Question on definition of transcendental numbers
WassilyDali
Def: A transcendental number (over Q) is a real (or complex) number r,
such that there does not exist a polynomial, with rational (equiv.
integral) coefficients, f(x) with f(r) = 0.
A polynomial (viewed has an "infinite" series) has, by definition, a
finite number of (nonzero) coefficients. My question is: Does there
necessarily exist an infinite series f(x) = sum_{k=0}^{\infty} a_k x^k
such that f(r) = 0 for every transcendental r?
For instance, Pi is transcendental over Q, however by the usual taylor
series for f(x) = sin x, we have an equation of the above form:
0 = Pi - Pi^3/3! + Pi^5/5! - . . .
What about e? With a little manipulation due to the issues with the
taylor series of log x, we can do the same:
f(x) = -log(1-x) + 1 has a well-defined taylor series, and then we just
plug in x = 1 - exp(1).
But of course e and Pi are very familiar numbers, and such
constructions of f(x) are unsurprisingly easy to find. Are there
general constructions for any transcendental number, or is there some
easier non-constructive way to see the existence of such an infinite
series f_r(x), with f_r(r)= 0, for a given transcendental r?
Virgil
"WassilyDali" <slach...@gmail.com> wrote:
> Corrections welcome.
>
> Def: A transcendental number (over Q) is a real (or complex) number r,
> such that there does not exist a polynomial, with rational (equiv.
> integral) coefficients, f(x) with f(r) = 0.
>
> A polynomial (viewed has an "infinite" series) has, by definition, a
> finite number of (nonzero) coefficients. My question is: Does there
> necessarily exist an infinite series f(x) = sum_{k=0}^{\infty} a_k x^k
> such that f(r) = 0 for every transcendental r?
Any power series that is zero on a set of values on a circle of
convergence with positive radius that is dense on the real line segment
of that circle will be identically zero everywhere.
Note that the set of transcendentals in any open real interval is dense
in that interval.
WassilyDali
thorough explication please.
> Any power series that is zero on a set of values on a circle of
> convergence with positive radius that is dense on the real line segment
> of that circle will be identically zero everywhere.
For a given transcendental number r, how do we know there exists a
power series that "is zero on a set of values on a circle of
convergence with positive radius."
>
> Note that the set of transcendentals in any open real interval is dense
> in that interval.
This follows by the Cantorian argument?
Rupert
WassilyDali wrote:
> Corrections welcome.
>
> Def: A transcendental number (over Q) is a real (or complex) number r,
> such that there does not exist a polynomial, with rational (equiv.
> integral) coefficients, f(x) with f(r) = 0.
>
> A polynomial (viewed has an "infinite" series) has, by definition, a
> finite number of (nonzero) coefficients. My question is: Does there
> necessarily exist an infinite series f(x) = sum_{k=0}^{\infty} a_k x^k
> such that f(r) = 0 for every transcendental r?
>
This is certainly true when r is real. Choose each a_i so that
sum_{k=0}^{i} a_k x^k is within 1/k of 0. It is probably true for
complex r as well.
Tonico
WassilyDali wrote:
> > Note that the set of transcendentals in any open real interval is dense
> > in that interval.
>
> This follows by the Cantorian argument?
Hi:
It all depends on what you call "a Cantorian argument" to, but the
answer is probably yes: the set of algebraic numbers is countable, and
thus the set of transcendental ones has to be uncountable ==> in any
open non-empty subset of the reals there HAS to be one transcendental
number ==> the trans. numbers are dense in the reals ( all with the
usual euclidean topology, of course).
Regards
Tonio
Mate
> WassilyDali wrote:
> > Corrections welcome.
> >
> > Def: A transcendental number (over Q) is a real (or complex) number r,
> > such that there does not exist a polynomial, with rational (equiv.
> > integral) coefficients, f(x) with f(r) = 0.
> >
> > A polynomial (viewed has an "infinite" series) has, by definition, a
> > finite number of (nonzero) coefficients. My question is: Does there
> > necessarily exist an infinite series f(x) = sum_{k=0}^{\infty} a_k x^k
> > such that f(r) = 0 for every transcendental r?
> >
>
> This is certainly true when r is real. Choose each a_i so that
> sum_{k=0}^{i} a_k x^k is within 1/k of 0. It is probably true for
> complex r as well.
>
You cannot be sure that the transcendental number is
in the (open) disc of covergence.
A more interesting question would be:
for t a transcendental number, does there exist
a nonzero entire function f with rational coefficients
such that f(t)=0?
Mate
Rupert
Mate wrote:
> Rupert wrote:
> > WassilyDali wrote:
> > > Corrections welcome.
> > >
> > > Def: A transcendental number (over Q) is a real (or complex) number r,
> > > such that there does not exist a polynomial, with rational (equiv.
> > > integral) coefficients, f(x) with f(r) = 0.
> > >
> > > A polynomial (viewed has an "infinite" series) has, by definition, a
> > > finite number of (nonzero) coefficients. My question is: Does there
> > > necessarily exist an infinite series f(x) = sum_{k=0}^{\infty} a_k x^k
> > > such that f(r) = 0 for every transcendental r?
> > >
> >
> > This is certainly true when r is real. Choose each a_i so that
> > sum_{k=0}^{i} a_k x^k is within 1/k of 0. It is probably true for
> > complex r as well.
> >
>
> You cannot be sure that the transcendental number is
> in the (open) disc of covergence.
In the real case I can ensure that the series does converge at the
transcendental number. I can't be sure that it's contained in an open
ball centred at the origin on which the series converges, no.
> A more interesting question would be:
> for t a transcendental number, does there exist
> a nonzero entire function f with rational coefficients
> such that f(t)=0?
>
That might be a better question, yes.
> Mate
Dave L. Renfro
> A more interesting question would be:
> for t a transcendental number, does there exist
> a nonzero entire function f with rational coefficients
> such that f(t)=0?
What follows is extracted from something I wrote about
three years ago. I thought I had already posted this
particular part, but a google search shows that I haven't.
Incidentally, the term "power series algebraic" is
one that I made up, but it may previously have been
used by others for this (or for some other) concept.
POWER SERIES ALGEBRAIC NUMBERS
One natural way to generalize the notion of an algebraic number
is to consider what I'll call "power series algebraic". These are
numbers that are the zeros of non-zero functions whose power series
expansions about 0 have rational coefficients and an infinite
radius of convergence. Adolf Hurwitz [5] proved in 1891 that
every complex number is power series algebraic.
This means that if we try to extend algebraic and transcendental
notions to "polynomials of infinite degree", we don't get
anything useful -- every complex number will be algebraic
in this extended sense.
The proof of the following stronger result is adapted
from Herzog [4].
THEOREM: Let E be a subset of the complex numbers such that
E has no (finite) limit point. Then there exists
a non-zero function f such that f(z) = 0 for all z
in E and f has a power series expansion about 0 with
rational coefficients and with an infinite radius
of convergence.
PROOF: In what follows, "SUM" represents a summation from n=0 to n=oo
and {Q} is the complex conjugate of Q. Let A(z) = SUM[a(n)*z^n]
be any entire function and let {A}(z) = SUM[{a(n)}*z^n]. Then
A(z)*{A}(z) is an entire function whose power series expansion
about z=0 has real coefficients. Let B(z) = SUM[b(n)*z^n] be
any entire function with real coefficients. Then there exists
an entire function C(z) = SUM[c(n)*z^n] such that C(z) is not
identically zero and the power series expansion of B(z)*C(z)
about z=0 has rational coefficients. We merely have to choose
c(0), c(1), c(2), ... successively so that for each n we have
-1/n! < c(n) < 1/n! (this is to ensure that C(z) is entire)
and the number b(0)*c(n) + b(1)*c(n-1) + ... + b(n)*c(0) is
rational. Note that c(0) cannot be chosen until we reach the
first non-vanishing coefficient among the b(n)'s, but after
this point we'll be able to choose the c(n)'s successively
regardless of whether the later values of the b(n)'s are zero.
To ensure that C(z) isn't identically zero, simply choose c(0)
different from zero.
To complete the proof, let A(z) be any non-zero entire
function that is zero on E, let B(z) be the function
A(z)*{A}(z), and let f(z) = B(z)*C(z) C(z) as constructed
above.
I believe Hurwitz [5] actually proved a stronger result in which
{z: f(z) = 0} = E and, in addition, the order of the zeros of f
at each point of E can be arbitrarily specified in advance. The
references below are re-discoveries and variations of Hurwitz's
result.
[1] David G. Cantor, "Solution to Monthly Problem #5898,
American Mathematical Monthly 81 (1974), 414-415.
[2] Michael Golomb, "Solution to Monthly Problem #5968",
American Mathematical Monthly 82 (1975), 1020.
[3] H. Graetzer, "Note on power series", Journal of the London
Mathematical Society 22 (1947), 90-92.
[MR 9,179d; Zbl 29.39203]
http://www.emis.de/cgi-bin/Zarchive?an=0029.39203
[4] Fritz Herzog, "Solution to Monthly Problem 4498", The American
Mathematical Monthly 60 (1953), 634.
[5] Adolf Hurwitz, "Ueber beständig convergirende Potenzreihen mit
rationalen Zahlencoefficienten und vorgeschriebenen Nullstellen",
Acta Mathematica 14 (1891), 211-215. [JFM 23.0250.01]
http://www.emis.de/cgi-bin/JFM-item?23.0250.01
[6] Robert Israel, sci.math post, 2 June 2002.
http://groups.google.com/group/sci.math/msg/47f0ed46057a3de9
[7] Joseph Lehner, "Note on power series with integral coefficients",
Journal of the London Mathematical Society 25 (1950), 279-282.
[MR 12,325e; Zbl 39.07902]
http://www.emis.de/cgi-bin/Zarchive?an=0039.07902
[8] Cornelis Gerrit Lekkerkerker, "On power series with integral
coefficients. I, II", Nederl. Akad. Wetensch. Proc. Ser. A 52
(1949), 740-746 and 1164-1174 (= Indagationes Mathematicae 11,
270-276 and 438-448). [MR 11,338b and 11,425e; Zbl 39.07901]
http://www.emis.de/cgi-bin/Zarchive?an=0039.07901
[9] Bill Taylor, sci.math post, 13 November 2001.
http://groups.google.com/group/sci.math/msg/dfffc4db4b60980d
[10] Arthur Geoffrey Walker, "Note on integral functions",
Journal of the London Mathematical Society 19 (1944), 106-107.
[MR 6,263a; Zbl 63.08134]
http://www.emis.de/cgi-bin/MATH-item?0063.08134
Dave L. Renfro