text:
# Which law states that "A good absorber is a good emitter"?

1. Stefan's law
3. Biot-Savart law
4. Bragg's Law

Option 2 : Kirchhoff's Radiation Law

## Detailed Solution

Kirchhoff's Law

For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity and absorptivity are equal. That means anybody absorbing heat well will also emit the heat energy well.

 Important Laws Details Stefan's Law Total radiant heat energy emitted from a surface is proportional to the fourth power of its absolute temperature Kirchhoff's Radiation Law For any arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to absorptivity. Biot-Savart Law It helps in calculating the strength of a magnetic field at various points. Bragg's Law It gives the angles for coherent and incoherent scattering from a crystal lattice.

# A 0.108 mol of sample gas has a volume of 0.505 L at 20°C. What is the pressure exerted by it?

1. 5.99 atm
2. 5.14 atm
3. 6 atm
4. 7 atm

Option 2 : 5.14 atm

## Detailed Solution

Given,

n= 0.108 mol

V= 0.505 L

T= 20 + 273= 293 K

R= 0.08206 L-atm/mol-K

To find P

According to the Ideal Gas Law,

PV= nRT

or, P= $$\frac{nRT}{V}$$

= $$\frac{0.108×0.08206×293}{0.505}$$

= 5.14 atm

# Which of the following is used in a fire alarm?

1. Photo cell
2. Water
3. Solar cell
4. None of the above

Option 1 : Photo cell

## Detailed Solution

The correct answer is Photo Cell.

• Photo Cell is used in fire alarm to detect the smoke.

Important Points

• A photoelectric smoke detector is characterized by its use of light to detect fire.
• Inside the alarm, there’s a light-sensing chamber.
• In this chamber, an LED light shoots a beam of light in a straight light across the chamber.
• The alarm detects smoke; when smoke enters the chamber, it deflects the LED light from the straight path into a photosensor in a different compartment in the same chamber.
• As soon as light beams hit this sensor, the alarm begins to sound.

# Newton’s first law is also known as ________.

1. Law of rotation
2. Law of displacement
3. Law of momentum
4.  Law of inertia

Option 4 :  Law of inertia

## Detailed Solution

CONCEPT:

There are three of Newton’s laws of motion.

• Newton’s first law is also known as the law of inertia. It states that ‘an object will continue to stay at rest and an object in motion will remain in motion in the same direction and speed unless an unbalanced force acts upon it’.
• Newton’s second law of motion states that ‘for a constant mass, force is equal to the product of mass and acceleration. Force is also equal to the change in momentum per change in time.’
• Newton’s third law of motion states that ‘for every action, there is an equal and opposite reaction.’

EXPLANATION:

• Newton’s first law is also known as the Law of inertia.

# Hydraulic machines work under the Principle of

1. Newton's Law
2. Joules Law
3. Pascal's Law
4. Floatation Law

Option 3 : Pascal's Law

## Detailed Solution

The correct answer is Pascal's Law.

• The hydraulic machine works on the principle of Pascal's Law.

Key Points

• Pascal's Law:
• "Any force applied to a confined fluid is transmitted uniformly in all directions throughout the fluid regardless of the shape of the container".
• The two types of hydraulic system:
• Open-loop hydraulic system
• Closed-loop hydraulic system
• In an open-loop system, when the actuating mechanism is idle, there will be fluid flow but no pressure.
• For a closed-loop system, when the pump operates there will be pressure in the fluids.
• The main applications of the hydraulic system are as follows:
• Industrial: Automated production lines, machine tool industries, Paper industries.
• Mobile hydraulics: Material handling equipment, commercial vehicles, tunnel boring equipment.
• Automobiles: It is used in the systems like breaks, shock absorbers, steering system.
• Marine applications: It mostly covers ocean-going vessels, fishing boats, and naval equipment.

• Newtons three laws are as follows:
• In the first law, an object will not change its motion unless a force acts on it.
• In the second law, the force on an object is equal to its mass times its acceleration.
• In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction
• Joule's law is the rate at which resistance in a circuit converts electric energy into heat energy.
• When any boat displaces a weight of water equal to its own weight, it floats. This is often called the "principle of flotation".

# Newton’s first law is also known as _______.

1. Law of rotation
2. Law of displacement
3. Law of momentum
4. None of the option is correct

Option 4 : None of the option is correct

## Detailed Solution

There are three Newton’s laws of motion.

• Newton’s first law is also known as the law of inertia. It states that ‘an object will continue to stay at rest and an object in motion will remain in motion in the same direction and speed unless an unbalanced force acts upon it’.
• Newton’s second law of motion states that ‘for a constant mass, force is equal to the product of mass and acceleration. Force is also equal to the change in momentum per change in time.
• Newton’s third law of motion states that ‘for every action, there is an equal and opposite reaction.

# Hydraulic press works on the principle of ________.

1. Hooke’s law
2. Newton’s third law of motion
3. Pascal’s law
4. Archimedes principle

Option 3 : Pascal’s law

## Detailed Solution

Explanation:

Hydraulic press:

• It is also called Hydraulic Jack or Hydraulic lift.
• The hydraulic press is a device used for lifting heavy weights by the application of a much smaller force.
• It is based on Pascal’s law.

Pascal's Law:

• Pascal's principle states that pressure applied to fluids is equally transmitted in all directionsto all parts of the fluid and to all parts of a container.
• The increase in pressure at one point of the enclosed liquid in an equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container, provided the effect of gravity is neglected.

Hooke’s law: According to this law, within the elastic limit, stress is proportional to the strain.

i.e. stress ∝ strain

or,

$$\frac{{stress}}{{strain}} = constant = E$$

Where E  = modulus of elasticity.

Newton's third law of motion

Newton's Third Law of Motion states that 'To every action, there is an equal and opposite reaction'.

Archimedes principle:

the principle of Archimedes states “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it.”

Thus, when a solid is fully immersed in a liquid, it loses weight which is equal to the weight of the liquid it displaces.

# Which law of thermodynamics defines the concept temperature?

1. First Law of Thermodynamics
2. Second Law of Thermodynamics
3. Zeroth Law of Thermodynamics
4. Third Law of Thermodynamics

Option 3 : Zeroth Law of Thermodynamics

## Detailed Solution

CONCEPT:

There are 4 laws to thermodynamics:

Zeroth law of thermodynamics:

• If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other.

First law of thermodynamics:

• Energy can neither be created nor destroyed. It can only change forms. In any process, the total energy of the universe remains the same.
• For a thermodynamic cycle, the net heat supplied to the system is equals the net work done by the system.

ΔQ = ΔU + ΔW

where ΔQ = change in heat, ΔU = change in internal energy and ΔW = change in work done

Second Law of Thermodynamics:

1. Clausius statement: It is impossible for a self-acting machine to transfer heat from a colder body to a hotter one without the aid of an external agency
2. Kelvin-Planck’s statement: It is impossible to design an engine that extracts heat and fully utilizes it into work without producing any other effect.

Third law of thermodynamics:

• As the temperature approaches absolute zero, the entropy of a system approaches a constant minimum.
• ΔST = 0K = 0

Where ΔS = change in entropy

EXPLANATION:

• The First Law of Thermodynamics tells us about the concept of internal energy. Therefore option 1 is incorrect.
• The Second Law of Thermodynamics tells us that some form of energy gets lost whenever energy is transferred or transformed. Therefore option 2 is incorrect.
• From above it is clear that the Zeroth Law of Thermodynamics defines the concept of temperature. Therefore option 3 is correct.
• The Third Law of Thermodynamics tells us about the concept of entropy. Therefore option 4 is incorrect.

# Dalton’s theory was based on the laws of:

1. Conservation of Momentum

2. Conservation of Mass

3. Conservation of Energy

4. Conservation of Charge

Option 2 :

Conservation of Mass

## Detailed Solution

The correct answer is Conservation of mass.

Key Points

• Dalton's theory was based on the law of conservation of mass and the law of constant composition.
• It is also known as the law of constant composition and the law of multiple proportions.
• Dalton’s atomic theory states that all matter is composed of atoms and all compounds are composed of atomic combinations in defined ratios.

• Conservation of momentum: Conservation of momentum, the general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.
• Conservation of Energy: The law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. In other words, we can say that energy can neither be created nor destroyed - only converted from one form of energy to another.
• Conservation of Charge: The law states that the total amount of electric charge in a system does not change with time.

# Working principle of washing machine is _________.

1. Pressure power
2. Dialysis
3. Centrifugation
4. Diffusion

Option 3 : Centrifugation

## Detailed Solution

• Washing machine works on the principle called Centrifugation.
• Centrifugation is a separation process which uses the action of centrifugal force to promote accelerated settling of particles in a solid-liquid mixture.

# Addition of the Avogadro's law into the combined gas law results in the development of the ________.

1. Ideal Gas Law
2. Real Gas Law
3. Henry's Law
4. Graham's Law

Option 1 : Ideal Gas Law

## Detailed Solution

• Ideal gas law is developed by the combination of the combined gas law and Avogadro's law.
• It is represented by,

PV= nRT, where, P is the pressure of the gas, V is its volume, T is its absolute temperature, R is the universal gas constant and n is the number of moles.

• It gives the idea regarding the behaviour of most of the gases under most of the possible conditions.
• Real gas law is associated with the real gases and they follow the Van Der Waal's Equation.
• Henry's law states that the amount of dissolved gas is proportional to its partial pressure in the gas phase.
• Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of the mass of its particles.

# According to Newton’s which law of motion is possible to swim?

1. First law
2. Second law
3. Third law
4. All options are correct.

Option 3 : Third law

## Detailed Solution

• According to Newton’s third law of motion is possible to swim.
• It states that every action must have an equal and opposite reaction.
• In swimming also when the hands and feet push against the water, the water pushes back on the swimmer and move the swimmer forwards.
 Newton’s law Description First law It states that an object will remain at in uniform motion in a straight line unless acted upon by an external force. Second law It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Third law It states that every action must have an equal and opposite reaction.

# On which of the following principles does a rocket in space work?

1. Principle of Conservation of Mass
2. Principle of Conservation of Speed
3. Principle of Conservation of Momentum
4. Principle of Conservation of Energy

Option 3 : Principle of Conservation of Momentum

## Detailed Solution

Concept:

• Momentum: The product of mass and velocity is called momentum or linear momentum.
• It is denoted by P and unit is kg⋅ m/s
• Momentum (P) = Mass (m) × velocity (V)
• Conservation of linear momentum: When there is no net external force on a system, then the total linear momentum of the system remains constant. This is called the conservation of linear momentum.

Explanation:

In the case of a rocket, we will take the rocket and its fuel as a system. There is no net force on this system, so the total momentum will be conserved.

• Thus the rockets work on the law of conservation of linear momentum is correct.
• The increased forward momentum of the rocket is equal but opposite in sign to the momentum of the ejected exhaust gases.

# Which law states that mass can neither be created nor destroyed in a chemical reaction?

1. Law of Reciprocal proportion
2. Law of Conservation of mass
3. Law of Constant proportions
4. Law of Conservation of energy

Option 2 : Law of Conservation of mass

## Detailed Solution

• Law of Conservation of mass-mass can neither be created nor destroyed in a chemical reaction. The mass of the system must remain constant over time.
• Law of Reciprocal proportion - If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine.
• Law of Constant proportions - It states that a chemical compound always contains exactly the same proportion of elements by mass.
• Law of Conservation of energy - Energy can neither be created nor be destroyed. The total energy of an isolated system remains constant.

# Time period of a pendulum will ______ at the Moon to that of on the Earth.

1. Remain same
2. Decrease
3. Increase
4. No option is correct

Option 3 : Increase

## Detailed Solution

The time period of a simple pendulum is given by 2π√l/g.

• Gravity on moon is one sixth of the gravity on the earth.
• As time period is inversely proportional to g, implies time period will increase.

# TV remote controls work on the principle of ________.

1. Bluetooth technology
2. Ultrasonic waves
3. Laser technology
4. Infrared waves

Option 4 : Infrared waves

## Detailed Solution

• Working principle of TV remote is based on Infrared waves, as a medium of communication.
• The wavelength of infrared waves ranges from 700 nm to 1 mm.
• The inventor of TV remote control: Eugene Polley.

# Which scientist gave 'laws of motion'?

1. Galileo
2. Newton
3. Einstein
4. Boyle

Option 2 : Newton

## Detailed Solution

CONCEPT:

• The ‘Laws of Motion’ were first compiled by Isaac Newton in 1687.
• There are three laws of motion given in the theory by Newton.
• According to Newton’s first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
• Newton’s second law of motion states that for a constant mass, force is equal to the product of mass and acceleration.
• Newton’s Third Law: To every action, there is always an equal (in magnitude) and opposite (in direction) reaction.

EXPLANATION:

• The ‘Laws of Motion’ were first compiled by Isaac Newton in 1687. So option 2 is correct.

# Which of the following is a device designed to protect electrical devices from voltage spikes?

2. Surge protector
3. Heat protector

Option 2 : Surge protector

## Detailed Solution

When voltage of the connection drastically increases to a very high value suddenly for a very short period of time is called a voltage spike.

A surge protector is a device that protects or shields your appliances against a voltage surge/ voltage spike

When the voltage increases beyond a certain level, the surge protector either blocks or transfers the excess energy to the ground. By doing this, it ensures that the appliances connected to it do not get affected or broken down due to a voltage surge.

# “When a card placed over a tumbler is flicked with the finger,  the coin placed over it falls in the tumbler”.

1. Law of conservation of energy
2. Newton’s third law of motion
3. Law of Inertia
4. Law of conservation of momentum

Option 3 : Law of Inertia

## Detailed Solution

• Sir Isaac Newton first presented his three laws of motion in 1686.
• His first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
• This concept is generally known as inertia.
• Extra facts:
• There are three Newton’s laws of motion.
• Newton’s first law is also known as the law of inertia. It states that ‘an object will continue to stay at rest and an object in motion will remain in motion in the same direction and speed unless an unbalanced force acts upon it’.
• Newton’s second law of motion states that ‘for a constant mass,  force is equal to the product of mass and acceleration. Force is also equal to the change in momentum per change in time.’
• Newton’s third law of motion states that ‘for every action,  there is an equal and opposite reaction.

# Which of the following is NOT true about nuclear fusion?

1. It produces much more energy than nuclear fission
2. The process is used for making nuclear bomb.
3. Electricity production through this process is commercially less viable than nuclear fission.
4. It is the source of energy in the stars.


text:
How do I apply MDS analysis on my data set?

Consider the following dataset (it is the emission probability matrix of a Hidden Markov Model):

    H4K12Ac H3K9Ac  H3K4me2 H3K4me3 PolII   H3K18Ac H3K4me1 H3K36me3 H3K27me3
1   0.00    0.00    0.00    0.00    0.00    0.00    0.00    0.00    0.00
2   0.93    0.95    0.86    0.96    0.53    0.46    0.61    0.06    0.00
3   0.08    0.12    0.19    0.70    0.12    0.01    0.00    0.09    0.04
4   0.19    0.17    0.53    0.82    0.10    0.23    0.98    0.11    0.04
5   0.00    0.00    0.01    0.00    0.00    0.00    0.87    0.00    0.01
6   0.00    0.00    0.03    0.00    0.03    0.94    0.99    0.00    0.01
7   0.69    0.46    0.27    0.00    0.12    0.88    0.98    0.05    0.00
8   0.01    0.00    0.00    0.00    0.00    0.59    0.00    0.00    0.00
9   0.12    0.03    0.00    0.02    0.05    0.33    0.87    0.90    0.00
10  0.00    0.00    0.00    0.00    0.00    0.00    0.00    0.66    0.00
11  0.97    0.01    0.00    0.00    0.00    0.01    0.32    0.29    0.01
12  0.03    0.00    0.00    0.00    1.00    0.05    0.15    0.14    0.09
13  0.00    0.00    0.00    0.00    0.00    0.01    0.03    0.00    0.87

So we have 13 states and each state can emit some mark/emission with a certain probability. Now I'd like to do some MDS analysis on those data (based on a paper I am reading). Let us consider the results (from a 51 state HMM)

Question For what I understand, the position of each point on the graph is decided based on the column vectors (ie the emission vector). Or is it the other way around, i.e. the position of each point is decided based on the row vector which represents a specific state?

Secondary question: What does it mean by the standard pairwise correlation coefficient and how can I compute using my dataset?

In your example, the author used the cmdscale function in Matlab. That particular function uses a distance matrix (which is symmetric and has the same labels for rows and columns), and you haven't generated that yet. Remember, MDS is used to generate a more interpretable 2- (or 3-) dimensional projection of an n-dimensional set of distances. The pairwise correlation coefficient they mention refers to the metric used to calculate those distances. You will have to decide if the correlation coefficient or another metric is suitable for determining the distance/dissimilarity between your items (given the number of zeroes in your dataset, correlation is probably not a good choice).

In R at least, the distance matrix can be generated on the rows of your input matrix using the dist function. Some R code that will produce a similar plot using Euclidean distance (and without the point colours) for your data would be:

# assume your data is a matrix called x
d = dist( x, method="euclidean" )
mds = cmdscale( d, k=2 ) # use 2 dimensions
plot( mds, pch=23 ) # 23 specifies diamond symbols
text( mds, rownames(mds), pos=1 ) # place text labels above points

The distance metric and aesthetics can be modified as needed.

• Thank you. I have a much better understanding now. Do you mind expanding on (given the number of zeroes in your dataset, correlation is probably not a good choice).? I don't have a statistics background so I am not sure why this is the case. Also, if I was to use correlation (just to see why its a bad idea), what would be a good formula to use? – masfenix Oct 24 '14 at 14:41
• It was more intuition that made me say that, I didn't have a formal reason in mind. Many zeroes will make a correlation much higher between the rows of your matrix than is probably indicated – scottzed Oct 24 '14 at 18:47
• (accidentally pressed enter above)... Also, the values aren't measurements, they're probabilities, so correlation doesn't make a lot of sense. You should submit a question asking what row-wise distance metric makes sense to use with a matrix of probabilities, I'm sure someone will have a better rationale than I can provide. – scottzed Oct 24 '14 at 18:55

In your document you have the vectors on the columns. In other words it means that each state (i.e. H4K12Ac) can be seen as a vector. A state is different from another ones by the fact that they have different probabilities of emissions.

So it's true what scottzed says:

...
d = dist( x, method="euclidean" )
mds = cmdscale( d, k=2 ) # use 2 dimensions
...

but the calculation of the matrix distance (d) relies on the column: at the position d[1,3] you will find the computed distance between H4K12Ac and H3K4me2. On the diagonal you will find:

d[i,i]=0

since a state has no differences with itself.

In other words the MDS is a way to plot the differences from the elements belonging to dimension n on 2 dimensions such that you can have an overview on the differences!

format: plain


text:
# T-join

## Packing T-joins ★★

Author(s): DeVos

\begin{conjecture} There exists a fixed constant $c$ (probably $c=1$ suffices) so that every graft with minimum $T$-cut size at least $k$ contains a $T$-join packing of size at least $(2/3)k-c$. \end{conjecture}

Keywords: packing; T-join


text:
# luogu2851 最少的硬币The Fewest Coins

### 题目描述

Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of coins he receives in change is minimized. Help him to determine what this minimum number is.

FJ wants to buy T (1 ≤ T ≤ 10,000) cents of supplies. The currency system has N (1 ≤ N ≤ 100) different coins, with values V1, V2, …, VN (1 ≤ Vi ≤ 120). Farmer John is carrying C1 coins of value V1, C2 coins of value V2, …., and CN coins of value VN (0 ≤ Ci ≤ 10,000). The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner (although Farmer John must be sure to pay in a way that makes it possible to make the correct change).

### 输入输出格式

#### 输入格式：

Line 1: Two space-separated integers: N and T.

Line 2: N space-separated integers, respectively V1, V2, …, VN coins (V1, …VN)

Line 3: N space-separated integers, respectively C1, C2, …, CN

#### 输出格式：

Line 1: A line containing a single integer, the minimum number of coins involved in a payment and change-making. If it is impossible for Farmer John to pay and receive exact change, output -1.

3 70
5 25 50
5 2 1

3

### 说明

Farmer John pays 75 cents using a 50 cents and a 25 cents coin, and receives a 5 cents coin in change, for a total of 3 coins used in the transaction.

×


text:
This topic is 4847 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Is it possible to combine multiple rotation and translation matrices in a specific order? For instance, say i want to first rotate, then translate, then rotate again, and then translate again. How do i do that? Can i take the first rotation matrix, add to it the first translation matrix, then multiply by the 2nd rotation matrix, then add the 2nd translation, and then multiply the coordinate matrix by the resulting matrix? Or do i need to apply each matrix seperately to the coordinates i want to transform (ie. (((coordMatrix * rot1)+trans1)*rot2)+trans2)? Thanks.

##### Share on other sites
Say u want to apply matrices A, B, and C to some pt P ... (in the order C,B,A)

newPt = A * (B * (C * P))

Due to the associative nature of matrix multiplication we can multiply all the matrices together first...

T = A * B * C;newPt = T * P;

##### Share on other sites
ooooh... i see my mistake now. So if you have some matrix, and you want to add a translation to it, you would multiply it by a translation matrix, instead of just adding the translation values to the 4th row? So just to confirm, say you have the following 2 matrices:

A:
1 0 0 0
0 cos(A) -sin(A) 0
0 sin(A) cos(A) 0
0 0 0 1

and

B:
1 0 0 T1
0 1 0 T2
0 0 1 T3
0 0 0 1

if you want to do the above rotation, then the translation, it would be B * A, and not just:

1 0 0 T1
0 cos(A) -sin(A) T2
0 sin(A) cos(A) T3
0 0 0 1

right?

edit: ack.. try to pretend that the columns of the matrices are lined up =P

[Edited by - HalcyonX on August 6, 2005 3:53:07 AM]

##### Share on other sites
Yep, but if you do the multiplication then you will find out that B*A actually equals the same thing as just putting the translation values in the 4th column. [smile] Whereas A*B would give you something different (aswell as a different transform)

##### Share on other sites
mm i see. Ok, thanks a lot!

##### Share on other sites
Quote:
 Original post by KuladusYep, but if you do the multiplication then you will find out that B*A actually equals the same thing as just putting the translation values in the 4th column. [smile] Whereas A*B would give you something different (aswell as a different transform)
Well, for homogenous matrices...

##### Share on other sites
Woah !! you mixed everything! To apply a matrix to a point you need to multiply the point by the matrix and not the oposite! For instance if you want to do A, B, c (in this order) to point p you should do: newP = P * A * B * C or, T = A * B * C, newP = P * T
How can you multiply a matrix by a point?? look:
Matrix (4X4)
Point (1X4)
Matrix * Point is legal only when the number columns in the Matrix (4) == Number of rows in the Point (1), which is illegal because 4 != 1, but the oposite is.

##### Share on other sites
Quote:
 Woah !! you mixed everything! To apply a matrix to a point you need to multiply the point by the matrix and not the oposite! For instance if you want to do A, B, c (in this order) to point p you should do: newP = P * A * B * C or, T = A * B * C, newP = P * THow can you multiply a matrix by a point?? look:Matrix (4X4)Point (1X4)Matrix * Point is legal only when the number columns in the Matrix (4) == Number of rows in the Point (1), which is illegal because 4 != 1, but the oposite is.
You can do it either way. Both p' = p*M and p' = M*p are valid - you just have to make sure your matrices are set up appropriately. The former is row-vector notation (1x4, like in D3D), and the latter is column-vector notation (4x1, like in OpenGL).

##### Share on other sites
I was thinking

[a b c d] [x][e f g h] [y][i j k l] [z][m n o p] [w]

as per jyk's post.

1. 1
2. 2
Rutin
19
3. 3
4. 4
5. 5

• 9
• 9
• 9
• 14
• 12
• ### Forum Statistics

• Total Topics
633302
• Total Posts
3011272
• ### Who's Online (See full list)

There are no registered users currently online

×


text:
Returning None or raising Exception

less than 1 minute read

During my development of toppra, a thing bugs me quite often: how to handle failure? The problem is that toppra is a fairly complicated algorithm with several hotspots for numerical instabilities. In another words, it fails often and in different places. For example, consider the below pseudo-code:

def solve_toppra(inputs):
ret = func1()
ret = func2(ret)
ret = func3(ret)
return ret

Now, the algorithm can fail at different points. As a result, there are a lots of error-checking stuffs in my implementation:

def solve_toppra(inputs):
ret = func1()
if ret is None:
return None
ret = func2(ret)
if ret is None:
return None
ret = func3(ret)
return ret

Now, the code is rather convoluted. The original version without error checking looks much cleaner and resemble the pseudo-code better. Furthermore, it is also very difficult to know why the algorithm fails, as the result of solve_toppra is None, conveying next to nothing information. Indeed, everytime toppra fails I have to look into the logs to figure which part has gone wrong, which is really tedious.

The solution http://www.cauldwell.net/patrick/blog/ThisIBelieveTheDeveloperEdition.aspx

Tags:

Categories:

Updated:


text:
# A "Sig Figs" Story

Author Unknown

A student once was given the assignment to make a cube of a metal with a mass of 83 grams. He knew the density of the metal was 8.67 g/cm3. Using his calculator, he found that the required volume would be 9.573 cm3. Taking the cube root of the volume gave him the length of each edge of the cube, 2.1233 cm. He drew up plans for the cube and took them to a local machine shop that his friend had used to do the same job a year earlier.

The foreman said "Yes, we can make this according to your specifications, but it will be expensive."

"That's okay," replied the student. "It's important."

He knew that his friend had paid $35.00 and he himself had been given a budget of$50.00 to get the job done.

He returned the next day, expecting to pick up the cube. "Sorry," said the foreman. "We're still working on it. Try next week."

Finally the day came, and our student got his cube. It looked very smooth and shiny and beautiful in its velvet lined case. Seeing it, our hero had a premonition of disaster and became a bit nervous, but he gathered up enough courage to ask for the bill. "$525.00 and cheap at the price. We had a terrible time getting it right... had to make three before we finally got it." "But -- but -- my friend paid only$35.00 for the same thing!"

"No. He wanted a cube 2.1 cm on an edge, and your specification called for 2.1233 cm. We had yours roughed out to 2.1 cm that afternoon, but it was the precision grinding and lapping to get it down to 2.1233 cm which took so long and cost the big money. The first one we made was 2.0895 cm on one edge when we finished, so we had to throw it away. The second one was closer, but still not what you specified. That's why the three tries."

"Oh."

MORAL: Calculators can increase speed, but never precision.


text:
# pijul remote default

This command sets the default remote repositories to use for the push and pull commands. This will set the default remote for both push and pull, unless changed in the command options.

## Usage

pijul remote default [-h | --help] [-V | --version] [--pull] [--push] <name>

## Options

• pull

When set, the default command will only set the default pull remote (and any other type set in the command flags)

• push

When set, the default command will only set the default push remote (and any other type set in the command flags)

• name

The name of the remote to set as the default. The remote must already be saved by pijul.


text:
There are two programs installed by the instructions in Download and Install. The older one Fokko is designed for doing single (perhaps large) computations: in the command line interface, the user can issue single commands (for example, “compute all of the Kazhdan-Lusztig polynomials for a split group of type E7”). This program was begun by Fokko du Cloux in 2004 and 2005. Since 2006 it has been developed primarily by Marc van Leeuwen, with contributions by Alfred Noel, Jeff Adams, and others.
The program atlas is designed for interactive exploration of real reductive Lie groups and their representations. It has been developed by Marc van Leeuwen since 2006. It is built on a programming language axis, also developed by Marc van Leeuwen. The atlas software invokes (invisibly to the user!) the libraries of Fokko to create mathematical objects; then the axis language offers standard programming tools (loops, variables, scripts, and so on) with which the user can manipulate them. An important part of atlas is the (constantly growing) library of predefined scripts, which can be used or modified to do interesting mathematical tasks. This library has been developed by Marc van Leeuwen, Jeff Adams, Annegret Paul, and others.


text:
# 如何替換插件的Div內容

"0 - String could not be parsed as XML"

$buffer = preg_replace('
([^]*?)<\/div>/','my custom content',$buffer);

Now in Vm3 I'm using following codes.

$docs = new DOMDocument(); libxml_use_internal_errors(true);$docs->resolveExternals = true;
$docs->loadHTML($buffer);
if($getRelatedProduct != '')$element = $docs->createElement('div',$getRelatedProduct);
$xpath = new DOMXPath($docs);
$query = '//div[@class="'.$class_name.'"]';
$entries =$xpath->query($query); foreach ($entries as $entry) {$entry->parentNode->replaceChild($element,$entry);
}
$docs->recover = true; libxml_clear_errors(); JResponse::setBody(htmlspecialchars_decode($docs->saveHTML()));

bcoz all the instance of the class needs to be replaced.

I know for testing the plugin required another Extension (VM3).its difficult for debuggers.But simply my requirement is simply replace all the class instance with mine.

is there any way to replace Div inside Div structure without breaking layouts using preg_replace ??

The Div Structure is

Related Products

Test Product 1Custom prototype for related products

I'm trying to replace "product-field product-field-type-R" it may comes more than one inside "product-fields`"

Hope someone will be able to figure out something :)


text:
• ### The Atmospheres of Earth-like Planets after Giant Impact Events(1401.1499)

Jan. 7, 2014 astro-ph.EP
It is now understood that the accretion of terrestrial planets naturally involves giant collisions, the moon-forming impact being a well known example. In the aftermath of such collisions the surface of the surviving planet is very hot and potentially detectable. Here we explore the atmospheric chemistry, photochemistry, and spectral signatures of post-giant-impact terrestrial planets enveloped by thick atmospheres consisting predominantly of CO2, and H2O. The atmospheric chemistry and structure are computed self-consistently for atmospheres in equilibrium with hot surfaces with composition reflecting either the bulk silicate Earth (which includes the crust, mantle, atmosphere and oceans) or Earth's continental crust. We account for all major molecular and atomic opacity sources including collision-induced absorption. We find that these atmospheres are dominated by H2O and CO2, while the formation of CH4, and NH3 is quenched due to short dynamical timescales. Other important constituents are HF, HCl, NaCl, and SO2. These are apparent in the emerging spectra, and can be indicative that an impact has occurred. The use of comprehensive opacities results in spectra that are a factor of 2 lower in surface brightness in the spectral windows than predicted by previous models. The estimated luminosities show that the hottest post-giant-impact planets will be detectable with near-infrared coronagraphs on the planned 30m-class telescopes. The 1-4um region will be most favorable for such detections, offering bright features and better contrast between the planet and a potential debris disk. We derive cooling timescales on the order of 10^5-10^6 Myrs, based on the modeled effective temperatures. This leads to the possibility of discovering tens of such planets in future surveys.
• ### SPT 0538-50: Physical conditions in the ISM of a strongly lensed dusty star-forming galaxy at z=2.8(1309.3275)

Sept. 12, 2013 astro-ph.CO
We present observations of SPT-S J053816-5030.8, a gravitationally-lensed dusty star forming galaxy (DSFG) at z = 2.7817, first discovered at millimeter wavelengths by the South Pole Telescope. SPT 0538-50 is typical of the brightest sources found by wide-field millimeter-wavelength surveys, being lensed by an intervening galaxy at moderate redshift (in this instance, at z = 0.441). We present a wide array of multi-wavelength spectroscopic and photometric data on SPT 0538-50, including data from ALMA, Herschel PACS and SPIRE, Hubble, Spitzer, VLT, ATCA, APEX, and the SMA. We use high resolution imaging from HST to de-blend SPT 0538-50, separating DSFG emission from that of the foreground lens. Combined with a source model derived from ALMA imaging (which suggests a magnification factor of 21 +/- 4), we derive the intrinsic properties of SPT 0538-50, including the stellar mass, far-IR luminosity, star formation rate, molecular gas mass, and - using molecular line fluxes - the excitation conditions within the ISM. The derived physical properties argue that we are witnessing compact, merger-driven star formation in SPT 0538-50, similar to local starburst galaxies, and unlike that seen in some other DSFGs at this epoch.
• ### Measurements of CO redshifts with Z-Spec for lensed submillimeter galaxies discovered in the H-ATLAS survey(1009.5983)

Aug. 17, 2012 astro-ph.CO
We present new observations from Z-Spec, a broadband 185-305 GHz spectrometer, of five sub-millimeter bright lensed sources selected from the Herschel Astrophysical Terahertz Large Area Survey (H-ATLAS) science demonstration phase (SDP) catalog. We construct a redshift finding algorithm using combinations of the signal-to-noise of all the lines falling in the Z-Spec bandpass to determine redshifts with high confidence, even in cases where the signal-to-noise in individual lines is low. We measure the dust continuum in all sources and secure CO redshifts for four out of five (z~1.5-3). In one source, SDP.17, we tentatively identify two independent redshifts and a water line, confirmed at z=2.308. Our sources have properties characteristic of dusty starburst galaxies, with magnification-corrected star formation rates of 10^(2-3) M_sun/yr. Lower limits for the dust masses (~a few 10^8 M_sun) and spatial extents (~1 kpc equivalent radius) are derived from the continuum spectral energy distributions, corresponding to dust temperatures between 54 and 69K. In the LTE approximation, we derive relatively low CO excitation temperatures (< 100 K) and optical depths (tau<1). Performing a non-LTE excitation analysis using RADEX, we find that the CO lines measured by Z-Spec (from J=4->3 to 10->9, depending on the galaxy) localize the best solutions to either a high-temperature / low-density region, or a low-temperature / high-density region near the LTE solution, with the optical depth varying accordingly. Observations of additional CO lines, CO(1-0) in particular, are needed to constrain the non-LTE models.
• ### The Detection of a Population of Submillimeter-Bright, Strongly-Lensed Galaxies(1011.1255)

Nov. 4, 2010 astro-ph.CO
Gravitational lensing is a powerful astrophysical and cosmological probe and is particularly valuable at submillimeter wavelengths for the study of the statistical and individual properties of dusty starforming galaxies. However the identification of gravitational lenses is often time-intensive, involving the sifting of large volumes of imaging or spectroscopic data to find few candidates. We used early data from the Herschel Astrophysical Terahertz Large Area Survey to demonstrate that wide-area submillimeter surveys can simply and easily detect strong gravitational lensing events, with close to 100% efficiency.
• ### Discovery of Lyman-alpha Pumped Molecular Hydrogen Emission in the Planetary Nebulae NGC 6853 and NGC 3132(astro-ph/0602547)

Feb. 24, 2006 astro-ph
We report the first observation of Ly$\alpha$ pumped molecular hydrogen emission lines in planetary nebulae. The H$_{2}$ emission observed in the ultraviolet spectra of NGC 6853 and NGC 3132 can be explained by excitation of vibrationally hot H$_{2}$ by Ly$\alpha$ photons. Constraints are placed on the nebular Ly$\alpha$ emission profile, as well as the molecular hydrogen temperature, column density and turbulent motion. These parameters are similar for the two nebulae, pointing to similar physical conditions in these objects. The ro-vibrational cascade following Ly$\alpha$ pumping is predicted to have low surface brightness signatures in the visible and near infrared.


text:
# Tag Info

### Optimize search for rational numbers on unit circle?

This method is called the Stereographic approach and can be used to generate arbitary rational triples without needing brute force searching. I was fortunate to have learned this trick off Minhyong ...
• 21.7k

### Optimize search for rational numbers on unit circle?

This is a compiled program to find all Pythagorean triples $(a,b,c)$ with $a^2 + b^2 = c^2$ and $a \leq b$ and where c lies between ...
• 98.3k
Accepted

### Memory leak with Mathematica Graph functions

Here is another possible way to eliminate the impact of history, etc. The following code aborts on my machine after printing to the console three times. ...
• 4,753
Accepted

### NDSolve very slow on 2D heat equation

The simplest solution I find is, add SolveDelayed -> True to NDSolve: ...
• 58.5k
Accepted

### Speed up replacement of very large alternatives expression

Use threaded rules instead of alternatives: instead of 1 | 2 -> 0 use {1 -> 0, 2 -> 0}. Combined with a ...
• 40.1k

### What is a good way to compute successive primorials with Mathematica?

Mathematica code is already provided in the The On-Line Encyclopedia of Integer Sequences, here http://oeis.org/A002110 oeis Sol 1 ...
• 33k
Accepted

### Optimize search for rational numbers on unit circle?

This is not the fastest solution in the world for finding rational points on the unit circle (takes 4.3 seconds on my laptop for denominators up to 100000), but at least it's short: ...
• 16.2k

### NDSolve very slow on 2D heat equation

Try FiniteElement method with NeumannValue instead of derivative-bc: ...
• 41.5k

### What is a good way to compute successive primorials with Mathematica?

Luckily Prime is Listable and we can make an elegant solution with FoldList: ...
• 7,754

### What is a good way to compute successive primorials with Mathematica?

Not sure if the following is considered a good way, but at least it gave results in realistic time frames and did not crash anything. Create the list of primes and then multiply the elements of the ...
• 9,067
Accepted

• 33k
Accepted

### NIntegrate: slow convergence error and speed efficiency

If we change the method of integration, we can obtain the NNt[1,1] which we see in the screenshot without any errors. The method is ...
• 9,067
1 vote

### Quickly summing matrix elements

One possibility is to use TensorContract/TensorProduct, but you need to also use Inactive to ...
• 126k
1 vote

### Efficient memory usage while building a large sparse matrix

Even though you have shared no details about f, as others have mentioned, I think I understand your problem. Consider this (note that ...
• 4,753
1 vote
Accepted

### How to refer to calling notebook instead of my personal include? / detector for high-memory-usage cells

The issue is that the default values EvaluationNotebook[] and EvaluationCell[] are evaluated when the function is defined, not ...
• 29.9k

Only top scored, non community-wiki answers of a minimum length are eligible


text:
# Reducing TSP to HAM-CYCLE to VERTEX-COVER to CLIQUE to 3 CNF-SAT to SAT

In Cormen's Algorithms book on NP-completeness they prove various problems are NP-complete by reducing a previously proved NP-complete problem (call $K$) to current problem (call $L$). Each proof involves some clever construction which reduces all instances of $K$ to few instances of $L$. Here is the proof order they follow. CIRCUIT-SAT, SAT, 3 CNF-SAT, CLIQUE, VERTEX-COVER, HAM-CYCLE, TSP. e.g. in reducing VERTEX-COVER to HAM-CYCLE they use a widget which does the trick.

After this previous question of mine, I think one can reduce back. i.e. one can reduce HAM-CYCLE to VERTEX-COVER problem. I tried searching web for such reductions, but most of the link return the normal reduction order. I'm interested to see if one can reduce in reverse order. i.e. TSP to HAM-CYCLE to VERTEX-COVER to CLIQUE to 3 CNF-SAT to SAT

I'm looking for reverse constructive proofs. I know all of these problems belong to NP-complete hence equivalent.

You don't have to give complete proof as an answer. Proof sketches are fine too. If you can point me where these proofs are available online, that's completely fine too. I'm just trying to lean how constructions are leveraged among problems that look so different on surface. Thanks!

• Well I found most reductions interesting because they are construction proofs. It's hard to think out-of-the-box constructions. So I wanted to see if reverse construction is possible. – Ankush Aug 17 '12 at 14:34
• What have you tried? All stated problems are $\mathsf{NP}$-Complete. Thus every problem in $\mathsf{NP}$ can be reduced to your problems (by definition of $\mathsf{NPC}$). Hint: Start reducing TSP to HAM-CYCLE. – Christopher Aug 17 '12 at 14:40
• @Chris As of now nothing. I mean till yesterday I was under impression that above isn't possible. Let me start with TSP to HAM-CYCLE :) – Ankush Aug 17 '12 at 15:13
• Note that not all NPC problems are equally hard, see e.g. weakly vs strongly NP-completeness. Therefore, not all reductions are equally simple; those from strong to weak problems have to be complex enough to prevent e.g. efficient approximations to carry over (unless P=NP). – Raphael Aug 18 '12 at 6:16
• I'd offer you to read Garey & Johnson book, they provide iff proof for some of problems, proof techniques are not easy and you can't expect to solve them yourself in few days. – user742 Aug 18 '12 at 18:32

As mentioned in one of the comments on the question, $3SAT$ to $SAT$ is trivial, an instance of $3SAT$ is already and instance of $SAT$, so there's no work needed at all.
To get from $VERTEX$ $COVER$ $(VC)$ to $CLIQUE$ we can do a couple of short jumps, that'll also put another problem in the loop. $VC$ is the dual of $INDEPENDENT$ $SET$ $(IS)$, that is, we can find a vertex cover of size $k$ in a graph $G$ iff we can find an independent set of size $n-k$ in $G$. Play around with this a little and you'll see why this is true. If we then take the edge-complement graph $\bar{G}$ of $G$, an independent set of size $t$ in $G$ is the same as a clique of size $t$ in $\bar{G}$, so that gets us to $CLIQUE$.


text:
# Friction: Static Cylinder And Ramp Problem: Again!

1. May 31, 2005

### afromanam

Thanks for the help,

i think i'm getting it, but i still have doubts, anyway, here's the problem:

A wedge of 15° is pushed under a tube of 50 kg as shown in fig 1. $$\mu_S = 0.20$$ in all the surfaces. Determine the required force P to move the wedge.

answer is $$P &= 283 N\leftarrow$$
but i get 377.73 N, please give some hints of what i'm doing wrong

what i'm doing is (please refer to attachments to see the fbd's)

$$\arctan\mu_S = \theta_S$$

$$\arctan0.20 = \theta_S$$

$$11.3 &= \theta_S$$

$$R &_B=\frac{W\sin101.3}{\sin52.4}$$

$$R &_B=607.09 N$$

and in the second system:

$$P &=\frac{607.09N\sin37.6}{sin78.7}$$

which yields:

$$P &=377.3N\leftarrow$$

#### Attached Files:

• ###### frprob2.GIF
File size:
4 KB
Views:
70
Last edited: May 31, 2005
2. May 31, 2005

### OlderDan

I have an idea on this, but I have not yet gotten it to work out to the given answer. See how this strikes you. I think your calculation is assuming that the maximum frictional force is applied at all surfaces. That is not likely the case here. For a very small angle, the normal force at the plane is going to be much greater than the normal force at the wall. The cylinder is going to slide up the wall, and roll up the plane. The torques have to be equal and opposite, so the vertical component of the force at the wall, which will be the maximaum friction force, must equal the friction force parallel to the plane, which will be less than the maximum friction force. This is going to reduce the force at B, and therefore reduce the force at C, resulting in less force at P. I believe the same will be true even at 15 degrees. See what you can do with this.

3. May 31, 2005

### afromanam

uhmm, i have a couple of questions

ok, let's say $$F &_A=F &_B^'$$

where $$F &_B^'$$ is the frictional force parallel to the plane

if my calculations are right $$F &_A = R &_A\sin 11.3$$

$$F &_A = 274.3 N\sin 11.3$$

$$F &_A = 53.74$$

ok, so if $$\frac{F &_B^'}{\mu_S} = 268.7 = N &_B^'$$

and $$R &_B^{'} = 274.06$$

because $$R &_B^{'} = \sqrt{N &_B^{'}^{2} + F &_B^{'}^{2}$$

that would make $$P &= 171.72 N$$

plz, any help!!!!

the other question, are the fbd right?

edit:

another thing i tried was to make the angle between $$W \ \mbox{and} \ R_B$$ 15 degrees in the first fbd, so that instead of having $$R_B$$ you have $$N &_B$$ directly and you can obtain $$F_B$$ and eventually $$R_B$$, but that didn't do it either.

Last edited: Jun 1, 2005
4. Jun 1, 2005

### OlderDan

Those numbers look too high to me. It works as I described earlier. I got the given answer. Your original free body diagrams are not correct because they assume maximum friction at all three surfaces. The direction of your B force is not correct. You need to equate the frictional force at A to the frictional force at B. The frictional force at A is 20% of the normal force at A. Here are my starting equations, where I have broken the normal and frictional forces at B into horizontal and vertical components. These are the forces acting on the cylinder.

$$\overrightarrow {F_A } = F_{AN} \widehat i - \mu F_{AN} \widehat j$$

$$\overrightarrow {F_B } = F_{BN} \left( { - \sin 15\widehat i + \cos 15\widehat j} \right) + \mu F_{AN} \left( { - \cos 15\widehat i - \sin 15\widehat j} \right)$$

Add the weight to these to get zero and solve for the normal force components. You should get

$$F_{AN} = 177.76$$

$$F_{BN} = 554.13$$

Plug those in to get the total force at B. The normal force at C must counter the vertical force on the wedge at B. The horizontal force at C is 20% of the normal force because the wedge is moving. Add the horizontal components at B and C, and you get the magnitude of P.

5. Jun 1, 2005

### afromanam

Thanks, thank you very much

once again, you saved me....thank you very much

$$F_B^{'}$$ which is parallel to the plane is 35.55, from that, everything was 'easy'.....

$$N_C = 526.05 N$$

$$\sum F_x = 0$$

$$F_N^{'}\cos15° + F_{BN}\sin15° + 0.2N_C - P = 0$$

$$P = 283.6 N$$

see you next problem....

6. Jan 29, 2010

### howellt

Here's one that I just saw on a Canadian First Class Operating Engineer Applied Mechanics exam:

A solid cylinder rests at the top of a ramp 1.2 meters high. Neglecting friction what is the coefficient of friction? How fast does the cylinder move down the ramp (in meters per second)?
Note: The radius of gyration is the same as the diameter of the solid cylinder.

I have no problem admitting that I had no idea where to start on this one (though I knew the note was a good place to start).
Any help with this question would be greatly appreciated!


text:
Illustration: Alan Stonebraker

Figure 1: Schematic representation of the variation of the energy gap (order parameter) plotted as function of the local angle circling the hole ($α$) and electron ($β$) Fermi surfaces of the material. Dashed line represents $Δ=0$. (a) Isotropic $s±$ state. (b) Anisotropic $s±$ state with nodes on $β$ pockets. (c) Same as (b) but for a state with marginal (“kissing”) nodes. (d) Same as (b) but for deep gap minima.


text:
0
Explain with an example Chomsky Normal Form

Mumbai university > Comp > SEM 4 > TCS

Marks: 5M

Year: Dec 2014

0
0
• A context-free grammar G = (V, Σ, R, S) is said to be in Chomsky Normal Form (CNF), if and only if every rule in R is of one of the following forms:

1. A → a, for some A ∈ V and some a ∈ Σ
2. A → BC, for some A ∈ V and B, C ∈ V \ {S}
3. S →є

For a Context Free Grammar to be in Chomsky Normal Form(CNF), the grammar should yield productions such as :

A => BC or

A => a

Where a is a terminal and A,B and C are non-terminals

We consider an example

S => bA | aB

A => bAA | aS | a

B => aBB | bS | b

For CNF, A => BC (two nonterminals)

Or

A => a (one terminal)

Consider S => bA

S => BA which is a CNF (substituting B => b)

Consider S => aB

S => AB which is CNF(substituting A => a)

Consider A => aS

Let $C_1$ => a

A => $C_1S$ which is CNF

Consider A =>bAA

A => BAA (substituting B => b)

Let $C_2$ => BA

A => $C_2A$ which is CNF

Consider B => bS

Let $C_3$ => b

B => $C_3S$ which is CNF

Consider B =>aBB

B => ABB (Substituting A =>a)

Let $C_4$ => AB

B => $C_4B$ which is CNF

0


text:
#### Vol. 11, No. 8, 2018

 Recent Issues
 The Journal About the Journal Editorial Board Subscriptions Editors’ Interests Scientific Advantages Submission Guidelines Submission Form Editorial Login Ethics Statement ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print) Author Index To Appear Other MSP Journals
On weak weighted estimates of the martingale transform and a dyadic shift

### Fedor Nazarov, Alexander Reznikov, Vasily Vasyunin and Alexander Volberg

Vol. 11 (2018), No. 8, 2089–2109
DOI: 10.2140/apde.2018.11.2089
##### Abstract

We consider weak-type estimates for several singular operators using the Bellman-function approach. In particular, we consider a concrete dyadic shift. We disprove the ${A}_{1}$ conjecture for those operators, which stayed open after Muckenhoupt and Wheeden’s conjecture was disproved by Reguera and Thiele.

##### Keywords
martingale transform, weak weighted estimate
##### Mathematical Subject Classification 2010
Primary: 42A45, 42A61, 42B20, 42B35, 42B37, 47A30
Secondary: 42A50, 49L20, 49L25


text:
Why Godel's theorem influences Hilbert's programme?

I have trouble understanding why it is being said that Godel's incompleteness theorem shows that Hilbert's programme is essentially impossible.

So, if I understand Godel's second incompleteness theorem, it says that sufficiently strong theorems such as Peano Arithmetic cannot prove its consistency within itself.

But isn't Hilbert's programme that we prove consistency using finitary means in metatheory? It seems to me that Godel's theorem implies that we cannot prove the consistency of the theory within the formal theory. But what about proving the consistency of the theory using metatheory (which is outside of the formal theory)?

I would appreciate your help and comments!

• For the purposes of Hilbert's programme, your metatheory would itself presumably be a formal theory. So, yes, you can prove the consistency of, say, PA in a metatheory, but how do you know your metatheory is consistent? The standard scenario in that case is proving the consistency of PA within ZFC, another formal theory. – Derek Elkins Oct 29 '18 at 21:46
• @DerekElkins But isn't the whole point of metatheory to be not formal theory, but, say, be intuitive reasoning about finite strings which we have intuition for because, for example, they could be checked in the real world by just writing these finite strings? – Daniels Krimans Oct 29 '18 at 21:48
• Arguably, but Gödel's theorem suggests that many of the formalizations of those "intuitions about finite strings" are either inconsistent or incomplete. That means that either we very much need to scale back what we "intuitively" believe about finite strings, or there are things about finite strings that we believe that are unformalizable. This is not an exhaustive listing of alternatives, but regardless, some serious reflection is needed. If you do consider things "true" which are verifiable in the real (physical) world, then you go beyond finitism to ultrafinitism. – Derek Elkins Oct 29 '18 at 21:56
• @DerekElkins I don't understand why do you say that we need formalizations of our intuitions? Formalizing some theory means that there is metatheory which talks about how it is formalized. That means if we formalize metatheory we need metametatheory, and so on. So in the end we still have something that relies purely on our intuition and is not formalized. Isn't that true? – Daniels Krimans Oct 29 '18 at 21:59
• I didn't say we needed to formalize our intuitions, just that the systems we usually present as formalizations of our intuitions of finite strings are, in fact, subject to Gödel's theorem. Formal systems that are conservative enough not to run afoul of Gödel are usually viewed as too weak to actually do math in as well as justify some of our other intuitions. One (overly colorful) way to frame this is that Hilbert thought he could battle ambiguities with infinities based in the "finite", but the battle lines are actually drawn within the "finite". Somewhere. – Derek Elkins Oct 29 '18 at 22:11

Short answer. Sure, we can prove that, e.g., Peano Arithmetic is consistent if we stand outside the theory and help ourselves to a sufficiently strong background theory to work in. But Gödel's second theorem tells us that this metatheory will have to be in some respects stronger than Peano Arithmetic itself to do the trick. Which probably isn't much help if we were worried about the consistency of Peano Arithmetic in the first place, for the worries will carry over to the stronger theory.

And the argument generalises. Hilbert's Program was to make it safe to play the game of infinitary mathematics (e.g. even wildly infinitary full set theory) by standing outside the theory and using uncontroversial weak mathematics to at least prove the consistency of the infinitary theory. Gödel tells us this can't be done. A consistency proof for a theory $$T$$ requires more oomph than is available in $$T$$, so we certainly can't prove $$T$$'s consistency in a much weaker theory.

Long answer: Not here but in Ch. 37 (in particular) of the second edition of my Introduction to Gödel's Theorems (should be in any academic library)! Which shows the story is a bit more complicated in interesting ways, though the short take-home message is much the same.

• Thanks for the answer. ''Gödel's second theorem tells us that the metatheory will have to be in some respects stronger than Peano Arithmetic''. How do you conclude that it has to be stronger? Why weaker one would not suffice? I think I have read that Godels theorem can be proven in PRA metatheory which is weaker than PA – Daniels Krimans Oct 30 '18 at 2:00
• @DanielsKrimans He is talking about proving the consistency of PA, not the incompleteness theorem. – spaceisdarkgreen Oct 30 '18 at 7:24
• Careful! Yes, a very weak background theory -- acceptable even to a finitist -- is enough to prove e.g. (G): PA cannot prove Con(PA). There's no wriggle room. We have to accept (G). But now what does (G) tell us? That PA can't prove Con(PA). It requires something stronger than PA (at least in some respects) to prove the consistency of PA. Likewise, that weak background proves (G'): ZFC cannot prove Con(ZFC). So it requires something stronger than ZFC to prove the consistency of ZFC. So a finitistically acceptable metatheory can't prove the consistency of ZFC. – Peter Smith Oct 30 '18 at 7:26
• @DanielsKrimans You indeed seem a bit confused about Hilbert's program. Read plato.stanford.edu/entries/hilbert-program – Peter Smith Oct 30 '18 at 7:49
• @DanielsKrimans See e.g. William Tait's "Finitism" Journal of Philosophy, 1981, 534-556 (available through your uni's Jstor), reprinted in a follow-up paper in Tait's book The Provenance of Pure Reason (OUP 2005) – Peter Smith Oct 31 '18 at 8:27


text:
# Protect fragile variable in file path. Case of \ifthenelse in \includegraphics{}

In this script:

\documentclass{article}
\usepackage{ifthen}
\usepackage{graphicx}
\newcommand{\myfigure}[1]{%
\def\mydirectory{\ifthenelse{\equal{#1}{first}}{dir1}{dir2}}

my directory is \mydirectory

\begin{figure}
%    \includegraphics[width=0.45\textwidth]{\protect\mydirectory/mypic.pdf}%
\includegraphics[width=0.45\textwidth]{example-image-a}%
\caption{picture in \protect\mydirectory}
\end{figure}

}

\begin{document}
\myfigure{first}
\end{document}

I \protect the fragile variable \mydirectory in \caption - see comments to How to protect a fragile command in a moving argument? The case of macro containing \ifthenelse in \caption.

However, when I provide:

\includegraphics[width=0.45\textwidth]{\mydirectory/mypic.pdf}

the classical symptoms of a fragile \ifthenelse appear:

! Undefined control sequence.
<argument> \equal
{first}{first}

that are not cured even by \protect:

\includegraphics[width=0.45\textwidth]{\protect\mydirectory/mypic.pdf}

How to protect a fragile variable in a file path?

• Ifthenelse is not expandable. You can't use it in this place. – Ulrike Fischer Oct 15 '18 at 11:29
• I think it should by now be clear to you that \ifthenelse doesn't work by pure expansion. A method with \ifstreq similar to one I suggested for another case will work flawlessly. – egreg Oct 15 '18 at 11:41
• This can not work. – David Carlisle Oct 15 '18 at 12:28

I would not use ifthenelse but if you use it (or any other test mechanism) you want \mydirectory to expand to the directory name, not to a complicated sequence of assignments and tests that may eventually typeset a directory name.

so

\newcommand{\myfigure}[1]{%
\def\mydirectory{\ifthenelse{\equal{#1}{first}}{dir1}{dir2}}

can not possibly work, you would need

\newcommand{\myfigure}[1]{%
\ifthenelse{\equal{#1}{first}}%
{\def\mydirectory{dir1}}%
{\def\mydirectory{dir2}}%
}

so that the definition of \mydirectory is either dir1 or dir2.

• As you define \mydirectory in {\def\mydirectory{dir1}} will not \mydirectory live only inside the scope {}? – Viesturs Oct 15 '18 at 12:36
• @Viesturs no: {}: that delimit macro arguments do not form a group. – David Carlisle Oct 15 '18 at 12:37
• Is it advisable to use \newcommand in place of \def? – Viesturs Oct 15 '18 at 14:25


text:
2015
04-16

# Genetics

A colony of alien bacteria has recently been discovered close to a crater in New Mexico. Dr. Poucher is in charge of the scientific team at the ICPC BioLab committed to the study of the alien DNA structure. We briefly sketch their discoveries here.
Alien DNA molecules have the structure of a circular sequence. Each sequence is composed of nucleotides.There are 26 different types of nucleotides, and each of them can occur in two faces. It is very important to remark that in any given alien DNA molecule, every nucleotide either does not appear at all or appears exactly twice (hence, the length of a DNA molecule is an even integer between 2 and 52). In case a nucleotide occurs twice, each occurrence can be of either type independently. Alien bacteria have two types of extremities,which in the technical biological jargon are referred to as arms and legs. A major discovery of Dr. Poucher’s team is a method to determine the exact number of arms and legs of a bacterium by examining its DNA structure.
Here we represent each nucleotide as a letter of the alphabet. We refer to the different nucleotides as a,A, . . . z, Z, where the lowercase and uppercase forms of a letter represent the two possible faces a nucleotide may appear with; we shall also use a/A, b/B, . . . z/Z to refer to a nucleotide in either face.To determine the number of extremities, Dr. Poucher starts by initializing two counters of arms and legs to zero, and then proceeds to perform a number of surgeries, transforming a DNA sequence into another one. After each transformation, you may need to increase some of the counters, depending on the type of surgery applied. When the empty sequence of nucleo- tides (which will be denoted by ) has been reached,the number of extremities of the original molecule has been found. The possible surgeries are:

1. Eliminate consecutive instances of a given nucleotide appearing with opposite faces. The number of arms and legs is preserved. For example: aBbCaC -> aCaC by eliminating Bb. Another example: DeHhEd -> eHhE by eliminating dD. Remember that DNA structure is circular, so in our representation as a string the last and first letters are connected.

2. Eliminate consecutive nucleotides appearing with the same face. Add one to the number of arms. For example: BBcgCg -> cgCg by eliminating BB. Another example: xabyyaBX -> xabaBX by eliminating yy.

3. Eliminate a sequence of four nucleotides formed by two different nucleotides that appear alternately where different occurrences of the same nucleotide have opposite faces. Add one to the number of legs.For example: dcDCefFe -> efFe, by eliminating dcDC. Another example: cmNMnC -> cC by eliminating mNMn.

4. Cut and paste, the most sophisticated procedure.

First, a nucleotide is selected, for instance a=A, and the DNA sequence is chopped into two linear chains such that the nucleotide appears once in each of them.

Second, if both occurrences of a=A are of the same face, one of the chains is "inverted" by reversing the sequ- ence and changing the face of every nucleotide in the chain.

Then, the chains are combined by concatenating the subsequence occurring before a with the sub-sequence occurring after A, and the subsequence occurring after a with the sub-sequence occurring before A.

Finally, two new a=A nucleotides are added to close the chain into a circular shape. The face of the new nucleotides are the same if the original pair of nucleotides selected had the same face, and is different otherwise.

Formally, suppose you select the nucleotide a=A, and further assume for the moment that it appears both times with the face a (A). The cut and paste surgery turns sequences of the form (respectively ) into (respectively ). On the other hand, if nucleo-tide a=A appears with its two different faces, the surgery turns sequences of the form into . S1, S2, S3 and S4 are arbitrary sub-chains (possibly empty). In both cases the original circular chain was chopped into and .

For example (see the figure below): starting with the sequence BacDcAbD, we can get chains BacDc and AbD. Then, merging at nucleotide a=A we get the sequence cDca’BbDA’ where a’ and A’ represent the new a=A nucleotides. Here, S1 = B, S2 = cDc, S3 = ; and S4 = bD.

Another example: take the same DNA sequence BacDcAbD, and cut to get the chains DBac and DcAb; paste nucleotide c=C (in this case you need to reverse one chain, for example BaCd) to get the sequence cDBadcBa. Here, S1 = DBa, S2 = ;, S3 = D and S4 = Ab.

This surgery does not modify the number of arms or legs, but can be used cleverly in combination with the previous surgeries to reduce the size of the DNA molecule and finish the calculation.

However, alien bacteria do not present both arms and legs at the same time. This is due to the fact that,in their early development, a leg, in the presence of one or more arms, becomes two arms. Because of the above, the end result is either a number of arms or a number of legs, but not both at the same time. In order to avoid expensive surgical procedures, Dr. Poucher has hired you to write a program that computes the number of arms and legs a bacterium will develop, given its DNA sequence. It is guaranteed that the result is determined uniquely by the original string, regardless of the particular sequence of surgeries applied.

Each test case consists of a string of even length between 2 and 52, inclusive, representing the DNA structure of an alien bacterium. All characters are letters. There will be one case per line in the input. The last line contains the word \END" and must not be processed.

Each test case consists of a string of even length between 2 and 52, inclusive, representing the DNA structure of an alien bacterium. All characters are letters. There will be one case per line in the input. The last line contains the word \END" and must not be processed.

rkrk
abcdeABCDE
shcoOCfFHS
END

1 arm
2 legs
none
1

1. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

2. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n

3. 可以根据二叉排序树的定义进行严格的排序树创建和后序遍历操作。如果形成的排序树相同，其树的前、中、后序遍历是相同的，但在此处不能使用中序遍历，因为，中序遍历的结果就是排序的结果。经在九度测试，运行时间90ms，比楼主的要快。


text:
Keywords = extreme point
Number of Articles: 2
##### 1. On certain subclasses of univalent $p$-harmonic mappings

Volume 41, Issue 2, April 2015, Pages 429-451

J. Qiao; J. Chen; M. Shi

##### 2. The starlikeness, convexity, covering theorem and extreme points of p-harmonic mappings

Volume 38, Issue 3, September 2012, Pages 581-596

Qiuhong Luo; Xiantao Wang


text:
# How do you integrate 1 / (x^5)?

Nov 9, 2016

$- \frac{1}{4 {x}^{4}} + c$

#### Explanation:

Integrate using the $\textcolor{b l u e}{\text{power rule for integration}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\int \left(a {x}^{n}\right) \mathrm{dx} = \frac{a}{n + 1} {x}^{n + 1}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

Express $\frac{1}{x} ^ 5 = {x}^{-} 5$

$\Rightarrow \int {x}^{-} 5 \mathrm{dx} = \frac{1}{- 5 + 1} {x}^{- 5 + 1} + c$

$= - \frac{1}{4} {x}^{-} 4 + c = - \frac{1}{4 {x}^{4}} + c$

where c is the constant of integration.


text:
# IBPS Clerk Quantitative Aptitude Questions with Answers Practice online test 45

Description: free IBPS Clerk Quantitative Aptitude Questions with Answers Practice test 45 for IBPS Clerk Preliminary and Main online test Prepare bank Clerk banking mock exams adda

1 .
A man has Rs. 972 in the denominations of one-rupee, two-rupee, five-rupee and ten-rupee. If there are equal no. of coins of each denomination, what is the total number of coins that he has?
 216 232 224 236 212
2 .
A alone can do a piece of work in 6 days. B can do the same piece of work in 8 days. A and B signed to do it for `3200.They completed the work in 3 days with the help of C. How much is to be paid to C?
 Rs. 380 Rs. 600 Rs. 400 Rs. 420 Rs. 450
3 .
Mr Basu's annual income is Rs. 5,10,000. He spends 28% of his monthly income on paying bills, 15% on household items and 8% he donates to an NGO. If thr ee-fifths of the r emaining amount heinvests in share market, what is the amount left with him every month?
 Rs. 5234 Rs. 7109 Rs. 6610 Rs. 8330 Rs. 8789
4 . Refer to the table below and answer the given questions:

Number of employees in five companies during five years

The number of employees in company T in 2008 is what per cent of the number of employees in company R in 2010?
 74$1 \over 4$ 66$3 \over 4$ 68$3 \over 4$ 72$3 \over 4$ 64$1 \over 4$
5 . Refer to the table below and answer the given questions:

Number of employees in five companies during five years

What is the difference between the total number of employees in company S in 2008 and 2010 together and the total number of employees in company Q in 2007 and 2009 together?
 134 138 144 132 128
6 . Refer to the table below and answer the given questions:

Number of employees in five companies during five years

What is the ratio of the total number of employees in company P in 2008 and 2009 together to that of the total number of employees in company Q in 2006 and 2010 together?
 73 : 84 71 : 86 73 : 88 71 : 88 73 : 86
7 . Refer to the table below and answer the given questions:

Number of employees in five companies during five years

The number of employees in company R in 2008 is approximately what per cent less than the number of employees in company S in 2006?
 36 27 33 24 21
8 . Refer to the table below and answer the given questions:

Number of employees in five companies during five years

What is the average number of employees in company P in 2006, 2007 and 2010 together?
 244 236 242 238 232
9 . What will come in place of question mark (?) in the given questions?

11 9 15 41 159 ?
 785 789 797 769 771


text:
There's a Hole in my Model...

When you assume, you make an ass of U and Me, or so they say.

Yesterday, I was alerted to a discovery that has the potential to challenge some of our assumptions, by somebody who wanted to know what I thought about it. To be honest, it presents something of a puzzle.

The discovery in question was a black hole. Well 'so what?', you might be tempted to ask; after all, black holes are hardly an Earth-shattering (pun intended) revelation. So what's all the hoo-ha about?

Well, it isn't the black hole itself that's remarkable. What's really remarkable is how far away it is, and its size. To show just how remarkable, I'm going to break with tradition somewhat and go over some previously-covered ground.

We've talked a fair bit about the Big Bang in quite a few other posts. We've noted that, in the popular understanding, the Big Bang is a theory dealing with the beginning of the universe. Despite the fact that this model is still talked about as the standard for the evolution of the universe, it hasn't actually been that for several decades, not least because of some fatal flaws, which we've looked at in some detail in previous posts. I'll include links to these posts at the bottom, along with any other whose subject matter crops up during the course of what follows, as these observations would take us too far afield for our purposes today. What I will do is to note that the Big Bang as we use it now doesn't actually deal with the beginning of anything. These days, the term 'Big Bang' is really nothing more than the name we give to the observed fact that the universe is expanding and was therefore smaller in the past.

It does, however, still contain some implications, not least in terms of what we observe when we look back through time at the light coming to us from the earliest times we can observe. It's certainly the case that the popular conception isn't far off the mark in some respects, and observations in the last few decades have reinforced some of them.

One of the things that all of our models have, aside from a Big Bang, is a time when our universe was extremely hot and dense. We've previously discussed the earliest observations we have of the cosmos, in a post about biological evolution in the context of entropy. In that outing, we noted that:
...one of our best current sources of information about the early universe is the cosmic microwave background radiation (CMBR), discovered serendipitously by Penzias and Wilson in 1964. What the CMBR actually represents is not, as I've seen suggested, the glow of the big bang, but the photons that come to us from the 'last scattering surface'. This is going to serve as a useful pointer to what we're discussing here, so it's worth spending a little time on.

For about the first 380,000 years or so after the Planck time - a theoretical construct dealing with the smallest useful amount of time after the 'beginning' of expansion - the cosmos was opaque to photons. The reason for this is that, after the onset of expansion, the cosmos was extremely hot and dense. So hot, in fact, that it was basically a white-hot plasma of photons and ionised hydrogen, which consists of protons and free electrons. Because of the abundance of free electrons at such temperatures, the distance that photons could travel freely was massively restricted due to Compton scattering.

Now, anybody who likes to smell nice knows what happens when a body of particles expands, because we've all felt the deodorant can cool down as we let out the smellies. The same thing happens with the cosmos. As expansion continues, the cosmos cools. Prior to a certain low temperature, electrons remain free, not bound to protons, because of a quirk of entropy, namely that, in the environment they find themselves in, there's nowhere for energy to go to become unavailable. As the termperature gets below about 4,000 Kelvin, however, something interesting happens; there's somewhere for energy to go, so the free electrons begin to bind to protons, forming the first neutral atoms. The reason they do this is because it's energetically favourable for them to do so. This is just another way of saying that, once they can shed some energy, their lowest available energy state consists of being bound to protons.

As an aside, this is where the CMBR comes to us from. By the time the temperature gets as low as 3,000K, most of the free electrons have become bound, and the photons from the last bit of Compton scattering are now free to travel through the cosmos, and it's these photons that we detect as the CMBR. Due to the expansion of the cosmos, these photons are hugely red-shifted, meaning that their wavelengths are stretched out, until their temperature is about 2.7K, or -270 degrees Celsius, a smidge less than three degrees above absolute zero.
Here's one of the latest images from the Planck satellite, the most recent generation of satellites imaging the CMBR.

So, as we can see, barring some variation, representing temperature differences of a few ten-thousandths of a degree between the hottest (red) spots and the coldest (blue) spots, it's pretty homogeneous.

Let's jump forward now, and deal with what's actually been discovered, and then we'll look at why it's a problem, as well as looking at some potential solutions.

The object that's been observed is a quasar, or 'quasi-stellar object'. A quasar is an active galactic nucleus consisting of a central black hole with an accretion disc of gas. As the gas spirals into the black hole at huge velocities, it emits enormous amounts of electromagnetic radiation. The name itself is a contraction of 'quasi-stellar radio source', as their earliest discoveries were as sources of radio waves which, in the visible part of the electromagnetic spectrum, looked an awful lot like stars. In any event the electromagnetic radiation they emit has huge luminosity, with the brightest ones giving off on the order of $1 \times 10^{40}$ W of radiation across the spectrum. When you compare that with the estimated luminosity of the Milky Way at around $1 \times 10^{36}$ W*, that's a future so bright that you've gotta wear shades. Or, at least, you would, if this were all visible light, and if we weren't talking about the past, rather than the future.

So, this quasar, known affectionately to its friends as $J1342+0928$, comes to us with a redshift of $z=7.54$. "Fascinating!" you might say. "What does that mean?" Well, as we've discussed previously, the universe is expanding. Because it's expanding at approximately the same rate everywhere - known as metric expansion - this means that objects further away are receding from us at a greater rate.

As we know from earlier discussions, when something is moving away from us, the light we receive from it is shifted toward the red end of the frequency spectrum, as the light-waves we receive get stretched out, making them look redder than their intrinsic colour. Thus, the further away something is, the faster it recedes, and the more its light is shifted toward the red.

What this means is that we can tell by how far the light has shifted from its intrinsic colour how far away it is, as in the image on the left from the European Space Agency.

This is very similar to what we all know as the Doppler Effect, in which passing sirens have their sound frequency shifted as they pass you, and it's what we now know as Hubble's Law, after Edwin Hubble, the American astronomer who first discovered extra-galactic sources and the expansion of the universe in the 1920s.

In this, what that redshift figure is telling us is that this quasar is a very, very long way away. In fact, it's at a comoving distance with Earth of just a smidgen over 29 Gly (giga-lightyears, or billion lightyears). That figure in turn means that, when the light we see from it left its source, it was about 13.1 Gly from us.

Now, this quasar is big. Really big. Not on the scale of those really bright ones discussed earlier, but still absolutely huge, on the order of 800 million times the mass of the sun, and with a luminosity of about $4 \times 10^{14}$ the luminosity of the sun.

And this is where the problems lie. We know, or we think we know, how black holes form. Now, it's certainly true that stars with larger mass exhaust their fuel much quicker than stars with low mass. Indeed, there's a direct proportionality to the mass of the star and the speed with which it runs out of fuel. Going back to the Planck image above, we know that the universe was pretty isotropic (the same in all directions) to within a few ten-thousandths of a degree, and we know that temperature relates to density. How, then, can we have a black hole of 800 million solar masses within only a few hundred million years (about 690 million, but who's counting)? In fact, although the CMBR comes to us from just shy of 690 million years before, that CMBR is itself part of a process, and electron bonding would have been continuing for some time afterwards, which means this is even closer than that.

This is a puzzle, and no mistake.

Now, there are some possible ways to resolve this. The first is that this is the merger of several smaller black holes that formed fairly early on after the last scattering. This seems statistically unlikely, though we can't rule it out as yet and, since this is the first observation of a large black hole at such a distance, and since this comes from a survey of only a very small portion of the sky, it may be that we find that this isn't anything like as uncommon as we think. It does seem to put a bit of a spanner in the works in terms of obtaining such high density from such a smooth distribution only a short time before, so this is going to be fruitful ground for research, not only in further observing this quasar itself, but also in the search for similar objects at commensurate redshifts.

Another possibility, but possibly a somewhat more troubling one in terms of our models, is that this started forming during the rebonding epoch, although how it would do this without leaving any evidence in the CMBR creates even more problems.

And the final possibility, though not one I'd give a huge amount of credence to, is that this black hole actually formed while the universe was still a plasma. Again, I'd expect evidence of this to be pretty obvious in the CMBR, so it causes more problems than it solves.

We can't really rule out either of these two scenarios, despite the lack of observed evidence in the CMBR, until we rule out the possibility that we've actually observed and corrected for it without realising it. It's certainly been the case in the past that announcements have been made of this or that thing in the CMBR when we've failed to correct for something that accounts for it, like that BICEP2 discovery that hit the world with some fanfare a few years ago, so it would be hubris to suppose that there was no possibility that we've over-corrected, or that some other assumption has been making an ass of you and me.

As always, thanks for reading, and a special thanks to my donors and patrons. You are awesome.

Corrections, nits and crits always gratefully received.

Arxiv pre-print server upload of the full paper.
Nature publication of the paper.
All Downhill From Here Evolution and entropy, featuring a potted history of the universe.

It Wasn't Big, and it Didn't Bang Before the Big Bang Part I
You Must Be Off Your Brane! Before the Big Bang Part II
The Certainty of Uncertainty Before the Big Bang Part III
Scale Invariance and the Cosmological Constant Dark matter, dark energy, and the Lambda term in General Relativity


text:
# Thermodynamics - Ideal Diatomic Gas

by ChronicQuantumAddict
Tags: diatomic, ideal, thermodynamics
 Share this thread:
 P: 39 The question is as follows: An ideal diatomic gas for which cv = (3/5)*R, where R is the ideal gas constant, and occupies a volume V= 2 meters^3, at a Pressure Pi= 5 atm and temperature Ti = 20 degrees Celcius (= 293 K). The gas is compressed to a final pressure of Pf = 12 atm. Calculate the final volume Vf, the final temperature Tf, the work done dW, the heat transferred dQ, and the change in Internal energy dU for the following: a). a reversible isothermal compression b). a reversible adiabatic compression i know in the adiabatic process, dQ = 0, and in the isothermal dT = 0. Where do i take it from here?
 P: 39 anyone please?
Sci Advisor
HW Helper
P: 6,679
 Quote by ChronicQuantumAddict The question is as follows: An ideal diatomic gas for which cv = (3/5)*R, where R is the ideal gas constant, and occupies a volume V= 2 meters^3, at a Pressure Pi= 5 atm and temperature Ti = 20 degrees Celcius (= 293 K). The gas is compressed to a final pressure of Pf = 12 atm. Calculate the final volume Vf, the final temperature Tf, the work done dW, the heat transferred dQ, and the change in Internal energy dU for the following: a). a reversible isothermal compression b). a reversible adiabatic compression i know in the adiabatic process, dQ = 0, and in the isothermal dT = 0. Where do i take it from here?
The first one is fairly straightforward. Think PV=nRT. If T is constant, what does that tell youi about PV at all times? That should give you Vf. What is U at intial and final volumes? What is W in terms of P and $\Delta V$? That should give you W. What is the relationship between U, Q and W? That will give you Q.

For the second part, it is more complicated. You have to use the adiabatic relationship:

$$PV^\gamma = constant$$
where $\gamma = C_p/C_v$

What is $\gamma$ here?
What is the expression for W?
How is W related to Q and U?

AM

 P: 39 Thermodynamics - Ideal Diatomic Gas i am stuck with the internal energy for part a) isothermal process. here is what i've done so far. since its isothermal: P1*V1 = P2*V2 and thus, V2= (P1/P2)*V1 = 0.8333333 m^3 For isothermal, Ti = Tf = T = constant = 293.15 K W = -nRT*(Integral: Pi to Pf) (dP/P) = -nRT*ln(P2/P1) = nRT*ln(P1/P2) thus W = -2.134 * 10^6 J, work done on the system Now, for U i am stuck. i keep reading in the book, that for an ideal gas U only depends on T, and they give me this formula. U - U_0 = dU = Cv*dT, where Cv is the specific heat at constant volume. but volume is not constant in this problem, and also, this leads to an integral with dT and for isothermal dT = 0 If this is true, then the integral yields a constant U and that means U_final = U_initial, but that still doesnt tell me what U is. Anything further? where am i going wrong?
 P: 39 please help, need to turn this in tomorrow and im stuck
Sci Advisor
HW Helper
P: 6,679
 Quote by ChronicQuantumAddict i am stuck with the internal energy for part a) isothermal process. here is what i've done so far. since its isothermal: P1*V1 = P2*V2 and thus, V2= (P1/P2)*V1 = 0.8333333 m^3 For isothermal, Ti = Tf = T = constant = 293.15 K W = -nRT*(Integral: Pi to Pf) (dP/P) = -nRT*ln(P2/P1) = nRT*ln(P1/P2) thus W = -2.134 * 10^6 J, work done on the system Now, for U i am stuck. i keep reading in the book, that for an ideal gas U only depends on T, and they give me this formula. U - U_0 = dU = Cv*dT, where Cv is the specific heat at constant volume. but volume is not constant in this problem, and also, this leads to an integral with dT and for isothermal dT = 0 If this is true, then the integral yields a constant U and that means U_final = U_initial, but that still doesnt tell me what U is. Anything further? where am i going wrong?
U is the internal energy due to kinetic energy of the molecules. If you keep the volume constant and increase the temperature, the gas does no work. So the increase in internal energy is equal to the heat input. Since the heat flow is just the heat capacity of the gas at constant volume multiplied by the change in temperature, $dU = nC_vdT => U = \int dU = \int nC_vdt = nC_vT$.

So $U = nC_vT$ (T in K).

AM
 P: 39 understood now, thanks very much AM

 Related Discussions Introductory Physics Homework 0 Introductory Physics Homework 2 Introductory Physics Homework 0 Introductory Physics Homework 9 Classical Physics 11


text:
# Velocity of particle constant under null initial magnetic force

I have got a doubt on the exact application of Newton's first, and third, law.

Let us for example suppose that a particle, electrically charged with charge $q$, is moving at the instant $t=0$ with velocity $\mathbf{v}(0)$ through a uniform magnetic field $\mathbf{B}$ and a uniform electric field $\mathbf{E}$, such that they are orthogonal to each other, as shown in the figure below. The force acting on the particle is $\mathbf{F}(t)=q\mathbf{E}+q\mathbf{v}(t)\times \mathbf{B}$ and, if $\mathbf{v}(0)=\|\mathbf{B}\|^{-2}\mathbf{E}\times \mathbf{B}$, we have $\mathbf{F}(0)=\mathbf{0}$ and therefore the acceleration is $\mathbf{a}(0)=\mathbf{0}$ at $t=0$.

$\hskip2in$

I intuitively am sure that, if $\mathbf{F}(0)=\mathbf{0}$, in this case nothing modifies the velocity and therefore, for all $t$, $\mathbf{v}(t)\equiv\mathbf{v}(0)$.

Nevetheless, I have got a mathematical problem. Of course, at any time $\tilde{t}$, if $\mathbf{v}(\tilde{t})=\|\mathbf{B}\|^{-2}\mathbf{E}\times \mathbf{B}$, we have $\mathbf{a}(\tilde{t})=\mathbf{0}$, but how can we mathematically be sure that, if $\mathbf{v}(0)=\|\mathbf{B}\|^{-2}\mathbf{E}\times \mathbf{B}$ (and therefore $\mathbf{a}(0)=\mathbf{0}$) the velocity $\mathbf{v}(\tilde{t})$ still is $\mathbf{v}(0)$ if $\tilde{t}>0$ belongs to a neighbourhood of $0$? In fact, if $\mathbf{a}(0)=\mathbf{0}$, the usual continuity assumptions do not guarantee at all that $\mathbf{a}(t)=\mathbf{0}$ in a neighbourhood of $0$ and, while velocity is "istantly constant" at $t=0$ in the sense that $\mathbf{a}(0)=\mathbf{0}$, I am not sure how we can mathematically prove that it does not change. I heartily thank you for any answer.

• You have a guess for the trajectory, right? Just substitute it into the equation of motion and see whether it satisfies it. If it does, there's nothing else to say. Dec 23, 2015 at 12:14
• You have the differential equation of motion in the OP. Solve that equation in the general form, then plug in the initial value. Dec 23, 2015 at 12:14
• @EmilioPisanty The equation(s) of motion determine one and only one solution, which, therefore, is the constant $\mathbf{v}\equiv\mathbf{v}(0)$. Thank you so much!!! Dec 23, 2015 at 13:45
• As an aside, you may notice that $F(v)=0$ has multiple solutions (as F is independent of the y-component/magnetic-field-parallel-component of v). Dec 23, 2015 at 15:09
• I should also suggest changing the title to something that will be easier to find for others with a similar question in the future. I'm not really sure what that title would be, though. Jan 4, 2016 at 13:59

$$\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \begin{pmatrix} C_1 \cos (\frac{q B t}{m} - \alpha) \\ C_2 \\ C_1 \sin (\frac{q B t}{m} - \alpha) + \frac{E}{B} \end{pmatrix}$$
where $\alpha$, $C_1$, $C_2$ are arbitrary constants. I leave it to the reader to realize what happens at certain initial conditions.