#NewCommander

[bubbloquacious]

3, 20 from math asks?

From the End of year math asks.

3; What math concept did you struggle the hardest with this year?

Linear optimization 😭😭 I can't seem to get the simplex method to stick, and I need to in order to turn in some exercises haha. If anyone wants to explain it to me at length I would be all ears, this is a cry for help.

20; Have you discovered any cool LaTeX tricks this year?

I figured out how to get a nice よ into TeX as a symbol for the Yoneda embedding! :)

\font\maljapanese=dmjhira at 2.5ex \newcommand{\yo}{{\textrm{\maljapanese\char"0H}}}

Also on this StackExhange post I found a way to make underlined text be interrupted by descenders.

\usepackage[outline]{contour} \usepackage{ulem} \normalem % use classical emph

\newcommand \myul[4]{% \begingroup% \renewcommand \ULdepth {#1}% \renewcommand \ULthickness {#2}% \contourlength{#3}% \uline{\phantom{#4}}\llap{\contour{white}{#4}}% \endgroup% }

\newcommand{\iul}[1]{\myul{1.75pt}{0.5pt}{1pt}{#1}}

Now all I have to do is figure out how to put a monospace font in a Tumblr post lol

[indirect]

omggg — \newcommand{\femb0t}{ (@__femb0t)

[lexiconarchitect]

Let's explore a new commander!

[babyawacs]

@new_narrative they use these as bargain factor but at thispoint the russians call for a peacedeal withhout able to retreat face savingly iftheir newcommander triestosave russianblood itis right steps in impossible situation

@new_narrative they use these as bargain factor but at thispoint the russians call for a peacedeal withhout able to retreat face savingly iftheir newcommander triestosave russianblood itis right steps in impossible situation

@new_narrative they use these as bargain factor but at thispoint the russians call for a peacedeal withhout able to retreat face savingly iftheir newcommander triestosave russianblood itis right steps in impossible situation I am Christian KISS BabyAWACS – Raw Independent Sophistication #THINKTANK + #INTEL #HELLHOLE #BLOG https://www.BabyAWACS.com/ Inquiry PHONE / FAX +493212 611 34 64 Helpful?…

[max1461]

\newcommand{\shit}[1]{pissing and shitting user onrtrp into the sewers and hitting them with #1 kg frying pan}

\leq and \geq are two of my acquaintances

[murilo-jeep-tijuca]

21 96437-7419 @jeepcommanderclub with 🔰 Novo Jeep Commander - O novo SUV de 7 lugares da Jeep no Brasil 🇧🇷 - Jeep Commander Limited T270 1.3 turboflex AT6 - R$ 199.990 - Jeep Commander Overland T270 1.3 turboflex AT6 R$ 219.990 - Jeep Commander Limited TD380 2.0 diesel AT9 R$ 259.990 - Jeep Commander Overland TD380 2.0 diesel AT9 R$ 279.990 Também temos descontos para cnpj ➖➖➖➖➖➖➖➖➖➖➖➖➖ #Jeep #JeepCommander #JeepLovers #JeepLover #Commander #NewCommander #NewJeepCommander #AllNewCommander #4x4 #Makehistory #OlllllllO #like4like #offroad #auto #autoshow #jeeplove #pictureoftheday #photoofday #jeepnation #car #jeepexperience #carsofinstagram #follow4follow #instacars #jeeplife #jeepcommanderclub #jeepgrandcommander #meridian #grandcommander (em Jeep | Brasil) https://www.instagram.com/p/CULHQZJFNJ4/?utm_medium=tumblr

[spiritsoulandbody]

APRIL 14, 2022 MAUNDY THURSDAY John 13 A New Commandment I Give You

https://youtu.be/oOmea43M9HE Read the full article

[iramos68]

A new commandment I give to you, that you love one another; as I have loved you, that you also love one another. By this all will know that you are My disciples, if you have love for one another.” John 13:34‭-‬35 #Agape #Love #BrotherlyLove #NewCommandment #❤️

[indirect]

https://t.co/Vp1IsClNeQ — \newcommand{\femb0t}{ (@__femb0t)

[nelcedar]

I am giving you a new commandment, that you love one another. Just as I have loved you, so you too are to love one another. John 13:34 AMP https://bible.com/bible/1588/jhn.13.34.AMP #TCInspire #tcinspire #TCI #newcommandment (at Pokuase Hills) https://www.instagram.com/p/COM2L-hh0gV/?igshid=17bpx3vrx9nho

[jameszeeshan]

Maundy Thursday A new command I give you: Love one another. As I have loved you, so you must love one another. By this everyone will know that you are my disciples, if you love one another. #HolyWeek #MaundyThursday #LastSupper #NewCommand #instaZJ #tumblrZJ #pinterestZJ #tweetZJ #linkedinZJ #quoteZJ https://www.facebook.com/570586976426543/posts/1562847983867099/?d=n&substory_index=0

[olehswift4]

\documentclass[12pt]{article}

\usepackage{pstricks}

\usepackage{pst-all}

\newcommand{\wgt}{\boldsymbol{\bar{w}}}

\usepackage{amsmath,amssymb}

\usepackage{verbatim}

\usepackage{color}

\usepackage{amsmath}

\usepackage{amssymb}

\usepackage{amsthm}

\newtheorem{theorem}{Theorem}

\newtheorem{corollary}{Corollary}[theorem]

\newtheorem{lemma}[theorem]{Lemma}

\renewcommand{\qedsymbol}{$\blacksquare$}

\DeclareMathOperator{\E}{\mathbb{E}}

\definecolor{orange}{RGB}{255,57,0}

\title{Variances convergence proof for negative weights}

\author{o.gorodnitskii@skoltech.ru}

\date{LANL, January-February 2017}

\usepackage{natbib}

\usepackage{graphicx}

\begin{document}

\maketitle

\begin{theorem} Consider reweighted min-sum algorithm with parametrization of messages as below:\\

\begin{equation}

m_{i \rightarrow j}^{m} = \frac{1}{2} a_{i \rightarrow j}^{m}x_j^2 + b_{i \rightarrow j}x_j

\end{equation}

\begin{equation}

A_{i \slash j}^{t} = \left[ J_{ii} + \sum_{k \in N(i)}{c_{ki} a_{k \rightarrow i}^{t - 1}} \right] - a_{j \rightarrow i}^{t - 1}

\end{equation}

\begin{equation}

B_{i \slash j}^{t} = \left[ h_{i} - \sum_{k \in N(i)}{c_{ki} b_{k \rightarrow i}^{t - 1}} \right] - b_{j \rightarrow i}^{t - 1}

\end{equation}

\begin{equation}

a_{i \rightarrow j}^{t} := \frac{-\big( \frac{J_{ij}}{c_{ij}} \big)^2 }{A_{i \slash j}^{t}} \ , a_{i \rightarrow j}^{0} = 0

\end{equation}

\begin{equation}

b_{i \rightarrow j}^{t} := \frac{B_{i \slash j}^{t} \big( \frac{J_{ij}}{c_{ij}} \big) }{A_{i \slash j}^{t}} \ , b_{i \rightarrow j}^{0} = 0

\end{equation}\bigskip

If $\forall i,j$ weights $c_{ij}$ are initialized as follows:

$$c_{ij} = -2(n + 1) \cdot \max \{ \frac{J_{ij}^{2}}{J_{ii}}, \frac{|J_{ij}|}{\sqrt{J_{ii}}} \}$$

then sequence $a_{i \rightarrow j}^{t}$ (and variances $\text{Var}_{ij}^{t} = \frac{1}{a_{i \rightarrow j}^{t}}$) converges.\\

\end{theorem}

Proof comes below after additional lemmas.\\

\begin{lemma} If $c_{ij} < 0 \ \forall \ i,j$ then:

\begin{equation}

A_{i \slash j}^{t} \geq J_{ii} > 0

\end{equation}

\begin{proof}

$a_{i \rightarrow j}^{t} \leq 0 \ \forall t > 0$ (Lemma 15 \cite{ruoz13}). From (2) and $c_{ij} < 0$ comes the statement.

\end{proof}

\end{lemma}

\begin{lemma}

If $c_{ij} < 0 \ \forall \ i,j$ then:

\begin{equation}

a_{i \rightarrow j}^{t} \geq \frac{-\big( \frac{J_{ij}}{c_{ij}} \big)^2}{J_{ii}}

\end{equation}

and\\

\begin{equation}

A_{i \slash j}^{t} \leq \left[ J_{ii} + \sum_{k \in N(i)}{|c_{ki}| \frac{\big( \frac{J_{ki}}{c_{ki}} \big)^2}{J_{kk}}} \right] + \frac{\big( \frac{J_{ji}}{c_{ji}} \big)^2}{J_{jj}}\\

\end{equation}\\

\begin{proof}

From $A_{i \slash j}^{t} > J_{ii}$ and (4) comes (7). From (7) and (2) comes (8).

\end{proof}

\end{lemma}

\begin{corollary}

If $c_{ij} < 0$ sequences $\{a_{i \rightarrow j}^{t}\}$ and $\{A_{i \slash j}^{t}\}$ are bounded from below and above.\\

\end{corollary}

\begin{lemma}

Consider even t ($t = 0 \ \text{mod} \ 2$). If $c_{ij} < 0$ then:

\begin{equation}

A_{i \slash j}^{t + 1} < A_{i \slash j}^{t + 3}

\end{equation}

\begin{equation}

A_{i \slash j}^{t + 2} > A_{i \slash j}^{t + 4}

\end{equation}

\begin{proof} Mathematical induction on t:\\

\begin{itemize}

\item The statement is true for t = 0.

\item Let it be true $\forall t < t^{*} > 0$.

\item From (2):

\begin{equation}

A_{i \slash j}^{t^{*} + 1} = \left[ J_{ii} + \sum_{k \in N(i)}{|c_{ki}| \frac{\big( \frac{J_{ki}}{c_{ki}} \big)^2}{A_{k \slash i}^{t^{*}}}} \right] + \frac{\big( \frac{J_{ji}}{c_{ji}} \big)^2}{A_{j \slash i}^{t^{*}}}

\end{equation}

\begin{equation}

A_{i \slash j}^{t^{*} + 3} = \left[ J_{ii} + \sum_{k \in N(i)}{|c_{ki}| \frac{\big( \frac{J_{ki}}{c_{ki}} \big)^2}{A_{k \slash i}^{t^{*} + 2}}} \right] + \frac{\big( \frac{J_{ji}}{c_{ji}} \big)^2}{A_{j \slash i}^{t^{*} + 2}}

\end{equation}

\item By the induction hypothesis for $t^{*} - 2$ from (10):\\

\begin{equation}

A_{i \slash j}^{t^{*}} > A_{i \slash j}^{t^{*} + 2}

\end{equation}

\item From (11),(12),(13):\\

\begin{equation}

A_{i \slash j}^{t^{*} + 1} < A_{i \slash j}^{t^{*} + 3}

\end{equation}

\item From (14) and equations for $A_{i \slash j}^{t^{*} + 2}$ and $A_{i \slash j}^{t^{*} + 4}$, similar to (11), (12):\\

\begin{equation}

A_{i \slash j}^{t^{*} + 2} > A_{i \slash j}^{t^{*} + 4}

\end{equation}

\end{itemize}

\end{proof}

\end{lemma}

\begin{corollary}

$\{A_{i \slash j}^{2t + 1}\}$ forms monotonically increasing sequence and $\{A_{i \slash j}^{2t + 2}\}$ forms monotonically decreasing sequence for $t \geq 0$

\end{corollary}

\begin{corollary}

Sequence $\{A_{i \slash j}^{2t + 1}\}, t \geq 0$ converges;

\begin{proof}

$\{A_{i \slash j}^{2t + 1}\}$ is bounded from above monotonically increasing sequence $\rightarrow$ it converges.

\end{proof}

\end{corollary}

\begin{corollary}

Sequence $\{A_{i \slash j}^{2t + 2}\}, t \geq 0$ converges;

\end{corollary}

\begin{theorem}

Sequence $\{A_{i \slash j}^{t}\}$ converges under particular choice of weights $c_{ij} < 0$.

\begin{proof}

Mathematical induction on t, to prove $|A_{i \slash j}^{t + 1} - A_{i \slash j}^{t}| \leq (\frac{1}{2})^{t}, \forall t > 0$:

\begin{itemize}

\item for t = 1:

\begin{equation}

|{A_{i \slash j}^{2} - {A_{i \slash j}^{1}| = \left| \sum_{k \in N(i)}{\frac{|c_{ki}|}{J_{kk}} \frac{J_{ki}^2}{c_{ki}^2}} + \frac{J_{ji}^2}{c_{ji}^2} \cdot \frac{1}{J_{jj}} \right| = \left| \sum_{k \in N(i)}{\frac{J_{ki}^2}{J_{kk}} \frac{1}{|c_{ki}|}} + \frac{J_{ji}^2}{c_{ji}^2} \cdot \frac{1}{J_{jj}} \right|

\end{equation}

Initialize $c_{ij}$ as:

\begin{equation}

c_{ij} = -2(n + 1) \cdot \max \{ \frac{J_{ij}^{2}}{J_{ii}}, \frac{|J_{ij}|}{\sqrt{J_{ii}}} \}

\end{equation}

Then:

\begin{equation}

|A_{i \slash j}^{2} - A_{i \slash j}^{1}| = \left| \sum_{k \in N(i)}{\frac{J_{ki}^2}{J_{kk}} \frac{1}{|c_{ki}|}} + \frac{J_{ji}^2}{c_{ji}^2} \cdot \frac{1}{J_{jj}} \right| \leq \frac{1}{2}

\end{equation}

\item Induction hypothesis: $|A_{i \slash j}^{t + 1} - A_{i \slash j}^{t}| \leq (\frac{1}{2})^{t}, \forall t \leq t^{*} $

\item For $t = t^{*} + 1$:

$$|A_{i \slash j}^{t^{*} + 2} - A_{i \slash j}^{t^{*} + 1}| = \left| \sum_{k \in N(i)}{\frac{|c_{ki}|}{c_{ki}^2} \cdot {J_{ki}^2} \frac{|A_{k \slash i}^{t^{*} + 1} - A_{k \slash i}^{t^{*}}|}{A_{k \slash i}^{t^{*} + 1}A_{k \slash i}^{t^{*}}}} + \frac{J_{ji}^2}{c_{ji}^2} \cdot \frac{|A_{j \slash i}^{t^{*} + 1} - A_{j \slash i}^{t^{*}}|}{A_{j \slash i}^{t^{*} + 1}A_{j \slash i}^{t^{*}}} \right| \leq $$

Because $A_{i \slash j}^{t} \geq J_{ii} \ \forall i,j$ and (17):

$$\leq \left( \sum_{k \in N(i)}{\frac{1}{2(n + 1)J_{kk}}} + \frac{1}{2(n + 1)J_{jj}} \right) \cdot \left( \frac{1}{2} \right)^{t^{*}} \leq$$

If we normalize matrix in a way $J_{ii} = 1, \ \forall i$:

\begin{equation}

\leq \frac{1}{2} \cdot \left( \frac{1}{2} \right)^{t^{*}} = \left( \frac{1}{2} \right)^{t^{*} + 1}

\end{equation}

\end{itemize}

\end{proof}

\end{theorem}

\begin{corollary}

If $c_{ij} < 0 \ \forall i,j$ initialized according to (17), sequence $a_{i \rightarrow j}^{t}$ (and variances $\text{Var}_{ij}^{t} = \frac{1}{a_{i \rightarrow j}^{t}}$) converges.\\

\end{corollary}

\bibliographystyle{plain}

\bibliography{references}

\end{document}

[transgenderer]

as i work through a problem set i add more commands and operators to the start of my latex file to make writing certain things faster (which, btw, is the key to LaTeX), so by the end it gets...kind of out of hand

\DeclareMathOperator{\R}{\mathbb{R}}

\DeclareMathOperator{\g}{\gamma}

\DeclareMathOperator{\B}{\beta}

\DeclareMathOperator{\e}{\epsilon}

\newcommand{\pma}[1]{\begin{pmatrix}

#1

\end{pmatrix}}

\newcommand{\lr}[1]{\langle#1\rangle}

\newcommand{\pp}[2]{\frac{\partial#1}{\partial#2}}

\newcommand{\bp}[1]{\bigg(#1\bigg)}

\DeclareMathOperator{\G}{\Gamma}

\newcommand{\dd}[2]{\frac{d #1}{d #2}}

\DeclareMathOperator{\X}{\mathcal{X}}

\DeclareMathOperator{\grad}{\text{grad}}

\DeclareMathOperator{\dv}{\text{div}}

\DeclareMathOperator{\la}{\vartriangle}

\newcommand{\D}[1]{\frac{D}{d#1}}

\DeclareMathOperator{\ep}{\epsilon}

\DeclareMathOperator{\w}{\omega}

(i forgot i already defined epsilon, so when it threw a "command already defined" error i changed the name. woops)

[goedel-rhymes-with-turtle]

I'm part of the tiny minority that hates \langle and \rangle. My .tex template has two \newcommand lines, one for when I need it to be a group given by generators and one for when I need it to be an inner product, so that I never have to read \langle or \rangle in the body of my document. I appreciate most silliness in math and in Latex (I unsuccessfully tried to get people to use the halloweenmath package and the one that puts coffee stains on the page) but for some reason I read langle jee rangle and I am instantly fuming.