bio | website | mathematik.uni-wuerzburg.de/… |
---|---|---|
location | Würzburg | |
age | 49 | |
visits | member for | 3 years, 10 months |
seen | 20 hours ago | |
stats | profile views | 2,382 |
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 |