# Sequence behavior

• March 20th 2011, 08:29 AM
Darkprince
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!!
• March 20th 2011, 08:36 AM
chisigma
Quote:

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

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

c) somewhat else...

Kind regards

$\chi$ $\sigma$
• March 20th 2011, 08:47 AM
Darkprince
What do you mean? My approach is completely wrong? Why in b the recursive relation is 3xn/(x(n)^2+2n)?
• March 20th 2011, 09:00 AM
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
• March 20th 2011, 09:44 AM
chisigma
Quote:

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...

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

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

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

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

http://digilander.libero.it/luposabatini/MHF112.bmp

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

Kind regards

$\chi$ $\sigma$
• March 20th 2011, 11:16 AM
Darkprince
Thank you very much for the help :)
• March 20th 2011, 02:24 PM
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!
• March 20th 2011, 02:48 PM
chisigma
Quote:

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 $x_{0}>0$ the sequence converges to 1, if $x_{0}<0$ the sequence converges to -1 and if $x_{0}=0$ the sequence is 'all zeroes'...

Kind regards

$\chi$ $\sigma$
• March 20th 2011, 02:56 PM
Darkprince
Thank you very much Chi Sigma, appreciate all your help :)
• March 21st 2011, 11:17 AM
Darkprince
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?