text:
No 'She-cession' for Top Female Disruptors of 2020
Lisa Buyer
It might have been the year of the she-cession, but it was also the year of female disruptors creating positive vibrations for women around the world. Diamonds are a girl's best friend. Not anymore. Female disruptors think less about the glitz and more about equal pay, equal rights, respect, the gender gap, and breaking the glass ceiling.
Females Took a Hit in 2020
As if 2020 COVID-19 pandemic did not disrupt the world enough; females took a hit in almost every category. There's even a new term called the 'she-cession'.
According to the World's Women, 2020: Trends and Statistics the progress towards equal power and equal rights for women remains elusive. No country has achieved gender equality, and the COVID-19 crisis threatens to erode the limited gains that have been made.
The Biggest Issues for Women in 2021
Unpaid domestic and care work holding women back – In 2020, only 47% of women of working age participated in the labor market, compared to 74% of men – a gender gap that has remained relatively constant since 1995.
No cracks in the glass ceiling – Only 18% of enterprises surveyed had a female Chief Executive Officer in 2020. Among Fortune 500 corporations only 7.4%, or 37 Chief Executive Officers, were women.
Women's participation in education on the rise worldwide – A plus is the world has made substantial progress in achieving universal primary education, with girls and boys participating equally in primary education in most regions.
Violence against women and girls remains a global issue – Most disturbing during COVID-19 lockdowns, many women and girls have been isolated in unsafe environments where they are at heightened risk of experiencing intimate partner violence. Around one-third of women worldwide have experienced physical and/or sexual violence by an intimate partner; 18% have experienced such violence in the past 12 months. In the most extreme cases, violence against women is lethal: globally, an estimated 137 women are killed by their intimate partner or a family member every day.
The editorial team behind Female Disruptors and the women featured have one thing in common, they are change agents. Ready to shake it up and make the future of females empowered and improved, even during a pandemic.
In 2020 we featured women disrupting and pivoting in the business world and beyond despite the challenges. If you're looking for inspiration on how to make a shift, change course, thrive, and survive – we've got you. Meet Female Disruptors most likely to succeed in 2021.
Power in the Palm of Her Hands: Nicole Balistreri
Nicole Balistreri has boosted her career with a recent move to Social Media Manager at USCCA – a membership organization that provides self-defense education, training, and legal protection for responsible gun owners. Some key advice from Nicole is that you can teach skills but not passion. "Never pass up someone for a role just because they lack specific skills if they are passionate. You can teach skills, but you can't force someone to care."
Meet Skin On Sundays Jessica Lakritz
Meet Jessica Lakritz, who combines the two worlds of poetry and the human body with Skin on Sundays. Some key advice from Jessica is that momentum comes and goes. "You have to keep pushing forward even when there is none. There are lots of times I have wondered, what am I doing, does anyone even see or care about my art, etc. when the momentum slows or stops. I guess I had to teach myself how to deal with that and have come to understand that you can still be disciplined even when you don't feel motivated, and that's how you succeed. Do it even when you don't feel like it."
Dr. Kim Chronister: Tik Tok Famous Therapist
Mental health therapist, entrepreneur, influencer. Dr. Kim Chronister is spreading her advice all over Instagram and TikTok, and she might be the epitome of a Female Disruptor. A clinical psychologist, Dr. Kim is also a social media star. Key advice from Kim is to avoid burnout with self-care. Kim says that we need to feel good about ourselves, but that takes work. For instance, she schedules massages, workouts, and time with friends to avoid burnout. According to her, that makes her a better model for her clients and employees.
How you are balancing work/life as a result of the 2020 pandemic impact?
As a result of the 2020 pandemic, my workload has increased due to the large impact it has had on people's mental health. As a psychologist, I've seen an increase in substance use relapses, relationship conflicts, breakups, and depressive episodes with my clients. That has led me to work longer hours. However, the work gives me meaning and purpose which makes it pleasurable for the most part. The key for me is to increase my self-care as my workload increases. For example, if I'm working thirty percent harder, then my self-care (time with friends, massages, movie time, stretching, workouts, etc.) increases by thirty percent.
What tips or advice can you share with other women?
Increase your self-care time even when you think you don't have enough time for anything. Just when you think you can't add anything more to your plate is when it's time to put more relaxation slots into your schedule. Also, studies show that we can ride the high of looking forward to an event or a getaway for up to three months on average. Schedule something at least every three months so that there is some sort of pleasurable event you're looking forward to so that it gets you through the challenges of work, relationship, or financial stressors.
Sarah Axtell: Create Boundaries for Yourself
Sarah Axtell is an entrepreneur at heart. As an adolescent, she ran a busy boat-cleaning business during the summer. Throughout naturopathic medical school, she had a strong vision of what she wanted her future clinic to be like. But that drive and ambition didn't come without challenges. She and her daughters live by Michelle Obama's quote, "When they go low, we go high." When the world seems like it is in turmoil, Sarah Axtell said this quote guides her family to be their best selves.
I make time every single day for exercise. It is an act of self-care and self-love. I also prioritize sleep. After I put my kids to bed, I go straight to bed. There is always the temptation to go back downstairs after putting them to bed to get just a little more work done, but I resist the urge, and every morning when my alarm goes off at 5 am, I am happy I went to bed early!
Make time for yourself. That may mean getting up at 5 am to fit in exercise if you are an early bird. It may mean letting the dishes soak in the sink while you take a bath after dinner. It may mean calling your best friend for a laugh, a cry, or simply a connection with another woman who "gets it." Self-care can also mean making time to cook a nourishing meal for yourself. Figure out what nourishes you and do it daily.
Beth Fiedler: Positivity Matters for Entrepreneurs
Beth Fiedler is curious and determined, two qualities that helped her start her own businesses. A Toronto native, she runs an NLP training program. Some key advice from beth is to follow the 20/80 rule. "I used to put in 80% effort to realize 20% results. Now I focus on 20% efforts to realize 80% results. This principle helps me enjoy a work-life balance for inner peace and harmony."
I realize that there are things I can't change. For example, the weather here in Ontario is going to be grey and chilly in winter. We have no say on the lockdown. In order to maintain work/life balance, I've been starting each day with a clear intention and commitment to stay positive and hopeful. Yes, it needs conscious effort. Unexpected emotion will come up as I'm just human. It's okay and I let it go. I forgive myself for the weak moments and promise myself not to make excuses to stay frustrated or depressed. When I feel stuck, I will get up and make some physical moves. This simple act is very effective to clear my head. When I get scared, I often remind myself that FEAR stands for:
Let go of things that don't serve us anymore.
The world is a different place for all of us. Look into possibilities and opportunities.
We always have at least three choices when we look deeper, not just yes and no or black and white.
The Marketing Whisperer: Sherron Washington
The founder and CEO of the marketing and communication company the P3 Solution and creator of social media campaign #YourWorthIsValued, Sherron Washington is an accomplished professional, with a Master's Degree in Communications. Juggling her careers as a marcomms strategist, speaker, trainer, and author, Sherron is a true female disruptor who's set on changing the world. To women who want to start their own business, Sherron advises the first thing they do is get help. Before even deciding on a name or a logo, aspiring entrepreneurs should consult friends, colleagues, and experts in the field. "No matter how good an idea might sound, there is always something to be learned from outside perspectives."
Shifting my focus inward, I have spent these days getting to know myself fully and learning how to obtain happiness in this crazy world. I have kept with my understand that life doesn't require balance, life requires synergy, and knowing this has helped me to pivot well and create an entirely new lifestyle brand called the SW Life.
1. Be kind and patient with yourself and how you navigate through your own life. Everyone's process is different so don't feel a need to be rushed.
2. Work on things in bite-sized portions, to reduce feelings of anxiety.
3. Celebrate Everything no matter the size. It is important that you factor in some feel-good time.
Alexis Dean: We All Need to Stand for Equality
Alexis Dean's goal seems simple: support female entrepreneurs. That's why she founded The Dovetail Community in 2017, a global organization on a mission to advance female entrepreneurs through a powerful online membership and mastermind community, and unforgettable shared experiences. Alexis Dean's advice for the next generation is that entrepreneurship is a long game. It takes effort and drives. A business doesn't just grow overnight, but the journey is worth it.
The pandemic gave me the time and space to recognize how little I used to focus on self-care. Now I find myself making a conscious effort to take a full day (or two) off every week, and really disconnect. I'm no longer working late hours and going to events on weeknights and weekends. Instead, I'm spending the time connecting with family and friends out in the woods! I've also refocused on some goals that have nothing to do with my work. This year I ran two solo half marathons, took up a new exercise class (The Class), and got really into stand-up paddleboarding!
My advice for other women is to take the time to check in with yourself and give yourself permission to have and do what you really want. Take up a new hobby, make time for adventures, do what brings you joy. Work and family life will always be there, but now we have the gift of fewer in-person work and social obligations- let's take advantage of it!
Jessica Rafaeil Helps Black-Owned Businesses and Entrepreneurs Get More Than Press
Having the guts to switch paths. An unapologetically big personality. Successfully launching her own business. Compassionate. These are some of the things that make Jessica Rafaeil a Female Disruptor. Jessica is a Public Relations Specialist who collaborates with The Buyer Group and her own business, More Than Press. Key advice from Jessica is to never stop learning. "Continuing to educate yourself and learn new things is so important in a world where things can shift and change with the click of a button. Also, apply that to your relationships with people. We can all learn something from one another, no matter our rank."
I've been working remotely for a few years, so that's an adjustment I made a while back. However, the pandemic itself has added loaded stress and now clients are working from home – which makes the workload feel even more "round the clock." At the beginning of 2020, I decided to pick one day (usually a Saturday or Sunday) as what I've dubbed my "veg out day." This is the day that I allow myself to do absolutely whatever I want. If I want to stay in bed and rest all day – I do it. If I want to go out and explore the city or spend some time outdoors – I do it. It's the one day I don't allow work or any of life's stresses to interfere with what I want.
I've also tried to make it a habit to regularly "detox" from digital devices by putting my phone on Do Not Disturb in the evenings and on weekends. At night I try to read and/or write (by hand on paper, not on my laptop). This helps me relax and unload, and I find that it helps me sleep better as well (something I've struggled with pre-COVID).
Own your time. It's important to handle your work, but apply time management to your free time as well. I love the saying "Just because you have free time, doesn't mean you're available."
Ending Period Shame With FemTech: Ruby Raut
CEO and co-founder of UK start-up WUKA (Wake Up Kick-Ass), Ruby Raut, causes a significant disruption in the world of FemTech with a new period product that confronts period shame head-on with a focus on sustainability and intersectionality. "Millions of girls and women don't have access to safe, menstrual hygiene, meaning girls missing education, and risking infections – and reusable period pants could solve this."
Katie Bulmer: Your Passion Has Value
Katie Bulmer took years to figure out what she wanted to do with her life — it turns out, the answer is she wants to help other young women figure that out themselves. Motivational speaker, author of "Sorority Girls can Change the World" and the voice behind "Truth for your Twenties," this Chattanooga gal draws on her own life experiences to make a difference. Katie is learning to change her internal rhetoric to something more appreciative of the time and effort she puts into her business. She makes sure she thanks herself for the work she does and congratulates herself on her accomplishments. Each night before she goes to bed, she thinks about the three things she's going to accomplish the next day. Being able to check off the "Big 3" at the end of the day makes her feel accomplished and relieved.
How Can You Support Women in 2021?
As we head into another year, Female Disruptors to lift up and feature females who are helping the future of females improve equality, decrease the gender gap, and add more seats to the boardroom.
Investing in diversity and inclusion (D&I) within organizations
Supporting women-led businesses
Supporting the leadership development of women throughout their career journeys
Ensuring the organization is following D&I best practices
"If the She-cession is to be combated, and we are not to lose years of hard-fought, precious progress for women's equality, we must be intentional. We must look very carefully at the decisions we make over the next year, and we must commit to supporting women — or risk losing them," notes Sarah Alter, president, and CEO of the Network of Executive Women, a learning and leadership community representing 12,400 members in 22 regional groups in the United States and Canada.
Diamonds are a girl's best friend. We can buy them ourselves.
Do you know a female disruptor who should be featured? Let us know by nominating here.
Featured Image Credit: Jessica Lakritz @Skinonsundays
Looking for the latest Social PR trends or digital life hacks? Meet Lisa Buyer, a subject matter expert in public relations/social media and a recovering digital life junkie. She is the author of Digital Detox Secrets and Social PR Secrets with the foreword by Guy Kawasaki. Lisa brings 20+ years of PR experience always staying ahead of the curve as an early adapter integrating PR, SEO, and social media. Clients include Fortune 500 companies, start-ups, and niche brands. Industries include technology, marketing, real estate, health, wellness, and yoga.
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Q: What are these atg.servlet.jsp.ByteContentBuffer entries in my dynamo.log? In my ATG dynamo.log I am seeing thousands of entries like:
atg.servlet.jsp.ByteContentBuffer@ed62b0(mBlockCount=4, mBlockSize=1024, mCurOffset=0, mOwner=null, mReader=null, mMaxContentSize=-1, , mInInitOrAdd=false, mBookmarkCount=0, [], , mBlocks=[[B@1eab262) dumping bookmarks, cur offset = 0, owner = null

What are they and how do I turn them off?  I know it has something to do with the content of the JSP (BodyContent?) but not sure of the purpose of the logging.

A: apparently, there is a piece of code in your project that prints the object of ByteContentBuffer, which it's toString() method returns the output printed in your log.
It would be very useful if you shared the 50 lines where this code is in the middle of.

redpajama_set_name: RedPajamaStackExchange


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\section*{Introduction}
Zariski predicted, and proved in characteristic $0$ in \cite{Z}, that any integral algebraic variety X
equipped with a dominant morphism $v : Spec(V) \to X$ from a valuation ring $V$ can be desingularized
along $V$ : there should exist a proper birational map ${\tilde X} \to  X$ for which the lift ${\tilde v} : Spec(V) \to {\tilde X}$
of $v$ supplied by the valuative criterion of properness would factor through the regular locus of $\tilde X$.
This local form of resolution of singularities remains open in positive and mixed characteristic, and
implies that every valuation ring V should be a filtered direct limit of regular rings.  A filtered direct limit (in other words a filtered colimit) is a limit indexed by a small category that is filtered (see \cite[002V]{SP} or \cite[04AX]{SP}). A filtered  union is a filtered direct limit in which all objects are subobjects of the final colimit, so that in particular all the transition arrows are monomorphisms. There exists several nice extensions of Zariski's Uniformization Theorem as for example recently the result of   B. Antieau, R. Datta \cite[Theorem 4.1.1]{AD}, which says that every perfect valuation ring of characteristic $p>0$ is a filtered union of its smooth ${\bf F}_p$-subalgebras. This result is an application of \cite[Theorem 1.2.5]{T} which relies on some results from \cite{J}. Also E. Elmanto and M. Hoyois proved that an absolute integrally closed valuation ring of residue field of characteristic $p>0$ is a filtered union of its regular finitely generated  ${\bf Z}$-subalgebras (see \cite[Corollary 4.2.4]{AD}).

The goal
of this paper is to establish
a kind of Zariski's Uniformization Theorem in characteristic $p>0$ and dimension one.

\begin{Theorem}\label{T} Let $V$ be a one dimensional valuation ring containing a perfect field $F$ of characteristic $p>0$, $k$ its residue field,   $\Gamma$ its  value group and $K$ its fraction field.  Then the following statements hold
\begin{enumerate}

\item  if $k\subset V$ and $K=k(x)$ for some   system of algebraically independent  elements  $x=(x_1,\ldots,x_r)\in V^r$ over $k$ such that   $\Gamma=\oplus_{i=1}^{r-1} {\bf Z}\val(x_i)$
then
 $V$ is a filtered
 union of  its polynomial $k$-subalgebras, in particular of its smooth $F$-subalgebras,

\item if $\Gamma$ is free of rank $r$,  $k\subset V$ and  $x=(x_1,\ldots,x_r)\in V^r$ is a system of elements such that  $\Gamma=\oplus_{i=1}^r {\bf Z}\val(x_i)$ and $K/k(x)$ is algebraic
  then $V$ is  a filtered
union of its smooth $F$-subalgebras,  in particular of its smooth ${\bf F}_p$-subalgebras,

\item if  $\Gamma$ is free of rank $r$, $k/F$ is a field extension of finite type and $\trdeg_FK=\trdeg_Fk+r$,
then $V$ is a filtered union of its smooth $F$-subalgebras, in particular of its smooth ${\bf F}_p$-subalgebras.
\end{enumerate}
\end{Theorem}

The proof is given in Corollaries \ref{C}, \ref{co1},   \ref{co2}.  The idea of the proof of Theorem \ref{T} (1)
is to see that $W=V\cap k(x)$ is a filtered  union of its localizations of polynomial $k$-subalgebras and then to reduce to show that  the immediate extension $W\subset V$ is  a filtered
union of its smooth $W$-subalgebras. This is done somehow in \cite[Proposition 18]{P} when  $k\supset {\bf Q}$. If char $k>0$ this does not work  (see e.g. \cite[Remark 6.10]{Po1}), the reason being that in general a valuation ring could have a so called defect as was noted by Ostrowski. The
 Generalized Stability Theorem of F.\ V.\ Kuhlmann \cite[Theorem 1.1]{K1} says in particular that $W$ is defectless (see Corollary \ref{ku}) and so we may show (1) using Lemma \ref{l1}. For  Theorem \ref{T} (2) we see that $W\subset V$ is dense and we use a N\'eron-Schappacher Theorem (see e.g. \cite[Theorem 4.1]{Po1}) or
 a kind of General N\'eron Desingularization (see Lemma \ref{k'}) which allow us to handle dense extensions.

 When $V$ contains $\bf Q$ then an immediate algebraic extension of valuation rings $V\subset V'$  is dense by Ostrowski's Defektsatz \cite[Sect. 9, No 57]{O}. It follows that
   $V\subset V'$ is  a filtered
direct limit of   smooth $V$-algebras (see e.g. \cite[Proposition 9]{P}). The converse is also mainly true as shows the following result (see Theorem \ref{t}).

\begin{Theorem}\label{T1}  Let  $ V'$ be an  immediate algebraic  extension  of a valuation ring $V$,
$\hat V$ the completion of $V$ and $K$, ${\hat K}$ the fraction fields of $V$, ${\hat V}$. Assume that   $\trdeg_K{\hat K}\geq 1$,  ${\hat K}/K$ is separable,  $V'$ is a filtered direct limit of smooth $V$-algebras and either $V$ is Henselian, or $\dim V=1$. Then $V\subset V'$ is dense.
\end{Theorem}

We owe thanks to F.\ V.\ Kuhlmann   who hinted us some mistakes in some preliminary forms of the paper. Also we  owe thanks to K\k{e}stutis \v{C}esnavi\v{c}ius  who hinted us some mistakes in the preliminary forms of Theorem \ref{ky}.
\vskip 0.5 cm

\section{The Defect}
Let $V$ be a  valuation ring with value group $\Gamma$,  $K$ its fraction field and its valuation $\val:K^*=K\setminus \{0\}\to \Gamma$. Let $F$ be a finite field extension of $K$, $v_i$, $1\leq i\leq r$ the valuations of $F$ extending $\val$ and $V_i$ be the valuation rings of $F$ defined by $v_i$. Let $e_i, f_i$ be the ramification index respectively the degree of the residue field extension of $V_i$ over $V$. It is well known that
$$[F:K]\geq \sum_{i=1}^r e_if_i.$$
 If this inequality is equality for all finite (resp. finite separable) field  extensions $F$ of $V$ we say that $V$ is {\em defectless} (resp. {\em separably defectless}, see \cite{K} for  details and examples).

We remind the following  important result for the existence of defectless valuation rings.

\begin{Theorem} (\cite[Theorem 2,  VI, (8.5)]{Bou})\label{b} Let $F$ be a finite  field extension of $K$ and $B$ be the integral closure of $V$ in $F$. Then the above inequality is equality if and only if $B$ is a free $V$-module.
\end{Theorem}

\begin{Lemma}\label{s}  Let $V$ be a  valuation ring  and $S$ a multiplicative closed system from $V$. If $V$ is defectless (resp. separably defectless) then the valuation ring $S^{-1}V$ is defectless (resp. separably defectless) too.
\end{Lemma}
\begin{proof} Assume $V$ is defectless (the proof in the separably defectless case is similar). Let $F$ be a finite field extension of the fraction field $K$ of $V$ (and of  $S^{-1}V$ too) and $B$, $B'$ be the integral closures of $V$, respectively  $S^{-1}V$ in $F$.
By Theorem \ref{b} we have to show that $B'$ is a free  $S^{-1}V$-module. As $V$ is defectless by the quoted theorem we see that $B$ is a free $V$-module and so  $S^{-1}B$ is a free  $S^{-1}V$-module.

We claim that $B'\cong  S^{-1}B$ which is enough. Indeed, let $z\in F$ be integral over $S^{-1}V$. We have
$$z^e+\sum_{i=0}^{e-1}(a_i/s)z^i=0$$
for some $e\in {\bf N}$, $a_i\in V$ and $s\in S$. Then
$$(sz)^e+\sum_{i=0}^{e-1}s^{e-i-1}a_i(sz)^i=0$$
and so $sz\in B$, which shows our claim.

\hfill\ \end{proof}

Next it is  very useful the following  particular form of the Generalized Stability Theorem of F. V. Kuhlmann (see \cite[Theorem 1.1]{K1}, \cite[Theorem 5.1]{K}).

\begin{Theorem}\label{k}  Let $V\subset V'$ be an extension of   valuation rings   with the same residue field, $\Gamma\subset \Gamma'$ its value group extension and $K\subset K'$ its fraction field extension.  Assume  $K'/K$ is a finite type field extension and  $\Gamma'/\Gamma$ is a finitely generated   free $\bf Z$-module  of rank   $\trdeg K'/K $. If $V$ is defectless (resp. separably defectless) then $V'$ is  defectless (resp. separably defectless) too.
\end{Theorem}

\begin{Corollary}\label{ku}  Let $V$ be a  valuation ring of characteristic $p>0$  with  finitely generated value group $\Gamma$ and fraction field $K$. Assume  $V$ contains its residue field $k$ and $K=k(y)$ for some elements $y=(y_1,\ldots,y_r)$ of $V$ such that $\val(y_i)$, $1\leq i\leq r$ is a basis of the free $\bf Z$-module $\Gamma$.
Then $V$ is  defectless.
\end{Corollary}
For the proof note that $k$ is defectless for the trivial valuation, $y$ is algebraically independent over $k$ by \cite[Theorem 1, (10.3)]{Bou} and apply the above theorem to $k\subset V$.

\begin{Proposition}(Kuhlmann, \cite[Theorem 1.1]{K1})  Let $V$ be a  valuation ring,  ${\frk m}$ its maximal ideal and $X=(X_1,\ldots,X_n)$ some variables. If  $V$ is defectless  then the valuation ring $V_1=V[X]_{{\frk m} V[X]}$ is defectless too.
\end{Proposition}
\begin{proof} Applying induction on $n$ we reduce to the case $n=1$. Let $\Gamma$ be the value group of $V$ and set $\Gamma'=\Gamma\oplus {\bf Z}\gamma$ where $\gamma$ is taken to be positive but $<$ than all positive elements of $\Gamma$ and let $w$ be the valuation on $K(X)$ extending $\val$ by $X\to \gamma$. By Theorem \ref{k} we see that $W$, the valuation ring of $w$, is defectless. Note that ${\frk m} W$ is a prime ideal of $W$ and $W_{{\frk m} W}\cong V_1$ is defectless by Lemma \ref{s}.
\hfill\ \end{proof}

An inclusion $V \subset V'$ of valuation rings is an \emph{immediate extension} if it is local as a map of local rings and induces equalities between the value groups and the residue fields of $V$ and $V'$.  It is dense if for any $x'\in V'$ and $\gamma\in \Gamma$ there exists $x\in V$ such that $\val(x-x')>\gamma$.

Let $\lambda$ be a fixed limit ordinal  and $v=\{v_i \}_{i < \lambda}$ a sequence of elements in $V$ indexed by the ordinals $i$ less than  $\lambda$. Then $v$ is \emph{pseudo convergent} if

$\val(v_{i} - v_{i''} ) < \val(v_{i'} - v_{i''} )     \ \ \mbox{for} \ \ i < i' < i'' < \lambda$
(see \cite{Kap}, \cite{Sch}).
A  \emph{pseudo limit} of $v$  is an element $x \in V$ if

$ \val(x -  v_{i}) = \val(v_{i} - v_{i'})) \ \ \mbox{for} \ \ i < i' < \lambda$. We say that $v$  is
\begin{enumerate}
\item
\emph{algebraic} if some $f \in V[T]$ satisfies $\val(f(v_{i})) < \val(f(v_{i'}))$ for large enough $ i < i' < \lambda$;

\item
\emph{transcendental} if each $f \in V[T]$ satisfies $\val(f(v_{i})) = \val(f(v_{i'}))$ for large enough $i < i' < \lambda$.
\end{enumerate}
When for any $\gamma\in \Gamma$ it holds $\val(v_{i} - v_{i'})>\gamma$ for  $ i < i' < \lambda$
large we call $v$ {\em fundamental}.
\begin{Lemma}\label{l0}
 Let $V\subset V'$ be an immediate extension of one dimensional valuation rings and $K\subset K'$ their fraction field extension. If $V$ is separably defectless and $K'/K$ is separable and algebraic then $ K'/K$ is dense. Moreover, if $K'/K$ is not separable but  $V$ is defectless then $K'/K$ is still dense.
 \end{Lemma}
 \begin{proof} We may reduce to the case when $K'/K$ is finite separable. Let $K^h$, $K'^h$ be the Henselizations of $K$ respectively $K'$ and $L=K^h(K')\subset K'^h$. By \cite[Theorem 2.3]{K} we have $K^h$ separably defectless and so we have $[L:K^h]=e_{L/K^h}f_{L/K^h}=1$ because there exists an unique extension of the valuation of $V$ to $K^h$. Thus $L\subset K^h$ and $K\subset K^h$ is dense because $\dim K=1$. The second statement goes similarly.
\hfill\ \end{proof}

 \begin{Example} We consider  \cite[Example 3.1.3]{Po1} inspirated by \cite{O}. Let $k$ be a field
of characteristic $p > 0$, $X$ a variable, $\Gamma = {\bf Q}$ and $K$ the fraction field of the group
algebra $k[\Gamma]$, that is the rational function in $\{X^q\}_{q\in {\bf Q}}$. Let $P$ be the field of all formal
sums $z =
\sum_{n\in {\bf N}} a_n X^{\gamma_n}$ where $(\gamma_n)_{n\in {\bf N}}$ is a monotonically increasing sequence from
$\Gamma$ and $a_n \in k$. Set $\val(z) = \gamma_s$, where $s = \min \{n \in {\bf N} : a_n \not= 0\}$ if $ z \not= 0$ and let $V'$
be the valuation ring defined by $\val : P^*\to \Gamma$.

Let $\rho_n=(p^{n+1}-1)/(p-1)p^{n+1}$,
$$y=-1+\sum_{n\geq 0} (-1)^nX^{\rho_n}$$
and $a_i=-1+\sum_{0\leq n\leq i} (-1)^nX^{\rho_n}$.
We have $1+\rho_n =p\rho_{n+1}$ for $n\geq 0$ and $p\rho_0=1$ and  y is a pseudo limit of the pseudo convergent
sequence $a = (a_i)_{i\in {\bf N}}$, which has no pseudo limit in $K$. Then $x$ is a root of the separable
polynomial $g=Y^p+XY+1\in K[Y]$ and the algebraic separable extension
$V = V'\cap K\subset V'\cap K(y)$ is not dense.
 The above lemma cannot be applied because $V$ is not separably defectless.
\end{Example}

\begin{Lemma}\label{l1}
 Let $V\subset V'$ be an  immediate extension of one dimensional valuation rings, $K\subset K'$ their fraction field extension and $x\in V'$ a transcendental element over $K$. Assume $V$ is separably defectless and $x$ is a pseudo limit of a pseudo convergent sequence $a$ over $V$ which has no pseudo limit in $V$. Then $a$ is transcendental.
 \end{Lemma}

\begin{proof} Let $a=(a_j)_{j<\lambda}$ and assume  $a$ is algebraic. Let $h\in V[X]$ be a primitive polynomial of minimal degree among the polynomials $f\in V[X]$ such that $\val(f(a_j))<\val(f(a_{j+1})) $ for all $j<\lambda$.
We may take $h$ separable. Indeed, assume that $h$ is not separable. We have $\val(h(x))>\val(h(a_j))$ for $j<\lambda$ large. Choose $b\in V$ such that $\val(bx)>\val(h(x))$ and set $h'=h+bX$. Note that $\val(x)=\val(a_j)$ for $j$ large, which implies $\val(ba_j)=\val(bx)>\val(h(x))>\val(h(a_j))$ for $j$ large. It follows that $\val(h'(a_j))=\val(h(a_j))$ and we may replace $h$ by $h'$, which is separable.

 By \cite[Theorem 3]{Kap} there exists a finite separable immediate extension $L=K(z)$ of $K$ such that $h(z)=0$ and  $z$ is a pseudo limit of $a$. By Lemma \ref{l0} $L$ is dense over $K$ and so $a$ has a pseudo limit in $K$ (see e.g. \cite[Lemma 2.5]{Po1}), which is false.
\hfill\ \end{proof}

\begin{Proposition}\label{p0}
 Let $V\subset V'$ be an  immediate extension of one dimensional valuation rings and $K\subset K'$ their fraction field extension. Assume $V$ is separably defectless and $K'=K(x)$ for some $x\in K'$ which is  transcendental over $V$.  Then $V'$ is a filtered  union of localizations of its polynomial $V$-subalgebras in one variable.
 \end{Proposition}

\begin{proof} Then $x$ is a pseudo limit of a pseudo convergent sequence $a$ over $V$ which has no pseudo limit in $V$ by \cite[Theorem 1]{Kap}. But $a$ is transcendental by Lemma \ref{l1} and using \cite[Lemma 3.2]{Po} or  \cite[Lemma 15]{P} we are done.
\hfill\ \end{proof}

\begin{Corollary} \label{C} Let $V$ be a one dimensional valuation ring, $K$ its fraction field, $k$ its residue field and $\Gamma$ its value group. Assume $k\subset V$ and $K=k(x)$ for some   system of algebraically independent  elements  $x=(x_1,\ldots,x_r)\in V^r$ over $k$ such that either $\Gamma=\oplus_{i=1}^r {\bf Z}\val(x_i)$, or  $\Gamma=\oplus_{i=1}^{r-1} {\bf Z}\val(x_i)$. Then $V'$ is a filtered  union of localizations of its polynomial $V$-subalgebras in $r$ variables.
\end{Corollary}
\begin{proof} The first case is a consequence of \cite[Theorem 1, VI, (10.3)]{Bou} and \cite[Lemma 4.6]{Po} (see also \cite[Lemma 26 (1)]{P}). In the second case we see that $V$ is an immediate extension of $W=V\cap k(x_1,\ldots,x_{r-1})$. As above $W$ is a filtered  union of localizations of its polynomial $V$-subalgebras in $(r-1)$ variables. By Corollary \ref{ku} we have $W$ defectless and using Proposition  \ref{p0} we are done.
\hfill\ \end{proof}

\vskip 0.5 cm

\section{Valuation rings as limits of smooth algebras}

\begin{Proposition}\label{p2}
Let $V\subset V'$ be an  immediate extension of one dimensional valuation rings and  $K\subset K'$ their fraction field extension.   Assume $V$ is separably defectless and $K'/K$ is  algebraic separable.
 Then $V'$ is a filtered  union of its smooth $V$-subalgebras.
\end{Proposition}
\begin{proof}
By Lemma \ref{l0} the extension $V\subset V'$ is dense.  Then by a N\'eron-Scappacher Theorem (see e.g. \cite[Theorem 4.1]{Po1}) we see  that $V'$ is a filtered direct limit of smooth $V$-algebras.
 \hfill\ \end{proof}

\begin{Corollary}\label{C'} Let $V$ be a one dimensional valuation ring of characteristic $p>0$  with a free (over $\bf Z$) value group $\Gamma$ of rank $r$ and fraction field $K$. Assume  $V$ contains its residue field $k$ and  $x=(x_1,\ldots,x_r)\in V^r$ is a system of elements such that  $\Gamma=\oplus_{i=1}^r {\bf Z}\val(x_i)$ and $K/k(x)$ is algebraic
 separable.
Then $V$ is a filtered union of its smooth $k$-subalgebras.
\end{Corollary}
\begin{proof}  By Corollary \ref{ku} we see that $W=V\cap k(x)$ is defectless and so  $V$ is a filtered  union of its smooth $W$-subalgebras using the above proposition.   But $W$ is a filtered union of its smooth $k$-algebras  (see \cite[Theorem 1, VI, (10.3)]{Bou}, \cite[Lemma 4.6]{Po},  \cite[Lemma 26 (1)]{P}), which is enough.
 \hfill\ \end{proof}

We need the following lemma   (\cite[Lemma 7]{P} which is an extension of
 \cite[Proposition 3]{KPP}, and \cite[Proposition 5]{P1}).

\begin{Lemma} \label{k'}
For a commutative diagram of ring morphisms

\xymatrix@R=0pt{
& B \ar[rd] & & && B \ar[dd]^{b \, \mapsto\, a} \ar[rd] & \\
A \ar[rd] \ar[ru] & & V & \mbox{that factors as follows} & A \ar[ru]\ar[rd] & & V/a^3V \\
& A' \ar[ru] & & && A'/a^3A' \ar[ru] &
}
\noindent with $B$ finitely presented over $A$, a $b \in B$ that is standard over $A$ (this means a special element from the ideal $H_{B/A}$ defining the non smooth locus of $B$ over $A$, for details see for example \cite[Lemma 4]{P}), and a nonzerodivisor $a \in A'$ that maps to a nonzerodivisor in $V$ that lies in every maximal ideal of $V$,
there is a smooth $A'$-algebra $S$ such that the original diagram factors as follows:

\hskip 4 cm\xymatrix@R=0pt{
& B \ar[rdd] \ar[rrd] & & \\
A \ar[rd] \ar[ru] & &  & V. \\
& A' \ar[r]  & S \ar[ru] &
}

 \end{Lemma}

In fact the separability condition is not necessary in Corollary \ref{C'}.

\begin{Corollary}\label{co1} Let $V$ be a one dimensional valuation ring of characteristic $p>0$  with a free (over $\bf Z$) value group $\Gamma$ of rank $r$ and fraction field $K$. Assume  $V$ contains its residue field $k$ and  $x=(x_1,\ldots,x_r)\in V^r$ is a system of elements such that  $\Gamma=\oplus_{i=1}^r {\bf Z}\val(x_i)$ and $K/k(x)$ is algebraic.
Then $V$ is a filtered union of its smooth $k$-subalgebras.
\end{Corollary}
\begin{proof}
As in Corollary \ref{C'} we see that $W=V\cap k(x)$ is defectless and so the algebraic extension $W\subset V$ is dense by Lemma \ref{l0}.
Let $E\subset V$ be a finitely generated $F$-subalgebra and $w:E\to V$ its inclusion. Assume $E=F[Y]/I$, for $Y=(Y_1,\ldots,Y_n)$.   Using \cite[Lemma 1.5]{S} it is enough to show that $w$ factors through a smooth $F$-algebra. Note that $K/F$ is separable because $F$ is perfect. Thus $E/F$ is separable and
 $w(H_{E/F})\not =0$, let us assume that  $ w( H_{E/F})V \supset zV$ for some $z \in W $, $z\not =0$. Replacing $z$ by a power of it we may assume that $z=\sum_i^s b_ib'_i$ for some $b_i=\det(\partial f_{ij}/\partial Y_{j_i})$ for some systems of polynomials $f_i$ from $I$ and $b''_i\in F[Y]$ which kills $I/(f_i)$.
Similarly as in  \cite[Lemma 4]{KPP} we may
assume that we can take $s=1$, that is   for some polynomials $ f = (f_1,\ldots,f_r)$
from $I$, we have  $z\in NM E$ for some $N\in ((f):I)$ and a $r\times r$-minor $M$ of the Jacobian matrix $(\partial f_i/\partial Y_j)$ (since $V$ is a valuation ring this reduction is much easier). Thus we may assume $z$ is standard over $F$,  which is necessary later to apply Lemma \ref{k'}.

 Set $E'=E\otimes_FW$ and let $w':E'\to V$ be the map induced by $w$. We have $w'(H_{E'/W})\supset zV$ and note that
$w'$ factors modulo $z^3$ through the smooth $W/z^3W$-algebra  $W/z^3W\cong V/z^3V$ since $W\subset V$ is dense. Then
 using Lemma \ref{k'} we see that $w'$ factors through a smooth $W$-algebra.  Since $W$ is a filtered direct limit of smooth $k$-algebra as in Corollary \ref{C'} we see that $w$ factors through a smooth $k$-algebra even $F$-algebra. Actually, we may apply as in Proposition \ref{p2} a variant of N\'eron-Schappacher Theorem to get $V$ as a filtered union of its smooth $W$-subalgebras and we are done.
 \hfill\ \end{proof}

\begin{Corollary}\label{co2} Let $V$ be a one dimensional valuation ring containing a perfect field $F$ of characteristic $p>0$,  $\Gamma$ its value group, $K$ its fraction field and $k$ its residue field. Assume $\Gamma$ is free of rank $r$, $k/F$ is a field extension of finite type and $\trdeg_FK=\trdeg_Fk+r$.
Then $V$ is a filtered union of its smooth $F$-subalgebras, in particular of its smooth ${\bf F}_p$-algebras.
\end{Corollary}

\begin{proof} Let $E\subset V$ be a finitely generated $F$-subalgebra and $w:E\to V$ its inclusion. Using \cite[Lemma 1.5]{S} it is enough to show that $w$ factors through a smooth $F$-algebra. Using Lemma \ref{k'}  we may replace $V$ by its completion as in \cite[Proposition 9]{P}. So we may assume $V$ is Henselian.  A lifting of $k$ to $V$ could be done when $V$ is Henselian and char\ $k=p=0$ (see \cite[Theorem 2.9]{v}) but the proof goes in the same way when $p>0$ and  $k$ is separably generated over $F$. In particular the lifting could be done when $k/F$ is of finite type because $F$ is perfect. Thus we may assume $k\subset V$ and $\trdeg_kK=r$. Choose some elements  $x=(x_1,\ldots,x_r)\in V^r$  such that  $\Gamma=\oplus_{i=1}^r {\bf Z}\val(x_i)$. Then $W=V\cap k(x)\subset V$ is algebraic and we may proceed as in Corollary \ref{co1}.
 \hfill\ \end{proof}

 \section{Algebraic pseudo convergent sequences}

 Let  $ V'$ be an  immediate extension  of a valuation ring $V$, $K\subset K'$ their fraction field extension, $v= (v_j)_{j<\lambda}$  a pseudo-convergent sequence in $V$ which is not fundamental and has  a pseudo limit $x$ in  $ V'$, but having no pseudo limit in $K$.  Suppose that  $K'=K(x)$.  We need the following result \cite[Theorem 1.8]{KC}.

 \begin{Theorem}(Kuhlmann-\' Cmiel)\label{kc}  Assume that $x$ is not a root of an irreducible polynomial $f\in K[X]$ and either  $\dim V=1$, or $V$ is Henselian. Then
for all $z\in V$ there is a  $\nu<\lambda$ such that $\val(f(z))<\val(f(v_j))$ for all $\nu<j<\lambda$.
 \end{Theorem}

\begin{Theorem} \label{ky}  Assume  $x$ is transcendental over $K$ and  either  $\dim V=1$, or $V$ is Henselian.
 Then the following statements  are equivalent:
 \begin{enumerate}

 \item for every  polynomial $f\in V[X]$ with $f(x)\not =0$ there exists a $y\in V$ such that $\val(f(y))= \val(f(x))$,
\item for every  polynomial $f\in V[X]$ with $f(x)\not =0$   there exists a $y\in V$ such that $\val(f(y))\geq \val(f(x))$
 \item $v$ is transcendental,
 \item $V'$ is a filtered  union of its localizations of polynomial $V$-subalgebras in one variable.
 \end{enumerate}

\end{Theorem}

\begin{proof} First assume that (4) holds and let $f$ be as in (1).  Let $d\in V$, $z\in V'$ be such that $f(x)=dz$ and $zz'=1$ for some $z'\in V'$. Then the solution $x,z,z'$ in $V'$ of the polynomials $F_1=f-dZ, F_2=ZZ'-1 \in V[X,Z,Z']$ must be contained in a localization of a polynomial $V$-subalgebra $C=V[u']_{P(u')}$ of $V'$, where $P\in V[U]$ and  $u'\in V'$ is  transcendental over $V$. Choose a $u\in V$ such that $u\equiv u'$ modulo ${\frk m}'$ the maximal ideal of $V'$. Then $P(u)$ is a unit in $V$ and the map $\rho:C\to V$ given by $u'\to u$ is a retraction of $V\subset C$ and $y=\rho(x), \rho(z), \rho(z')$ is a solution of $F_1,F_2$ in $V$, and so $\val(f(y))=\val(d)=\val(f(x))$.

Clearly (2) follows from (1) and assume that (2) holds but $v$ is algebraic. This is not possible as F.\ V.\ Kuhlmann said, the proof being done in Theorem \ref{kc} (see also \cite[Lemma 5.4]{KV}). Let  $h\in V[T]$ be a polynomial with  minimal degree among the polynomials $f\in V[T]$ such that
 $\val(f(v_i))<\val(f(v_j))$ for large $i<j<\lambda$. Then $h$ is irreducible (see the proof of \cite[(II,4), Lemma 12]{Sch}). Note that $\val(h(x)-h(v_j))\geq \val(x-v_j)=\val(v_{j+1}-v_j)$. As the last term is increasing we must have $\val(h(x))\geq \val(h(v_j))$ for all $j$.   By Theorem \ref{kc} for every $z\in V$ we have $\val(h(z))<\val(h(v_j))$ for $j$ large. Thus
   $h$ fails the condition (2) and so $v$ cannot be algebraic.

If $v$ is transcendental then $V'$ is  a filtered  union of its localizations of polynomial $V$-subalgebras in one variable by \cite[Lemma 3.2]{Po} (see also \cite[Lemma 15]{P}).
\hfill\ \end{proof}

 \begin{Corollary} In the assumptions of the above theorem, if $v$ is algebraic  then $V'$ is not  a filtered  union of its localizations of polynomial $V$-subalgebras in one variable.
 \end{Corollary}

 \begin{Lemma} \label{r2}  In the assumptions of the above theorem, assume that $V$ is Henselian and  there exists an immediate extension of valuation rings $V'\subset V_1$ such that $V_1$ is a filtered direct limit of smooth $V$-algebras. Then   for every  polynomial $f\in V[X]$ such that $f(x)\not =0$   there exists a $y\in V$ such that $\val(f(y))= \val(f(x))$.
\end{Lemma}
\begin{proof}  Fix $f\in V[X]$. Let $d\in V$, $z\in V'$ be such that $f(x)=dz$ and $zz'=1$ for some $z'\in V'$. Since $V_1$ is a filtered direct limit of smooth $V$-algebras, the solution $x,z',z'$ in $V'$ of the polynomials $F_1=f-dZ, F_2=ZZ'-1 \in V[X,Z,Z']$ must  come from  a solution $\bar x$, $\bar z$, ${\bar z}'$ in a smooth $V$-algebra $C$. But $V$ is Henselian and so there exists  a retraction  $\rho:C\to V$ of $V\subset C$. Thus $y=\rho({\bar x}), \rho({\bar z}), \rho({\bar z}')$ is a solution of $F_1,F_2$ in $V$, and so $\val(f(y))=\val(d)=\val(f(x))$.
\hfill\ \end{proof}

\begin{Lemma} \label{r3}  In the assumptions of the above theorem, assume that $\dim V=1$ and  there exists an immediate extension of valuation rings $V'\subset V_1$ such that $V_1$ is a filtered direct limit of smooth $V$-algebras. Let  $\hat V$  be the completion of $V$ and $K$, $\hat K$ the fraction fields of $V$,  $\hat V$. Suppose that ${\hat K}/K$ is separable.  Then   for every  polynomial $f\in V[X]$ with $f(x)\not=0$ there exists a $y\in V$ such that $\val(f(y))= \val(f(x))$,
\end{Lemma}
\begin{proof}  Let     ${\hat V}_1$ be the completion of $V_1$ and $K_1$ the fraction field of $V_1$. Then $ {\hat V}_1\cap {\hat K}={\hat V}$. Note that  $K_2={\hat K}(V_1)$ is a localization of ${\hat K}\otimes_K K_1$ since ${\hat K}/K$ is separable. Then $V_2={\hat V}_1\cap K_2$ is  a localization of  ${\hat V}\otimes_V V_1$ and so $V_2$ is a filtered direct limit of smooth $\hat V$-algebras. As $\hat V$ is Henselian we get $\val(f(z))=\val(f(x))$ for some $z\in {\hat V}$ using the above lemma. Choose $y\in V$ such that $\val(y-z)>\val(f(x))$. Then $\val(f(y))=\val(f(x))$.
\hfill\ \end{proof}

\begin{Theorem}\label{t}  Let  $ V'$ be an  immediate algebraic  extension  of a valuation ring $V$,
$\hat V$ the completion of $V$ and $K$, ${\hat K}$ the fraction fields of $V$, ${\hat V}$. Assume that   $\trdeg_K{\hat K}\geq 1$,  ${\hat K}/K$ is separable, there exists an immediate extension of valuation rings $V'\subset V_1$ such that $V_1$ is a filtered direct limit of smooth $V$-algebras and either $V$ is Henselian, or $\dim V=1$. Then $V\subset V'$ is dense.
\end{Theorem}

 \begin{proof}
 Assume that $V\subset V'$ is not  dense and let $x\in V'$ which is not in $\hat V$. Then $x$ is a pseudo limit of a pseudo convergent sequence $v$ which is not fundamental and has no pseudo limits in $V$ by \cite[Theorem 1]{Kap}. Note that $v$ is algebraic because $x$ is algebraic over $K$.  Choose a transcendental $t\in {\hat V}$ over $K$. Multiplying $t$ with a constant of $V$ of high enough value we may assume that $x+t$ is still a pseudo limit of $v$.

 Let     ${\hat V}_1$ be the completion of $V_1$  and set $V_2={\hat V}_1\cap {\hat K}(V_1)$. As in the proof of the above lemma we see that $V_2$ is a  filtered direct limit of smooth $\hat V$-algebras. Thus  for every  polynomial $f\in V[X]$ such that $f(x)\not =0$   there exists a $y\in V$ such that $\val(f(y))= \val(f(x+z))$ using Lemmas \ref{r2}, \ref{r3}.  By Theorem \ref{ky} applied to $x+z$ we get $v$ transcendental, which is false.
\hfill\ \end{proof}

\vskip 0.5 cm

redpajama_set_name: RedPajamaArXiv


Hunting Season is always a special time of the year. Whether you are planning on waiting patiently in a deer stand to bag that trophy buck or sitting knee deep in the marsh hoping to bag a band. Being able to take advantage of all the hunting seasons is why The Waterfowl Hunter has put together the hunting dates for Missouri.
These dates serve as a general guide for more specific hunting visit Missouri Dept of Conservation.


text:
Zebulon Alexander Terry (June 17, 1891 – March 14, 1988) was a professional baseball player who played infielder in the Major Leagues from - for the Boston Braves, Chicago Cubs, Chicago White Sox, and Pittsburgh Pirates. He played college baseball at Stanford University.

Terry made his big-league debut on April 12, 1916, starting at shortstop for the White Sox against the Detroit Tigers in a game at Comiskey Park. He went hitless with a walk, but teammates didn't fare much better that day against Detroit's Harry Coveleski, who pitched a three-hit shutout.

Sparingly used by the Sox, and hitting just .190 as a rookie, Terry ended up changing teams in 1918, 1919 and 1920. He found his greatest success with the Cubs, getting most of his 605 career hits for them. In the 1922 season, his last in the majors, Terry led the National League in sacrifice hits with 22. Terry announced his "voluntary retirement" from baseball to pursue real estate in the Los Angeles area in early 1923

Born in Texas, he attended Long Beach Polytechnic High School in California, which by one recent count has produced 19 Major League Baseball players, including Tony Gwynn and brother Chris Gwynn, Milton Bradley, Rocky Bridges and Chase Utley.

References

External links

1891 births
1988 deaths
Major League Baseball infielders
Baseball players from Texas
Boston Braves players
Chicago Cubs players
Chicago White Sox players
Pittsburgh Pirates players
Stanford Cardinal baseball players
Los Angeles Angels (minor league) players
Long Beach Polytechnic High School alumni

redpajama_set_name: RedPajamaWikipedia


Our latest case study features a project we completed with TGW Ltd, one of the world's leading warehousing automation companies. The case study highlights how our meeting room and collaboration technology expertise helped the company improve productivity.
The project included the fit out of the company's new meeting rooms. The meeting rooms are in high demand at the company's new headquarters, so we wanted to make them as simple to use as possible.
This seems to have worked as some of the early feedback from employees includes "Many people have commented how easy the technology is to operate; it helps rather than hinders meetings. Video conferencing is a much better, more effective experience".


text:
Q: Authenticate an AWS SQS scaler in Keda I have a Keda deployment that I've been trying to get to work for about a month now. At the moment, my scaler looks like this:
apiVersion: keda.k8s.io/v1alpha1
kind: ScaledObject
metadata:
  name: {service-name}-scaler
spec:
  scaleTargetRef:
    deploymentName: {service-name}
    containerName: {service-name}
  pollingInterval: 30
  cooldownPeriod:  600
  minReplicaCount: 0
  maxReplicaCount: 10
  triggers:
  - type: aws-sqs-queue
    authenticationRef:
      name: keda-trigger-authentication
    metadata:
      queueURL: https://sqs.ap-northeast-1.amazonaws.com/{AWS ID}/{Queue-name}
      queueLength: "1"
      awsRegion: "ap-northeast-1"
      identityOwner: pod

The associated trigger authentication and secret are:
apiVersion: v1
kind: Secret
metadata:
  name: keda-secrets
data:
  AWS_ACCESS_KEY_ID: {base64-encoded-string}
  AWS_SECRET_ACCESS_KEY: {base64-encoded-string}
  KEDA_ROLE_ARN: {base64-encoded-string}

---

apiVersion: keda.k8s.io/v1alpha1
kind: TriggerAuthentication
metadata:
  name: keda-trigger-authentication
spec:
  env:
  - parameter: awsRegion
    name: AWS_REGION
  - parameter: awsAccessKeyID
    name: AWS_ACCESS_KEY_ID
  - parameter: awsSecretAccessKey
    name: AWS_SECRET_ACCESS_KEY
  - parameter: awsRoleArn
    name: KEDA_ROLE_ARN
  secretTargetRef:
  - parameter: awsRoleArn
    name: keda-secrets
    key: KEDA_ROLE_ARN

I understand that the KEDA_ROLE_ARN value is repeated here; I left both for debugging purposes. The order of deploying this is as follows:

*

*Install common environment variables (this is where the AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY and KEDA_ROLE_ARN values are stored. The AWS_ACCESS_KEY_ID and AWS_SECRET_ACCESS_KEY values are listed as AWS_ACCESS_KEY_ID_ASSUME and AWS_SECRET_ACCESS_KEY_ASSUME respectively in the file and will assume their appropriate values on the container. Again, these are duplicated for debugging purposes. I would prefer to use these values rather than a separate secret.
Install Keda pods with Helm

*Deploy the keda-secrets secret and the keda-trigger-authentication trigger authentication

*Deploy the container that should be scaled. This is where the AWS_ACCESS_KEY_ID_ASSUME value will assume the name of AWS_ACCESS_KEY_ID and the AWS_SECRET_ACCESS_KEY_ASSUME value will assume the name of AWS_SECRET_ACCESS_KEY and where the AWS_REGION value is defined.

*The scaled object is deployed

For some reason, I keep getting an error from AWS when the scaler attempts to scale saying that there are no credential providers in the chain. It appears that the AWS credentials are not being sent. What am I doing wrong here?

A: I will show you two ways to successfully scale deployment based on AWS SQS
First way : Using AWS IAM role attached to node
If your IAM role (node role) has permission to SQS then accessing SQS becomes easier you just have to change identityOwner: pod field to identityOwner: operator so that KEDA can use node role to access AWS SQS
Sample ScaledObject file with SQS trigger
apiVersion: keda.sh/v1alpha1
kind: ScaledObject
metadata:
  name: aws-sqs-queue-scaledobject
  namespace: default
spec:
  scaleTargetRef:
    name: test-deployment
  minReplicaCount: 0
  maxReplicaCount: 2
  triggers:
  - type: aws-sqs-queue
    metadata:
      queueURL: https://sqs.us-east-1.amazonaws.com/3243234432432/Queue
      queueLength: "5"
      awsRegion: "us-east-1"
      identityOwner: operator

Second way: Using IAM user
In this approach, we need to create below objects

*

*Create IAM user in AWS.

*Create secret in Kubernetes.

*Create TriggerAuthentication in Kubernetes.

*Create scaledObject in Kubernetes.

Create IAM user and give SQS permissions to this IAM user.
first encode IAM user Access Key and Secret key using base64 which will be required while creating Kubernetes secret.
Create secret
apiVersion: v1
kind: Secret
metadata:
  name: test-secrets
  namespace: default
data:
  AWS_ACCESS_KEY_ID: <base64-encoded-key>
  AWS_SECRET_ACCESS_KEY: <base64-encoded-secret-key>

Create TriggerAuthentication this will be used in scaledObject
apiVersion: keda.sh/v1alpha1
kind: TriggerAuthentication
metadata:
  name: keda-trigger-auth-aws-credentials
  namespace: default
spec:
  secretTargetRef:
  - parameter: awsAccessKeyID     # Required.
    name: test-secrets            # Required.
    key: AWS_ACCESS_KEY_ID        # Required.
  - parameter: awsSecretAccessKey # Required.
    name: test-secrets            # Required.
    key: AWS_SECRET_ACCESS_KEY    # Required.

Create scaledObject to map keda with deployment you want to scale based on SQS trigger
apiVersion: keda.sh/v1alpha1
kind: ScaledObject
metadata:
  name: aws-sqs-queue-scaledobject
  namespace: default
spec:
  scaleTargetRef:
    name: test-deployment
  minReplicaCount: 0
  maxReplicaCount: 2
  triggers:
  - type: aws-sqs-queue
    authenticationRef:
      name: keda-trigger-auth-aws-credentials
    metadata:
      queueURL: https://sqs.us-east-1.amazonaws.com/012345678912/Queue
      queueLength: "5"
      awsRegion: "us-east-1"


safe and helpful ways to improve your cancer care.
and treatments used to help people with cancer.
out what works best for patients with cancer.
and how to give it.
treatment still works after a long period.


text:
Netflix's "I Am Mother" deals in a slow burn of unanswered questions and half truths, bound to keep you engaged. (Illustration by Claire Maske, Skidmore College)
Screens x June 25, 2019
'I Am Mother' Is A Sci-fi Thriller as Thoughtful as it Is Frightening
The latest Netflix original sci-fi film is a glowing example of the genre at its best.
By Logan Floyd, Mercyhurst University
Recent years have arguably seen a noticeable change in science fiction titles, with movies such as 2016's "Arrival," 2017's "Blade Runner 2049" and 2018's "Annihilation" mostly abandoning whirlwind blockbuster action in favor of slower, more thoughtful pacing and an emphasis on exploring concepts. Now, in 2019, Netflix's newest original sci-fi film, "I Am Mother," is continuing this trend, providing an unhurried narrative that still manages to be intense, thought-provoking and often deeply disturbing.
"I Am Mother" takes place in a future where an event — never revealed to the audience — has wiped out the human race, resulting in the activation of a hidden bunker designed to repopulate humanity. The bunker's automated systems fashion a lone robot called Mother (voiced by Rose Byrne, performed by Luke Hawker), who draws a single human embryo from a vast storage area, grows it in an incubator and begins to raise it as her child.
The child, simply known as Daughter (Clara Rugaard), grows up with Mother acting as both her parent and her teacher. While Mother is surprisingly warm, gentle and affectionate for a machine, she still puts firm boundaries in place, forbidding Daughter from leaving the bunker due to the world outside being contaminated.
Although Daughter loves her mechanical guardian deeply, as a teenager she comes to question Mother's warnings, particularly after a mouse manages to find its way into the bunker. Despite Daughter's protests, Mother swiftly incinerates the animal, citing the risk it poses due to outside contamination. Daughter, however, wonders if the mouse is or is not proof that life can exist beyond the bunker's walls.
Season 3 of 'Stranger Things' Delivers on Visuals and Tears
When Daughter's growing suspicions lead her to explore the bunker's airlock, she is startled by an injured woman (Hilary Swank) who suddenly appears outside, begging for help. Daughter convinces the woman to don a hazmat suit before reluctantly allowing her inside, hiding her from Mother.
The woman's arrival marks a major shift in both the story and in Daughter's life. Before, Daughter had been told she was the last human and that it would be her responsibility to assist with raising her future "brothers and sisters." Now, as the woman begins to tell her of other survivors, Daughter is forced to confront the possibility that Mother has either been wrong in her assumptions or — perhaps even worse — lying.
Adding to Daughter's growing doubts are the woman's claims that the outside world is completely clean. The real threat, she insists, are hostile robots who will attack humans on sight — robots that supposedly resemble Mother.
Still, Daughter is not certain she can trust the woman, who appears desperate and may not be telling the whole truth. Then again, this woman does not seem to be contaminated, whereas the gunshot wound she has reportedly suffered from the robots is quite real.
"I Am Mother" is a film that plays with the audience's expectations, challenging its watchers, much like Daughter, to question what has been assumed and what is actually true.
Even though Mother is initially presented as a compassionate caregiver, both Daughter and the viewer must soon consider whether she is truly capable of feeling emotion or if she is, in fact, a cold, calculating machine whose real motives remain hidden behind an artificial, motherly facade.
What exactly is it Mother wants — to reestablish humanity? If so, will this be a humanity with self-determination or one which Mother herself will guide? What degree of power will she have? What of the other robots the woman speaks of? If they are real, is Mother aware of them? Does she wish to protect Daughter from them, or is she working with them? Is there a darker plan for humanity at play?
7 Reasons to Watch 'Neon Genesis Evangelion' on Netflix
Much like the protagonist, the audience is presented with a number of disturbing possibilities, not just with Mother. The woman is a stranger and could be making up any number of things simply to bring Daughter to her side. Did she really have a brother, as she claims? Is there such a place as "the mine," where other human survivors can be found? What if this is really all an elaborate cover story, meant to lure the unwary?
The woman is also armed. She has the first real gun Daughter has ever seen. Does the woman only carry it for protection, or does she have more sinister intentions in mind? Is she an innocent survivor, or could she be a bandit, a murderer? Did her gunshot wound come from hostile robots, or from other humans?
"I Am Mother" raises all of these questions, both directly or implicitly, although, ultimately, it is up to Daughter to find the answers. But the more Daughter starts to dig, the darker the secrets she uncovers.
On one level, the film can be said to represent that time in adolescence where one realizes the sources of information they have placed their trust in, including their own parents, may not have been as reliable as they first thought. Daughter, the very foundation of her beliefs shaken by the woman's revelations, comes to understand how uncertain and terrifying reality can be.
'Dead to Me' Is a Dark, Cutting Comedy About Grief and Murder
On another level, "I Am Mother" critiques humanity's increasing reliance on machines. In this instance, a machine is placed in charge of the destiny of the entire human race, starting with Daughter. Mother is given full reign over Daughter's life and education from the time she is a baby up to the film's present.
Even though Mother claims she is programmed only to care for humans and never to harm them, the woman's intense fear of the robot, coupled with her stories of machines gunning down humans in cold blood, gradually begins to make Mother's presence feel profoundly ominous. Mother slowly comes to seem less like Daughter's guardian and more like her warden.
"I Am Mother" does not need to rely on special effects to frighten its viewers. The film's thrills come primarily from the tension it builds through sudden twists and turns, as unexpected happenings and alarming discoveries keep the audience ever in a state of uneasy anticipation.
Perhaps even more horrifying is simply the vision of the future the film supplies: the human race under grave threat, dependent on a machine that may yet prove to be its doom. Like all great science fiction, "I Am Mother" offers a window into what could be — and perhaps what already is.
Culture /// Screens
'Tales of the City' Conveys the Importance of a Chosen Family as a Safe Haven For the LGBTQ+ Community
Does 'When They See Us' Make You Uncomfortable? Good.
'Hemlock Grove' Is One of the Best Horror Series that Netflix Has to Offer
In 'See You Yesterday,' Spike Lee Does It Again!
Season 4 of 'Lucifer,' Now Raunchier As a Netflix Program, Has Found Its Home
The 3 Best Episodes from Seasons 1 & 2 of Netflix's 'Easy'
Top 6 Underrated Must-Watch Netflix Originals
Everything You Need to Know About 'The Crown' Season 3
Why Fans Were so Adamant 'One Day at a Time' Get a Second Chance
Netflix's 'The Perfection' Is Anything But Perfect
'My Next Guest Needs No Introduction with David Letterman' Is a Blueprint for Interviewers

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Bootstrap Hamburger icon animation in navbar-header Good day everyone,
How to make menu Hamburger button animate to X in bootstrap navbar-fixed-top?

              <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar">
                Toggle navigation

              </button>
               Logo

Menu1

Menu2

Menu3

I was trying to plug in all possible examples, but I am missing the point. Any ideas?
Many thanks.

A: You asked for how to animate the transition, I think jQuery would be great for this.

$(".navbar-toggle").click(function () {
  $(".open").toggle(500).siblings().toggle(500);
});
.close {
 display:none;
}
<!DOCTYPE html>

        <meta charset="utf-8">
        <meta http-equiv="X-UA-Compatible" content="IE=edge">
        <meta name="viewport" content="width=device-width, initial-scale=1">

        <meta name="description" content="">
        <meta name="author" content="">
        <link rel="icon" href="../../favicon.ico">

        <title>Test</title>

        <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" integrity="sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u" crossorigin="anonymous">

        <link href="../../assets/css/ie10-viewport-bug-workaround.css" rel="stylesheet">

        <link href="style.css" rel="stylesheet">

              <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar">

                  Toggle navigation

                  X

              </button>
               Logo

Menu1

Menu2

Menu3

A: First of all, you need create additional markup for close button
<button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar">
            Toggle navigation

               x

          </button>

So, after that you need to update your css:
 .navbar-toggle .burger{
  display: none;
 }
 .navbar-toggle .close{
   display: block;
 }

 .navbar-toggle.collapsed .burger{
   display: block;
 }
 .navbar-toggle.collapsed .close{
   display: none;
 }

P.S. also you can easily add animations between default and .collapsed classes
JSFiddle example


Working at UMass Amherst is like living in a city: the campus feels geographically expansive, it's hard to communicate across boroughs, information gets lost, and you find yourself sticking to the same familiar locations. But within this city, there are always opportunities to seize if you know how to find them; we apply this same analogy to locating on-campus funding for your teaching and research development.
Identify three approaches you can use to find on-campus funding for teaching and research.
Describe the benefits of participating in on-campus fellowships that build community and offer funding for teaching or research.


We're delighted you've chosen to spend your break in the Roi Morvan country. You made a great decision!
After booking your accommodation, why not take a look at the best restaurants in our region, note down the addresses of some of the leisure facilities and guided tours, and to download our diary! Enjoy your holidays!


Ready to Book and Pay Online Now?
Can't find your preferred tour dates? Give us a call and we'll add it for you so you can start inviting other groups to join you.
Looking for big groups discounts for company & school outings? Give us a call to request for a quote.
San Antonio, Zambales is the camping capital of the Philippines since 2009. Since then, the number of visitors has been steadily increasing year after year, rivaling other 'best beaches' destinations like Puerto Galera and Boracay in terms of tourist arrivals. One of the best kept secrets of Luzon Island.
Pundaquit, San Antonio is famous for its 4 coves and 1 island. The 4 coves are named Anawangin, Nagsasa, Talisayin & Silanguin. The island is called Capones Island, home of the antique Spanish era lighthouse that looks out to guard the West Philippine Sea.
Beach Camping is the only form of accommodation in the coves and there is no direct road access to the camp grounds. The lack of electricity and cellphone signal adds to the allure of getting 'trapped' in an island with only a few people to spend quality time with. In the tradition of Survivor Philippines, the night is lit with tiki torches (suló), bonfires and the night sky. A very special and unique experience like no other in the world.
Pine Tree Forest provides cover for tent shelters and produces that unique swooshing sound inside it. It's weird enough to see pine forests in tropical islands, and it's even more weird that they line up along the white sand beaches. Surely a sight and sound to behold.
Bangka is the only means of transportation to the coves. Bangka is the name of the traditional and ancient Filipino mode of sea transportation. Experience our culture and enjoy our company.


Wasps are going in and out of an air vent on my house but I can't see the nest can you help?
Yes a residual insecticide dust can be applied to the air vent and the wasps carry it into the nest which then destroys the wasp colony.
Birds are roosting on the roof of my house but I don't want any harm to come to them, how can I get rid of them?
The parts of the roof they are roosting on can be fitted with bird spikes which stop birds landing and roosting.
Why is payment made quarterly for routines ?
It can be arranged to be taken monthly, however as standard it is set up quarterly with two routine visits taking place each period; 8 each year.
Is it true if you have mice you don't have rats?
No, not necessarily, if there is enough food such as cereals they will quite happily live alongside each other.
We try to minimise changes to our pricing however impacts outside our influence such as fuel, chemicals, bait, and labour can increase price over time. When signing a contract with us, we commit to keeping that price in place for a minimum of two years.
Do we use humane traps?
Humane, means causing as little distress to the pest as possible and this is right at the heart of our business. The poisons we use actually put the animal to sleep without causing pain or stress. Live traps are available however, are far more expensive as they need to be monitored daily they also don't get rid of the problem as it is illegal to re-home a pest on someone else land.
Is there a weekend call out charge?
No, we charge the same prices throughout the week. We've built our business on being there for our customers when they need us, so we don't believe we should pass on an extra cost to our clients.
My house has been treated for rats or mice but I am still hearing them, why?
The anticoagulant poison that is used can take up to 5 days to work. It also may be that you have not found the source of your problem, a site visit would allow us to see the issue in question and advise any alternative methods.
I have rats in my garden but I am afraid my dog might get to the poison?
The poison is put in tamper proof bait boxes and are fixed down so non target species can't get to the poison.
How do I know if you've caught the moles?
As standard we offer to show you the moles caught as guarantee for work carried out successfully.
What damage can mice or rats cause?
Whats the difference between a clean up (one -off treatment) and a contract?
A clean up is when we come in and get rid of the pest and then remove all bait boxes and treatment measures. A Contract is when we clear -up the infestation, but then return regularly to ensure that the outbreak does not occur again.
All birds are protected. Proofing measures such as spikes and nets must be used in the first instance.
Why am I now getting moles?
I would suggest there has either been a recent population increase in your area, or the soil quality has improved and is now housing a higher quantity of worms.


text:
"Just being in New York is an incredible benefit of attending Juilliard! So many orchestras come here. When the London Phil was in town, their tuba player gave a master class at Juilliard and their music director conducted our orchestra — and I performed in the concert. As a sophomore I got to play with the Juilliard Orchestra at Carnegie Hall. It was an amazing experience!" — Ibanda Ruhumbika, BM '12, Tuba
Degrees & Diplomas
See Application Requirements See Application Req.
Core Experiences
Performing before a live audience is an essential part of the Juilliard experience. Whether you are an instrumentalist, singer, or composer, you will have frequent opportunities to share your art with the public. Most performances take place in one of Juilliard's superb auditoriums, but you will also have occasion to appear elsewhere at Lincoln Center, including David Geffen and Alice Tully halls, homes of the New York Philharmonic and the Chamber Music Society; at other venues in New York City, such as Carnegie Hall, where the Juilliard Orchestra performs annually, the Blue Note Jazz Club, where Jazz students enjoy regular gigs, and the Museum of Modern Art, where the New Juilliard Ensemble appears in the Summergarden series; or on national and international tours.
Classroom Studies
The Music Division's main areas of classroom instruction—Ear Training, Music Theory and Analysis, Music History, and Keyboard Studies—complement your private lessons and performance experiences to broaden your understanding of the foundations of music and to provide a context for the music literature you study and perform. Juilliard's Center for Innovation in the Arts equips students with skills to compose and perform music using new technology, and offers hands-on experience with state of the art equipment.
Whatever your discipline, your development as an artist will be deepened by a foundation in the liberal arts, which provides the humanistic, ethical, social, critical, and aesthetic background essential to personal development and professional excellence. All undergraduate degree students take classes in literature, philosophy, history, social sciences, arts, and languages – studies that will help you develop a deeper understanding of yourself and the complex world in which you live. Through your coursework in liberal arts, you will refine your skills in reading, writing, speaking, and critical thinking, and learn to communicate with greater clarity and effectiveness.
In the changing cultural landscape, performers must think more creatively about their careers than ever. While some pursue positions in traditional settings, others are exploring new ways to create opportunities that are both creatively satisfying and financially sustainable. The Alan D. Marks Center for Career Services and Entrepreneurship encourages leadership, initiative, and entrepreneurial thinking, enabling young performing artists to develop a more expansive view of success in the 21st century. It teaches students the skills essential for success in today's entrepreneurial environment, with workshops in public speaking and business planning, panel discussions with industry leaders, and entrepreneurship residencies. It also facilitates more than 450 freelance performance opportunities for Juilliard students.


text:
Q: How to find overlapping / redundant nodes in a tree I have a list of nodes in a big hierarchy. How do I efficiently determine which of my nodes are descendants of any others in my list?
Specifically this is a list of ASP.NET controls. My code examines each of these nodes and their descendants, so I am trying to eliminate redundancy.
A simple example is this tree:
         Parent
         /    \
    Gladis    Arthur
    /    \
 Jon      Sam

My list contains { Parent, Jon }.
I would like to write an algorithm that reduces the list to just Parent based on the fact that the Parent node contains Jon as a descendant.
Here is code:
    public class Node
    {
        public Node(string name_, params Node[] children_)
        {
            this.children = children_;
        }
        public string name;
        public Node[] children;
    }

    public class NodeAnalysis
    {
        public Node node;
        public Node containingAncestor = null;
    }

    static void Main(string[] args)
    {
        Node jon = new Node("jon");
        Node parent = new Node("parent",
            new Node("Gladis",
                jon,
                new Node("Sam")),
            new Node("Arthur")
        );

        List<NodeAnalysis> results = new List<NodeAnalysis>();
        results.Add(new NodeAnalysis{ node = parent });
        results.Add(new NodeAnalysis{ node = jon });

        foreach (NodeAnalysis item in results)
        {
            // ??? populate item.containingAncestor
        }
    }

I can't think of an efficient algorithm to accomplish this. I cannot control the order in which the nodes are added to the list. It seems like it could be optimized by examining both the tree and the relationships I've already identified in the list as I traverse it.
**EDIT: .parent is available in the Node struct. It's easier to discover ancestral relationship using .parent.

A: Here's one approach:
public static void RemoveDescendants(IList<Node> list)
{
    for (int index = 0; index < list.Count; index++)
        RemoveDescendants(list[index], list);
}
private static void RemoveDescendants(Node node, IList<Node> list)
{
    foreach (var child in node.children)
    {
        list.Remove(child);
        RemoveDescendants(child, list);
    }
}


Void-off-Course: August 31 @ 2:53 am (ET) August 30 11:53 pm (PT) to August 31 4:33 am (ET) 1:33 am (PT) Pisces is a fortunate sign for the Void off Course moon.
House(s): 5 th & 6th: 5th house of creativity, fun, children and pets: 6th house of health, service and detail.
Opportunity Period: August 29 6:21 pm (ET) 3:21 (PT) to August 31 4:33 am (ET) 1:33 am (PT): Good for meditation, artistic activities and being helpful to others.
In the 5th house: We may find we are spending more time with kids and pets… It' s a time to have some fun, be entertained, or be entertaining… This could be a challenging time for those who enjoy sticking to a routine, and for those who love to fun it may be more than they usually handle. A healthy balance between work (duty) and play is strongly encouraged.
In the 6th house: Our well-being may be on our mind and our health may be in focus. We could be tempted to do more for less or find we are in a position to be of service to someone. Some of us can see all the small details now, don't get overwhelmed, just do what's needed.
The full moon will emphasize the traits mentioned above.
Folk should be on the watch for deception… from without as well as from within.
It may be easier to full into a day dreaming world, especially if reality is offering a harsh alternative, take care when choosing the method of "escape".
Those with addictions challenges may be especially challenged now, finding a safe place is recommended.
Medical issue's and or appointments regarding the feet may be on the agenda.
Those with a low tolerance to prescription medications are advised to care over the next few days when taking any, negative or unusual reactions to such, may occur.
Other's may express unusually selfish behavior.
The moon rules a water sign, (Cancer), Pisces is a fellow water sign, and a duel one… Emotional situations could be on the agenda.


GAZA CITY – The Israeli lethal attacks against Gaza – described as a "Holocaust" by a top Israeli army official – killed as many as toddlers and innocent civilians, and turned the tiny territory into a moonscape with destroyed homes and scorched earth; but it breathed life into long-dormant unity between the rival brothers Hamas and Fatah.
"Grief has united us all," Kamal, a Fatah member, told IslamOnline.net Wednesday, March 5.
Kamal did not think twice and rushed to the home of a Hamas neighbor to condole with his family after he was killed in one of the ongoing Israeli missile attacks on the Gaza Strip.
Now he says he will provide for the family of the Hamas victim.
"Aren't we brothers? It is a duty not a favor to solace one another in such hard times. My heart breaks for my (Hamas) neighbor," said Kamal, his eyes red and brimming with pain.
Five days of Israeli air strikes and ground incursions in Gaza have killed at least 123 Palestinians, mostly women and children, and translated into action threats by Israeli Deputy Defense Minister Matan Vilnai to turn Gaza into a "bigger Holocaust."
More than 350 Palestinians, mostly civilians, were wounded in the offensive with many of them were left die as Israel impeded ambulances.
Tuesday night, Israeli occupation troops shot dead a Palestinian baby girl and a senior Hamas member during an incursion into Gaza.
In a spontaneous moving scene, dozens of Hamas and Fatah supporters were seen over the past few days sitting shoulder-to-shoulder to offer their condolences to the families of the raid victims.
The buried the ghost of their bitter disunity and trampled upon simmering political rivalries between their leaders.
In the Hamas stronghold of Gaza, the Fatah supporters shrugged off moles who routinely report to the leadership in Ramallah about Fatahists who "switch allegiance."
The heart-wrenching scenes that keep unfolding from the Gaza Strip have made national unity at the top of the ladder of priorities.
"It is high time we united and let bygones be bygones," Aziz, a Fatah supporter, said after offering his heartfelt condolences to the family of a Hamas victim in Gaza.
Abu-Ghaleb Al-Banna, a Hamas supporter, said the Israeli onslaught is a bittersweet incident.
"It broke our hearts, but united us," he said.
The rift between Hamas and Fatah widened last June after Abbas had sacked a national unity government headed by Hamas leader Ismail Haniyeh after Hamas seized control of the Gaza Strip.
Citing confidential documents, the US magazine Vanity Fair revealed that the Bush administration had drawn up plans to topple the ruling Hamas after its 2006 parliamentary victory with the help of Fatah strongman Mohammad Dahlan, prompting Hamas to take a preemptive action by taking over the Gaza Strip.
Hamas leader Khaled Meshaal renewed calls on Saturday for a national unity government and unconditional, open dialogue with Fatah to tackle all sticking points.
In the West Bank, the green flags of Hamas were seen mingling with the yellow ones of Fatah in mass marches in Nablus, Tulkarm and Bethlehem against the Israeli incursions.
"Unity, unity…Hey Fatah and Hamas!" shouted a group of students. "What are you waiting for? Come on and shake hands."
Is Israel Using Prohibited Weapons?


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Bitcoin: Be prepared for the rise of digital currency, adopters say
14 Feb 2014 / CBC.ca – Each day, Lorne Lantz buys his lunch with Bitcoin. Some say the burgeoning digital currency has no real value — but his sandwiches say otherwise.
Bitcoins and other "cryptocurrency" have started to rise from the fringes of the internet to the cusp of mainstream use — a remarkable breakthrough for a currency which made its online debut only four years ago.
When they first began moving online, Bitcoins could buy you almost nothing. Now, there's almost nothing they can't buy. From songs to survival gear, cars to consumer goods, hard currency to hard drugs — retailers are rushing to adopt the virtual currency for all its perks. Bitcoin ATMs are popping up in major cities like Toronto, Ottawa and Vancouver.
READ MORE: Why are Canadians buying Bitcoin?
"It solves so many big problems in the financial industry," said Lantz, a recent DeGroote School of Business graduate who is now an engineer working on financial startups in Silicon Valley. "We're moving towards everything being done digitally."
Advocates describe Bitcoin as the foundation of an almost utopian economy: one with no borders, no change fees, no closing hours, and no one to dictate what you can and can't do with your money.
The currency's detractors say Bitcoin in and of itself has no value – rather that it just functionally meets certain payment and monetary needs — which has attracted investors who then give Bitcoin real financial value through speculation. The Conservative government also singled out the seedier side of the currency in its 2014 federal budget this week, pledging to ensure it can't be used to launder money or finance terrorism.
But early adopters in the Hamilton area say digital currency is just going to keep growing…… Read more
http://www.cbc.ca/news/canada/hamilton/news/bitcoin-be-prepared-for-the-rise-of-digital-currency-adopters-say-1.2537391
On February 14, 2014 / companies, exchange, finance, mining, politics


You recently got into a fistfight with your ex-significant other. The police arrived and now you have been charged with domestic battery.
But your relationship ended some time ago. Is it still domestic battery?
The answer is yes. A recent Illinois Supreme Court decision held that the domestic battery law places no time limit on a dating relationship. (See People v Gray.
In Illinois, a person commits domestic battery if he or she knowingly, without legal justification and by any means, causes bodily harm to or makes insulting or provoking physical contact with any family or household member. A family or household member includes persons who have or have had a dating or engagement relationship. This definition does not include a casual acquaintanceship or ordinary fraternization between two people in a business or social context. But it can include your ex-partner from several years ago.
If you have been charged with a crime, contact an experienced criminal law attorney immediately. An attorney can review your case for its best possible defense. Perhaps you never you were just friends with the alleged victim, or you acted in self defense. Even if the evidence against you is overwhelming, an attorney who is respected in the court house may be able to negotiate a more favorable plea agreement than you could on your own.
See also: Illinois Domestic Battery Law.
SEARCH INCIDENT TO ARREST: HOW FAR CAN IT GO?


text:
Joost Swarte (1947) is a cartoonist, illustrator and graphic designer. He studied industrial design in Eindhoven, but abandoned his studies after three years and started drawing comic strips. His comic books and books he has illustrated have been translated into various languages. He is best known for his 'clear-line' style of drawing and coined the term to describe the defining characteristic of his work: a refined drawing style with clear outlines filled with bright blocks of colour.
Over the years his work shifted from comic strips (Charlie Hebdo, Metal Hurlant, Raw Magazine) to illustrations (for the Humo, Vrij Nederland, The New Yorker magazines), posters, stamps, lettering design and free-form graphics. He also occasionally designs furniture and other objects. In 1992 he initiated the Stripdagen, a biennial international comic event held in Haarlem. In 1996 he presented his design for the Toneelschuur, a new theatre in Haarlem. His plans were elaborated in partnership with Mecanoo architects and the theatre opened in 2003. He also designs leaded and stained glass windows, the largest of which (± 100 m²) was installed in the Palace of Justice in Arnhem in 2004. He was also closely involved in the conception and design of the Hergé Museum in Louvain-la-Neuve (Belgium).
Joost Swarte has won numerous awards for his work, including, in 2012, the Marten Toonder Award, an oeuvre award presented to a cartoonist who has made an outstanding contribution to Dutch comics.
Als de blaadjes vallen
The title of Joost Swarte's Miffy statue is Als de blaadjes vallen [When the leaves fall]. 'Miffy in the autumn. Winter is still to come. But after that there is always the spring.'
more about Joost Swarte