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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT 256 8 "Calculus" }{TEXT 271 37 "
\n\nMathematics and Modeling with Maple" }}{PARA 19 "" 0 "" {TEXT -1 
28 "\nDaniel D. Warner\n\nFall 2000" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT 
-1 15 "Modeling Change" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 18 "Drugs i
n the Body\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 19 "Novocaine Example
 \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 20 "Initial Condition:  " }{XPPEDIT 18 0 "u[0
] = 500" "6#/&%\"uG6#\"\"!\"$+&" }{TEXT -1 20 ".\nRecursion Formula:" 
}{XPPEDIT 18 0 "u[n] = u[n-1]-.2*u[n-1];" "6#/&%\"uG6#%\"nG,&&F%6#,&F'
\"\"\"\"\"\"!\"\"F,*&$\"\"#!\"\"F,&F%6#,&F'F,\"\"\"F.F,F." }{TEXT -1 
22 ".\nDefine the function " }{XPPEDIT 18 0 "u(n);" "6#-%\"uG6#%\"nG" 
}{TEXT -1 60 "  directly from the difference equation.  \nThis is call
ed a " }{TEXT 257 9 "recursive" }{TEXT -1 13 " definition.\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "u := proc(n) if n<=0 then 50
0 else u(n-1) - 0.2*u(n-1) fi end;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 21 "\nWe can now evaluate " }{XPPEDIT 18 0 "u;" "6#%\"uG" }{TEXT 
-1 23 " for various values on " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT 
-1 20 " simply by entering " }{XPPEDIT 18 0 "u(n);" "6#-%\"uG6#%\"nG" 
}{TEXT -1 6 ".    \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "u(0
); u(1); u(2); u(3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "\nWe can \+
generate a list of points for " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT 
-1 50 " taking on the values from 0 to 25 by  entering .\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "[seq([n,u(n)], n=0..25)];" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 152 "\nWe can avoid printing out the e
ntire list, but save it for future use by assigning the list to a vari
able and by terminating the command with a colon.\n" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 32 "pts := [seq([n,u(n)], n=0..25)]:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "\nWe can plot the list of points u
sing the plot command.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 
"plot(pts,style=point);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "\nUpon
 reflection, we see that we can write " }{XPPEDIT 18 0 "u(n);" "6#-%\"
uG6#%\"nG" }{TEXT -1 6 " in a " }{TEXT 258 11 "closed form" }{TEXT -1 
4 " as " }{XPPEDIT 18 0 "u(n) = 500*.8^n;" "6#/-%\"uG6#%\"nG*&\"$+&\"
\"\")$\"\")!\"\"F'F*" }{TEXT -1 38 ".  \nWhat happens if we simply rep
lace " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 4 " by " }{XPPEDIT 18 
0 "t;" "6#%\"tG" }{TEXT -1 9 " and let " }{XPPEDIT 18 0 "t;" "6#%\"tG
" }{TEXT -1 27 " take on arbitrary values?\n" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 49 "u := proc(t) 500*(0.8)^t end;\nplot(u(t),t=0..25
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "\nWe can display both plot
s by giving each one a name, supressing any output, \nand displaying t
hem with the plots[display] command.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 79 "P1 := plot(pts,style=point):\nP2 := plot(u(t),t=0..25
):\nplots[display](\{P1,P2\});" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 
20 "Amoxicillin Example\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "
restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "The patient's kidney
s eliminate half the drug every hour, and \nthe patient takes a 500 mg
 pill every 4 hours.\n\nInitial Condition " }{XPPEDIT 18 0 "u[0] = 500
;" "6#/&%\"uG6#\"\"!\"$+&" }{TEXT -1 23 ". \n\nRecurrence formula\n" }
{XPPEDIT 18 0 "u[1] = 500+.5^4*u[0];" "6#/&%\"uG6#\"\"\",&\"$+&\"\"\"*
&$\"\"&!\"\"\"\"%&F%6#\"\"!F*F*" }{TEXT -1 12 "  =  531.25\n" }
{XPPEDIT 18 0 "u[2] = 500+.5^4*u[1];" "6#/&%\"uG6#\"\"#,&\"$+&\"\"\"*&
$\"\"&!\"\"\"\"%&F%6#\"\"\"F*F*" }{TEXT -1 16 "  =  533.203125\n" }
{XPPEDIT 18 0 "u[3] = 500+.5^4*u[2];" "6#/&%\"uG6#\"\"$,&\"$+&\"\"\"*&
$\"\"&!\"\"\"\"%&F%6#\"\"#F*F*" }{TEXT -1 18 "  =  533.3251953 \n" }
{TEXT 259 6 " . . ." }{TEXT -1 1 "\n" }{XPPEDIT 18 0 "u[n] = 500+.5^4*
u[n-1];" "6#/&%\"uG6#%\"nG,&\"$+&\"\"\"*&$\"\"&!\"\"\"\"%&F%6#,&F'F*\"
\"\"!\"\"F*F*" }{TEXT -1 13 "\n\nNow define " }{XPPEDIT 18 0 "u(n);" "
6#-%\"uG6#%\"nG" }{TEXT -1 62 " using the initial conditions and the r
ecurrence formula.\n   \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
61 "u := proc(n) if n<=0 then 500 else 500 + 0.5^4*u(n-1) fi end;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "\nDouble check the four values tha
t we already know.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "u(0
); u(1); u(2); u(3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "\nPlot th
e amount of amoxicillin in the patient's body after taking 25 pills ov
er a 6 day period.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "pts
 := [seq([n,u(n)],n=0..24)]:\nplot(pts, style=point);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 48 "\nThe values appear to have rapidly appro
ached a " }{TEXT 260 8 "limiting" }{TEXT -1 35 " value.  Let's check a
 few values.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "u(10); u(
15); u(20);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "\nTo 10 digits th
ese are the same.  But if we increase our precision to 25 digits we wi
ll see\nsome small differences.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "Digits := 25;\nu(10); u(15); u(20);\nDigits := 10;" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 234 "\nOf course, while these differ
ences are interesting from a mathematical point of view, differences\n
that are less than one hundredth of a milligram would be of no practic
al significance.\n\nSuppose that we assume that there is a number, " }
{XPPEDIT 18 0 "L;" "6#%\"LG" }{TEXT -1 78 ", which remains the same fo
r each iteration.\nThen we would have the relation  " }{XPPEDIT 18 0 "
L = 500+.5^4*L;" "6#/%\"LG,&\"$+&\"\"\"*&$\"\"&!\"\"\"\"%F$F'F'" }
{TEXT -1 41 ".  We can simply solve this equation for " }{XPPEDIT 18 
0 "L;" "6#%\"LG" }{TEXT -1 19 " \nand learn that   " }{XPPEDIT 18 0 "L
 = 500/(1-.5^4);" "6#/%\"LG*&\"$+&\"\"\",&\"\"\"F'*$$\"\"&!\"\"\"\"%!
\"\"F/" }{TEXT -1 49 "    =  533.3333333.  \n\nGiven a recursion formu
la " }{XPPEDIT 18 0 "u[n] = F(u[n-1]);" "6#/&%\"uG6#%\"nG-%\"FG6#&F%6#
,&F'\"\"\"\"\"\"!\"\"" }{TEXT -1 4 ", a " }{TEXT 263 10 "rest point" }
{TEXT -1 16 " is any number, " }{XPPEDIT 18 0 "x;" "6#%\"xG" }{TEXT 
-1 12 ", for which " }{XPPEDIT 18 0 "x = F(x);" "6#/%\"xG-%\"FG6#F$" }
{TEXT -1 102 ".\n  \nAlthough this was a simple equation which was eas
y to do by hand, we can also solve it\nusing the " }{TEXT 264 5 "solve
" }{TEXT -1 86 " command in  Maple. This can be helpful when the equat
ion is complicated.\nSimply type " }{TEXT 265 5 "solve" }{TEXT -1 56 "
 along with the equation and the variable to solve for\n." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "solve(L = 500 + 0.5^4*L,L);\n" }}}}
{SECT 1 {PARA 5 "" 0 "" {TEXT -1 12 "Exercise 17\n" }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
138 "Suppose that 600 mg of an antibiotic are administered every three
 hours, and\nthat the body eliminates 40% of the antibiotic in one hou
r.\n\n" }{TEXT 266 1 "a" }{TEXT -1 79 ".  Model the maximum level of t
he antibiotic with a difference equation.\n\n     " }{XPPEDIT 18 0 "u[
0] = 600;" "6#/&%\"uG6#\"\"!\"$+'" }{TEXT -1 7 " \n     " }{XPPEDIT 
18 0 "u[n] = 600+.4^3*u[n-1];" "6#/&%\"uG6#%\"nG,&\"$+'\"\"\"*&$\"\"%!
\"\"\"\"$&F%6#,&F'F*\"\"\"!\"\"F*F*" }{TEXT -1 6 ".    \n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 267 2 "b." }{TEXT -1 171 "  Does \+
the maximum value stabilize?  If so what is this \"stable\" value.\n\n
     First, define the difference equation in Maple and determine the \+
first few numerical values.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 61 "u := proc(n) if n<=0 then 600 else 600 + 0.4^3*u(n-1) fi end;" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "u(0); u(1); u(2); u(3);" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "\n    Now determine any rest poi
nts.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "solve(L = 600 + 0
.4^3*L,L);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 " \n   The number 6
41.0256410 is a rest point and it certainly appears that the maximum c
oncentration of the antibiotic will\n   quickly \"stabilize\" at this \+
value. \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 268 2 "c." }
{TEXT -1 94 "  Produce a graph to depict the maximum amount of the ant
ibiotic in the first 12 hours.  \n    " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "pts := [seq([n,u(n)],n=0..4)]:\nplot(pts, style=point
);" }}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 25 "Patterns of Accumulation
\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 11 "Summation.\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 66 "To sum the odd integers from 3 to 19 in Maple use the sum
 command\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Sum(2*k+1, k=
1..9) = sum(2*k+1, k=1..9);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "
\nThe capital S in Sum tells Maple to simply write the summation in st
andard mathematical\nnotation.  The second sum, with the lower case s,
 tells Maple to do the calculation.\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 18 "A sum of the form " }{XPPEDIT 18 0 "Sum(a^k,k = 0 .. n) =
 1+a+a^2;" "6#/-%$SumG6$)%\"aG%\"kG/F);\"\"!%\"nG,(\"\"\"\"\"\"F(F0*$F
(\"\"#F0" }{TEXT -1 3 " + " }{TEXT 262 8 ". . . + " }{XPPEDIT 18 0 "a^
n;" "6#)%\"aG%\"nG" }{TEXT -1 76 " is called a finite geometric series
.\nIf we ask Maple to evaluate it we get\n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 26 "S := sum(a^k,k = 0 .. n); " }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 35 "\nThis is not exactly correct.  If  " }{XPPEDIT 18 0 "
a = 1;" "6#/%\"aG\"\"\"" }{TEXT -1 8 ", then  " }{XPPEDIT 18 0 "S = n+
1;" "6#/%\"SG,&%\"nG\"\"\"\"\"\"F'" }{TEXT -1 51 ".   The formula is c
orrect for all other values of " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT 
-1 2 ".\n" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 31 "Amoxicillin Exampl
e Revisited.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 "We now return to the Amoxicilli
n example.  Remember that the patient's kidneys eliminate\nhalf the dr
ug every hour, and  that the patient takes a 500 mg pill every 4 hours
.\n\nInitial Condition " }{XPPEDIT 18 0 "u[0] = 500;" "6#/&%\"uG6#\"\"
!\"$+&" }{TEXT -1 23 ". \n\nRecurrence formula\n" }{XPPEDIT 18 0 "u[1]
 = 500+.5^4*u[0];" "6#/&%\"uG6#\"\"\",&\"$+&\"\"\"*&$\"\"&!\"\"\"\"%&F
%6#\"\"!F*F*" }{TEXT -1 5 "  =  " }{XPPEDIT 18 0 "500*(1+.5^4);" "6#*&
\"$+&\"\"\",&\"\"\"F%*$$\"\"&!\"\"\"\"%F%F%" }{TEXT -1 23 "  =  500(1 \+
+ 0.0625)  \n" }{XPPEDIT 18 0 "u[2] = 500+.5^4*u[1];" "6#/&%\"uG6#\"\"
#,&\"$+&\"\"\"*&$\"\"&!\"\"\"\"%&F%6#\"\"\"F*F*" }{TEXT -1 5 "  =  " }
{XPPEDIT 18 0 "500*(1+.5^4+.5^8);" "6#*&\"$+&\"\"\",(\"\"\"F%*$$\"\"&!
\"\"\"\"%F%*$$\"\"&!\"\"\"\")F%F%" }{TEXT -1 27 "  =  500(1 + 0.0625 +
 0.062" }{XPPEDIT 18 0 "5^2;" "6#*$\"\"&\"\"#" }{TEXT -1 3 ") \n" }
{XPPEDIT 18 0 "u[3] = 500+.5^4*u[2];" "6#/&%\"uG6#\"\"$,&\"$+&\"\"\"*&
$\"\"&!\"\"\"\"%&F%6#\"\"#F*F*" }{TEXT -1 4 "  = " }{XPPEDIT 18 0 "500
*(1+.5^4+.5^8+.5^12);" "6#*&\"$+&\"\"\",*\"\"\"F%*$$\"\"&!\"\"\"\"%F%*
$$\"\"&!\"\"\"\")F%*$$\"\"&!\"\"\"#7F%F%" }{TEXT -1 27 "  =  500(1 + 0
.0625 + 0.062" }{XPPEDIT 18 0 "5^2;" "6#*$\"\"&\"\"#" }{TEXT -1 8 " + \+
0.062" }{XPPEDIT 18 0 "5^3;" "6#*$\"\"&\"\"$" }{TEXT -1 3 ")\n " }
{TEXT 261 6 " . . ." }{TEXT -1 1 "\n" }{XPPEDIT 18 0 "u[n] = 500+.5^4*
u[n-1];" "6#/&%\"uG6#%\"nG,&\"$+&\"\"\"*&$\"\"&!\"\"\"\"%&F%6#,&F'F*\"
\"\"!\"\"F*F*" }{TEXT -1 4 "  = " }{XPPEDIT 18 0 "500*Sum(.625e-1^k,k \+
= 0 .. n);" "6#*&\"$+&\"\"\"-%$SumG6$)$\"$D'!\"%%\"kG/F-;\"\"!%\"nGF%
" }{TEXT -1 17 " . \n \nNow define " }{XPPEDIT 18 0 "u(n);" "6#-%\"uG6
#%\"nG" }{TEXT -1 35 " in tems of this summation formula." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "u := proc(n) 500*sum(0.0625^k,k=0..
n) end;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "\nDouble check the fou
r values that we already know.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "u(0); u(1); u(2); u(3);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 13 "\nNow examine " }{XPPEDIT 18 0 "u(10);" "6#-%\"uG6#\"#5" 
}{TEXT -1 2 ".\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "u(10);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "\nNote that the values of  " }
{XPPEDIT 18 0 "u(3);" "6#-%\"uG6#\"\"$" }{TEXT -1 5 " and " }{XPPEDIT 
18 0 "u(10);" "6#-%\"uG6#\"#5" }{TEXT -1 477 " differ from the previou
s calculation by 2 units in the last digit.\nThis is an example of rou
nd-off  error.  When calculations are done in two different ways using
  a fixed\nnumber of decimal digits, the different calculations will c
ause rounding to occur at different times.\nThis will typically produc
e slightly different results.  Rounding error is another reason to per
form\ncalculations with more digits than the minimum necessary to refl
ect the accuracy of the measurements." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 97 "\nPlot the amount of amoxicillin in the patient's body af
ter taking 25 pills over a 6 day period.\n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 55 "pts := [seq([n,u(n)],n=0..24)]:\nplot(pts, style=po
int);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "\nWe can also define " }
{XPPEDIT 18 0 "u(n);" "6#-%\"uG6#%\"nG" }{TEXT -1 40 " in terms of the
 closed form solution .\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
82 "u := proc(n) 500*((1-0.0625^(n+1))/(1-0.0625)) end;\nu(0); u(1); u
(2); u(3); u(10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "\nThe close
d form solution yields results that are identical to those with the su
mmation formula,\nbecause Maple uses the closed form solution to calcu
late the sum.\n" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 21 "Piecewise Fu
nctions.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Piecewise functions can be define
d in Maple using the piecewise command.\nFor a simple example suppose \+
that " }{XPPEDIT 18 0 "f(x) = 4;" "6#/-%\"fG6#%\"xG\"\"%" }{TEXT -1 5 
" for " }{XPPEDIT 18 0 "x < -3;" "6#2%\"xG,$\"\"$!\"\"" }{TEXT -1 13 "
, then 2 for " }{XPPEDIT 18 0 "x < 2;" "6#2%\"xG\"\"#" }{TEXT -1 20 ",
 and\nfinally 3 for " }{XPPEDIT 18 0 "2 <= x;" "6#1\"\"#%\"xG" }{TEXT 
-1 38 ".  We define the function as follows.\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 43 "f := proc(x) piecewise(x<-3,4,x<2,2,3) end;" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "\nTyping the expression " }
{XPPEDIT 18 0 "f(t);" "6#-%\"fG6#%\"tG" }{TEXT -1 93 " causes Maple to
 display the piecewise function the way it is usually\nprinted in math
 books.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(t);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "\nWe plot " }{XPPEDIT 18 0 "f(x);" 
"6#-%\"fG6#%\"xG" }{TEXT -1 33 " as we would any other function.\n" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(f(x),x=-5..5,y=0..5);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "\nFor a slightly more intere
sting example, consider a function that decays exponentially, then gro
ws\nexponentially,  and then begins to decay again.\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "g := proc(x) piecewise(x<3,5*(0.585
)^x,x<6,(1.71)^(x-3),5*(0.7)^(x-6)) end;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "plot(g(x),x=0..10,y=0..5);" }}}}{SECT 1 {PARA 5 "" 0 
"" {TEXT -1 21 "Exercise 17 Revisited" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "\nRe
call Exercise 17 from Section 1.1.  For the three hours after the firs
t dose the amount of the antibiotic\ndecays exponentially.  The closed
 form solution is " }{XPPEDIT 18 0 "u[0] = 600;" "6#/&%\"uG6#\"\"!\"$+
'" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[n] = .4^n*u[0];" "6#/&%\"uG6#
%\"nG*&)$\"\"%!\"\"F'\"\"\"&F%6#\"\"!F-" }{TEXT -1 65 ".  However this
 only holds\nuntil the next dose which occurs when " }{XPPEDIT 18 0 "n
 = 3;" "6#/%\"nG\"\"$" }{TEXT -1 16 ".  We start the " }{XPPEDIT 18 0 
".4^n;" "6#)$\"\"%!\"\"%\"nG" }{TEXT -1 54 " decay process all over wi
th an initial \ncondition of " }{XPPEDIT 18 0 "600+.4^3*600;" "6#,&\"$
+'\"\"\"*&$\"\"%!\"\"\"\"$\"$+'F%F%" }{TEXT -1 171 ".  The effect of t
wo pills can be described by a piecewise function.\nNote that when dec
ay process starts over after the second pill, we must adjust the expon
ent and write\n" }{XPPEDIT 18 0 ".4^(n-3);" "6#)$\"\"%!\"\",&%\"nG\"\"
\"\"\"$!\"\"" }{TEXT -1 45 " in order to multiply by the correct facto
r.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "u := proc(n) piece
wise(n<=0, 600, n<3, 600*0.4^n, \n                      (600 + 600*0.4
^3)*0.4^(n-3)) end;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "\nWe can p
lot the effect of these two pills over a 12 hour period.\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "pts := [seq([n,u(n)],n=0..12)]:\npl
ot(pts, style=point);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "\nWe can
 extend this piecewise function to handle three pills by adding anothe
r term.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "u := proc(n) \+
piecewise(n<=0, 600, n<3, 600*0.4^n, \n     n<6, (600 + 600*0.4^3)*0.4
^(n-3), \n     (600 + 600*0.4^3 + 600*0.4^6)*0.4^(n-6)) end;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "\nOver a 12 hour period we now get
 the following plot.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "p
ts := [seq([n,u(n)],n=0..12)]:\nplot(pts, style=point);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 187 "This app
roach will get extremely awkward if we want to consider the effect of \+
taking a pill\nevery three hours for several days.  However, there is \+
another way to look at what's happening." }}{PARA 0 "" 0 "" {TEXT -1 
100 "\nNotice that if we multiply out the last term in the definition \+
of the piecewise solution we get\n\n  " }{XPPEDIT 18 0 "(600+600*.4^3+
600*.4^6)*.4^(n-6) = 600*.4^(n-6)+600*.4^(n-3)+600*.4^n;" "6#/*&,(\"$+
'\"\"\"*&\"$+'F'*$$\"\"%!\"\"\"\"$F'F'*&\"$+'F'*$$\"\"%!\"\"\"\"'F'F'F
')$\"\"%!\"\",&%\"nGF'\"\"'!\"\"F',(*&\"$+'F')$\"\"%!\"\",&F;F'\"\"'F=
F'F'*&\"$+'F')$\"\"%!\"\",&F;F'\"\"$F=F'F'*&\"$+'F')$\"\"%!\"\"F;F'F'
" }{TEXT -1 121 "\n \nThis means that we can also think of the total a
mount in the body as the sum of the amount\nremaining from each pill. \+
 " }{TEXT 282 116 "Each term in the sum is just the closed form soluti
on shifted to\nstart at the time the corresponding pill was taken." }
{TEXT -1 90 "  We can write one piecewise function which\nrepresents a
ny shifted solution.  We will let " }{TEXT 270 1 "n" }{TEXT -1 34 " de
note the current hour and let  " }{TEXT 269 1 "h" }{TEXT -1 46 "  deno
te the\nhour at which the pill is taken.\n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 53 "v := proc(n,h) piecewise(n<h, 0, 600*0.4^(n-h)) end
; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 "\nNow the complete regimen
 is represented by the sum over all the prescribed pills and the\ntime
s they are to be taken. For 13 pills taken at 3 hour intervals over a \+
period of 48 hours\nwe have\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 95 "u := proc(n) sum(v(n,3*k),k=0..12) end;\npts := [seq([n,u(n)],n=
0..48)]:\nplot(pts, style=point);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
95 "\nSince we have a closed form solution, we can also plot the funct
ion as a continuous function, " }{XPPEDIT 18 0 "u(t);" "6#-%\"uG6#%\"t
G" }{TEXT -1 3 " .\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plo
t(u(t),t=0..48, 0..700);" }}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 27 "A \+
Model for Natural Growth\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 18 "The
 Rumor Example\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 405 "\nIn this model we consider \+
the spread of a rumor.  Our idea is that the number of people who know
\nthe rumor at the next time step is the sum of those who know it at t
he present time plus a small percentage\nof the product of those who k
now it with those who don't.  The idea of this product is that it will
 measure\nthe number of possible interactions betweeen those who know \+
the rumor and those who don't.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 73 "u := proc(n) if n<=0 then b else u(n-1) + a*u(n-1)*(P
op - u(n-1)) fi end;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "\nLet's a
ssume that the population, " }{XPPEDIT 18 0 "Pop;" "6#%$PopG" }{TEXT 
-1 62 ", has been normalized to 1, and the proportionality constant, \+
" }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 17 ", \nis 0.05.\n     " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "a := 0.05;\nPop := 1;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "\nNow set the size of the group wh
o initially knows the rumor to 1% of the total population.\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "b := 0.01;\npts := [seq([n,u
(n)],n=0..10)];\nplot(pts, style=point);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 170 "\nThis calculation takes quite a bit of time, and the ru
mor has not even spread to 2% of the population.\nThe problem is that \+
our recursive definition causes this function " }{TEXT 272 5 "u(n) " }
{TEXT -1 70 "to be evaluated many many\ntimes.  We can fix this by tel
ling Maple to " }{TEXT 273 8 "remember" }{TEXT -1 76 " the values that
 it has alreaady calculated.\nSimply add the Maple statement " }{TEXT 
274 17 "option remember; " }{TEXT -1 28 "to the function definition.\n
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "u := proc(n) option rem
ember; if n<=0 then b \n     else u(n-1) + a*u(n-1)*(Pop - u(n-1)) fi \+
end;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "a := 0.05;\nPop := \+
1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "b := 0.01;\nu(100);\n
pts := [seq([n,u(n)],n=0..200)]:\nplot(pts, style=point);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 18 "\nWhen we evaluate " }{TEXT 275 5 "u(n) \+
" }{TEXT -1 175 "with new values of its parameters we must first tell \+
it to forget all the\nvalues that it remebered from the calculations w
ith the old parameters.  The Maple command\nis called " }{TEXT 276 6 "
forget" }{TEXT -1 2 ".\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
103 "readlib(forget);\nforget(u);\nb := 0.04;\nu(100);\npts := [seq([n
,u(n)],n=0..200)]:\nplot(pts, style=point);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 96 "forget(u);\nb := 0.07;\nu(100);\npts := [seq([n,u(
n)],n=0..200)]:\nplot(pts, style=point);forget(u);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 95 "forget(u);\nb := 0.1;\nu(100);\npts := [seq
([n,u(n)],n=0..200)]:\nplot(pts, style=point);forget(u);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 58 "\nWe can solve the recurrence formula for
 its rest points.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "solv
e(x = x + a*x*(Pop-x),x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "\nCa
n you explain what happens when  " }{XPPEDIT 18 0 "b = 0;" "6#/%\"bG\"
\"!" }{TEXT -1 7 "  and  " }{XPPEDIT 18 0 "b = 1;" "6#/%\"bG\"\"\"" }
{TEXT -1 4 "?\n  " }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 18 "The Yeast \+
Example\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "restart;\nreadl
ib(forget);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "YeastData :
= [[10,9.6],[20,29.0],[30,71.1],[40,175],[50,351],\n              [60,
513],[70,594],[80,640],[90,656],[100,662]];" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 28 "plot(YeastData,style=point);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 96 "u := proc(n) option remember; if n<=0 then b \+
\n     else u(n-1) + r*u(n-1) - p*(u(n-1))^2 fi end;" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 100 "\nTo get started, let's estimate that the long
 term population will be 690 and that the growth\nrate, " }{XPPEDIT 
18 0 "r;" "6#%\"rG" }{TEXT -1 20 ", is about 0.1.  \n  " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "b := 9.6;\nr := 0.1;\np := r/690;
\nu(10);\nP1 := plot(YeastData,style=point):\npts := [seq([n,u(n)],n=0
..100)]:\nP2 := plot(pts):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 99 "\nThis isn't too bad a fit, but there are a coupl
e of points we can improve on.\nFirst, we're using  " }{XPPEDIT 18 0 "
u(0) = 9.6;" "6#/-%\"uG6#\"\"!$\"#'*!\"\"" }{TEXT -1 34 ", instead of \+
the given fact that  " }{XPPEDIT 18 0 "u(10) = 9.6;" "6#/-%\"uG6#\"#5$
\"#'*!\"\"" }{TEXT -1 41 ".\nSecond, we simply guessed the value of " 
}{TEXT 283 1 "r" }{TEXT -1 33 ", but we could use the values of\n" }
{XPPEDIT 18 0 "u(10);" "6#-%\"uG6#\"#5" }{TEXT -1 5 " and " }{XPPEDIT 
18 0 "u(20);" "6#-%\"uG6#\"#?" }{TEXT -1 175 " to get a more precise e
stimate. \n\nTo get the more precise estimates, assume that initially \+
the yeast simply\ngrows exponentially.  Then we would have the closed \+
form solution\n" }{XPPEDIT 18 0 "u(n) = u[0]*(1+r)^n;" "6#/-%\"uG6#%\"
nG*&&F%6#\"\"!\"\"\"),&\"\"\"F,%\"rGF,F'F," }{TEXT -1 35 ".  This lead
s to the two equations\n" }{XPPEDIT 18 0 "u[0]*(1+r)^10 = 9.6;" "6#/*&
&%\"uG6#\"\"!\"\"\"*$,&\"\"\"F)%\"rGF)\"#5F)$\"#'*!\"\"" }{TEXT -1 7 "
  and  " }{XPPEDIT 18 0 "u[0]*(1+r)^20 = 29.0;" "6#/*&&%\"uG6#\"\"!\"
\"\"*$,&\"\"\"F)%\"rGF)\"#?F)$\"$!H!\"\"" }{TEXT -1 47 ". \nWe can sol
ve these by hand, first eliminate " }{XPPEDIT 18 0 "u[0];" "6#&%\"uG6#
\"\"!" }{TEXT -1 37 " from the second equation, solve\nfor " }{TEXT 
284 1 "r" }{TEXT -1 32 ", then substitute that value of " }{TEXT 285 
1 "r" }{TEXT -1 36 " in the first equation to determine " }{XPPEDIT 
18 0 "u[0];" "6#&%\"uG6#\"\"!" }{TEXT -1 51 ".\nOr we can simply ask M
aple to do the work.\n      " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 94 "r := 'r';\nb := 'b';\neq1 := b*(1+r)^10 = 9.6;\neq2 := b*(1+r)^2
0 = 29.0;\nsolve(\{eq1,eq2\},\{b,r\});" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 95 "\nThis seems like a lot more answers than we want.  Howev
er, there are only two\nsolutions where " }{TEXT 286 1 "r" }{TEXT -1 
4 " is " }{TEXT 287 3 "not" }{TEXT -1 59 " a complex number, and of th
ese only one of the solutions,\n" }{XPPEDIT 18 0 "r = .11689;" "6#/%\"
rG$\"&*o6!\"&" }{TEXT -1 79 "  has a positive value.  The  correspondi
ng value for the yeast population\nat  " }{XPPEDIT 18 0 "n = 0;" "6#/%
\"nG\"\"!" }{TEXT -1 5 "  is " }{XPPEDIT 18 0 "b = 3.1779;" "6#/%\"bG$
\"&z<$!\"%" }{TEXT -1 26 ".\n\nUsing these values for " }{TEXT 288 1 "
b" }{TEXT -1 5 " and " }{TEXT 289 1 "r" }{TEXT -1 75 ", and the same l
ong term population, 690, \nwe get the following fit.\n      " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "forget(u);\nb := 3.1779;\nr
 := 0.11689;\np := r/690;\nu(10);\npts := [seq([n,u(n)],n=0..100)]:\nP
2 := plot(pts):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 157 "\nThis is a big improvement near the beginning of the da
ta, and\nwe can improve the fit at the other end by lowering our estim
ate\nof the long term population.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 102 "forget(u);\np := r/680;\nu(10);\npts := [seq([n,u(n)
],n=0..100)]:\nP2 := plot(pts):\nplots[display](P1,P2);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 308 "\nThis is good enough for now, but notic
e that we don't really\nhave a good fit where the function becomes con
cave down.\nIn fact, the data looks like it might eventually get large
r than\nour model predicts.  We will learn some other techniques later
\nthat can further improve the fitted parameters of our model.\n" }}}}
{SECT 1 {PARA 5 "" 0 "" {TEXT -1 21 "A Chaotic Population\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 28 "If the initial growth rate, " }{TEXT 290 1 "r" }{TEXT -1 
75 ", is large, then our difference equation can\nhave some surprising
 results.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "u := proc(n)
 option remember; if n<=0 then b else r*u(n-1) - p*(u(n-1))^2 fi end;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "b := 100;\nr := 4;\np :
= 0.005;\nseq(round(u(n)),n=0..11); " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "pts := [seq([n,u(n)],n=0..50)]:\nplot(pts);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "\nLooking at things over a longer
 period dosn't show any improvement\nin the regularity of the results.
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "pts := [seq([n,u(n)],
n=0..100)]:\nplot(pts);" }}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 21 "Inv
estigating Limits\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 24 "Limits fro
m Closed Form\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 264 "\nMaple can compute limits usi
ng the limit command.\nJust as in the sum command, using Limit with a \+
capital L\ncauses Maple to write the mathematical expression without\n
evaluating it.  Using limit with a lower case l, causes\nMaple to eval
uate the limit -- if it can.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 55 "Limit((1/2)^n,n=infinity) = limit((1/2)^n,n=infinity);;" }}}}
{SECT 1 {PARA 5 "" 0 "" {TEXT -1 29 "Rest Points and Limit Points\n" }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "restart;\nreadlib(forget):
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "\nLet's investigate the limit
 points and rest points of the following difference equation\n   " }
{XPPEDIT 18 0 "u[n] = 2*u[n-1]-u[n-1]^2;" "6#/&%\"uG6#%\"nG,&*&\"\"#\"
\"\"&F%6#,&F'F+\"\"\"!\"\"F+F+*$&F%6#,&F'F+\"\"\"F0\"\"#F0" }{TEXT -1 
8 "   and  " }{XPPEDIT 18 0 "u[0] = b;" "6#/&%\"uG6#\"\"!%\"bG" }
{TEXT -1 3 ". \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "u := pr
oc(n) option remember; if n<=0 then b else 2*u(n-1) - u(n-1)^2 fi end;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(x=2*x-x^2,x);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "\nThe rest points are zero and on
e.  Zero is clearly a limit point for  " }{XPPEDIT 18 0 "b = 0;" "6#/%
\"bG\"\"!" }{TEXT -1 40 ", and\n one is clearly a limit point for " }
{XPPEDIT 18 0 "b = 1;" "6#/%\"bG\"\"\"" }{TEXT -1 40 ".  But what happ
ens for other values of " }{TEXT 291 1 "b" }{TEXT -1 2 "?\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "b := 0.1;\nseq(u(n),n=0..7);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "forget(u);\nb := 0.5;\n
seq(u(n),n=0..7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "forget
(u);\nb := 0.8;\nseq(u(n),n=0..7);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 21 "\nIt appears that for " }{TEXT 277 2 "0<" }{XPPEDIT 18 0 "b <= \+
1;" "6#1%\"bG\"\"\"" }{TEXT -1 44 " we have a limit point and the valu
e is 1.\n " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "forget(u);\nb
 := 1.1;\nseq(u(n),n=0..7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
38 "forget(u);\nb := 1.5;\nseq(u(n),n=0..7);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 38 "forget(u);\nb := 1.9;\nseq(u(n),n=0..7);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "forget(u);\nb := 2;\nseq(u(n
),n=0..7);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "\nIt appears that o
ne is the limit point for all values of " }{TEXT 278 1 "b" }{TEXT -1 
13 " such that 0<" }{TEXT 279 1 "b" }{TEXT -1 30 "<2.\nZero is a limit
 point for " }{TEXT 280 1 "b" }{TEXT -1 11 "=0 and for " }{TEXT 281 1 
"b" }{TEXT -1 4 "=2.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "f
orget(u);\nb := 2.1;\nseq(u(n),n=0..7);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "forget(u);\nb := 2.5;\nseq(u(n),n=0..7);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "forget(u);\nb := 3.0;\nseq(u(n),n=0
..7);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "\nFor values of " }
{TEXT 292 1 "b" }{TEXT -1 68 " larger than 2, it looks like the values
 tend to\nnegative infinity.\n" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 
16 "Proving a Limit\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "v :
= proc(n) if n<=0 then 1 else cos(v(n-1))/n^2 fi end;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "pts := seq(evalf(v(k),4),k=0..9);" 
}}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 43 "Continuous and Piecewise Con
tinuous Models\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 19 "Painless Dent
istry\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "\nStarting with the discrete Novoc
aine model we created a continuous model \nby substituting " }{TEXT 
293 1 "t" }{TEXT -1 5 " for " }{TEXT 294 1 "n" }{TEXT -1 123 " in the \+
closed form representation of the model.\nThis continuous model predic
ts the same values as the discrete model\nwhen " }{TEXT 295 1 "t" }
{TEXT -1 81 " is an integer, and it provides \"reasonable\" looking an
swers for\nother values of " }{TEXT 296 3 "t, " }{TEXT -1 110 "but at \+
this point in time, we don't have any physical\nreason to justify this
 as the correct continuous model.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 50 "N := proc(t) 0.8^t * 500 end;\nplot(N(t), t=0..24);" 
}}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 22 "Amoxicillin Revisited\n" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 54 "\nThe Amoxicilin example led to the discrete model
\n\n   " }{XPPEDIT 18 0 "u[n] = 500*(1-.625e-1^(n+1))/(1-.625e-1);" "6
#/&%\"uG6#%\"nG*(\"$+&\"\"\",&\"\"\"F*)$\"$D'!\"%,&F'F*\"\"\"F*!\"\"F*
,&\"\"\"F*$\"$D'!\"%F3F3" }{TEXT -1 69 " .\n\nWe can also turn this in
to a continuous function by\nsubstituting " }{TEXT 297 1 "t" }{TEXT 
-1 5 " for " }{TEXT 298 1 "n" }{TEXT -1 5 ".\n   " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 80 "A := proc(t) 500*(1.0 - 0.0625^(t+1)) / (1.0
 - 0.0625) end;\nplot(A(t), t=0..24);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 20 "\nHowever, this is a " }{TEXT 299 9 "very poor" }{TEXT 
-1 74 " model for the amount of Amoxicillin in the\nblood.  In the dis
crete model " }{XPPEDIT 18 0 "u[n];" "6#&%\"uG6#%\"nG" }{TEXT -1 183 "
 represents the amount on Amoxicillin \nimmediately after taking a pil
l.  Over the next four hours the amount is\nreduced by the kidneys.  \+
\n\nFollowing the Novacaine model we can define " }{XPPEDIT 18 0 "h(t)
;" "6#-%\"hG6#%\"tG" }{TEXT -1 8 ".\n      " }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 47 "h := proc(t) 500*0.5^t end;\nplot(h(t), t=0..4);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "\nAt " }{XPPEDIT 18 0 "t = 4;" "
6#/%\"tG\"\"%" }{TEXT -1 141 ", the patient takes a second pill which \+
adds 500 mg to the amount of\nAmoxicillin in the patients blood.  This
 can be represented by defining " }{XPPEDIT 18 0 "h(t);" "6#-%\"hG6#%
\"tG" }{TEXT -1 29 " as\na piecewise function.\n   " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 87 "h := proc(t) piecewise(t<4, 500*0.5^t, 53
1.25*0.5^(t-4)) end;\nh(t);\nplot(h(t), t=0..8);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 60 "\nIt's interesting to look at a small table of val
ues around " }{XPPEDIT 18 0 "t = 4;" "6#/%\"tG\"\"%" }{TEXT -1 5 ".\n \+
  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 169 "[3.0,h(3.0)];[3.9,h(
3.9)];[3.99,h(3.99)];[3.999,h(3.999)];[3.9999,h(3.9999)];\n[4.0,h(4.0)
];[4.0001,h(4.0001)];[4.001,h(4.001)];[4.01,h(4.01)];[4.1,h(4.1)];\n[5
.0,h(5.0)];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "\nMaple can handle
 left and right sided limits, and \nit will indicate when limits are n
ot defined.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "limit(h(t)
,t=4,left);\nlimit(h(t),t=4,right);\nlimit(h(t),t=4);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 141 "\nIt's possible for a function to have a
 limit at a point\nwhere it is undefined.  These points are called hol
es\nor removable discontinuities.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 171 "f := proc(x) (x^3-8)/(x-2) end;\n[1.9,f(1.9)];[1.99,
f(1.99)];[1.999,f(1.999)];[1.9999,f(1.9999)];\n[2.0001,f(2.0001)];[2.0
01,f(2.001)];[2.01,f(2.01)];[2.1,f(2.1)];\n[2,f(2)];" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 62 "limit(f(t),t=2,left);\nlimit(f(t),t=2,rig
ht);\nlimit(f(t),t=2);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 
"plot(f(t),t=-3..3);\nf(2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "\n
Maple has a plot option, called " }{TEXT 300 7 "discont" }{TEXT -1 96 
" which will first attempt\nto find all discontinuities, so that the j
umps are correctly plotted.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 31 "plot(h(t),t=0..8,discont=true);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 452 "\nA better approach to the Amoxicillin model is to defin
e one function\nwhich follows the Novacaine model for the amount of am
oxicillin from\none pill.  However, add a parameter to this function s
o that it can be\nshifted to account for the time the pill was taken. \+
 The amount of Amoxicillin\nat any time is just the sum of the amounts
 remaining from each pill that\nhas been taken.  This leads to the fol
lowing model for 6 pills taken at\n4 hour intervals.\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "H := proc(t,h) piecewise(t<h, 0, 5
00*0.5^(t-h)) end;\nA := proc(t) sum(H(t,4*k),k=0..5) end;\nplot(A(t),
t=0..24);" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 20 "Sine-based Sequenc
e\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(sin(2*Pi/x),x=-2
..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "limit(sin(2*Pi/x),
x=0,right);" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 23 "Continuous Yeast
 Model\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "\nLet's revisit the yeast model to
 see if there's a continuous model\nlurking in the background.\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "YeastData := [[10,9.6],[20,
29.0],[30,71.1],[40,175],[50,351],\n              [60,513],[70,594],[8
0,640],[90,656],[100,662]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
54 "P1 := plot(YeastData,style=point):\nplots[display](P1);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "\nNow consider the " }{TEXT 301 8 
"logistic" }{TEXT -1 10 " function\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "y := proc(t) a/(1+b*c^t) end;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 40 "assume(0<c,c<1);\nlimit(y(t),t=infinity);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "\nThis tells us that we should set
 " }{TEXT 302 1 "a" }{TEXT -1 88 " to our best estimate of the long te
rm \nyeast population.  Let's estimate it to be 680.\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "a := 680;\ny(10);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 22 "\nNow we can determine " }{TEXT 303 1 "b" }
{TEXT -1 5 " and " }{TEXT 304 1 "c" }{TEXT -1 13 " by forcing  " }
{XPPEDIT 18 0 "y(t);" "6#-%\"yG6#%\"tG" }{TEXT -1 67 " to match two sp
ecific data points.\nThis is usually referred to as " }{TEXT 305 13 "i
nterpolating" }{TEXT -1 77 " the data at the selected points.  Two\npo
ints that might be interesting are  " }{XPPEDIT 18 0 "t = 10;" "6#/%\"
tG\"#5" }{TEXT -1 7 "  and  " }{XPPEDIT 18 0 "t = 50;" "6#/%\"tG\"#]" 
}{TEXT -1 4 ".\n  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "solve
(\{y(10)=9.6, y(50)=351\},\{b,c\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 117 "\nAgain the mathematical problem has lots of answers, but ther
e is only one\nanswer that satisfies our condition that  " }{XPPEDIT 
18 0 "0 < c;" "6#2\"\"!%\"cG" }{TEXT -1 7 "  and  " }{XPPEDIT 18 0 "c \+
< 1;" "6#2%\"cG\"\"\"" }{TEXT -1 4 ".  \n" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 76 "b := 205.17;\nc := 0.89783;\nP2 := plot(y(t),t=0..1
00):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 148 "
\nMeasuring the fit.  We can calculate the differences between the dat
a values and the\nfunction values.  These differences are called the r
esiduals.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "res := [seq(
YeastData[k,2]-y(YeastData[k,1]),k=1..nops(YeastData))];" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 347 "\nHere are three different ways to recor
d the size of these residuals.\nWe can look at the sum of the absolute
 values of the residuals,\n the maximum of the absolute values of the \+
residuals, or \nthe sum of the squares of the residuals.  \n\nThis las
t measure, usually referred to as the \"sum of squared errors\"\nturns
 out to be one of the most useful.\n " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "sum(res[k],k=1..nops(res));" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 32 "sum(abs(res[k]),k=1..nops(res));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "max(seq(abs(res[k]),k=1..nops(res))
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sum(res[k]^2,k=1..nop
s(res));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{SECT 0 
{PARA 3 "" 0 "" {TEXT -1 16 "Measuring change" }}{SECT 0 {PARA 4 "" 0 
"" {TEXT -1 30 "Analyzing a Discrete Function\n" }}{SECT 1 {PARA 5 "" 
0 "" {TEXT -1 17 "Richer or Broker\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "\nWe \+
want to look at the differences in prices of a stock from day to day.
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "FwdDiff := proc(Y) lo
cal k; [seq(Y[k]-Y[k-1], k=2..nops(Y))] end:#Define the FwdDiff functi
on." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 236 "DiffTable := proc(X
,Y) local k, fd, sd;\n             fd := FwdDiff(Y);\n             sd \+
:= FwdDiff(fd);\n             linalg[transpose](convert([X,Y,[op(fd), \+
\"-\"],[op(sd),\"-\",\"-\"]],array));\n             end:# Define the D
iffTable function" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "a := \+
[58.33, 59.04, 61.10, 62.27, 61.62, 59.26,\n       58.31, 57.92, 57.86
, 58.00, 59.11, 62.33, \n       67.84, 68.55, 69.96, 71.37, 71.96, 72.
15];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "pts := [seq([k-1, a
[k]],k=1..nops(a))]:\nplot(pts,style=point);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 37 "DiffTable([seq(k-1,k=1..nops(a))],a);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 164 "P1 := plot(pts,style=point)
:\nfd := FwdDiff(a);\npts := [seq([k-1,fd[k]],k=1..nops(fd))]:\nP2 := \+
plot(pts, style=point, symbol=box, color=blue):\nplots[display](P1,P2)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 140 "sd := FwdDiff(fd);\np
ts := [seq([k-1,sd[k]],k=1..nops(sd))]:\nP3 := plot(pts, style=point, \+
symbol=circle, color=green):\nplots[display](P2,P3);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 83 "pts := [seq([k-1, a[k]],k=1..nops(a))]:\n
plot(pts,style=point, view=[4..11,57..65]);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 118 "b := [seq(100-a[k],k=1..nops(a))];\npts := [seq([
k-1, b[k]],k=1..nops(b))]:\nplot(pts,style=point, view=[4..11,35..43])
;" }}{PARA 13 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 34 "Constructi
ng Models from Patterns\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 27 "Sun \+
Microsystem's Revenues\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r
estart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "FwdDiff := proc(
Y) local k; \n   [seq(Y[k]-Y[k-1], k=2..nops(Y))]\nend:#Define the Fwd
Diff function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 280 "DiffTabl
e := proc(XY) local k, X, Y, fd, sd;\n   X  := [seq(XY[k,1], k=1..nops
(XY))];\n   Y  := [seq(XY[k,2], k=1..nops(XY))];\n   fd := FwdDiff(Y);
\n   sd := FwdDiff(fd);\n   linalg[transpose](convert([X,Y,[op(fd), \"
-\"],[op(sd), \"-\", \"-\"]],array));\nend:# Define the DiffTable func
tion" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "SSE := proc(f,Xd,Y
d) local k;\n   sum((Yd[k]-f(Xd[k]))^2, k=1..nops(Xd));\nend:# Define \+
the Sum of Squared Errors." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
132 "LinReg := proc(Xd,Yd) local eq;\n   eq := stats[fit,leastsquare[[
x,y], \n      y=a*x+b, \{a,b\}]]([Xd,Yd]); \n   unapply(rhs(eq),x);\ne
nd:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "\nThe Revenue data\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "SunRev := [ 210.0,  538.0, \+
1052.0, 1765.0, 2466.0, 3221.0,\n           3589.0, 4309.0, 4690.0, 59
02.0, 7095.0];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "Xd := [s
eq(85+k, k=1..nops(SunRev))]:\npts := [seq([Xd[k], SunRev[k]], k=1..no
ps(SunRev))]:\nplot(pts,style=point, color=red);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 97 "fd := FwdDiff(SunRev);\na0 := sum(fd[k],k=1..
nops(fd))/nops(fd);\nb0 := solve(SunRev[1]=a0*86+b,b);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "DiffTable(pts);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 192 "a0 := (SunRev[11] - SunRev[1])/10;\nb0 := \+
solve(SunRev[1]=a0*86+b,b);\nf0 := t -> a0*t+b0;\nP1 := plot(pts,style
=point, color=red):\nP2 := plot(f0(t),t=86..96,color=green):\nplots[di
splay](P1,P2);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "f1 := L
inReg(Xd,SunRev);\nP3 := plot(f1(t),t=86..96,color=blue):\nplots[displ
ay](P1,P3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "SS0 := SSE(f
0,Xd,SunRev);\nSS1 := SSE(f1,Xd,SunRev);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 73 "f := x -> 3*x - 5;\npts := [seq([2*k-1,f(2*k-1)],k=1.
.8)]:\nDiffTable(pts);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 19 "Bouncing Golf Ball\n" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 25 "\nPreliminary definitions." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 98 "FwdDiff := proc(Y) local k; \n   [seq(Y[k]-Y[k
-1], k=2..nops(Y))]\nend:#Define the FwdDiff function." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 280 "DiffTable := proc(XY) local k, X, \+
Y, fd, sd;\n   X  := [seq(XY[k,1], k=1..nops(XY))];\n   Y  := [seq(XY[
k,2], k=1..nops(XY))];\n   fd := FwdDiff(Y);\n   sd := FwdDiff(fd);\n \+
  linalg[transpose](convert([X,Y,[op(fd), \"-\"],[op(sd), \"-\", \"-\"
]],array));\nend:# Define the DiffTable function" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 112 "SSE := proc(f,Xd,Yd) local k;\n   sum((Yd[k]
-f(Xd[k]))^2, k=1..nops(Xd));\nend:# Define the Sum of Squared Errors.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "LinReg := proc(Xd,Yd) \+
local eq;\n   eq := stats[fit,leastsquare[[x,y], \n      y=a*x+b, \{a,
b\}]]([Xd,Yd]); \n   unapply(rhs(eq),x);\nend:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 186 "QuadReg := proc(Xd,Yd) local k, eq;\n   eq :=
 stats[fit,leastsquare[[x,y], \n      y=a*x^2+b*x+c, \{a,b,c\}]]([Xd,Y
d]); \n   unapply(rhs(eq),x);\nend:# Define the Quadratic Regression f
unction" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "\nNow consider the dat
a from the picture of the bouncing golf ball.\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 249 "Golf := [1.7, 4.3, 6.4, 8.2, 9.7, 10.9, 11.7,
 12.3, 12.6, 12.5, 12.2, 11.6, 10.7,\n         9.5, 7.9, 5.9, 3.6, 0.9
];\nXd  := [seq(k-1,k=1..nops(Golf))];\npts := [seq([Xd[k], Golf[k]], \+
k=1..nops(Golf))]:\nplot(pts,style=point, symbol=circle, color=red);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 221 "\nThe plot looks a lot like th
e photo.  The y-values are the same, but the x-values are\nnot the sam
e.  How are the x-values related to the horizontal values in the photo
?\n\nLet's fit a least squares quadratic to this data.\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "g := QuadReg(Xd,Golf);\nP1 := plot
(pts,style=point, symbol=circle, color=red):\nP2 := plot(g(x),x=0..17,
color=blue):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 88 "\nThis fit looks really good.  Let's check the first and second
 differences of the data.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "DiffTable(pts);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 158 "\nThe sec
ond differences are all small negative numbers, which are fairly close
 together.\nThe first differences are steadily getting smaller. Let's \+
plot them.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 210 "fd := FwdD
iff(Golf):\nP1 := plot([seq([k-1,fd[k]],k=1..nops(fd))],style=point,co
lor=red,symbol=box):\nsd := FwdDiff(fd):\nP2 := plot([seq([k-1,sd[k]],
k=1..nops(sd))],style=point,color=blue):\nplots[display](P1,P2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 208 "\nThe first differences of the da
ta almost lie on a straight line, and the second differences\nof the d
ata are almost constant.  Let's examine the first and second differenc
es of the\nfitted quadratic function, " }{XPPEDIT 18 0 "g(x);" "6#-%\"
gG6#%\"xG" }{TEXT -1 4 ".  \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 225 "gd := [seq(g(k),k=0..17)]:\nfd := FwdDiff(gd):\nP3 := plot([seq
([k-1,fd[k]],k=1..nops(fd))],style=point,symbol=box):\nsd := FwdDiff(f
d):\nP4 := plot([seq([k-1,sd[k]],k=1..nops(sd))],style=point,color=blu
e):\nplots[display](P3,P4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 229 "
\nThe picture is nearly identical.  The first differences seem to lie \+
on a straight line\nwith a negative slope, and the second differences \+
appear to be constant.\n\nLet's look at the first differences for anot
her quadratic function\n" }{XPPEDIT 18 0 "f(x) = 3*x^2-16*x+5;" "6#/-%
\"fG6#%\"xG,(*&\"\"$\"\"\"*$F'\"\"#F+F+*&\"#;F+F'F+!\"\"\"\"&F+" }
{TEXT -1 5 ".\n   " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "f :=
 x -> 3*x^2-16*x+5;\nP1 := plot(f(x),x=-1..7):\nfd := [seq([k,f(k+1)-f
(k)], k=0..7)];\nP2 := plot(fd,style=point):\nplots[display](P1,P2);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 "\nIt certainly appears that th
e first differences lie on a straight line with\na positive slope.  In
 fact the slope appears to be 6.\nLet's check the Difference Table.\n
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "pts := [seq([x,f(x)],x=
0..7)];\nDiffTable(pts);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "\nYe
s, the first differences are increasing linearly.\nCan we provide some
 algebraic confirmation?  Let Maple help us\nlook at the formula for \+
" }{XPPEDIT 18 0 "f(x+1)-f(x);" "6#,&-%\"fG6#,&%\"xG\"\"\"\"\"\"F)F)-F
%6#F(!\"\"" }{TEXT -1 4 ".\n  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 27 "f(x+1) - f(x);\nsimplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 157 "\nVoila!  The formula for the first difference is the formula \+
for a straight\nline with slope 6.\n\nWhat happens if we do this for a
 general quadratic function?\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 61 "g := proc(x) a*x^2 + b*x + c end;\ng(x+1) - g(x);\nsimplify(%)
;" }}}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 47 "Divided Differences and A
verage Rate of Change\n" }}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 16 "Ponder
osa Pines\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "\nPreliminary definitions" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "FwdDiff := proc(Y) local k; \+
\n   [seq(Y[k]-Y[k-1], k=2..nops(Y))]\nend:#Define the FwdDiff functio
n." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 280 "DiffTable := proc(XY
) local k, X, Y, fd, sd;\n   X  := [seq(XY[k,1], k=1..nops(XY))];\n   \+
Y  := [seq(XY[k,2], k=1..nops(XY))];\n   fd := FwdDiff(Y);\n   sd := F
wdDiff(fd);\n   linalg[transpose](convert([X,Y,[op(fd), \"-\"],[op(sd)
, \"-\", \"-\"]],array));\nend:# Define the DiffTable function" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "SSE := proc(f,Xd,Yd) local \+
k;\n   sum((Yd[k]-f(Xd[k]))^2, k=1..nops(Xd));\nend:# Define the Sum o
f Squared Errors." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "LinRe
g := proc(Xd,Yd) local eq;\n   eq := stats[fit,leastsquare[[x,y], \n  \+
    y=a*x+b, \{a,b\}]]([Xd,Yd]); \n   unapply(rhs(eq),x);\nend:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "QuadReg := proc(Xd,Yd) loca
l k, eq;\n   eq := stats[fit,leastsquare[[x,y], \n      y=a*x^2+b*x+c,
 \{a,b,c\}]]([Xd,Yd]); \n   unapply(rhs(eq),x);\nend:# Define the Quad
ratic Regression function" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
94 "DivDiff := proc(X,Y) \n   local k; \n   [seq((Y[k]-Y[k-1])/(X[k]-X
[k-1]),\n   k=2..nops(X))]\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
26 "\nThe Ponderosa Pine data.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 158 "Girth :=     [53.0, 60.0, 63.0, 72.0, 78.0, 88.0, 10
0.0, 119.0, 122.0, 129.0]; \nBoardFeet := [19.0, 26.0, 32.0, 48.0, 61.
0, 86.0, 128.0, 214.0, 232.0, 275.0];" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 80 "pts := [seq([Girth[k], BoardFeet[k]], k=1..nops(Girth
))]:\nplot(pts,style=point);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 196 "P1 := plot(pts,style=point):\nf1 := LinReg(Girth,BoardFeet);\nP
2 := plot(f1(x),x=53..129,color=blue):\nf2 := QuadReg(Girth,BoardFeet)
;\nP3 := plot(f2(x),x=53..129,color=red):\nplots[display](P1,P2,P3);" 
}{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "\nTo the eye, th
e quadratic looks like a much better fit.\nWhat do the first differenc
es tell us?\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "fd := FwdD
iff(BoardFeet):\npts := [seq([Girth[k],fd[k]],k=1..nops(fd))]:\nplot(p
ts,style=point);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
19 "\nThis is certainly " }{TEXT 306 3 "not" }{TEXT -1 337 " what we w
ould expect to see from the first differences\nof a quadratic.  What's
 wrong?\n\nThe problem is that the data are not equally spaced along t
he x-axis.  We\ncan't ignore that because a large difference in the y-
values may not be all\nthat large when we consider the difference in t
he x-values.  The way to\naccount for this is to use " }{TEXT 307 19 "
divided differences" }{TEXT -1 47 "  which are computed\naccording to \+
the formula  " }{XPPEDIT 18 0 "(y[n+1]-y[n])/(x[n+1]-x[n]);" "6#*&,&&%
\"yG6#,&%\"nG\"\"\"\"\"\"F*F*&F&6#F)!\"\"F*,&&%\"xG6#,&F)F*\"\"\"F*F*&
F16#F)F.F." }{TEXT -1 4 ".\n  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 100 "dd := DivDiff(Girth,BoardFeet):\npts := [seq([Girth[k],dd[k]]
,k=1..nops(dd))]:\nplot(pts,style=point);" }{TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 62 "\nThe first divided differences definitel
y show a linear trend." }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 22 "Darin
g Turnpike Drive\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "\nPreliminary definitions" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "DivDiff := proc(X,Y) \n   \+
local k; \n   [seq((Y[k]-Y[k-1])/(X[k]-X[k-1]),\n   k=2..nops(X))]\nen
d:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "\nThe Turnpike Data\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "Time     := [0, 0.19, 0.35,
 0.55, 0.77, 1.10, 1.43,  1.75,  2.00 ];\nDistance := [0, 13.0, 26.0, \+
42.5, 60.0, 85.0, 110.0, 135.0, 150.0]; " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 62 "\nPlot the average speed (in miles per hour) for each str
etch.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "dd := DivDiff(T
ime,Distance);\npts := [seq([Time[k],dd[k]],k=1..nops(dd))]:\nplot(pts
,0..2,55..85,style=point);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 27 "\nNow consider the function " }{XPPEDIT 18 0 "f(x) = 2*x*
cos(2.5*x);" "6#/-%\"fG6#%\"xG*(\"\"#\"\"\"F'F*-%$cosG6#*&$\"#D!\"\"F*
F'F*F*" }{TEXT -1 145 ".  We are interested\nin examining the average \+
rate of change of this function over different\nintervals.  First cons
ider intervals of length 1.\n  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 52 "f := proc(x) 2*x*cos(2.5*x) end;\nplot(f(x),x=-1..6);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "\nConsider the average rate of cha
nge over intervals of length 1\nfor " }{TEXT 311 1 "x" }{TEXT -1 22 " \+
ranging from 1 to 5.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "X
d := [seq(k,k=1..5)];\nYd := [seq(f(Xd[k]),k=1..5)]:\nevalf(Yd,5);\ndd
 := evalf(DivDiff(Xd,Yd),5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "
\nNow consider intervals of length 0.1.  Also display the function and
\nthe divided differences together.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 226 "Xd := [seq((1.0+0.1*k),k=0..40)]:\nYd := [seq(f(Xd[k
+1]),k=0..40)]:\ndd := DivDiff(Xd,Yd):\npts := [seq([Xd[k],dd[k]],k=1.
.nops(dd))]:\nP2 := plot(pts,-1..6,style=point,color=blue):\nP1 := plo
t(f(x),x=-1..6):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 3 "\n  " }{XPPEDIT 18 0 "Delta*g/(Delta*x) = (g(b)-g(a))/(b-a
);" "6#/*(%&DeltaG\"\"\"%\"gGF&*&F%F&%\"xGF&!\"\"*&,&-F'6#%\"bGF&-F'6#
%\"aGF*F&,&F/F&F2F*F*" }{TEXT -1 20 " \n\nThe linear case\n " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "g := proc(x) m*x+k end;\n(g(
b)-g(a))/(b-a);\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 12 "Secant line\n" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 44 "\nSuppose that we have an arbitrary function " }
{XPPEDIT 18 0 "f(x);" "6#-%\"fG6#%\"xG" }{TEXT -1 77 ", and suppose th
at\nwe want the straight line that passes through the points (" }
{XPPEDIT 18 0 "a,f(a);" "6$%\"aG-%\"fG6#F#" }{TEXT -1 7 ") and\n(" }
{XPPEDIT 18 0 "b,f(b);" "6$%\"bG-%\"fG6#F#" }{TEXT -1 106 ").  What's \+
the equation of that straight line?\n\nHere's what we can learn from M
aple.  First consider the \n" }{TEXT 308 16 "slope intercept " }{TEXT 
-1 5 "form " }{XPPEDIT 18 0 "m*x+k;" "6#,&*&%\"mG\"\"\"%\"xGF&F&%\"kGF
&" }{TEXT -1 4 ".\n  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "so
lve(\{m*a+k=f(a),m*b+k=f(b)\},\{m,k\});" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 24 "\nNote that the slope is " }{XPPEDIT 18 0 "(f(b)-f(a))/(b
-a);" "6#*&,&-%\"fG6#%\"bG\"\"\"-F&6#%\"aG!\"\"F),&F(F)F,F-F-" }{TEXT 
-1 8 " .\n\nThe " }{TEXT 309 11 "point slope" }{TEXT -1 9 " form is " 
}{XPPEDIT 18 0 "m*(x-a)+k;" "6#,&*&%\"mG\"\"\",&%\"xGF&%\"aG!\"\"F&F&%
\"kGF&" }{TEXT -1 3 ".\n " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
45 "solve(\{m*(a-a)+k=f(a),m*(b-a)+k=f(b)\},\{m,k\});" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 163 "\nThe point slope form is a little easie
r to remember.  Of course, both the point\nslope form and the slope in
tercept form have the same slope, and the slope is\nthe " }{TEXT 316 
22 "average rate of change" }{TEXT -1 4 " in " }{XPPEDIT 18 0 "f(x);" 
"6#-%\"fG6#%\"xG" }{TEXT -1 17 " on the interval " }{TEXT 317 6 "(a, b
)" }{TEXT -1 43 ".\n\nThis is so important that we define the " }
{TEXT 318 5 "avgRC" }{TEXT -1 10 " function." }}{PARA 0 "" 0 "" {TEXT 
-1 2 "  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "avgRC := proc(f
,a,b)\n  simplify((f(b)-f(a))/(b-a))\nend:" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 22 "\nFinally there is the " }{TEXT 315 9 "two point" }{TEXT 
-1 6 " form " }{XPPEDIT 18 0 "alpha*(x-a)+beta*(b-x);" "6#,&*&%&alphaG
\"\"\",&%\"xGF&%\"aG!\"\"F&F&*&%%betaGF&,&%\"bGF&F(F*F&F&" }{TEXT -1 
4 "\n.  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "solve(\{alpha*(
a-a)+beta*(b-a)=f(a),alpha*(b-a)+beta*(b-b)=f(b)\},\{alpha,beta\});" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "\nThe two point form is usually \+
written   " }{XPPEDIT 18 0 "f(b);" "6#-%\"fG6#%\"bG" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "(x-a)/(b-a)+f(a);" "6#,&*&,&%\"xG\"\"\"%\"aG!\"\"F',&%
\"bGF'F(F)F)F'-%\"fG6#F(F'" }{TEXT -1 1 " " }{XPPEDIT 18 0 "(b-x)/(b-a
);" "6#*&,&%\"bG\"\"\"%\"xG!\"\"F&,&F%F&%\"aGF(F(" }{TEXT -1 92 " .   \+
 \nWritten this way it is easy to see that the function takes on the c
orrect \nvalues at " }{XPPEDIT 18 0 "x = a;" "6#/%\"xG%\"aG" }{TEXT 
-1 8 "  and   " }{XPPEDIT 18 0 "x = b;" "6#/%\"xG%\"bG" }{TEXT -1 66 "
.  However, the value of the slope is not \nquite as obvious.\n\nThe \+
" }{TEXT 312 11 "point slope" }{TEXT -1 155 " form is the most conveni
ent way to represent the secand line.\nWe can easily define a routine \+
for computing the secant line given the function\nand the two " }
{TEXT 313 1 "x" }{TEXT -1 9 "-values.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 133 "SecantLine := proc(f,a,b) \n   local slope,fa,x;\n  \+
 fa := simplify(f(a));\n   slope := avgRC(f,a,b);\n   unapply(slope*(x
-a)+fa,x);\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "f := x \+
-> 2*x*cos(2.5*x);\nplot(f(x),x=-1..6,-12..12);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 58 "g := SecantLine(f,1,5);\nplot([f(x),g(x)],x=-1
..6,-12..12);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 167 "Xd := [se
q(1+0.1*(k-1),k=1..41)]:\npts := [seq([Xd[k],avgRC(f,Xd[k],Xd[k]+0.1)]
,k=1..41)]:\nP1 := plot(f(x),x=-1..6):\nP2 := plot(pts,style=point):\n
plots[display](P1,P2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 13 "Secant Movie\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart;\nwith(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "avgRC := proc(f,a,b)\n  simplify((f
(b)-f(a))/(b-a))\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "
SecantLine := proc(f,a,b) \n   local slope,fa,x;\n   fa := simplify(f(
a));\n   slope := avgRC(f,a,b);\n   unapply(slope*(x-a)+fa,x);\nend:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 801 "SecantMovie := proc(f,n,
a,b,x1,x2,y1,y2)\n   local sl,pt,tt,dt,rho,k,h,slope,P1,P2,P3,P4;\n   \+
sl := array(1..n);\n   pt := array(1..n);\n   tt := array(1..n);\n   d
t := 0.5;\n   rho := exp(ln(0.001)/(n-1));\n   for k from 1 to n do\n \+
     h := rho^(k-1)*(b-a);\n      slope := (f(a+h)-f(a))/h;\n      sl[
k] := SecantLine(f,a,a+h);\n      pt[k] := [[a,f(a)],[(a+h),f(a+h)]];
\n      tt[k] := cat(\"Slope: \",convert(evalf(slope,4),string));\n   \+
od;   \n   P1 := display(seq(plot(sl[k](x),x=x1..x2,y1..y2),k=1..n),in
sequence=true):\n   P2 := display(seq(plot(pt[k],style=point,symbol=ci
rcle),k=1..n),insequence=true):\n   P3 := animate(f(x),x=x1..x2,y=y1..
y2,color=black,frames=n):\n   P4 := display(seq(textplot([a+dt,f(a)-dt
,tt[k]],align=RIGHT),k=1..n),insequence=true):\n   display(P1,P2,P3,P4
,view=[x1..x2,y1..y2]);\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "f := x -> 2*x*cos(2.5*x);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "SecantMovie(f,50,1,5,-1,6,-10,10);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 39 "SecantMovie(f,50,1.37,5.6,-1,6,-10,10);" 
}}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 46 "Average Rate of Change and Se
lected Exercises\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "avgRC := proc(f,a,b)\n
  simplify((f(b)-f(a))/(b-a))\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "f := x -> 2*x*cos(2.5*x);\navgRC(f,1,5);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 19 "\nExercises 12 & 14\n" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 75 "f := x -> m*x+b;\navgRC(f,x,x+h);\nf := x -
> a*x^3+b*x^2+c*x+d;\navgRC(f,x,h);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 27 "\nExercises 16, 18, and 20.\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 61 "f := x -> -3*x + 5;\nplot([f(x),avgRC(f,x-0.5,x+0.5)]
,x=0..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "f := x -> sin(
x);\nplot([f(x),avgRC(f,x-0.5,x+0.5)],x=0..6);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 56 "f := x -> 2^x;\nplot([f(x),avgRC(f,x-0.5,x+0.5
)],x=0..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 
0 {PARA 4 "" 0 "" {TEXT -1 15 "Rate of Change\n" }}{SECT 1 {PARA 5 "" 
0 "" {TEXT -1 18 "Surfing in Bombay\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "\nPre
liminary definitions" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "Fw
dDiff := proc(Y) \n   local k; \n   [seq(Y[k]-Y[k-1], k=2..nops(Y))] \+
\nend:#Define the FwdDiff function." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 112 "SSE := proc(f,Xd,Yd) local k;\n   sum((Yd[k]-f(Xd[k]
))^2, k=1..nops(Xd));\nend:# Define the Sum of Squared Errors." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 161 "SinReg2 := proc(w,Xd,Yd) lo
cal eq;\n   eq := stats[fit,leastsquare[[x,y],\n      y = a*cos(w*x)+b
*sin(w*x)+d,\n      \{a,b,d\}]]([Xd,Yd]);\n   unapply(rhs(eq),x);\nend
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 240 "sdCount := proc(y) lo
cal fd,sd,ct,sg,k;\n   fd := FwdDiff(y);\n   sd := FwdDiff(fd);\n   ct
 := 0;\n   sg := sd[1];\n   for k from 2 to nops(sd) do\n      if sg*s
d[k] < 0 then\n         sg := sd[k];\n         ct := ct+1;\n      fi;
\n   od;\n   ct;\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3120 
"SinReg1 := proc(wlo,whi,Xd,Yd) \n   local k,w1,w2,w3,w4,rs1,rs2,rs3,r
s4,\n      SS1,SS2,SS3,SS4,ww,rs,SS,q,a,b,c,ans,gf;\n   userinfo(3,Sin
Reg,`wlo,whi:`,wlo,whi);\n   w1  := wlo;\n   rs1 := SinReg2(w1,Xd,Yd);
\n   SS1 := SSE(rs1,Xd,Yd);\n   w4  := whi;\n   rs4 := SinReg2(w4,Xd,Y
d);\n   SS4 := SSE(rs4,Xd,Yd);\n   gf  := evalf((sqrt(5)-1)/2);\n   w3
  := w1 + gf*(w4-w1);\n   rs3 := SinReg2(w3,Xd,Yd);\n   SS3 := SSE(rs3
,Xd,Yd);\n   w2  := w1 + gf*gf*(w4-w1);\n   rs2 := SinReg2(w2,Xd,Yd);
\n   SS2 := SSE(rs2,Xd,Yd);\n   userinfo(3,SinReg,\"w1,w2,w3,w4:\",w1,
w2,w3,w4);\n   userinfo(3,SinReg,\"SSE:\",SS1,SS2,SS3,SS4);\n   q  := \+
x -> a*x^2 + b*x + c;\n   for k from 1 to 25 do\n      if ((SS2 < SS1)
 and (SS2 < SS3) and (SS3 < SS4)) then\n         w4  := w3;\n         \+
rs4 := rs3;\n         SS4 := SS3;\n         ans := solve(\{q(w1)=SS1, \+
q(w2)=SS2, q(w3)=SS3\},\{a,b,c\});\n         ww  := -subs(ans,b)/(2*su
bs(ans,a));\n         if ((w1 < ww) and (ww < w2)) then\n            w
3  := w2;\n            rs3 := rs2;\n            SS3 := SS2;\n         \+
   w2  := ww;\n            rs2 := SinReg2(w2,Xd,Yd);\n            SS2 \+
:= SSE(rs2,Xd,Yd);\n         elif ((w2 < ww) and (ww < w3)) then\n    \+
        w3  := ww;\n            rs3 := SinReg2(w3,Xd,Yd);\n           \+
 SS3 := SSE(rs3,Xd,Yd);s;\n         else\n            w3  := w2;\n    \+
        rs3 := rs2;\n            SS3 := SS2;\n            w2  := w1 + \+
gf*(w3-w1);\n            rs2 := SinReg2(w2,Xd,Yd);\n            SS2 :=
 SSE(rs2,Xd,Yd);\n         fi;\n      elif (SS2 < SS1) and (SS3 < SS2)
 and (SS3 < SS4) then\n         w1  := w2;\n         rs1 := rs2;\n    \+
     SS1 := SS2;\n         ans := solve(\{q(w2)=SS2, q(w3)=SS3, q(w4)=
SS4\},\{a,b,c\});\n         ww  := -subs(ans,b)/(2*subs(ans,a));\n    \+
     if ((w2 < ww) and (ww < w3)) then\n            w2  := ww;\n      \+
      rs2 := SinReg2(w2,Xd,Yd);\n            SS2 := SSE(rs2,Xd,Yd);\n \+
        elif ((w3 < ww) and (ww < w4)) then\n            w2  := w3;\n \+
           rs2 := rs3;\n            SS2 := SS3;\n            w3  := ww
;\n            rs3 := SinReg2(w3,Xd,Yd);\n            SS3 := SSE(rs3,X
d,Yd);\n         else\n            w2  := w3;\n            rs2 := rs3;
\n            SS2 := SS3;\n            w3  := w2 + gf*gf*(w4-w2);\n   \+
         rs3 := SinReg2(w3,Xd,Yd);\n            SS3 := SSE(rs3,Xd,Yd);
\n         fi;\n      elif SS2 < SS3 then\n         w4  := w3;\n      \+
   rs4 := rs3;\n         SS4 := SS3;\n         w3  := w2;\n         rs
3 := rs2;\n         SS3 := SS2;\n         w2  := w1 + gf*gf*(w4-w1);\n
         rs2 := SinReg2(w2,Xd,Yd);\n         SS2 := SSE(rs2,Xd,Yd);\n \+
     else\n         w1  := w2;\n         rs1 := rs2;\n         SS1 := \+
SS2;\n         w2  := w3;\n         rs2 := rs3;\n         SS2 := SS3;
\n         w3  := w1 + gf*(w4-w1);\n         rs3 := SinReg2(w3,Xd,Yd);
\n         SS3 := SSE(rs3,Xd,Yd);\n      fi;\n      userinfo(3,SinReg,
\"w1,w2,w3,w4:\",w1,w2,w3,w4);\n      userinfo(3,SinReg,\"SSE:\",SS1,S
S2,SS3,SS4);\n      if (abs(w4-w1) < 0.001*w4) then\n         break;\n
      fi;\n   od;\n   SS := min(SS1,SS2,SS3,SS4);\n   if (SS1 = SS) th
en\n      rs1;\n   elif (SS2 = SS) then \n      rs2;\n   elif (SS3 = S
S) then\n      rs3;\n   else\n      rs4;\n   fi;\nend:# Define the Sin
usoidal Regression function" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1064 "SinReg := proc(Xd,Yd) \n   local k,cnt,w0,Sk,SS,kk,rs,A,B,a,b,c,
d;\n   cnt := sdCount(Yd);\n   userinfo(3,SinReg,`Count:`,cnt);\n   if
 cnt < 2 then\n      cnt := 2;\n   fi;\n   #  We assume that the data \+
covers at least a quarter of the period.\n   w0 := evalf(0.5*Pi/abs(Xd
[nops(Xd)]-Xd[1]));\n   kk := 1;\n   rs := SinReg2(kk*w0,Xd,Yd);\n   S
S := SSE(rs,Xd,Yd);\n   userinfo(3,SinReg,`w, SS:`,w0, SS);\n   for k \+
from 2 to 2*cnt+2 do\n      rs := SinReg2(k*w0,Xd,Yd);\n      Sk := SS
E(rs,Xd,Yd);\n      userinfo(3,SinReg,`w, SS:`,k*w0, Sk);\n      if Sk
 < SS then\n         kk := k;\n         SS := Sk;\n      fi;\n   od;\n
   if kk = 1 then\n      rs := SinReg1(w0,3*w0,Xd,Yd);\n   elif kk = 2
*cnt+2 then\n      rs := SinReg1(2*cnt*w0,(2*cnt+2)*w0,Xd,Yd);\n   els
e\n      rs := SinReg1((kk-1)*w0,(kk+1)*w0,Xd,Yd);\n   fi;\n   userinf
o(3,SinReg,`rs`,rs(x));\n   A := op(1,op(1,rs(x)));\n   B := op(1,op(2
,rs(x)));\n   a := sqrt(A^2+B^2);\n   b := op(1,op(1,op(2,op(1,rs(x)))
));\n   c := arctan(A,B);\n   d := op(3,rs(x));\n   unapply(a*sin(b*x+
c)+d,x);\nend:# Define the Sinusoidal Regression function" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 310 13 "\nThe Tide dat" }{TEXT -1 2 "a\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "Hour  := [  6,   7,   8,   \+
9,   10,   11,   12,   13,   14,   15,   16,  17,  18,  19,  20];\nDep
th := [2.5, 2.7, 4.5, 7.5, 11.1, 14.5, 17.0, 18.0, 17.2, 15.0, 11.7, 8
.0, 4.9, 2.9, 2.5]; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "pts
 := [seq([Hour[k],Depth[k]],k=1..nops(Hour))]:\nplot(pts,style=point);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Bombay := SinReg(Hour,D
epth);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "P1 := plot(pts,s
tyle=point,color=blue):\nP2 := plot(Bombay(x),x=6..20,2..20):\nplots[d
isplay](P1,P2);\nSS5 := SSE(Bombay,Hour,Depth);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 1 "\n" }{TEXT 314 11 "Zooming In\n" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 172 "pt := [[11,Bombay(11)]];\na  := 8;\nb  := 14
;\nP1 := plot(pt,view=[8..14,6..18],style=point):\nP2 := plot(Bombay(x
),x=a..b):\nplots[display](P1,P2);\n(Bombay(b)-Bombay(a))/(b-a);" }}
{PARA 13 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 175 "pt := [[11,Bombay(11)]];\na  := 10;\nb  := 12;\nP1 := plot(pt,v
iew=[10..12,11..17],style=point):\nP2 := plot(Bombay(x),x=a..b):\nplot
s[display](P1,P2);\n(Bombay(b)-Bombay(a))/(b-a);" }}{PARA 13 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 187 "pt := [[11
,Bombay(11)]];\na  := 10.9;\nb  := 11.1;\nP1 := plot(pt,view=[10.9..11
.1,14.2..14.8],style=point):\nP2 := plot(Bombay(x),x=a..b):\nplots[dis
play](P1,P2);\n(Bombay(b)-Bombay(a))/(b-a);" }}{PARA 13 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 194 "pt := [[11,Bomba
y(11)]];\na  := 10.99;\nb  := 11.01;\nP1 := plot(pt,view=[10.99..11.01
,14.49...14.56],style=point):\nP2 := plot(Bombay(x),x=a..b):\nplots[di
splay](P1,P2);\n(Bombay(b)-Bombay(a))/(b-a);" }}{PARA 13 "" 0 "" 
{TEXT -1 0 "" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 20 "Bombay Secant M
ovie\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart;\nwith(plo
ts):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "avgRC := proc(f,a,b
)\n  simplify((f(b)-f(a))/(b-a))\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 133 "SecantLine := proc(f,a,b) \n   local slope,fa,x;\n  \+
 fa := simplify(f(a));\n   slope := avgRC(f,a,b);\n   unapply(slope*(x
-a)+fa,x);\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 801 "Secant
Movie := proc(f,n,a,b,x1,x2,y1,y2)\n   local sl,pt,tt,dt,rho,k,h,slope
,P1,P2,P3,P4;\n   sl := array(1..n);\n   pt := array(1..n);\n   tt := \+
array(1..n);\n   dt := 0.5;\n   rho := exp(ln(0.001)/(n-1));\n   for k
 from 1 to n do\n      h := rho^(k-1)*(b-a);\n      slope := (f(a+h)-f
(a))/h;\n      sl[k] := SecantLine(f,a,a+h);\n      pt[k] := [[a,f(a)]
,[(a+h),f(a+h)]];\n      tt[k] := cat(\"Slope: \",convert(evalf(slope,
4),string));\n   od;   \n   P1 := display(seq(plot(sl[k](x),x=x1..x2,y
1..y2),k=1..n),insequence=true):\n   P2 := display(seq(plot(pt[k],styl
e=point,symbol=circle),k=1..n),insequence=true):\n   P3 := animate(f(x
),x=x1..x2,y=y1..y2,color=black,frames=n):\n   P4 := display(seq(textp
lot([a+dt,f(a)-dt,tt[k]],align=RIGHT),k=1..n),insequence=true):\n   di
splay(P1,P2,P3,P4,view=[x1..x2,y1..y2]);\nend:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 51 "Bombay := x -> 7.78*sin(0.47146*x +1.688) + 10
.186;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "SecantMovie(Bombay
,35,11,18,6,20,0,25);" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT -1 13 "Tangen
t Line\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 167 "TangentLine := proc(f,a)\n \+
  local slope,fa;\n   fa := simplify(f(a));\n   slope := simplify(D(f)
(a));\n   unapply(slope*(x-a)+fa,x);\nend:# Define the TangentLine fun
ction" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Bombay := x -> 7.7
80117521*sin(.4714658670*x+1.688424783)+10.18610980;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 144 "f := TangentLine(Bombay,11);\npt := [[11
,Bombay(11)]];\nP1 := plot(pt,style=point):\nP2 := plot([Bombay,f],8..
14,color=red):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 186 "Hour  := [  6,   7,   8,   9,   10,   11,   12,   13
,   14,   15,   16,  17,  18,  19,  20];\nDepth := [2.5, 2.7, 4.5, 7.5
, 11.1, 14.5, 17.0, 18.0, 17.2, 15.0, 11.7, 8.0, 4.9, 2.9, 2.5]; " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "pts := [seq([Hour[k],Depth[
k]],k=1..nops(Hour))]:\nP1 := plot(pts,style=point):\nP2 := plot([Bomb
ay,f],6..20,2..22,color=red):\nplots[display](P1,P2);" }}}}}{SECT 0 
{PARA 4 "" 0 "" {TEXT -1 29 "Rate of Change as a Function\n" }}{SECT 
1 {PARA 5 "" 0 "" {TEXT -1 16 "Plotting Slopes\n" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 55 "avgRC := proc(f,a,b)\n  simplify((f(b)-f(a))/(b-a))\nend:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 261 "Xd := [0,Pi/6,Pi/4,Pi/3,Pi/
2,2*Pi/3,3*Pi/4,5*Pi/6,Pi,\n       7*Pi/6,5*Pi/4,4*Pi/3,3*Pi/2,5*Pi/3,
7*Pi/4,11*Pi/6,2*Pi];\npts := [seq([Xd[k],sin(Xd[k])],k=1..nops(Xd))]:
\nP1 := plot(pts,style=point,symbol=circle):\nP2 := plot(sin(x),x=0..2
*Pi):\nplots[display](P1,P2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 116 "h := 0.25;\nZd := [seq(evalf(avgRC(sin,Xd[k]-h,Xd[k]+h),4),k=1.
.nops(Xd))];\nfp := [seq(D(sin)(Xd[k]),k=1..nops(Xd))];" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "sin(Pi/3);\nD(sin)(Pi/3);\n\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Bombay := x -> 7.78*sin(.471
*x+1.69)+10.2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "pts := [s
eq([k,D(Bombay)(k)],k=0..12)]:\nplot(pts,style=point);\n\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "plot([Bombay(x),D(Bombay)(x)],x=0..
12);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "f := x -> x^2;\nplo
t([f(x),D(f)(x)],x=-2..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
48 "g := x -> 3 + x^2;\nplot([g(x),D(g)(x)],x=-2..2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "f := x -> abs(5 - x^2);\nD(f);\nplo
t([f(x),D(f)(x)],x=-4..4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "plot(f(x),x=-4..4,-5..10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 26 "plot(f(x),x=2..2.5,-1..1);" }}}}{SECT 1 {PARA 5 "" 0 "" {TEXT 
-1 46 "Average Rate of Change and Selected Exercises\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 55 "avgRC := proc(f,a,b)\n  simplify((f(b)-f(a))/(b-a))
\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "\nExercises 12, 18, and
 20.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "f := x -> -3*x + \+
5;\nplot([f(x),avgRC(f,x-0.5,x+0.5),D(f)(x)],x=0..6);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot((avgRC(f,x-0.5,x+0.5) - D(f)(x
)),x=0..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "f := x -> si
n(x);\nplot([f(x),avgRC(f,x-0.5,x+0.5),D(f)(x)],x=0..6);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot((avgRC(f,x-0.5,x+0.5) - D(f)(x
)),x=0..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "f := x -> 2^
x;\nplot([f(x),avgRC(f,x-0.5,x+0.5),D(f)(x)],x=0..2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot((avgRC(f,x-0.5,x+0.5) - D(f)(x
)),x=0..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "2 5 2" 0 }
{VIEWOPTS 1 1 0 1 1 1803 }