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!!
... and that can be written as...
The function is represented here...
If we define 'attractive fixed point' a value of x for which crosses the x axis with negative slope, then has two attractive fixed points in . Because is an 'odd function' we can analyse what does happen for . The point isn't an attractive fixed point and the initial value will produce the 'all zero sequence'. Setting the positive attractive fixed point, any which satisfies the condition [the 'red line' in the figure...] will produce a sequence monotonically convergent at . If the first iteration will produce an between 0 and 1 and after that the sequence converges monotonically at ...
... more exact is to say that if the sequence converges to 1, if the sequence converges to -1 and if the sequence is 'all zeroes'...
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?