text:
## Why not crushed by air?

1. The problem statement, all variables and given/known data
It has been estimated that the mass of nitrogen alone in the column of atomsphere above each square metre of the Earth's surface amounts to about 8 tonnes. Why aren't we crushed by this?

2. Relevant equations
none

3. The attempt at a solution
Air in the atomsphere is not just nitrogen and they are dense as the question suggested. Are we are not crushed by them essentially because of Newton's third law?

We are surrounded by dense air in ground level. What kind of molcules are they? They surround us and collide with our bodies and make us not as cold if without them.

The different densities of air all exert forces to the next layer and due to the third law there is an upward force from the succeeding layers. These layers does the same and you get somewhat of a cancellation of forces on each layer. Hence equilibrium is reached for each layer as there is no net force and we are not crushed by the upper level air. So a cushioning effect is passed down through the layers.

The image tries to illustrate this.
Attached Thumbnails

 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 We can't be crushed by earth's atmosphere because air spreads out equally. The only way you could be crushed by air is if you were in a tank and someone kept pumping in air.
 Recognitions: Gold Member Science Advisor Staff Emeritus The air pressure at the surface of the earth is 14.7 psi. This force does not act in a single direction (downwards) -- it acts in all directions. In other words, the air pushes on your left side with the same force that it pushes on your right side. The force is balanced on both sides, so you don't go sliding sideways. The same is true of the forces in any two directions, including up and down. This is easily understood when you realize that pressure is caused by molecules of air moving in all directions, randomly, imparting forces upon things they strike from any direction. Water is essentially imcompressible, meaning that it does not change volume much with changing pressure. Thus, we stay the same size even when the air pressure changes due to weather or when we walk up hills. - Warren

## Why not crushed by air?

 Quote by chroot The air pressure at the surface of the earth is 14.7 psi. This force does not act in a single direction (downwards) -- it acts in all directions. In other words, the air pushes on your left side with the same force that it pushes on your right side. The force is balanced on both sides, so you don't go sliding sideways. The same is true of the forces in any two directions, including up and down. This is easily understood when you realize that pressure is caused by molecules of air moving in all directions, randomly, imparting forces upon things they strike from any direction. Water is essentially imcompressible, meaning that it does not change volume much with changing pressure. Thus, we stay the same size even when the air pressure changes due to weather or when we walk up hills. - Warren
But the net force of the air in the atomsphere is towards the earth because of gravity. I was trying to illustrate the net effect with the arrows. Is my diagram correct? If so it seems that we are not crushed by air above us because the air the surrounds us cusions the above air. Correct? And the air that surrounds us dosen't crush us because of the reason you suggested.
 Recognitions: Gold Member Science Advisor Staff Emeritus The net force due to the air pressure is zero, because the air pushes on you from every direction. I cannot understand your 'cushioning' concept, but I can tell you immediately that it's not useful. - Warren
 Have you ever tried to suck all the air out of a plastic coke bottle? Yup, it gets crushed. Our bodies would too if we weren't filled with air and liquid.

 Quote by chroot The net force due to the air pressure is zero, because the air pushes on you from every direction. I cannot understand your 'cushioning' concept, but I can tell you immediately that it's not useful. - Warren
Imagine most of the nitrogen is on the top few layers on my diagram. It's dense as specified in the question - although maybe not because it is spread out over a large volume. Gravity acts it towards the earth. So why doesn't it all come down and crush us all at once? The reason is as I suggested in my previous post.
 Recognitions: Gold Member Science Advisor Staff Emeritus No. No offense, but your notion is nonsense. We understand gas very well via a theory known as statistical mechanics, which is derived immediately from Newton's laws of motion. The reason the gas doesn't all come down at once is: Thermal energy The individual gas particles have thermal energy, and move about randomly. They rise and fall in ballistic trajectories, just like baseballs (except there are many, many collisions). - Warren

 Quote by chroot No. No offense, but your notion is nonsense. We understand gas very well via a theory known as statistical mechanics, which is derived immediately from Newton's laws of motion. The reason the gas doesn't all come down at once is: Thermal energy The individual gas particles have thermal energy, and move about randomly. They rise and fall in ballistic trajectories, just like baseballs (except there are many, many collisions). - Warren
So you think the gravitational force on the air molecules are neglible compared to their random forces from thermal energy. Although gravity is still a factor because air dosen't just escape into space. However hydrogen does because it's even lighter. So there is still a net force toward the surface of the earth from the atomsphere. I am talking in a very macroscopic way.

Is that the reason why some bollons rise but some don't. The ones that do are filled mostly with hydrogen.

As for the original question, the answers in the book suggested: "While the atmosphere exerts a pressure on us, we exert an equal pressure back." The pressure on us must be very small because the effect of gravity is tiny. The overriding force are random forces due to thermal motion as you say. That is why they don't crush us because it's also good that they don't escape into space as would be predicted without gravity.

 Quote by pivoxa15 Is that the reason why some bollons rise but some don't. The ones that do are filled mostly with hydrogen.
Helium actually, but that's just bifurcating bunnies...

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 Quote by pivoxa15 So you think the gravitational force on the air molecules are neglible compared to their random forces from thermal energy. Although gravity is still a factor because air dosen't just escape into space. However hydrogen does because it's even lighter. So there is still a net force toward the surface of the earth from the atomsphere. I am talking in a very macroscopic way.
Of course there is a force on the earth's surface due to the weight of the atmosphere. But there's no net force on you due to atmospheric pressure, since the air is pressing on you from all sides. (Actually, there is a small upward buoyant force due to the change in pressure with height--the upward buoyant force on you will equal the weight of the air you displace.)

 Is that the reason why some bollons rise but some don't. The ones that do are filled mostly with hydrogen.
The reason some balloons rise is because they are light enough (they are filled with a gas less dense than air) that the buoyant force from the surrounding air is greater than the weight of the balloon plus its contents.

 As for the original question, the answers in the book suggested: "While the atmosphere exerts a pressure on us, we exert an equal pressure back."
That's not a very good answer, since that statement would be true whether we were crushed or not! (Imagine an elephant stepping on a peanut. The peanut exerts the same force on the elephant that the elephant exerts on the peanut--but the peanut is surely crushed.)
 The pressure on us must be very small because the effect of gravity is tiny. The overriding force are random forces due to thermal motion as you say. That is why they don't crush us because it's also good that they don't escape into space as would be predicted without gravity.
The pressure on you is not small--it's about 14.7 lbs per square inch (at sea level). But we are perfectly capable of handling it. (For one thing, we wisely keep our lungs full of air at the same pressure as the outside air.)

 Quote by Doc Al The pressure on you is not small--it's about 14.7 lbs per square inch (at sea level). But we are perfectly capable of handling it. (For one thing, we wisely keep our lungs full of air at the same pressure as the outside air.)
But the net pressure is virtually 0 as net force is 0.

But imagine a person standing on the ground. He is more exposed to air on top of him than from below him so there is a net force downwards from the atomsphere? And if you factor in gravity than that is another tiny bit more net force directed towards your feet.

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 Quote by pivoxa15 But the net pressure is virtually 0 as net force is 0.
Sorry, but this is incorrect. The fact that the net force might be zero tells you nothing about the pressure. Recall my example of the elephant crushing the peanut. The net force on the peanut is zero, but the pressure is very high. Pressure is a force per unit area.

Further, air pressure is not zero!

 But imagine a person standing on the ground. He is more exposed to air on top of him than from below him so there is a net force downwards from the atomsphere?
That would make some sense if your feet made an airtight seal with some surface, but that's unlikely to be the case. Rather than a person standing, imagine a suction cup stuck on a smooth surface. It sticks to the surface because air pressure pushes it. When you break the seal, air (at atmospheric pressure) gets inside and neutralizes the net force from the air.

Note that the orientation of the suction cup and surface is irrelevant: the surface can be horizontal, vertical, sideways, upside down--it doesn't matter. Air pressure acts in all directions.
 And if you factor in gravity than that is another tiny bit more net force directed towards your feet.
I don't know what you mean by "factor in gravity"; gravity is the source of air pressure.

 Quote by Doc Al Sorry, but this is incorrect. The fact that the net force might be zero tells you nothing about the pressure. Recall my example of the elephant crushing the peanut. The net force on the peanut is zero, but the pressure is very high. Pressure is a force per unit area. Further, air pressure is not zero!
Force is a vector but area is a scalar. So if we assume net force is 0 than how can there be nonzero pressure=force/area?

Or are you saying even though if you add all the forces up, it equals 0, the area these forces act on may not be equal. i.e. There are two forces 8N and -8N. The 8N force act on 2m^2 of the surface only whereas the -8N force act on 3m^2 of the surface only. In this way net pressure is nonzero.

But the mouse could also get completely crushed even if total force and pressure on it is 0. i.e. there could be opposite force and pressure (hence net 0) on top of it and from the bottom. The top exerted by the elephant and the bottom by the ground.

 Quote by Doc Al That would make some sense if your feet made an airtight seal with some surface, but that's unlikely to be the case. Rather than a person standing, imagine a suction cup stuck on a smooth surface. It sticks to the surface because air pressure pushes it. When you break the seal, air (at atmospheric pressure) gets inside and neutralizes the net force from the air. Note that the orientation of the suction cup and surface is irrelevant: the surface can be horizontal, vertical, sideways, upside down--it doesn't matter. Air pressure acts in all directions.
Those things are amazing. There was a time when I didn't know how these thigns worked and thought that new physics was needed to explain them. Then I realised it was only the air around us that was causing it to suck.

 Quote by Doc Al I don't know what you mean by "factor in gravity"; gravity is the source of air pressure.
But a weak source? I was trying to say that on the surface of the earth, the net force of air is towards the earth is not 0 but not much higher either.

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 Quote by pivoxa15 Force is a vector but area is a scalar. So if we assume net force is 0 than how can there be nonzero pressure=force/area?
Area is a vector quantity. Pressure is a scalar.

Note: If the edges of the soles of your shoes make airtight seals with the floor below them, then there will be a net downward force (because there's no air below your feet) of about 1000-2000lbs.

 Quote by Gokul43201 Area is a vector quantity. Pressure is a scalar. Note: If the edges of the soles of your shoes make airtight seals with the floor below them, then there will be a net downward force (because there's no air below your feet) of about 1000-2000lbs.
Force is a vector quantity and if area is a vector quantitiy than Pressure=Force/Area meaning a vector divide another vector. How does that make sense

Recognitions:
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$$P = \frac{\vec{F} \cdot \vec{A}}{|A|^2}$$
$$\vec{dF}=P~\vec{dA}$$


text:
SOLVING ABSOLUTE VALUE EQUATIONS

If you're going through this entire online Algebra I course,
then you already know that I love WolframAlpha.

Well, WolframAlpha allows you to create widgets to share its amazing computational power!

See that ‘WolframAlpha widget’ in the right-hand column?
It has a starting equation already in place, which happens to be an example from this page.
Go ahead and press the ‘Submit’ button—and behold the power of WolframAlpha!
You'll see a graph (for slightly more advanced users), the solutions,
and a number line visualization of the solutions.

Next, start playing!
Type in any absolute value equation or inequality;
I like to press ‘Enter’ instead of clicking the ‘Submit’ button.
(There is a vertical bar ‘ | ’ for absolute value somewhere on your keyboard—look around!)

You can even copy-and-paste from the randomly-generated exercises and worksheets.

|x| = 3       |x| < 3       |x| <= 3       |x| > 3       |x| >= 3

Feel free to move the widget around to keep it nearby while you're reading the lesson.
When you're done exploring, just close the widget window.
Have fun!

## WolframAlpha Widget

Now, let's talk about the concepts involved in solving absolute value equations:

THEOREM solving absolute value equations
Let $\,x\in\mathbb{R}\,$, and let $\,k\ge 0\,$.   Then, $$|x| = k\ \ \text{ is equivalent to }\ \ x = \pm k$$
TRANSLATING THE THEOREM

Recall first that normal mathematical conventions dictate that   ‘$\,|x| = k\$’   represents an entire class of sentences,
including the members   ‘$\,|x| = 2\$’,   ‘$\,|x| = 5.7\$’,   and   ‘$\,|x| = \frac{1}{3}\,$’.
The variable $\,k\$ changes from sentence to sentence, but is constant within a given sentence.

Also recall that   ‘$\,x=\pm k\$’   is a shorthand for   ‘$\,x = k\ \text{ or }\ x = -k\$’.

When you see a sentence of the form $\,|x| = k\$, here's what you should do:

• Check that $\,k\,$ is a nonnegative number (zero, or greater than zero).
• The symbol $\,|x|\,$ represents the distance between $\,x\,$ and $\,0\,$.
• Thus, you want the numbers $\,x\,$, whose distance from $\,0\,$ is $\,k\$.
• You can walk from $\,0\,$ in two directions: to the right, or to the left.
Walk to the right a distance $\,k\$, and you get to the number $\,k\$.
Walk to the left a distance $\,k\$, and you get to the number $\,-k\$.
• Thus,   $\,|x| = k\$   is equivalent to   $\,x = k\ \text{ or }\ x = -k\$, which goes by the shorthand   $\,x=\pm k\$.
• Equivalent sentences are completely interchangeable, and you can use whichever is easiest to work with.
In this case, you're getting rid of the troublesome absolute value in exchange for a less-troublesome ‘plus or minus’ sign.

Recall that   ‘$\,\iff \,$’   is a symbol for ‘is equivalent to’.

The power of the sentence-transforming tool   ‘$\,|x| = k \iff x = \pm k\$’
goes far beyond solving simple sentences like $\,|x| = 5\,$!

Since $\,x\,$ can be any real number, you should think of $\,x\,$ as merely representing
the stuff inside the absolute value symbols.
Thus, you could think of rewriting the tool as:   ‘$\,|\text{stuff}| = k \iff \text{stuff} = \pm k\$’

See how this idea is used in the following examples:

EXAMPLE:
Solve: $|2 - 3x| = 7$
Solution:
Write a nice, clean list of equivalent sentences:
 $|2 - 3x| = 7$ (original equation) $2-3x = \pm 7$ (check that $\,k\ge 0\,$; use the theorem) $2-3x = 7\ \text{ or }\ 2-3x = -7$ (expand the plus/minus) $-3x = 5\ \text{ or }\ -3x = -9$ (subtract $\,2\,$ from both sides of both equations) $\displaystyle x = -\frac{5}{3}\ \text{ or } x = 3$ (divide both sides of both equations by $\,-3\,$)
It's a good idea to check your solutions:

$|2 - 3(-\frac{5}{3})|\ \overset{\text{?}}{=}\ 7$,     $|2 + 5| = 7$     Check!

$|2 - 3(3)|\ \overset{\text{?}}{=}\ 7$,     $|2 - 9| = 7$     Check!
EXAMPLE:
Solve: $5 - 2|3 - 4x| = -7$
Solution:
To use the theorem, you must have the absolute value all by itself on one side of the equation.
Thus, your first job is to isolate the absolute value:
 $5 - 2|3 - 4x| = -7$ (original equation) $-2|3 - 4x| = -12$ (subtract $\,5\,$ from both sides) $|3 - 4x| = 6$ (divide both sides by $\,-2\,$) $3 - 4x = \pm 6$ (check that $\,k\ge 0\,$; use the theorem) $3 - 4x = 6\ \text{ or }\ 3 - 4x = -6$ (expand the plus/minus) $-4x = 3\ \text{ or }\ -4x = -9$ (subtract $\,3\,$ from both sides of both equations) $\displaystyle x = -\frac{3}{4}\ \text{ or }\ x = \frac{9}{4}$ (divide both sides of both equations by $\,-4\,$)
EXAMPLE:
Solve: $|3x + 1| = -5$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative?   NO!
No matter what number you substitute for $\,x\,$,
the left-hand side of the equation will always be a number that is greater than or equal to zero.
Therefore, this sentence has no solutions. It is always false.
Master the ideas from this section

When you're done practicing, move on to:
Solving Absolute Value Inequalities Involving ‘Less Than’

Solve the given absolute value equation.
Write the result in the most conventional way.

For more advanced students, a graph is displayed.
For example, the equation $\,|2 - 3x| = 7\,$
is optionally accompanied by the graph of $\,y = |2 - 3x|\,$ (the left side of the equation, dashed green)
and the graph of $\,y = 7\,$ (the right side of the equation, solid purple).
In this example, you are finding the values of $\,x\,$ where the green graph intersects the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.

CONCEPT QUESTIONS EXERCISE:

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# Calling All Fall 2012 Consortium Applicants!

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Intern
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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31 Jan 2012, 17:02
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KUDOS
I interviewed at Darden, Yale, Tuck, and Wharton, and am happy to share my experience if anyone needs it! Good idea Senator, thanks for taking the lead.
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31 Jan 2012, 17:10
1
KUDOS
Here to help. Have interviewed with:

Tuck - on campus - 2nd year (self initiated)
Johnson - phone - not an adcom but someone who works with adcom
Anderson - off campus - alum (Boston)
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01 Feb 2012, 08:33
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KUDOS
kawasaki wrote:
gablaze23 wrote:
rackbar wrote:
Yes, I checked their website when I applied and the applicant-initiated interview was listed as N/A for consortium applicants. Link: tuck[dot]dartmouth[dot]edu/admissions/how-to-apply/important-dates/

Thus, I thought they only invited consortium candidates. Based on what I'm reading on this forum though, it seems their website probably wasn't updated to reflect the new consortium interview policy. Looks like no Tuck for me this year

I applied to Tuck without self-initiating an interview as well. I did explain it in my "Why Tuck" essay though. I don't think all hope is lost. I spoke to a student who was accepted without self-initiating an interview. He said Tuck will call around 5 weeks before the decision deadline if they are interested

Dont lose hope!! Even though I did not self-initiate an interview, I ranked Tuck as #1 in my rankings and showed deep interest. "Why Tuck" was probably the best essay I wrote. Very curious to see if they will call applicants like us.

Are there any Tuck applicants here who did not self-initiate an interview? Would love to hear from you guys!

Guilty as charged. I think I may not make the cut cause I did not rank them 1st, I am still expectant though that it'll work out.

I was in the same boat, and had them down in my rankings as well. They called me for an interview anyways however.
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01 Feb 2012, 15:38
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Just wanted to post something to be the 1000th post .
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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01 Feb 2012, 17:59
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Got an invite from Tuck about an hour ago. For the "conspiracy theorists" out there, I ranked them #1
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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02 Feb 2012, 14:41
1
KUDOS
Hi everyone. I just had my first interview (!) with Johnson. It was a Skype interview with an adcom member. I was a bit nervous but the interview was very conversational. Went overtime (interview lasted almost an hour) but she said it was okay since we had a good conversation (Hope that's a good sign!)

Why MBA, why now?
Why Cornell?
Greatest failure.
Weaknesses.
Strengths.
What other schools did you apply to?
How would you prepare for the quantitative portion of MBA? (Notes about my lack of quant classes and Verbal stronger than Quant GMAT from those who reviewed my application) I mentioned about possibly taking courses at the local university but not wanting my core courses to become redundant since a current student mentioned it happening to him. She mentioned the option of enrolling in MBAMath. com to prepare.
How would you contribute to the school?
Any questions?

So I'm planning to email my interviewer thanking her - any tips about what the contents should involve? And when? Current day? Next day?

Also, she told me that Consortium applicants should receive a decision by Feb. 21
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02 Feb 2012, 17:14
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KUDOS
Decided to come out from the shadows! I got invited to interview at Yale even though I ranked them last (5th). I'm actually really bummed because I like the program so much more than I did when I submitted the app. Does anyone know WHY the Consortium would provide the rankings to schools prior to admissions decisions? Ugggh, probably going to get dinged, but really hope not.
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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03 Feb 2012, 22:16
1
KUDOS
TheSenator wrote:
"Thank you for applying to the Tuck School of Business through the Consortium for Graduate Study in Management. We have established an account for you in our secure application system. Please follow the directions below to access your account:

......

Admission decisions will be available at 5:00PM EST on Friday, February 13, 2012. On that date, you will see a link to view your online decision letter. To ensure that your User Name and Password are working properly, please log on to your account prior to the notification date. The Consortium will announce fellowship awards in March."

Tick Tock Tick Tock Tick Tock

February 13 isn't a Friday, it's a Monday. Hmmmm.
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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06 Feb 2012, 10:48
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KUDOS
TheSenator wrote:
kawasaki wrote:
You are definitely in an enviable spot Senator with all the admits. Just curious, when push comes to shove and all else being equal what school would you MOST likely matriculate in?

Why thank you kawasaki. IDK, what would you do? I could use advice right now.

kawasaki, I'll be making Senator come to Johnson with me. I may be making you come as well. I'll need some gmatclub roomies out there in Ithaca, haha.
_________________

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06 Feb 2012, 11:36
1
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mixedin86 wrote:
TheSenator wrote:
kawasaki wrote:
You are definitely in an enviable spot Senator with all the admits. Just curious, when push comes to shove and all else being equal what school would you MOST likely matriculate in?

Why thank you kawasaki. IDK, what would you do? I could use advice right now.

kawasaki, I'll be making Senator come to Johnson with me. I may be making you come as well. I'll need some gmatclub roomies out there in Ithaca, haha.

Haha... Cornell is definitely a fine choice, and Senator would be one heck of a buddy on campus...I love the focus on Leadership in Cornell and Tuck.
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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06 Feb 2012, 12:57
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KUDOS
gablaze23 wrote:
TheSenator wrote:
kawasaki wrote:
Senator, with all the conversations we've had on this forum, I am pretty sure that you'll make the right decision cause you know what you want. Would the $$play a part in your decision?$$\$ will play a part. How big of a part when it comes down to it... I'm not sure. I think by the end of next week I'll have it narrowed down to three schools, that I really want to go to.

The decision could be made by the end of this week though. We. Shall. See. #fingerscrossed

Good luck man!! Not that you need any of it, but I am sure you will be offered tons of $$everywhere. Now here's a question to all the other gmatclubbers around here. How much would free$$ play a part in your decision on where to matriculate?

The decision will be really simple for me. I only applied to schools that I was really interested in attending. I also ranked schools in the exact order of my interest...ie, I didn't try to "game the system". As a result, if I do get the fellowship, I will attend the school that gives it to me

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07 Feb 2012, 11:44
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TheSenator wrote:
First chart compares total intake from last year vs. just gmatclub this year. The bar graph is just gmatclub cgsm applicants. It's a good barometer of where you fit.

We should also mention that gmat cgsm applicants probably represent the best of the cgsm pool, the average cgsm applicant to Tuck doesn't have anywhere near a 693 GMAT.
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08 Feb 2012, 07:13
1
KUDOS
Just got the admit email from Emory, anyone looking to interview feel free to PM if you have any questions.

PS ranked them 3/5
Senior Manager
Status: MBAing!!!!
Joined: 24 Jun 2011
Posts: 297
Location: United States (FL)
Concentration: Finance, Real Estate
GPA: 3.65
WE: Project Management (Real Estate)
Followers: 6

Kudos [?]: 52 [1] , given: 56

Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

### Show Tags

11 Feb 2012, 09:42
1
KUDOS
I'm happy to report that I had my interview at Johnson this past Friday. I arrived on Thursday night and stayed at the Courtyard Ithaca, this hotel is highly recommended, they shuttled me around for free. They picked me up at the airport; they drove me to Sage Hall, and took me back to the airport after the interview and info session. The campus is beautiful, and I loved the old world feel. My interview was at 11:15am with a recent grad that is going into consulting full time this summer. We talked for about 45 minutes; other than the traditional questions, she stressed on teamwork dynamics and pressed on the issue of leadership style. Follow up questions are always the hardest, so be on point. After a smooth but intense interview I went to a Q&A held by Consortium students at Johnson that lasted around an hour. It was very informative and I was happy to interact with current students and a couple of members of this forum iamvolitant and treydingo. I think I will be very happy spending 2 years at Johnson, I felt the passion of the consortium members and I hope I get the call from Johnson.
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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16 Feb 2012, 08:14
1
KUDOS
wow- i wish i would have found this forum sooner. I feel like i've been going through this process all alone. it would have been great to have your insights. I applied to 5 consortium schools- Haas, Yale, USC, UCLA and Emory. So far I interviewed with Yale and have recieved interview requests from USC and Emory. No word from Haas and UCLA... and i've all but given up at this point. I've also applied to Owen, and am considering applying to Wisconsin- as I am a wisconsin resident.

I didn't realize the ranking order was so important, I do know someone who went through the process last year, he got into all three of his schools but was awarded the the fellowship to his 3rd choice.

At this point i just want tMarch to come so that i can sleep normally again.
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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16 Feb 2012, 13:03
1
KUDOS
Sure- He applied to Dartmouth, Yale and CMU (Tepper) he was admitted to all three, but the first two opted not to award fellowship. So he went to Tepper.

I'll go ahead and update my stats but here's the basics: 690 GMAT, 3.45 GPA- Dual Degree- Business Admin and Spanish Language from a state school, 5years work exp in nonprofit management.
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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16 Feb 2012, 15:40
1
KUDOS
mooze wrote:
Opened my email, I see Yale Shool of Management in big fat letters,...then look at the subject line...some marketing newletter. Dagger.

Now I know how it feels to get dumped and the girl who dumped you turn around and include you on a mass email with stories of how great her life is and pictures from her Cancun trip....stupid b****, take me out of your f'ing contacts!!... Whoops, having a bit of a moment... Away, I got the same Yale email.

Posted from my mobile device
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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20 Feb 2012, 09:16
1
KUDOS
Just got a nice phonecall from Johnson this morning.
I'm IN!
:D
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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20 Feb 2012, 11:16
1
KUDOS
good to know method! Today I talked to a girl who I met visiting Stern in the fall - she said her Ross interviewer was stressful, but she thinks her experience was the exception and that most people had good experiences. Good luck- hope you get in!

Vamos azul indeed!
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Re: Calling All Fall 2012 Consortium Applicants! [#permalink]

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20 Feb 2012, 12:46
1
KUDOS
gablaze23 wrote:
kheti wrote:
Any new haas or ucla news from anyone?

Posted from my mobile device

Got an UCLA interview invite on Thursday... still no word from Haas the fact that UCLA came so late encourages me, since they're both public cali schools, i would imagine their process may be similar.

Wow! Glad to hear they have started rolling their invites.

They offered in person or locally with Alumni... they stressed that there is no preference given to coming there. My guess is that they do this because they know they have less time to complete them in than other schools.
Re: Calling All Fall 2012 Consortium Applicants!   [#permalink] 20 Feb 2012, 12:46

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# Calling All Fall 2012 Consortium Applicants!

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text:
Prealgebra 2e

# Practice Test

Prealgebra 2ePractice Test

### Practice Test

626.

For the polynomial $8y4−3y2+18y4−3y2+1$

1. Is it a monomial, binomial, or trinomial?
2. What is its degree?

In the following exercises, simplify each expression.

627.

$(5a2+2a−12)+(9a2+8a−4)(5a2+2a−12)+(9a2+8a−4)$

628.

$(10x2−3x+5)−(4x2−6)(10x2−3x+5)−(4x2−6)$

629.

$(−34)3(−34)3$

630.

$n·n4n·n4$

631.

$(10p3q5)2(10p3q5)2$

632.

$(8xy3)(−6x4y6)(8xy3)(−6x4y6)$

633.

$4u(u2−9u+1)4u(u2−9u+1)$

634.

$(s+8)(s+9)(s+8)(s+9)$

635.

$(m+3)(7m−2)(m+3)(7m−2)$

636.

$(11a−6)(5a−1)(11a−6)(5a−1)$

637.

$(n−8)(n2−4n+11)(n−8)(n2−4n+11)$

638.

$(4a+9b)(6a−5b)(4a+9b)(6a−5b)$

639.

$56585658$

640.

$(x3·x9x5)2(x3·x9x5)2$

641.

$(47a18b23c5)0(47a18b23c5)0$

642.

$24r3s6r2s724r3s6r2s7$

643.

$8y2−16y+204y8y2−16y+204y$

644.

$(15xy3−35x2y)÷5xy(15xy3−35x2y)÷5xy$

645.

$4−14−1$

646.

$(2y)−3(2y)−3$

647.

$p−3·p−8p−3·p−8$

648.

$x4x−5x4x−5$

In the following exercises, factor the greatest common factor from each polynomial.

649.

$80a3+120a2+40a80a3+120a2+40a$

650.

$−6x2−30x−6x2−30x$

651.

According to www.cleanair.org, the amount of trash generated in the US in one year averages out to $112,000112,000$ pounds of trash per person. Write this number in scientific notation.

652.

Convert $5.25×10−45.25×10−4$ to decimal form.

In the following exercises, simplify, and write your answer in decimal form.

653.

$(2.4×108)(2×10−5)(2.4×108)(2×10−5)$

654.

$9×1043×10−19×1043×10−1$

655.

A hiker drops a pebble from a bridge $240240$ feet above a canyon. The polynomial $−16t2+240−16t2+240$ gives the height of the pebble $tt$ seconds a after it was dropped. Find the height when $t=3.t=3.$


text:
In an FWF-funded project, our group brings the "black-box" approach of quantum information theory into space and time. In an nutshell, we study the implications of symmetries (via group representation theory) on (semi-)device-independent scenarios. Here, we describe some of the background and ideas of the project.

### Black boxes: foundations and applications

Quantum cryptography and the generation of random numbers are two of the most exciting applications of quantum information theory (QIT). Ideally, we would like to base the security of such protocols on as few assumptions as possible. Device-independent [5, 6, 7] and semi-device-independent [8, 9] protocols offer a kind of "gold standard" towards this goal. For example, there are protocols that allow two parties (say, Alice and Bob) to establish a provably secure cryptographic key if only two assumptions are satisfied:

• the no-signalling principle: there is no instantaneous information transfer between Alice and Bob;
• violation of a Bell inequality [10].

In principle, we can guarantee that there is no signalling by placing Alice and Bob at spacelike separation, such that special relativity forbids information transfer on the relevant time scales. The violation of a Bell inequality is a statistical property which Alice and Bob can observe in their recorded data. No assumptions on the devices have to be made (they could have been bought from the eavesdropper) - we do not even have to assume the validity of quantum theory! In principle, all components of the experiment are modelled as "black boxes" [10]: unknown physical systems with some specific observable statistical input-output behavior. It is remarkable that such black-box descriptions are sufficient to guarantee the functionality and security of some quantum information-theoretic protocols.

Apart from protocols, this black-box picture has proven extremely useful for the study of the foundations of quantum physics. For example, we can ask about the set of correlations that any physical theory admits in a causal scenario as the one in the figure above. What possible probability tables p(a,b|x,y) (also called "behaviors" or "correlations") can we generate? Bell's theorem [10] tells us that there are correlations that can be generated via quantum physics in this causal scenario, but not via classical physics, as testified by a violation of a Bell inequality. For example, correlations that arise from certain measurements on a singlet state have this property. Despite these difference, classical and quantum correlations satisfy one common constraint. These are the no-signalling conditions [10]: Bob's choice of input ("choice of measurement") has no impact on the marginal probabilities of Alice's outputs ("outcomes") and vice versa:

In a relativistic universe, these conditions are crucial: if they were violated, we could construct a "Bell telephone" that allows us to transmit informtion faster than light. This is impossible in classical and quantum physics. However, quantum physics does not give us the most general non-signalling correlations that satisfy this important constraint: there are conceivable correlations that are non-signalling, but that violate Bell inequalities by more than any quantum state. An example is given by "Popescu-Rohrlich" ("PR-box") [10] correlations: Alice and Bob both choose a single bit as input, i.e. x,y∈{0,1}, and they each obtain either +1 or -1 as outputs, i.e. a,b∈{-1,+1}. The PR-box probabilities are

That is, the outcomes are either perfectly anticorrelated (if Alice and Bob choose inputs "1") or otherwise perfectly correlated. This probability table satisfies the no-signalling conditions, but it cannot be realized in a Bell scenario within quantum physics.

The situation can be illustrated as follows:

The set of classical correlations C is a strict subset of the set of quantum correlations Q, which in turn is a strict subset of the set of non-signalling correlations NS. Both C and NS are polytopes (for any fixed number of parties, inputs and outputs), but Q is not. The set of quantum correlations Q sits strangely "in between" the most restrictive (classical) and most general (non-signalling) possibilities. "Why" does nature admit the set of correlations Q, and not a smaller or larger set? Instead of simply postulating the validity of quantum theory, can we perhaps find physical principles that determine Q? The last few years have seen tremendous research efforts on this question, and some fascinating partial results, but the general problem is still open (see, however, our other research ares: "Reconstructions of Quantum Theory").

Our research project builds on these previous findings, but studies one additional and surprisingly insightful aspect: the role of space and time.

### Putting the black boxes into space and time

In the context of an FWF-funded project, our group is reconsidering the "black-box" approach - not only for Bell scenarios as sketched above, but also for more general settings, such as semi-device-independent prepare-and-measure scenarios or other causal structures. As we have seen above, the inputs and outputs of black boxes are typically considered to be abstract labels, for example single bits. However, in many (if not all) experimentally relevant scenarios, the inputs are spatiotemporal quantities: for example, directions of a magnetic field, the duration of a pulse, or the the angle of a polarizer.

In other words, inputs and/or outputs of the black boxes are often quantities that break a spatiotemporal symmetry: physical systems that carry group representations. Do these symmetries impose additional constraints on the correlations? Do they yield additional information that can play a similar role as the (ultimately spatiotemporally motivated) no-signalling principle - perhaps information that we can use as well-motivated assumptions in semi-device-independent scenarios? These are some of the questions that we address in our project. Our aim is two-fold:

1. To analyze how the structures of spacetime and quantum theory mutually constrain each other. This is a foundational question motivated by pure curiosity, where we aim for insights into the logical structure of the physical world.
2. To use the additional structures and constraints, via group representation theory, to construct novel device- or semi-device-independent protocols, or to improve the analysis of assumptions that underly, for example, the analysis of experimentally relevant "nonlocality witnesses".

In preliminary work [1], we have given a proof of principle that natural assumptions on the response of a physical system to rotational symmetries allows us to certify nonlocality in ways that would not otherwise be possible. Furthermore, we have shown that a partial characterization of quantum correlations in terms of symmtries is possible, which we describe next.

### Exact characterization of the (2,2,2)-quantum correlations

Consider a Bell scenario where Alice and Bob each obtain one of two possible outcomes, a,b∈{-1,+1}, and each of them can choose between two possible inputs. However, now we assume that Alice and Bob reside in d-dimensional space (where d≥2), and that their inputs are spatial directions: unit vectors on which they can act with the rotation group SO(d). For example, we could have d=3, and the choice of inputs could be given by setting a magnetic field (or its inhomogeneity) in a specific direction. Now we make one specific assumption and a non-assumption:

• We do not assume that quantum theory holds. That is, we base our proofs and calculations only on probability theory, and we demand that the probabilities respond to rotations in a consistent way.
• We assume that "the probabilities transform locally like vectors". This physics jargon means mathematically that the local outcome probabilities are affine-linear in the rotations RA, RB that Alice and Bob perform locally to choose their inputs.

In our paper [1], we prove that these conditions uniquely characterize the set of quantum correlations Q as depicted in the figure above -- in the special case of (2,2,2)-correlations, i.e. if we have 2 parties (Alice and Bob), each having a choice of 2 inputs and obtaining one of 2 possible outputs:

Theorem. Every correlation that can be obtained under these two assumptions is a quantum correlation, i.e. can be realized within quantum theory. Furthermore, all (2,2,2)-quantum correlations can be realized under these two assumptions (restricting Alice and Bob to two possible input directions) if we supplement Alice and Bob with shared randomness.

For example, consider the case d=2 depicted above. In this case, the input (unit vectors) can be parametrized by single angles. We can think of Alice (and Bob) choosing his (her) input by rotating a polarizer by a certain angle α (resp. β), obtaining outcome -1 if the particle passes the polarizer and +1 otherwise, yielding correlations p(a,b|α,β),  The set of probability tables that can be realized in this scenario, assuming that the probabilities transform like vectors, that Alice and Bob are restricted to two possible choices of angles, and that Alice and Bob have a supply of shared randomness, is exactly the set Q of (2,2,2)-quantum correlations.

### Towards semi-device-independent protocols

This fundamental result rests on a particular assumption: that outcome probabilities transform according to a certain representation of the symmetry group. We think that assumptions of this form can, in some cases, be used as natural replacements of ad-hoc assumptions in semi-device-independent protocols.

For example, prepare-and-measure scenarios can be used to generate certified random numbers, but they typically work under certain formal assumptions like Hilbert space dimension bounds. We believe that such dimension bounds can be replaced by physically better motivated assumptions on the response of the physical systems to spatiotemporal symmetries, such as theory-independent energy bounds.

### Space, time and quantum theory

How are the structures of spacetime and quantum theory related? While this is one of the motivating questions for this ongoing project, we have previously [2, 3, 4] studied further aspects of this question. Please see "General Probabilistic Theories" for more information.

### References [our group]

[1] A. J. P. Garner, M. Krumm, and M. P. Müller, Semi-device-independent information processing with spatiotemporal degrees of freedom, Phys. Rev. Research 2, 013112 (2020). arXiv:1907.09274

[2] A. J. P. Garner, M. P. Müller, and O. C. O. Dahlsten, The complex and quaternionic quantum bit from relativity of simultaneity on an interferometer, Proc. R. Soc. A 473, 20170596 (2017). arXiv:1412.7112

[3] P. A. Höhn and M. P. Müller, An operational approach to spacetime symmetries: Lorentz transformations from quantum communication, New J. Phys. 18, 063026 (2016); in NJP's "Highlights of 2016" collection. arXiv:1412.8462

[4] M. P. Müller and Ll. Masanes, Three-dimensionality of space and the quantum bit: an information-theoretic approach, New J. Phys. 15, 053040 (2013); in NJP's "Highlights of 2013" collection. arXiv:1206.0630

### References [other authors]

[5] A. Acín, N. Brunner, N. Gisin, S. Massar, S. Pironio, and V. Scarani, Device-Independent Security of Quantum Cryptography against Collective Attacks, Phys. Rev. Lett. 98, 230501 (2007). arXiv:quant-ph/0702152

[6] J. Barrett, L. Hardy, and A. Kent, No Signaling and Quantum Key Distribution, Phys. Rev. Lett. 95, 010503 (2005). arXiv:quant-ph/0405101

[7] R. Colbeck, Quantum And Relativistic Protocols For Secure Multi-Party Computation, PhD thesis, University of Cambridge, 2006. arXiv:0911.3814

[8] M. Pawłowski and N. Brunner, Semi-device-independent security of one-way quantum key distribution, Phys. Rev. A 84, 010302(R) (2011). arXiv:1103.4105

[9] H. Li, Z. Yin, Y. Wu, X. Zou, S. Wang, W. Chen, G. Guo, and Z. Han, Semi-device-independent random-number expansion without entanglement, Phys. Rev. A 84, 034301 (2011). arXiv:1108.1480

[10] N. Brunner, D. Cavalcanti, S. Pironio, V. Scarani, and S. Wehner, Bell nonlocality, Rev. Mod. Phys. 86, 419 (2014). arXiv:1303.2849


text:
# Curve Fitting problem MATLAB

I have a system of differential equations with some unknown parameters and I need to find the optimal parameter set that fits best with the data I have.

For each parameter set choice (using for loops), my code solves the system of DE numerically, then extracts from the time vector a time subvector that agrees with data time values. The next step requires to extract a new output subvector for each of the variables that match with the time subvector (in order to compute the error).

Here is how my code is looking:

OO=[];
y2=(y(:,2))';
t1=t';
FF=[];
y1=(y(:,1))';

global "parameters and other global variables"

for parameter_1
for parameter_2
[t,y]=ode45('system',[time interval], [initial conditions]);
time; % "time" is a code that extracts the time subvector

for i=1:length(time)
for j=1:length(t1)
if time(i)==t1(j);
FF(end+1) = y1(j);
end
end
end

for k=1:length(time)
for m=1:length(t1)
if time(k)==t1(m);
OO(end+1) = y2(m);
end
end
end

end
end

At this point the code is not over, but I would be expecting it to return the OO vector and the FF vector for the last parameter set choice of the loop. However when the code runs (without any errors) it produces vectors OO and FF that are longer than vector time: length(time) < {length(OO),length(FF)}.

• Welcome to SciComp.SE. It is not clear what your question is. What is the problem that you have? Jul 1 '15 at 21:20
• Hi. Thanks for your reply. When I run the code. the dimension of the OO and FF vector is not the same as the time vector. My time vector has dimension 111. I need my OO and FF vector to be the same. Jul 1 '15 at 22:33
• The length of the time subvector is not given, and the double loops you use will, for the pathological case, produce a vector of length l=length(time)*length(t1). The length of FF and OO can be of any length between 0 and my l defined before depending on the level of agreement. Maybe you could draw a picture of what you're trying to accomplish? Jul 1 '15 at 23:52
• I use ode45 to solve the system of DE over the time interval t=[0 110] However MATLAB, when computing numerically the solution, generates a discrete time interval whose elements are not equally spaced. I have a code where MATLAB picks 111 elements of the vector t (closest to integer time values). So the length of "time" is 111. When I run the above code. The length of OO and FF is 221 (they must be of same length as time) For the picture, after getting OO and FF, I want to compute the error between the simulated model and data. The lengths of the data vectors are also 111. Jul 2 '15 at 1:23
• Finally, I get a vector whose entries are errors for each parameter set. I want to find the minimum of this vector and the only output that i want to see from the code is the choice of parameter set that minimizes the error. It will be my baseline choice for the model. Jul 2 '15 at 1:24

This is a more general nonlinear optimization problem than would usually be described as simply "curve fitting". I suggest you approach it as such. You will need to define a few things:

1. The parameters you are optimizing.
2. The quantity to be minimized.
3. Helper functions to simplify things.

Matlab has tools that can help and it will help you to break the problem into a couple parts as follows:

### Main script

clear

% load the curve and time values you want the ODE to fit
[ tReference, yReference ] = loadKnownValues();

% load the initial conditions for the ODE:

% initial guess for parameter values (a column vector)
parametersGuess = [ 1; 6.83; ... ];

% define an objective function for the optimization
% I will assume you want to minimize the sum of the errors squared
% lsqnonlin is in the Optimization toolbox
%
% You could also use fminsearch. In that case you would want to use the
% objective function F = @(param) sum( objectiveFunction(param).^2 );
objectiveFunction = @(parameters) yReference ...
- solveODE(tReference,initialConditions,parameters);

optimizedParameters = lsqnonlin(objectiveFunction,parametersGuess);

This main script relies on a few other functions that do specific things. Obviously you will need to load in the curve you're trying to fit and the initial conditions for the ODE (loadKnownValues() and loadInitialConditions()). You will also need a function that solves the ODE given a set of initial conditions and parameters and returns the values at the desired times. Those first two should be easy, the last one can be done as follows:

### solveODE.m

function yVals = solveODE(tReference,initialConditions,parameters)
% yVals = solveODE(tReference,initialConditions,parameters)
% Solves the ODE for your problem and with the given initialConditions
% and parameters and returns the values at the specified times in
% tReference. The first time in tReference must correspond to the
% initial time.

% function that returns the right hand side of the ODE,
% use the parameters in the input vector "parameters"
rightHandSide = @(t,y) ... ;

[~, yVals] = ode45(rightHandSide, tReference, initialConditions);

Obviously I can't comment on the convergence or performance of the optimization without knowing a lot more about the system, but this should get you started.

• Hi Doug, I decided to give your code a try. In my code, I had to "manually" zoom into the 5-dimensional grid (5 parameter space). I am also not satisfied with the results. parameters are very much correlated. I ran your code but I think I have a tolerance problem. The code stops after 1 second and says that the initial point is a local minimum. values of yReference are between 0 and 1 and so errors should be of the order 0.001 or less. I tried changing Tolerance but I couldn't manage. How should I proceed? @User1 Jul 30 '15 at 16:51
• i used options=optimset('TolX',a,'TolFun',b) and I am letting a and b be of very small orders e.g. 10^-10 and it still takes one second. I also defined a lower and upper bound for each of the parameters to increase the grid search. Nothing is helping @Doug Jul 30 '15 at 17:27
• If you are getting "The code stops after 1 second and says that the initial point is a local minimum" from lsqnonlin(), this is commonly due to a discontinuous objective function. A common case is where the residual function samples some data using nearest-neighbor interpolation, e.g. Residual = @(t) y - interp1(tdat,ydat,t,'nearest'); In this case, changing from 'nearest' to 'linear' will fix the problem, allowing lsqnonlin() to "see" small changes in t. Oct 6 '15 at 1:32


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Creating anonymous functions?

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Is it possible to create functions without names, just like you can create objects without names? So that you can have an array of functions without real names that clutter up the global (or any) namespace.
void func() { /*blah*/ }

// instead I want something like:

void (*pFunc)() = ???

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If I understand your question, then the answer is no, you can''t. But if you are worrying about cluttering the global namespace, then just put your functions in a seperate namespace. Is that a problem?

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Well, technically, yes it is possible to do so at runtime. Allocate some memory and fill it with the machine code for your functions, and create an array that contains function pointers that point to the various function entry points.

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Hmm... how would I go about doing that? I suppose it''s not directly supported by the language, so it wouldn''t be pretty, I guess?

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There are some things to keep in mind when creating such code. It must be relocatable code, since you don''t know before hand where it will absolutely be in memory. Or if is isn''t relocatable, then you must patch it up which would be much more trouble than its worth. If the only reason you want to do this is to not pollute the global namespace I wouldn''t do this. Just create a namespace for them.

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C and C++ do not treat functions as first-class data.

You can malloc() a bunch of memory and generate machine code into this memory; this is not at all supported by the language and will look different for each type of CPU architecture you intend to run on (PPC, x86, Itanium, etc).

THEN you have to make sure that you flush/invalidate the D-cache and I-cache for the malloc-ed area, else bad things might happen when you try to jump to it :-) Some OS-es also require that you map the area "executable" which it might not be by default. These two operations are OS dependent (Windows, MacOS, Linux, etc).

Thus, you have the cross section of possible CPUs and possible OSes to deal with. Delicious!

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Your best bet is to use a static function (or unnamed namespace for you C++ people). The name will be local to this translation unit, so you''re only cluttering the current namespace.

As for the filling an array with machine instructions, there''s a reason the standard doesn''t guarantee you can cast a function pointer to a void pointer. It either cannot work or is very difficult on some systems. Of course, that''s a matter of portability. It''ll rule out some platforms and may require significant work to port it to others. It''s a trade off.

Another option would be to interface with a language with first class functions. For example, from what I can gather, Haskell (with the FFI adendum) allows conversion of its functions to C functions at runtime. This may or may not be wise. (using multiple languages isn''t something to take lightly, but shouldn''t be ruled out)

Also, if you''re using gcc and don''t care about other compilers, you can make functions local to a function. Thus, you only clutter the namespace within the function (not a big deal).

Depending on why you need this, you might even be able to find another, simpler solution.

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Yeah I''ll probably go with an unnamed namespace.

Too bad lambda functions aren''t supported in C++ though.

I basically need to run (and possibly create) particular functions runtime, so I need to map functions with strings. Since I won''t ever call a function explicitly in code I don''t need or want them to have names.

Does gcc support nested functions?

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quote:
Original post by eighty
Yeah I''ll probably go with an unnamed namespace.

Too bad lambda functions aren''t supported in C++ though.

I basically need to run (and possibly create) particular functions runtime, so I need to map functions with strings. Since I won''t ever call a function explicitly in code I don''t need or want them to have names.

Check out Boost Lambda (http://www.boost.org). It''s fairly limited and not nearly as flexible as lambda constructs in (mostly) functional languages, but if your needs are simple, it might be what you''re after.

quote:

Does gcc support nested functions?

Gcc does, but not g++, AFAIK. It only works as an extension to the C language.

Given your needs for lambda constructs and nested functions, I''m wondering if you''re using the right language for the job though...

Hope this helps.

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I haven''t really tried it yet, but the boost library implements lambda functions: http://www.boost.org/libs/lambda/doc/index.html Boost contains some other useful stuff too. I don''t know if it''s suited for your problem, though.

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quote:
Original post by Anonymous Poster
Check out Boost Lambda (http://www.boost.org). It's fairly limited and not nearly as flexible as lambda constructs in (mostly) functional languages, but if your needs are simple, it might be what you're after.

Interesting... I'll check it out.

quote:

Gcc does, but not g++, AFAIK. It only works as an extension to the C language.

Hm... I was under the impression gcc == g++...?

Thanks for the replies.

[edited by - eighty on July 9, 2003 2:38:38 AM]

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quote:
Original post by eighty
Hm... I was under the impression gcc == g++...?

Most (all?) extensions are language specific. AFAIK, all the languages supported by GCC are compiled to a Lisp-like intermediate code by the language specific frontend. The resulting code is then compiled by the compiler backend. Therefore, the C++ frontend will likely not be able to handle a C extension, and vice-versa.

I could be wrong of course: I don''t know GCC''s internals, so feel free to correct me.

Hope this helps.

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Umm... I meant I thought gcc was the same thing as g++. Just another name for it. I guess I was wrong.

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gcc used to stand for GNU C Compiler. Now, it stands for GNU Compiler Collection, and includes the C compiler (gcc), the C++ compiler (g++), the pascal compiler, as well as a couple others (I forget exactly which ones).

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C, C++, Object C, Fortran, Java, and Ada are supported by gnu, and there are many other front-ends available.

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gcc will call the correct back end for the job. You need C++? It''ll send your code to g++. You need Fortran? g77 handles it. Etc.

As noted, the best solution may be to use another language. OCaml might be a good place to start (good reputation for speed, supports procedural, OO, and functional paradigms (with a focus on functional))

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# PyMKS Introduction¶

In this short introduction, we will demonstrate the functionality in PyMKS. We will quantify microstructures using 2-point statistics, predict effective properties using homogenization and predict local properties using localization. If you would like more technical details about any of these methods please see the theory section.

In [1]:

import pymks

import warnings
warnings.filterwarnings('ignore')

%matplotlib inline

## Quantify Microstructures using 2-Point Statistics¶

Lets make two dual-phase microstructures with different morphologies.

In [2]:

from pymks.datasets import make_microstructure
import numpy as np

X_1 = make_microstructure(n_samples=1, grain_size=(25, 25))
X_2 = make_microstructure(n_samples=1, grain_size=(95, 15))

X = np.concatenate((X_1, X_2))

Throughout PyMKS X is used to represent microstructures. Now that we have made the two microstructures, lets take a look at them.

In [3]:

from pymks.tools import draw_microstructures

draw_microstructures(X)

We can compute the 2-point statistics for these two periodic microstructures using the correlate function from pymks.stats. This function computes all of the autocorrelations and cross-correlation(s) for a microstructure. Before we compute the 2-point statistics, we will discretize them using the PrimitiveBasis function.

In [4]:

from pymks import PrimitiveBasis
from pymks.stats import correlate

p_basis = PrimitiveBasis(n_states=2, domain=[0, 1])
X_corr = correlate(X, p_basis, periodic_axes=[0, 1])

Let’s take a look at the two autocorrelations and the cross-correlation for these two microstructures.

In [5]:

from pymks.tools import draw_correlations

print(X_corr[0].shape)

draw_correlations(X_corr[0])

(101, 101, 3)

In [6]:

draw_correlations(X_corr[1])

2-Point statistics provide an object way to compare microstructures, and have been shown as an effective input to machine learning methods.

## Predict Homogenized Properties¶

In this section of the intro, we are going to predict the effective stiffness for two-phase microstructures using the MKSHomogenizationModel, but we could have chosen any other effective material property.

First we need to make some microstructures and their effective stress values to fit our model. Let’s create 200 random instances 3 different types of microstructures, totaling to 600 microstructures.

In [7]:

from pymks.datasets import make_elastic_stress_random

grain_size = [(47, 6), (4, 49), (14, 14)]
n_samples = [200, 200, 200]

X_train, y_train = make_elastic_stress_random(n_samples=n_samples, size=(51, 51),
grain_size=grain_size, seed=0)

Once again, X_train is our microstructures. Throughout PyMKS y is used as either the property, or the field we would like to predict. In this case y_train is the effective stress values for X_train. Let’s look at one of each of the three different types of microstructures.

In [8]:

draw_microstructures(X_train[::200])

The MKSHomogenizationModel uses 2-point statistics, so we need to provide a discretization method for the microstructures by providing a basis function. We will also specify which correlations we want.

In [9]:

from pymks import MKSHomogenizationModel

p_basis = PrimitiveBasis(n_states=2, domain=[0, 1])
homogenize_model = MKSHomogenizationModel(basis=p_basis, periodic_axes=[0, 1],
correlations=[(0, 0), (1, 1), (0, 1)])

Let’s fit our model with the data we created.

In [10]:

homogenize_model.fit(X_train, y_train)

Now let’s make some new data to see how good our model is.

In [11]:

n_samples = [10, 10, 10]
X_test, y_test = make_elastic_stress_random(n_samples=n_samples, size=(51, 51),
grain_size=grain_size, seed=100)

We will try and predict the effective stress of our X_test microstructures.

In [12]:

y_pred = homogenize_model.predict(X_test)

The MKSHomogenizationModel generates low dimensional representations of microstructures and regression methods to predict effective properties. Take a look at the low-dimensional representations.

In [13]:

from pymks.tools import draw_components_scatter

draw_components_scatter([homogenize_model.reduced_fit_data[:,:2],
homogenize_model.reduced_predict_data[:,:2]],
['Training Data', 'Test Data'])

Now let’s look at a goodness of fit plot for our MKSHomogenizationModel.

In [14]:

from pymks.tools import draw_goodness_of_fit

fit_data = np.array([y_train, homogenize_model.predict(X_train)])
pred_data = np.array([y_test, y_pred])

draw_goodness_of_fit(fit_data, pred_data, ['Training Data', 'Test Data'])

Looks good.

The MKSHomogenizationModel can be used to predict effective properties and processing-structure evolutions.

## Predict Local Properties¶

In this section of the intro, we are going to predict the local strain field in a microstructure using MKSLocalizationModel, but we could have predicted another local property.

First we need some data, so let’s make some.

In [15]:

from pymks.datasets import make_elastic_FE_strain_delta

X_delta, y_delta = make_elastic_FE_strain_delta()

Once again, X_delta is our microstructures and y_delta is our local strain fields. We need to discretize the microstructure again, so we will also use the same basis function.

In [16]:

from pymks import MKSLocalizationModel

p_basis = PrimitiveBasis(n_states=2)
localize_model = MKSLocalizationModel(basis=p_basis)

Let’s use the data to fit our MKSLocalizationModel.

In [17]:

localize_model.fit(X_delta, y_delta)

Now that we have fit our model, we will create a random microstructure and compute its local strain field, using finite element analysis. We will then try and reproduce the same strain field with our model.

In [18]:

from pymks.datasets import make_elastic_FE_strain_random

X_test, y_test = make_elastic_FE_strain_random()

Let’s look at the microstructure and its local strain field.

In [19]:

from pymks.tools import draw_microstructure_strain

draw_microstructure_strain(X_test[0], y_test[0])

Now let’s pass that same microstructure to our MKSLocalizationModel and compare the predicted and computed local strain field.

In [20]:

from pymks.tools import draw_strains_compare

y_pred = localize_model.predict(X_test)
draw_strains_compare(y_test[0], y_pred[0])

The MKSLocalizationModel can be used to predict local properties and local processing-structure evolutions.
In [ ]:


text:
sage: H = SymmetricGroup(4)

Then G will be represented as permutations of 1, 2, ..., 8. H.direct_product(H) is a tuple, the first entry of which is the actual direct product; the other entries give inclusions of the two factors into the product and projections, likewise.


text:
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# Some people form a joint account for one year with the condi

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Some people form a joint account for one year with the condi [#permalink]

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03 Sep 2013, 08:03
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Some people form a joint account for one year with the condition that every month each member deposits an amount equal to the number of members in the account in that month. Also, the person who withdraws from the account before the end of the year get his/her amount at the end of the year. After 6 months, 1/4th of the people withdraw and 1/3rd of the remaining withdraw after 3 months. At the end of the year there is an amount of Rs. 4860 in the account before paying of the withdrawn amounts. Find out the number of people in the beginning of the year.

A. 12
B. 24
C. 36
D. 48
E. 60
[Reveal] Spoiler: OA

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Last edited by Bunuel on 03 Sep 2013, 08:08, edited 1 time in total.
Renamed the topic and edited the tags.
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Re: Some people form a joint account for one year with the condi [#permalink]

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03 Sep 2013, 18:41
Qoofi wrote:
Some people form a joint account for one year with the condition that every month each member deposits an amount equal to the number of members in the account in that month. Also, the person who withdraws from the account before the end of the year get his/her amount at the end of the year. After 6 months, 1/4th of the people withdraw and 1/3rd of the remaining withdraw after 3 months. At the end of the year there is an amount of Rs. 4860 in the account before paying of the withdrawn amounts. Find out the number of people in the beginning of the year.

A. 12
B. 24
C. 36
D. 48
E. 60

Need the solution, cant reach to B as the solution
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Re: Some people form a joint account for one year with the condi [#permalink]

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03 Sep 2013, 20:42
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N people in the start.

first 6 months there would 6.N.N money deposited.
1/4 members are gone so 3/4 left
Then next 3 month 3.3N/4.3N/4

then 1/3 of remaining gone = 3N/4*2/3 = N/2
so money deposited 3.N/2.N/2

total = 6N^2+27/16N^2+3/4N^2 = 4860
solve for N = 24
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Re: Some people form a joint account for one year with the condi [#permalink]

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04 Sep 2013, 00:30
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Let x be the number of people in the beginning.
Amount deposited for 6 months = 6*x*x = 6x^2

Number of people for the next 3 months = x -x/4 = 3x/4

Amount deposited for the next 3 months = 3 * (3x/4)^2

Number of people for the last 3 months = 3x/4 - (1/3 * 3x/4) = x/2

Amount deposited for the last 3 months = 3* (x/2)^2

Total amount = 6x^2 + 27x^2/16 + 3x^2/4 = 4860

x= 24 (option B)
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Re: Some people form a joint account for one year with the condi [#permalink]

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14 Apr 2015, 05:37
Qoofi wrote:
Let x be the number of people in the beginning.
Amount deposited for 6 months = 6*x*x = 6x^2

Number of people for the next 3 months = x -x/4 = 3x/4

Amount deposited for the next 3 months = 3 * (3x/4)^2

Number of people for the last 3 months = 3x/4 - (1/3 * 3x/4) = x/2

Amount deposited for the last 3 months = 3* (x/2)^2

Total amount = 6x^2 + 27x^2/16 + 3x^2/4 = 4860

x= 24 (option B)

Why are we multiplying things two times(^2)?
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Re: Some people form a joint account for one year with the condi [#permalink]

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15 Apr 2015, 01:00
ssriva2 wrote:
Why are we multiplying things two times(^2)?

Because at the end of each month, each person deposits an amount equal to the number of people. For example, if there are 24 people, each would deposit $24 (or whatever currency the question is asking), meaning a total of 24*24 dollars is deposited. So if you start out with $$n$$ people: > After month 1: ($$n$$ x $$n$$) deposited = $$n^2$$ > After month 2: +$$n^2$$ deposited > After month 3: +$$n^2$$ deposited > After month 4: +$$n^2$$ deposited > After month 5: +$$n^2$$ deposited > After month 6: +$$n^2$$ deposited ** Here, 1/4 of people leave and you are left with $$\frac{3}{4}n$$ people, each depositing $$\frac{3}{4}n$$ dollars ** > After month 7: +($$\frac{3}{4}n$$) x ($$\frac{3}{4}n$$) deposited = $$\frac{9}{16}n^2$$ > After month 8: +$$\frac{9}{16}n^2$$ deposited > After month 9: +$$\frac{9}{16}n^2$$ deposited ** Here, 1/3 of the remaining people leave and you are left with $$(\frac{2}{3})(\frac{3}{4}n)$$ people, each depositing $$(\frac{2}{3})(\frac{3}{4}n)$$ dollars ** > After month 10: + ($$\frac{2}{3})(\frac{3}{4}n$$) x $$(\frac{2}{3})(\frac{3}{4}n)$$ = $$\frac{1}{4}n^2$$ > After month 11: +$$\frac{1}{4}n^2$$ > After month 12: +$$\frac{1}{4}n^2$$ Put it all together and for the 12-month period, you get: $$6(n^2) + 3(\frac{9}{16}n^2) + 3 (\frac{1}{4})n^2 = 4860$$ Finally, solve to get: $$n=24$$ Answer: B Re: Some people form a joint account for one year with the condi [#permalink] 15 Apr 2015, 01:00 Similar topics Replies Last post Similar Topics: 9 This year, x people won an Olympic medal for water competitions. One 7 16 Sep 2015, 07:16 4 In a joint family there are 3 couples. The ages (all in years) of the 2 06 Nov 2014, 20:56 11 At the end of year X, automobile installment credit accounte 5 11 Nov 2013, 13:31 John deposits$5000 in a savings account for 5 years 1 12 Jan 2012, 18:41
5 A committee of 3 has to be formed randomly from a group of 6 people. 6 29 Dec 2009, 20:56
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text:
# Elliptic Torque

The above image shows a dam and an outlet pipe. All dimensions are given in figure itself. The bold segment indicates an elliptical plate acting as valve. It is inclined with the horizontal plane at an angle $$\alpha$$.

The torque necessary to hold the valve in place can be expressed as $$\tau =\kappa \rho g{ h }^{ \eta }\cfrac { { R }^{ \mu } }{ \sin ^{ \lambda }{ \alpha } }$$ where $$\kappa , \lambda , \mu , \eta$$ are non negative numerical constants.

Find value of $$\kappa + \lambda + \mu + \eta$$.

×


text:
$$\require{cancel}$$

# 5.27 Effects of Tidal Forces

The tidal force is a universal consequence of Newton's law of gravity, and we can see its effects throughout the universe. The force that causes the Earth's oceans to rise and fall also operates elsewhere in the Solar System and beyond. Large objects in close proximity exert the strongest tidal forces, but any time two large objects orbit each other closely, tides are important. This is true of planets, stars, and even entire galaxies!

Tidal locking results in the Moon rotating about its axis in about the same time it takes to orbit the Earth. Except for libration effects, this results in it keeping the same face turned towards the Earth, as seen in the figure on the left. (The Moon is shown in polar view, and is not drawn to scale.) If the Moon didn't spin at all, then it would alternately show its near and far sides to the Earth while moving around our planet in orbit, as shown in the figure on the right. Click here for original source URL.

The same tidal force that stretches a satellite also tends to slow its rotation until the longest axis of the satellite lines up with the planet. Just as the Earth's rotation is slowing due to the Moon's tidal force on it, the Moon's rotation has slowed until it is locked into this position. This is why most satellites, like the Moon, face toward their planet - they are tidally locked"" in that orientation. We always see the same face of the Moon from the Earth because the Moon's rotation period is the same as the time it takes to complete one orbit around the Earth. Another way of saying this is that the Moon is in a 1:1 spin orbit resonance — the ratio of its rotational (spin) period to its orbital period is 1 to 1. Examples of this are common in our Solar System. Pluto and Charon are tidally locked to each other. Mercury's eccentric orbit prevents it from being in a 1:1 spin orbit resonance. Instead, it's in a 3:2 resonance — in other words, Mercury's day is two-thirds as long as its year.

Closer to the Roche limit the body (an exoplanet) is deformed by tidal forces. Click here for original source URL.

If a satellite (or a passing body) comes very close to a planet, the tidal forces can be destructive. The closer two objects are in space, the stronger the gravity between them, and the stronger the tidal force. So the closer an object comes to a planet, the more it's stretched. Within a certain distance called the Roche limit, stretching forces can break it apart. That's why we don't see satellites orbiting too close to planets — instead, we see ring systems within a certain distance of the planet. These rings are the remnants of bodies that were broken up by tidal forces.

Even when there's no water to respond to tidal forces, the solid mass of a planet feels the stress caused by these forces. In fact, tidal forces can heat the interior of a satellite in an elliptical orbit. In that case, the satellite comes closer to the planet during one part of its orbit and there it's subjected to strong stretching forces. As it moves away from the planet the stress is partly released and the body relaxes back toward a more spherical shape. This continual flexing of the satellite creates heat through internal friction, in the same way that if you flex a tennis ball enough times it becomes warm. This effect is called tidal heating. The more elliptical the orbit, the stronger the tidal heating. Jupiter's large Galilean satellites experience this kind of heating — enough to produce extensive volcanism on Io, and possibly create liquid-water oceans beneath the surface of Europa.


text:
# Regular Expression to Context-Free Grammar

Anyone knows if there is an algorithm for directly write the context-free grammar that generates a given regular expression?

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## migrated from cstheory.stackexchange.comJan 20 '13 at 9:01

This question came from our site for theoretical computer scientists and researchers in related fields.

Do you want the CFG that generates the valid regular expression strings like "a*(a|b)", or do you want the CFG that generates strings like "aaaaab" given the regular expression "a*(a|b)"? –  Alex ten Brink Jan 20 '13 at 10:14

I assume you want to get a grammar that generates the same language as the given regular expression.

You can achieve that by the following steps:

1. Translate the regular expression into an NFA.
2. Translate the NFA into a (right-)regular grammar.

Both translations are standard and covered in basic textbooks on formal languages and automata. Note that any regular grammar is also context-free.

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Yes, better than my answer. –  Hendrik Jan Jan 21 '13 at 0:39

Yes. I give the high-level answer, without many details.

First you have to parse the expressions. That can be done using a simple recursive decent parser. Several examples on the web.

Then you should add "semantic" rules to the parser, when returning from the recursion. Those are standard in any formal language theory course. If $S_1$ and $S_2$ are non-terminals that generate expressions $E_1$ and $E_2$ then we can generate $E_1+E_2$ by $S$ and the rules $S\to S_1$, $S\to S_2$. We can generate concatenation $E_1 E_2$ by $S$ and the rule $S\to S_1 S_2$. $E_1*$ by $S$ and the rules $S\to S_1 S$, $S\to \lambda$. Assuming we choose fresh nonterminals each time.

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