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jmanton.wordpress.com | ||
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extremal010101.wordpress.com
With Alexandros Eskenazis we posted a paper on arxiv "Learning low-degree functions from a logarithmic number of random queries" exponentially improving randomized query complexity for low degree functions. Perhaps a very basic question one asks in learning theory is as follows: there is an unknown function $latex f : \{-1,1\}^{n} \to \mathbb{R}$, and we are... |
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almostsuremath.com
I will give a proof of the measurable section theorem, also known as measurable selection. Given a complete probability space $latex {(\Omega,\mathcal F,{\mathbb P})}&fg=000000$, we denote the projection from $latex {\Omega\times{\mathbb R}}&fg=000000$ by $latex \displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\pi_\Omega\colon \Omega\times{\mathbb R}\rightarrow\Omega,\smallskip\\ &\displaystyle\pi_\Omega(\omega,t)=\omega. \end{array} &fg=000000$ By definition, if $latex {S\subseteq\Omega\times{\mathbb R}}&fg=000000$ then, for every $latex {\omega\in\pi_\Omega(S)}&fg=000000$, there... |
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nhigham.com
The spectral radius $latex \rho(A)$ of a square matrix $latex A\in\mathbb{C}^{n\times n}$ is the largest absolute value of any eigenvalue of $LATEX A$: $latex \notag \rho(A) = \max\{\, |\lambda|: \lambda~ \mbox{is an eigenvalue of}~ A\,\}. $ For Hermitian matrices (or more generally normal matrices, those satisfying $LATEX AA^* = A^*A$) the spectral radius is just... |
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theproofangel.wordpress.com
Do you believe that standards of English are declining, and as a result the end of the world is nigh? If you do, or if you are a disciple of Gwynne's Grammar, you might find this post enlightening. As the writer says, English is more flexible & robust than some people think. It can be... |
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