So now I wonder if there's an x such that the countable sequence
x, e^x, e^(e^x), e^(e^(e^x)), ... are *all* transcendental.
Chaitin's constant is an explicit example.
"Adam P. Goucher" <apgoucher@gmx.com> wrote:
I had a simpler example in mind, namely one that follows from *only* the following facts about algebraic numbers:
-- The algebraic numbers form a field of characteristic zero; -- x and exp(x) cannot both be algebraic.
As did I. The solution I was thinking of is the unique real solution to x = exp(-x). Approximately 0.5671432.
Very nice example. It does, however, make the (perfectly valid, but unnecessary) assumption that the reals are (analytically) complete. An analysis-free solution to the problem is simply: 2 + log(2) whose image under exp is: 2 * exp(2) both of which are transcendental.
Decades ago, by the same reasoning, I realized that the unique real solution to x = cos(x) must be transcendental. Of course that means cos(cos(x)), cos(cos(cos(x))), cos(cos(cos(cos(x)))), etc., are all also transcendental, since they're all the same number. Approximately 0.7390851.
(Puzzle: Is the solution to x = cos(x) still transcendental if you do it in degrees instead of radians?)
Transcendental, by Gelfond-Schneider. If it were algebraic, then x = cos(pi x / 180) = Re(exp(pi i / 180) ^ x) would be transcendental. Sincerely, Adam P. Goucher