I've read in an article of this site that if you consider the sequence defined for $n \in \mathbb{N}$ by $$ u_0=0 \text{ and }u_{n+1}=\left(u_n\right)^2+\frac{1}{4} $$ Then the sequence $\displaystyle \left(u_n\right)_{n \in \mathbb{N}}$ converges to $1/2$, no problem so far. However, it also explains that the sequence for $\alpha>0$ given by $$ v_0=0 \text{ and }v_{n+1}=\left(v_n\right)^2+\frac{1}{4}+\alpha $$ is a divergent sequence but if we let $N$ be the number of iteration needed so that $v_N>2$,it was written that $$ N\left(\alpha\right)\underset{(0^{+})}{\sim}\frac{\pi}{\sqrt{\alpha}} $$ Is there a way to show this equivalence ? Here is a link to the article

  • 1
    $\begingroup$ "I've read in an article of this site" – what about adding a link to that article, so that we know the context? $\endgroup$ – Martin R Apr 10 '18 at 13:01
  • 1
    $\begingroup$ The article can be found at pdfs.semanticscholar.org/dbed/…. $\endgroup$ – Ewan Delanoy Apr 10 '18 at 14:38
  • $\begingroup$ Isn' there already a proof in the article? $\endgroup$ – Seewoo Lee Apr 12 '18 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.