text:
#### Question 1:

Is displacement a scalar quantity?

No, displacement is a vector quantity because it has magnitude as well as direction.

#### Question 2:

State whether distance is a scalar or a vector quantity.

Distance is a scalar quantity; it has magnitude but no specific direction.

#### Question 3:

Change the speed of 6 m/s into km/h.

We have to convert 6 m/s into km/h. So,

#### Question 4:

What name is given to the speed in a specified direction?

Velocity is the name given to the speed of a body in a particular direction.

#### Question 5:

Give two examples of bodies having non-uniform motion.

Two examples of non-uniform motion:
(a) Motion of a bus on a curved road
(b) Motion of a bee flying randomly in air.

#### Question 6:

Name the physical quantity obtained by dividing 'Distance travelled' by 'Time taken' to travel that distance.

Speed is obtained by dividing 'distance travelled' by 'time taken' to travel that distance.

#### Question 7:

What do the following measure in a car?

(a) Speedometer
(b) Odometer

(a) Speedometer of a car is used to measure the instantaneous speed of the car.
(b) Odometer of a car is used to record and measure the overall distance travelled by the car.

#### Question 8:

Name the physical quantity which gives us an idea of how slow or fast a body is moving.

Speed is the physical quantity that gives the idea of how slow or fast a body is moving.

#### Question 9:

Under what conditions can a body travel a certain distance and yet its resultant displacement be zero?

When the body comes back to its starting point, its resultant displacement is zero. It is because it has covered a certain distance in due course of time; however, there is no difference between the initial and final positions.

#### Question 10:

In addition to speed, what else should we know to predict the position of a moving body?

Besides speed, we should know the direction to predict the position of a moving body.

#### Question 11:

When is a body said to have uniform velocity?

When a body covers equal distances in equal intervals of time in a particular direction, it is said to have uniform velocity.

#### Question 12:

Under which condition is the magnitude of average velocity equal to average speed?

The magnitude of average velocity is equal to average speed when an object covers equal distances in equal intervals of time in a particular direction.

#### Question 13:

Which of the two can be zero under certain conditions : average speed of a moving body or average velocity of a moving body?

Average velocity of a moving body can be zero. This is because the net displacement for a given time interval can be zero.

#### Question 14:

Give one example of a situation in which a body has a certain average speed but its average velocity is zero.

Motion of a boy from his home to shop and back to home is an example of a situation in which a body has a certain average speed but its average velocity is zero.

#### Question 15:

What is the acceleration of a body moving with uniform velocity?

The acceleration of a body moving with uniform velocity is zero.

#### Question 16:

What is the other name of negative acceleration?

Retardation is the other name for negative acceleration.

#### Question 17:

Name the physical quantity whose SI unit is :

(a) m/s
(b) m/s2

(a) Speed or velocity is expressed in m/s.
(b) Acceleration or retardation is expressed in m/s2.

#### Question 18:

What type of motion is exhibited by a freely falling body?

A freely falling body exhibits uniform accelerated motion.

#### Question 19:

What is the SI unit of retardation?

The S.I. unit of retardation is m/s2.

#### Question 20:

(a) Displacement is a __________ quantity, whereas distance is a __________ quantity.
(b) The physical quantity that includes both the speed and direction of the motion of a body is called its __________.
(c) A motorcycle has a steady __________ of 3 m/s2. This means that each __________ its __________ increases by __________.
(d) Velocity is the rate of change of __________. It is measured in __________.
(e) Acceleration is the rate of change of __________. It is measured in __________.

(a) Displacement is a vector quantity, whereas distance is a scalar quantity.
(b) The physical quantity that includes both the speed and direction of the motion of a body is called its velocity.
(c) A motorcycle has a steady acceleration of 3 m/s2. This means that each second its velocity increases by 3 m/s.
(d) Velocity is the rate of change of displacement. It is measured in m/s.
(e) Acceleration is the rate of change of velocity. It is measured in m/s2.

#### Question 21:

What type of motion, uniform or non-uniform, is exhibited by a freely falling body? Give reason for you answer.

A freely falling body exhibits non-uniform motion. Its velocity increases at a constant rate, so it shows uniformly accelerated motion.

#### Question 22:

State whether speed is a scalar or a vector quantity. Give reason for your choice.

Speed is a scalar quantity. This is because it has magnitude, but it does not specify direction. It is the distance travelled by a body per unit time.

#### Question 23:

Bus X travels a distance of 360 km in 5 hours whereas bus Y travels a distance of 476 km in hours. Which bus travels faster?

We should first consider Bus X.
Distance travelled (d1) = 360 km
Time taken (d1) = 5 hr
So, we can calculate the speed as:

Thus, the speed of Bus X is 72 km/h.
Similarly,
For Bus Y:
Distance travelled
Time taken (t2) = 7 hr
So, we can calculate the speed as:

Thus, the speed of Bus Y is 68 km/h.
Speed of Bus X is more than that of Bus Y. Hence, Bus X travels faster.

#### Question 24:

Arrange the following speeds in increasing order (keeping the least speed first):

(i) An athlete running with a speed of 10 m/s.
(ii) A bicycle moving with a speed of 200 m/min.
(iii) A scooter moving with a speed of 30 km/h.

We have three different moving bodies. To compare their speeds, we need to express them in similar S.I. units.
(i) An athlete running with a speed of 10 m/s.
(ii) A bicycle moving with a speed of 200 m/min. Converting into S.I. units, the speed will be:

(iii) A scooter moving with a speed of 30 km/h. Converting into S.I. units, the speed will be:

On arranging the speeds in ascending order, we get:
(ii) < (iii) < (i)

#### Question 25:

(a) Write the formula for acceleration. Give the meaning of each symbol which occurs in it.
(b) A train starting from Railway Station attains a speed of 21 m/s in one minute. Find its acceleration.

(a) Acceleration is the change in velocity per unit time; it is a vector quantity.

The above expression can also be written as:

Where,
a = Acceleration
= Final velocity
= Initial velocity
= Time taken

(b) We will find the value of uniform acceleration.
Initial velocity (u) 0 m/s
Final velocity (v) = 21 m/s
Time taken (t) = 60 s
Acceleration:
$a\mathit{=}\frac{v\mathit{-}u}{t}$
Putting the values in the above equation, we get:

#### Question 26:

(a) What term is used to denote the change of velocity with time?
(b) Give one word which means the same as 'moving with a negative acceleration'.
(c) The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Give reason for your answer.

(a) Acceleration is used to denote the change in velocity with time.
(b) Retardation is the same as ‘moving with a negative acceleration'.
(c) Displacement is a vector quantity, so it can be zero for two reasons:
1. When the body doesn't move at all: In this case, the distance travelled will also be zero.
2. When the body comes back to its initial position: In this case, the distance travelled is non-zero, but the displacement is zero.

#### Question 27:

A snail covers a distance of 100 metres in 50 hours. Calculate the average speed of snail in km/h.

Figure

Distance (d) = 1000 m = 0.1 km
Time (t) = 50 hr
So, we can calculate the speed as:

Average speed for the entire journey:

#### Question 28:

A tortoise moves a distance of 100 metres in 15 minutes. What is the average speed of tortoise in km/h?

Distance (d) = 0.1 km

So, we can calculate the speed as:

Average speed for the entire journey:

#### Question 29:

If a sprinter runs a distance of 100 metres in 9.83 seconds, calculate his average speed in km/h.

Distance (d) = 100 m
Time (t) = 9.83 s
So, we can calculate the speed as:

Average speed for the entire journey:

Now, we can convert it in km/h as:

#### Question 30:

A motorcyclist drives from place A to B with a uniform speed of 30 km h−1 and returns from place B to A with a uniform speed of 20 km h−1. Find his average speed.

We have to find the average velocity of the entire journey. For this, we have the following information:
Speed from A to B = (v1) = 30 m/s
Let the distance from A to B be  (d).
Also, let the time taken to travel from A to B be (t1).

We have:
${t}_{1}=\frac{d}{30}$
Speed from B to A (v2) = 20 m/s
Let the time taken to travel from B to A be(t2).
Thus, we have:

Total distance travelled is 2d.
Therefore,

On putting the values to obtain the average speed of the motorcyclist, we get:

#### Question 31:

A motorcyclist starts from rest and reaches a speed of 6 m/s after travelling with uniform acceleration for 3 s. What is his acceleration?

We have to find the value of uniform acceleration.
Initial velocity (u) = 0 m/s
Final velocity (v) = 6 m/s
Time taken (t) = 3s
Acceleration:

Put the values in the above equation to obtain the value of acceleration.

#### Question 32:

An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the speed of sound as 1100 km/h at the aircraft's altitude, how long will it take to reach the 'sound barrier'?

Final velocity, v = 1100 km/h = 1100 × 5/18 = 305.55 m/s
Initial velocity, u = 600 km/h = 600 × 5/18 = 166.66 m/s
Acceleration = 10 km/h per second = 10 × 5/18 = 2.77 m/s2
Time taken by body,
v = u + at
t = (v - u)/a
t  = (305.55 -166.66 )/2.77
t = 50.14 sec
t ≃ 50 sec

#### Question 33:

If a bus travelling at 20 m/s is subjected to a steady deceleration of 5 m/s2, how long will it take to come to rest?

We have to find the time taken to reach the given final velocity.
We have:
Initial velocity (u) = 20 m/s
Final velocity  (v) = 0 m/s
Acceleration for the entire journey (a) = –5 ms2.
Let the time taken be (t).
We can calculate the time taken using the first equation of motion.

Time taken by the bus to come to rest:

#### Question 34:

(a) What is the difference between 'distance travelled' by a body and its 'displacement'? Explain with the help of diagram.
(b) An ant travels a distance of 8 cm from P to Q and then moves a distance of 6 cm at right angles to PQ. Find its resultant displacement.

(a)

 Distance Displacement 1. Distance has only magnitude, with no specified direction. 1. Displacement has magnitude as well as direction. 2. It is a scalar quantity. 2. It is a vector quantity. 3. Two different distances can be added directly. 3. We have to follow the vector addition method to add displacements.

One difference in diagrammatic form is as follows:

Here the curved line is the distance traveled and the straight line is the magnitude of the displacement.

(b) We have to find the resultant displacement from the given diagram:

We have:
PQ = 8 cm and QR = 6 cm
Resultant displacement:

The direction of this displacement is from P to R. If θ is the angle made by PR with PQ then,'

This is the angle made by the resultant with PQ.

#### Question 35:

Define motion. What do you understand by the terms 'uniform motion' and 'non-uniform motion'? Explain with examples.

A body is said to be in motion when its position changes continuously with respect to a stationary point taken as the reference point.

Uniform motion: A body is said to be in uniform motion if it travels equal distances in equal intervals of time in a particular direction, no matter how small these time intervals are.
For example, a car running at a constant speed of 10 m/s towards east will cover the equal distance of 10 m every second towards east, so its motion will be uniform.

Non-uniform motion: A body is said to be in non-uniform motion if it travels unequal distances in equal intervals of time.
For example, motion of a freely falling ball from the roof of a tall building.

#### Question 36:

(a) Define speed. What is the SI unit of speed?
(b) What is meant by (i) average speed, and (ii) uniform speed?

(a) Speed of a body is the distance travelled by it per unit time. The S.I. unit of speed is m/s.
(b) (i) Average speed of a body is the total distance travelled by it divided by the total time taken by it to cover the given distance.

(ii) A body has a uniform speed if it travels equal distance in equal intervals of time, no matter how small these time intervals are.

#### Question 37:

(a) Define velocity. What is the SI unit of velocity?
(b) What is the difference between speed and velocity?
(c) Convert a speed of 54 km/h into m/s.

(a) Velocity of a body is the distance travelled by it per unit time in a given direction. The S.I. unit of velocity is m/s. It is a vector quantity.

(b) (i) Speed is a scalar quantity, whereas velocity is a vector quantity.
(ii) Speed of a body is the distance travelled by it per unit time, whereas the velocity of a body is the distance travelled by it per unit time in a given direction.
(iii) Speed is always positive, while velocity can be both positive and negative, depending upon the direction.

(c) We have to convert 54 km/h into m/s.

#### Question 38:

(a) What is meant by the term 'acceleration'? State the SI unit of acceleration.
(b) Define the term 'uniform acceleration'. Give one example of a uniformly accelerated motion.

(a) Acceleration of a body is defined as the rate of change of its velocity with respect to time. It is a vector quantity.    The S.I. unit of acceleration is  (m/s2) .
(b) A body has uniform acceleration if it travels in a straight line and its velocity increases by equal amounts in equal intervals of time.
For example, a freely falling body has uniform acceleration.

#### Question 39:

The distance between Delhi and Agra is 200 km. A train travels the first 100 km at a speed of 50 km/h. How fast must the train travel the next 100 km, so as to average 70 km/h for the whole journey?

We have the following data to find the speed for the second part of the journey:
Total distance to be travelled by train (D) = 200 km
Average speed required (vavg) = 70 km/hr
Time required for the entire journey (T):

For the first part of the trip:
Distance covered (d1) = 100 km
Speed for this part of journey (v1) = 50 km/hr
Time taken for the first part of journey:

For the second part of the trip,
Distance covered (d2) = 100 km
Time taken for the second part of journey:

Speed of the train for the second part of the journey:

#### Question 40:

A train travels the first 15 km at a uniform speed of 30 km/h; the next 75 km at a uniform speed of 50 km/h; and the last 10 km at a uniform speed of 20 km/h. Calculate the average speed for the entire train journey.

(i) In the first case, the train travels at a speed of 30 km/h for a distance of 15 km.
We can find the time as:

(ii) In the second case, the train travels at a speed of 50 km/h for a distance of 75 km.
We can find the time as:

(iii) In the third case, the train travels at a speed of 20 km/h for a distance of 10 km.
We can find the time as:

Total distance covered:
= (15 + 75 + 10) km
= 100 km
Total time taken = (0.5 + 1.5 + 0.5) km
= 2.5
Therefore,
Now, put the values to get the average speed.

#### Question 41:

A car is moving along a straight road at a steady speed. It travels 150 m in 5 seconds:

(a) What is its average speed?
(b) How far does it travel in 1 second?
(c) How far does it travel in 6 seconds?
(d) How long does it take to travel 240 m?

(a) We have:
Distance (d) = 150 m
Time (t) = 5s
So, we can calculate average the speed as:

Average speed for the entire journey:

(b) We have to calculate the distance travelled in 1s.
Distance = (Speed) (Time)
Distance travelled in one second:
=(30) (1) m

(c) We have to calculate the distance travelled in 6 s.
Distance = (Speed) (Time)
Distance travelled in one second:
= (30) (6) m

(d) We have:
Distance (d) = 240 m
Speed (v) = 30 m/s
We can find the time as:

#### Question 42:

A particle is moving in a circular path of radius r. The displacement after half a circle would be:

(a) 0
(b) πr
(c) 2r
(d) 2πr

(c) Displacement is the difference between the final and initial position of a body. It is a vector quantity and is independent of the path taken. So, for the movement of half of a circle, the displacement is 2r, where r is the radius of the circular path.

#### Question 43:

The numerical ratio of displacement to distance for a moving object is :

(a) always less than 1
(b) equal to 1 or more than 1
(c) always more than 1
(d) equal to 1 or less than 1

(d), i.e., Equal to 1 or less than 1.
Displacement is always smaller than or equal to displacement.

#### Question 44:

A boy is sitting on a merry-go-round which is moving with a constant speed of 10 m s−1. This means that the boy is :

(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

(c) Acceleration is the rate of change of velocity, and the velocity of the merry-go-round is changing with respect to time. Thus, it will move in an accelerated motion.

#### Question 45:

In which of the following cases of motion, the distance moved and the magnitude of displacement are equal?

(a) if the car is moving on straight road
(b) if the car is moving on circular road
(c) if the pendulum is moving to and fro
(d) if a planet is moving around the sun

(a) The magnitude of displacement is equal to the distance travelled by a body when it travels in a straight line.

#### Question 46:

The speed of a moving object is determined to be 0.06 m/s. This speed is equal to :

(a) 2.16 km/h
(b) 1.08 km/h
(c) 0.216 km/h
(d) 0.0216 km/h

(c) We can convert 0.06 m/s as:

So, the answer is 0.216 km/h.

#### Question 47:

A freely falling object travels 4.9 m in 1st second, 14.7 m in 2 nd second, 24.5 m in 3rd second, and so on. This data shows that the motion of a freely falling object is a case of :

(a) uniform motion
(b) uniform acceleration
(c) no acceleration
(d) uniform velocity

(b) The displacement of the body in equal interval of time is unequal, but acceleration is constant. The acceleration will thus be uniform, so the answer is (b).

#### Question 48:

When a car runs on a circular track with a uniform speed, its velocity is said to be changing. This is because :

(a) the car has a uniform acceleration
(b) the direction of car varies continuously
(c) the car travels unequal time intervals.
(d) the car travels equal distances in unequal time intervals

(d) When a car runs on a circular track, its velocity changes continuously, as its direction keeps changing.

#### Question 49:

Which of the following statement is correct regarding velocity and speed of a moving body?

(a) velocity of a moving body is always higher than its speed
(b) speed of a moving body is always higher than its velocity
(c) speed of a moving body is its velocity in a given direction
(d) velocity of a moving body is its speed in a given direction

(d) Velocity is a vector quantity having a magnitude and a specific direction. So, velocity is nothing but speed in a particular direction.

#### Question 50:

Which of the following can sometimes be 'zero' for a moving body?

(i) average velocity
(a) only (i)

(ii) distance travelled
(b) (i) and (ii)

(iii) average speed
(c) (i) and (iv)

(iv) displacement
(d) only (iv)

(c) Distance is the length of the actual path covered by a moving body. This implies that the distance travelled by a moving body and also its average speed can never be zero at any point of time. However, it is possible for the displacement as well as average velocity for the body to be zero at an instance during the motion of the particle.

#### Question 51:

When a car driver travelling at a speed of 10 m/s applies brakes and brings the car to rest in 20 s, then retardation will be:

(a) + 2 m/s2
(b) − 2 m/s2
(c) − 0.5 m/s2
(d) + 0.5 m/s2

(d) The term “retardation” means negative acceleration.
Initial velocity = 10 m/s
Final velocity = 0 m/s
Time taken = 20 s

#### Question 52:

Which of the following could not be a unit of speed?

(a) km/h
(b) s/m
(c) m/s
(d) mm s−1

(b) Speed is the distance travelled by a moving body per unit time.

#### Question 53:

One of the following is not a vector quantity. This one is :

(a) displacement
(b) speed
(c) acceleration
(d) velocity

(b) Vector quantities have magnitude as well as direction, and they obey the laws of vector addition.

#### Question 54:

Which of the following could not be a unit of acceleration?

(a) km/s2
(b) cm s−2
(c) km/s
(d) m/s2

(c) Acceleration is defined as the rate of change of velocity.

#### Question 55:

A body is moving along a circular path of radius R. What will be the distance travelled and displacement of the body when it completes half a revolution?

We have to analyse the distance and displacement of a body that has covered half the perimeter of a circle.
Distance travelled in half a rotation of a circular path is equal to the circumference of semi-circle.
Distance travelled = πR
Displacement is calculated from the initial and final positions of a body. It is independent of the path covered. So, displacement is the diameter of the semi-circle.
Hence, displacement is 2R, where R is the radius of the circular path.

#### Question 56:

If on a round trip you travel 6 km and then arrive back home :

(a) What distance have you travelled?
(b) What is your final displacement?

(a) Distance travelled is the actual path covered.
Total distance covered in this case due to going and coming back:
2d = 6 km
is the distance of one-side journey.

(b) Displacement is calculated from the initial and final positions of a body. It is independent of the path covered. So, displacement in this case is 0 because the initial and final positions are the same.

#### Question 57:

A body travels a distance of 3 km towards East, then 4 km towards North and finally 9 km towards East.
(i) What is the total distance travelled?
(ii) What is the resultant displacement?

1-22-57
(i) Distance travelled is the actual path covered.
Total distance travelled:
= (3 + 4 + 9) km
= 16 km

(ii) The body travels a total distance of 12 km east, which means towards the x-axis, on a Cartesian plane. It travels a distance of 4 km in north direction, which means towards y-axis.
Resultant displacement:

#### Question 58:

A boy walks from his classroom to the bookshop along a straight corridor towards North. He covers a distance of 20 m in 25 seconds to reach the bookshop. After buying a book, he travels the same distance in the same time to reach back in the classroom. Find (a) average speed, and (b) average velocity, of the boy.

(a)    Distance travelled is the length of the actual path covered.
Total distance covered in going and coming back:
= (20 + 20) m
= 40 m
Total time taken, = (25 + 25) s
= 50 s
So, we can calculate average speed as:

Average speed for the entire journey:

(b) Average velocity is zero, as displacement is zero because the boy arrives at the initial point.

#### Question 59:

A car travels 100 km at a speed of 60 km/h and returns with a speed of 40 km/h. Calculate the average speed for the whole journey.

In the first case, the car travels at a speed of 60 km/h for a distance of 100 km.
Thus,

In the second case, the car travels at a speed of 40 km/h for a distance of 100 km.
Thus,

Total time taken:

Total distance travelled = 200 km
We can calculate average speed as:

Average speed for the entire journey:

#### Question 60:

A ball hits a wall horizontally at 6.0 m s−1. It rebounds horizontally at 4.4 m s−1. The ball is in contact with the wall for 0.040 s. What is the acceleration of the ball?

We have to find the value of uniform acceleration.
We have:
Initial velocity (u) = 6 m/s
Final velocity is in opposite direction to that of initial velocity (v) = –4.4 m/s
Time taken (t) = 0.04 s
Acceleration:

Put the values in the above equation to get the value of acceleration.


text:
## Quantum graphs, convex bodies, and a century-old problem of Minkowski

Series
Analysis Seminar
Time
Wednesday, October 30, 2019 - 1:55pm for 1 hour (actually 50 minutes)
Location
Skiles 005
Speaker
Yair Shenfeld – Princeton University – yairs@princeton.eduhttps://www.yairshenfeld.com/
Organizer
Galyna Livshyts

That the ball minimizes surface area among all sets of fixed volume, was known since antiquity; this is equivalent to the fact that the ball is the unique set which yields equality in the isoperimetric inequality. But the isoperimetric inequality is only a very special case of quadratic inequalities about mixed volumes of convex bodies, whose equality cases were unknown since the time of Minkowski. This talk is about these quadratic inequalities and their unusual equality cases which we resolved using degenerate diffusions on the sphere. No background in geometry will be assumed. Joint work with Ramon van Handel.

format: plain


text:
# Natural language processing (NLP): text vectorization and bag of words approach

###### 04.02.2019 - Jay M. Patel - Reading time ~4 Minutes

Let us consider a simple task of classifying a given email into spam or not spam. As you might have already guessed, this is an example of a simple binary classification problem with target or label values being 0 or 1, within the larger field of supervised machine learning (ML).

However, we still need to convert non-numerical text contained in emails into numerical features which can be fed into a ML algorithm of our choice. This process is known as “text vectorization”, and there are many different ways of doing it, from simple approach of count based vectorization, discussed here, to more sophisticated term frequency-inverse document frequency (tf-idf) and word embedding and neural network methods such as Glove, word2vec etc.

### Introduction to bag of words based text vectorization

In this model, any text is represented as the bag or a multiset of its words (known as “tokens”) by disregarding grammar, punctuation and word order but keeping multiplicity of individual words/tokens. Splitting individual sentences into it’s constituent words or tokens is referred to as “tokenization”.

As an example, let us consider the sentence below

Sam likes to watch baseball. Christine likes baseball too.

Let us remove punctuation and tokenize it.

"Sam","likes","to","watch","baseball","Christine","likes","baseball","too"

In a bag of words representation, we can convert the list above into a dictionary, with keys being the words, and values being the number of occurrences.

BoW_dict = {"Sam":1,"likes":2,"to":1,"watch":1,"baseball":2,"Christine":1,"too":1}

For our purposes, we are more interested in getting a list of “word frequencies” from the above dictionary which serves as a sparse vector representation of the underlying sentence, with number of dimensions being equal of total number of unique words in the corpus. As you might have already noticed, this term frequency vector does not preserve order of the words in the sentence. However, for many text classification tasks this bag of words model works pretty satisfactorily.

vect = [1, 2, 1, 1, 2, 1, 1]

### Count vectorization with scikit-learn.

Let us take a block of text below from a random page on open learn.

text = '''As a psychologist and coach, I have long been interested in the power of stories to convey important lessons. Some of my earliest     recollections were fables, stories involving animals as protagonists. Fables are an oral story-telling tradition typically involving animals, occasionally humans or forces of nature interacting with each other. The best-known fables are attributed to the ancient Greek storyteller Aesop, although there are fables found in almost all human societies and oral traditions making this an almost universal feature of human culture. A more modern example of an allegorical tale involving animals is George Orwell’s Animal Farm , in which animals rebel against their human masters only to recreate the oppressive system humans had subjected them to. Fables work by casting animals as characters with minds that are recognisably human. This superimposition of human minds and animals serves as a medium for transmitting and teaching moral and ethical values, in particular to children. It does this by drawing on the symbolic connection between nature and society and the dilemmas that this connection can be used to represent. Fables have survived into the modern age, although their role seems to have waned under the emergence of other storytelling media. Animals are, thus, still a feature of modern day stories (e.g. David Attenborough’s wildlife programmes, stories about animals in the news, and and CGI-heavy children’s cartoons involving cute animal characters). Our more immediate and visceral connection with nature and with the different type of animal species seems to have been reduced to being observers rather than participants. Wild animals in ancient times were revered as avatars of ancient gods but also viewed with fear. By and large, our modern position as observer in relation to animal stories is that we remotely view animals in peril from climate change on television and social media news coverage, but we may remain disconnected to the fundamental moral lessons their stories could teach us today.'''

Let us remove punctuations from the text using regex function.

import re
def preprocessor(text):

text = re.sub('<[^>]*>', '', text)
emoticons = re.findall('(?::|;|=)(?:-)?(?:\)|\(|D|P)',
text)
text = (re.sub('[\W]+', ' ', text.lower()) +
' '.join(emoticons).replace('-', ''))
return text

After preprocessing, the text looks like shown below:

Now, this is ready to be fed into scikit learn’s count vectorizer which will convert it into bag of words.

from sklearn.feature_extraction.text import CountVectorizer
import numpy as np
import pandas as pd

cv = CountVectorizer(analyzer='word')
#cv.get_feature_names()

You can intuitively see the sparse vectors by converting the cv into a pandas dataframe.

dtm = cv.transform([text])
pd.DataFrame(dtm.toarray(), columns=cv.get_feature_names())

You can directly use the vectors created above for text classification. However, there are some disadvantages with using these count vectors directly and in the next post we will talk about term frequency-inverse document frequency (tf-idf) and discuss more robust approaches and preprocessing techniques which we can use instead.


text:
# Trade Count Time Series

Is historical information on the counts of trades in single stocks, futures, options etc. available somewhere for download or purchase? If not, which ways can you think of to gather it?

-

## 2 Answers

Interactive Brokers provides it as a field called count.

In this page of the IB API Reference Guide count is described as follows

When TRADES historical data is returned, represents the number of trades that occurred during the time period the bar covers.

-
 Ok, so IB provides this data. Nice. This is not an option if you do not trade via IB, but for me and, I believe, many others this is ok. – Konsta Sep 23 '12 at 12:41 You asked if the data was "available somewhere for download or purchase." You don't have to make any trades in order to get the data although you may have to pay something like 10 or 20 dollars a month. You may even be able to get the data with a paper trading account for free. – user508 Sep 23 '12 at 14:50 I am undecided which answer to accept. Since I believe that IB offers a lot of asset classes and exchanges and the count is supposedly then available for all history, I opt for this one. – Konsta Sep 24 '12 at 22:40

Its not available for free anywhere that I know of. Your only option is to purchase tick data for the instruments you're interested in and then count the ticks per trading session (or whichever timeframe you want to use).

EDIT: This link might be interesting for you regarding how to get market data.

http://www.quantshare.com/sa-426-6-ways-to-download-free-intraday-and-tick-data-for-the-us-stock-market

-
 Yes, I stumbled upon this link yesterday and figured that simply loading tick data might give a hint - how to interpret ticks? It is not the BBO I get, but the prices of trades that happened, right? – Konsta Sep 23 '12 at 12:37 @Konsta You said you are trying to count trades, right? Each tick is one trade. So for a daily bar, just count the total number of ticks that happen in that day. Note that this is different from the total volume, which would be the sum of the trade size of each tick (like 300 shares, etc). – Paccc Sep 23 '12 at 14:08


text:
# Covering of uncountable cardinal

Let $$\alpha$$ be an uncountable cardinal number and $$x$$ a countable set of countable subsets of $$\alpha$$ with $$\alpha = \bigcup \bigcup x$$. How can be shown that there exists a countable subset $$y$$ of $$\alpha$$ such that $$\alpha = \bigcup y$$?

• Hint: take $y = \cup x$ . – Shervin Sorouri Dec 19 '19 at 13:27
• Where do we use that $\alpha$ is a cardinal? And why is $\cup x$ countable? – Fblthp Dec 19 '19 at 14:15
• Any countable union of countable sets is countable by the axiom of choice. So $\cup x$ is countable. The fact that $\alpha$ is a cardinal is not used here. – Shervin Sorouri Dec 19 '19 at 14:28
• Does this still hold if we consider ZF without the axiom of choice? – Fblthp Dec 19 '19 at 14:31
• @Shervin Without choice your suggestion does not work. – Andrés E. Caicedo Dec 19 '19 at 15:00

If $$x$$ is a set of ordinals, then $$\bigcup x=\sup x$$, and if $$x$$ is a set of sets of ordinals, then $$\bigcup\bigcup x=\sup\{\sup y\mid y\in x\}$$. So in other words, you statement can be recast in the following way:

Suppose that there is a countable family of sets $$y_n\subseteq\alpha$$ such that $$\alpha=\sup\left(\bigcup_{n<\omega}y_n\right)$$, then there is a countable $$y\subseteq\alpha$$ such that $$\alpha=\sup y$$.

There are two options:

1. There is some $$y\in x$$ such that $$\sup y=\alpha$$. Then we are done, since $$y$$ is countable set by definition.

2. Otherwise, let $$\alpha_n=\sup y_n$$ for $$y_n\in x$$, then $$y=\{y_n\mid n\in\omega\}$$, or equivalently $$y=\{\bigcup z\mid z\in x\}$$, is such that $$\bigcup y=\alpha$$ and $$y$$ is countable since there is a surjection from $$x$$ onto $$y$$. (Note that this holds even without the Axiom of Choice, since a surjection from a countable set is finite or countable even in ZF.)

Finally, note that this is in fact equivalent to $$\alpha$$ having a countable cofinal set, since in that case taking a countable cofinal subset $$y=\{y_n\mid n<\omega\}$$ and taking $$x=\{\{y_n\}\mid n<\omega\}$$ works.

Asaf's is clearly the right way of phrasing things. Anyway, let's clean this up a bit.

Enumerate $$x$$, say $$x=\{x_n\}_n$$. If any of its members is cofinal in $$\alpha$$, you are done - the set you want is any member of $$x$$ that is cofinal in $$\alpha$$. Otherwise, let $$\alpha_n:=\sup x_n$$ and note that $$\{\alpha_k:k\in\omega\}$$ is cofinal in $$\alpha$$, since $$\bigcup\bigcup x\subseteq\sup_k\alpha_k$$.

For instance, (without choice) we could have $$\omega_1$$ be singular and let $$x$$ be countable and cofinal in $$\omega_1$$. In this case we already have $$\bigcup x=\omega_1$$.


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > KEYWORD > SMOOTHED ANALYSIS:
Reports tagged with Smoothed Analysis:
TR05-063 | 24th June 2005
Bodo Manthey, Rüdiger Reischuk

#### Smoothed Analysis of the Height of Binary Search Trees

Revisions: 2

Binary search trees are one of the most fundamental data structures. While the
height of such a tree may be linear in the worst case, the average height with
respect to the uniform distribution is only logarithmic. The exact value is one
of the best studied problems in average case ... more >>>

TR06-023 | 7th February 2006
Xi Chen, Xiaotie Deng, Shang-Hua Teng

#### Computing Nash Equilibria: Approximation and Smoothed Complexity

By proving that the problem of computing a $1/n^{\Theta(1)}$-approximate Nash equilibrium remains \textbf{PPAD}-complete, we show that the BIMATRIX game is not likely to have a fully polynomial-time approximation scheme. In other words, no algorithm with time polynomial in $n$ and $1/\epsilon$ can compute an $\epsilon$-approximate Nash equilibrium of an \$n\times ... more >>>

TR07-039 | 27th March 2007
Bodo Manthey, Till Tantau

#### Smoothed Analysis of Binary Search Trees and Quicksort Under Additive Noise

Revisions: 1

Binary search trees are a fundamental data structure and their height
plays a key role in the analysis of divide-and-conquer algorithms like
quicksort. Their worst-case height is linear; their average height,
whose exact value is one of the best-studied problems in average-case
complexity, is logarithmic. We analyze their smoothed height ... more >>>

TR13-008 | 7th January 2013


text:
# To calculate: The vertices and foci of the conic section: x29 + y24=1

To calculate: The vertices and foci of the conic section:
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

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Step 1

Formula Used: Eccentricity

Step 2

Calculation: Compare the given equation with the equation of an ellipse and we get:

Now,

Vertices are:

Now, put the values of a and b to get the vertice of the conic section and we get: Now,

Thus, the vertices and foci of the conic section are:


text:
Categories

# Log4j XML Configuration

Note to self – don’t forget this!

For some reason, I can’t seem to retain a key piece of information – how to configure the logging level per-package in the log4j XML configuration. I’m putting this snippet here to cater for the likely event that I’ll forget it again. It’s all to do with the category element, as it happens.

<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE log4j:configuration SYSTEM "log4j.dtd">

<log4j:configuration xmlns:log4j="http://jakarta.apache.org/log4j/">

<appender name="ConsoleAppender" class="org.apache.log4j.ConsoleAppender">
<layout class="org.apache.log4j.SimpleLayout"/>
</appender>

<category name="org.apache">
<priority value="warn"/>
</category>

<root>
<priority value="debug" />
<appender-ref ref="ConsoleAppender"/>
</root>

</log4j:configuration>


text:
# Angles in triangle

The triangle is ratio of the angles β:γ = 6:8. Angle α is 40° greater than β. What are the size of angles of the triangle?

Correct result:

β =  82 °
γ =  42 °
α =  56 °

#### Solution:

$\beta =\frac{6\cdot \frac{180-40}{\frac{2\cdot 6}{8}+1}}{8}=4{2}^{\circ }$
$\gamma =\frac{180-40}{\frac{2\cdot 6}{8}+1}=5{6}^{\circ }$

We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!

Showing 0 comments:

Tips to related online calculators
Do you have a system of equations and looking for calculator system of linear equations?
See also our trigonometric triangle calculator.

#### You need to know the following knowledge to solve this word math problem:

We encourage you to watch this tutorial video on this math problem:

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text:
## anonymous one year ago Find cos theta if sin theta =-5/13 and tan theta>0

1. freckles

$\sin^2(x)+\cos^2(x)=1 \\ (\sin(x))^2+(\cos(x))^2=1$

2. freckles

replace sin(x) with -5/13

3. freckles

and solve for cos(x)

4. freckles

since tangent is positive and sine is negative then cosine is also negative.

5. freckles

$(\frac{-5}{13})^2+(\cos(x))^2=1$ do you know how to solve this equation for cos(x)?

6. anonymous

Isolate cos x

7. freckles

yes that is what we need to do


text:
77 views

Consider the following closed loop control system

where $G\left ( s \right )=\dfrac{1}{s\left ( s+1 \right )}$ and $C\left ( s \right )=K\dfrac{s+1}{s+3}$. If the steady state error for a unit ramp input is $0.1$, then the value of $K$ is ______________.


text:
# Variable selection for an exploratory multilevel categorical model with weak priors

I am conducting an exploratory study of a linguistic outcome with 4 unordered categories (logit link). There are two group-level effects (varying-intercept terms) and about 30 population-level covariates, for a total of 150 parameters. N is 2,220. I use weak N(0, 2.5) priors for all di-/polychotomous covariates, and equivalently weak normal priors for quantitative covariates. The group-level SD’s have exponential(2) priors and are not subject to variable selection because they are of inherent interest. The purpose of my priors is just to help the model converge and to rule out infinite/preposterous values. I am basically trying to fit a model that produces the same estimates as a frequentist model, without the computational and convergence problems.

I presently use brms to do all fitting.

Many covariates will likely not make the “final” the model due to excessively noisy/overfit coefficients, however, due to my exploratory emphasis I don’t want those coefficients shrunk to zero by strong priors. Rather, I want to witness every weak and unreliable signal of potential effect and make at least a descriptive comment on whether it accords with theory, before proceeding to remove the associated covariate.

I am now trying to decide on a reasonable method of variable selection that is not too likely to attract serious methodological criticism. The projection predictive approach doesn’t seem an option because it appears to be intended for parsimonious prediction rather than exploratory investigation. It is also not available for categorical responses.

I’m presently aware of two fairly appealing options (but feel free to suggest others!). Each of them was originally developed for frequentist analysis. Below, in outline, are my “improvised quasi-frequentist” and “improvised Bayesian” implementations of each procedure:

1a. Hosmer et al’s “Purposeful Selection” (2013: 89–124), improvised quasi-frequentist version:

1. Start with a null model containing only the fixed and group-level intercepts.

2. Fit 30 “univariate” models, adding to the null model each of the 30 population-level covariates in turn, noting each one that improves “Deviance” at p < 0.25 in a likelihood-ratio test (LRT).

3. Fit a “full model” containing every population-level covariate that was significant at p < 0.25.

4. Rank the population-level covariates by significance in the LRT.

5. One at a time, delete every population-level covariate that is neither significant at p < 0.05 nor a confounder, i.e. whose removal doesn’t result in a change of \geq 20% in any of the remaining coefficients.

6. After removing all non-significant non-confounders, go through each covariate that failed to make the cut in Step 2, one at a time, adding it to the model and checking whether it’s significant at p < 0.05 conditional on the other covariates in the model.

7. Once no more covariates can be added or removed, check for non-linear and interaction effects among the remaining covariates.

This procedure relies heavily on the LRT and “Deviance”, i.e. minus twice the in-sample log score calculated using the posterior means of the parameters. This quantity is not quite the true Deviance because it is not based solely on the likelihood, but it should be very close due to the weak priors.

1b. Hosmer et al’s “Purposeful Selection”, improvised Bayesian version:

Like 1a except we use elpd_loo comparisons instead of LRT p-values. In Step 2, our criterion is whether elpd_loo improves at all. In Steps 5 and 6, our criterion is whether elpd_loo improves by 4 or more. The confounding criterion is unchanged.

2a. Dunkler et al’s “Augmented Backward Elimination” (2014), improvised quasi-frequentist version:

1. Start with a “full model” containing the group-level intercepts and all population-level covariates of interest.

2. Rank the population-level covariates by p-value and start eliminating those not significant at p < 0.2 that are not confounders.

3. Once no more covariates can be eliminated, check for non-linear and interaction effects among the remaining covariates.

These authors don’t specify which significance test to use, but I would use the LRT based on “Deviance” here as well.

2b. Dunkler et al’s “Augmented Backward Elimination” (2014), improvised Bayesian version:

Like 2a except we use elpd_loo comparisons instead of “Deviance”-based LRT p-values. Simply, main effects are removed from the full starting model if they are not confounders and their removal improves elpd_loo by any amount, no matter how small.

In addition to the time saving of not having to compute LOO posteriors, the quasi-frequentist approaches are arguably more faithful to the original frequentist procedures because they approximate the p-value criteria more closely. AFAIK, with priors this weak, -2*elpd_loo is pretty much equivalent to AIC, so any improvement in elpd_loo that results from adding a parameter should imply a p-value \leq 0.157 in a LRT. This is much more stringent than the p < 0.25 criterion Hosmer et al use in their Step 2, and it is also more stringent than Dunkler et al’s backward-elimination criterion of p > 0.2 for removing parameters. I’m not sure how to go about relaxing it for better equivalence to these frequentist criteria. One idea might be to include a number of pure noise covariates in the model. Then, any covariate of interest whose addition/removal yielded an elpd_loo change within the same range as that observed for the pure noise covariates would be declared inconsequential/non-significant.

My main questions are the following:

1. Do you think either of the two selection philosophies (Hosmer et al 2013 or Dunkler et al 2014) is more likely than the other to attract serious methodological criticism?
2. Whichever of the two selection philosophies I choose, do you think I could get away with using the quasi-frequentist version of it?
3. Might there be some third approach that is better suited for my exploratory aims than these two?

Thanks in advance.

References
Hosmer, David W., Stanley Lemeshow, and Rodney X. Sturdivant. 2013. Applied logistic regression (3rd ed.). Hoboken, N.J.: Wiley.

Dunkler, Daniela, Max Plischke, Karen Leffondré, and Georg Heinze. 2014. “Augmented Backward Selection: A Pragmatic and Purposeful Way to Develop Statistical Models”. PLoS ONE 2014; 9 (11)

Whether it’s exploratory or not, the prior should reflect what you know, and not your utility/loss in decision making. If you are worried that the prior you’ve chosen to present your prior knowledge might be misspecified, you can do prior sensitivity checking (see, e.g. [2107.14054] Detecting and diagnosing prior and likelihood sensitivity with power-scaling). With 30 covariates, you probably have some prior knowledge how likely it is that all of them are relevant or just some or whether they are all relevant but correlating (Piranha theorem states that you can’t have many big effects unless the covariates are correlating [2105.13445] The piranha problem: Large effects swimming in a small pond, so you can include that assumption in your prior).

If the model can predict with less covariates, you have learned something about those covariates, and thus it’s fine for exploratory investigation, too.

It is, see, e.g. [2109.04702] Latent space projection predictive inference
However, it might be possible that the current projpred package doesn’t work yet with unordered categorical response.

Your N is quite big compared to the number of covariates, so it’s likely that quite many approaches would work for ordering the covariates in some relevance order, but this depends also on signal-noise-ratio, so can’t be sure. Any approach you choose, you can probably improve it with use of reference model approach even if you would not do the projection, see [2004.13118] Using reference models in variable selection

I recommend checking if you can formulate your problem as in [2109.04702] Latent space projection predictive inference

2 Likes

Thank you the exhaustive answer, Aki. Much appreciated. That is a wealth of new literature to go through. Your input will definitely help me make a better-informed choice.

Indeed, it says Error in get(extend_family_specific, mode = "function") : object 'extend_family_categorical' of mode 'function' was not found. Otherwise I would certainly give it a try out of pure academic curiosity.

1 Like


text:
## The Annals of Probability

### Poly-logarithmic localization for random walks among random obstacles

#### Abstract

Place an obstacle with probability $1-\mathsf{p}$ independently at each vertex of $\mathbb{Z}^{d}$, and run a simple random walk until hitting one of the obstacles. For $d\geq2$ and $\mathsf{p}$ strictly above the critical threshold for site percolation, we condition on the environment where the origin is contained in an infinite connected component free of obstacles, and we show that the following path localization holds for environments with probability tending to 1 as $n\to\infty$: conditioned on survival up to time $n$ we have that ever since $o(n)$ steps the simple random walk is localized in a region of volume poly-logarithmic in $n$ with probability tending to 1. The previous best result of this type went back to Sznitman (1996) on Brownian motion among Poisson obstacles, where a localization (only for the end point) in a region of volume $t^{o(1)}$ was derived conditioned on the survival of Brownian motion up to time $t$.

#### Article information

Source
Ann. Probab., Volume 47, Number 4 (2019), 2011-2048.

Dates
Revised: July 2018
First available in Project Euclid: 4 July 2019

https://projecteuclid.org/euclid.aop/1562205700

Digital Object Identifier
doi:10.1214/18-AOP1300

Mathematical Reviews number (MathSciNet)
MR3980914

Zentralblatt MATH identifier
07114710

#### Citation

Ding, Jian; Xu, Changji. Poly-logarithmic localization for random walks among random obstacles. Ann. Probab. 47 (2019), no. 4, 2011--2048. doi:10.1214/18-AOP1300. https://projecteuclid.org/euclid.aop/1562205700

#### References

• [1] Antal, P. (1995). Enlargement of obstacles for the simple random walk. Ann. Probab. 23 1061–1101.
• [2] Antal, P. and Pisztora, A. (1996). On the chemical distance for supercritical Bernoulli percolation. Ann. Probab. 24 1036–1048.
• [3] Astrauskas, A. (2007). Poisson-type limit theorems for eigenvalues of finite-volume Anderson Hamiltonians. Acta Appl. Math. 96 3–15.
• [4] Astrauskas, A. (2008). Extremal theory for spectrum of random discrete Schrödinger operator. I. Asymptotic expansion formulas. J. Stat. Phys. 131 867–916.
• [5] Athreya, S., Drewitz, A. and Sun, R. (2017). Random walk among mobile/immobile traps: A short review. Preprint. Available at arXiv:1703.06617.
• [6] Biskup, M. and König, W. (2016). Eigenvalue order statistics for random Schrödinger operators with doubly-exponential tails. Comm. Math. Phys. 341 179–218.
• [7] Biskup, M., König, W. and dos Santos, R. S. (2018). Mass concentration and aging in the parabolic Anderson model with doubly-exponential tails. Probab. Theory Related Fields 171 251–331.
• [8] Bolthausen, E. (1994). Localization of a two-dimensional random walk with an attractive path interaction. Ann. Probab. 22 875–918.
• [9] Chayes, J. T., Chayes, L. and Newman, C. M. (1987). Bernoulli percolation above threshold: An invasion percolation analysis. Ann. Probab. 15 1272–1287.
• [10] Donsker, M. D. and Varadhan, S. R. S. (1975). Asymptotics for the Wiener sausage. Comm. Pure Appl. Math. 28 525–565.
• [11] Donsker, M. D. and Varadhan, S. R. S. (1979). On the number of distinct sites visited by a random walk. Comm. Pure Appl. Math. 32 721–747.
• [12] Fiodorov, A. and Muirhead, S. (2014). Complete localisation and exponential shape of the parabolic Anderson model with Weibull potential field. Electron. J. Probab. 19 no. 58, 27.
• [13] Fukushima, R. (2009). From the Lifshitz tail to the quenched survival asymptotics in the trapping problem. Electron. Commun. Probab. 14 435–446.
• [14] Gärtner, J., König, W. and Molchanov, S. (2007). Geometric characterization of intermittency in the parabolic Anderson model. Ann. Probab. 35 439–499.
• [15] Gärtner, J. and Molchanov, S. A. (1990). Parabolic problems for the Anderson model. I. Intermittency and related topics. Comm. Math. Phys. 132 613–655.
• [16] Grenkova, L. N., Molčanov, S. A. and Sudarev, J. N. (1983). On the basic states of one-dimensional disordered structures. Comm. Math. Phys. 90 101–123.
• [17] Grimmett, G. R. and Marstrand, J. M. (1990). The supercritical phase of percolation is well behaved. Proc. Roy. Soc. London Ser. A 430 439–457.
• [18] Hall, R. R. (1992). A quantitative isoperimetric inequality in $n$-dimensional space. J. Reine Angew. Math. 428 161–176.
• [19] Kesten, H. and Zhang, Y. (1990). The probability of a large finite cluster in supercritical Bernoulli percolation. Ann. Probab. 18 537–555.
• [20] König, W. (2016). The Parabolic Anderson Model: Random Walk in Random Potential. Birkhäuser/Springer, Cham.
• [21] König, W., Lacoin, H., Mörters, P. and Sidorova, N. (2009). A two cities theorem for the parabolic Anderson model. Ann. Probab. 37 347–392.
• [22] Lacoin, H. and Mörters, P. (2012). A scaling limit theorem for the parabolic Anderson model with exponential potential. In Probability in Complex Physical Systems. Springer Proc. Math. 11 247–272. Springer, Heidelberg.
• [23] Lawler, G. F. and Limic, V. (2010). Random Walk: A Modern Introduction. Cambridge Studies in Advanced Mathematics 123. Cambridge Univ. Press, Cambridge.
• [24] Lee, S. (1997). The power laws of $M$ and $N$ in greedy lattice animals. Stochastic Process. Appl. 69 275–287.
• [25] Martin, J. B. (2002). Linear growth for greedy lattice animals. Stochastic Process. Appl. 98 43–66.
• [26] Povel, T. (1999). Confinement of Brownian motion among Poissonian obstacles in ${\mathbf{R}}^{d},d\ge3$. Probab. Theory Related Fields 114 177–205.
• [27] Sidorova, N. and Twarowski, A. (2014). Localisation and ageing in the parabolic Anderson model with Weibull potential. Ann. Probab. 42 1666–1698.
• [28] Sznitman, A.-S. (1990). Lifschitz tail and Wiener sausage. I, II. J. Funct. Anal. 94 223–246, 247–272.
• [29] Sznitman, A.-S. (1991). On the confinement property of two-dimensional Brownian motion among Poissonian obstacles. Comm. Pure Appl. Math. 44 1137–1170.
• [30] Sznitman, A.-S. (1993). Brownian asymptotics in a Poissonian environment. Probab. Theory Related Fields 95 155–174.
• [31] Sznitman, A.-S. (1993). Brownian survival among Gibbsian traps. Ann. Probab. 21 490–508.
• [32] Sznitman, A.-S. (1996). Brownian confinement and pinning in a Poissonian potential. I, II. Probab. Theory Related Fields 105 1–29, 31–56.
• [33] Sznitman, A.-S. (1997). Fluctuations of principal eigenvalues and random scales. Comm. Math. Phys. 189 337–363.
• [34] Sznitman, A.-S. (1998). Brownian Motion, Obstacles and Random Media. Springer, Berlin.
• [35] van der Hofstad, R., Mörters, P. and Sidorova, N. (2008). Weak and almost sure limits for the parabolic Anderson model with heavy tailed potentials. Ann. Appl. Probab. 18 2450–2494.
• [36] Whittington, S. G. and Soteros, C. E. (1990). Lattice animals: Rigorous results and wild guesses. In Disorder in Physical Systems. 323–335. Oxford Univ. Press, New York.


text:
Next: Summary Up: THEORY Previous: Elastic specular reflectivity estimation

## Elastic parameter estimation

Once we have obtained the elastic reflectivity, , we can consider how to invert for three elastic parameters at each subsurface location.

We begin by making the linear Bortfeld approximation (Bortfeld, 1961, p. 489) to the nonlinear Zoeppritz Rpp coefficient (Aki and Richards, 1980, p. 153). This means that we assume that changes in elastic parameters are linearly related to reflectivity. In practice, this is a good assumption for elastic parameter changes less than about 20% and specular angles less than about 30 ( is a normal incidence reflection). By elastic parameters, we mean the change in elastic property divided by the average elastic property at any given location. In other words, the parameters we estimate are relative changes and are not absolute magnitudes of properties. Hence we will quote dimensionless values of the parameters, like +10%, instead of +0.2 g/cc (density) or +300 m/s (velocity).

The linearized Bortfeld approximation to can be written as

 (4)

where the are known coefficients of specular angle (mathematically determined by choice of elastic parameterization), and is the set of three elastic parameters to be found. Generally, P is related to compressional wave properties of the earth, S to shear wave properties, and D to density. At a single location we have Rpp defined for several values: as many as there are offsets in the recorded data (recall the constant offset summations). The Bortfeld approximation above can then be written as a linear matrix equation

 (5)

where is the matrix of the coefficients , is the parameter vector , and is the corresponding vector of Rpp values,

 (6)

If our original data contain 60 offsets (as in a 120 trace per shot standard marine survey), then we will have 60 values, will be a 60x3 matrix, will be the 3x1 parameter vector, and will be the 60x1 Rpp (data) vector. This is clearly an overdetermined problem since we are using 60 data equations to solve for only 3 parameters. Hence we solve the matrix solution by least squares with a technique known as Singular Value Decomposition (SVD), which decomposes the system into an orthogonal reference frame and reconstructs an optimal solution by eigenvector methods (Lines and Treitel, 1984). We use the spectrum of the 3 eigenvalues to determine the stability of the solution and to estimate the confidence in the inversion result, as discussed later.

Next: Summary Up: THEORY Previous: Elastic specular reflectivity estimation
Stanford Exploration Project
12/18/1997


text:
# Triangular Numbers.....Part-2

Number Theory Level pending

Let $$t_{n}$$ denote the $$n^{th}$$ triangular number. Let $$t_{a},t_{b},t_{c}$$ be three successive triangular numbers such that their sum is perfect square.Find the minimum possible value of $$a+b+c$$.

Note- A triangular number is a number in the form of $$\frac{n(n+1)}{2}$$ where $$n\ge1$$.

×


text:
# Analytical continuation of Taylor series

Let us expand $\sqrt{1-x^2}$ in Taylor series...

Images:

### I get an error when

I get an error when evaluating this code, saying that ** should be used for exponentiation instead of ^.  If I change the carets on lines 7 and 29 to double stars, then the errors go away, even though I can see several other uses of the caret in the code . . . strange.

### Youp  - and I think this

Youp  - and I think this error was absent at the time I posted this interact.
Anyways - I have changed all ^ to ** and it it fine ;-)


text:
Skip Nav Destination

# Flow Induced Vibration of Power and Process Plant Components: A Practical Workbook

By
M. K. Au-Yang, Ph.D., P.E.
M. K. Au-Yang, Ph.D., P.E.
Search for other works by this author on:
ISBN-10:
0791801667
No. of Pages:
494
Publisher:
ASME Press
Publication date:
2001

Although not the most costly, acoustically induced vibration is probably one of the most common vibration problems in the power and process industries. Vessels, piping systems, valve cavities, heat exchanger internals, ducts and many other components are potential resonators in which standing waves can form, while fans, pumps, valves, elbows, obstruction and discontinuities in flow channels, or even the addition or removal of heat all have the potential to excite these standing waves. Once the resonant conditions are met, the resulting sound intensity in most cases will require remedial action. In some cases, acoustic excitation can cause rapid fatigue failure of piping welded points, valve internal parts and other components.

The first requirement in acoustically induced vibration analysis is to calculate the velocity of sound in the fluid media. The velocity of sound in air at 68 deg. F (20 deg. C) and one atmospheric pressure is 13,500 in/s or 343 m/s. The velocity of sound at other temperatures can be readily calculated from the equation,
$c=(∂p∂ρ)s=γpρ=γGT∝Tγ=Cp∕Cv$
where T is the absolute temperature in either deg. R or deg. K, depending on which unit system is used. At atmospheric pressure and 0 deg. C (32 deg. F) the velocity of sound in water is 55,288 in/s or 1,404 m/s. The velocity of sound in water, steam or water-steam mixture at any given temperature and pressure combination can be calculated from information given in the ASME Steam Table (1979), as outlined in Examples 12.2 and 12.3. A table of velocity of sound in water at selected values of temperatures and pressures are given in Chapter 2, Table 2.1.
In a heat exchanger, the velocity of sound in the direction transversal to the tube bundle axis is decreased by the presence of the tubes. If c0 is the velocity without the tubes and c is the velocity in the presence of tubes, then
$c=c01+σ$
where σ is the ratio of the heat exchanger internal volume occupied by the tubes to the total volume.
This content is only available via PDF.


text:
# Why don't charges move transverse to an EM wave?

Image we have an ultra-high intensity, ultra low frequency laser, with wattage on the order of terawatts and a wavelength on the order of a lightsecond. We rotate it that the electric field component is oriented on the $\hat z$ axis, then fire it at a macroscopic block with a positive electric charge. Because of the low frequency the block will experience an electric that doesn't immediately change direction, and because of the high intensity the field will be very strong. So from this naive understanding of classical physics, the block will briefly levitate.

Except this blatantly contradicts both QM (Compton scattering) and multiple macroscopic experiments (like solar sails), which both say that the block will be pushed in the direction of the laser. What assumptions in the original problem are missing/wrong?

-


text:
2:41 AM
@PauloCereda I found my old self-written manual for ghostscript.yaml
You coul add it here, if you want:
2

I have the arara rule ghostscript.yaml written by @clemens and @esdd and some lints added by myself, which converts pdf-files into graphic files, like png, using ghostscript. I set up TeXLive 2018. My old arara rules do not work any more. I tried the rule converter rc.jar without succes. What ...

%### ghostscript.yaml ###
% arara: pdflatex
% arara: ghostscript: {  resolution: 300, device: pngalpha  }

%#######################################
%# Parameters by default:
%# { program: gs / gswin32c ,  resolution: 300  , device: png16m ,
%#   alphabits: 4 , allpages: false ,
%#   options: -q -dNOPAUSE -dBATCH  -dEPSCrop }
%#
%# resolution: <number>  (300 [dpi] by default)
%#
%# allpages: false (1st page as PNG)/ true (all pages as seperatly PNGs)
%#
%# alphabits: <number> (4 by default)
%#
%# device (examples):

4 hours later…
6:39 AM
@UlrikeFischer nothing at all:-) actual reason is to learn some German language and culture

6:57 AM
@DavidCarlisle I thought the z is perhaps some special english idiom that I don't know as I can't find the z on my keyboard ;-) I wonder how much German culture you can learn in a park which is mostly about "back to wilderness and nature".

7:40 AM
@UlrikeFischer day trips to Cologne and Kommern Freilicht Museum as far as I can see and some lessons at a Language school somewhere around there...

7:51 AM
@DavidCarlisle we were quite regularly in Kommern when we still lived in Bonn and Siegburg and we are still member of their "Förderverein". That's quite nice if you are interested in history.

8:25 AM
@DavidCarlisle I just ran that graphicspath with utf8 in latex-dev and it seems to work ...

@UlrikeFischer in a case when it wouldn't find the file normally? What's Frank changed.....

@DavidCarlisle I didn't check on a real case, only if you still get an "invalid utf8"

@UlrikeFischer ah OK, I thought you meant \graphicspath{a/b/c} worked instead of \graphicspath{{a/b/c}}

9:14 AM
I'm looking at the PGF manual, and I was hoping someone could clarify one of the examples for me. I'm not sure what is going on here.
\def\a{bottom}
\def\b{\a}
\def\c{\b}
\pgfkeys{/key1/.initial=\c}
\pgfkeys{/key2/.initial/.expand once=\c}
\pgfkeys{/key3/.initial/.expand twice=\c}
\pgfkeys{/key4/.initial/.expanded=\c}
\def\a{{\ttfamily\string\a}}
\def\b{{\ttfamily\string\b}}
\def\c{{\ttfamily\string\c}}
\begin{tabular}{ll}
Key 1:& \pgfkeys{/key1} \\
Key 2:& \pgfkeys{/key2} \\
Key 3:& \pgfkeys{/key3} \\
Key 4:& \pgfkeys{/key4}
\end{tabular}
This is the Key handler hkeyi/.expand once=<value> section.
It's probably suitable as a question. Should I ask it as a question?

9:49 AM
@FaheemMitha if I make that into a document I get what I would have expected from the names of the .zzzz functions:
\documentclass{article}

\usepackage{pgfkeys}

\begin{document}

\def\a{bottom}
\def\b{\a}
\def\c{\b}
\pgfkeys{/key1/.initial=\c}
\pgfkeys{/key2/.initial/.expand once=\c}
\pgfkeys{/key3/.initial/.expand twice=\c}
\pgfkeys{/key4/.initial/.expanded=\c}
\def\a{{\ttfamily\string\a}}
\def\b{{\ttfamily\string\b}}
\def\c{{\ttfamily\string\c}}
\begin{tabular}{ll}
Key 1:& \pgfkeys{/key1} \\
Key 2:& \pgfkeys{/key2} \\
Key 3:& \pgfkeys{/key3} \\
Key 4:& \pgfkeys{/key4}
\end{tabular}
\end{document}
@FaheemMitha \c, \b, \a, bottom

@DavidCarlisle I'm not saying there is anything wrong.
I just don't completely understand what is happening.
Is the TeX book an understandable/simple guide to TeX expansion? I realise I don't really understand that either.

@FaheemMitha as I say the output is as I'd expect from names like .expand once even without looking what pgfkeys is really doing.

@DavidCarlisle Yes, I realise that.

@FaheemMitha you haven't said what you don't understand, were you expecting a different outcome (basically you haven't asked a question that can be answered, just posted some code:-)

@DavidCarlisle Yes, I was writing a question. But it's not complete.
Like I said, I realised I don't really understand how expansion works.
So I was looking at some documentation.

9:57 AM
@DavidCarlisle quack

But to start with, would I be correct to think that after the macro \c was redefined by \def\c{{\ttfamily\string\c}}, when the key executed the code, it woudl be executing the new macro?
Also, it's a bit confusing to have a macro redefined in terms of itself.

@FaheemMitha \c is used after \string, so it won't be expanded.

@egreg Oh. Hmm. One more thing I didn't know.

@FaheemMitha Of course, you should not redefine \c at all.

I don't understand. First, isn't that code redefining the macro \c as well as the others? Sure looks like it.

10:02 AM
@FaheemMitha Probably I'd choose \edef\c{{\noexpand\ttfamily\string\c}} so the new meaning of \c doesn't depend on the value of \escapechar at call time.
@FaheemMitha I mean, in a “real” document.

@egreg Sorry, that's a bit over my head.
@egreg Oh, right. Sure.
I realise this is just for illustrative purposes.

@FaheemMitha Try \def\c{{\ttfamily\string\c}}\c \escapechar=-1 \c

@egreg By try it, you mean feed it to the interpreter?

@FaheemMitha Yes
@FaheemMitha Next, try \edef\c{{\noexpand\ttfamily\string\c}}\c \escapechar=-1 \c

@egreg It complains about \c being undefined. Not surprisingly.
@egreg Pretty much the same thing.
Ok, just to be clear, calling \def a second time overwrites the old defn. I know that much.

10:07 AM
@FaheemMitha Try this:
\documentclass{article}

\begin{document}

{\def\c{{\ttfamily\string\c}}\c \escapechar=-1 \c}

{\edef\c{{\noexpand\ttfamily\string\c}}\c \escapechar=-1 \c}

\end{document}}

Now, presumably it uses the old defn when it does so, in this case.

10:35 AM
Actually, let's back up to the beginning of that code.
So, in:
\pgfkeys{/key1/.initial=\c}
Am I correct to think that the value saved from the rhs is simply the text \c.
So what it expands to in the future will depend on whatever it is defined to expand to at that time?

@FaheemMitha Yes

10:59 AM
@egreg Excellent. Thank you for the confirmation.
Also, \def\a{{\ttfamily\string\a}} clearly overwrites the previous definition of -a.
However, I'm not sure what the resulting macro is.
At first glance it would appear that the previous defn simply gets overwritten and forgotten, but clearly that does not happen.
According to en.wikibooks.org/wiki/TeX/string, \string simply converts it its corresponding string, as one would expect.
So, naively, it looks like \a simply expands to the printed representation of \a. But that is also clearly wrong.
Because that would render the exercise pointless, and the entire output would be simply \c.
Sorry, I should have written above - \string simply converts the following token to its corresponding string.

@FaheemMitha the macro is defined as you show, {\ttfamily\string\a} which if used typesets the string \a in monospace font, the previous definition of \a is lost in the current scope.
@FaheemMitha yes that does happen.

@DavidCarlisle Oh.
In that case, why don't all those strings output \c?

@FaheemMitha which, in your pgf example or egreg's edef version?

@DavidCarlisle My PGF example. Which is straight from the documentation, to be clear.
So in that sense it's not my example.

@FaheemMitha because the first key is defined to be \c but the second key is defined to be \b the third key is defined to be \a and the fourth key is defined to be bottom

11:13 AM
@DavidCarlisle Oh, I see. The expansion occurs at the time the keys are .initialized.
So not at the time they are run? Or I guess the expansion would continue at the time they were run, if they weren't fully expanded?
BTW, is it possible in TeX to actually define a macro in terms of itself?
That doesn't seem like a great idea, but I also don't see how it would work.

@FaheemMitha \def\z{\z} but it loops forever
@FaheemMitha that's the natural reading of \pgfkeys{/key2/.initial/.expand once=\c} that the key is initially defined with the first expansion of \c.

@DavidCarlisle That's what I thought. I suppose TeX will obediently try to expand \z in terms of the current definition of \z, so recursion without end.
@DavidCarlisle Ok. I guess this didn't quite register with me. I suppose I thought it would happen when the key was executed.

@FaheemMitha \def\z{\z}  is literally without end, it's tail recursion elimiation means you will have to kill the program. \def\z{.\z}  will run for a while then overflow the call stack and \def\z{\z.}  will run for a while then overflow the input buffer

But I see that the documentation does say the first expansion is saved.
@DavidCarlisle So there is no useful way to call a TeX macro in terms of itself?

@FaheemMitha I haven't seen the doc or that form before I just guessed what it was doing from the name

11:25 AM
@DavidCarlisle Ok.

@FaheemMitha sure but like any recursion in any language, you need some end condition somewhere

@DavidCarlisle So that's possible?
Adding an end condition, I mean.

@FaheemMitha \def\z{\advance\count0 1 \ifnum\count0<100 \z \else stop\fi}

@DavidCarlisle Yes, I see. So it prints 1 to 99?

@FaheemMitha it could do that but as written it just calls itself 100 times (if count0 starts at 0) and then typesets stop (so count0 will be 100 but if you did \the\count0,\z instead of just \z in the recurse case it would print all the values

11:32 AM
@DavidCarlisle Ok, right.

11:46 AM
@DavidCarlisle fixed point. :)

12:36 PM
@PauloCereda \def\duck{\def\duck{dinner}\duck}\duck

@DavidCarlisle ooh I see what you did there
@DavidCarlisle it's mean yet clever :)

@PauloCereda and tasty

@DavidCarlisle oh no
@DavidCarlisle ^^

@PauloCereda wise advice, shame it never worked and no text ever appeared.
Jan 12 '16 at 12:15, by David Carlisle

12:51 PM
oh no
@DavidCarlisle oh no
<3

1:13 PM
Does anyone know why @marmot was suspended?

@Raaja best not to discuss these things in public chat (I didn't know he was but if you scroll up this chat a few days you will see some discussions.....)

@DavidCarlisle ok I am sorry, I will refrain from repeating this in the future. Thanks for the reference.

1:33 PM
@Raaja To stop escalating heated discussions on the site, a few users have been temporarily suspended to have a pause from this for a few days. Not meaning a bad thing from a single one or another user but to stop a fight here, so actually not a name needed.

1:56 PM
@StefanKottwitz Thanks for the reply. Just to be clear for me, so is it common that both the end of parties will be suspended? is it the usual way?

@Raaja Nothing is common (luckily) as it's rare but that can happen if it wasn't possible to figure out who's fault it is (or if it's just one causing it) and when it has to be stopped and that wasn't possible by simply asking for it.

@Raaja Each case is handled on its own merits; there isn't really 'usual'

@StefanKottwitz @JosephWright Thanks for the clarification.
@StefanKottwitz @JosephWright [personal opinion] I think nothing is usual way or moderating is not fair to both the parties. Of course, the discussion will indeed end with the statement "it is what it is". Sometimes, it is worthwhile to sit in the shoes of the parties to get a different perpsective, I think. Nevertheless, thanks again for the clarification.

2:16 PM
Hello everyone! I want to make citations without adding a references section at the end
\bibliographystyle{named}
\vspace{10000pt}
\bibliography{references}
Am I doing this correctly?
(lol)

2:41 PM
@MeetTaraviya the \vspace  is weird, I'd remove that, otherwise your markup looks OK if you have a references.bib and run bibtex, but that will make a references section.

Hi @DavidCarlisle I know how to use bibtex, I am trying to make in-text citations without creating a references section at the end
Hence the vspace
Its sends the references section the hell out of my page

@MeetTaraviya the vspace will do nothing it will just force a page break then be discarded. But if you don't want a reference section what do your citations refer to? If you just want it to say [Smith] then you can write [Smith] and don't need \cite at all.
@MeetTaraviya something very strange happening if it has that effect.

It's a single page poster

@MeetTaraviya and we were supposed to be able to tell that from the supplied code?

> According to British tabloid The Mirror Doctor Who’s twelfth season will include an episode in which Mary Shelley is inspired to write Frankenstein after meeting the Cybermen.

2:46 PM
I am not trying to get help, I am just sharing something that I ended up doing after spending hours of tring to do it the "proper" way

@MeetTaraviya so why ask Am I doing this correctly? ???

You're an avid reader, but you missed the lol at the end

@MeetTaraviya I saw it but read what you posted from the viewpoint of someone who isn't you, it looks like a question about a document, not a statement about a poster.
@MeetTaraviya in general chat is a bad way of sharing things as it just scrolls up and essentially no one ever looks at the history. If you do want to share some technique it's better to ask a question and then post an answer.

@DavidCarlisle ^^ total of videos might take a while

3:14 PM
I don't understand what Boris is getting at in the last two lines of this answer:
23

Suppose you want to define a command \foo2. You cannot do this because 2 is not a letter. However, this construction works: \csname foo2\endcsname. Sometimes this is useful, e.g. when you need a series of commands, \foo1, \foo2, etc (another way is to use roman numerals). Another example, sup...

> Another example, suppose you want to define a series of commands like \endsection, \endsubsection, etc. Then you can use a loop with \expandafter\def\csname end#1\csname...

3:28 PM
@FaheemMitha You could set them up by passing a list section,subsection,subsubsection, Etc.

3:52 PM
@JosephWright Ok, but shouldn't it be \expandafter\def\csname end#1\endcsname...?

Hmmm...

@FaheemMitha yes (it is now)
@Werner scroll up an hour or two

@DavidCarlisle OK. Thank you.

4:08 PM
@FaheemMitha Yes

@JosephWright Ok, thanks.

4:31 PM
@DavidCarlisle Thanks.

@AlanMunn oh

5:55 PM
Does \edef leave parameter tokens (e.g. #1) alone, as in not expand them? I suppose I could experiment, but it's easier to ask.
I can't imagine what it would do with it. After all, it still needs the parameter(s) even after expansion...
Also, TeX doesn't care if I stick the definition and execution of \defs inside \defs, correct?

@TeXnician a reply will come soon, hopefully. :)

@PauloCereda Take your time. I have quite a lot to reply as well :)

@TeXnician I blame the island. :)
@DavidCarlisle Now I decided to do some research on ω-languages because of our talk yesterday. :)

6:13 PM
Is there some simple way of telling TeX not to expand anything from here to there?

@FaheemMitha \unexpanded{..............} (etex)

@DavidCarlisle Oh. But I don't see people using it.
That \noexpand thing is very popular.

@DavidCarlisle bah extensions

Doing a search.

@FaheemMitha there are not many times it is needed, also of course etex is still (after all these decades) not as well known as tex

6:14 PM
@DavidCarlisle Does it work with LuaTeX too?

@FaheemMitha yes

@DavidCarlisle Ok, thank you.
@DavidCarlisle Trying to find a definition. But I suppose the argument needs to be balanced text, right?

@FaheemMitha yes, texdoc etex to see the original spec for \unexpanded (page 14)

@DavidCarlisle Ok.
Yes, that's what I was just looking at.
> \unexpanded<general text>.
The expansion is the token list <balanced text>.
That's the definition?

@FaheemMitha yes exactly. It just returns its argument

6:20 PM
@DavidCarlisle Oh. So like verbatim?

@PauloCereda ?

@FaheemMitha well \detokenize is more like verbatim unexpanded is more like putting \noexpand before each

@DavidCarlisle I forget what \detokenize does, but ok.

@FaheemMitha specified in the same document as that's an etex command as well

6:23 PM
Oh, here's the definition.
> • \detokenize<general text>.
The expansion is a list of character tokens representing the token list
<balanced text>. As with the lists of character tokens produced by TEX’s
\the and ε-TEX’s \readline, these tokens have category 12 (‘other’),
except that the character code 32 gets category 10 (‘space’).

@FaheemMitha so \detokenize is like putting \string in front of each token, instead of \noexpand

@DavidCarlisle Yes, I see. So it takes them out of action, so to speak.

@PauloCereda You shouldn't. Maybe the warm temperatures will never return from there ;)

@PhelypeOleinik See email

@JosephWright ^^ :-)

6:36 PM
@DavidCarlisle ooh

@DavidCarlisle :)
@PauloCereda Did you see the secret message?

@JosephWright no... :(

@JosephWright Replied :D
@DavidCarlisle Should I "oh no" to that?

@PhelypeOleinik And actioned

@PhelypeOleinik oh no

6:43 PM
@PhelypeOleinik Welcome to the team :)

@JosephWright Thanks :D

@PhelypeOleinik did you see the deleted message?

@DavidCarlisle Barely. You deleted before I was able to read.

2 days ago, by David Carlisle
@FaheemMitha no it's just that Ulrike is the newest member of the l3 team so she inherits the blame for all reported bugs.

@DavidCarlisle ooh a secret
@DavidCarlisle ooh
@JosephWright a secret!

6:45 PM
@PhelypeOleinik @UlrikeFischer will be pleased that is no longer true:-)

@DavidCarlisle woo

@DavidCarlisle oh no? :-)

@PhelypeOleinik oh yes :)
@PhelypeOleinik congrats

@PauloCereda Thanks :-)

@PhelypeOleinik secrety secrets :)

6:49 PM
@DavidCarlisle and what the second newest member of the team inherit?

@PhelypeOleinik welcome and you could ask Frank for a shirt (but in view of my experiences with the brasilian post system you probably won't get it before christmas):
Oct 17 '18 at 12:14, by Ulrike Fischer

@UlrikeFischer das Bär
@UlrikeFischer my tee is somewhere. :)

@DavidCarlisle that I got before I was in the team, that was my ticket.

@UlrikeFischer /pdfmark then :-)

6:53 PM

@DavidCarlisle buh. I want to keep the blame, that's more fun.
2

@UlrikeFischer Thanks! It would be a nice christmas present :-)

@PhelypeOleinik But you are going to pay the customs anyway. :)

@PauloCereda That is if someone at the customs doesn't decide the shirt is illegal, for some reason :P
@PauloCereda Perhaps (and that is a very long perhaps) I'll move to Europe for my PhD, then I'll ask for one :-)

@PhelypeOleinik No. Correios is now in charge of any international parcel that arrives in Brazilian soil, and they charge R\$ 15,00 for releasing it.
@PhelypeOleinik Yay! Yay! Yay!
@PhelypeOleinik congrats!

6:58 PM
@PauloCereda Hm... Less than I would've guessed :-)

@PhelypeOleinik Still, you are paying for something others sent you. :)

@PauloCereda As I said, a very long perhaps.
@PauloCereda Not surprising

@PhelypeOleinik not at all. :)

7:34 PM
@PhelypeOleinik we should update the getRandBlame routine now. :)

7:49 PM
@PauloCereda I'd propose weighing the distribution based on the time the subject is part of the Team :-)

I wouldn't mind a TeX t-shirt.

@PauloCereda Sorry you got beaten to be first Brazilian on the team ...

8:07 PM
@PauloCereda here's a suggested implementation echo Phelype

@DavidCarlisle My life has no meaning anymore ;-(

8:23 PM
@UlrikeFischer you can still be blamed for anything wrong with hyperref or the oberdiek packages, and of course anything in miktex

1 hour later…
9:26 PM
@marmot Hi, The Physics SE website is impressive: I don't see a serene climate but many heavy and rude comments. I don't even know how to write a question anymore. There is no union, there is no humanity, most of them are cold and without a bit of heart. They vote against me or even eliminate the good questions, just for theoretical clarification. But what is this site for?

10:13 PM
@StefanKottwitz @JosephWright Good evening. Excuse me for this request. I don't want to go into the merits of your decisions and personal situations. I do not understand marmot should be suspended. I personally absolutely do not see anything that has done bad. Why do good and competent people always have to flee? For some there were very good reasons for marmot I do not understand. Excuse me, but I believe in the friendship of all of us. Good evening

@Sebastiano Can we please leave this discussion out of chat.
2

@Sebastiano If you wish to discuss this, meta is the right place. However, note that there are limits on what the moderators can say in public

My phone isn't really up to it, but I heard there was a moon landing, or something....

@JosephWright <3
@DavidCarlisle ooh smiley thingy

10:33 PM
@AlanMunn Yes, of course.
@JosephWright I'm sorry, Joseph, but I'm sorry a friend's leaving. I have never entered into discussions that remain personal between you. Sorry for my sensitivity.


text:
×

# Sheokand and String(SHKSTR) TLE

 0 While solving this question I first used a Map (Key: Character, Value: TrieNode) to solve the problem. Considering O(1) insertion and lookup, my code should not have resulted in TLE but when I used an array of length 26 to store trie nodes, the code got accepted. So should I avoid using maps or dictionaries in CP and why? Is this because its time complexity for insertion and lookups is not exactly O(1)? Solution using Map: 30 points Solution using array of length 26: 100 points asked 12 Jun '18, 11:20 3★baap_42 1●1 accept rate: 0%

 0 The $O(1)$ doesn't mean it is fast enough.... The time taken by earth for one revolution around the sun is $O(1)$ infact everything periodic is O(1)... As the time took by earth to revolve around sun won't change by the "current year" and hence it is $O(1)=constant$. Even time taken for rotation is also $O(1)$. So understand there is a difference between both O(1)'s. Both are constant.. but one constant is large enough... And another is not.. Hence my advice is not to use map set unnecessarily... Use them whenever they are required... Map and sets are meant when "n" is large enough so u need something that doesn't take more time with more "n" and hence maps and sets were useful.. but for 26 characters it is not useful at all answered 12 Jun '18, 12:43 1.4k●1●9 accept rate: 25%
 0 Consider using even faster IO for both reading and writing. I submitted your code with my even faster IO. Your code with my IO. Notice the difference in speed. I haven't gone through your code, but I solved this using Map(Character, TrieNode) myself. You can have a look at my soln here code. Have added useful messages. Let me know if you face issues! answered 12 Jun '18, 14:02 60●3 accept rate: 25% 1 @l_returns also has a valid point. Some questions really need to you squeeze the bajezuz out of the hardware. CHSIGN in prev month competition required a lot of optimisation (not related to core logic). You can continue using maps. However if you notice that 1 or 2 are TLE and AC ones are close to Time Limit, consider using alternatives like arrays (or perhaps a better logic <- No shit @kukreja_vlk) (12 Jun '18, 14:08)
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question asked: 12 Jun '18, 11:20

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last updated: 12 Jun '18, 14:08


text:
Johansens Cointegration test - VARselect output showing all Lag 1's

I am running a cointegration test on 4 input variables over 1 year and when I run the VARselect it outputs:

AIC(n)  HQ(n)  SC(n) FPE(n)
1      1      1      1

Input - VARselect(Data, lag.max = 10, type = "const")

This is an issue as the minimum K for the ca.jo in RStudio is 2.

I have looked for answers on this site and other papers but cant get a clear answer (or perhaps understand them!).

Question - Can anyone advise how I interpret this or is there a means to choosing a correct lag (say 2)? When I run the ca.jo with 2 or say 4 lags it gives me different hypothesis outcomes.

• Do you understand why ca.jo requires a minimum K of 2? Nov 25 at 12:00
• It is because VARselect does not try to output a result that would necessarily fit into ca.jo. It finds a "best" model, and if that happens to contain 1 lag, then that is it. Might it be that ca.jo only considers a special case of $K\geq2$ while in the general case $K=1$ is also permissible (i.e. it is not incorrect mathematically)? I used to know these things but have forgotten... Nov 26 at 8:15


text:
# Quark Matter 2018

May 13 – 19, 2018
Venice, Italy
Europe/Zurich timezone
The organisers warmly thank all participants for such a lively QM2018! See you in China in 2019!

## Centrality dependence study of nuclear modification factor of electrons from heavy-flavour hadron decay in p-Pb collisions with ALICE at the LHC

May 15, 2018, 5:00 PM
2h 40m
First floor and third floor (Palazzo del Casinò)

### First floor and third floor

#### Palazzo del Casinò

Poster Open heavy flavour

### Speakers

Sudipan De (Indian Institute of Technology Indore (IITI)) Shingo Sakai (University of Tsukuba (JP))

### Description

The goal of relativistic heavy ion collider experiments is to explore the properties of the strongly interacting matter produced with very high temperature and energy density, conditions under which the formation of a Quark-Gluon Plasma (QGP) is expected. Heavy quarks, i.e. charm and beauty are sensitive probes of the QGP as they are produced in the initial stages of the collision and witness the entire evolution of the system. Measurements in p-Pb collisions help understanding Cold Nuclear Matter (CNM) effects such as the modification of the nuclear Parton Distribution Function (nPDF) with respect to the expectation from proton PDF, parton momentum ($k_{\rm T}$) broadening from soft scattering processes and parton energy loss in nuclear matter. Studies of heavy-flavour production in different centrality intervals can provide information on the dependence of CNM effects on the collision geometry and on the density of final-state particles. In addition, these allow us to study the interplay between the hard and soft processes in heavy-flavour production. In this contribution, we will present the $p_{T}$-differential cross-section of heavy-flavour decay electrons for different centrality intervals in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV at mid-rapidity ($|\eta| <$ 0.8). The Time Projection Chamber (TPC) was used to identify the electrons in 2 $< p_{T} <$ 8 GeV/c and the Electromagnetic Calorimeter (EMCal) was used to extend the $p_{T}$ range up to 16 GeV/c. The nuclear modification factor, $Q_{pPb}$, ratio of transverse momentum spectra in p-Pb collisions and the corresponding cross section in pp collisions scaled by the nuclear overlap function, and the central-to-peripheral ratio $Q_{cp}$, ratio of transverse momentum spectra measured at central collisions to that from peripheral collisions, scaled by the corresponding nuclear overlap functions, will be presented for different centrality intervals of p-Pb events.

Content type Experiment ALICE Presenter name already specified

### Primary author

Sudipan De (Indian Institute of Technology Indore (IITI))


text:
Publication
Title
An econometric property of the g-index
Author
Abstract
 Let X = (x1, , xN) and Y = (y1, , yN) be two decreasing vectors with positive coordinates such that (representing e.g. citation data of articles of two authors or journals with the same number of publications and the same number of citations (in total)). It is remarked that if the Lorenz curve L(X) of X is above the Lorenz curve L(Y) of Y, then the g-index g(X) of X is larger than or equal to the g-index g(Y) of Y. We indicate that this is a good property for so-called impact measures which is not shared by other impact measures such as the h-index. If L(X) = L(Y) and we prove that g(X) g(Y). We can even show that g(X) > g(Y) in case of integer values xi and yi and we also investigate this property for other impact measures.
Language
English
Source (journal)
Information processing and management. - Oxford
Publication
Oxford : 2009
ISSN
0306-4573
Volume/pages
45:4(2009), p. 484-489
ISI
000267170100008
Full text (Publisher's DOI)
UAntwerpen
 Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address


text:
# Body Mass Index

Not Reviewed
Equation / Last modified by KurtHeckman on 2018/07/17 13:30
BMI =
Tags:
Rating
ID
KurtHeckman.Body Mass Index
UUID
69ac8c5d-3abb-11e3-bfbe-bc764e049c3d

The Body Mass Index (BMI) calculator computes one's body mass index using their weight and height.

INSTRUCTIONS: Choose your preferred weight and height units and enter the following:

• (h) This is the person's height.
• (w) This is the person's weigth.

BMI: The calculator returns the Body Mass Index (BMI).

Related Calculators:

#### The Math / SCIENCE

The body mass index (BMI), or Quetelet index, is a measure of relative size based on the mass and height of an individual.  The formula for BMI is as follows:

BMI = height / weight2

where:

The calculator makes the unit conversions and then computes the index in kilograms and meters.  It also rounds to the nearest tenth.

Wikipedia (wikipedia.org/wiki/Boddy_mass_index) list the following interpretations for BMI.

General Interpretation:

CategoryBMI (kg/m2)BMI Prime

fromtofromto
Very severely underweight
15
0.60
Severely underweight15160.600.64
Underweight1618.50.640.74
Normal (healthy weight)18.5250.741.0
Overweight25301.01.2
Obese Class I (Moderately obese)30351.21.4
Obese Class II (Severely obese)35401.41.6
Obese Class III (Very severely obese)40
1.6

International Interpretations:

The Hospital Authority of Hong Kong recommends the use of the following BMI ranges:[17]

CategoryBMI (kg/m2)

fromto
Underweight
18.5
Normal Range18.523
Overweight—At Risk2325
Overweight—Moderately Obese2530
Overweight—Severely Obese30

Japan Society for the Study of Obesity (2000):[18]

CategoryBMI (kg/m2)

fromto
Low
18.5
Normal18.525
Obese (Level 1)2530
Obese (Level 2)3035
Obese (Level 3)3540
Obese (Level 4)40

Sinagpore

Health RiskBMI (kg/m2)
Risk of developing problems such as nutritional deficiency and osteoporosisunder 18.5
Low Risk (healthy range)18.5 to 23
Moderate risk of developing heart disease, high blood pressure, stroke, diabetes23 to 27.5
High risk of developing heart disease, high blood pressure, stroke, diabetesover 27.5

The index was devised by Adolphe Quetelet during the course of developing what he called "social physics", between 1830 and 1850.[1] The BMI for a person is defined as their body mass divided by the square of their height—with the value universally being given in units of kg/m2; however, vCalc allows the user to use other units via the pull-down menu.

# Reference

• Wikipedia - http://en.wikipedia.org/wiki/Body_mass_index


text:
frudawski

# ciespec2Y

The ciespec2Y function calculates the integrated photometric value Y of a V(\lambda) weighted spectral power distribution. Any errors in the data set or in results generated from the reference data or the Lighting Toolbox are not in the liability of the CIE nor me, see licence.

Y = ciespec2Y(lam,spec,k)

Where:

Note: alternatively one can use the function ciespec2unit with reference ‘y’ or ciespec2xyz.

Examples

Determine the photometric value from the spectral power distribution of standard illuminant ‘A’:

lam = 360:830;
spec = ciespec(lam,'A');
Y = ciespec2Y(lam,spec)

Result:

Y = 7.3689e+06

Determine the photometric values from the spectral power distributions of standard illuminant ‘D55’ and ‘D65’:

lam = 360:830;
spec = ciespec(lam,{'D55','D65'});
Y = ciespec2Y(lam,spec)

Y =
7.2173e+06
ISO/CIE 11664-1:2019(E): Colorimetry - Part 1: CIE standard colorimetric observers. Commission International de l’Éclairage (CIE), Vienna Austria, 2019.