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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 46 "Sequences -- Classifyinn
g 2-Cycles -- Answer 3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 74 "The cell below finds the equilibrium points and the \+
2-cycles for the model" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "
" 0 "" {TEXT -1 50 "R(p) = 0.5 + 0.0015 p - 0.000002166667 p (p - 500)
" }}{PARA 258 "" 0 "" {TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 13 "f
(p) = R(p) p" }}{PARA 260 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" 
{TEXT -1 18 "p(n + 1) = f(p(n))" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 262 "" 0 "" {TEXT 256 16 "Evaluate it now." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "R := p -> \+
0.5 + 0.0015 * p - 0.000001166667 * p * (p - 500):\n\nf := p -> R(p) *
 p:\n\nsolve(f(p) = p);\n\nsolve(f(f(p)) = p);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Notice that the equi
librium points are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 257 13 "0.285.7142713" }{TEXT -1 4 " and" }}{PARA 263 "" 0 "" 
{TEXT -1 11 "1499.999647" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 108 "and the 2-cycle points that are positive, and rea
l (have no imaginary part),  and not equilibrium points are" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 11 "1353.969201" }
{TEXT -1 4 " and" }}{PARA 264 "" 0 "" {TEXT -1 11 "1600.388102" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "The next
 cell evaluates the derivative of the function  f(f(p)) atone of these
 two 2-cycle points.  " }{TEXT 259 16 "Evaluate it now." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 42 "subs(p = 1353.969201, diff(f(f((p))), p));" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
119 "Notice that the absolute value of the derivative at this 2-cycle \+
point is less than one, so this 2-cycle is attracting." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 "The next cell gives
 an example with an initial condition so high that the population drop
s below the \"threshold\" and dies out.  " }{TEXT 260 16 "Evaluate it \+
now." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 292 "InitialValue := 2000:\n\n  pop :=\n     proc(n)\n   \+
     if n = 1 then InitialValue else f(pop(n-1))\n        fi\n     end
:\n\nfor k from 1 to 20 \n      do printf(`%4d %10.4f`, k, pop(k));\n \+
     print();\n      od;\n\nplot([seq([n, pop(n)], n = 1 .. 20)], \n  \+
      year = 1 .. 20, title = `Population`);" }}}}{MARK "5 0 0" 246 }
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