1. ## Sequence behavior

Let x(n+1)= (3*x(n)) / (x(n))^2+2 n belongs in N, xo>=0 is some starting value

a) Show that if the sequence converges to a finite limit then this is 0 or 1.
b) Show that if 0<x(n)<1 then 0<x(n)<x(n+1)<1 and determine the limit behavior of the sequence when 0<x0<1.
c) Show that if 1<x(n)<2 then 1<x(n+1)<x(n) and determine the limit behavior of the sequence when 1<x0<2
d) What happens for x0>=2?

My work and if someone can check it and correct me if I am anywhere wrong.

a) If x(n) tends to K in R then so does x(n+1) so K=3K/K^2+2 so K=0 or K=1 or K=-1 but we cross out K=-1 since x0>=0 and so ...>x2>x1>x0 (x0>0) or all x(n)=0 (x0=0)

b) x(n+1)-x(n)=(x(n)*(1-(x(n))^2)) / ((x(n))^2+2) which is >0 if 0<x(n)<1 and so x(n+1)>x(n). So the sequence is an increasing one with 0<x(n)<x(n+1)<1, since the only positive finite limit is 1.

When 0<x0<1 our sequence is increasing and so 1>...>x2>x1>x0>0. It is bounded above by 1 since is the only positive finite limit.

c) If 1<x(n)<2 then x(n+1)-x(n)<0 so x(n)>x(n+1) so our sequence is decreasing but >1 since is the only positive finite limit and so 1<x(n+1)<x(n)

When 1<x0<2 then 1<.....<x2<x1<x0 our sequence is decreasing and bounded below by 1.

d) If x0=2 then all x(n) are zero so we have a sequence full of zeros.
If xo>2 we observe that x1<x0 but 1>...>x3>x2>x1 and our sequence is bounded above by 1 since is the only possible finite limit.

This is my work. Could anyone please check it and confirm my work? Thanks in advance, appreciate it!!

2. Originally Posted by Darkprince
Let x(n+1)= (3*x(n)) / (x(n))^2+2 n belongs in N, xo>=0 is some starting value...
There is an essential detail: the recursive relation is...

a) $\displaystyle \displaystyle x_{n+1}= \frac{3\ x_{n}}{x^{2}_{n} +2}$

b) $\displaystyle \displaystyle x_{n+1}= \frac{3\ x_{n}}{x^{2}_{n} +2n}$

c) somewhat else...

Kind regards

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

3. What do you mean? My approach is completely wrong? Why in b the recursive relation is 3xn/(x(n)^2+2n)?

4. Oh sorry, now I understand. I meant n belongs in N in my first sentence. So the sequence is just x(n+1)= (3*x(n)) / (x(n))^2+2

5. Originally Posted by Darkprince
Oh sorry, now I understand. I meant n belongs in N in my first sentence. So the sequence is just x(n+1)= (3*x(n)) / (x(n))^2+2
... so the recursive relation is...

$\displaystyle \displaystyle x_{n+1}= \frac{3\ x_{n}}{x^{2}_{n} + 2}$ (1)

... and that can be written as...

$\displaystyle \displaystyle \Delta_{n}= x_{n+1}-x_{n} = \frac{x_{n}- x^{3}_{n}}{x^{2}_{n}+2}= f(x_{n})$ (2)

The function $\displaystyle f(x)$ is represented here...

If we define 'attractive fixed point' a value of x for which $\displaystyle f(x)$ crosses the x axis with negative slope, then $\displaystyle f(x)$ has two attractive fixed points in $\displaystyle x= \pm 1$. Because $\displaystyle f(x)$ is an 'odd function' we can analyse what does happen for $\displaystyle x_{0}>0$. The point $\displaystyle x=0$ isn't an attractive fixed point and the initial value $\displaystyle x_{0}=0$ will produce the 'all zero sequence'. Setting $\displaystyle x^{*}=1$ the positive attractive fixed point, any $\displaystyle x_{0}$ which satisfies the condition $\displaystyle |f(x_{0})|< |x_{0}-x^{*}|$ [the 'red line' in the figure...] will produce a sequence monotonically convergent at $\displaystyle x=1$. If $\displaystyle x_{0}>2$ the first iteration will produce an $\displaystyle x_{1}$ between 0 and 1 and after that the sequence converges monotonically at $\displaystyle x=1$...

Kind regards

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

6. Thank you very much for the help

7. Sorry Chi Sigma for bothering again. I understood your logic. My solution was correct? I mean I prefer to work on my approach, so if you can tell me if it was correct or if I have to add anything I would greatly appreciate it! Thank you!

8. Originally Posted by Darkprince
Sorry Chi Sigma for bothering again. I understood your logic. My solution was correct? I mean I prefer to work on my approach, so if you can tell me if it was correct or if I have to add anything I would greatly appreciate it! Thank you!
Your work was correct!... a minor detail is in the point a)...if the sequence converges to a finite limit then this is 0 or 1...

... more exact is to say that if $\displaystyle x_{0}>0$ the sequence converges to 1, if $\displaystyle x_{0}<0$ the sequence converges to -1 and if $\displaystyle x_{0}=0$ the sequence is 'all zeroes'...

Kind regards

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

9. Thank you very much Chi Sigma, appreciate all your help

10. Hello Chi Sigma, sorry for bothering again. I want to ask if there is anything more to add to my answers for part b and c so I can deduce that my series is converging in that cases. I mean the pattern is obvious but is there something to add up to my answers to explain them better mathematically?