{VERSION 5 0 "IBM INTEL NT" "5.0" } {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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 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 } {CSTYLE "" -1 256 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 24 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 0 "" }{TEXT 257 21 "Numeric al Integration" }}{PARA 0 "" 0 "" {TEXT -1 213 "Here we try to evaluat e definite integrals using numerical methods. This is particularly us eful when we can find no antiderivative of the integrand and therefore cannot apply the fundemental theorem of calculus. " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 502 "The following sections contain several Maple commands in a row. To execute all of the comman d in a \"cell\" just hit \"enter\" on any line in the cell. Maple wil l then execute all of the commands in sequence starting from the first one in the cell. To create a cell with several commands you must hit \"shift\" and \"enter\" at the same time. This goes to the next lin e but does not execute the command. When Maple executes the commands \+ in sequence, it prints the results from every command ending with a " }{TEXT 258 10 "semi-colon" }{TEXT -1 60 " but does not print the resul ts from commands ending with a " }{TEXT 259 7 "colon. " }{TEXT -1 12 " Finally the " }{TEXT 261 1 "#" }{TEXT -1 105 " symbol, and everything \+ after it, is ignored by Maple. I use this to make comments about some commands. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "For the first 3 examples that follow, we are integrating " } {XPPEDIT 18 0 "f(x) = sqrt(1-x^2);" "6#/-%\"fG6#%\"xG-%%sqrtG6#,&\"\" \"F,*$F'\"\"#!\"\"" }{TEXT -1 128 " from x=0 to x=1. From geometry w e know that this is one quarter the area of a circle with radius 1. I e. The exact answer is " }{XPPEDIT 18 0 "Pi/4;" "6#*&%#PiG\"\"\"\"\"%! \"\"" }{TEXT -1 50 ". In a sense, we are approximating the value of \+ " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 1 " " }{TEXT -1 3 " " }}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 39 " Left and Right Endpoint Approxim ations" }}{PARA 0 "" 0 "" {TEXT -1 65 "The following sequence of comma nds generates an approximation to " }{XPPEDIT 18 0 "int(f(x),x = a .. \+ b);" "6#-%$intG6$-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 48 " using lef t and right endpoint approximations. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "with(student): #This loads the \"student\" library containing \"leftb ox\" and \"rightbox\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "f \+ := x -> sqrt(1 - x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#% \"xG6\"6$%)operatorG%&arrowGF(-%%sqrtG6#,&\"\"\"F0*$)9$\"\"#F0!\"\"F(F (F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 430 "a := 0: # the left limit of integration\nb := 1: # the right limit of integration\nn := 20: # the number of subintervals\ndeltax := (b-a)/n: # the interval \+ width\nxk := k -> a + k*deltax: #this functions finds the k'th x valu e\nyk := k -> f(xk(k)): # this function finds the k'th y value\nleftb ox(f(x),x=a..b,n); #this is a special plotting function\nLeftApprox : = evalf(deltax * sum(yk(k),k=0..n-1)); #left endpoint approximation\n " }{TEXT -1 0 "" }{MPLTEXT 1 0 76 "MapleArea := evalf(int(f(x),x=a..b) ); #Exact Answer as determined by Maple\n" }{TEXT -1 0 "" }{MPLTEXT 1 0 395 "LeftError := MapleArea - LeftApprox; #The error using the l eft endpoint approximation\nrightbox(f(x),x=a..b,n); #this is a specia l plotting function \nRightApprox := evalf(deltax * sum(yk(k),k=1..n)) ; #right endpoint approximation\nMapleArea := evalf(int(f(x),x=a..b)); #Exact Answer as determined by Maple\nRightError := MapleArea - Righ tApprox; #The error using the right endpoint approximation" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "69-%'CURVESG6&7Y7$$\"\"! 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"Curve 14" "Curve 15" "Curve 16" "Curve 17 " "Curve 18" "Curve 19" "Curve 20" "Curve 21" }}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,RightApproxG$\"+*>i6d(!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*MapleAreaG$\"+N;)R&y!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+RightErrorG$\"*O%>GG!#5" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "# end of this section" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 6 "Notice" }{TEXT -1 213 ", if the function is decreasing (o r increasing) over x=a to x=b then the exact answer lies between the l eft and right endpoint approximations. This provides one way of determ ining the error in the approximation. " }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 22 "Midpoint Approximation" }}{PARA 0 "" 0 "" {TEXT -1 65 "Th e following sequence of commands generates an approximation to " } {XPPEDIT 18 0 "int(f(x),x = a .. b);" "6#-%$intG6$-%\"fG6#%\"xG/F);%\" aG%\"bG" }{TEXT -1 34 " using the midpoint approximation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 72 "with(student): # This loads the \"student\" library containing \"middlebox\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := x -> sqrt(1-x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf* 6#%\"xG6\"6$%)operatorG%&arrowGF(-%%sqrtG6#,&\"\"\"F0*$)9$\"\"#F0!\"\" F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 458 "a := 0: # the l eft limit of integration\nb := 1: # the right limit of integration\nn := 20: # the number of subintervals\ndeltax := (b-a)/n: # the interv al width\nxk := k -> a + k*deltax:\nxmidk := k -> (xk(k-1) + xk(k))/2: \nymidk := k -> f(xmidk(k)):\nyk := k -> f(xk(k)):\nmiddlebox(f(x),x=a ..b,n);\nMidApprox := evalf(deltax * sum(ymidk(k),k=1..n));\nMapleArea := evalf(int(f(x),x=a..b));\nMidError := MapleArea - MidApprox; #The \+ error using the Midpoint method. " }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "69-%)POLYGONSG6$7&7$$\"\"!F)F(7$F($\"+7X(o***!#57 $$\"+++++]!#6F+7$F/F(-%&COLORG6&%$RGBG$\"\"(!\"\"$\"\"*F9F7-F$6$7&F27$ F/$\"+Q`$=(**F-7$$\"+++++5F-F@7$FCF(F3-F$6$7&FE7$FC$\"+;uc@**F-7$$\"++ +++:F-FJ7$FMF(F3-F$6$7&FO7$FM$\"+QVoX)*F-7$$\"+++++?F-FT7$FWF(F3-F$6$7 &FY7$FW$\"+QweV(*F-7$$\"+++++DF-Fhn7$F[oF(F3-F$6$7&F]o7$F[o$\"+iAW9'*F -7$$\"+++++IF-Fbo7$FeoF(F3-F$6$7&Fgo7$Feo$\"+i,9d%*F-7$$\"+++++NF-F\\p 7$F_pF(F3-F$6$7&Fap7$F_p$\"+4\"[-F*F-7$$\"+++++SF-Ffp7$FipF(F3-F$6$7&F [q7$Fip$\"+%\\L>0*F-7$$\"+++++XF-F`q7$FcqF(F3-F$6$7&Feq7$Fcq$\"+]z&)*z )F-7$$F0F-Fjq7$F]rF(F3-F$6$7&F^r7$F]r$\"+iA-6&)F-7$$\"+++++bF-Fcr7$Ffr F(F3-F$6$7&Fhr7$Ffr$\"+)3M:=)F-7$$\"+++++gF-F]s7$F`sF(F3-F$6$7&Fbs7$F` s$\"+(\\Zi!yF-7$$\"+++++lF-Fgs7$FjsF(F3-F$6$7&F\\t7$Fjs$\"+\"Gx\"ytF-7 $$\"+++++qF-Fat7$FdtF(F3-F$6$7&Fft7$Fdt$\"+c')[()oF-7$$\"+++++vF-F[u7$ F^uF(F3-F$6$7&F`u7$F^u$\"+JBh>jF-7$$\"+++++!)F-Feu7$FhuF(F3-F$6$7&Fju7 $Fhu$\"+ysK^cF-7$$\"+++++&)F-F_v7$FbvF(F3-F$6$7&Fdv7$Fbv$\"+$=H7%[F-7$ $\"+++++!*F-Fiv7$F\\wF(F3-F$6$7&F^w7$F\\w$\"+Q5n*z$F-7$$\"+++++&*F-Fcw 7$FfwF(F3-F$6$7&Fhw7$Ffw$\"+/'[?A#F-7$$\"\"\"F)F]x7$F`xF(F3-%'CURVESG6 &7Y7$F(F`x7$$\"3emmm;arz@!#>$\"3e(=:\"QTi(***!#=7$$\"3[LL$e9ui2%F[y$\" 3!e2E0a)o\"***F^y7$$\"3nmmm\"z_\"4iF[y$\"3)f\\s^f/2)**F^y7$$\"3[mmmT&p hN)F[y$\"3)37fk0E]'**F^y7$$\"3CLLe*=)H\\5F^y$\"3R\\N+$H'zW**F^y7$$\"3g mm\"z/3uC\"F^y$\"3^l4$yi$*=#**F^y7$$\"3%)***\\7LRDX\"F^y$\"3%p#zm\"3WR *)*F^y7$$\"3]mm\"zR'ok;F^y$\"3iaa\"Q\\n/')*F^y7$$\"3w***\\i5`h(=F^y$\" 3\"Rj1r%eUA)*F^y7$$\"3WLLL3En$4#F^y$\"3_.Ayu2Py(*F^y7$$\"3qmm;/RE&G#F^ y$\"3wIBk#=x`t*F^y7$$\"3\")*****\\K]4]#F^y$\"31q9\"H%H@#o*F^y7$$\"3$** ****\\PAvr#F^y$\"3B_^A([sOi*F^y7$$\"3)******\\nHi#HF^y$\"3m:#e7,zAc*F^ y7$$\"3jmm\"z*ev:JF^y$\"3)>#\\;*p8A]*F^y7$$\"3?LLL347TLF^y$\"3pW!GHZL` U*F^y7$$\"3,LLLLY.KNF^y$\"3#=1W(>`Yb$*F^y7$$\"3w***\\7o7Tv$F^y$\"3!QZN cQ$eo#*F^y7$$\"3'GLLLQ*o]RF^y$\"3=`?'yv9l=*F^y7$$\"3A++D\"=lj;%F^y$\"3 ]kxU3;t!4*F^y7$$\"31++vV&R5j')F^y7$$\"3&em;zRQb@&F^y$\"3e0&=5t$=K&)F^y7$$\"3\\** *\\(=>Y2aF^y$\"3ld&yZ.e=T)F^y7$$\"39mm;zXu9cF^y$\"3]/X@$oS\\F)F^y7$$\" 3l******\\y))GeF^y$\"3k!yShX>b7)F^y7$$\"3'*)***\\i_QQgF^y$\"35nMlUm1rz F^y7$$\"3@***\\7y%3TiF^y$\"3bX$z*QqP8yF^y7$$\"35****\\P![hY'F^y$\"3QXh Iao;GwF^y7$$\"3kKLL$Qx$omF^y$\"3q\"p\\+YH?X(F^y7$$\"3!)*****\\P+V)oF^y $\"378IrHv-`sF^y7$$\"3?mm\"zpe*zqF^y$\"3QynnKd;iqF^y7$$\"3%)*****\\#\\ 'QH(F^y$\"3G\"QTZ_=5%oF^y7$$\"3GKLe9S8&\\(F^y$\"3a%Rp()p\"*)>mF^y7$$\" 3R***\\i?=bq(F^y$\"3=&F^y7$$\"3=LLe9tOc()F^y$\"3PnBuRIq H[F^y7$$\"3u******\\Qk\\*)F^y$\"3+@I:/fPhWF^y7$$\"3CLL$3dg6<*F^y$\"3au )*4GH?')RF^y7$$\"3ImmmmxGp$*F^y$\"3s!RL6R._\\$F^y7$$\"3A++D\"oK0e*F^y$ \"3!yTAE(=!f'GF^y7$$\"3C+++]oi\"o*F^y$\"3?UJ1O#=K]#F^y7$$\"3A++v=5s#y* F^y$\"37vO>U4Dt?F^y7$$\"3;+D1k2/P)*F^y$\"3mLW(eL^zz\"F^y7$$\"35+]P40O \"*)*F^y$\"31H.f%oH+Z\"F^y7$$\"3k]7.#Q?&=**F^y$\"3G#>z^*=&RF\"F^y7$$\" 31+voa-oX**F^y$\"3(pV\\8H')3/\"F^y7$$\"3[\\PMF,%G(**F^y$\"3?&=hJ*p=ltF [yFbx-%'COLOURG6&F6$\"*++++\"!\")F(F(-%&STYLEG6#%%LINEG-%*THICKNESSG6# \"\"#-%+AXESLABELSG6$Q\"x6\"Q!Fejl-%%VIEWG6$;F(F`x%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "C urve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" "Curve 8" "Curve 9" "C urve 10" "Curve 11" "Curve 12" "Curve 13" "Curve 14" "Curve 15" "Curve 16" "Curve 17" "Curve 18" "Curve 19" "Curve 20" "Curve 21" }}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*MidApproxG$\"+rkdjy!#5" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%*MapleAreaG$\"+N;)R&y!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)MidErrorG$!(O[f*!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "# end of Midpoint approximation method section" }}}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "Trapezoidal Approximation" }} {PARA 0 "" 0 "" {TEXT -1 65 "The following sequence of commands genera tes an approximation to " }{XPPEDIT 18 0 "int(f(x),x = a .. b);" "6#-% $intG6$-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 37 " using the Trapezoid al approximation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := x -> sqrt(1-x^2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&a rrowGF(-%%sqrtG6#,&\"\"\"F0*$)9$\"\"#F0!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 487 "a := 0: # the left limit of integration\n b := 1: # the right limit of integration\nn := 20: # the number of su bintervals\ndeltax := (b-a)/n: # the interval width\nxk := k -> a + k *deltax: #this functions finds the k'th x value\nyk := k -> f(xk(k)): # this function finds the k'th y value\nTrapApprox := evalf(deltax/2 * (yk(0) + sum(2*yk(k),k=1..n-1) + yk(n))); \nMapleArea := evalf(int( f(x),x=a..b)); #Exact Answer\nTrapError := MapleArea - TrapApprox; \+ #The error using Trapazoid Method" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %+TrapApproxG$\"+*>i6#y!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*Maple AreaG$\"+N;)R&y!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*TrapErrorG$\" )O%>G$!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "There is no easy way to graph the boxes associated with the trapazoidal method. Sorry." }} {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 263 7 "Notice:" }{TEXT -1 155 " \+ The error using the trapazoidal method is smaller than that for the le ft and right endpoint approximations but not as good as the midpoint a pproximation. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "# end of \+ Trapazoidal Method Section" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 56 "E rror Bounds for Midpoint and Trapezoidal approximations" }}{PARA 0 "" 0 "" {TEXT -1 217 "It is certainly useful to know how accurate an appr oximation is. With this goal, we look at the error bounds associated \+ with the midpoint and trapezoidal approximations. Consider the typical problem of approximating " }{XPPEDIT 18 0 "int(f(x),x = a .. b);" "6# -%$intG6$-%\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 98 " using n equally s paced subintervals. Furthermore, assume that f''(x) is continous on [a ,b]. Let " }{XPPEDIT 18 0 "E[M];" "6#&%\"EG6#%\"MG" }{TEXT -1 59 " re presents the error using the midpoint approximation and " }{XPPEDIT 18 0 "E[T];" "6#&%\"EG6#%\"TG" }{TEXT -1 110 " represents the error us ing the trapazoidal approximation. Bounds on these erros may then be \+ calculated from " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 16 "Formula (1) " }{XPPEDIT 18 0 "abs(E[M]) <= (b-a)^3*k [2]/(24*n^2);" "6#1-%$absG6#&%\"EG6#%\"MG*(,&%\"bG\"\"\"%\"aG!\"\"\"\" $&%\"kG6#\"\"#F.*&\"#CF.*$%\"nGF5F.F0" }{TEXT -1 10 " , where " } {XPPEDIT 18 0 "k[2];" "6#&%\"kG6#\"\"#" }{TEXT -1 48 " is the maximum \+ value of | f''(x) | on [a,b]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 4 "and " }}{PARA 0 "" 0 "" {TEXT -1 1 " " } {TEXT 266 11 "Formula (2)" }{TEXT -1 3 " " }{XPPEDIT 18 0 "abs(E[T]) <= (b-a)^3*k[2]/(12*n^2);" "6#1-%$absG6#&%\"EG6#%\"TG*(,&%\"bG\"\"\"% \"aG!\"\"\"\"$&%\"kG6#\"\"#F.*&\"#7F.*$%\"nGF5F.F0" }{TEXT -1 10 " , \+ where " }{XPPEDIT 18 0 "k[2];" "6#&%\"kG6#\"\"#" }{TEXT -1 46 " is the maximum value of | f''(x) | on [a,b]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "In finding these error bounds the fi rst goal is to find " }{XPPEDIT 18 0 "K[2];" "6#&%\"KG6#\"\"#" }{TEXT -1 221 " (the maximum of the absolute value of the second derivative o f f on [a,b]). If we cannot find an exact value for this number, it su ffices to approximate it as long as our approximation is bigger than t he actual number. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 45 "The question of accuracy comes in two forms: " }}{PARA 0 "" 0 "" {TEXT -1 104 "(1) Given f(x), a, b, and n, what is the maxi mum error that can occur with our approximation technique?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 149 "(2) Given f(x), \+ a, b, and an error bound, how many subintervals should we use so that \+ our approximation stays within this error of the exact answer? " }} {PARA 0 "" 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 100 "In the example that follow, we will look at these two questions using the tr apezoidal approximation." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 264 12 "Example (1) " }}{PARA 0 "" 0 "" {TEXT -1 58 "What is the maximum e rror that can occur by approximating " }{XPPEDIT 18 0 "int(1/x,x = 1 . . 2);" "6#-%$intG6$*&\"\"\"F'%\"xG!\"\"/F(;F'\"\"#" }{TEXT -1 53 " usi ng the trapezoidal method with 10 subintervals ? " }}{PARA 0 "" 0 "" {TEXT -1 209 "The first goal is to find the maximum of | f''(x) | on [ 1,2]. We can do this and analytically and determine the maximum is 2. However, we can also arrive at this conclusion by plotting f''(x) ov er [1,2] by " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f := x -> 1/x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*&\"\" \"F-9$!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot(a bs(diff(f(x),x,x)), x=1..2);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$\"\"\"\"\"!$\"\"#F*7$$\"3hmm;arz@5 !#<$\"3+uU*Q\"zru=F07$$\"3OL$e9ui2/\"F0$\"3ai\"z3['3u\"F07$$\"3YLL$3En$47F0$\"3smr$*y2sI6F07$$\"3 cmmT!RE&G7F0$\"3)eQy\"4+ky5F07$$\"3)*****\\K]4]7F0$\"3-Ibj$[mP-\"F07$$ \"3))****\\PAvr7F0$\"3#y_=3b'\\B(*!#=7$$\"3/++]nHi#H\"F0$\"3_jmtFio7$$\"3-+]7=lj;9F0$\"3= z\"\\vlZ[.(Fio7$$\"3&***\\PaR/<*p\\vt'Fio7$$\"3GLLe9Ege9F0$ \"3_3:[%GU\\W'Fio7$$\"3WL$eR\"3Gy9F0$\"3W(zQ/xw4>'Fio7$$\"3mmmT5k]*\\ \"F0$\"3!)\\>iG(z<$fFio7$$\"3em;zRQb@:F0$\"3ufl,JAkxcFio7$$\"3%)**\\(= >Y2a\"F0$\"3I'zI8+.\"oaFio7$$\"3imm\"zXu9c\"F0$\"3MG2*4,+n=VFio7$$\"3')****\\P+V)o\"F0$\" 3-crVhL4bTFio7$$\"3im;zpe*zq\"F0$\"3Q&pZZrFR,%Fio7$$\"3)*****\\#\\'QH< F0$\"3/F*H.o=o'QFio7$$\"37L$e9S8&\\9xd`C$Fio7$$\"3_ mmTvJga=F0$\"3T**zG6[GNJFio7$$\"3KL$e9tOc(=F0$\"3e%[$fZ^)4.$Fio7$$\"3' ******\\Qk\\*=F0$\"3hHXqQW=RHFio7$$\"3@LL3dg6<>F0$\"3!o(z'=4t%QGFio7$$ \"3_mmmw(Gp$>F0$\"3[P'RqNdAv#Fio7$$\"3-+]7oK0e>F0$\"35')))>'fOTm#Fio7$ $\"3-+](=5s#y>F0$\"3%\\;*y%e#G$e#Fio7$F+$\"3++++++++DFio-%'COLOURG6&%$ RGBG$\"#5!\"\"$F*F*F^[l-%+AXESLABELSG6$Q\"x6\"Q!Fc[l-%%VIEWG6$;F(F+%(D EFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curv e 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Alright, we now have that " }{XPPEDIT 18 0 "k[2];" "6#&%\"kG6#\"\"#" }{TEXT -1 6 " from " } {TEXT 267 12 "Formula (2) " }{TEXT -1 74 "is 2. Plugging this and a= 1, b=2, n=10, into the same formula yeilds " }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 45 "MaxError := evalf(((2-1)^3 * 2)/(12*(10)^2)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)MaxErrorG$\"+nmmm;!#7" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 270 23 "Answer to Example (1): " }{TEXT -1 87 "The maximum error in using the trapezoidal metho d with 10 subintervals is 0.001666667. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "# end of example 1" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 268 12 "Example (2) " }}{PARA 0 "" 0 "" {TEXT -1 112 "What is t he minimum number of subintervals that should be used in order to get a trapazoidal approximation to " }{XPPEDIT 18 0 "int(1/x,x = 1 .. 2) " "6#-%$intG6$*&\"\"\"F'%\"xG!\"\"/F(;F'\"\"#" }{TEXT -1 37 " with an \+ error less than or equal to " }{XPPEDIT 18 0 "10^(-12);" "6#)\"#5,$\"# 7!\"\"" }{TEXT -1 2 "? " }}{PARA 0 "" 0 "" {TEXT -1 49 "We'll use the \+ result from the first example that " }{XPPEDIT 18 0 "k[2];" "6#&%\"kG6 #\"\"#" }{TEXT -1 4 " in " }{TEXT 269 12 "Formula (2) " }{TEXT -1 38 " is 2 and set the error bound equal to " }{XPPEDIT 18 0 "10^(-12)" "6#) \"#5,$\"#7!\"\"" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {XPPEDIT 18 0 "abs(E[T]) <= (b-a)^3*k[2]/(12*n^2);" "6 #1-%$absG6#&%\"EG6#%\"TG*(,&%\"bG\"\"\"%\"aG!\"\"\"\"$&%\"kG6#\"\"#F.* &\"#7F.*$%\"nGF5F.F0" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "10^(-12)" "6# )\"#5,$\"#7!\"\"" }{TEXT -1 31 " solving this equation for " } {XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 9 " yields " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "solve( ((2-1)^3 * 2)/(12*n^2) = 10^(-12),n) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,$*$-%%sqrtG6#\"\"'\"\"\"#!'++]\" \"$,$F$#\"'++]F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Now we need t o choose the positive answer and we get " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "n := evalf(500000/3 * sqrt(6));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG$\"+1H[#3%!\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "Since n must be an integer (it is, after all, the number of su bintervals) we choose the first integer above this." }}{PARA 0 "" 0 " " {TEXT -1 1 " " }{TEXT 271 23 "Answer to Example (2): " }{TEXT -1 51 " In order to ensure an error less than or equal to " }{XPPEDIT 18 0 " 10^(-12)" "6#)\"#5,$\"#7!\"\"" }{TEXT -1 81 ", you must use at least \+ 408,249 subintervals in the trapezoidal approximation. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "# end of this section" }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 26 "Assignmen t: Approximating " }{XPPEDIT 18 0 "ln(3);" "6#-%#lnG6#\"\"$" }}{PARA 0 "" 0 "" {TEXT -1 269 "We know that we can type ln(3) into a calculat or or Maple and get an answer. In this assignment, we go through the \+ process of determining this value in much the same way that a calculat or or computer would. First, the Fundemental Theorem of Calculus ass ures us that " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "int(1/x,x = 1 .. 3) = \+ ln(3)-ln(1);" "6#/-%$intG6$*&\"\"\"F(%\"xG!\"\"/F);F(\"\"$,&-%#lnG6#F- F(-F06#F(F*" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "ln(3)-0 = ln(3);" "6#/ ,&-%#lnG6#\"\"$\"\"\"\"\"!!\"\"-F&6#F(" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 30 "Therefore, we can approximate " }{XPPEDIT 18 0 "ln(3 );" "6#-%#lnG6#\"\"$" }{TEXT -1 61 " by numerically integrating the le ft hand definite integral. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 272 11 "Problem (1)" }{TEXT -1 13 " App roximate " }{XPPEDIT 18 0 "ln(3);" "6#-%#lnG6#\"\"$" }{TEXT -1 261 " b y using the midpoint approximation to the definite integral with 50 s ubintervals. What is the maximum error possible with this approximat ion? What is the actual error? (You can find this by assuming Maple \+ gives the exact solution when you type ln(3)). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 273 11 "Problem (2)" }{TEXT -1 13 " Approximate " }{XPPEDIT 18 0 "ln(3);" "6#-%#lnG6#\"\"$" }{TEXT -1 263 " by using the trapezoidal approximation to the definite integ ral with 50 subintervals. What is the maximum error possible with th is approximation? What is the actual error? (You can find this by ass uming Maple gives the exact solution when you type ln(3)). " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 11 "Problem \+ (3)" }{TEXT -1 68 " What is the minimum number of subintervals require d to approximate " }{XPPEDIT 18 0 "ln(3);" "6#-%#lnG6#\"\"$" }{TEXT -1 37 " with an error less than or equal to " }{XPPEDIT 18 0 "10^(-6); " "6#)\"#5,$\"\"'!\"\"" }{TEXT -1 68 " using the midpoint approximati on to the definite integral? Note: " }{XPPEDIT 18 0 "10^(-6)" "6#)\"# 5,$\"\"'!\"\"" }{TEXT -1 14 " = 0.000001. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 11 "Problem (4)" }{TEXT -1 68 " Wh at is the minimum number of subintervals required to approximate " } {XPPEDIT 18 0 "ln(3);" "6#-%#lnG6#\"\"$" }{TEXT -1 38 " with an error \+ less than or equal to " }{XPPEDIT 18 0 "10^(-6)" "6#)\"#5,$\"\"'!\"\" " }{TEXT -1 69 "using the trapazoidal approximation to the definite i ntegral? Note: " }{XPPEDIT 18 0 "10^(-6)" "6#)\"#5,$\"\"'!\"\"" } {TEXT -1 12 " = 0.000001" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 276 11 "Problem (5)" }{TEXT -1 144 " Generalize the infor mation found above and describe which method (trapezoidal or midpoint) does a better job of approximating ln(3) and why. " }}}}{MARK "6" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }