{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 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 "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 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 16 "Dr. Wei-Chi Yang" }} {PARA 257 "" 0 "" {TEXT -1 21 "www.radford.edu/wyang" }}{PARA 0 "" 0 " " {TEXT -1 104 "Given arbitrary irrationals, a=1/arctan(101), and b=1 /arctan(100). Find a rational between 'a' and 'b'." }}{PARA 0 "" 0 "" {TEXT -1 128 "Remark: We note the function f below. When x gets large, arctan(x) gets close to Pi/2. So a and b should be close to each othe r." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f:=x->1/arctan(x);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot(f(x),x=-10..10,y=-10. .10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=20;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "a:=evalf((1/arctan(101)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "b:=evalf((1/arctan(100))) ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "We want to find the " } {TEXT 256 8 "smallest" }{TEXT -1 81 " possible integer n so that (b-a) >1/n, which exists by the Archemedian Principle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve((b-a)>1/x,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(x>1/(b-a),x);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 9 "n:=24609;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 194 "W e consider the set S=\{x is integer and x/n >a\}. Since S is unbounded above so S is non-empty. Also, since S is bounded below, S has the sm allest member. We want to find such member as follows. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve((x/n)>a,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "m:=15766;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(m/n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf((m/n)-a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "ev alf((m/n)-b);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf(((m+ 1)/n)-a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf(((m+1)/n )-b);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf(((m-1)/n)-a) ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Remark: We note that the in teger m indeed is the smallest possible integer such that m/n >a since (m-1)/n