psi20 Posted July 24, 2006 Share Posted July 24, 2006 Putting powers upon powers aren't taught in school. But this problem is interesting. The problem is to find x. x^x^x^x^x^x... = 2. Find x. This was in the book The Art of the Infinite. Now the question is whether or not there is such a number x. If there was, it'd have to be greater than 1. But it'd have to be less than 2. The idea is that because x^x^x^x^x... = 2, it is the same as x^x^x^x... = 2 and so you can have x^2 = 2, and x is the square root of 2. Now the steps seemed logical to me, but I was skeptical so I tried to check it on the calculator. But x can't be the square root of 2 because this diverges. I tried it on the calculator different ways, pressing Ans ^ Ans and then going step by step, but it still showed that it diverged. So the question still remains. What is x? I played on the calculator a bit and saw something interesting. sqr rt 2 to sqrrt 2 to sqr rt 2. cube root of 3 of to the cube root of 3 to the cube root of 3 to the cube root of 3. and so on. Interesting to look at, but I didn't make much sense of it. Oh well. Link to comment Share on other sites More sharing options...

CPL.Luke Posted July 25, 2006 Share Posted July 25, 2006 and what were all of these roots tell you? other than that you can raise the square root of two to the square root of two? I don't think that there is any algebraic solution too your equation, I think you would just need a computer run numbers through till it found one that worked Link to comment Share on other sites More sharing options...

shmoe Posted July 25, 2006 Share Posted July 25, 2006 I tried it on the calculator different ways, pressing Ans ^ Ans and then going step by step, but it still showed that it diverged. This isn't what x^x^x... represents. It's the limit of the sequence x^x, x^(x^x), x^(x^(x^x)),... what you are doing on your calculator appears to be x^x, (x^x)^(x^x), ((x^x)^(x^x))^((x^x)^(x^x)),... which is something different. With x=sqrt(2) you'll get: x=1.414213562.... x^x=1.632526919... x^(x^x)=1.760839556... You can keep going, it will be approaching 2 but of course doing a finite number of terms proves nothing at all. You should prove (when x=sqrt(2)) that a) this sequence is increasing b) this sequence is bounded above by 2 (induction for both). This will prove the limit does in fact exist, you can then do some manipulations (combined with continuity of the exponential) to show this limit is 2. Link to comment Share on other sites More sharing options...

psi20 Posted July 25, 2006 Author Share Posted July 25, 2006 Ah cool. Is the method to find x described in my initial post a valid one? Link to comment Share on other sites More sharing options...

matt grime Posted July 25, 2006 Share Posted July 25, 2006 What method? You went from x^x^x^x..=2 to x^2=2 without justifying why this is true (it is: you are just raising x to the same power, but you didn't say this explicitly enough, I feel). Link to comment Share on other sites More sharing options...

jordan Posted July 25, 2006 Share Posted July 25, 2006 matt, would you be able to show what an explicit proof for this problem looks like? Link to comment Share on other sites More sharing options...

matt grime Posted July 25, 2006 Share Posted July 25, 2006 the method has been given. Let f_n(x) be the result of x^x^x^..^x, with n terms, and let f(x) be the limit as n tends to infinity of f_n(x). This will not a priori exist for all x, but that is immaterial at this stage. Now, assume that there is a solution to f(x)=2, if there is, then it must be root 2, since x^f(x)=x^2, and x^f(x)=f(x)=2. It remains to show that f_n(sqrt(2)) actually converges which is done using shmoe's hints. Link to comment Share on other sites More sharing options...

Primarygun Posted July 30, 2006 Share Posted July 30, 2006 Did someone mention this method? It suits the name of the book "The Art of the Infinite" let the left-hand side of the equation be y, and since there are infinite x, do the "order" again. so we have the yth power of x = 2. As y is originally equal to 2. We have x being the root of 2 Link to comment Share on other sites More sharing options...

matt grime Posted July 30, 2006 Share Posted July 30, 2006 Yes, that has been mentioned (by the OP, by me and by shmoe), and doesn't address the fact that it assumes that the limit exists in the first place. You also need to prove that fact. After all, if I assume there is a largest integer N, then N^2 is larger than N but N is the largest, hence N^2=N so N=0 or 1, and since 1 is larger than 0 there are no integers larger than 1. See what problems you have if you assume a false premise? Link to comment Share on other sites More sharing options...

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