functional iteration convergence

Question

functional iteration sequence $x_{n+1} = 2 - (1+c)x_n + cx_{n}^3$ will converge for some values of c to $ \alpha = 1$ for what values of c this sequence will converge?

My attempt to solve this was using contractive mapping theorem First finding a closed set $ D = [1-a, 1+b] $ and show that $ f: D \rightarrow D $ and also $ |f^{'}(x)| < 1$ then by the contractive mapping theorem this series will converge to a unique fixed point in $D$ which is 1.

$f^{'}(x) = -1 - c + 3cx^2$ is for some values of c always less than 1 but for some values not. I'm not sure how to define c value.

Answer 1

When $c=0$ then $x_{n+1}=2-x_n$ and except when $x_0=1$ the sequence will alternate between values $\{x_0,2-x_0\}$ and won't be convergent.

So let assume $c\neq 0$ and study $f'(x)=3cx^2-1-c$

• $f'(x)<1\implies 3cx^2-1-c<1\implies c<\dfrac{2}{3x^2-1}=1-3(x-1)+o(x-1)$

Remember that we want also that $x$ belongs to a neighbourhood of $1$, therefore $c\le 1$

• $f'(x)>-1\implies 3cx^2-1-c>-1\implies (3x^2-1)c>0\implies c>0$

So at least for $f$ to be contractive we need $c\in(0,1]$.

https://www.desmos.com/calculator/g8fqi7qo0y

But as Paresseux noticed, the choice of $x_0$ is also very prominent, indeed the conditions above can be expressed in term of $x$:

• $f'(x)<1\implies x^2<\frac{c+2}{3c}$
• $f'(x)>-1\implies x^2>\frac{1}{3}$

Which you can verify are the regions in orange on the diagram.

If some $x_{n0}$ falls in this region then convergence is ensured for the following terms, but still there are some $x_0$ outside this region for which it is possible that a subsequent term falls in the region and the sequence will converge anyway.

Try for instance $c=0.4$ and $x_0=0.2$

f(0.2);    1.7232
f(%);      1.634280604
f(%);      1.457989590
f(%);      1.198533584
f(%);      1.010722110
f(%);      0.997994028
f(%);      1.000406020

And this is a much difficult issue to predict for which initial values $x_0(c)$ the sequence will always be convergent.

Answer 2

Except when $c=0$, the convergence depends largely on the initial value $x_0$.
For example, if $c \ne 0$ and $|x_0|> \max(100/|c|,5)$, you can prove easily that by induction that: $$|x_{n+1}|>2|x_n|$$ which means the sequence doesn't converge.

Answer 3

You get as progression of the distance to the fixed point $$ \frac{x_{n+1}-1}{x_n-1}=c(x_n^2+x_n)-1. $$ To get convergence you need the right side to be smaller than $1$ in absolute value. This gives $$ 0<c(x^2+x)\le 2\iff \frac14<\left(x+\frac12\right)^2<\frac2c+\frac14 $$ To get $x=α=1$ inside the contracting interval one needs $0<c<1$. The interval is then $$ 0<x<\sqrt{\frac2c+\frac14}-\frac12=1+\frac{4(1-c)}{\sqrt{8c+c^2}+3c} $$