I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
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.
Labels:
Math
Subscribe to:
Post Comments (Atom)
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).