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stancioff
Charter Member
4 posts
Mar-29-01, 00:41 AM (EST)    "Responding to x^x^x=3"

 I can't seem to figure out how to respond directly to a topic in the arithmetic/algebra column, so I'll say it here. I remeber being given this problem in the following form. Solve: x^(x^(x^...) = 2 The elippsis indicate an infinite nesting of the exponents. A cute solution is to note that the exponent of the lowest x is the entire expression itself. i.e.in x^<(x^(x^...> the square bracket encloses an expression which is the same as the original expression. So that the exponent on x is 2 or x^2 = 2 and x= sqrt(2). This is cute but to challenge the claim made in the page I read try this if you replace the right side with any value a, then the solution of x^(x^(x... = a would be x = a^(1/a). Now try this: a=4. Go ahead, try it.

Subject     Author     Message Date     ID RE: Responding to x^x^x=3 stancioff Mar-30-01 1 RE: Responding to x^x^x=3 alexb Mar-31-01 2 RE: Responding to x^x^x=3 stancioff Mar-31-01 3 RE: Responding to x^x^x=3 alexb Mar-31-01 4 RE: Responding to x^x^x=3 stancioff Mar-31-01 5 RE: Responding to x^x^x=3 alexb Apr-01-01 6

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stancioff
Charter Member
4 posts
Mar-30-01, 08:38 PM (EST)    1. "RE: Responding to x^x^x=3"
In response to message #0

 I'm correcting typos from the first post. When I typed square brackets, they turned into angle brackets (less than or greater than) in the posted message, which makes it look like I was trying to do some kind of inequality. Now I see, above this message box, that should use square brackets to express angle brackets. I hope the intent can be understood. Please let me know if I can further clarify it. I'm new to this kind of exchange.Paul alexb
Charter Member
672 posts
Mar-31-01, 00:36 AM (EST)    2. "RE: Responding to x^x^x=3"
In response to message #1

 Paul, thank you for your message. It would be great if you could just repost it using curly braces instead of the brackets. Note also that you misstated the problem in the subject line. It'should have been xx3 = 3.. Lastly, the connection you alude to between the two problems is not altogether obvious. Could you please elucidate with a few lines.Do not worry about the current thread. I shall remove it (which I can do) as soon as you posted an updated message (which I can't do for you.)Thank you,Alexander Bogomolny stancioff
Charter Member
4 posts
Mar-31-01, 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 alexb
Charter Member
672 posts
Mar-31-01, 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 xx2 = 2 and xx4 = 4 have the same solution 21/2 = 41/4 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 stancioff
Charter Member
4 posts
Mar-31-01, 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 xx2 = 2 and >xx4 = 4 have the >same solution 21/2 = 41/4 >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 a1/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 alexb
Charter Member
672 posts
Apr-01-01, 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 a1/a (There is no need to use the tag here. It is only useful (or so I believe) wherever there are spaces (like in xx3 = 3).)>Thanks for your feedback. The pleasure was mine.Alexander Bogomolny

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