# Find a general solution for this recurrence: $a_n = \sqrt{a_{n-1}a_{n-2}}$

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.

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