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Question about exponents and i .

  1. Jun 28, 2012 #1
    how would I simplify or figure out [itex] i^{i^{i..}} [/itex]
    this keep going an infinite tower of i's
     
    Last edited: Jun 28, 2012
  2. jcsd
  3. Jun 28, 2012 #2
    (a^b)^c = a^(bc)

    i^i^i^... = i^(i*i*i*...)
    [tex]= \lim_{n→∞}i^{(n\,i)} = \lim_{n→∞}(i^{\,i})^{n}[/tex]
    [tex]i^{\,i} = e^{-\pi/2}[/tex]
    [tex]\lim_{n→∞}(i^{\,i})^{n} = \lim_{n→∞}\frac{1}{e^{n\pi/2}} = 0[/tex]
     
  4. Jun 28, 2012 #3
    why does (a^b)^c equal a^(bc)
    x^x^x does not equal x^(x^2)
     
  5. Jun 28, 2012 #4

    chiro

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    Hey cragar.

    In terms of this I don't think you are going to get a closed-form solution but there is a way to go from say i^i using eulers identity.

    To do this we use i^i = e^(i * ln(i)) = e^(itheta) where theta = ln(i). We use the principle branch for ln(i), which gives cos(ln(i)) + isin(ln(i)) = e^(-pi/2) = 0.207879576.

    Another way is to use i^i = ((-1)^(1/2))^i = e^(1/2*i^2*pi) = e^(-pi/2), since i has principal argument of pi/2 (cos(pi/2) + isin(pi/2) = i).

    Applying this again gives us (e^(-pi/2))^i = e^(-ipi/2) = cos(-pi/2) + isin(-pi/2) = -i. But now are back to the imaginary i in negative form!

    I'm guessing (but you will have verify) that we will eventually get back to i after doing this process again (kind of like a spinor).

    Because of this, the value in the infinite limit will not exist in the sense of something fixed and something that converges to a definitive value.
     
  6. Jun 28, 2012 #5

    Mute

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    (a^b)^c = a^(bc), but that is not what you have. You have a^(b^c), which is not equal to a^(bc). Exponentiation is not associative.

    As for the actual limit, the result is apparently quoted on this wikipedia page: http://en.wikipedia.org/wiki/Super-exponentiation#Extension_to_complex_bases

    There is a formula for a number x raised to itself infinitely many times in terms of the Lambert-W function (which is also quoted on that page), but there does not appear to be a proved extension for complex numbers.
     
    Last edited: Jun 28, 2012
  7. Jun 28, 2012 #6

    haruspex

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    If the limit exists, it is a solution of iz = z.
    Writing z = re:
    r π cos(θ) = 2θ
    r π sin(θ) = -2 ln(r)
    The solution at http://en.wikipedia.org/wiki/Super-e..._complex_bases [Broken] does indeed satisfy these equations.
     
    Last edited by a moderator: May 6, 2017
  8. Jun 29, 2012 #7

    HallsofIvy

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    This is true but cragar i talking about [itex]a^{b^c}[/itex]

     
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