"For Step 1. you have proven that x<1, but "Obviously, that limit is b^(1/b)" seams to be weak. You'd probably should show that there is such n1 that f(n1) is in \epsilon range from e and for any n2>n1 and so on..."I take that back. It was at some point earlier. I just wanted you not to use something like f(n+1) = x^f(n), but there is no point not to.So if b is in (0,e] then the solution is b^1/b. (e, infty) no solution thus the original equation has no solution. Cool.Thanks

- alexandreC
**Posts:**678**Joined:**

(hang on a second... final draft being prepared.)

Last edited by alexandreC on July 16th, 2004, 10:00 pm, edited 1 time in total.

alexandreC,I guess the confusion comes from the fact that (a^b^c)^d = a^(b^c*d) and not a^b^c^d. (Hint: a^b^c^d = a^(b^(c^d)) )So if you take the power 1/x on both sides at best you get 2^(1/x) = 2^(1/x) which is OK but proves nothing.

- alexandreC
**Posts:**678**Joined:**

alexanderC,I have a proble with the line 3. Do I read well a^(1/b)=a^(-b)? Then for a=b=2 we will have sqrt(2)=1/4!!!. Cool, but wrong.

Hi, if I can helpLet's try to go back to the original problemIf the equation is the following ^x^x = 2, then the solution is clearly 2^(1/2)If the equation is x^x^x^x^x^x^x^x^x^x^x^x^x^x^x^x^x^x^....until infinity = 2, I will say that there is in theory no solution to the problem.For me,if 0<x<1, it will converge to 1if x=1, the solution is obviously 1if x>1, it will divergeIF it can be helpful

mat,For me = 1.76084 not 2!!!A proof for this problem was posted on July 15. Dose anybody know a shorter one?

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