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Limit of a shrinking function



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Let Tk = kth triangular number = sum(i=1,k) { i } = k(k+1)/2

y(x) = prod(k=1,inf) { x/2k } = lim(k->inf) { xk/2Tk } = lim(k->inf) e[k ln(x)]/e[Tk ln(2)]

     (1)   y(x) = lim(k->inf)  e[k ln(x) - Tk 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) - Tk ln(2)] } ]

For finite x>0, lim(k->inf) { xk/2Tk } exists and is zero, since Tk 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) - Tk 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.


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I have a strong feeling, and I'm working on a proof, that


both answers are 1.



(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/(2k) }

(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 xk but decreases as 1/(2k). 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/(2k), as well. So for infinitely large x, ln z also goes strongly to zero.

limx->inf { ln z(x) } = 0 and so

(5) limx->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.


(6) limx->0 { z(x) } = 1


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