Question: Consider $E_n\in(0,1)$ and $E_n\to0,\text{as }n\to\infty$

Assuming that $E_{n+1} = E_n E_{n-1}$, show that, for a real $C$ independent of $n$, $$E_{n+1} \leq C{E_n}^\phi$$ where $\phi=\frac{1+\sqrt{5}}{2}$, the Golden Ratio

My attempt so far:

We can take logarithms of both side to see that

\begin{align*}\log(E_{n+1}) &= \log(E_n E_{n-1})\\ &=\log(E_n)+\log(E_{n-1})\end{align*}

We can see that this fulfils the conditions of the Fibonacci Sequence, $F_{n+1} = F_{n}+F_{n-1}$

Therefore, we can assume that $$\frac{\log(E_{n+1})}{\log(E_n)}\to \phi,\quad\text{as }n\to\infty$$

We can then say that \begin{align*}\lim_{n\to\infty}\left|\frac{\log(E_{n+1})}{\log(E_n)}-\phi\right|&=0\\ \lim_{n\to\infty}\left|\log(E_{n+1})-\phi\log(E_n)\right|&=0\\ \lim_{n\to\infty}\left|\log(E_{n+1}) -\log({E_n}^\phi)\right|&=0\\ \lim_{n\to\infty}\left|E_{n+1} - {E_n}^\phi\right|&=1\end{align*}

I'm not entirely sure that the last line of this is correct - I have attempted to get rid of the logarithms as I would have done in an equation with no limits or absolute values in. Could someone let me know if I have done anything wrong please. I am also struggling to continue from this point so any guidance would be much appreciated too.

  • 1
    I don't think that you can replace $\log(E_{n+1})-\log(E_n^\phi)$ like that. – Simply Beautiful Art Dec 5 '17 at 14:35
  • 1
    I also don't think that you can multiply both sides by $\log(E_n)$ near the bottom, since $\lim_{n\to\infty}\log(E_n)=\log(0^+)=-\infty$. – Simply Beautiful Art Dec 5 '17 at 14:37
  • @SimplyBeautifulArt Thanks for your input, how would you propose I tackle this question then? – lioness99a Dec 5 '17 at 14:39

Hint : The following functional equation : $$f(n+1)=f(n)f(n-1)$$ Have to solution : $$f(n)=e^{c_1F_n+c_2L_n}$$ Where $F_n$ are the Fibonnaci number , $L_n$ the Lucas number and $c_1$,$c_2$ are arbitrary parameters.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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