
Store







CTK Exchange

stancioff
Charter Member
4 posts 
Mar3101, 09:11 AM (EST) 

3. "RE: Responding to x^x^x=3"
In response to message #2

Forgive the typos please, and sorry about the subject line. My intention was an infinite nesting of the exponents. The problem is to solve f(x) = x^{x^{x^{x^...ad infinitum...}}} = a. What makes it different from your x^x^3=3 is that there is no last exponent in the nesting. Otherwise the two have some similar features in their solutions. My problem has the solution x=a^(1/a) but with some caveats. When a friend originally gave me the problem he had it with a=2. I figured it out when I realized that the exponent on the lowest x in the expression on the left side is the just the left side itself. Since the expression = 2, the exponent must be 2. Or x^2=2 and x=2^(1/2). Fooling around with the generalization, f(x)=a, with the solution x=a^(1/a), I tried a=4. Which would have the solution x=4^1/4 = 2^1/2. i.e. f(x)= 2 and f(x)= 4 have the same solution. f(2^(1/2))=2 and f(4^(1/4)) = 4, then 2=4 and I'm the pope, and everything else is true. Actually, the problem only works for values of a between 1/e and e. (ln(a)<1.) And the infinite nested series only converges for values of x between 1/e^e and e^1/e. I always thought it was a fun problem. (Anytime you can prove that both of us are the pope, it's fun)Paul 

Alert  IP 
Printerfriendly page  Edit 
Reply 
Reply With Quote  Top 




alexb
Charter Member
672 posts 
Mar3101, 11:26 AM (EST) 

4. "RE: Responding to x^x^x=3"
In response to message #3

>Forgive the typos please, and sorry >about the subject line. Do not worry about that. I have pointed this out because the two problems are in fact very different. True, they lead to the same kind of solutions and both deal with repeated exponents. However, the problem I discussed in the text, has solutions for any (positive) right hand side. >Fooling around with the generalization, >f(x)=a, with the solution x=a^(1/a), >I tried a=4. Which would >have the solution x=4^1/4 = >2^1/2. i.e. f(x)= 2 and >f(x)= 4 have the same >solution. f(2^(1/2))=2 and f(4^(1/4)) = >4, then 2=4 and >I'm the pope, and everything >else is true. This is a curious observation indeed. And I gather you are not the Pope. But again, your remark underscores the difference between the two problems. Two eqautions x^{x2} = 2 and x^{x4} = 4 have the same solution 2^{1/2} = 4^{1/4} without engendering a contrudiction. > Actually, >the problem only works for >values of a between 1/e >and e. (ln(a)<1.) And the >infinite nested series only converges >for values of x between >1/e^e and e^1/e. Yes, this is true. I recollect seeing this discussed in some book, but can't now place it. Do you have a reference? > I >always thought it was a >fun problem. Absolutely. Alexander Bogomolny


Alert  IP 
Printerfriendly page  Edit 
Reply 
Reply With Quote  Top 




stancioff
Charter Member
4 posts 
Mar3101, 00:01 AM (EST) 

5. "RE: Responding to x^x^x=3"
In response to message #4

>Do not worry about that. I >have pointed this out because >the two problems are in >fact very different. True, they >lead to the same kind >of solutions and both deal >with repeated exponents. However, the >problem I discussed in the >text, has solutions for any >(positive) right hand side. It was the similarities that drew my attention to this problem in your forum. The differences of course make all the difference. > >>Fooling around with the generalization, >>f(x)=a, with the solution x=a^(1/a), >>I tried a=4. Which would >>have the solution x=4^1/4 = >>2^1/2. i.e. f(x)= 2 and >>f(x)= 4 have the same >>solution. f(2^(1/2))=2 and f(4^(1/4)) = >>4, then 2=4 and >>I'm the pope, and everything >>else is true. > >This is a curious observation indeed. >And I gather you are >not the Pope. But again, >your remark underscores the difference >between the two problems. Two >eqautions x^{x2} = 2 and >x^{x4} = 4 have the >same solution 2^{1/2} = 4^{1/4} >without engendering a contrudiction. The two problems are ever so similar in the region of convergence. The top exponent in the finite problem has the same value as the infinite nested series. And if 2=4 then not only am I the pope, but I am also not the pope. > >> Actually, >>the problem only works for >>values of a between 1/e >>and e. (ln(a)<1.) And the >>infinite nested series only converges >>for values of x between >>1/e^e and e^1/e. > >Yes, this is true. I recollect >seeing this discussed in some >book, but can't now place >it. Do you have a >reference? Sorry, I only found these results while fooling around with the problem. I can't even remember who posed the problem to me initially. I do recall that he posed it as a curiosity with a=2 and I then noticed the a=4 and a=2 results, which then led me to look into the convergence issues. First I noticed that a^{1/a} reaches a maximum at a=e, which then led to finding the convergence region. I have not seen this problem otherwise. > >> I >>always thought it was a >>fun problem. > >Absolutely. Thanks for your feedback. > >Alexander Bogomolny


Alert  IP 
Printerfriendly page  Edit 
Reply 
Reply With Quote  Top 




alexb
Charter Member
672 posts 
Apr0101, 00:12 AM (EST) 

6. "RE: Responding to x^x^x=3"
In response to message #5

>It was the similarities that drew >my attention to this problem >in your forum. The >differences of course make all >the difference. I was just reacting to your introductory remark in the very first post: "I remember being given this problem in the following form. " >Sorry, I only found these results >while fooling around with the >problem. I can't even >remember who posed the problem >to me initially. It must be in one of the books in my own library. Sooner or later I am bound to bump into it again. I shall post then the reference. >First I noticed that a^{1/a} (There is no need to use the <nobr> tag here. It is only useful (or so I believe) wherever there are spaces (like in x^{x3} = 3).) >Thanks for your feedback. The pleasure was mine. Alexander Bogomolny 

Alert  IP 
Printerfriendly page  Edit 
Reply 
Reply With Quote  Top 



You may be curious to visit the old CTK Exchange archive.
Front page
Contents
Copyright © 19962018 Alexander Bogomolny
[an error occurred while processing this directive]

Advertise
