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Sequence based on sequential square root function

  1. Feb 14, 2006 #1
    Express each term of the sequence [tex]\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, ...\}[/tex] as a power of 2.

    I found [itex]\{2^{\frac{1}{2}}, 2^{\frac{3}{4}}, 2^{\frac{7}{8}}, ...\}[/itex] but I can't get the formula for it so I can find it's limit.
     
  2. jcsd
  3. Feb 14, 2006 #2

    Hurkyl

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    Try a few more then!

    Or, it might be better just to write down a way to compute the next term from the current term.
     
  4. Feb 14, 2006 #3
    If I consider just the fractional exponents as a sequence on its own, there is no common difference or common ratio so I'm stuck in this aspect.
     
  5. Feb 14, 2006 #4

    Hurkyl

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    Do you at least recognize the pattern?
     
  6. Feb 14, 2006 #5
    The numerator and denominator go up by 2, 4, 8, 16, etc., or powers of two.
     
    Last edited: Feb 14, 2006
  7. Feb 14, 2006 #6
    I see... so it'd be like [2^(n) - 1]/2^n ?
     
  8. Feb 14, 2006 #7

    Hurkyl

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    It seems to work for the first three!
     
  9. Feb 15, 2006 #8

    VietDao29

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    Be careful with the parentheses. It should read:
    2 ^ ((2 ^ (n) - 1) / 2 ^ n), or
    [tex]2 ^ {\frac{2 ^ {n} - 1}{2 ^ n}}[/tex]
    :)
     
  10. Feb 15, 2006 #9

    Hurkyl

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    Well, I thought he was writing down the formula for the exponent, not the term of his sequence!
     
  11. Feb 15, 2006 #10
    Yeah, I meant it only as the exponent.
     
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