X^X^X^... = 2Date: 05/17/2000 at 20:13:49 From: Dominic Subject: limits Solve for x: x^x^x^... = 2 I tried to use limits, tried substituting various values, and tried graphing the function on the graphing calculator, but got no solution. Help! Date: 05/17/2000 at 23:07:30 From: Doctor Douglas Subject: Re: limits Hi Dominic, Thanks for submitting your question to Ask Dr. Math. This is a very interesting problem. Before answering it, let's make sure that we mean the same thing by x^x^x^... My convention is that we do all powers in the exponent before raising a base to that exponent, meaning, for example, that: 4^3^2 = 4^(3^2) = 4^9 = 262144 Now let's see if we can make progress on this problem. Let's let: y = x^x^x^... provided that this limit exists (assume it does) Then: y = 2 of course we know y; what's x? But we also see that x^(y) = 2 seems like magic, the tower collapses instantly! x^2 = 2 x = sqrt(2) Check this on a calculator, and see if it converges. (According to my calculator, it does.) - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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