An old conjecture
Andrew Usher
(x,y), x and y are either both algebraic or both transcendental. What
additional conditions are needed for the converse of this?
Certainly some are, as we could otherwise pick any random set of
points as the 'curve'. Let's say that the curve shall be everywhere
continuous and piecewise analytic; is that enough?
How about the logarithmic spiral defined by x = 2^t cos(2pi t), y =
2^t sin(2pi t)? If t is rational, x,y are both algebraic, if t is
algebraic irrational, x,y are both transcendental (Gelfond-Schneider),
but if t is transcendental, I don't know if we can say anything. I
proposed this problem to a professor of math in college and he could
not answer it.
Andrew Usher
Jon Slaughter
This made me wonder about the polynomials with algebraic coefficients and
transcendental coefficients.... are these of any use?
A N Niel
<c1e08214-c875-4812...@m26g2000yqb.googlegroups.com>,
Andrew Usher <k_over...@yahoo.com> wrote:
I'm not sure I understand the question.
x = 2^t, y = e^t; continuous analytic curve ??
x = 2 t, y = e t; continuous analytic curve ??
x = t, y = 4; continuous analytic curve ??
Andrew Usher
> I'm not sure I understand the question.
You haven't.
> x = 2^t, y = e^t; continuous analytic curve ??
> x = 2 t, y = e t; continuous analytic curve ??
> x = t, y = 4; continuous analytic curve ??
The question was about curves that DO satisfy the property that (x,y)
are either both algebraic or both not. Curves that do not, such as
your examples, are irrelevant. Again, I asked whether there might be
such curves that are analytic but not algebraic.
Andrew Usher
Robert Israel
> Any algebraic curve in the plane has the property that for any point
> (x,y), x and y are either both algebraic or both transcendental.
Perhaps you mean an algebraic curve defined by a polynomial
with algebraic coefficients. It's certainly not true if you allow
transcendental coefficients.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Andrew Usher
<isr...@math.MyUniversitysInitials.ca> wrote:
> On Fri, 1 Jan 2010 20:57:37 -0800 (PST), Andrew Usher wrote:
> > Any algebraic curve in the plane has the property that for any point
> > (x,y), x and y are either both algebraic or both transcendental.
>
> Perhaps you mean an algebraic curve defined by a polynomial
> with algebraic coefficients. It's certainly not true if you allow
> transcendental coefficients.
Well, yeah, that should have been obvious.
Andrew Usher