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Question

let us consider the following quadratic map:

$$s_{n}=s_{n-1}²+c$$ $$(*)$$

There is several papers disscuting the dynamics of (*). I want to know the behavior of this map for $c=-2$ and I am asking if this map is equivalent to the logistic map $$s_{n}=rs_{n-1}(1-s_{n-1})$$ for some $0≤r≤4$.

Empy2 · Accepted Answer · 2014-10-18T18:34:04.747

1

$rs_n=rs_{n-1}(r-rs_{n-1})$, let $t_n=-rs_n$,
so $t_n+r/2=(t_{n-1}+r/2)^2-r^2/4+r/2$, let $u_n=t_n+r/2$
and $u_n=u_{n-1}^2-(r^2-2r)/4$.
So $r^2-2r=8$,$(r-1)^2=9$,$r=4$
$u_n=2-4s_n$, $u_n=u_{n-1}^2-2$,$s_n=4s_{n-1}(1-s_{n-1})$

edited Oct 18 '14 at 18:34

answered Oct 18 '14 at 16:56

Empy2

50,853

@ Michael: Can you specify the homeomorphism between them – DER Oct 18 '14 at 17:34
$u_n=2s_n-1$, then $u_n=u_{n-1}^2-2$ and $s_n=2s_{n-1}(s_{n-1}-1)$ – Empy2 Oct 18 '14 at 17:38
@ Michael: Thank you very much. – DER Oct 18 '14 at 17:41
@ Michael: there is a typo in the question: $s_{n}=rs_{n-1}(1-s_{n-1})$ rathar than $s_{n}=rs_{n-1}(s_{n-1}-1)$ – DER Oct 18 '14 at 18:09
@DER I have adjusted my answer – Empy2 Oct 18 '14 at 19:42

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