Is the Recursive Series Convergent?

[boyo]

Homework Statement

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

Homework Equations

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

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.

Thanks in advance.

 

[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 )

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.

Hint: try calculating the limit first …

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

 

[boyo]

Thanks for the reply, tiny-tim.

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

^a_n+1 = L
^a_n = 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.

 

[tiny-tim]

Hi boyo!

(what happened to that α i gave you? )

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.

Well, if α > 1/4, there isn't a limit, is there?

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

there may, for example, be a number for which a_n 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, {a_n} is monotonic, and goes towards L₁ or L₂ (and does it fast enough to get there! ).

 

[boyo]

a_n 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.

 

[tiny-tim]

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

 

[boyo]

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

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

Thanks again.

 

[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 a_n heading as n increases?

 

[boyo]

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

 

[tiny-tim]

Well, if a_n+1 > a_n in the middle region, then so long as a_n+1 stays in the middle region, that means that {a_n} 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.