$ \color{goldenrod}{\text{Th}} $: For any $ \sigma \in S_n $, we have $ \text{ord}(\sigma) \leq (e^{\frac{1}{e}})^n \text{ } ( \leq 1.445^n ). $ $ \color{goldenrod}{\text{Pf}} $: Let $ \sigma = \sigma_1 \ldots \sigma_k $ be decomposition into disjoint cycles, and $ | \sigma_i | =: \ell_i $. Now $ \text{ord}(\sigma) $ $ = \text{LCM}(\ell_1, \ldo... Read more 06 Aug 2021 - 1 minute read
$ \color{goldenrod}{\text{Th}}$: Take a continuous map $ [a,b] \overset{f}{\to} \mathbb{R} $, such that $ f’’ $ exists and is $ > 0 $ on $ (a,b) $. Then $ f((1-t)a+tb) < (1-t)f(a) + tf(b) $ for all $ t \in (0,1) $. $ \color{goldenrod}{\text{Pf}}$: LHS is just $ f(x) $ evaluated at $ (1-t)a + tb $. RHS is just “line” $ f(a) + \left( \dfr... Read more 06 Aug 2021 - 1 minute read
Consider $ f : \mathbb{R} \to \mathbb{R} $ with $ f(x) = 0 $ at irrational $ x $ and $ f(\frac{p}{q}) = \frac{1}{q} $ at rationals. This is Thomae’s function. Btw when we say “a rational $ \frac{p}{q} $” it will be implicit that $ p, q $ are integers, $ q > 0 $ and the fraction is in reduced form. $ f $ is discontinuous at rationals. ... Read more 05 Aug 2021 - 1 minute read
$\color{goldenrod}{\text{Th}}$: Let $ (g(k)) $ be a sequence in a complete normed space, and $ (a_k) $ a sequence in $ \mathbb{R}_ {\geq 0} $ decreasing to $ 0 $. Suppose seq $ G(k) := g(1) + \ldots + g(k) $ is bounded. Then $ \displaystyle \sum_{k=1}^{\infty} a_k g(k) $ converges. $\color{goldenrod}{\text{Pf}} $: To get rid of $ g(k) $s, we ca... Read more 05 Aug 2021 - 2 minute read
Consider functions $ f_n (x) = \frac{1}{1+n^2 x^2} $. Pointwise limit $ \displaystyle f(x) = \lim_{n\to\infty} f_n (x) $ is $ 1 $ at $ 0 $ and $ 0 $ at every non-zero $ x $ (esp discontinuous). Plotting in geogebra shows the graphs of $ f_n $ getting ‘‘pinched”. Read more 04 Aug 2021 - less than 1 minute read