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endow
endow, v. (ɛnˈdaʊ) Also 7–8 indow. [f. en- prefix1 + F. douer:—L. dōtāre, f. dōt-em dowry. In legal AF. (15th c.) endouer.] 1. trans. † a. To give a dowry to (a woman) (obs.). b. To provide dower for (a widow). Formerly const. of.1535 Act 27 Hen. VIII, c. 10 §7 Suche woman shalbe endowed of as muche...
Oxford English Dictionary
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endow
endow/ɪnˈdau; ɪn`daʊ/ v1 [Tn, Tn.pr]~ sb/sth (with sth) give money, property, etc to provide a regular income for (eg a school, a college) (经常性)资助, 捐助(如学校等) endow a bed in a hospital 资助医院的一个床位(即经常资助一名住院病人的全部医疗费用).2 [Tn.pr usu passive 通常用於被动语态]~ sb with sth provide sb naturally with (any good quality...
牛津英汉双解词典
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San Francisco Ballet Receives $60 Million Gift to Endow New Work ...
San Francisco Ballet today announced a historic $60 million gift—the largest single gift ever given to the company, and what SF Ballet believes to be one of the largest gifts given to an American ballet company—to ensure the organization's capacity to create new works and acquire masterpieces, and to bolster its vision of revolutionizing ballet.
www.sfballet.org
What is the meaning of "endow with the Euclidean norm" From the paper titled, _Distributed Asynchronous Deterministic and Stochastic Gradient Optimization Algorithms_ , the author defines: !Definition Can someone ex...
To add to @Peyton: Regard $H$ as all $L$ tuples, whose $i$th component is chosen from the real vector space $H_i$. Then for $h \in H$ $$\Vert h \Vert = \Vert (h_1, \ldots, h_L)\Vert = \sqrt{\Vert h_1 \Vert^2 + \ldots \Vert h_L \Vert^2}$$ where the norm of each $h_i$ is from its respective space. Tho...
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Inner product and unit vector $u_1 = (1, -1)'$ and $u_2 = (1, 1)'$ are two vector of $R^2$. Endow $R^2$ with an inner product such that $||u_1|| = 1$ and $||u_2|| = 1$. Well, honestly, I don't completely understand w...
An inner product is a function of two vectors into $\Bbb R$ that satisfies certain properties. You are asked to find a function $f((a,b),(c,d))$ that satisfies these. You can't have $u_1=1$ because $u_1$ and $1$ are different kinds of things. You can have $f(u_1,u_1)=1$, which is what you want. The ...
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Discrete subspace of $\mathbb{N}^\mathbb{N}$ Endow $\mathbb{N}$ with the discrete topology. Does $\mathbb{N}^\mathbb{N}$ contain a discrete subspace of size $2^{\aleph_0}$?
Since $\mathbb{N}$ is separable (with the discrete topology), we have that $\mathbb{N}^{{\mathbb{N}}}$ is separable with the product topology. Therefore, any subspace is separable, so there is no discrete subspace of cardinal $2^{\aleph_{0}}$
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$X=${(x,y)$\in \mathbb R^2: 2x^2+3y^2=1$}. Endow $\mathbb R^2$ with discrete topology, and $X$ with the subspace topology. Let $X={(x,y)\in \mathbb R^2: 2x^2+3y^2=1}$. Endow $\mathbb R^2$ with discrete topology, and $...
You need to be careful with applying a polynomial saying it is continuous - the topology (and thus the notion of continuity) is very different from the usual one. It is a good exercise to try and show that a continuous map from a discrete space to $\mathbb{R}$ (with the usual topology) must be const...
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How does one endow $\mathbb GF(p)^{n}$ with a field structure? I've tried to endow $\mathbb{F}^{n}$, where $\mathbb{F}=\mathbb GF(p)$ with a field structure, but I was not able to do it. Could you please help me with ...
So, if you start with ${\Bbb F_p}^n$ and you want to endow it with a field structure you can do it in the following two steps:
* Pick a monic irreducibe
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Is $[0,1]$ a topological group? Can one endow the unit interval $[0,1]$ with a group operation to make it a topological group under its natural Euclidean topology?
No. A topological group is homogeneous, and $[0,1]$ is not, since it has the two endpoints. (An open neighborhood of one of the endpoints, like $[0,1/2)$, is not homeomorphic to any open neighborhood of an interior point via a homeomorphism mapping $0$ to the interior point.)
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What is the Pontryagin dual of the rationals? Endow the rational numbers (or any global field) with the discrete topology, what will be the (compact) Pontryagin dual of the additive group and of the multiplicative gro...
The dual of the additive group $\mathbf Q$ with the discrete topology is $A_\mathbf Q/\mathbf Q$, where $A_\mathbf Q$ is the adele ring of $\mathbf Q$ (viewed as an additive group). See a proof here. The standard topology on $A_\mathbf Q$ makes it locally compact and $\mathbf Q$ (embedded diagonally...
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An example of a compact multiplicatively unbounded ring My teacher asked me to build an associative topological Hausdorff compact ring $R$ with $1$, which is multiplicatively unbounded. That means there is a neighborh...
To make the question answered I note that a required example was given by Uri Bader here.
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Making $\mathbb R^n$ a vector space over $\mathbb R$ of dimension $1$ Is it possible to endow $V_{\mathbb R}=\mathbb R^n$ with $+$ and $\cdot$ operations in such a way that with those operations, the vector space $(V_...
Of course this is possible. Choose a bijection $b: \mathbb R^n \to \mathbb R$ and define $\lambda \cdot v = b^{-1}(\lambda b(v))$ and $v+w=b^{-1}(b(v)+b(w))$.
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Inducing a comodule structure on Hom If C is an R-coalgebra and M is an R-module... then is it possible to endow $Hom_{_RMod}(C,M)$ or $Hom_{_RMod}(M,C)$ with the strucutre of a C-comodule?
_Hint._ We have a canonical span $\hom_R(M,C) \to \hom_R(M,C \otimes_R C) \leftarrow \hom_R(M,C) \otimes_R C$. The right arrow is an isomorphism when $M$ is finitely generated projective (first check $M=R$, then direct sums, then direct summands). I don't see such a construction for $\hom_R(C,M)$.
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An example of compact diffeomorphism group? Let $X$ be a compact smooth manifold. Then we can endow the set of diffeomorphisms of $X$ with $C^1$ topology. Are there any examples when $\mathrm{Diff}(X)$ is compact (as ...
It is very far from being compact, its Lie algebra is the set of vector fields which is infinite dimensional if $dim>0$ so it is an infinite dimensional Lie group. <
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Ideal contained in annihilator problem Let $M$ be an $A$-module, and $I\subseteq \operatorname{Ann}(M)$ be an ideal. Why do we can endow $M$ in a natural way with the structure of an $A/I$-module, and why do $M\simeq ...
Note that $M$ being an $A$-module is equivalent to $M$ being an abelian group and giving a ring map $\varphi:A\rightarrow \mbox{End}(M)$. Then $\mbox{Ann}(M)=ker(\varphi)$ and so if $I\subset \mbox{Ann}(M)$ our ring map factors through the quotient $A/I\rightarrow \mbox{End}(M)$. This is the natural...
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