$$x_{n+1}=x_{n}\mu \left( 1-x_{n}\right)$$ this has a 'STABLE' period 1 orbit with 2 fixed points upto a certain parameter value,its a fact.

$$x_{n+2}=x_{n+1}\mu \left( 1-x_{n+1}\right)$$

For the next iteration,it would have a 'STABLE' period 2 orbit with 4 fixed points for a certain range of parameter values

Calculation of fixed points for period 2 orbit equation yields:

$$x_{n+2} = \mu^2 x_n(1-x_n)(1-\mu x_n(1-x_n))$$

So,to gather the fixed points,we set:

$$x_{n+2} =x_n$$


$$x_n^* = \mu^2 x_n^*(1-x_n^*)(1-\mu x_n^*(1-x_n^*))$$,where $$x_n^*$$ are the fixed points

In order to solve the above expression, my book assumes that 2 of the roots are 'already known' i.e they are the fixed points of the period 1 orbit and hence continues to find the other 2 roots,thereby obtaining all 4 roots of the period 2.

Why would fixed points of the period 1 orbit be the same as the fixed points of the period 2 orbit?I'm trying to visualize this in terms of Poincare' maps but it doesn't make sense why the above assumption is true.

(Just for reference,I add that the fixed points of the period 1 orbit are 0 ,(1-1/mu))

  • 1
    $\begingroup$ If $x_{n+1} = f(x_n) = x_n$, then $x_{n+2} = f(x_{n+1}) = f(x_n)$ (since $x_{n+1} =x_n$) $= x_{n+1}$ (by definition of $x_{n+1}$) $= x_n$ (again since $x_{n+1} =x_n$). $\endgroup$ Mar 27, 2013 at 16:12

2 Answers 2


If $x_{n+1}$ is a function of $x_n$ and $x_{n+1}=x_n$ then $x_{n+2}=x_{n+1}=x_n$ so it is an order $2$ fixed point.


How fixed point coincide with period 2 orbit Here one can see that as parameter changes, then points of period 2 repelling orbit come closer each other and in one point coincide.

In that point 3 points are the same : 2 points of period 2 orbit and 1 point of period 1 orbit.

It is geometric explanation.



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