>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