8,984 reputation
13246
bio website mathematik.uni-wuerzburg.de/…
location Würzburg
age 49
visits member for 3 years, 10 months
seen 20 hours ago
I'm a professor of mathematics at the University of Würzburg (Germany). For further infos see my web page.

Aug
26
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
The discriminant of $\phi_n(x,\alpha)$, with respect to $x$, seems to be a polynomial in $\alpha$ with all coefficients of the same sign (or $0$). If one can prove that, this would be the missing piece.
Aug
26
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
another fix
Aug
26
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
simplified the partial proof, fixed small errors
Aug
25
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
added images which illustrate the argument
Aug
25
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
added images which illustrate the argument
Aug
25
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
added 585 characters in body
Aug
25
answered Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Aug
25
awarded  Enlightened
Aug
24
reviewed Approve Is this an instance of any existing convex pentagonal tilings?
Aug
23
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Indeed, in order to approximately compute $\int_0^1\text{exp}(-x-1/x)dx$, Raabe (in 1836) computes approximately the four roots of $\phi_6(x)$ with $x>1$, and uses a theorem of Fourier (a refinement of Descartes for intervals) to show that there are no more roots.
Aug
21
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
added 307 characters in body
Aug
21
revised Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
edited body
Aug
21
answered Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Aug
21
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
Did a recent comment break the view? Comments don't render in a readable form anymore.
Aug
20
comment Finding the inertia group
@Pablo: The only thing I see: Modulo $5$ $H$ factors into degree $5$ irreducibles, and modulo $19$ it factors into degree $4$ irreducibles, so $H$ need to be irreducible.
Aug
19
comment Why does $d^n \exp(-x-x^{-1})/(dx)^n$ only have $n$ positive real zeroes?
The expected number of roots holds even up to $n=6000$ via Decartes' Rule. This is a little surprising as Descartes only gives an upper bound. While Faa di Bruno allows to compute messy expressions for the coefficients of $\phi_n$, I don't believe that one can decide the signs of them in order to apply Descartes. Experimentally, the first somewhat less than $n$ coefficients have constant sign, the somewhat less than the last $n$ coefficients alternate, and around the mid the behavior is somewhat irregular (I didn't see a good pattern).
Aug
19
comment Is there an algorithm to test whether a vector is an eigenvector of a power of a matrix?
@Hurkyl: And how do you determine the possibilities of $s$? Even for matrices of size $2$, $s$ can be any positive integer.
Aug
19
comment Is there an algorithm to test whether a vector is an eigenvector of a power of a matrix?
@Stefan: In general $\alpha$ won't be rational, so $g(T)$ normally won't have rational coefficients. Of course if $f(T)$ happens to have a rational root $\alpha$, then $g=G$ and things are a little easier.
Aug
19
comment Is there an algorithm to test whether a vector is an eigenvector of a power of a matrix?
I enhanced my answer to cover the case $k=\mathbb Q$.
Aug
19
revised Is there an algorithm to test whether a vector is an eigenvector of a power of a matrix?
added 1033 characters in body