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Math Help - infinite exponentials

  1. #1
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    Exclamation infinite exponentials

    Hi all,
    A friend asked me if i could sove Ln(x)=-x in exact form.
    The solution i gave was x=e^(-e)^(-e)^(-e)^...
    Ln(x)=Ln[e^(-e)^(-e)^...]=(-e)^(-e)^(-e)^...=-x

    Whilst exploring this answer i have had difficulty believing this to be well defined or even correct.
    A similar problem is x^x^x^x....=2 (or 4)
    Which can be solved as x^2=2 x=root 2
    or x^4=4 x=root 2 also

    I would apreciate any thoughts on this and infinite exponention. All proofs are welcome.
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  2. #2
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    Quote Originally Posted by AndrewAllen View Post
    Hi all,
    A friend asked me if i could sove Ln(x)=-x in exact form.
    The solution i gave was x=e^(-e)^(-e)^(-e)^...
    Ln(x)=Ln[e^(-e)^(-e)^...]=(-e)^(-e)^(-e)^...=-x

    Whilst exploring this answer i have had difficulty believing this to be well defined or even correct.
    First, note that "a^b^c" must be interpreted as a^(b^c) because (a^b)^c= a^(bc) which gives nothing new. If x=e^(-e)^(-e)^(-e)^... then, taking e to the negative power of each side, e^-x= e^-(e^(-e)^(-e)^(-e)^...)= x (since the right side is an infinite tower of "-e"s still). Solve the equation e^-x= x, which is, of course, the same as -x= ln(x). There is no algebraic way to solve such an equation but numerical methods give x as approximately 0.55

    A similar problem is x^x^x^x....=2 (or 4)
    Which can be solved as x^2=2 x=root 2
    or x^4=4 x=root 2 also

    I would apreciate any thoughts on this and infinite exponention. All proofs are welcome.
    For the second, if there were an x satisfying x^x^x^x...= 2, then taking x to the power of each side, x^(x^x^x^x...)= x^2 or x= x^2. No, that is NOT "x^2= 2 x= root 2". It is x^2- x= x(x- 1)= 0. That has roots 0 and 1. But 0^0 is not defined and 1^(1^(...)) is 1 no matter how many powers you take, certainly not 2. There is no x satisfying that equation.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post


    For the second, if there were an x satisfying x^x^x^x...= 2, then taking x to the power of each side, x^(x^x^x^x...) = x^2 or x= x^2. No, that is NOT "x^2= 2 x= root 2". It is x^2- x= x(x- 1)= 0. That has roots 0 and 1. But 0^0 is not defined and 1^(1^(...)) is 1 no matter how many powers you take, certainly not 2. There is no x satisfying that equation.
    But isn't the red just 2 so x^2 = 2?

    This problem is a famous problem dating back to Euler 1778. Here's a website with more information

    Hyperpower Function
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  5. #5
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    Quote Originally Posted by AndrewAllen View Post
    Hi all,
    A friend asked me if i could sove Ln(x)=-x in exact form.
    The solution i gave was x=e^(-e)^(-e)^(-e)^...
    Ln(x)=Ln[e^(-e)^(-e)^...]=(-e)^(-e)^(-e)^...=-x

    Whilst exploring this answer i have had difficulty believing this to be well defined or even correct.
    [snip]
    \ln x = -x \Rightarrow x = e^{-x} \Rightarrow x e^x = 1 \Rightarrow x = W(1) where W is the famous Lambert W-function:

    http://www.cs.uwaterloo.ca/research/tr/1993/03/W.pdf

    Lambert W function - Wikipedia, the free encyclopedia

    Lambert W-Function -- from Wolfram MathWorld
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