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Let $x, x_0inmathbb{R}^n$ be two vectors satisfying $$frac{|x|_1}{|x|_2}leqfrac{|x_0|_1}{|x_0|_2}.$$$| cdot|_1$ and $| cdot|_2$ are the $ell_1$ and $ell_2$ norm in $mathbb{R}^n$, respectively. Suppose...
Asked on 12/01/2020 by Yiming Xu
0 answerSuppose that $(E,H)$ is a rigged (infinite dimensional, separable) Hilbert space, i.e. $H$ is a Hilbert space, and $E$ is a Fréchet space, equipped with a continuous...
Asked on 11/30/2020
2 answerCall a finite subset $S$ of the plane with an even number of points an odd Jackson set, if there is an $Asubset mathbb R^2$ such that ...
Asked on 11/30/2020 by domotorp
0 answerI am currently trying to build the derivatives of $$f(x) = frac{1}{e^x+e^{-x}}.$$It is fairly straightforward to obtain$$ frac{d^n f}{dx^n} = frac{P_n(e^x)}{e^{(n-1)cdot x} (e^x+e^{-x})^{n+1}}, $$where ...
Asked on 11/28/2020 by tobias
1 answerI'm not sure this question is research level question. Sorry in advance. Hypothesis$k$ is a commutative ring.$A$ is an augmented $k$-algebra.$A^e$ is defined as the ...
Asked on 11/28/2020 by TTip
1 answerLet $T$ be a triangle in $mathbb{R}^2$ defined by $y = alpha x$, $y = beta$ and $x = gamma$ where$alpha, beta, gamma...
Asked on 11/27/2020 by Johnny T.
1 answerThe unitary group, $U(n)$, acts transitively on the Grassmann manifold $X = Gr(2, C^n)$. The isotropy group is $H = U(2)times U(n-2)$, i.e. the group elements...
Asked on 11/26/2020 by Norman Goldstein
1 answerAs every MO user knows, and can easily prove, the inverse of the matrix $begin{pmatrix} a & b \ c & d end{pmatrix}$ is $dfrac{1}{ad - bc} begin{pmatrix}...
Asked on 11/21/2020 by Frank Thorne
9 answerSuppose $X$ is a complex projective variety with a model $X_mathbb{Q}$ defined over the rational numbers. Then there is a rational de Rham lattice $H^k_{dR}(X_mathbb{Q}, mathbb{Q})subset H^k(X,...
Asked on 11/19/2020 by Dmitry Vaintrob
0 answerGiven a graph $G=(V,E)$, each edge $e$ has $k$ costs $c_i(e)$, $1le ile k$. Correspondingly, a path $P$ is also characterized by $k$...
Asked on 11/18/2020 by lchen
2 answerGet help from others!
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