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Tower of sqrt(2)

  1. Jun 10, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    Let [itex]a_1=\sqrt{2}[/itex], [itex]a_2=(\sqrt{2})^{a_1}[/itex],... [itex]a_{n+1}=(\sqrt{2})^{a_n}[/itex]. Show that a_n-->2. you may use any relevant fact from calculus.

    3. The attempt at a solution
    I noticed that [itex]\ln(a_1)=\ln(\sqrt{2})=\ln(2)/2[/itex], [tex]\ln(a_2)=\ln((\sqrt{2})^{\ln(2)/2})=(\ln(2)/2)^2[/tex], ..., [itex]\ln(a_n)=(\ln(2)/2)^n[/itex]. But ln(2) < 2. So ln(2)/2 < 1, and thus ln(a_n)-->0 <==> a_n-->1. What's wrong?
     
    Last edited: Jun 10, 2007
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  3. Jun 10, 2007 #2

    Hurkyl

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    [tex]\log a_2 = \log (\sqrt{2})^{a_1} [/tex]

    not

    [tex]\log a_2 = \log (\sqrt{2})^{\log a_1} [/tex]
     
  4. Jun 10, 2007 #3

    matt grime

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    we know a_n= (sqrt(2))^{a_{n-1}}

    eugh.

    Suppose there is a limit. Let that limit be a. Can you show from this that a=2? I.e. what equality does a satisfy, and thus what is a (by inspection).

    Now we sjust need to show that a_n converges to anything? What are the only things we know converge for sure? increasing bounded above sequences, or decreasing bounded below ones.
     
  5. Jun 10, 2007 #4
    Did someone say... proof by induction!?:biggrin:
     
  6. Jun 10, 2007 #5

    NateTG

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    One way to do this is to look at the sequence:
    [tex]b_n=2-(a_n)[/tex]
    So if [itex]b_n \rightarrow 0[/itex] you're good. Maybe you can look at:
    [tex]\frac{b_n}{b_{n+1}}[/tex].
     
  7. Jun 10, 2007 #6

    quasar987

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    Nice. Suppoes a_n-->a. Then [itex]a_{n+1}=\sqrt{2}^{a_n}\rightarrow a[/itex]. But also, [itex]\sqrt{2}^{a_n}\rightarrow \sqrt{2}^a[/itex]. So [itex]\sqrt{2}^a=a[/itex]. So a=2.

    Did you follow through with that idea? Cuz it did cross my mind, but I rejected it quickly because that [tex]\frac{b_n}{b_{n+1}}[/tex] fraction is not too cooperative as far as i can see.
     
    Last edited: Jun 11, 2007
  8. Jun 10, 2007 #7

    quasar987

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    How could we show the sequence is bounded though? :/
     
  9. Jun 10, 2007 #8

    NateTG

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    [tex]b_{n+1}=2-a_{n+1}=2-\sqrt{2}^{2-b_n}=2-\frac{2}{\sqrt{2}^{b_n}}[/tex]
    I'm pretty sure it's possible to show that
    [tex]0 < \frac{2-\frac{2}{\sqrt{2}^{b_n}}}{b_n} < 1[/tex]
    if [itex]b_n \in (0,1)[/tex]
     
    Last edited: Jun 10, 2007
  10. Jun 10, 2007 #9

    Hurkyl

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    One method is guessing what the bound is, and then proving that the sequence never exceeds the bound.
     
  11. Jun 11, 2007 #10

    matt grime

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    That doesn't follow at all. in fact you appear to be supposing the answer, unless it is just a typo, which seems most likely. If a_n converges to a, then a satisfies a=sqrt(2)^a.
     
  12. Jun 11, 2007 #11
    I'm thinking induction-- which would just be what Hurkyl hinted at.
    [tex]a_1 = \sqrt{2}^{\sqrt{2}} < 2[/tex] => true for n=1.
    Assume true for n, i.e. [tex]a_n = \sqrt{2}^{a_n -1} < 2[/tex]... and show that this implies n+1 true, which would be something like:
    [tex]a_{n+1} = \sqrt{2}^{a_n} < \sqrt{2}^{2} = 2[/tex]
    Or something.
    And it's a monotone increasing sequence, so...
     
    Last edited: Jun 11, 2007
  13. Jun 11, 2007 #12

    quasar987

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    Yeah I realized induction does the trick at work today :smile:


    (typo corrected in post #6)
     
  14. Jun 11, 2007 #13

    Office_Shredder

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    That's the point though..... IF an is a convergent sequence, by algebra of limits, it must converge to 2 by the argument given. This is always a good way to start, because often to show a sequence converges, it's best to know what it converges to. For example, in this case if we didn't have the number 2, it would be difficult to pull the number out of mid-air and show it's an upper bound of the sequence (which it is regardless of whether we've verified it's going to be the limit, so there's no circular logic); so starting by showing it converges to something and then finding the limit would have been a much harder way to progress
     
  15. Jun 12, 2007 #14

    quasar987

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    What guarantees us once we're at

    [tex]a^{1/a}=2^{1/2}[/tex]

    that a=2 is the only solution?
     
  16. Jun 12, 2007 #15

    morphism

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    It's not. a=4 works too.

    I guess you're going to need to use some bounds.
     
  17. Jun 12, 2007 #16

    quasar987

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    How do we methodically find all the solutions to this?
     
  18. Jun 13, 2007 #17

    NateTG

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    [tex]\frac{d}{da}a^{\frac{1}{a}}=\frac{1-\ln a}{a^2}a^{\frac{1}{a}}[/tex]
    Since the derivative only changes sign once for [itex]a>0[/itex], those are the only positive real solutions. I expect that there is an infinite number of complex solutions.
     
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