text:
Q: Integration with textstyle fractions limits I am trying to write an integration question like below with amsmath package:
\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\,\mathrm{d}x

However, the two fractions limits appear to be too small. I would like to do something like this:
\displaystyle\int_{\textstyle\frac{\pi}{4}}^{\textstyle\frac{\pi}{2}}\sin x\,\mathrm{d}x

Is there an efficient way to set all the integration limits in the document to \textstyle?

A: The only sensible way is to use \tfrac. Note the inconsistency in a common situation either with default or using \textstyle:
    \documentclass{article}
    \usepackage[utf8]{inputenc}
    \usepackage[T1]{fontenc}
    \usepackage{mathtools}

    \begin{document}

    Default:
    \[  \int_{k\frac{\pi}{4}}^{k\frac{\pi}{2}}\sin x\,\mathrm{d}x \]%
    \vskip 2ex
    With \verb+\tfrac+:
      \[ \int_{k\tfrac{\pi}{4}}^{k \tfrac{\pi}{2}}\sin x\,\mathrm{d}x  \]%
    \vskip 2ex
    With \verb+\textstyle+:
    \[  \int_{\textstyle k\frac{\pi}{4}}^{\textstyle  k\frac{\pi}{2}}\sin x\,\mathrm{d}x \]%

    \end{document}

A: You asked,

Is there an efficient way to set all the integration limits in the document to \textstyle?

When TeX is in displaymath mode, \int produces a "large" integral symbol and the upper and lower limits of integration are typeset in scriptstyle. What you're encountering is that the numerators and denominators inside these limits are typeset in scriptscriptstyle if \frac is used.
To change this default behavior globally, i.e., to switch from scriptstyle to textstyle in the limits of integration on a document-wide basis, would require major surgery on TeX's math innards and is not to be undertaken lightly. And I would hate to call such coding efficient. Moreover, you will probably not like the resulting look if the limits contain simple numbers and letters (since these will now be set in textstyle, i.e., the same size as applies for \sin x\,\mathrm{d}x.
You could use \tfrac explicitly -- \tfrac is short for \textstyle\frac-- to get scriptsize- rather than scriptscriptsize-sized numerator and denominator terms; the resulting look is also shown in @Bernard's earlier answer. Alternatively, you could use "inline-style" fraction notation, i.e., write \pi/4 and \pi/2 in the limits; these terms will be set in scriptstyle automatically.

\documentclass{article}
\usepackage{amsmath} % for \text and \tfrac macros
\begin{document}
\[
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\,\mathrm{d}x    % \frac
\quad\text{vs.}\quad
\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\sin x\,\mathrm{d}x  % \tfrac
\quad\text{vs.}\quad
\int_{\pi/4}^{\pi/2}\sin x\,\mathrm{d}x                    % inline-style
\]
\end{document}

redpajama_set_name: RedPajamaStackExchange


text:
RIORDAN, RICK
From the creator of PERCY JACKSON, bestselling author Rick Riordan, comes a brand-new adventure, inspired by Jules Verne's Twenty Thousand Leagues Under the Sea.
Ana Dakkar is a freshman at Harding-Pencroft Academy, a five-year high school that graduates the best marine scientists, naval warriors, navigators, and underwater explorers in the world. Ana's parents died while on a scientific expedition two years ago, and the only family's she's got left is her older brother, Dev, also a student at HP. Ana's freshman year culminates with the class's weekend trial at sea, the details of which have been kept secret.
She only hopes she has what it'll take to succeed. All her worries are blown out of the water when, on the bus ride to the ship, Ana and her schoolmates witness a terrible tragedy that will change the trajectory of their lives.
But wait, there's more.
The professor accompanying them informs Ana that their rival school, Land Institute, and Harding-Pencroft have been fighting a cold war for a hundred and fifty years. But now the heat is on and the freshman are in danger of becoming fish food. In a race against deadly enemies, Ana will make amazing friends and astounding discoveries about her heritage as she puts her leadership skills to the test for the first time.
Rick Riordan's trademark humour, fast-paced action, and wide cast of characters are on full display in this undersea adventure.
Rick Riordan has now sold an incredible 180 million copies of his books worldwide.
'This is the stuff of legends' - The Guardian on Percy Jackson
One boy and his toy are about to change everything... Jack loves his childhood toy, Dur Pig. DP has always been there for him, through good and bad.Until one Christmas Eve, something terrible happens - DP is lost. But Christmas Eve is a night for miracles and lost causes, a night when all things can come to life... even toys.And Jack's newest toy - the Christmas Pig (DP's annoy...
DIARY OF A WIMPY KID: BIG SHOT BOOK 16
KINNEY, JEFF
In Big Shot, book 16 of the Diary of a Wimpy Kid series from #1 international bestsetlling author Jeff Kinney, Greg Heffley and sports just don't mix.After a disastrous field day competition at school, Greg decides that when it comes to his athletic career, he's officially retired. But after his mom urges him to give sports one more chance, he reluctantly agrees to sign up for ...
SOMETHING I SAID
SMITH BAILEY, BEN
From comedian, actor, rapper and screenwriter Ben Bailey Smith comes the biggest middle grade debut of the year - a blazingly funny, big-hearted story about family, friendship and how far one boy will go to get a laugh. For thirteen-year-old Carmichael Taylor, life is one big joke - in a good way. He just can't understand why no one else seems to find everything as funny as he ...
COUNTING ON KATHERINE
BECKER, HELAINE
Winner of the information book category of the UKLA Book Awards 2020. Katherine Johnson is the mathematical genius who helped make the historic Apollo 11 moon landings possible and made sure that Apollo 13 returned home safely when the mission was in critical danger. As a child, Katherine loved to count.She counted the steps on the road, the number of dishes and spoons she wash...
A TALE OF MAGIC...
COLFER, CHRIS
A new adventure series set in the Land of Stories universe from the No. 1 New York Times bestselling author, Chris Colfer, perfect for new and old fans! For fans of The School for Good and Evil and Pages & Co. When Brystal Evergreen stumbles across a secret section of the library, she discovers a book that introduces her to a world beyond her imagination and learns the impossib...
CRAFT, JERRY
Winner of the Newbery Medal, Coretta Scott King Author Award, and Kirkus Prize for Young Readers' Literature! Perfect for fans of Raina Telgemeier and Gene Luen Yang, New Kid is a timely, honest graphic novel about starting over at a new school where diversity is low and the struggle to fit in is real, from award-winning author-illustrator Jerry Craft. Seventh grader Jordan Ban...
LA ÚLTIMA DESCENDIENTE
UN SECRETO EN LAS PROFUNDIDADES DEL OCÉANOUNA BATALLA BAJO EL MAR PARA DESCUBRIR LA VERDADLa Academia Harding-Pencroft no es un instituto normal: de sus clases salen los mejores exploradores submarinos y los navegantes más prestigiosos del mundo, que compiten en condiciones extremas divididos entrelas casas Delfín, Cefalópodo, Orca y Tiburón. Ana Dakkar, una brillante estudiant...
THE TOWER OF NERO
LA TORRE DE NERON
BURNING MAZE, THE
FROM 12 YEARSThe third book in the latest series from international bestselling author, Rick Riordan He was a God once. Until he was cast out his father, Zeus. Now, he's an awkward teenager. Called Lester. The only way out is a series of scary and dangerous trials, of course. For his third trial, Apollo must journey through the Labyrinth to free an Oracle who only speaks in puz...
LA PROFECÍA OSCURA
PERCY JACKSON AND THE SEA OF MONSTERS (FILM)
FROM 12 YEARS ...
Vista previa: DAUGHTER OF THE DEEP

redpajama_set_name: RedPajamaCommonCrawl


text:
Mayor says mound will be demolished, hill developed [LINK/Lead]
Anniston Star – Mayor says mound will be demolished hill developed
Oxford may be planning to remove a controversial pile of stones from a hill behind the Oxford Exchange shopping center, according to a story in a national newspaper.
A story that appeared on the Web site of The New York Times on Friday said Mayor Leon Smith plans to demolish the stone pile, which may have been built a thousand years ago by American Indians, and make it the site of a hotel, restaurant or health clinic.
When Scholarship and Tribal Heritage Face Off Against Commerce [LINK/Lead]
Oxford Journal – When Scholarship and Tribal Heritage Face Off Against Commerce – NYTimes.com
OXFORD, Ala. — Overlooking the Interstate and an outdoor shopping mall here stands a sad little hill, bald but for four bare trees and a scattering of stones.
Bob Farley for The New York Times
Harry O. Holstein, the archaeology professor who tried to protect the stone mound.
That the stones are there is beyond argument. But everything else about them — whether somebody put them there, how long they have been there and what should be done with them — became a matter of fierce debate last summer and has continued to yield surprising twists into recent weeks.
The latest episode in the very long history of the Oxford stones began last June, when an excavator showed up on the hill. The city was planning for the construction of a Sam's Club nearby and intended to use dirt from the hill for the area where the store would sit.
Then a local archaeology professor began making phone calls.
The professor, Harry O. Holstein of nearby Jacksonville State University, had concluded that a stone mound at the top of the hill was constructed by American Indians more than a thousand years ago, and in 2003 he recorded it in a state archaeological registry. The possibility of its being destroyed, Dr. Holstein said, made him sick.
"I'm not against development," he said. "But some things should just be saved."


@dianamarisol_ messaged one of our homies out of the wood work, first without proper greeting. Straight asking for a ride on the private jet. Talk about wasting no time. I wonder if her boyfriend who's she's been with for over a year knows about her hoeism. TagTeam, find me this mans Instagram.
Diana take this L you nut drainer.


text:
Q: Join only the lowest row with the nearest row from another table SQL I have two table in Teradata like below:
Table A
A|B|ID|C |D
1|2|11|hh|
1|2|31|zz|ss
1|2|21|ee|nn

Table B
A|B|ID|C |D |dat
1|2|10|yy|tt|'2021-01-01'
1|2|07|tt|uu|'2021-01-01'
1|3|12|ee|oo|'2021-01-02'

output
A|B|ID|C |D
1|2|11|hh|yy
1|2|21|ee|hh
1|2|31|zz|ee

Explanation
The goal is to calculate column D value so, we start with rows from table A (which is a with query in fact)
for the same A and B we look for the anterior row in ID values and we get the value of C which will be the D value of current row,
Till we come to the lowest row in ID value (11 in the example) in this case we look in the table B for the max(dat) for the same A and B ( if there is multiple rows with the same max(dat) we get the row contain the max(ID) with id < id of the current row (11 in the example)
I tried (the query below) to get the result with the Lag function but I can't do the join for only the lowest id from table A with the row of table B.
SELECT
a.A,
a.B,
a.ID,
a.C,
case when lag(a.A,1)over(partition by a.A,a.B order by a.id) isnot null then
lag(a.D,1)over(partition by a.A,a.B order by a.id) else b.c end as D

from
table_a a
left join ( select A,B,C,id,dat
from
table_b) b
on a.A=b.A
and a.B = b.B
and a.id > b.id

qualify row_number () over(partion by b.A,b.B order by b.dat desc,b.id desc) = 1

A: Non-equi joins will have bad performance, even worse when they require row_number.
Instead of joining combine both tables using UNION ALL, then sort the data the required way and use LAG to get the previous row's value.
select a, b, id, c
-- find the previous NOT NULL value, prefer table_a over table_b
  ,coalesce(lag(case when x = 1 then c end ignore nulls) over (partition by A, B order by id)
           ,lag(case when x = 2 then c end ignore nulls) over (partition by A, B order by id)) as D
from
 ( -- combine both table into one, x indicates where a row is coming from
   select 1 as x, a, b, id, c
   from table_a
   union all
   select 2 as x, a, b, id, c
   from table_b
) as dt
-- now remove all rows from table_b
qualify x = 1


text:
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Show: Season - Playoffs
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view top 50...
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This was a great review of the past tense system and its many applications in modern speech. After reading through this chapter, I began to understand concepts like all continuous forms feature some form of verb 'to be" plus the 'ing" form of verb. If the verb "to be" is in the present, it will be the present continuous. If the verb 'to be' is in the past....which means it will be the past continuous. As a native english speaker these concepts are such that I use quite regularly, yet becoming extremely aware through these diagnostics is a always a pleasantly exciting effect.


text:
\section{Introduction}

This paper addresses the following question.
{\it What is the aspect ratio of
  the shortest smooth paper Moebius band?\/}
Let's state the basic question more precisely.
Given $\lambda>0$, let
\begin{equation}
  \label{quotient}
  M_{\lambda}=([0,1] \times [0,\lambda])\!/\sim, \hskip 30 pt
  (x,0) \sim (1-x,\lambda)
\end{equation}
denote the standard flat Moebius band of width $1$ and
height $\lambda$.  This Moebius band has aspect ratio $\lambda$.
Let $S \subset \R_+$ denote the set of values of
$\lambda$ such that there is a smooth \footnote{
The smoothness requirement (or some
suitable variant) is necessary in order
to have a nontrivial problem.
Given any $\epsilon>0$, one
can start with the strip $[0,1] \times [0,\epsilon]$
and first fold it (across vertical folds) so that it
becomes, say, an $(\epsilon/100) \times \epsilon$ ``accordion''. One can
then easily twist this ``accordion'' once around
in space so that it makes a Moebius band.  The corresponding
map from $M_{\epsilon}$ is an isometry but it cannot be approximated by smooth
isometric embeddings.}
isometric embedding $I: M_{\lambda} \to \R^3$.
The question above asks for the quantity
\begin{equation}
  \lambda_0=\inf S.
\end{equation}
The best known result, due to Halpern and Weaver
[{\bf HW\/}], is that
\begin{equation}
  \lambda_0 \in [\pi/2,\sqrt 3].
  \end{equation}
In \S 14 of their book,
{\it Mathematical Omnibus\/}
[{\bf FT\/}],
Fuchs and Tabachnikov give a beautiful
exposition of the problems and these bounds.
This is where I learned about the problem.

The lower bound is local in nature and does not
see the difference between immersions and
embeddings. Indeed, in
[{\bf FT\/}], a sequence of
immersed examples whose aspect ratio tends to $\pi/2$ is given.
The upper
bound comes from an explicit construction.  The left side of
Figure 1.1 shows $M_{\sqrt 3}$, together with a certain
union of {\it bends\/} drawn on it.  The right
side shows the nearly embedded paper Moebius
band one gets by folding this paper model up
according to the bending lines.

\begin{center}
\resizebox{!}{1.5in}{\includegraphics{fig1.eps}}
\newline
    {\bf Figure 1.1:\/} The conjectured optimal paper Moebius band
\end{center}

The Moebius band just described is degenerate: It coincides as a set with the
equilateral triangle $\Delta$  of semi-perimeter $\sqrt 3$.
However, one can choose any $\epsilon>0$ and
find a nearby smoothly embedded image of
$M_{\sqrt 3+\epsilon}$ by a process of rounding out the
folds and slightly separating the sheets.  Halpern and
Weaver conjecture that $\lambda_0=\sqrt 3$,
so that the triangular example is the best one can do.

The Moebius band question in a sense goes back a long
time, and it is related to many
topics.  The early paper [{\bf Sa\/}] proves rigorously that
smooth paper Moebius bands exist.  (See [{\bf HF\/}] for
a modern translation to english.)  The paper [{\bf CF\/}] gives a general
framework for considering the differential
geometry of developable surfaces.
Some authors have discussed optimal shapes for
Moebius bands from other perspectives, e.g.
algebraic or physical.
 See, e.g. [{\bf MK\/}] and [{\bf S1\/}].
The Moebius band question has connections to origami.
See e.g. the beautiful
examples of isometrically embedded flat tori
[{\bf AHLM\/}].  It is also related to the main
optimization question from geometric knot theory:
What is the shortest
piece rope one can use to tie a given knot?
See e.g. [{\bf CKS\/}].

In this paper we improve the lower bound.  Because we do not want
to worry about possible pathologies, we assume explicitly that the
smooth extension of our map $I$ to a neighborhood of
$M_{\lambda}$ is regular in the sense that the differential
$dI$ is $1$-to-$1$ everywhere.  This regularity will
help us when we make polygonal approximations.
We don't want to worry about
any funny behavior at the boundary of $M_{\lambda}$.

\begin{theorem}[Main]
  \label{main}
An embedded paper Moebius band must have aspect ratio at least
$$\lambda_1=
\frac{2 \sqrt{4-2 \sqrt{3}}+4}{\sqrt[4]{3} \sqrt{2}+2 \sqrt{2 \sqrt{3}-3}}
=1.69497...$$
\end{theorem}
This value $\lambda_1$ arises naturally in a geometric optimization
problem involving trapezoids.  To quantify the way that our result
improves over the previous lower bound, we note that
$\lambda_1> \sqrt 3-(1/26)$
whereas $\pi/2 < \sqrt 3 - (4/26)$.

The proof of the Main Theorem has $2$ ideas, which we now explain.
Being a ruled surface, $I(M_{\lambda})$ contains a continuous
family of line segments which have their endpoints
on $\partial I(M_{\lambda})$.  We call these line segments
{\it bend images\/}.  Say that a $T$-{\it pattern\/} is a pair of
perpendicular coplanar bend images.
The $T$-pattern looks somewhat like the two vertical
and horizontal segments on the right side of
Figure 1.1 except that the two segments are
disjoint in an embedded example.
Here is our first idea.

\begin{lemma}
  \label{T}
  An embedded paper Moebius band of aspect ratio
  less than $7 \pi/12$ contains a $T$-pattern.
\end{lemma}

Note that $7 \pi/12>\sqrt 3$, so Lemma
\ref{T} applies to the examples of interest to us.
Lemma \ref{T} relies crucially on the
embedding property.
The immersed examples in
[{\bf FT\/}] do not have these $T$-patterns.

The two bend images comprising the $T$-pattern
divide $I(M)$ into two halves.  Our second
idea is to observe that the image
$I(\partial M_{\lambda})$ makes a loop which
hits all the vertices of the $T$-pattern.
When we compare the relevant portions of
$\partial M_{\lambda}$ with a polygon made from
the vertices of the $T$-pattern, we
get two constraints which lead naturally
to the lower bound of $\lambda_1$.

Since we can almost effortlessly give {\it some\/}
improvement to the lower bound
 using Lemma \ref{T}, we do this now.
The convex hull of the $T$-pattern contains a
triangle of base at least $1$ and height at least $1$.
Such a triangle has semi-perimeter at
least $\phi=1.61...$, the golden ratio.  Hence
$\lambda_0 \geq \phi$.  The argument which
gives $\lambda_0 \geq \lambda_1$, the
constant in Theorem \ref{main}, is similar
in spirit but more elaborate.

It would have been nice if the existence
of a $T$-pattern forced $\lambda>\sqrt 3$ on its
own.  This is not the case.  In Figure 2.1
we show half of an immersed paper Moebius
band of aspect ratio less than $\sqrt 3$
that has a $T$-pattern.  I did not try
to formally prove that these exist, but
using my computer program I can construct
them easily. In a sequel paper I explain
how to reduce the main conjecture above to
some finite dimensional calculations (which
currently are too hard for me to perform.)

  This paper is organized as follows.
  In \S 2 we introduce polygonal paper Moebius
  bands and some basic geometric
  objects associated to them.  We then
prove Lemma \ref{T} for polygonal Moebius bands.
In \S 3 we give the argument involving the
vertices of the $T$-pattern and thereby
prove Theorem \ref{main} for polygonal paper Moebius bands.
In the brief \S 4 we explain
why the results in the polygonal case
imply the same results in the smooth case.

I would like to thank Dan Cristofaro-Gardiner,
Dmitry Fuchs,
Steve Miller, and Sergei Tabachnikov for
helpful discussions about this
problem. I would especially like to thank Sergei
for telling me about the problem and pointing
me to his book with Dmitry.
I would also like to
acknowledge the support of the Simons
Foundation, in the form of a 2020-21 Simons
Sabbatical Fellowship, and also the
support of the Institute for Advanced Study, in the
form of a 2020-21 membership funded by a grant
from the Ambrose Monell Foundation.

\newpage

\section{Existence of the T Pattern}
\label{basic}

\subsection{Polygonal Moebius Bands}
\label{basic00}

\noindent
{\bf Basic Definition:\/}
Say that a {\it polygonal Moebius band\/} is a
pair ${\cal M\/}=(\lambda,I)$ where
$I: M_{\lambda} \to \R^3$ is a map which is
an isometry on each triangle.
 We insist that the vertices of
these triangles all lie on $\partial M_{\lambda}$,
as in Figure 1.1.   In this paper we will
work entirely with polygonal Moebius bands.

At the end of the paper, we will include
a short chapter which explains why
the results in the polygonal case
imply the results in the smooth case.
\newline
\newline
{\bf Associated Objects:\/}
Let $\delta_1,...,\delta_n$ be the successive triangles
of $\cal M$.
\begin{itemize}
\item The {\it ridge\/} of $\delta_i$ is
edge of $\delta_i$ that is contained in
$\partial M_{\lambda}$.
\item The {\it apex\/} of $\delta_i$ to be the
  vertex of $\delta_i$ opposite the ridge.
\item A {\it bend\/}
  is a line segment
  of $\delta_i$ connecting the apex to
  a ridge point.
\item A {\it bend image\/} is the image of a bend under $I$.
\item A {\it facet\/} is the image of some $\delta_i$ under $I$.
\end{itemize}
We always represent $M_{\lambda}$ as
a parallelogram with top and bottom
sides identified.  We do this by
cutting $M_{\lambda}$ open at
a bend. See Figure 2.1 below.
\newline
\newline
{\bf The Sign Sequence:\/}
Let $\delta_1,...,\delta_n$ be the
triangles of the triangulation associated
to $\cal M$, going from bottom to top
in $P_{\lambda}$.
We define $\mu_i=-1$ if
$\delta_i$ has its ridge on the left edge
of $P_{\lambda}$ and $+1$ if the ridge is on the right.
The sequence for the example in Figure 1.1 is
$+1,-1,+1,-1$.
\newline
\newline
{\bf The Core Curve:\/}
There is a circle $\gamma$ in $M_{\lambda}$ which
stays parallel to the boundary and
exactly $1/2$ units away.  In Equation
\ref{quotient}, this circle is the
image of $\{1/2\} \times [0,\lambda]$
under the quotient map.  We call
$I(\gamma)$ the {\it core curve\/}.

The left side of Figure 2.1 shows $M_{\lambda}$ and the pattern of bends.
The vertical white segment is the bottom half of $\gamma$.
The right side of Figure 2.1 (which has been magnified to
show it better) shows $I(\tau)$ where $\tau$ is the colored half of $M_{\lambda}$.
All bend angles are $\pi$ and the whole picture is planar.
  The colored curve on the right is
  the corresponding half of the core curve.
  Incidentally, for $\tau$ we have
    $L+R=1.72121...<\sqrt 3$.
\begin{center}
\resizebox{!}{3.1in}{\includegraphics{example2.eps}}
\newline
    {\bf Figure 2.1:\/} The bend pattern and the bottom half of the image
\end{center}

\noindent
    {\bf The Ridge Curve:\/}
    We show the picture first, then explain.

\begin{center}
\resizebox{!}{3.4in}{\includegraphics{example3.eps}}
\newline
    {\bf Figure 2.2:\/} Half 2x core curve (red/blue) and
    half ridge curve (black).
    \end{center}

    Let $\beta_b$ be the bottom edge of the
    parallelogram representing $M_{\lambda}$.
We normalize so that $I$ maps the left
vertex of $\beta_b$ to $(0,0,0)$ and
the right vertex to $(B,0,0)$, where
$B$ is the length of $\beta_b$.
Let $E_1,...,E_n$ be the successive
edges of the core curve, treated as vectors.
Let
\begin{equation}
  \label{switch}
  \Gamma_i'=2\mu_i E_i, \hskip 30 pt i=1,...,n.
\end{equation}
Let $\Gamma$ be the
curve whose initial vertex is $(B,0,0)$
and whose edges
are $\Gamma_1',...,\Gamma_n'$.
Here $\mu_1,...,\mu_n$ is the sign sequence.

$\Gamma$ has length $2\lambda$,
connects $(B,0,0)$ to $(-B,0,0)$,
and is disjoint from the open unit ball.  The lines extending the sides of $\Gamma$
are tangent to the unit sphere.
We rotate so that $\Gamma$ contains $(0,T,0)$ for
some $T>1$.  If we cone $\Gamma$ to the
origin, we get a collection $\Delta_1,...,\Delta_n$
of triangles, and $\Delta_i$ is the translate of
$\mu_i I(\delta_i)$ whose apex is at the origin.
In particular, the vectors pointing to the vertices
of $\Gamma$ are parallel to the corresponding bend images.
Figure 2.2 shows the portion of the
ridge curve (in black) associated to the example in Figure 2.1.
We have also scaled the core curve by $2$ and translated it to
show the relationships between the two curves.

\subsection{Geometric Bounds}

While we are in the neighborhood, we re-prove the lower bound
from  [{\bf FT\/}].  The proof in [{\bf FT\/}] is somewhat similar, though
it does not use the ridge curve.
Let $\lambda$ be the aspect of
the polygonal Moebius band
$\cal M$ and let $\Gamma$ be the
associated ridge curve.
Let $f: \R^3-B^3 \to S^2$ be orthogonal
projection.  The map $f$ is arc-length decreasing.  Letting
$\Gamma^*=f(\Gamma)$, we have
$|\Gamma^*|<|\Gamma|=2\lambda$.
Since $\Gamma^*$ connects a point on $S^2$ to
its antipode,
$|\Gamma^*| \geq \pi$. Hence
$\lambda>\pi/2$.

Now we use the same idea in a different way.

\begin{lemma}
  \label{geom1}
  Suppose $\cal M$ has aspect ratio less than
  $7 \pi/12$.  Then the ridge curve
  $\Gamma$ lies in the open slab bounded by
  the planes $Z=\pm 1/\sqrt 2$.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We divide $\Gamma$ into halves. One half
goes from $(B,0,0)$ to $(0,T,0)$ and
the second half goes from $(0,T,0)$ to
$(-B,0,0)$.  Call the first half $\Gamma_1$.
Suppose that $\Gamma_1$ intersects the plane $Z=1/\sqrt 2$.
Then the spherical projection $\Gamma_1^*$
goes from $A=(1,0,0)$ to some unit vector
$B=(u,v,1/\sqrt 2)$ to $C=(0,1,0)$.
Here $u^2+v^2=1/2$.
The shortest
path like this is the geodesic bigon connecting
$A$ to $B$ to $C$.  Such a bigon has length at least
$$\arccos(A \cdot B) + \arccos(B \cdot C)=
\arccos(u)+\arccos(v) \geq^* 2 \arccos(1/2)= 2 \pi/3.$$
The starred inequality comes from
the fact that the minimum, subject to the
constraint $u^2+v^2=1/2$, occurs at $u=v=1/2$.

But then $\Gamma$ has length at least $2 \pi/3 + \pi/2 = 7\pi/6$.
This exceeds twice the
aspect ratio of $\cal M$. This is a contradiction.
The same argument works if
$\Gamma_1$ hits the plane $Z=-1/\sqrt 2$.
Likewise the same argument works with
the second half of $\Gamma$ in place of
the first half.
$\spadesuit$  \newline

\begin{corollary}
  \label{perp0}
  Suppose $\cal M$ has aspect ratio less than $7 \pi/12$.
  Let $\beta_1^*$ and $\beta_1^*$ be two perpendicular bend
  images.  Then a plane parallel to both $\beta_1^*$ and
  $\beta_2^*$ cannot contain a vertical line.
\end{corollary}

{\bf {\medskip}{\noindent}Proof: }
Every bend image is parallel to
some vector from the origin to a point of $\Gamma$.
By the previous result, such a vector make an angle
of less than $\pi/4$ with the $XY$-plane.
Hence, all bend images make angles of
less than $\pi/4$ with the $XY$-plane.
Suppose our claim is false.   Since
$\beta_1^*$ and $\beta_2^*$ are perpendicular
to each other, one of them must
make an angle of at least $\pi/4$ with the
$XY$-plane.  This is a contradiction.
$\spadesuit$  \newline

\subsection{Perpendicular Lines}

As a prelude to the work in the next section,
we prove a few results about lines and planes.
Say that an {\it anchored line\/} in $\R^3$ is
a line through the origin. Let
$\Pi_1$ and $\Pi_2$ be planes through the origin in
$\R^3$.

\begin{lemma}
  \label{perp}
  Suppose that $\Pi_1$ and $\Pi_2$ are not perpendicular.
The set of
perpendicular anchored lines
$(L_1,L_2)$ with $L_j \in P_j$ for
$j=1,2$ is diffeomorphic to a circle.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
For each anchored line $L_1 \in \Pi_1$ the line
$L_2 = L_1^{\perp} \cap \Pi_2$ is the unique
choice anchored line in $\Pi_2$
which is perpendicular to $L_1$.
The line $L_2$ is a smooth function of $L_1$.
So, the map $(L_1,L_2) \to L_1$ gives a diffeomorphism
between the space of interest to us and a circle.
$\spadesuit$  \newline

A {\it sector\/} of the plane $\Pi_j$ is a set linearly
equivalent to the union of the $(++)$ and $(--)$ quadrants
in $\R^2$.  Let $\Sigma_j \subset \Pi_j$ be a sector.  The boundary
$\partial \Sigma_j$ is a union of two anchored lines.

\begin{lemma}
  \label{perp2}
  Suppose (again) that the planes
  $\Pi_1$ and $\Pi_2$ are not perpendicular.  Suppose also that
no line of $\partial \Sigma_1$ is
perpendicular to a line of $\partial \Sigma_2$.
Then the set of perpendicular
pairs of anchored lines $(L_1,L_2)$ with
$L_j \in \Sigma_j$ for $j=1,2$ is either
empty or diffeomorphic to a closed line segment.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
Let $S^1$ denote the set of perpendicular pairs
as in Lemma \ref{perp}.  Let $X \subset S^1$
denote the set of those pairs with $L_j \in \Sigma_j$.
Let $\pi_1$ and $\pi_2$ be the two diffeomorphisms
from Lemma \ref{perp}.   The set of anchored
lines in $\Sigma_j$ is a line segment and hence
so is its inverse image $X_j \subset S^1$
under $\pi_j$.  We have $X=X_1 \cap X_2$.
Suppose $X$ is nonempty.  Then some
$p \in X$ corresponds to a pair of
lines $(L_1,L_2)$ with at most one
$L_j \in \partial \Sigma_j$.  But then we
can perturb $p$ slightly, in at least
one direction, so that the corresponding
pair of lines remains in $\Sigma_1 \times \Sigma_2$.
This shows that $X_1 \cap X_2$, if nonempty,
contains more than one point. But then the
only possibility, given that both
$X_1$ and $X_2$ are segments, is that their
intersection is also a segment.
$\spadesuit$  \newline

\subsection{The Space of Perpendicular Pairs}

We prove the results in this section more generally for
piecewise affine maps $I: M_{\lambda} \to \R^3$ which
are not necessarily local isometries.
The reason for the added generality is that it
is easier to make perturbations within this
category.
Let $\X$ be the space of such maps which
also satisfy the conclusion of Corollary
\ref{perp0}. (In this section we will not
use this property but in the next section we will.)
So, $\X$ includes all (isometric)
polygonal Moebius bands of aspect ratio less
than $7\pi/12$.
The notions of bend images and facets makes
sense for members of $\X$.

\begin{lemma} The space $\X$ has a dense set $\Y$
  which consists of members such that no two
  facets lie in perpendicular planes and no two special bend images
  are perpendicular.
      \end{lemma}

      {\bf {\medskip}{\noindent}Proof: }
   One can start with any member of $\X$ and postcompose the
   whole map with a linear transformation arbitrarily close to the identity
so as to get a member of $\Y$.  The point is that we just need to
    destroy finitely many perpendicularity relations.
    $\spadesuit$  \newline

Let $\gamma$ be center circle of $M_{\lambda}$.
We can identify the space of bend images of $\cal M$ with $\gamma$:
The bends and bend-images are in bijection, and each
bend intersects $\gamma$ once.
The space of ordered pairs of unequal bend images can be
identified with $\gamma \times \gamma$ minus the diagonal.
We compactify this space by
adding in $2$ boundary components.  One of the boundary
components comes from approaching the main diagonal
from one side and the other comes from approaching the
diagonal from the other side.  The resulting space
$A$ is an annulus.    For the rest of the section we choose a member of $\Y$ and
    make all definitions for this member.

\begin{lemma} $\cal P$ is a
  piecewise smooth $1$-manifold in $A$.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We apply Lemma \ref{perp2} to the planes through the origin
parallel to the facets and to the anchored lines
parellel to the bend images within the facets.
(Within a single facet the bend images
and the corresponding anchored lines are in smooth bijection.)
By Lemma \ref{perp2},
the space $\cal P$ is the union of finitely many
smooth connected arcs.  Each arc corresponds
to an ordered pair of facets which contains at least
one point of $\cal P$.
Each of these arcs has two endpoints.  Each endpoint
has the form $(\beta^*_1,\beta^*_2)$ where exactly
one of these bend images is special.
Let us say that $\beta^*_1$ is special.
Then $\beta^*_1$ is the edge between two
consecutive facets, and hence
$(\beta^*_1,\beta^*_2)$ is the endpoint of
exactly $2$ of the arcs.
Hence the arcs fit together to make a piecewise
smooth $1$-manifold.
$\spadesuit$  \newline

A component of $\cal P$ is {\it essential\/} if it
separates the boundary components of $A$.

\begin{lemma}
$\cal P$ has an odd number of essential components.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
An essential component, being embedded, must represent
a generator for the first homology
$H_1(A)= \Z$.  By duality, a transverse arc running
from one boundary component of $A$ to the other intersects
an essential component an odd number of times and
an inessential component an even number of times.
Let $a$ be such an arc.  As we move
along $a$ the angle between the corresponding bends
can be chosen continuously so that it starts at $0$
and ends at $\pi$.  Therefore, $a$ intersects
$\cal P$ an odd number of times.  But this means that
there must be an odd number of essential components of $\cal P$.
$\spadesuit$  \newline

\subsection{The Main Argument}

Now we prove Lemma \ref{T}.
Say that a member of $\X$ is {\it good\/} if
(with respect to this member) there is a
path connected subset  ${\cal K\/} \subset {\cal P\/}$ such
that both $(\beta_1^*,\beta_2^*)$ and
$(\beta_2^*,\beta_1^*)$ belong to $\cal K$ for
some pair $(\beta_1^*,\beta_2^*)$.

\begin{lemma}
  If $\cal M$ is good then $\cal M$ has a $T$-pattern.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
      Each pair $(\beta_1^*,\beta_2^*)$
      in $\cal P$ determines a unique pair of parallel planes $(P_1,P_2)$
      such that $\beta_j^* \subset P_j$ for $j=1,2$.  By hypothesis, members
      These planes do not contain vertical lines.
      Hence, our
      planes intersect the $Z$-axis in single and continuously
      varying points.  As we move
      along $\cal K$ these planes exchange places and so do
      their $Z$-intercepts.
      So, at some instant, the planes coincide and give us a $T$-pattern.
      $\spadesuit$  \newline

      \begin{lemma}
        A dense set of members of $\X$ are good.
      \end{lemma}

      {\bf {\medskip}{\noindent}Proof: }
      Let $\Y$ be the dense subset of $\X$ considered in the previous section.
      Relative to any member of $\Y$, the space $\cal P$ is a piecewise
      smooth $1$-manifold of the annulus $A$ with an odd number of essential components.
      The involution $\iota$,
      given by
      $\iota(p_1,p_2)= (p_2,p_1)$, is a continuous involution
      of $A$ which preserves $\cal P$ and permutes the essential components.
      Since there are an odd number
      of these, $\iota$ preserves some essential component of $\cal P$.
      But then this essential component contains our set $\cal K$.
      $\spadesuit$  \newline

      Now we know that there are $T$-patterns for members of a dense subset of $\X$.
      By compactness and continuity, every member of $\X$ has a
      $T$-pattern.  By Corollary \ref{perp0}, $\X$ contains all
      (ordinary) embedded polygonal
      Moebius bands of aspect ratio less than $7\pi/12$.
      Hence all such embedded Moebius bands contain $T$-patterns.
      This completes the proof of
      Lemma \ref{T}.

\newpage

\section{The Aspect Ratio Bound}

\subsection{Constraints coming from the T Pattern}

Let $\cal M$ be a polygonal Moebius band of
aspect ratio $\lambda<7\pi/12$.  We keep
the notation from the previous chapter.

      Let $\beta_1$ and $\beta_2$ be two bends whose corresponding
      images $\beta_1^*=I(\beta_1)$ and $\beta_2^*=I(\beta_2)$ form a $T$-pattern.
      Since these segments do not intersect, we can label so
      that the line extending $\beta_2^*$ does not intersect
      $\beta_1^*$.  We cut $M_{\lambda}$ open along
      $\beta_1$ and treat $\beta_1$ as the bottom edge.
      We now set $\beta_b=\beta_1$ and $\beta_t=\beta_2$ and (re)normalize as in
      \S \ref{basic00}.
      So, $\beta_b^*$ connects $(0,0,0)$ to $(B,0,0)$, and
      $\beta_t^*$ is a translate of the segment connecting
      $(0,0,0)$ to $(0,T,0)$.
      Here $B$ and $T$ are the lengths of these segments.

      The left side of Figure 3.1 shows $M_{\lambda}$.  Reflecting
      in a vertical line, we normalize so that $L_1 \geq R_1$.
      This means that $L_2 \geq R_2$.  The right side of
      Figure 3.1 shows that $T$ pattern, and the corresponding
      images of the sets on the left under the isometry $I$.
      The wiggly curves we have drawn do not necessarily lie in the
      $XY$-plane but their endpoints do.

\begin{center}
\resizebox{!}{2.1in}{\includegraphics{trap1.eps}}
\newline
    {\bf Figure 3.1:\/} A Paper Moebius band interacting with the $T$-pattern.
\end{center}
There is some $\epsilon$ such that the vertical distance  (i.e.,
the difference in the $y$-coordinates) from the white vertex
to the yellow vertex is $T/2+\epsilon$ and the vertical distance from the
white vertex to the green vertex is $T/2-\epsilon$.

Looking at the picture, and using the fact that geodesics in the
Euclidean plane are straight lines, we get the following constraints:
\begin{equation}
  \label{constraint1}
R_1+R_1 \geq T, \hskip 10 pt
\end{equation}
\begin{equation}
  \label{constraint2}
  L_1+L_2 \geq \sqrt{B^2+(T/2-\epsilon)^2}+\sqrt{B^2+(T/2+\epsilon)^2} \geq 2\sqrt{B^2+T^2/4}.
\end{equation}

\subsection{An Optimization Problem}
\label{opto}

We are done with the Moebius band. We just have a
  parallelogram as on the left side of Figure 3.1 which satisfies the
  constraints in Equations \ref{constraint1} and \ref{constraint2} and
  we want to minimize $L_1+L_2+R_1+R_2$.
  Let $L=(L_1+L_2)/2$ and $R=(R_1+R_2)/2$.
  If we replace $L_1,L_2$ by $L,L$ and
  $R_1,R_2$ by $R,R$ the constaints are still satisfied
  and the sum of interest is unchanged.
  The constraints now become:
  \begin{equation}
    \label{const}
    2R \geq T, \hskip 30 pt L \geq \sqrt{B^2+T^2/4}.
  \end{equation}
  We show that $L+R \geq \lambda_1$, the constant from
  the Main Theorem, which means
    $$\lambda = \frac{1}{2} (L_1+L_2+R_1+R_2) = L+R \geq \lambda_1.$$
  So, showing that $L+R \geq \lambda_1$ finishes the proof of
  the Main Theorem.

  Let $b$ and $t$ respectively denote the slopes of the sides
  labeled $B$ and $T$ in Figure 3.2. Let $S=L+R$.
  Since $L \geq R$ we have $b \geq t$.  In Figure 3.2 we
  depict the case when $b>0$ and $t<0$.  We mention without proof that
  this must happen when $S<\sqrt 3$. (Our sequel paper has a proof.)
\begin{center}
\resizebox{!}{1.5in}{\includegraphics{trap2.eps}}
\newline
    {\bf Figure 3.2:\/} The basic trapezoid.
\end{center}
Since $L+R=S$ and $L-R=b-t$, and by the Pythagorean Theorem,
\begin{equation}
  \label{relations}
  L=\frac{S+b-t}{2}, \hskip 10 pt
  R=\frac{S-b+t}{2}, \hskip 10 pt
  B=\sqrt{1+b^2}, \hskip 10 pt
  T=\sqrt{1+t^2}.
\end{equation}
Plugging these relations into
Equation \ref{const}, we get $S \geq f(b,t)$ and $S \geq g(b,t)$ where
$$f(b,t)= b - t + T, \hskip 20 pt
g(b,t)=- b +t + \sqrt{4B^2+T^2}.$$

Let $\phi=\max(f,g)$ and let $D$ be the domain
where $b \geq t$.  We have $S \geq \phi$.  To finish the
proof of the Main Theorem we just need to show
that $\min_D \phi =\lambda_1$.  This is what we do.

\begin{lemma}
  $\phi$ achieves its minimum at a point in
  the interior of $D$, and at this point we have $f=g$.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We note $3$ properties of our functions:
\begin{enumerate}
\item $f(0,-1/\sqrt 3)=g(0,-1/\sqrt 3)=\sqrt 3$.  Hence $\phi(0,-1/\sqrt 3)=\sqrt 3$.
\item On $\partial D$ we have $g(b,b)=B \sqrt 5 \geq \sqrt 5$.
  Hence $\min_{\partial D} \phi > \sqrt 3$.
\item  As $b^2+t^2 \to \infty$ in $D$, we have $f,g \to +\infty$.
\end{enumerate}
From these properties, we see that the global minimum of
$\phi$ is achieved at some point in the interior of $D$.
Next, we compute
$$\frac{\partial f}{\partial t}= -1+\frac{t}{T}<0, \hskip 20 pt
\frac{\partial g}{\partial t} =1+\frac{t}{4B^2+T^2}>0.$$
Hence the gradients of $f$ and $g$ are never zero, and
are never positive multiples of each other.  For this reason,
  any point where $\phi$ achieves a global minimum
  lies on the set where $f=g$.
  $\spadesuit$  \newline

  Setting $f=g$ and solving for $b$,
we get
  \begin{equation}
  b=\beta(t)=
 \frac{ t^3 - T^3 - 3 t}{ 3 t^2-1}.
\end{equation}
There no solutions to $f=g$ when $t=+1\sqrt 3$ and Item 1 above
(or a direct calculation) shows that
$\beta(-1/\sqrt 3)=0$.  In particular,
$\beta$ has a removable singularity at $t=-1/\sqrt 3$ and
hence is smooth on the domain
$$D^*=(-\infty,1/\sqrt 3).$$
When $t>1/\sqrt 3$ we have $b=\beta(t)<0<t$.
Hence, only $t \in D^*$ corresponds to points in $D$.
So, we just need show $\min_{D^*} \phi^*=\lambda_1$, where
\begin{equation}
 \phi^*(t)=f(\beta(t),t)=
  \frac{2T (t^2-tT - 1)}{3 t^2-1}
\end{equation}

As $t \to -\infty$ or $t \to 1/\sqrt 3$ we have $\phi^*(t) \to +\infty$.
Hence $\phi^*$ achieves its minimum on $D^*$ at some point where
$d\phi^*/dt=0$.
This happens only at
$$t_0=-\sqrt{\frac{2}{\sqrt 3}-1}= -.39332...$$
We check that $\phi^*(t_0)=\lambda_1$.

\section{Polygonal Approximation}

In this short chapter we
explain why the results in the polygonal
case imply the results in the smooth case.
There is some subtlety here because the
obvious method of polygonal approximation (described below)
does not quite yield
the polygonal Moebius bands that we have been working with.
We need to broaden the definition.

A $K$-{\it bilipschitz\/} map between metric spaces
is one which distorts distances by a
factor of at most $K$. A $1$-bilipschitz map is an isometry.
Say that a $K$-{\it bilipschitz polygonal
  Moebius band\/} is the same kind of map as
a polygonal Moebius band, except that the restriction
of the map to each triangle
is linear and $K$-bilipschitz.
All the arguments in this paper work the same way for
$K$-bilipschitz polygonal Moebius bands as
they do for $1$-bilipschitz polygonal Moebius bands,
except that the bound $\lambda>\lambda_1$
in the Main Theorem
is replaced by
$\lambda>\lambda_1/K$.  Lemma \ref{T}
has a similar easy modification.

The following result justifies the claim that
``polygonal implies smooth'',

\begin{lemma} Let
  $I': M_{\lambda} \to \R^3$ be a smooth
  paper Moebius band.  There is a sequence
of $K_n$-bilipschitz polygonal Moebius bands
$I_n: M_{\lambda} \to \R^3$ with
$K_n \to 1$, so that:
\begin{enumerate}
\item For each $p \in M_{\lambda}$ we have
  $\|I'(p)-I_n(p)\|<K_n-1$.
\item If $I_n$ is linear about $p$ then
  the plane tangent to
  $I_n(M_{\lambda})$ at $I_n(p)$ makes an angle
  less than $K_n-1$ with the tangent plane
  to $I'(M_{\lambda})$ at $I'(p)$.
\end{enumerate}
$I_n$ is an embedding for $n$ large provided that $I'$ is an embedding.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We take a large finite mesh $\beta_0,...,\beta_n$
of bends in $M_{\lambda}$, with $\beta_0$ being the
first bend and $\beta_n$ being the last.  The
union $$\partial M_{\lambda} \cup \beta_0 \cup ... \cup \beta_n$$ is a union
of $n$ trapezoids, some possibly degenerate -- i.e. triangles.
For each nondegenerate trapezid we insert one of the two diagonals.
This gives us a triangulation of $M_{\lambda}$ of the kind considered
above.  The restriction of $I'$
to $\beta_0 \cup ... \cup \beta_n$
gives us exactly enough information to
extend linearly to all of
$M_{\lambda}$ by making the map linear on each
triangle of the triangulation.
Call the resulting map $I_n$.

Given the compactness of $M_{\lambda}$ and the
smoothness of $I'$, the map
$I_n$ is a $K_n$-bilipschitz polygonal
Moebius band with $K_n \to 1$.
Because $I'$ and $I_n$ agree on a set which
is becoming dense in $M_{\lambda}$ as
$n \to \infty$, these maps
satisfy Condition 1 above.  Now we
justify Condition 2.  Given the
regularity of $I'$, the convex hull
of the images $I(\beta_j)$ and $I(\beta_{j+1})$ will be
nearly flat, in the sense that the $4$ faces of
this tetrahedron will be all nearly parallel to
each other, and all faces be nearly parallel to the
tangent planes to $I'(M_{\lambda})$ at nearby points.

The final statement about the embedding follows from
the $C^1$ nature of the approximation.
$\spadesuit$  \newline

\newpage

\section{The Range of Possibilities}

\subsection{The Main Result}

We state our result for immersed paper Moebius bands of
aspect ratio less than $\sqrt 3$.  Thanks to Lemma \ref{T},
these results apply to any embedded examples, should they exist.
Figure 4.1 shows a copy of Figure 3.1, with an extra vector
$(x,y)$ added in.  This vector points from the right endpoint
of the horizontal bend image to the midpoint of the vertical bend image.

\begin{center}
\resizebox{!}{2in}{\includegraphics{Pix/trap1B.eps}}
\newline
    {\bf Figure 4.1:\/} An elaboration of Figure 3.1.
\end{center}

Let $\Delta$ be the (shaded) convex hull of the $T$-pattern $I(T) \cup I(B)$.
Let $S_j=L_j+R_j$ for $j=1,2$.  The aspect ratio of
our paper Moebius band is $\lambda=(S_1+S_2)/2$.  Theorem \ref{main}
gives a bound on this average.  Now we give a bound on each sum
$S_j$ separately, and we also bound the geometry of $\Delta$.

\begin{theorem}
  \label{fine}
  \label{half}
  For any immersed paper Moebius with a $T$-pattern and
  aspect ratio less than $\sqrt 3$, the following is true.
  \begin{enumerate}
  \item $S_j \geq \sqrt 3 - \frac{1}{3}b(1-2b)$ for $j=1,2$.
  \item $x<1/18$ and $|y|<1/30$.
    \item $\Delta$ is a triangle with minimum angle greater than $\pi/4$.
  \end{enumerate}
\end{theorem}

The function $f(b)=\frac{1}{3}b(1-2b)$ has maximum value $1/24$.
Hence we have $S_j>\sqrt 3-(1/24)$ for $j=1,2$.  The corresponding
bound for $\lambda$ is only slightly weaker than the one in
Theorem \ref{main}.

\subsection{The Range of Slopes}

We continue with the notation from the previous chapter.
In particular, let $f$ and $g$ be the two constraint
functions from \S \ref{opto} and let
$\phi=\max(f,g)$.
Again, $$f(b,t)= b - t + T, \hskip 20 pt
g(b,t)=- b +t + \sqrt{4B^2+T^2}.$$
Here $T=\sqrt{1+t^2}$ and $B=\sqrt{1+b^2}$.

Let $\Omega$ be the set of values $(b,t)$ satisfying
$\phi(b,t)<\sqrt 3$.    We plot $\Omega$ in
Figure 4.1.  These are the possible values
$(b,t)$ of slopes of bends which can participate
in a $T$-pattern when the aspect ratio is less
than $\sqrt 3$.

\begin{center}
\resizebox{!}{4in}{\includegraphics{Pix/contour1.eps}}
\newline
    {\bf Figure 4.1:\/} The range of slopes.
\end{center}

 We have
added in some points and lines to help frame $\Omega$,
and in particular we have placed it inside a
trapezoid which tightly hugs it.
All the lines drawn touch $\partial \Omega$ except
the red one, which lies just barely above $\Omega$.
The lines of slope $2/3$ and $4/3$ through
$(0,-1/\sqrt 3)$ are tangent to $\Omega$.
The right vertex of $\Omega$ is $(a,-a/2)$ where
$a=(\sqrt{27}-\sqrt{11})/4=.4698...$

\begin{lemma}
  $\Omega$ is a subset of the trapezoid indicated in Figure 4.1.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We compute
$$\frac{\partial f}{\partial t}=-1+\frac{t}{T}<0, \hskip 20 pt
\frac{\partial g}{\partial t}=1 + \frac{t}{5+5b^2+t^2}>0$$
This calculation shows that the sets
 $f=\sqrt 3$ and $g=\sqrt 3$ are
  graphs of smooth functions.  Call these graphs $\Gamma_f$ and $\Gamma_g$.
  Let $a=(\sqrt{27}-\sqrt{11})/4$.  We check algebraically that
  $\Gamma_f$ and $\Gamma_g$ intersect only at
  $(0,-1/\sqrt 3)$ and $(a,-a/2)$.  Moreover, we check
  that $\Gamma_g$ lies below $\Gamma_f$ at $b=-1,1$.
  Hence $\Gamma_g$ lies below $\Gamma_f$ for all $b \not \in [0,a]$.
  Given our derivative calculation, this shows that $\Omega$
  lies in the strip between the lines $b=0$ and $b=a$.
  Now we consider the $3$ non-vertical constraint lines:
  \begin{itemize}
\item We show that $\Omega$ lies below the
line $t=(2/3)b-(1/2).$
Since $\partial g/\partial t>0$ it suffices to prove
that the restriction of $g$ to our line is greater than $\sqrt 3$.
The minimum of $g(b,(2/3) b - 1/2)$ occurs  at
$b=(39 + \sqrt{8151})/520$
and the value there is about $\sqrt 3+.00002$.

\item We show that $\Omega$ lies above the line
$t=(2/3)b-1/\sqrt 3$.  Since
  $\partial f/\partial t<0$ it suffices to show that
  $f(b,(2/3)b-1/\sqrt 3)>\sqrt 3$ for $b>0$.  We check that
  this is the case.  The infimum occurs at $b=0$ and the value is $\sqrt 3$.

\item A similar argument, with $g$ in place of $f$, shows that
  $\Omega$ lies below the line $t=(4/3)t - (1/\sqrt 3)$.
  \end{itemize}
This completes the proof.
$\spadesuit$  \newline

Here is a corollary.

\begin{theorem}
  \label{perpendicular}
  If $\cal M$ is an immersed paper Moebius band with a
  $T$-pattern and aspect ratio less than $\sqrt 3$ then
  $\cal M$ has at least $2$ bends that are perpendicular to the
  boundary of the Mobius band.
\end{theorem}

{\bf {\medskip}{\noindent}Proof: }
One thing we notice right away is $(b,t) \in \Omega$
implies that $b>0$ and $t<0$.
That is, the bends have opposite slopes.
Applying the Intermediate Value Theorem, we
see that there are two bends that have slope $0$.
More geometrically, these bends are perpendicular
to the boundary of the Moebius band.
$\spadesuit$  \newline

\subsection{Another Constraint}

We first establish another
constraint imposed by Figure 3.1.

\begin{lemma}
  \label{const3}
  $B^2 - L_j^2 + (T-R_j)^2 \leq 0$ for $j=1,2$.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We take the case
$j=1$.  The case $j=2$ is the same, except that
the vector $(x,y)$ is replaced
by the vector $(x,-y)$.  We set $L=L_1$, etc.

Let
\begin{equation}
  \label{THETA}
  \Theta=(B/2+x,y).
\end{equation}
Let $\Gamma$ be the portion of the ridge curve corresponding to $\tau$.
Here $\Gamma$ connects $(B,0,0)$ to $(0,T,0)$.
Let $\mu_1,...,\mu_n$ be the associated sign sequence.
The $(-)$ signs corresponding to the left
side of the trapezoid $\tau$ and the $(+)$ signs correspond to the right.
Let $M_{\pm} \subset \{1,...,n\}$ denote those
indices $i$ such that $\mu_i = \pm 1$.
Let $\Gamma_1',...,\Gamma_n'$ denote the edges of $\Gamma$.
Let
\begin{equation}
  L_{\rm str\/}=\sum_{i \in M_-} \Gamma'_i, \hskip 30 pt
  R_{\rm str\/}=\sum_{i \in M_+} \Gamma'_i.
\end{equation}
It follows from the definitions that
\begin{equation}
  \label{midpoint}
  \|L_{\rm str}\/\| \leq L, \hskip 12 pt
  \|R_{\rm str\/}\| \leq R, \hskip 12 pt
  R_{\rm str\/}+L_{\rm str\/}=(-B,T,0), \hskip 12 pt
  R_{\rm str\/}-L_{\rm str\/}=2\Theta.
\end{equation}
See Equation \ref{switch} for the justification of the
last inequality. Define
\begin{equation}
\label{POINT}
p=(B,0,0)+R_{\rm str\/} = (0,T,0) - L_{\rm str\/}
=\bigg(\frac{B}{2},\frac{T}{2},0\bigg)+ \Theta.
\end{equation}

Let $\pi_k$ be projection onto the $k$th coordinate.
We have
$\pi_1(p)=B+x$. Also, $$\pi_2(p) \leq \|R_{\rm str\/}\| \leq R<1 \leq T.$$
Set $\zeta=(0,T,0)$. By the Pythagorean Theorem:
\begin{equation}
  \label{PY}
  \|(B,R,0)-\zeta\| \leq \|(B+x,\|R_{\rm str\/}\|,0)-\zeta\| \leq \|p-\zeta\|=\|L_{\rm str\/}\| \leq L.
\end{equation}
Hence $\|(B,R)-(0,T)\| \leq L$.  Squaring both sides and rearranging, we get the
advertised constraint.
$\spadesuit$  \newline

\subsection{Proof of Statement 1}

We pick an index $j=1,2$ and then set
    $L=L_j$, etc.
Plugging in Equation \ref{relations} into the constraint from
Lemma \ref{const3}, we see that
this constraint is equivalent to
    $$2+b^2+t^2+bT-tT - S(b-t+T) \leq 0.$$
    Rearranging this expression, we see that we must have
    \begin{equation}
      S \geq \psi(b,t):=\frac{2+b^2+t^2+bT-tT}{b-t+T}.
    \end{equation}
Hence
    \begin{equation}
      \label{better}
      S+ \frac{b(1-2b)}{3} - \sqrt 3 \geq \widehat \psi(b,t):=\psi(b,t)+ \frac{b(1-2b)}{3} - \sqrt 3.
    \end{equation}
    Hence, it suffices to prove that $\min_{\Omega} \widehat \psi=0$,
    where $\Omega$ is as in Figure 5.2.

    The expression $b-t+T$ is positive when $b>t$, which we have
    on $\Omega$.  Using the solvable expression trick on
    $(b-t+T)\widehat \psi$, we find that
        $\widehat \psi(b,t)=0$ only if $P(b,t)=0$, where $P(b,t)$ is the following polynomial:
    \small{
    $$4 b^6-8 b^5 t-16 b^5+20 b^4 t+12 \sqrt{3} b^4+12 b^4-24 \sqrt{3} b^3 t-8 b^3 t-24 \sqrt{3} b^3- 8 b^3+$$
      $$9 b^2 t^2+12 b^2 t+30 \sqrt{3} b^2 t-12 \sqrt{3} b^2+59 b^2+18 b t^3-42 b t-12 \sqrt{3} b+27 t^2+18 \sqrt{3} t+9$$}
    It suffices to prove that $P>0$ on $\Omega$.

    The expression for $P$ is a monster but it is cubic in $t$.
    Let $Z$ be the segment which is the subset of the line
    \begin{equation}
      \label{subt}
      t=(2/3)b-(1/\sqrt 3).
    \end{equation}
    parametrized by $b \in (0,1/2)$.  Given the description of $\Omega$ in Figure 3.1, every point
        of $\Omega$ can be reached from a point of $Z$ by travelling upwards on a
    vertical ray.  So, to prove that $P>0$ on $\Omega$ it suffices to prove that
    $P'''>0$ when $b>0$ and that $P,P',P''>0$ on $Z$.  Here we are taking
    derivatives with respect to $t$ but then we will use Equation
    \ref{subt} to express the answer in terms of $b$.

    We compute $P'''=108b$, which is certainly positive when $b>0$.  We then
    compute other derivatives and restrict to $Z$:
    $$P''=54 - 36 \sqrt 3 b + 90b^2, \hskip 20 pt
    P'=b(12  + 12 b+ 28 b^2 - 24 \sqrt 3 b^2) + b^4(20 - 8 b).$$
    \begin{equation}
      \label{expression}
      (3/4)P=b^2(21-12 \sqrt 3+18 b - 10\sqrt 3 b + 12 b^2 - 8 \sqrt 3 b^2 - 2 b^2) + b^5(2 \sqrt 3 -b)
    \end{equation}
    On $(0,1/2)$,
    every linear polynomial in sight is positive and, as can be checked by the
    quadratic formula, every quadratic polynomial is positive as well.
    Hence, the expressions in Equation \ref{expression}
    are all positive on $(0,1/2)$.  This completes the proof.

    \subsection{Proof of Statement 2}

    We can normalize by an isometry so that $y \geq 0$.

    \begin{lemma}
      $x<1/18$.
    \end{lemma}

    {\bf {\medskip}{\noindent}Proof: }
    Let $L_2'$ denote the length of the segment in Figure 4.1 joining the endpoints (yellow, black) of
    $I(L_2)$.  Define $R_1'$ similarly.
    From the Pythagorean Theorem,
    \begin{equation}
      \label{py0}
      L_2 \geq L_2'=\sqrt{(B+x)^2+(T/2+y)^2} \geq \sqrt{(B+x)^2+T^2/4}.
      \end{equation}
    Also, we have
    $R_1 \geq R_1'= T/2+y \geq T/2.$
    Define the function
    $$f(b,t,x) = T/2+\sqrt{(B+x)^2+T^2/4}- \sqrt 3.$$
    Since $L_2+R_1<\sqrt 3$, we can prove that $x<1/18$ by showing
    that $f>0$ on $\Omega$.

Every point in $\Omega$ can be reached by following a downward-pointing vertical
ray which starts on the segment
 \begin{equation}
  \label{subt2}
 Z=\{(b,t)|\ t=(2/3)b-(1/2), b \in I\}, \hskip 30 pt I=(0,1/2).
  \end{equation}
 Since (clearly) $\partial f/\partial t<0$, we just need to prove
 that $\phi>0$ on $[0,1/2]$, where
\begin{equation}
  \phi(b)=f(b,(2/3)b - (1/2),1/18).
\end{equation}
We check that $\phi(0)>0$.  So, we just have to show
that $\phi$ does not vanish on $[0,1/2]$.  The expression for $\phi$
is an iterated radical.  An expression of
the form $u+v \sqrt w$ vanishes only if $u^2-v^2w$ vanishes.
Using this fact $3$ times, to eliminate the iterated radical,
we find that $\phi(b)=0$ only if the following
polynomial vanishes.
$$
379204871936 - 2821217402880 b - 3788174241792 b^2 +
59974706921472 b^3 - $$
$$ 81516306161664 b^4 - 11284439629824 b^5 +
30126667530240 b^6 - 2821109907456 b^8
$$
Any number of methods -- e.g., Sturm Sequences  -- shows
that this does not happen.  The closest root is $0.624325...$.
$\spadesuit$  \newline

\begin{lemma}
  $|y|<1/30$.
\end{lemma}

{\bf {\medskip}{\noindent}Proof: }
We keep the notation from the previous lemma.  We have
$$R_1-y \geq R_1'-y=T/2 \hskip 30 pt \Longrightarrow \hskip 30 pt
y \leq R_1-(T/2).$$
Since $x \geq 0$, Equation \ref{py0} says that
\begin{equation}
  \label{L2b}
L_2 \geq \sqrt{B^2 + (T/2+y)^2}.
\end{equation}
Since $L_2+R_1<\sqrt 3$, we have
$$y \leq (R_1+L_2) - L_2 - (T/2) < \sqrt 3 - \sqrt{B^2+(T/2+y)^2} - (T/2).$$
The last inequality comes from Equation \ref{L2b}.

Define
\begin{equation}
  h(b,t,y)=  \sqrt{B^2+(T/2+y)^2} + (T/2) + y -\sqrt 3
\end{equation}
To finish our proof it suffices to prove that
$h(b,t,1/30)>0$ on $\Omega$.  The argument is just as in the previous case,
except that now the final polynomial is
$$-300 b^4-40 \sqrt{3} b^2+1600 b^2-600 b+80 \sqrt{3}-79.$$
This polynomial has two negative roots and two non-real roots.
$\spadesuit$  \newline

\subsection{Proof of Statement 3}

Since $|y|<T/2$, the convex hull $\Delta$ is a triangle.  Now we consider the angles.
To make the numbers easier, all we use is that $x<1/18$ and $|y|<1/8$.

 The (vertical) base of $\Delta$ has length $T=\sqrt{1+t^2}$.
    The (horizontal) altitude of $\Delta$ has length $B=\sqrt{1+b^2}+x$.
    Let $\alpha$ be the foot of the altitude and
    let $\beta$ be the midpoint of the base.  We have
    $|t|<1/\sqrt 3$ on $\Omega$.
        Hence $\Delta$ has base length less than $5/4$.
        Since $\|\alpha-\beta\|=|y|<1/8$, the foot $\alpha$
    is at most $3/4$ from either the top or bottom vertex.
    The altitude has length at least $1$.
    Hence, both top and bottom angles exceed $$\tan^{-1}(4/3)>\pi/4.$$

    Now we consider the angle at the left vertex.
        The function $f(b,t)=(B+\frac{1}{18})/T$ is maximized at the
        right vertex of $\Omega$ and its value there is $1.129...<1.13$.
        Hence, the altitude/base ratio
        of $\Delta$ is at most $1.13$.  At the same time, the base length
        is at least $1$ and (again) $\|\alpha-\beta\|<1/8$.
        Hence, the left angle of $\Delta$ is at least
        $$\tan^{-1}\bigg(\frac{3/8}{1.13}\bigg) +
        \tan^{-1}\bigg(\frac{5/8}{1.13}\bigg)=(1/4) \times 3.30...>\pi/4.$$
        This completes the proof.

    \newpage

\section{References}

\noindent
[{\bf AHLM\/}] A. Malaga, S. Lelievre, E. Harriss, P. Arnoux, \newline
{\it ICERM website\/}:
https://im.icerm.brown.edu/portfolio/paper-flat-tori/  (2019)
\newline
\newline
[{\bf CF\/}] Y. Chen and E. Fried,
{\it Mobius bands, unstretchable material sheets and developable surfaces\/},
Proceedings of the Royal Society A, (2016)
\newline
\newline
[{\bf CKS\/}] J. Cantarella, R. Kusner, J. Sullivan, {\it On the
minimum ropelength of knots and links\/},  Invent. Math. {\bf 150\/} (2) pp 257-286 (2003)
\newline
\newline
[{\bf FT\/}], D. Fuchs, S. Tabachnikov, {\it Mathematical Omnibus: Thirty Lectures on Classic Mathematics\/}, AMS 2007
\newline
\newline
[{\bf HF\/}, D.F. Hinz, E. Fried, {\it Translation of Michael Sadowsky's paper
'An elementary proof for the existence of a developable MÖBIUS band and the attribution
of the geometric problem to a variational problem'\/}. J. Elast. 119, 3–6 (2015)
\newline
\newline
[{\bf HW\/}], B. Halpern and C. Weaver,
{\it Inverting a cylinder through isometric immersions and embeddings\/},
Trans. Am. Math. Soc {\bf 230\/}, pp 41.70 (1977)
\newline
\newline
[{\bf MK\/}] L. Mahadevan and J. B. Keller,
{\it The shape of a Mobius band\/}, Proceedings of the Royal Society A (1993)
\newline
\newline
[{\bf Sa\/}], M. Sadowski, {\it Ein elementarer Beweis für die Existenz eines
abwickelbaren MÖBIUSschen Bandes und die Zurückführung des geometrischen
 Problems auf einVariationsproblem\/}. Sitzungsberichte der Preussischen Akad. der Wissenschaften, physikalisch-mathematische Klasse 22, 412–415.2 (1930)
\newline
\newline
[{\bf S1\/}] G. Schwarz, {\it A pretender to the title ``canonical Moebius strip''\/},
Pacific J. of Math., {\bf 143\/} (1) pp. 195-200, (1990)
\newline
\newline
[{\bf S2\/}] R. E. Schwartz, {\it The $5$ Electron Case of Thomson's Problem\/},
Journal of Experimental Math, 2013.
\newline
\newline
    [{\bf W\/}] S. Wolfram, {\it The Mathematica Book, 4th Edition\/},  Wolfram Media and Cambridge University Press (1999).

\end{document}

Sadowsky M.. Hinz DF, Fried E. (doi:10.1007/s10659-014-9490-5)3. Sadowsky M. 1929 Die Differentialgleichungen des MÖBIUSschen Bandes, Jahresbe

redpajama_set_name: RedPajamaArXiv


Full Time Entertainment and Events Supervisor at Six Flags, Inc.
Summary of Job:This position is responsible for supervising the daily operation of the Entertainment department, assisting with Resort-wide events and working with the Operations management team to improve the overall operation of the park.
* Model and enforce all park policies and procedures and maintain all Six Flags standards.
* Minimum High School diploma or equivalent. Bachelor's degree preferred.
* Possess a minimum of three years of Supervisor experience.
* Minimum three years of experience with live shows and event production preferred.
* Strong knowledge of all Microsoft applications.
* Leadership ability with a team oriented approach.
* Excellent communication and presentation skills, both written and verbal.
* Strong motivational, leadership, flexible and team building skills with a "can-do" attitude.
* Good organizational skills, creative, outgoing, detail oriented, and self-motivator.
* Ability to work a flexible schedule including nights, weekends and holidays.


Thanks to the Friends of the Kingston Public Library we're excited to announce our newest discounted museum pass to the Peabody Essex Museum! One of the nation's largest museums, PEM features art and culture from New England and around the world.
You can see what's currently on exhibit here.
To book a discounted pass, visit our museum pass reservation site.
To see the rest of our museum passes, visit the museum passes section of our website.
MeeScan – A New Way to Check Out!
Try our new self-checkout service – MeeScan.
There's no need to wait in line at the circulation desk – users can check out items in the privacy of their own devices, on their own time, from anywhere in the library. In addition to a quicker, simpler, and more private experience, meeScan will help you stay green and keep track of your checkouts by sending you an email receipt for every transaction.
More Than Just Books at the Kingston Public Library!
Check out some of our great new services!
The Kingston Public Library History Room offers several resources for anyone interested in the history of Kingston.
Stop by and check out our new Teen Room – we've got something for teens of all interests!
Want to stream the big game but the power is out? Can't get Wi-Fi service on vacation? Using up too much of your data plan while on the road? The Kingston Public Library has the solution – mobile Wi-Fi hotspots.
Kingston's "library of things" continues to expand with more items you may not expect to find at a library. In addition to a great selection of board games and launch pads – easy to use preloaded tablets – the Kingston Public Library now offers card holders the chance to check out mobile Wi-Fi hotspots. Hotspots are ideal for staying connected during vacations, road trips, long commutes, power outages and more.
With over eight hours of battery life and support for up to ten devices at 4G speeds, your friends and family can stay connected too. Whether it's your laptop, tablet, phone, or gaming console, mobile Wi-Fi hotspots will help keep you and your family connected – and keep your data usage down – no matter where you go.
For more information on mobile Wi-Fi hotspots, or to explore our other collections, contact the library at 781-585-0517 or visit www.kingstonpubliclibrary.org.
You can now access the New York Times from the library or at home. Visit our Print Media Page for links and instructions.
On July 1st, we're introducing a change to our eBook and eAudiobook collection that we're pretty excited about. We have offered eBooks and eAudiobooks on a number of platforms in recent years. On July 1st, we'll be streamlining those services a bit and building up the services that work best for Kingston library users.
Axis 360 will be discontinued, but you will find even more books and video on Overdrive and Hoopla in the coming months. Of course, you can continue to stream and download digital music from Hoopla.
We're excited about making things a little simpler for you! And remember, we can also help you find digital research resources. If you want help with any of this, drop in to see us or give us a call at 781-831-6272.
For additional information, visit the Massachusetts Board of Library Commissioners Website.


Actress Sofia Vergara is the subject of a shortform series co-produced with Fusion that will debut this summer exclusively on Snapchat.
The six-episode series "Vergaraland" is billed as a tongue-in-cheek look at the thesp's career through the perspective of her son, Manolo González Vergara. The project is being produced in conjunction with Sofia Vergara's talent management and entertainment marketing firm, Latin World Entertainment.
Fusion, a joint venture of ABC News and Univision Communications, aims to reach a millennial audience — and sees Snapchat as a key content-distribution partner. Fusion was among media companies that launched on the Snapchat Discover service in January. The Discover service lets partners post video and content "stories" that are available for 24 hours; ordinarily, messages posted between Snapchat users vanish after 10 seconds or less.
"Vergaraland," which will be shot to fit Snapchat's vertical video format, joins Fusion's other original series for the service including "Outpost," "Weird Threads," and "Off the Record," among others.
For Snapchat, Fusion produces minute-long news updates, "News in a Snap," and also has created content for the platform in several languages including Portuguese, French and Spanish.


text:
Tomás Rojas Gómez (* 12. Juni 1980 in Veracruz, Mexiko) ist ein mexikanischer Boxer im Superfliegengewicht.

Profikarriere
1996 begann Rojas seine Profikarriere. Am 20. September 2010 boxte er gegen Kōhei Kōno um den Weltmeistertitel des Verbandes WBC und siegte durch einstimmige Punktrichterentscheidung. Diesen Titel verteidigte er insgesamt zweimal und verlor ihn im August des darauffolgenden Jahres an Suriyan Sor Rungvisai nach Punkten.

Weblinks

Boxer (Mexiko)
Weltmeister (Boxen)
Mexikaner
Geboren 1980
Mann

redpajama_set_name: RedPajamaWikipedia


How do I remove the calendar from the left side of the page?
Is there a way to get a member's activity history, like when was the member added to a certain group and when removed and readded etc..
Are there ever any templates in your gallery for special dates in the Spring and Summer, such as, Easter, Mother's Day, Father's Day?
Is it possible to change the format of the sent date when viewing sent email campaigns?


text:
HBO Max abruptly pulls 200 Sesame Street episodes from its catalog
By David Osiecki / TV / August 21, 2022 August 29, 2022 / Leave a Comment / 2 minutes of reading
According to Variety, about 200 episodes of Sesame Street have been taken off HBO Max, which could be devastating for children and parents. According to Variety, the number of Sesame Street episodes on the service has decreased from about 650 to 456, and many of the episodes that have been dropped are from the program's first few seasons.
HBO and Sesame Street have been partners for many years; in 2019, WarnerMedia (now Warner Bros. Discovery) signed a significant five-year deal to stream new episodes of the program exclusively, and Sesame Street debuted on HBO in 2016.
A spokesperson for HBO Max, Chris Willard, told The Verge that "Sesame Street is and has always been an important part of television culture and a crown jewel of our preschool offering." In addition to the nearly 400 episodes of the most recent and classic seasons that are still available on Cartoonito on HBO Max, which will premiere this fall, "We are committed to continuing to bring Sesame Street into families' homes.
Older Sesame Street episodes are being canceled, and Warner Bros. Discovery also intends to remove 36 titles from HBO Max, according to NPR. Along with some Sesame Street specials, the Elmo-hosted late-night program, and HBO Max Originals like Infinity Train and Summer Camp Island, this includes.
The new app that will replace HBO Max and Discovery Plus will debut in the summer of 2019. HBO Max explained the decision to pull 36 shows in a statement to Variety: "As we work toward bringing our content catalogs together under one platform, we will be making changes to the content offering available on both HBO Max and Discovery Plus." "That will involve some content being deleted from both platforms."
Warner Bros. Discovery has been discontinuing content at an alarming rate; most recently, Batgirl was put on hold, and at least six Warner Bros. films were removed from HBO Max. Due to the Warner Bros. Discovery megamerger's overlap earlier this week, the company cut staff at HBO Max.


Are you looking for an activity to do with your class that combines both written and spoken language but has a creative element, too?
If so, try this lesson on invention which challenges students to think creatively as well as linguistically and just may improve the world in the process.
Ask your students think of something they could not live without. After a minute or two, break your class into groups to share those most important items. While in those groups, ask your students to tell the others why they need this item, what they would do without it, and how they could make it better. Your students may say that they could not live without a cell phone or laptop computer, but what about life's more basic items? Could they live without a bed? The light bulb? A refrigerator? We often do not realize how many modern inventions we find irreplaceable. Have your groups make a list of all the modern conveniences they use on a daily basis and would not want to live without.
Once your students are thinking of those important items in their lives, challenge them to think beyond their daily limits. If they could invent anything, what would it be? What would it do? Why would they want it? Give your students some time to think and make notes about their invention, and then divide your class into groups to talk about what they would create. Encourage students to ask questions of one another to refine their ideas. You should allow students to take notes on their classmates' comments and ideas so they can use them later when refining their inventions.
Now that your students have inventions that they can really picture, it is time to introduce those inventions to the world. Each student will be responsible for creating an advertisement for his or her invention. Help your students get ideas by making old catalogs and magazines available to them. They can look through the magazines for style and content of advertisements and model their own advertisements off those in the magazines. Once your class is familiar with these ads, have each person create an ad for his or her invention. The ad should include a picture of the product, a description of it, the creator's name and some information about the product. Once everyone in class has completed his or her invention, compile them into a class catalogue. You can reproduce a few copies for your class and then break them into groups to look at their classmate's products in the catalogue.
After the groups have had a chance to read about their classmates' inventions, explain to your class that they will make a commercial for one of the products. Each group should select one product from those created by its group members. As a group, they should then write a commercial for the product. Encourage them to be creative and engaging. You may want to show them classic commercials in English like Wendy's "Where's the Beef?" or other classic commercial such as Dunkin' Donut's "Time to Make the Donuts" or Coke's commercial starring Mean Joe Green. Stress that effective commercials appeal to the viewer's emotion as well as his logic.
Once the commercials are written, give your students a chance to perform their commercials in front of the class. Stress to your class that everyone should participate in the commercial. Once all the groups have performed, take a vote in class for the most popular product. Either by a show of hands or by ballot, have your class vote for the product they would be most likely to purchase. Recognize the inventor of the winning product with a certificate of creativity and/or a prize.
To follow up your lesson on inventions, you can use the Newberry Award winning book The 21 Balloons by William Pene Dubois. In this book, a balloonist travels around the world via a series of original inventions. The main character, a retired schoolteacher, decides to spend one year in a hot air balloon of his original design. Though he fails in his attempt, he is rescued to an island whose inhabitants are inventors in their own right. Your students will enjoy the straightforward language and creative inventions, some real and some imagined, presented in this novel. Another follow up activity would be researching an inventor who made a significant impact on the world. You may want students to research the inventor of that item they said they could not live without, or you may want your students to focus on more classic examples such as Thomas Edison. In either case, give your students an opportunity to present their findings to the class.
People everywhere everyday strive to make the world a better place in which to live. Give your students a hand at this goal by asking them to invent their own creation.
You will be surprised at the creativity your students show when given the chance. Who knows, maybe one day one of their inventions will take the rest of the world by storm.


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    [ValidateNotNullOrEmpty()][Parameter (Mandatory = $True)][string]$SubscriptionName,
    [ValidateNotNullOrEmpty()][Parameter (Mandatory = $False)][string]$ResourceGroupName = "IoTJourney",
    [ValidateNotNullOrEmpty()][Parameter (Mandatory = $False)][string]$DeploymentName = "WarmStorage-AzureStreamAnalyticsAndSql",
    [ValidateNotNullOrEmpty()][Parameter (Mandatory = $False)][bool]$AddAccount = $true
)
PROCESS
{
    $ErrorActionPreference = "Stop"

    $ScriptsRootFolderPath = Join-Path $PSScriptRoot -ChildPath "..\"
    $ModulesFolderPath = Join-Path $PSScriptRoot -ChildPath "..\..\Modules"

    Push-Location $ScriptsRootFolderPath
        .\Init.ps1
    Pop-Location

    Load-Module -ModuleName ResourceManager -ModuleLocation $ModulesFolderPath

    if($AddAccount)
    {
        Add-AzureAccount
    }

    Select-AzureSubscription $SubscriptionName

    $deploymentInfo = $null
    Invoke-InAzureResourceManagerMode ({
        $deploymentInfo = Get-AzureResourceGroupDeployment -ResourceGroupName $ResourceGroupName  -Name $DeploymentName
    })

    #simulator settings
    $simulatorSettings = @{
        'Simulator.EventHubNamespace'= $deploymentInfo.Outputs["serviceBusNamespaceName"].Value;
        'Simulator.EventHubName' = $deploymentInfo.Outputs["eventHubName"].Value;
        'Simulator.EventHubSasKeyName' = $deploymentInfo.Outputs["sharedAccessPolicyName"].Value;
        'Simulator.EventHubPrimaryKey' = $deploymentInfo.Outputs["sharedAccessPolicyPrimaryKey"].Value;
        'Simulator.EventHubTokenLifetimeDays' = ($deploymentInfo.Outputs["messageRetentionInDays"].Value -as [string]);
    }

    Write-SettingsFile -configurationTemplateFile (Join-Path $PSScriptRoot -ChildPath "..\..\..\..\src\Simulator\ScenarioSimulator.ConsoleHost.Template.config") `
                       -configurationFile (Join-Path $PSScriptRoot -ChildPath "..\..\..\..\src\Simulator\ScenarioSimulator.ConsoleHost.config") `
                       -appSettings $simulatorSettings
}

redpajama_set_name: RedPajamaGithub


There is just some produce that screams summer: Stone fruits, strawberries, and especially sweet, juicy fresh corn! There's nothing like it.
The best thing about perfectly ripe summer corn is that it's so sweet it can be eaten raw. And even though the kernels are crisp and crunchy without being cooked, they're tender and juicy on the inside. Corn makes a perfect summer salsa. I love this sweet raw corn salsa recipe with chips of course, but it's also super delicious over grilled fish, chicken or steak.
This is the perfect complement to chips, and makes an amazing summer topping for fish, chicken, or steaks.
Step 1: Drizzle the corn with the lemon juice, using about ½ tablespoon per cob. Let them sit for a few minutes, while you prepare the remaining ingredients.
Step 2: Add the tomatoes, onion, cilantro, and jalapeño to a medium-sized mixing bowl.
Step 3: Use a chef's knife to cut the lemon-coated corn kernels off of their cobs, adding them to the bowl with the other ingredients as you go. Use the back of the knife to scrape as much of the corn "juice" as possible into the bowl as well.
Step 4: Add the sugar and season to taste with the salt.
How to Build a Better Cheese BoardBeer & Cheese Pairings: What Goes With What?


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text:
Russia's State Duma Approves Bill to Ban 'LGBT propaganda'
by David November 24, 2022 · Gay News / Gay Pride / Gay Rights / The Advocate
Author: Uliana Pavlova
(CNN) — Russia's lower house of parliament passed in the third reading amendments to a law on so-called "LGBT propaganda" on Thursday, expanding liability to all ages.
The discriminatory law proposes to ban all Russians from promoting or "praising" homosexual relationships or publicly suggesting that they are "normal."
The original version of the law adopted in 2013 banned "propaganda of non-traditional sexual relations" among minors. The new iteration would apply the law to adults as well.
Individuals who spread or attempt to distribute what the bill calls "LGBT propaganda" will be fined up to 400,000 roubles ($6,600). Legal entities can be fined up to 5 million rubles ($82,100). Foreigners can be arrested for up to 15 days or deported, according to the bill.
It will now be forwarded to the Federation Council, Russia's upper house of Parliament, before being signed into law by Russian President Vladimir Putin.
The European Court of Human Rights ruled in 2017 that Russia's so-called "gay propaganda law" is discriminatory, promotes homophobia, and violates the European Convention on Human Rights.
The court found that the law "served no legitimate public interest," rejecting suggestions that public debate on LGBT issues could influence children to become homosexual, or that it threatened public morals.
Homosexuality was decriminalized in Russia in 1993, but homophobia and discrimination is still rife. It is ranked 46th out of 49 European countries for LGBTQ+ inclusion by watchdog ILGA-Europe.
Related: Russia Lawmakers Begin Steps to Toughen 'Gay Propaganda' Law
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RESISTANCE WATCH--Organizers of the Women's March on Washington—which took place the day after Donald Trump's inauguration and inspired solidarity marches in hundreds of cities across the globe—are planning a protest against the National Rifle Association for July 14, in response to a video published to the NRA's Facebook page last month.
This time around, the two-day 18-mile march will begin Friday at NRA headquarters in Fairfax, Virginia, and culminate with a rally in front of the Department of Justice Saturday morning. Organizers are promoting the march on social media with #NRA2DOJ and details are available on the group's Facebook event.
"We know that we are not safe," the group stated. "But we will not be intimidated into silence."
In the minute-long video that provoked the march, conservative talk-radio host Dana Loesch pushes the false narrative that protesters are violent, in an effort to encourage viewer support for the NRA and gun ownership. Loesch says U.S. media and protest organizers encourage demonstrators "to smash windows, burn cars, shutdown interstates and airports, bully and terrorize the law-abiding..."
"Recent actions by the NRA demonstrate not only a complete disregard for the lives of black and brown people in America—your fellow citizens—but appear to be a direct endorsement of violence against these citizens exercising their constitutional right to protest….
The advertisement released by the NRA is a direct attack on people of color, progressives and anyone who exercises their First Amendment right to protest. At a time when our nation is seeing a rise in racially charged incidents and violence motivated by hate speech, it is unconscionable for a powerful organization like yours to unashamedly peddle an 'us versus them' narrative. You are calling for our grassroots, nonviolent resistance movement to be met with violence."
The NRA responded three days later, with a longer video titled "We Don't Apologize For Telling the Truth," starring NRA TV Live host Grant Stinchfield. In the response video, Stinchfield repeatedly attacks the "lying media" and "violent left," and addresses Mallory directly. "I'm talking to you Tamika Mallory," he says, dismissing her concerns, and those of activists and politicians who have publicly condemned the Loesch video.


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text:
Discuss all things related to the pop culture of the 2010\'s. Politics & Religious topics are a seperate sub-board, and can ONLY appear on that board.
This is an index of topics from the The 2010s topic on the inthe00s
The 2010s 2017 Archive: (December) (November) (October) (September) (August) (July) (June) (May) (April) (March) (February) (January)
Why Does Nobody Trick-or-Treat Anymore? by #Infinity - 10/31/2017
2017 World Series! by TheReignMan99 - 10/30/2017
Jersey shore is back Nov 27 by John Titor - 10/30/2017
Is disco/funk revival the "post-grunge" of the 2010s? by bchris02 - 10/28/2017
My predictions for the next couple of years by HeyJealousy - 10/28/2017
Is 2017 a core 10s year? by questoftheninja - 10/27/2017
Adobe Flash Will Be Killed by 2020 by Shemp97 - 10/26/2017
When do you think music will be completely late 10s? by Charmed17 - 10/25/2017
Mid 2010s by Wobo - 10/25/2017
When did the Mac fad peak? Are Macs still popular? by bchris02 - 10/25/2017
Windows 7 by bchris02 - 10/25/2017
Taylor Swift about to release her "Born This Way"? by Fearsword - 10/24/2017
I'm noticing a more macho,right, testosterone culture coming in! by Charmed17 - 10/24/2017
George Michael heading for posthumous number one by Philip Eno - 10/24/2017
Do you think Adult Contemporary music will make a comeback? by mxcrashxm - 10/23/2017
2018 predictions? by HeyJealousy - 10/23/2017
2018 might actually be great! by Charmed17 - 10/22/2017
Looks Like Roman Polanski is Next by LyricBoy - 10/21/2017
The Most Memorable Year Of The 2010s by Bab2781 - 10/20/2017
Is Nicki Minaj the most consistent artist this decade? by bchris02 - 10/20/2017
So is 2010 & 2011 at iconic status yet? by John Titor - 10/19/2017
The 2020s will be a wild but fun decade by Charmed17 - 10/19/2017
Top 10 Songs (2010) by mxcrashxm - 10/17/2017
Will the 2020s be backlash of 10s or embrace it. by Charmed17 - 10/17/2017
Is anyone else resentful they had to be young in the 2010s? by nintieskid999 - 10/17/2017
YouTube's days appear to be numbered by LooseBolt - 10/16/2017
Mario Odyssey receives perfect 10 from EDGE: 'Spiritual successor to Mario 64' by 2001 - 10/15/2017
I'm Just Gonna Be a Pretentious Jerk by BornIn86 - 10/15/2017
US adult obesity rate hits 40% — and shows no signs of slowing down by 2001 - 10/15/2017
Platinum hair: a rising trend? by HeyJealousy - 10/14/2017
When did the late 10s transition start? by Charmed17 - 10/14/2017
Kim Kardashian : "Selfies are dead" by John Titor - 10/14/2017
Liam Gallagher gets to number one on his own in the UK charts by Philip Eno - 10/13/2017
Avatar: The start of the '10s for movies? by bchris02 - 10/12/2017
Aspects of 2010's culture you liked the most by Looney Toon - 10/11/2017
2008-2011 is way better than 2012-2017!!? by Charmed17 - 10/11/2017
To the people who are leaving this site. by John Titor - 10/10/2017
What year do you think trap rap will fade? by Charmed17 - 10/09/2017
Something about 2010s culture by mxcrashxm - 10/09/2017
Are the 2010s aimless by Charmed17 - 10/08/2017
I'm really starting to feel some distance from the aughts by bchris02 - 10/07/2017
Mainstream music is at an all time low! by Charmed17 - 10/07/2017
Why do people say it's harder to date now? by Charmed17 - 10/07/2017
Anyone for the new Blade Runner sequel by Philip Eno - 10/06/2017
Harvey Weinstein, the harassment allegations, is going to trial by Philip Eno - 10/06/2017
Is pop culture taking a hit from staleness? by Charmed17 - 10/06/2017
Is Sonic the Hedgehog officially out of its "dark age"? by Zelek3 - 10/06/2017
When people say you look younger than your age by Brian06 - 10/06/2017
AIM to shut down in December by bchris02 - 10/06/2017
I'm leaving for good by 80sfan - 10/06/2017
What defines pop culture? by Charmed17 - 10/05/2017
Will this era the Donald Trump's era end up being an in between era? by Charmed17 - 10/04/2017
Jared Leto to star in Hugh Hefner biopic by Philip Eno - 10/04/2017
Pop culturally and musically this is the worse decade since the 40s by Charmed17 - 10/03/2017
What year will most likely be seen as the last real 10s year by Charmed17 - 10/03/2017
9/11 or Oklahoma Bombing of this Decade? by Wobo - 10/03/2017
Another member leaving... by HazelBlue99 - 10/03/2017
TRL Is Back today by Charmed17 - 10/02/2017
50 People Were Murdered In Las Vegas Last Night by BornIn86 - 10/02/2017
Tom Petty Heart Attack by LyricBoy - 10/02/2017
I think culture will shift somewhere in the beginning of 2018! by Charmed17 - 10/01/2017


text:
Toast the Ghost
Standing in a cemetery across from a distillery doesn't really sound like the type of tour introduction one might expect, but that's exactly how our recent visit to The Glenrothes began. Ronnie Cox, Brands Heritage Director, stood amidst the headstones memorializing denizens of the village of Rothes and told us a ghost story.
The distillery was founded in 1879, and the building that now houses the big-bellied pot stills producing this magical elixir was once its malting floor. That malt house was converted to house the stills in 1979.
At the turn of the 20th century, an orphan boy named Biawa "Byeway" Makalaga was found under a bush in Africa by Colonel Grant of Rothes. Grant brought "Byeway" back to Scotland to be raised in his hometown. The young man grew up to be the Colonel's helper and a well-known figure in the village, playing on the local football team. "Byeway" passed away in 1972, seven years later when the former malting floor was renovated and the new stills installed there were two reports of his ghost appearing in the still house.
The distillery brought in a local professor to investigate, he found a disruption in an energy line that was caused by the new stills installation, this led him to walk to a specific gravestone across the road, 70+ yards from the still house, and after a brief moment of conversation he returned and said the situation had been resolved amicably. The gravestone was that of Biawa "Byeway" Makalaga, and his ghost was not seen again.
The legend of The Glenrothes continues, and why not pick up a couple of bottles so that next time you want a fine Scotch Whisky you can raise your glass and "toast the ghost", in the tradition of Rothes, Scotland.
Glenrothes Sherry Cask Reserve Single Malt Scotch Whisky Speyside
Glenrothes Peated Cask Reserve Single Malt Scotch Whisky Speyside
1999 Signatory 'Glenrothes' 16yr Single Malt Whisky