# Recursive logistic map vs continuous logistic function

1. Sep 5, 2013

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. Sep 6, 2013

### pasmith

It depends on what you mean by "continuous version". Given a suitable $f(x)$, you can either turn it into a continuous-time system by setting
$$\dot x = f(x)$$
or a discrete-time system by setting
$$x_{n+1} = f(x_n).$$
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 $t = nh$ and $x_n = x(nh)$ for some $h > 0$ then one can approximate $\dot x(t)$ by $(x_{n+1} - x_n)/h$ to obtain
$$x_{n+1} = x_n + hf(x_n).$$
But this is not the same recurrence relation as $x_{n+1} = f(x_n)$.
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