Recursive series convergance

1. Nov 20, 2009

boyo

1. The problem statement, all variables and given/known data
Given a recursive series I'm asked to determine for which values of $$\alpha , x$$ the series will be a convergent series.
I'm also asked to calculate the limit, according to those values.
Given values: $$\alpha, x$$ being non-negative

2. Relevant equations
I am introduced to a recursive series:
$$a_{n+1} = a^{2}_{n} + \alpha ,$$
$$a_{1} = x$$

3. The attempt at a solution
Usually I'm introduced to a less generalized series (in which $$a_{1}$$ is given a real value) but now I find myself awfully confused and cannot determine a good starting point.

I appreciate any kind of help on this.

2. Nov 20, 2009

tiny-tim

Welcome to PF!

Hi boyo! Welcome to PF!

(have an alpha: α and try using the X2 and X2 tags just above the Reply box )
Hint: try calculating the limit first …

then, starting at x, see whether the series goes towards that limit :tongue2:, or shoots off in the wrong direction!

3. Nov 21, 2009

boyo

However, while trying to calculate the limit, in the following manner

an+1 = L
an = L

and coming up with the limit calculation:

$$L = L^2 - \alpha \longrightarrow L^2 - L - \alpha = 0$$

I end up with:

$$L_{1,2} = \frac{1}{2}\pm\sqrt{\frac{1}{4}-\alpha}$$

I can't seem to understand where x takes place here, since it doesn't appear in the L1 or L2. Furthermore I can't come up with any conclusion about $$\alpha$$ from it.

4. Nov 21, 2009

tiny-tim

Hi boyo!

(what happened to that α i gave you? )
Well, if α > 1/4, there isn't a limit, is there?

x is your starting-point … if an = x, an+1 might be higher or lower than x …

there may, for example, be a number for which an always increases on one side of that number, and decreases on the other side.

So one idea would be to prove that, with that starting-point, {an} is monotonic, and goes towards L1 or L2 (and does it fast enough to get there! ).

5. Nov 21, 2009

boyo

an is not x, but a1 is.

So $$a_{1} = x , a_{2} = x^2 + \alpha , x_{3} = (x^2 + \alpha)^2 + \alpha$$
and so on...

I remember from the limit calculation (because I am looking for values in which it will converge) that $$0\leq\alpha \leq \frac{1}{4}$$ but because there is some complex exponential growing, I cannot determine for which values the expression (x')2 will be less than x' (that is: x' = some long exponential value which results in being a fracture).

So basically still in a dead end for me.

6. Nov 21, 2009

tiny-tim

Try this: for what values of an is an+1 > an ?

7. Nov 21, 2009

boyo

it leads to a solution of an x2 < 0 template, that is:
$$\frac{1}{2} - \sqrt{\frac{1}{4} - \alpha}< a_{n} < \frac{1}{2} + \sqrt{\frac{1}{4} - \alpha}$$

but that still leaves me with the unknown relation between alpha and an.

Thanks again.

8. Nov 21, 2009

tiny-tim

ok, so there are three regions, left middle and right (the one above is the middle) …

in each of those three regions, in which direction is an heading as n increases?

9. Nov 21, 2009

boyo

Uh.... I'm freshed out of ideas.
I'm so sorry!!!

10. Nov 21, 2009

tiny-tim

Well, if an+1 > an in the middle region, then so long as an+1 stays in the middle region, that means that {an} is monotone increasing in that region, and it must converge (why? ), so it will converge to the top of the region, ie 1/2 + √(1/4 - α).

Now fill out the proof for that case (the middle region), and then try the other two cases.