Vtyjkyuk

I'm stating the problem that I'm stuck with, along with the progress that I've made.

Problem : The iteration defined by $x_k+1=frac12left(x_k^2+cright)$, where $0<c<1$,
has two fixed points $xi_1$ and $xi_2$, where $0 < xi_1 < 1 < xi_2$ . Show that
$$x_k+1-xi_1=frac12left(x_k+xi_1right)left(x_k-xi_1right), ,,k=0,1,2,dots$$
and deduce that $lim_k to inftyx_k=xi_1$ if $0 leq x_0<xi_2$. How does the
iteration behave for other values of $x_0$?

I have progressed in the following way :

The fixed point iteration is given by
beginequationtag1
x_k+1=frac12(x_k^2+c)
endequation

If $xi_1$ is a fixed point, then
beginequationtag2
xi_1=frac12(xi_1^2+c)
endequation

$(1)-(2)$ gives, for $k=0,1,2,dots$,
beginalign*
x_k+1-xi_1&=frac12(x_k^2+c)-frac12(xi_1^2+c)\
&=frac12left(x_k^2-xi_1^2right)\
&=frac12left(x_k+xi_1right)left(x_k-xi_1right)
endalign*

Now,
beginalign*
x_k+1-xi_1&=frac12left(x_k+xi_1right)left(x_k-xi_1right)\
&=frac12left(x_k+xi_1right)frac12left(x_k-1+xi_1right)left(x_k-1-xi_1right)\
&=left(frac12right)^2 left(x_k+xi_1right)left(x_k-1+xi_1right)left(x_k-1-xi_1right)\
&=left(frac12right)^3 left(x_k+xi_1right)left(x_k-1+xi_1right)left(x_k-2+xi_1right)left(x_k-2-xi_1right)\
&dots dots dots dots dots dots\
&=left(frac12right)^k+1 left(x_0-xi_1right) prodlimits_i=0^k left(x_i+xi_1right)
endalign*

How can I ensure that this quantity goes to $0$ as $k to infty$? Can I show that $x_i+xi_1$ will eventually be $<2$? Also, what happens to the situation when $x_0 notin [0,xi_2)$? Any help would be much appreciated. Thank you.

Hints. You have to find those fixed points, given $x_k+1=f(x_k)=frac12left(x^2_k+cright)$, where $f(x)=frac12left(x^2+cright)$ of course, fixed points $f(xi)=xi$ are those satisfying $xi^2-2xi+c=0$ or $xi_1=1-sqrt1-c$ and $xi_2=1+sqrt1-c$.

One thing to observe is that $f'(x)=x geq0, forall xgeq0$ so the function is ascending. As a result
$$0leq x_0<xi_2 Rightarrow 0leq x_1=f(x_0) leq f(xi_2)=xi_2$$
and by induction
$$0leq x_k leq xi_2, forall k$$
which makes $(x_k)$ a bounded sequence when $x_0 in left[0,xi_2right)$.

Another thing to observer is that

• $f(x)geq x iff x in left[0, 1-sqrt1-cright] cup left[1+sqrt1-c,inftyright)$


• $f(x)< x iff x in left(1-sqrt1-c,1+sqrt1-cright)$

As a result:

• $x_0 in left[0, 1-sqrt1-cright] Rightarrow x_1=f(x_0)geq x_0$ and by induction $x_k+1geq x_k$, making $(x_k)$ increasing and bounded, thus it has a limit.


• $x_0 in left[1+sqrt1-c,inftyright) Rightarrow x_1=f(x_0)geq x_0$ and by induction $x_k+1geq x_k$, making $(x_k)$ increasing, but it's not bounded anymore.


• $x_0 in left(1-sqrt1-c,1+sqrt1-cright) Rightarrow x_1=f(x_0)leq x_0$ and by induction $x_k+1leq x_k$, making $(x_k)$ decreasing and bounded, thus it has a limit.

Hint. Note that $0<xi_1 = 1 - sqrt1-c<1<xi_2 = 1 + sqrt1-c$. Show by induction that

i) if $x_0in [0,xi_1)$ then $x_k<x_k+1<xi_1$.

ii) if $x_0in (xi_1,xi_2)$ then $xi_1<x_k+1<x_k$.

iii) if $x_0in (xi_2,+infty)$ then $xi_2<x_k<x_k+1$.

Then recall that a monotone bounded sequence has a finite limit and any finite limit $L$ of the recurrence $x_k+1=frac12left(x_k^2+cright)$ satisfies the equation $L=frac12left(L^2+cright)$ that is $Linxi_1,xi_2$.