Determine Value of Constant for Iterative Convergence The question I am trying to answer is stated as follows:
The iteration
$x_{n+1} = 2 - (1+c)x_n + cx_n^3$
will converge to $\alpha = 1$ for some values of $c$ (provided that $x_0$ is sufficiently close to $\alpha$). Find the values of $c$ for which this is true. For what values of $c$ will the convergence be quadratic?
In approaching this problem, I have assumed that it is some variant of Newton's method. A large part of this assumption comes from the fact that we are both using $x$ rather than $y$, and that convergence is assumed so long as our initial guess is "sufficiently close". At first, I tried exploring with other basic iterative methods that aren't root finding (such as Euler's method) with the consideration that perhaps $c$ represented some step size. Attempts at that had failed.
If I can show that this is some variant of Newton's method, I think that quadratic convergence would come automatically. However, that makes me think that that wouldn't be the case. If it was, why would they ask? In that case, it could be some other method with super-linear convergence, but gains quadratic convergence with stricter values of $c$. The last thing here that throws me off is, expanding the parenthesis, I get a leading term of $-x_n$. I'm not sure how to approach this either.
I appreciate any help or suggestions!
 A: Since you're wondering whether $x_n\to1$ it may make things more transparent if you set $x_n=y_n+1$ and check whether $y_n\to0$ (just because "small" can be easier to see than "close to $1$"). You get $$y_{n+1}=(c-1)y_n+cy_n^2.$$
Now regardless of $c$, if $y_n$ is small enough then $cy_n^2$ is even smaller. When could you conclude that $(c-1)y_n$ is smaller than $y_n$?
Oops That's all wrong. I misread the original recurrence. So substitute $x_n=y_n+1$ into the actual recurrence. You get $$y_{n+1}=Ay_n+By_n^2+Cy_n^3,$$where $A$, $B$, and $C$ are certain constants depending on $c$. As above, if $y_n$ is small enough then the last two terms are even smaller, so you need to figure out what valuues of $c$ will make $Ay_n$ even smaller, given that $y_n$ is small.
A: As always with iterative systems  $x_{n+1}=f(x_n)$ you get convergence around the fixed point $x_*$ if $|f'(x_*)|<1$. In this case,
$$
f(x)=2-(1+c)x+cx^3\implies f'(x)=-(1+c)+3cx^2,\quad f'(1)=-1+2c
$$
Thus for $c\in (0,1)$ you get a locally contractive iteration around $x_*=1$.
