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Wonderful exponent tower property!

  1. Jan 25, 2006 #1
    Hi dudes, don't be put off by the clumsy notation here.

    I was wondering about these particular exponent towers and this curious property of theirs...

    Let p be a positive integer. Then the exponent tower, composed of p+1 parts each of value p^(1/p), equals p.

    e.g. for p=2.
    tower part = 2^(1/2)
    (2^(1/2))^ ((2^(1/2))^(2^(1/2)))=2
    bah, this looks clumsy, but it's concise written by hand, i.e. a 3 part exponent tower.

    Anyway, I heard that it's difficult to prove any particular case for p, let alone the general case. I had a go myself for case p=2. I set x equals exponent tower and tried to show x=2, but I got nowhere.

    Can anyone post the easiest proof for case p=2?, or any other case?
     
  2. jcsd
  3. Jan 25, 2006 #2

    StatusX

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    Well, that isn't true as you've written it, since the only number you can raise [itex]\sqrt{2}[/itex] to to get 2 is 2 (did I mention 2?). However, if you write it like this:

    [tex]((\sqrt{2})^{\sqrt{2}})^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2}\cdot\sqrt{2}} = (\sqrt{2})^2=2[/tex]

    Then it is true, and it is clear how this extends to the general case.
     
  4. Jan 25, 2006 #3
    Doops, well that's another of reality's amazing mysterys unweaved. Feels like the time I discovered santa claus didn't really exist.
     
  5. Jan 25, 2006 #4

    Hurkyl

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    And FYI, a power tower looks like this:

    [tex]a^{a^{a^\ldots}}[/tex]

    and not

    [tex]((a^a)^a)^\ldots[/tex]

    A quick google search reveals that people seem to use exponent tower to mean the same thing... so it sounds like whoever posed the problem to you has their terms wrong. :frown:
     
  6. Feb 18, 2006 #5

    arildno

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    Dearly Missed

    However, the power tower SEQUENCE of square roots of 2 does converge to 2.

    Here's a proof:
    1. The sequence is bounded above by 2. This is seen in that each member of the sequence must be less than the number where the last square root of 2 is replaced by a 2. That number is easily sen to be 2.

    2. The sequence is increasing, by 1., it must therefore converge to some number x.

    3. x must satisfy the equation:
    [tex]x=(\sqrt{2})^{x}[/tex]
    This equation has two solutions; x=2 and x=4
    Since 2 is the lesser upper bound, the tower converges to 2, rather than to 4
     
    Last edited: Feb 18, 2006
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