Transcendental numbers: yet another classification - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:40:15Z http://mathoverflow.net/feeds/question/59972 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59972/transcendental-numbers-yet-another-classification Transcendental numbers: yet another classification To be cont'd 2011-03-29T14:29:44Z 2011-03-30T12:02:03Z <p>Let $\mathbb{A^+}$ be the set of non-negative <a href="http://en.wikipedia.org/wiki/Algebraic_number" rel="nofollow">algebraic numbers</a>. 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 $\alpha \in \mathbb{R}, \alpha \geq 0$ <em>extra-algebraic</em> 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 <a href="http://en.wikipedia.org/wiki/Gelfond%25E2%2580%2593Schneider_constant" rel="nofollow">$2^\sqrt2$</a> which are extra-algebraic but not algebraic and more by the <a href="http://en.wikipedia.org/wiki/Gelfond%25E2%2580%2593Schneider_theorem" rel="nofollow">Gelfond–Schneider theorem</a>). We call $\beta \in \mathbb{R}, \beta > 0$ <em>extra-transcendental</em> if it is not extra-algebraic. Candidates for examples of extra-transcendental numbers are <a href="http://en.wikipedia.org/wiki/Gelfond%2527s_constant" rel="nofollow">$e^\pi$</a> and $e^\frac{-\pi}{2}$. </p> <p><strong>Question</strong>:</p> <ol> <li>Do extra-transcendental numbers exist?</li> <li>Is $\mathbb{R^+} - \mathbb{A}_E$ uncountable? </li> </ol> <p>Many thanks.</p> http://mathoverflow.net/questions/59972/transcendental-numbers-yet-another-classification/59976#59976 Answer by Emil Jeřábek for Transcendental numbers: yet another classification Emil Jeřábek 2011-03-29T15:05:36Z 2011-03-30T12:02:03Z <p>$\mathbb P$ 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.</p> <p>For a concrete example, the <a href="http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem" rel="nofollow">Lindemann–Weierstrass theorem</a> implies that $e$ is extra-transcendental.</p> <p>EDIT: To tie up a loose end, every nonzero $f\in\mathbb P$ has only <em>finitely</em> 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 $U$.</p> <p>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. </p> <p>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.</p> <p>I guess that one could also prove a variant of <a href="http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs" rel="nofollow">Decartes' rule of signs</a> for these generalized polynomials along similar lines.</p>