What am I missing? Don't you want intersection rather than union? On Wed, Aug 12, 2015 at 2:42 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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