Most proofs using this sort of counting argument are actually constructive, if you unwind them, including this one. You can explicitly enumerate the algebraics, so you can explicitly enumerate the numbers of the form log^n(x), where x is algebraic, and the exponent n indicates function iteration, not powers of log(x). Let {a_i} be such an enumeration, and then define T as the number whose decimal expansion has 3 in the nth place unless a_i has a 3 in the nth place, in which case it has 6. Then all exponentiates of T are transcendental. I agree that while this is technically constructive, it would be nice to have a more "natural" explicit example. "Everybody knows" that e, or pi, would be such an example, but that doesn't mean we can prove it. Andy On Wed, Aug 12, 2015 at 3:49 PM, Dan Asimov <asimov@msri.org> wrote:
Nice existence proof, Andy.
I wonder if it's possible to prove that some explicit x works. Maybe x = e ? Why did God make it so hard to prove transcendence?
—Dan
On Aug 12, 2015, at 12:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Aug 12, 2015 at 12:50 PM, Dan Asimov <dasimov@earthlink.net> wrote:
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.
Doesn't a simple counting argument prove this?
There are countably many x that are algebraic There are countably many x such that e^x is algebraic There are countably many x such that e^(e^x) is algebraic ...
and the union of countably many countable sets is countable.
So all but countably many x are the x you're looking for.
I'm working only in the reals here, so that the exponential function is injective.
Andy
These are the first few fixed points of f(z) = exp(z) in the complexes: (copied from a post by "Joffan", Physics Forums, January 6, 2012):
0.318131505 ± i * 1.337235701 2.06227773 ± i * 7.588631178 2.653191974 ± i * 13.94920833 3.020239708 ± i * 20.27245764 3.287768612 ± i * 26.5804715 3.498515212 ± i * 32.88072148 3.672450069 ± i * 39.17644002 3.820554308 ± i * 45.4692654
Would a fixed point of exp(z) be transcendental?
—Dan
On Aug 12, 2015, at 7:56 AM, Rich <rcs@xmission.com <mailto:rcs@xmission.com>> wrote:
Take x as the root of x e^x = 1 near .567143. Then log(x), x, and e^x are all transcendental.
------ Quoting Adam P. Goucher <apgoucher@gmx.com>:
Nice. 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.
I believe the following extension puzzle has no such elementary solution:
"Find x such that x, exp(x) and exp(exp(x)) are all transcendental."
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