|
     
|
Question: Tower of Powers Given a list of integers, we can create a TOWER OF POWERS, for example, 3^5^7^14^2^... and so on. You have to use each number exactly once in the tower of powers. The tower associates right-to-left, that is, the expression a^b^c should be interpreted as a^(b^c), with the natural extension for a longer tower of powers. Suppose we create a tower of powers with the integers 2, 3, ... n. Which arrangement of integers creates a tower with the largest value? The smallest value? Prove your result. Answer: Given the tower a1 ^ (a2 ^ ....^ (b ^ (c ^ D))...) where D is either the continuation of the tower or 1 (in which case c is the last term of the tower), does the tower become larger or smaller by switching b and c? it is sufficient to concentrate on b^(c^D) as making this larger will make the tower larger. b^(c^D) > c^(b^D) (*) when Dlog(c) + log(log(b)) > Dlog(b) + log(log(c)) (by taking logs twice) when Dlog(c) - log(log(c)) > Dlog(b) - log(log(b)) let f(x) = Dlog(x) - log(log(x)) so f'(x) = 1/x(D-1/log(x)) f'(x) is positive when D > 1/log(x) remembering that b and c are integers greater than or equal to 2, f'(x) is only negative when D=1 and x=2 (x>=2). therefore (*) is always true for c>b unless b&c are the last two terms of the tower and one of them is 2. When one of them is 2 you need to check whether 2^p > p^2, p >= 3. This is always the case apart from p=3 and p=4. (2^3 < 3^2 and 2^4 = 4^2). Essentially by switching terms in any given tower the maximum will be 2^(3^(4^...^n)..) apart from the case n = 3 in which case its 3^2 The minimum will be given by n^(...(4^(2^3))) apart from the case n=3 which will be 2^3. OR The virtually same proof typed in LaTeX |
