Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have $z_{n+1}=\frac{z_{n}^2}{1+cz_{n}}$ where $c>1$ and $z_{1}>0$. What can we say about $z_{n}$? Can we find an explicit formula? Can we at least get an approximation of the form $c_{1}a_{1}(n)+c_{2}a_{2}(n)+\mathcal O(a_{3}(n))$ for some constants $c_i$ and some sequences $a_{i}$ with decreasing order of magnitude?

Edited: the index of the sequence in $\mathcal O$.

share|cite|improve this question
Note that $\frac{z_{n}}{z_{n+1}}=\frac{1+c z_n}{z_n}=c+\frac1{z_n}>c>1$, hence the sequence is strictly decreasing. Ultimately, $z_n\to 0$ with $z_{n+1}\approx z_n^2$. Letting $a=\lim_{n\to\infty}\frac{\ln z_n}{2^n}$ (if it exists?) it seems reasonable to compare $z_n$ with $\exp(2^na)$. By the way, note that $c_1a_1(n)+c_2a_2(n)+{\mathcal O}(a_1(n))=c_2a_2(n)+{\mathcal O}(a_1(n))$. – Hagen von Eitzen Sep 15 '12 at 12:04

Introduce $\gamma=\gamma(c,z_1)$ with $0\lt\gamma\lt1$, as $$ \gamma=z_1^{1/2}\cdot\prod_{n\geqslant1}(1+cz_n)^{-1/2^n}. $$ Then $z_n^{1/2^n}\to\gamma$ decreasingly, and, considering $\delta\lt\gamma$ defined by $\delta=\gamma^3/4$, $$ z_n=\gamma^{2^n}+c\cdot\delta^{2^n}+o(\delta^{2^n}). $$ To prove this, one can start from the formula $$ z_n^{1/2^n}=\gamma\cdot\prod_{k\geqslant n+1}(1+cz_k)^{1/2^k}. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.