Discuss the convergence of $ \left \{ a_n \right\} $ where $ a_{n+1}=\frac{a_0}{2}+\frac{a_n^2}{2},n\geq 1 $

Question

Let $$ a_{n+1} = \dfrac{a_0}{2} + \dfrac{a_n^2}{2} $$ where $ a_1 = \dfrac{a_0}{2} $ and $ n\geq 1 $

Discuss the convergence of $ \left\{a_n\right\} $

Answer 1

As hints Potato, I wrote the following Pascal program.

program Sequence;
 uses
   Crt;
const
  a0=1;
var
  an:Double;
begin
clrscr;
an:=a0/2;
repeat
an:=a0/2+sqr(an)/2;
writeln(an);
until keypressed;
end.

The obtained experimental evidence suggests the complex behavior of the sequence, like the sequence generated by the logistic map. Here are some of my intuitions.

\begin{array}{rl} a_0 & \mbox{Behavior of the sequence} \\ 0 & \mbox{Stabilizes at } 0 \\ 1 & \mbox{Monotonically increases and relatively slowly converges to } 1 \\ 2 & \mbox{Monotonically increases and very quickly “converges” to } \infty \\ 1/2 & \mbox{Monotonically increases and very quickly converges to } 0.2928932188\dots \\ 1/3 & \mbox{Monotonically increases and very quickly converges to } 0.1835034190\dots \\ -1 & \mbox{Alternately and very quickly converges to }–0.4142135623\dots \\ -2 & \mbox{Alternately converges to }– 0.7320508075\dots \\ -1/2 & \mbox{Alternately converges to }– 0.2247448713\dots \\ - 3 & \mbox{Alternately and very very slowly converges to } –1 \\ -4 & \mbox{Stabilizes at a cycle (0,-2) of length } 2 \\ -8 & \mbox{Stabilizes at } 4 \\ -5 & \mbox{Very slowly “converges” to a cycle} \\ & (-2.414088914\dots, 0.4139131902\dots,-2.4143379354\dots,0.414513833\dots) \mbox{ of length } 4 \\ -6 & \mbox{Chaotic?} \end{array}

Answer 2

A more coherent presentation is to define $(a_n)_{n\geqslant0}$ by $a_0=0$ and $a_{n+1}=u_\alpha(a_n)$ for every $n\geqslant0$, for some $\alpha$, where $u_\alpha:x\mapsto\frac12\alpha+\frac12x^2$. (This replaces $a_0$ in the question by $\alpha$ but yields the same sequence $(a_n)_{n\geqslant1}$.) As with every recursive sequence, the first thing to do is to draw on a same picture the graph of the functions $u_\alpha$ and the diagonal which is the graph of the function $x\mapsto x$. This picture then allows to determine the asymptotics of $(a_n)$.

Assume first that $\alpha\gt0$ and that $\alpha$ is so large that $u_\alpha(x)\gt x$ for every $x$, then $a_n\to\infty$ (and $(a_n)$ is increasing). This happens for every $\alpha\gt1$. If $\alpha=1$, $x=1$ is the unique fixed point of $u_\alpha$, hence $a_n\to1$ (and $(a_n)$ is increasing). Lowering again the value of $\alpha$, $u_\alpha(x)=x$ has two solutions, one of them $x_\alpha\lt1$, hence $a_n\to x_\alpha$ (and $(a_n)$ is increasing). This happens for every $\alpha$ in $[0,1]$, and $x_\alpha=1-\sqrt{1-\alpha}$. (You might want to compare with some of the numerical values @AlexRavsky computed.)

When $\alpha\lt0$, $u_\alpha(x)=x$ has two solutions $x_\alpha=1-\sqrt{1-\alpha}$ and $z_\alpha=1+\sqrt{1-\alpha}$, such that $x_\alpha\lt0$ and $z_\alpha\gt2$, and $a_1\lt0$ hence the graph of $u_\alpha$ on $(-\infty,0)$ becomes relevant, where $u_\alpha$ is decreasing, and it becomes less simple to establish the asymptotics of $(a_n)$.

Nevertheless, if $\alpha\gt-3$, the fixed point $x_\alpha$ is stable since $|u'_\alpha(x_\alpha)|\lt1$ hence the convergence to $x_\alpha$ is possible while if $\alpha\lt-3$, $x_\alpha$ is unstable hence there is no convergence to $x_\alpha$. Note that in the regime $\alpha\lt-3$, $(a_n)$ may nevertheless converge for some exceptional values of $\alpha$. For example, if $\alpha=-8$, the positive fixed point of $u_\alpha$ is $z_\alpha=4$ and $a_2=z_\alpha$ hence $a_n\to4$, while if $\alpha$ is just below $-8$, $a_2$ is just above $z_\alpha$ hence $a_n\to\infty$. Likewise, if $\alpha\approx-4.99038$, the positive fixed point of $u_\alpha$ is $z_\alpha=1+\sqrt{1-\alpha}\approx3.44753$ and $a_4=z_\alpha$ hence $a_n\to z_\alpha$.

Answer 3

Let $z_n = \frac{a_n}{2}$. Then $$ z_{n+1} = c + z_n^2$$ where $c = \frac{a_0}{4}$

The set of $c$ for which the sequence $(z_n)$ is bounded is the Mandelbrot set. This indicates that for complex $c$ the question is hopeless. For real $c$ the iteration is conjugate to that of the logistic maps $\lambda x (1-x)$ for $\lambda \in [1,4]$, which includes the chaotic regime in addition to analyzable cases.

http://en.wikipedia.org/wiki/Mandelbrot_set#Basic_properties
http://en.wikipedia.org/wiki/Logistic_map