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Recursive series convergance

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a recursive series I'm asked to determine for which values of [tex]\alpha , x [/tex] the series will be a convergent series.
    I'm also asked to calculate the limit, according to those values.
    Given values: [tex]\alpha, x[/tex] being non-negative


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

    3. The attempt at a solution
    Usually I'm introduced to a less generalized series (in which [tex] a_{1}[/tex] 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.
     
  2. jcsd
  3. Nov 20, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi boyo! Welcome to PF! :smile:

    (have an alpha: α and try using the X2 and X2 tags just above the Reply box :wink:)
    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! :rolleyes:
     
  4. Nov 21, 2009 #3
    Thanks for the reply, tiny-tim.

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

    an+1 = L
    an = L

    and coming up with the limit calculation:

    [tex]L = L^2 - \alpha \longrightarrow L^2 - L - \alpha = 0 [/tex]

    I end up with:

    [tex]L_{1,2} = \frac{1}{2}\pm\sqrt{\frac{1}{4}-\alpha} [/tex]

    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 [tex]\alpha [/tex] from it.
     
  5. Nov 21, 2009 #4

    tiny-tim

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    Hi boyo! :smile:

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

    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! :rolleyes:). :smile:
     
  6. Nov 21, 2009 #5
    an is not x, but a1 is.

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

    I remember from the limit calculation (because I am looking for values in which it will converge) that [tex]0\leq\alpha \leq \frac{1}{4}[/tex] 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.
     
  7. Nov 21, 2009 #6

    tiny-tim

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    Try this: for what values of an is an+1 > an ? :smile:
     
  8. Nov 21, 2009 #7
    it leads to a solution of an x2 < 0 template, that is:
    [tex]\frac{1}{2} - \sqrt{\frac{1}{4} - \alpha}< a_{n} < \frac{1}{2} + \sqrt{\frac{1}{4} - \alpha}[/tex]

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

    Thanks again.
     
  9. Nov 21, 2009 #8

    tiny-tim

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    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? :smile:
     
  10. Nov 21, 2009 #9
    Uh.... I'm freshed out of ideas.
    I'm so sorry!!! :blushing:
     
  11. Nov 21, 2009 #10

    tiny-tim

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    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? :wink:), 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. :smile:
     
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