What is the solution for x in the infinite exponent tower x^x^x^...=2?

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Homework Help Overview

The problem involves solving the equation for x in the context of an infinite exponent tower, specifically x^{x^{x^{...}}} = 2. This falls under the subject area of mathematical analysis and exponentiation, particularly focusing on the behavior of infinite sequences and limits.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore various values for x, such as 1 and \sqrt{2}, and discuss their implications on the convergence of the infinite exponent tower. Some participants question the assumptions made regarding convergence and divergence, particularly for values greater than 1.

Discussion Status

The discussion includes multiple interpretations of the problem, with some participants suggesting that \sqrt{2} may be a solution while others argue that it diverges. There is an ongoing exploration of the conditions under which the infinite exponent tower converges or diverges, and some participants have offered insights into the nature of the sequence.

Contextual Notes

Participants note that the problem could be interesting to consider in the context of complex numbers, and there are references to the limitations of calculators in evaluating such expressions. The discussion also highlights the importance of distinguishing between different forms of exponentiation in the context of the problem.

  • #31
saltydog said:
Thanks Daniel,
However, my understanding is that the hyperexponent function above converges only when x is in some interval. Can you please tell me how to incorporate math symbols into my posting so I can be more specific? I have Mathematica. Can I export data from there?

Thanks,
SD

Ok Daniel,
I checked out the FAQ and found LaTeX. Nice feature here!
SD
 
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  • #32
I still don't get how this works. I believe you that the answer is 2^1/2 but I still don't know why. Is is calcuable using a calcuator or only provable through algebra? Could someone post a sollution?
 
  • #33
Physics is Phun said:
I still don't get how this works. I believe you that the answer is 2^1/2 but I still don't know why. Is is calcuable using a calcuator or only provable through algebra? Could someone post a sollution?

If you mean how do you find that the answer is sqrt(2), look at my post above. If you mean how could it be that a number greater than 1 converges in an infinite tower, which at least I found strange, just think of it like this: Sqrt(2)^2 is 2, so sqrt(2)^x for any x less than 2 will be less than 2. This means that sqrt(2)^sqrt(2)<2, and so then sqrt(2)^(sqrt(2)^sqrt(2))<2, and so on.
 
  • #34
Though I am still alittle quesitonable about sqrt 2 as an answer, I think I understand the reason the answer is correct.
 

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