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\begin{document}
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%Slide 1
\title{{\LARGE\bf The Logistic Flow}\newline (Continuous)\newline \newline}
\author{Tom Carter
\newline
\newline
\newline
\tthdump{\href{http://astarte.csustan.edu/\~tom/SFI-CSSS}{http://astarte.csustan.edu/\~\ tom/SFI-CSSS}}
%%tth:\href{http://astarte.csustan.edu/~tom/SFI-CSSS}{http://astarte.csustan.edu/\~tom/SFI-CSSS}
\vfill
Complex Systems Summer School
\newline
}
\date{June, 2008}
\maketitle
%Slide 2
%\sectionhead{Our general topics:}
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%\pagedone
\sectionhead{Logistic flow \ldots}
We all know that the discrete logistic map
\begin{eqnarray*}
P_{n+1} & = & r P_n (1 - P_n)
\end{eqnarray*}
exhibits interesting behavior of various sorts for various values of the parameter $r$, including chaos, etc.
\vspace{1cm}
\
\centerline{\includegraphics[width = 6.5in]{logistic-pic}}
What kind of behavior can we expect for the continuous version of a logistic flow:
\begin{eqnarray*}
\frac{dP}{dt} & = & r P (1 - P) \ \ \ ?
\end{eqnarray*}
Note that this is a non-linear ODE, but fortunately we can actually integrate \ldots
\begin{eqnarray*}
\frac{dP}{dt} & = & r P (1 - P) \\
\\
\frac{dP}{P (1 - P)} & = & r dt \\
\end{eqnarray*}
Thus:
\begin{eqnarray*}
\int \frac{dP}{P (1 - P)} & = & \int r dt \\
\\
\int \frac{dP}{P (1 - P)} & = & r t + c_1 \\
\end{eqnarray*}
By partial fractions, we have:
\begin{eqnarray*}
\int \frac{dP}{P} + \int \frac{dP}{(1 - P)} & = & r t + c_1 \\
\\
\log (P) - \log (1 - P) & = & r t + c_1 \\
\\
\log ( \frac{P}{1 - P}) & = & r t + c_1 \\
\\
\frac{P}{1 - P} & = & e^{r t + c_1} \\
\\
\frac{P}{1 - P} & = & c_2 e^{r t} \\
\\
P & = & ( 1 - P ) c_2 e^{r t} \\
\end{eqnarray*}
\pagedone
And thus:
\begin{eqnarray*}
P & = & ( 1 - P ) c_2 e^{r t} \\
\\
P & = & c_2 e^{r t} - P c_2 e^{r t} \\
\\
P + P c_2 e^{r t} & = & c_2 e^{r t} \\
\\
P ( 1 + c_2 e^{r t}) & = & c_2 e^{r t} \\
\end{eqnarray*}
giving us:
\begin{eqnarray*}
P & = & \frac { c_2 e^{r t}} { 1 + c_2 e^{r t} } \\
\end{eqnarray*}
and, dividing top and bottom by $c_2 e^{r t}$, and simplifying, we have:
\begin{eqnarray*}
P & = & \frac {1} { 1 + c e^{-r t} } \\
\end{eqnarray*}
\pagedone
This function just gives us the classic logistic/sigmoid curve:
\vspace{1cm}
\
\centerline{\includegraphics[width = 5in]{Logistic-curve}}
and changes in $c$ and $r$ make minor changes in the behavior near $0$ \ldots
The difference between the behavior of the discrete and continuous logistic functions
can give us some idea of the significance of working in the discrete regime . . .
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\end{document}
%% \item Oh, all right. Since we're here, let's use Stirling's approximation
%% $$n! \approx \sqrt{2\pi}\ n^ne^{-n}\sqrt{n}.$$
%% to calculate:
%%
%% $$P(S_n = k) = \frac{n!}{(\frac{n + k}{2})!
%% \ (\frac{n - k}{2})!\ 2^n} $$
%%
%% \begin{eqnarray*} & \approx & \frac{\sqrt{2\pi}\ n^ne^{-n}\sqrt{n}}
%% {\sqrt{2\pi}\ (\frac{n + k}{2})^{\frac{n + k}{2}}
%% e^{-\frac{n + k}{2}}\sqrt{\frac{n + k}{2}}
%% \ \sqrt{2\pi}\ (\frac{n - k}{2})^{\frac{n - k}{2}}
%% e^{-\frac{n - k}{2}}\sqrt{\frac{n - k}{2}}\ 2^n} \\
%% \\
%% & = & \frac{n^n\sqrt{n}}
%% {\sqrt{2\pi}\ (\frac{n + k}{2})^{\frac{n + k}{2}}
%% \sqrt{\frac{n + k}{2}}
%% \ \ (\frac{n - k}{2})^{\frac{n - k}{2}}
%% \sqrt{\frac{n - k}{2}}\ 2^n} \\
%% \\
%% & = & \frac{n^n\sqrt{n}}
%% {\sqrt{2\pi}\ (n + k)^{\frac{n + k}{2}}
%% \sqrt{\frac{n + k}{2}}
%% \ \ (n - k)^{\frac{n - k}{2}}
%% \sqrt{\frac{n - k}{2}}} \\
%% \\
%% & = & \frac{n^n\sqrt{\frac{4n}{(n+k)(n-k)}}}
%% {\sqrt{2\pi}\ (n + k)^{\frac{n + k}{2}}
%% \ (n - k)^{\frac{n - k}{2}}}
%% \end{eqnarray*}
%%
%% \pagedone
%%
%% \begin{eqnarray*}
%% & = & \frac{e^{n \ln(n)}\sqrt{\frac{4n}{(n + k)(n - k)}}}
%% {\sqrt{2\pi}\ e^{\frac{n + k}{2}\ln(n + k)}
%% \ e^{\frac{n - k}{2}\ln(n - k)}} \\
%% \\
%% & = & \frac{1}{\sqrt{2\pi}}e^{n \ln(n)-\frac{n + k}{2}\ln(n + k)
%% - \frac{n - k}{2}\ln(n - k)} \\
%% & & * \sqrt{\frac{4n}{(n + k)(n - k)}}
%% \end{eqnarray*}
%% so
%%
%% $\ln(P(s_n = k)) $
%% \begin{eqnarray*}
%% & \approx & \ln(\frac{1}{\sqrt{2\pi}})
%% + n \ln(n)-\frac{n + k}{2}\ln(n + k)\\
%% & & - \frac{n - k}{2}\ln(n - k)
%% + \frac{1}{2}\ln\left(\frac{4n}{(n + k) (n - k)}\right) \\
%% \\
%% & = & \ln(\frac{1}{\sqrt{2\pi}})
%% + n \ln(n)-\frac{n + k}{2}\ln(n + k)\\
%% & & - \frac{n - k}{2}\ln(n - k)
%% + \frac{1}{2}(\ln(4)) + \frac{1}{2}\ln(n)\\
%% & & - \frac{1}{2}\ln((n + k)(n - k))
%% \end{eqnarray*}
%%
%% \pagedone
%% \begin{eqnarray*}
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \frac{2n + 1}{2} \ln(n)-\frac{n + k + 1}{2}\ln(n + k)\\
%% & & - \frac{n - k + 1}{2}\ln(n - k) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \frac{2n + 1}{2}(\ln(n)
%% - \frac{n+k+1}{2n + 1}\ln(n + k) \\
%% & & - \frac{n-k+1}{2n + 1}\ln(n - k)) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \frac{1}{\frac{2}{2n + 1}}(\ln(n)
%% - \frac{n+k+1}{2n + 1}\ln(n + k) \\
%% & & - \frac{n-k+1}{2n + 1}\ln(n - k)) \\
%% \end{eqnarray*}
%%
%% so $ \lim_{n \to \infty} \ln(P(S_n = k)) $
%% \begin{eqnarray*}
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% - \lim_{n \to \infty}\frac{1}{\frac{4}{(2n+1)^2}}
%% (\frac{1}{n} \\
%% & & + \frac{2k + 1}{(2n + 1)^2}\ln(n+k)
%% - \frac{n + k + 1}{(2n + 1)(n + k)} \\
%% & & - \frac{2k - 1}{(2n + 1)^2}\ln(n-k)
%% - \frac{n - k + 1}{(2n + 1)(n - k)})
%% \end{eqnarray*}
%%
%% \pagedone
%%
%% \begin{eqnarray*}
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{1}{4}
%% (\frac{(2n+1)^2}{n} \\
%% & & + (2k + 1)\ln(n+k)
%% - \frac{(n + k + 1)(2n + 1)}{(n + k)} \\
%% & & - (2k - 1)\ln(n-k)
%% - \frac{(n - k + 1)(2n + 1)}{(n - k)}) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{1}{4}
%% (\frac{(2n+1)^2}{n} \\
%% & & - \frac{(n - k + 1)(2n + 1)}{(n - k)}
%% - \frac{(n + k + 1)(2n + 1)}{(n + k)} \\
%% & & + (2k + 1)\ln(n+k) - (2k - 1)\ln(n-k)) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{2n+1}{4}
%% (\frac{(2n+1)}{n} \\
%% & & - \frac{n - k + 1}{(n - k)}
%% - \frac{n + k + 1}{(n + k)}) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{2n+1}{4}
%% (\frac{(2n+1)(n-k)(n+k)}{n(n-k)(n+k)} \\
%% & & - \frac{n(n-k+1)(n+k) + n(n-k)(n+k+1)}{n(n-k)(n+k)})
%% \end{eqnarray*}
%%
%% \pagedone
%%
%% \begin{eqnarray*}
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{2n+1}{4}
%% (\frac{(2n+1)(n^2 - k^2)}{n^3 - nk^2} \\
%% & & - \frac{(n-k+1)(n^2+nk) + (n^2-nk)(n+k+1)}{n^3 - nk^2}) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{2n+1}{4} \\
%% & & (\frac{2n^3 - 2nk^2 + n^2 - k^2}{n^3 - nk^2} \\
%% & & - \frac{n^3 +n^2k - kn^2 -nk^2 +n^2 +nk}{n^3 - nk^2} \\
%% & & - \frac{n^3 +n^2k + n^2 -n^2k -nk^2 -nk}{n^3 - nk^2}) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}})
%% + \lim_{n \to \infty}\frac{2n+1}{4}
%% (\frac{- n^2 - k^2 -2nk}{n^3 - nk^2}) \\
%% \\
%% & = & \ln(\frac{2}{\sqrt{2\pi}}) - \frac{1}{2}
%% \end{eqnarray*}
%%
%%