BMAD 63 Report post Posted April 2, 2018 Find the limit of x^(x/2)^(x/4)^(x/8)^(x/16)^(x/32).... (a) as x goes to infinity (b) as x goes to zero Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted April 3, 2018 Which way should we evaluate this? [ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ].... or x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ] Quote Share this post Link to post Share on other sites
0 BMAD 63 Report post Posted April 3, 2018 2 hours ago, bonanova said: Which way should we evaluate this? [ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ].... or x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ] The first case Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted April 3, 2018 I have a strong feeling, and I'm working on a proof, that Spoiler both answers are 1. Spoiler Consider: (1) z(x) = x ^ (x/2) ^ (x/4) ^ (x/8) ^ (x/16/) ^ (x/32) ^ .... (evaluated left to right) (2) Let y(x) = Prod_{(}_{k=1, inf) } { x/(2^{k}) } (3) Then z(x) = x ^{y(x)}. = e ^{y(x) ln(x)} Question 1: What is the behavior of z(x) as x increases without bound? (4) From (3), ln z = y ln x. From (2) y increases as x^{k} but decreases as 1/(2^{k}). For any finite x, the product terms in y individually go exponentially with k to zero. Thus y itself is clearly zero for any finite x. From (4), ln z increases more than y by a factor of ln x. But ln x is dominated by 1/(2^{k}), as well. So for infinitely large x, ln z also goes strongly to zero. lim_{x}_{->inf} { ln z(x) } = 0 and so (5) lim_{x}_{->inf} { z(x) } = 1. Question 2: What is the behavior of z(x) as x decreases to zero? From (2), y(0) = 0 by inspection. From (3) z(0) = 0^{0}, which is indeterminate without the knowledge that y(x) goes much more strongly to 0 than x does, deciding in favor of the exponent. Thus (6) lim_{x}_{->0} { z(x) } = 1 Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted April 10, 2018 Spoiler Let T_{k} = k_{th} triangular number = sum(i=1,k) { i } = k(k+1)/2 y(x) = prod_{(}_{k}_{=1,inf)} { x/2^{k} } = lim_{(}_{k}_{->inf)} { x^{k}/2^{T}_{k} } = lim_{(}_{k}_{->inf)} e^{[}^{k}^{ ln(}^{x}^{)]}/e^{[}^{T}_{k}^{ ln(2)]} (1) y(x) = lim_{(}_{k}_{->inf)} e^{[}^{k}^{ ln(}^{x}^{) - }^{T}_{k}^{ ln(2)]} Then if the continued exponent is evaluated left to right, z(x) = x ^ (x/2) ^ (x/4) ^ (x/8) ^ (x/16) ^ (x/32) ^ .... = x ^ y(x) = e ^ [ y(x) ln(x) ] (2) z(x) = e ^ [ ln(x) lim_{(}_{k}_{->inf)} { e^{[}^{k}^{ ln(}^{x}^{) - }^{T}_{k}^{ ln(2)]} } ] For finite x>0, lim_{(}_{k}_{->inf)} { x^{k}/2^{T}_{k} } exists and is zero, since T_{k} dominates k, and y(x) = 0, and z(x) = 1. Otherwise, from equation (2) z(x) is indeterminate: z(x) is 1 for finite x, but goes with x to infinity for finite k. The problematic term is k ln(x) - T_{k} ln(2). Since there is no satisfactory extension of L'Hopital's rule to multiple variables, we would have to arbitrarily assign a relation between x and k to get a definitive answer. If we let x = k, for example, we still get the above result. But if k increased more slowly, say letting x = ln(k), then z(x) diverges to infinity. Similarly different behaviors occur when x goes to zero. Quote Share this post Link to post Share on other sites
Find the limit of
x^(x/2)^(x/4)^(x/8)^(x/16)^(x/32)....
(a) as x goes to infinity
(b) as x goes to zero
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