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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 46 "Experiments on Limits and
 Linear Approximation" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "        \+
                                                     Version .7 Maple \+
VR7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "These packages need to be \+
loaded first." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "with(plots): with(
linalg): with(plottools):" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 22 " Li
mits and Graphs in " }{XPPEDIT 18 0 "R^3;" "6#*$)%\"RG\"\"$\"\"\"" }
{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Look at the graph \+
below. Perhaps try changing the value of eps and rotating the graph." 
}}{PARA 0 "" 0 "" {TEXT 279 63 "What does this say about the limit as \+
(x,y) approaches (0,0) of" }{TEXT -1 1 " " }{XPPEDIT 18 0 "sin((x-y)^2
)/(x^2+y^2);" "6#*&-%$sinG6#*$),&%\"xG\"\"\"%\"yG!\"\"\"\"#F+F+,&*$)F*
F.F+F+*$)F,F.F+F+F-" }{TEXT -1 1 "?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "eps:=.5;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot3d(sin((x-y)^2)/
(x^2+y^2),x=-eps..eps,y=-eps..eps,axes=framed);" }}}{PARA 3 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "The animation belo
w shows the y= kx sections of this graph for various values of k." }}
{PARA 0 "" 0 "" {TEXT -1 87 "To play the animation, click on the graph
 and then choose play from the animation menu." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 272 64 "Why does the animation tell you something
 about the above limit?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "animate(
subs(y=k*x,sin((x-y)^2)/(x^2+y^2)),x=-eps..eps,k=-3..3);" }}}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 44 "Some Linear Approximations for a function
 F:" }{XPPEDIT 18 0 "R^2;" "6#*$)%\"RG\"\"#\"\"\"" }{TEXT -1 2 "->" }
{XPPEDIT 18 0 "R^2;" "6#*$)%\"RG\"\"#\"\"\"" }{TEXT -1 1 "." }}{EXCHG 
{PARA 0 "" 0 "" {TEXT 256 147 "Try changing the scale below - at least
 for the values scale =.1,.5 and 1. \n(You need to re-execute the line
s which follow after you change scale.)" }}{PARA 0 "" 0 "" {TEXT -1 
76 "The pictures compare the image of a certain square under the funct
ion F with" }}{PARA 0 "" 0 "" {TEXT -1 80 "the image of the square und
er the linear approximation obtained at the center of" }}{PARA 0 "" 0 
"" {TEXT -1 12 "the square.\n" }{TEXT 258 12 "Compare the " }{TEXT 
257 8 "accuracy" }{TEXT 259 68 " of linear approximation as a function
 of the size of the rectangle." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 
"Defining F:" }{XPPEDIT 18 0 "R^2;" "6#*$)%\"RG\"\"#\"\"\"" }{TEXT -1 
2 "->" }{XPPEDIT 18 0 "R^2;" "6#*$)%\"RG\"\"#\"\"\"" }{TEXT -1 4 " by \+
" }{XPPEDIT 18 0 "F(x,y) = (exp(x)+y, exp(y)+x);" "6#/-%\"FG6$%\"xG%\"
yG6$,&-%$expG6#F'\"\"\"F(F.,&-F,6#F(F.F'F." }{TEXT -1 42 ". DF is its \+
derivative at the point (x,y)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "F
:=(x,y)->[1-exp(x)-y,1-exp(y)-x]; DF:=jacobian(F(x,y),[x,y]);\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "We will be linearizing F at the p
oint p0. And scale controls the size of the rectangle we work with." }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "scale:=1;\np0:=[.5*scale,.5*scale]
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "An easy way to draw a square
 with corner (0,0) and (scale,scale)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 87 "gr1:=conformal(z,z=0..scale*(1+I)):\ngr1a:=conformal(z,z=0..scal
e*(1+I),color=magenta):\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "This
 picture shows the original rectangle together with its image under F.
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "F_gr:=transform(F):\ndisplay([g
r1,F_gr(gr1)],title=`Image of Rectangle under F`);\n" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 121 "Below is the corresponding picture under the l
inearization of F at p0.\nDon't worry necessarily about the Maple code
, but " }}{PARA 0 "" 0 "" {TEXT -1 88 "     F0= the image of the point
 p0 under F.\n     L=derivative DF at p0 (a 2 by 2 matrix)" }}{PARA 0 
"" 0 "" {TEXT -1 43 "     L1=the linear approximation of F at p0" }}
{PARA 0 "" 0 "" {TEXT -1 55 "     L2=a different form of L1 for Maple \+
syntax reasons" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 251 "F0:=F(op(p0));\n
L:=map(evalf,subs(\{x=p0[1],y=p0[2]\},eval(DF)));\nL1:=convert(evalm(L
 &*vector([x-p0[1],y-p0[2]])+F0),list);\nL2:=unapply(L1,[x,y]);\nL2_gr
:=transform(L2):\ndisplay([gr1a,L2_gr(gr1a)],title=`Image of Rectangle
 under Linear Approximation`);\n\n" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 28 "Linearization at the Origin." }}{EXCHG {PARA 0 "" 0 "" {TEXT 
264 75 "Try changing the scale below - at least for the values scale =
.1,.5 and 1. " }}{PARA 0 "" 0 "" {TEXT 268 52 "Why do the linearized p
ictures look the way they do?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "
Defining F:" }{XPPEDIT 18 0 "R^2;" "6#*$%\"RG\"\"#" }{TEXT -1 2 "->" }
{XPPEDIT 18 0 "R^2;" "6#*$%\"RG\"\"#" }{TEXT -1 4 " by " }{XPPEDIT 18 
0 "F(x,y) = (exp(x)+y, exp(y)+x);" "6#/-%\"FG6$%\"xG%\"yG6$,&-%$expG6#
F'\"\"\"F(F.,&-F,6#F(F.F'F." }{TEXT -1 42 ". DF is its derivative at t
he point (x,y)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "F:=(x,y)->[1-exp
(x)-y,1-exp(y)-x]; DF:=jacobian(F(x,y),[x,y]);\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 100 "We will be linearizing F at the point p0. And sca
le controls the size of the rectangle we work with." }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 22 "scale:=0.1;\np0:=[0,0];" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 65 "An easy way to draw a square with corner (0,0) and (sca
le,scale)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "gr1:=conformal(z,z=0.
.scale*(1+I)):\ngr1a:=conformal(z,z=0..scale*(1+I),color=magenta):\n" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "This picture shows the original
 rectangle together with its image under F." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 81 "F_gr:=transform(F):\ndisplay([gr1,F_gr(gr1)],title=`I
mage of Rectangle under F`);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
121 "Below is the corresponding picture under the linearization of F a
t p0.\nDon't worry necessarily about the Maple code, but " }}{PARA 0 "
" 0 "" {TEXT -1 88 "     F0= the image of the point p0 under F.\n     \+
L=derivative DF at p0 (a 2 by 2 matrix)" }}{PARA 0 "" 0 "" {TEXT -1 
43 "     L1=the linear approximation of F at p0" }}{PARA 0 "" 0 "" 
{TEXT -1 55 "     L2=a different form of L1 for Maple syntax reasons" 
}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 251 "F0:=F(op(p0));\nL:=map(evalf,sub
s(\{x=p0[1],y=p0[2]\},eval(DF)));\nL1:=convert(evalm(L &*vector([x-p0[
1],y-p0[2]])+F0),list);\nL2:=unapply(L1,[x,y]);\nL2_gr:=transform(L2):
\ndisplay([gr1a,L2_gr(gr1a)],title=`Image of Rectangle under Linear Ap
proximation`);\n\n" }}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 
3 "" 0 "" {TEXT -1 19 "Another function F:" }{XPPEDIT 18 0 "R^2;" "6#*
$%\"RG\"\"#" }{TEXT -1 2 "->" }{XPPEDIT 18 0 "R^2;" "6#*$%\"RG\"\"#" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "" 0 "" {TEXT 260 150 "Try changing th
e scale below - at least for the values scale =1, .5, .1, .01. \n(You \+
need to re-execute the lines which follow after you change scale.)" }}
{PARA 0 "" 0 "" {TEXT -1 76 "The pictures compare the image of a certa
in square under the function F with" }}{PARA 0 "" 0 "" {TEXT -1 80 "th
e image of the square under the linear approximation obtained at the c
enter of" }}{PARA 0 "" 0 "" {TEXT -1 12 "the square.\n" }{TEXT 262 12 
"Compare the " }{TEXT 261 8 "accuracy" }{TEXT 263 68 " of linear appro
ximation as a function of the size of the rectangle." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 55 "F:=(x,y)->[x^2-y^2,2*x*y]; DF:=jacobian(F(x,y),[x,y]);\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "We will be linearizing F at the p
oint p0. And scale controls the size of the rectangle we work with." }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "scale:=1;\np0:=[.5*scale,.5*scale]
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "An easy way to draw a square
 with corner (0,0) and (-scale,-scale)." }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 89 "gr1:=conformal(z,z=0..scale*(-1-I)):\ngr1a:=conformal(z,z=0..s
cale*(-1-I),color=magenta):\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "
This picture shows the original rectangle together with its image unde
r F." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "F_gr:=transform(F):\ndispla
y([gr1,F_gr(gr1)],title=`Image of Rectangle under F`);\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 121 "Below is the corresponding picture under
 the linearization of F at p0.\nDon't worry necessarily about the Mapl
e code, but " }}{PARA 0 "" 0 "" {TEXT -1 88 "     F0= the image of the
 point p0 under F.\n     L=derivative DF at p0 (a 2 by 2 matrix)" }}
{PARA 0 "" 0 "" {TEXT -1 43 "     L1=the linear approximation of F at \+
p0" }}{PARA 0 "" 0 "" {TEXT -1 55 "     L2=a different form of L1 for \+
Maple syntax reasons" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 251 "F0:=F(op(p
0));\nL:=map(evalf,subs(\{x=p0[1],y=p0[2]\},eval(DF)));\nL1:=convert(e
valm(L &*vector([x-p0[1],y-p0[2]])+F0),list);\nL2:=unapply(L1,[x,y]);
\nL2_gr:=transform(L2):\ndisplay([gr1a,L2_gr(gr1a)],title=`Image of Re
ctangle under Linear Approximation`);\n\n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 44 "Some Line
ar Approximations for a function F:" }{XPPEDIT 18 0 "R^2;" "6#*$)%\"RG
\"\"#\"\"\"" }{TEXT -1 2 "->" }{XPPEDIT 18 0 "R^3;" "6#*$)%\"RG\"\"$\"
\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 76 "The pictures comp
are the image of a certain square under the function F with" }}{PARA 
0 "" 0 "" {TEXT -1 80 "the image of the square under the linear approx
imation obtained at the center of" }}{PARA 0 "" 0 "" {TEXT -1 12 "the \+
square.\n" }}{PARA 0 "" 0 "" {TEXT 277 86 "Try changing the scale belo
w - at least for the values scale =2, 1, .5, .25,  and .1. " }{TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 12 "Compare the " }{TEXT 274 8 "ac
curacy" }{TEXT 276 68 " of linear approximation as a function of the s
ize of the rectangle." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "F:=
(x,y)->[sin(x)*cos(y),sin(x)*sin(y),cos(x)]; DF:=jacobian(F(x,y),[x,y]
);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "We will be linearizing F
 at the point p0. And scale controls the size of the rectangle we work
 with." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "scale:=1;\np0:=[.5*scale,
.5*scale];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "An easy way to draw
 a square with corner (0,0) and (scale,scale)." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 102 "gr1:=conformal(z,z=0..scale*(1+I)):\ngr1a:=conformal
(z,z=0..scale*(+1+I),color=magenta):\ndisplay(gr1);\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 37 "This picture shows the image under F." }}
{PARA 0 "" 0 "" {TEXT -1 83 "The option scaling=constrained means to c
hoose the same scale along all three axes." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 107 "F_gr:=transform(F):\ndisplay(F_gr(gr1),title=`Image \+
of Rectangle under F`,axes=framed,scaling=constrained);\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 117 "Below is the picture of the image of the
 rectangle under the linearization of F at p0 together with the actual
 image." }}{PARA 0 "" 0 "" {TEXT -1 50 "Don't worry necessarily about \+
the Maple code, but " }}{PARA 0 "" 0 "" {TEXT -1 88 "     F0= the imag
e of the point p0 under F.\n     L=derivative DF at p0 (a 2 by 2 matri
x)" }}{PARA 0 "" 0 "" {TEXT -1 43 "     L1=the linear approximation of
 F at p0" }}{PARA 0 "" 0 "" {TEXT -1 55 "     L2=a different form of L
1 for Maple syntax reasons" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 277 "F0:=
F(op(p0));\nL:=map(evalf,subs(\{x=p0[1],y=p0[2]\},eval(DF)));\nL1:=con
vert(evalm(L &*vector([x-p0[1],y-p0[2]])+F0),list);\nL2:=unapply(L1,[x
,y]);\nL2_gr:=transform(L2):\ndisplay([F_gr(gr1),L2_gr(gr1a)],title=`I
mage and its Linear Approximation`,axes=framed,scaling=constrained);\n
\n" }}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 256 "" 0 "" 
{TEXT -1 68 "You could also try the above without the scaling=constrai
ned option." }}{PARA 257 "" 0 "" {TEXT -1 34 "Can you explain what hap
pens then?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1
 0 0" 62 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 
}