{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 "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 
0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 }
{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 "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 "Normal" -1 256 1 {CSTYLE "" -1 -1 "Tim
es" 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 0 "" 0 "" {TEXT -1 83 "We want to estimate sin(1)
 by using the Maclaurin Series of sin(x) within 10^(-3). " }}{PARA 0 "
" 0 "" {TEXT -1 91 "Recall the Taylor's Theorem, we need the remainer \+
R_n(1) less than or equal 10^(-3), where " }}{PARA 256 "" 0 "" {TEXT 
-1 33 "R_n(x)=(sin^(n+1)(c)/(n+1)!)*x^n," }}{PARA 0 "" 0 "" {TEXT -1 
296 "where sin^(n+1)(x) refers to the (n+1)-th derivative of sin(x), a
nd c is in the interval of (0,x). since since |(sin^(n+1)(c)| is less \+
than or equal to 1, we want to find a postive integer n so that 1/(n+1
)! is less than or equal to 10^(-3), which is the same as solving n so
 that (n+1)! >= 1000." }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Digits:=20;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%'DigitsG\"#?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "g:=n->(n+1)!-1000;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gGf*6#%\"nG6\"6$%)operatorG%&arrow
GF(,&-%*factorialG6#,&9$\"\"\"F2F2F2\"%+5!\"\"F(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 36 "evalf(g(5));evalf(g(6));evalf(g(7));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$!$!G\"\"!" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"%SS\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"&?$R
\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 252 "But n=6 for g does not \+
work because the degree 6 Maclaurin polynomial for sin(x) does not exi
st. So we need to take n=7. This means that if use degree 5 Maclaurin \+
polynomial for sin(x), we will be able to approximate sin(1) within th
e error of 10^(-3)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "tayl
or( sin(x), x=0, 7);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++%\"xG\"\"\"F
%#!\"\"\"\"'\"\"$#F%\"$?\"\"\"&-%\"OG6#F%\"\"(" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 23 "p5:=convert(%,polynom);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#p5G,(%\"xG\"\"\"*&#F'\"\"'F'*$)F&\"\"$F'F'!\"\"*&#F'
\"$?\"F'*$)F&\"\"&F'F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 
"a:=evalf(subs(x=1,p5));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\"5
nmmmmmmm;%)!#?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "b:=evalf(
sin(1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG$\"5l1l*y![)4ZT)!#?
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "a-b;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$\"2-g,xe=o&>!#?" }}}}{MARK "11" 0 }{VIEWOPTS 1 1 0 
3 2 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }