Recursive logistic map vs continuous logistic function

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  • #1
nomadreid
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Main Question or Discussion Point

how is the logistic function characterized by the differential equation
df(x)/dx = f(x)(1-f(x))
[with solution f(x)=1/(1+e-x), but this is irrelevant to the question]
the continuous version of the logistic map, given by the recursive function:
xn+1 = xn(1-xn)?
It would seem to me that, in order for the limit of the latter, as n goes to zero, to go to the former, you would need the latter to look like this:
xn+1-xn = xn(1-xn)

A second question: usually the logistic map is given by
xn+1 = r.xn(1-xn) for some real r.
when taking the continuous version, does the r survive as
df(x)/dx = r.f(x)(1-f(x))?

Thanks.
 

Answers and Replies

  • #2
pasmith
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how is the logistic function characterized by the differential equation
df(x)/dx = f(x)(1-f(x))
the continuous version of the logistic map, given by the recursive function:
xn+1 = xn(1-xn)?
It depends on what you mean by "continuous version". Given a suitable [itex]f(x)[/itex], you can either turn it into a continuous-time system by setting
[tex]
\dot x = f(x)
[/tex]
or a discrete-time system by setting
[tex]
x_{n+1} = f(x_n).
[/tex]
But it is not always the case that the two will behave similarly.

A different idea is the approximation of a differential equation by a recurrence relation. If one sets [itex]t = nh[/itex] and [itex]x_n = x(nh)[/itex] for some [itex]h > 0[/itex] then one can approximate [itex]\dot x(t)[/itex] by [itex](x_{n+1} - x_n)/h[/itex] to obtain
[tex]x_{n+1} = x_n + hf(x_n).[/tex]
But this is not the same recurrence relation as [itex]x_{n+1} = f(x_n)[/itex].
 

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