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Recursive logistic map vs continuous logistic function

  1. Sep 5, 2013 #1
    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.
     
  2. jcsd
  3. Sep 6, 2013 #2

    pasmith

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    Homework Helper

    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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