abs( a(n)-0)<10^(-4)." }{TEXT 200 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 17 "a:=n->(5^n)/(n!);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#>I\"aG6\"f*6#I\"nGF%F%6$I)operatorGF%I&arro wGF%F%*&)\"\"&9$\"\"\"-I*factorialGI*protectedGF36#F/!\"\"F%F%F%" } {TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 28 "b:=proc (x,eps) a(x)-eps end;" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#>I\"bG6\"f*6$I\"xGF%I$epsGF%F%F%F%,&-I\"aGF%6#9$\"\"\"9 %!\"\"F%F%F%" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 21 "evalf(b(10,10^(-4)));" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 " " {XPPMATH 20 "6#$\"+bW/\"p#!\"*" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 " > " 0 "" {MPLTEXT 1 208 21 "evalf(b(20,10^(-4)));" }{MPLTEXT 1 208 0 " " }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$!+]c4!3'!#9" }{TEXT 20 0 "" }}} {EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 30 "fsolve(b(x,10^(-4)),x,10 ..20);" }{MPLTEXT 1 208 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#$\"+yL %G$>!\")" }{TEXT 20 0 "" }}}{EXCHG {PARA 207 "> " 0 "" {MPLTEXT 1 208 30 "fsolve(b(x,10^(-7)),x,20..40);" }{MPLTEXT 1 208 0 "" }}{PARA 208 " " 1 "" {XPPMATH 20 "6#$\"+kXZ(R#!\")" }{TEXT 20 0 "" }}}{EXCHG {PARA 212 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 10 "Method 2. " }{TEXT 200 0 "" }}}{EXCHG {PARA 212 "> " 0 "" {MPLTEXT 1 0 22 "solve(a(n)=epsilon,n);" }{MPLTEXT 1 0 0 "" }}{PARA 208 "" 1 "" {XPPMATH 20 "6#-I'RootOfG6$I*protectedGF&I(_syslibG6\"6#,& )\"\"&I#_ZGF%!\"\"*&I(epsilonGF(\"\"\"-I*factorialGF&6#F-F1F1" }{TEXT 20 0 "" }}}{EXCHG {PARA 212 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 133 "Note. In this case, Maple solved the \+ equation in real numbers. If you know how to solve the equation in int egers, please let me know." }}}{EXCHG {PARA 206 "" 0 "" {TEXT 200 0 "" }}}{EXCHG {PARA 205 "" 0 "" {TEXT 206 149 "Exercise 3. Use the techni que above to find the smallest integer so that n!/n^n is less than a g iven epsilon. Repeat Exercise 2 when epsilon=10^(-5)." }{TEXT 206 0 "" }}}{PARA 213 "" 0 "" {TEXT 212 0 "" }}{PARA 214 "" 0 "" {TEXT 213 0 " " }}{PARA 214 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 15 10 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }