Find a general solution for this recurrence: $$a_n = \sqrt{a_{n-1}a_{n-2}}$$ when $a_1 = 2$, $a_2 = 8$.

My attempt to solve it:

This recurrence isn't a regular one. In order to solve it, I have tried to count the first elements in this recurrence, which are $$a_1 = 2, a_2 = 8, a_3 = 4, a_4 = \sqrt{32}, a_5 = \sqrt{512} ...,$$ but I didn't find any way to proceed.

Any help will be very appreciated.

  • 3
    $\begingroup$ Try $a_n$ = $2^{b_n}$ $\endgroup$ – DanielV Jan 11 '19 at 20:06
  • $\begingroup$ and what is $b_n$? $\endgroup$ – Robo Yonuomaro Jan 11 '19 at 20:10
  • $\begingroup$ I mean how can I find a general solution for $b_n$? $\endgroup$ – Robo Yonuomaro Jan 11 '19 at 20:16
  • $\begingroup$ @RoboYonuomaro using the given recurrence formula for $a_n$. $\endgroup$ – Scientifica Jan 11 '19 at 20:18
  • $\begingroup$ First write the recurrence among $b_n, b_{n-1}, b_{n-2}$ using $\endgroup$ – Will Jagy Jan 11 '19 at 20:18

Hint: By taking the log (with respect to any base) of both sides, we can rewrite the recurrence as $$ \log(a_n) = \frac 12[\log(a_{n-1}) + \log(a_{n-2})] $$ So, the sequence $b_n := \log(a_n)$ satisfies a linear recurrence.


Taking logs, and letting $b_n = \log a_n$, this becomes $b_n =\frac12(b_{n-1}+b_{n-2}) $.

The characteristic polynomial is $x^2-\frac12 x-\frac12 = 0$ which has roots $x =\dfrac{\frac12\pm\sqrt{\frac14+2}}{2} =\dfrac{\frac12\pm\sqrt{\frac94}}{2} =\dfrac{\frac12\pm\frac32}{2} =\dfrac{2, -1}{2} =1, -\frac12 $ so the solutions are $b_n = 1$ and $b_n = (-1/2)^n $.

As a check $\frac12(b_{n-1}+b_{n-2}) =\frac12((-1/2)^{n-1}+(-1/2)^{n-2}) =\frac12(-1/2)^{n-2}(-\frac12+1) =\frac12(-1/2)^{n-2}(\frac12) =\frac14(-1/2)^{n-2} =(-1/2)^{n} $.

Therefore $b_n = u+v(-1/2)^{n}$ for any reals $u$ and $v$, so, $a_n = rs^{(-1/2)^{n}}$ where $r$ and $s$ are positive reals (actually, you can let them be any types that you can raise to a real power).


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