I need to solve the recurrence relation: $$ a_n = (a_{n-1})^{3}a_{n-2}$$

I tried guessing or changing variables but that didn't work out. I'm also not sure if you could solve this with polynoms for homogeneous recurrence relations.


closed as off-topic by Namaste, José Carlos Santos, Shailesh, JonMark Perry, J. M. is a poor mathematician Sep 4 '17 at 1:39

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If $a_{m}=0$ for some $m$ then $a_n=0$ for all $n\geq m$ (this means that $a_0=0$ or $a_1=0$).

Now assume that $a_n\not=0$ for all $n$. Then solve the linear recurrence $$x_n = 3x_{n-1}+x_{n-2}$$ where $x_n=\ln(|a_n|)$. Since the characteristic equation $\lambda^2-3\lambda-1=0$, has two distinct solutions, $s_1=(3+\sqrt{13})/2$, $s_2=(3-\sqrt{13})/2$, then the general solution is $$x_n=c_1 s_1^n+c_2 s_2^n$$ where the constants $c_1$, $c_2$ have to be determined by using the initial conditions $x_0$ and $x_1$. Note that the sign of $a_n$ depends on the sign of $a_0a_1$.

Can you take it from here?

  • $\begingroup$ Removed the -1 I had put when you had the condition $a_n>0$. $\endgroup$ – Hellen Sep 3 '17 at 14:43
  • $\begingroup$ Yes, thank you. Great answer. $\endgroup$ – Gabi G Sep 3 '17 at 14:44
  • $\begingroup$ No, I am putting it back. The absolute value $|a_n|$ is completely unnecessary. $\endgroup$ – Hellen Sep 3 '17 at 14:47
  • $\begingroup$ @Hellen I do not agree with you! $a_n$ can be negative. The sign of $a_n$ depends on the sign of the product $a_0a_1$. Take for example the solution $-1,1,-1,-1,1,-1,\dots$. $\endgroup$ – Robert Z Sep 3 '17 at 14:52
  • $\begingroup$ I never said $a_n$ cannot be negative. I said that the absolute value is unnecessary. $\endgroup$ – Hellen Sep 3 '17 at 14:52

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