Jump to content
BrainDen.com - Brain Teasers
  • 1

Limit of a shrinking function


BMAD
 Share

Question

4 answers to this question

Recommended Posts

  • 0

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/(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.

Thus

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

 

Link to comment
Share on other sites

  • 1
Spoiler

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.

 

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...