This is elementary, but I found it somewhat surprising. Define $$ a_n = \frac{1+ a_{n-2}}{\sqrt{1+a_{n-1}}} \;,$$ where $a_1$ and $a_2$ are constants. For example, here is a plot for $a_1=5$ and $a_2=3$:

It appears that $\lim_{n\to\infty} a_n \approx 1.61803$, which is close to the golden ratio $\phi$. The limit appears to be independent of the two initial values $a_1$ and $a_2$, which is the aspect I found surprising. Of course $\frac{1+\phi}{\sqrt{1+\phi}}=\phi$, so it is not so surprising that the limit is $\phi$.

Q. For which values of $a_1$ and $a_2$ is $\lim_{n\to\infty} a_n = \phi$?

  • 2
    $\begingroup$ Well, you've said that the final result is independent of $ a_1, a_2 $, so what are you trying to ask? $\endgroup$ – user122283 Mar 20 '14 at 2:15
  • $\begingroup$ Since you want to show that $a_n \to \phi$, perhaps if you set $a_n = b_n+\phi$ you can show that $b_n \to 0$ using your identity for $\phi$. $\endgroup$ – marty cohen Mar 20 '14 at 2:24
  • 1
    $\begingroup$ I think you answered your own question. If there is a limit, then it is the golden ratio and this is independent of the initial values. $\endgroup$ – Claude Leibovici Mar 20 '14 at 4:08
  • $\begingroup$ @SanathDevalapurkar: "appears to be independent of $a_1,a_2$." I don't have a proof that it is. And I am not certain this is true for all $a_1,a_2$. $\endgroup$ – Joseph O'Rourke Mar 20 '14 at 10:01
  • $\begingroup$ it is true for all non negative $a_1,a_2$ $\endgroup$ – chenbai Mar 20 '14 at 12:37

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.