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Infinite Exponent towers

  1. Jan 28, 2005 #1
    Infinite Exponent "towers"

    Please help me solve this problem; I dont even know how to start...

    Solve for x: [tex] x^{x^{x^{x^{...}}}}=2 [/tex]
    Last edited: Jan 28, 2005
  2. jcsd
  3. Jan 28, 2005 #2
    if x=1, x^x^x^x.... =1
    if x >1, x^x^x^x.... = infinity
    so x is undefine
  4. Jan 28, 2005 #3


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    You can say that the equation does not admit a real solution...


    P.S.The problem would be interesting to consider and solve in [itex] \mathbb{C} [/itex]... :wink:
  5. Jan 28, 2005 #4


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    [tex] x^{x^{x^{x^{...}}}}=2 [/tex]

  6. Jan 28, 2005 #5


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    Let the iterated exponent (a teteration) be called LHS (left hand side)

    Then LHS=y=2
    but the iterated exponent is also equal to LHS so that:

    but y=2
    so that:



    Yea, I know it's hard to grasp. I need to work on it too.

  7. Jan 28, 2005 #6
    So far, I got that

    if x=1, my LHS=1
    if 0<x<1, LHS converges to 1
    if x>1, LHS diverges

    I plugged it in on a calculator and it divirged into infinity...

    Go for it, I'll think about that, too.
    Last edited: Jan 28, 2005
  8. Jan 28, 2005 #7


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    Try to see whether it verifies the equation...

    Last edited: Jan 28, 2005
  9. Jan 28, 2005 #8
    It doesn't, [tex]\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}[/tex] does infact come very close to 2 (I think exactly, dont have my 89 with me...), however. But as soon as more terms are piled on, it spirals into infinity.
  10. Jan 28, 2005 #9
    This is obvious for the expression [tex]((x^x)^x)^x...[/tex]
    but how did you show convergence for
    For the first, you get the sequence 2^{-1}, 2^{-1/2}, 2^{-1/4},...,2^{-1/2^i} but for the second the form gets ugly:
    2^{-1}, 2^{-1/2}, {sqrt{2}/2}^sqrt{2}, {1/2}^{{sqrt{2}/2}^sqrt{2}}, ...
    My forays with Windows calculator are giving me oscillations, so I can't be sure it doesn't converge to something less than 1.
    Last edited: Jan 28, 2005
  11. Jan 28, 2005 #10
    I'm sorry, the correct expression for me would have been 0<x<1 will converge to 1
  12. Jan 28, 2005 #11
    2^{-1} is less than 1. :) But I'm not sure the power tower sequence it generates converges to 1, as I can't find a general ith term for the sequence.
  13. Jan 28, 2005 #12


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    [tex] x^{x^{x^{x^{...}}}}=a [/tex]

    Has the solution

    [tex] x=a^{\frac{1}{a}} [/tex]

    Iteration & logarithmation to show it...

  14. Jan 28, 2005 #13
    I see where you are coming from, but I cant see it working the the equation...

    SQRT [2]^SQRT [2]^SQRT [2]=2, but as soon as more SQRT [2]s are stacked, it flies off the mark...
  15. Jan 28, 2005 #14


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    No,it doesn't,trust me...Do you approximate results (intermediary) ??If so,then that's why it may jump over 2...

  16. Jan 28, 2005 #15
    Hmm. Mathworld has an expression for the general solution of the infinite power tower at http://mathworld.wolfram.com/PowerTower.html , but its not as simple as a^{1/a}. However, the equation seems to verify that the power tower of sqrt(2) converges to 2.
    Regarding dex, it's true. sqrt(2) is an irrational number so it can't be rationed about like a finite decimal on a calculator. :)
    Last edited: Jan 28, 2005
  17. Jan 28, 2005 #16
    I idnt approximate. I got 2 for a stack of 3 sqrt twos, but as I did [tex]x^{x^{x^{x^{...}}}}[/tex], it went to infinity.


    Can you show me how your solution was attained?

    I see the link, I'll go check it out.
  18. Jan 28, 2005 #17
    are you having a bad day?? ......I'll show you if x=2^1/2, then x^x^x^x >2,

    let x=2^1/2

    x^x^x^x=x^(x*x*x) = x^(2*x) = 2^(1/2*2)^x = 2^x >2
  19. Jan 28, 2005 #18


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    vincent: a^a^a^a means




    Or, written flatly, it's x^x^x^x := x^(x^(x^x)))
    Last edited: Jan 28, 2005
  20. Jan 28, 2005 #19
    This step is wrong. We're finding the result of x^(x^(x^...)), not ((x^x)^x)^...
    Ie., (sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2,
    but there is no similar way to simplify sqrt(2)^(sqrt(2)^sqrt(2)).
  21. Jan 28, 2005 #20
    so the expression is:
    i was keep doing

    idoit me
    for your [tex]
    it really depend on how ppls read.....
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