I have the following iterative process
$$a\_n=a\_{n-1}(1-\phi(a\_{n-1})),\quad 0< a\_0<1,$$
where $\phi(x)$ is a continuous increasing function, $\phi(0)=0$, and if $x\in(0,1)$ then $0< \phi(x)<1$. For instance, if $\phi(x)=x$, then this is just the logistic map.


Clearly, $a\_n$ converges to $0$ as $n\to\infty$ for any function $\phi$, but the convergence rate depends on $\phi$.


My question is how we can estimate the convergence rate. Are there any standard methods which can be used here?


Thanks.



Convergence rate of an iterative process
The standard techniques is as follows. At first, we guess the growth rate by replacing the discrete dynamical system to ODE: $a\_n-a\_{n-1}=-a\_{n-1}\phi(a\_{n-1})$ is replaced to $A'(n)=-A(n)\phi(A(n))$, where function $A(n)$ imitates $a\_n$. Nex, we solve this equation, rewriting it as $g(A(n))'=1$, where $g'(t)=-1/(t\phi(t))$. Our guess is that $g(A(n))$ behaves like $n$ (since it derivative behaves as 1), and the same for $g(a\_n)$. For proving this guess rigorously we introduce a new sequence $b\_n=g(a\_n)$ and write down the formula $b\_n-b\_{n-1}=g(a\_n)-g(a\_{n-1})=(a\_n-a\_{n-1})g'(s)$ for some $s$ between $a\_{n-1}$ and $a\_n$. Next, $(a\_n-a\_{n-1})g'(s)=\frac{a\_{n-1}\phi(a\_{n-1})}{s\phi(s)}$. Assume that $\phi(x):\phi(y)$ tends to 1 when $x/y$ tends to 1 and $x,y$ tend to $+0$. Then we may conclude that our last fraction tends to $1$, and by usual Stolz theorem conclude that $b\_n$ behaves like $n$.


Example: $\phi(x)=x$, $g(t)=1/t$, so we consider the sequence $b\_n=1/a\_n$. We get $b\_{n}-b\_{n-1}=\frac{a\_{n-1}-a\_{n}}{a\_na\_{n-1}}=\frac{a\_{n-1}}{a\_{n}}$, what clearly tends to 1, hence $b\_n\sim n$. 


