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{SECT 0 {PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 256 41 "Section 5.4: A \+
String in a Viscous Medium" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT 257 30 "Maple Packages for Section 5.4" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 15 "with(PDEtools):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 80 "The variation on the wave equation we consider in this se
ction is the following:" }}{PARA 0 "" 0 "" {TEXT -1 29 " \+
" }{XPPEDIT 18 0 "diff(u,`$`(x,2))-k*diff(u,t)-g = diff
(u,`$`(t,2));" "6#/,(-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#\"\"\"*&%\"kGF.-
F&6$F(%\"tGF.!\"\"%\"gGF4-F&6$F(-F*6$F3F-" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 " \+
u(t, 0) = 0 = u(t, L)" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 59 " \+
u( 0, x) = F(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
" }{XPPEDIT 18 0 "u[t];" "6#&%\"uG6#%\"tG" }
{TEXT -1 17 " ( 0 , x) = G(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 333 "This problem has a physical realization.
It models a string for which there is a downward pull, perhaps gravit
y. Also, there is a force pulling in the opposite direction from the v
elocity. Thus, if the vibration of the string occurs in a viscous medi
um, there is a resistance proportioned to the velocity of the motion o
f the string." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 286 "If you were in a Mathematics office or an Engineering of
fice and stepped out into the hall from your workplace, stopped the fi
rst person you saw, and asked them how to solve this PDE, they would r
espond something such as, \"Separate variables. Isn't that how you do \+
all those problems?\" " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 208 "I say to you: \"Surely. Separation of variables is \+
one of the main ideas of these notes.\" So, let's solve the equation a
s the person we might stop in the hall would. Let's use the techniques
of Fourier Series." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 73 "To be specific, we will get solutions that we can graph
for the equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 26 " " }{XPPEDIT 18 0 "diff(w,`$`(x,
2))-diff(w,t)/5-32 = diff(w,`$`(t,2));" "6#/,(-%%diffG6$%\"wG-%\"$G6$%
\"xG\"\"#\"\"\"*&-F&6$F(%\"tGF.\"\"&!\"\"F4\"#KF4-F&6$F(-F*6$F2F-" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 " \+
w(t, 0) = 0 = w(t, " }{XPPEDIT 18
0 "Pi;" "6#%#PiG" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 61 " \+
w( 0, x) = sin(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
" }{XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG" }
{TEXT -1 14 " ( 0 , x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 151 "Recognize
that this is a non-homogeneous problem. The -32 term makes it so. We \+
must separate out the steady state solution and the transient solution
. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "Th
e steady state solution v(x) is independent of time. Thus, that soluti
on will satisfy the equation" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+
" }{XPPEDIT 18 0 "diff(v(x),`$`(x,2));" "6#-%%diffG6$-%\"vG6#%
\"xG-%\"$G6$F)\"\"#" }{TEXT -1 30 " - 32 = 0, with v(0) = 0 = v(" }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 ""
{TEXT -1 58 "We solve this second order ordinary differential equation
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "dsolve(\{diff(v(x),x,x)
=32,v(0)=0,v(Pi)=0\},v(x));" }}}{PARA 0 "" 0 "" {TEXT -1 42 "We find t
hat the steady state solution is " }{XPPEDIT 18 0 "v(x) = 16*x^2-16*Pi
*x;" "6#/-%\"vG6#%\"xG,&*&\"#;\"\"\"*$F'\"\"#F+F+*(F*F+%#PiGF+F'F+!\"
\"" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "v:=x->
16*x^2-16*Pi*x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Next we s
olve the transient equation. The transient equation is" }}{PARA 0 ""
0 "" {TEXT -1 24 " " }{XPPEDIT 18 0 "diff(w,`$`
(x,2))-diff(w,t)/5 = diff(w,`$`(t,2));" "6#/,&-%%diffG6$%\"wG-%\"$G6$%
\"xG\"\"#\"\"\"*&-F&6$F(%\"tGF.\"\"&!\"\"F4-F&6$F(-F*6$F2F-" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 " \+
w(t, 0) = 0 = w(t, " }{XPPEDIT 18 0 "Pi;
" "6#%#PiG" }{TEXT -1 1 ")" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA
0 "" 0 "" {TEXT -1 68 " w( \+
0, x) = sin(x) - v(x)" }}{PARA 0 "" 0 "" {TEXT -1 44 " \+
" }{XPPEDIT 18 0 "w[t]" "6#&%\"wG6#%\"tG"
}{TEXT -1 14 " ( 0 , x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 119 "How do we solve this transient equation? Just \+
like the hall-walker said: we separate variables. This leads to equati
ons" }}{PARA 0 "" 0 "" {TEXT -1 44 " X''/X = (T'' +
T '/5 )/T." }}{PARA 0 "" 0 "" {TEXT -1 128 "In the usual manner, we ar
rive at two ordinary differential equations, the solutions of which le
ad to the solutions for the PDE:" }}{PARA 0 "" 0 "" {TEXT -1 19 " \+
X '' = " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 20 " \+
X, with X(0) = X( " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 " ) = \+
0" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
T '' + T'/5 = " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }
{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 56 "We've seen the X equation enough to know all solutions: \+
" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 3 " = " }{XPPEDIT
18 0 "-n^2;" "6#,$*$%\"nG\"\"#!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18
0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 10 "(x) = sin(" }{XPPEDIT 18 0 "
n*Pi*x/L;" "6#**%\"nG\"\"\"%#PiGF%%\"xGF%%\"LG!\"\"" }{TEXT -1 10 ") =
sin(n " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT -1 24 "). Check this by \+
asking:" }}{PARA 0 "" 0 "" {TEXT -1 7 "(1) Is " }{XPPEDIT 18 0 "X[n];
" "6#&%\"XG6#%\"nG" }{TEXT -1 5 "'' = " }{XPPEDIT 18 0 "lambda;" "6#%'
lambdaG" }{TEXT -1 2 " " }{XPPEDIT 18 0 "X[n];" "6#&%\"XG6#%\"nG" }
{TEXT -1 2 " ?" }}{PARA 0 "" 0 "" {TEXT -1 7 "(2) Is " }{XPPEDIT 18 0
"X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 15 "(0) = 0 ? Is " }{XPPEDIT 18
0 "X[n];" "6#&%\"XG6#%\"nG" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "Pi;" "6#%#
PiG" }{TEXT -1 7 ") = 0 ?" }}{PARA 0 "" 0 "" {TEXT -1 41 "The answer t
o the three questions is yes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 76 "For each n, we solve the T equation. Be s
ure we know what the T equation is:" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+
T '' + T'/5 = " }{XPPEDIT 18 0 "-n^2;" "6#,$*$%
\"nG\"\"#!\"\"" }{TEXT -1 3 " T." }}{PARA 0 "" 0 "" {TEXT -1 141 "Sinc
e this is a second order equation in T, we expect two constants. These
will be constants that we will determine by some Fourier process. " }
}{PARA 0 "" 0 "" {TEXT -1 94 " How many terms shall we do? Let's c
all the number of terms we choose to do by the name N." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "N:=5;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 89 "for n from 1 to N do\n dsolve(diff(T(t),t,t)+diff(T
(t),t)/5=-n^2*T(t),T(t));\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 109 "It would be nice to
have a good formula for those trig terms. I need a formula that repre
sents this sequence:" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+
" }{XPPEDIT 18 0 "sqrt(99),sqrt(399),sqrt(899),sqrt(1599),sqrt(2499);
" "6'-%%sqrtG6#\"#**-F$6#\"$*R-F$6#\"$**)-F$6#\"%*f\"-F$6#\"%*\\#" }
{TEXT -1 7 ", ... ." }}{PARA 0 "" 0 "" {TEXT -1 3 " " }}{PARA 0 ""
0 "" {TEXT -1 237 "The question is: how can I write that multiple of t
inside those sine or cosine function as a function of n. (Such questi
ons are often on IQ tests.) Here's a way to find the answer. I'll solv
e the equation again without saying what n is." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 53 "dsolve(diff(T(t),t,t)+diff(T(t),t)/5=-n^2*T(t),T
(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 133 "It is clear what the terms are now, isn't it? Using th
e Euler identity relating exponential and trig functions, each T is of
the form" }}{PARA 0 "" 0 "" {TEXT -1 13 " " }{XPPEDIT 18
0 "T[n](t) = A[n]*exp(-t/10)*cos(t*sqrt(100*n^2-1)/10)+B[n]*exp(-t/10)
*sin(t*sqrt(100*n^2-1)/10);" "6#/-&%\"TG6#%\"nG6#%\"tG,&*(&%\"AG6#F(\"
\"\"-%$expG6#,$*&F*F0\"#5!\"\"F7F0-%$cosG6#*(F*F0-%%sqrtG6#,&*&\"$+\"F
0*$F(\"\"#F0F0F0F7F0F6F7F0F0*(&%\"BG6#F(F0-F26#,$*&F*F0F6F7F7F0-%$sinG
6#*(F*F0-F=6#,&*&FAF0*$F(FCF0F0F0F7F0F6F7F0F0" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 68 "I can now make the general solution for t
he PDE. I hope you can too." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "w:=(t,x)->\n sum(A[n]*exp(-t/10)*c
os(t*sqrt(100*n^2-1)/10)*sin(n*x),n=1..N)+\n sum(B[n]*exp(-t/10)*sin(t
*sqrt(100*n^2-1)/10)*sin(n*x),n=1..N);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 162 "We check that this \+
is a general solution. Ask: does it satisfy the PDE and the boundary c
onditions? (The initial conditions will determine the values of A and \+
B.)" }}{PARA 0 "" 0 "" {TEXT -1 8 "The PDE:" }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 61 "simplify(diff(w(t,x),x,x)-diff(w(t,x),t)/5-diff(w(t
,x),t,t));" }}}{PARA 0 "" 0 "" {TEXT -1 20 "Boundary conditions:" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "w(t,0); w(t,Pi);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 57 "It
remains to determine the A's and B's. Observe what is " }{XPPEDIT 18
0 "u(0,x);" "6#-%\"uG6$\"\"!%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0
"u[t](0,x);" "6#-&%\"uG6#%\"tG6$\"\"!%\"xG" }{TEXT -1 1 "." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "w(0,x);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 13 "D[1](w)(0,x);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 101 "DerivTerm:=collect(collect(collect(collect(collect(%
,sin(x)),sin(2*x)),sin(3*x)),sin(4*x)),sin(5*x));" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 247 "It seems c
lear that we can get the A's through a simple Fourier Series. Having t
he A's we can get the B's. We do that here. This is the first time in \+
this problem to use the initial conditions.\n\nRecall that the solutio
n for the original problem is" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+
u(t, x) = v(x) + w(t, x)." }}{PARA 0 "" 0 "" {TEXT -1 42 "Thus, u(0,
x) = v(x) + w(0, x) and " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diff
G6$%\"uG%\"tG" }{TEXT -1 8 "(0,x) = " }{XPPEDIT 18 0 "diff(w,t);" "6#-
%%diffG6$%\"wG%\"tG" }{TEXT -1 6 "(0,x)." }}{PARA 0 "" 0 "" {TEXT -1
17 "This means that " }}{PARA 0 "" 0 "" {TEXT -1 3 " " }{XPPEDIT
18 0 "A[n] = int((f(x)-v(x))*sin(n*x),x)/int(sin(n*x)^2,x);" "6#/&%\"A
G6#%\"nG*&-%$intG6$*&,&-%\"fG6#%\"xG\"\"\"-%\"vG6#F1!\"\"F2-%$sinG6#*&
F'F2F1F2F2F1F2-F*6$*$-F86#*&F'F2F1F2\"\"#F1F6" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 " " }
{XPPEDIT 18 0 "(sqrt(100*n^2-1)*B[n]-A[n])/10;" "6#*&,&*&-%%sqrtG6#,&*
&\"$+\"\"\"\"*$%\"nG\"\"#F,F,F,!\"\"F,&%\"BG6#F.F,F,&%\"AG6#F.F0F,\"#5
F0" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "int(g(x)*sin(n*x),x)/int(sin(n
*x)^2,x());" "6#*&-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6#*&%\"nGF,F+F,
F,F+F,-F%6$*$-F.6#*&F1F,F+F,\"\"#-F+6\"!\"\"" }{TEXT -1 2 " ." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "We comput
e" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "fmv:=x->sin(x)-v(x);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "for n from 1 to N do\n A
[n]:=int(fmv(x)*sin(n*x),x=0..Pi)/int(sin(n*x)^2,x=0..Pi);\nod;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g:=x->0;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
136 "for n from 1 to N do\n B[n]:=(10*int(g(x)*sin(n*x),x=0..Pi)/int
(sin(n*x)^2,x=0..Pi)+A[n])\n /sqrt(100*n^2-1);\nod;\nn:
='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 105 "We now have the transient solution completely determin
ed. We create the solution to the original problem:" }}{PARA 0 "" 0 "
" {TEXT -1 53 " u(t,x) = w(t,x) + v(x)."
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "u:=(t,x)->w(t,x)+v(x);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 97 "It seems appropriate to check that this is really the solution.
We check the boundary conditions." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 16 "u(t,0); u(t,Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 17 "We \+
check the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "simplify(d
iff(u(t,x),x,x)-diff(u(t,x),t)/5-32-diff(u(t,x),t,t));" }}}{PARA 0 ""
0 "" {TEXT -1 107 "We check the initial conditions. For each one, we d
raw graphs to see how close we are to the initial value." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "plot([sin(x),u(0,x)],x=0..Pi,color=
[black,red]);" }}}{PARA 0 "" 0 "" {TEXT -1 107 "We draw a graph to see
how close the initial velocity is to zero. Off-set the graph so that \+
it can be seen." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plot(eval
(subs(t=0,diff(u(t,x),t)))+0.001,x=0..Pi,y=-1..1);" }}}{PARA 0 "" 0 "
" {TEXT -1 261 "We now show an animation. Can you predict the movement
? It should be a vibrating string; the first derivative term is a reta
rding force so the oscillations should decrease in size; and the downw
ard force should pull the string down to that hanging, steady state" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "with(plots):\nanimate(u(t,x
),x=0..Pi,t=0..20);" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 371 "In this Section, we have made the physi
cal model more interesting. We assumed that the string lies in a visco
us medium and is subject to a constant force pulling it downward. This
made the partial differential equation more interesting. There were m
ore terms to consider than the simple wave equation. To solve the prob
lem, we used the methods of separation of variables." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT 258 16 "Unassisted Map
le" }}{PARA 0 "" 0 "" {TEXT -1 267 "Because the problem of this sectio
n has been a wave equation problem and because we did not use d'Alembe
rt's techniques, we wonder what Maple will suggest when unassisted. Wi
ll it suggest a variation on d'Alembert's Method, or will it suggest s
eparation of variables?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 102 "Note that at the beginning, we called up the packag
e PDEtools. We make a generic u and define the PDE." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 7 "u:='u';" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 56 "PDE:=diff(u(t,x),x,x)-diff(u(t,x),t,t)-diff(u(t,x),t)
=0;" }}}{PARA 0 "" 0 "" {TEXT -1 49 "Look to see what form Maple gets \+
for a solutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "ans := p
dsolve(PDE);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "build(ans);
" }}}{PARA 0 "" 0 "" {TEXT -1 297 "This process provides solutions as \+
exponentials. Most important, it suggests that one can obtain a soluti
on to this problem by performing a d'Alembert type transformation to g
et a different second order partial differential equation. Going in th
ese directions might be pursued at a different time." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 50 "EMAIL: herod@math.gatech.edu or jhero
d@tds.net" }}{PARA 0 "" 0 "" {TEXT -1 38 "URL: http://www.math.gatech.
edu/~herod" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT
-1 36 "Copyright \251 2003 by James V. Herod" }}{PARA 256 "" 0 ""
{TEXT -1 19 "All rights reserved" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
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33 1 1 }