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xorshammer.com | ||
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pfzhang.wordpress.com
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| | | | | Consider a monic polynomial with integer coefficients: $latex p(x)=x^d + a_1 x^{d-1} + \cdots + a_{d-1}x + a_d$, $latex a_j \in \mathbb{Z}$.The complex roots of such polynomials are called algebraic integers. For example, integers and the roots of integers are algebraic integers. Note that the Galois conjugates of an algebraic integer are also algebraic integers.... | |
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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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jmanton.wordpress.com
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| | | | | If $latex Y$ is a $latex \sigma(X)$-measurable random variable then there exists a Borel-measurable function $latex f \colon \mathbb{R} \rightarrow \mathbb{R}$ such that $latex Y = f(X)$. The standard proof of this fact leaves several questions unanswered. This note explains what goes wrong when attempting a "direct" proof. It also explains how the standard proof... | |
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mikespivey.wordpress.com
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| | | It's fairly well-known, to those who know it, that $latex \displaystyle \left(\sum_{k=1}^n k \right)^2 = \frac{n^2(n+1)^2}{4} = \sum_{k=1}^n k^3 $. In other words, the square of the sum of the first n positive integers equals the sum of the cubes of the first n positive integers. It's probably less well-known that a similar relationship holds... | ||
