2
$\begingroup$

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.

$\endgroup$
  • 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
11
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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).

$\endgroup$

Your Answer

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

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