Given the recursive sequence :
$$x_{n+1}=e^{x_{n}x_{n-1}}-1$$
Find the order of convergence if the sequence converges to $0$. I tried getting the $x_n$ sequences in the left hand side but couldn't since we need to do $\ln$ and we get
$$\ln(x_{n+1})=\ln(e^{x_{n}x_{n-1}}-1).$$
What can we do when we have $e$'s and such in the sequences ?
Thank you in advance.
The sequence is essentially identical to the well-known sequence $$y_{n+1} = y_n y_{n-1}$$ and the rate of convergence is identical. The analysis follows below.
We first ensure that the exercise is meaningful. Specifically, we show that there exists $\rho > 0$ such that the sequence $$x_{n+1} = e^{x_n x_{n+1}} - 1$$ is convergent and $$x_n \rightarrow 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}$$ for all choices of $$\{x_1, x_2\} \subset (0,\rho).$$ It is straightforward to verify that if $x_0 > 0$ and $x_1>0$, then $x_n > 0$ for all $n$. By the mean value theorem of differentiation there exists at least one $\xi_n \in (0, x_n x_{n-1})$ such that $$x_{n+1} = e^{x_n x_{n-1}} - 1 = \frac{e^{x_n x_{n-1}} - 1}{x_n x_{n-1}} x_n x_{n-1} = e^{\xi_n} x_n x_{n-1}.$$ It follows that if $\{x_{n-1},x_n\} \subset (0, \rho)$ for a particular choice of $\rho$, then $$0 < x_{n+1} < e^\rho \rho^2.$$ We want $x_{n+1} < \rho$ and so we now choose any $\rho > 0$ such that $\rho e^\rho \leq 1$. The largest acceptable $\rho$ is $$\rho = W(1) \approx 0.567143$$ where $W$ is the so-called Lambert W-function. It follows that if $$\{x_{n-1},x_n\} \subset (0, \rho)$$ then $$0 < x_{n+1} < \rho e^{\rho} x_n \leq x_n.$$ We conclude that the sequence is (strictly) positive and (strictly) decreasing provided that $x_2 < x_1$ and $\{x_1,x_2\} \subset (0, \rho)$. It follows that the sequence is convergent. It follows from the continuity of the natural exponential function that the limit must be a nonnegative solution of the nonlinear equation $$x = e^{x^2} - 1.$$ There are two solutions of this equation namely $x=0$ and $x \approx 0.75$. We can discard $x \approx 0.75$ because it is well outside the largest relevant interval $(0,W(1))$. We conclude that the limit of the sequence is $0$.
We now turn towards the problem of determining the order of convergence. We have already made the critical observation. Specifically, we have expressed the recursion as $$x_{n+1} = e^{\xi_n} x_n x_{n-1}$$ where $$\xi_n \in (0,x_{n-1} x_n).$$ There are now two auxiliary sequences that are relevant. They are $$y_{n+1} = y_n y_{n-1}$$ and $$ z_{n+1} = e^\rho z_n z_{n-1}.$$ We use $\{x_1, x_2\}$ to initialize the auxiliary sequences, i.e., we choose $$y_1 = z_1 = x_1. \quad y_2 = z_2 = x_2.$$ We claim that $$\forall k \in \mathbb{N} \: : \: 0 < y_k \leq x_k \leq z_k.$$ This is trivially true for $$k \in \{1,2\}.$$ Suppose for a moment these inequalities are satisfied for all $k = n-1$ and $k=n$. Then we have $$ 0 < y_{n+1} = y_n y_{n-1} \leq x_n x_{n-1} < x_{n+1} = e^{\xi_n} x_n x_{n-1} < e^\rho x_n x_{n-1} < e^\rho z_n z_{n-1} = z_{n+1}.$$ By induction on $n$ we conclude that $$\forall n \in \mathbb{N} \: : \: 0 < y_n \leq x_k \leq z_n.$$ It follows that $\{x_n\}_{n=1}^\infty$ tends to zero at least as fast as $\{z_n\}_{n=1}^\infty$, but no faster than $\{y_n\}_{n=1}^\infty$. Finally, we note that it is possible to obtain an explicit formula for $y_n$ and $z_n$. We observe that if $t_n = \log y_n$, then $$t_{n+1} = t_n + t_{n-1}$$ is the Fibonacci sequence and if $s_n = e^\rho z_n$, then $$s_{n+1} = s_n s_{n-1}.$$