1. ## Alternating Sequence

Fix $\displaystyle \alpha>1$, take $\displaystyle x_1 > \sqrt{\alpha}$ and define

$\displaystyle x_{n+1} = \frac{\alpha+x_n}{1+x_n} = x_n + \frac{\alpha-x_n^2}{1+x_n}$

I know this sequence should alternate about $\displaystyle \sqrt{\alpha}$. I.E., the odd terms are greater than $\displaystyle \sqrt{\alpha}$ and the even terms are less than $\displaystyle \sqrt{\alpha}$. However, I cannot seem to show this.

I've tried to use the fact that $\displaystyle \alpha>1$ and $\displaystyle \sqrt{\alpha} < x_n$ but I can't seem to get anything to work.

I would appreciate a hint very much.

2. You may consider a sequence
$\displaystyle y_n=x_n-\sqrt{\alpha}$

and

$\displaystyle y_n \rightarrow 0.$

Now you may examine

$\displaystyle y_{n+1}-y_n$

how it behaves around zero.

3. Originally Posted by Haven
Fix $\displaystyle \alpha>1$, take $\displaystyle x_1 > \sqrt{\alpha}$ and define

$\displaystyle x_{n+1} = \frac{\alpha+x_n}{1+x_n} = x_n + \frac{\alpha-x_n^2}{1+x_n}$

I know this sequence should alternate about $\displaystyle \sqrt{\alpha}$. I.E., the odd terms are greater than $\displaystyle \sqrt{\alpha}$ and the even terms are less than $\displaystyle \sqrt{\alpha}$. However, I cannot seem to show this.

I've tried to use the fact that $\displaystyle \alpha>1$ and $\displaystyle \sqrt{\alpha} < x_n$ but I can't seem to get anything to work.

I would appreciate a hint very much.
Problems of this type have been solved a lot of times, recently in...

http://www.mathhelpforum.com/math-he...it-170087.html

The recursive relation can be written as...

$\displaystyle \displaystyle \Delta_{n}= x_{n+1}- x_{n} = \frac{\alpha-x_{n}^{2}}{1+x_{n}} = f(x_{n})$ (1)

Supposing $\displaystyle \alpha>0$ there is an 'attractive fixed point' at $\displaystyle x_{0}=\sqrt{\alpha}$ and the condition for monotonic convergence at $\displaystyle x_{0}$ is that...

$\displaystyle |f(x)| \le |x_{0}-x|$ (2)

It is easy to verify that the condition (2) is satisfied for $\displaystyle 0<\alpha<1$ so that in this case the sequence will be increasing if $\displaystyle x_{1}< \sqrt{\alpha}$ and decreasing if $\displaystyle x_{1}> \sqrt{\alpha}$. For $\displaystyle \alpha>1$ the sequence remains convergent but is 'oscillating'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$