Inspired by Otto Bretscher and Abhay Tiwari

Calculus Level pending

x x x x . . x n = 2 \huge x^{x^{x^{x^{.^{.^{x^{n}}}}}}}=2

The above power tower can be viewed as a recurrence relation: a 0 = n , a k + 1 = x a k a_0=n, a_{k+1}=x^{a_{k}} for k 0 k \ge 0 and lim k a k = 2 \displaystyle\lim_{k \to \infty}a_k=2 .

It is known that n n is a constant and x = 2 x=\sqrt{2} is the solution of the above equation. What is the largest range for n n ?

Inspiration and more inspiration .

n ( , ) n \in (-\infty, \infty ) n = 2 n=\sqrt{2} n ( , 0 ) n \in (-\infty, 0) n ( , 4 ) n \in (-\infty, 4) n ( 0 , ) n \in (0, \infty) n ( 2 , 2 ) n \in (-\sqrt{2}, \sqrt{2}) n ( 0 , 4 ) n \in (0, 4) n ( 0 , 2 ) n \in (0, \sqrt{2})

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1 solution

Chan Lye Lee
May 9, 2016

Here are the graphs showing iteration of a k a_k for (i) n < 2 n <2 , (ii) 2 < n < 4 2 <n<4 and (iii) n > 4 n>4 respectively.

We can get the idea from here that the lim k a k = 2 \displaystyle \lim_{k \to \infty} a_k=2 for all n < 4 n<4 .

Any good algebraic proof?

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