No. $M\mathrel{:=}k(x,y)$ has a $k$-automorphism $\sigma:x\mapsto 1/x,\,y\mapsto 1/y$, of order 2. Let $G\mathrel{:=}\langle\sigma\rangle$, and put $L\mathrel{:=}M^{G}$, the fixed field. The elements $x+1/x$ and $y+1/y$ of $L$ are algebraically independent over $k$, hence $L$ has transcendency degree 2. However, no non-constant polynomial $g\in k[x,y]$ can ...


The irrationality measure of the Champernowne constant $C_b$ in base $b>2$ is exactly $b$.


A quick thought for your Question 2. If the limit for $\xi_f$ exists, then as you noted we'll get $\xi_f ^2 - 1 = f(\xi_f +1)$. If we define $g(x) = f(x+1) - x^2 +1 = f(x+1) - (x+1)^2 + 2(x+1)$, then we have $x^2 - 1 = f(x+1)$ iff $g(x) = 0$. So your second question more or less reduces to saying ``we have a function $g$, and we would like to know when the ...


This recent result Root separation for trinomials by Koiran also uses Baker's Theorem. In my PhD thesis, we use this result to demonstrate a polynomial time algorithm to isolate real roots of integer trinomials.


Regarding the expression of $f(1)$ in terms of other known constants: $$f(1) = -e\left(\gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!} \right) \qquad\quad \tag{$*$} \label{id}$$ $$\qquad\qquad\qquad= -e \left(\gamma - 1 + \frac1{4} - \frac1{18} + \frac1{96} - \frac1{600} + \cdots \right)$$ where $e = \sum_{n\geq0} \frac1{n!} \approx 2.718$ and $\gamma \...

Only top voted, non community-wiki answers of a minimum length are eligible