$\newcommand{\set}[1]{\left\{#1\right\}}$ I have been asked to solve the following problem: Let $\set{x_n}_{n=1}^\infty$ be a real sequence defined by $$x_{n+1}=\frac{C}{2}+\frac{x_n^2}{2},$$ with $x_1=\frac{C}{2}$, where $C$ is a constant. Try to show that

  1. If $C>1$, then $\set{x_n}_{n=1}^\infty$ is divergent;
  2. If $0< C\leq 1$, then $\set{x_n}_{n=1}^\infty$ is convergent;
  3. If $-3\leq C < 0$, then $\set{x_n}_{n=1}^\infty$ is convergent;

For the case of $C< -3$, when is $\set{x_n}_{n=1}^\infty$ divergent?

the original link were here: A Problem about Limits of Sequence

  • 2
    $\begingroup$ First check: What could be the limit if there is one? $\endgroup$ – Hagen von Eitzen Sep 24 '12 at 6:31

For (1), observe that if the sequence has a limit $x$, then $x$ must satisfy the equation $$x=\frac{C}{2}+\frac{x^2}{2}\;,$$ which can be written $x^2-2x+C=0$. Completing the square, we see that $(x-1)^2+C-1=0$, so $(x-1)^2=1-C$; what does this tell you about $C$?

Added: Here’s a bit more.

When $C=-8$, $x_0=-4$, and $x_n=4$ for all $n\ge 1$, so the sequence converges. Now suppose that $C<-8$, and let $a=|C/2|-4>0$. Then $x_0=C/2=-(4+a)$, so


In general, if $x_n\ge 4+ka$ for some positive integer $k$, then

$$x_{n+1}\ge -(4+a)+\frac12(4+ka)^2=\frac12k^2a^2+(4k-1)a+4>4+(4k-1)a\ge 4+3ka\;,$$

so by induction $x_n\ge 4+3^na$ for $n\in\Bbb N$, and $\langle x_n:n\in\Bbb N\rangle$ diverges to $\infty$.

If $C=-4$, $x_0=-2$, $x_1=-2+\frac42=0$, $x_2=-2+\frac02=-2$, and clearly $x_n=-1-(-1)^n$, so the sequence diverges by oscillating between $-2$ and $0$. This suggests that we should examine the case $-8<C<-4$, i.e., $-4<x_0<-2$.

For what it’s worth, numerical evidence rather strongly suggests that $\liminf_n x_n=x_0$ and $\limsup_nx_n=x_1$ for $C$ in this interval. It is at least true that $x_0\le x_n\le x_1$ for all $n$ when $C$ is in this interval. Indeed, for all $n>0$ we have $$x_n-x_0=\frac{x_{n-1}^2}2\ge 0$$ and hence $x_n\ge x_0$. If $-4<x_0<-2$, then $$-1<1+\frac{x_0}2<0\;,$$ and $x_1=x_0\left(1+\dfrac{x_0}2\right)$, so $0<x_1<|x_0|=-x_0$.

Thus, if $x_n\le x_1$, then $$x_1-x_{n+1}=x_0+\frac{x_0^2}2-\left(x_0+\frac{x_n^2}2\right)=\frac12\left(x_0^2-x_n^2\right)\ge 0\;,$$

and by induction $x_n\le x_1$ for all $n$.

Suppose that $x_n=x_0+\epsilon$ for some small $\epsilon>0$. Then $$x_{n+1}=x_0+\frac{x_n^2}2=x_0+\frac12(x_0+\epsilon)^2=x_0+\frac{x_0^2}2+\epsilon x_0+\frac{\epsilon^2}2=x_1+\epsilon\left(x_0+\frac{\epsilon}2\right)\;,$$ so if $x_0$ is a cluster point of $\langle x_n:n\in\Bbb N\rangle$, then so is $x_1$, and the sequence diverges. However, I have yet been able to show that $x_0$ is a cluster point of the sequence.

  • $\begingroup$ the (1) and (2) are relatively easy to show, the (3) and the last discussion is my point of asking!! $\endgroup$ – van abel Sep 24 '12 at 8:24
  • $\begingroup$ and the initial data is that $x_1=C/2$, sorry about it! $\endgroup$ – van abel Sep 24 '12 at 8:26
  • $\begingroup$ @vanabel: That makes all the difference in the world; now we can actually think about the rest of the question! Please edit the question to add that information. $\endgroup$ – Brian M. Scott Sep 24 '12 at 8:27
  • $\begingroup$ how about to consider the even and odd subsequence ? Which will lead a proof of 3). $\endgroup$ – van abel Oct 6 '12 at 3:11

For $C<-3$, consider the following two cases:

  1. $C=-4$ the sequence is divergent.

  2. $C=-8$ the sequence is constant 4 except the first term.

I can't figure out the answer for $C<-3$.


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