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For a small constant $\varepsilon>0$ and positive integer $n$, we are given a sequence of $(0,1)$-reals $\left\{a_0,a_1,\dots\right\}$, with $a_0=n^{-\varepsilon}$ and $$ a_{i+1}=\left(\frac{a_i}{1-a_i}\right)^2. $$ The question is about determining the minimal integer $i$ (or some reasonable lower-bound) such that $a_i$ drops below $\frac{1}{n}$. Clearly, the limit (or fixed point of the function) is $0$, and the rate of convergence seems to be quadratic.

Intuitively, by $\left(\frac{a}{1-a}\right)^2<\frac{a^2}{1-2a}\sim a^2$, I suspect the result to be close to $-\log_{2}\varepsilon$, at least for big enough $n$, however I would need somehow more precise statement.

Matjaž Krnc
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1 Answers1

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Since $a_{i+1} =\left(\dfrac{a_i}{1-a_i}\right)^2 $, $\dfrac{a_{i+1}}{a_i} =\dfrac{a_i}{(1-a_i)^2} $.

Since $\dfrac1{1-x} \le 1+2x $ for $1 \le 1+x-2x^2 $ or $x \le \frac12$, if $a_i \le\frac15 < \frac12$,

$\begin{array}\\ a_{i+1} &=\dfrac{a_i^2}{(1-a_i)^2}\\ &\le a_i^2(1+2a_i)^2 \\ &= a_i^2(1+4a_i+4a_i^2)\\ &= a_i^2(1+4a_i(1+a_i))\\ &= a_i^2(1+5a_i)\\ &\le 2a_i^2\\ \text{so}\\ a_{i+k} &\le (2a_i)^{2^{k-1}}\\ \end{array} $

Note: The exponent of $2$ may be not quite right, but the exponent of $a_i$ probably is correct.

marty cohen
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    Thank you for an answer; I believe your arguments gives a bound of $2^{2^{k+1}}\cdot a_i^{2^k}$. For the sake of formal correctness, I think your statements could be improved in a couple of details... I'll fix them in the evening if you don't want to do it yourself by then. – Matjaž Krnc Jan 13 '18 at 11:38
  • I was rushed when I did it, so I am not surprised if there are minor errors. Make any corrections you want. – marty cohen Jan 13 '18 at 22:55