The sequence $(x_n)$ is given recursively by $x_{0}=0$, $x_{1}=1$, $ x_{n+1}=x_{n} \sqrt{x^{2}_{n-1}+1}+x_{n-1}\sqrt {x^{2}_{n}+1} , \ \ \ n \geq 1 $

Question

The sequence $(x_n)$ is given recursively by $x_{0}=0$, $x_{1}=1$, $$ x_{n+1}=x_{n} \sqrt{x^{2}_{n-1}+1}+x_{n-1}\sqrt {x^{2}_{n}+1} , \ \ \ n \geq 1$$ Find $x_{n}$

I attempted to do a substitution for $x_{n-1}=\tan(u)$ and $x_{n} =\tan(v)$ and I obtained $$x_{n+1}=\frac{sin(v)+ sin(u)}{cos(v)cos(u)}$$.

This is what I have done so far. Would someone help me out with that. Thank you !

Answer 1

By setting $x_n=\sinh u_n$ as suggested by Ron Gordon, we get $$ \sinh u_{n+1} = \sinh u_n \cosh u_{n-1} + \sinh u_{n-1}\cosh u_n = \sinh(u_n+u_{n-1}) \tag{1}$$ from which $u_{n+1}=u_n+u_{n-1}$ follows from the injectivity of the $\sinh $ function.
Since $u_0=0$ and $u_1=\log(1+\sqrt{2})$, we have $$ u_n = F_n\cdot\log(\sqrt{2}+1) \tag{2}$$ $$ x_n = \color{red}{\frac{1}{2}\left[(\sqrt{2}+1)^{F_n}-(\sqrt{2}-1)^{F_n}\right]}\tag{3}$$ where $F_n$ is the $n$-th Fibonacci number.