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Recursive sequence - show it does not converge to zero

  1. Jul 10, 2012 #1
    Hi guys,

    I'm new here at this forum, but I don't understand this problem.

    Let a1 be a positive real number. Define a sequence an recursively by a(n+1) = (an)^2 - 1. Show that an does not converge to zero.

    (Is there a1 such that the sequence an converges to some non-zero value?)

    I'm not sure if this requires something with epsilon and delta.
     
  2. jcsd
  3. Jul 10, 2012 #2

    micromass

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    Can you prove that the limit of the sequence must be a fixed point of [itex]f(x)=x^2-1[/itex]? That is, it satisfies [itex]x=x^2-1[/itex].
     
  4. Jul 10, 2012 #3
    Well I have this:

    an = (an)^2 - 1
    0 = (an)^2 - an - 1
    an = 1 +/- sqrt (5)/2

    I'm not sure what this tells me though
     
  5. Jul 10, 2012 #4

    micromass

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    See my post 2. Prove that [itex]x=\lim_n a_n[/itex]satisfies [itex]x=x^2-1[/itex].
     
  6. Jul 10, 2012 #5
    a1 = real number > 0

    a2 = (a1)^2 - 1 = (a1+1)(a1 - 1)

    a3 = (a2)^2 - 1 = (a2 + 1)(a2-1)

    a(n+1) = (an)^2 - 1 = (an+1)(an-1). This can only converge to 0 if and only if an equals either +1 or -1. However, that seems contradictory, doesn't it? How can something converge to 0 if it must converge to 1 or -1?
     
    Last edited: Jul 10, 2012
  7. Jul 10, 2012 #6
    a1 = real number > 0

    a2 = (a1)^2 - 1 = (a1+1)(a1 - 1)

    a3 = (a2)^2 - 1 = (a2 + 1)(a2-1)

    a(n+1) = (an)^2 - 1 = (an+1)(an-1). This can only converge to 0 if and only if an equals either +1 or -1. However, that seems contradictory, doesn't it? How can something converge to 0 if it must converge to 1 or -1?
     
  8. Jul 10, 2012 #7
    Here is what I am thinking:

    An+1=an^2-1
    L=L^2-1
    0=L^2-L-1
    L=1+\- sqrt 5/2 if An converges

    Therefore if it does converge it does not
    converge to zero

    Is that enough?
     
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