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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 299 
"" 0 "" {TEXT -1 25 "Lesson 20:  Taylor Series" }}}{SECT 0 {PARA 4 "" 
0 "" {TEXT -1 32 "Definition of Taylor polynomials" }}{EXCHG {PARA 0 "
" 0 "Taylor polynomial" {TEXT -1 31 " Suppose the nth derivative of " 
}{XPPEDIT 18 0 "y = f(x)" "6#/%\"yG-%\"fG6#%\"xG" }{TEXT -1 15 " is de
fined at " }{XPPEDIT 18 0 " x = a" "6#/%\"xG%\"aG" }{TEXT -1 16 " . Th
en the nth " }{TEXT 262 17 "Taylor polynomial" }{TEXT -1 10 " for f at
 " }{XPPEDIT 18 0 "x = a" "6#/%\"xG%\"aG" }{TEXT -1 23 " is defined as
 follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "p[n](x) = sum((D@@i)(f)(a)/i!*(x-a)^i,i=0..n);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-&%\"pG6#%\"nG6#%\"xG-%$sumG6$*&*&---
%#@@G6$%\"DG%\"iG6#%\"fG6#%\"aG\"\"\"),&F*F;F:!\"\"F6F;F;-%*factorialG
6#F6F>/F6;\"\"!F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 
262 16 "Taylor remainder" }{TEXT -1 24 " function is defined as " }
{XPPEDIT 19 1 "R[n](x) = abs(f(x) - p[n](x))" "6#/-&%\"RG6#%\"nG6#%\"x
G-%$absG6#,&-%\"fG6#F*\"\"\"-&%\"pG6#F(6#F*!\"\"" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}{PARA 7 "" 1 "" {TEXT -1 32 "Warning, premature en
d of input\n" }}}{EXCHG {PARA 0 "" 0 "taylor" {TEXT -1 17 "There is a \+
word, " }{TEXT 260 6 "taylor" }{TEXT -1 141 ", in the Maple vocabulary
 already which compute Taylor polynomials. Suppose we want the 11 th T
aylor polynomial of the sin function at x = 0." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 29 "p11 := taylor(sin(x),x=0,12);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$p11G+1%\"xG\"\"\"F'#!\"\"\"\"'\"\"$#F'\"$?\"\"
\"&#F)\"%S]\"\"(#F'\"'!)GO\"\"*#F)\")+o\"*R\"#6-%\"OG6#F'\"#7" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 193 "p11 is not actually a polynomial \+
because of the term at the end which is used to signal which polynomia
l is represented (in case some of the coefficients are 0). We can conv
ert to a polynomial." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "p11
 := convert(p11,polynom);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$p11G,.
%\"xG\"\"\"*&#F'\"\"'F'*$)F&\"\"$F'F'!\"\"*&#F'\"$?\"F')F&\"\"&F'F'*&#
F'\"%S]F'*$)F&\"\"(F'F'F.*&#F'\"'!)GOF')F&\"\"*F'F'*&#F'\")+o\"*RF'*$)
F&\"#6F'F'F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "The Taylor polyn
omials are usually good approximations to the function near a. Let's p
lot the first few polynomials for the sin function at x =0. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "sinplot := plot(sin,-Pi..2*P
i,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 217 "tays:= \+
plots[display](sinplot):\nfor i from 1 by 2 to 11 do\ntpl := convert(t
aylor(sin(x), x=0,i),polynom):\ntays := tays,plots[display]([sinplot,p
lot(tpl,x=-Pi..2*Pi,y=-2..2,\ncolor=black,title=convert(tpl,string))])
 od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plots[display]([tay
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45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" "Curve
 6" "Curve 7" "Curve 8" "Curve 9" "Curve 10" "Curve 11" "Curve 12" "Cu
rve 13" }}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 38 "Animation of Taylor \+
series convergence" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 241 "The Taylor \+
Polynomials gradually converge to the Taylor Series which is a represe
ntation of the original function in some interval of convergence. In t
his section, we'll see with our own eyes how this convergence takes pl
ace in an animation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 149 "First we define a function and a generic Taylor polyno
mial. Then we define some constant so that our graph desplays for a =<
 x =< b, and c =<y =< d. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}{PARA 7 "
" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "f := x -> sin(x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG%$sinG" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 43 "on := x -> piecewise( x <0, 0, x < 1, 1,1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#onGR6#%\"xG6\"6$%)operatorG%&arrowG
F(-%*piecewiseG6'29$\"\"!F12F0\"\"\"F3F3F(F(F(" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 32 "a := -6: b := 6: c := -3: d:= 4:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "g := (x,k) -> convert(series(f(x), \+
x, k), polynom):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 522 "displa
y( plot( f(x), x = a..b, y = c..d, thickness = 3, color = blue), anima
te( on(t-1)*g(x,2) , x = a..b, t = 0..7, view = c..d, color = cyan), a
nimate( on(t-2)*g(x,4) , x = a..b, t = 0..7, view = c..d, color = cora
l),animate( on(t-3)*g(x,6) , x = a..b, t = 0..7, view = c..d, color = \+
green), animate( on(t-4)*g(x,8) , x = a..b, t = 0..7, view = c..d, col
or = violet), animate( on(t-5)*g(x,10) , x = a..b, t = 0..7, view = c.
.d, color = red), animate( on(t-6)*g(x,12) , x = a..b, t = 0..7, view \+
= c..d, color = coral) ); " }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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{TEXT -1 24 "Taylor remainder theorem" }}{EXCHG {PARA 297 "" 0 "" 
{TEXT -1 158 "Theorem: (Taylor's remainder theorem) If the (n+1)st der
ivative of f is defined and bounded in absolute value by a number M in
 the interval from a to x, then " }{XPPEDIT 18 0 "R[n](x) <= M/(n+1)!*
abs(x-a)^(n+1)" "6#1-&%\"RG6#%\"nG6#%\"xG*(%\"MG\"\"\"-%*factorialG6#,
&F(F-F-F-!\"\")-%$absG6#,&F*F-%\"aGF2,&F(F-F-F-F-" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 168 "This theorem is essential when you are u
sing Taylor polynomials to approximate functions, because it gives a w
ay of deciding which polynomial to use. Here's an example." }}{PARA 
298 "" 0 "" {TEXT -1 47 "Problem Find the 2nd Taylor polynomial p[2] o
f " }{XPPEDIT 18 0 "f(x) = ln(x)*sin(exp(x))+1" "6#/-%\"fG6#%\"xG,&*&-
%#lnG6#F'\"\"\"-%$sinG6#-%$expG6#F'F-F-F-F-" }{TEXT -1 4 " at " }
{XPPEDIT 18 0 "x = 1" "6#/%\"xG\"\"\"" }{TEXT -1 140 ". Plot both the \+
polynomial and f on the interval [.5,1.5]. Determine the maximum error
 in using p[2] to approximate ln(x) in this interval. " }}{PARA 0 "" 
0 "" {TEXT 266 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 31 "f := x -> ln(x)*sin(exp(x))+1; " }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)operatorG%&arrowGF(,&*&
-%#lnG6#9$\"\"\"-%$sinG6#-%$expGF0F2F2F2F2F(F(F(" }}}{EXCHG {PARA 0 ">
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "p[2] := x -> sum((D@@i)(f)(1
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{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,($\"
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0 "" 0 "eyeball" {TEXT -1 169 "In order to use Taylor's remainder theo
rem, we need to find a bound M on the 3rd derivative of the function f
. In this case, we could just plot the third derivative and " }{TEXT 
262 7 "eyeball" }{TEXT -1 28 " an appropriate value for M." }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 26 " plot((D@@3)(f),.5..1.5) ;" }}{PARA 13 "" 
1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$\"3++++++++
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](=5s#y9F-$\"35y$)o;OJ%o'Fbq7$$\"3++++++++:F-$\"3U\\J'o5mh&pFbq-%'COLO
URG6&%$RGBG$\"#5!\"\"$\"\"!Fa[lF`[l-%+AXESLABELSG6$Q!6\"Fe[l-%%VIEWG6$
;$\"\"&F_[l$\"#:F_[l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
20 "We could use M = 75." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "
M := 75;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG\"#v" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 17 "So the remainder " }{XPPEDIT 18 0 "R2 =  \+
abs(f(x) - p2(x)) " "6#/%#R2G-%$absG6#,&-%\"fG6#%\"xG\"\"\"-%#p2G6#F,!
\"\"" }{TEXT -1 15 " is bounded by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "M/3!*(1.5-1)^3;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#$\"+++]i:!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
125 "We can see from the plot of f and the polynomial that the actual \+
error is never more than about .1 on the interval [.5,1.5]. " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 265 15 "Another example" }{TEXT -1 2 ": \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 298 "" 0 "" {TEXT -1 118 "Wh
ich Taylor polynomial would you use to approximate the sin function on
 the interval from -Pi to Pi to within 1/10^6?" }}{PARA 0 "" 0 "" 
{TEXT -1 11 " Solution: " }}{PARA 0 "" 0 "" {TEXT -1 100 "Well, 1 is a
 bound on any derivative of the sin on any interval. So we need to sol
ve the inequality " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "ineq \+
:= 1/n!*Pi^n <= 1/10^6;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ineqG1*&
)%#PiG%\"nG\"\"\"-%*factorialG6#F)!\"\"#F*\"(+++\"" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 123 "for n. Solve will not be much help here because \+
of the factorial, but we can find the smallest n by running through a \+
loop." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "n := 1: while eval
f(1/n!*Pi^n) > 1/10^6 do n := n+1 od: print (`take n to be `,n);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$%.take~n~to~be~G\"#<" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 35 "(seq(evalf( 1/n!*Pi^n) ,n=15..20));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6($\"+\\``\">#!#9$\"+'fpII%!#:$\"+=S0_z!
#;$\"+]_*yQ\"F+$\"+1H%[H#!#<$\"+23t/O!#=" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 
"t17 := convert(taylor(sin(x),x=0,18),polynom);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$t17G,4%\"xG\"\"\"*&#F'\"\"'F'*$)F&\"\"$F'F'!\"\"*&#F
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "plot(t17,x=-Pi..Pi);" }}{PARA 13 "
" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7en7$$!3*****4
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F37$$\"3!y32s$*Qxc\"F*$\"3-o8)=E`*****F37$$\"3&R90-3LQj\"F*$\"3M#*4z;%
Q,)**F37$$\"35+K?Bs#**p\"F*$\"3W!ygr%=u;**F37$$\"39YK*)Gjak<F*$\"35G>3
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.+s2N#*F37$$\"3kYuyfix%4#F*$\"3=zq$>iZ$e')F37$$\"3T4q))fu.GAF*$\"38P'y
sI2o\"zF37$$\"3y?w'**=&>gBF*$\"3OA9l!3AF/(F37$$\"3!*>3a=Wj\"[#F*$\"3DY
03x&Q38'F37$$\"3A?c*)yu\"3i#F*$\"3O1Ky1q_v\\F37$$\"3-i-HnWIXFF*$\"3A\"
HvYio*fQF37$$\"3u=RWhN.yGF*$\"3\"osVj\"R=0EF37$$\"3ss(*RSA20IF*$\"3AZS
vRu'4O\"F37$$\"3!)***\\/l#fTJF*$\"3!>#pim.$fb#F--%'COLOURG6&%$RGBG$\"#
5!\"\"$\"\"!Fj]lFi]l-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6$;$!+aEfTJ!\"
*$\"+aEfTJFh^l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" }}}}{PARA 5 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 40 "Looks pretty much like the sin function.
" }}}}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 267 8 "Problems" }}{EXCHG 
{PARA 298 "" 0 "" {TEXT -1 20 "Exercise: Show that " }{XPPEDIT 18 0 "c
os(x)" "6#-%$cosG6#%\"xG" }{TEXT -1 41 " is approximated to within 7 d
ecimals by " }{XPPEDIT 18 0 "1-x^2/2!+x^4/4!-x^6/6!+x^8/8!" "6#,,\"\"
\"F$*&%\"xG\"\"#-%*factorialG6#F'!\"\"F+*&F&\"\"%-F)6#F-F+F$*&F&\"\"'-
F)6#F1F+F+*&F&\"\")-F)6#F5F+F$" }{TEXT -1 14 " for all x in " }
{XPPEDIT 18 0 "[-Pi/4,Pi/4]" "6#7$,$*&%#PiG\"\"\"\"\"%!\"\"F)*&F&F'F(F
)" }{TEXT -1 2 " ." }}{PARA 298 "" 0 "convert(..,polynom)" {TEXT -1 
25 "Exercise: Use taylor and " }{TEXT 260 19 "convert(..,polynom)" }
{TEXT -1 187 " to compute and plot, on the interval specified, the fir
st few taylor polynomials of the following functions. Observe the conv
ergence of the polynomials to the function and make comments." }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "f(x)=ln(x)" "6#/-%\"fG6
#%\"xG-%#lnG6#F'" }{TEXT -1 32 " at x=1, on the interval [-1,3]." }
{MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "f(x)=1/(1-x)" "6#/-
%\"fG6#%\"xG*&\"\"\"F),&F)F)F'!\"\"F+" }{TEXT -1 31 " at x =0 on the i
nterval [-2,2]" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "f(x) =  arctan(x)" "6
#/-%\"fG6#%\"xG-%'arctanG6#F'" }{TEXT -1 33 " at x = 0 on the interval
 [-2..2]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 298 "" 0 "" {TEXT 
-1 108 "Exercise: Write a procedure to compute sin(x) for any x by usi
ng p[5]. restricted to the interval [0,Pi/4]. " }}{PARA 0 "" 0 "" 
{TEXT 268 19 "Outline of solution" }{TEXT -1 35 ": If x is negative, r
eplace x with " }{XPPEDIT 18 0 "-x" "6#,$%\"xG!\"\"" }{TEXT -1 21 " an
d use the oddness " }{XPPEDIT 18 0 "sin(x) = -sin(-x)" "6#/-%$sinG6#%
\"xG,$-F%6#,$F'!\"\"F," }{TEXT -1 101 " property. If x is greater than
 or equal to 2*Pi, then replace x with x-2*Pi and use the periodicity \+
" }{XPPEDIT 18 0 "sin(x) = sin(x-2*Pi)" "6#/-%$sinG6#%\"xG-F%6#,&F'\"
\"\"*&\"\"#F+%#PiGF+!\"\"" }{TEXT -1 87 " . Repeat this step until [0,
 2*Pi ). If Pi/4 < x < Pi/2 , then use the trig indentity " }{XPPEDIT 
18 0 "sin(x) = 1-sin(Pi/2-x)" "6#/-%$sinG6#%\"xG,&\"\"\"F)-F%6#,&*&%#P
iGF)\"\"#!\"\"F)F'F0F0" }}{PARA 0 "" 0 "" {TEXT -1 16 "and approximate
 " }{XPPEDIT 18 0 "sin(Pi/2-x)" "6#-%$sinG6#,&*&%#PiG\"\"\"\"\"#!\"\"F
)%\"xGF+" }{TEXT -1 4 " by " }{XPPEDIT 18 0 "p[5](Pi/2-x)" "6#-&%\"pG6
#\"\"&6#,&*&%#PiG\"\"\"\"\"#!\"\"F,%\"xGF." }{TEXT -1 5 ". If " }
{XPPEDIT 18 0 "Pi/2 < x " "6#2*&%#PiG\"\"\"\"\"#!\"\"%\"xG" }{TEXT -1 
5 " and " }{XPPEDIT 18 0 "x< Pi" "6#2%\"xG%#PiG" }{TEXT -1 7 ", then \+
" }{XPPEDIT 18 0 "sin(x) = sin(Pi-x)" "6#/-%$sinG6#%\"xG-F%6#,&%#PiG\"
\"\"F'!\"\"" }{TEXT -1 5 ". If " }{XPPEDIT 18 0 "Pi < x " "6#2%#PiG%\"
xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x< 2*Pi" "6#2%\"xG*&\"\"#\"\"
\"%#PiGF'" }{TEXT -1 7 ", then " }{XPPEDIT 18 0 "sin(x) = - sin(x - Pi
)" "6#/-%$sinG6#%\"xG,$-F%6#,&F'\"\"\"%#PiG!\"\"F." }{TEXT -1 2 ". " }
}}{EXCHG {PARA 0 "" 0 "Digits" {TEXT -1 78 "Exercise: Find the smalles
t n such that the nth Taylor polynomial p[n](x) for " }{XPPEDIT 18 0 "
y = exp(x) " "6#/%\"yG-%$expG6#%\"xG" }{TEXT -1 3 "at " }{XPPEDIT 18 
0 "x = 0 " "6#/%\"xG\"\"!" }{TEXT -1 31 " approximates exp(x) to withi
n " }{XPPEDIT 18 0 "10^(-12) " "6#)\"#5,$\"#7!\"\"" }{TEXT -1 38 "for \+
x in [0,1]. (You will want to set " }{TEXT 260 6 "Digits" }{TEXT -1 
37 " equal to 15 in order to do this one." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 26 "or so to do this problem.)" }}}}{MARK "2 7 1 0" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }