Calculating the convergence order I have the iterative formula: 
$$x_{n+1}=x_{n-1}\cdot (x_n)^2$$
How can i calculate the convergence's order when the series is not a constant but converge. I know that i have to substitute $x_n=\epsilon_n+c$ but how can i find this constant $c$ if I'm using the right method ?

The full exercise is:

  
*
  
*A. Let $x_{n+1}=(x_n)^3$, find the numbers that the series could convege to.
  
*B. Find the the starting points for which the series will converge and the points for which the series will not converge, and prove it, and for the starting points for which the series will converge find the value to which the series will converge. 
  
*C. find the convergence order for which the series in A converge.
  
*D. You're given an iterative formula: 
  $x_{n+1}=x_{n-1}\cdot(x_n)^2$ (not same series as A), calculate the convergence's order when the series is not a constant but converges.
  
*E. You're given an iterative formula:
  $x_{n+1}=x_n\cdot(x_{n-1})^2$
  calculate the convergence's order when the series is not a constant but converges.



*

*In D+E, I know that i have to substitute $x_n=\epsilon_n+c$ and then express $x_{n+1}$ and $x_{n-1}$ with the relationship with $\epsilon$ but how can i find this constant c if I'm using the right method ?

*In A i found that i have three values: $-1,0,1$, both $0$ and $1$ give a constant series, but $-1$ gives a divergent series.

*In B i did not understand what i have to do exactly.

*In C I found that the convergence's order for 1 is first order and for 0 is 3rd.


Can you give tips and methods for solving.
 A: On the original question, part d)
Consider the logarithm of the sequence, $u_k=\ln(x_k)$. Then
$$
u_{n+1}=2u_n+u_{n-1}
$$
is a linear recursion formula with characteristic roots $1\pm\sqrt2$. If the recursion is not locked in to the smaller root $1-\sqrt2=-\frac1{1+\sqrt2}$, $$u_n=A(1+\sqrt2)^n+B(1-\sqrt2)^n$$ will have a component that diverges with the larger root $1+\sqrt2>2$. 

If the initial values $x_0,x_1$ are smaller than $1$, the coefficients that are a solution of
\begin{align}
\ln|x_0|&=A+B\\
\ln|x_1|&=A(1+\sqrt2)+B(1-\sqrt2)
\\
\ln|x_1|+(\sqrt2-1)\ln|x_0|&=2\sqrt2A
\end{align}
will have $A<0$, so that the divergence of that term leads to the convergence of the $(x_n)$ sequence to zero with a convergence rate $1+\sqrt2$.

On your responses to the full exercise


*

*a) that is correct, also for the other iterations

*b) for the iteration in a) it is relatively easy to see that you get convergence to $0$ for $|x_0|<1$ and divergence to infinity for $|x_0|>1$.

*c) the only point to consider is $0$, there is no sequence that converges to one of $\pm 1$, except the constant one.

*d) you get a curve in the set of starting point pairs $(x_0,x_1)$ from where you get convergence to $\pm 1$ corresponding to the condition $A=0$ (and probably needing $x_0x_1>0$). Else see above.

*e) $q^2-q-2=0$ has roots $2$ and $-1$. So you get for instance directly that $(x_{n+1}x_n)=(x_nx_{n-1})^2$
A: Excellent. You know that you have to substitute $x_n=\epsilon_n+c$.
You don't need to know $c$ - it is the value that your sequence will converge to.
Instead, you want to know about the relationship between $\epsilon_{n+1}$ and $\epsilon_{n}$
Since $x_n=\epsilon_n+c$, we can also say that $x_{n+1}=\epsilon_{n+1}+c$ and that $x_{n-1}=\epsilon_{n-1}+c$
Substituting those into $x_{n+1}=x_{n-1}*(x_n)^2$ gives us:
$\epsilon_{n+1}+c=\left(\epsilon_{n-1}+c\right)*\left(\epsilon_{n}+c\right)^2$
$\epsilon_{n+1}+c=\left(\epsilon_{n-1}+c\right)*\left(\epsilon_{n}^2+2\epsilon_{n}+c^2\right)$
$\epsilon_{n+1}+c=\epsilon_{n-1}\epsilon_{n}^2+2\epsilon_{n-1}\epsilon_{n}+\epsilon_{n-1}c^2+ \epsilon_{n}^2c+2\epsilon_{n}c+c^3$
We can simplify this by saying that any powers of $\epsilon$ are so small that they can be discarded.
$\epsilon_{n+1}+c=\epsilon_{n-1}c^2+2\epsilon_{n}c+c^3$
We can also note that $c$ is the value that the sequence converges to, so $c=c^3$
$\epsilon_{n+1}+c^3=\epsilon_{n-1}c^2+2\epsilon_{n}c+c^3$
and eliminate $c^3$ from each side
$\epsilon_{n+1}=\epsilon_{n-1}c^2+2\epsilon_{n}c$
This is a recursive formula with auxiliary equation $\lambda^2-2c\lambda-c^2=0$
Roots are $\lambda=\frac{2c \pm \sqrt{4c^2+4c^2}}{2}=c\left(1\pm\sqrt 2\right)$
So $\epsilon_n=Ac^n\left(1+\sqrt 2\right)^n+Bc^n\left(1-\sqrt 2\right)^n$
We must have $A=0$ or the values of $\epsilon$ would not tend to zero.
So for large $n$ we have $\epsilon_n=B\left(1-\sqrt 2\right)^n$
Thus $\frac{\epsilon_{n+1}}{\epsilon_n}=\frac{c^{n+1}\left(1 -\sqrt 2\right)^{n+1}}{c^n\left(1 -\sqrt 2\right)^{n}}$
Or $\frac{\epsilon_{n+1}}{\epsilon_n}=c\left(1 -\sqrt 2\right)$
