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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 442 "MAPLE WORKSHEET FOR THE A
NIMATED DISPLAY OF TAYLOR POLYNOMIALS\n\nBased on a posting to sci.mat
h.symbolic on 15 February 1994 by Helmer Aslaksen (mathelmr@nusunix.nu
s.sg)\n\nModified by Douglas Meade (meade@math.scarolina.edu) to work \+
for an arbitrary function (not just sin).\n\nEXAMPLES:\n  taylor_anima
te( sin(x), x=0, x=-5..5, frames=10 );\n  taylor_animate( exp(x), x=2,
 x= 0..5, frames=5 );\n  taylor_animate( ln(x), x=1, x=0.1..3, frames=
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be? (b) what do you think the 'limit' of this power series will be?" }
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Y5Fbo7$FB$!+(\\j%\\\"*F47$FE$!+DJ(*zzF47$FH$!+**>nHpF47$FK$!+T;zxfF47$
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\"+jo1,IF47$Fip$\"+3$*zkLF47$F\\q$\"+\"Rxer$F47$F_q$\"+TIKbSF47$Faq$\"
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pHcF47$F`r$\"+_(G,%fF47$Fcr$\"+$3-JE'F47$Ffr$\"+%[f1h'F47$Fir$\"+eg=+q
F47$F\\s$\"+1#\\jX(F47$F^s$\"+Qpd8!)F47$Fas$\"+QC<>()F47$Fds$\"+k?=P'*
F47$Fgs$\"+zZJ&3\"Fbo7$Fjs$\"+f*))zC\"Fbo7$F]t$\"+fyWm9Fbo7$F`t$\"+Iyl
f<Fbo7$Fct$\"+WZ$=:#Fbo7$Fft$\"+`;htEFbo7$Fit$\"+4oSjLFboFjtFau-%+AXES
LABELSG6$%\"xG%!G-%%VIEWG6$;F,$\"#DF.%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 
2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "Example: \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "taylor((x)^(1/3),x=8,3)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++,&%\"xG\"\"\"\"\")!\"\"*$)F'#F&
\"\"$F&\"\"!,$*&\"#CF(F'F+F&F&,$*&\"$w&F(F'F+F(\"\"#-%\"OG6#F&F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "taylor(x^(1/3),x=a,3);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#++,&%\"xG\"\"\"%\"aG!\"\"*$)F'#F&\"\"$
F&\"\"!,$*&F&F&*&F,F&)F'#\"\"#F,F&F(F&F&,$*&F&F&*&\"\"*F&)F'#\"\"&F,F&
F(F(F3-%\"OG6#F&F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "p2:=c
onvert(%,polynom);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#p2G,(*$)%\"aG
#\"\"\"\"\"$F*F**(F+!\"\"F(#!\"#F+,&%\"xGF*F(F-F*F**(\"\"*F-F(#!\"&F+F
0\"\"#F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "P2:=proc(x,a) a
^(1/3)+(x-a)/(3*a^(2/3))-(x-a)^2/(9*a^(5/3)) end; " }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#P2Gf*6$%\"xG%\"aG6\"F)F),(*$)9%#\"\"\"\"\"$F/F/*&F
.F/*&,&9$F/F-!\"\"F/F-#!\"#F0F/F/*&#F/\"\"*F/*&F3\"\"#F-#!\"&F0F/F5F)F
)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "P2(x,a);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#,(*$)%\"aG#\"\"\"\"\"$F(F(*(F)!\"\"F&#!\"#F),&%
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" {MPLTEXT 1 0 8 "P2(x,8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)\"
\")#\"\"\"\"\"$F(F(*(\"#C!\"\",&%\"xGF(F&F,F(F&F'F(*(\"$w&F,F-\"\"#F&F
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{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrow
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0 31 "R2:=proc(x,a) f(x)-P2(x,a) end;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R2Gf*6$%\"xG%\"aG6\"F)F),&
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{MPLTEXT 1 0 8 "R2(x,8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**$)%\"xG
#\"\"\"\"\"$F(F(*$)\"\")F'F(!\"\"*(\"#CF-,&F&F(F,F-F(F,F'F-*(\"$w&F-F0
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(abs(R2(x,8)),x=7..9,y=0..0.000
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G6$7Y7$$\"\"(\"\"!$\"3Y4^b?nhKE!#@7$$\"3Rmm;arz@q!#<$\"3s[gCLWAfCF-7$$
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&zF1$\"3yv/6))3a$)>!#D7$$\"3KLL$3#G,**zF1$\"3)y'>k*fP)>B!#I7$$\"3sKLez
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$\"3OuQK25b:6F-7$$\"3Jnm;9@BM))F1$\"3U+ZtP;\\48F-7$$\"3#RLLLbdQ())F1$
\"3BmD0?4_+:F-7$$\"30++DOl5;*)F1$\"3.BKd0AFB<F-7$$\"3r)*****p`KO*)F1$
\"3#\\#zpo@2P=F-7$$\"3;***\\P?Wl&*)F1$\"31'fy;Rac&>F-7$$\"3e**\\(=5s#y
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5!\"\"$F*F*Fc]l-%+AXESLABELSG6$Q\"x6\"Q\"yFh]l-%%VIEWG6$;F(Fh\\l;Fc]l$
\"\"%!\"%" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Cu
rve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "Remark: Therefore, we \+
see that |R2(x,8)|<0.0003, when x is in [7,9]. Since the highest point
 of happens at x=7, to be exact, we get the following:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "evalf(abs(R2(7,8)));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"'hKE!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "diff(R2(x,8),x$3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#,$*(\"#5\"\"\"\"#F!\"\"%\"xG#!\")\"\"$F&" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 161 "Now, let's verify if the Remainder's theorem holds, i.e.
 If we pick any point x not equal to x0=8, then we can find a c in the
 interval (x,x0) or (x0,x) so that " }}{PARA 258 "" 0 "" {TEXT -1 29 "
R2(x)=(R2'''(c)/3!)*(x-x0)^3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 48 "g:=proc(a,x) (diff(R2(x,8),x$3))/3!*(a-8)^3 end;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"gGf*6$%\"aG%\"xG6\"F)F)*(-%%diffG6$-%#R2G6$9%
\"\")-%\"$G6$F1\"\"$\"\"\"-%*factorialG6#F6!\"\",&9$F7F2F;F6F)F)F)" }}
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F(F,F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Let's solve the zeros \+
for the equation below. Assume x=7.5 (We can pick any point x as long \+
as x is not equal to 8)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 
"fsolve(R2(7.5,8)=g(7.5,x),x,7..9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#$\"+qs\"G(y!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot(\{
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$FfoFe[l7$F[pFe[l7$F`pFe[l7$FepFe[l7$FjpFe[l7$F_qFe[l7$FdqFe[l7$FiqFe[
l7$F^rFe[l7$FcrFe[l7$FhrFe[l7$F]sFe[l7$FbsFe[l7$FgsFe[l7$F\\tFe[l7$Fat
Fe[l7$FftFe[l7$F[uFe[l7$F`uFe[l7$FeuFe[l7$FjuFe[l7$F_vFe[l7$FdvFe[l7$F
ivFe[l7$F^wFe[l7$FcwFe[l7$FhwFe[l7$F]xFe[l7$FbxFe[l7$FgxFe[l7$F\\yFe[l
7$FayFe[l7$FfyFe[l7$F[zFe[l7$F`zFe[l7$FezFe[l-Fjz6&F\\[lF`[lF][lF`[l-%
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LTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1
" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "We can rewrite di
ff(R2(x,8),$3) as function  F  below and write  (R2(x,8)*3!)/(x-8)^3 a
s function G below. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "F:=x->diff(R2(x,8),x$3);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FGf*6#%\"xG6\"6$%)operatorG%&arrow
GF(-%%diffG6$-%#R2G6$9$\"\")-%\"$G6$F2\"\"$F(F(F(" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 5 "F(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*(
\"#5\"\"\"\"#F!\"\"%\"xG#!\")\"\"$F&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "G:=x->R2(x,8)*3!/(x-8)^3;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"GGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*(-%#R2G6$9$\"
\")\"\"\"-%*factorialG6#\"\"$F2,&F0F2F1!\"\"!\"$F(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "G(x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$*(\"\"'\"\"\",**$)%\"xG#F&\"\"$F&F&*$)\"\")F+F&!\"\"*(\"#CF0,&
F*F&F/F0F&F/F+F0*(\"$w&F0F3\"\"#F/F+F&F&F3!\"$F&" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 34 "evalf(subs(x=7,F(x)));evalf(G(7));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+\\tdl?!#7" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"'dz:!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "The green is
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\{F(x),G(x)\},x=7..9,thickness=2);" }}{PARA 13 "" 1 "" {GLPLOT2D 287 
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w$\"3mwd@eL1:7F-7$Fcw$\"3+t!eZ-%f*>\"F-7$Fhw$\"3?<_%pt(o$=\"F-7$F]x$\"
3MWV^q4ko6F-7$Fbx$\"3aC-BST_`6F-7$Fgx$\"3pQYxenzQ6F-7$F\\y$\"3\")zhs63
\\D6F-7$Fay$\"302)Q>>-06\"F-7$Ffy$\"3!R'H!zuFt4\"F-7$F[z$\"3O1Zd0m^$3
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[l-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$Q\"x6\"Q!Fael-%%VIEWG6$;F(Fez%(D
EFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curv
e 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 465 "So The Largran
ge Remainder Theorem actually says that if we pick any point  x  in (7
,9) except x=8, we can find a c so that G(x)=F(c). Since the green is \+
F, the red is G, we start with any point (x,y) on G (red). Then If x<8
, then we can find the corresponding point (by going horizontal direct
ion) on F so that F(c)=G(x). Similarly, if x>8, we start with any poin
t (x,y) on G, we can find the corresponding point on F so that F(c)=G(
x). So if we pick x=7.5, we see" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(G(7.5));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"'$*4:!\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "fsolve(F(x)=.150993e-2,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+J-\"G(y!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}}{MARK "49 1 0" 0 }{VIEWOPTS 1 1 0 3 2 1804 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }