text:
Chapter 4:- Control Volume Analysis Using Energy¶

Example 4.1 Page no-125

In [1]:
# Given:-
# At inlet 1:-
p1= 7.0                                        # pressure in bar
T2= 200.0                                      # temperature in degree celcius
m1dot= 40.0                                    # mass flow rate in kg/s

# At inlet 2:-
p2= 7.0                                        # pressure in bar
T2= 40.0                                       # temperature in degree celcius
A2= 25.0                                       # area in cm^2

# At exit:-
p3= 7.0                                        # pressure in bar
AV3= 0.06                                      # Volumetric flow rate through wxir in m^3/s

# From table A-3
v3 = (1.108)*(10**(-3))                        # specific volume at the exit in m^3/kg

# from table A-2
v2= (1.0078)*(10**(-3))                        # specific volume in state 2 in m^3/kg

# Calculation:-
m3dot= AV3/v3                                  # mass flow rate at exit
m2dot = m3dot-m1dot                            # mass flow rate at inlet 2
V2= (m2dot*v2)/(A2*(10**(-4)))

# Results:-
print '-> The mass flow rate at the inlet 2 is', round(m2dot,2),' kg/s.'
print '-> The mass flow rate at the exit is', round(m3dot,2),' kg/s.'
print '-> The velocity at the inlet is ', round(V2,2),'m/s.'

-> The mass flow rate at the inlet 2 is 14.15  kg/s.
-> The mass flow rate at the exit is 54.15  kg/s.
-> The velocity at the inlet is  5.7 m/s.

Example 4.3 Page no-135

In [2]:
# Given:-
p1= 40.0                                     # pressure in bar
T1= 400.0                                    # temperature in degree celcius
V1= 10.0                                     # velocity m/s

# At exit:-
p2= 10.0                                     # pressure in bar
V2= 665.0                                    # velocity in m/s
mdot= 2.0                                    # mass flow rate in kg/s

# From table A-4
h1= 3213.6                                   # snpecific enthalpy in kJ/kg
v2 = 0.1627                                  # specific volume at the exit in m^3/kg

# Calculation:-
h2 = h1 + ((V1**2-V2**2)/2)/1000             # snpecific enthalpy in kJ/kg
A2=(mdot*v2)/V2                              # Exit area

# Results:-
print '-> The exit Area of the nozzle is', round(A2,4),'m^2.'

-> The exit Area of the nozzle is 0.0005 m^2.

Example 4.4 Page no-138

In [3]:
# Given:-
m1dot = 4600.0                                    # mass flow rate in kg/h
Wcvdot= 1000.0                                    # turbine power output in kv
p1= 60.0                                          # pressure in bar
T1=400.0                                          # temperature in degree celc
V1= 10.0                                          # velocity in m/s

# At exit:-
p2= 0.10                                          # pressure in bar
q2= 0.90                                          # quality
V2= 50.0                                          # velocity in m/s

# From table A-2 and A-3:-
h1= 3177.2                                        # specific enthalpy at inlet in kJ/kg
hf2= 191.83
hg2= 2584.63

# Calculation:-
h2 = hf2+q2*(hg2-hf2)                             # specific enthalpy at exit in kJ/kg
Qcvdot = Wcvdot + m1dot*((h2-h1)+(V2**2- V1**2)/(2*1000))/3600

# Results:-
print '-> The rate of heat transfer between the turbine and surroundings is',round(Qcvdot,2),'kW.'

-> The rate of heat transfer between the turbine and surroundings is -61.39 kW.

Example 4.5 Page no-140

In [4]:
# Given:-
p1=1.00                                         # pressure in bar
t1= 290.00                                      # temperature in kelvin
A1= 0.1                                         # area in m^2
V1= 6.00                                        # velocity in m/s

# At exit:-

p2=7.00                                         # pressure in bar
t2= 450.00                                      # temperature in kelvin
V2= 2.00                                        # velocity in m/s
Qcvdot= -180.0                                  # heat transfer rate in kJ/min
R= 8.314                                        # universal gas constant in SI units

# from table A-22

h1= 290.16                                      # specific enthalpy in kJ/kg
h2= 451.8                                       # specific enthalpy in kJ/kg

# Calculations:-

v1 =  (R*1000*t1)/(28.97*p1*10**5)              # specific volume
mdot=(A1*V1)/v1                                 # mass flow rate
Wcvdot = Qcvdot/60 + mdot*((h1-h2)+(V1**2-V2**2)/(2*1000))

# Results:-

print '-> The power input to the compressor is ',round(Wcvdot,2),'kW.'

-> The power input to the compressor is  -119.52 kW.

Example 4.6 Page no-141

In [5]:
# Given:-
# At Entry:=
t1=20.0                                          # Temperatue in deg celcius
p1=1.0                                           # pressure in atm
AV1= 0.1                                         # volumetric flow rate in litre/s
D1=2.5                                           # Diameter of th hose in cm

# At Exit:=
t2=23.0                                          # temperatuer in deg celcius
p2=1.0                                           # pressure in atm
V2=50.0                                          # Velocity in m/s
Z2=5.0                                           # elevation in m
g= 9.8                                           # acceleration due to gravity in m/s^2

# from table A-2 and A-19:-

v= (1.0018)*((10.0)**(-3))                       # specific volume in m^3/kg
c= 4.18

# Calculation:-
mdot = (AV1/1000)/v                              # mass flow rate in kg/s
V1= (AV1/1000)/(3.14*(D1/(2*100))**2)            # Entry velocity in m/s
deltah = c*(t2-t1)+v*(p2-p1)
Wcvdot= ((mdot*10)/9)*(-deltah+(V1**2-V2**2)/(2*1000)+g*(0-Z2)/1000)

# Results:-
print '-> The power input to the motor is', round(Wcvdot,2),'KW.'

-> The power input to the motor is -1.53 KW.

Example 4.7:- Page no-144

In [6]:
# Given:-
# Entering:-
p1=0.1                                        # pressure in bar
x1= 0.95                                      # Quality
p2= 0.1                                       # pressure in bar
t2= 45.0                                        # temperature in deg celcius
t3=20.0                                         # temperature of cooling entry in deg cel
t4=35.0                                         # temperature of cooling exit

# From table A-3
hf= 191.53                                    # Enthalpy in KJ/kg
hg= 2584.7                                    # Enthalpy in KJ/kg
h2=188.45                                     # Assumption at states 2,3 and 4, h is approx equal to hf(T), in kJ/kg
deltah4_3= 62.7                               # Assumption 4, in kJ/kg

# Calculations:-
h1= hf + x1*(hg-hf)
ratio= (h1-h2)/(deltah4_3)
QRate= (h2-h1)                                # Part B

# Results:-
print '-> The rate of the mass flow rate of the cooling water to the mass flow rate of the condenstaing stream is (m3dot/m1dot)',round(ratio,2)
print '-> The rate of energy transfer from the condensing steam to the cooling water of the steam passing through the condenser is',round(QRate,2),'kJ/kg.'

-> The rate of the mass flow rate of the cooling water to the mass flow rate of the condenstaing stream is (m3dot/m1dot) 36.31
-> The rate of energy transfer from the condensing steam to the cooling water of the steam passing through the condenser is -2276.59 kJ/kg.

Example 4.8:- Page no-146

In [7]:
# Given:-
T1 = 293.0                                            # In kelvin
P1= 1.01325 * (10**5)                                 # In pascal
V1max= 1.3                                            # maximum velocity of entering air in m/s
T2max= 305.0                                          # maximum temperature at the exit in kelvin
pec= -80.0                                            # power received by electronic components in watt
Pf= -18.0                                             # Power received by fan in watt
R= 8.314                                              # Universal gas constant
M= 28.97*(10**(-3))                                   # Molar mass of air in kg
Qcvdot=0                                              # Heat transfer from the outer surface of the electronics enclosure to the surroundings is negligible.
Cp= 1.005*(10**3)                                     # in j/kg*k
pi=3.14

# Calculations:-

Wcvdot = pec +Pf                                      # total electric power provided to electronic components and fan in watt
mdotmin=  (-Wcvdot)/(Cp*(T2max-T1))                   # minimum mass flow rate
v1= ((R/M)*T1)/P1                                     # specific volume
A1min = (mdotmin*v1)/V1max
D1min = (4*A1min/(pi))**(0.5)

# Results:-
print '-> The smallest fan inlet diameter is',round(D1min*100,2),'cm.'

-> The smallest fan inlet diameter is 8.13 cm.

Example 4.9 Page no-148

In [8]:
# Given:-
P1 = 20.0                             # pressure in supply line in bars
P2 = 1.0                              # exhaust pressure in bar
T2 = 120.0                            # exhaust temperature in degree celcius

# from table A-3 at 20 bars
hf1 = 908.79                        # Enthalpy in kj/kg
hg1 = 2799.5                        # Enthalpy in kj/kg

# from table A-4, at 1 bar and 120 degree celcius
h2 = 2766.6                         # in kj/kg
h1 = h2                             # from throttling process assumption

# Calculations:-
x1 = (h1-hf1)/(hg1-hf1)

# Results:-
print '-> The quality of the steam in the supply line is',round(x1,2)

-> The quality of the steam in the supply line is 0.98

Example 4.10 Page no-150

In [9]:
# Given:-
P1 = 1.0                      # pressure of industrial discharge in bar
T1 = 478.0                    # temperature of industrial discharge in kelvin
m1dot = 69.78                 # mass flow rate of industrial discharge in kg/s
T2 = 400.0                    # temperature of exit products from steam generator in kelvin
P2 = 1.0                      # pressure of exit products from steam generator in bar
P3 = 0.275                    # pressure of water stream entering the generator in Mpa
T3 = 38.9                     # temperature of water stream entering the generator in degree celcius
m3dot = 2.079                 # mass flow rate of water stream entering in kg/s
P5 = 0.07                     # exit pressure of the turbine in bars
x5 = 0.93                     # quality of turbine exit

# Part (a)
m2dot = m1dot                 # since gas and water streams do not mix
m5dot = m3dot                 # --DO

# from table A-22, A-2 and A-3:-
h1 = 480.3                    # in kj/kg
h2 = 400.98                   # in Kj/kg
h3 = 162.9                    # assumption: h3 = hf(T3), units in Kj/kg
hf5 = 161.0                   # in kj/kg
hg5 = 2571.72                 # in kj/kg

# Part (b)
P4 = P3                       # from the assumption that there is no pressure drop for water flowing through the steam generator
T4 = 180                      # in degree celcius

# Calculations:-
h5 = hf5 + x5*(hg5-hf5)
Wcvdot = m1dot*h1 + m3dot*h3 - m2dot*h2 - m5dot*h5
h4 = h3 + (m1dot/m3dot)*(h1 -h2)    # from steady state energy rate balance
# interpolating in table A-4, with these P4 and h4
# Results:-
print '-> The power developed by the turbine is ',round(Wcvdot,2),'kJ/s.'
print '-> Turbine inlet temperature is',round(T4,2),'degree celcius.'

-> The power developed by the turbine is  877.84 kJ/s.
-> Turbine inlet temperature is 180.0 degree celcius.

Example 4.11 Page no-153

In [10]:
# Given:-
V = 0.85                               # volume of tank in m^3
T1 = 260.0                             # initial temperature of the tank in degree celcius
X1 = 0.7                               # initial quality

# from table A-2
uf1 = 1128.4                           # in kg/kg
ug1 = 2599.0                           # in kg/kg

vf1 = 1.2755e-3                        # in m^3/kg
vg1 = 0.04221                          # in m^3/kg

# for final state, from table A-2,
u2 = 2599.0                             # units in KJ/kg
v2 = 42.21e-3                           # units in m^3/Kg
he = 2796.6                             # units in KJ/kg

# Calculations:-
u1 = uf1 + X1*(ug1-uf1)                 # in kj/kg
v1 = vf1 + X1*(vg1-vf1)                 # in m^3/kg
m1 = V/v1                               # initial mass in kg
m2 =  V/v2                              # final mass in kg
U2 = m2*u2                              # final internal energy in KJ
U1 = m1*u1                              # initial internal energy in KJ
Qcv = (U2-U1) - he*(m2-m1)

# Results:-
print '-> The amount of heat transfer is',round(Qcv,2),'KJ.'

-> The amount of heat transfer is 14162.16 KJ.

Example 4.12 Page no-155

In [11]:
# Given:-
Pv = 15.0                                            # pressure in the vessel in bar
Tv = 320.0                                           # temperature in the vessel in degree celcius
Vt = 0.6                                             # volume of a tank in m^3
Tt = 400.0                                           # temperature in the tank in degree celcius when the tank is full

# Since the tank is initially empty:-
m1 = 0
u1 = 0

# From table A-4, at 15bar and 400 degree celcius:-
v2 = 0.203                                           # Volume in m^3/kg
m2 = Vt/v2                                           # mass within the tank at the end of the process in kg
hi = 3081.9                                          # in kj/kg
u2 = 2951.3                                          # in kj/kg

# Calculations:-
deltaUcv = m2*u2-m1*u1
Wcv = hi*(m2-m1)-deltaUcv

# Results:-
print '-> The amount of work developed by the turbine is ',round(Wcv,2),'kJ.'

-> The amount of work developed by the turbine is  386.01 kJ.


text:
# 15.4 Pendulums  (Page 4/7)

 Page 4 / 7

## Measuring the torsion constant of a string

A rod has a length of $l=0.30\phantom{\rule{0.2em}{0ex}}\text{m}$ and a mass of 4.00 kg. A string is attached to the CM of the rod and the system is hung from the ceiling ( [link] ). The rod is displaced 10 degrees from the equilibrium position and released from rest. The rod oscillates with a period of 0.5 s. What is the torsion constant $\kappa$ ?

## Strategy

We are asked to find the torsion constant of the string. We first need to find the moment of inertia.

## Solution

1. Find the moment of inertia for the CM:
${I}_{\text{CM}}=\int {x}^{2}dm={\int }_{\text{−}L\text{/}2}^{+L\text{/}2}{x}^{2}\lambda dx=\lambda {\left[\frac{{x}^{3}}{3}\right]}_{\text{−}L\text{/}2}^{+L\text{/}2}=\lambda \frac{2{L}^{3}}{24}=\left(\frac{M}{L}\right)\frac{2{L}^{3}}{24}=\frac{1}{12}M{L}^{2}.$
2. Calculate the torsion constant using the equation for the period:
$\begin{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{\kappa }};\hfill \\ \hfill \kappa & =\hfill & I{\left(\frac{2\pi }{T}\right)}^{2}=\left(\frac{1}{12}M{L}^{2}\right){\left(\frac{2\pi }{T}\right)}^{2};\hfill \\ & =\hfill & \left(\frac{1}{12}\left(4.00\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(0.30\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}\right){\left(\frac{2\pi }{0.50\phantom{\rule{0.2em}{0ex}}\text{s}}\right)}^{2}=4.73\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\text{.}\hfill \end{array}$

## Significance

Like the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the period.

## Summary

• A mass m suspended by a wire of length L and negligible mass is a simple pendulum and undergoes SHM for amplitudes less than about $15\text{°}$ . The period of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$ , where L is the length of the string and g is the acceleration due to gravity.
• The period of a physical pendulum $T=2\pi \sqrt{\frac{I}{mgL}}$ can be found if the moment of inertia is known. The length between the point of rotation and the center of mass is L .
• The period of a torsional pendulum $T=2\pi \sqrt{\frac{I}{\kappa }}$ can be found if the moment of inertia and torsion constant are known.

## Conceptual questions

Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.

A pendulum clock works by measuring the period of a pendulum. In the springtime the clock runs with perfect time, but in the summer and winter the length of the pendulum changes. When most materials are heated, they expand. Does the clock run too fast or too slow in the summer? What about the winter?

The period of the pendulum is $T=2\pi \sqrt{L\text{/}g}.$ In summer, the length increases, and the period increases. If the period should be one second, but period is longer than one second in the summer, it will oscillate fewer than 60 times a minute and clock will run slow. In the winter it will run fast.

With the use of a phase shift, the position of an object may be modeled as a cosine or sine function. If given the option, which function would you choose? Assuming that the phase shift is zero, what are the initial conditions of function; that is, the initial position, velocity, and acceleration, when using a sine function? How about when a cosine function is used?

## Problems

What is the length of a pendulum that has a period of 0.500 s?

Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?

24.8 cm

What is the period of a 1.00-m-long pendulum?

How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

4.01 s

The pendulum on a cuckoo clock is 5.00-cm long. What is its frequency?

Two parakeets sit on a swing with their combined CMs 10.0 cm below the pivot. At what frequency do they swing?

1.58 s

(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is $9.79\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ is moved to a location where the acceleration due to gravity is $9.82\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ . What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.

A pendulum with a period of 2.00000 s in one location ( $g=9.80{\text{m/s}}^{2}$ ) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?

$9.82002\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$

(a) What is the effect on the period of a pendulum if you double its length? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?

what is imaginary mass and how we express is
what is imaginary mass how we express it
Yash
centre of mass is also called as imaginary mass
Lokmani
l'm from Algeria and fell these can help me
Many amusement parks have rides that make vertical loops like the one shown below. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b) The car goes over the top at slower than this speed?
how can I convert mile to meter per hour
1 mile * 1609m
Boon
hey can someone show me how to solve the - "Hanging from the ceiling over a baby bed ...." question
i wanted to know the steps
Shrushti
sorry shrushti..
Rashid
which question please write it briefly
Asutosh
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 312 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Note that Mars has an acceleration of gravity of 3.7 m/s2 .
what is summit
Asutosh
highest point on earth
Ngeh
पृथवी को इसके अक्ष पर कितने कोणीय चाल से घूमाऐ कि भूमधय पे आदमी का भार इसके वासतविक भार से 3/5अधिक हो
best
Murari
At a post office, a parcel that is a 20.0-kg box slides down a ramp inclined at 30.0° 30.0° with the horizontal. The coefficient of kinetic friction between the box and plane is 0.0300. (a) Find the acceleration of the box. (b) Find the velocity of the box as it reaches the end of the plane, if the length of the plane is 2 m and the box starts at rest.
As an IT student must I take physics seriously?
yh
Bernice
hii
Raja
IT came from physics and maths so I don't see why you wouldn't
conditions for pure rolling
Md
the time period of jupiter is 11.6 yrs. how far is jupiter from the sun. distance of earth from rhe sun is 1.5*10 to the power 11 meter.
lists 5 drawing instruments and their uses
that is a question you can find on Google, anyway of top of my head, compass, ruler, protractor, try square, triangles.
Rongfang
A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of radius 2R will be
2F
Jacob
it will be doubled
kelvin
double
Devesh
The difference between vector and scaler quantity
vector has both magnitude & direction but scalar has only magnitude
Manash
my marunong ba dto mag prove ng geometry
ron
how do I find resultant of four forces at a point
Inusah
use the socatoa rule
kingsley
draw force diagram, then work out the direction of force.
Rongfang
In a closed system of forces... Summation of forces in any direction or plane is zero... Resolve if there is a need to then add forces in a particular plane or direction.. Say the x direction... Equate it tk zero
Jacob
define moment of inertia
it is the tendency for a body to continue in motion if is or continue to be at rest if it is.
prince
hay
Khlil
what is Euler s theorem

format: plain


text:
# Blog

## Many Metrics. Such Data. Wow.

Geoffrey Bilder – 2014 February 24

[

Crossref Labs loves to be the last to jump on an internet trend, so what better than than to combine the Doge meme with altmetrics?

Note: The API calls below have been superceeded with the development of the Event Data project. See the latest API documentation for equivalent functionality

Want to know how many times a Crossref DOI is cited by the Wikipedia?

http://det.labs.crossref.org/works/doi/10.1371/journal.pone.0086859

Or how many times one has been mentioned in Europe PubMed Central?

http://det.labs.crossref.org/works/doi/10.1016/j.neuropsychologia.2013.10.021

Or DataCite?

http://det.labs.crossref.org/works/doi/10.1111/jeb.12289

## Background

Back in 2011 PLOS released its awesome ALM system as open source software (OSS). At Crossref Labs, we thought it might be interesting to see what would happen if we ran our own instance of the system and loaded it up with a few Crossref DOIs. So we did. And the code fell over. Oops. Somehow it didn’t like dealing with 10 million DOIs. Funny that.

But the beauty of OSS is that we were able to work with PLOS to scale the code to handle our volume of data. Crossref contracted with Cottage Labs  and we both worked with PLOS to make changes to the system. These eventually got fed back into the main ALM source on Github. Now everybody benefits from our work. Yay for OSS.

So if you want to know technical details, skip to Details for Propellerheads. But if you want to know why we did this, and what we plan to do with it, read on.

## Why?

There are (cough) some problems in our industry that we can best solve with shared infrastructure. When publishers first put scholarly content online, they used to make bilateral reference linking agreements. These agreements allowed them to link citations using each other’s proprietary reference linking APIs. But this system didn’t scale. It was too time-consuming to negotiate all the agreements needed to link to other publishers. And linking through many proprietary citation APIs was too complex and too fragile. So the industry founded Crossref to create a common, cross-publisher citation linking API. Crossref has since obviated the need for bilateral linking arrangements.

So-called altmetrics look like they might have similar characteristics. You have ~4000 Crossref member publishers and N sources (e.g. Twitter, Mendeley, Facebook, CiteULike, etc.) where people use (e.g. discuss, bookmark, annotate, etc.) scholarly publications. Publishers could conceivably each choose to run their own system to collect this information. But if they did, they would face the following problems:

• The N sources will be volatile. New ones will emerge. Old ones will vanish.
• Each publisher will need to deal with each source’s different APIs, rate limits, T&Cs, data licenses, etc. This is a logistical headache for both the publishers and for the sources.
• If publishers use different systems which in turn look at different sources, it will be difficult to compare results across publishers.
• If a journal moves from one publisher to another, then how are the metrics for that journal’s articles going to follow the journal? This isn’t a complete list, but it shows that there might be some virtue in publishers sharing an infrastructure for collecting this data. But what about commercial providers? Couldn’t they provide these ALM services? Of course - and some of them currently do. But normally they look on the actual collection of this data as a means to an end. The real value they provide is in the analysis, reporting and tools that they build on top of the data. Crossref has no interest in building front-ends to this data. If there is a role for us to play here, it is simply in the collection and distribution of the data.

## No, really, WHY?

Aren’t these altmetrics an ill-conceived and meretricious idea? By providing this kind of information, isn’t Crossref just encouraging feckless, neoliberal university administrators to hasten academia’s slide into a Stakhanovite dystopia? Can’t these systems be gamed?

FOR THE LOVE OF FSM, WHY IS CROSSREF DABBLING IN SOMETHING OF SUCH QUESTIONABLE VALUE?

takes deep breath. wipes spittle from beard

These are all serious concerns. Goodhart’s Law and all that… If a university’s appointments and promotion committee is largely swayed by Impact Factor, it won’t improve a thing if they substitute or supplement Impact Factor with altmetrics. Amy Brand has repeatedly pointed out, the best institutions simply don’t use metrics this way at all (PowerPoint presentation). They know better.

But yes, it is still likely that some powerful people will come to lazy conclusions based on altmetrics. And following that, other lazy, unscrupulous and opportunistic people will attempt to game said metrics. We may even see an industry emerge to exploit this mess and provide the scholarly equivalent of SEO. Feh. Now I’m depressed and I need a drink.

So again, why is Crossref doing this? Though we have our doubts about how effective altmetrics will be in evaluating the quality of content, we do believe that they are a useful tool for understanding how scholarly content is used and interpreted. The most eloquent arguments against altmetrics for measuring quality, inadvertently make the case for altmetrics as a tool for monitoring attention.

Critics of altmetrics point out that much of the attention that research receives outside of formal scholarly communications channels can be ascribed to:

• Puffery. Researchers and/or university/publisher “PR wonks” over-promoting research results.
• Innocent misinterpretation. A lay audience simply doesn’t understand the research results.
• Deliberate misinterpretation. Ideologues misrepresent research results to support their agendas.
• Salaciousness. The research appears to be about sex, drugs, crime, video games or other popular bogeymen.
• Neurobollocks. A category unto itself these days.

In short, scholarly research might be misinterpreted. Shock horror. Ban all metrics. Whew. That won’t happen again.

Scholarly research has always been discussed outside of formal scholarly venues. Both by scholars themselves and by interested laity. Sometimes these discussions advance the scientific cause. Sometimes they undermine it. The University of Utah didn’t depend on widespread Internet access or social networks to promote yet-to-be peer-reviewed claims about cold fusion. That was just old-fashioned analogue puffery. And the Internet played no role in the Laetrile or DMSO crazes of the 1980s. You see, there were once these things called “newspapers.” And another thing called “television.” And a sophisticated meatspace-based social network called a “town square.”

But there are critical differences between then and now. As citizens get more access to the scholarly literature, it is far more likely that research is going to be discussed outside of formal scholarly venues. Now we can build tools to help researchers track these discussions. Now researchers can, if they need to, engage in the conversations as well. One would think that conscientious researchers would see it as their responsibility to remain engaged, to know how their research is being used. And especially to know when it is being misused.

That isn’t to say that we expect researchers will welcome this task. We are no Pollyannas. Researchers are already famously overstretched. They barely have time to keep up with the formally published literature. It seems cruel to expect them to keep up with the firehose of the Internet as well.

Which gets us back to the value of altmetrics tools. Our hope is that, as altmetrics tools evolve, they will provide publishers and researchers with an efficient mechanism for monitoring the use of their content in non-traditional venues. Just in the way that citations were used before they were distorted into proxies for credit and kudos.

We don’t think altmetrics are there yet. Partly because some parties are still tantalized by the prospect of usurping one metric for another. But mostly because the entire field is still nascent. People don’t yet know how the information can be combined and used effectively. So we still make naive assumptions such as “link=like” and “more=better.” Surely it will eventually occur to somebody that, instead, there may be a connection between repeated headline-grabbing research and academic fraud. A neuroscientist might be interested in a tool that alerts them if the MRI scans in their research paper are being misinterpreted on the web to promote neurobollocks. An immunologist may want to know if their research is being misused by the anti-vaccination movement. Perhaps the real value in gathering this data will be seen when somebody builds tools to help researchers DETECT puffery, social-citation cabals, and misinterpretation of research results?

But Crossref won’t be building those tools. What we might be able to do is help others overcome another hurdle that blocks the development of more sophisticated tools; getting hold of the needed data in the first place. This is why we are dabbling in altmetrics.

Wikipedia is already the 8th largest referrer of Crossref DOIs. Note that this doesn’t just mean that the Wikipedia cites lots of Crossref DOIs, it means that people actually click on and follow those DOIs to the scholarly literature. As scholarly communication transcends traditional outlets and as the audience for scholarly research broadens, we think that it will be more important for publishers and researcher to be aware of how their research is being discussed and used. They may even need to engage more with non-scholarly audiences. In order to do this, they need to be aware of the conversations. Crossref is providing this experimental data source in the hope that we can spur the development of more sophisticated tools for detecting and analyzing these conversations. Thankfully, this is an inexpensive experiment to conduct - largely thanks to the decision on the part of PLOS to open source its ALM code.

## What Now?

Crossref’s instance of PLOS’s ALM code is an experiment. We mentioned that we had encountered scalability problems and that we had resolved some of them. But there are still big scalability issues to address. For example, assuming a response time of 1 second, if we wanted to poll the English-language version of the Wikipedia to see what had cited each of the 65 million DOIs held in Crossref, the process would take years to complete. But this is how the system is designed to work at the moment. It polls various source APIs to see if a particular DOI is “mentioned”. Parallelizing the queries might reduce the amount of time it takes to poll the Wikipedia, but it doesn’t reduce the work. Another obvious way in which we could improve the scalability of the system is to add a push mechanism to supplement the pull mechanism. Instead of going out and polling the Wikipedia 65 million times, we could establish a “scholarly linkback” mechanism that would allow third parties to alert us when DOIs and other scholarly identifiers are referenced (e.g. cited, bookmarked, shared). If the Wikipedia used this, then even in an extreme case scenario (i.e. everything in Wikipedia cites at least one Crossref DOI), this would mean that we would only need to process ~ 4 million trackbacks.

The other significant advantage of adding a push API is that it would take the burden off of Crossref to know what sources we want to poll. At the moment, if a new source comes online, we’d need to know about it and build a custom plugin to poll their data. This needlessly disadvantages new tools and services as it means that their data will not be gathered until they are big enough for us to pay attention to. If the service in question addresses a niche of the scholarly ecosystem, they may never become big enough. But if we allow sources to push data to us using a common infrastructure, then new sources do not need to wait for us to take notice before they can participate in the system.

Supporting (potentially) many new sources will raise another technical issue- tracking and maintaining the provenance of the data that we gather. The current ALM system does a pretty good job of keeping data, but if we ever want third parties to be able to rely on the system, we probably need to extend the provenance information so that the data is cheaply and easily auditable.

Perhaps the most important thing we want to learn from running this experimental ALM instance is: what it would take to run the system as a production service? What technical resources would it require? How could they be supported? And from this we hope to gain enough information to decide whether the service is worth running and, if so, by whom. Crossref is just one of several organizations that could run such a service, but it is not clear if it would be the best one. We hope that as we work with PLOS, our members and the rest of the scholarly community, we’ll get a better idea of how such a service should be governed and sustained.

### Warning, Caveats and Weasel Words

The Crossref ALM instance is a Crossref Labs project. It is running on R&D equipment in a non-production environment administered by an orangutang on a diet of Redbulls and vodka.

### So what is working?

The system has been initially loaded with 317,500+  Crossref DOIs representing publications from 2014. We will load more DOIs in reverse chronological order until we get bored or until the system falls over again.

We have activated the following sources:

• PubMed
• DataCite
• PubMedCentral Europe Citations and Usage
• We have data from the following sources but will need some work to achieve stability:

• Wikipedia
• CiteULike
• Reddit
• Some of them are faster than others. Some are more temperamental than others. WordPress, for example, seems to go into a sulk and shut itself off  after approximately 1,300 API calls.

In any case, we will be monitoring and tweaking the sources as we gather data. We will also add new sources as we get requested API keys. We will probably even create one or two new sources ourselves. Watch this blog and we’ll update you as we add/tweak sources.

### Dammit, shut up already and tell me how to query stuff.

You can login to the Crossref ALM instance simply using a Mozilla Persona (yes, we’d eventually like to support ORCID too). Once logged-in, your account page will list an API key. Using the API key, you can do things like:

http://det.labs.crossref.org/api/v5/articles?ids=10.1038/nature12990

And you will see that (as of this writing), said Nature article has been cited by the Wikipedia article here:

http://en.wikipedia.org/wiki/HE0107-5240</a>

PLOS has provided lovely detailed instructions for using the APISo, please, play with the API and see what you make of it. On our side we will be looking at how we can improve performance and expand coverage. We don’t promise much- the logistics here are formidable. As we said above, once you start working with millions of documents, the polling process starts to hit API walls quickly. But that is all part of the experiment. We appreciate your helping us and would like your feedback. We can be contacted at:


text:
# Retrieving input from either GET/POST

Today, I created an input class with a method to check whether or not input via GET/POST exists, and another method to retrieve input from either GET/POST, without having to check both individually.

class input {
public static function exists($type = 'post') { switch ($type) {
case "post":
return (!empty($_POST)) ? true : false; break; case "get": return (!empty($_GET)) ? true : false;
break;
default:
return false;
break;
}

}

public static function get($data) { if (isset($_GET[$data])){ return$_GET[$data]; } else if (isset($_POST[$data])){ return$_POST[$data]; } else { return ''; } } }  Here is my form and example usage: <?php require_once('core/init.php'); if(input::exists()){ echo input::get('username'); echo input::get('password'); echo input::get('confirm_password'); } ?> <form method="POST" autocomplete="off">
  Username  <input type="text" name="username" id="username" />

  Password  <input type="password" name="password" id="password" />

  Confirm Password  <input type="password" name="confirm_password" id="confirm_password" />

 <input type="submit" />
 </form>  First of all, is it okay for me to access the methods in the way that I have? Is using statics in this way bad practice? Secondly, aside from what I asked above, are there any recommendations or improvements that you can suggest? ## 3 Answers I see a few small issues with the code: • Method names: particularly get is quite ambiguous, I'd use something more descriptive like 'get_element_data($element_name)' or something like that. As it is now it is obvious when used but it's not clear when looking just at the class itself.
• Instead of return (!empty($_GET)) ? true : false; you can actually use just return (!empty($_GET))
• You don't need to break after return. Some might, however, argue that using it anyway is a good practice that prevents errors. So choose for yourself.

Other than that, I don't see any big problems with it.

• Brilliant! Thanks for your help :). Regarding your third point, I think I'll carry on using the break just to build up the habit so that I don't forget when it actually is needed. – user21611060 Dec 11 '14 at 23:30

has a native GLOBAL variable $_REQUEST that you can check to see if $_GET or $_POST was assigned. By the same one you can retrieve the values from that: <?php$name = isset($_REQUEST['name'])?: false; //this will check whether the sent data was GET or POST ?>  Also, you do not need to create a class to solve this problem. This one (there are several that do the same) https://github.com/auraphp/Aura.Http can be a good solution for you. I recommend that you read about HTTP RFC. • But doesn't the$_REQUEST take the $_COOKIE values too? – user21611060 Dec 15 '14 at 17:43 • No no. I guess you are new to OOP and you are getting out of the way like me on past =) . First, avoid static methods always you can. They are inside OOP, but are not good for maintainability, testability and another good things.... Cookie are just cookies, Request are just Requests. Everyone in the right place, ok? It is not good to "mix" things in OOP, search for "Single Responsibility Principle" and you will understand what I mean. – Daniel Lima Dec 15 '14 at 17:50 Just a note: In your input class you made a function. exists($type = 'post') that make difference between get and post methods. But you made just a get($name) function that use $_GET and $_POST arrays. I think you should make different functions for get and post methods or just add a secound parameter ($type like exists() function) to your get function to be clear everyone what your function does.

When you use your function(s) you have to know which method do you use because the exists() function need it as parameter.


text:
# Welcome to StackEdit!

Hello, I am your first Markdown document within StackEdit1. Don't delete me, I can be helpful. I can be recovered anyway in the Utils tab of the Settings dialog.

## Documents

StackEdit stores your documents in the browser local storage, which means all your documents are automatically saved locally and are accessible offline.

#### Create a document

You can create a new document by clicking the button in the navigation bar. This will switch from the current document to the new one.

#### Switch to another document

You can list all your local documents and switch from one to another by clicking the button in the navigation bar.

#### Rename a document

You can rename the current document by clicking the document title in the navigation bar.

#### Delete a document

You can delete the current document by clicking the button in the navigation bar.

## Synchronization

#### Import a document

You can import a document from the Cloud by going to the Google Drive or the Dropbox sub-menu and by clicking Import from.... Once imported, your document will be automatically synchronized with the Google Drive / Dropbox file.

#### Export a document

You can export any document by going to the Google Drive or the Dropbox sub-menu and by clicking Export to.... Even if your document is already synchronized with Google Drive or Dropbox, you can export it to a another location. StackEdit can synchronize one document with multiple locations.

#### Synchronize a document

If you just have modified your document and you want to force the synchronization, click the button in the navigation bar.

NOTE: The button is disabled when:

• you are offline,
• or the document is not synchronized with any location,
• or the document has not been modified since the last synchronization.

#### Manage document synchronization

Since one document can be synchronized with multiple locations, you can list and manage synchronized locations by clicking Manage synchronization in the menu. This will open a dialog box allowing you to add or remove synchronization links that are associated to your document.

NOTE: If you delete the file from Google Drive or from Dropbox, the document will no longer be synchronized with that location.

## Publication

Once you are happy with your document, you can publish it on different websites directly from StackEdit. As for now, StackEdit can publish on Blogger, Dropbox, Gist, GitHub, Google Drive, Tumblr, WordPress and on any SSH server.

#### Publish a document

You can publish your document by going to the Publish on sub-menu and by choosing a website. In the dialog box, you can choose the publication format:

• Markdown, to publish the Markdown text on a website that can interpret it (GitHub for instance),
• HTML, to publish the document converted into HTML (on a blog for instance),
• Template, to have a full control of the output.

NOTE: The default template is a simple webpage that wraps your document in HTML format. You can customize it in the Publish tab of the Settings dialog.

#### Update a publication

After publishing, StackEdit will keep your document linked to that publish location so that you can update it easily. Once you have modified your document and you want to update your publication, click on the button in the navigation bar.

NOTE: The button is disabled when:

• you are offline,
• or the document has not been published anywhere.

#### Manage document publication

Since one document can be published on multiple locations, you can list and manage publish locations by clicking Manage publication in the menu. This will open a dialog box allowing you to remove publication links that are associated to your document.

NOTE: In some cases, if you remove the file from the website or the post from the blog, the document will no longer be published on that location.

## Markdown Extra

StackEdit supports Markdown Extra, which extends Markdown syntax with some nice features.

### Tables

Markdown Extra has a special syntax for tables:

Item Value
Computer $1600 Phone$12
Pipe 1 You can specify column alignment with one or two colons: Item Value Qty Computer1600 5
Phone $12 12 Pipe$1 234

### Definition Lists

Markdown Extra has a special syntax for definition lists too:

Term 1
Term 2
Definition A
Definition B
Term 3

Definition C

Definition D

part of definition D

### Fenced code blocks

GitHub's fenced code blocks are also supported with Prettify syntax highlighting:

// Foo
var bar = 0;

NOTE: To use Highlight.js instead of Prettify, just configure the Markdown Extra extension in the Settings dialog.

### Special Attributes

With Markdown Extra, you can specify class and id attributes on headers and fenced code blocks just like this:

var foo = bar;

Then you can create cross-references like this: beginning of the document.

### Footnotes

You can create footnotes like this2.

You can insert a table of contents using the marker [TOC]:

### MathJax

You can render LaTeX mathematical expressions using MathJax, as on math.stackexchange.com:

The Gamma function satisfying Γ(n)=(n1)!nN$\Gamma(n) = (n-1)!\quad\forall n\in\mathbb N$ is via through the Euler integral

Γ(z)=0tz1etdt.

• about Markdown Extra extension here,
• about Prettify syntax highlighting here.
• about Highlight.js syntax highlighting here.

1. StackEdit is a free, open-source Markdown editor based on PageDown, the Markdown library used by Stack Overflow and the other Stack Exchange sites.
2. Here is the text of the footnote.

### Welcome document

Welcome to StackEdit! {#welcome} ===================== Hello, I am your first Markdown document within **StackEdit**[^stackedit]. Don't delete me, I can be helpful. I can be recovered anyway in the Utils tab of the Settings dialog. ---------- Documents --------- **StackEdit** stores your documents in the browser local storage, which means all your documents are automatically saved locally and are accessible offline. #### Create a document You can create a new document by clicking the button in the navigation bar. This will switch from the current document to the new one. #### Switch to another document You can list all your local documents and switch from one to another by clicking the button in the navigation bar. #### Rename a document You can rename the current document by clicking the document title in the navigation bar. #### Delete a document You can delete the current document by clicking the button in the navigation bar. ---------- Synchronization --------------- **StackEdit** can be combined with **Google Drive** and **Dropbox** to have your documents centralized in the *Cloud*. The synchronization mechanism will take care of uploading your modifications or downloading the latest version of your documents. #### Import a document You can import a document from the *Cloud* by going to the Google Drive or the Dropbox sub-menu and by clicking Import from.... Once imported, your document will be automatically synchronized with the **Google Drive** / **Dropbox** file. #### Export a document You can export any document by going to the Google Drive or the Dropbox sub-menu and by clicking Export to.... Even if your document is already synchronized with **Google Drive** or **Dropbox**, you can export it to a another location. **StackEdit** can synchronize one document with multiple locations. #### Synchronize a document Once your document is linked to a **Google Drive** or a **Dropbox** file, **StackEdit** will periodically (every 3 minutes) synchronize it by downloading/uploading any modification. Any conflict will be detected, and a local copy of your document will be created as a backup if necessary. If you just have modified your document and you want to force the synchronization, click the button in the navigation bar. > **NOTE:** The button is disabled when: > > - you are offline, > - or the document is not synchronized with any location, > - or the document has not been modified since the last synchronization. #### Manage document synchronization Since one document can be synchronized with multiple locations, you can list and manage synchronized locations by clicking Manage synchronization in the menu. This will open a dialog box allowing you to add or remove synchronization links that are associated to your document. > **NOTE:** If you delete the file from **Google Drive** or from **Dropbox**, the document will no longer be synchronized with that location. ---------- Publication ----------- Once you are happy with your document, you can publish it on different websites directly from **StackEdit**. As for now, **StackEdit** can publish on **Blogger**, **Dropbox**, **Gist**, **GitHub**, **Google Drive**, **Tumblr**, **WordPress** and on any SSH server. #### Publish a document You can publish your document by going to the Publish on sub-menu and by choosing a website. In the dialog box, you can choose the publication format: - Markdown, to publish the Markdown text on a website that can interpret it (**GitHub** for instance), - HTML, to publish the document converted into HTML (on a blog for instance), - Template, to have a full control of the output. > **NOTE:** The default template is a simple webpage that wraps your document in HTML format. You can customize it in the Publish tab of the Settings dialog. #### Update a publication After publishing, **StackEdit** will keep your document linked to that publish location so that you can update it easily. Once you have modified your document and you want to update your publication, click on the button in the navigation bar. > **NOTE:** The button is disabled when: > > - you are offline, > - or the document has not been published anywhere. #### Manage document publication Since one document can be published on multiple locations, you can list and manage publish locations by clicking Manage publication in the menu. This will open a dialog box allowing you to remove publication links that are associated to your document. > **NOTE:** In some cases, if you remove the file from the website or the post from the blog, the document will no longer be published on that location. ---------- Markdown Extra -------------- **StackEdit** supports **Markdown Extra**, which extends **Markdown** syntax with some nice features. ### Tables **Markdown Extra** has a special syntax for tables: Item | Value --------- | ----- Computer | \$1600 Phone | \$12 Pipe | \1 You can specify column alignment with one or two colons: | Item | Value | Qty | | :-------- | ------:| :--: | | Computer | \1600 | 5 | | Phone | \$12 | 12 | | Pipe | \$1 | 234 | ### Definition Lists **Markdown Extra** has a special syntax for definition lists too: Term 1 Term 2 : Definition A : Definition B Term 3 : Definition C : Definition D > part of definition D ### Fenced code blocks **GitHub**'s fenced code blocks are also supported with **Prettify** syntax highlighting:  // Foo var bar = 0;  > **NOTE:** To use **Highlight.js** instead of **Prettify**, just configure the Markdown Extra extension in the Settings dialog. ### Special Attributes With **Markdown Extra**, you can specify class and id attributes on headers and fenced code blocks just like this: ##### Header example {#my-header}  {#my-id .my-class} var foo = bar;  Then you can create cross-references like this: [beginning of the document](#welcome). ### Footnotes You can create footnotes like this[^footnote]. [^footnote]: Here is the *text* of the **footnote**. ### Table of contents You can insert a table of contents using the marker [TOC]: [TOC] ### MathJax You can render *LaTeX* mathematical expressions using **MathJax**, as on [math.stackexchange.com][1]: The *Gamma function* satisfying $\Gamma(n) = (n-1)!\quad\forall n\in\mathbb N$ is via through the Euler integral $$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,.$$ > **NOTE:** You can find more information: > > - about **Markdown** syntax [here][2], > - about **Markdown Extra** extension [here][3], > - about **Prettify** syntax highlighting [here][4]. > - about **Highlight.js** syntax highlighting [here][5]. [^stackedit]: StackEdit is a free, open-source Markdown editor based on PageDown, the Markdown library used by Stack Overflow and the other Stack Exchange sites. [1]: http://math.stackexchange.com/ [2]: http://daringfireball.net/projects/markdown/syntax "Markdown" [3]: https://github.com/jmcmanus/pagedown-extra "Pagedown Extra" [4]: https://code.google.com/p/google-code-prettify/ [5]: http://softwaremaniacs.org/soft/highlight/en/

### test

this is a test blog


text:
# Dimension of symmetric and skew symmetric bilinear forms

1. Oct 11, 2007

### quasar_4

Given the vector space consisting of all bilinear forms of a vector space V (let's call it B) it's very easy to prove that B is the direct sum of two subspaces, the subspace of symmetric and the subspace of skew symmetric bilinear forms. How would one go about determining the dimension of these spaces without any definition of bases? Or is that the only way to go about it?

2. Oct 11, 2007

### HallsofIvy

Staff Emeritus
Think of it in terms of matrices. If V has dimension n, so that every vector in V can be written as an n-tuple of numbers, every bilinear form on V can be associated with an n by n matrix: The vector space of all such bilinear forms is the same as the dimension of all such matrices: n2. Now look at the set of symmetric bilinear forms- that is equivalent to the space of all symmetric n by n matrices. What dimension does that have? Well, we can divide a n by n matrix into 3 parts: the main diagonal, all entries above the main diagonal and all entries below the main diagonal. Obviously, there are n entries on the main diagonal. That leaves n2- n entries not on the main diagaon, (n2- n)/2 are above the main diagonal and (n2- n)/2 are below the main diagonal. Now for symmetric matrices, we can choose any numbers we want for the main diagonal: n choices. If we then choose numbers for the region above the main diagonal, that's another (n2-n)/2 choices but, because the matrix is symmetric, the entries below the main diagonal must be the same: no more choices. We are free to choose entries for n+ (n2- n)/2= (2n+ n2- n)/2= (n2+ n)/2 places independently. That's the dimension of the space of all symmetric n by n matrices and so for all symmetric bilinear forms on an n dimensional vector space.
For the space of all anti-symmetric bilinear forms, you can do either of two things:
Since the space of all bilinear forms is the direct sum of those two subspaces, the sum of the two dimensions must be n2. Or, look at the number of independent entries: now the main diagonal must be all 0s- no choices there. But again, once you have chosen numbers for the region above the main diagonal, the entries below the main diagonal are fixed: they must be the negative of those above the main diagonal. Do you see how many independent choices you have now? Those two caculations should be the same.

3. Oct 12, 2007

### quasar_4

oh ok. So this is actually quite a lot like the problem you do in elementary linear algebra where you show that the space of nxn matrices is the direct sum of symmetric and skew symmetric. Actually, it seems about exactely the same, just that you have to think of the bilinear forms as matrices... which is a little awkward in a sense, because they're not quite the same as linear transformations. But this should work. Thanks. :)


text:
What is Schwarz inequality

1. Jul 23, 2014

Greg Bernhardt

Definition/Summary

the Schwarz inequality (also called Cauchy–Schwarz inequality and Cauchy inequality) has many applications in mathematics and physics.

For vectors $a,b$ in an inner product space over $\mathbb C$:

$$\|a\|\|b\| \geq |(a,b)|$$

For two complex numbers $a,b$ :
$$|a|^2|b|^2 \geq |ab|^2$$

In bra-ket notation, which is commonly used in Quantum Mechanics:
$$| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2$$

For functions $a, b$ which maps a real numbers into complex: $x \in \mathbb{R}$, $a(x), \: b(x) \in \mathbb{C}$:
$$\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a^*bdx \right|^2 ,$$
if the integrals have finite values.

Equations

The Schwarz inequality may be written in a number of equivalent ways:

$$\|a\|\|b\| \geq |(a,b)|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$

$$|a|^2|b|^2 \geq |ab|^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$$

$$| \langle a | a \rangle |^2| \langle b | b \rangle |^2 \geq | \langle a | b \rangle |^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$$

$$\int |a (x)|^2 dx\int |b(x)|^2 dx \geq \left| \int a(x)^*b(x) \, dx \right|^2 ~~~(4)$$

Extended explanation

Theorem:

If a and b are vectors in an inner product space X over $$\mathbb C[/itex], then [tex]|(a,b)| \leq \|a\|\|b\|$$

where the norm is the standard norm on an inner product space.

Proof #1:

Let t be an arbitrary complex number.

$$0 \leq (a+tb,a+tb)=\|a\|^2+t(a,b)+t^*(b,a)+|t|^2\|b\|^2$$

$$=\|a\|^2+2\mbox{ Re}(t(a,b))+|t|^2\|b\|^2$$

The inequality is obviously satisfied when the real part of t(a,b) is non-negative, so we can only learn something interesting when it's negative. Let's choose Arg t so that it is

$$=\|a\|^2-2|t||(a,b)|+|t|^2\|b\|^2$$

Now let's choose |t| so that it minimizes the sum of the last two terms. (This should give us the most interesting result).

$s=|t|$, $A=\|b\|^2$, $B=2|(a,b)|$

$$f(s)=As^2-Bs$$

$$f'(s)=2As-B=0 \implies s=\frac{B}{2A} = \frac{|(a,b)|}{\|b\|^2}$$

$$f''(s)=2A>0$$

Continuing with this value of |t|...

$$=\|a\|^2-2\frac{|(a,b)|}{\|b\|^2}|(a,b)|+\frac{|(a,b)|^2}{\|b\|^4}\|b\|^2$$

$$=\|a\|^2-\frac{|(a,b)|^2}{\|b\|^2}$$

Thus, $0 \leq \|a\|^2\|b\|^2 - |(a,b)|^2$

Proof #2:

A proof for Ineq(3) (the last inequality in the "equations" section) will now be given, but first some basic properties of complex numbers...

$$z = \text{Re}z + i \text{Im}z$$
$$z^* = \text{Re}z - i \text{Im}z$$
$$|z|^2 = z^*z \geq 0$$
$$|zy|^2 = (zy)^*zy$$

Now let $c(x) = a(x) + d \, b(x)$, $d \in \mathbb{C}$.

$$\int |c(x)|^2 dx \geq 0$$

$$\int |a(x)|^2dx + d^*d \int |b(x)|^2dx + d \int a(x)^*b(x) dx + d^* \int a(x)b(x)^* dx \geq 0$$

if the integrals have finite value, this inequality must hold for all $d$, we can choose:

$$d = - \left( \int b(x)^*a(x) dx \right) / \left( \int |b(x)|^2 dx \right)$$

$$\int |a(x)|^2 dx \int |b(x)|^2 dx \geq \int a(x)^*b(x)dx \int a(x)b(x)^*dx = \left| \int a(x)^*b(x)dx \right|^2$$

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!


text:
# Carmichael Number

I am little bit confused with the definition of Carmichael Number

Wikipedia(http://en.wikipedia.org/wiki/Carmichael_number) saying that

Carmichael Number is a composite number satisfies

$b^{n}\equiv b \mod {n}$

for all integers $1<b<n$

Wolfram Mathworld(http://mathworld.wolfram.com/CarmichaelNumber.html) says that

Carmichael Number is a composite number satisfies

$a^{n-1}\equiv 1\mod {n}$

for all integers $1<a<n$

Which one of the above is correct ?

Doubt arosed because if $b^{n}\equiv b \mod {n}$ and if $b$ is not divisible by $n$ then its not necessary that $b^{n-1}\equiv b\mod {n}$

example : $3^{561} \equiv3\mod{561}$ but $3^{560}\not \equiv 1 mod{561}$

$$a^{n-1} \equiv 1 \pmod n$$
for every choice of $a$ satisfying $(a,n) = 1$ with $1 < a < n$"


text:
# Robust Covariance Estimation for Approximate Factor Models.

## Author(s): Fan, Jianqing; Wang, Weichen; Zhong, Yiqiao

To refer to this page use: http://arks.princeton.edu/ark:/88435/pr17g5h
DC FieldValueLanguage
dc.contributor.authorFan, Jianqing-
dc.contributor.authorWang, Weichen-
dc.contributor.authorZhong, Yiqiao-
dc.date.accessioned2021-10-11T14:17:36Z-
dc.date.available2021-10-11T14:17:36Z-
dc.date.issued2019-01en_US
dc.identifier.citationFan, Jianqing, Wang, Weichen, Zhong, Yiqiao. (2019). Robust Covariance Estimation for Approximate Factor Models.. Journal of econometrics, 208 (1), 5 - 22. doi:10.1016/j.jeconom.2018.09.003en_US
dc.identifier.issn0304-4076-
dc.identifier.urihttp://arks.princeton.edu/ark:/88435/pr17g5h-
dc.description.abstractIn this paper, we study robust covariance estimation under the approximate factor model with observed factors. We propose a novel framework to first estimate the initial joint covariance matrix of the observed data and the factors, and then use it to recover the covariance matrix of the observed data. We prove that once the initial matrix estimator is good enough to maintain the element-wise optimal rate, the whole procedure will generate an estimated covariance with desired properties. For data with only bounded fourth moment, we propose to use adaptive Huber loss minimization to give the initial joint covariance estimation. This approach is applicable to a much wider class of distributions, beyond sub-Gaussian and elliptical distributions. We also present an asymptotic result for adaptive Huber's M-estimator with a diverging parameter. The conclusions are demonstrated by extensive simulations and real data analysis.en_US
dc.format.extent5 - 22en_US
dc.languageengen_US
dc.language.isoen_USen_US
dc.relation.ispartofJournal of econometricsen_US
dc.rightsAuthor's manuscripten_US
dc.titleRobust Covariance Estimation for Approximate Factor Models.en_US
dc.typeJournal Articleen_US
dc.identifier.doidoi:10.1016/j.jeconom.2018.09.003-
pu.type.symplectichttp://www.symplectic.co.uk/publications/atom-terms/1.0/journal-articleen_US

Files in This Item:
File Description SizeFormat
Robust covariance estimation for approximate factor models.pdf2.07 MBAdobe PDF


text:
# 1.3: Binomial Coefficients

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Recall the appearance of Pascal's Triangle in example 1.2.6. If you have encountered the triangle before, you may know it has many interesting properties. We will explore some of these here.

You may know, for example, that the entries in Pascal's Triangle are the coefficients of the polynomial produced by raising a binomial to an integer power. For example,

$(x+y)^3=1\cdot x^3+3\cdot x^2y+ 3\cdot xy^2+1\cdot y^3$

and the coefficients 1, 3, 3, 1 form row three of Pascal's Triangle. For this reason the numbers $$n\choose k$$ are usually referred to as the binomial coefficients.

Theorem 1.3.1: Binomial Theorem

$$(x+y)^n={n\choose 0}x^n+{n\choose 1}x^{n-1}y+ {n\choose 2}x^{n-2}y^2+\cdots+{n\choose n}y^n= \sum_{i=0}^n {n\choose i}x^{n-i}y^i$$

Proof

We prove this by induction on $$n$$. It is easy to check the first few, say for $$n=0,1,2$$, which form the base case. Now suppose the theorem is true for $n-1$, that is,

$$(x+y)^{n-1} =\sum_{i=0}^{n-1} {n-1\choose i}x^{n-1-i}y^i.$$

Then

$$(x+y)^n=(x+y)(x+y)^{n-1}=(x+y)\sum_{i=0}^{n-1} {n-1\choose i}x^{n-1-i}y^i.$$

Using the distributive property, this becomes

\eqalign{ x\sum_{i=0}^{n-1} {n-1\choose i}x^{n-1-i}y^i+ y&\sum_{i=0}^{n-1} {n-1\choose i}x^{n-1-i}y^i\cr &=\sum_{i=0}^{n-1}{n-1\choose i}x^{n-i}y^i+ \sum_{i=0}^{n-1} {n-1\choose i}x^{n-1-i}y^{i+1}.\cr}

These two sums have much in common, but it is slightly disguised by an "offset'': the first sum starts with an $$x^ny^0$$ term and ends with an $$x^{1}y^{n-1}$$ term, while the corresponding terms in the second sum are $$x^{n-1}y^{1}$$ and $$x^{0}y^{n}$$. Let's rewrite the second sum so that they match:

\eqalign{ \sum_{i=0}^{n-1}{n-1\choose i}&x^{n-i}y^i+ \sum_{i=0}^{n-1} {n-1\choose i}x^{n-1-i}y^{i+1}\cr &=\sum_{i=0}^{n-1}{n-1\choose i}x^{n-i}y^i+ \sum_{i=1}^{n} {n-1\choose i-1}x^{n-i}y^{i}\cr &={n-1\choose 0}x^n+\sum_{i=1}^{n-1}{n-1\choose i}x^{n-i}y^i+ \sum_{i=1}^{n-1} {n-1\choose i-1}x^{n-i}y^{i}+{n-1\choose n-1}y^{n}\cr &={n-1\choose 0}x^n+ \sum_{i=1}^{n-1}({n-1\choose i}+{n-1\choose i-1})x^{n-i}y^{i}+ {n-1\choose n-1}y^{n}.\cr }

Now we can use theorem 1.2.7 to get:

\eqalign{ {n-1\choose 0}x^n+ &\sum_{i=1}^{n-1}({n-1\choose i}+{n-1\choose i-1})x^{n-i}y^{i}+ {n-1\choose n-1}y^{n}.\cr &={n-1\choose 0}x^n+ \sum_{i=1}^{n-1}{n\choose i}x^{n-i}y^{i}+ {n-1\choose n-1}y^{n}.\cr &={n\choose 0}x^n+ \sum_{i=1}^{n-1}{n\choose i}x^{n-i}y^{i}+ {n\choose n}y^{n}\cr &=\sum_{i=0}^{n}{n\choose i}x^{n-i}y^{i}.\cr }

At the next to last step we used the facts that $${n-1\choose 0}={n\choose 0}$$ and $${n-1\choose n-1}={n\choose n}$$.

$$\square$$

Here is an interesting consequence of this theorem: Consider

$$(x+y)^n=(x+y)(x+y)\cdots(x+y).$$

One way we might think of attempting to multiply this out is this: Go through the $$n$$ factors $$(x+y)$$ and in each factor choose either the $$x$$ or the $$y$$; at the end, multiply your choices together, getting some term like $$xxyxyy\cdots yx= x^iy^j$$, where of course $$i+j=n$$. If we do this in all possible ways and then collect like terms, we will clearly get

$$\sum_{i=0}^n a_ix^{n-i}y^i.$$

We know that the correct expansion has ${n\choose i}=a_i$; is that in fact what we will get by this method? Yes: consider $$x^{n-i}y^i$$. How many times will we get this term using the given method? It will be the number of times we end up with $$i$$ $$y$$-factors. Since there are $$n$$ factors $$(x+y)$$, the number of times we get $$i$$ $$y$$-factors must be the number of ways to pick $$i$$ of the $$(x+y)$$ factors to contribute a $$y$$, namely $$n\choose i$$. This is probably not a useful method in practice, but it is interesting and occasionally useful.

Example $$\PageIndex{1}$$:

Using this method we might get

$$(x+y)^3 = xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy$$

which indeed becomes $$\displaystyle x^3+3x^2y+3xy^2 + y^3$$ upon collecting like terms.

The Binomial Theorem, 1.3.1, can be used to derive many interesting identities. A common way to rewrite it is to substitute $$y=1$$ to get

$$(x+1)^n=\sum_{i=0}^n {n\choose i} x^{n-i}.$$

If we then substitute $$x=1$$ we get

$$2^n=\sum_{i=0}^n{n\choose i},$$

that is, row $$n$$ of Pascal's Triangle sums to $$2^n$$. This is also easy to understand combinatorially: the sum represents the total number of subsets of an $n$-set, since it adds together the numbers of subsets of every possible size. It is easy to see directly that the number of subsets of an $$n$$-set is $$2^n$$: for each element of the set we make a choice, to include or to exclude the element. The total number of ways to make these choices is $$2\cdot2\cdots2=2^n$$, by the multiplication principle.

Suppose now that $$n\ge 1$$ and we substitute $$-1$$ for $$x$$; we get

\eqalignno{ (-1+1)^n&=\sum_{i=0}^n{n\choose i}(-1)^{n-i}.& (1.3.1)\cr }

The sum is now an alternating sum: every other term is multiplied by $$-1$$. Since the left hand side is $0$, we can rewrite this to get

\eqalignno{ {n\choose 0}+{n\choose 2}+\cdots&={n\choose 1}+{n\choose 3}+\cdots.& (1.3.2)\cr }

So each of these sums is $2^{n-1}$.

Another obvious feature of Pascal's Triangle is symmetry: each row reads the same forwards and backwards. That is, we have:

Theorem 1.3.3

\displaystyle {n\choose i}={n\choose n-i}$. Proof This is quite easy to see combinatorially: every$i$-subset of an$n$-set is associated with an$(n-i)$-subset. That is, if the$n$-set is$A$, and if$B\subseteq A$has size$i$, then the complement of$B$has size$n-i$. This establishes a 1–1 correspondence between sets of size$i$and sets of size$n-i$, so the numbers of each are the same. (Of course, if$i=n-i$, no proof is required.)$\square$Note that this means that the Binomial Theorem, 1.3.1, can also be written as $$(x+y)^n=\sum_{i=0}^n {n\choose n-i}x^{n-i}y^{i}.$$ or $$(x+y)^n=\sum_{i=0}^n {n\choose i}x^{i}y^{n-i}.$$ Another striking feature of Pascal's Triangle is that the entries across a row are strictly increasing to the middle of the row, and then strictly decreasing. Since we already know that the rows are symmetric, the first part of this implies the second. Theorem 1.3.4 For$1\le i\le \lfloor{n\over 2}\rfloor$,$\displaystyle {n\choose i}>{n\choose i-1}. Proof This is by induction; the base case is apparent from the first few rows. Write \eqalign{ {n\choose i}&={n-1\choose i-1}+{n-1\choose i}\cr {n\choose i-1}&={n-1\choose i-2}+{n-1\choose i-1}\cr } Provided that1\le i\le \lfloor{n-1\over 2}\rfloor$, we know by the induction hypothesis that $${n-1\choose i}>{n-1\choose i-1}.$$ Provided that$1\le i-1\le \lfloor{n-1\over 2}\rfloor$, we know that $${n-1\choose i-1}>{n-1\choose i-2}.$$ Hence if$2\le i\le \lfloor{n-1\over 2}\rfloor$, $${n\choose i}>{n\choose i-1}.$$ This leaves two special cases to check:$i=1$and$i\ge \lfloor{n-1\over 2}\rfloor+1$. These are left as an exercise.$\square\$


text:
## Illinois Journal of Mathematics

### Homomorphic images of pro-nilpotent algebras

George M. Bergman

#### Abstract

It is shown that any finite-dimensional homomorphic image of an inverse limit of nilpotent not-necessarily-associative algebras over a field is nilpotent. More generally, this is true of algebras over a general commutative ring $k$, with “finite- dimensional” replaced by “of finite length as a $k$-module.”

These results are obtained by considering the multiplication algebra $M(A)$ of an algebra $A$ (the associative algebra of $k$-linear maps $A → A$ generated by left and right multiplications by elements of $A$), and its behavior with respect to nilpotence, inverse limits, and homomorphic images.

As a corollary, it is shown that a finite-dimensional homomorphic image of an inverse limit of finite-dimensional solvable Lie algebras over a field of characteristic 0 is solvable.

It is also shown by example that infinite-dimensional homomorphic images of pro-nilpotent algebras can have properties far from those of nilpotent algebras; in particular, properties that imply that they are not residually nilpotent.

Several open questions and directions for further investigation are noted.

#### Article information

Source
Illinois J. Math., Volume 55, Number 3 (2011), 719-748.

Dates
First available in Project Euclid: 29 May 2013

https://projecteuclid.org/euclid.ijm/1369841782

Digital Object Identifier
doi:10.1215/ijm/1369841782

Mathematical Reviews number (MathSciNet)
MR3069281

Zentralblatt MATH identifier
1285.17001

#### Citation

Bergman, George M. Homomorphic images of pro-nilpotent algebras. Illinois J. Math. 55 (2011), no. 3, 719--748. doi:10.1215/ijm/1369841782. https://projecteuclid.org/euclid.ijm/1369841782

#### References

• H. Bass, Algebraic K-theory, W. A. Benjamin, New York–Amsterdam, 1968.
• G. M. Bergman, An invitation to general algebra and universal constructions, Henry Helson, Berkeley, CA, 1998. Available at http://math.berkeley.edu/~gbergman/245.
• G. M. Bergman, Continuity of homomorphisms on pro-nilpotent algebras, Illinois J. Math. 55 (2011), 749–770.
• G. M. Bergman and A. O. Hausknecht, Cogroups and co-rings in categories of associative rings, Mathematical Surveys and Monographs, vol. 45, Amer. Math. Soc., Providence, RI, 1996.
• G. M. Bergman and N. Nahlus, Homomorphisms on infinite direct product algebras, especially Lie algebras, J. Algebra 333 (2011), 67–104. Available at http://math. berkeley.edu/~gbergman/papers.
• G. M. Bergman and N. Nahlus, Linear maps on $k^I$, and homomorphic images of infinite direct product algebras, J. Algebra 356 (2012), 257–274. Available at http:// math.berkeley.edu/~gbergman/papers.
• K. Bowman and D. A. Towers, On almost nilpotent-by-abelian Lie algebras, Linear Algebra Appl. 247 (1996), 159–167.
• A. L. S. Corner, Three examples of hopficity in torsion-free abelian groups, Acta Math. Acad. Sci. Hungar. 16 (1965), 303–310.
• J. D. Dixon, M. P. F. du Sautoy, A. Mann and D. Segal, Analytic pro-$p$ groups, 2nd ed., Cambridge Studies in Advanced Mathematics, vol. 61, Cambridge University Press, Cambridge, 1999.
• L. Henkin, A problem on inverse mapping systems, Proc. Amer. Math. Soc. 1 (1950), 224–225.
• G. Higman and A. H. Stone, On inverse systems with trivial limits, J. London Math. Soc. 29 (1954), 233–236.
• V. A. Hiremath, Hopfian rings and Hopfian modules, Indian J. Pure Appl. Math. 17 (1986), 895–900. Available at http://www.new.dli.ernet.in/rawdataupload/upload/insa/INSA_2/20005a4f_895.pdf.
• K. H. Hofmann and S. A. Morris, The Lie theory of connected pro-Lie groups. A structure theory for pro-Lie algebras, pro-Lie groups, and connected locally compact groups, EMS Tracts in Mathematics, vol. 2, European Mathematical Society, Zürich, 2007.
• N. Jacobson, Lie algebras, Dover, New York, 1979.
• T. Y. Lam, A first course in noncommutative rings, Graduate Texts in Mathematics, vol. 131, Springer, New York, 1991.
• T. Y. Lam, Lectures on modules and rings, Graduate Texts in Mathematics, vol. 189, Springer, New York, 1999.
• S. Lang, Algebra, 3rd ed., Addison-Wesley, Reading, 1993; reprinted as Graduate Texts in Mathematics, vol. 211, Springer, New York, 2002.
• S. Mac Lane, Categories for the working mathematician, Graduate Texts in Mathematics, vol. 5, Springer, New York, 1971.
• E. Sąsiada and P. M. Cohn, An example of a simple radical ring, J. Algebra 5 (1967), 373–377.
• R. D. Schafer, An introduction to nonassociative algebras, Dover, New York, 1995.
• J.-P. Serre, Galois cohomology, Springer, Berlin, 1997.
• A. M. Slin'ko, The equivalence of certain nilpotencies of right-alternative rings, Algebra Logika 9 (1970), 342–348 (in Russian).
• K. Varadarajan, Hopfian and co-Hopfian objects, Publ. Mat. 36 (1992), 293–317.
• W. C. Waterhouse, An empty inverse limit, Proc. Amer. Math. Soc. 36 (1972) 618.


text:
# UVA 11396：Claw Decomposition(二分图判定)

A claw is defined as a pointed curved nail on the end of each toe in birds, somereptiles, and some mammals. However, if you are a graph theory enthusiast, youmay understand the following special class of graph as shown in the followingfigure by the word claw.If you are more concerned about graph theory terminology, you may want todefine claw as K1,3.Lets leave the definition for the moment and come to the problem. You aregiven a simple undirected graph in which every vertex has degree 3. You are tofigure out whether the graph can be decomposed into claws or not.Just for the sake of clarity, a decomposition of a graph is a list of subgraphs such that each edgeappears in exactly one subgraph in the list.

Input

There will be several cases in the input file. Each case starts with the number of vertices in the graph,V (4 ≤ V ≤ 300). This is followed by a list of edges. Every line in the list has two integers, a and b,the endpoints of an edge (1 ≤ a, b ≤ V ). The edge list ends with a line with a pair of ‘0’. The end ofinput is denoted by a case with V = 0. This case should not be processed.

Output

For every case in the input, print ‘YES’ if the graph can be decomposed into claws and ‘NO’ otherwise.

Sample Input

4
1 2
1 3
1 4
2 3
2 4
3 4
0 0
6
1 2
1 3
1 6
2 3
2 5
3 4
4 5
4 6
5 6
0 0
0

Sample Output

NO
NO

AC代码：

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<vector>
#include<string.h>
#include<queue>
using namespace std;
int mapp[500][500];
int vis[500],n;
bool dfs()
{
queue<int>q;
memset(vis,0,sizeof(vis));
q.push(0);
vis[0]=1;
while(!q.empty())
{
int p=q.front();
q.pop();
for(int i=0; i<n; i++)
if(mapp[p][i])
{
if(vis[i]==0)
{
vis[i]=-vis[p];
q.push(i);
}
else if(vis[i]==vis[p])
return false;
}
}
return true;
}
int main()
{
while(~scanf("%d",&n)&&n)
{
int a,b;
memset(mapp,0,sizeof(mapp));
while(~scanf("%d%d",&a,&b)&&(a||b))
mapp[a-1][b-1]=mapp[b-1][a-1]=1;
printf(dfs()?"YES\n":"NO\n");
}
return 0;
}


text:
1. ## subspaces

Hi,

How can I show that th set of all solutions to Ax = b where A is mxn, is not a subspace of R^n if b <> 0.

If Ax = 0, (b = 0), the set of solutions IS a subspace. What difference does it make whether b is 0 or not?

Thanks!

2. Originally Posted by kpizle
Hi,

How can I show that th set of all solutions to Ax = b where A is mxn, is not a subspace of R^n if b <> 0.

If Ax = 0, (b = 0), the set of solutions IS a subspace. What difference does it make whether b is 0 or not?

Thanks!

EDIT: Moving question to more appropriate forum
Suppose $x$ is a solution to this and $\lambda$ is a scalar, then $A(\lambda x)=\lambda b$, so $\lambda x$ is not in the set of solutions unless $b=0$, but if the set of solutions is a subspace it would have to be.

CB