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!!

Darkprince
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 20032011 16:16

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 20032011 21:29
No one?

ghostwalker
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 21032011 08:00
Note: I'm somewhat rusty on what's acceptable.
That doesn't seem very rigorous, with chunks missing. Only one error I can see; bar that, as a sketch it would be fine, but....
(Original post by Darkprince)
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.
I'd prefer better justification on x(n+1)<1, as you haven't shown that it goes to a finite limit at this stage.
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.
And that's the error that would make me think you've not worked through all the details, and were being a bit sloppy.
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.Last edited by ghostwalker; 21032011 at 08:10. Reason: Spelling & more 
Darkprince
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 21032011 14:41
How should I explain more that the sequence converges to that limit? Thanks for the answer!!! Appreciate it!!

DFranklin
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 21032011 15:19
You need to show that if x(n) < 1 then x(n+1) is also less than 1. (And similarly for the > 1 case).

Darkprince
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 21032011 15:30
We know from part a that the only possible finite limits is 1 and 0. So if x(n)>0, the series converges to 1. if x(n)=0 it converges to zero so for my two cases x0 is larger than zero, so the sequence converges to 1. Isn't this enough? What should I say more? Thanks in advance, appreciate all the help!

DFranklin
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 21032011 16:06
(Original post by Darkprince)
We know from part a that the only possible finite limits is 1 and 0. So if x(n)>0, the series converges to 1. if x(n)=0 it converges to zero so for my two cases x0 is larger than zero, so the sequence converges to 1. Isn't this enough? What should I say more? Thanks in advance, appreciate all the help! 
ghostwalker
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 21032011 16:59
(Original post by Darkprince)
We know from part a that the only possible finite limits is 1 and 0. So if x(n)>0, the series converges to 1. if x(n)=0 it converges to zero so for my two cases x0 is larger than zero, so the sequence converges to 1. Isn't this enough? What should I say more? Thanks in advance, appreciate all the help!
Other than that, note DFranklin's comments; he's more one the ball with the methodology etc. than I am (I haven't used this stuff in over 30 years!).Last edited by ghostwalker; 21032011 at 17:16. 
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 21032011 18:00
Thank you guys, what would be the actual calculation in my case?

ghostwalker
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 21032011 18:33
(Original post by Darkprince)
Thank you guys, what would be the actual calculation in my case? 
Darkprince
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 21032011 18:55
Yes i understand but when doing some calculations then it is obvious that the sequence converges! You meant numerical calculations?

ghostwalker
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 21032011 19:13
(Original post by Darkprince)
Yes i understand but when doing some calculations then it is obvious that the sequence converges! You meant numerical calculations?
This together with the fact that you've already proved that x(n+1) > x(n) is what shows convergence.
You're not interested in a specific numerical value as such.Last edited by ghostwalker; 21032011 at 19:16. 
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 21032011 19:31
Oh, so I just have to say that if x(n)<1 then x(n+1)<(3*1)/(1^2+2)<1!
That's correct since I used the recurrance formula, right? 
ghostwalker
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 21032011 21:57
(Original post by Darkprince)
Oh, so I just have to say that if x(n)<1 then x(n+1)<(3*1)/(1^2+2)<1!
That's correct since I used the recurrance formula, right?
Since this is dragging on, here's my thinking:
x(n+1) = 3x(n)/[(x(n))^2+2] = 1  [(x(n))^23x(n)+2]/[(x(n))^2+2]
= 1  [(x(n))2)(x(n)1)]/[(x(n))^2+2]
Then in the interval (0,1) your fraction works out as postive, and since it's being subtracted the whole thing is less than 1. 
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 21032011 22:07
Thank you very much! Appreciate it!

Darkprince
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 24032011 17:17
Ghostwalker sorry for bothering again! I transformed my recursive relation to the one you guided me to do but as you said before it is 1 something positive! Shouldn't I also prove that that something positive is less than one? So 1  (something smaller than one) so the whole thing would be larger than one??

DFranklin
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 24032011 17:21
If I want to prove something is smaller than 1, then showing it is 1  "something positive" is *exactly* what I need to do...

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 24032011 17:36
Yes correct! I don't know what was I thinking! I am gonna have a final look at my solution and if anything appears I'll let you know. Thanks for all the help, heavily appreciate it!!

Darkprince
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 24032011 18:41
Ok just reviewed my solution. I was able to prove that if 0<x(n)<1 then 0<x(n)<x(n+1)<1 and that if 1<x(n)<2 then 1<x(n+1)<x(n).
Then for determining the limit behavior of the sequence when 0<x(0)<1 I said that we have an increasing sequence bounded above by 1 (since 0<x(n)<x(n+1)<1 ), so the series converges to 1 by Monotone Sequence Theorem and 1 is the only possible finite limit for this case.
For determining the limit behavior of the sequence when 1<x(0)<2 I said that we have a decreasing sequence bounded below by 1 ( since 1<x(n+1)<x(n)) so it converges to 1 by Monotone Sequence Theorem and 1 is the only possible finite limit for this case.
This sounds better and more mathematically correct? Do you have anything to add? Thanks again for all the help!!Last edited by Darkprince; 24032011 at 19:44. 
DFranklin
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 24032011 21:37
You seem to be saying that if x_n is monotonically increasing and bounded above by 1, then x_n must tend to 1. This isn't true. (e.g. x_n = 1/n). All you know is that x_n tends to a limit. You then need to argue (using the first part) that that limit must be 1.
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