If given the iterative method:
$$x_{n+1}=e^{x_n x_{n-1}}-1$$
we need to find the rate of convergence if given that it converges to zero.
can I do the following:
$$x_{n+1}+1=e^{x_n x_{n-1}}$$
$$\ln(x_{n+1}+1)=\ln(e^{x_n x_{n-1}})$$
$${x_n x_{n-1}}=\ln(x_{n+1}+1)$$
$${x_n}={\varepsilon_n} +c$$ where c is zero.
and then with the help of Taylor series:
$$\ln(\varepsilon_{n+1}+1)= \ln(1)+{\varepsilon_{n+1}}\ln'(1)+...∼ \varepsilon_{n+1}$$
then I can get that:
$$\varepsilon_{n+1} ∼ {\varepsilon_{n}} {\varepsilon_{n-1}}$$
for all n we define $$|\varepsilon_n|=\delta_n$$ and then we get $$(1) \delta_{k+1}∼|A|\delta_{k}\delta_{k-1}$$ then we define m=rate and guess $$\delta_{n+1}∼\lambda \delta_{n}^m$$ for $$\lambda>0$$ and $$1 \le m \le 2$$ and then we substitute n= k-1, k and get $$\delta_{k+1}∼\lambda \delta_{k}^m, \delta_{k}∼\lambda \delta_{k-1}^m$$ $$\delta_{k+1}∼\lambda \delta_{k}^m∼\lambda(\lambda\delta_{k-1}^m)^m=\lambda^{m+1}\delta_{k-1}^{m^2}$$ then we substitute in (1) and get $$\lambda^{m+1}\delta_{k-1}^{m^2}∼|A|(\lambda\delta_{k-1})\delta_{k-1}$$ we can pick $$\lambda=|A|^\frac{1}{m}$$ $$\lambda^{m+1}\delta_{k-1}^{m^2}∼\lambda^m(\lambda \delta_{k-1}^{m})\delta_{k-1}$$ $$\delta_{k-1}^{m^2}∼\delta_{k-1}^{m+1}$$ then we just solve $$m^2=m+1$$ and get $$m=\frac{1+\sqrt{5}}{2} $$