## Monday, August 13, 2007

### And I Will Raise You Up

Given that x^(x^(x^(x^(x^(x^(x^(x^(x...)))...) = 2, solve for x.

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Given that x^(x^(x^(x^(x^(x^(x^(x^(x...)))...) = 2, solve for x.

Labels:
Math

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Limit X^(infinity)=2 X-> 1

ReplyDeleteIn other words, the infinate root of 2 is a hair larger than 1.

Mr Don, of course, is wrong: there's no particular reason that the exponent need by infinite.

ReplyDeleteIndeed, if x^(x^(x^(x^(x^(x^(x^(x^(x...)))...) is going to have any meaning at all, then we can solve the equation by substitution:

x^(x^(x^(x^(x^(x^(x^(x^(x...)))...) = 2

x^(x^(x^(x^(x^(x^(x^(x^(x^(x...)))...)) = x^2

so x^2 = 2, and x=\pm\sqrt{2}.

But, of course, \sqrt{2} > 1, so the limit goes to \infty if taken in the usual metric, whereas negative numbers aren't easily raised to irrational exponents (the function is multivalued without a good choice for what value it should take). Which is to say: x ought to equal \sqrt{2}, but only with a careful definition for the infinite exponent.

I'm not sure I want to get involved in this debate.

ReplyDeleteOh. With such a lucid explanation, how can anyone dispute the correct answer?

ReplyDeletehaha, you know he's hardcore if he writes his answer in pseudo-latex code. mr. don, I detected some sarcasm...

ReplyDeleteI'm sorry I was out of town for this one, I would have liked to weigh in. theo is right.

x^(x^(x...))...) = 2

x^(x^(x^(x...))...) = x^2

but the left sides of those two equations are identical, so x^2 = 2. thus, x = sqrt(2).