tag:blogger.com,1999:blog-15628310.post6844697347038129051..comments2021-01-21T13:53:28.761-05:00Comments on Question of the day: And I Will Raise You UpAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-15628310.post-52547851345836930182007-08-14T23:49:00.000-04:002007-08-14T23:49:00.000-04:00haha, you know he's hardcore if he writes his answ...haha, you know he's hardcore if he writes his answer in pseudo-latex code. mr. don, I detected some sarcasm...<BR/><BR/>I'm sorry I was out of town for this one, I would have liked to weigh in. theo is right.<BR/><BR/>x^(x^(x...))...) = 2<BR/>x^(x^(x^(x...))...) = x^2<BR/><BR/>but the left sides of those two equations are identical, so x^2 = 2. thus, x = sqrt(2).Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-38253171562784513572007-08-14T10:23:00.000-04:002007-08-14T10:23:00.000-04:00Oh. With such a lucid explanation, how can anyone...Oh. With such a lucid explanation, how can anyone dispute the correct answer?Mr. Donhttps://www.blogger.com/profile/04844019232053683131noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-50566507863113556002007-08-14T09:28:00.000-04:002007-08-14T09:28:00.000-04:00I'm not sure I want to get involved in this debate...I'm not sure I want to get involved in this debate.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-89097249173259218322007-08-13T17:13:00.000-04:002007-08-13T17:13:00.000-04:00Mr Don, of course, is wrong: there's no particular...Mr Don, of course, is wrong: there's no particular reason that the exponent need by infinite.<BR/><BR/>Indeed, 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:<BR/><BR/>x^(x^(x^(x^(x^(x^(x^(x^(x...)))...) = 2<BR/><BR/>x^(x^(x^(x^(x^(x^(x^(x^(x^(x...)))...)) = x^2<BR/><BR/>so x^2 = 2, and x=\pm\sqrt{2}.<BR/><BR/>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.Theohttps://www.blogger.com/profile/03344294173628793721noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-58793462335692518302007-08-13T12:11:00.000-04:002007-08-13T12:11:00.000-04:00Limit X^(infinity)=2 X-> 1In other words, the infi...Limit X^(infinity)=2 X-> 1<BR/><BR/>In other words, the infinate root of 2 is a hair larger than 1.Mr. Donhttps://www.blogger.com/profile/04844019232053683131noreply@blogger.com