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"" -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 }} {SECT 0 {EXCHG {PARA 258 "" 0 "" {TEXT -1 0 "" }{TEXT 258 9 "Su Nguyen " }}{PARA 261 "" 0 "" {TEXT 260 0 "" }{TEXT 261 8 "Ways of " }{TEXT 299 10 "ESTIMATING" }{TEXT 300 1 " " }{XPPEDIT 271 0 "Pi;" "6#%#PiG" } {TEXT 259 0 "" }}{PARA 258 "" 0 "" {TEXT 257 9 "Math 107B" }}{PARA 261 "" 0 "" {TEXT -1 0 "" }{TEXT 262 0 "" }{TEXT 263 0 "" }{TEXT 264 13 "June 14, 2002" }}{PARA 256 "" 0 "" {TEXT -1 12 "\nThe number " } {XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 78 " has been the subject of a great deal of mathematical (and popular) folklore. " }}{PARA 256 "" 0 "" {TEXT -1 51 "It's been worshipped, maligned, and misunderstood. \+ " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 101 " has recently (Sep. 20, 1999) been computed to a world record 206,158,430, 208 decimal digits (2^36). " }}{PARA 256 "" 0 "" {TEXT -1 30 "There ar e many algorithms for " }{TEXT 295 10 "estimating" }{TEXT -1 1 " " } {XPPEDIT 302 0 "Pi" "6#%#PiG" }{TEXT -1 59 ", ranging from the Monte C arlo (by generating random points" }}{PARA 256 "" 0 "" {TEXT -1 108 " \+ and testing whether or not they fall inside the unit circle. Pi is app roximately 4*nHits / nTrials) to the " }}{PARA 256 "" 0 "" {TEXT -1 85 "Continued Fraction method. The remarkable result is that the proba bility is directly " }{TEXT 296 7 "related" }{TEXT -1 8 " to the " } {TEXT 297 9 "value of " }{XPPEDIT 298 0 "Pi" "6#%#PiG" }{TEXT -1 2 ". \+ " }}{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 26 "S ince the actual value of " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 84 " has been calculated to be equal to 3.1415926535897932384626433832 79502884197......." }}{PARA 256 "" 0 "" {TEXT -1 93 "(and has been com puted using more powerful hardware and software rather than using Mapl e 7), " }}{PARA 256 "" 0 "" {TEXT -1 34 "the purpose of this assignmen t is " }{TEXT 285 3 "NOT" }{TEXT 286 34 " to calculate the actual valu e of " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {TEXT -1 20 "Therefore, the main " }{TEXT 290 8 "PURPOSES" } {TEXT -1 25 " of this assignment are: " }}{PARA 265 "" 0 "" {TEXT -1 33 " " }{TEXT 287 3 "(1)" }{TEXT -1 74 " Combine the uses and capabilities of Maple 7 and mathematics to e stimate " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 2 ". " }}{PARA 265 "" 0 "" {TEXT -1 33 " " }{TEXT 288 3 " (2)" }{TEXT -1 93 " Use several different methods (ranging from the le ast accurate to the more precise ones) to " }{TEXT 291 8 "estimate" } {TEXT -1 12 " the number " }{XPPEDIT 293 0 "Pi" "6#%#PiG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 33 " \+ " }{TEXT 294 4 "(3) " }{TEXT -1 89 "Involve Maple programming and grap hics tools to better visualize the materials presented." }}{PARA 265 " " 0 "" {TEXT -1 33 " " }{TEXT 289 3 "( 4)" }{TEXT -1 66 " Maple will output the differences between the estim ated value of " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 25 " and the \+ actual value of " }{XPPEDIT 292 0 "Pi" "6#%#PiG" }{TEXT -1 8 " itself. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 270 47 "Calculating PI using Circumference and Diameter" }{TEXT -1 0 " " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 79 " is def ined as the ratio of a circle's circumference C to its diameter d = 2r . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "P i;" "6#%#PiG" }{TEXT -1 22 " = C / d = C / 2r. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 62 " is equal to 3.141592653589793238462643383279502884197.......\n " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 193 "restart;\nwith(plottool s):\nc := circle([0,0], 1, color=blue, linestyle=3, thickness=2):\nj : = line([0,0], [.95, .3], color=red, linestyle=1, thickness=2):\nplots[ display](c,j,scaling=CONSTRAINED);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6&7U7$ $\"\"\"\"\"!$F*F*7$$\"+8q9@**!#5$\"+OBL`7F/7$$\"+6;$eo*F/$\"+t))*o[#F/ 7$$\"+f[w(H*F/$\"+FbC\"o$F/7$$\"++o1j()F/$\"+Tn`<[F/7$$\"+V*p,4)F/$\"+ CD&y(eF/7$$\"+tio*G(F/$\"+g5ZXoF/7$$\"+'*)RUP'F/$\"+HC80xF/7$$\"+^zEe` F/$\"+a#zKW)F/7$$\"+=HzdUF/$\"+B0F[!*F/7$$\"+Q*p,4$F/$\"+l^c5&*F/7$$\" +UJ\"Q(=F/$\"+3D(G#)*F/7$$\"+D>0zi!#6$\"+%Gn-)**F/7$$!+m>0ziFaoFbo7$$! +YJ\"Q(=F/$\"+2D(G#)*F/7$$!+U*p,4$F/$\"+j^c5&*F/7$$!+8HzdUF/$\"+E0F[!* F/7$$!+YzEe`F/$\"+d#zKW)F/7$$!+$*)RUP'F/$\"+JC80xF/7$$!+$G'o*G(F/$\"+] 5ZXoF/7$$!+]*p,4)F/$\"+9D&y(eF/7$$!+0o1j()F/$\"+Kn`<[F/7$$!+i[w(H*F/$ \"+>bC\"o$F/7$$!+8;$eo*F/$\"+l))*o[#F/7$$!+9q9@**F/$\"+IBL`7F/7$$!\"\" F*$!+:w1-T!#>7$$!+8q9@**F/$!+QBL`7F/7$$!+6;$eo*F/$!+t))*o[#F/7$$!+f[w( H*F/$!+FbC\"o$F/7$$!+,o1j()F/$!+Sn`<[F/7$$!+Y*p,4)F/$!+?D&y(eF/7$$!+xi o*G(F/$!+c5ZXoF/7$$!+())RUP'F/$!+PC80xF/7$$!+RzEe`F/$!+i#zKW)F/7$$!+1H zdUF/$!+H0F[!*F/7$$!+M*p,4$F/$!+m^c5&*F/7$$!+QJ\"Q(=F/$!+4D(G#)*F/7$$! +%)=0ziFao$!+&Gn-)**F/7$$\"+2?0ziFao$!+%Gn-)**F/7$$\"+]J\"Q(=F/$!+2D(G #)*F/7$$\"+Y*p,4$F/$!+i^c5&*F/7$$\"+%\"CG-%$IntG6$,$*&\"\"\"F**$-%%sqrtG6#,&F*F**$)%\"xG\" \"#F*!\"\"F*F4\"\"%/F2;\"\"!F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "#Evaluate the integeral from 0 to 1\nC := int( 4 * ( 1 / sqrt( 1-x^2) ), x=0..1 ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG,$%#PiG \"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "#Pi is equal to th e circumference/diameter, displayed 100 digits\nevalf(C/2, 100); " }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"_qoq6U`#[.G')**3iG1k\"yI#fW\\(4#e5v $*Rpr>%)G]zKQVEYQKz*e`EfTJ!#**" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 273 34 "Estimating PI using simple methods" }}{PARA 256 "" 0 "" {TEXT -1 83 "This section will fo cus on some of the calculated values that are closely resemble " }} {PARA 256 "" 0 "" {TEXT -1 16 "to the value of " }{XPPEDIT 18 0 "Pi" " 6#%#PiG" }{TEXT -1 50 ". It will then output the difference between ea ch " }}{PARA 256 "" 0 "" {TEXT -1 21 "calculated value and " } {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "#Set Maple to disp lay 20 digits\nDigits := 20; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'DigitsG\"#?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 174 "#A small fraction that is almost close to Pi\nP := 7/2; \+ \nP := evalf(P);\n\n#Display the difference between calcula ted value and Pi\nevalf( Pi-P ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG#\"\"(\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"PG$\"5+++++++++N!#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!4:w1-TYtSe$ !#>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 169 "#A smaller fraction that is close to Pi\nP := 355/113; \nP := evalf(P);\n\n #Display the difference between calculated value and Pi\nevalf( Pi-P ) ; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG#\"$b$\"$8\" " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"54I#)RN?HfTJ!#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!.C1*=knE!#>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 178 "#A number with exponent that is very close to Pi\nP \+ := 43^(7/23); \nP := evalf(P);\n\n#Display the difference \+ between calculated value and Pi\nevalf( Pi-P ); " }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG*$)\"#V#\"\"(\"#B\"\"\"" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"5E^`Hy_)R:9$!#>" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#$\"0fsRo!3!G&!#>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 176 "#A natural log (ln) number that is close to Pi\nP := ln(2198) / sqrt(6); \nP := evalf(P);\n\n#Display the difference be tween calculated value and Pi\nevalf( Pi-P ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,$*&-%#lnG6#\"%)>#\"\"\"-%%sqrtG6#\"\" 'F+#F+F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"5-$=3]%\\VfTJ!#> " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "#A ln and sqrt number that is close to P i\nP := ln(2625374126407687744) / sqrt(163); \nP := evalf(P);\n\n#Dis play the difference between calculated value and Pi\nevalf( Pi-P ); \+ " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,$*&-%#lnG6#\"4Wx o2k7u`i#\"\"\"-%%sqrtG6#\"$j\"F+#F+F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"5*H6?(>P]%>K$!#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!49*y S2OQ_.=!#>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 389 "#Using equat ions to calculated a value that is close to Pi\na := 0: \+ \nb := 1:\nc := 1 / sqrt(2): \nd := 1 / 4: \ne := 1: \na := b: \nb := (b + c) / 2: \nc := sqrt(c * a): \nd := d - (e * (b - a) * (b \+ - a)): \ne := 2 * e: \nf := (b * b) / d: \ng := ((b + c) * (b + c)) / \+ (4 * d);\ng := evalf(g);\n\n#Display the difference between calculated value and Pi\nevalf( abs(Pi-g) ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG,$*&,(#\"\"\"\"\"#F)*&#F)\"\"%F)-%%sqrtG6#F*F)F)* &F(F))F*#\"\"$F-F)F)F*,&F,F)*$),&#!\"\"F*F)*&F,F)F.F)F)F*F)F:F:F," }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG$\"5#[#o@_]#z09$!#>" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"2.*\\inIS85!#>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}}{SECT 1 {PARA 263 "" 0 "" {TEXT 272 37 "Estimating PI by Archimedes Algorithm" }}{PARA 256 "" 0 "" {TEXT 274 0 "" }{TEXT 275 0 "" }{TEXT -1 0 "" }{TEXT 276 0 "" }{TEXT 277 0 "" }{TEXT -1 0 "" }{TEXT 278 36 "This section will use Archimede s alg" }{TEXT 282 0 "" }{TEXT 281 47 "orithm to provide successive app roximations to " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT 279 2 ". " }} {PARA 256 "" 0 "" {TEXT 283 56 "Archimedes obtained the first rigorous approximation of " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT 280 28 " by i nscribing on a circle. " }}{PARA 256 "" 0 "" {TEXT -1 123 "As shown in the graph below, if we divide the the circle into 10 triangles, each \+ angle theta of the triangle is 36 degrees." }}{PARA 256 "" 0 "" {TEXT -1 21 "We use the formula: t" }{TEXT 301 18 "an^-1 (36) = 1 / x" } {TEXT -1 64 ", where x is the opposite side of the angle theta (36 deg rees). " }}{PARA 256 "" 0 "" {TEXT -1 118 "If we sum up all the x valu es of the 10 triangles, we would get the approximation to the circumfe rence of the circle. " }}{PARA 256 "" 0 "" {TEXT -1 5 "Then " } {XPPEDIT 308 0 "Pi" "6#%#PiG" }{TEXT -1 4 " is " }{TEXT 307 12 "approx imated" }{TEXT -1 17 " by the formula: " }{XPPEDIT 18 0 "Pi;" "6#%#PiG " }{TEXT -1 22 " = C / d = C / 2r. " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 680 "#Graph the example of a circle of radius 1 circumscribed an octagon.\nrestart;\nwith(plots): \nwith(plottools):\none_poly := [[0,0],[0,1],[1,1] ,[1,0]]:\npolygonpl ot(one_poly,axes=boxed):\ne := circle([0,0], 1, color = blue, linestyl e=3, thickness=1):\nf := line([0,0], [-.31,.95], color=red, linestyle= 1, thickness=2):\ng := line([0,0], [.31,.95], color=red, linestyle=1, \+ thickness=2):\nh := line([0,0], [.81,.58], color=red, linestyle=1, thi ckness=2):\ni := line([0,0], [1,0], color=red, linestyle=1, thickness= 2):\nngon := n -> [seq([ cos(2*Pi*i/n), sin(2*Pi*i/n) ], i = 1..n)]:\n display( ([ polygonplot(ngon(10)), textplot([0,0,`Octagon`]) ], color= yellow),e,f,g,h,i,scaling=CONSTRAINED );" }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n" }}{PARA 7 "" 1 "" {TEXT -1 43 "Warning, the name arrow has been redefined\n" }} {PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6+-%)POLYGONSG6#7, 7$$\"+V*p,4)!#5$\"+CD&y(eF*7$$\"+Q*p,4$F*$\"+l^c5&*F*7$$!+Q*p,4$F*F07$ $!+V*p,4)F*F+7$$!\"\"\"\"!$F;F;7$F6$!+CD&y(eF*7$F3$!+l^c5&*F*7$F.FA7$F (F>7$$\"\"\"F;F<-%%TEXTG6$7$F0zi!#6$\"+%Gn-)**F*7$$!+m>0ziF\\qF]q7$$!+YJ\" Q(=F*$\"+2D(G#)*F*7$$!+U*p,4$F*$\"+j^c5&*F*7$$!+8HzdUF*$\"+E0F[!*F*7$$ !+YzEe`F*$\"+d#zKW)F*7$$!+$*)RUP'F*$\"+JC80xF*7$$!+$G'o*G(F*$\"+]5ZXoF *7$$!+]*p,4)F*$\"+9D&y(eF*7$$!+0o1j()F*$\"+Kn`<[F*7$$!+i[w(H*F*$\"+>bC \"o$F*7$$!+8;$eo*F*$\"+l))*o[#F*7$$!+9q9@**F*$\"+IBL`7F*7$F9$!+:w1-T!# >7$$!+8q9@**F*$!+QBL`7F*7$$!+6;$eo*F*$!+t))*o[#F*7$$!+f[w(H*F*$!+FbC\" o$F*7$$!+,o1j()F*$!+Sn`<[F*7$$!+Y*p,4)F*$!+?D&y(eF*7$$!+xio*G(F*$!+c5Z XoF*7$$!+())RUP'F*$!+PC80xF*7$$!+RzEe`F*$!+i#zKW)F*7$$!+1HzdUF*$!+H0F[ !*F*7$$!+M*p,4$F*$!+m^c5&*F*7$$!+QJ\"Q(=F*$!+4D(G#)*F*7$$!+%)=0ziF\\q$ !+&Gn-)**F*7$$\"+2?0ziF\\q$!+%Gn-)**F*7$$\"+]J\"Q(=F*$!+2D(G#)*F*7$$\" +Y*p,4$F*$!+i^c5&*F*7$$\"+ \+ " 0 "" {MPLTEXT 1 0 483 "EstPI1 := proc(n)\n #Declare and intialize l ocal variables\n local c, num; \n #Find the artan of (360 \+ degrees/n triangles) \n c := evalf( arctan(Pi/n) );\n #Num equal \+ to (c * n triangles) \n num := evalf(c*n);\n #Return the calculated \+ value of Pi\n return num; \nend proc:\n\n#Assume there \+ are 1000 to 1010 triangles inscribed on a circle\nfor i from 1000 to 1 010 do\n #Call procedure EstPI1 with i between 1000 and 1010 \n Est PI1(i); \nend do;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# $\"+>BeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+SBeTJ!\"*" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+fBeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+!Q#eTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+- CeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+@CeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+TCeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#$\"+iCeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+#[#eTJ!\"*" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+-DeTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+ADeTJ!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "#Display the difference between calculated value and Pi\nevalf( \+ Pi - EstPI1(1010) ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"&K,\"!\"* " }}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 284 38 "Estimating PI using Monte Carlo Method" }}{PARA 256 "" 0 "" {TEXT -1 64 "This section will use probability theory to estimate the number " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 41 ", the area o f a circle with unit radius. " }}{PARA 256 "" 0 "" {TEXT -1 40 "This m ethod produce an approximation of " }{XPPEDIT 18 0 "Pi" "6#%#PiG" } {TEXT -1 40 " by generating random points and testing" }}{PARA 256 "" 0 "" {TEXT -1 65 " whether or not those random points fall inside the \+ unit circle. " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 37 " is approxi mately 4*nHits / nTrials. " }}{PARA 256 "" 0 "" {TEXT -1 97 "The progr am will calculate the distance from the circle center to coordinates ( x,y) to determine " }}{PARA 256 "" 0 "" {TEXT -1 70 "whether (x,y) is \+ inside the circle or not. It counts the total points " }}{PARA 256 "" 0 "" {TEXT -1 65 "land inside the circle and divided by the number of iterations. " }}{PARA 256 "" 0 "" {TEXT -1 23 "The result will be the " }{TEXT 304 13 "approximation" }{TEXT -1 4 " to " }{XPPEDIT 305 0 "P i" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 57 "#Include the plots packages\nwith(plots): \+ " }}{PARA 7 "" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 566 "# Setting up the graph ready to be plotted\nPIPix := proc(pts::list, Est Pie::float)\n #Declare local variables to be used in this section\n \+ local CirclePlot, Plot1, Plot2, PTitle, PlotPoint; \n PTitle := cat( \"The Estimate of PI is: \", convert(EstPie, string)); \n CirclePlot \+ := plot(\{sqrt(1-x*x), -1*sqrt(1-x*x)\}, x=-1..1, color=green,\n \+ title = PTitle); \n Plot1 := plot([[0,1], [1,1]]); \n P lot2 := plot([[1,1], [1,0]]); \n PlotPoint := plot(pts, color = blue ); \n display( \{CirclePlot, Plot1, Plot2, PlotPoint\}, scaling=CONS TRAINED ); \nend proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 119 "RandNum := proc()\n #Generate and return a random number between 0 and 1\n return evalf(rand() / 10^12); \nend proc: \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 439 "DrawPi := proc(Points::integer)\n \+ local x, y, i, pts, Circ, pi, CirclePlot, PlotPoint, Plot1, Plot2 ; \+ \n pts := [] ; Circ := 0 ; \n\n #Begin of for loop from 1 to Points \n for i from 1 to Points do\n x := RandNum() ; y := RandNum() ; \+ \n pts := [op(pts), [x,y]] ; \n if ( (x*x)+(y*y) ) <= 1 then Cir c := Circ + 1 ; end if ; \n end do ; \n\n #Evalulate the value pf pi \n pi := evalf(4 * Circ / Points) ; \n\n PIPix(pts, pi); \nend proc: \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "#Estimated the valu e of Pi at 500 points\nDrawPi(500); " }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6+-%'CURVESG6$7_o7$$!\"\"\"\"!$F*F*7$$!3-n;Hd Nvs**!#=$!3aIk@d'**oP(!#>7$$!3/MLe9r]X**F/$!3^xD,?$RD/\"F/7$$!3/,](=ng 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1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Cur ve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" }}}}{EXCHG {PARA 264 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 259 "" 0 "" {TEXT -1 60 "Es timating PI by adding up the first MAX terms of the series" }}{PARA 256 "" 0 "" {TEXT -1 64 "This section will adds up the first MAX terms of the series to " }{TEXT 303 11 "approximate" }{TEXT -1 1 " " } {XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 256 "" 0 "" {TEXT -1 94 "Since the series is alternating, the result alternately g ives the upper and lower estimate to " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 2 ". " }}{PARA 256 "" 0 "" {XPPEDIT 256 0 "Pi;" "6#%#PiG" } {TEXT -1 60 " = 4 - (4/3) + (4/5) - (4/7) + ...- (-1)^n [4/(2n+1)] + . ..," }}{PARA 256 "" 0 "" {TEXT -1 68 "Noted that the sign changes from positive to negative or vice versa " }}{PARA 256 "" 0 "" {TEXT -1 41 "and the denominator is incremented by 2. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 764 "EstPI2 := proc(n) \n#Declare and initialize local variables\nlocal sgn, dnm, term, sm, c ount;\nsgn := 1: \ndnm := 1:\nterm := 0:\nsm := 0.0 :\n\n#Begin for loop from 0 to n, the number of iterations\nfor count \+ from 0 to n do\n #Denominator equal 2*count + 1 \n dnm := 2*coun t + 1:\n #Term equal sign (+ or -) * 4/denominator \n term := sgn * (4.0/dnm):\n #Sum = sum + term \n sm := sm + term:\n #Swit ch the sign after each term \n sgn := -sgn: \+ \nend do;\n\n#Return the sum, which is the caculated value of Pi \nre turn sm; \nend proc:\n\n#Assume i is from 100 to 105 iterations\nfor i from 10000 to 10005 do\n #Calculated the value of \+ Pi from 100 to 105 iterations \n EstPI2(i); \nend do;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+7EpTJ!\"*" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#$\"+UE\\TJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#$\"+#f#pTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+iE\\TJ!\"*" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+sDpTJ!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+#o#\\TJ!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "#Display the difference between the calculated value and Pi.\n evalf ( Pi - EstPI2(105) ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\" (_PV*!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 259 "" 0 "" {TEXT -1 66 "Estimating PI using the Double Exponent ial Random Variables Method" }}{PARA 256 "" 0 "" {TEXT -1 64 "This sec tion will simulate Double Exponential Random Variables. " }}{PARA 256 "" 0 "" {TEXT -1 87 "First, we calculate the cdf (cummulative distribu tion function) and inverse cdf of the " }}{PARA 256 "" 0 "" {TEXT -1 102 "Double Exponential distribution 0.5e^(-0.5(x^2+y^2)) dxdy, and th en generate uniform random variables " }}{PARA 256 "" 0 "" {TEXT -1 81 "and plug them in. It will generate pairs of Double Exponential ra ndom variables " }}{PARA 256 "" 0 "" {TEXT -1 8 "and thes" }{TEXT 269 0 "" }{TEXT 267 0 "" }{TEXT 268 0 "" }{TEXT 265 0 "" }{TEXT 266 0 "" } {TEXT -1 56 "e are plugged into 2*exp( (-.5*((x^2)+(y^2))+|x|+|y|) ) \+ " }}{PARA 256 "" 0 "" {TEXT -1 107 "as x and y arguments of the functi on. Finally, it will\ncalculate the average over all the iterations as an " }{TEXT 306 13 "approximation" }{TEXT -1 4 " of " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "RandNum := proc()\n #Gener ates a random number between 0 and 1\n return evalf(rand() / 10^12); \+ \nend proc:" }{TEXT -1 1 "\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 791 "EstPI4 := proc(n)\n#Declare and intialize local variables\nlo cal i, sm, a, b, x, y, doubleexp; \nsm := 0;\n\n#Begin for loop from \+ 1 to n, the number of iterations\nfor i from 1 to n do \n \+ #Assign a random number to variables a and b\n a := RandNum(); \+ \n b := RandNum(); \n \n #Using if then else statement f or appropriate calculation\n if a <= .5 then x := log(2*a); else x := -log(2-2*a); end if; \n if b <= .5 then y := log(2*b); else y := -lo g(2-2*b); end if;\n doubleexp := evalf( 2 * exp( (-.5 * ((x*x) + (y*y ))) + abs(x) + abs(y) ) );\n sm := evalf(doubleexp + sm);\nend do; \+ \n#End of the for loop\n\n#Return sum / n, whic h is the calculated value of Pi\nreturn (sm/n); \nen d proc: \n#End the procedure\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "#Assume the number of iterations is 1000\nEstPI4(1000); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+'fyl7$!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "#Di splay the difference between the caculated value and Pi\nevalf( Pi - E stPI4(1000) ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!(Hu9&!\"*" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}}{SECT 1 {PARA 259 "" 0 "" {TEXT -1 38 "Estimating PI using Continued Fractio n" }}{PARA 256 "" 0 "" {TEXT -1 59 "This section use the Continued Fra ction method to estimate " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 28 ". We assumed that the value " }}{PARA 256 "" 0 "" {TEXT -1 3 "of \+ " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 47 " is known (3.1415...), \+ then we can approximate " }{XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 37 " using the continued fraction method." }}{PARA 256 "" 0 "" {XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 45 " = a1 + 1 / ( a2+1/(a3+1/(a4+1/a5+. .....) ) )" }}{PARA 256 "" 0 "" {XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 45 " = 3 + 1 / ( 7 + 1/(15 + 1/(1+1/292+....) ) )" }}{PARA 256 "" 0 "" {TEXT -1 82 "Continued fraction provide a series of \"best\" esti mates for an irrational number. " }}{PARA 256 "" 0 "" {TEXT -1 96 "The \"best\" approximation of a given order, is [3,7,15,1,292,1,1,1,2,1,3 ,1,14,2,1,1,2,2,2,2,...] " }}{PARA 256 "" 0 "" {TEXT -1 99 "The very l arge term 292 means that the convergent: [3,7,15,1] = [3,7,16] = 355/1 13 = 3.14159292... " }}{PARA 256 "" 0 "" {TEXT -1 122 "is an extremely good approximation. The first few convergents are 22/7, 333/106, 355/ 113, 103993/33102, 104348/33215, ... " }}{PARA 256 "" 0 "" {TEXT -1 46 "A nice expression for the third convergent of " }{XPPEDIT 256 0 "P i" "6#%#PiG" }{TEXT -1 14 " is given by: " }}{PARA 256 "" 0 "" {XPPEDIT 256 0 "Pi" "6#%#PiG" }{TEXT -1 42 " = 2[1,1,1,3,32] = 355/113 = 3.14159292..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 541 "#Generate the \"best\" approximation of a giv en order\nGenerateNum := proc (x, n) \n#Declare and initialize local variables\nlocal i, num, a; \nnum := evalf(x, 1000);\na : = [seq(0, i=1..n)];\n\n#Begin the for loop from 1 to n\nfor i from 1 t o n do \n #Take the floor or num and insert into array a of e lement i\n a[i] := floor(num); \n num := evalf (1 / (num - a[ i]), 1000);\nend do; \n#End of for loop\n#Return \+ the array a\nreturn a; \n\nend proc:\n#End of proce dure \n " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 409 "#Give an example of the \"best\" approximation of a given order\n GenerateNum(Pi, 10); \n#Assign array d to store the number s generated by GenerateNum\nd := GenerateNum(Pi, 10); \n#Display th e size of array d\nnops(d); \n#Display the 3rd ele ment stored in array d\nd[3]; \n#Display up to \+ the 10th fractions that was calculated to Pi\nconvergs( d(10) ); \+ " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7,\"\"$\"\"(\"#:\"\"\"\"$#HF' F'F'\"\"#F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG7,\"\"$\"\"(\"#: \"\"\"\"$#HF)F)F)\"\"#F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"#:" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6$\"\"##\"#A\"\"(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"$#\"$L$\"$1\" " }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"%#\"$b$\"$8\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"&#\"'$*R5\"&-J$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"'#\"'[V5\"&:K$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"(#\"'T $3#\"&P$)\"'\"Ql#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"#5#\"(3k9\"\"'8\\O" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 563 "EstPI5 := proc (n)\n#Declare and initialize local va riables\nlocal arysze, b, d, num; \nnum := 0;\n#Assign array d to store the numbers generated by GenerateNum\nd := GenerateNum(Pi, n); \+ \n#Assign arysze to store the size of array d\narysze := nops(d); \+ \n\n#Begin for loop from arysze to 2 by decrease -1 each tim e\nfor b from arysze by -1 to 2 do \n num := evalf( 1 / (d[b] + num ) , 1000);\nend do; \n#End of for loop\n\n#Return the calculated value of Pi\nreturn (3+num); \nend proc: \n#End of procedure \n" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 81 "#Set the iterations to 75 times, display 100 digits \nevalf( EstPI5(75), 100 ); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\" _qR7&zUEy#y1jK;viS;yI#fW\\(4#e5v$*Rpr>%)G]zKQVEYQKz*e`EfTJ!#**" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 168 "#The continued fraction met hod proved to be the most accurate \n#method of all of the above metho ds as a way get an approximation to Pi.\nevalf( (Pi - EstPI5(75)), 100 );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\";He;9*)*p?gbtX5\"!#**" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}}{MARK "5" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }