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thatsmaths.com | ||
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dominiczypen.wordpress.com
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| | | | | Let $latex \omega$ denote the first infinite cardinal - that is, the set of non-negative integers. Let $latex p_0 = 2$ be the smallest prime number, and let $latex (p_n)_{n\in\omega}$ enumerate all prime numbers in ascending order. Let $latex \mathcal{U}$ be a free ultrafilter on $latex \omega$. We consider the field $latex F = \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{\mathcal... | |
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fabricebaudoin.blog
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| | | | | In this section, we consider a diffusion operator $latex L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i (x)\frac{\partial}{\partial x_i}, $ where $latex b_i$ and $latex \sigma_{ij}$ are continuous functions on $latex \mathbb{R}^n$ and for every $latex x \in \mathbb{R}^n$, the matrix $latex (\sigma_{ij}(x))_{1\le i,j\le n}$ is a symmetric and non negative matrix. Our... | |
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almostsuremath.com
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| | | | | Given a sequence $latex {X_1,X_2,\ldots}&fg=000000$ of real-valued random variables defined on a probability space $latex {(\Omega,\mathcal F,{\mathbb P})}&fg=000000$, it is a standard result that the supremum $latex \displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle X\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\},\smallskip\\ &\displaystyle X(\omega)=\sup_nX_n(\omega). \end{array} &fg=000000$ is measurable. To ensure that this is well-defined, we need to allow X to have values in $latex... | |
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nhigham.com
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| | | A Householder matrix is an $latex n\times n$ orthogonal matrix of the form $latex \notag P = I - \displaystyle\frac{2}{v^Tv} vv^T, \qquad 0 \ne v \in\mathbb{R}^n. $ It is easily verified that $LATEX P$ is orthogonal ($LATEX P^TP = I$), symmetric ($LATEX P^T = P$), involutory ($LATEX P^2 = I$ that is, $LATEX P$ is... | ||
