text:
This paper deals with the strongly nonstationary squeeze of an oil film at the interface between the chain pin and pulley in chain belt continuously variable transmission. We concentrate on the squeeze motion as it occurs as soon as the pin enters the pulley groove. The duration time to complete the squeeze process compared with the running time the pin takes to cover the entire arc of contact is fundamental to understand whether direct asperity-asperity contact occurs between the two approaching surfaces to clarify what actually is the lubrication regime (elastohydrodynamic lubrication (EHL), mixed, or boundary) and to verify if the Hertzian pressure distribution at the interface can properly describe the actual normal stress distribution. The Hertzian pressure solution is usually taken as a starting point to design the geometry of the pin surface; therefore, it is of utmost importance for the designers to know whether their hypothesis is correct or not. Taking into account that the traveling time, the pin spends in contact with the pulley groove, is of about 0.01 s, we show that rms surface roughness less than $0.1 μm$, corresponding to values adopted in such systems, guarantees a fully lubricated EHL regime at the interface. Therefore, direct asperity-asperity contact between the two approaching surfaces is avoided. We also show that the Hertzian solution does not properly represent the actual pressure distribution at the pin-pulley interface. Indeed, after few microseconds a noncentral annular pressure peak is formed, which moves toward the center of the pin with rapidly decreasing speed. The pressure peak can grow up to values of several gigapascals. Such very high pressures may cause local overloads and high fatigue stresses that must be taken into account to correctly estimate the durability of the system.

1.
Carbone
,
G.
,
Mangialardi
,
L.
, and
Mantriota
,
G.
, 2002, “
Fuel Consumption of a Mid Class Vehicle With Infinitely Variable Transmission
,”
SAE Transactions 2001 - Journal of Engines
, Section 3, Vol.
110
(3), pp.
2474
2483
.
2.
Carbone
,
G.
,
Mangialardi
,
L.
,
Mantriota
,
G.
, and
Soria
,
L.
, 2004, “
Performance of a City Bus Equipped With a Toroidal Traction Drive
,”
IASME Transactions
,
1
(
1
), pp.
16
23
.
3.
Mantriota
,
G.
, 2005, “
Fuel Consumption of a Vehicle With Power Split CVT System
,”
Int. J. Veh. Des.
0143-3369,
37
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4
), pp.
327
342
.
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Simons
,
S. W. H.
,
Klaassen
,
T. W. G. L.
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Steinbuch
,
M.
,
Veenhuizen
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P. A.
, and
Carbone
,
G.
, “
Shift Dynamics Modelling for Optimization Variator Slip Control in a Push-Belt CVT
,”
Int. J. Veh. Des.
0143-3369,
48
(
1-2
), pp.
45
64
.
5.
Carbone
,
G.
,
Mangialardi
,
L.
, and
Mantriota
,
G.
, 2005, “
The Influence of Pulley Deformations on the Shifting Mechanisms of MVB-CVT
,”
ASME J. Mech. Des.
0161-8458,
127
, pp.
103
113
.
6.
Carbone
,
G.
,
Mangialardi
,
L.
,
Bonsen
,
B.
,
Tursi
,
C.
,
Veenhuizen
,
P. A.
, and
Dynamics
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C. V. T.
, 2007, “
Theory and Experiments
,”
Mech. Mach. Theory
0094-114X,
42
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4
), pp.
409
428
.
7.
Srnik
,
J.
, and
Pfeiffer
,
F.
, 1997, “
Dynamics of CVT Chain Drives: Mechanical Model and Verification
,” ASME Paper No. DETC97/VIB-4127.
8.
Saito
,
T.
, 2007, “
Simulation of Stress on Elements of CVT Metal Pushing V-Belt Under Transient Operating Conditions
,”
Proceedings of the International Congress on Continuously Variable and Hybrid Transmissions
, Yokohama, Paper No. 20074544.
9.
Kanehara
,
S.
,
Fujii
,
T.
, and
Kitagawa
,
T.
, 1994, “
A Study of a Metal Pushing V-Belt Type CVT-Part 3: What Forces Act on Metal Blocks?
,” SAE Paper No. 940735.
10.
Carbone
,
G.
,
Scaraggi
,
M.
, and
Mangialardi
,
L.
, “
EHL-Squeeze at Pin-Pulley Interface in CVTs: Influence of Lubricant Rheology
,”
Tribol. Int.
0301-679X, submitted.
11.
Christensen
,
H.
, 1962, “
The Oil Film in a Closing Gap
,”
Proc. R. Soc. Lond. A Math. Phys. Sci.
,
266
, pp.
312
328
. 0080-4630
12.
Lee
,
K. M.
, and
Cheng
,
H. S.
, 1973, “
The Pressure and Deformation Profiles Between Two Normally Approaching Lubricated Cylinders
,”
ASME J. Lubr. Technol.
0022-2305,
95
(
3
), pp.
308
317
.
13.
Safa
,
M. M. A.
, and
Gohar
,
R.
, 1986, “
Pressure Distribution Under a Ball Impacting a Thin Lubricant Layer
,”
Trans. ASME, J. Tribol.
,
108
, pp.
372
376
. 0742-4787
14.
Yang
,
P.
, and
Wen
,
S.
, 1991, “
Pure Squeeze Action in an Isothermal Elastohydrodynamic Lubricated Spherical Conjunction. Part 1: Theory and Dynamic Load Results
,”
Wear
0043-1648,
142
, pp.
1
16
.
15.
Yang
,
P.
, and
Wen
,
S.
, 1991, “
Pure Squeeze Action in an Isothermal Elastohydrodynamic Lubricated Spherical Conjunction. Part 2: Constant Speed and Constant Load Results
,”
Wear
0043-1648,
142
, pp.
17
30
.
16.
Wong
,
P. L.
,
Lingard
,
S.
, and
Cameron
,
A.
, 1992, “
The High Pressure Impact Microviscometer
,”
STLE Tribol. Trans.
1040-2004,
35
(
3
), pp.
500
508
.
17.
Dowson
,
D.
, and
Wang
,
D.
, 1994, “
An Analysis of the Normal Bouncing of a Solid Elastic Ball on an Oily Plate
,”
Wear
0043-1648,
179
, pp.
29
37
.
18.
,
R.
, and
Höglund
,
E.
, 1994, “
Elastohydrodynamic Lubrication at Pure Squeeze Motion
,”
Wear
0043-1648,
179
, pp.
39
43
.
19.
,
R.
, and
Höglund
,
E.
, 1995, “
Numerical Simulation of a Ball Impacting and Rebounding a Lubricated Surface
,”
Trans. ASME, J. Tribol.
0742-4787,
117
, pp.
94
102
.
20.
Chu
,
H. M.
,
Li
,
W. L.
, and
Chen
,
M. D.
, 2006, “
Elastohydrodynamic Lubrication of Circular Contacts at Pure Squeeze Motion With Non-Newtonian Lubricants
,”
Tribol. Int.
0301-679X,
39
, pp.
897
905
.
21.
Guo
,
F.
,
Kaneta
,
M.
,
Wang
,
J.
,
Nishikawa
,
H.
, and
Yang
,
P.
, 2006, “
Occurrence of a Noncentral Dimple in Squeezing EHL Contacts
,”
ASME J. Tribol.
0742-4787,
128
, pp.
632
640
.
22.
Hamrock
,
B. J.
, 1994,
Fundamentals of Fluid Film Lubrication
,
McGraw-Hill
,
New York
.
23.
Dowson
,
D.
, 1995, “
Elastohydrodynamic and Micro-Elastohydrodynamic Lubrication
,”
Wear
0043-1648,
190
, pp.
125
138
.
24.
Moore
,
A. J.
, 1997, “
The Behaviour of Lubricants in Elastohydrodynamic Contacts
,”
Proc. Inst. Mech. Eng., Part J: J. Eng. Tribol.
1350-6501,
211
, pp.
91
106
.
25.
Carbone
,
G.
, and
Scaraggi
,
M.
, 2008, Dipartimento di Ingegneria Meccanica e Gestionale, Politecnico di Bari, Confidential Report No. 3.
26.
Johnson
,
K. L.
, 1985,
Contact Mechanics
,
Cambridge University Press
,
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.
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Maugis
,
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,
Contact Adhesion and Rupture of Elastic Solids
(
Springer Series in Solid State Sciences
),
Springer-Verlag
,
Berlin
.


text:
Javascript is required
1
JEE Main 2021 (Online) 17th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
The velocity of a particle is v = v0 + gt + Ft2. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is :
A
v0 + g + f
B
v0 + $${g \over 2}$$ + $${F \over 3}$$
C
v0 + 2g + 3F
D
v0 + $${g \over 2}$$ + F
2
JEE Main 2021 (Online) 17th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to $${{81} \over {100}}$$ of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms$$-$$2)
A
2.50 ms$$-$$1
B
3.0 ms$$-$$1
C
2.0 ms$$-$$1
D
3.50 ms$$-$$1
3
JEE Main 2021 (Online) 17th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is :
A
$${{4\alpha \beta } \over {(\alpha + \beta )}}{t^2}$$
B
$${{2\alpha \beta } \over {(\alpha + \beta )}}{t^2}$$
C
$${{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}$$
D
$${{\alpha \beta } \over {4(\alpha + \beta )}}{t^2}$$
4
JEE Main 2021 (Online) 16th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
A mosquito is moving with a velocity $$\overrightarrow v = 0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k$$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?
A
$${\tan ^{ - 1}}\left( {{\sqrt {85} } \over 6}\right)$$ from y-axis
B
$${\tan ^{ - 1}}\left( {{5 \over 2}} \right)$$ from y-axis
C
$${\tan ^{ - 1}}\left( {{2 \over 3}} \right)$$ from x-axis
D
$${\tan ^{ - 1}}\left( {{5 \over 2}} \right)$$ from x-axis
Policy

format: plain


text:
# Happy maths 1: numbers  (Page 5/5)

 Page 5 / 5

There is a scene in the movie that has a jailertelling his officers,

“Half of you go this side. Half of you go that side. Rest ofyou, follow me!”

Zzero: Tell me Eka, how many people followed the jailer?
Eka: Zzero, you must be joking!

Zzero: No one followed the jailer. One half plusanother half of the same thing makes a full!

Sankhya and Ganith had to share a huge aloo Paratha for lunch. Theyboth liked aloo Paratha very much. “Let’s cut it into two. Each of uscan have one part,” said Ganith. “You cut it, Sankhya.”

Sankhya was going to fold the parathainto two, when her friends Faisal andJohn walked in. “We want a piece too!”they said.

Sankhya and Ganith would have had $\frac{1}{2}$ a paratha each. Now they would have tohave $\frac{1}{4}$ of a paratha.

So they had to divide one paratha into four equal pieces. After Sankhya had cut theparatha neatly into four pieces, John saw a sugarcane piece in the kitchen. “I’ll eatthe sugarcane. I don’t want the paratha,” he decided.

Sankhya, Ganith and Faisal had a quarter of theparatha each. And they cut up John’s quarter intothree. What do you think they did with that!

## Fractions are nothing but parts of a whole.

Try this:

Take a sheet of notepaper, fold it in hald, and colour the top flap. You have colured half of the sheet.

It means you have coloured 1 part of two ewual parts that make up a whole. ( $\frac{1}{2}$ ) Now fold over in half again and colour the top in a different colour. You have colored 1 part of 4.

Fold the sheet over another time. Colour the top using anothercolour pencil. Open the sheet out fully and see whatfraction you have coloured.

## Shapely numbers

Sankhya watched as Amma made a beautiful rangoli. First she made many dots. Then she joined the dots to make designs.

“Amma. How do you know how many dots to make?”

Amma mumbled something, because she really did not have an answer.

But Zzero has discovered a way to make shapes out of numbers.

Take any number and its next number. (Two numbers that have a differenceof 1 are called consecutive numbers.)

Multiply them and divide the answer by 2.

You can make a triangular number grid with this new answer.

$6\times 7$ divided by 2 = 21.

You can form a triangle with 21 dots, starting with 6 dots at the bottom, reducing one dot as you go higher, till you have just onedot on top.

Or you can start with one dot on top, and keep increasing one dot in the next line as you go down.

Can you tell how many dots you need to make an 8 - decker triangle with dots?

36.
$\frac{8\times 9}{2}=36$ .

Try this:

Make grids using dots. Then draw funny figures.

## Let's fly

Zzero: Eka, let’s fly!
Eka: Zzero, we are not birds. And we are not pilots. Sohow can we fly?

Zzero: Let’s make paper planes and fly them. Listen friends, let’splay detectives!

Write a coded message using numbers on a piece of paper. Make this into a paper plane.Send it zooming in the class.

The person on whom the plane lands has to decode the message.The numbers must represent alphabets.

The numbers that make up one word can be put together with hyphens.

Some scratched busily on papers with their pencils. Soon there were differentkinds of planes flying in class.

Can you decode some of these messages that ‘landed’?

13-29 / 15-29-25-9! (Notice that all the numbers are ODD)

GO HOME! (Starting with 1 for A, 3 for B…….51 for Z)

49-29-41 / 1-35-9 /13-29-29-7.

YOU ARE GOOD.

51-51-9-35-29 /17-37 / 1 /15-9-35-29.

ZZERO IS A HERO

28-42-26-4-10-36-38 / 2-36-10 / 12-42-28.

NUMBERS ARE FUN. (2 for A, 4 for B…..52 for Z)

## Conclusion

Sankhya and Ganith have made friends with numbers. Sometimes the ‘friends’ behave oddly!But most of the time, they are good! Look at the numbers around you.You may find something interesting about them.

My name is Gopalji Srivastava. I am in class 5 and never miss a movie by Amitabh Bachchan and Kajol. I also enjoy badmintonand the Bhangra.

Thank you for buying this book. My friends and I will get to read many more books in our library because you bought this book.

Mala Kumar is a journalist, writer and editor based in Bangalore. Her stories for children have won awards from Children’s Book Trust. She discovered her love forteaching while conducting non-formal workshops in Mathematics in schools, using the day’s newspaper instead of text-books.

Angie is a graphic designer and in her spare time loves to keep busy with ceramic. Upesh is an animator who collects graphic novels and catches up with odd films in hisspare time. Together they form ‘The Other Design Studio’.

This is a Mathematics book with a difference. There are more stories here than problems!So read the stories, take in the mixture of facts and fiction and enjoy teasing your brain.

## Titles in this series

Happy Maths 1 : Numbers
Happy Maths 2 : Shapes and Data
Happy Maths 3 : Measurements
Happy Maths 4 : Time and Money

Our books are available in English, Hindi, Tamil, Telugu, Kannada, Marathi, Gujarati, Bengali, Punjabi, Urdu and Oriya.

Pratham Books is a not-for-profit publisher that produces high-quality and affordable children’s books in Indian languages.

Age Group: 11 - 14 years
Happy Maths - 2 Shapes and Data (English)
MRP: Rs. 25.00

what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
how did you get the value of 2000N.What calculations are needed to arrive at it
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text:
# Annoying space before the first item in an enumeration which starts a theorem

Consider the following example:

\documentclass{article}
\newtheorem{Theorem}{Theorem}
\begin{document}
\begin{Theorem}
On the other hand, this one has only one claim.
\end{Theorem}
\begin{Theorem}
\begin{enumerate}
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\end{document}

Typesetting it results in:

Notice there is extra horizontal space before the first item in the list of the second theorem.

Is there a sensible way to get rid of that?

Ideally, the solution would work with amsthm, thmtools, enumitem.

N.B.: this is related to Is it possible to skip the first line in a theorem environment? but not quite the same.

• @Martin: the plus signs meant something different than your commas :) – Mariano Suárez-Álvarez May 5 '11 at 9:22
• Sorry, I didn't saw that. The main edit was the formatting of the package name with back-ticks . Feel free to put the +s back in. – Martin Scharrer May 5 '11 at 9:24

Two variants of a solution using the enumitem package. In the first case, two enumerate environments were used; the first one uses leftamargin=* to suppress the undesired horizontal space; the second enumerate uses resume to keep the numbering but preserves the standard value for \leftmargin; the vertical spacing between the two environments was also corrected. In the second case, the option leftmargin=* was also used, but this time only one enumerate environment was used so the horizontal space will be removed for all the items.

\documentclass{article}
\usepackage{enumitem}

\newtheorem{Theorem}{Theorem}

\begin{document}

\begin{Theorem}
On the other hand, this one has only one claim.
\end{Theorem}

\begin{Theorem}
\begin{enumerate}[leftmargin=*]
\item First claim
\end{enumerate}\vspace*{-2\partopsep}
\begin{enumerate}[resume]
\item Second claim
\end{enumerate}
\end{Theorem}

\begin{Theorem}
On the other hand, this one has only one claim.
\end{Theorem}

\begin{Theorem}
\begin{enumerate}[leftmargin=*]
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}

\end{document}

The result:

You can use a \@labels trick (as in Phil Hirschhorn's answer to “Avoiding a line break at the beginning of an enumerate”). Here's the result:

There are two ways you can use this. The first is by manually putting a \thmenumhspace after \begin{enumerate} inside a theorem:

\documentclass{article}
%\usepackage{amsthm,enumitem,thmtools}% works with or without
\newtheorem{Theorem}{Theorem}
\makeatletter
\newlength{\@thlabel@width}%
\makeatother
\begin{document}
\begin{Theorem}
1. First claim 2. Second claim.
\end{Theorem}
\begin{Theorem}
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\end{document}

With a little more work, this can be automated, hopefully without any side effects (this time, you need amsthm for it to work correctly):

\documentclass{article}
\usepackage{amsthm,enumitem,thmtools}% needs amsthm
\newtheorem{Theorem}{Theorem}
\makeatletter
\let\original@item\item
\newlength{\@thlabel@width}%
\newcommand{\RedefineItem}{%
\settowidth{\@thlabel@width}{\itshape1.}%
}
\let\original@Theorem\Theorem
\def\Theorem{
\@ifnextchar[{\Theorem@Opt}
{\Theorem@NoOpt}
}
\def\Theorem@Opt[#1]{%
\@ifnextchar\begin{\RedefineItem\original@Theorem[#1]}%
{\original@Theorem[#1]}%
}
\def\Theorem@NoOpt{%
\@ifnextchar\begin{\RedefineItem\original@Theorem}%
{\original@Theorem}%
}
\makeatother
\begin{document}
\begin{Theorem}
1. First claim. 2. Second claim.
\end{Theorem}
\begin{Theorem}
\begin{enumerate}
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\begin{Theorem}
This one has two claims.
\begin{enumerate}
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\begin{Theorem}[Title]
\begin{enumerate}
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\begin{Theorem}[Title]
This one has two claims.
\begin{enumerate}
\item First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\end{document}

• From where comes the dimension -1em? To me it is not enough. How to determine the exactly value to use to have the same output as with no enumerate? – Sigur Apr 8 '15 at 17:52
• @Sigur: I've edited the answer; it should now have the exact right amount of space whatever the settings are. – Philippe Goutet Apr 10 '15 at 16:00
• I saw that you edited only the automated code. But I'm using the simplest one since I'm using it only once. I tried the same value as \baselineskip (in my case, 1.5em) and it looks right but I'm not sure. – Sigur Apr 10 '15 at 17:35
• @Sigur: I've also adapted the manual code. The exact value cannot be 1.5em: it depends on the leftmargin, labelsep, itemindent and the size of the enumerate label. – Philippe Goutet Apr 10 '15 at 19:37
• Nice! Since I'm using the label style (i) and so on in upright font, should I use \upshape(i) instead of \itshape1.? – Sigur Apr 10 '15 at 20:03

You can hack a bit the first \item as follows:

\documentclass{article}
\newtheorem{Theorem}{Theorem}
\newcommand*\fixitem {\item[]%
\refstepcounter{enumi}\hskip-\leftmargin\labelenumi\hskip\labelsep}
\begin{document}
\begin{Theorem}
On the other hand, this one has only one claim.
\end{Theorem}
\begin{Theorem}
\begin{enumerate}
\fixitem First claim
\item Second claim
\end{enumerate}
\end{Theorem}
\end{document}

and the result is:

I found this trick that might help:

\documentclass{article}
\newtheorem{Theorem}{Theorem}
\begin{document}
\begin{Theorem}
On the other hand, this one has only one claim.
\end{Theorem}
\begin{Theorem}
\mbox{}
\begin{itemize}
\item[(i)] item one.
\item[(ii)] item two.
\item[(iii)]  item three.
\end{itemize}
\end{Theorem}
\end{document}

• That is mentioned in the question I linked to, but does not do what I want because it starts a new line. I would like the list to start right beside the Theorem 2, but without the unnecessary space. – Mariano Suárez-Álvarez May 5 '11 at 9:41
• How is it unnessary? That is the default space between items in a list, that has nothing to do with theorems. If you whant to be able to configure this, see the enumitem package – daleif May 5 '11 at 9:43
• @daleif: well, I consider it unnecessary to add extra horizontal space between the end of the Theorem 2 text and the label of the first item in the list: it is inb that way that it is unnecessary! Also «see the enumitem package» does in no way help me... I have already seen it —I have been using for years now…— and I would not be asking if I knew how to achieve the effect I want to achieve. If you read the other answers, you'll notice that it is quite not a matter of using the please-mariano option to enumitem`, too. – Mariano Suárez-Álvarez May 5 '11 at 17:01
• @mariano ahh, then I misunderstood, I was thinking vertical – daleif May 5 '11 at 22:28


text:
# Common Signatures for Images

Some modules can be used for more than one task (e.g., image classification modules tend do to some feature extraction on the way). Therefore, each module provides (1) named signatures for all the tasks anticipated by the publisher, and (2) a default signature output = m(images) for its designated primary task.

## Image Feature Vector

### Usage summary

An image feature vector is a dense 1-D tensor that represents a whole image, typically for classification by the consumer model. (Unlike the intermediate activations of CNNs, it does not offer a spatial breakdown. Unlike image classification, it discards the classification learned by the publisher model.)

A module for image feature extraction has a default signature that maps a batch of images to a batch of feature vectors. It can be used like so:

  module_spec = hub.load_module_spec("path/to/module")
height, width = hub.get_expected_image_size(module_spec)
images = ...  # A batch of images with shape [batch_size, height, width, 3].
module = hub.Module(module_spec)
features = module(images)   # A batch with shape [batch_size, num_features].

It also defines the corresponding named signature.

### Signature specification

The named signature for extracting image feature vectors is invoked as

  outputs = module(dict(images=images), signature="image_feature_vector",
as_dict=True)
features = outputs["default"]

The input follows the general convention for input of images.

The outputs dictionary contains a "default" output of dtype float32 and shape [batch_size, num_features]. The batch_size is the same as in the input, but not known at graph construction time. num_features is a known, module-specific constant independent of input size.

These feature vectors are meant to be usable for classification with a simple feed-forward classifier (like the pooled features from the topmost convolutional layer in a typical CNN for image classification).

Applying dropout to the output features (or not) should be left to the module consumer. The module itself should not perform dropout on the actual outputs (even if it uses dropout internally in other places).

The outputs dictionary may provide further outputs, for example, the activations of hidden layers inside the module. Their keys and values are module-dependent. It is recommended to prefix architecture-dependent keys with an architecture name (e.g., to avoid confusing the intermediate layer "InceptionV3/Mixed_5c" with the topmost convolutional layer "InceptionV2/Mixed_5c").

## Image Classification

### Usage summary

Image classification maps the pixels of an image to linear scores (logits) for membership in the classes of a taxonomy selected by the module publisher. This allows consumers to draw conclusions from the particular classification learned by the publisher module, and not just its underlying features (cf. Image Feature Vector).

A module for image feature extraction has a default signature that maps a batch of images to a batch of logits. It can be used like so:

  module_spec = hub.load_module_spec("path/to/module")
height, width = hub.get_expected_image_size(module_spec)
images = ...  # A batch of images with shape [batch_size, height, width, 3].
module = hub.Module(module_spec)
logits = module(images)   # A batch with shape [batch_size, num_classes].

It also defines the corresponding named signature.

### Signature specification

The named signature for extracting image feature vectors is invoked as

  outputs = module(dict(images=images), signature="image_classification",
as_dict=True)
logits = outputs["default"]

The input follows the general convention for input of images.

The outputs dictionary contains a "default" output of dtype float32 and shape [batch_size, num_classes]. The batch_size is the same as in the input, but not known at graph construction time. num_classes is the number of classes in the classification, which is a known constant independent of input size.

Evaluating outputs["default"][i, c] yields a score predicting the membership of example i in the class with index c.

It depends on the underlying classification whether these scores are meant to be used with softmax (for mutually exclusive classes), sigmoid (for orthogonal classes), or something else. The module documentation should describe this, and refer to a definition of the class indices.

The outputs dictionary may provide further outputs, for example, the activations of hidden layers inside the module. Their keys and values are module-dependent. It is recommended to prefix architecture-dependent keys with an architecture name (e.g., to avoid confusing the intermediate layer "InceptionV3/Mixed_5c" with the topmost convolutional layer "InceptionV2/Mixed_5c").

## Image input

This is common to all types of image modules and image signatures.

A signature that takes a batch of images as input accepts them as a dense 4-D tensor of dtype float32 and shape [batch_size, height, width, 3] whose elements are RGB color values of pixels normalized to the range [0, 1]. This is what you get from tf.images.decode_*() followed by tf.image.convert_image_dtype(..., tf.float32).

A module with exactly one (or one principal) input of images uses the name "images" for this input.

The module accepts any batch_size, and correspondingly sets the first dimension of TensorInfo.tensor_shape to "unknown". The last dimension is fixed to the number 3 of RGB channels. The height and width dimensions are fixed to the expected size of input images. (Future work may remove that restriction for fully convolutional modules.)

Consumers of the module should not inspect the shape directly, but obtain the size information by calling hub.get_expected_image_size() on the module or module spec, and are expected to resize input images accordingly (typically before/during batching).

For simplicity, TF-Hub modules use the channels_last (or NHWC) layout of Tensors, and leave it to TensorFlow's graph optimizer to rewrite to channels_first (or NCHW) if needed. It has been doing that by default since TensorFlow version 1.7.


text:
[

# [

[ Mathematical Institute

24-29 St Giles’
Oxford OX1 3LB
UK
20 October 200626 January 2007
20 October 200626 January 2007
20 October 200626 January 2007
###### Abstract

We describe a simple criterion for showing that a group has Serre’s property FA. By exhibiting a certain pattern of finite subgroups, we show that this criterion is satisfied by and when .

property FA
automorphisms of free groups
–trees
\proceedingstitle

The Zieschang Gedenkschrift \conferencestart5 September 2007 \conferenceend8 September 2007 \conferencenameConference in honour of Heiner Zieschang \conferencelocationToulouse, France \editorMichel Boileau \givennameMichel \surnameBoileau \editorMartin Scharlemann \givennameMartin \surnameScharlemann \editorRichard Weidmann \givennameRichard \surnameWeidmann Configurations of torsion and Property ]A condition that prevents groups
from acting nontrivially on trees Martin R Bridson]Martin R Bridson \givennameMartin R \surnameBridson \urladdrhttp://www.maths.ox.ac.uk/ bridson \dedicatoryDedicated to the memory of Heiner Zieschang \volumenumber14 \issuenumber \publicationyear2008 \papernumber08 \startpage129 \endpage133 \doi \MR \Zbl \arxivreference \subjectprimarymsc200020E08 \subjectprimarymsc200020F65 \published29 April 2008 \publishedonline29 April 2008 \proposed \seconded \corresponding \version \makeautorefnamethmaTheorem \makeautorefnamelemmaaLemma

{asciiabstract}

We describe a simple criterion for showing that a group has Serre’s property FA. By exhibiting a certain pattern of finite subgroups, we show that this criterion is satisfied by Aut(F_n) and SL(n,Z) when n¿=3.

{htmlabstract}

We describe a simple criterion for showing that a group has Serre’s property FA. By exhibiting a certain pattern of finite subgroups, we show that this criterion is satisfied by Aut(F¡sub¿n¡/sub¿) and SL(n,¡b¿Z¡/b¿) when n≥ 3.

An –tree is a geodesic metric space in which there is a unique arc connecting each pair of points. A group is said to have property if for every action of by isometries on an –tree, the fixed point set is nonempty. Serre’s property FA is similar except that one considers only actions on simplicial trees. A group has FA if and only if it cannot be expressed as a nontrivial amalgamated free product or HNN extension.

###### Lemma 1.

Let be a group that is generated by the union of the subsets . If has property for all , then also has property .

###### Proof.

Let be connected subsets of an –tree. It is not difficult to see that if is nonempty for , then is nonempty (cf Serre [7, p 65]). Setting proves the lemma, since is assumed to be nonempty and . ∎

Every finite group has because the circumcentre of any –orbit in an –tree will be a fixed point.

###### Corollary 1 (The Triangle Criterion).

If is generated by and is finite for , then has property .

Let denote the automorphism group of the free group of rank .

###### Theorem 1.

If then and satisfy the Triangle Criterion and hence have property .

J-P Serre [7] was the first to prove that has FA if , and his argument shows that these groups actually have . Our argument is very similar to his except that he exploited the pattern of nilpotent subgroups rather than finite ones. In the light of a theorem of J Tits [8], Serre’s argument shows that all subgroups of finite index in have FA. In contrast, there is a subgroup of finite index that does not have property FA (see McCool [6]), while it is unknown if has such subgroups when . O Bogopolski [1] was the first to prove that has FA. M Culler and K Vogtmann [5] gave a short proof based on their idea of “minipotent” elements.

The obvious appeal of \fullrefmain lies in the final phrase, but the stronger fact that these groups satisfy the Triangle Criterion is useful in my work on fixed point theorems for actions of automorphism groups of free groups on higher-dimensional CAT spaces [2]. One can extend the theorem in various ways (cf \fullrefvariation) but I shall not present the details here as to do so would obscure the simple and transparent proof that has property FA, which is the main point of this note. I hope that it is a proof that Zieschang would have enjoyed.

#### Acknowledgements

This research was supported by Fellowships from the EPSRC and by a Royal Society Wolfson Research Merit Award.

## 1 Generating \rm{Aut}(Fn) and \rm{SL}(n,Z) by finite subgroups

We assume that and fix a basis of . For , let be the automorphism of that sends to and fixes the other basis elements. J Nielsen proved that is generated by the right Nielsen transformations and the involutions .

Let be the group generated by permutations111We shall write to denote the transposition of and . of . Conjugation by a permutation sends to and to . Therefore is generated by and . In particular is generated by and the subgroup generated by and the . (The action of on the abelianisation of gives a epimorphism , and the image of under this map is the group of monomial matrices.)

We write and for the subgroups corresponding to the sub-basis . Let , let and , and note that each is an involution. Define

 A1={εn,η}∪Σn−2,  A2={θ},  A3={τ}.
###### Lemma 1.1.

is generated by .

###### Proof.

Conjugating by we get , which conjugates to and to . Thus and are in the subgroup generated by the ; hence and are too. We already noted that is generated by and . ∎

###### Lemma 1.2.

The groups are finite.

###### Proof.

and commute with the involutions and , and has order , so . As , we have . And . ∎

These lemmas prove that satisfies the Triangle Criterion if , and by taking the images of the under the natural map we see that does too. When is odd, we obtain the corresponding result for by replacing the image of each by ; let denote the image of modified in this manner.

When is even we need to adjust the a little more. Let and note that is generated by the image of and the subgroup corresponding to the sub-basis . If then the groups and remain finite if we add to . Thus the sets , , demonstrate that satisfies the Triangle Criterion.

###### Remark.

If then by modifying the sets slightly one can also show that , the inverse image in of , satisfies the Triangle Criterion.

### 1.1 The geometry of the Hij

It would be unfair of me to leave the reader to guess the origin of the finite subgroups used in the above proof, so let me explain the geometry behind the construction.

Any finite subgroup of can be realised as a group of basepoint-preserving isometries of a graph of Euler characteristic . \fullreffigure222I am grateful to Karen Vogtmann for producing this figure. below gives such realisations for the groups . An important point is that if then cannot be realised as a group of symmetries of . I wanted to obtain the generating set that proved useful in my work with Karen Vogtmann [4]. Thus, starting with the rose and the graph for , I looked for a third graph where could be realised together with a symmetry intertwining and .

## 2 Variations on the theme

I have concentrated on configurations of finite subgroups in this note but \fullreflem1 can also be applied to situations where the subgroups are infinite. For example, if lies in the commutator subgroup of its centralizer, then will be nonempty whenever acts by isometries on an –tree. (This is a special instance of a general fact about semisimple actions on CAT spaces [3].) By exploiting such facts in conjunction with \fullreflem1 one can prove, for example, that the mapping class group of a surface of genus at least 3 has property , a result first proved in [5].

One can also strengthen \fullreflem1 using an argument due to J-P Serre [7, p 64]: it suffices to require that the have and, in any action of on an –tree, that have a fixed point, for every and . To see this, one reduces to the case and argues that if the fixed point sets of and did not intersect then the point of closest to would be fixed by all with and , which is a contradiction.

A quite different strengthening begins with the observation that the behaviour of convex sets described in the proof of \fullreflem1 is a manifestation of the fact that trees are –dimensional objects. A suitable version of Helly’s Theorem provides constraints on the way in which convex sets can intersect in higher-dimensional CAT spaces, and by applying these constraints to the fixed point sets of finite subgroups one can prove far-reaching generalisations of \fullrefmain; this is the theme of [2].

## References

• [1] O V Bogopol’skiĭ, Arborial decomposability of groups of automorphisms of free groups, Algebra i Logika 26 (1987) 131–149, 271 \xoxMR964922
• [2] M R Bridson, Helly’s theorem, CAT spaces, and actions of automorphism groups of free groups, preprint (2007)
• [3] M R Bridson, A Haefliger, Metric spaces of nonpositive curvature, Grundlehren series 319, Springer, Berlin (1999) \xoxMR1744486
• [4] M R Bridson, K Vogtmann, Automorphisms of automorphism groups of free groups, J. Algebra 229 (2000) 785–792 \xoxMR1769698
• [5] M Culler, K Vogtmann, A group-theoretic criterion for property , Proc. Amer. Math. Soc. 124 (1996) 677–683 \xoxMR1307506
• [6] J McCool, A faithful polynomial representation of , Math. Proc. Cambridge Philos. Soc. 106 (1989) 207–213 \xoxMR1002533
• [7] J-P Serre, Trees, Springer Monographs in Mathematics, Springer, Berlin (2003) \xoxMR1954121  Translated from the French original by John Stillwell, Corrected 2nd printing of the 1980 English translation
• [8] J Tits, Systèmes générateurs de groupes de congruence, C. R. Acad. Sci. Paris Sér. A-B 283 (1976) Ai, A693–A695 \xoxMR0424966
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text:
# cabal-bounds: A command line program for managing the bounds/versions of the dependencies in a cabal file.

[ bsd3, development, library, program, utils ] [ Propose Tags ]

A command line program for managing the bounds/versions of the dependencies in a cabal file.

cabal-bounds is able to do these things with the bounds of the dependencies in the cabal file:

• drop them

• update them by the library versions of the current cabal build

• update them by the library versions of a haskell platform release

• update them by the library versions specified by a file

• dump the libraries/dependencies and their lower bound versions from the cabal file(s) into a file

Installation

You have to ensure, that the Cabal library of cabal-bounds matches the one used by the cabal binary:

$cabal --version cabal-install version 1.18.0.2 using version 1.18.1 of the Cabal library $ cabal install --constraint="Cabal == 1.18.1" cabal-bounds

If you update the cabal binary and the used Cabal library changes, then you have to rebuild cabal-bounds.

Issues

Perhaps the currently most annoying thing is, that you have to live with the reformating of your cabal file done by the pretty printer of the Cabal library.

To reformat your cabal file without changing any bounds you can call cabal-bounds with the name of a section that isn't present in the cabal file:

$cabal-bounds drop --executable=blub myproject.cabal [Skip to Readme] Versions [faq] 0.1.2, 0.1.3, 0.1.4, 0.1.5, 0.1.6, 0.1.7, 0.1.8, 0.1.9, 0.1.10, 0.1.11, 0.1.12, 0.1.13, 0.1.14, 0.1.15, 0.1.16, 0.2, 0.2.1, 0.3, 0.4, 0.4.1, 0.5, 0.6, 0.7, 0.8, 0.8.1, 0.8.2, 0.8.3, 0.8.4, 0.8.5, 0.8.6, 0.8.7, 0.8.8, 0.8.9, 0.9, 0.9.1, 0.9.2, 0.9.3, 0.9.4, 0.9.5, 1.0.0, 1.0.1, 1.0.2, 1.0.3, 1.0.4, 1.0.5, 1.1.0, 1.2.0, 1.3.0, 1.4.0, 1.5.0, 1.6.0, 2.0.0, 2.1.0, 2.1.1, 2.2.0, 2.2.1 (info) CHANGELOG base (>=3 && <5), Cabal (>=1.18.0 && <1.23), cabal-bounds, cabal-lenses (>=0.4.4 && <0.5), cmdargs (>=0.10.5 && <0.11), either (>=4.1.1 && <4.4), lens (>=4.0.1 && <4.9), strict (>=0.3.2 && <0.4), transformers (>=0.3.0.0 && <0.5), unordered-containers (>=0.2.3.3 && <0.3) [details] BSD-3-Clause Daniel Trstenjak daniel.trstenjak@gmail.com Utils, Development head: git clone https://github.com/dan-t/cabal-bounds by DanielTrstenjak at Fri Apr 3 15:36:46 UTC 2015 NixOS:2.2.1 cabal-bounds 17904 total (830 in the last 30 days) 2.0 (votes: 1) [estimated by rule of succession] λ λ λ Docs available Last success reported on 2015-04-03 ## Modules [Index] • CabalBounds ## Downloads #### Maintainer's Corner For package maintainers and hackage trustees ## Readme for cabal-bounds-0.9.3 [back to package description] # cabal-bounds A command line program for managing the bounds/versions of the dependencies in a cabal file. cabal-bounds is able to do these things with the bounds of the dependencies in the cabal file: • drop them • update them by the library versions of the current cabal build • update them by the library versions of a haskell platform release • update them by the library versions specified by a file • dump the libraries/dependencies and their lower bound versions from the cabal file(s) into a file # Example: Raise the Upper Bounds If you have several cabalized projects, then it can be quite time consuming to keep the bounds of your dependencies up to date. Especially if you're following the package versioning policy, then you want to raise your upper bounds from time to time, to allow the building with newer versions of the dependencies. cabal-bounds tries to automate this update process to some degree. So a typical update process might look like: # update the version infos of all libraries$> cabal update

# drops the upper bound of all dependencies in 'myproject.cabal', most likely you want to ignore 'base'
$> cabal-bounds drop --upper --ignore=base myproject.cabal # create a cabal sandbox for building of 'myproject'$> cabal sandbox init

# build 'myproject'
$> cabal install # update the upper bound of all dependencies in 'myproject.cabal' by the cabal build information$> cabal-bounds update --upper --ignore=base myproject.cabal dist/dist-sandbox-*/setup-config

# Example: Update Bounds by Haskell Platform

Ensuring that your project builds with the current haskell platform - or perhaps the last two ones - can make it, especially for beginners, a lot easier to build your project.

cabal-bounds supports the updating of the bounds by the library versions of a specific haskell platform release.

To update the bounds to the haskell platform 2013.2.0.0:

$> cabal-bounds update --haskell-platform=2013.2.0.0 myproject.cabal  There're two additional symbolic names for specifying a haskell platform release: current and previous. So one use case might be to initialize the bounds to library versions used by a haskell platform release, test if your project builds and works with these, and then raise the upper bounds to the newest available versions: # intialize the bounds to the previous haskell platform release$> cabal-bounds update --ignore=base --haskell-platform=previous myproject.cabal

# build and test the project

# initialize the lower bounds of libraries not present in the haskell platform
$> cabal-bounds update --lower --missing --ignore=base myproject.cabal dist/dist-sandbox-*/setup-config # drop the upper bounds to test your project with the newest available library versions$> cabal-bounds drop --upper --ignore=base myproject.cabal

# build and test the project

# set the upper bounds to the ones used in the current build
$> cabal-bounds update --upper --ignore=base myproject.cabal dist/dist-sandbox-*/setup-config  If you specify a haskell platform release and a setup config file at once, then the setup config library verions are only used for the libraries not present in the haskell platform release. # Example: Update Bounds by File It's also possible to update the bounds by library versions specified in a file: $> cabal-bounds update --fromfile=libs.hs myproject.cabal

The libs.hs file has to be of the format:

[ ("libA", [0,2,1]), ("libB", [2,1]), ("libC", [1]) ]

If you specify a library file, a haskell platform release and a setup config file at once, then first the haskell platform libraries and versions are considered, then the library file and at the end the setup config file.

The library file can be created by the dump command:

$> cabal-bounds dump --output=libs.hs myproject.cabal  The dump command will dump dependencies with their lower bound version. The command can take multiple cabal files. If the same dependencies is present in multiple files, then the lowest lower bound version is taken. # Example: Bound Changes The => shows what the result is of the operation for every dependency. Left is the dependency before calling the command, right the one after calling. $> cabal-bounds drop ...
lens >=4.0.1 && <4.1   =>   lens

$> cabal-bounds drop --upper ... lens >=4.0.1 && <4.1 => lens >=4.0.1  If the cabal build (the setup-config) uses lens 4.1.2, then the results of the update command would be: $> cabal-bounds update ...
lens >=4.0.1 && <4.1   =>   lens >=4.1.2 && <4.2
lens                   =>   lens >=4.1.2 && <4.2

$> cabal-bounds update --lower ... lens >=4.0.1 && <5 => lens >=4.1.2 && <5 lens >=4.0.1 && <4.1 => lens >=4.1.2 lens <4.1 => lens >=4.1.2 lens => lens >=4.1.2$> cabal-bounds update --upper ...
lens >=4.0.1 && <4.1   =>   lens >=4.0.1 && <4.2
lens >=4.0.1           =>   lens >=4.0.1 && <4.2
lens                   =>   lens <4.2

You can also specify which component of the version number should be updated:

$> cabal-bounds update --lowercomp=minor ... lens >=4.0.1 && <4.1 => lens >=4.1.2$> cabal-bounds update --lowercomp=major2 ...
lens >=4.0.1 && <4.1   =>   lens >=4.1

$> cabal-bounds update --lowercomp=major1 ... lens >=4.0.1 && <4.1 => lens >=4 && <4.1$> cabal-bounds update --uppercomp=minor ...
lens >=4.0.1 && <4.1   =>   lens >=4.0.1 && <4.1.3

$> cabal-bounds update --uppercomp=major2 ... lens >=4.0.1 && <4.1 => lens >=4.0.1 && <4.2$> cabal-bounds update --uppercomp=major1 ...
lens >=4.0.1 && <4.1   =>   lens >=4.0.1 && <5

# Options

You can restrict the modification to certain sections in the cabal file by specifing the type and the name of the section:

• --library
• --executable=name
• --testsuite=name
• --benchmark=name

If you omit these options, then all sections are considered and modified.

You can also restrict the modification of dependencies by specifing which dependencies should only or shouldn't be modified:

• --only=name
• --ignore=name

You can also only update the dependencies without a bound by specifying the --missing flag. If you omit these options, then all dependencies are considered and modified.

All options taking a name can be specified multiple times: e.g. --executable=exe1 --executable=exe2 or --ignore=base --ignore=whatever

Please consult cabal-bounds --help for a complete list of options.

# Installation

You have to ensure, that the Cabal library of cabal-bounds matches the one used by the cabal binary:

$> cabal --version cabal-install version 1.18.0.2 using version 1.18.1 of the Cabal library$> cabal install --constraint="Cabal == 1.18.1" cabal-bounds

If you update the cabal binary and the used Cabal library changes, then you have to rebuild cabal-bounds.

# Issues

Perhaps the currently most annoying thing is, that you have to live with the reformating of your cabal file done by the pretty printer of the Cabal library.

To reformat your cabal file without changing any bounds you can call cabal-bounds with the name of a section that isn't present in the cabal file:

\$> cabal-bounds drop --executable=blub myproject.cabal


text:
# Trig function of arc trig functions and the reverse

#### Philosophaie

I know the sin(arccos(x)) = (1-x^2)^0.5

I was wondering what some of the others are:

cos(arcsin(X))
tan(arcsin(X))
tan(arccos(x))
sin(arctan(x))
cos(arctan(x))

also the reverse:

arcsin(cos(x))
arcsin(tan(X))
arccos(Sin(X))
arccos(tan(X))
arctan(sin(X))
arctan(cos(X))

#### micromass

Let me do two:

$$\cos(arcsin(x))=\sqrt{1-\sin(arcsin(x))}=\sqrt{1-x^2}$$

and

$$\tan(arccos(x))=\sqrt{\frac{1}{\cos^2(arccos(x))}-1}=\sqrt{\frac{1}{x^2}-1}$$

I'll let you find out the other ones...

#### gsal

are these always possible? I mean take

arccos(tan(x))

for example, the cosine of an angle is always between 0 and 1, and so, the argument to the arccos function should be a number between 0 and 1...but the tangent of an angle can get pretty large...so, I think these is no solution here...same for others.

#### LCKurtz

Homework Helper
Gold Member
Let me do two:

$$\tan(arccos(x))=\sqrt{\frac{1}{\cos^2(arccos(x))}-1}=\sqrt{\frac{1}{x^2}-1}$$
Not if arccos(x) is in the second quadrant.

#### HallsofIvy

Homework Helper
As long as your "angles" are in the first quadrant (so you don't have multi-value problems), you can get all of those formulas by constructing an appropriate right triangle.

For example, to get sin(arctan(x)), imagine a right triangle with "opposite side" x and "near side" 1 (so that the tangent of the angle opposite side "x" is x/1= x and the angle is arctan(x)). By the Pythagorean theorem, it will have "hypotenuse" $\sqrt{x^2+ 1}$. Sine is "opposite side over hypotenuse" so $sin(arctan(x))= \frac{x}{\sqrt{x^2+ 1}}$.

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text:
# Properties

 Label 2001.1.i.b Level 2001 Weight 1 Character orbit 2001.i Analytic conductor 0.999 Analytic rank 0 Dimension 2 Projective image $$D_{4}$$ CM disc. -23 Inner twists 4

# Related objects

## Newspace parameters

 Level: $$N$$ = $$2001 = 3 \cdot 23 \cdot 29$$ Weight: $$k$$ = $$1$$ Character orbit: $$[\chi]$$ = 2001.i (of order $$4$$ and degree $$2$$)

## Newform invariants

 Self dual: No Analytic conductor: $$0.998629090279$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(i)$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Projective image $$D_{4}$$ Projective field Galois closure of 4.0.116116029.1

## $q$-expansion

The $$q$$-expansion and trace form are shown below.

 $$f(q)$$ $$=$$ $$q$$ $$+ ( 1 - i ) q^{2}$$ $$+ i q^{3}$$ $$-i q^{4}$$ $$+ ( 1 + i ) q^{6}$$ $$- q^{9}$$ $$+O(q^{10})$$ $$q$$ $$+ ( 1 - i ) q^{2}$$ $$+ i q^{3}$$ $$-i q^{4}$$ $$+ ( 1 + i ) q^{6}$$ $$- q^{9}$$ $$+ q^{12}$$ $$+ 2 i q^{13}$$ $$+ q^{16}$$ $$+ ( -1 + i ) q^{18}$$ $$+ i q^{23}$$ $$+ q^{25}$$ $$+ ( 2 + 2 i ) q^{26}$$ $$-i q^{27}$$ $$-i q^{29}$$ $$+ ( -1 - i ) q^{31}$$ $$+ ( 1 - i ) q^{32}$$ $$+ i q^{36}$$ $$-2 q^{39}$$ $$+ ( -1 - i ) q^{41}$$ $$+ ( 1 + i ) q^{46}$$ $$+ ( 1 + i ) q^{47}$$ $$+ i q^{48}$$ $$+ q^{49}$$ $$+ ( 1 - i ) q^{50}$$ $$+ 2 q^{52}$$ $$+ ( -1 - i ) q^{54}$$ $$+ ( -1 - i ) q^{58}$$ $$-2 i q^{59}$$ $$-2 q^{62}$$ $$-i q^{64}$$ $$- q^{69}$$ $$+ ( -1 + i ) q^{73}$$ $$+ i q^{75}$$ $$+ ( -2 + 2 i ) q^{78}$$ $$+ q^{81}$$ $$-2 q^{82}$$ $$+ q^{87}$$ $$+ q^{92}$$ $$+ ( 1 - i ) q^{93}$$ $$+ 2 q^{94}$$ $$+ ( 1 + i ) q^{96}$$ $$+ ( 1 - i ) q^{98}$$ $$+O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q$$ $$\mathstrut +\mathstrut 2q^{2}$$ $$\mathstrut +\mathstrut 2q^{6}$$ $$\mathstrut -\mathstrut 2q^{9}$$ $$\mathstrut +\mathstrut O(q^{10})$$ $$2q$$ $$\mathstrut +\mathstrut 2q^{2}$$ $$\mathstrut +\mathstrut 2q^{6}$$ $$\mathstrut -\mathstrut 2q^{9}$$ $$\mathstrut +\mathstrut 2q^{12}$$ $$\mathstrut +\mathstrut 2q^{16}$$ $$\mathstrut -\mathstrut 2q^{18}$$ $$\mathstrut +\mathstrut 2q^{25}$$ $$\mathstrut +\mathstrut 4q^{26}$$ $$\mathstrut -\mathstrut 2q^{31}$$ $$\mathstrut +\mathstrut 2q^{32}$$ $$\mathstrut -\mathstrut 4q^{39}$$ $$\mathstrut -\mathstrut 2q^{41}$$ $$\mathstrut +\mathstrut 2q^{46}$$ $$\mathstrut +\mathstrut 2q^{47}$$ $$\mathstrut +\mathstrut 2q^{49}$$ $$\mathstrut +\mathstrut 2q^{50}$$ $$\mathstrut +\mathstrut 4q^{52}$$ $$\mathstrut -\mathstrut 2q^{54}$$ $$\mathstrut -\mathstrut 2q^{58}$$ $$\mathstrut -\mathstrut 4q^{62}$$ $$\mathstrut -\mathstrut 2q^{69}$$ $$\mathstrut -\mathstrut 2q^{73}$$ $$\mathstrut -\mathstrut 4q^{78}$$ $$\mathstrut +\mathstrut 2q^{81}$$ $$\mathstrut -\mathstrut 4q^{82}$$ $$\mathstrut +\mathstrut 2q^{87}$$ $$\mathstrut +\mathstrut 2q^{92}$$ $$\mathstrut +\mathstrut 2q^{93}$$ $$\mathstrut +\mathstrut 4q^{94}$$ $$\mathstrut +\mathstrut 2q^{96}$$ $$\mathstrut +\mathstrut 2q^{98}$$ $$\mathstrut +\mathstrut O(q^{100})$$

## Character Values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/2001\mathbb{Z}\right)^\times$$.

 $$n$$ $$553$$ $$668$$ $$1132$$ $$\chi(n)$$ $$i$$ $$-1$$ $$-1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1172.1
 1.00000i − 1.00000i
1.00000 1.00000i 1.00000i 1.00000i 0 1.00000 + 1.00000i 0 0 −1.00000 0
1931.1 1.00000 + 1.00000i 1.00000i 1.00000i 0 1.00000 1.00000i 0 0 −1.00000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char. orbit Parity Mult. Self Twist Proved
1.a Even 1 trivial yes
23.b Odd 1 CM by $$\Q(\sqrt{-23})$$ yes
87.f Even 1 yes
2001.i Odd 1 yes

## Hecke kernels

This newform can be constructed as the kernel of the linear operator $$T_{2}^{2}$$ $$\mathstrut -\mathstrut 2 T_{2}$$ $$\mathstrut +\mathstrut 2$$ acting on $$S_{1}^{\mathrm{new}}(2001, [\chi])$$.


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Synopsis: Reluctant swimmers

The swimming motion of organisms in complex fluids is hindered by elasticity.

Microorganisms routinely live in complex fluids such as mud, intestinal fluid, and mucus. In this world of varying rheological properties, organisms propagate with a motion analogous to swimming. Think of spermatozoa propelled by undulations of flagella. Researchers have known that the properties of a fluid strongly affect how the organisms swim in it—the beat of a spermatozoon’s flagella, for example, goes from sinusoidal in an “ordinary” Newtonian fluid to asymmetric in a viscoelastic fluid, a medium that shows both solid- and fluid-like behavior—but quantitative studies have been scarce.

In a paper in Physical Review Letters, Xiaoning Shen and Paulo E. Arratia, at the Department of Mechanical Engineering and Applied Mechanics, University of Pennsylvania, report usage of video microscopy and velocimetry methods to measure body kinematics and swimming dynamics of a species of roundworm in a range of shear viscosities and viscoelasticities. They find that the fluid elasticity hinders swimming speed as well as efficiency. This result is surprising, if only because myriad biological processes occur in such fluids. Examples range from sperms in the female reproductive tract to allergens in the respiratory passage. The authors find that the roundworm’s swimming speed is reduced by up to $35%$ compared to that in ordinary fluids. This work provides a valuable experimental benchmark for what appears to be a ubiquitous phenomenon. – Sami Mitra

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Viewpoint: Particles Move to the Beat of a Microfluidic Drum

A thin vibrating plate can organize microscopic particles within a liquid into different patterns, an effect like that observed in 18th century studies of musical instruments. Read More »


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Welcome to Windows 7 Forums. Our forum is dedicated to helping you find support and solutions for any problems regarding your Windows 7 PC be it Dell, HP, Acer, Asus or a custom build. We also provide an extensive Windows 7 tutorial section that covers a wide range of tips and tricks.

# Windows 7: How to get the current date automatically?

 19 Jan 2015 #1 sujith123 64 bit 3 posts How to get the current date automatically? Hi All, I have the following Copy Command which copies files from one location to another XCOPY D:\Test C:\Test /i Now i would like to have the files copied from D to C everyday based on the date for today, i googled around and found this one which is working fine XCOPY D:\Test C:\Test /D:01-19-2015. The thing here is i want the date to be automated as i would like to schedule a job for this one which would copy the files daily automated . Can someone please help me with this? PLease let me know if you have any questions. Thanks My System Specs
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 19 Jan 2015 #2 Bugija w7x64 2 posts Hi sujith try this xcopy "D:\Test" "C:\Test %DATE:~7,2% %DATE:~4,2% %DATE:~-4%" /y /i /h /s /e /s /e - recursive copy, including copying empty directories. /h - copy system and hidden files. /y - don't prompt before overwriting existing files. My System Specs
 19 Jan 2015 #3 Pyprohly Windows 10, Windows 8.1 Pro, Windows 7 Professional, OS X El Capitan 722 posts In a house Hello Sujith, In batch scripts there is a predefined environmental variable that expands to the current date. It is %DATE%. The variable includes an abbreviated weekday name at the beginning, so some string slicing needs to take place before implementing the variable into your command. Code: Xcopy D:\Test C:\Test /d:%DATE:~4% EDIT: The above will only work if your locale is set to display dates in the form of MM/DD/YYYY An alternate way of doing things, that does not require the %DATE% variable, would be to loop off the results of Forfiles. Code: Forfiles /p D:\Test /d 0 /c "XCOPY @PATH C:\Test" Now, to schedule a batch file--or any application/executable really--you'd use Task Scheduler. The following is a command that will create your described task to be scheduled for you. Execute the command in an elevated command prompt. Code: At 00:00 /every:M,T,W,TU,F,S,SU "C:\path\to\batch\script.bat" This will run the batch program for every day of the week at midnight. My System Specs
 .
 20 Jan 2015 #4 sujith123 64 bit 3 posts Hi Guys, Thanks a lot for your reply, I made it work using the following code set curr_date=%DATE:~4,2%/%DATE:~7,2%/%DATE:~10,4% & XCOPY C:\Test D:\Test /d:%curr_date% My System Specs
 20 Jan 2015 #5 Pyprohly Windows 10, Windows 8.1 Pro, Windows 7 Professional, OS X El Capitan 722 posts In a house Um... Have you tested that, Sujith? You might find it does not do what you want. In batch, if you're going to assign a variable and use the it immediately on the same line, you have to enable delayed expansion, or the variable will expand to an empty string. So the actual Xcopy command that you are triggering each time is: Code: XCOPY C:\Test D:\Test /d: Which means to append files from C:\Test to D:\Test, only if the source file is newer than than the destination file. 　 But you asked for... Code: XCOPY C:\Test D:\Test /d:01/21/2015 (The date should vary and be the current date^^) Which says to copy files changed on or after 01/21/2015 (the current date etc.). Sujith, to avoid an empty string value for "/d:", you should either put both of your commands on two separate lines: Code: set curr_date=%DATE:~4,2%/%DATE:~7,2%/%DATE:~10,4% XCOPY C:\Test D:\Test /d:%curr_date% Or enable and use delayed expansion, like as follows: Code: setlocal EnableDelayedExpansion set curr_date=%DATE:~4,2%/%DATE:~7,2%/%DATE:~10,4% & XCOPY C:\Test D:\Test /d:!curr_date! But because you are situated in the USA (where dates are in the form of MM/DD/YYYY), Sujith, I suggest simply using: Code: XCOPY D:\Test C:\Test /d:%DATE:~4% My System Specs
 21 Jan 2015 #6 sujith123 64 bit 3 posts Hi Pyprohly, Thanks a lot for taking your valuable time and replying to me...i was using my old method in the batch file it was running fine but i was using it in some sql job and it was failing for no reason, but now i changed the code of my batch file to be using this in the end /d:%DATE:~4% and even my job is also running fine, Thanks a lot for your help. My System Specs
 21 Jan 2015 #7 Pyprohly Windows 10, Windows 8.1 Pro, Windows 7 Professional, OS X El Capitan 722 posts In a house You're most welcome. And speaking of welcome, welcome to SF. Be sure to make a visit should you need any assistance at all (regarding Windows that is ). My System Specs


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# Probability Problem

Hi from Melbourne.

I have a rudimentary probability question that you will probably find amusing but hopefully you can also help me!

I am studying construction planning and this question is taken from an old exam paper.

## Homework Statement

"Risk analysis: At 5% level of significance, what is the probability of completing the project at or earlier than the employer's desired duration in the question above.

The question above requires that project is completed 7 days earlier than the normal duration, μ

Where:
Mean, μ = 50
Required duration, D = 43
Variance, σ² = 3.06

## Homework Equations

I assume Z = D - μ / √σ²

## The Attempt at a Solution

I thought the answer would be z=43-50/√3.06
This gives 4.001, would this mean there is a less than 1% probability of the project being completed in 43 days?

On further reading I also discovered that a 5% level of significance is the same as a 95% confidence interval? If this is the case, doesnt this make the z value 1.645? In which case none of the variables for the equation are unknown.

As you can see I have got myself very confused :

Any and all help will be greatly appreciated. I'll have this forum open all day (and night)!

Mark44
Mentor
Hi from Melbourne.

I have a rudimentary probability question that you will probably find amusing but hopefully you can also help me!

I am studying construction planning and this question is taken from an old exam paper.

## Homework Statement

"Risk analysis: At 5% level of significance, what is the probability of completing the project at or earlier than the employer's desired duration in the question above.

The question above requires that project is completed 7 days earlier than the normal duration, μ

Where:
Mean, μ = 50
Required duration, D = 43
Variance, σ² = 3.06

## Homework Equations

I assume Z = D - μ / √σ²
You need parentheses here.

z = (x - μ )/ σ

## The Attempt at a Solution

I thought the answer would be z=43-50/√3.06
You need parentheses here, too, otherwise this will be interpreted as 43 - (50/√3.06
).

The answer to the question is not a value of z - it's the probability of completing the project at or earlier than the employer's desired duration (43 days).
This gives 4.001, would this mean there is a less than 1% probability of the project being completed in 43 days?
Wouldn't this give you a z value of -4.001? Where are you getting a probability of 1%?
On further reading I also discovered that a 5% level of significance is the same as a 95% confidence interval?
Yes.
If this is the case, doesnt this make the z value 1.645? In which case none of the variables for the equation are unknown.
The mean length of jobs is 50 days, with a standard deviation of 3.06 days. What fraction of jobs are done in 43 days or fewer? To answer this question look at the standard normal distribution to see how much area under the curve corresponds to your (corrected) z value.