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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Calculating Pi" }}{PARA 0 "" 0 "
" {TEXT -1 37 "by Peter Stone, Nanaimo, B.C., Canada" }}{PARA 0 "" 0 "
" {TEXT -1 19 "Version:  28.3.2007" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 1 ";" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 12 "Ca
lculating " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 67 " using regular
 polygons with a progressively larger number of sides" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}{PARA 0 "" 0 "" {TEXT -1 34 "The met
hod we use for calculating " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 
118 " involves calculating the perimeter of regular polygons inscribed
 in a circle of radius 1. We can get an estimate for " }{XPPEDIT 18 0 
"2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 90 " by calculating the per
imeter of such a regular polygon when the number of sides is large." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Suppose \+
that we know the length of one side of a regular polygon with " }
{TEXT 269 1 "n" }{TEXT -1 228 " sides.\nThen consider the regular poly
gon with twice as many sides formed by constructing two new sides in p
lace of each previous side by adding an additional vertex mid-way arou
nd the arc between two adjacent existing vertices." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "In the picture, the line \+
segment DE represents one original side and B is the new vertex." }}
{PARA 0 "" 0 "" {TEXT -1 88 "O is the centre of the polygon and A is a
 point diametrically opposite the new vertex B." }}{PARA 0 "" 0 "" 
{TEXT -1 32 "DE intersects AB in the point C." }}{PARA 13 "" 1 "" 
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1&444444***F*7$7$Fcp$\"1/+++++D**F*F`gl7$F^gl7$FipFgp7$7$F]qFIFjgl7$Ff
gl7$$\"1%************G\"F*$\"1/+++++5**F*7$7$F]qF`dlF^hl7$F\\hl7$Fcq$!
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jsF`am7$Fbam7$$\"1;THN#)eq()F*F[`l7$7$Fa`lFitFdam7$Fham7$F\\alFj`l7$7$
FbalF`alFjam7$7$FbalF_[m7$F\\[mF`v7$7$F\\[mF\\v7$$\"1.++++++()F*$\"1#*
************[F*7$7$Fc\\mF[wFbbm7$Fhbm7$Fj]mFd]m7$F\\bm7$Fhal$!1]w6%HN#
)o$F*7$7$F\\blF0F\\cm7$F`cm7$$\"1+++++v$\\*F*Fdbl7$7$FjblF<Fbcm7$7$Fgc
lF<7$F\\dlFjcl7$7$F`dlFKFicm7$F[dm7$FjdlFhdl7$7$F^elFWF]dm7$F_dm7$Fiel
Fgel7$7$F]flFaoFadm7$Fcdm7$F]flFefl7$7$F]flFioFedm7$Fgdm7$$\"1%444444*
**F*$\"1Q!444444%F\\o7$7$FgglFcpFidm7$F_em7$FahlF_hl7$7$F`dlF]qFaem7$F
cem7$F_ilF]il7$7$FgclFgqFeem7$Fgem7$F[jlFiil7$7$F_jlFcrFiem7$F[fm7$Fba
lFc[m-%'COLOURG6&%$RGBG\"\"\"\"\"!Fcfm-F$6$7$7$$!\"\"FcfmFcfm7$$\"1+++
7y1rqF*F[gm-F_fm6&FafmFcfmFcfmFcfm-F$6$7$7$FcfmFcfmFjfmF]gm-F$6$7$Fbgm
7$F[gm$!1+++7y1rqF*F]gm-F$6$7$7$$FbfmFcfmFcfmFjfmF]gm-F$6$7$F\\hmFfgmF
]gm-F$6$7$FgfmF\\hmF]gm-F$6$7$FjfmFfgmF]gm-%%TEXTG6$7$$!$3\"!\"#$\"\"$
F]im%\"AG-Fhhm6$7$$\"$3\"F]imF^im%\"BG-Fhhm6$7$$\"#xF]imFjim%\"DG-Fhhm
6$7$Fjim$!#uF]im%\"EG-Fhhm6$7$$!\"(F]im$!\"&F]im%\"OG-Fhhm6$7$$\"#jF]i
mFhjm%\"CG-%*AXESTICKSG6$FcfmFcfm-%*AXESSTYLEG6#%%NONEG-%+AXESLABELSG6
$%!GF[\\n-%(SCALINGG6#%,CONSTRAINEDG-%%VIEWG6$%(DEFAULTGFc\\n" 1 2 0 
1 10 0 2 9 1 1 1 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" 
"Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" 
"Curve 10" "Curve 11" "Curve 12" "Curve 13" "Curve 14" }}}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "Let the length DE of
 the original " }{TEXT 270 1 "n" }{TEXT -1 18 "-sided polygon be " }
{XPPEDIT 18 0 "s[n];" "6#&%\"sG6#%\"nG" }{TEXT -1 52 ", and let the le
ngth DB = BE of one side of the new " }{XPPEDIT 18 0 "2*n" "6#*&\"\"#
\"\"\"%\"nGF%" }{TEXT -1 18 "-sided polygon be " }{XPPEDIT 18 0 "s[2*n
];" "6#&%\"sG6#*&\"\"#\"\"\"%\"nGF(" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 88 "Angle ADB is a right-angle (why?), so the area of trian
gle ADB can be expressed both as " }}{PARA 256 "" 0 "" {TEXT -1 2 "  \+
" }{XPPEDIT 18 0 "1/2;" "6#*&\"\"\"F$\"\"#!\"\"" }{TEXT -1 1 " " }
{TEXT 260 1 "." }{TEXT -1 6 " (AB) " }{TEXT 261 1 "." }{TEXT -1 17 " (
DC)   and as   " }{XPPEDIT 18 0 "1/2;" "6#*&\"\"\"F$\"\"#!\"\"" }
{TEXT -1 1 " " }{TEXT 262 1 "." }{TEXT -1 6 " (AD) " }{TEXT 263 1 "." 
}{TEXT -1 6 " (DB)." }}{PARA 0 "" 0 "" {TEXT -1 24 "Since AB = 2, this
 gives" }}{PARA 256 "" 0 "" {TEXT -1 2 "2 " }{TEXT 265 1 "." }{TEXT 
-1 13 " DC =   (AD) " }{TEXT 264 1 "." }{TEXT -1 6 " (DB) " }}{PARA 0 
"" 0 "" {TEXT -1 3 "or " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "s[n] = AD*` .`* s[2*n]" "6#/&%\"sG6#%\"nG*(%#ADG\"\"\"%#~.GF*&F%
6#*&\"\"#F*F'F*F*" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 9 "that \+
is, " }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "s[n]/s[2*n] \+
= AD;" "6#/*&&%\"sG6#%\"nG\"\"\"&F&6#*&\"\"#F)F(F)!\"\"%#ADG" }{TEXT 
-1 13 " ------- (i)." }}{PARA 0 "" 0 "" {TEXT -1 69 "On the other hand
, applying Pythagoras' theorem to triangle ADB gives" }}{PARA 256 "" 
0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "AD^2 = 4-s[2*n]^2;" "6#/*$%#ADG\"
\"#,&\"\"%\"\"\"*$&%\"sG6#*&F&F)%\"nGF)F&!\"\"" }{TEXT -1 14 " -------
 (ii)." }}{PARA 0 "" 0 "" {TEXT -1 49 "Eliminating AD from equations (
i) and (ii)  gives" }}{PARA 256 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 
0 "s[n]^2 = s[2*n]^2*(4-s[2*n]^2);" "6#/*$&%\"sG6#%\"nG\"\"#*&&F&6#*&F
)\"\"\"F(F.F),&\"\"%F.*$&F&6#*&F)F.F(F.F)!\"\"F." }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "Writing  \+
" }{XPPEDIT 18 0 "x = s[2*n]^2;" "6#/%\"xG*$&%\"sG6#*&\"\"#\"\"\"%\"nG
F+F*" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "S = s[n]^2;" "6#/%\"SG*$&%\"
sG6#%\"nG\"\"#" }{TEXT -1 9 ", we get " }}{PARA 256 "" 0 "" {TEXT -1 
2 "  " }{XPPEDIT 18 0 "x*(4-x) = S;" "6#/*&%\"xG\"\"\",&\"\"%F&F%!\"\"
F&%\"SG" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 34 "which is a qu
adratic equation for " }{TEXT 271 1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 74 "We can solve this quadratic equation by completing t
he square as follows. " }}{PARA 0 "" 0 "" {TEXT -1 35 " ( <=>  means \+
\"is equivalent to\" ) " }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "x*(4-x) = S" "6#/*&%\"xG\"\"\",&\"\"%F&F%!\"\"F&%\"SG" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 
6 "<=>   " }{XPPEDIT 18 0 "4*x-x^2 = S" "6#/,&*&\"\"%\"\"\"%\"xGF'F'*$
F(\"\"#!\"\"%\"SG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 256 "" 0 "" {TEXT -1 6 "<=>   " }{XPPEDIT 18 0 "x^2-4*x = -S" "6
#/,&*$%\"xG\"\"#\"\"\"*&\"\"%F(F&F(!\"\",$%\"SGF+" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 6 "<=>   " 
}{XPPEDIT 18 0 "x^2-4*x+4 = 4-S" "6#/,(*$%\"xG\"\"#\"\"\"*&\"\"%F(F&F(
!\"\"F*F(,&F*F(%\"SGF+" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 256 "" 0 "" {TEXT -1 6 "<=>   " }{XPPEDIT 18 0 "(x-2)^2 = 4-
S" "6#/*$,&%\"xG\"\"\"\"\"#!\"\"F(,&\"\"%F'%\"SGF)" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 6 "<=>   " 
}{XPPEDIT 18 0 "x-2 = -sqrt(4-S)" "6#/,&%\"xG\"\"\"\"\"#!\"\",$-%%sqrt
G6#,&\"\"%F&%\"SGF(F(" }{TEXT -1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 2 
"  " }}{PARA 256 "" 0 "" {TEXT -1 6 "<=>   " }{XPPEDIT 18 0 "x = 2-sqr
t(4-S)" "6#/%\"xG,&\"\"#\"\"\"-%%sqrtG6#,&\"\"%F'%\"SG!\"\"F." }{TEXT 
-1 1 " " }}{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
40 "We have to take the minus sign because  " }{XPPEDIT 18 0 "x = s[2*
n]^2;" "6#/%\"xG*$&%\"sG6#*&\"\"#\"\"\"%\"nGF+F*" }{TEXT -1 24 " is cl
early less than 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 14 "We now have   " }{XPPEDIT 18 0 "s[2*n]^2 = 2-sqrt(4-s[n]^
2);" "6#/*$&%\"sG6#*&\"\"#\"\"\"%\"nGF*F),&F)F*-%%sqrtG6#,&\"\"%F**$&F
&6#F+F)!\"\"F5" }{TEXT -1 6 "   or " }}{PARA 256 "" 0 "" {TEXT -1 4 " \+
   " }{XPPEDIT 18 0 "s[2*n] = sqrt(2-sqrt(4-s[n]^2));" "6#/&%\"sG6#*&
\"\"#\"\"\"%\"nGF)-%%sqrtG6#,&F(F)-F,6#,&\"\"%F)*$&F%6#F*F(!\"\"F6" }
{TEXT -1 1 "." }}{PARA 256 "" 0 "" {TEXT -1 1 " " }{TEXT 268 11 "_____
______" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 67 "Let's start with a regular polygon with 4 sides, tha
t is, a square." }}{PARA 0 "" 0 "" {TEXT -1 47 "By drawing a picture y
ou can easily check that " }{XPPEDIT 18 0 "s[4] = sqrt(2);" "6#/&%\"sG
6#\"\"%-%%sqrtG6#\"\"#" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }{XPPEDIT 18 0 "s[8] = sqrt(2-s
qrt(4-2));" "6#/&%\"sG6#\"\")-%%sqrtG6#,&\"\"#\"\"\"-F)6#,&\"\"%F-F,!
\"\"F2" }{TEXT -1 13 " , that is,  " }{XPPEDIT 18 0 "s[8] = sqrt(2-sqr
t(2));" "6#/&%\"sG6#\"\")-%%sqrtG6#,&\"\"#\"\"\"-F)6#F,!\"\"" }{TEXT 
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 
"Since " }{XPPEDIT 18 0 "s[8]^2 = 2-sqrt(2);" "6#/*$&%\"sG6#\"\")\"\"#
,&F)\"\"\"-%%sqrtG6#F)!\"\"" }{TEXT -1 10 ", we have " }{XPPEDIT 18 0 
"s[16] = sqrt(2-sqrt(4-(2-sqrt(2))));" "6#/&%\"sG6#\"#;-%%sqrtG6#,&\"
\"#\"\"\"-F)6#,&\"\"%F-,&F,F--F)6#F,!\"\"F5F5" }{TEXT -1 6 "  or  " }
{XPPEDIT 18 0 "s[16] = sqrt(2-sqrt(2+sqrt(2)));" "6#/&%\"sG6#\"#;-%%sq
rtG6#,&\"\"#\"\"\"-F)6#,&F,F--F)6#F,F-!\"\"" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "In a similar wa
y we get   " }{XPPEDIT 18 0 "s[32] = sqrt(2-sqrt(2+sqrt(2+sqrt(2))));
" "6#/&%\"sG6#\"#K-%%sqrtG6#,&\"\"#\"\"\"-F)6#,&F,F--F)6#,&F,F--F)6#F,
F-F-!\"\"" }{TEXT -1 3 ",  " }{XPPEDIT 18 0 "s[64] = sqrt(2-sqrt(2+sqr
t(2+sqrt(2+sqrt(2)))));" "6#/&%\"sG6#\"#k-%%sqrtG6#,&\"\"#\"\"\"-F)6#,
&F,F--F)6#,&F,F--F)6#,&F,F--F)6#F,F-F-F-!\"\"" }{TEXT -1 14 " ,  and s
o on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 90 "As the number of sides n tends to inf
inity, the n sided regular polygon tends to a circle." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 119 "It follows that the p
erimeter of the n-sided polygon approaches the circumference of a circ
le of unit radius, which is " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%
#PiGF%" }{TEXT -1 1 "." }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "limit(2^n*s
[n],n = infinity) = 2*Pi;" "6#/-%&limitG6$*&)\"\"#%\"nG\"\"\"&%\"sG6#F
*F+/F*%)infinityG*&F)F+%#PiGF+" }}{PARA 0 "" 0 "" {TEXT -1 22 "In orde
r to calculate " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 76 " using t
his limit, we need to perform repeated applications of the function " 
}{XPPEDIT 18 0 "f(x) = 2+sqrt(x);" "6#/-%\"fG6#%\"xG,&\"\"#\"\"\"-%%sq
rtG6#F'F*" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 79 "The standar
d way of doing this from a programming point of view is to set up a " 
}{TEXT 259 8 "for loop" }{TEXT -1 12 " as follows." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "We can calculate " }
{XPPEDIT 18 0 "s[64];" "6#&%\"sG6#\"#k" }{TEXT -1 12 " as follows." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
80 "t := 2:\nn := 3:\nfor k from 1 to n do\n   t := 2+sqrt(t);\nend do
:\nsqrt(2-sqrt(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*$-%%sqrtG6#,&
\"\"#\"\"\"*$-F%6#,&F(F)*$-F%6#,&F(F)*$-F%6#,&F(F)*$-F%6#F(F)F)F)F)F)F
)F)!\"\"F)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 56 "In order to use this to obtain a numerical estimate for " }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 49 " we need to multiply this \+
by 32 and put in a few " }{TEXT 0 5 "evalf" }{TEXT -1 3 "'s." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "t
 := 2:\nn := 3:\nfor k from 1 to n do\n   t := evalf(sqrt(t)+2);\nend \+
do:\nevalf((2^(n+2))*sqrt(2-sqrt(t)));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"+87LSJ!\"*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 33 "To get a more accurate value for " }{XPPEDIT 18 0 "Pi;
" "6#%#PiG" }{TEXT -1 26 " we need to do two things:" }}{PARA 15 "" 0 
"" {TEXT -1 9 "Increase " }{TEXT 272 1 "n" }{TEXT -1 1 "." }}{PARA 15 
"" 0 "" {TEXT -1 61 "Increase the number of digits being used for the \+
computation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 168 "Digits := 30:\nt := 2:\nn := 22:\nfor k from 1 to
 n do\n   t := evalf(2+sqrt(t));\nend do:\ntemp := evalf((2^(n+2))*sqr
t(2-sqrt(t)));\nDigits := 15:\nevalf(temp);\nDigits := 10:" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%%tempG$\"?gGrA//w)y*e`EfTJ!#H" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#$\"0z*e`EfTJ!#9" }}}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 42 "Comparing with the Maple's calculati
on of " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 42 " we see that this
 is correct to 15 digits." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf[15](Pi);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"0z*e`EfTJ!#9" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 ";" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Tasks" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "(a) Find suitable choic
es of " }{TEXT 0 6 "Digits" }{TEXT -1 5 " and " }{TEXT 0 1 "n" }{TEXT 
-1 48 " to use in the loop of the last section so that " }{XPPEDIT 18 
0 "Pi;" "6#%#PiG" }{TEXT -1 31 " is given correct to 20 digits." }}
{PARA 0 "" 0 "" {TEXT -1 66 "_________________________________________
_________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 66 "_______________________________________________________
___________" }}{PARA 0 "" 0 "" {TEXT -1 89 "(b) How many sides does th
e final approximating polygon for your calculation in (a) have?" }}
{PARA 0 "" 0 "" {TEXT -1 65 "_________________________________________
________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 47 "The final approximating polygon has      sides." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "________
__________________________________________________________" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 110 "(c) How can you \+
determine that your value in (a) is correct to 20 digits without using
 Maple's calculation of " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 "
?" }}{PARA 0 "" 0 "" {TEXT -1 65 "____________________________________
_____________________________" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 66 "_______________________________________________
___________________" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 
0 "" 0 "" {TEXT -1 16 "Code for picture" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 465 "with(plots):\nrt := ev
alf(sqrt(0.5)):\npA := [-1,0]:\npB := [1,0]:\npD := [rt,rt]:\npE := [r
t,-rt]:\npO := [0,0]:\nlines := plot([[pA,pD],[pO,pD],[pO,pE],[pB,pD],
\n[pB,pE],[pA,pB],[pD,pE]],color=black):\ncrcle := implicitplot(x^2+y^
2=1,x=-1..1,y=-1..1):\ntxt := textplot([[-1.08,0.03,`A`],[1.08,0.03,`B
`],\n[0.77,0.77,`D`],[0.77,-0.74,`E`],[-0.07,-0.05,`O`],\n[0.63,-0.05,
`C`]]):\ndisplay([crcle,lines,txt],axes=NONE,\n        scaling=CONSTRA
INED,tickmarks=[0,0],labels=[``,``]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "4 0 0" 0 }{VIEWOPTS 1 1 
0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }