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I have been asked to prove that for $r>4$, the discrete dynamic system $x_{n+1}=r x_n(1-x_n)$, $x_0$ given has a periodic point of period 3. Nevertheless, I am totally lost. I do not know even how to start. If somebody knows any reference where this fact is proven or could provide a proof I would be very thankful.

X1921
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I use $f_a(x)=ax(1-x)$, then $a>1$ implies $$ \frac{d}{dx}f_a^3(0)=(f_a'(0))^3=a^3>1, $$ so that close to $x=0$ one has $f(x)>x$. As $f(1)=0<1$ is below the diagonal, there has to be an intercept point in-between.

One now has to exclude the 1-periodic point $1-\frac1a$, which for $a>4$ is closer to $1$ than to $0$ which should not be identical to the first crossing which is closer to $0$.

More precisely, for $a>4$ at the middle of the interval $$ f_a^3(\frac12)=f_a^2(\frac a4)=f_a(\frac{a^2(4-a)}{16})<0<\frac12. $$ Thus there is a 3-periodic point in $(0,\frac12)$. One finds that $f_a^3(\frac12)<\frac12$ already for $a>3.832$ and generally a triple point for $a\ge3.828427124746191$.

Lutz Lehmann
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  • I named wrongly the variables. In my previous descrption, $a=r$. I do not know if may be this could confuse you when you say $a=r/4$... Or, if it is not the case, could you explain your proof more in detail? Thank you. – X1921 May 20 '17 at 17:42
  • It is just the intermediate value theorem applied to the function $h(x)=f_a^3(x)-x$. – Lutz Lehmann May 20 '17 at 17:57
  • But I am looking for period 3 points which are not 1,2 periodic. Of course, 1-1/r is always a fixed point. – X1921 May 20 '17 at 19:15
  • There are no 2-periodic points that also are 3-periodic. But you are right, one has to exclude the stationary points which amounts to discuss the real roots of the degree 6 polynomial (a*x)^6 - (3*a + 1)*(a*x)^5 + (3*a^2 + 4*a + 1)*(a*x)^4 - (a^3 + 5*a^2 + 3*a + 1)*(a*x)^3 + (2*a^3 + 3*a^2 + 3*a + 1)*(a*x)^2 - (a^3 + 2*a^2 + 2*a + 1)*(a*x) + a^2 + a + 1. – Lutz Lehmann May 20 '17 at 19:50