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nickhar.wordpress.com
| | mathematicaloddsandends.wordpress.com
8.6 parsecs away

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| | I recently came across this theorem on John Cook's blog that I wanted to keep for myself for future reference: Theorem. Let $latex f$ be a function so that $latex f^{(n+1)}$ is continuous on $latex [a,b]$ and satisfies $latex |f^{(n+1)}(x)| \leq M$. Let $latex p$ be a polynomial of degree $latex \leq n$ that interpolates...
| | mikespivey.wordpress.com
7.6 parsecs away

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| | The Riemann zeta function $latex \zeta(s)$ can be expressed as $latex \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$, for complex numbers s whose real part is greater than 1. By analytic continuation, $latex \zeta(s)$ can be extended to all complex numbers except where $latex s = 1$. The power sum $latex S_a(M)$ is given by $latex S_a(M) =...
| | mkatkov.wordpress.com
7.6 parsecs away

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| | For probability space $latex (\Omega, \mathcal{F}, \mathbb{P})$ with $latex A \in \mathcal{F}$ the indicator random variable $latex {\bf 1}_A : \Omega \rightarrow \mathbb{R} = \left\{ \begin{array}{cc} 1, & \omega \in A \\ 0, & \omega \notin A \end{array} \right.$ Than expected value of the indicator variable is the probability of the event $latex \omega \in...
| | elenasquareeyes.com
21.5 parsecs away

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| Didn't realise I hadn't done a masterpost for the 2024 edition of the A-Z Blogging Challenge until just now so thought I'd better get on that. In 2024 the theme was Disney Channel Original Movies - or soem random movies available on Disney+ when there wasn't a DCOM starting with the right letter. Sign-Up Post...