Find $u_n$, where $u_0 = 2, u_{n+1} = 3u_n+\sqrt{8u_n^2-1}$

Question

Let $u_0 = 2, u_{n+1} = 3u_n+\sqrt{8u_n^2-1}$. Find $u_n$.

I didn't see an easy way of finding the closed form for $u_n$ since it is not a linear recurrence. What other method can we use to solve it?

Answer 1

Using the substitution $u_n = \frac{\cosh(t_n)}{\sqrt{8}}$, we get $$ \frac{\cosh(t_{n+1})}{\sqrt{8}} = \frac{3\cosh(t_n)}{\sqrt{8}}+\sqrt{\cosh^2(t_n)-1} $$ Multiplying this with $\sqrt{8}$ and using the identity $\cosh^2(t)-\sinh^2(t)=1$, we get $$ \cosh(t_{n+1}) = 3\cosh(t_n)+\sqrt{8}\sinh(t_n) $$ As $(3)^2-(\sqrt{8})^2 = 1$, there is a number $d$ with $\cosh(d)=3$ and $\sinh(d)=\sqrt{8}$. Note: $d=\ln (3+\sqrt{8})$. With this $d$ and with the addition theorem of hyperbolic functions, we get $$ \cosh(t_{n+1}) = \cosh(d)\cdot\cosh(t_n) + \sinh(d)\cdot\sinh(t_n) = \cosh(t_n+d) $$ This means $$ t_{n+1} = t_n + d $$ or $$ t_n = t_0 + nd $$ from which the closed form can easily be obtained.