TITLE: How fast does the sequence $y_t$ defined by $y_{t+1}=y_t(1-y_t)$ decay to zero?
QUESTION [10 upvotes]: The question is in the title; I'm looking for the exact decay rate. Naturally, assume the starting point $y_0$ belongs to $(0,1)$. 
This is motivated by one of the answers to a previous question of mine.

REPLY [2 votes]: I think it is clear that the slowest way this sequence can decay is by taking $y_0 = \frac{1}{2}$. This is because $x -x^2$ take maximum value $\frac{1}{4}$ on $(0,1)$
and $x - x^2$ is increasing on $(0,\frac{1}{2})$. Hence, whatever value of 
$y_0$ you choose, you have $y_ 1 \leq \frac{1}{4}$, and then $y_{2} \leq \frac{3}{16}$,
$y_{3} \leq \frac{39}{256}$, etc, and by choosing $y_0 = \frac{1}{2}$, you get the unique
slowest decreasing sequence. 
It is worth noting that the sequence $(y_t)$ is the sequence obtained by 
applying Newton's method for solving $f(y) = 0$, where $f(y) = e^{\frac{-1}{y}}$
(to extend $f$ to a continuous function on $[0,1]$, we have to define 
$f(0) =0$, and there is clearly no other value of $y \in [0,1]$ for which $f(y) =0$).