Limit of a shrinking function

[BMAD]

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

[bonanova]

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.

 

[bonanova]

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) .... ] ] ] ] ]

[BMAD]

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

[bonanova]

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) = 0⁰, 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