# Purely Transcendental

• Aug 12th 2007, 07:46 AM
ThePerfectHacker
Purely Transcendental
Given an element $\displaystyle x\in \mathbb{R}$ having a property that no non-constant polynomial in $\displaystyle \mathbb{Q}$ has $\displaystyle x$ as a zero, we call it "transcendental". For example, $\displaystyle \pi \in \mathbb{R}$ is an example.

However, if we allow "infinite polynomials", i.e. power series then it is no longer transcendental. Because $\displaystyle \pi$ is a zero of $\displaystyle f(x)=x - \frac{x^3}{3!}+\frac{x^5}{5!}-...$.

My question is: Is there an element in $\displaystyle \mathbb{R}$ which is purely transcendental? Meaning it has no rational power series even.
• Aug 12th 2007, 08:11 AM
Rebesques
What do you mean by "rational power series"? Rational coefficients? :confused:
• Aug 12th 2007, 08:41 AM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
What do you mean by "rational power series"? Rational coefficients? :confused:

Yes
• Aug 12th 2007, 08:52 AM
Rebesques
Like $\displaystyle e^{1/\pi}=\sum_n \frac{1}{n!\pi^n}$ ? :eek:
• Aug 12th 2007, 11:12 AM
CaptainBlack
Quote:

Originally Posted by Rebesques
Like $\displaystyle e^{1/\pi}=\sum_n \frac{1}{n!\pi^n}$ ? :eek:

He is interested in those numbers, if such exist, which are not the root of any
rational power series:

$\displaystyle f(x)=\sum_{n=0}^{\infty} a_n x^n,\ a_n \in \bold{Q}$

RonL
• Aug 12th 2007, 11:41 AM
topsquark
Quote:

Originally Posted by CaptainBlack
He is interested in those numbers, if such exist, which are not the root of any
rational power series:

$\displaystyle f(x)=\sum_{n=0}^{\infty} a_n x^n,\ n \in \bold{Q}$

RonL

Correct me if I'm wrong, but we should be able to represent any real number as a decimal, yes? Which means that all real numbers, including the transcendentals can be expressed as $\displaystyle \sum_{n = 0}^{\infty}a_n(10)^n$ for some choice of $\displaystyle \{ a_n \}$. (This argument would seem to depend on the Axiom of Choice?) So my stab at the answer is that there is no such transcendental number.

-Dan
• Aug 12th 2007, 12:48 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Correct me if I'm wrong, but we should be able to represent any real number as a decimal, yes? Which means that all real numbers, including the transcendentals can be expressed as $\displaystyle \sum_{n = 0}^{\infty}a_n(10)^n$ for some choice of $\displaystyle \{ a_n \}$. (This argument would seem to depend on the Axiom of Choice?) So my stab at the answer is that there is no such transcendental number.

-Dan

$\displaystyle x=\sum_{n = 0}^{\infty}a_n(10)^n$ does not guarantee that

$\displaystyle \sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}$

for some $\displaystyle a_0, a_1, .. \in \bold{Q}$

RonL
• Aug 12th 2007, 05:25 PM
Rebesques
I wonder where he comes up with those :confused:

Ok, suppose such a number x exists -- not a root of any power series with rational coefficients. We will prove Hacker's questions are transcedental. :D

The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.

"Quod erat demonstratum"
• Aug 12th 2007, 07:45 PM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.

How do we know that given any $\displaystyle r\in \mathbb{R}$ there exists $\displaystyle q_i\in \mathbb{Q}$ so that $\displaystyle r=q_0+q_1x+q_2x^2+...$, i.e. is spams?
• Aug 12th 2007, 09:24 PM
CaptainBlack
Quote:

Originally Posted by Rebesques
I wonder where he comes up with those :confused:

Ok, suppose such a number x exists -- not a root of any power series with rational coefficients. We will prove Hacker's questions are transcedental. :D

The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.

"Quod erat demonstratum"

It is obvious that the set of numbers which are roots of some rational power series has the same cardinality as $\displaystyle \bold{R}$, but that does not prove tht the set exhausts $\displaystyle \bold{R}$.

It is also obvious tha the set is dense in $\displaystyle \bold{R}$, as it inherits this propety from the set of algebraic numbers which it contains.

RonL
• Aug 13th 2007, 04:26 AM
topsquark
Quote:

Originally Posted by CaptainBlack
$\displaystyle x=\sum_{n = 0}^{\infty}a_n(10)^n$ does not guarantee that

$\displaystyle \sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}$

for some $\displaystyle a_0, a_1, .. \in \bold{Q}$

RonL

Ah well. I guess when I'm off, I'm way off. :D

-Dan
• Aug 13th 2007, 05:14 PM
Rebesques
Quote:

How do we know (...) spans?

Quote:

It is also obvious tha the set is dense in , as it inherits this propety from the set of algebraic numbers which it contains.

You are both right, I mistakingly took it for a maximal set. :o