Transcendental numbers: yet another classification

2011-03-29T14:29:44Z

Let $\mathbb{A^+}$ be the set of non-negative algebraic numbers. Consider the set of "polynomials" : $$\mathbb{P} = \lbrace a_0 + a_1x^{r_1} + a_2x^{r_2} + a_3x^{r_3} +\cdots + a_nx^{r_n}| a_0, a_i, r_i \in \mathbb{A}, r_i > 0, i= 1,2,\cdots,n\rbrace$$ We call $\alpha \in \mathbb{R}, \alpha \geq 0$ extra-algebraic if there exists a polynomial in $\mathbb{P}$ satisfying $f(\alpha)=0$. Denote the set of all extra-algebraic numbers by $\mathbb{A}_E$. So, $\mathbb{A} \subset \mathbb{A}_E$.(The strict inclusion is because of numbers like $2^\sqrt2$ which are extra-algebraic but not algebraic and more by the Gelfond–Schneider theorem). We call $\beta \in \mathbb{R}, \beta > 0$ extra-transcendental if it is not extra-algebraic. Candidates for examples of extra-transcendental numbers are $e^\pi$ and $e^\frac{-\pi}{2}$.

Question:

Do extra-transcendental numbers exist?
Is $\mathbb{R^+} - \mathbb{A}_E$ uncountable?

Many thanks.

Answer by Emil Jeřábek for Transcendental numbers: yet another classification

2011-03-29T15:05:36Z

$\mathbb P$ is countable. Moreover, any $f\in\mathbb P$ is analytic, hence it has only countably many zeros. Thus $\mathbb A_E$ is countable, and in particular, extra-transcendental reals exist, and $\mathbb R^+\smallsetminus\mathbb A_E$ has the power of continuum.

For a concrete example, the Lindemann–Weierstrass theorem implies that $e$ is extra-transcendental.

EDIT: To tie up a loose end, every nonzero $f\in\mathbb P$ has only finitely many positive real roots. Since $f(x)$ is eventually dominated by its nonzero term with the highest exponent, the roots are bounded. Similarly, $f(x)$ is dominated by the term with the smallest exponent when $x\to0+$, hence the roots are bounded away from $0$, i.e., they are contained in a compact subset of $(0,+\infty)$. However, choosing a branch of logarithm makes $f$ holomorphic in $U=\mathbb C\smallsetminus(-\infty,0]$, therefore it can have only finitely many roots in any compact subset of $U$.

In fact, if $f(x)=a_0x^{r_0}+a_1x^{r_1}+\cdots+a_nx^{r_n}$ (with $r_i\in\mathbb R$ pairwise distinct, $a_i\in\mathbb R\smallsetminus\{0\}$), then $f$ has at most $n$ positive real roots.

We can prove this by induction on $n$. If $n=0$, then $f(x)=a_0x^{r_0}$ has no positive real root. Assume the statement holds for $n-1$. Put $g(x)=a_0+a_1x^{r_1-r_0}+a_2x^{r_2-r_0}+\cdots+a_nx^{r_n-r_0}=f(x)/x^{r_0}$. Then every positive root of $f$ is also a root of $g$. Moreover, between each two consecutive roots of $g$, there is a root of its derivative $g'$. Since $g'$ has at most $n$ nonzero terms (the derivative of the constant $a_0$ vanishes), it has at most $n-1$ positive real roots by the induction hypothesis, thus $f$ has at most $n$ such roots.

I guess that one could also prove a variant of Decartes' rule of signs for these generalized polynomials along similar lines.

Transcendental numbers: yet another classification - MathOverflow

Transcendental numbers: yet another classification

Answer by Emil Jeřábek for Transcendental numbers: yet another classification