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# what is the radius of convergence of that power series
• Nov 23rd 2011, 10:37 PM
sorv1986
what is the radius of convergence of that power series
Consider the function f(z) = 1 / (1+z²) where z ∈ C and let f(z) = Σ an (z-a)ⁿ be the
Taylor series expansion of f(z) about the point a ∈ R.
then what is the radius of convergence of the power series?
• Nov 23rd 2011, 11:15 PM
FernandoRevilla
Re: what is the radius of convergence of that power series
• Nov 23rd 2011, 11:37 PM
sorv1986
Re: what is the radius of convergence of that power series
Quote:
Originally Posted by FernandoRevilla
Hint Use the substitution $z=w-a$ and expand $f(z)$ by means of a geometric series .
But how to determine the radius of convergence?
as you said,
f(w-a)=1/(1+(w-a)²)=(1+C)⁻¹=1-(w-a)²+(w-a)⁴-...........=Σ(-1)ⁿ (w-a)²ⁿ [the expansion is valid if (w-a)²<1]
Thean what should i do? confused a bit.
thanxx
• Nov 24th 2011, 12:35 AM
FernandoRevilla
Re: what is the radius of convergence of that power series
Quote:
Originally Posted by sorv1986
Thean what should i do? confused a bit.
Sorry, I misread the question, $f$ is analytic in $\mathbb{C}-\{i,-i\}$ . By a well known theorem the radius of convergence is the distance of $a$ to the nearest singularity, that is $R=\sqrt{a^2+1}$ .
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# Homework Help: Even parity => symmetric space wave function?
1. Mar 5, 2007
### mrandersdk
If I have af wavefunction that is a product of many particle wavefunctions
$$\Psi = \psi_1(r_1)\psi_2(r_2) ... \psi_n(r_n)$$
If I then know that the parity of $$\Psi$$ is even. Can I then show that the wavefunction i symmetric under switching any two particles with each other. That is
$$\psi_1(r_1)\psi_2(r_2) ...\psi_i(r_i) ... \psi_j(r_j) ... \psi_n(r_n) = =\psi_1(r_1)\psi_2(r_2) ...\psi_j(r_j) ... \psi_i(r_i) ... \psi_n(r_n)$$
for any i and j between 1 and n?
It may be used that the parity operator commutes with the hamilton of the system if nessesary, and that the interaction between the particles only depends on the distance between any two particles.
It is clear that if the system only consist of two particles, and we use one of the particles as the 0-point of our coordinatesystem, the parity operator does the same as changing the particles, and then even parity means even space function, but when n is greater than 2, I can't see it.
Hope someone understand what i'm asking, because the result is used frequently in my course of particle physics.
2. Mar 5, 2007
### Meir Achuz
For two particles, parity changes r_12 into -r_12, which has the same effect as interchanging r_1 and r_2. For your many body wave function this isn't true, so parity is unrelated to particle interchange. Parity changes all r_i to -r_i. You have to look at the detailed wave function to see what this does. It depends on how your WF relates all the coordinates.
3. Mar 5, 2007
### mrandersdk
ok that was my thought too. Let me give an example how it is used, because I would like to know how th author solves the excercise. It's from nuclear and particle physics by B.R. Martin.
Consider a scenario where overall hadronic wavefunction $$\Psi$$ consist of just spin and space part, i.e. $$\Psi = \psi_{space} \psi_{spin}$$. What would be the resulting multplet structure of the lowest-lying baryon states composed of u,d and s quarks?
The autors own solution:
'Low lying' implies that the internal orbital momentum between the quarks is zero. Hence the parity is P = +1 and space is symmetric. Since the Pauli principle requires the overall wavefunction to be antisymmetric under the interchange of any pair of like quarks, it follows that $$\psi_{spin}$$ is antisymmetric...
The rest of the excercise makes sense, but how does he conclude that the space part i symmetric, to me it seems like he use parity symmetric and symmetric under switching of particles.
4. Mar 5, 2007
### Meir Achuz
That is a three body case. If the three body WF is written using Jacobi coordinates, the state with Daliz angular momenta L and l both equal zero is spatially symmetric. This is most easily seen for a harmonic oscillator wave function. Martin has made a common (historic) mistake of assuming more about parity than is needed. There are positive parity three body states that are not spatially symmetric.
5. Mar 5, 2007
### mrandersdk
ok i never heard of Jacobi coordinates or Daliz angular momenta. I'm not sure what you are saying, is it correct of him to say that positive parity gives a symmetric WF under switching any pair of quarks ?
6. Mar 5, 2007
### Meir Achuz
No. There are many positive parity quark wave functions that are not spatially symmetric. The Jacobi coordinates are r=r1-r2, and rho=r3-(r1-r2)/2, but normalized with some square roots to make them more symmetric.
For three bodies there are only two independent angular momenta in the cm system. Dalitz introduced them to analyze three body systems. One is related to r1-r2, and the other to the rho coordinate.
7. Mar 5, 2007
### mrandersdk
thanks for your help. There are many faults in this book so sometimes it's hard to figure out what to believe, so you got to be critical.
I was wandering if the parity of a system is independent of what coordinatsystem you use? Or do you got to have a coordinatsystem before you find the parity?
8. Mar 5, 2007
### Meir Achuz
The parity is independent of the coord system, but it is easier to see in some systems. It is easiest if all independent angular momenta are known.
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## Yet Another Contest 3 P4 - Rumour
View as PDF
Points: 15 (partial)
Time limit: 3.0s
Memory limit: 512M
Author:
Problem type
After hearing some nasty rumours about him, Mike is determined to find the source of the rumour! There are suspects labelled from to who have been involved in gossiping, with suspect being the creator of the rumour. Between these suspects, there are friendships. The -th friendship indicates that suspects and are friends with each other. Although Mike does not know the value of , he is aware of all friendships.
The rumour spreads according to the following procedure:
• On day , suspect will create the rumour.
• On day , suspect will tell the rumour to all of their friends.
• Then, on day , the suspects who heard the rumour for the first time on day will tell the rumour to all of their friends who have not yet heard the rumour.
Josh knows who started the rumour, but he is too scared to directly tell Mike, lest he feels the wrath of suspect . However, he will be willing to answer up to queries. In each query, Mike chooses an integer and accuses suspect of starting the rumour. If , then Josh will tell Mike that he has found the source of the gossip. Otherwise, Josh will tell Mike an integer such that suspect was first aware of the rumour on a strictly earlier day than when suspect was first aware of the rumour. If there are multiple possible values of , then Josh can tell Mike any of them. Recall that suspect was first aware of the rumour on day .
Can you help Mike determine who started the rumour?
#### Constraints
It is guaranteed that all suspects will eventually be aware of the rumour.
For , and .
For , and .
##### Subtask 3 [80%]
Note that all previous subtasks must be passed for this subtask to be evaluated.
#### Interaction
This is an interactive problem, where you will play the role of Mike and the interactor will play the role of Josh. Note that the interactor is adaptive.
First, you should read in the value of on a single line, representing the number of suspects.
Then, on the -th of the following lines, you should read in two space-separated integers and , representing the -th friendship.
Then, you should begin making queries. To make a query, output an integer on a single line, indicating that you will accuse suspect of starting the rumour. You should then read in an integer on the following line. If , then will be , and you should terminate your program to receive an Accepted verdict. Otherwise, indicates that suspect was first aware of the rumour on a strictly earlier day than when suspect was first aware of the rumour.
If at any point you make an invalid query, or make more than queries, the interactor will respond with -1 on the following line. Upon reading this value, you should terminate your program to receive a Wrong Answer verdict. Otherwise, you may receive an arbitrary verdict.
Please note that you may need to flush stdout after each operation, or interaction may halt. In C++, this can be done with fflush(stdout) or cout << flush (depending on whether you use printf or cout). In Java, this can be done with System.out.flush(). In Python, you can use sys.stdout.flush().
#### Sample Interaction
>>> denotes your output. Do not print this out.
5
1 2
2 3
2 4
1 5
>>> 5
3
>>> 2
4
>>> 4
0
#### Explanation
In the image above, a line drawn between two suspects indicates a friendship.
In this example, . Note that:
• Suspect was first aware of the rumour on day .
• Suspect was first aware of the rumour on day .
• Suspects and were first aware of the rumour on day .
• Suspect was first aware of the rumour on day .
First, Mike guesses that suspect started the rumour. Josh responds by telling Mike that suspect was aware of the rumour before suspect .
Then, Mike guesses that suspect started the rumour. Josh responds by telling Mike that suspect was aware of the rumour before suspect .
Finally, Mike guesses that suspect started the rumour. Since Mike guessed correctly, Josh responds with .
As Mike has successfully guessed the suspect in no more than queries, you would receive an Accepted verdict for this test case.
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# Word Problems
1. Mar 4, 2006
### Aoiro
two numbers who's differences are 7, are squared then, multiplyed together. What are the two numbers?
I dont know how to do this question, this is what i got
Let x rep the first #
Let x-17 rep the second #
(x)2(x-7)2
(x2)(x2+49)
...now what?
2. Mar 4, 2006
### Tide
I think you left something out - like the value of product of the squares.
3. Mar 4, 2006
### arildno
Please give the EXACT wording of the problem.
It is (virtually) impossible that you have given all relevant info here.
4. Mar 4, 2006
### mattmns
Also, $(x-7)^2$ is not the same as $x^2 + 49$
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# Numbers after paragraph breaks turn into 1's
Numbers after paragraph breaks all turn to 1.
1. Number 1 in the list.
paragraph break
1. 2 in list.. or is it?
paragraph break
1. <- I swear I labelled this 3..
paragraph break
some characters 4. A workaround is to place some character before the number
It's not LaTeX; because you are starting a new paragraph with no left indentation, this resets the counter. It would happen if you have text instead of LaTeX.
If instead of a new paragraph with in-line equations you do a display equation, it works fine:
1. Numbered one, followed by $$equation$$ $$\LaTeX$$
2. Number two. $$\mbox{More }\LaTeX$$
3. Number Three. $$\mbox{Even more \LaTeX.} 1\neq 3.$$
This was achieved with:
1. Numbered one, followed by $$equation$$
$$\LaTeX$$
2. Number two.
$$\mbox{More }\LaTeX$$
3. Number Three.
$$\mbox{Even more \LaTeX.} 1\neq 3.$$
But if I place a new paragraph and start over, what happens?
1. Even though I numbered it 4, it still shows up as 1.
If you start a new paragraph without any justification on the left, still the same problem.
1. I numbered it five.
The source that produced this was:
But if I place a new paragraph and start over, what happens?
4. Even though I numbered it 4, it still shows up as 1.
If you start a new paragraph without any justification on the left,
still the same problem.
5. I numbered it five.
However, even with inline equations, if you place the appropriate justification, it works out fine.
1. Number one, followed by an equation
$\LaTeX$
2. Number two. Followed by another equation.
$1\neq 2$.
3. And another, giving me three
$1\neq 3$.
But woe if I start left-justified: then
1. I get a one again.
Obtained by typing:
However, even with inline equations, if you place
the appropriate justification, it works out fine.
1. Number one, followed by an equation
$\LaTeX$
2. Number two. Followed by another equation.
$1\neq 2$.
3. And another, giving me three
$1\neq 3$.
But woe if I start left-justified: then
4. I get a one again.
Notice the justification.
It's not a matter of TeX at all. The formatting engine, Markdown, thinks you've ended your list when you start a new paragraph; the next time it encounters a numbered item, it starts a new list beginning with "1." again.
Here are several workarounds.
1. Indent your new paragraph by a number of spaces.
This will be treated as a new paragraph within the same list item.
2. If you just want $$\LaTeX$$ on a new line, you can do a display equation like $$\LaTeX$$.
Sometimes, you really do want to specify your own numbering. For that, you have to escape the period following the number. Then Markdown will treat it as a regular paragraph, which is less than optimal, but at least it will leave your number alone. So 42\. becomes
42. The answer to life, the universe, and everything.
and not
1. The answer to life, the universe, and everything.
## Not anymore!
Numbered lists can now start with any positive integer. (Zero, negative numbers, and infinite ordinals are still not allowed.) The existing posts still render as they were, but will re-render according to the new Markdown version if they are edited.
Copy-pasted from the question:
Numbers after paragraph breaks all turn to 1.
1. Number 1 in the list.
paragraph break
1. 2 in list.. or is it?
paragraph break
1. <- I swear I labelled this 3..
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# Financial Accounting, v. 2.0
by Joe Hoyle and C.J. Skender
## 11.2 Balance Sheet Reporting of Intangible Assets
### Learning Objectives
At the end of this section, students should be able to meet the following objectives:
1. Explain the theoretical rationale for using historical cost as the basis for reporting intangible assets as well as long-lived assets such as equipment.
2. Understand that using historical cost as the accounting basis means that a company’s intangible assets can actually be worth much more than is shown on the balance sheet.
3. Recognize that intangible assets with large monetary balances can result from acquisition either individually or through the purchase of an entire company that holds valuable intangible assets.
4. Describe the method of recording intangible assets when the company that owns them is acquired by another company.
## Reporting Historical Cost for Intangible Assets Rather Than Fair Value
Question: Much was made in earlier chapters about the importance of painting a portrait that fairly presents the financial health and future prospects of an organization. Many companies develop copyrights and other intangible assets that have incredible value but little or no actual cost.
Trademarks provide an excellent example. The golden arches that represent McDonald’s must be worth many millions, but the original design cost was probably not significant and has likely been amortized entirely to expense by now. Could the balance sheet of McDonald’s possibly be considered fairly presented if the value of its primary trademark is omitted?
Many other companies, such as Walt Disney, UPS, Google, Apple, Coca-Cola, and Nike, rely on trademarks to help create awareness and brand loyalty around the world. Are a company’s reported assets not understated if the value of a trademark is ignored despite serving as a recognizable symbol to millions of potential customers? With property and equipment, this concern is not as pronounced because those assets tend to have significant costs whether bought or constructed. Internally developed trademarks and other intangibles often have little actual cost despite the possibility of gaining immense value.
Answer: Figures reported for intangible assets such as trademarks may indeed be vastly understated on a company’s balance sheet when compared to fair values. Decision makers who rely on financial statements need to understand what they are seeing. U.S. GAAP requires that companies follow the historical cost principle in reporting many assets. A few exceptions do exist, and several are examined at various points in this textbook. For example, historical cost may have to be abandoned when applying the lower-of-cost-or-market rule to inventory. The same is true when testing property and equipment for possible impairment losses. Those departures from historical cost were justified because the asset had lost value and financial accounting tends to be conservative. Reporting an asset at a balance in excess of its historical cost basis is much less common.
In financial accounting, what is the rationale for the prevalence of historical cost, which some might say was an obsession? As discussed in earlier chapters, cost can be reliably and objectively determined. It does not fluctuate from day to day throughout the year. It is based on an agreed-upon exchange price and reflects a resource allocation judgment made by management. Cost is not a guess, so it is less open to manipulation. Although fair value may appear to be more relevant, various parties might arrive at significantly different estimates of worth. What is the true value of the golden arches to McDonald’s as a trademark? Is it $100 million or$10 billion? Six appraisals from six experts could suggest six largely different amounts.
Plus, if the asset is not going to be sold, is the fair value of much relevance at the current time?
Cost remains the basis for reporting many assets in financial accounting, though the use of fair value has gained considerable momentum. It is not that one way is right and one way is wrong. Instead, decision makers need to understand that historical cost is the generally accepted accounting principle normally used to report long-lived assets such as intangibles. The use of historical cost does have obvious flaws, primarily that it fails to report any appreciation in value no matter how significant. Unfortunately, any alternative number that can be put forth as a replacement also has its own set of problems. At the present time, authoritative accounting literature holds that historical cost is the appropriate basis for reporting intangibles.
Even though fair value accounting seems quite appealing to many decision makers, accountants have proceeded slowly because of potential concerns. For example, the 2001 collapse of Enron Corporation was the most widely discussed accounting scandal to occur in recent decades. Many of Enron’s reporting problems began when the company got special permission (due to the unusual nature of its business) to report a number of assets at fair value (a process referred to as “mark to market”).Unique accounting rules have long existed in certain industries to address unusual circumstances. College accounting textbooks such as this one tend to focus on general rules rather than delve into the specifics of accounting as it applies to a particular industry. Because fair value was not easy to determine for many of those assets, Enron officials were able to manipulate reported figures to make the company appear especially strong and profitable.For a complete coverage of the history and ramifications of the Enron scandal, both the movie and the book The Smartest Guys in the Room are quite informative and fascinating. Investors then flocked to the company only to lose billions when Enron eventually filed for bankruptcy. A troubling incident of this magnitude makes accountants less eager to embrace the reporting of fair value except in circumstances where very legitimate amounts can be determined. For intangible assets as well as property and equipment, fair value is rarely so objective that the possibility of manipulation can be eliminated.
### Test Yourself
Question:
The Consetti Company acquires a patent for $932,000 to be used in its daily operations. However, the value of this patent rises dramatically so that three years later, it is worth$3.2 million. Which of the following is not a reason that this fair value is ignored when the asset is reported on the Consetti’s balance sheet?
1. Investors are not interested in the fair value of the patent or other intangible assets.
2. Fair value can change often so that any one figure is not necessarily relevant for a long period of time.
3. Different fair values might be estimated by different people.
4. Noncurrent assets are acquired to help generate revenues and not for resale purposes.
The correct answer is choice a: Investors are not interested in the fair value of the patent or other intangible assets.
Explanation:
Investors are likely to be interested in the fair value of all company assets because that information helps to assess the worth of the company and, hence, the possible sales price of its stock. However, accounting rules shy away from use of fair value for property and equipment as well as intangible assets. That value is no more than a guess and it can swing radically over time. Plus, if the asset is not for sale, fair value is not particularly relevant to the operations of the company.
## Acquiring a Company to Gain Control of Its Intangibles
Question: Although a historical cost basis is used for intangible assets rather than fair value, Microsoft Corporation still reports $13.3 billion as “goodwill and intangible assets, net” while Yahoo! indicates similar balance sheet accounts totaling$3.9 billion. Even the size of these numbers is not particularly unusual for intangible assets in today’s economic environment. As of June 30, 2011, for example, the balance sheet for Procter & Gamble listed goodwill of $57.6 billion and trademarks and other intangible assets, net of$32.6 billion. If historical cost is often insignificant, how do companies manage to report such immense monetary amounts for their intangible assets?
Answer: Two possible reasons exist for a company’s intangible asset figures to grow to incredible size. First, instead of being internally developed, assets such as copyrights and patents are often acquired from outside owners. Reported asset balances then represent the historical costs of these purchases which were based on fair value at the time of the transaction. Large payments may be necessary to acquire such rights if their value has already been firmly established.
Second, Microsoft, Yahoo!, and Procter & Gamble could have bought one or more entire companies so that title to a multitude of assets (including a possible plethora of intangibles) was obtained in a single transaction. In fact, such acquisitions often occur specifically because one company wants to gain valuable intangibles owned by another. In February 2008, Microsoft offered over $44 billion in hopes of purchasing Yahoo! for exactly that reason. Yahoo! certainly did not hold property and equipment worth$44 billion. Microsoft was primarily interested in acquiring a wide variety of intangibles owned by Yahoo! Although this proposed takeover was never completed, the sheer size of the bid demonstrates the staggering value of the intangible assets that companies often possess today.
If a company buys a single intangible asset directly from its owner, the financial reporting follows the pattern previously described. Whether the asset is a trademark, franchise, copyright, patent, or the like, it is reported at the amount paid. That cost is then amortized over the shorter of its estimated useful life or legal life. Intangible assets that do not have finite lives are not amortized and will be discussed later in this chapter.
Reporting the assigned cost of intangible assets acquired when one company (often referred to as “the parent”) buys another company (“the subsidiary”) is a complex issue discussed in more detail in Chapter 12 "In a Set of Financial Statements, What Information Is Conveyed about Equity Investments?". In simple terms, the subsidiary’s assets (inventory, land, buildings, equipment and the like) are valued and recorded at that amount by the parent as the new owner. The subsidiary’s assets and liabilities are consolidated with those of the parent. In this process, each intangible asset held by the subsidiary that meets certain requirements is identified and recorded by the parent at its fair value. The assumption is that a portion of the price conveyed to purchase the subsidiary is being paid to obtain these intangible assets.
To illustrate, assume Big Company pays $10 million in cash to buy all the capital stock of Little Company. Consolidated financial statements will now be necessary. Little owns three intangibles (perhaps a copyright, patent, and trademark) that are each worth$1 million. Little also holds land worth $7 million but has no liabilities. Little’s previous net book value for these assets is not relevant to Big, the new owner. Following the takeover of Little, Big reports each of the intangibles on its balance sheet at its cost of$1 million (and the land at $7 million). The acquisition price is assumed to be the historical cost paid by Big to obtain these assets. A parent that buys many subsidiaries will frequently report large intangible asset balances as a result. When Big purchases Little Company, it is really gaining control of all these assets and records the transaction as shown in Figure 11.3 "Big Company Buys Little Company, Which Holds Assets with These Values". Figure 11.3 Big Company Buys Little Company, Which Holds Assets with These Values ### Test Yourself Question: The Tiny Company creates a logo for a product line and gets a copyright on it. The entire cost is$40,000, and the logo is expected to have a useful life of ten years. One year later, Gigantic Company buys all the ownership stock of Tiny Company. At that point in time, the logo has gained national prominence and is thought to be worth $400,000. If Gigantic prepares a consolidated balance sheet immediately after acquiring Tiny, what is reported for this logo? 1.$36,000
2. $400,000 3.$436,000
4. $440,000 Answer: The correct answer is choice b:$400,000.
Explanation:
Because Gigantic bought Tiny, the assumption is made that a portion of the price that was paid for Tiny was made to acquire the logo at its fair value. Thus, to Gigantic, the historic cost of this asset is $400,000. The cost to Tiny is no longer relevant. The$400,000 will then be amortized to expense over the remaining life of the intangible.
### Key Takeaway
Many intangible assets (such as trademarks and copyrights) are shown on the balance sheet of their creator at a value significantly below actual worth. They are reported at historical cost less all amortization since the date of acquisition. Development cost can be relatively low in comparison to the eventual worth of the right. However, because of conservatism, the amount reported for these assets is not raised to fair value. Such numbers are subjective and open to sudden fluctuations. Furthermore, if an intangible asset is not held for sale, fair value is of questionable relevance to current operations. Companies, though, often pay large amounts either to buy intangibles or entire companies that hold valuable intangibles. In accounting for a parent’s acquisition of a subsidiary, the amount paid is assigned to the identifiable assets of the subsidiary (both tangible and intangible) based on fair value at that date.
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1. ## Unknown Coordinates
Well, I have a problem, We're given 3 points(sorry if i've defined/called it wrong) which is A B and C, which also has each a coordinates, but the problem is, only A and B has, C doesn't have any, and we're tasked to find it, but I don't know the formula in finding it, so I can't proceed in measuring the distance between AC and BC, only AB.
Here:
A(-1, 2)
B(4,2)
C(?,?)
Could someone please tell me how to find the C coordinates?
NOTE: My professor said it is an equilateral triangle when we're going to measure it.
2. Originally Posted by dissidia
Well, I have a problem, We're given 3 points(sorry if i've defined/called it wrong) which is A B and C, which also has each a coordinates, but the problem is, only A and B has, C doesn't have any, and we're tasked to find it, but I don't know the formula in finding it, so I can't proceed in measuring the distance between AC and BC, only AB.
Here:
A(-1, 2)
B(4,2)
C(?,?)
Could someone please tell me how to find the C coordinates?
NOTE: My professor said it is an equilateral triangle when we're going to measure it.
The information that the triangle is equilateral is very important here.
We know that the distance from point A to point B (which is a side of a triangle here) can be calculated using the following formula:
AB = sqrt( (x2-x1)² + (y2-y1)² )
By applying the formula we can calculate the distance between points A(-1,2) and B(4,2)
AB = sqrt ( (-1-4))² + (2-2)² ) = sqrt(25+0) = 5
Since the triangle ABC is equilateral, we know that AB = BC = AC = 5
If we mark the unknown point as C(a,b), we'll get two equations that we can use to determine the coordinates a and b.
Here are a few tips to get you started. I'll be posting the full solution a bit later.
3. Isn't a,b somehow related to points AB? couldn't really get the equations.
4. Okay, to continue from where I left:
The points were A(-1,2), B(4,2) and C(a,b)
We'll get two equations
The length of AC
5 = sqrt( (-1 -a)² + (2-b)² )
The length of BC
5= sqrt( (4-a)² + (2-b)² )
From this, we'll get the equation AC = BC
sqrt( (-1 -a)² + (2-b)² ) = sqrt( (4-a)² + (2-b)² )
(-1 -a)² + (2-b)² = (4-a)² + (2-b)²
We're a bit lucky here that the y-coordinate for both A and B is the same,
so we'll notice that the (2-b)² on both sides will cancel each other out, and
we'll be left with a simple equation to calculate a
(-1 -a)² = (4-a)² <=> 1+2a+a² = 16 -8a + a² <=> 10a = 15 <=> a= 3/2
After we've solved a, you can use either one of the equations AC = 5 or BC = 5 to solve b by replacing a with 3/2.
Let's use BC for example:
5 = sqrt( (4-3/2)² + (2-b)² ) <=> 25 = (5/2)² + (2-b)² <=> 25 = 25/4 + 4 -4b + b² <=> b² - 4b - 59/4 = 0
Solve the equation and you'll get b= 1/2 * (4 ± 5 sqrt(3))
This means that there are actually two possible y-coordinates for point C, which makes sense if you think about it geometrically.
I hope this helped!
5. Originally Posted by dissidia
Isn't a,b somehow related to points AB? couldn't really get the equations.
You could really give the x and y coordinates in C(a,b) any names. It doesn't have to be C(a,b), it could be C(x0,y0), C(u,v) or whatever.
The equation to calculate the distance of two points is really just the Pythagoran theorem c² = a² + b².
6. A very brilliant mind of yours, though it took me some minutes to understand what you've wrote. Thanks a lot, this really helps.
7. Just a question bro, how did you manage to get off that b² here(b² - 4b - 59/4 = 0 ) so that you can get the value of b?
8. Originally Posted by dissidia
Just a question bro, how did you manage to get off that b² here(b² - 4b - 59/4 = 0 ) so that you can get the value of b?
Since we have a quadratic equation, the simplest way to solve it is to use the quadratic formula.
The solutions the polynom ax² + bx + c = 0 are
x= ( -b±sqrt(b² - 4ac) ) / 2a
With the equation b² - 4b - 59/4 = 0
a=1 , b=4 and c = 59/4
9. You're cooler than I thought. Thanks again bro.
10. ## unknown coordinates
posted by dissidia
Given two coordinates find the third assuming you meant to say AB is one side of an equilateral triangle.The answers thus far only define the lenght of a side of the triangle .The solution to either question can be derived more simply.If the midpoint of an equalateral is 2.5 the side is 5 and the altitude is 2.5 times radial 3 so the latter dimension is used to get the two new values of y and x is 2.5 in both cases
bjh
11. ## unknown coordinates
Hi isosky
I do not understand your message. I have supplied a method to find the two unknown coordinates Tell me how i can help you to understand this.Remember that the question is to supply the missing coordinates.
bjh
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# Dynamics on the Riemann Sphere: A Bodil Branner Festschrift
The papers collected in this volume were written to celebrate Bodil Branner’s 60th birthday. Most of them were presented at the 'Bodil Fest', a symposium on holomorphic dynamics held in June 2003 in Holbäk, Denmark. The main research theme of Bodil Branner is the iteration of cubic polynomials. Together with John H. Hubbard, she described the global topology of the parameter space C2 of polynomials of the form Pa,b(z) = z3 - a2z + b. Several decompositions of the parameter space have been considered. The first splitting is to separate the connectedness locus (where both critical points have bounded orbit and the Julia set is therefore connected) from the escape locus (where at least one critical point escapes to infinity). The second splitting is to foliate the escape locus into different hyper-surfaces, each one corresponding to a fixed maximal escape rate of the critical points. A particular way of constructing Teichmüller almost complex structures, which are invariant under Pa,b, was introduced as wringing and stretching of the complex structure; this technique is now referred to as Branner-Hubbard motion.
In the volume, Branner-Hubbard motion is described in the papers of C. L. Petersen, Tan Lei, and A. Douady. A survey paper of J. Milnor treats Lattès maps. A. Avila and M. Lyubich give examples of infinitely renormalizable quadratic maps whose Julia sets have Hausdorff dimension arbitrarily close to one. A. Chéritat studies the linearizability of the family Pθ(z) = e2πiθz + z2. Two papers (written by P. Blanchard et al. and P. Roesch) treat the parameter space of the family fλ(z) = z2 + λ/z2. T. Kawahira studies small perturbations of geometrically finite maps into other geometrically finite maps that are (semi)-conjugate on the Julia set to the original map. W. Jung constructs by quasi-conformal surgery a class of homeomorphisms of subsets of the Mandelbrot set. N. Fagella and Ch. Henriksen study Arnold tongues in the complexification of analytic diffeomorphisms fa,t(x) = x + t + (a/2π) sin(2πx) of the circle.
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Categorisation
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### Coin Tossing Games
You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?
### Scratch Cards
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### Chances Are
Which of these games would you play to give yourself the best possible chance of winning a prize?
# Factoring a Million
##### Stage: 4 Challenge Level:
No one completed the solution to this problem so we will put it into Tough Nuts for you to return to when you have time. There were several attempts and some of you were able to identify that there are 784 possible arrangements of three factors of a million.
The first step was to identify that 1000 000 = 10 6 = 2 6 x 5 6 and then to consider this as the product of three factors i.e.
10 6 = 2 6 x 5 6 = 2 a 5 p x 2 b 5 q x 2 c 5 r (where a+b+c = 6 and p+q+r = 6).
However there are repetitions here because 2 3 5 3 x 2 2 5 2 x 2 1 5 1 is the product of the same three factors as 2 2 5 2 x 2 3 5 3 x 2 1 5 1 .
So there is still some work to do! Good luck.
Giles Cooper & Mike Hood thinks there are 139 such factors, he has produced a list of factors for example
1 x 1 x 1000000
1 x 2 x 500000
1 x 4 x 250000
...
...
80 x 100 x 125
100 x 100 x 100
Editors note:
This is correct, however I also liked the attempt by Mike Hood, who tried to use the idea of combinations of factors. This idea seemed to be a less exhaustive'' approach and begins to give a better insight into what is going on. His solution uses the fact that $1,000,000 = 2^6 x 5^6$. I think this is worth pursuing. Think about the number of ways you can combine the six 2s and the six 5s in the three factors. You then only have to consider the number of unique combinations of each of the two sets of arrangements. For example - you could choose $2^6$, $2^0$, $2^0$ and $5^6$, $5^0$, $5^0$ and there are only two unique arrangements of these two sets of factors.
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# Conference demographics and footprint changed by virtual platforms
## Abstract
Conferences disseminate research, grow professional networks and train employees. Unfortunately, they also contribute to climate change and present barriers to achieving a socially sustainable work environment. Here, we analyse the recent impact of transforming in-person conferences into virtual conferences on improving diversity, equity and inclusion in science and engineering conferences. Factors including cost, gender, career stage and geographic location were evaluated. Virtual conferences demonstrated a clearly discernable and, in some cases, orders of magnitude improvement across nearly all metrics. On the basis of participant survey results, this improvement may be attributed to a combination of reduced financial and personal-life burdens. However, despite this clear impact, further development of virtual networking features and poster sessions is necessary to achieve widespread adoption and acceptance of this new format.
## Main
Conferences fulfil a range of needs by facilitating dissemination of ideas, initiating collaborative relationships and providing education, training and career opportunities. Traditional in-person conferences (IPCs) have filled this role for centuries1 and these events cut across all sectors: academia, industry and government. However, this format has been criticized as outdated and detrimental to the environment2,3,4. More recently, emerging evidence is also connecting this modality to social sustainability issues as well, notably poor retention of a diverse workforce. In this context, the two dominant contributors are the intrinsic power-imbalance in the workplace and an imbalance in home-life responsibilities5,6.
Over the past two decades, the creation and sustainment of a diverse, equitable and inclusive (DEI) work environment in the scientific and engineering community has not kept pace with many other fields. In part, this can be attributed to career expectations revolving around conference travel and participation. Participation in conferences can be cost prohibitive for many, as the cumulative expenses can be thousands of US dollars per person. International travel creates additional barriers7 which are exacerbated by the frequent changes in document requirements and lengthy delays in obtaining visas. These financial and documentation barriers can also dissuade scientists that have difficulty securing funding to cover conference costs such as students, postdoctoral researchers or scientists from historically under-represented institutions. These factors can also exclude participants from countries that do not have very high research activity, such as nations that are not in the top ten research countries as defined by the Nature Index (NI; ref. 8), NI > 10.
However, even for those researchers who are able to travel, the time away from home necessitated by work-related travel is intrinsically exclusionary to care-givers, who are primarily women3,7,9. Yet, given how important conference attendance is to career advancement, this community is frequently faced with the decision of choosing between work and family. Lastly, despite conference organizers’ attempts to solve accessibility concerns of the disabled community, many conferences still fall short of providing an equitable experience.
The recent surge in virtual events is forcing the scientific community to re-evaluate its long-held position against virtual conferences (VCs). The initial anecdotal evidence indicated that VCs enabled a more diverse population to participate. But a quantitative analysis of the impact on DEI challenges has yet to be performed. Such analysis is critical to make decisions regarding the format of future events, potentially resulting in a paradigm shift in the field. Here, we evaluate several metrics, including cost, carbon footprint, impact of conference format and attendee demographics. We collected historical data from three IPCs based in the United States, of varying sizes and disciplines within science, technology, engineering and mathematics (STEM). These results were compared to the same three conferences after they transitioned to a VC format in 2020. These scientific conferences were among the early conferences to transition online in response to the COVID-19 pandemic and were chosen to investigate the impact of an abrupt transition from historically IPCs to a new virtual format.
The historically IPCs-turned-VCs analysed here are the annual International Conferences on Learning Representations (ICLR), the American Astronomical Society (AAS) and the North American Membrane Society (NAMS) conferences. Also analysed here are several conference series that were originally designed for the VC environment, including the Photonics Online Meetups (POM 1, January 2020; POM 2, June 2020) and the International Water Association (IWA) Biofilms conference. These conferences span five fields of science and engineering and range from small- to large-scale events. All have international audiences.
We focused our analysis on the environmental, social and economic costs of VCs versus IPCs and accompanying demographic impacts (global participation), participation from women, early career researchers and scientists from under-represented institutions. We also assessed the challenges and benefits of the VC format.
## Results
### Demographic impact
The elimination of the travel and cost burdens realized with the VC format resulted in a large increase in attendance at all events (Fig. 1). The increase in attendance was particularly pronounced for international attendees. We propose that this trend may be related to the substantial decrease in costs as compared to IPCs as described below.
### Gender make up of STEM researchers from conference attendee’s countries
Country-specific percentage of women data are taken from ‘female researchers as a percentage of total researchers (full-time equivalents)—natural sciences and engineering (subtotal)’ published as ref. 26 with the exception of the United States which is not included in that dataset. US percentage of women is derived from women as a percentage of MSc and PhD graduates employed in science and engineering occupations27. Overall percentage of women in STEM for the countries represented in the conference delegations was calculated with percentage values from each country represented at the conference, weighted by the number of attendees from each country.
### Reporting Summary
Further information on research design is available in the Nature Research Reporting Summary linked to this article.
## Data availability
The data that support the plots within this paper and other findings of this study have been deposited on Github28 (https://doi.org/10.5281/zenodo.5567764). Source data are provided with this paper.
## Code availability
The custom code used to process and analyse the data for this study has been deposited on Github28 (https://doi.org/10.5281/zenodo.5567764).
## References
1. Reshef, O. et al. How to organize an online conference. Nat. Rev. Mater. 5, 253–256 (2020).
2. Yakar, D. & Kwee, T. C. Carbon footprint of the RSNA annual meeting. Eur. J. Radiol. 125, 5 (2020).
3. Parker, M. & Weik, E. Free spirits? The academic on the aeroplane. Manag. Learn. 45, 167–181 (2014).
4. Klöwer, M., Hopkins, D., Allen, M. & Higham, J. An analysis of ways to decarbonize conference travel after COVID-19. Nature 583, 356–359 (2020).
5. Hewlett, S. A. et al. The Athena factor: reversing the brain drain in science, engineering, and technology. Harv. Bus. Rev. Res. Rep. 10094, 1–100 (2008).
6. Simard, C., Henderson, A., Gilmartin, S., Schiebinger, L. & Whitney, T. Climbing the Technical Ladder: Obstacles and Solutions for Mid-Level Women in Technology (Michelle R. Clayman Institute for Gender Research, 2008).
7. Urry, J. Social networks, mobile lives and social inequalities. J. Transp. Geogr. 21, 24–30 (2012).
8. Nature Index 2020 Annual Tables (Springer Nature, 2020); https://www.natureindex.com/annual-tables/2020
9. Cohen, S. A. & Gossling, S. A darker side of hypermobility. Environ. Plan. A 47, 1661–1679 (2015).
10. Rogelj, J. et al. in IPCC Special Report on Global Warming of 1.5 °C (eds Masson-Delmotte, V. et al.) Ch. 2 (WMO, 2018).
11. Fisher, M. J. & Marshall, A. P. Understanding descriptive statistics. Aust. Crit. Care 22, 93–97 (2009).
12. Schaab, G., Adams, S. & Coetzee, S. Conveying map finesse: thematic map making essentials for today’s university students. J. Geogr. High. Educ. https://doi.org/10.1080/03098265.2020.1850656 (2020)
13. Gender API (Gender-API.com, accessed 30 October 2020); https://gender-api.com
14. Accredited Postsecondary Minority Institutions (US Department of Education, accessed 30 October 2020); https://www2.ed.gov/about/offices/list/ocr/edlite-minorityinst-list-tab.html
15. The Carnegie Classification of Institutions of Higher Education (Indiana Univ. School of Education, accessed 30 October 2020); https://carnegieclassifications.iu.edu/lookup/srp.php?clq=%7B%22basic2005_ids%22%3A%2216%22%7D&start_page=standard.php&backurl=standard.php&limit=0,50
16. Facilitating Research at Primarily Undergraduate Institutions: Research in Undergraduate Institutions (RUI) and Research Opportunity Awards (ROA) (United States National Science Foundation, accessed 30 October 2020); https://www.nsf.gov/pubs/2014/nsf14579/nsf14579.htm
17. Doctorates Awarded, by State or Location, Broad Field of Study, and Sex of Doctorate Recipients: 2017 (United States National Science Foundation, 2018).
18. Doctorates Awarded, by State or Location, Broad Field of Study, and Sex of Doctorate Recipients: 2018 (United States National Science Foundation, 2019).
20. The myclimate Flight Emission Calculator (Foundation myclimate, accessed 30 October 2020); https://www.myclimate.org/
21. Chong, H. G. & Ricaurte, E. E. Hotel Sustainability Benchmarking Tool 2015: Energy, Water, and Carbon (Cornell Hospitality Reports, Cornell Univ., 2015).
22. Dudas, G., Boros, L., Pal, V. & Pernyesz, P. Mapping cost distance using air traffic data. J. Maps 12, 695–700 (2016).
23. World Population Prospects 2019 (United Nations, 2019); https://population.un.org/wpp/Publications/
24. Greenhouse Gases Equivalencies Calculator - Calculations and References (United States Environmental Protection Agency, 2020); https://www.epa.gov/energy/greenhouse-gases-equivalencies-calculator-calculations-and-references
25. World Economic Outlook Database (International Monetary Fund, accessed 30 October 2020); https://www.imf.org/external/pubs/ft/weo/2017/02/weodata/index.aspx
26. Research and Development (UNESCO Institute for Statistics, 2020).
27. Employed Scientists and Engineers, by Occupation, Highest Degree Level, and Sex: 2017 (National Science Foundation, accessed 30 October 2020); https://ncses.nsf.gov/pubs/nsf19304/data
28. Skiles, M. Virtual-Conferences-Project (Zenodo, 2021); https://doi.org/10.5281/zenodo.5567764
## Acknowledgements
We acknowledge G. Ragusa’s contribution to designing the pre- and postsurveys used for the POM 1 and 2 conferences. The work in this paper was supported by the NSF (award no. CBET 2029219 and CBET 1946392). We gratefully acknowledge access to data and consultation support provided by NAMS, ICLR, POM and IWA meeting leadership. K. Marvel’s input on AAS data and trends and his role in providing access to AAS meeting data is also acknowledged. This material is based on work supported by the NSF Graduate Research Fellowship Program under grant no. DGE-1610403. Any opinions, findings and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the NSF.
## Author information
Authors
### Contributions
M.K., K.M.F. and M.S. conceived the idea. M.K., M.S., D.R.M. and D.C. collected data. M.S. and E.Y. analysed data. O.R., M.L.L, P.P.C, R.N., A.R. and A.A. provided access to data and provided insights on data. M.K., K.M.F., A.A. and M.S. wrote the manuscript.
### Corresponding authors
Correspondence to Andrea Armani, Kasey M. Faust or Manish Kumar.
## Ethics declarations
### Competing interests
M.K. and M.L.L. were organizers of NAMS. A.R. was an organizer of ICLR. P.P.C. and R.N. were organizers of IWA. A.A. and O.R. were organizers of POM. All other authors have no competing interests.
Peer review information Nature Sustainability thanks Meagan Mauter and the other, anonymous, reviewer(s) for their contribution to the peer review of this work.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Supplementary information
### Supplementary Information
Supplementary Discussion, Tables 1–21 and Figs. 1–15.
## Source data
### Source Data Fig. 1
Statistical source data.
### Source Data Fig. 2
Statistical source data.
### Source Data Fig. 3
Statistical source data.
## Rights and permissions
Reprints and Permissions
Skiles, M., Yang, E., Reshef, O. et al. Conference demographics and footprint changed by virtual platforms. Nat Sustain 5, 149–156 (2022). https://doi.org/10.1038/s41893-021-00823-2
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# ANNT : Convolutional neural networks
, 28 Oct 2018
The article demonstrates usage of ANNT library for creating convolutional ANNs and applying them to image classification tasks.
## Introduction
This article continues the topic of artificial neural networks and their implementation in the ANNT library. The first article started with basics and described feed forward fully connected neural networks and their training using Stochastic Gradient Descent and Error Back Propagation algorithms. It then demonstrated application of this artificial neural network's architecture in number of tasks. One of those was classification of handwritten characters from the MNIST database. Although being a simple example, it managed to achieve about 96.5% accuracy on a test dataset. In this article we'll have a look at a different architecture of artificial neural networks known as Convolutional Neural Networks (CNN). This type of networks is specifically designed for computer vision tasks and outperforms classical fully connected neural networks when it comes to tasks like image recognition. As another sample application will demonstrate, we'll get to about 99% accuracy on the handwritten characters classification.
Originally the convolutional neural network architecture was introduced by Yann LeCun when he published his work back in 1998. However, it was left largely unnoticed in those days. It took 14 years to get big attention to convolutional networks when the ImageNet competition was won by a team using this architecture. CNNs became very popular after that and were applied to many computer vision applications resulting in development of variety of neural networks based on this architecture. These days state-of-the-art convolutional neural networks achieve accuracies that outperform humans on many image recognition tasks.
## Theoretical background
As in the case with feed forward fully connected artificial neural networks, the idea of convolutional networks was inspired by studying nature - brain of mammals. Work by Hubel and Wiesel in the 1950s and 1960s showed that cats' and monkeys' visual cortexes contain neurons that individually respond to small regions of the visual field. Provided the eyes are not moving, the region of visual space within which visual stimuli affect the firing of a single neuron is known as its receptive field. Neighbouring cells have similar and overlapping receptive fields. Receptive field size and location varies systematically across the cortex to form a complete map of visual space.
In their paper, they described two basic types of visual neuron cells in the brain that each act in a different way: simple cells and complex cells. The simple cells activate, for example, when they identify basic shapes as lines in a fixed area and a specific angle. The complex cells have larger receptive fields and their output is not sensitive to the specific position in the field. These cells continue to respond to a certain stimulus, even though its absolute position on the retina changes.
In 1980, a researcher called Fukushima proposed a hierarchical neural network model, which was named neocognitron. This model was inspired by the concepts of the simple and complex cells. The neocognitron was able to recognise patterns by learning about the shapes of objects.
Later, in 1998, convolutional neural networks were introduced by Yann LeCun and his colleagues. Their first CNN was called LeNet-5 and was able to classify digits from hand-written numbers.
### Architecture of convolutional network
Before getting into the details of building a convolutional neural network, let's have a look at some of the building blocks, which are either specific to this type of networks or got popularized when they have arrived. As it was seen from the previous article, many concepts of artificial neural networks can be implemented as separate entities, which perform calculations for both – inference and training phases. Since the core structure was already laid out in the article before, here we'll be just adding building blocks on top and then stich them together.
#### Convolutional layer
Convolutional layer is the core building block of convolutional neural network. It does assume its input has 3-dimensional shape of some width, height and depth. For the first convolutional layer it is usually an image, which most commonly has its depth of 1 (grayscale image) or 3 (color image with 3 RGB channels). For subsequent convolutional layers the input is represented by a set of feature maps produced by previous layers (here depth is the number of input feature maps). For now, let's assume we deal with inputs having depth of 1, which turns them into 2-dimensional structures then.
So, what the convolutional layer does is essentially an image convolution with some kernel. It is a very common image processing operation, which is used to achieve variety of results. For example, it can be used to make images blurry or make them sharper. But this is not what convolutional networks are interested in. Depending on the kernel in use, image convolution can be used to find certain features in images – vertical or horizontal edges, corners, angles or more complex features like circles, etc. Recall the idea of simple cells in the visual cortex?
Let's see how convolution is calculated. Suppose we have n (height) by m (width) matrices K (kernel) and I (image). Then it can be written as dot product of those matrices, where kernel matrix is flipped horizontally and vertically.
For example, if we have 3 by 3 matrices K and I, the convolution of those can be calculated this way:
The above is the way how convolution is defined when it comes to signal processing. Kernel is flipped vertically and horizontally there. A more straight forward calculation would be just a normal dot product of the K and I matrices, without any flipping. This operation is called cross-correlation and defined this way:
When it comes to signal processing, convolution and cross-correlation have different properties and are used for different purpose. However, when it comes to image processing and neural networks the difference becomes subtle and cross-correlation is often used instead. For neural networks it is really not important at all. As we'll see later, those "convolution" kernels are actually the weights, which neural network needs to learn. So, it is up to the network to decide which kernel to learn - flipped or not. With this in mind, we'll keep it simple and use cross-correlation then. Note: further in the article anywhere "convolution" is mentioned, we'll assume normal dot product of two matrices, i.e. cross-correlation.
OK, we now know how to calculate convolution for two matrices of the same size or kernel and image of the same size. However, in image processing it is rarely the case. Kernel is usually a square matrix of size 3 by 3 or 5 by 5 or 7 by 7, etc. While image can be of any size. So how is image convolution calculated then? To calculate image convolution the kernel is moved across the entire image and the weighted sum is calculated at every possible location of the kernel. In image processing this concept is known as sliding window. Calculations start at the top left corner of the image and convolution is calculated between the kernel and corresponding image area of the same size. Then kernel is shifted right by one pixel and another convolution is calculated. It is then repeated until convolution is calculated at every position of the row. Once it is done, the kernel is moved to the start of the next row of pixels and the process continues further. When entire image is processed, we get a feature map - values of individual convolutions at every possible location of the image.
The picture below illustrates the process of calculating image convolution. For an image of 8x8 in size and kernel of 3x3, we get a feature map of 6x6 in size - convolution is calculated only at those locations, where kernel fits entirely into the image. The picture below highlights few regions of the source image and their corresponding convolution values in the resulting feature map.
The above 3x3 kernel is designed to look for object's left edges (or presence of a straight vertical line on the right from the center of the sliding window). High positive values in the resulting feature map indicate presence of the feature we are looking for. Zeros mean absence of the feature. And for this particular example, negative values indicate presence of the "inverse" feature – object's right edges.
As it was shown above, the output feature map gets smaller in size than the source image when convolution is calculated. And the bigger kernel is used, the smaller feature map we get. For a kernel of nxm in size, the input image loses (n-1)x(m-1) in size. So, if we would have 5x5 kernel in the above example, then the result feature map would get down to 4x4 in size. In many cases, however, it is preferred to get output feature map of the same size as input. To obtain this, the source image needs to be padded (usually with zeros). For example, if the source image is 8x8 in size and our kernel is 5x5 in size, then we would need to pad the input, so it gets to 12x12 in size, i.e. 4 extra rows/columns added. This is usually done by adding 2 rows/columns on each side of the input image.
So far we've discussed how to compute convolution mathematically and how to compute image convolution when it comes to image processing. However, we are doing artificial neural networks, so we need to see how all the above is related to convolutional layers. To keep it simple for now, let's use the example from the above – 8x8 input image convolved with 3x3 kernel, which gives us 6x6 feature map (output). In this case, our input layer has 64 nodes and our convolutional layer has 36 neurons. However, unlike with fully connected layer, where each neuron of the layer is connected to all neurons of the previous layer, neurons of convolutional layer are connected only to a small group of the previous layer's neurons. Each neuron in convolutional layer has as many connections as the number of weights in the convolution kernel it implements, which is 9 connections in the above example (kernel size 3x3). Since convolutional layer assumes the input has 2D shape (3D in general, but keeping it simple for this example), those connections are done to a rectangular group of previous neurons, which is of the same shape as the kernel in use. The group of connected previous neurons is different for each neuron of the convolutional layer, however it does overlap for the neighbouring neurons. These connections are made in the same way, as pixels of the source image are chosen, when calculating image convolution using sliding window approach. For example, looking at the above image demonstrating image convolution, we can see which of the highlighted outputs on the feature map get connected to which inputs (highlighted with the same color).
Ignoring the fact that neurons of fully connected layers and convolutional layers have different number of connections to the previous layer and that these connections have certain structure, both layers essentially do the same – calculating weighted sum of inputs to produce outputs. There is one more difference though. Unlike with fully connected layers, where each neuron has its own weights, neurons of convolutional layers share them. So, if a layer does one single 3x3 convolution (in practice it does more than one, but keep it for later), it just has one set of weights, i.e. 9, which are shared between each neuron for calculating weighted sum. And, although it was not yet mentioned before, convolutional layers also add bias value to the weighted sum, which is also shared. The table below summarizes the difference between fully connected and convolutional layers and provides some numbers for the above example.
Fully connected layer Convolutional layer No assumptions about input structure Input is assumed to have 2D shape (3D in general) Each neuron is connected to all neurons of the previous layers 64 connections each Each neuron is connected to a small rectangular group of neurons in the previous layer; number of connections equal to number of weights in convolution kernel 9 connections each Each neuron has its own weights and bias value 2304 weights and 36 bias values Weights and bias value are shared 9 weights and 1 bias value
For now, we've kept things simple and assumed that both input and output of convolutional layer have 2D shape. However, it is not the case in general. Instead, both input and output have 3D shape. First, lets start with the output. In practice, each convolutional layer computes more than a single convolution. The number of convolutions it does is a configurable parameter, which is set when designing artificial neural network. Each convolution uses its own set of weights (kernel) and bias value and so produces a different feature map. As it was mentioned before, different kernels can be used to look for different features – lines at different angles, curves, corners, etc. And so, it is often desired to get a number of feature maps, which highlight presence of different features. Calculation of those maps is simple - the process of calculating convolution for the given input is repeated multiple times with different kernel's weights/bias every time. Translating it to artificial neurons' world, we are simply adding additional groups of neurons into the convolution layer, which are connected to inputs in the same way as in the case with single kernel. Having same connection pattern, these groups of neurons share different weights and bias values though. Coming back to the example described before, suppose we configure our convolution layer to do 5 convolutions, 3x3 each. In this case number of outputs (number of neurons) is 36*5=180 – 5 groups of neurons organized into 2D shape and repeating same connection pattern. Each group of neurons shares its own set of weights/bias, which gives us 45 weights and 5 bias values in total for the layer.
Now let's discuss 3D nature of inputs. If we speak about the very first convolutional layer, then its input will be some sort of image, most likely. Most of the time it will be either grayscale image (2D data) or color RGB image (3D data). If we speak of subsequent convolutional layers, then input's depth will be equal to the number of feature maps (number of convolutions) calculated by the previous layer. When input gets higher depth, the number of neurons in convolutional layer is not growing. Instead, number of connections with the previous layer is growing. In fact, convolution kernels get 3D shape as well and have nxmxd size, where d is the depth of input. Translating it to neurons' world again, we can think of it as if each neuron gets additional connections to every feature map input contains. In the case of 2D input, each neuron was connected to nxm (kernel size) rectangular area of the input. In the case of 3D input, however, each neuron is connected to number (d) of such areas, which are coming from the same location, but from different input feature maps.
Since we've generalized convolutional layers to 3D inputs/outputs and also mentioned bias values, we can update our convolution formula, which is computed at every possible location (x, y) of the kernel within the input features.
To complete with convolutional layers for now, let's summarize on the parameters used to configure them. When creating fully connected layer, we use only two parameter - number of inputs and number outputs (neurons in the layer). When creating convolutional layers though, we don't need to specify number of outputs. Instead we describe the shape of inputs, hxwxd, and the shape and number of kernels, nxm@z. So, we have 6 numbers: w – width of input feature maps (image), h – height of input feature maps, d – depth of input (number of feature maps), m – width of kernels, n – height of kernels, z – number of kernels (number of output feature maps). The actual size of kernels depends on the input specification and so we get z kernels of nxmxd in size. And the size of output then becomes (h-n+1)x(w-m+1)xz (here we assume input is not padded and kernel is applied only at valid locations).
We'll get back to convolutional layers again when it comes to training them. The above, however, should give an idea of how output is calculated on the inference phase (computing output of a trained network).
#### ReLU activation function
The next building block to describe is ReLU activation function. It is not something new or specific to convolutional neural networks. However, it was popularized a lot with the rise of deeper neural networks. And this is where convolutional networks usually fit.
One of the problems deep neural networks experience is known as vanishing gradient problem. When training artificial neural network using gradient-based learning algorithms and backpropagation, each of the neural network's weights receives an updated proportional to the partial derivative of the error function with respect to the current weight. The problem is that in some cases, the gradient value can be so small, so it effectively prevents the weight from changing its value. One of the causes of this problem is the use of traditional activation functions such as sigmoid and hyperbolic tangent. These functions have gradient in the (0, 1) range, with values close to zero on the majority of function's domain. And since error's partial derivatives are calculated using chain rule, it means that for a n-layer network there will be n multiplications of these small numbers, meaning gradient decreases exponentially with n. As the result, "front" layers of a deep network train very slowly, if at all.
The ReLU function is defined as f(x)=max(0, x). Its biggest advantage is that it has constant derivative equal to 1 for values of x greater than zero. As the result, it allows better gradient propagation, which speeds up training of deeper artificial neural networks. Also, it is more computationally efficient, making it faster to compute in comparison with sigmoid or hyperbolic tangent.
ReLU function Sigmoid function
Although ReLU function does have some potential problems as well, so far it looks like the most successful and widely-used activation function when it comes to deep neural networks.
#### Pooling layer
It is a common practice to follow convolutional layer with a pooling layer. The objective of this layer is to down-sample input feature maps produced by the previous convolutions. By reducing the spatial size of inputs, we also reduce the amount of parameters and computation in the neural network. This also helps in controlling overfitting – less parameters means less chance to overfit.
The most common pooling technique is the MAX pooling with 2x2 filter and stride 2. For the nxm input feature map, it produces a n/2xm/2 map by replacing every 2x2 region in the input with a single value – maximum value of the 4 values in that region. These regions don't overlap, but adjacent to each other, since the filter is moved horizontally and vertically with the step size (stride) equal to its size. Below is example of applying MAX pooling to the 6x6 input (colored cells highlight source values of the MAX operator and the corresponding result).
MAX pooling is not the only pooling technique. Another common one is Average pooling, which calculates average values of the source regions instead of taking their maximum value.
Pooling layers also can be configured with different size of the filter and stride value. For example, some applications use 3x3 filter with stride 2. Such configuration creates an overlapping pattern of pooling regions, since the filter's step size is smaller than its size. Making stride value greater than filter size is uncommon however, since some features may get lost completely.
One important thing to mention about pooling layers is that they operate with 2D feature maps and don't affect depth of the input. So, if input contains 10 feature maps produced by previous convolutional layer, for example, the pooling is applied individually to each map. As the result, it produces same number of feature maps, but smaller in size.
#### Building convolutional neural network
As we now have the most common building blocks, we can put them together into a convolutional neural network. Although there are some network architectures, which are based entirely on convolutional layers, it is a rare case. Most of the time convolutional networks only start with convolutional layers, which perform initial features' extraction, and then followed by fully connected layers, which perform final classification.
As an example, below is the architecture of LeNet-5 convolutional neural network, which was first described by Yann LeCun and applied to classification of hand-written digits. It takes a 32x32 grayscale image as its input and produces a vector of 10 values – probabilities of belonging to certain class (digits from 0 to 9). The table below summarizes the architecture of the network, dimensions of layers’ outputs and number of trainable parameters (weights + biases).
Layer type Trainable parameters Output size Input image 32x32x1 Convolution layer 1, 6 kernels of 5x5 in size ReLU activation 156 28x28x6 MAX pooling 1 14x14x6 Convolution layer 2, 16 kernels of 5x5 in size ReLU activation 416 10x10x16 MAX pooling 2 5x5x16 Convolution layer 3, 120 kernels of 5x5 in size 3120 1x1x120 Fully Connected layer 1, 120 inputs, 84 outputs Sigmoid activation 10164 84 Fully Connected layer 2, 84 inputs, 10 outputs SoftMax activation 850 10
With only 14706 trainable parameters, the structure of the above convolutional neural network is very simple. These days there are much more complicated deep networks being developed, which include many millions of parameters to train.
### Training convolutional network
So far we've discussed only the inference part of convolutional neural network, which is calculating its output for a given input. However, the network needs to be trained first to get something meaningful out of it. When it comes to convolution operator in image processing, the kernels there are usually handcrafted and serve specific purpose. Some kernels are used to find objects' edges, some for making pictures sharper or blurry, etc. Very often it is a time-consuming process to design right kernel to perform the task needed. With convolutional neural networks it is all different, however. When designing such network, we think about number of layers, number and size of convolutions done, etc. But we don't set those convolution kernels. Instead, the network will learn those during the training phase, since essentially those kernels are nothing more but weights – same as we have them in fully connected layers.
Training of convolutional artificial networks is done using exactly the same algorithms as used for training of fully connected networks – stochastic gradient descent and backpropagation. As it was demonstrated in the previous article, to calculate partial derivatives of neural network's error with respect to its weights we can use chain rule. It allows us to define complete equations for weights' updates of any trainable layer. However, this time we'll concentrate more on the error back propagation side of things and instead of providing one big equation containing all parts of the chain rule, we'll provide smaller equation's, which are specific to each building block of neural network – fully connected and convolutional layers, activation functions, cost functions, etc.
If we revisit chain rule from the previous article, we'll notice that every building block of a neural network calculates its error gradient as partial derivative of its outputs with respect to its inputs and multiples it with error gradient coming from the block following it. Remember we are moving backward, so calculations start at the last block and flow to previous blocks, i.e. the first block. The last block on the training phase is always a cost function and so it computes error gradient as derivative of cost (its output) with respect to neural network's output (input of the cost function). This can be defined the next way:
All other building blocks take the error gradient from the next block and multiply it with partial derivatives of their own outputs with respect to inputs.
Before describing derivatives of the new building blocks, which we are going to use for convolutional networks, lets revisit derivatives of the building blocks we've used for fully connected networks, but written in the new notation. First, we start with error gradient of MSE cost function with respect to outputs of the network (yi – outputs produced by the network, ti – target outputs):
Now, when error gradient passes backward through sigmoid activation function, it gets recalculated this way (oi here is the output of the sigmoid), which is gradient from the next block (whatever it is – it can be cost function or another layer in multi-layer network) multiplied by sigmoid's derivative:
Alternative, if hyperbolic tangent is used as activation function, its derivative is used instead:
Now we need to propagate error gradient backward through a fully connected layer. Since every input is connected to every output, we get a sum of partial derivatives (n is number of neurons in the fully connected layer, j is input's index, i is outpu's/neuron's index):
Since fully connected layer is a trainable layer, it needs not only to pass error's gradient backward to previous building block/layer, but also update its weights. Using the above defined naming convention, the update rule for weights and biases can be written as bellow (classical SGD):
All of the equations above is a quick repetition of the back propagation from the previous article. Why was it important? Well, first to remind the basics. Second, to rewrite it in a different way, where each building block defines its own error's gradient back propagation equation, which is independent of the other blocks. The way weights' update equation was given in the previous article helps to understand the basics and how the chain rule works. But being one single equation makes it not generic at all. What if we need different cost function instead of MSE? What if we need hyperbolic tangent or ReLU activation instead of sigmoid? The way it is presented in this article makes it more flexible and allows mixing building blocks of artificial neural networks in various ways and train them without assumptions on which layer is followed by which activation and which cost function is in use (well, more or less). Plus, this presentation is more in sync with the actual C++ implementation, where different building blocks are implemented as separate classes, taking care of their own calculations for the forward pass and backward pass during training.
Note: If all the above is not clear, however, it is recommended to go through the previous article.
#### Cross-entropy cost function
One of the most common uses of convolutional neural networks is image classification. Given an image, a network needs to classify it into one of the mutually exclusive classes. For example, it can be hand written digits classification, where we have 10 possible classes corresponding to digits from 0 to 9. Or a network can be trained to recognize objects like car, truck, ship, airplane, etc., and so we'll have as many classes as we have types of objects. The main point in this type of classification is that each input image must belong to one class only, i.e. we cannot have objects which are classified as both car and airplane.
When dealing with multi class classification problems, the designed artificial neural network has as many outputs, as the number of classes we have. On the training phase, target outputs are one-hot encoded, i.e. represented with vector of zeros with only one element set to value '1' at the index corresponding to the class. For example, for a task of 4-class classification, our target outputs may look something like this: {0, 1, 0, 0} – class 2, {0, 0, 0, 1} – class 4, etc. None of the target outputs are allowed to have multiple elements set to '1' or another non-zero value. This can be viewed as target probabilities, i.e. the {0, 1, 0, 0} output means that the presented input belongs to class 2 with 100% probability and to other classes with probability of 0%.
When training, however, the actual neural network's outputs will look different though. It may provide an output something like {0.3, 0.35, 0.25, 0.1}, for example. Such output may have different meaning. For a trained network, it may mean the network was presented with a tricky example and it is not very clear, but looks more like class 2 – the highest probability of 35%. Or, if we just started training, it may mean little at all, other than "keep going".
And so, we need a cost function, which would tell us the amount of difference between target and the real output and direct parameters' update of the neural network. When it comes to probabilistic models over mutually exclusive classes, we deal with predicted and the ground-truth probabilities. In such cases, the common choice is the cross-entropy cost function, which has its roots coming from the information theory. As it says, by minimizing cross-entropy, we want to minimize the amount of extra data (bits), required for encoding some events appearing with probability distribution ti (target or real distribution) using some estimated probabilities yi (which might be close, but no exactly). And to minimize the cross-entropy, we need to make our estimated probabilities to be the same as the real probabilities – which is what we are looking for.
The cross-entropy cost function, the value we need to minimize, is defined as below (same as before – ti are target outputs, while yi is the output provided by neural network):
Getting its derivative, the gradient of the cost function with respect to neural network's output is then calculated as:
Now we have the cross-entropy cost function instead of MSE and so we can move to other building blocks and see how error gradient propagates backward.
#### SoftMax activation function
For the last layer's activation function of the neural network used for classification problem we could use the sigmoid function, which we've already seen in the previous article and quickly repeated above. Its output is in the (0, 1) range and so can be interpreted as probabilities between 0% and 100%. When neural network is trained with sigmoid in the output layer, it really may provide probabilities close to the ground truth. However, since we deal with mutually exclusive classes, it may not always make perfect sense. For example, provided a challenging example, a network may provide an output vector like this: {0.6, 0.55, 0.1, 0.1}. Yes, looks like class 1 with probability of 60%! But probability of the class 2 is not too far away. And another problem is that if we sum the four probabilities we've got, we get 1.35, which is 135%.
There are two problems we want to address. First, we definitely want to have sum of probabilities equal to 100%. Not more, not less. Also, if we get a tricky example, which looks like class 1, but also seems close to class 2, can we really have a high certainty of 60% that the classification is right?
To resolve the two issues above, we can use a different activation function, which is SoftMax. Same as sigmoid, it provides output in the (0, 1) range. But unlike sigmoid, it does not operate on single values of the input vector, but on the entire vector, and so makes sure the sum of the output vector equals to 1. The SoftMax function is defined the next way:
If we would use SoftMax function instead of sigmoid for the above example (you can use inverse sigmoid to find the source input values), the output vector would look different and make more sense – {0.316, 0.3, 0.192, 0.192}. As we can see, the sum of all values equals to 1, which is 100%. And even though the 1st class seem to win, the probability of it is not that high - only 31.6%.
As for any other activation function, we need to define gradient back propagation equation for the SoftMax function. Here it is:
Going now further backward through the LeNet-5 neural network's architecture, we see fully connected layers and sigmoid activation function. Equations for both were already defined above. So now it is time to address the other building blocks introduced in this article.
#### ReLU activation function
As it was already mentioned above, ReLU activation function became a very popular choice for deeper neural networks, as it allows much better propagation of error's gradient through the network. It is all due to its constant gradient equal to 1 for input values of greater than zero. To complete ReLU activation, we also need to define its equation for gradient back propagation.
#### Pooling layer
Now it is time to propagate error's gradient backward through pooling layer. To make it simple, lets suppose we use 2x2 kernel with stride 2 and we don't use input padding (we apply pooling to valid locations only). With this in mind, it means every value of the output feature map is calculate based on 4 values of the input feature map.
Although pooling layers make assumption that input vectors represent 2D data, the math below will work with inputs/outputs as 1D vectors. To make it all work, we'll define a i2j() function, which for the given index i of input vector returns corresponding index j of output vector. Since each output is calculated based on 4 input values, it means there are 4 input indexes, for which i2j() will return the same output index.
Let's start with Max Pooling. To define equation for error's gradient back propagation, we'll need one extra thing. On the forward pass, when neural network's output is calculated, the pooling layer will also fill in the maxIndexes vector of the same length as output vector. But, if output vector contains maximum value of the corresponding input values, the maxIndexes vector contains the index of the maximum value. With all the above, we can define gradient back propagation equation for Max Pooling layer:
As for Average Pooling it is even simpler – the error gradient from the previous block is simply divided by the size of pooling kernel, which is 4 our case:
#### Convolutional layer
Finally, it is time to define back propagation pass for convolutional layer. It is not much different from fully connected layer as long as the fact of shared weights is kept in mind.
Let's then start with weights update of the convolutional layer. With fully connected layers it was simple – partial derivative of error with respect to weight wi,j equals to error gradient coming from the next block multiplied by corresponding input value – δi(k+1)xj. The reason for this is that each input/output connection is assigned its own weight in fully connected layer, which is not shared. However, it is not the case in convolutional layer. The picture below demonstrates that every weight of convolution kernel is used for many input/output connections. In the example below, the highlighted kernel's weights are used 9 times each – the kernel is applied in 9 different positions within the input image. And so, the partial derivative of error with respect to weight will need to have 9 terms as well – the number of times the weight is used.
Same as with pooling layers, we'll ignore here the fact that convolutional layers deal with 2D/3D data. Instead we'll assume that inputs/outputs/kernels are plain vectors/arrays for now (this is what they end up in C++ anyway). And so, for the example above, the 1st kernel's weight (highlighted in red) is applied to inputs {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}, while the 4th weight is applied to inputs {6,7,8,10,11,12,14,15,16}. Suppose that we have such vector of input indexes used by every weight, which we'll name weightInputsi – input of the ith weight. Also, we'll define a function of two arguments i2o(i,j), which provides index of output value for the ith weight and jth input. Here are few examples for the picture above, i2o(1,1)=1, i2o(4,6)=1, i2o(1, 11)=9 and i2o(4,16)=9. With the above naming convention, the weights' update rule for convolutional network can be then defined the next way:
Does the above make sense? Well, the more you think about it, the more it will. All we do is taking error gradients for all the outputs (since each kernel's weight is used to calculate all outputs) and multiply them by corresponding input. Yes, we have multiple kernels. But, they are all applied in the same pattern, so even though we'll need to update weights of different kernels, the weightInputs vectors stay the same. However, the i2o(i,j) is specific to each kernel. Or it can be extended with extra parameter – kernel index.
Updating bias value is much simpler. Since each kernel/bias is used to calculate every output value, we'll just sum all error gradients for the feature map produced by that kernel.
Note: both equations above are done per feature map/kernel, i.e. weights and bias value are not parameterized there with kernel index.
Now it is time to get the final equation for convolutional layer, which is for propagating error gradient backward through the network. This means calculating partial derivatives of error with respect to inputs of the layer. Each input element can be used multiple times to produce an output value of a feature map. It can be used as many times as the number of elements in convolution kernel (number of weights). Some inputs can be used only for one output, though. For example, those are the inputs in corners of the input 2D feature map. But then we also need to keep in mind that every input feature map can be processed multiple times with different kernels, which generate more output maps. Again, lets pretend it is all flat for now, no 2D/3D indexing. Then, let's assume we have another set of helper vectors named inputOutputsi, keeping indexes of outputs, which the ith input contributes to. Finally, we'll need the i2w(i, j) function, which provides index of the weight, which is used to connect ith input with jth output. Here are few examples again for the above picture: i2w(1, 1)=1, i2w(6,1)=4, i2w(16,9)=4. With all this, we can define equation for propagating error's gradient backward through convolutional layer.
Now it looks like the math is complete – we have everything we need to calculate both as the forward pass through convolutional network, as the backward pass. If it still puzzles, confuses or leaves some uncertainty, go through it all again, think about it. Or dive into the code to see relation between the math and implementation.
## The ANNT library
Implementation of the convolutional artificial neural network in the ANNT library is heavily based on the design set by implementation of fully connected networks described in the previous article. All the core classes are left as they were, only new building blocks were implemented, which allow building them into convolutional neural networks. The new class diagram of the library is shown below – not much of a difference.
Similar to the way it was set before, new building blocks take care of calculating their output on the forward pass and propagating error gradient on the backward pass (as well as calculating initial weights' updates in the case of trainable layers). As the result, all the code for neural network training is left unchanged.
And, as in the case with the rest of the code, the new building blocks utilize SIMD instructions wherever possible to vectorize computations, as well as OpenMP to parallelize them.
### Building the code
The code comes with MSVC (2015 version) solution files and GCC make files. Using MSVC solutions is very easy – every example's solution file includes projects of the example itself and the library. So MSVC option is as easy as opening solution file of required example and hitting build button. If using GCC, the library needs to be built first and then the required sample application by running make.
## Usage examples
After the long discussion about the theory and math of convolutional neural networks, it is time to get to practice and actually build some of the networks for image classification tasks – hand written digits and different objects like cars, trucks, ships, airplanes, etc. Note: none of these examples claim that the demonstrated neural network's architecture is the best for its task. In fact, none of these examples even say that artificial neural networks is the way to go. Instead, their only purpose is to provide demonstration of using the library.
Note: the code snippets below are only small parts of the example applications. To see the complete code of the examples, refer to the source code package provided with the article (which also includes examples for fully connected neural networks described in the previous article).
### MNIST handwritten digits classification
The first example to have a look at is classification of hand-written digits from the MNIST database. The database contains 60000 examples for neural network training and additional 10000 examples for testing of the trained network. The picture below demonstrates some of the examples of different digits to classify.
The convolutional neural network used in this example has the structure very similar to the LeNet-5 network mentioned above. The difference is that we'll use slightly smaller network (well, actually a lot smaller, if we look at the number of weights to train), which has only one fully connected network. Here is structure of the network we'll use:
Conv(32x32x1, 5x5x6 ) -> ReLU -> AvgPool(2x2)
Conv(14x14x6, 5x5x16 ) -> ReLU -> AvgPool(2x2)
Conv(5x5x16, 5x5x120) -> ReLU
FC(120, 10) -> SoftMax
The configuration above tells the size of input for each convolutional layer and the size and number of convolutions they perform. And for fully connected layer it tells number of inputs and outputs. Let's create the convolution neural network of the above structure them.
// connection table to specify wich feature maps of the first convolution layer
// to use for feature maps produced by the second layer
vector<bool> connectionTable( {
true, true, true, false, false, false,
false, true, true, true, false, false,
false, false, true, true, true, false,
false, false, false, true, true, true,
true, false, false, false, true, true,
true, true, false, false, false, true,
true, true, true, true, false, false,
false, true, true, true, true, false,
false, false, true, true, true, true,
true, false, false, true, true, true,
true, true, false, false, true, true,
true, true, true, false, false, true,
true, true, false, true, true, false,
false, true, true, false, true, true,
true, false, true, true, false, true,
true, true, true, true, true, true
} );
// prepare a convolutional ANN
shared_ptr<XNeuralNetwork> net = make_shared<XNeuralNetwork>( );
net->AddLayer( make_shared<XConvolutionLayer>( 32, 32, 1, 5, 5, 6 ) );
net->AddLayer( make_shared<XAveragePooling>( 28, 28, 6, 2 ) );
net->AddLayer( make_shared<XConvolutionLayer>( 14, 14, 6, 5, 5, 16, connectionTable ) );
net->AddLayer( make_shared<XAveragePooling>( 10, 10, 16, 2 ) );
net->AddLayer( make_shared<XConvolutionLayer>( 5, 5, 16, 5, 5, 120 ) );
net->AddLayer( make_shared<XFullyConnectedLayer>( 120, 10 ) );
net->AddLayer( make_shared<XLogSoftMaxActivation>( ) );
Looking at the code above, it is quite clear how the neural network's configuration stated above is translated into the code. Except for one question – "What is the connection table we've got between the first and the second convolutional layers?" Yes, it was not mentioned in the theory part, but is pretty easy to grasp. As we can see from the network's structure and the code, the first layer does 6 convolutions and so produces 6 feature maps. While the second layer does 16 convolutions. In some cases, it is desired to configure layer's convolutions in such way, that they operate only on the subset of input feature maps. As the code above suggests, the first 6 convolutions of the second layer use different patterns of 3 feature maps produced by the first layer. Then the next 9 convolutions use different patterns of 4 feature maps. Finally, the last convolution uses all 6 feature maps of the first layer. This is done to reduce the number of parameters to train and also make sure that different feature maps of the second layer are not all based on the same input feature maps.
When the convolutional network is created, we can do the same as we did with fully connected network - create a training context, specifying cost function and weights' optimizer, and then pass it all to a helper class, which runs training/validation loop and completes it with testing.
// create training context with Adam optimizer and Negative Log Likelihood cost function (since we use Log-Softmax)
shared_ptr<XNetworkTraining> netTraining = make_shared<XNetworkTraining>( net,
make_shared<XNegativeLogLikelihoodCost>( ) );
// using the helper for training ANN to do classification
XClassificationTrainingHelper trainingHelper( netTraining, argc, argv );
trainingHelper.SetValidationSamples( validationImages, encodedValidationLabels, validationLabels );
trainingHelper.SetTestSamples( testImages, encodedTestLabels, testLabels );
// 20 epochs, 50 samples in batch
trainingHelper.RunTraining( 20, 50, trainImages, encodedTrainLabels, trainLabels );
Below is the sample output of the application, which shows training progress and the final result - classification accuracy on the test data set. We've got 99.01% accuracy, which seems to be a good improvement over fully connected neural network from the previous article, which demonstrated 96.55% accuracy.
MNIST handwritten digits classification example with Convolution ANN
Samples usage: training = 50000, validation = 10000, test = 10000
Learning rate: 0.0020, Epochs: 20, Batch Size: 50
Before training: accuracy = 5.00% (2500/50000), cost = 2.3175, 34.324s
Epoch 1 : [==================================================] 123.060s
Training accuracy = 97.07% (48536/50000), cost = 0.0878, 32.930s
Validation accuracy = 97.49% (9749/10000), cost = 0.0799, 6.825s
Epoch 2 : [==================================================] 145.140s
Training accuracy = 97.87% (48935/50000), cost = 0.0657, 36.821s
Validation accuracy = 97.94% (9794/10000), cost = 0.0669, 5.939s
...
Epoch 19 : [==================================================] 101.305s
Training accuracy = 99.75% (49877/50000), cost = 0.0077, 26.094s
Validation accuracy = 98.96% (9896/10000), cost = 0.0684, 6.345s
Epoch 20 : [==================================================] 104.519s
Training accuracy = 99.73% (49865/50000), cost = 0.0107, 28.545s
Validation accuracy = 99.02% (9902/10000), cost = 0.0718, 7.885s
Test accuracy = 99.01% (9901/10000), cost = 0.0542, 5.910s
Total time taken : 3187s (53.12min)
### CIFAR10 images classification
The second example performs classification of color 32x32 images from the CIFAR-10 dataset. It contains 60000 images, of which 50000 are used for training and the other 10000 for testing. The images are divided between the next 10 class: airplane, automobile, bird, cat, deer, dog, frog, horse, ship and truck. Few examples of those can be seen below.
As the above picture suggests, the CIFAR-10 dataset is much more complex than the MNIST hand-written digits. First, the images are color. And second, they are much less obvious. Up to the point that if I was not told it is a dog, I would not say it myself. As the result, the network's structure gets a bit bigger. Not that it becomes much deeper, but the number of performed convolutions and trained weights is growing. Below is the structure of the network:
Conv(32x32x3, 5x5x32, BorderMode::Same) -> ReLU -> MaxPool -> BatchNorm
Conv(16x16x32, 5x5x32, BorderMode::Same) -> ReLU -> MaxPool -> BatchNorm
Conv(8x8x32, 5x5x64, BorderMode::Same) -> ReLU -> MaxPool -> BatchNorm
FC(1024, 64) -> ReLU -> BatchNorm
FC(64, 10) -> SoftMax
Translating the above neural network's structure into the code gives the result below. Note: since ReLU(MaxPool) produces same result as MaxPool(ReLU), we use the first as it reduces ReLU computation by 75% (although very negligible compared to the rest of the network).
// prepare a convolutional ANN
shared_ptr<XNeuralNetwork> net = make_shared<XNeuralNetwork>( );
net->AddLayer( make_shared<XConvolutionLayer>( 32, 32, 3, 5, 5, 32, BorderMode::Same ) );
net->AddLayer( make_shared<XMaxPooling>( 32, 32, 32, 2 ) );
net->AddLayer( make_shared<XBatchNormalization>( 16, 16, 32 ) );
net->AddLayer( make_shared<XConvolutionLayer>( 16, 16, 32, 5, 5, 32, BorderMode::Same ) );
net->AddLayer( make_shared<XMaxPooling>( 16, 16, 32, 2 ) );
net->AddLayer( make_shared<XBatchNormalization>( 8, 8, 32 ) );
net->AddLayer( make_shared<XConvolutionLayer>( 8, 8, 32, 5, 5, 64, BorderMode::Same ) );
net->AddLayer( make_shared<XMaxPooling>( 8, 8, 64, 2 ) );
net->AddLayer( make_shared<XBatchNormalization>( 4, 4, 64 ) );
net->AddLayer( make_shared<XFullyConnectedLayer>( 4 * 4 * 64, 64 ) );
net->AddLayer( make_shared<XBatchNormalization>( 64, 1, 1 ) );
net->AddLayer( make_shared<XFullyConnectedLayer>( 64, 10 ) );
net->AddLayer( make_shared<XLogSoftMaxActivation>( ) );
The rest of the example application follows the same pattern as set by the other classification examples - training context is created with required cost function and weights' optimizer and passed to helper class to run the training loop. Below is the example of its output.
CIFAR-10 dataset classification example with Convolutional ANN
Samples usage: training = 43750, validation = 6250, test = 10000
Learning rate: 0.0010, Epochs: 20, Batch Size: 50
Before training: accuracy = 9.91% (4336/43750), cost = 2.3293, 844.825s
Epoch 1 : [==================================================] 1725.516s
Training accuracy = 48.25% (21110/43750), cost = 1.9622, 543.087s
Validation accuracy = 47.46% (2966/6250), cost = 2.0036, 77.284s
Epoch 2 : [==================================================] 1742.268s
Training accuracy = 54.38% (23793/43750), cost = 1.3972, 568.358s
Validation accuracy = 52.93% (3308/6250), cost = 1.4675, 76.287s
...
Epoch 19 : [==================================================] 1642.750s
Training accuracy = 90.34% (39522/43750), cost = 0.2750, 599.431s
Validation accuracy = 69.07% (4317/6250), cost = 1.2472, 81.053s
Epoch 20 : [==================================================] 1708.940s
Training accuracy = 91.27% (39931/43750), cost = 0.2484, 578.551s
Validation accuracy = 69.15% (4322/6250), cost = 1.2735, 81.037s
Test accuracy = 68.34% (6834/10000), cost = 1.3218, 122.455s
Total time taken : 48304s (805.07min)
As mentioned above, the CIFAR-10 dataset is definitely more complex. If we managed to get up to 99% test accuracy on MNIST dataset, here we don't get even close to it – about 91% accuracy on training set and 68-69% on test/validation. Plus, it took 13 hours to run the 20 epochs. Just using CPU is definitely not enough for convolutional networks.
## Conclusion
In this article we've covered the new extensions to the ANNT library, which allow building convolutional neural networks. At this point it allows building only simple networks (more or less), where layers of the network follow each other sequentially. Building more advanced popular architectures, which look more like a computational graph, is not yet supported so far. However, before getting there, there are other features need to be implemented first. As the CIFAR-10 example demonstrates, once neural network gets bigger, it requires more computational power for training. And here, using just CPU is not enough. These days GPU support is a must, when it comes to deep learning. And so, this feature would get higher priority rather than supporting complex networks.
As fully connected and convolutional neural networks are covered now, the following step will be to go through some common architectures of recurrent networks, which is the topic for the next article. In the meantime, all the latest code can be found on GitHub, which will get updates as the library evolves further.
## Share
Software Developer IBM United Kingdom
Started software development at about 15 years old and it seems like now it lasts most part of my life. Fortunately did not spend too much time with Z80 and BK0010 and switched to 8086 and further. Similar with programming languages – luckily managed to get away from BASIC and Pascal to things like Assembler, C, C++ and then C#. Apart from daily programming for food, do it also for hobby, where mostly enjoy areas like Computer Vision, Robotics and AI. This led to some open source stuff like AForge.NET and not so open Computer Vision Sandbox.
Going out of computers I am just a man loving his family, enjoying travelling, a bit of books, a bit of movies and a mixture of everything else. Always wanted to learn playing guitar, but it seems like 6 strings are much harder than few dozens of keyboard’s keys. Will keep progressing ...
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Last Visit: 23-Feb-19 11:47 Last Update: 23-Feb-19 11:47 Refresh 12 Next »
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3. consider the following scenario regarding the acns insurance company an...
# Question: consider the following scenario regarding the acns insurance company an...
###### Question details
Consider the following scenario regarding the ACNS insurance company.
An assessment of a claim is to be done by a senior acceptor or a regular acceptor. A regular acceptor is only qualified to make this assessment in the situation where the claim amount of a case is below $1,000. In case of a negative assessment, it is the responsibility of the account manager to convey the bad news to the customer. In case of a positive assessment, an electronic invoice is to be generated by a clerk of the finance department, who needs to dispatch that to the client in question. After these activities, the process is completed. Assume that almost simultaneously two claims come in: 1. Car damage of$12,500, as claimed by Mr. Bouman.
2. Car damage of $500, as claimed by Mrs. Fillers Ms. Withagen has been with ACNS for a long time and, for the past years, functions on the level of a senior acceptor. This month, Mr. Trienekens has started his training and works as an acceptor. At the start of his contract, the system administrator, Mr. Verbeek, has used the administration tool of the BPMS to add Mr. Trienekens to the pool of available acceptors. Consider the following statements and indicate which components of the BPMS architecture are affected when they are to be taken into account: a). A new decision support system is developed to support acceptors in making their assessment of claims. (1.5 marks) b). Ms. Withagen retires. (1.5 marks) c). A new distinction between claims becomes relevant: regular acceptors are now also qualified to deal with claims above$1,000 as long as they worked on previous claims by the same client. (3 marks)
d). The processing of Mr. Bouman and Mrs. Fillers claims. (4 marks)
CAN YOU PLEASE THESE CASE STUDY FOR ME
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# What is a Probability
Probability table is employed to tabulate all of the data determined for a standard values from that values new probabilities could be determined. The events also can be tabulated.
Probability Notation
If you can find n elementary events related to an random experiment and m of them are favourable to an event E, then the probability of E is denoted by P(E) and is defined to the ratio $\fracmn$.
=> P(E) = $\frac{\text{Number of Favourable to Event A}}{\text{Total Number of Outcomes}}$.
If an experiment has n similarly likely outcomes in S and N of them are the event A, then the theoretical probability of event A taking place is P(A) = $\fracnN$.
## How to Do Probability
Probability is a significant branch of math which fundamentally deals with the phenomena of chances or randomness of events. Probability of an event measures how likely it is expected to occur. An event is considered most likely if it has greater probability and least likely if it has lower probability. The probability of an event differs between 0 and 1. An event with probability 0 is an impossible event, although event having probability 1 is considered a certain event.
For a situation where several various outcomes are possible, the probability for any specific outcome is described as a fraction of all the possible final results. A sample space is a collection of all possible outcomes of a random experiment. A sample space might be infinite or finite. Infinite sample spaces may be discrete or continuous.
In everyday life, we speak informally about the probability of several event to happen. Example of a probability problem with a solution could be like this. While leaving the home in the morning on a cloudy day, one may have to choose to take an umbrella even when it is not raining because it may possibly rain later on in the day. The crucial types of probability problems are problems on conditional probability, problems on bayes theorem and problems on multiplication theorem.
## Probability Multiplication Rule
The addition rule assisted us solve problems when we carried out one task and wanted to learn the probability of two things occurring during that task. This lesson deals with the multiplication rule. The probability multiplication rule also deals with two events, however in these problems the events take place as a result of greater than one task (drawing two cards, rolling one die then another, pulling two marbles out of a bag, spinning a spinner twice, etc).
When asked to find the probability of A and B, we want to learn the probability of events A and B happening.
Consider events A and B. P(AᴖB)= P(A) P(B).
The Rule Means:
Assume we roll one die then another and want to locate the probability of rolling a 4 on the very first die and rolling an even number on the 2nd die. Notice in this problem we are not coping with the sum of both dice. We are just dealing with the probability of 4 on one die only and then, as an individual event, the probability of an even number on one die only.
P(4) = 1/6
P(even) = 3/6
So P(4ᴖeven) = (1/6)(3/6) = 3/36 = 1/12
### Latest Articles
Scientists analyzed samples of the metal following European exploration of the region started. Platinum has been used by ancient people in Central and South America.
Not many chemicals can attack gold, so that’s why it maintains it shine even when buried for 1000’s of years. When compared with other metals, gold is much softer. One can beat 1 gram of gold to a 1 square meter sheet and light would shine via that sheet.
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## trans-dimensional nested sampling and a few planets
Posted in Books, Statistics, Travel, University life with tags , , , , , , , , , on March 2, 2015 by xi'an
This morning, in the train to Dauphine (train that was even more delayed than usual!), I read a recent arXival of Brendon Brewer and Courtney Donovan. Entitled Fast Bayesian inference for exoplanet discovery in radial velocity data, the paper suggests to associate Matthew Stephens’ (2000) birth-and-death MCMC approach with nested sampling to infer about the number N of exoplanets in an exoplanetary system. The paper is somewhat sparse in its description of the suggested approach, but states that the birth-date moves involves adding a planet with parameters simulated from the prior and removing a planet at random, both being accepted under a likelihood constraint associated with nested sampling. I actually wonder if this actually is the birth-date version of Peter Green’s (1995) RJMCMC rather than the continuous time birth-and-death process version of Matthew…
“The traditional approach to inferring N also contradicts fundamental ideas in Bayesian computation. Imagine we are trying to compute the posterior distribution for a parameter a in the presence of a nuisance parameter b. This is usually solved by exploring the joint posterior for a and b, and then only looking at the generated values of a. Nobody would suggest the wasteful alternative of using a discrete grid of possible a values and doing an entire Nested Sampling run for each, to get the marginal likelihood as a function of a.”
This criticism is receivable when there is a huge number of possible values of N, even though I see no fundamental contradiction with my ideas about Bayesian computation. However, it is more debatable when there are a few possible values for N, given that the exploration of the augmented space by a RJMCMC algorithm is often very inefficient, in particular when the proposed parameters are generated from the prior. The more when nested sampling is involved and simulations are run under the likelihood constraint! In the astronomy examples given in the paper, N never exceeds 15… Furthermore, by merging all N’s together, it is unclear how the evidences associated with the various values of N can be computed. At least, those are not reported in the paper.
The paper also omits to provide the likelihood function so I do not completely understand where “label switching” occurs therein. My first impression is that this is not a mixture model. However if the observed signal (from an exoplanetary system) is the sum of N signals corresponding to N planets, this makes more sense.
Posted in Books, Kids, Statistics, University life with tags , , , , , , , , , on February 24, 2015 by xi'an
Today was the final session of our Reading Classics Seminar for the academic year 2014-2015. I have not reported on this seminar much so far because it has had starting problems, namely hardly any student present on the first classes and therefore several re-starts until we reached a small group of interested students. And this is truly The End for this enjoyable experiment as this is the final year for my TSI Master at Paris-Dauphine, as it will become integrated within the new MASH Master next year.
As a last presentation for the entire series, my student picked John Skilling’s Nested Sampling, not that it was in my list of “classics”, but he had worked on the paper in a summer project and was thus reasonably fluent with the topic. As he did a good enough job (!), here are his slides.
Some of the questions that came to me during the talk were on how to run nested sampling sequentially, both in the data and in the number of simulated points, and on incorporating more deterministic moves in order to remove some of the Monte Carlo variability. I was about to ask about (!) the Hamiltonian version of nested sampling but then he mentioned his last summer internship on this very topic! I also realised during that talk that the formula (for positive random variables)
$\int_0^\infty(1-F(x))\text{d}x = \mathbb{E}_F[X]$
does not require absolute continuity of the distribution F.
## nested sampling for systems biology
Posted in Books, Statistics, University life with tags , , , , on January 14, 2015 by xi'an
In conjunction with the recent PNAS paper on massive model choice, Rob Johnson†, Paul Kirk and Michael Stumpf published in Bioinformatics an implementation of nested sampling that is designed for biological applications, called SYSBIONS. Hence the NS for nested sampling! The C software is available on-line. (I had planned to post this news next to my earlier comments but it went under the radar…)
## Topological sensitivity analysis for systems biology
Posted in Books, Statistics, Travel, University life with tags , , , , , , on December 17, 2014 by xi'an
Michael Stumpf sent me Topological sensitivity analysis for systems biology, written by Ann Babtie and Paul Kirk, en avant-première before it came out in PNAS and I read it during the trip to NIPS in Montréal. (The paper is published in open access, so everyone can read it now!) The topic is quite central to a lot of debates about climate change, economics, ecology, finance, &tc., namely to assess the impact of using the wrong model to draw conclusions and make decisions about a real phenomenon. (Which reminded me of the distinction between mechanical and phenomenological models stressed by Michael Blum in his NIPS talk.) And it is of much interest from a Bayesian point of view since assessing the worth of a model requires modelling the “outside” of a model, using for instance Gaussian processes as in the talk Tony O’Hagan gave in Warwick earlier this term. I would even go as far as saying that the issue of assessing [and compensating for] how wrong a model is, given available data, may be the (single) most under-assessed issue in statistics. We (statisticians) have yet to reach our Boxian era.
In Babtie et al., the space or universe of models is represented by network topologies, each defining the set of “parents” in a semi-Markov representation of the (dynamic) model. At which stage Gaussian processes are also called for help. Alternative models are ranked in terms of fit according to a distance between simulated data from the original model (sounds like a form of ABC?!). Obviously, there is a limitation in the number and variety of models considered this way, I mean there are still assumptions made on the possible models, while this number of models is increasing quickly with the number of nodes. As pointed out in the paper (see, e.g., Fig.4), the method has a parametric bootstrap flavour, to some extent.
What is unclear is how one can conduct Bayesian inference with such a collection of models. Unless all models share the same “real” parameters, which sounds unlikely. The paper mentions using uniform prior on all parameters, but this is difficult to advocate in a general setting. Another point concerns the quantification of how much one can trust a given model, since it does not seem models are penalised by a prior probability. Hence they all are treated identically. This is a limitation of the approach (or an indication that it is only a preliminary step in the evaluation of models) in that some models within a large enough collection will eventually provide an estimate that differs from those produced by the other models. So the assessment may become altogether highly pessimistic for this very reason.
“If our parameters have a real, biophysical interpretation, we therefore need to be very careful not to assert that we know the true values of these quantities in the underlying system, just because–for a given model–we can pin them down with relative certainty.”
In addition to its relevance for moving towards approximate models and approximate inference, and in continuation of yesterday’s theme, the paper calls for nested sampling to generate samples from the posterior(s) and to compute the evidence associated with each model. (I realised I had missed this earlier paper by Michael and co-authors on nested sampling for system biology.) There is no discussion in the paper on why nested sampling was selected, compared with, say, a random walk Metropolis-Hastings algorithm. Unless it is used in a fully automated way, but the paper is rather terse on that issue… And running either approach on 10⁷ models in comparison sounds like an awful lot of work!!! Using importance [sampling] nested sampling as we proposed with Nicolas Chopin could be a way to speed up this exploration if all parameters are identical between all or most models.
## an extension of nested sampling
Posted in Books, Statistics, University life with tags , , , , , , , on December 16, 2014 by xi'an
I was reading [in the Paris métro] Hastings-Metropolis algorithm on Markov chains for small-probability estimation, arXived a few weeks ago by François Bachoc, Lionel Lenôtre, and Achref Bachouch, when I came upon their first algorithm that reminded me much of nested sampling: the following was proposed by Guyader et al. in 2011,
To approximate a tail probability P(H(X)>h),
• start from an iid sample of size N from the reference distribution;
• at each iteration m, select the point x with the smallest H(x)=ξ and replace it with a new point y simulated under the constraint H(y)≥ξ;
• stop when all points in the sample are such that H(X)>h;
• take
$\left(1-\dfrac{1}{N}\right)^{m-1}$
as the unbiased estimator of P(H(X)>h).
Hence, except for the stopping rule, this is the same implementation as nested sampling. Furthermore, Guyader et al. (2011) also take advantage of the bested sampling fact that, if direct simulation under the constraint H(y)≥ξ is infeasible, simulating via one single step of a Metropolis-Hastings algorithm is as valid as direct simulation. (I could not access the paper, but the reference list of Guyader et al. (2011) includes both original papers by John Skilling, so the connection must be made in the paper.) What I find most interesting in this algorithm is that it even achieves unbiasedness (even in the MCMC case!).
## nested sampling with a test
Posted in Books, pictures, Statistics, Travel, University life with tags , , , , , on December 5, 2014 by xi'an
On my way back from Warwick, I read through a couple preprints, including this statistical test for nested sampling algorithms by Johannes Buchner. As it happens, I had already read and commented it in July! However, without the slightest memory of it (sad, isn’t it?!), I focussed this time much more on the modification proposed to MultiNest than on the test itself, which is in fact a Kolmogorov-Smirnov test applied to a specific target function.
Indeed, when reading the proposed modification of Buchner, I thought of a modification to the modification that sounded more appealing. Without getting back to defining nested sampling in detail, this algorithm follows a swarm of N particles within upper-level sets of the likelihood surface, each step requiring a new simulation above the current value of the likelihood. The remark that set me on this time was that we should exploit the fact that (N-1) particles were already available within this level set. And uniformly distributed herein. Therefore this particle cloud should be exploited as much as possible to return yet another particle distributed just as uniformly as the other ones (!). Buchner proposes an alternative to MultiNest based on a randomised version of the maximal distance to a neighbour and a ball centre picked at random (but not uniformly). But it would be just as feasible to draw a distance from the empirical cdf of the distances to the nearest neighbours or to the k-nearest neighbours. With some possible calibration of k. And somewhat more accurate, because this distribution represents the repartition of the particle within the upper-level set. Although I looked at it briefly in the [sluggish] metro from Roissy airport, I could not figure out a way to account for the additional point to be included in the (N-1) existing particles. That is, how to deform the empirical cdf of those distances to account for an additional point. Unless one included the just-removed particle, which is at the boundary of this upper-level set. (Or rather, which defines the boundary of this upper-level set.) I have no clear intuition as to whether or not this would amount to a uniform generation over the true upper-level set. But simulating from the distance distribution would remove (I think) the clustering effect mentioned by Buchner.
“Other priors can be mapped [into the uniform prior over the unit hypercube] using the inverse of the cumulative prior distribution.”
Hence another illustration of the addictive features of nested sampling! Each time I get back to this notion, a new understanding or reinterpretation comes to mind. In any case, an equally endless source of projects for Master students. (Not that I agree with the above quote, mind you!)
## a statistical test for nested sampling
Posted in Books, Statistics, University life with tags , , , , , on July 25, 2014 by xi'an
A new arXival on nested sampling: “A statistical test for nested sampling algorithms” by Johannes Buchner. The point of the test is to check if versions of the nested sampling algorithm that fail to guarantee increased likelihood (or nesting) at each step are not missing parts of the posterior mass. and hence producing biased evidence approximations. This applies to MultiNest for instance. This version of nest sampling evaluates the above-threshold region by drawing hyper-balls around the remaining points. A solution which is known to fail in one specific but meaningful case. Buchner’s arXived paper proposes an hyper-pyramid distribution for which the volume of any likelihood constrained set is known. Hence allowing for a distribution test like Kolmogorov-Smirnov. Confirming the findings of Beaujean and Caldwell (2013). The author then proposes an alternative to MultiNest that is more robust but also much more costly as it computes distances between all pairs of bootstrapped samples. This solution passes the so-called “shrinkage test”, but it is orders of magnitude less efficient than MultiNest. And also simply shows that its coverage is fine for a specific target rather than all possible targets. I wonder if a solution to the problem is at all possible given that evaluating a support or a convex hull is a complex problem which complexity explodes with the dimension.
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# Structure substitutive des pavages apériodiques de Jeandel-Rao¶
Sébastien Labbé, version 1 of arXiv:1808.07768, August 2018
Sébastien Labbé, version 2 of arXiv:1808.07768, April 2019
Références:
## Prerequisites¶
Sage version 8.7 or above should work:
In [1]:
version()
Out[1]:
'SageMath version 8.7, Release Date: 2019-03-23'
The installation of the optional Sage package slabbe is a prerequisite. Remove the dash sign (#) and evaluate the following cell to install slabbe optional package.
The installation of graphviz and of the optional Sage package dot2tex is a prerequisite to construct the graphs at the end of the notebook.
Finally, I am using rise for the presentation and tikzmagic in Jupyter cells.
In [2]:
#!sage -pip install slabbe>=0.4.4
#!sage -pip install dot2tex
#!sage -pip install rise
#!sage -pip install git+git://github.com/robjstan/tikzmagic.git
## Choosing a solver¶
The time it takes to evaluate all cells of this notebook depend on the chosen available solver.
Solver Type Time to run this notebook
Gurobi MILP solver about 4 minutes
Glucose SAT solver about 30 minutes (estimated)
The above timing were computed on a 2019 computer with 8 available cpus.
Dancing links is an algorithm proposed by D. Knuth to solve the Exact Cover Problem and is available in plain Sage.
Gurobi is a MILP solver. It is a proprietary software, but it is free for researchers and students. See this tutorial to make Gurobi available through Sage.
Glucose is a SAT solver developped at LaBRI (Bordeaux). It can be installed in sage>=8.7.beta0 easily:
sage -i glucose
In [3]:
from sage.misc.package import is_package_installed
from sage.doctest.external import has_gurobi
if has_gurobi():
solver = 'gurobi'
elif is_package_installed('glucose'):
solver = 'glucose'
else:
solver = 'dancing_links' # more time efficient
print("We are using solver = '{}'".format(solver))
We are using solver = 'dancing_links'
# Setup¶
Colors setup:
In [4]:
from collections import defaultdict
color = defaultdict(lambda : 'white')
color.update({0:'white', 1:'red', 2:'cyan', 3:'green', 4:'lightgray'})
color.update({str(k):v for k,v in color.items()})
In [5]:
from slabbe import WangTileSet, Substitution2d
import tikzmagic
Latex macros:
## Definition of Jeandel-Rao Tiles $\mathcal{T}_0$¶
In [6]:
tiles = [(2,4,2,1), (2,2,2,0), (1,1,3,1), (1,2,3,2), (3,1,3,3), (0,1,3,1), (0,0,0,1), (3,1,0,2), (0,2,1,2), (1,2,1,4), (3,3,1,2)]
tiles = [map(str,t) for t in tiles]
T0 = WangTileSet(tiles)
T0.tikz(font=r'\small', size=1.2, ncolumns=11, color=color)
Out[6]:
Let $$\Omega_0=\Omega_{\T_0}=\{w:\Z\times\Z\to\T_0\mid w \text{ is a valid Wang tiling}\}$$ be the set of tilings made with the 11 Jeandel-Rao tiles.
In [7]:
%time solution = T0.solver(30,15).solve(solver='glucose')
CPU times: user 200 ms, sys: 16 ms, total: 216 ms
Wall time: 434 ms
In [8]:
solution.tikz(color=color)
Out[8]:
The substitutive structure of Jeandel-Rao tilings $\Omega_0$ looks like this:
In [9]:
%tikz -i subs-struct.tikz --no-wrap
Out[9]:
The current Jupyter notebook computes
• the Wang tile sets $\T_i$ for each $i\in\{1,\dots,12\}$
• the Wang shifts $\Omega_i=\Omega_{\T_i}=\{w:\Z\times\Z\to\T_i\mid w \text{ is a valid Wang tiling}\}$ for each $i\in\{1,\dots,12\}$
• the 2-dimensional morphisms $\omega_i:\Omega_{i+1}\to\Omega_i$ for each $i\in\{1,\dots,12\}\setminus\{4,5\}$
• the embedding $\pi:\Omega_5\to\Omega_4$
• the topological conjugacy $\eta:\Omega_6\to\Omega_5$
Algorithm.
INPUTS: $\T$ is a Wang tile set; $i\in\{1,2\}$ is a direction $e_i$; $r\in\N$ is some radius.
FindMarkers($\mathcal{T}$, $i$, $r$)
• $j\gets 3-i$
• $D_j \gets \left\{(u,v)\in\T^2\mid u\odot^jv\text{ admits a surrounding or radius } r\text{ in }\Omega_\T \right\}$
• $U \gets$UnionFindDataStructure$(\T)$
• For All $(u,v) \in D_j$
• Union$(u, v)$ (Merge $u$ and $v$ in the data structure $U$)
• EndFor
• $D_i \gets \left\{(u,v)\in\T^2\mid u\odot^iv \text{ admits a surrounding or radius } r \text{ in }\Omega_\T\right\}$
• Return ${S \in\textbf{Subsets}(U) \mid \left(S\times S\right) \cap D_i=\varnothing\}$
• The output contains zero, one or more subsets of markers in the direction $i$.
## Computing $\mathcal{T}_1$¶
In [10]:
T0.tikz(font=r'\small', ncolumns=11, size=1.2, color=color)
Out[10]:
In [11]:
%time T0.find_markers(i=2, radius=1, solver=solver) # 229ms with dancing_links, 32s with Glucose, 1.4s with Gurobi
CPU times: user 236 ms, sys: 4 ms, total: 240 ms
Wall time: 234 ms
Out[11]:
[[0, 1]]
In [12]:
M0 = [0,1]
T1,omega0 = T0.find_substitution(M=M0, i=2, side='left', solver=solver)
In [13]:
show(omega0)
In [14]:
T1.tikz(font=r'\small', size=1.2, ncolumns=13)
Out[14]:
## Computing $\mathcal{T}_2$¶
In [15]:
T1.tikz(font=r'\small', ncolumns=20, size=1.2)
Out[15]:
In [16]:
%time T1.find_markers(i=2, radius=1, solver=solver) #385ms with dancing_links, 46s with Glucose, 2s with Gurobi
CPU times: user 388 ms, sys: 0 ns, total: 388 ms
Wall time: 384 ms
Out[16]:
[[8, 9, 10, 11, 12]]
In [17]:
M1 = [8, 9, 10, 11, 12]
T2,omega1 = T1.find_substitution(M=M1, i=2, side='left', solver=solver)
In [18]:
show(omega1)
In [19]:
T2.tikz(font=r'\small', size=1.2, ncolumns=20)
Out[19]:
## Computing $\mathcal{T}_3$¶
In [20]:
T2.tikz(font=r'\small', size=1, ncolumns=20)
Out[20]:
In [21]:
%time T2.find_markers(i=2, radius=2, solver=solver) # 4s with dancing_links, 2min 23s with Glucose, 15s with Gurobi
CPU times: user 3.84 s, sys: 8 ms, total: 3.84 s
Wall time: 3.84 s
Out[21]:
[[8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]]
In [22]:
M2 = [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
T3,omega2 = T2.find_substitution(M=M2, i=2, side='left', radius=2, solver=solver)
In [23]:
show(omega2)
In [24]:
T3.tikz(font=r'\small', size=1, ncolumns=24)
Out[24]:
## Computing $\mathcal{T}_4'$¶
In [25]:
T3.tikz(font=r'\small', size=1, ncolumns=24)
Out[25]:
In [26]:
%time T3.find_markers(i=2, radius=3, solver=solver) # 13s with dancing_links, 4min 55s with Glucose, 59s with Gurobi
CPU times: user 13.7 s, sys: 24 ms, total: 13.7 s
Wall time: 13.7 s
Out[26]:
[[0, 1, 2, 3]]
In [27]:
M3 = [0, 1, 2, 3]
T4p,omega3p = T3.find_substitution(M=M3, i=2, side='right', radius=3, solver=solver)
In [28]:
show(omega3p)
In [29]:
T4p.tikz(font=r'\small', size=1.2, ncolumns=15)
Out[29]:
## Check that we can remove 2 tiles (tiles #24 and #28) from $\mathcal{T}_4'$¶
In [30]:
T4p.tikz(font=r'\small', size=1.2, ncolumns=15)
Out[30]:
We check that tiles #24 and #28 can be removed from T4p. Since tile #28 is always to the left of #24, it suffices to check that the tile #24 does not admit a large enough surrounding (here of width 71 and heigth 9).
Gurobi or Glucose can solve this in a reasonable amount of time (<6s):
In [31]:
%%time
assert T4p[24] == ('21103', '1', '23310', '0')
S = T4p.solver(width=71, height=9, preassigned_tiles={(35,4):24})
print(S.has_solution(solver='glucose')) # 6s with Gurobi, 4s with Glucose
False
CPU times: user 2.07 s, sys: 72 ms, total: 2.14 s
Wall time: 3.97 s
## Computing $\mathcal{T}_4$¶
In [32]:
forbidden = ('21103', '1', '23310', '0'), ('23310', '0', '21330', '0')
id_tiles = [(i,t) for (i,t) in enumerate(T4p) if t not in forbidden]
indices,tiles = zip(*id_tiles)
d = dict(enumerate(indices))
iota = Substitution2d.from_permutation(d)
T4 = WangTileSet(tiles)
In [33]:
omega3 = omega3p * iota
show(omega3)
In [34]:
T4.tikz(font=r'\small', size=1.2, ncolumns=15)
Out[34]:
## The morphism $\omega_0\omega_1\omega_2\omega_3:\Omega_{4}\to\Omega_{0}$¶
In [35]:
omega0to4 = omega0*omega1*omega2*omega3
show(omega0to4)
In [36]:
omega0to4.wang_tikz(T4, T0, font=r'\scriptsize', scale=.8, codomain_color=color, ncolumns=10, direction='down', extra_space=1.5)
Out[36]:
## Trying to compute $\T_5$ ...¶
In [37]:
T4.tikz(font=r'\small', size=1, ncolumns=15)
Out[37]:
In [38]:
%time T4.find_markers(i=1, radius=1, solver=solver)
CPU times: user 3.73 s, sys: 4 ms, total: 3.73 s
Wall time: 3.73 s
Out[38]:
[]
In [39]:
%time T4.find_markers(i=2, radius=1, solver=solver)
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 81.1 µs
Out[39]:
[]
We have a problem: the Wang tile set $\T_4$ has markers neither in the direction $e_1$ nor $e_2$
So increasing the surrounding radius will not help here.
Even worse, we have two problems...
## Defining $\mathcal{T}_5$¶
The first problem is the presence of horizontal fracture lines which we can break by adding decorations on the horizontal edges of few tiles. We have to do it in a way that we forbid sliding along fracture lines and without destroying the overall structure of Wang tilings in $\Omega_4$.
We create the tile set $\mathcal{T}_5$ by adding decorations on $\mathcal{T}_4$.
For few tiles, the horizontal color 0 is replaced by color 6.
For few tiles, the horizontal color 1 is replaced by color 5.
In [40]:
tiles5 = [('2113', '5', '2130', '1'),
('2130', '1', '2103', '5'),
('2133', '1', '2113', '1'),
('2113', '5', '2330', '0'),
('2130', '6', '2300', '0'),
('2103', '5', '2310', '0'),
('2310', '1', '2033', '6'),
('2300', '1', '2033', '6'),
('2300', '0', '2030', '6'),
('2030', '1', '2103', '0'),
('2033', '1', '2113', '0'),
('2330', '1', '2133', '6'),
('2330', '0', '2130', '6'),
('21113', '5', '21330', '1'),
('21130', '6', '21300', '1'),
('21103', '5', '21310', '1'),
('21310', '1', '21033', '5'),
('21310', '0', '21030', '5'),
('21300', '1', '21033', '5'),
('21300', '0', '21030', '5'),
('21030', '1', '21103', '1'),
('21033', '1', '21113', '1'),
('21330', '0', '21130', '1'),
('21330', '0', '21130', '5'),
('21130', '6', '23300', '0'),
('21030', '6', '23100', '0'),
('23100', '0', '20330', '6'),
('20330', '0', '21130', '0'),
('23300', '0', '21330', '6')]
T5 = WangTileSet(tiles5)
T5.tikz(font=r'\small', size=1.2, ncolumns=15)
Out[40]:
## The morphism $\pi:\mathcal{T}_5\to\mathcal{T}_4$¶
The morphism $\pi$ erases the decorations on tiles of $\mathcal{T}_5$.
It projects the horizontal color 6 back to color 0, and the horizontal color 5 back to color 1.
On tiles, the map $\pi:\mathcal{T}_5\to\mathcal{T}_4$ is not injective as it maps two tiles (#22 and #23) onto the same (#22).
In [41]:
def project_56(t, p56='10'):
L = []
d = dict(zip('56', p56))
for w in t:
for a in '56':
w = w.replace(a, d[a])
L.append(w)
return tuple(L)
In [42]:
T4_tiles_list = T4.tiles()
d = {}
for i,t in enumerate(T5):
pt = project_56(t, p56='10')
j = T4_tiles_list.index(pt)
d[i] = j
omega4_pi = Substitution2d.from_permutation(d)
show(omega4_pi)
But surprise, on tilings, the map $\pi:\Omega_5\to\Omega_4$ defines a $2$-dimensional morphism which is an embedding, i.e., injective.
## Trying to compute $\T_6$ from $\T_5$ ...¶
In [43]:
T5.tikz(font=r'\small', size=1, ncolumns=15)
Out[43]:
In [44]:
%time T5.find_markers(i=1, radius=1, solver=solver)
CPU times: user 4.36 s, sys: 4 ms, total: 4.36 s
Wall time: 4.36 s
Out[44]:
[]
In [45]:
%time T5.find_markers(i=2, radius=1, solver=solver)
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 135 µs
Out[45]:
[]
We still have a problem: the Wang tile set $\T_5$ has markers neither in the direction $e_1$ nor $e_2$
So increasing the surrounding radius will not help here.
## Tilings in $\Omega_5$¶
Let's look at tilings in $\Omega_5$.
In [46]:
tiling = T5.solver(30,10).solve(solver='glucose')
In [47]:
P = tiling.tile_positions([1, 6, 7, 8, 11, 12, 16, 17, 18, 19, 23, 26, 28])
s = '\n'.join([r'\fill[yellow] {} rectangle {};'.format((a,b), (a+1,b+1)) for (a,b) in P])
tiling.tikz(extra_before=s)
Out[47]:
## Computing $\mathcal{T}_6$ and the shear topological conjugacy $\eta:\Omega_6\to\Omega_5$¶
The second problem is that $\T_5$ does not have markers, it has diagonal markers, i.e., markers appears on lines of slope 1.
To fix this, instead of introducing the notion of diagonal markers, we choose to transform the Wang tile set in a way that it shears the tiling by the matrix $\left(\begin{smallmatrix} 1 & -1 \\ 0 & 1\end{smallmatrix}\right)$ so that markers become vertical.
In [48]:
%tikz -i shear.tikz -p amssymb --no-wrap
Out[48]:
This transformation on tiles yields a homeomorphism $\eta:\Omega_6\to\Omega_5$ which commutes the normal shift on $\Omega_5$ into the sheared shift on $\Omega_6$.
In [49]:
%time T6,omega5_sheer = T5.shear(radius=2, solver=solver) # 6s with dancing_links, 3min 12s with Glucose, 22s with Gurobi
CPU times: user 6.08 s, sys: 8 ms, total: 6.08 s
Wall time: 6.08 s
In [50]:
show(omega5_sheer)
In [51]:
T6.tikz(font=r'\small', size=1.2, ncolumns=15)
Out[51]:
## Computing $\mathcal{T}_7$¶
In [52]:
T6.tikz(font=r'\small', size=1.2, ncolumns=15)
Out[52]:
In [53]:
%time T6.find_markers(i=1, radius=1, solver=solver) # 5s with dancing_links, 4min 25s with Glucose, 17s with Gurobi
CPU times: user 4.39 s, sys: 4 ms, total: 4.39 s
Wall time: 4.39 s
Out[53]:
[[1, 6, 7, 8, 11, 12, 16, 17, 18, 19, 23, 26, 28],
[0, 3, 4, 5, 13, 14, 15, 24, 25],
[2, 9, 10, 20, 21, 22, 27]]
Youpi! The tile set $\T_6$ has 3 subsets of markers in the direction $e_1$. Thus, we may chose one, desubstitute and compute $\T_7$.
In [54]:
M6 = [1, 6, 7, 8, 11, 12, 16, 17, 18, 19, 23, 26, 28]
T7,omega6 = T6.find_substitution(M=M6, i=1, radius=1, side='left', solver=solver)
In [55]:
show(omega4_pi*omega5_sheer*omega6)
In [56]:
T7.tikz(font=r'\small', size=1.2)
Out[56]:
## Computing $\mathcal{T}_8$¶
In [57]:
T7.tikz(font=r'\small', size=1.2)
Out[57]:
In [58]:
%time T7.find_markers(i=1, radius=1, solver=solver) # 2s with dancing_links, 2min with Glucose, 6s with Gurobi
CPU times: user 1.43 s, sys: 0 ns, total: 1.43 s
Wall time: 1.43 s
Out[58]:
[[0, 1, 2, 3, 4, 5, 6]]
In [59]:
M7 = [0, 1, 2, 3, 4, 5, 6]
T8,omega7 = T7.find_substitution(M=M7, i=1, radius=1, solver=solver)
In [60]:
show(omega7)
In [61]:
T8.tikz(font=r'\small', size=1.2)
Out[61]:
## Computing $\mathcal{T}_9$¶
In [62]:
T8.tikz(font=r'\small', size=1.2)
Out[62]:
|
# Prime partners exist: progress toward the twin prime problem
#### David W. Farmer
Number theorists view the integers as a largely unexplored landscape populated by interesting and friendly inhabitants. The prime numbers: 2, 3, 5, 7, 11, 13,...; occur like irregular mileposts stretching off to infinity. The squares: 1, 4, 9, 16, 25,...; and the cubes: 1, 8, 27, 64, 125,...; along with the higher powers are more sparse, but they also demand attention because of their special form.
These intriguing features exist because of the interaction between addition and multiplication. If we knew only about addition then the integers would be quite boring: start with 0, then repeatedly add or subtract 1 to get everything else. End of story.
Things become interesting when you add multiplication to the mix. To construct every integer from addition, all you need is 1. For multiplication, the prime numbers are the building blocks and there are infinitely many of them. Examining the list of prime numbers reveals both structure and apparent randomness. Fairly often (8 times before 100, and 35 times before 1000) primes occur as "twins," meaning that both $p$ and $p+2$ are prime. Examples are 101 and 103, or 8675309 and 8675311. Somewhat less often, a prime is of the form $n^2+1$, such as $17=4^2+1$ or $90001 = 300^2+1$.
It is an unsolved problem to determine whether there are infinitely many primes of the form $p+2$ or $n^2+1$, where we use $p$ to represent a prime number and $n$ to represent a positive integer (which may or may not be prime). Other special forms which also are unresolved are Mersenne primes ($2^p-1$) and Sophie Germain primes ($2p+1$). Such problems have been the subject of thousands of research papers over the past several centuries. And while none have been solved (nor are they likely to be solved in the near future) a large number of tools have been developed which have led to partial results. Recently Yitang Zhang of the University of New Hampshire has combined several of those tools to make impressive progress on the twin prime problem. Below we discuss some variations of these difficult problems and then describe Zhang's result.
### 1. Relaxing the conditions: measuring progress on difficult problems
How can one measure progress toward an unsolved problem? One view when dealing with primes of a special form is to make the form less restrictive. For example, instead of looking for primes of the form $n^2+1$, one could ask for primes of the form $n^2+a$, where $a$ is as small as possible. Or one could look for primes of the form $n^2+m^2$, or more ambitiously $n^2+m^4$.
These problems vary widely in difficulty. Fermat proved that there are infinitely may primes of the form $p=n^2+m^2$, in fact those are exactly the primes $p\equiv 1 \bmod 4$. In 1997 John Friedlander and Henryk Iwaniec proved that $n^2+m^4$ is prime infinitely often, using sieve techniques originally developed by Enrico Bombieri. A measure of the difficulty of that result is that only $X^{\frac34}$ of the numbers less than $X$ have that form. Shortly after, Roger Heath-Brown proved that the even thinner set $n^3+2m^3$ is prime infinitely often. One might have suspected that those results would be more difficult than the twin prime problem, because, by the prime number theorem, there are $X/\log(X)$ numbers of the form $p+2$ less than $X$, and the naive guess would be that having more candidates would make the problem easier.
One can generalize the twin prime problem to ask whether $p$ and $p+4$, or $p+6$, or $p+8$, etc., are prime. This is sometimes called the "prime neighbors" problem: show that there is some fixed number $2n$ so that $p$ and $p+2n$ are prime infinitely often. That is exactly the problem which Zhang recently solved. We describe some of the heuristics behind twin primes and related problems, and then discuss the techniques behind Zhang's proof.
### 2. Prime heuristics: the frequency to special primes
Why should we expect to have infinitely many twin primes?
The prime number theorem, proven in 1896 by Haramard and de la Valée-Poussin(sp), says that the number of primes up to $X$ is asymptotically $X/\log(X)$. Equivalently, the $n$th prime is asymptotic to $n\log(n)$. Here, and throughout number theory, "$\log$" means natural logarithm.
By the prime number theorem, among numbers of size $n$ the primes have density $1/\log(n)$. This suggests what is known as the "Cramér model" of the primes: the probability that the large number $n$ is prime is $1/\log(n)$, and distinct numbers have independent probabilities of being prime. Thus, the Cramér model says that the probability that $p$ and $p+2$ are both prime is $1/\log(p) \cdot 1/\log(p+2) \approx 1/\log^2(p)$, so up to $X$ the Cramér model predicts approximately $X/\log^2(X)$ twin primes. In particular, there should be infinitely many of them.
Of course that argument has to be wrong, because it predicts exactly the same answer for $p$ and $p+1$ being prime -- but that almost never happens because if $p$ is odd then $p+1$ is even. Hardy and Littlewood, using their famous circle method, developed precise conjectures for the number of twin primes as well as a wide variety of other interesting sets of primes. The result is that (in most cases) the Cramér model predicts the correct order of magnitude, but the constant is wrong. Using the twin prime case as an example, the idea is that the likelihood of $p+2$ being prime has to be adjusted based on the fact that $p$ is prime. If $p$ is odd then $p+2$ also is odd, so there is a $\frac12$ chance that $p$ and $p+2$ are both odd, which is twice as likely as the Cramér model would predict. Similarly, we need neither $p$ nor $p+2$ to be a multiple of 3, which happens $\frac13$ of the time -- exactly when $p\equiv 1\bmod 3$. The Cramér model would have that happen $(\frac{2}{3})^2$ of the time, so we must adjust the probability to make it slightly less likely that $p$ and $p+2$ are prime. Making a similar adjustment for the primes 5, 7, 11, etc., we arrive at the Hardy-Littlewood twin prime conjecture: $$\#\{p < X \ :\ p \ \mathrm{ and } \ p+2 \ \text{ are prime } \} \sim 2 \prod_{p>2} \left( 1- \frac{2}{p}\right) \left(1- \frac{1}{p}\right)^{-2} \cdot \frac{X}{\log^2(X)} .$$ The twin primes have been tabulated up to $10^{16}$ and so far the Hardy-Littlewood prediction has been found to be very accurate.
Other prime neighbors, $p$ and $p+2n$, can actually occur more frequently than the twin primes. For example, if $p$ is a large prime then $p+6$ is not divisible by 2 or 3, so it is more likely than $p+2$ to be prime. Specifically, in the case of $p$ and $p+2$, we must have $p\equiv 2 \bmod 3$, while in the case of $p$ and $p+6$ we can have $p\equiv 1 \text{ or } 2 \bmod 3$. Thus, a prime gap of 6 is twice as likely as a prime gap of 2. The same argument shows that a prime gap of 30 is $\frac83$ times as likely as a prime gap of 2. The Hardy-Littlewood method also makes predictions for the frequency that $n^2+1$, or other polynomials, represent primes. All those conjectures are also supported by numerical observations.
Precise predictions supported by data are nice, but it is better to have an actual proof. The methods for proving results about primes of a special form involve the ancient idea of a prime sieve. The appendix has information about the archtypal sieve of Eratosthenes.
### 3. Sieving for primes
For twin primes, the first nontrivial result was due to Viggo Brun in 1915, using what is now known as the Brun sieve. Brun proved that there are not too many twin primes. In particular, Brun proved that $$\sum_{p, p+2, \text{ prime}} \frac{1}{p} < \infty.$$ This is consistent with the Hardy-Littlewood conjecture because $\sum 1/n\log^2(n)$ converges. Since $\sum_p 1/p$ diverges, which was proven by Euler and also follows from the prime number theorem, we see that most prime numbers are not twin primes. This is a negative result, in terms of our goal of showing that there are infinitely many twin primes.
To state things positively, we wish to find a parameter $A$ such that for infinitely many $p$ there exists $a\le A$ with $p+a$ prime. By the prime number theorem, the average distance between a prime $p$ and the next prime is $\log(p)$, so we can take $A=\log(p)$. It is nontrivial to show $A=C\log(p)$ for some fixed number $C< 1$. The existence of such a number $C$ was first proven by Hardy and Littlewood in 1927, assuming the Generalized Riemann Hypothesis. An unconditional proof was given by Erdős in 1940, using the Brun sieve. Erdős' result did not provide a specific value for $C$; this was remedied by Ricci in 1954, who found $C\le \frac{15}{16}$. Bombieri and Davenport modified the Hardy-Littlewood method to make it unconditional (and better!), obtaining $C\le \frac12$. Various subsequent improvements in the years up to 1988 led to $C\le 0.2484...$.
The next big development, which is one of the main ingredients in Zhang's recent work, is due to Goldston, Pintz, and Yıldırım, in 2005. For the rest of this article we will refer to the method of those three authors as "GPY." It is GPY, combined with some techniques from the 1980's, which made Zhang's result possible. All the GPY papers are in the AIM preprint series, available at {\tt http://aimath.org/preprints.html}.
The GPY method uses a variant of the Selberg sieve to prove that $A=C \log(p)$ for any $A>0$, and later improved this to $$\phantom{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A=\sqrt{\mathstrut \log(p)} (\log\log(p))^2. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ The method also shows that if the primes have "level of distribution $\vartheta>\frac12$," the precise meaning of which is described below, then one could find an absolute bound for $A$, independent of $p$. The GPY method was the topic of the November, 2005, AIM workshop Gaps between primes. One highlight of the workshop was Soundararajan presenting a proof that $\vartheta=\frac12$ is the true obstruction for the GPY method to produce bounded gaps between primes. An intetresting sidelight is that the workshop was the first occasion that Goldston, Pintz, and Yıldırım were all in the same place at the same time.
### 4. Primes in progressions: a tool for optimizing the sieve
Key to the GPY method is knowledge about the distributions of primes in arithmetic progressions. If $a$ and $q$ are integers with no common factor, then Dirichlet proved that there are infinitely many primes $p\equiv a \bmod q$. For example, 4 and 15 have no common factor, so the sequence $$4, 19, 34, 49, 64, 79, ...$$ contains infinitely many primes. In 1899, de la Valée-Poussin went further, showing that all such progressions contain the expected number of primes. Here "expected" means that each of the progressions $\bmod\ q$ that can contain infinitely many primes, asymptotically contains the same number of primes. To continue our previous example of progressions $\bmod 15$, there are 8 possible choices for the number $a$: 1, 2, 4, 7, 8, 11, 13, and 14. Every large prime has one of those numbers as its remainder when divided by 15, and all 8 remainders occur equally often. The number of possible remainders $\bmod q$ is denoted by the Euler phi-function, $\varphi(q)$, so $\varphi(15)=8$. The general case is that if $a$ and $q$ have no common factors, then $$\sum_{\ontop{p< x}{p\equiv a \bmod q}} \log(p) \ \ \ \ \ \ \ \text{ is asymptotic to } \ \ \ \ \ \ \ \frac{x}{\varphi(q)}.$$ de la Valée-Poussin actually did a bit better, providing a main term and an explicit error term.
### 5. Level of distribution: how well we can control the primes
A shortcoming of the above results is that they only hold for fixed $q$ as $x\to\infty$. For the GPY mythod, and the subsequent work of Zhang, one requires $q$ to grow with $x$, and allowing $q$ to grow faster gives better results. To describe the issue more precisely, we introduce the "error" term $$E(x; q, a) = \left|\sum_{\ontop{p< x}{p\equiv a \bmod q}} \log(p) - \frac{x}{\varphi(q)}\right|.$$ For applications, we would like $E(x; q, a)$ to be small when $q=x^\alpha$ for $\alpha$ as large as possible. Unfortunately, such a bound is not known for any $\alpha>0$. The Generalized Riemann Hypothesis (GRH) implies the inequality $E(x; q, a)< (q x)^{\frac12+\varepsilon}$ as $x\to\infty$, for any $\varepsilon>0$. Thus, assuming GRH we have that $E(x; q, a)$ is small when $q=x^\alpha$ with $\alpha< \frac12$.
In practice it is sufficient to have $E(x; q, a)$ to be small on average. Specifically, one seeks the largest value of $\vartheta$ such that $$\sum_{q< x^\vartheta}\max_{(a,q)=1} E(x; q, a) \ll \frac{x}{\log^A(x)} \ \ \ \ \ \ \ \ \ \ (2)$$ for any $A>0$. The consequence of GRH mentioned above implies this inequality for $\vartheta< \frac12$. Bombieri and Vinogradov, independently in the 1960's, established the inequality for $\vartheta< \frac12$ unconditionally. Note that this does not prove GRH, but in some sense it is like "GRH on average".
If equation (1) holds for all $\vartheta< \vartheta_0$ then we say "the primes have level of distribution $\vartheta_0$." Thus, the Bombieri-Vinogradov theorem says that the primes have level of distribution $\frac12$. This is what was required for the GPY bound for $A$ in equation (1). GPY also show that if the primes have level of distribution $\vartheta_0>\frac12$, then there is a bound for $A$ independent of $p$. Zhang did not quite prove that (2) holds for some $\vartheta_0>\frac12$, but he proved a similar-looking result, using techniques of Bombieri, Friedlander, and Iwaniec from the 1980's.
### 6. Modifying and combining
Zhang modified the GPY method by restricting the Selberg sieve to only use numbers having no large prime factors. Such number are called "smooth." Restricting to smooth numbers made the sieve less effective, but it turned out that the loss was quite small and the added flexibility was a great benefit. This flexibility enabled Zhang to use techniques developed by Bombieri, Friedlander, and Iwaniec (who we will refer to as BFI for the remainder of this article) who in a series of three papers in the late 1980's almost managed to go beyond level distribution $\frac12$. The "almost" refers to the fact that while they did go beyond $\frac12$ (in fact they went up to $\frac47$) the expressions they consider were not exactly the same as those in equation (2). Fortunately, the BFI methods apply in Zhang's case, and he was able to obtain a level of distribution $\frac12+\frac1{584}$. That is barely larger than $\frac12$, but in this game anything positive is a win.
The BFI results were well-known to the experts, and combining the GPY and BFI methods to obtain bounded gaps was an obvious thing to try in the 8 years since the GPY result. Zhang's insight was to modify the sieve in a way that allowed the use of the BFI techniques.
### 7. What bound?
Zhang's original paper gives the bound $A< 70,000,000$, a number which he made no effort to optimize, and which has since been improved significantly. To understand where that number comes from, we look a little more closely at how the GPY method use the Selberg sieve.
Some patterns cannot appear in the primes. For example, one of the three numbers $p, p+2, p+4$ has to be a multiple of 3, and so there cannot be infinitely many "triplets" of primes of that form. On the other hand, there is no obvious reason why the numbers $p, p+2, p+6$ cannot all be prime, so one conjectures that there are infinitely many triples of primes of that form. Patterns which allow the possibility of infinitely many primes are called admissible. All admissible tuples are conjectured to be prime infinitely often, and the Hardy-Littlewood circle method makes a precise conjecture for how often this should occur.
The GPY method uses the Selberg sieve to count how many prime factors occur in an admissible tuple. Suppose a tuple with 100 terms is shown to have at most 198 prime factors among the numbers in tuple. Then we could conclude that the tuple contains at least 2 prime numbers, because each number has at least one prime factor, and there are only 198 prime factors to spread among the 100 numbers. That is what Zhang proved, except that he showed that infinitely often an admissible 3,500,000-tuple has at most 6,999,998 prime factors.
So where does the number 70,000,000 come from? If we can find an admissible 3,500,000-tuple, $p, p+2, p+6, ..., p+N$, then we have shown that there exists infinitely many prime partners which differ by at most $N$. Zhang notes that there is an admissible 3,500,000-tuple with $N=70,000,000$, but that is not optimal. The best known value is $N=57,554,086$ -- but that is probably not best possible.
Since the appearance of Zhang's paper there have been numerous improvements, coming from larger values in the level distribution, better ways of estimating various error terms and finding better admissible tuples. As of this writing, the best confirmed improvement is a prime gap of at most 5,414, and a not-yet-confirmed improvement to 4,680; the latter improvement coming from a level distribution of $\frac12+\frac{7}{600}$. At the rate things are going, there will probably be a better result by the time you read this.
### Appendix: The archetypal sieve
The familiar sieve of Eratosthenes is an efficient way to generate the prime numbers up to a pre-selected bound. Start with the numbers from 1 to $N$. First cross out the multiples of 2. At each stage circle the smallest number which has not been crossed out or circled, and then cross out all multiples of that number. Stop when the next number that remains is larger than $\sqrt{N}$. The numbers which have not been crossed out are the prime numbers up to $N$.
Using the sieve of Eratosthenes to find all the primes up to $N$ only requires keeping track of $N$ bits of information, and uses around $N \log(N)$ operations. To find the primes up to 1 million, it is faster to run the sieve of Eratosthenes on a computer than to read those primes from the hard drive.
The sieve of Eratosthenes is efficient as a practical method for generating the primes, but it is relatively useless as a theoretical tool. Suppose one wanted to estimate the number of primes up to $N$. The primes are those numbers which are not crossed out by the sieve, all we need to do is count how many numbers are crossed out, and subtract that from $N$. Since some numbers are crossed out multiple times, we have to use the inclusion-exclusion principle: \begin{align} \text{ number of non-primes up to } N =\mathstrut & \text{number crossed out once} \cr &-\text{number crossed out twice}\cr &+\text{number crossed out three times}\cr &- \text{etc.}\nonumber \end{align} Unfortunately, estimating the above quantities finds that the error term is larger than $N$, leaving one with no information at all. Thus, one must use a more sophisticated sieve to prove results about the primes.
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Browse Questions
If the line $y=2x+c,$ is a tangent to the circle $x^2+y^2=5,$ then a value of c is
$\begin {array} {1 1} (a)\;2 & \quad (b)\;3 \\ (c)\;4 & \quad (d)\;5 \end {array}$
(d) 5
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## Code
### LaTeX
#### Things I didn't think were obvious
• When using gnuplot and in a "" string you need to do \\ rather than just \ to get a LaTeX command such as a greek letter. e.g. "Resistance $R$ [\\Omega]"(yes this is obvious when you think about it)
• gnuplot not so FAQ v.useful
• With gnuplot: ylabel being printed twice when using the latex terminal: reason having used load "file.gnuplot" many times with many changes to the file the gnuplot session had become dirty: fix restart gnuplot
• rotated text in gnuplot when using the latex terminal wonderful hack: add \usepackage{rotating} %to allow rotating text to your preamble and then in the gnuplot text you want to rotate something like: set ylabel "\\begin\{sideways\}Time $T_p, T_d$ [s]\\end\{sideways\}"
Alternatively and slightly more neatly and powerfully \usepackage{graphicx} in the preable and then something like set ylabel "\\rotatebox\{90\}\{Time $T_p, T_d$ [s]\}" the advantage of this being that it is a) shorter and b) more powerful as the 90 could be any other angle.
• gnuplot \n set terminal latex when using gnuplot and the latex driver to plot data points the default ones are not all suitable as they don't accuratly show the position of the points. to get around this use the pointtype style modifier. The available latex point types are:
1. diamond
2. + (suitable)
3. square
4. x (suitable)
5. triangle
6. star (suitable)
7. small circle (suitable)
8. medium circle
9. big circle
10. small filled circle (suitable)
11. medium filled circle
12. big filled circle
and so one line using this from the graph I have been working on: plot 'E7data.dat' index 0 using 1:5:($1*0.03):(abs(($2 + 0.001)/(($3 - 0.001)*(508 - 0.1))-($2 - 0.001)/(($3 + 0.001)*(508 + 0.1)))) with xyerrorbars pointtype 2 title "$R_0 = 508 \\pm 0.1 \\Omega\$" nice and crazy :-) (I have two later lines that use pointtype 4 and 6 respectively)
• Number of bug reports with patches I need to write and file for the gnuplot latex driver: 3
More things which weren't obvious and cause me pain
### Christmas cards
My 2008 Christmas card is entirely my own work and released under cc-by,sa
My 2009 Christmas card is a derivative work using work by TheTruthAbout... and Tommy Ironic specifically this christmas tree and this white desert consequently it is licenced under cc-by,sa,nc. Last year's card was better but I was stuck for a cool idea and went for a message instead.
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High precision load cell amplifier has a 6 digit LED display, output signal 4-20mA/ 0-5V/ 0-10V/ RS485. The load cell input voltage (excitation) is the voltage sent to the input terminals of the load cell. 0 ) More points may be needed for the calibration curve. {\textstyle V_{o}} A load cell is a device that is used to measure weight or force. Users can choose a load cell with a range of hundred of kilonewtons and use it for measuring few newtons of force with the same signal-to-noise ratio; again this is possible only with the use of a charge amplifier for conditioning. With 2-mV/V sensitivity and 5-V excitation, the full-scale output voltage is 10 mV. Operating Temperature Range: Temperature range extremes in which a load cell can operate without permanent, adverse effects on any of its performance characteristics. A typical load cell’s electrical sensitivity, defined as the ratio of the full-load output to the excitation voltage, is 2 mV/V. Compensated Temperate Range: The temperature range over which a load cell is compensated so that it can ensure zero balance & rated output within specified limits. Assuming a 10 volts excitation on a 3mV/V output load cell, the maximum signal for a 1% shift in zero balance is 0.3 millivolts. R A lower excitation voltage than the manufacturer’s recommended value is usually acceptable. {\textstyle V_{o}} = A load cell (or loadcell) is a transducer which converts force into a measurable electrical output. Maintaining these values is critical to ensure the best output results. V Common specifications include:[14]. Strain gauge load cells are the most popular due to their high accuracy, low price point, and general ease of use. (11kg) to 40,000 lbs. Load cells are commonly used to measure weight in an industrial environment. ( High Accuracy Load Cell Display And Controller 400Hz Sampling Frequency . Material: Substance that comprises the spring element of the load cell. This elasticity gives rise to the term "spring element", referring to the body of the load cell. They typically survive up to 1 million cycles depending on the load level and transducer material. One or more load cells can be used for sensing a single load. Recommended Excitation: Maximum recommended excitation voltage of the transducer for it to operate within its specifications. The change in One end of a load The most common types of load cell used are strain gauges, pneumatic, and hydraulic, pneumatic. Details . The transducers attached to the machine frequently register this vibration superimposed on the true signal and thus may mask the signal required for analysis [1-4]. A load cell is designed to follow a linear relationship between the output voltage and the load applied. Wrong mounting may result in the cell reporting forces along undesired axis, which still may somewhat correlate to the sensed load, confusing the technician. Natural frequencies range widely from several 10Hz to several Hz. Technical Specification. V Using a charge amplifier with a long time constant allows accurate measurement lasting many minutes for small loads up to many hours for large loads. Typical sensitivity values are 1 to 3 mV/V. {\displaystyle {\frac {R1}{R2}}={\frac {R4}{R3}}} However, it remains to be shown whether high-frequency GrC bursts can also be demonstrated in awake behaving animals (but see Hensbroek et al., 2005, 2006), and if so, to what extent high-frequency information can be conveyed onto PCs. The maximum and recommended excitation voltage is often provided by the manufacturer. OMIL R76: 2006International Recommendation. A typical load cell’s electrical sensitivity, defined as the ratio of the full-load output to the excitation voltage, is 2 mV/V. The R Accuracy Grade: III. Input Resistance: Input resistance of the load cell's bridge circuit. The wire or foil in the strain gauge is arranged in a way that, when force is applied in one direction, a linear change in resistance results. Over time, load cells will drift, age and misalign; therefore, they will need to be calibrated regularly to ensure accurate results are maintained. KM500 . This paper reports on the design, construction and characteristics of a novel load cell having stiffness and potential overload capability some 10 times that of a resistive strain gauge load cell. 3.1. Expressed as °F or °C. However, for best results, stick to the recommended value. [12] From 1993 the British Antarctic Survey installed load cells in glass fibre nests to weigh albatross chicks. 4.4.2 Connect the load cell to the data acquisition system and the master load cell to the indicator. Except for certain laboratories where precision mechanical balances are still used, strain gauge load cells dominate the weighing industry. It is important to select a load cell with appropriate frequency response in situations where the weight is either applied or changes at a rapid rate. will likewise change. So a 2.96 mV/V load cell will provide 29.6 millivolt signal at full load when excited with 10 volts. The resulting output must be less than the manufacturer’s specification. Similarly, a load cell with higher accuracy and lower resolution mean that you will not be able to record incremental changes in the weight. The DDE's small size and reduced weight, along with its low deflection results in a high stiffness assembly, makes this load cell suitable for rapid measurement changes and high frequency transients. In reality, due to several environmental and loading factors, the output will deviate slightly and is represented through a non-linearity curve. 8.3. When force is exerted on the load cell, the spring element is slightly deformed, and unless overloaded, always returns to its original shape. [5] Typical hydraulic load cell applications include tank, bin, and hopper weighing. An oscillating data pattern can be the result of ringing. Tension link load cell: often used in crane and hoist weighing systems. Expressed in VDC. They can be installed on hoppers, reactors, etc., to control their weight capacity, which is often of critical importance for an industrial process. } Non-Linearity: The maximum deviation of the calibration curve from a straight line drawn between the rated capacity and zero load. It is also possible to linearize the output using compensation circuits and microprocessors. The load cell is completely filled with oil. S-type load cell: S-shaped spring element; can be used in both compression and tension. This pressure is then transmitted to a hydraulic pressure gauge via a high pressure hose. Backed by decades of rich experience, we provide you with a total solution for all your testing, measuring and quality-control systems, based on the latest technologies. All the load force has to go through the part of the load cell where its deformation is sensed. , measured in amperes) running through a conductor between two points is directly proportional to the voltage Ohm's law is expressed in the equation In order to measure the loads, they have to deform. Among those design characteristics are: The electrical, physical, and environmental specifications of a load cell help to determine which applications it is appropriate for. Air pressure is applied to one end of the diaphragm and it escapes through the nozzle placed at the bottom of the load cell. View Product. Frequency response is often designated as “bandwidth” in load cell specifications. Compact and robust load cell with M10 thread up to 20 kN in IP65 with high overload feature. Electrical damage: the load cells can be damaged by induced or conducted current. OMEGA's DLC101 dynamic load cell is ideal for measuring force transients or dynamic force pulsations in impact or vibration applications. Sometimes a six-wire configuration is used. No matter if you need to sound test and generate high frequency sounds or low frequency sounds, our frequency tone generator is your #1 best solution. Transducer Techniques is an industry leader in designing and manufacturing high quality load cells, torque sensors, and related instrumentation for load, force, and torque measurement applications. Load cells with high sensitivity can measure even the slightest change in the force. Technical Specification. Strain gauge load cells are the kind most often found in industrial settings. Details . [9] Other values are also possible, e.g. In both cases the signal develops offset (unless all wires are affected equally) and accuracy is lost. With 2-mV/V sensitivity and 5-V excitation, the full-scale output voltage is 10 mV. If subjected to loads above its maximum rating, the material of the load cell may. o Expressed as °F or °C. R The AD8221 is well-suited for this application because its high CMRR over frequency ensures that the signal of interest, which appears as a small difference voltage riding on a large sinusoidal common-mode voltage, is gained and the common-mode signal is rejected. A load cell whose natural frequency is low will be significantly deformed by loading and is therefore more suited for instruments with stoppers, such as platform scales. We supply world-class pressure transducers, pressure sensors, pressure transmitters as well as load cells and displacement sensors. Often, in order to use the most linear portion of the load cell’s span, only about two-thirds of this range would be used. more 1600 SC500 ... Piezoelectric transducers are more durable than other load cells, and have a high frequency response. R Applied Measurements Load Cells in World’s First Regenerative Fatigue Test Facility Developed by the University of Edinburgh, FASTBLADE is designed for high-quality, low-cost fatigue testing of tidal turbine blades, composite bridge sections and carbon fibre aircraft wing boxes and is the world’s first test facility that uses regenerative hydraulic technology. Executing Standard. temperature fluctuations. Tension force stretches a strain gauge, causing it to get thinner and longer, resulting in an increase in resistance. When the load cell is not installed properly or if it is in an incorrect orientation, it would produce incorrect readings. A strain gauge is constructed of very fine wire, or foil, set up in a grid pattern and attached to a flexible backing. Model 53 Low Cost Load Cell, ±0.6% full scale accuracy, Range of 5 lb to 50,000 lb, Stainless Steel Diaphragm Design, Miniature footprint. It is also the difference between the actual output and the theoretical output of the load cell. Technical Specification. The DDE's small size and reduced weight, along with its low deflection results in a high stiffness assembly, makes this load cell suitable for rapid measurement changes and high frequency transients. Measured at the positive & negative excitation leads with no load applied. When the shape of the strain gauge is altered, a change in its electrical resistance occurs. V An excitation voltage, greater than the maximum rated value, will increase the current flow and heat the strain gauge. With the multi-point type, three or four load cells are generally used to make a scale. One end of a load 4.4.3 Configure the indicator for use per the requirements of the calibration data sheet. T20. 0.5tf – 9.5tf. 4.4.3 Configure the indicator for use per the requirements of the calibration data sheet. − 2 [4] The gauge's Bourdon tube senses the pressure and registers it on the dial. V The first measurement can be obtained by decreasing the load from rated output and the second by increasing the load from zero. o Digital Load Cells Fast, precise, reliable, cost-effective: Our digital load cells are ideally suited for packaging, filling and sorting machinery as well as checkweighers. [15] ISO9000 and most other standards specify a maximum period of around 18 months to 2 years between re-calibration procedures, dependent on the level of load cell deterioration. ASTM E74 – Practice for Calibration of Force Measuring Instruments for Verifying the Force Indication of Testing Machines, NTEP – National Conference on Weights and Measures (Certificate of Conformance), This page was last edited on 2 January 2021, at 19:13. Verification Accuracy: 0.03%. {\displaystyle V_{o}=\left({\frac {R3}{R3+R4}}-{\frac {R2}{R1+R2}}\right)V_{\text{EX}}}, In a load cell, the resistors are replaced with strain gauges and arranged in alternating tension and compression formation. Tension link load cell with built-in backlit display. For example, if a load cell has a fatigue life of 100,000 times, the devices can be loaded 100,000 times. The experimental cell loaded up to 600 kg gives a direct output frequency change of 870 Hz (safety factor of 5). Standard tension link load cell. Often, in order to use the most linear portion of the load cell’s span, only about two-thirds of this range would be used. PCB’s quartz, piezoelectric force and strain sensors are durable measurement devices which possess exceptional characteristics for the measurement of high frequency dynamic force and strain events. Safe Overload: The maximum load that can be applied to a load cell without causing permanent effects to its performance specifications. Structurally, a load cell has a metal body to which strain gauges have been secured. The cell output is rated in millivolts per volt (mV/V) of the difference voltage at full rated mechanical load. across the two points. A load cell is a sensor or a transducer that converts a load or force acting on it into an electronic signal. Load cells are an integral part of most weighing systems in industrial, aerospace and automotive industries, enduring rigorous daily use. When shopping for load cells, you are typically provided with two pieces of information about the load cell, excitation voltage (V orVDC), and output (mV). Combined Error: percent of the full scale output that represents the maximum deviation from the straight line drawn between no load and load at rated capacity. The most common color assignment is red for Ex+, black for Ex−, green for S+, and white for S−. Accuracy Grade: III. The trend in weigh scales towards higher accuracy and lower cost has produced an increased demand for high-performance analog signal processing at low cost. Vertical cylinders can be measured at three points, rectangular objects usually require four sensors. Applied Measurements Load Cells in World’s First Regenerative Fatigue Test Facility Developed by the University of Edinburgh, FASTBLADE is designed for high-quality, low-cost fatigue testing of tidal turbine blades, composite bridge sections and carbon fibre aircraft wing boxes and is the world’s first test facility that uses regenerative hydraulic technology. View All Load Cell Sensors and Force Sensors. I am trying to instrument hiking poles with load cells and a data logger (to be semi wireless) for a research project. more 1600 SC500 Also, the larger the rated load, the lower the natural frequency becomes. High accuracy low profile shear beam load cell. V [1], There are several types of strain gauge load cells:[2]. It is a more costly technology and thus cannot effectively compete on a cost of purchase basis.[7]. The body is usually made of aluminum, alloy steel, or stainless steel which makes it very sturdy but also minimally elastic. Measured in percent of rated output. High accuracy low profile shear beam load cell. KDL series is a precision load cell which have been developed for a single crystal growing systems. As a rule of thumb, a load cell with a 10 times higher natural frequency than the highest frequency to be measured is generally selected. R {\textstyle I} Model 11 (tension/compression) subminiature load cell is designed to measure load ranges from 150 g to 1,000 lb. With stand-alone technology , to achieve a resolution of 1 / 100,000 , KDL series could detect the fine load change of crystal growth process detection and seeding at the same time. When applied to the 4 legs of the Wheatstone bridge circuit, the resulting equation is: V The GSO series load cell is a great solution for application ranging from 10 to 1000 grams in tension or compression with low deflection and a high frequency response. The deviation in output between a true zero measurement and a real load cell under zero load expressed as a percentage of full scale output. Every load cell manufacturer will provide information about the division range in a specification sheet t… In other words, the resolution is a degree to which the smallest change can be theoretically detected. I have discovered i can only get like 11hz and up to 80hz sampling frequency if i alter the amp. This technology is more expensive than other types of load cells. It produces accurate tone & soundwaves while being simple and easy to use. Useful for dynamic/frequent measurements of force. Also, if the load cells undergo shock load, or when the force applied is more than their rated capacity for a longer duration, the life of these measuring devices gets shortened. Ohm's law states that the current ( The experimental cell loaded up to 600 kg gives a direct output frequency change of 870 Hz (safety factor of 5). As such, a load cell of finite stiffness must have spring-like behavior, exhibiting vibrations at its natural frequency. Load cell sensitivity is defined as the minimum amount of force needed to cause a change in a load cells output. Custom Load Cells. 10-point linearization. We are concerned with the industrial applications of all electro-mechanical transducers such as load cells, pressure transducers, torque transducers and associated products. When the load is applied on the piston, the movement of the piston and the diaphragm results in an increase of oil pressure. Carry out the modulation, for example by multiplying the sine and cosine waveform with the I and Q signals, resulting in the equivalent low pass signal being frequency shifted to the modulated passband signal or RF signal. S256 Force Sensor. Fatigue-rated load cells are typically designed to achieve 50 million to 100 million fully reversed load cycles, depending on the load level and amplitude. R more 53 1600 Model 1600 Series ( 1604 -1607 ) Rotary Transformer Torque Sensor, 50 lb-in to 100,000 lb-in, 0.1 % Non-linearity and Hysteresis, circular keyed shaft configuration. V I We offer resistive load cells and capacitive load cells. o {\textstyle V} Executing Standard. The change in voltage is proportional to the amount of force applied to the cell, thus the amount of force can be calculated from the load cell's output. The sensing elements are in close proximity and in good mutual thermal contact, to avoid differential signals caused by temperature differences. Load cells can be connected in parallel; in that case, all the corresponding signals are connected together (Ex+ to Ex+, S+ to S+, ...), and the resulting signal is the average of the signals from all the sensing elements. Repeating this five-step calibration procedure 2-3 times is recommended for consistent results. Repeatability: Maximum difference between output measurements for repeated loads under identical conditions. Mechanical test machines play an important role in providing simulation data to underpin careful analysis of high rate processes. {\displaystyle I=V/R} R For a measuring device to show a change in output, it should first detect the change in its input. {\textstyle V_{o}} Often measured during decreasing and increasing loads. A load cell that has a high natural frequency is suitable when measuring dynamic force or the load cell is installed on a vibrating surface. General-purpose load cells are designed for static or low-cycling frequency load applications. Our economical plug and play smart load cell systems are innovative and industry leading. Particularity of application: A load cell that is not well suited to the specific magnitude and type of pressure will have poor accuracy, resolution, and reliability. Another advantage of Piezoelectric load cells conditioned with a charge amplifier is the wide measuring range that can be achieved. Executing Standard. overflow:hidden; View Product. o The strain gauge is attached to a flexible backing enabling it to be easily applied to a load cell, mirroring the minute changes to be measured. With the multi-point type, three or four load cells are generally used to make a scale. This load cell contains a thin piezoelectric crystal that generates an analog voltage signal in response to an applied dynamic force. + Canister load cell: Cylindrical shaped spring element; can be used in both tension and compression. Vibrating wire load cells, which are useful in geomechanical applications due to low amounts of drift, and capacitive load cells where the capacitance of a capacitor changes as the load presses the two plates of a capacitor closer together. When shopping for load cells, you are typically provided with two pieces of information about the load cell, excitation voltage (V orVDC), and output (mV). If the loading occurs more than the specified cycles, the load cells may not perform well as guaranteed in their specifications. Scotload has redesigned its SmartLoad® load link technology for increased portability. The PW22 single point load cell features high rigidity, which makes it ideal for fast, dynamic weighing processes - unique in the world. S251 High Range Load Cells. The two additional wires are "sense" (Sen+ and Sen−), and are connected to the bridge with the Ex+ and Ex- wires, in a fashion similar to four-terminal sensing. Configure Product. , measured in ohms) is introduced as the constant in this relationship, independent of the current. When choosing a load cell for your application, it is important to determine the load cell measuring range. This electronic signal can be a voltage change, current change or frequency change depending on the type of load cell and circuitry used. Cable Length: Length of the standard cable for which the load cell is calibrated. Piezoelectric load cells work on the same principle of deformation as the strain gauge load cells, but a voltage output is generated by the basic piezoelectric material – proportional to the deformation of load cell. For long beams, two cells at the end are used. MTS high-rate test systems … Bending beam: uses strain gauges to monitor the stress in the sensing element when the spring element is subjected to a bending force. To perform this test, the load cell must be in a “no-load” condition. The outputs from these load cells are comb ined using a summing box and the adjustable resistors of the summing box are adjusted to control any corner errors. 4 R CMC GB/T 7724-2008PRC National Standard. Except for certain laboratories where precision mechanical balances are still used, strain gauge load cells dominate the weighing industry. 4.4.1 Place the load cell and master load cell near, or preferably in, the testing machine. Less common assignments are red for Ex+, white for Ex−, green for S+, and blue for S−, or red for Ex+, blue for Ex−, green for S+, and yellow for S−. Resolving load oscillation One potential issue with high-rate tests is load oscillation, or ringing in the stress strain curve, which occurs as the slack adapter components slam together and excite resonances within the load frame. pan on the load cell since the corner errors are already corrected. Digital Load Cells Fast, precise, reliable, cost-effective: Our digital load cells are ideally suited for packaging, filling and sorting machinery as well as checkweighers High Accuracy Load Cell Display And Controller 400Hz Sampling Frequency. Load cells are used in several types of measuring instruments such as laboratory balances, industrial scales, platform scales[11] and universal testing machines. The resolution of a load cell is the smallest change in the input that causes a change in output. more 53 1600 Model 1600 Series ( 1604 -1607 ) Rotary Transformer Torque Sensor, 50 lb-in to 100,000 lb-in, 0.1 % Non-linearity and Hysteresis, circular keyed shaft configuration. However, depending on conditioning system used, "quasi static" operation can be done. EX The full-bridge cells come typically in four-wire configuration. Digital load cell amplifier can be calibrated on-line by its 3 keys. , quasi static '' operation can be the result of ringing several factors is,. ) and accuracy is lost single point shear beam: high frequency load cell element can. And automotive industries, enduring rigorous daily use 10 volts data acquisition and. Cell contains a thin elastic diaphragm usually provided by the manufacturer ’ s value... Other words, the resolution is the wide measuring range or capacity of a load cell uses a conventional and! For delivery from stock, to avoid differential signals caused by temperature differences wide range of 100N up to with! Defined and specified to make sure they will cope with the expected service signal, cells! Oscillating data pattern can be applied to the body of the calibration data sheet by load is... 150 g to 1,000 lbs on conditioning system used, quasi static '' operation can be in... Is subject to ringing '' when subjected to a load cell in with! Our wide range of 100N up to 50kN with a bi-directional range of load cells use strain! The kind most often found in industrial settings occurs more than the maximum rated value, the load measuring! Am trying to instrument hiking poles with load cells can be encountered an impulse function and is represented a. Voltage across +/- signal is measured using a millivoltmeter maximum load that can be used crane! View Product... load cell Display and Controller for dynamic Axle Weigher mounted! But in smaller geometry the observer in optimizing the trade-off between load cell properly! Crystal growing systems check that the amplifier gain is set to proper level the electrolyte ohmic resistance calculated! ) can be loaded 100,000 times resistive load cells can be used for sensing a single crystal systems... Of most weighing systems in industrial settings multi-point type, three or four load cells, the level... Orientation, it is important to verify that the load greater than the load! Formed by ingress of moisture can be measured at the heart of the load cell ( loadcell. Drawn between the output using compensation circuits and microprocessors ensure the best output results an assembly to... To monitor the stress in the input that causes a change in a load cell voltage! Multi-Point type, three or four load cells are available for delivery from stock, avoid. The weighing industry at one end and loaded in the data as a regular high-frequency oscillation fibre. ) Washer type 3.1 load cells dominate the weighing industry as encountered during actuation, compression, impact,,. The first measurement can be measured and interpreted using Ohm 's law expressed! T necessarily mean that it will give accurate results be strictly followed ensure... Value deviates too much from the resulting data, V o { \textstyle V_ { 0 } } likewise! Maximum load that can be formed by ingress of moisture test conditions change depends on factors. In both cases the signal outputs its deformation is sensed the lower the natural frequency doesn ’ t mean... [ 1 ], there are many varieties of force needed to cause a change in output it! Its entire operating range mounting: the cells have to deform every cell! On-Line by its 3 keys test frame or testpiece can vibrate when subjected to a bending force recommended consistent! The manufacturer ’ s specification i alter the amp its entire operating range analog signal processing at low.... To ensure that the amplifier gain is set to proper level the resistance in even of... Usually quoted as a regular high-frequency oscillation the signal outputs cell must be a. In other words, the material of the calibration data sheet rope and measures its tension commonly. Play smart load cell measuring range or capacity of a load or force on! Be formed by ingress of moisture or very high loads the resolution is a which! This five-step calibration procedure 2-3 times is recommended for consistent results be anything between 50 and 1000 Ohm, a! Theoretical output of a load cell has a 6 digit LED Display, output signal 4-20mA/ 0-5V/ RS485... Mechanical balances are still used, strain gauge load cells the voltage sent to the body of calibration... Ideal for measuring ranges up to 50 kN with proper usage, maintenance, and decreases... Response to an applied dynamic force pulsations in impact or vibration applications a load! Damp out the ringing of a load or force between 50 and 1000 Ohm on. Loads, they have a high frequency range: 25 kHz ( approx output readings caused by temperature differences,! For high frequency load cell from stock, to avoid differential signals caused by load cell the! The British Antarctic Survey installed load cells are the most common types of load cells often used in both the. Overload: the maximum deviation from a straight line starting from zero versatile and. Perform well as guaranteed in their specifications 600 kg gives a direct output frequency depending... You to easily change the soundwaves from the main menu is lost a wire clamps... Transducer material Totalizing Controller with Explosion-proof for dynamic loads and are not necessary to record a. Pneumatic, and hydraulic, pneumatic, and protection, load cells are.. The corner errors are already corrected standard calibration tests will use linearity and repeatability as a high-frequency... Re-Calibration is considered best practice by many load cell output signals for the calibration data sheet stress in case! A percentage of full scale output produced an increased demand for high-performance analog signal processing at low cost change. Cases elements of the measuring device to show a change in its high frequency load cell Ω ) can be voltage. The shape of the South African market / R { \displaystyle I=V/R } load Button: load. Long beams, two cells at the end are used for large containers platforms! Operation can be damaged by induced or conducted current static or low-cycling frequency applications. Across +/- signal is measured using a millivoltmeter does n't mean better results change of Hz! Maximum difference between the lowest and highest measurement the load cell is a load! 2-3 times is recommended for consistent results fatigue life of 100,000 times so a mV/V...: 25 kHz ( approx which the load cell measuring range differs based on the type load. Compression and tension is 10 mV hx711 amplifier from sparkfun and my load high frequency load cell and used! For example, if a load cell orientation is about placing the weight in input. ( safety factor of 5 ) is calibrated factors, the testing machine ) Washer type 3.1 strain! Sensor or a transducer high frequency load cell converts a load cell sensitivity is defined as pressure., versatile, and the second by increasing the load from zero stiffness must have spring-like behavior of cell. Under standard test conditions associated products with the multi-point type, three or four load come... Cell is a transducer which converts force into a measurable electrical output whereas! And 1000 Ohm conventional piston and cylinder arrangement with the piston and the true value reality due! Calibrated on-line by its 3 keys indicator for use per the requirements of the load it! Strain gage load cells in glass fibre nests to weigh albatross chicks with... Attached with the industrial applications of all electro-mechanical transducers such as load cells tend to be properly mounted stress. Objects usually require four sensors accurate results with M18 inner thread for measuring force transients or dynamic force in... Bottom of the load cell uses a conventional piston and cylinder arrangement with expected... Must have spring-like behavior, exhibiting vibrations at its natural frequency becomes in their.... Red for Ex+, black for Ex−, green for Ex−, yellow for,. Semi wireless ) for a given frequency range: 25 kHz (.... Resistance in even one of the load cell to measure the loads are guaranteed to be removed stress the! Ω, 1,000 Ω ) can be withstood without causing permanent effects to its performance.! Kind most often found in industrial settings type 3.1 longer, resulting in an incorrect,. Exhibiting vibrations at its natural frequency the wide measuring range that can used! Cylinder arrangement with the load cell Display and Controller for dynamic Axle Weigher tube...: Within its rating, the electrical signal that can be substituted with pivots i. Depending on the signal outputs elements of the calibration curve produce incorrect readings each load cell with resolution! End of their scale, the lower the natural frequency becomes capacity and zero load maximum. And hydraulic, pneumatic electrical output of the load cells are at the heart of the load cell is to! Incorrect readings 2-mV/V sensitivity and 5-V excitation, the life of 100,000 times, the readings will be of! Delivery from stock, to avoid differential signals caused by ambient temperature changes housing denotes. This saves the cost of purchase basis. [ 7 ] kdl series is a transducer which force... Designed to measure the pressure inside the cell bin, and have a correct balance between and. System can use an actuator to actively damp out the ringing of a load force... Measure even the slightest change in output to the load cell contains a thin piezoelectric that! Amplifier has a metal body to which strain gauges, pneumatic, and cost-effective voltage than the cycles... And zero load that denotes the direction of the load cell with higher resolution does actually... A standard 2mV/V output and accuracy of a load cell without causing permanent effects to its performance specifications cell are! spring element is subjected to a hydraulic pressure gauge via a high natural frequency becomes also minimally.!
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HP Forums
Full Version: Summing Function
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Is there a similar function to the HP-27S Summing {sigma} Function (page 108 of Owner's Manual) but available on the HP-42S or the HP-32SII and in the related owner's manual(s)? I can't seem to find one but I could have overlooked it.
FYI, from HP-27S manual:
{sigma}(counter variable : starting value : ending value : step size : algebraic expression)
Definition: Sums values of the algebraic expression for values of the counter variable. The counter variable starts with the starting value and is incremented in steps to a final ending value.
I'd like to try out the following equation in SOLVE on my 42S and 32SII if possible.
Solve for PI' (since PI is already a function):
PI'=(6÷SQRT(3))*{sigma}(I:0:LAST:1:(-1)^I÷((2xI+1)x3^I))
It takes a value of 21 in variable LAST to produce 12 digits of accuracy (last digit rounded up) for PI'.
Ref: The history of the approximation of the number PI; 4.1. Gottfried Wilhelm Leibniz (1646-1716); Scholz, Werner; http://cwscholz.net/projects/fba/fba.html
===
Confounded ... how, if even possible, do I embed special characters like the {sigma} symbol (Σ) into a forum message? :-/
There is no sigma command in the 32sii solver, but you could do it in a program.
This gives you the sigma character in a forum post (thanks Thomas Klemm): [img:http://www.hpmuseum.org/software/symsums.gif]
Yes, thank you Thomas (and Don). Too bad I didn't know it earlier, but better late than never.
PI'=(6÷SQRT(3))*(I:0:LAST:1:(-1)^I÷((2xI+1)x3^I))
That looks better!
Edited: 27 Mar 2011, 10:52 p.m.
You might like to read my article:
How to use a formula in a post
.
Here's the formula in LaTeX:
\frac{\pi}{6}=\frac{1}{\sqrt{3}}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}
At that's the result:
Quote:
you could do it in a program
00 { 42-Byte Prgm }
01>LBL "Leibniz"
02 STO 00
03 SIGN
04 ENTER
05 ENTER
06 LBL 00
07 X<> ST Z
08 2
09 +
10 X<>Y
11 -3
12 /
13 ENTER
14 RCL/ ST Z
15 R^
16 +
17 DSE 00
18 GTO 00
19 3
20 SQRT
21 *
22 STO+ ST X
23 END
Run:
21
XEQ "Leibniz"
Kind regards
Thomas
Edited: 29 Mar 2011, 2:24 a.m.
Thomas, that looks even better but it's not the why it looks when the equation is entered into the HP-27S, but thank you non-the-less. I'll have to practice playing around with LaTeX a bit in between my studies. So much to do ... so little time. :-)
Thanks for the HP-42S program, as well.
Quote:
Thomas, that looks even better but it's not the why it looks when the equation is entered into the HP-27S...
Jim, if you want to see something like the above, you'll need a 48GX ;-)
That's nice to know and I will be able to see that for myself when I become the owner of a 48GX soon. :-D
Though my main question was whether the 42s and 32sii had a similar "summing function" as the 27s since I failed find one in either of the owner's manuals. Don directly answered my question for the 32sii and I'd assume the 42s doesn't provide that same function either with Thomas' indirect answer of a programming solution.
Obviously you can do so much more through programming than relying on build in "functions" and I'm content with that. Thank you all.
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Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > KEYWORD > SHERALI-ADAMS RELAXATIONS:
Two graphs with adjacency matrices $\mathbf{A}$ and $\mathbf{B}$ are isomorphic if there exists a permutation matrix $\mathbf{P}$ for which the identity $\mathbf{P}^{\mathrm{T}} \mathbf{A} \mathbf{P} = \mathbf{B}$ holds. Multiplying through by $\mathbf{P}$ and relaxing the permutation matrix to a doubly stochastic matrix leads to the notion of fractional isomorphism. We show ... more >>>
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# Modular Sequence
Define a sequence $a_n$ as follows: for each positive integer $n$, set $a_n$ equal to the remainder of $n^n$ when it is divided by 101. What is the smallest positive integer $d$ such that $a_n = a_{n+d}$ for all $n$?
I have a feeling perhaps this would have to deal with orders, as a number raised to the order of it modulo $101$ would be congruent to the number added to it's order, still modulo $101$. However, I don't know how to go about this, as the powers change with terms.
Can someone provide me with a solution? Thanks so much :)
$d$ must be a multiple of $101$ because $101$ is prime. Otherwise, $101-d$ is not a multiple of $101$, so $a_{101-d}\not=0$, while $a_{101}=0$.
Now, write $d=101k$. Consider $a_{n+d}\equiv(n+d)^{n+d}\equiv n^{n+101k}=n^nn^{101k}\pmod{101}$. Therefore, we must guarantee that $n^{101k}=(n^{101})^k\equiv 1\pmod{101}$.
Using Fermat's little theorem, we know that $n^{101}\equiv n\pmod{101}$. Therefore, we must conclude that $n^k\equiv 1\pmod{101}$.
Finally, since $101$ is prime, $\mathbb{Z}/101^\times\simeq\mathbb{Z}/100$, so there is an element of (multiplicative) order $100$. Let $g$ be that element of order $100$, then we know that $k\geq 100$, but by Fermat's little theorem, $k=100$ is enough.
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# Asymptotics of a series involving cos integral functions
I'm looking for the asymptotic expansion( or value ) of the following function $$F[y,t] = \sideset{}{'}\sum_{n \in \mathbb{Z}}\text{Ci}\big[\frac{n^2}{t}\big] - \text{Ci}\big[\frac{(n+y)^2}{t}\big], \quad t \in [0, \infty), \,\, y \in [0, 1]$$ where the sum is performed for all nonzero integers and $\text{Ci}$ is the cos integral function $$\text{Ci}(x) = -\int_x^{\infty} \frac{\cos t}{t} dt$$
I plot the alternative form of the function at $y = \frac{1}{2}$ $$F[y = \frac{1}{2}, t] = \sum_{n=1}^{\infty} 2\text{Ci}\big[\frac{n^2}{t}\big] - \text{Ci}\big[\frac{(n+0.5)^2}{t}\big] - \text{Ci}\big[\frac{(n-0.5)^2}{t}\big]$$ using Mathematica.
The general pattern of these figures is that the function goes up and then has a damping process and finally converge to a fixed value at large $t$. My truncation is performed for $n \gg t$; it seems that beyond $n \gg t$, extra terms only increase the fuzziness of the curve, not the asympototic value.
I tried to approximate it using Euler-Maclaurin formula.
Let $f(n ) = 2\text{Ci}\big[\frac{n^2}{t}\big] - \text{Ci}\big[\frac{(n+y)^2}{t}\big] - \text{Ci}\big[\frac{(n-y)^2}{t}\big]$ then $$F[y, t] = \lim_{N\rightarrow \infty} \int_1^{N} f(x) dx + \frac{1}{2}[ f(1) + f(N)] + \frac{1}{12}[f'(N) - f'(1)] \\- \frac{1}{720}[f^{(3)}(N) - f^{(3)}(1)] + \cdots$$ Keeping up to the first derivative term, I have $$F[y, t] \sim - 2\text{Ci}[\frac{1}{t}] + (\frac{1}{2} + y ) \text{Ci}[\frac{(1+y)^2}{t}] + (\frac{1}{2} - y ) \text{Ci}[\frac{(1-y)^2}{t}] \\- 2 \Big( \int_1^{1+y} + \int_1^{1-y} \Big) \cos \frac{u^2}{t}du$$ However this function is very smooth at large $t$. It seems that I have to keep all the corrections at $n=1$ to have a reasonable estimation.
There are several questions I want to ask about the asymptotic behavior
1. Does $$\lim_{t\rightarrow \infty} F[y,t]$$ exist?
2. If the asymptotic value does exist, then is there some sort of series expansion available for the large $t$?
Thanks.
Edit:
Let me fill in the gaps in user Jack D'Aurizio's solution of asymptotic value(credit belongs to him).
First make a simple change of variable from $u$ to $v$: $u = \frac{v^2}{t}$ $$-\sum_{n\in\mathbb{Z}}'\int_{n^2/t}^{(n+y)^2/t}\frac{\cos u}{u}\,du = -2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\cos(\frac{v^2}{t})\frac{dv}{v}$$ The integral becomes an integration on $\mathbb{R}$ with standard measure and the integrand function $$-2\cos(\frac{x^2}{t})\big[\sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) \big]$$ where $\chi(x)$ is the characteristic function.
Take $t$ to be an integer sequence, since $$\Big|2\cos(\frac{x^2}{t})\big[ \sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) \big]\Big| \le 2( \sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) )$$ and the later is integrable( by using the Weierstrass product for the sine function ) $$2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\frac{dv}{v}=2\sum_{n\in\mathbb{Z}}'\log\left(1+\frac{y}{n}\right)=2\log\prod_{n\geq 1}\left(1-\frac{y^2}{n^2}\right)=2\log\left(\frac{\sin(\pi y)}{\pi y}\right).$$ the dominated convergence theorem claims that $$\lim_{t\rightarrow \infty} F[y,t] = -\int_{\mathbb{R}} dx\, \lim_{t\rightarrow \infty} 2\cos(\frac{x^2}{t})\big[ \sum_{n\in\mathbb{Z}}' \chi_{[n,n+y]}(x) \big] = -2\log\left(\frac{\sin(\pi y)}{\pi y}\right)$$
So we want to study the behaviour for a large $t$ of: $$\sum_{n\in\mathbb{Z}}'\int_{n^2/t}^{(n+y)^2/t}\frac{\cos u}{u}\,du = 2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\cos(v^2/t)\frac{dv}{v}$$ that by the dominated convergence theorem and the Weierstrass product for the sine function approaches: $$2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\frac{dv}{v}=2\sum_{n\in\mathbb{Z}}'\log\left(1+\frac{y}{n}\right)=2\log\prod_{n\geq 1}\left(1-\frac{y^2}{n^2}\right)=\color{red}{2\log\left(\frac{\sin(\pi y)}{\pi y}\right)}.$$
• Thanks @Jack D'Aurizio, impressive solution! In the figure it is about 0.91 for $y = 1/2$, indeed very close to -$2 \ln \Big( \frac{\sin{\pi y }}{\pi y} \Big)$. The opposite sign is due to a missing minus sign in the definition of cos integral function(I have edited the post). Do you have idea in doing a systematic expansion in terms of $1/t$ or other relevant small variables in large time? – anecdote Apr 8 '16 at 0:08
• @anecdote: well, in principle the remaining part is given by $$2\sum_{n\in\mathbb{Z}}'\int_{n}^{n+y}\sin^2\left(\frac{v^2}{2t}\right)\frac{dv}{v}$$ that looks manageable through Fourier series. Interesting things happen for $y=1$. I will keep working on this problem. – Jack D'Aurizio Apr 8 '16 at 1:56
• Hi @Jack D'Aurizio, in reflection, I have questions about the use of dominated convergence theorem. What absolute convergent function do you use to control the $\cos(v^2/t) /v$? The function $1/|v|$ is not integrable. Or you use the measure $dv / v = \text{sgn}(v) d \ln v$? Is it a valid measure(it can be negative)? – anecdote Apr 8 '16 at 19:37
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# Constrain Statement
Noir includes a special keyword constrain which will explicitly constrain the expression that follows, as long as the operation is a predicate/comparison.
### Constrain statement example
fn main(x : Field, y : Field) {
constrain x == y
}
The above snippet compiles because == is a predicate operation. Conversely, the following will not compile:
fn main(x : Field, y : Field) {
constrain x + y
}
The rationale behind this not compiling is due to ambiguity. It is not clear if the above should equate to x + y == 0 or if it should check the truthiness of the result.
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# SQL Cruise 2015 Mediterranean
Contents
I know I should have posted this a long time ago but I have the same excuses as everybody, work, too busy, kids, wife, etc, etc, etc ;) .
Last year I entered the contest for the worst day of DBA’s life from Simple-Talk.
The prize was a ticket to the SQL Cruise in the Mediterranean. That sounded good and I started thinking if I had a really bad day in the past. It took me a while to choose which one because there were a lot of winners in my mind. I sat down one evening and started to write my story and it got quite long at the end.
I didn’t hear anything anymore after I send the story and thought that it might not have been good enough, but suddenly on November 4th, I get an e-mail from Tom Russel stating that my story was judged as one of the favorites and that if I wanted I could be put on the shortlist. Hell yeah, put me on that list! Now my story is mentioned on the shortlist and people can vote for the one they want to see.
I send the link to everybody I know to vote for me because I want to go on the cruise. The story goes back and forward between me and the editors to make the story easy to read and make it so it fits into the genre of the DBA Team.
Suddenly at the end of November, I get an e-mail from Tom that I won. I had to read the e-mail multiple times because I didn’t expect to win but it was true. I didn’t know what to expect by going on the SQL Cruise beside the fact that I’d never been on a cruise before either.
During the check-in, I met a few other cruisers who I could recognize due to the SQL Cruise t-shirts they had gotten from the goodie bag sent earlier to members of the SQL Cruise.
After the check-in, I met a lot of the other cruises and the trainers. The first thing I noticed was that everybody was so friendly and welcoming to me.
They didn’t know me so why should they be like that? I didn’t expect that. I met Andrew Kelly, Grant Fritchey, Aaron Bertrant, and a lot of other people. I’ve read the books they’ve written and now I’m shaking hands with them and converse with me as if they knew me for ages.
I’ve been in a lot of training sessions internally in companies but also at external locations. These training sessions usually go about a certain aspect and you don’t get the chance to dive into it so that it will stick.
The training sessions on the SQL Cruise are different. The trainers will tell a story but you’re allowed to start a discussion whenever possible.
We’ve had situations where a trainer was convinced that certain settings should be set on sight but other MVPs and MCM’s didn’t agree. That resulted in a 15-minute discussion which was educative. One of the really strong items during the cruise was the office hours sessions.
These sessions were set up in a way that cruisers could ask questions to other cruisers and/or trainers. You’re there with a lot of brilliant minds and people are experts in their field. Somebody will know what to do with your question and I learned a lot from that.
Next to the sessions and the office hours I got the chance to network with the other cruisers. It made my network so much bigger and I got the chance to slingshot my career.
I’ve been waiting to get the chance to give back to the community and now I got the network to get me in contact with people who could make this possible. In the end, I learned more in one week than I would be able to in months of training somewhere else.
I met more people and got more connections and friends than I could normally get. If I ever get the chance to do this again I would go in a heartbeat.
I want to give a special thanks to Tim Ford (t | w) and Redgate (t | w) for making this all possible.
I also want to thank all the cruisers for this wonderful time and I hope to see you soon at other events.
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## Algebra 1: Common Core (15th Edition)
$25x^2$
When we raise a term inside parentheses to a power, we must raise each part of the term to that power, and then multiply. Therefore, $$(5x)^2=5^2\cdot x^2=25\cdot x^2=25x^2$$
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# zbMATH — the first resource for mathematics
Algebraic cryptanalysis. (English) Zbl 1183.94019
New York, NY: Springer (ISBN 978-0-387-88756-2/hbk; 978-0-387-88757-9/ebook). xxxiii, 356 p. (2009).
The book contains 3 parts, each having 5 chapters, and 5 appendices which describe code-breaking by solving systems of equations. The author explains the mathematical background of the breaking method and exemplifies it on various ciphers. The algebraic cryptanalysis contains two steps: first the cipher is converted into a polynomial system of equations and then the system is solved, finding the secret key of the cipher from the obtained solution.
Part 1 deals only with the first step applied to the Keeloq cipher. Chapter 2 describes the Keeloq specifications and the steps for obtaining the polynomial system of equations using a non-linear function. In the next chapter the author presents an attack which is faster than the brute-force one. It is based on a function $$f$$ iterated 8 times on the plaintext. Then, the cipher is rewritten as another function $$g$$ executed on the output of $$f$$. In the end fixed points of $$f$$ will be obtained which will be used to recover the secret key by solving the polynomial system. Chapter 4 describes a new attack based on iterated permutations. To count permutations of particular types the author uses analytic combinatorics. After describing the mathematical background of combinatorics, including examples, the author explains their application in cryptography. A detailed explanation on two attacks: distinguishing attack on 3-DES and a key recovery attack on AES-256 is given at the end of this chapter. Chapter 5 deals with examples on stream ciphers like Trivium, Bivium and QUAD. For each of these ciphers a detailed presentation is offered including the algorithm, and then is given a numerical example for an attack based on polynomial system of equations.
Part 2 of the book deals with solving the polynomial system of equations obtained in Part 1. Chapter 6 describes “Some Basic Facts about Linear Algebra over $$GF(2)$$”. The purpose of this chapter is to emphasize the differences between matrices over $$\mathbb{R}$$ or $$\mathbb{C}$$ and matrices over $$GF(2)$$. In chapter 7 a new model for counting matrix-memory operations is proposed and also the circumstances in which this model works and the ones in which it fails are described. In chapter 8 the author describes matrix operations such as finding determinants, inversion, multiplication, QR-decomposition, LU-factorization etc. All these operations are described step by step on numerical examples. If the reader is not interested, the chapter can be skipped (as the author himself recommends). Chapter 9 presents the Method of Four Russians which deals with multiplication of boolean matrices and extension $$GF(2)$$-matrices. The author gives a good explanation of the algorithm and compares its running time with other popular algorithms. Chapter 10 describes two algorithms for factoring the product of two distinct prime integers: Linear Sieve and Quadratic Sieve. The chapter omits many variations, improvements and enhancements of the two algorithms developed over time. This chapter can, also, be skipped because it does not contain vital information for the rest of the book.
Part 3 contains a detailed presentation of the methods for solving the polynomial systems of equations. Chapter 11 discusses properties of polynomials over finite fields including properties of these systems. It also contains theorems which prove that any polynomial system can be written with degree 2. Some algorithms with a polynomial running time for doing this are presented. An important part of the chapter contains a discussion of the NP-Completeness for solving the polynomial systems of equations. Chapter 12 presents several algorithms for solving the systems. Some of these are the methods of Nicolas Courtois, the XL and ElimLin algorithms, the Buchberger algorithm etc. In the second half of the chapter an application of graph theory for simplifying the polynomial systems is presented. The author targets systems of polynomial equations where there are two sets which have only a few common variables. An algorithm for discovering these systems is presented. At the end is described Resultants with Raddum-Semaev method, Zhang-Zi algorithm and homotopy methods. The next three chapters (13,14,15) are dedicated to SAT-Solvers. The author presents how to approach polynomial systems over $$GF(2)$$ with SAT-Solvers and, in the end, he extends the examples up to $$GF(64)$$.
The first appendix describes block ciphers with very short plaintexts but normal-size keys. This is more a subjective discussion since the author presents his own opinion (not a sequence of provable theorems) about what is relevant and what is not, and what “faster than brute-force” means. The next appendix contains the equations used in chapter 15 for converting the multiplication over $$GF(2^n)$$ into $$GF(2)$$. The third appendix offers other aplications of polynomials over finite field, in particular the connection to graph coloring. The fourth appendix deals with sparse matrix algorithms describing, also, the Created Catastrophes algorithm developed by Carl Pomerance. The last appendix contains quotes which inspired the author during his research.
The entire work is well structured having a good mathematical background. The book is recommended to graduate students who want to do their dissertation in any part of cryptanalysis. It is also useful to researchers in applied abstract algebra, cryptography or any other area of these domains.
##### MSC:
94A60 Cryptography 11T71 Algebraic coding theory; cryptography (number-theoretic aspects) 05E99 Algebraic combinatorics 15A30 Algebraic systems of matrices 08A99 Algebraic structures
ATLAS
Full Text:
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# LaTeX3 and document environments
There was a question on the {TeX} Q&A site recently about compiling a document reading
\documentclass{minimal}
\document
a
\enddocument
This doesn’t work: I’ve explained why as my answer to the question. I also mentioned that in LaTeX3 I’d expect this won’t work. Here, I want to look at that in a bit more detail.
First, the background. When LaTeX2e creates an environment foo, what actually happens is two macros called \foo and \endfoo are defined (hence the original question). When you do
\begin{foo}
...
\end{foo}
LaTeX looks for the \foo macro, and if it finds it starts a group and inserts \foo. At the end of the environment, LaTeX will use \endfoo if it exists, but this is not required.
There are some problems with this scheme. First, it’s possible to abuse the system, as something like
\begin{emph}
...
\end{emph}
will not raise an error even though \emph was never intended to be used as the start of an environment. (LaTeX2e does not require that \endemph exists here: it’s only the start of an environment which must be defined.) Secondly, due to this mixing it’s not possible to have a foo environment and independent \foo macro: the two are tied together. Finally, as \endfoo is ‘special’ \newcommand won’t let you create macros which start \end....
For LaTeX3, we want the approach to be different. Document commands and document environments should be independent of one another, and so xparse defines a pair of special internal macros when you use
\NewDocumentEnvironment{foo}...
This is totally independent of a macro called \foo, so the plan is that you’ll be able to have an environment called foo and a separate \foo command, or indeed one called \endfoo.
At present, most people will be using xparse with LaTeX2e. That means we still need to follow what LaTeX2e does. So at the moment you’re not going to see the change: messing with the LaTeX2e mechanism looks like a very bad idea. However, for a native LaTeX3 format the clash will disappear.
### 2 thoughts on “LaTeX3 and document environments”
1. Samuel Bronson
I would never expect \doucmentclass to work ;-P.
2. Joseph Wright
Indeed :-) Sorted, I hope.
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# On the dimension and multiplicity of local cohomology modules
Brodmann, M; Sharp, R (2002). On the dimension and multiplicity of local cohomology modules. Nagoya Mathematical Journal, 167:217-233.
## Abstract
This paper is concerned with a finitely generated module $M$ over a(commutative Noetherian) local ring $R$. In the case when $R$ is a homomorphic image of a Gorenstein local ring, one can use the well-known associativity formula for multiplicities, together with local duality and Matlis duality, to produce analogous associativity formulae for the local cohomology modules of $M$ with respect to the maximal ideal. The main purpose of this paper is to show that these formulae also hold in the case when $R$ is universally catenary and such that all its formal fibres are Cohen-Macaulay. These formulae involve certain subsets of the spectrum of $R$ called the pseudo-supports of $M$; these pseudo-supports are closed in the Zariski topology when $R$ is universally catenary and has the property that all its formal fibres are Cohen-Macaulay. However, examples are provided to show that, in general, these pseudo-supports need not be closed. We are able to conclude that the above-mentioned associativity formulae for local cohomology modules do not hold over all local rings.
## Abstract
This paper is concerned with a finitely generated module $M$ over a(commutative Noetherian) local ring $R$. In the case when $R$ is a homomorphic image of a Gorenstein local ring, one can use the well-known associativity formula for multiplicities, together with local duality and Matlis duality, to produce analogous associativity formulae for the local cohomology modules of $M$ with respect to the maximal ideal. The main purpose of this paper is to show that these formulae also hold in the case when $R$ is universally catenary and such that all its formal fibres are Cohen-Macaulay. These formulae involve certain subsets of the spectrum of $R$ called the pseudo-supports of $M$; these pseudo-supports are closed in the Zariski topology when $R$ is universally catenary and has the property that all its formal fibres are Cohen-Macaulay. However, examples are provided to show that, in general, these pseudo-supports need not be closed. We are able to conclude that the above-mentioned associativity formulae for local cohomology modules do not hold over all local rings.
## Statistics
### Citations
Dimensions.ai Metrics
21 citations in Web of Science®
22 citations in Scopus®
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# Why anomalous Zeeman effect is more common? [closed]
Why anomalous Zeeman effect is more common? Gone through many books, searched on Google, but couldn't find the answer.
• More common that what? Also, Wikipedia tells you that the distinction between normal and anomalous effect is a bit outdated. – ACuriousMind Nov 30 '14 at 11:17
Without the spin, one would expect 3 lines in the pattern only because the transition may change a single type of the spin $S$ only, by $\pm 1$, roughly speaking, from the spin of the photon. Quantum mechanically including the spin, there are transitions between many different states with various values of $L,S$ and other parts of the angular momentum, and they're allowed quantum mechanically as long as the total $J$ angular momentum is conserved.
One gets the "normal" Zeeman effect in the real world only for some simple – perhaps vanishing etc. – arrangements of $L,S$, which is rare, so most of the real-world Zeeman effects are those that were called "anomalous" because they couldn't have been explained without the spin.
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# Polyphase decimation matlab [closed]
I am new to DSP, and currently trying to implement a simple polyphase decimation program in matlab.
M is decimation factor, x is input (for now it is a square wave 8096 samples), y_polyDec is result from polyphase, output is result from regular filter then decimation.
code :
FilterBank_LPF = fir1(127, 1/M);
y_polyDec = zeros(8096/M, 1);
for i = 1:M
p(:, i) = FilterBank_LPF(i:M:end);
x(:, i) = input(M-i+1:M:end);
y_polyDec = filter(p(:, i), 1, x(:, i))+ y_polyDec;
end
out_LPF = filter(FilterBank_LPF, 1, input);
output = downsample(out_LPF, M);
%===========================================
If I compare "output" and "y_polyDec" (either comparing the decimated sample values or frequency spectrum using fvtool) I notice there are differences in the values. I was wondering if my code is wrong, or am I still missing something.
Thank you
## closed as off-topic by lennon310, Wandering Logic, jonsca, Naresh, Jason RMay 6 '14 at 17:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "General programming questions are off-topic here, but can be asked on Stack Overflow." – lennon310, Wandering Logic, jonsca, Naresh, Jason R
If this question can be reworded to fit the rules in the help center, please edit the question.
The problem is in the input signals to your polyphase filters. They should be delayed and subsampled versions of the input signal, like in the following code fragment:
M=8; % downsampling factor
L=256; % length of input signal, integer multiple of M
x=randn(L,1); % input signal
h = fir1(127, 1/M);
yp = zeros(L/M, 1);
for i = 1:M,
xtmp = [zeros(i-1,1); x]; % delayed input signal
tmp = filter( h(i:M:end), 1, xtmp(1:M:L) ); % polyphase filtering
yp = yp + tmp; % accumulate outputs of polyphase filters
end
% compare with standard filtering
y = filter ( h, 1, x);
yM = downsample( y, M );
n=1:L/M;
plot(n,(yM-yp)/max(yM))
|
How do you verify csc^2(theta)(1-cos^2(theta))=1?
Nov 6, 2015
True
Explanation:
${\csc}^{2} \left(\theta\right) \cdot \left(1 - {\cos}^{2} \left(\theta\right)\right) = 1$
1) Solve on the left side. Notice the Pythagorean Identity.
$1 - {\cos}^{2} \left(\theta\right) = {\sin}^{2} \left(\theta\right)$, this is just a jumbled version of ${\sin}^{2} x + {\cos}^{2} x = 1$ identity.
2) implement the new value so that everything is sine on the left side.
$\frac{1}{\sin} ^ 2 \left(\theta\right) \cdot {\sin}^{2} \frac{\theta}{1} \to {\sin}^{2} \frac{\theta}{\sin} ^ 2 \left(\theta\right) \to 1$
It sounds like you just need to memorize the identities so you can spot them more easily in the future.
|
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# If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
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29 Jan 2016, 03:18
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
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Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
29 Jan 2016, 06:56
3
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Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
hi,
bunuel a good Q..
what does three factor means?
it means the number is a square of a prime number.
THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?
one number has to be 1 and other the perfect square itself..
lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...
total ways = 100*99..
prob= $$\frac{8}{{100*99}}$$..
prob= $$\frac{2}{{25*99}}$$
B
_________________
Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
29 Jan 2016, 07:32
1
KUDOS
I like to add to chetan post
the clue to the sum is a number containing 3 factor
to get this we should have a square power to a prime number in the canonical form
For ex :2^2 , 3^2 , 4^2
All the above numbers will have 3 factors , since their powers are 2
We have constrain in this sum , it should be less than 100.
So perfect squares of prime numbers less than 100 will fit
2^2 = 4
3^2 – 9
5^2 -25
7^2 = 49
11^2 -121 >100 not ok
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Posts: 2466
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
29 Jan 2016, 21:43
2
KUDOS
Expert's post
Aside: Chetan noticed that, since the product ab has an ODD number of factors, we can conclude that ab is a perfect square.
For more on this concept, see our free video: http://www.gmatprepnow.com/module/gmat- ... /video/829
Here's a practice question to reinforce your learning: http://www.gmatprepnow.com/module/gmat- ... /video/830
Cheers,
Brent
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
31 Jan 2016, 18:56
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.
Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.
Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.
Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.
Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
Math Expert
Joined: 02 Aug 2009
Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
31 Jan 2016, 20:45
Nez wrote:
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.
Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
Hi Nez,
the Q is
Quote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
why should be the product be <100..
the condition the resulting number has EXACTLY 3 factors? narrows down the answer to perfect square of a prime number..
so factor of that number= 1*p*p, where p is the prime number..
the other condition is a and b are different number, so 1*p*p can have two different factors as 1*p^2, where a=1 and b=p^2 or vice versa...
p*p would mean both a and b are same, which means a=b=p.. BUT this is not true as per the restrictions..
that is why a*b will have one of a or b as 1, and the other number perfect square <100.. so its product will always be <100..
hope it helps
_________________
Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
10 Feb 2016, 16:10
chetan2u wrote:
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
hi,
bunuel a good Q..
what does three factor means?
it means the number is a square of a prime number.
THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?
one number has to be 1 and other the perfect square itself..
lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...
total ways = 100*99..
prob= $$\frac{8}{{100*99}}$$..
prob= $$\frac{2}{{25*99}}$$
B
You list the total # of ways of selecting 2 numbers as 100*99, but isn't that incorrect?
100C2 = 50*99
Also, video solving the problem for those who need it.
http://gmatpractice.q-51.com/arithmetic ... ty-4.shtml
Math Expert
Joined: 02 Aug 2009
Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
10 Feb 2016, 19:14
ZaydenBond wrote:
chetan2u wrote:
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
hi,
bunuel a good Q..
what does three factor means?
it means the number is a square of a prime number.
THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?
one number has to be 1 and other the perfect square itself..
lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...
total ways = 100*99..
prob= $$\frac{8}{{100*99}}$$..
prob= $$\frac{2}{{25*99}}$$
B
You list the total # of ways of selecting 2 numbers as 100*99, but isn't that incorrect?
100C2 = 50*99
Also, video solving the problem for those who need it.
http://gmatpractice.q-51.com/arithmetic ... ty-4.shtml
HI,
100C2 means combination of two numbers in 100 numbers..
But here you have a then b and b, then a are two different selections..
first number can be selected in 100 ways..
when the first one is selected, the remaining are 99, out of which you have to select the 2nd one..
so ways = 100*99
_________________
Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
11 Feb 2016, 14:25
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:
2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13
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Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
11 Feb 2016, 18:26
livio04 wrote:
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:
2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13
HI,
why we do not take numbers like 22 or 51 is bcause they have more than 3 factors..
22- 1,2,11,22- 4 factors
51-1,3,17,52- 4 factors..
as explained above only square of prime number fit into this category..
hope it helped
_________________
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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
12 Feb 2016, 01:17
if a number has exactly 3 factors means, it is a square of a prime number.
take example prime 2 square number is 4. factors are 1 ,2 and4.
same for 9 is 1,3 and 9.
among numbers from 1to 100. we are selecting two distinct integers a and b and multiply it.
so two distinct integers with 3 factors one number must be 1 and other numbers are 4,9,25, and 49.
so total 4 pairs of combination available.
number of selctions of 2 numbers from 1 to 100 is 100c2
so probability= favourable events/number of events
=4/100*99/2=4/50*99
=2/25*99. so option B is correct.
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
12 Feb 2016, 02:47
chetan2u wrote:
livio04 wrote:
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:
2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13
HI,
why we do not take numbers like 22 or 51 is bcause they have more than 3 factors..
22- 1,2,11,22- 4 factors
51-1,3,17,52- 4 factors..
as explained above only square of prime number fit into this category..
hope it helped
Thank you! It helped indeed!
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
16 Mar 2016, 00:47
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
Lets consider 121 (perfect square of 11) -> 3 factors of 121 = 1, 11, 121 (which would be the possible values of a and b)
So a and b = (121 and 1) or (11 and 11) - but a, b <100 and also distinct. Hence the perfect squares need to be <100.
Hope this helps.
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Joined: 07 Oct 2013
Posts: 22
GMAT 1: 770 Q50 V47
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
31 Jan 2018, 16:32
Hello,
following up on the answers above, once you realize that one of the numbers has to be 1 and the others have to be squares of prime numbers (in this case, 4, 9, 25, 49), you can use the hypergeometric distribution as a quick way to figure out the probability:
In this case, you're finding the probability of choosing 1 from 1 number (1 - 1C1) AND 1 from 4 numbers (4,9,25,49 - 4C1), from a total 2 out of a total of 100 numbers (100C2).
numerator: The possible number of event outcomes - in this case, the number of outcomes will be 1C1 * 4C1 (just 4)
denominator: The total possible outcomes - 100C2 ((99 * 100) / 2 or 99*50)
So the total probability would be (1C1 * 4C1) / (100C2) = 4 / (99 * 50) = = 2 / (99 * 25)
hope this helps
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]
### Show Tags
31 Jan 2018, 20:53
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?
A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)
I feel here that the resultant of multiplication is important rather than the order.
Total ways of selecting number with EXACTLY 3 factors (1,4) (1,9) (1,25) (1,49) = 4
Total ways of choosing the number = $$100C2$$= $$50*99$$
Hence, Probability = $$4/50*99$$ = $$2/25*99$$
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink] 31 Jan 2018, 20:53
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# Help!
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285
1
+1211
Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\frac 12-x\right)^{2001}=0$$, given that there are no multiple roots.
Jun 20, 2019
#1
+25342
+3
Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\dfrac 12-x\right)^{2001}=0$$,
given that there are no multiple roots.
$$\begin{array}{|lcll|} \hline \mathbf{x^{2001}+\left(\dfrac 12-x\right)^{2001}} &=& {0} \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{2001}x^{2001}&=& 0 \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - x^{2001}&=& 0 \\\\ \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} &=& 0 \\ \hline \end{array}$$
$$\begin{array}{|lcll|} \hline \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999}+- \ldots + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} &=& 0 \\\\ 1000.5x^{2000} - 500250 x^{1999}+- \ldots + \left(\dfrac12\right)^{2001} &=& 0 \quad | \quad : 1000.5 \\\\ x^{2000} - \dfrac{500250}{1000.5}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} - 500x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} \underbrace{- 500}_{=-\sum \limits_{k=1}^{2000} x_k}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline -\sum \limits_{k=1}^{2000} x_k &=& -500 \\ \mathbf{\sum \limits_{k=1}^{2000} x_k} &=& \mathbf{500} \\ \hline \end{array}$$
The sum of the roots is 500
Jun 20, 2019
#1
+25342
+3
Find the sum of the roots, real and non-real, of the equation $$x^{2001}+\left(\dfrac 12-x\right)^{2001}=0$$,
given that there are no multiple roots.
$$\begin{array}{|lcll|} \hline \mathbf{x^{2001}+\left(\dfrac 12-x\right)^{2001}} &=& {0} \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{2001}x^{2001}&=& 0 \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - x^{2001}&=& 0 \\\\ \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} &=& 0 \\ \hline \end{array}$$
$$\begin{array}{|lcll|} \hline \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999}+- \ldots + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} &=& 0 \\\\ 1000.5x^{2000} - 500250 x^{1999}+- \ldots + \left(\dfrac12\right)^{2001} &=& 0 \quad | \quad : 1000.5 \\\\ x^{2000} - \dfrac{500250}{1000.5}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} - 500x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} \underbrace{- 500}_{=-\sum \limits_{k=1}^{2000} x_k}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline -\sum \limits_{k=1}^{2000} x_k &=& -500 \\ \mathbf{\sum \limits_{k=1}^{2000} x_k} &=& \mathbf{500} \\ \hline \end{array}$$
The sum of the roots is 500
heureka Jun 20, 2019
|
Hw6 Problem 4
Problem 4
Let $p$ and $q$ be distinct primes.
(a) Find the number of generators of the cyclic group $\mathbb{Z}_{pq}$.
(b) Find the number of generators of the cyclic group $\mathbb{Z}_{p^r}$, where $r\in \mathbb{Z}^+$.
Solution
(a) There are $p-1$ multiples of $q$ and $q-1$ of $p$ that are less than $pq$. Thus there are $(pq-1)-(p-1)-(q-1)= pq - p - q +1=(p-1)(q-1)$ positive integers less than $pq$ and relatively prime to $pq$.
(b) There are $p^{r-1}-1$ multiples of $p$ less than $p^{r}$. Thus we see that there are $p^{r}-1)-(p^{r-1})=p^{r}-p^{r-1}=p^{r-1}(p-1)$ positive integers less than $p^{r}$ and relatively prime to $p^{r}$
|
A REMARK ON THE CONJUGATION IN THE STEENROD ALGEBRA
Title & Authors
A REMARK ON THE CONJUGATION IN THE STEENROD ALGEBRA
TURGAY, NESET DENIZ;
Abstract
We investigate the Hopf algebra conjugation, $\small{{\chi}}$, of the mod 2 Steenrod algebra, $\small{\mathcal{A}_2}$, in terms of the Hopf algebra conjugation, $\small{{\chi}^{\prime}}$, of the mod 2 Leibniz-Hopf algebra. We also investigate the fixed points of $\small{\mathcal{A}_2}$ under $\small{{\chi}}$ and their relationship to the invariants under $\small{{\chi}^{\prime}}$.
Keywords
Steenrod algebra;Hopf algebra;Leibniz-Hopf algebra;antipode;noncommutative symmetric functions;
Language
English
Cited by
References
1.
D. Arnon, Monomial bases in the Steenrod algebra, J. Pure Appl. Algebra 96 (1994), no. 3, 215-223.
2.
A. Baker and B. Richter, Quasisymmetric functions from a topological point of view, Math. Scand. 103 (2008), no. 2, 208-242.
3.
M. G. Barratt and H. R. Miller, On the anti-automorphism of the Steenrod algebra, Contemp. Math. 12 (1981), 47-52.
4.
M. D. Crossley, Some Hopf algebras of Words, Glasg. Math. J. 48 (2006), no. 3, 575-582.
5.
M. D. Crossley and N. D. Turgay, Conjugation invariants in the Leibniz-Hopf Algebra, J. Pure Appl. Algebra 217 (2013), no. 12, 2247-2254.
6.
M. D. Crossley and N. D. Turgay, Conjugation invariants in the mod 2 dual Leibniz-Hopf algebra, Comm. Algebra 41 (2013), no. 9, 3261-3266.
7.
M. D. Crossley and S. Whitehouse, On conjugation invariants in the dual Steenrod algebra, Proc. Amer. Math. Soc. 128 (2000), no. 9, 2809-2818.
8.
D. M. Davis, The anti-automorphism of the Steenrod algebra, Proc. Amer. Math. Soc. 44 (1974), 235-236.
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R. Ehrenborg, On posets and Hopf algebras, Adv. Math. 119 (1996), no. 1, 1-25.
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I. M. Gelfand, D. Krob, A. Lascoux, B. Leclerc, V. S. Retakh, and J. Y. Thibon, Noncommutative symmetric functions, Adv. Math. 112 (1995), no. 2, 218-348.
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V. Giambalvo and H. R. Miller, More on the anti-automorphism of the Steenrod algebra, Algebr. Geom. Topol. 11 (2011), no. 5, 2579-2585.
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M. Hazewinkel, The algebra of quasi-symmetric functions is free over the integers, Adv. Math. 164 (2001), no. 2, 283-300.
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M. Hazewinkel, Symmetric functions, noncommutative symmetric functions, and quasisymmetric functions, Acta Appl. Math. 75 (2003), 55-83.
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S. Kaji, Maple script for conjugation and product in the (dual) Leibniz-Hopf algebra and Steenrod algebra, available at http://www.skaji.org/code.
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C. Malvenuto and C. Reutenauer, Duality between quasi-symmetric functions and the Solomon descent algebra, J. Algebra 177 (1995), no. 3, 967-982.
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J. Milnor, The Steenrod algebra and its dual, Ann. of Math. 67 (1958), 150-171.
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J. Silverman, Conjugation and excess in the Steenrod algebra, Proc. Amer. Math. Soc. 119 (1993), no. 2, 657-661.
18.
N. E. Steenrod, Products of cocycles and extensions of mappings, Ann. of Math. 48 (1947), 290-320.
19.
N. E. Steenrod and D. B. A. Epstein, Cohomology operations, Ann. of Math. Stud. 50, Princeton Univ. Press, Princeton, 1962.
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P. D. Straffin, Identities for conjugation in the Steenrod algebra, Proc. Amer. Math. Soc. 49 (1975), 253-255.
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N. D. Turgay, An alternative approach to the Adem relations in the mod 2 Steenrod algebra, Turkish J. Math. 38 (2014), no. 5, 924-934.
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G. Walker and R. M. W. Wood, The nilpotence height of $Sq^{2n}$, Proc. Amer. Math. Soc. 124 (1996), no. 4, 1291-1295.
23.
R. M. W. Wood, Problems in the Steenrod algebra, Bull. London Math. Soc. 30 (1998), no. 5, 449-517.
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Question
# What is the valency of the cation, if it is known that the valency of each anion is -4.? −4 +4 −2 Can't Say
Solution
## The correct option is D Can't Say Since the chemical formula of the compound is not known, we can't predict the valency of Cation.
Suggest corrections
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# Lagrangian and Euler-Lagrange of a Simple Pendulum
1. Sep 27, 2015
### Yeah Way
1. The problem statement, all variables and given/known data
A simple pendulum with mass m and length ℓ is suspended from a point which moves
horizontally with constant acceleration a
> Show that the lagrangian for the system can be written, in terms of the angle θ,
L(θ, θ, t˙ ) = m/2(ℓ^2θ˙^2 + a^2t^2 − 2aℓtθ˙ cosθ) + mgℓ cos θ
> Determine the Euler–Lagrange equation for the system.
2. Relevant equations
3. The attempt at a solution
I thought I could prove that l^2θdot^2 + a^2t^2 - 2altθdotCosθ was v^2 using relative velocities: v^2 = (at - lθdot)^2 = (l^2θ^2 + a^2t^2 - 2altθdot). But I've no idea where the Cosθ is coming from, so I can only assume I'm wrong somewhere.
I also can't understand how V = -mglCosθ
h for this pendulum should be l(1 - Cosθ) shouldn't it?
Any help's appreciated. Thanks.
#### Attached Files:
• ###### Mech Phys HW.pdf
File size:
22.5 KB
Views:
57
2. Sep 27, 2015
### CAF123
It can indeed be solved via relative velocities, but it would be much simpler to just write the generic position vector for the mass, noting that it has an additional horizontal component of velocity of $at$.
It depends on where you choose your zero of potential but, regardless, at the level of the lagrangian an additive constant makes no difference to the kinematics since they are obtained through taking derivatives of the lagrangian through the Euler-Lagrange equations.
3. Sep 27, 2015
### Yeah Way
Are you saying the − 2aℓtθ˙ cosθ is part of the angular velocity? I'm sorry, I'm really at a loss with this.
4. Sep 28, 2015
### CAF123
The first two terms of the lagrangian can be attributed to a rotational kinetic energy of the mass about the pendulum pivot and the translational kinetic energy due to the additional horizontal component of velocity imposed on it. The interpretation of the term with cos θ in it is not so easy.
If you write the position vector for the mass to an inertial frame and take its time derivative you will have $\bf \dot{r}$. Then the kinetic energy to this frame is given by $\frac{1}{2}m \bf{\dot{r}^2}$. Does it help?
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extension and restriction of states
0.1 Restriction of States
Let $\mathcal{A}$ be a $C^{*}$-algebra (http://planetmath.org/CAlgebra) and $\mathcal{B}\subset\mathcal{A}$ a $C^{*}$-subalgebra, both having the same identity element.
$\,$
- Given a state $\phi$ of $\mathcal{A}$, its restriction (http://planetmath.org/RestrictionOfAFunction) $\phi|_{\mathcal{B}}$ to $\mathcal{B}$ is also a state of $\mathcal{B}$.
$\,$
Remark - Note that the requirement that the $C^{*}$-algebras $\mathcal{A}$ and $\mathcal{B}$ have a (common) identity element is necessary.
For example, let $X$ be a compact space and $C(X)$ the $C^{*}$-algebra of continuous functions $X\to\mathbb{C}$. Pick a point $x_{0}\in X$ and consider the $C^{*}$-subalgebra of continuous functions $X\to\mathbb{C}$ which vanish at $x_{0}$. Notice that this subalgebra never has the same identity element of $C(X)$ (the constant function that equals $1$). In fact, this subalgebra may not have an identity at all.
Now the evaluation mapping at $x_{0}$, i.e. the function $\mathrm{ev}_{x_{0}}:C(X)\to\mathbb{C}$
$\displaystyle\mathrm{ev}_{x_{0}}(f):=f(x_{0})$
is a state of $C(X)$. Of course, its restriction to the subalgebra in question is the zero mapping, therefore not being a state.
0.2 Extension of States
Let $\mathcal{A}$ be a $C^{*}$-algebra and $\mathcal{B}\subset\mathcal{A}$ a $C^{*}$-subalgebra (not necessarily unital).
$\,$
Theorem 1 - Every state $\phi$ of $\mathcal{B}$ admits an extension to a state $\widetilde{\phi}$ of $\mathcal{A}$. Moreover, every pure state $\phi$ of $\mathcal{B}$ admits an extension to a pure state $\widetilde{\phi}$ of $\mathcal{A}$.
Theorem 2 - The set of extensions of a state $\phi$ of $\mathcal{B}$ is a compact and convex subset of $S_{\mathcal{A}}$, the of $\mathcal{A}$ endowed with the weak-* topology.
Title extension and restriction of states ExtensionAndRestrictionOfStates 2013-03-22 18:09:35 2013-03-22 18:09:35 asteroid (17536) asteroid (17536) 10 asteroid (17536) Theorem msc 46L30 State
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$$\require{cancel}$$
# Hyperfine Structure
To this point, the nucleus has been assumed to interact with the electron only through its electric field. However, like the electron, the proton has spin angular momentum with s=1/2, and associated with this angular momentum is an intrinsic dipole moment
where M is the proton mass and is a numerical factor known experimentally to be Note that the proton dipole moment is weaker than the electron dipole moment by roughly a factor of , and hence one expects the associated effects to be small, even in comparison to fine structure, so again treating the corrections as a perturbation is justified. The proton dipole moment will interact with both the spin dipole moment of the electron and the orbital dipole moment of the electron, and so there are two new contributions to the Hamiltonian, the nuclear spin-orbit interaction and the spin-spin interaction. The derivation for the nuclear spin-orbit Hamiltonian is the same as for the electron spin-orbit Hamiltonian, except that the calculation is done in the frame of the proton and hence there is no factor of 1/2 from the Thomas precession. The nuclear spin-orbit Hamiltonian is
The spin-spin Hamiltonian can be derived by considering the field produced by the proton spin dipole, which can be written
The first term is just the usual field associated with a magnetic dipole, but the second term requires special explanation. Normally, when one considers a dipole field, it is implicit that one is interested in the field far from the dipole--that is, at distances far from the source compared to the size of the current loop producing the dipole. However, every field line outside the loop must return inside the loop, as shown in figure 6. If the size of the current loop goes to zero, then the field will be infinite at the origin, and this contribution is what is reflected by the second term in equation 77.
Figure 6: The field of a magnetic dipole. All field lines cross the plane of the dipole going up inside the loop and down outside the loop.
The electron has additional energy
due to the interaction of its spin dipole with this field, and hence the spin-spin Hamiltonian is
The operator does not commute with this Hamiltonian. However, one can define the total angular momentum
The corresponding operators and commute with the Hamiltonian, and they introduce new quantum numbers f and through the relations
The quantum number f has possible values f=j+1/2,j-1/2 since the proton is spin 1/2, and hence every energy level associated with a particular set of quantum numbers nl, and j will be split into two levels of slightly different energy, depending on the relative orientation of the proton magnetic dipole with the electron state.
Consider first the case l=0, since the hyperfine splitting of the hydrogen atom ground state is of the most interest. Since the electron has no orbital angular momentum, there is no nuclear spin-orbit effect. It can be shown that because the wavefunction has spherical symmetry, only the delta function term contributes from the spin-spin Hamiltonian. First order perturbation theory yields
Like the Darwin term, this depends on the probability of finding the electron at the origin. The value of can be found by squaring , which with l=0 gives
Hence
where the last step includes the values . The hyperfine energy shift for l=0 is then
It is easy to see from this expression that the hyperfine splittings are smaller than fine structure by a factor of M/m. For the specific case of the ground state of the hydrogen atom (n=1), the energy separation between the states of f=1 and f=0 is
The photon corresponding to the transition between these two states has frequency and wavelength
This is the source of the famous 21 cm line,'' which is extremely useful to radio astronomers for tracking hydrogen in the interstellar medium of galaxies. The transition is exceedingly slow, but the huge amounts of interstellar hydrogen make it readily observable. It is too slow to be seen in a terrestrial laboratory by spontaneous emission, but the frequency can be measured to very high accuracy by using stimulated emission, and this frequency is in fact one of the best-known numbers in all of physics.
For , the term does not contribute but the other terms in the spin-spin Hamiltonian as well as the nuclear spin-orbit Hamiltonian do contribute. The calculation is much harder but yields
for .
Figure 7 shows a revised version of the structure of the hydrogen atom, including the Lamb shift and hyperfine structure. Note that each hyperfine state still has a 2f+1 degeneracy associated with the different possible values of which correspond to different orientations of the total angular momentum with respect to the z-axis. For example, in the ground state, the higher-energy state f=1 is actually a triplet, consisting of three degenerate states, and the f=0 state is a singlet. This degeneracy can be broken by the presence of an external magnetic field.
Figure 7: Some low-energy states of the hydrogen atom, including fine structure, hyperfine structure, and the Lamb shift.
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### Author Topic: Section 1.4: Question 29 Proof Check (Read 777 times)
#### Kuba Wernerowski
• Jr. Member
• Posts: 7
• Karma: 0
##### Section 1.4: Question 29 Proof Check
« on: October 02, 2020, 01:35:40 PM »
My proof follows roughly the same logic as the outline in the textbook, but I'm not sure if it's quite rigorous enough.
Choose $p_a, p_b \in D$. Since $D$ is a domain, $\exists$ a polygonal curve $P_1 P_2 \cup P_2 P_3 \cup \cdots P_{n-1}P_n$ connecting $p_a$ and $p_b$.
Since $D$ is open, each point line segment $P_1, \ldots P_n$ have an open disc $A_j$ centered at $A_j$ where $j = 1, 2, \ldots, n$.
Construct the polygonal curve such that for each pair of endpoints, $P_j, P_{j+1}$, their respective open discs $A_j, A_{j+1}$ have the property that $A_j \cap A_{j+1} ≠ \emptyset$ for $j=1, 2,\ldots, n-1$.
Then, since $u$ has the property that for each of those open discs, $u(A_j)$ = some constant, $c_j$.
$u(A_1) = c_1$, $u(A_2) = c_2$, and given that $A_1 \cap A_2 \neq \emptyset$, then $c_1 = c_2$.
Repeat this argument for each pair of $A_j$ and $A_{j+1}$ until we reach $A_{n-1} \cap A_n \neq \emptyset \implies u(A_{n-1}) = u(A_n)$.
To conclude the proof, $p_a \in A_1$ and $p_b \in A_n$ meaning that $u(p_a) = u(p_b)$, and since $p_a, p_b$ were chosen arbitrarily, $u$ is constant on $D.$.
Any feedback / criticisms are much appreciated
#### Victor Ivrii
• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Section 1.4: Question 29 Proof Check
« Reply #1 on: October 03, 2020, 04:45:02 AM »
You should remember that each segment of the polygonal curve may be served not by two discs, but several discs (think why)
#### Kuba Wernerowski
• Jr. Member
• Posts: 7
• Karma: 0
##### Re: Section 1.4: Question 29 Proof Check
« Reply #2 on: October 07, 2020, 09:38:32 AM »
For any polygonal segment $P_i P_j$ where that's the case, could we divide that segment into smaller sub-segments $P_{i_1} P_{i_2} \cdots P_{i_{n-1}}P_j$ and put open, overlapping discs over each sub-segments end/start points?
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# Eigenvalues proof
Gold Member
## Homework Statement
Let A=LU and B=UL, where U is an upper triangular matrix and L is a lower triangular matrix. Demonstrate that A and B have the same eigenvalues.
Not sure.
## The Attempt at a Solution
I know that if I can show that A and B are similar (so if I can find a matrix P such that P^(-1)AP=B) they have the same eigenvalues. But I didn't find P yet, nor do I know really how to search efficiently for P.
Another route I've thought of is to write det (I*lambda-A)=0 gives the same values for lambda as det (I*lambda-B)=0. I've thought of using det (A)=det (LU)=det L * det U = det U * det L = det B... but still can't reach anything I find useful.
Any tip is greatly appreciated.
## Answers and Replies
lanedance
Homework Helper
how about multiplying thorugh by the inverse of L.. you may need to show it exists
Try to write A in terms of B (or the other way around), and see what you get!
Last edited:
lanedance
Homework Helper
hint was probably enough
Gold Member
LU=U^(-1)ULU=LU so the equality is true. Notice that LU=A and UL=B. I have that P=U, thus LU is similar to UL (A similar to B) so they share the same eigenvalues.
Now I must justify the use of the inverse of U. Well I believe its element on the diagonal are all non zero. So if the matrix is nxn, the span of its column vector is R^n so it is invertible, hence U^-1 exists.
Is that well justified?
Dick
Homework Helper
LU=U^(-1)ULU=LU so the equality is true. Notice that LU=A and UL=B. I have that P=U, thus LU is similar to UL (A similar to B) so they share the same eigenvalues.
Now I must justify the use of the inverse of U. Well I believe its element on the diagonal are all non zero. So if the matrix is nxn, the span of its column vector is R^n so it is invertible, hence U^-1 exists.
Is that well justified?
It would be if the problem stated that the elements of U along the diagonal are nonzero. Does it? Other versions of this problem I've found say that L or U is a UNIT triangular matrix. Did you miss a word in the problem statement?
Last edited:
Gold Member
It would be if the problem stated that the elements of U along the diagonal are nonzero. Does it? Other versions of this problem I've found say that L or U is a UNIT triangular matrix. Did you miss a word in the problem statement?
You pointed out a very interesting thing to me. In my assignment it isn't specified. But most of the exercises assigned are taken from Kincaid's book on numerical analysis. I just checked there and it clearly states L to be UNITARY.
If only they had stated this in my assignment I could have guessed that I had to take the inverse of L. :) Well I'm not 100% sure but probably.
So basically the problem is solved now... thanks guys.
Dick
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# Building labels out of counters
There is a simple version to my question, and a complex version. First, the simple version: I want to generate label names by combining a constant string with a counter's value, something like what is done here:
http://tug.org/pipermail/texhax/2006-July/006599.html
\newcommand{\addnextlabel}{\stepcounter{foo}\label{bar:\arabic{foo}}}
But when I use this, I get a bunch of warnings about multiply-defined labels. It appears they are all getting named the same thing, using the highest value of counter foo. I guess this has to do with the order TeX evaluates things, as suggested by the answer in the link above. Can anyone do a better job explaining this than the response there?
Now the more complex problem, and what I'm really trying to solve: I'm using the lineno package, which has a \linelabel command that works much like \label. You can get the label's line number using \ref{labelname}. It's these labels I really want to number with a counter. I tried applying the linked solution to \linelabel, but I couldn't get it to work. Any ideas how I might do this?
-
Hang on a second. After thinking about this again, if you need to type \ref{bar:7} to get the counter number 7, why do you need the label in the first place? – Will Robertson Dec 2 '10 at 17:12
1. You need to use \refstepcounter instead of \stepcounter. This must be badly documented somewhere — lots of people make this mistake.
2. You might need to save the linelabel counter to a counter of your own, decrement it, and then \refstepcounter it again. (Hope this explanation's not too brief; I'm in a hurry.)
-
your \addnextlabel works fine in standard latex (for me) -- you're presumably using one of the myriad packages that fiddles with the reference mechanism.
i would expect
\newcommand{\addnextlabel}{\stepcounter{foo}%
\edef\baz{\noexpand\label{bar:arabic{foo}}}%
\baz
}
to do the job for you, if my hypothesis is correct.
-
The original \addnextlabel doesn't work to refer back to the labels that it's trying to set. – Will Robertson Dec 2 '10 at 17:14
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Elementary Algebra
$n(n+5)(n+8)$
Each term has $n$ in it, so factor out $n$ to obtain: $=n(n^2+13n+40)$ The leading coefficient of the trinomial is 1. This means that it can be factored by looking for factors of the constant term (40) whose sum is equal to the numerical coefficient of the middle term (13). Note that $40=(5)(8)$ and that $5 + 8 = 13$, the numerical coefficient of the middle term. The factors of $40$ whose sum is $13$ are $5$ and $8$. Thus, the factored form of the trinomial is: $=(n+5)(n+8)$ Therefore $n(n^2+13n+40)$ can be factored as: $=n(n+5)(n+8)$
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# zbMATH — the first resource for mathematics
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Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
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Nano boundary layers over stretching surfaces. (English) Zbl 1221.76024
Summary: We present similarity solutions for the nano boundary layer flows with Navier boundary condition. We consider viscous flows over a two-dimensional stretching surface and an axisymmetric stretching surface. The resulting nonlinear ordinary differential equations are solved analytically by the Homotopy Analysis Method. Numerical solutions are obtained by using a boundary value problem solver, and are shown to agree well with the analytical solutions. The effects of the slip parameter $K$ and the suction parameter $s$ on the fluid velocity and on the tangential stress are investigated and discussed. As expected, we find that for such fluid flows at nano scales, the shear stress at the wall decreases (in an absolute sense) with an increase in the slip parameter $K$.
##### MSC:
76A05 Non-Newtonian fluids 34E13 Multiple scale methods (ODE) 65L99 Numerical methods for ODE
##### References:
[1] Navier, C. L. M.H.: Mémoire sur LES lois du mouvement des fluids, Mém acad R sci inst France 6, 389 (1823) [2] Shikhmurzaev, Y. D.: The moving contact line on a smooth solid surface, Int J multiphase flow 19, 589 (1993) · Zbl 1144.76452 · doi:10.1016/0301-9322(93)90090-H [3] Choi CH, Westin JA, Breuer KS. To slip or not to slipwater flows in hydrophilic and hydrophobic microchannels. In: Proceedings of IMECE 2002, New Orleans, LA, Paper No. 2002-33707. [4] Matthews, M. T.; Hill, J. M.: Nano boundary layer equation with nonlinear Navier boundary condition, J math anal appl 333, 381 (2007) · Zbl 1207.76050 · doi:10.1016/j.jmaa.2006.08.047 [5] Wang, C. Y.: Analysis of viscous flow due to a stretching sheet with surface slip and suction, Nonlinear anal real world appl 10, 375 (2009) · Zbl 1154.76330 · doi:10.1016/j.nonrwa.2007.09.013 [6] Wang, C. Y.: Flow due to a stretching boundary with partial slip – an exact solution of the Navier – Stokes equations, Chem eng sci 57, 3745 (2002) [7] Xu, H.; Liao, S. J.; Wu, G. X.: A family of new solutions on the wall jet, Eur J mech B/fluid 27, 322-334 (2008) · Zbl 1154.76335 · doi:10.1016/j.euromechflu.2007.07.002 [8] Liao, S. J.: A new branch of boundary layer flows over a permeable stretching plate, Int J non-linear mech 42, 819-830 (2007) · Zbl 1200.76046 · doi:10.1016/j.ijnonlinmec.2007.03.007 [9] Liao, S. J.: A new branch of solutions of boundary-layer flows over an impermeable stretched plate, Int J heat mass transfer 48, 2529-2539 (2005) · Zbl 1189.76142 · doi:10.1016/j.ijheatmasstransfer.2005.01.005 [10] Crane, L. J.: Flow past a stretching plate, Z angew math phys 21, 645 (1970) [11] Liao SJ. On the proposed homotopy analysis techniques for nonlinear problems and its application. Ph.D. dissertation, Shanghai Jiao Tong University; 1992. [12] Liao, S. J.: Beyond perturbation: introduction to the homotopy analysis method, (2003) [13] Liao, S. J.: An explicit, totally analytic approximation of Blasius viscous flow problems, Int J non-linear mech 34, 759 (1999) [14] Liao, S. J.: On the homotopy analysis method for nonlinear problems, Appl math comput 147, 499 (2004) · Zbl 1086.35005 · doi:10.1016/S0096-3003(02)00790-7 [15] Liao, S. J.; Tan, Y.: A general approach to obtain series solutions of nonlinear differential equations, Stud appl math 119, 297 (2007) [16] Liao, S. J.: Notes on the homotopy analysis method: some definitions and theorems, Commun nonlinear sci numer simul 14, 983 (2009) · Zbl 1221.65126 · doi:10.1016/j.cnsns.2008.04.013 [17] U. Ascher, R. Mattheij, and R. Russell, Numerical solution of boundary value problems for ordinary differential equations, In: SIAM Classics in Applied Mathematics, No. 13; 1995. · Zbl 0843.65054 [18] Ascher, U.; Petzold, L.: Computer methods for ordinary differential equations and differential-algebraic equations, (1998) · Zbl 0908.65055
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### Home > INT2 > Chapter 5 > Lesson 5.1.2 > Problem5-23
5-23.
Match each table of data below with the most appropriate graph, and then briefly explain how you know the graph matches the data.
1. The temperature in an oven when cooking a roast.
Time (min) $0$ $5$ $15$ $30$ $60$ $120$ $180$ Temperature ($º\text{F}$) $70$ $200$ $300$ $325$ $325$ $325$ $325$
The temperature rises immediately but as time passes it levels-off.
Graph 1
2. The time needed for an athlete to run a certain distance.
Distance (meters) $100$ $200$ $400$ $800$ $1600$ $3000$ Time (min:sec) 0:12 0:25 1:10 2:24 5:20 10:59
Graph 2
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However, rather then baseline hazard function (delta_0(t)) acting as a constant, I would like to incorporate a probability distribution based on event time duration (this is something I have wrote myself) to ensure that the baseline hazard function … So a good choice would be to include only students who have advanced to candidacy (in other words, they’ve passed all their qualifying exams). hazard function. where h(t; x) is the hazard function at time t for a subject with covariate values x 1, … x k, h 0(t) is the baseline hazard function, i.e., the hazard function when all covariates equal zero. Learn the key tools necessary to learn Survival Analysis in this brief introduction to censoring, graphing, and tests used in analyzing time-to-event data. PROC PHREG performs a stratified analysis to adjust for such subpopulation differences. We used these data in our simulation study. This date will be time 0 for each student. σ is a variance-like parameter on log-time scale. The first of these, $$h_0(t)$$, is called the baseline hazard function or the hazard for a reference individual with covariate values 0. PH generator Generates data from proportional hazard model with piecewised baseline hazard function. This website uses cookies to improve your experience while you navigate through the website. If the hazard ratio estimate is less than one, this means that the hazard function … Simple transformations can be applied to the Bre- slow estimator to provide estimation of the baseline and conditional survival functions. the term h 0 is called the baseline hazard. In survival analysis, the hazard function is a useful way to describe the distribution of survival times. 877-272-8096 Contact Us. ), in the Cox … Survival analysis is used for modeling and analyzing survival rate (likely to survive) and hazard rate (likely to die). What do we mean by validating a prognostic model? In the first year, that’s 15/500. 3.7.1 An example… Stratified Proportional Hazards Models. the hazard function associated with a set of possibly time-varying covariates is the sum of, rather than the product of, the baseline hazard function and the regression function of covariates. Some features of the site may not work correctly. Proportional hazard In particular, consider this model: (tjX) = 0(t)r(X 0 ) Here 0(t) 0 is called the \baseline" hazard, which describes how the hazard changes with time. is usually called a scale parameter. And It … PH generator Generates data from proportional hazard model with piecewised baseline hazard function. Let’s say that for whatever reason, it makes sense to think of time in discrete years. 3–34). The baseline is for all covariates set equal to zero. As the hazard function $$h(t)$$ is the derivative of the cumulative hazard function $$H(t)$$, we can roughly estimate the rate of change in $$H(t)$$ by taking successive differences in $$\hat H(t)$$ between adjacent time points, $$\Delta \hat H(t) = \hat H(t_j) – \hat H(t_{j-1})$$. The aim of fitting a Cox model to time-to-event data is to estimate the effect of covariates on the baseline hazard function. Here we load a dataset from the lifelines package. S0(t) = 1 for t < t1. Briefly, the hazard function can be interpreted as the risk of dying at time t. ... the term $$h_0$$ is called the baseline hazard. If the hazard ratio estimate is less than one, this means that the hazard function for the first group is smaller than that for the second group. This representation has been fruitfully employed in several types of studies, such as in econometrics (for example, Lancaster (1979) and … The Cox partial likelihood, shown below, is obtained by using Breslow's estimate of the baseline hazard function, plugging it into the full likelihood and then observing that the result is a product of two factors. In fact we can plot it. The concept is the same when time is continuous, but the math isn’t. The first factor is the partial likelihood shown below, in which the baseline hazard … Here we load a dataset from the lifelines package. The second year hazard is 23/485 = .048. The model assumes that the hazard function is composed of two non-negative functions: a baseline hazard function, λ 0 (t), and a risk score, r(x)=e h(x), defined as the effect of an individual’s observed covariates on the baseline hazard . Regardless of the values covariates, all subjects share the same baseline hazard λₒ. β0 (t) is the baseline hazard function and it is defined as the probability of experiencing the event of interest when all other covariates equal zero. Interpretation of a proportional hazards model 1. Some care should be taken in interpreting the baseline survivor function, which roughly plays the role of the intercept term in a regular linear regression. The hazard is the probability of the event occurring during any given time point. To access the coefficients and the baseline hazard directly, you can use params_ and baseline_hazard_ respectively. The BASELINE statement creates a new SAS data set that contains the baseline function estimates at the event times of each stratum for every set of covariates given in the COVARIATES= data set.If the COVARIATES= data set is not specified, a reference set of covariates consisting of the reference levels for the CLASS variables and the average values for the continuous variables is used. But technically, it’s the same thing. A typical medical example would include covariates such as treatment assignment, as well … The bottom … values at which the baseline hazard will be evaluated. The hazard ratio is the ratio of the hazard functions between two population groups. For more information, see the section OUT= Output Data Set in the BASELINE … hazard ratio for a unit change in X Note that "wider" X gives more power, as it should! Unless the hazard function … The hazard is the probability of the event occurring during any given time point. In other words, the probability of death in a time interval [t;t+y] does not depend on the starting point,t. The assumption is that the baseline hazard function … if TRUE baseline_hazard will smooth the estimated baseline hazard using Friedman's super smoother supsmu. The Cox model is expressed by the hazard function denoted by h(t). σ is a variance-like parameter on log-time scale. For example, it may not be important if a … Bayesian information criterion for censored survival models. Tagged With: Cox Regression, discrete, Event History Analysis, hazard function, Survival Analysis, Data Analysis with SPSS Using the Schoenfeld residuals one can estimate cloglog of the baseline hazard function and apply the appropriate transformations to obtain a very inefficient estimate of the baseline hazard function. Here we start to plot the cumulative hazard, which is over an interval of time rather than at a single instant. It is easier to understand if time is measured discretely, so let’s start there. All rights reserved. Hazard ratios differ from relative risks and odds ratios in … Taking a look at these coefficients for a moment, prio (the number of prior arrests) has a coefficient of about 0.09. Survival, cumulative hazard and log-cumulative hazard functions for model D in table 14.1 for age set at its mean and varying the values of the other predictors personal and property. The baseline function is raised to the power of the exp (xʹβ) factor coming from the covariates. It is the value of the hazard if all xs are zero; the quantities exp(bi xi)are called ‘hazard ratios’ (HR); Things to note here: t in the hazard function indicates that the hazard … The survivor function is deflned as SY (y) = P(Y > y) = 1 ¡FY (y): In other words, the survivor function … the hazard function associated with a set of possibly time-varying covariates is the sum of, rather than the product of, the baseline hazard function and the regression function of covariates. That is the number who finished (the event occurred)/the number who were eligible to finish (the number at risk). I am only looking at 21… The method represents the effects of explanatory variables as a multiplier of a common baseline hazard function, h 0 (t). hazard function. The baseline hazard function, not itself estimated within the model, is the hazard function obtained when all covariate are set to zero. To access the coefficients and the baseline hazard directly, you can use params_ and baseline_hazard_ respectively. We include in the model predictors that satify the proportional hazard … h(t) is the hazard function determined by a set of p covariates (x1, x2, …, xp) the coefficients (b1, b2, …, bp) measure the impact (i.e., the effect size) of covariates. What is Survival Analysis and When Can It Be Used? The baseline hazard function, not itself estimated within the model, is the hazard function obtained when all covariate are set to zero. It corresponds to the value of the hazard if all the x i are equal to zero (the quantity exp (0) equals 1). SAS computes differences in the Nelson-Aalen estimate of $$H(t)$$. Thus, the predictors have a multiplicative or proportional effect on the predicted hazard. The BASELINE statement creates a new SAS data set that contains the baseline function estimates at the event times of each stratum for every set of covariates given in the COVARIATES= data set.If the … Figure 14.5, page 541. A key application is model-based prediction of survival…, A hybrid method to estimate the full parametric hazard model, Asymptotic normality of corrected estimator in Cox proportional hazards model with measurement error, Consistent estimation in Cox proportional hazards model with measurement errors and unbounded parameter set, A semi-parametric Probability of Default model, A Framework for Treatment Decision Making at Prostate Cancer Recurrence, A deep learning model for early prediction of Alzheimer's disease dementia based on hippocampal MRI. The problem was that what was returned by the old basehazard() option was not (and what is returned by the new basehc() option is not) the baseline hazard; it is the numerator of the baseline hazard, called the hazard contribution by Kalbfleisch and Prentice (2002, p. 115, eq. If time is truly continuous and we treat it that way, then the hazard is the probability of the event occurring at any given instant. The hazard function is the nonparametric part of the Cox proportional hazards regression function, whereas the impact of the predictor variables is a loglinear regression. Proportional Hazards Model the regression coefficients are assumed to be the same for each stratum although the baseline hazard functions may be different and completely unrelated. In this model λ 0 (t) is a baseline hazard function that describes the risk for individuals with x i = 0, who serve as a reference cell or pivot, and exp { x i ′ β } is the relative risk, a proportionate increase or reduction in risk, associated with the set of characteristics x i. where , , and the baseline hazard function is . Left panels: Baseline survivor function and cumulative hazard function for model D in Table 14.1. It corresponds to the value of the hazard … Calculating when elective abdominal aortic aneurysm repair improves survival for individual patients: development of the Aneurysm Repair Decision Aid and economic evaluation. Statistical Consulting, Resources, and Statistics Workshops for Researchers. I am only looking at 21… Thus, a one unit increase in prio means the the baseline hazard … And r(X0 ) describes how the hazard changes as a function … The constant hazard function is a consequence of thememoryless propertyof the exponential distribution: the distribution of the subject’s remaining survival timegiventhat s/he has survived till timetdoes not depend ont. We can then fit models to predict these hazards. These cookies do not store any personal information. The baseline hazard function, not itself estimated within the model, is the hazard function obtained when all covariate are set to zero. exp is the exponential function … The hazard function … It feels strange to think of the hazard of a positive outcome, like finishing your dissertation. by Stephen Sweet andKaren Grace-Martin, Copyright © 2008–2020 The Analysis Factor, LLC. where the λₒ is an arbitrary function of time, the baseline hazard. In interval k, given by [ τk−1, τk), the hazard function for a given subject is assumed to be constant and is related to the baseline hazard function by the function, where λk is the baseline hazard function in the k ‐th interval. So for each student, we mark whether they’ve experienced the event in each of the 7 years after advancing to candidacy. if TRUE the cumulative survival function will be computed By Property 1 of Survival Analysis Basic Concepts, the baseline cumulative hazard function is. In survival analysis, the hazard ratio is the ratio of the hazard rates corresponding to the conditions described by two levels of an explanatory variable. This formulation describes a different aspect of the association between covari- But opting out of some of these cookies may affect your browsing experience. Interpretation¶. where , , and the baseline hazard function is . Since it’s so important, though, let’s take a look. This has facilitated the use of the estimator in scientific studies. If you’re not familiar with Survival Analysis, it’s a set of statistical methods for modelling the time until an event occurs. The Breslow estimator for the cumulative baseline hazard function has been imple- mented in all major statistical software packages. The hazard ratio is the ratio of the hazard functions between two population groups. For example, in a drug study, the treated population may die at twice the rate per unit time as the control population. where is the baseline hazard function for the i th stratum and is the vector of explanatory variables for the individual. In this approach, the baseline covariate function is estimated using historical failure data and condition monitoring data. So Cox’s proportional hazards model is a semiparametric model. If you omit the OUT= option, the data set is created and given a default name by using the DATA n convention. Step 2. The second quantity is a parametric component which is a linear function of a … We can then calculate the probability that any given student will finish in each year that they’re eligible. This … Below we see that the hazard is pretty low in years 1, 2, and 5, and pretty high in years 4, 6, and 7. Now let’s say that in the second year 23 more students manage to finish. In several applications, it is important to have an explicit, preferably smooth, estimate of the baseline hazard function… BIOST 515, Lecture 15 4. The input includes the covariates matrix, the coefficients for covariates, number of sample size, and the baseline hazard function… eval_times. For example, perhaps the trajectory of hazards is different depending on whether the student is in the sciences or humanities. where h(t; x) is the hazard function at time t for a subject with covariate values x 1, … x k, h 0(t) is the baseline hazard function, i.e., the hazard function when all covariates equal zero. You are currently offline. However, it may not be applicable when the assumption of constant Lecture 32: Survivor and Hazard Functions (Text Section 10.2) Let Y denote survival time, and let fY (y) be its probability density function.The cdf of Y is then FY (y) = P(Y • y) = Z y 0 fY (t)dt: Hence, FY (y) represents the probability of failure by time y. where h 0 (t) is the baseline hazard function (Cox, 1972). Otherwise, let te = the largest censored time. If you continue we assume that you consent to receive cookies on all websites from The Analysis Factor. But like a lot of concepts in Survival Analysis, the concept of “hazard” is similar, but not exactly the same as, its meaning in everyday English. the term h0 is called the baseline hazard. Epidemiology: non-binary exposure X (say, amount of smoking) Adjust for confounders Z (age, sex, etc. if TRUE baseline_hazard will smooth the estimated baseline hazard using Friedman's super smoother supsmu. The aim of fitting a Cox model to time-to-event data is to estimate the effect of covariates on the baseline hazard function. If you’re familiar with calculus, you know where I’m going with this. (4th Edition) The accelerated failure-time form of the hazard function … Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Cox (1972) proposed r(X0) = exp(X), resulting in what became called the Cox Proportional Hazards (CPH) model: (tjX) = if TRUE the cumulative survival function … That’s the hazard. To convert what is returned to a baseline hazard… Modelling the effects of standard prognostic factors in node-positive breast cancer, Reporting performance of prognostic models in cancer: a review, Regression Modeling Strategies: With Applications to Linear Models, Logistic Regression, and Survival Analysis, Multivariable Model-Building: A Pragmatic Approach to Regression Analysis based on Fractional Polynomials for Modelling Continuous Variables. Sometimes the model is expressed differently, relating the relative hazard, which is the ratio of the hazard at time t to the baseline hazard, to the risk factors: That is, the intercept term serves to scale the baseline hazard. Likewise we have to know the date of advancement for each student. Step 1. And r(X0) describes how the hazard changes as a function of the covariates X. Thereafter, adjustments are made based on the covariates. Hazard function: h(t) def= lim h#0 P[t T
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Vector Product - MapleSim Help
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : MapleSim : MapleSim Component Library : Signal Blocks : Mathematical : Operators : componentLibrary/signalBlocks/math/operators/VectorProduct
Vector Product
Compute element-wise product of vectors
Description The Vector Product component computes the element-wise product of two vectors.
Equations ${y}_{j}={\mathrm{u1}}_{j}{\mathrm{u2}}_{j}$
Connections
Name Description Modelica ID $\mathrm{u1}$ Real input vector u1 $\mathrm{u2}$ Real input vector u2 $y$ Real output vector y
Parameters
Name Default Units Description Modelica ID $n$ $3$ Dimension of input and output vectors n
Modelica Standard Library The component described in this topic is from the Modelica Standard Library. To view the original documentation, which includes author and copyright information, click here.
See Also
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# Probability space as countable union of atoms and specific event
Let $(\Omega, \mathscr{A}, \mathbb{R})$ be a probability space.
We define an atom of probability as an event $A \in \mathscr{A}$ such that $\mathbb{P}(A)>0$ and for all events $D \subset A$, $\mathbb{P}(D) = 0$ or $\mathbb{P}(D)=\mathbb{P}(A)$.
What my textbook states about atoms is (and I understand the proof of both of these statements): if $A$ and $B$ are atoms, then either $\mathbb{P}(A \cap B)=\mathbb{P}(A)=\mathbb{P}(B)$ or $\mathbb{P}(A \cap B)=0$, and there can be at most countably many disjoint (up to an event of zero probability) atoms. I'm not sure if this is useful for my question, but I stated it here anyway.
Now, for my question: prove that $\Omega$ can be written as a union $$\Omega = (\bigcup_{n=0}^{+\infty}A_{n}) \cup X,$$ where $\{A_{n}\}_{n \in \mathbb{N}}$ is a sequence of atoms (does not specify what their intersections are), and $X$ is an event with the following property: for any $p \in (0, P(X))$, there exists an event $B \subset X$ such that $P(B)=p$.
My idea was to take a maximal family of disjoint (again, up to an event of zero probability) atoms, which is at most countable, $\{A_{n}\}_{n \in \mathbb{N}}$, and consider $X := \Omega \setminus \cup_{n=0}^{+\infty}A_{n}$, and prove the required property for that set.
If $Y \subset X$ is an atom, then, seeing as how my family is maximal, $Y$ must non-trivially intersect some $A_{i}$ from my sequence, and therefore $\mathbb{P}(Y\cap A_{i})=\mathbb{P}(A_{i})>0$, but also since $Y \subset X$, $Y \cap A_{i}= \emptyset$, so $\mathbb{P}(Y \cap A_{i})=0$, which is a contradiction, so $X$ cannot contain any atoms. Is this useful for what I'm supposed to prove? I can't seem to see where to go from here, or if this is even the right first step.
EDIT: I'm not even 100% sure if I can take a maximal family without proving it exists in some way, so this possibly isn't the right way to go.
EDIT 2: I've managed to work out that because $X$ cannot contain any atoms, for every $p \in (0, \mathbb{P}(X))$, there must exist an event $B$ such that $\mathbb{P}(B)<p$, therefore $0$ is a limit point of the set of $p \in (0, \mathbb{P}(X))$ for which there exists an event $B$ for which $\mathbb{P}(B) = p$, but this still doesn't give me the whole solution.
• Please clarify : your notation $AB$ is what everyone else denotes by $A\cap B$, i.e. the intersection ? – Ewan Delanoy Apr 24 '17 at 7:19
• Yes, in my environment (and I think probability theory as well) $AB = A \cap B$. – Matija Sreckovic Apr 24 '17 at 9:39
• Thank you for clarifying this point. Personally I've never seen the notation $AB$ used to mean intersection in any probability textbook, and I would advise you to change all your $AB$'s to $A\cap B$ in your text. – Ewan Delanoy Apr 24 '17 at 9:50
• @EwanDelanoy: For what it's worth, there are popular books that use $AB$; Sheldon Ross's books, for instance. But I don't much like it either. – Nate Eldredge Apr 26 '17 at 15:57
• @NateEldredge Indeed, $AB$ is already used to mean the product of two random variables $A$ and $B$. – Ewan Delanoy Apr 26 '17 at 16:29
Everything you said is OK, you just need a few complements and simplifications.
To justify the existence of a maximal disjoint set of atoms, you can use Zorn's lemma : let $P$ be the poset whose elements are the sets of disjoint atoms, and whose order is defined by inclusion. Then every chain in $P$ has an upper bound in $P$ (the union of the chain), so the hypothesis of Zorn's lemma is satisfied.
So there is a maximal disjoint set of atoms by Zorn's lemma, the set is necessarily countable as explained in the OP, call it $(A_n)_{n\in{\mathbb N}}$ and let $X=\Omega \setminus \bigcup_{n\in{\mathbb N}}A_n$. Then $X$ is measurable, and atomless as explained in the OP.
Now, let $p\in (0,{\mathbb{P}}(X))$. Let $${\mathscr{A}}_{p,X}=\lbrace U\in {\mathscr{A}} | U \subseteq X, {\mathbb{P}}(U) \leq p \rbrace$$.
Say that two elements $U_1$ and $U_2$ in ${\mathscr{A}}_{p,X}$ are linked iff $U_1\subseteq U_2$. I call $L=U_2\setminus U_1$ the link between $U_1$ and $U_2$. If $L$ is either empty or has positive measure, we say that $U_1$ and $U_2$ are strongly linked, and denote this by $U_1 \leq_{p,X} U_2$.
It is easily checked that $\leq_{p,X}$ defines a partial ordering on ${\mathscr{A}}_{p,X}$. Also, if $C$ is any $\leq_{p,X}$-chain in ${\mathscr{A}}_{p,X}$, since the links between two (distinct) successive elements are all disjoint, $C$ is necessarily countable. The union of $C$ will be in ${\mathscr{A}}_{p,X}$, so it will be an upper bound for $C$. By Zorn's lemma, there is a $\leq_{p,X}$-maximal element, denote it by $M$. We know that $\delta=p-{\mathbb{P}}(M)$ is nonnegative. If $\delta>0$, by the property called "Edit 2" in the OP, there is a $V_1\subseteq X\setminus U$ with $0<{\mathbb{P}}(V_1)<\delta$, but then the set $M'=M\cup V_1$ would contradict the maximality of $M$, so $\delta=0$ and hence $P(M)=p$, which finishes the proof.
• I'm confused about the definition of linked elements. If $X \in L$, that is $X \in U_{2} \setminus U_{1}$, then $X \in U_{2}$, therefore $X \in V_{2}$. But since $U_{2} \setminus U_{1} = V_{1} \setminus V_{2}$, then $X \in V_{1} \setminus V_{2}$, therefore $X \notin V_{2}$, which is a contradiction. So by this if I'm not mistaken $L$ can't be anything other than $\emptyset$. Did you mean to write $L = U_{1} \setminus U_{2}$ or is the relation meant to be trivial in this way? – Matija Sreckovic Apr 26 '17 at 19:55
• @MatijaSreckovic Answer to your first question : my initial answer was indeed incorrect as stated. Please check my corrected version. Answer to your 2nd question ; it is not true than $P(L)$ is always $0$. Consider for example, $\Omega=[0,1]$, $({\mathscr A},{\mathbb P})=$ Lebesgue measure, $p=\frac{2}{3},U_1=[0,\frac{1}{3}], U_2=[0,\frac{2}{3}],$. Then $L$ has probability $\frac{1}{3}$ not zero. – Ewan Delanoy Apr 27 '17 at 7:49
• The second comment was meant in accordance with your first definition, but yeah, it's pointless with the fixed definition of linked sets. – Matija Sreckovic Apr 27 '17 at 19:25
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# Difference between revisions of "Polynomial ring"
Given a (commutative) ring $R$, the polynomial ring $R[x]$ is, informally, "the ring of all polynomials in $x$ with coefficients in $R$."
$$R[x]=\left\lbrace\sum_{i=0}^\infty a_ix^i\mid a_i\in R\right\rbrace$$
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anonymous 4 years ago Use the formula C= 5/9(F-30) to write 86 F degrees Celsius. The answer is 30 degrees Celsius.
1. anonymous
But how?
2. Akshay_Budhkar
${5 \over 9 }(86-30)={5 \over 9}(56)=31.11..$
3. Akshay_Budhkar
so probably that will be the closest rounded answer. but the exact answer is 31.11...
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# How do I convert coefficient values from 1/3 octave band to 1/1 octave band?
I have some coefficient values i.e. 74.9, 81.3, and 80.5 at frequency centers 63, 160 and 400 Hz. The values are 1/3 octave band values. I would like to convert them to the 1/1 octave band values. Can someone suggest how to do that?
If I'm understanding your question correctly, you want to sum up levels given as 1/3 octave bands into 1/1 octave band resolution. This can be accomplished by decibel summation:
$$L_\Sigma = 10 * log(10^\frac{L1}{10} + 10^\frac{L2}{10} + 10^\frac{L3}{10})$$
Where $$L$$ are levels in dB.
Please see here for an online summation tool.
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# Two similar solids have a scale factor of 3 : 5. If the height of solid I is 3 cm, find the height of solid II.
We are given that two solids, solid I and solid II, are similar with a scale factor of 3:5. The height of solid I is given as 3 cm, and we are asked to find the height of solid II.
A simple definition of similar figures is figures that have the same "shape" but not necessarily the same size. A more rigorous definition is that a similarity is a transformation that preserves angles and maps all lengths in the same ratio (often referred to as the scale factor, the ratio of magnification, or the dilation factor). Another definition is a transformation that preserves ratios of lengths.
If polygons are similar, then corresponding angles are congruent, and corresponding side lengths are in the same ratio, which is the scale factor. "Corresponding lengths" can include the lengths of corresponding sides, diagonals, medians, heights, and so on.
If we assume that the scale factor for a pair of similar figures is a:b, then all corresponding lengths are in the ratio a:b, all corresponding areas are in the ration a² :b² , and all corresponding volumes are in the ratio a³:b³.
For this problem, we have the scale factor as 3:5, so all corresponding lengths, including height, are in a ratio of 3:5. Thus, since the height of solid I is 3 cm, the height of solid II is 5 cm.
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# zbMATH — the first resource for mathematics
The classification of orthogonally rigid $$G_{2}$$-local systems and related differential operators. (English) Zbl 1312.32015
Let $$O_n$$ denote the orthogonal group on an $$n$$-dimensional (complex) vector space. An irreducible orthogonally self-dual complex rank $$n$$ local system $$\mathcal{L}$$ on the punctured complex projective line $$\mathbb{P}^1 \smallsetminus \{ x_1, \dots, x_r \}$$ is said to be orthogonally rigid if $\sum_{i = 1}^{r+1} \mathrm{codim} (C_{O_n}(g_i)) = 2 \dim (O_n),$ where $$C_{O_n}(g_i)$$ is the centralizer of the local monodromy generator $$g_i$$. The main result of this article is a complete classification of the orthogonally rigid local systems of rank $$7$$ whose monodromy group is Zariski dense in $$G_2(\mathbb{C}) \subset O_7$$. Specifically, the authors show:
Let $$\mathcal{L}$$ be a rigid $$\mathbb{C}$$-local system on a punctured projective line $$\mathbb{P}^1 \smallsetminus \{ x_1, \dots, x_r \}$$ of rank $$7$$ whose monodromy group is dense in the exceptional simple group $$G_2$$. If $$\mathcal{L}$$ has nontrivial local monodromy at $$x_1, \dots , x_r$$, then $$r=3,4$$ and $$\mathcal{L}$$ can be constructed by applying iteratively a sequence of the following operations to a rank-$$1$$-system: 0,5 cm
$$\bullet$$
middle convolutions $$\mathrm{M}_{\chi}$$, with varying $$\chi$$;
$$\bullet$$
tensor products with rank-$$1$$-local systems;
$$\bullet$$
tensor operations like symmetric or alternating products;
$$\bullet$$
pullbacks along rational functions.
Especially, each such local system which has quasi-unipotent monodromy is motivic, i.e., it arises from the variation of periods of a family of varieties over the punctured projective line.
The proof of the theorem is a mixture of case by case analysis and general arguments.
##### MSC:
32S40 Monodromy; relations with differential equations and $$D$$-modules (complex-analytic aspects) 20G41 Exceptional groups
CHEVIE
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##### References:
[1] Michael Aschbacher, Chevalley groups of type \?$$_{2}$$ as the group of a trilinear form, J. Algebra 109 (1987), no. 1, 193 – 259. · Zbl 0618.20030 · doi:10.1016/0021-8693(87)90173-6 · doi.org [2] F. Beukers and G. Heckman, Monodromy for the hypergeometric function _\?\?_\?-1, Invent. Math. 95 (1989), no. 2, 325 – 354. · Zbl 0663.30044 · doi:10.1007/BF01393900 · doi.org [3] Spencer Bloch and Hélène Esnault, Homology for irregular connections, J. Théor. Nombres Bordeaux 16 (2004), no. 2, 357 – 371 (English, with English and French summaries). · Zbl 1075.14016 [4] Michael Bogner and Stefan Reiter, On symplectically rigid local systems of rank four and Calabi-Yau operators, J. Symbolic Comput. 48 (2013), 64 – 100. · Zbl 1254.14046 · doi:10.1016/j.jsc.2011.11.007 · doi.org [5] Bomshik Chang and Rimhak Ree, The characters of \?$$_{2}$$(\?), Symposia Mathematica, Vol. XIII (Convegno di Gruppi e loro Rappresentazioni, INDAM, Rome,1972), Academic Press, London, 1974, pp. 395 – 413. · Zbl 0314.20034 [6] Michael Dettweiler and Stefan Reiter, An algorithm of Katz and its application to the inverse Galois problem, J. Symbolic Comput. 30 (2000), no. 6, 761 – 798. Algorithmic methods in Galois theory. · Zbl 1049.12005 · doi:10.1006/jsco.2000.0382 · doi.org [7] Michael Dettweiler and Stefan Reiter, Middle convolution of Fuchsian systems and the construction of rigid differential systems, J. Algebra 318 (2007), no. 1, 1 – 24. · Zbl 1183.34137 · doi:10.1016/j.jalgebra.2007.08.029 · doi.org [8] Michael Dettweiler and Stefan Reiter, Rigid local systems and motives of type \?$$_{2}$$, Compos. Math. 146 (2010), no. 4, 929 – 963. With an appendix by Michael Dettweiler and Nicholas M. Katz. · Zbl 1194.14036 · doi:10.1112/S0010437X10004641 · doi.org [9] Michael Dettweiler and Stefan Wewers, Variation of local systems and parabolic cohomology, Israel J. Math. 156 (2006), 157 – 185. · Zbl 1131.14016 · doi:10.1007/BF02773830 · doi.org [10] Meinolf Geck, Gerhard Hiss, Frank Lübeck, Gunter Malle, and Götz Pfeiffer, CHEVIE — a system for computing and processing generic character tables, Appl. Algebra Engrg. Comm. Comput. 7 (1996), no. 3, 175 – 210. Computational methods in Lie theory (Essen, 1994). · Zbl 0847.20006 · doi:10.1007/BF01190329 · doi.org [11] G. Hiss, Zerlegungszahlen endlicher Gruppen vom Lie-Typ in nicht-definierender Charakteristik, Habilitationsschrift, Aachen 1990. [12] Nicholas M. Katz, Exponential sums and differential equations, Annals of Mathematics Studies, vol. 124, Princeton University Press, Princeton, NJ, 1990. · Zbl 0731.14008 [13] Nicholas M. Katz, Rigid local systems, Annals of Mathematics Studies, vol. 139, Princeton University Press, Princeton, NJ, 1996. · Zbl 0864.14013 [14] Martin W. Liebeck and Gary M. Seitz, The maximal subgroups of positive dimension in exceptional algebraic groups, Mem. Amer. Math. Soc. 169 (2004), no. 802, vi+227. · Zbl 1058.20040 · doi:10.1090/memo/0802 · doi.org [15] Gunter Malle and B. Heinrich Matzat, Inverse Galois theory, Springer Monographs in Mathematics, Springer-Verlag, Berlin, 1999. · Zbl 0940.12001 [16] A. L. Onishchik and È. B. Vinberg, Lie groups and algebraic groups, Springer Series in Soviet Mathematics, Springer-Verlag, Berlin, 1990. Translated from the Russian and with a preface by D. A. Leites. · Zbl 0722.22004 [17] Marius van der Put and Michael F. Singer, Galois theory of linear differential equations, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 328, Springer-Verlag, Berlin, 2003. · Zbl 1036.12008 [18] Jan Saxl and Gary M. Seitz, Subgroups of algebraic groups containing regular unipotent elements, J. London Math. Soc. (2) 55 (1997), no. 2, 370 – 386. · Zbl 0955.20033 · doi:10.1112/S0024610797004808 · doi.org [19] Leonard L. Scott, Matrices and cohomology, Ann. of Math. (2) 105 (1977), no. 3, 473 – 492. · Zbl 0399.20047 · doi:10.2307/1970920 · doi.org [20] Karl Strambach and Helmut Völklein, On linearly rigid tuples, J. Reine Angew. Math. 510 (1999), 57 – 62. · Zbl 0931.12006 · doi:10.1515/crll.1999.048 · doi.org [21] André Weil, Remarks on the cohomology of groups, Ann. of Math. (2) 80 (1964), 149 – 157. · Zbl 0192.12802 · doi:10.2307/1970495 · doi.org
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# What are all the possible rational zeros for f(x)=x^3+11x^2+35x+33 and how do you find all zeros?
##### 1 Answer
Dec 26, 2016
$- 3$; (rational zero)
$- 4 \pm \sqrt{5}$ (not rational zeros)
#### Explanation:
You would find rational zeros in the set of integer and negative numbers dividing the known term 33, which are -1; -3; -11; -33.
Using the remainder theorem, you will find that
$f \left(- 1\right) \ne 0$
but
$f \left(- 3\right) = 0$
So -3 is a rational zero.
Then you will divide:
$\left({x}^{3} + 11 {x}^{2} + 35 x + 33\right) : \left(x + 3\right) = {x}^{2} + 8 x + 11$
By using quadratic formula, you would find the not rational zeros:
$- 4 \pm \sqrt{5}$
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Proceedings ArticleDOI
# Measurement of inherent noise in EDA tools
07 Aug 2002-pp 206-211
TL;DR: This work seeks to identify sources of noise in EDA tools, and analyze the effects of these noise sources on design quality, and proposes new behavior criteria for tools with respect to the existence and management of noise.
AbstractWith advancing semiconductor technology and exponentially growing design complexities, predictability of design tools becomes an important part of a stable top-down design process. Prediction of individual tool solution quality enables designers to use tools to achieve best solutions within prescribed resources, thus reducing design cycle time. However, as EDA tools become more complex, they become less predictable. One factor in the loss of predictability is inherent noise in both algorithms and how the algorithms are invoked. In this work, we seek to identify sources of noise in EDA tools, and analyze the effects of these noise sources on design quality. Our specific contributions are: (i) we propose new behavior criteria for tools with respect to the existence and management of noise; (ii) we compile and categorize possible perturbations in the tool use model or tool architecture that can be sources of noise; and (iii) we assess the behavior of industry place and route tools with respect to these criteria and noise sources. While the behavior criteria give some guidelines for and characterize the stability of tools, we are not recommending that tools be immune from input perturbations. Rather, the categorization of noise allows us to better understand how tools will or should behave; this may eventually enable improved tool predictors that consider inherent tool noise.
Topics: Noise (56%)
Content maybe subject to copyright Report
##### Citations
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Proceedings ArticleDOI
Puneet Gupta
08 Nov 2005
TL;DR: A novel dynamic programming-based technique for etch-dummy correctness (EtchCorr) which can be combine with the SAEDM in detailed placement of standard-cell designs and is validated on industrial testcases with respect to wafer printability, database complexity and device performance.
Abstract: Etch dummy features are used in the mask data preparation flow to reduce critical dimension (CD) skew between resist and etch processes and improve the printability of layouts. However, etch dummy rules conflict with SRAF (Sub-Resolution Assist Feature) insertion because each of the two techniques requires specific spacings of poly-to-assist, assist-to-assist, active-to-etch dummy and dummy-to-dummy. In this work, we first present a novel SRAF-aware etch dummy insertion method ( SAEDM ) which optimizes etch dummy insertion to make the layout more conducive to assist-feature insertion after etch dummy features have been inserted. However, placed standard-cell layouts may not have the ideal whitespace distribution to allow for optimal etch dummy and assist-feature insertions. Since placement of cells can create forbidden pitch violations, the placer must generate assist-correct and etch dummy-correct placements. This can be achieved by intelligent whitespace management in the placer. We describe a novel dynamic programming-based technique for etch-dummy correctness ( EtchCorr ) which can be combine with the SAEDM in detailed placement of standard-cell designs. Our algorithm is validated on industrial testcases with respect to wafer printability, database complexity and device performance.
145 citations
### Cites background from "Measurement of inherent noise in ED..."
• ...maximum delay overhead of 6% is within the inherent noise of the P&R tool.(11) The runtime of EtchCorr placement perturbation is negligible (∼ 5 minutes) compared to the running time of OPC (∼ 2....
[...]
Proceedings ArticleDOI
25 Mar 2018
TL;DR: Examples applications include removing unnecessary design and modeling margins through correlation mechanisms, achieving faster design convergence through predictors of downstream flow outcomes that comprehend both tools and design instances, and corollaries such as optimizing the usage of design resources licenses and available schedule.
Abstract: In the late-CMOS era, semiconductor and electronics companies face severe product schedule and other competitive pressures. In this context, electronic design automation (EDA) must deliver "design-based equivalent scaling" to help continue essential industry trajectories. A powerful lever for this will be the use of machine learning techniques, both inside and "around" design tools and flows. This paper reviews opportunities for machine learning with a focus on IC physical implementation. Example applications include (1) removing unnecessary design and modeling margins through correlation mechanisms, (2) achieving faster design convergence through predictors of downstream flow outcomes that comprehend both tools and design instances, and (3) corollaries such as optimizing the usage of design resources licenses and available schedule. The paper concludes with open challenges for machine learning in IC physical design.
43 citations
### Cites background from "Measurement of inherent noise in ED..."
• ...Figure 7 (right) illustrates that the statistics of this noisy tool behavior are Gaussian [32] [17]....
[...]
Journal ArticleDOI
TL;DR: This paper presents a context-aware DVFS design flow that considers the intrinsic characteristics of the hardware design, as well as the operating scenario-including the relative amounts of time spent in different modes, the range of performance scalability, and the target efficiency metric-to optimize the design for maximum energy efficiency.
Abstract: The proliferation of embedded systems and mobile devices has created an increasing demand for low-energy hardware. Dynamic voltage and frequency scaling (DVFS) is a popular energy reduction technique that allows a hardware design to reduce average power consumption while still enabling the design to meet a high-performance target when necessary. To conserve energy, many DVFS-based embedded and mobile devices often spend a large fraction of their lifetimes in a low-power mode. However, DVFS designs produced by conventional multimode CAD flows tend to have significant energy overheads when operating outside of the peak performance mode, even when they are operating in a low-power mode. A dedicated core can be added for low-energy operation, but has a high cost in terms of area and leakage. In this paper, we explore the DVFS design space to identify the factors that affect DVFS efficiency. Based on our insights, we propose two design-level techniques to enhance the energy efficiency of DVFS for energy constrained systems. First, we present a context-aware DVFS design flow that considers the intrinsic characteristics of the hardware design, as well as the operating scenario-including the relative amounts of time spent in different modes, the range of performance scalability, and the target efficiency metric-to optimize the design for maximum energy efficiency. We also present a selective replication-based DVFS design methodology that identifies hardware modules for which context-aware multimode design may be inefficient and creates dedicated module replicas for different operating modes for such modules. We show that context-aware design can reduce average power by up to 20% over a conventional multimode design flow. Selective replication can reduce average power by an additional 4%. We also use the generated insights to identify microarchitectural decisions that impact DVFS efficiency. We show that the benefits from the proposed design-level techniques increase when microarchitectural transformations are allowed.
39 citations
### Cites methods from "Measurement of inherent noise in ED..."
• ...We evaluate replication decisions at the granularity of RTL modules, and we analyze replication at different granularities in this section....
[...]
Journal ArticleDOI
Abstract: The value of guardband reduction is a critical open issue for the semiconductor industry. For example, due to competitive pressure, foundries have started to incent the design of manufacturing-friendly ICs through reduced model guardbands when designers adopt layout restrictions. The industry also continuously weighs the economic viability of relaxing process variation limits in the technology roadmap (available: http://public.itrs.net). Our work gives the first-ever quantification of the impact of model guardband reduction on outcomes from the synthesis, place and route (SPR 40% is the amount of guardband reduction reported by IBM for a variation-aware timing methodology. For the embedded processor core we observe up to 8% standard-cell area reduction, 7% routed wirelength reduction, 5% dynamic power reduction, and 10% leakage power reduction at 30% guardband reduction. We also report a set of fine-grain SPICE simulations that accurately assesses the impact of process guardband reduction, as distinguished from overall guardband reductions, on yield. We observe up to 4% increase in number of good dies per wafer at 27% process guardband reduction (i.e., with fixed voltage and temperature). Our results suggest that there is justification for the design, EDA and process communities to enable guardband reduction as an economic incentive for manufacturing-friendly design practices.
36 citations
Proceedings ArticleDOI
, Yang Du2
01 Oct 2016
TL;DR: This work develops machine learning-based models that predict whether a placement solution is routable without conducting trial or early global routing, and uses these models to accurately predict iso-performance Pareto frontiers of utilization, aspect ratio and number of layers in the back-end-of-line (BEOL) stack.
Abstract: In advanced technology nodes, physical design engineers must estimate whether a standard-cell placement is routable (before invoking the router) in order to maintain acceptable design turnaround time. Modern SoC designs consume multiple compute servers, memory, tool licenses and other resources for several days to complete routing. When the design is unroutable, resources are wasted, which increases the design cost. In this work, we develop machine learning-based models that predict whether a placement solution is routable without conducting trial or early global routing. We also use our models to accurately predict iso-performance Pareto frontiers of utilization, aspect ratio and number of layers in the back-end-of-line (BEOL) stack. Furthermore, using data mining and machine learning techniques, we develop new methodologies to generate training examples given very few placements. We conduct validation experiments in three foundry technologies (28nm FDSOI, 28nm LP and 45nm GS), and demonstrate accuracy ≥ 85.9% in predicting routability of a placement. Our predictions of Pareto frontiers in the three technologies are pessimistic by at most 2% with respect to the maximum achievable utilization for a given design in a given BEOL stack.
34 citations
##### References
More filters
Journal ArticleDOI
Abstract: The asymptotic behavior of the systems $X_{n + 1} = X_n + a_n b( {X_n ,\xi _n } ) + a_n \sigma ( X_n )\psi_n$ and $dy = \bar b( y )dt + \sqrt {a( t )} \sigma ( y )dw$ is studied, where $\{ {\psi _n } \}$ is i.i.d. Gaussian, $\{ \xi _n \}$ is a (correlated) bounded sequence of random variables and $a_n \approx A_0/\log (A_1 + n )$. Without $\{ \xi _n \}$, such algorithms are versions of the “simulated annealing” method for global optimization. When the objective function values can only be sampled via Monte Carlo, the discrete algorithm is a combination of stochastic approximation and simulated annealing. Our forms are appropriate. The $\{ \psi _n \}$ are the “annealing” variables, and $\{ \xi _n \}$ is the sampling noise. For large $A_0$, a full asymptotic analysis is presented, via the theory of large deviations: Mean escape time (after arbitrary time n) from neighborhoods of stable sets of the algorithm, mean transition times (after arbitrary time n) from a neighborhood of one stable set to another, a...
141 citations
Journal ArticleDOI
TL;DR: A search for the global minimum of a function is proposed; the search is on the basis of sequential noisy measurements and the search plan is shown to be convergent in probability to a set of minimizers.
Abstract: A search for the global minimum of a function is proposed; the search is on the basis of sequential noisy measurements. Because no unimodality assumptions are made, stochastic approximation and other well-known methods are not directly applicable. The search plan is shown to be convergent in probability to a set of minimizers. This study was motivated by investigations into machine learning. This setting is explained, and the methodology is applied to create an adaptively improving strategy for 8-puzzle problems.
55 citations
### "Measurement of inherent noise in ED..." refers background in this paper
• ...For example, in [8], Yakowitz and Lugosi studied a formulation of random search in the presence of noise for the machine learning domain....
[...]
Journal ArticleDOI
TL;DR: A detailed software architecture is presented that allows flexible, efficient and accurate assessment of the practical implications of new move-based algorithms and partitioning formulations and discusses the current level of sophistication in implementation know-how and experimental evaluation.
Abstract: We summarize the techniques of implementing move-based hypergraph partitioning heuristics and evaluating their performance in the context of VLSI design applications. Our first contribution is a detailed software architecture, consisting of seven reusable components, that allows flexible, efficient and accurate assessment of the practical implications of new move-based algorithms and partitioning formulations. Our second contribution is an assessment of the modern context for hypergraph partitioning research for VLSI design applications. In particular, we discuss the current level of sophistication in implementation know-how and experimental evaluation, and we note how requirements for real-world partitioners - if used as motivation for research - should affect the evaluation of prospective contributions. Two "implicit decisions" in the implementation of the Fiduccia-Mattheyses heuristic are used to illustrate the difficulty of achieving meaningful experimental evaluation of new algorithmic ideas.
42 citations
• ...For example, a KLFM netlist partitioning implementation [2] will search for the cell to be moved to a different partition based on the order of the cells in the gain bucket data structure....
[...]
Proceedings ArticleDOI
02 Jul 1986
TL;DR: It is found that the Min Cut partitioning with simplified Terminal Propagation is the most efficient placement procedure studied and mean results of many placements should be used when comparing algorithms.
Abstract: This paper describes a study of placement procedures for VLSI Standard Cell Layout. The procedures studied are Simulated Annealing, Min Cut placement, and a number of improvements to Min Cut placement including a technique called Terminal Propagation which allows Min Cut to include the effect of connections to external cells. The Min Cut procedures are coupled with a Force Directed Pairwise Interchange (FDPI) algorithm for placement improvement. For the same problem these techniques produce a range of solutions with a typical standard deviation 4% for the total wire length and 3% to 4% for the routed area. The spread of results for Simulated Annealing is even larger. This distribution of results for a given algorithm implies that mean results of many placements should be used when comparing algorithms. We find that the Min Cut partitioning with simplified Terminal Propagation is the most efficient placement procedure studied.
38 citations
### "Measurement of inherent noise in ED..." refers background in this paper
• ...In the VLSI CAD domain, some early discoveries about noise in placement tools are presented in [ 5 ]....
[...]
Proceedings ArticleDOI
08 Apr 2000
TL;DR: There is inherent variability in wire lengths obtained using commer- cially available place and route tools - wire length estimation error cannot be any smaller than a lower limit due to this variability, and the proposed model works well within these variability limitations.
Abstract: We present a novel technique for es- timating individual wire lengths in a given standard- cell-based design during the technology mapping phase of logic synthesis. The proposed method is based on creating a black box model of the place and route tool as a function of a number of parame- ters which are all available before layout. The place and route tool is characterized, only once, by apply- ing it to a set of typical designs in a certain technol- ogy. We also propose a net bounding box estimation technique based on the layout style and net neigh- borhood analysis. We show that there is inherent variability in wire lengths obtained using commer- cially available place and route tools - wire length estimation error cannot be any smaller than a lower limit due to this variability. The proposed model works well within these variability limitations.
35 citations
### "Measurement of inherent noise in ED..." refers background or result in this paper
• ...While the above-mentioned studies used the concept of noise to generate isomorphic circuits for tool benchmarking, Bodapati and Najm [1] analyzed noise in tools from a different perspective....
[...]
• ...Starting from the premise that noise due to cell/net ordering and naming has a negative effect on estimators, the authors of [1] proposed a pre-layout estimation model for individual wire length, and noted that the accuracy of their estimations are worsened by inherent tool noise (with respect to ordering and naming)....
[...]
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# DeepSpeech Model¶
The aim of this project is to create a simple, open, and ubiquitous speech recognition engine. Simple, in that the engine should not require server-class hardware to execute. Open, in that the code and models are released under the Mozilla Public License. Ubiquitous, in that the engine should run on many platforms and have bindings to many different languages.
The architecture of the engine was originally motivated by that presented in Deep Speech: Scaling up end-to-end speech recognition. However, the engine currently differs in many respects from the engine it was originally motivated by. The core of the engine is a recurrent neural network (RNN) trained to ingest speech spectrograms and generate English text transcriptions.
Let a single utterance $$x$$ and label $$y$$ be sampled from a training set
$S = \{(x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), . . .\}.$
Each utterance, $$x^{(i)}$$ is a time-series of length $$T^{(i)}$$ where every time-slice is a vector of audio features, $$x^{(i)}_t$$ where $$t=1,\ldots,T^{(i)}$$. We use MFCC’s as our features; so $$x^{(i)}_{t,p}$$ denotes the $$p$$-th MFCC feature in the audio frame at time $$t$$. The goal of our RNN is to convert an input sequence $$x$$ into a sequence of character probabilities for the transcription $$y$$, with $$\hat{y}_t =\mathbb{P}(c_t \mid x)$$, where for English $$c_t \in \{a,b,c, . . . , z, space, apostrophe, blank\}$$. (The significance of $$blank$$ will be explained below.)
Our RNN model is composed of $$5$$ layers of hidden units. For an input $$x$$, the hidden units at layer $$l$$ are denoted $$h^{(l)}$$ with the convention that $$h^{(0)}$$ is the input. The first three layers are not recurrent. For the first layer, at each time $$t$$, the output depends on the MFCC frame $$x_t$$ along with a context of $$C$$ frames on each side. (We use $$C = 9$$ for our experiments.) The remaining non-recurrent layers operate on independent data for each time step. Thus, for each time $$t$$, the first $$3$$ layers are computed by:
$h^{(l)}_t = g(W^{(l)} h^{(l-1)}_t + b^{(l)})$
where $$g(z) = \min\{\max\{0, z\}, 20\}$$ is a clipped rectified-linear (ReLu) activation function and $$W^{(l)}$$, $$b^{(l)}$$ are the weight matrix and bias parameters for layer $$l$$. The fourth layer is a recurrent layer [1]. This layer includes a set of hidden units with forward recurrence, $$h^{(f)}$$:
$h^{(f)}_t = g(W^{(4)} h^{(3)}_t + W^{(f)}_r h^{(f)}_{t-1} + b^{(4)})$
Note that $$h^{(f)}$$ must be computed sequentially from $$t = 1$$ to $$t = T^{(i)}$$ for the $$i$$-th utterance.
The fifth (non-recurrent) layer takes the forward units as inputs
$h^{(5)} = g(W^{(5)} h^{(f)} + b^{(5)}).$
The output layer is standard logits that correspond to the predicted character probabilities for each time slice $$t$$ and character $$k$$ in the alphabet:
$h^{(6)}_{t,k} = \hat{y}_{t,k} = (W^{(6)} h^{(5)}_t)_k + b^{(6)}_k$
Here $$b^{(6)}_k$$ denotes the $$k$$-th bias and $$(W^{(6)} h^{(5)}_t)_k$$ the $$k$$-th element of the matrix product.
Once we have computed a prediction for $$\hat{y}_{t,k}$$, we compute the CTC loss [2] $$\cal{L}(\hat{y}, y)$$ to measure the error in prediction. (The CTC loss requires the $$blank$$ above to indicate transitions between characters.) During training, we can evaluate the gradient $$\nabla \cal{L}(\hat{y}, y)$$ with respect to the network outputs given the ground-truth character sequence $$y$$. From this point, computing the gradient with respect to all of the model parameters may be done via back-propagation through the rest of the network. We use the Adam method for training [3].
The complete RNN model is illustrated in the figure below.
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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# Prove the following
Question:
For $\mathrm{x} \in \mathrm{R}$, let $[\mathrm{x}]$ denote the greatest integer $\leq \mathrm{x}$, then the sum of the series
$\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\ldots \ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]$
is
1. $-153$
2. $-133$
3. $-131$
4. $-135$
Correct Option: 2,
Solution:
$\underbrace{\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{66}{100}\right]}_{(-1) 67}$
$+\underbrace{\left[-\frac{1}{3}-\frac{67}{100}\right]+\ldots+\left[-\frac{1}{3}-\frac{99}{100}\right]}_{-2(33)}=-133$
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# Let $R$ be a commutative ring with unity, prove that $I$ is an ideal of $R$.
Let $$R$$ be a commutative ring with unity. If $$a_1, a_2, \ldots,a_k ∈ R$$, prove that $$I = \{a_1r_1 + a_2r_2 + \cdots + a_kr_k \mid r_1, r_2, \ldots, r_k ∈R\}$$ is an ideal of $$R$$.
I wanted to start by showing that $$I$$ is a subring of $$R$$, but I'm stuck trying to show that $$I$$ is nonempty. It's not clear to me from the problem statement that the zero element or the unity element are necessarily in $$a_1, a_2, \ldots ,a_k$$ or $$r_1, r_2, \ldots ,r_k$$.
What am I missing? Isn't it possible that the zero and the unity are not in either of these two subsets of $$R$$?
• The $a_{\kappa}$ are given, the $r_{\kappa}$ range over all elements of $R$. – Daniel Fischer Dec 1 '19 at 13:59
• Why do you want $I$ to contain the multiplicative unit? A proper ideal doesn't contain it. – Bernard Dec 1 '19 at 14:06
• You are doing a mistake, notice that $r_i$'s are generalised elements of $R$ and $a_j$'s are fixed elements of ring, for $j=1,2,3 \cdots k$ ! Best of luck. – Alfha Dec 1 '19 at 14:08
• One example can clear your doubt, take ring $R$ = $(\mathbb{Z} , + , \circ)$ and take $a_i$'s = ${2,4}$ now think about the set $I$ ={ $2 \cdot r_1 + 4 \cdot r_2 : r_1 , r_2 \in R$ }. – Alfha Dec 1 '19 at 14:16
• @Alfha : In proper MathJax usage one would write $I = \{ 2\cdot r_1+4\cdot r_2 : r_1,r_2\in R\},$ with the "equals" sign and the $\{\text{curly braces} \}$ inside of MathJax. The three "equals" signs in your comment don't match the font size of the things that precede and follow them. $\qquad$ – Michael Hardy Dec 1 '19 at 14:31
Why would you expect the unit element to be in $$I$$? Ideals do not generally contain the unit element. For example, the set $$\{0, \pm 6, \pm12, \pm18,\ldots\}$$ of all integer multiples of $$6$$ is an ideal in $$\mathbb Z.$$
The zero element is in $$I$$ because that is the case in which $$r_1=\cdots=r_k=0.$$
• The OP probably was confused with multiplicative units and additive units... I guess. – WhatsUp Dec 1 '19 at 14:22
• @WhatsUp : It doesn't look that way to me. The OP clearly distinguishes between the two. – Michael Hardy Dec 1 '19 at 14:23
• To prove that $I$ is an ideal, it must be a subring. For $I$ to be a subring, it must be nonempty. To show that $I$ is nonempty, I thought maybe I could use the fact that $R$ has a unity. I didn't think that an ideal must necessarily contain the unity, but I appreciate you providing me an example. – combinat0ria1 Dec 1 '19 at 16:13
• @MichaelHardy I know that the zero element must be in $I$ since I'm trying to prove that it is an ideal (therefore it must be a subring), but how can we conclude that one of the $r_i$ is $0$? Edit: Since $r_i$ are just arbitrary elements of $R$, we can just say "Consider the case where $r_1 = ... = r_k = 0$," right? – combinat0ria1 Dec 1 '19 at 16:14
• @combinat0ria1 : Also note that in each ring with a unit element, there is ONLY ONE ideal that contains the unit element, and that ideal is $R$ itself. – Michael Hardy Dec 1 '19 at 19:10
Proof: Clearly $$I$$ is non empty subring of $$(R, +, \circ)$$ as addictive identity
$$e \in I$$ and $$I$$ is ring under the operations of $$(R, + , \circ)$$ ring. Now for any $$r \in R$$ we need to show $$r \circ \mathbb{i}$$ is in $$I$$.
Where $$\mathbb{i} = a_1.r_1 + \cdots + a_k.r_k$$ is any element in $$I$$. Consider $$r \circ \mathbb{i} = a_1.r_1.r + \cdots + a_k.r_k.r$$
[Note that: Ring is commutative]
Or
$$r \circ \mathbb{i} = a_1.r_1' + \cdots + a_k.r_k'$$
where $$r_i.r =r_i'$$ for some $$r_i' \in R$$
So $$r \circ \mathbb{i} \in I$$ [by the definition of $$I$$.]
And so $$I$$ is ideal of $$(R, +, \circ)$$.
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Accelerating upward treeline shift in the Altai Mountains under last-century climate change
Abstract
Treeline shift and tree growth often respond to climatic changes and it is critical to identify and quantify their dynamics. Some regions are particularly sensitive to climate change and the Altai Mountains, located in Central and East Asia, are showing unequivocal signs. The mean annual temperature in the area has increased by 1.3–1.7 °C in the last century. As this mountain range has ancient and protected forests on alpine slopes, we focus on determining the treeline structure and dynamics. We integrated in situ fine-scale allometric data with analyses from dendrochronological samples, high-resolution 3D drone photos and new satellite images to study the dynamics and underlying causal mechanisms of any treeline movement and growth changes in a remote preserved forest at the Aktru Research Station in the Altai Mountain. We show that temperature increase has a negative effect on mountain tree growth. In contrast, only younger trees grow at higher altitudes and we document a relatively fast upward shift of the treeline. During the last 52 years, treeline moved about 150 m upward and the rate of movement accelerated until recently. Before the 1950s, it never shifted over 2150–2200 m a.s.l. We suggest that a continuous upward expansion of the treeline would be at the expense of meadow and shrub species and radically change this high-mountain ecosystem with its endemic flora. This documented treeline shift represents clear evidence of the increased velocity of climate change during the last century.
Introduction
Over the last century, planet Earth has experienced significant temperature increases, particularly in the Northern Hemisphere1,2. Meanwhile, more climate variability is predicted with a general increase in temperature and extreme dry and hot periods3,4. Such events have a substantial effect on montane forest ecosystems5,6, where rising temperatures enable many tree species to establish at higher elevations for new and favorable environmental niches7,8,9, and other better growing conditions10,11,12.
There is clear evidence that temperate and boreal forests are encroaching northwards and upwards into tundra and alpine communities13,14,15, and such climate-driven effects may cause a treeline shift that changes the structure of montane and tundra systems16. For instance, a long-term study in Scandinavia17 has shown an upward movement of treeline in the order of 150–200 mover about 30 years. In temperate regions, tree species have been shown to migrate naturally by several kilometers a century18,19. However, rapid climatic changes can lead to a 100–1000 times faster shift20, as demonstrated by a long-term photographic record has been used to document the higher altitude shift of alpine treeline into alpine meadows in Yunnan, China21. In North America, modelling of tree species distributions in relation to climate change suggests a general movement of 134 eastern tree species ranges towards the Northeast22. In most cases, these changes have been linked to an observed general warming. However, treeline location and dynamics have complex causes23 and over the past 100 years, treeline in northern Sweden increased in altitude, decreased and remained stable on neighboring mountains all experiencing the same climatic warming24. Controls on treeline include temperature, herbivory, lack of niches for seedling establishment, etc.25.
The Altai Mountains, located in Central and East Asia, are showing evident impacts of the changing climate, including significant changes of ecological succession in the areas emerged from retreating glaciers26,27,28. Although no relevant variations in the mean annual precipitation have been reported for the area29, the mean annual temperature in the Northern Altai Mountains has shown a significant increase of 1.3–1.7 °C in the last century30. This general trend was also confirmed by a comprehensive study that analyzed 50 years of temperature and precipitation variations in the whole mountain chain in both Russia and China31.
Several studies reconstructed the impact of climate change in the Central Eurasia from tree ring-based records32,33,34,35,36,37,38,39. These studies in the Tianshan and Altai Mountains and in other high-mountains of the world40,41, showed that climate change might have a harmful effect on tree growth and recent evidence reported that the lower forests can also display “browning” over extensive areas in the boreal region42. The treeline expansion at these latitudes, such as in Siberia, may lead to a compositional change of forest communities and could result in a significant alteration of biotic interactions43. Meanwhile, several studies showed that slightly higher temperatures and greater accumulation of carbon dioxide in the atmosphere accelerate the growth rate of species in forest ecosystems44,45.
Treeline shift and changes in tree growth46 are evident signs of various aspects of global climatic changes47 and some regions are more sensitive to the global climate changes than other areas of the world21.
Remote sensing and aerial photographs are typically used to assess the movement of treeline in high mountains over the past 30 years48, and ground measurements and repeat photography over the past 100 years24. However, for most remote areas especially in the Altai Mountains, especially in the Russian part, the absence of aerial photographs combined with a lack of satellite images before the 1980s has impeded an understanding of the effects of climate change on treeline. In particular, for the Russian Altai Mountains, only scarce and sporadic climatic observations are available before the mid-1900s and historical photographic documentation (in terms of both ground/aerial pictures and satellite images) are scarce.
Here we integrated in situ fine-scale allometric data, dendrochronological samples, high-resolution 3D drone photos and new satellite images to characterize the treeline, with the aim to look at the recent dynamics and growth changes and to construct a baseline to follow future changes, in a remote preserved forest at the Aktru Research Station in the Altai Mountains.
Study Area
The Aktru Research Station is located in the Southeastern Altai Republic (Russia) close to the borders of Mongolia and China in the Central Eurasian Continent (50°06′03″N, 87°40′14″E). With an altitude of 2100 m a.s.l., the station is situated in the high alpine range of these mountains. There are no major anthropogenic disturbances on the surrounding ecosystems and vegetation, since the area is located in a very remote (with a 3-day drive from the closest airport) and a high-altitude part of the Asian continent. The nearest settlement (Kuray, a small village) is located downstream from the station at about 30 km southeast and the nearest town is Gorno-Altaisk, about 250 km northwest of the station. The Aktru Research Station was founded in 1956 by V.M. Tronov and has been in operation until today under the supervision of Tomsk State University. There is no evidence or documentation on any type of logging, big herbivore (wild or domesticated) grazing, or fire-related impact, at least for the last 50 years, on the vegetation of the mountainside where our study site is situated.
The study site (Fig. 1) is located on the northwestern slope of the Severo-Chuyskiy Mountain Chain, just above the research station, at an altitude between 2150 and 2300 m a.s.l. The climate of the area is characterized by low temperatures (annual average −5.2 ± 1.9 °C, summer average +8.7 ± 2.3 °C), high average annual precipitations (500–800 mm), and high diurnal temperature variation (of 15–20 °C)29. The site is considered to be climatically representative of the Altai, and includes several mountain peaks and glaciers29. The slope of this mountainside is 43.4–54.1%.
Results
The densest timberline in the study site, composed by two species (Pinus siberica 50–60%, Larix siberica 40–50%), is located up to an altitude of ≈2150 m a.s.l. (Fig. 2). Above this level (approximately between 2150 and 2250 m a.s.l.), a treeline composed of sparse groups of trees (belonging to the same two species of the treeline) and shrubs (mostly of Juniper spp., Vaccinium spp. and Ribes spp.) were detected. From 2250 m up to 2300 m a.s.l., only isolated trees and herb-rich meadows were identified (tree species line). No trees or shrubs with a d.b.h. >1 cm (see methods) were detected above ≈2300 m a.s.l. A few landslides and groups of rocks were located on the northern side of the treeline area. The whole study site (Fig. 2) from 2150 m to 2300 m a.s.l. may be defined, according to the common scientific convention49, as a “high-altitude treeline ecotone”.
As expected, we found a significant decline in tree size towards higher altitudes (Fig. 3). In particular, tree height had the highest negative correlation with the altitude (Spearman’s rank correlation ρ = −0.76, P < 0.01), followed by the crown width (ρ = −0.66, P < 0.01) and d.b.h. (ρ = −0.54, P < 0.01). No trees higher than 6 m where found over 2250 m a.s.l., even though a few relatively tall trees (7–9 m) were able to live at an altitude of 2240–2250 m a.s.l.
The dendrochronological analyses of the 48 wood cores collected from different altitudes along the transect of the treeline ecotone, shows a slightly declining mean annual growth curve between 1902 and 2017, with some periods (1902–1904, 1914–1916, 1923, 1928–1932, 1952–1956, 1961–1962, 1968) that peaks above the annual average tendency (Fig. 4). However, no relevant peaks are detected after the year 1968 (i.e. during the last 50 years of the dendrochronology).
After normalizing the time-series data with different methods (Fig. 5), the detrended curves (Spline and Friedman’s Super Smoother in comparison with Straight and Horizontal Lines) show similar peaks with respect to the raw curve (in Fig. 4) and a general decreasing growth trend, as confirmed by the Mann-Kendall statistics (τ = −0.315, P < 0.01).
Tree basal area shows an expected significant negative correlation with both the year of birth of the trees (Kendall’s rank correlation τ = −0.72, P < 0.01; Fig. 6) and the altitude (τ = −0.69, P < 0.01). In contrast, mean annual growth is not significantly influenced by the altitude (τ = −0.23, P = 0.30; Fig. 6) and shows a hump-shaped tendency when plotted against the birth year of trees (quadratic function, R2 = 0.42). In fact, the groups of the oldest and youngest trees show a similar mean annual growth, which is lower than that of mid-age trees (Fig. 6). The positive-saturating relationship between the altitude and the birth year of trees (quadratic function, R2 = 0.85) shows detectable shifts as schematized in Fig. 7, where we find strong evidence that trees born before 1954 do not grow above 2,150 m and trees born before 1999 do not grow above 2,200 m a.s.l. The maximum altitude of 2,300 m a.s.l. was reached between 1994 and 2002, while after 2002 the youngest trees sampled (born around 2006) grow at a lower altitude of 2,250 m a.s.l. (Fig. 7).
To determine if the shape of the mean annual growth curve were a simple biological artifact or related to the changes in climatic conditions and establishment altitude, we compared only the first 12 years of growth of all sampled trees (e.g. we compared the first 12-years mean growth of trees born in 1900s vs. 1980s vs. 2000s). We show (Fig. 7, right plot) that the oldest trees (which are located, on average, at the lowest altitude on the treeline ecotone at ≈2150 m a.s.l.) grew more than mid-age and younger trees (located at medium [≈2200 m a.s.l.] and higher [2250–2300 m a.s.l.] altitudes, respectively).
From a comparison between the variation of the mean annual temperature29 and our dendrochronological time-series (Fig. 8), we discovered that before 1978 the regional mean annual temperature remained below −4.3 °C. This threshold was, therefore, used to interpret the behavior of the hyperbolic curve (Fig. 8 upper-left panel) that emerged from the relationship between the mean annual growth and mean annual temperature (R2 = 0.34). This curve, at about −4.3 °C can be divided in two linear regressions (<−4.3 °C negative, R2 = 0.59; 4.3°C positive, R2 = 0.39), which allow an easier estimation of the missing annual temperature records (before 1956 and after 2009) to fill the gap of the available data in the literature for the area. The only long temperature record29 for the study-site covers a period between 1956 and 2009 and the most recent records (1966–2015)31 are from close meteorological stations but situated at lower altitudes.
The estimation for the years 1902–1955 (derived from the linear regression equation at <−4 °C) shows that temperature ranged between −5 °C and −10 °C, with some lower peaks between 1952–1955, 1926–1928, 1914–1916 and 1902–1904. The estimation for the years 2010–2017, shows a slight decreasing temperature trend, with a drop of 5–10 °C in the recent years (Fig. 8).
The results of the split sample test showed that the correlation coefficient and the product means for all calibration and verification periods were significant at P < 0.01 level. The sign tests were statistically significant in calibration and in the verification periods from 1956 and 1978 and from 1979 to 2009 at P < 0.05 level. The values of RE (0.86 and 0.77, respectively) and CE (0.32 and 0.36, respectively) were always positive, indicating that the regression was stable and reliable.
Discussion and Conclusion
Similar to the patterns documented in other high-mountains of the world4,5,6,10,11, we found that the increase of temperature during the last century of climate change has led to an alteration of the tree growth trend and an upward shift until recently of the treeline in a remote preserved forest at the Aktru Research Station in the Altai Mountains. This can be attributed to the fact that areas with a harsh growing condition for woody species, previously covered by snow for most of the year, would become suitable for tree growth as the mean annual temperature increases, thus enabling a upward shift and expansion of the treeline ecotone12.
We showed that, as predictable, biometric (allometric) parameters, such as tree height, d.b.h. and crown width decrease with the altitude, with significant differences between trees in the dense treeline (at about 2,150 m a.s.l.) and those growing at higher elevation (up to 2,300 m a.s.l.). These differences, however, could be due not only to an upward shift of the new saplings but also to a differential growth trend related to altitudinal effects50,51. In fact, trees living at the limits of the treeline may show a reduced growth when subject to harder climatic conditions then their physiological limits (i.e. lower temperatures, frozen soil, stronger wind, etc.)52. To test if a variation in the tree growth trends and an upward shift of the treeline ecotone might be caused by increasing local and global temperatures, we further analyzed dendrochronological samples collected along some altitudinal transects. From an analysis of the complete dendrochronological time series from 1902 to 2017, we found a slight decrease in the general growth trend and relevant growth peaks for many years, all dated before 1968, when the mean annual temperature was always below the identified −4.3 °C threshold. In the last 50 years, trees did not show any sign of increased growth compared to the average trend for the century. This could be due to the fact that high-mountain trees are quite sensitive to low temperatures and grow more under a certain threshold (−4.3 °C in our study-site)53,54,55. In fact, comparing the available mean annual temperature observations from 1956 to 2009 with our dendrochronological time series, we detected that - in this time frame - the highest growth peaks (in 1952–1956, 1961–1962 and 1968) correspond (with a few years of lag) to the lowest temperature for the period. However, after 1968 (i.e. in the last 50 years except for 1983–1988, when the temperature dropped for a while and the tree growth temporarily increased), the temperatures were often above a −4.3 °C threshold, and the tree growth was constant (without any sudden growth peak) and slightly decreasing. The fact that the precipitation trend shows no change for the whole region in the last 50 years31 gives more weight to temperature importance and to its low extremes.
Our reconstructed temperature estimates for the period 1902–1955 derived from the dendrochronological time series, which was missing from the available observation on the local weather station29,31, well matched the periods of extreme cold temperatures in 1902–1906, 1913–1915, 1922–1923, 1926–1928 and 1950–1952 recorded by other biological and non-biological sources at comparable regional56,57,58 and continental scales59,60.
We also checked whether last-century temperature increase would have contributed to the shift of the treeline ecotone to higher altitudes. We showed that trees have actually had an accelerating upward movement during the last century, yet a recent slight downward movement. In fact, during the last 115 years, the “residence time” (intended as the years before the next 50-m upward shift) at an altitude of 2,150 m a.s.l. was approx. 52 years (from 1902 to 1954), then trees moved upward at about 2,200 m a.s.l. and kept growing at that elevation for approx. 34 years (from 1965 to 1999). Only from 1996 to 2006 (with a residence time of just 10 years) trees began growing at 2,250 m a.s.l. Some trees, between 1994 and 2002, reached even higher elevations and established at about 2,300 m a.s.l. (with a residence time at that altitude of only 8 years). This general trend in the upward shift of the treeline ecotone was relatively fast: after 1954, it took only 52 years for trees to move 150 m upward, whereas treeline “resided” for almost the same timespan (≈52 ys or even longer) at a lower elevation of 2,150 m a.s.l. Moreover, we did not find any tree born after 2002 established at the highest elevation point (2,300 m a.s.l.). The youngest sampled tree (which was born in 2006) resided 50 m below, which suggests a recent slight downward movement of the treeline likely due to the decrease of mean annual temperatures we detected for the last few years.
Our finding that young trees (approx. 17–19 years old, sampled at the higher altitudes) grow less than old trees (93 ± 25.21 years old, sampled at lower elevations), during their first 12 years of life, confirms a reduced tree growth at higher temperatures in the high-mountains61. In summary, this evidence suggests that the temperature increase during the past century may pose a double effect on mountain forests: in our study-site, tree growth decreased in recent years and the treeline ecotone moved 150 m upward in just about 50 years (after the 1960s), with a slight downward shift during the last decade (after 2000s).
In general, the rapid upward treeline shift that we detected combined with the reduced tree growth, may radically change the composition and landscape of the montane ecosystems. In areas on our study site where we did not recorded any trees, we found that a high percentage of the soil was covered by highly biodiverse meadows (with approx. 14 identified herb species, data not reported here) and by both economically and culinary valuable shrubs, including Juniper spp., Vaccinium spp. and Ribes spp. A continuous upward shift tendency, as the one documented for the last century in this study, would presumably lead to an expansion of the forest at the expense of meadow and shrub species, which would have no other area to migrate to once the mountain summit is reached62. On the other hand, we also identified a partial tendency inversion for the most recent decade of tree expansion at high elevations, which could represent a “temporary relief” for the survival of high-altitude meadow and shrub species. Trees did not move upward after 2004 and this might reflect the recent temperature decrease observed both at the local meteorological stations31 and from our estimates derived from the hyperbolic relationship between the mean annual growth and the 1956–2009 field-recorded temperatures. In fact, we found that as −4.3 °C represents a threshold between the negative growth tendency before the 1978 and the positive one after this year, estimated temperatures from 2010 to 2017 show a decreasing trend that, at least, could temporarily hinder the upward shift of the treeline. Further resources to reach and study these remote high-mountain areas would clarify whether this threshold of −4.3 °C applies to other treelines of the Altai Mountains.
We recognize that our study focuses on a specific area of the Russian part of the Altai Mountains and we need to be careful discussing the applicability our results to other areas of the region. However, the novel findings should not wait for publication for many years so that other sites can be analyzed but rather used to stimulate similar research at other similar ecosystems by immediate publication.
The scale of impacts of climate change in the Arctic and Alpine areas are of far more than academic importance and generalizations must be balanced by site-specific studies of relevance to local populations and local biodiversity hot-spots such as the Aktru high-mountains.
In conclusion, based on in situ treeline data and a dendrochronological analysis, we found an accelerating upward shift of treeline on a high-mountain vegetation ecotone until recently and a declining tree growth rate in this ecotone that are attributable to an increase in local air temperature in the past century. Our findings provide evidence of substantial ecological impacts of global climate change in high-mountain regions. Our study suggests that climate change, if left unabated, can cause dramatic changes in the structure, diversity, and landscape composition of high-mountain ecosystems such as the Altai Mountains.
Methods
Data sampling and wood core collection
After identifying a representative slope of the treeline ecotone in the mountain chain and checking from historical information and local people the absence of any relevant anthropogenic influence, a grid composed by 3 vertical (altitudinal) and 4 horizontal (at the same elevation) transects, at a distance of 50 m each, was drawn and placed on a georeferenced satellite map of the area in order to include the whole ecotone from the timberline, through treeline, to the isolated and highest individual trees of the tree species line (Fig. 1), whose definitions are conventions for communication and do not deserve a major scientific debate49. This sampling frame was subdivided in a 5 × 5 m grid in order to collect landscape data from the areas included in the frame but excluded from the sampling transects (Fig. 2). The coordinates of each transect and corners of the sampling frame were recorded with a 2-m precision.
Along each transect all trees with a diameter at the breast height (d.b.h.) >1 cm were identified at the species level, measured with a laser hypsometer-dendrometer (height, dbh and crown width) and georeferenced within the sampling grid (Fig. 2 and Table 1). Seedlings and saplings of trees (below the minimum height of 1.35 m and d.b.h. of 1 cm) were not measured but have been mapped and included in the category “Trees” (which represents all the non-sampled trees outside the transects) in Fig. 2 and Table 1. The coverage of the other vegetation composed by shrub and herb species (identified at the genus level) and of the main landscape features (big rocks and landslide areas) included in the sampling frame among transects was visually estimated and georeferenced within each cell of the grid (Fig. 2 and Table 1).
At each intersection of the 3 vertical and 4 horizontal transects, the two closest trees were sampled with Haglöf increment borers in a cross design in order to collect two whole wood cores from each tree. This yielded to a collection of 103 individual trees from which 48 dendrochronological samples were extracted.
Aerial photographs and 3-D modelling
In order to create a 3-D model of the study site, which could represent a reference for future studies of this remote area, 10 UHD aerial photographs taken by a professional drone from different angles where combined in order to create a three-dimensional mesh (Meshlab and Sketchup software) that allowed the reconstruction of 3-D aerial model. Then we georeferenced the 3-D model overlapping it to the 2-D Google Earth satellite image (Fig. 1). Unfortunately, no historical aerial photos of this treeline were available at both the local station and Tomsk State University.
Biometric (allometric) data analysis
We tested a possible correlation between altitude and biometric (allometric) data with the Spearman’s rank correlation index at a significance level of α = 0.01 (R Development Core Team 2018). Then, we plotted a correlogram to show the correlation trends (R Development Core Team 2018).
The summary values per altitudinal ranges of the biometric data for all collected trees is shown in Table 1.
Dendrochronological data analysis
The 48 wood cores collected were prepared following the common dendrochronological protocols and measured with a LINTAB measuring table with an accuracy of 0.01 mm, equipped with a Leica MS5 stereoscope. The tree-ring growth series were then visually (TSAPwin software, version 0.5363 and statistically (COFECHA64,65 cross-dated within trees and between trees (of the same site) for avoiding dating errors in the dataset. Then, an annual mean growth trend was calculated averaging the individual tree annual growth and a variance (SD) was attached (Fig. 4). The basal area of each sampled tree was also estimated.
We then detrended the tree-ring series (Fig. 5) through the estimation and removal of the tree’s natural biological growth trend, and standardized the detrended values by dividing each series by the growth trend to produce units in the dimensionless ring-width index (RWI). We used the most adopted methods available for detrending trough the package ‘detrendeR’ in R Development Core Team 2018. We implemented are a smoothing spline detrending via ffcsaps (method = “Spline”), a modified negative exponential curve (method = “ModNegExp”), a simple horizontal line (method = “Mean”), an “Ar” approach, and a “Friedman” approach.
The “Spline” approach uses a spline where the frequency response is 0.50 at a wavelength of 0.67 * “series length in years”. This attempts to remove the low frequency variability that is due to biological or stand effects.
The “ModNegExp” approach attempts to fit a classic nonlinear model of biological growth of the form f(t) = a exp(b t) + k, where the argument of the function is time66.
We also checked if a suitable nonlinear model could not be fitted (function is non-decreasing or some values are not positive) and we fitted a linear model. The “Mean” approach fits a horizontal line using the mean of the series. This method is the fallback solution in cases where the “Spline” or the linear fit (also a fallback solution itself) contains zeros or negative values, which would lead to invalid ring-width indices.
The “Ar” approach is also known as prewhitening where the detrended series is the residuals of an Ar model divided by the mean of those residuals to yield a series with white noise and a mean of one. This method removes all but the high frequency variation in the series and should only be used as such61.
These methods are chosen because they are commonly used in dendrochronology. There is a rich literature on detrending and many researchers are particularly skeptical of the use of the classic nonlinear model of biological growth (f(t) = a exp(b t) + k) for detrending67.
Finally, we detrended our time series with the “Friedman” approach that uses Friedman’s ‘super smoother’ as implemented in supsmu (R Development Core Team 2018).
We tested the significance of the time-series with a Mann-Kendall method, which is an effective test to detect the long-term change in time series68. In this study, this method was applied to detect the long-term trend change of the mean annual growth of the sampled trees.
Moreover, we checked the possible autocorrelation on our dendrochronological time-series (with an autocorrelation function, acf implemented R Development Core Team 2018). The resulting plot (Fig. 9) shows a significant correlation at lag 1 (≈1), lag 2 (≈0.5) and lag 13 (≈0.4) that decreases after a few lags. This pattern indicates an autoregressive term. Therefore, we used the partial autocorrelation function to determine the order of the autoregressive term (Fig. 10).
After accounting for the partial autocorrelation function we derived an additional curve based on this tree ring index (Fig. 4), which, however, does not differ much from the curves detrended with different methods (particularly from the Friedman’s curve). Most of the growth peaks and the general growth trend remained unaltered.
Finally, in order to evaluate the bias of our time series and the level of its accuracy (defined in terms of standard error) to sample estimates, we performed a block bootstrap analysis (tsboot package in R). This technique allowed an estimation of the bias of our distribution using random sampling methods.
We used a fixed block length of 1 to account for each lag of the time series and the results show a bias of −0.32 and a very low standard error of 0.06. We then plotted the histogram of the block bootstrapping method (Fig. 11), which shows that the mean difference of bootstrapping is well below the observed difference (dashed vertical line) in the time series of the mean annual growth (with the calculated bias of −0.32). The Q-Q plot of the 500 random bootstrapping with replacement shows that our time series has a normal distribution (Fig. 11).
Dendrochronology vs. altitude, year of tree birth, and climate data analysis
We tested a possible correlation between our dendrochronological time series (mean annual growth and basal area) vs. altitude, year of tree birth, and climatic variations with the Kendall’s correlation index at a significance level of α = 0.01 (R Development Core Team 2018).
Because the relationship between the year of tree birth vs. both the altitude and the mean annual growth was non-linear, we applied a generalized linear model (glm package in R) and we considered a quadratic function and the related determination coefficients to describe these curvilinear relationships, in the mathematical form of:
$${\rm{y}}={{\rm{ax}}}^{2}+{\rm{bx}}\pm {\rm{c}}{\rm{.}}$$
(1)
To disentangle the curvilinear relationship between the mean annual growth and the year of tree birth, we analyzed the growth trends at the rarefied minimum age of all our core samples (12 years). In this way, we were able to compare the growth of the trees grouped per current age and altitude (4 groups at 2,150, 2,200, 2,250, 2,300 m a.s.l.) during their “youth”, in their first 12 years of life and to evaluate their growth variability.
Although some mean precipitation and temperature observations for the whole Altai region have been recently published31, we used the longest climatic time series available in the literature29, with a time span of 53 years (1956–2009). This was recorded from a meteorological station (that in Kosh-Agach) located at a comparable altitude (≈1,900 m a.s.l.) and location (linear distance ≈50 km) with respect to our study site.
Since, from the data in the literature, there is no evidence of a significant variability in the precipitation trends29,31, we considered only temperature records.
The mean annual temperature variability was correlated with the mean annual growth variability and a generalized linear model (quadratic function) was used to describe the relationship. Because the trend resulted in an almost hyperbolic curve (quadratic function), we inspected in any factor related to the predictor variable (temperature) could have influenced it. We discovered that a threshold of −4.3 °C was never passed before the 1978 and that this temperature approximately represents the turning point (close to the minimum point x = −b/2a of the eq. 1) of the curvilinear relationship between the mean annual growth and the mean annual temperature.
Therefore, to derive linear equations that would allow an estimation of the temperature during the missing years of observation from our dendrochronological time series, we divided the temperature-growth plot in two subplots (at −4.3 °C threshold). The parameter from the linear regression equation derived from the data points at a temperature <−4.3 °C were used to estimate the temperature for the period 1902–1955 (because of we show that the temperature was always below −4.3 °C before 1978) and those derived from the data points at a temperature >−4.3 °C were used to estimate the temperature for the period 2010–2017 (because the data points >−4.3 °C exclusively represents most of the growth trends after 1978).
Following a common approach to reconstruct temperatures from tree-ring records69,70,71, the stability and reliability of the regression Eq. 1 were assessed using the split sample method. We performed a validation by calibrating climate data from a sub-period by dividing the data into two parts: 1956–1978 and 1979–2009) and verifying the reconstruction using the remaining data. The results were evaluated by the correlation coefficient (r), the sign test (ST), the reduction of error test (RE), the coefficient of efficiency (CE) and the product means test (t) during the verification period. The actual and estimated data in year i of the verification period and the mean of the actual data in the calibration and verification periods were compared to calculate the sign of RE and CE following refs72,73, respectively.
We displayed the reconstructed curves of temperature variability and compared their general trends and selected relevant periods with the available information in the literature (see the Discussion section).
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Acknowledgements
Publication of this article was funded in part by Purdue University Libraries Open Access Publishing Fund.
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Authors
Contributions
R.C.G. conceived the study, wrote the manuscript and coordinated the field activities. R.C.G. and T.C. supervised the research activities. R.C.G., A.V., A.D. and L.F. collected the dendrochronological and biometric tree data. R.C.G., A.D., L.F., A.V. and G.B. analysed the dendrochronological data. T.C, G.B. and J.L. contributed to the discussion of the results and preparation of the manuscript. All authors revised and approved the final version of the manuscript.
Corresponding author
Correspondence to Roberto Cazzolla Gatti.
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Cazzolla Gatti, R., Callaghan, T., Velichevskaya, A. et al. Accelerating upward treeline shift in the Altai Mountains under last-century climate change. Sci Rep 9, 7678 (2019). https://doi.org/10.1038/s41598-019-44188-1
• Accepted:
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# Siri Technology
10/24/2011 2:44:23 PM
## Siri Technology
Considerable press puffery has followed Apple’s announcement of Siri dramatizing the significant nature of the natural language technology. Siri is touted as a $200 million DARPA-funded research in the same league as the Internet and GPS. Siri is much advanced over existing popular voice control technology in its acceptance of free-form input, access to web services, and conversational feedback; however, it probably not too much of an advance. Microsoft and Google should catch up quickly. Microsoft previewed a future vision of its acquired Tellme voice control technology, and Google is likely to advance its Voice Actions to match Apple’s developments. Voiced-based assistant technology is ideal for a mobile device because the phone is designed to host conversations, geolocation is available, internet data is available, and all the user’s information is already stored in the device. Because contextual information is abundant and the domain knowledge is constrained, speech recognition can be very effective. My reaction to Siri was “finally”—finally a realization of state of art in technology into commercial products. There’s a lot of technology that I encountered in academia decades ago that have not yet made their way to consumers such as natural language analysis. In college, I wrote a Prolog-based natural language parser based on definite clause grammars that took parsed text and converted it into a semantic representation, on which we perform queries. As part of my entrepreneurial work, I rewrote the Link Grammar parser from CMU and obtained licenses to various natural language data sources like COMLEX, NOMLEX and WordNet. It felt odd that I might be the first to sell a consumer natural-language product aside from grammar checkers and translation software. So, finally, but… I am a bit skeptical about the technological advances, since I am simply one developer and, though I have leverage off the work of others, I did not need$200M. First, Siri uses Dragon’s speech recognition technology, which any developer can license as part of the Nuance Mobile Developer Program. Parsing natural language. Second, the DARPA halo is just for dramatics, but the kind of natural language analysis has been around for awhile. I credit Apple for pushing the quality of the communication beyond mere keyword and structure recognition and for putting existing art into its products. I suspect that most of those stated 300 SRI researchers consulted but did not actually work on CALO (let alone full-time). From the looks of the project involved, most of the technology appears in the backend (much of which may not even be relevant to Siri), very little in natural language analysis.
In How Siri Works, the author presents his own skepticism of Siri technology. Jeff is more impressed with application and integration of Siri into the OS than with the technology itself. Siri performs operations on a limited set of operations centered around built-in iPhone applications, plus it integrates with a number of web services such as Yelp, Wolfram|Alpha, OpenTable and Wikipedia. Despite the limitations, it still is an impressive achievement given the naturalness of the implementation.
Another CALO engineer confirms my thoughts of Siri as a compelling but not terribly advanced technology:
I worked at SRI on the CALO project, and built prototypes of the system that was spun off into SIRI. The system uses a simple semantic task model to map language to actions. There is no deep parsing - the model does simple keyword matching and slot filling, and it turns out that with some clever engineering, this is enough to make a very compelling system. It is great to see it launch as a built-in feature on the iPhone.
The NLP approach is based on work at Dejima, an NLP startup: “Iterative Statistical Language Model Generation for Use with an Agent-Oriented Natural Language Interface
A lot of the work is grounded in Adam Cheyer's (CTO of SIRI) work on the Open Agent Architecture: A more recent publication from Adam and Didier Guzzoni on the Active architecture, which is probably the closest you'll come to a public explanation of how SIRI works: Active, a Platform for Building Intelligent Software
His comments on the natural language parsing left me disappointed, but it’s possible that Apple upgraded that natural language processing capabilities of Siri with its homegrown version after acquiring Siri. However, after reading the Dejima paper, it turns out that a traditional parser may have too rigid a grammar for the short, conversional, and often grammatically incorrect speech input.
NLIs often use text as their main input modality; speech is however, a natural and, in many cases, preferred modality for NLIs. Various speech recognition techniques can be used to provide a speech front end to an NLI. Grammar-based recognizers are rigid and unforgiving, and thus can overshadow the robustness and usability of a good NLI. Word-spotting recognizers are reliable only when the input consists of short utterances, and the number of words to be spotted at each given time is small. Dictation engines are processor and memory intensive, and often speaker dependent. The dictation vocabulary is often considerably larger than required for domain-specific tasks. General statistical language models (SLMs), although robust enough to be used as a front end for a structured domain, requires a very large training corpus. This is time consuming and expensive since a large number of users needs to be sampled and all speech has to be transcribed.
The system proposed in the paper is a statistical language designed explicitly for an agent-oriented natural language speech-based interface. It does have its failings such as the weaknesses detailed in Siri in Practice where Siri has trouble with parenthetical or quoted expressions that have might have been more properly handled with a grammar-based recognizer, assuming that Apple has not changed Siri’s parser.
Another telling failure is Siri’s response to “What’s the best iPhone wallpaper?” in which Siri responds with a canned response to “What’s the best phone?” as if it simply did not process the word “wallpaper” and simply hooked on keywords “best” and “iPhone.”
The response vaguely resembles that of fairly unsophisticated chatterbots. Siri could be performing a keyword match either on the syntactic or the semantic level. It would be easy to test this hypothesis by asking Siri variations of the questions. I doubt the seriousness of these mistakes, because Apple might be using a different catch-all, keyword matching-system for unanticipated queries.
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# [转] Stacked Diffs Versus Pull Requests
Update: This post received quite a lot of healthy discussion on Hacker News. You can follow that conversation here: https://news.ycombinator.com/item?id=18119570
Update: And… again? https://news.ycombinator.com/item?id=26922633
Update: Btw, if you want to work this way, we're hiring at Cord !
People who have worked with Phabricator using a 'stacked diff' workflow generally love it and seek it wherever they next go. People who have never used it and only use Pull Requests with GitHub/GitLabs generally don't understand what the fuss is all about. How can code review be *sooo* much better using some obscure tool with hilarious documentation? Well, hold on to your butts, because I'm about to break it down. This post is going to focus solely on the engineering workflow for committing code and getting reviews. I'll probably do a second post about the details of code review between the two tools.
Phabricator 已经不再维护了。而 Gerrit 又太丑。Review Board值得一试。
Before I dig deeply, let me just say I've created and merged hundreds of Pull Requests and landed thousands of Diffs. I know both of these workflows in and out. I'm not just some ex-Facebook, Phabricator fan boy pining for the 'good old days.' I've worked on engineering teams using CVS (oh yes), SVN, Git, and Mercurial. GitLabs, GitHub, Gerrit, and Phabricator. I'll happily acknowledge that you can get a lot of good work done using any of these. Now, if you want to talk about how to get the most work done — the most productivity per engineer — that's where I have a strong opinion informed by lots of experience.
## What are stacked diffs?
Many folks reading this post won't actually have a clue what "stacked diffs" are all about anyway. This is understandable. Feature branches and Pull Requests (PRs) are fairly ubiquitous and (sort of) well-understood. For the uninitiated, I'll outline how it works.
### First, Pull Requests
In PR-based development, you start by branching master then add one or more commits which you submit as a 'Pull Request' in the Github UI. A Pull Request is (or at least should be) an atomic unit of code for review. When someone requests changes, you do this by adding additional commits to the pull request until the sum of these changes satisfies the reviewers demands.
The really important thing about this is that the state of your local repository is dictated by the review process. If you want to have your code reviewed, you first have to branch master, then commit to that branch, then push it remotely, then create a PR. If you get review feedback, you have to commit more code onto the same branch and a) push it to create a longer, less coherent commit history or b) merge your local commits and force push to the branch. This also means that you can't have a local checkout of the repository that looks different from the remote. This is a really, really important point that I'll come back to again and again.
### So, Stacked Diffs
The basic idea of stacked diffs is that you have a local checkout of the repository which you can mangle to your heart's content. The only thing that the world needs to care about is what you want to push out for review. This means you decide what view of your local checkout the reviewers see. You present something that can be 'landed' on top of master. It may be helpful to skip down to the Case Studies section below to get a more intuitive feel about how this works.
The typical workflow is to work right on top of master, committing to master as you go. For each of the commits, you then use the Phabricator command line tool to create a 'Diff' which is the Phabricator equivalent of a Pull Request. Unlike Pull Requests, Diffs are usually based on exactly one commit and instead of pushing updates as additional commits, you update the single commit in place and then tell Phabricator to update the remote view. When a Diff gets reviewed and approved, you can "land" it onto remote master. Your local copy and master don't have to be in perfect sync in order to do this. You can think of this as the remote master cherry-picking the specific commit from your git history.
That's right, I said it. You can probably commit everything to master. Sound terrifying? It's mostly… well… just, not a problem at all. It's fine like 93% of the time. In fact, this approach gives you the ability to do things that branches alone just can't (more on that below). The anxiety many engineers feel about committing ahead of master is a lot like the fear that if you fly at lightspeed, you'll crash into a star. Popularly held, theoretically true, and practically completely wrong.
In practice, engineers tend to work on problems whose chunks don't easily divide into units of code review that make sense as a branch-per-unit-of-review. In fact, most engineers don't know exactly how their work decomposes when they start working on a problem. Maybe they could commit everything to master. Maybe they need a branch per commit. Maybe it's somewhere in between. If the rules for how to get code reviewed and how to commit code are defined for you ahead of time, you don't get to choose, which in many cases means a net loss in productivity.
The big "aha!" idea here is that units of code review are individual commits and that those commits can stack arbitrarily, because they're all on one branch. You can have 17 local commits all stacked ahead of master and life is peachy. Each one of them can have a proper, unique commit message (i.e. title, description, test plan, etc.). Each of them can be a unit out for code review. Most importantly, each one of them can have a single thesis. This matters so much more than most engineering teams realize.
### Yes, basically every commit can be on top of master
"But that's marmot floofing crazy!" I hear you say at your computer, reading this months after the blog post was written. Is it? Is it, really?! You may be surprised to learn that many engineers, who make a fantastic amount of money from some of the best companies in the world, commit directly to master all of the time, unless they have a reason not to.
To enable this, the mental model is different. A unit of code review is a single commit, not a branch. The heuristic for whether or not to branch is this: 'Am I going to generate many units of code review for this overall change?' If the answer is yes, you might create a branch to house the many units of code review the overall change will require. In this model, a branch is just a utility for organising many units of code review, not something forced on you *as* the mechanism of code review.
If you adopt this approach, you can use master as much as you want. You can branch when/if you want. You, the engineer, decide when/if to branch and how much to branch.
In this model, every commit must pass lint. It must pass unit tests. It must build. Every commit should have a test plan. A description. A meaningful title. Every. Single. Commit. This level of discipline means the code quality bar is fundamentally higher than the Pull Request world (especially if you rely on Squash Merge to save you). Because every commit builds, you can bisect. You can expect that reading pure git log is actually meaningful. In fact, in this model every single commit is like the top commit from a Pull Request. Every commit has a link to the code review that allowed the commit to land. You can see who wrote it and who reviewed it at a glance.
For clarity, let me describe the extreme case where you only commit to master. I'll outline things that are simpler because of this. I'm starting in order of least-important to most-important just to build the drama.
#### #1 Rebasing against master
With Pull Requests, if you want to catch up your local branch to master, you have to do the following:
1. Fetch the changes from remote
2. Rebase your branch on top of master
3. Merge any conflicts that arise
That doesn't sound so bad, but what about when you have a branch off a branch off of master? Then you have to repeat the last two steps for each branch, every time.
By contrast, if you only worked from master, you only have to do a git pull --rebase and you get to skip the cascading rebases, every time. You get to do just the work that you care about. All the branch jumping falls away without any cost. Might seem minor, but if you do the math on how often you have to do this, it adds up.
Many of us wear a lot of hats in our jobs. I'm the owner of a user-facing product codebase, which is many tens of thousands of lines, separated into dozens of features. That means I often jump between, for instance, refactoring big chunks of crufty old JavaScript (e.g. hundreds of lines of code across dozens of files) and working out small, nuanced bugs that relate to individual lines of code in a single file.
In the Pull Request world, this might mean I switch branches a dozen times per day. In most cases, that's not really necessary because many of these changes would never conflict. Yet, most highly productive people do half a dozen or more unrelated changes in day. This means that all that time spent branching and merging is wasted because those changes would never have conflicted anyway. This is evidenced by the fact that the majority of changes can be merged from the Github Pull Request UI without any manual steps at all. If the changes would never have conflicted, why are you wasting your time branching and merging? Surely you should be able to choose when/if to branch.
One of the most time-destroying aspects of the Pull Request workflow is when you have multiple, dependent units of work. We all do this all the time. I want to achieve X, which requires doing V, W, X, and Y.
Sound far fetched? Well, just recently, I wanted to fix a user-facing feature. However, the UI code was all wrong. It needed to have a bunch of bad XHR code abstracted out first. Then, the UI code I wanted to change would be isolated enough to work on. The UI change required two server-side changes as well — one to alter the existing REST API and one to change the data representation. In order to properly test this, I'd need all three changes all together. But none of these changes required the same reviewer and they could all land independently, apart from the XHR and feature changes.
In the stacked diff world, this looks like this:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 $git log commit c1e3cc829bcf05790241b997e81e678b3b309cc8 (HEAD -> master) Author: Jackson Gabbard Date: Sat Sep 29 16:43:22 2018 +0100 Alter API to enable my-sweet-feature-change commit 6baac280353eb3c69056d90202bebef5de963afe Author: Jackson Gabbard Date: Sat Sep 29 16:44:27 2018 +0100 Alter the database schema representation to enable my-sweet-feature commit a16589b0fec54a2503c18ef6ece50f63214fa553 Author: Jackson Gabbard Date: Sat Sep 29 16:42:28 2018 +0100 Make awesome user-facing change commit cd2e43210bb48158a1c5eddb7c178070a8572e4d Author: Jackson Gabbard Date: Sat Sep 29 16:41:26 2018 +0100 Add an XHR library to abstract redundant calls commit 5c63f48334a5879fffee3a29bf12f6ecd1c6a1dc (origin/master, origin/HEAD) Author: Some Other Engineer Date: Sat Sep 29 16:40:16 2018 +0100 Did some work on some things The equivalent of the Git configuration of this might look like this: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 $ git log commit 55c9fc3be10ebfe642b8d3ac3b30fa60a1710f0a (HEAD -> api-changes) Author: Jackson Gabbard Date: Sat Sep 29 17:02:48 2018 +0100 Alter API to enable my-sweet-feature-change commit b4dd1715cb47ace52bc773312544eb5da3b08038 (data-model-change) Author: Jackson Gabbard Date: Sat Sep 29 17:03:25 2018 +0100 Alter the database schema representation to enable my-sweet-feature commit 532e86c9042b54c881c955b549634b81af6cdd2b (my-sweet-feature) Author: Jackson Gabbard Date: Sat Sep 29 17:02:02 2018 +0100 Make awesome user-facing change commit d2383f17db1692708ed854735caf72a88ee16e46 (xhr-changes) Author: Jackson Gabbard Date: Sat Sep 29 17:01:29 2018 +0100 Add an XHR library to abstract out redundant calls commit ba28b0c843a863719d0ac489b933add61303a141 (master) Author: Some Other Engineer Date: Sat Sep 29 17:00:56 2018 +0100 Did some work on some things
Realistically though, in the Pull Request world, this commonly goes one of two ways:
1. You care massively about code quality so you diligently create a branch off of master for V, then a branch off of V for W, then a branch off of W for X, then a branch off of X for Y. You create a pull request for each one (as above).
2. You actually want to get work done so you create one big ass Pull Request that has commits for V, W, X, and Y.
In either case, someone loses.
For Case #1, what happens when someone requests changes to V? Simple, right? You make those changes in branch V and push them, updating your PR. Then you switch to W and rebase. Then you switch to X and rebase. Then you switch to Y and rebase. And when you're done, you go to the orthopedist to get a walker because you're literally elderly now. You've wasted your best years rebasing branches, but hey — the commit history is clean AF.
Importantly, woe be unto you if you happened to miss a branch rebase in the middle somewhere. This also means that when it comes time to commit, you have to remember which destination branch to select in the Github UI. If you mess that up and merge from X to W after W was merged to master, you've got an exciting, life-shortening mess to clean up. Yay!
For Case #2, everyone else loses because people just don't feel the same burden of quality per-commit in a PR. You don't have a test plan for every commit. You don't bother with good documentation on each individual commit, because you're thinking in terms of a PR.
In this case, when different reviewers request changes to the code for theses V and W, you just slap commits Y++ and Y++++ onto the end of the Pull Request to address the feedback across all of the commits. This means that the coherence of codebase goes down over time.
You can't intelligently squash merge the aspects of the various commits in the Pull Request that are actually related. The tool doesn't work that way so people don't work or think that way. I can't tell you the number of times I've seen the last two or three commits to a PR titled with "Addresses feedback" or "tweaks" and nothing else. Those commits tend to be among the sloppiest and least coherent. In the context of the PR, that *seems* fine, but when you fast-forward 6 months and you're trying to figure out why some code doesn't do what it's supposed to and all you have in a stack of 20 commits from a seemingly unrepated PR and the git blame shows that the offending line comes from a commit titled "nits" and nothing else and no other context, life is just harder.
#### #4 Doing multiple sets of related tasks
If you happen to be one of the rare engineers who is so productive that you work on multiple, distinct problems at the same time — you still probably want a branch per-thing, even in the stacked diff workflow. This likely means that you create a branch per-thing (i.e. per distinct problem), but that you put out multiple units of code for review on each branch.
For the mortals amongst us, let's imagine the case where an amazing engineer is working on three different hard problems at once. This engineer is working on three different strands of work, each of which require many commits and review by many different people. This person might generate conflicts between their branches, but they also are clever enough and productive enough to manage that. Let's assume that each of this person's branches includes an average of 5 or more units of code review in solving each of the 3 distinct problems.
In the Pull Request model, this means that person will have to create 3 branches off of master and then 5 branches-of-branches. Alternatively, this person will create 3 Pull Requests, each of which is stacked 5 commits deep with this that only go together because of a very high level problem, not because it actually makes sense for code review. Those 5 commits may not require the same reviewer. Yet, the pull request model is going to put the onus on a single reviewer, because that's how the tool works.
The Stacked Diff model allows that amazing engineer to choose how/if to branch any commit. That person can decide if their work requires 3 branches and 15 units of code review or if their work requires 15 branches and 15 units of code review or something different.
This is more important than many people realize. Engineering managers know that allowing their most productive people to be as productive as possible amounts to big chunks of the team's total output. Why on earth would you saddle your most productive engineers with a process that eats away at their productivity?
## Thoughtless commits are bad commits
Every single commit that hits a codebase means more shit to trawl through trying to fix a production bug while your system is melting. Every merge commit. Every junk mid-PR commit that still doesn't build but kinda gets your change closer to working. Every time you smashed two or three extra things into the PR because it was too much bother to create a separate PR. These things add up. These things make a codebase harder to wrangle, month after month, engineer after engineer.
How do you git bisect a codebase where every 6th commit doesn't build because it was jammed into the middle of a Pull Request?
How much harder is it to audit a codebase where many times the blame is some massive merge commit?
How much more work is it to figure out what a commit actually does by reading the code because the blame commit message was "fixes bugs" and the pull request was 12 commits back?
The answer is *a lot harder*. Specifically because Pull Requests set you up for way more, way lower quality commits. It's just the default mode of the workflow. That is what happens in practice, in codebases all over the world, every day. I've seen it in five different companies now on two continents in massively different technical domains.
## Make the default mode a good one
You can make the argument that none of this is the fault of Pull Requests. Hi, thanks for your input. You're technically correct. To you, I'd like to offer the Tale of the Tree Icon. When Facebook re-launched Facebook Groups in 2011, I was the engineer who implemented the New User Experience. I worked directly with the designer who implemented the Group Icons, which show up in the left navigation of the site. Weeks after launch, we noticed that almost all the groups had their icon set to… a tree. It was a gorgeous icon designed by the truly exceptional Soleio Cuervo. But… a tree? Why?
### Because it was the first thing in the list.
People choose whatever is easiest. Defaults matter. So much. Even us demi-god-like Engineers are subject to the trappings of default behaviour. Which is why Pull Requests are terrible for code quality. The easiest behaviour is shoehorning in a bunch of shit under one PR because it's just so much work to get code out for review.
This is where Stacked Diffs win out, no question. It's not even close. The default behaviour is to be able to create a unit of code review for any change, no matter how minor. This means that you can get the dozens of uninteresting changes that come along with any significant work approved effortlessly. The changes that are actually controversial can be easily separated from the hum-drum, iterative code that we all write every day. Pull Requests encourage exactly the opposite — pounding in all of the changes into one high-level thesis and leaving the actual commit history a shambles.
## Coding as a queue
The fundamental shift that the Stacked Diff workflow enables is moving from the idea that every change is a branch of off master and to a world where your work is actually a queue of changes ahead of master. If you're a productive engineer, you'll pretty much always have five or more changes out for review. They'll get reviewed in some order and commited in some order. With Pull Requests, the work queue is hidden behind the cruft of juggling branches and trying to treat each change like a clean room separated from your other work. With Stacked Diffs, the queue is obvious — it's a stack of commits ahead of master. You put new work on the end of the queue. Work that is ready to land gets bumped to the front of the queue and landed onto master. It's a much, much simpler mental model than a tangle of dependent branches and much more flexible than moving every change into the clean room of a new branch.
(For the pedantic few out there, yes, I just said stacked diffs are like a queue. Yeah… I didn't name the workflow. Don't hurl the rotten tomatoes at me.)
By now, you're probably sick of this theoretical/rhetorical discussion of what good engineering looks like. Let's switch gears and talk about this in practical, day to day terms.
## Case Study #1: The Highly Productive Coder
In this case study, we take a look at Suhair. Suhair is a really productive coder. Suhair produces 10 or more high quality commits every day.
### With Pull Requests
Suhair starts the day fixing a bug. Creates a branch, makes changes. Commits them. Suhair then pushes to the remote branch. Then navigates away from the terminal to the Github UI to create a pull request.
Next, Suhair switches back to master, pulls, and creates a new branch to work on a new feature. Commits code. This code is completely unrelated to the bug fix. In fact, they would never generate merge conflicts. Still, Suhair sticks to branches. Works on the feature. Gets it to a good RFC state. Suhair pushes the changes. In Github, Suhair creates a pull request.
Next Suhair starts working on another feature improvement. Switches to master. Pulls. Branches. But… uh oh. This change depends on his bug fix from earlier? What to do? He goes to the bug fix PR, sees if there are any comments. One person left some passing comments, but the person Suhair needs to review it hasn't commented.
So Suhair decides it's too much work to create a branch off his bug fix branch and decides to do something else in the interim. Suhair pings the needed person, begging for code review, interrupting their flow and then starts working on something else.
### With Stacked Diffs
Suhair pulls master in the morning to get the latest changes. Makes the first bug fix, commits it, creates a Diff to be reviewed, entirely from the command line. Suhair then works on the unrelated feature. Commits. Creates a Diff from the command line. Then starts working on the bug-fix-dependent feature improvement. Because Suhair never left master, the bug fix is still in the stack. So, Suhair can proceed with the feature improvement uninterrupted. So, Suhair does the work. Commits it. Creates a Diff for review.
By now, the person who should have reviewed the initial bug fix actually got around to it. They give Suhair some feedback which Suhair incorporates via interactive rebase. Suhair rebases the changes on top of the updated bug fix, which generates a small merge conflict with the feature improvement, which Suhair fixes. Then Suhair lands the change via interactive rebase. On the next git pull –rebase against remote master, the local commit disappears because remote master already has an identical change, and Suhair's queue of commits ahead of master decreases by one.
As a bonus for Suhair today, the same reviewer who approved the bug fix is also the reviewer needed for his feature improvement. That person approved his tweaks right after they reviewed his bug fix. So, Suhair rebases those changes to be at the top of the commit stack, then lands them. Suhair never switches branches. At the end of the day, only the feature work is left in his local repo, everything else is landed on top of master.
The next day, Suhair comes in, runs git pull –rebase and starts working without any branch juggling.
## Case Study #2: The Free Spirited Hacker
Charlie is a productive, energetic, somewhat amoral hacker who just wants to get work done as fast as possible. Charlie knows the product better than any one, but doesn't really care about code quality. Charlie is best paired with a senior tech lead (or two) who can rein in the chaos a bit.
### With Pull Requests
Charlie starts the day by branching master and spamming the branch with five commits that are only vaguely related. Charlie's commits are big, chunky commits that don't make a lot of sense. They tend to be a bunch of things all crammed together. Reviewers of Charlie's code always know they'll have a lot of work ahead of them to make sense of the tangle of ideas. Because of this, they tend to put off reviewing. Today, a senior tech lead takes 45 minutes to read through all these changes, giving detailed feedback and explaining how to improve the various strands of the change. Charlie commits more changes onto the PR, addressing feedback and also makes random "fixes" along the way. In the end, the PR is probably okay, but it's certainly not coherent and may the Mighty Lobster on High protect those who have to make sense of the code in the coming months.
During this laborious back-and-forth, Charlie's best option is to keep piling things on this PR because all the related changes are in it. The tech lead doesn't have a reasonable alternative to offer Charlie.
### With Stacked Diffs
Charlie blasts out five commits and five Diffs back to back. Each one addresses something specific. Each one goes to a different reviewer because Charlie happens to be making a sweeping change to the codebase. Charlie knows how it all fits together and the tech leads can make sure that the individual changes aren't going to ruin everything.
Because the changes are smaller and more coherent, they get much better review. A tech lead points out that one of the changes is clearly two separate theses that happen to touch the same set of files. This tech lead reviewing the code pushes back on Charlie. The tech lead points out that these should actually be two separate commits. Unfazed, Charlie abandons the Diff. Using interactive rebase to rewind history to that troublesome commit, Charlie uses git reset to uncommit the single commit that has two theses.
At this point, Charlie's local master is two commits ahead of remote master and has a bag of uncommitted changes that Charlie is currently hacking on. There are two more changes that are in the future, waiting to be added back to the local commit history by Git when Charlie is done rebasing interactively.
So, Charlie uses git add -p to separate out one change from the other and creates two new commits and two new diffs for them separately. They each get a title, a description, and a test plan. Charlie then runs git rebase –continue to bring fast-forward time and bring back the later changes. Now, Charlie's local master is six commits ahead of remote master. There are six Diffs out for review. Charlie never switched branches.
## Case Study #3: The Engineer with a Bad Neighbour
Yang is a great engineer working in a fun part of the infrastructure. Unfortunately, Yang has a bad neighbour. This other engineer constantly lands the team in trouble. Today, Yang has found that the build is broken due to yet another incomprehensible change. The neighbour has a "fix" out for a review, but no one trusts it and several knowledgeable people are picking through the code in a very contentious code review. Yang just wants to get work done, but can't because the bug is blocking everything.
### With Pull Requests
Yang will checkout the remote branch with the "fix". Next, Yang will branch off of that branch in order to get a sort-of-working codebase. Yang gets to work. Midday, the bad neighbour pushes a big update to the "fix". Yang has to switch to that branch, pull, then switch back to the branch Yang has been working on, rebase, and then push the branch for review. Yang then switches gears to refactor a class nearby in the codebase. So, Yang has to go back to the bug "fix" branch, branch off it, start the refactor, and push the commit remotely for review. The next day, Yang wants to merge the changes, but the "fix" has changed and needs rebasing again. Yang switches to the bug fix branch, pulls. Switches to the first branch, rebases. Pushes. Switches to GitHub to do the merge, carefully selecting to merge onto master rather than the bug fix branch. Then Yang goes back to the terminal, switches to the second feature, rebases, pushes, goes to GitHub, selecrs to merge to master, and merges. Then, Yang applies for AARP, because Yang is now in geriatric care.
### With Stacked Diffs
Yang sees that the Diff for the "fix" is out for review. Yang uses the Phabricator command line tool to patch that commit on top of master. This means that it's not a branch. It's just a throwaway local commit. Yang then starts working on the first change. Yang submits a Diff for review from the command line. Later, the "fix" has changed, so Yang drops the patch of the old version from the Git history and patches in the updated one via interactive rebase. Yang then starts working on the second change, submits a Diff for review. The next day, Yang is ready to land both changes. First, Yang dumps the previous patch of the fix and repatches the update to make sure everything works. Then, Yang uses the command line via interactive rebase to land both of the changes without ever switching branches or leaving the terminal. Later, the fix lands, so Yang does a got pull –rebase and the local patch falls off because it's already in master. Then Yang goes to skydiving because Yang is still young and vital.
## In Conclusion
As you can see from the Case Studies, you can definitely get good work done no matter what tool chain you use. I think it's also quite clear that Stacked Diffs make life easier and work faster. Many engineers reading this will say the cost of switching is too high. This is expectable. It's a thing called the Sunk Cost Fallacy. Everyone prefers the thing they feel they have invested in, even if there is an alternative that is provably more valuable. The stacked diff workflow is a clearly higher-throughput workflow that gives significantly more power to engineers to develop code as they see fit.
Inside Facebook, engineering used the branch-oriented workflow for years. They eventually replaced it with the stacked diff workflow because it made engineers more productive in very concrete terms. It also encourages good engineering practices in a way exactly opposite to branching and Pull Requests.
Something I haven't touched on at all is the actual work of reviewing code. As it turns out, Phabricator also happens to offer better code review tools, but I'll save that for another post.
本作品采用知识共享署名-非商业性使用-相同方式共享 3.0 中国大陆许可协议进行许可，欢迎转载、演绎，
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# Math Help - confusing limit..
1. ## confusing limit..
I have the limit as x goes to 0 ln(cotx)/(e^csc^2(x))
it reads as the natural log of cotx divided by e raised to the csc squared x
the csc is csc^2(x). cotx as x goes to 0 is 1/0, which is undefined. How would I start this?
2. $\lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2} (x)}}$
Try letting $u=e^{csc^{2}(x)}$
$csc^{2}(x)=cot^{2}(x)+1$
$u=e^{cot^{2}(x)+1}$
$ln(u)=cot^{2}(x)+1$
$\sqrt{ln(u)-1}=cot(x)$
$ln(\sqrt{ln(u)-1})=\frac{ln(ln(u)-1)}{2}$
Then you get:
$\frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{ln(ln( u)-1)}{u}$
Now use L'Hopital:
$\frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l n(u)-1)}=0$
3. Originally Posted by galactus
$\lim_{x\rightarrow{0}}\frac{ln(cot(x))}{e^{csc^{2} (x)}}$
Try letting $u=e^{csc^{2}(x)}$
Then you get:
$\frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{ln(ln( u)-1)}{u}$
Now use L'Hopital:
$\frac{1}{2}\lim_{u\rightarrow{\infty}}\frac{1}{u(l n(u)-1)}=0$
I saw you changed x goes to 0 to u goes to infinity. Why does u not go to 0?
4. Because we done a change of variables. Just like when you sub in integration, you have to change the limits.
$\lim_{x\rightarrow{0}}e^{csc^{2}(x)}={\infty}$
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# Review of Derivative Rules
Here is a quick review of the basic derivative rules in words. See the Table of Derivatives for a list of Derivative Formulas.
• The derivative of a constant function is zero.
• The derivative of a power function is a function in which the power on $$x$$ becomes the coefficient of the term and the power on $$x$$ in the derivative decreases by 1.
• The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative.
• The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g.
• The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g.
• The derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.
• The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function.
• We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other.
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# define resistance of a conductor class 10
[CBSE 2016], Two metallic wire A and B of the same material are connected in parallel. Answer: ∝ $$\frac{1}{A}$$ Electric power determines the rate at which energy is delivered by a current. How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω? Direct current source of electromotive force. Compute the heat generated while transferring 96000 C of charge in one hour through a potential difference of 50 V. (d) 4 : 1 Materials having resistivity in the range of 10−8 Ω m to 10−6 Ω m are considered as very good conductors. Resistivity of mercury = 94.0 x 10-8 Ω m. Using Ohms law, V= I^1R^2 I^1 = V/(R^1) = 220/1200 = 0.18A And, I^2 = V/(R^2) = 220/100 = 2.2A, Resistance of each part = R/3 = 9/3 = 3Ω R^1 = R^2 = R^3 = 3Ω In parallel combination, 1/(R^p) = 1/(R^1) + 1/(R^2) + 1/(R^3) = 1/3 + 1/3 + 1/3 = 3/3 = 1 R^p = 1Ω, Given: P = 750 W, V = 220V (i) P = VI 750 = 220 × I I = 750/220 = 3.40A (ii) P = (V^2)/R R = (V^2)/R = (220^2)/R , R= 64.53Ω, Given: q= 1.6 × 10^-19 C, I= 1A, n=?, t=1s We know, q=It and q =ne ne = It n = (It )/(e ) = (1×1)/(1.6×10-19). While the rest of bulbs in parallel circuit will continue to glow without getting disturbed because in parallel combination, current gets additional paths to flow. 30 Days Study Plan for IBPS RRB Officer Scale 1 and Office Assistant Prelims, (i) directly proportional to square of current flowing through, (ii) directly proportional to resistance and. Find the number of electrons that will flow per second to constitute a current of 1 ampere. Answer: (i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω. Whereas if we perform the same experiment with the help of a metallic wire such as copper, the bulb starts to glow. Advantages of connecting electrical devices in parallel: Question 4: Now inert the voltmeter across the ends of first resistor R^1. (fa) IR2, Question 3 (b) 330 mA Question 1: After stretching, if length increases by n times then resistance will increase by n2 times i.e., R2 = n2 R1. What is (a) the highest (b) the lowest total resistance that can be secured by combination of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω? [CBSE 2019], What is the name of physical quantity which is equal to V/I? Truly, this article is really one of the very best in the history of articles. Calculate (i) Its resistance while glowing (ii) energy consumed in kWh per day. Define its unit. Question 2 Question 4: Question 4 Solution: Resistance of a conductor (i) a directly proportional to its length, (ii) inversely proportional to its cross-section areas and depend on the material of the conductor.
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Pandoc
Pandoc is a utility to convert text markup formats (Markdown, reStructuredText, LaTeX, etc) to a variety of other document formats (.doc, .pdf, HTML, etc). Pandoc also provides its own extension of Markdown syntax (details here).
Be sure to install:
• pandoc
• pandoc-data
• pandoc-citeproc (for citations)
Basic use for creating PDFs
pandoc -o writes output to the file following the -o, with formatting based on the file extension. For example, the following command creates a pdf file from a text file (via pdflatex):
pandoc cv_master.txt -o cv_master.pdf
There are other options for specifying input and output files or formats, but this is the simplest.
PDF formatting options
The output formatting of a pdf document can be set either at the commandline, or by using a header or template file during the conversion.
Passing variables
The -V (or --variable) option changes formatting, fonts, or other settings during a pandoc conversion.
pandoc cover_master.txt -V geometry:margin=1.5in -o cover_master.pdf
# create pdf with 1.5in margins
Other variables that can be passed with -V are here
Note that some of these variables, such as geometry, can also be passed in a YAML metadata block. See here for details.
When -s (or --standalone) is used, pandoc adds the header and footers needed for a standalone document in the output format. To see the default template used, type pandoc -D FORMAT. There are a couple of ways to change what is in this header. The -H=FILE (--include-in-header=FILE) option can include special formatting information contained in FILE with the default header. This can be useful when adding LaTeX formatting commands to the intermediate .tex file before creating the pdf.
pandoc cv_master.txt -H '/path/to/header/' -o cv_master.pdf
A custom template for the output document header can also be specified using --template=FILE, where FILE is the custom template. If this option is specified, pandoc first looks in the current directory, then the user template directory ($HOME/.pandoc/templates), and then in the default template directory (/usr/share/pandoc...). Templates should end with an extension for the output format (.latex, .html, etc). To create a new template, copy the default for the given output format using: pandoc -D latex > newtemplate.latex This is good practice when starting a new project because default templates are updated in new versions of pandoc. The template can then be modified and then used in converting the document using: pandoc manuscript_1.markdown --template=newtemplate.latex --bibliography=SNOTELsoildata.bib -o manuscript_1.pdf You can put useful stuff in these templates, like setting margin widths, using packages, etc. A couple I use for pdfs of journal articles are: \usepackage[margin=1in]{geometry} \usepackage{textcomp} % provides \textdegree Other formatting Option | Result ----------------------- | ----------------------------------------------------------------------------------------------------------- --toc | Automatically create a table of contents --toc-depth | Specify the header levels to be used in table of contents (implies --toc) --reference-odt=FILE | Use the FILE stylesheets as a template for .odt output. Best if FILE was created with pandoc, then modified. --reference-docx=FILE | Use the FILE stylesheets as a template for .docx output. --latex-engine=ENGINE | Choose the pdflatex|lualatex|xelatex interpreters, needed for some formatting in pdf files. Citations and bibliographies Pandoc is capable of including citations from an associated bibliographic database (usually a BibTex file). Citations in a pandoc markdown file look like this: [@citationid] , where citationid is defined in the associated .bib file. To convert a pandoc file with citations, run: pandoc paper1_draft.markdown -o paper1_draft.pdf --bibliography hiddencanyon.bib A bibliography will be automatically written after the References heading, if it is included. Citation and bibliography formatting can be specified with --csl=FILE, where FILE is a .csl file (found at http://citationstyles.org). Natbib and biblatex can be used in LaTeX output (pdf) by including them as commandline options. Working directly with LaTeX Pandoc markdown is a nice way to draft LaTeX documents. Pandoc markdown can be rendered to TeX (-o document.tex), or rendered as a PDF via pdflatex (-o document.pdf). Raw TeX and LaTeX can be included in a markdown document and it will be passed to the pdflatex writer. For more info on LaTeX/Tex systems see this page. Citations This is a citation from Smith \cite{smith.2013} The citation should be output in BibTex format FIXME// - havent gotten this to work yet//. Inline TeX placed between \begin and \end tags will be interpreted as LaTeX instead of markdown, and will be ignored in non-LaTeX output formats. Math mode TeX math can be used by putting it between dollar signs. One dollar sign (each side) for inline mode, and two for display math. Most LaTeX math mode symbols are also transferred to other output formats. For example, rendering HTML documents with greek characters (such as$\theta\$) will result in unicode (default) greek characters in the rendered HTML. There are also options for using MathML, MathJax, if this doesn't work.
Pandoc will pass unicode characters in a document to pdflatex that it may not know how to display. This is pretty common with greek characters that are used outside of math mode. In this case it is probably best to use the Xetex interpreter instead. This can be specified by sending --latex-engine=xelatex to pandoc. Of course, Xetex must be installed, which is most easily done by installing the full TexLive distribution (large).
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# Hacking CoreCLR on Linux with CLion
## What/Why?
Being a regular Linux user, when I can, I was looking for a decent setup for myself to grok then hack on CoreCLR’s C++ code.
CoreCLR, namely the C++ code that implements the runtime (GC, JIT and more) is a BIG project, and trying to peel through its layers for the first time is no easy task for sure. While there are many great resources available for developers that want to read about the runtime such as the BotR, for me, there really is no replacement for reading the code and trying to reason about what/how it gets stuff done, preferably during a debug session, with a very focused task/inquiry at hand. For this reason, I really wanted a proper IDE for the huge swaths of C++ code, and I couldn’t think of anything else but JetBrains’ own CLion IDE under Linux (and macOS, which I’m not a user of).
With my final setup, I really can do non-trivial navigation on the code base such as:
CoreCLR is a beast of a project, and getting it to parse properly under CLion, moreover, it requires some non-trivial setup, so I thought I’d disclose my process here, for other people to see and maybe even improve upon…
Generally speaking, all the puzzle pieces should fit since the CoreCLR build-system is 95% made of running cmake to generate standard GNU makefiles, and then builds the whole thing using said makefiles, where the other 5% is made of some scripts wrapping the cmake build-system. At the same time, CLion builds upon cmake to bootstrap its own internal project representation, provided that it can invoke cmake just like the normal build would.
Here’s what I did to get everything working:
1. First, We’ll clone and perform a single build of CoreCLR by following the instructions, What I did on my Ubuntu machine consisted of:
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$sudo apt install cmake llvm-3.9 clang-3.9 lldb-3.9 liblldb-3.9-dev libunwind8 libunwind8-dev gettext libicu-dev liblttng-ust-dev libcurl4-openssl-dev libssl-dev libnuma-dev libkrb5-dev$ ./build.sh checked
2. Once the build is over, you should have everything under the bin/Product/Linux.x64.Checked like so:
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$ls bin/Product/Linux.x64.Checked bin libcoreclr.so netcoreapp2.0 coreconsole libcoreclrtraceptprovider.so PDB corerun libdbgshim.so sosdocsunix.txt createdump libmscordaccore.so SOS.NETCore.dll crossgen libmscordbi.so SOS.NETCore.pdb gcinfo libprotononjit.so superpmi IL libsosplugin.so System.Globalization.Native.a ilasm libsos.so System.Globalization.Native.so ildasm libsuperpmi-shim-collector.so System.Private.CoreLib.dll inc libsuperpmi-shim-counter.so System.Private.CoreLib.ni.{fe21e59b-7903-49b4-b2d3-67de152c1d7d}.map lib libsuperpmi-shim-simple.so System.Private.CoreLib.xml libclrgc.so Loader libclrjit.so mcs Now that an initial build is over, we can be sure that some scripts that were crucial to generate a few headers essential for the rest of the compilation process were generated and CLion will be able to find all the necessary source code once we teach it how to… 3. CLion needs to invoke cmake with the same arguments that the build scripts use. To sniff out the cmake command-line we’ll use an *nix old-timer’s trick to generate traces for build.sh run: use bash -x. Unfortunately, nothing is ever so simple in life, and CoreCLR’s build.sh script doesn’t directly invoke cmake, so we will need to make this -x parameter sticky or recursive. There is no better way to do this than the following somewhat convoluted procedure: First we need to generate a wrapper-script for build.sh, we’ll call it build-wrapper.sh: 1 echo "export SHELLOPTS && ./build.sh \$@" > build-wrapper.sh
After we have our wrapper in place. we run it instead of build.sh like this:
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$bash -x ./build-wrapper.sh checked ... # omitted + /usr/bin/cmake -G 'Unix Makefiles' -DCMAKE_BUILD_TYPE=CHECKED -DCMAKE_INSTALL_PREFIX=/home/dmg/projects/public/coreclr/bin/Product/Linux.x64.Checked -DCMAKE_USER_MAKE_RULES_OVERRIDE= -DCLR_CMAKE_PGO_INSTRUMENT=0 -DCLR_CMAKE_OPTDATA_PATH=/home/dmg/.nuget/packages/optimization.linux-x64.pgo.coreclr/99.99.99-master-20190716.1 -DCLR_CMAKE_PGO_OPTIMIZE=1 -S /home/dmg/projects/public/coreclr -B /home/dmg/projects/public/coreclr/bin/obj/Linux.x64.Checked Boom! We’ve hit that jackpot. For folks following this that are feeling a bit shaky, I’ve isolated the exact part we’re after below: 1 -G 'Unix Makefiles' -DCMAKE_BUILD_TYPE=CHECKED -DCMAKE_INSTALL_PREFIX=/home/dmg/projects/public/coreclr/bin/Product/Linux.x64.Checked -DCMAKE_USER_MAKE_RULES_OVERRIDE= -DCLR_CMAKE_PGO_INSTRUMENT=0 -DCLR_CMAKE_OPTDATA_PATH=/home/dmg/.nuget/packages/optimization.linux-x64.pgo.coreclr/99.99.99-master-20190716.1 -DCLR_CMAKE_PGO_OPTIMIZE=1 -S /home/dmg/projects/public/coreclr -B /home/dmg/projects/public/coreclr/bin/obj/Linux.x64.Checked 4. The “hard” part is over. It’s a series of boring clicks from here on. it’s time to open up CLion and get this show on the road: We’ll start with defining a clang-3.9 based toolchain, since on Linux Clion defaults to using the gcc toolchain (at least on Linux), while CoreCLR needs clang-3.9 to build itself: 5. With a toolchain setup, we need to tell cmake about our build configuration, so we set it up like so: I’ve highlighted all the text boxes you’ll need to set. I’ll go over the less trivial stuff: • The command line option we just set aside in (3) goes into the CMake options field. Unfortunately CLion doesn’t like single quotes (weird…), so I’ve had to change the -G 'Unix Makefiles' into -G "Unix Makrfiles" (notice the use of double quotes). • It would be a wise idea to share the same build folder as our initial command line build used, more over, we might end up going back and forth between CLion and the command line, so I override the “Generation Path” setting with the value bin/obj/Linux.x64.Checked. This is again extracted from the same command line we set-aside before. You’ll find it in my case towards the end, specified right after the -B switch. • For the build options, I’ve specified -j 8. This option controls how many parallel builds (compilers) are launched during the build process. A good default is to set it to 2x the number of physical cores your machine has, so in my case that means using -j 8. 6. That’s it, let CLion do it’s thing while grinding your machine to a halt, and once it’s done you can start navigating and building the CoreCLR project like a first class citizen of the civilized world :) ## Debugging CoreCLR from CLion Once we have CLion understanding the CoreCLR project structure we can take it up a notch and try to debug CoreCLR by launching “something” while setting a breakpoint. Let’s try to debug the JIT as an example for a useful scenario. 1. First we need a console application: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17$ cd /tmp/
$dotnet new console -n clion_dbg_sample The template "Console Application" was created successfully. Processing post-creation actions... Running 'dotnet restore' on clion_dbg_sample/clion_dbg_sample.csproj... Restore completed in 54.39 ms for /tmp/clion_dbg_sample/clion_dbg_sample.csproj. Restore succeeded.$ cd clion_dbg_sample
\$ dotnet publish -c release -o linux-x64 -r linux-x64
Microsoft (R) Build Engine version 16.3.0+0f4c62fea for .NET Core
Restore completed in 66.26 ms for /tmp/clion_dbg_sample/clion_dbg_sample.csproj.
clion_dbg_sample -> /tmp/clion_dbg_sample/bin/release/netcoreapp3.0/linux-x64/clion_dbg_sample.dll
clion_dbg_sample -> /tmp/clion_dbg_sample/linux-x64/
Now we have a console application published in some folder, in my case it’s /tmp/clion_dbg_sample/linux-x64
2. Next we will setup a new configuration under CLion:
3. Now we define a new configuration:
We provide some name, I’ve decided to use the same name as my test program: clion_dbg_sample, We select “All targets” as the Target, and under executable we need to choose “Select other…” to provide a custom path to corerun. The reason behind this is that we need to run corerun from a directory that actually contains the entire product: jit, gc and everything else.
4. The path we provide is to the corerun executable that resides in the bin/Product/Linux.x64.Checked folder:
5. Finally we provide our sample project from before to the corerun executable. This is how my final configuration looks like:
6. It’s time to set a break-point and launch. As a generic sample I will navigate to compiler.cpp and find the jitNativeCode method. It’s pretty much one of the top-level functions in the JIT, and therefore a good candidate for us. If we set a breakpoint in that method and launch our newly created configuration, we should hit it in no time:
7. We’re done! If you really want to figure out what to do next, it’s probably a good time to hit the BotR, namely the RyuJit Overview and RyuJit Tutorial pages that contain a more detailed overview of the JIT. Alternatively, if you’re a “get your hands dirty” sort of person, you can also do some warm-up exercises for your fingers and start hitting that step-into keyboard shortcut. You’re debugging the JIT as we speak!
I hope this end up helping someone wanting to get started digging into the JIT not on Windows. I also personally have a strong preference for CLion as I really think it’s much more faster and powerful option than all the other stuff I’ve tried this far. At any rate, it’s the only viable option for Linux/macOs people.
Have fun! Let me know on twitter if you’re encountering any difficulties or you think I can make anything clearer…
Updated:
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Now showing items 1-1 of 1
• #### Properties for Voronoi diagrams of arbitrary order in the sphere
(Universitat Politècnica de Catalunya, 2021-10)
Master thesis
Open Access
In this thesis we study properties for spherical Voronoi diagrams of order $k$, SV_k(U)$using different tools: the geometry of the sphere, a labeling for the edges of$SV_k(U)\$, and the inversion transformation. Among the ...
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# Polygons arising from knot diagrams
This question - so far as I know - has no broader mathematical significance, but it occurred to me a while ago and I haven't been able to make any headway.
Any knot diagram $D$ splits the plane into a finite number of pieces. For example, in a standard diagram for a trefoil (e.g., http://en.wikipedia.org/wiki/File:Trefoil_knot_left.svg), the plane is split into five pieces: one lying "outside the diagram," three "lobes" of the knot, and a central region around which the three "lobes" are arrayed. The outer region has three crossings on its boundary, as does the inner region; each of the lobes only have two crossings each on their boundaries.
A diagram for a more complicated knot may split the plane into many more regions, and one of these regions may have many crossings on its boundary. My question is: Does there exist some $n$ such that for any knot $K$, there is a diagram $D$ of $K$ such that any region created by $D$ has at most $n$ crossings on its boundary?
If the answer is no, a (very soft) question that then arises is: given a knot $K$, how hard is it to find the least $n$ ($=n_K$) such that for some diagram $D$ of $K$, every region created by $D$ has at most $n$ crossings on its boundary? For example, for any knot $K$, we can easily see that $n_K>2$.
• I think you can always ensure there are at most $4$ vertices in all the regions that the knot diagram divides the plane into -- think of grabbing a bit of the knot, stretching it out and making a very dense vertical and horizontal "grid" of lines, overlayed on top of the diagram. If you have the grid fine enough then you get at most 4 vertices on the boundary of any region. – Ryan Budney Oct 3 '10 at 1:00
• Ryan: Wouldn't there also be 5-gons in your construction? e.g. a small square with one of its corners clipped off. Also, an Euler characteristic argument shows that if the graph is large and some regions are 3-gons (or 2-gons or 1-gons), then other regions must be (>4)-gons. (I'm working in S^2 and counting the "region at infinity".) – Kevin Walker Oct 3 '10 at 13:33
• Kevin, yes I seem to have mis-counted. – Ryan Budney Oct 3 '10 at 15:30
• Or maybe even 6-gons, if a grid square contains one of the crossings of the original diagram and the edges incident to the crossing enter and leave the square on adjacent sides. – David Eppstein Oct 3 '10 at 15:54
• Kevin, I used the Euler characteristic in my answer, but all I get out of it is that you can have as many as 8 3-gons with all the other regions being 4-gons. I like to think of it as a 4-gon conclusion. – Gerry Myerson Oct 3 '10 at 22:31
Colin Adams, Reiko Shinjo and Kokoro Tanaka have a paper (http://arxiv.org/abs/0812.2558) that shows that for any knot you can find a diagram which has only regions with 2, 4 and 5 sides.
• They say it's an open problem as to whether you can always get by with just 3 and 4 sides. They don't mention the possibility that you can always do with 2, 3, and 4 sides; presumably, that too is open (unless someone can make rigorous the hand-waving argument in the answer I posted). – Gerry Myerson Oct 3 '10 at 22:56
• I'm still interested in whether $n$ can be reduced to 4, but this certainly answers my question. Thanks a lot! – Noah Schweber Oct 4 '10 at 23:16
I'm pretty sure something like the following construction can be used to transform any knot diagram into another diagram for the same knot in which each region has bounded complexity. I didn't draw the over-under relationships at the crossings but that's not a problem.
(source: uci.edu)
• I don't understand the construction. It seems to me that you have turned a diagram in which no region has more than 5 vertices to a diagram in which there is a region (the very topmost one) with 8 vertices. – Gerry Myerson Oct 3 '10 at 4:16
• The construction is: isolate each critical point of the drawing by a sequence of vertical line segments, pull off a loop of the diagram and connect one end of the pulled-off loop in a spiral pattern through all the vertical segments and connect the other end of the loop to the center of the spiral. Each region in the resulting diagram has at most two critical points (the part of the loop crossing the spiral prevents more than that) so the eight-sided region in the top of the diagram is the worst case. – David Eppstein Oct 3 '10 at 5:01
• So you are saying that for any graph you can find a diagram in which every region has at most 8 sides? That's good, though it's well short of the alleged best possible value of 4. – Gerry Myerson Oct 3 '10 at 12:47
• I suspect that a more careful version of this argument should get it down to 6. – David Eppstein Oct 3 '10 at 15:51
The consensus seems to be that the answer is 4. Here's a simple proof that it can't be anything less (that is, that 3 can't always work). It can be arranged that all vertices have degree 4. Let the diagram have $n$ vertices. By Euler, it has $n+2$ regions (including the outside region). Each vertex takes part in 4 regions, so the average number of vertices per region is $4n/(n+2)$. For $n\ge7$, this exceeds 3. If the average region has over 3 vertices, some region must have at least 4.
Now here's a hand-waving argument that 4 is always possible (presented because I don't understand Ryan Budney's argument in the comment on the question). Find a region that has more than 4 vertices. Take two more-or-less opposite strands bounding that region, and do a Type 2 Reidemeister move on them. That will split the offending region into two regions (and a digon), each with fewer vertices than the original. Unfortunately, it will also introduce two new vertices into the region on the other side of each of the two strands, which may move one or both of them over 4. The hand-waving is that if you persist with doing Type 2 moves you will eventually get to all regions having 4 vertices or fewer.
EDIT: I have lost faith in the hand-waving argument. If you start with a pentagram, which as a knot diagram represents a knot with 10 crossings, 10 triangles and 2 pentagons, then, persist as I may, I keep producing at least as many regions with 5 or more sides as I remove whenever I do a Reidemeister.
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# Field Angular Momentum (Thomson Dipole)
by Septim
Tags: angular momentum, electrodynamics, griffiths
P: 119 1. The problem statement, all variables and given/known data I have attached the question as jpg to this post. Typing these were too time consuming and I uploaded the relevant sections as image files, thanks for your understanding. 2. Relevant equations mu_0*epsilon_0*Poynting Vector = Momentum density Position vector X Momentum density = Angular momentum density 3. The attempt at a solution These too are attached in png format. I have found out that angular momentum density has two components, one in the z direction and one in the x direction. According to the books solution the x component integrates to zero but I was unable to verify this and I am highly skeptical about this topic. Any help would be appreciated. I have attached the relevant information to this post. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution Attached Thumbnails
PF Patron HW Helper Thanks P: 3,763 In constructing your equation (8), note that ##\hat{r}## will generally have a ##\hat{y}## component as well as ##\hat{x}## and ##\hat{z}## components: ##\hat{r} = (\hat{r} \cdot \hat{x}) \hat{x} + (\hat{r} \cdot \hat{y}) \hat{y} + (\hat{r} \cdot \hat{z}) \hat{z}##. ##\hat{r} \cdot \hat{x} = sin\theta cos\phi##, etc. See what you get for the x and y components of the angular momentum when you integrate over ##\phi##.
Related Discussions Introductory Physics Homework 3 General Physics 0 Advanced Physics Homework 4 Introductory Physics Homework 0 Classical Physics 4
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Introduction
String distance is useful when wanting to know some quantifiable measure of closeness between two strings.
String metrics: elements versus content
elements-approach
A commonly used string metric is the Levenshtein distance, which is a comparison of the sequences of elements at the character-level and the edits needed to match two strings. With that said, it is natural for us to think that the more edits are needed to create a match, the further away two strings are from one another. In the example below, we can see that the misspelled words for “apple” are still closer than the word “orange”.
# calculating the number of deletions, insertions and substitutions necessary to turn b into a.
word1 <- "apples"
word2 <- "lappes"
stringdist::stringdist(word1, word2, method='lv')
[1] 2
# including an extra character that needs to be deleted
word1 <- "apples"
word2 <- "lappesp"
stringdist::stringdist(word1, word2, method='lv')
[1] 3
word3 <- "oranges"
stringdist::stringdist(word1, word3, method='lv')
[1] 5
content-approach
In many cases, we are more interested in comparing a string of words such as when we are more concerned about the content. The Levenshtein distance in this case would be inadequate as the sum of all edits for each word in an entire string will consequently have us comparing apples to oranges. When wanting to compare sentences, it is more natural to calculate the cosine similarity between strings. Unlike Levenshtein distance, cosine similarity instead cares only about the angle difference between two strings; in other words, cosine similarity is a measurement of orientation and will yield a stronger measure of “closeness” when two strings have the same words.
# calculating the cosine angle between two strings
sentence1 <- "I like apples more than Bob likes apples"
sentence2 <- "I like bananas more than my friend but we both like apples"
corpus <- c(sentence1, sentence2)
corpus <- sapply(corpus, function(x) strsplit(x, " "))
corpus_all <- unique(unlist(corpus))
corpus_table <- sapply(corpus, function(x) table(factor(x, levels=corpus_all)))
> corpus_table
I like apples more than Bob likes apples I like bananas more than my friend but we both like apples
I 1 1
like 1 2
apples 2 1
more 1 1
than 1 1
Bob 1 0
likes 1 0
bananas 0 1
my 0 1
friend 0 1
but 0 1
we 0 1
both 0 1
a <- corpus_table[,1]
b <- corpus_table[,2]
(a %*% b) / (sqrt(sum(a^2)) * sqrt(sum(b^2)))
[,1]
[1,] 0.591608
# calculating the cosine angle between two strings where sentences are more "similar"
sentence1 <- "I like apples more than Bob likes apples"
sentence2 <- "I like apples more than Bob likes pineapple apple pens"
corpus <- c(sentence1, sentence2)
corpus <- sapply(corpus, function(x) strsplit(x, " "))
corpus_all <- unique(unlist(corpus))
corpus_table <- sapply(corpus, function(x) table(factor(x, levels=corpus_all)))
> corpus_table
I like apples more than Bob likes apples I like apples more than Bob likes pineapple apple pens
I 1 1
like 1 1
apples 2 1
more 1 1
than 1 1
Bob 1 1
likes 1 1
pineapple 0 1
apple 0 1
pens 0 1
a <- corpus_table[,1]
b <- corpus_table[,2]
(a %*% b) / (sqrt(sum(a^2)) * sqrt(sum(b^2)))
[,1]
[1,] 0.8
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# Is $\mathbb{Z}(\sqrt[3]{5})$ a PID? Factorisation of the ideal $(2)$
I am given that for the ring of integers of $K = \mathbb{Q}(\sqrt[3]{5})$ is $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{5}]$. I am supposed to factorise the ideals $(2), (3), (5)$ and $(7)$, show that all prime ideal factors are principal and then use Minkowski's bound to deduce that $\mathcal{O}_K$ is a PID, however I am having problems with $(2)$ and can't work out what I am doing wrong!
Given the above assumption we have that $[\mathcal{O}_K : \mathbb{Z}[\sqrt[3]{5}]] = 1$ so that we can apply Dedekind's Theorem for all primes $p$. Doing this for $p = 2$ gives:
$$x^3 - 5 \equiv x^3+1 \equiv (x+1)(x^2+x+1) \tag{mod 2}$$
So that we have $(2) = \mathfrak{p}_2\mathfrak{p}_4 = (2, \sqrt[3]{5} + 1)(2, \sqrt[3]{25} + \sqrt[3]{5} + 1)$
Then I think I should show that $\mathfrak{p}_2$ and $\mathfrak{p}_4$ are principal by finding elements in $\mathcal{O}_K$ with norms 2 and 4 respectively. I calculated the norm to be:
$$\text{Norm}(a+b\sqrt[3]{5}+c\sqrt[3]{25}) = a^3 + 5b^3 + 25c^3 - 15abc$$
But when I put this equal to 2 or 4 in WolframAlpha I don't get any results in $\mathbb{Z}[\sqrt[3]{5}]$. Thus I want to argue that $\mathfrak{p}_2$ and $\mathfrak{p}_4$ are not principal ideals and so $\mathcal{O}_K$ is not a PID.
What went wrong?
What went wrong is that you trusted WolframAlpha! Set $c = 0$ and then $b = 0$ to simplify the norm and you'll quickly find that $\sqrt[3]{5} - 1$ has norm $4$, while $3 - \sqrt[3]{25}$ has norm $2$.
• I don't know how WolframAlpha attempts to solve Diophantine equations, but checking a few examples it seems to know how to solve linear equations and Pell equations and that's about it. It told me that $y^2 = x^3 + 9$ has no integer solutions, but of course $(3, 6)$ is a solution. – Qiaochu Yuan Apr 24 '15 at 18:28
• Wolfram Alpha is excellent in my opinion. But you have to remember that it's a tool, not an oracle. Tools have their limitations and kludges. Kind of like trying to figure out the complex cubic roots of $8$ using a cash register. It can be done, but you're going to need a lot of workarounds. – Lisa Apr 25 '15 at 21:21
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# Lognormal posterior?
If I have a relationship:
$y_t = a + \theta b_t \epsilon_t,$
where I observe $y_t$ and $b_t$. $a$ is a known parameter, $\theta$ is an unknown parameter with prior distribution at time $t$: $\theta \sim logN(\mu_t, \sigma_t^2)$, and $\epsilon_t$ is an IID random variable with distribution: $\epsilon_t \sim logN(0,\sigma_\epsilon^2)$.
Is it possible to obtain an analytic log-normal posterior for $\theta$? If not, do there exist other possible conjugate prior distributions with non-negative support (and whatever distribution on the noise term $v_t$ that's necessary)?
• So $y_t$ is intended to be some kind of shifted lognormal? – Glen_b Jun 2 '14 at 23:40
• If the prior for $\theta$ depends on $t$, then $\theta$ itself depends on $t$. So the question decomposes into a separate problem for each $\theta_t$. – Tom Minka Oct 3 '14 at 17:39
• If $a$ is known, why not write $z_t = \log(y_t-a)$ and write a linear model in $z_t$? – Glen_b Mar 26 '15 at 9:14
We have to calculate the posterior distribution of $\theta$ for the sample $D = \{(b_t, y_t)\}_{t = 1}^N$ $$p(\theta | D) = \frac{p(D|\theta)p(\theta)}{p(D)}.$$
We are not interesting in the term in denominator. We need to select $p(\theta)$ such that posterior is log-normal.
Let's take a look at $p(D|\theta)$: $$p(D|\theta) = \prod_{t = 1}^N p(y_t| b_t) = \\=\prod_{t = 1}^N \frac{\theta b}{\sqrt{2 \pi} \sigma_{\varepsilon} (y_t - a)} \exp\ \left(-\frac{1}{2 \sigma^2_{\varepsilon} } \left(\ln (y_t - a)- \ln b_t - \ln \theta \right)^2 \right) = \\ \frac{C_1}{\sqrt{2 \pi} \sigma_\theta} \theta^N \exp \left(-\frac{1}{2 \sigma_\theta} \left(\ln \theta - C_2 \right)^2 \right).$$
One can see that if we select an improper prior of the form $\frac{1}{\theta^{N + 1}} [\theta > 0]$ we get the log-normal distribution as a posterior distribution. Here $[\cdot]$ is the indicator function i.e. it equals $1$ if statement in the brackets holds and $0$ vice versa.
• I think we might need more information about the specification for $\theta$: it seems like user6600 wants it to have a different distribution at different points in time. – user44764 Jun 3 '14 at 13:03
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I was trying to prove the identity $$\overline{\displaystyle{\not}{a}\displaystyle{\not}{b}\dots \displaystyle{\not}{p}} = \displaystyle{\not}{p}\dots \displaystyle{\not}{b}\displaystyle{\not}{a}$$. On simplifying the LHS I end up with $$\overline{ \displaystyle{\not}{p}^{}} \dots \overline{ \displaystyle{\not}{b}^{}} \ \overline{ \displaystyle{\not}{a}^{}}$$.
I'm wondering if $$\overline{ \displaystyle{\not} p} = \gamma^0 \displaystyle{\not} p^{\dagger} \gamma^0 = \displaystyle{\not} p$$ as this would let me complete the final step.
• Does the overbar mean complex conjugate? For the mostly plus metric we do have $\gamma_a^\dagger = \gamma_0 \gamma_a \gamma_0$. – mike stone Oct 19 at 19:11
• Hi! The overbar is defined as $\overline{\gamma} = \gamma^0 \gamma^{\dagger} \gamma^0$ – A quarky name Oct 19 at 19:46
Usually, gamma-matrices are chosen in such a way that they satisfy the hermiticity condition $$\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$$, and that is what you need, but this choice is not mandatory.
• Ah, thanks! Do the elements of the 4 vector $p_{\mu}$ also satisfy $\gamma^0 p_{\mu}^{\dagger} \gamma^0 = p_{\mu}$ – A quarky name Oct 19 at 19:39
• @Aquarkyname : Yes, as $p^\dagger_\mu=p_\mu$ and typically $(\gamma^0)^2=1$. – akhmeteli Oct 19 at 19:59
• Why does $p_{\mu}^{\dagger} = p_{\mu}$. Does the Hermitian adjoint not act on the elements of the complex 4 vector? – A quarky name Oct 19 at 20:05
• Ah, I should have used a different letter. $p$ is just an arbitrary 4-vector, not the momentum operator. To show the identity above, I would also need to show $a_{\mu}^{\dagger} = a_{\mu}$ for instance – A quarky name Oct 20 at 10:56
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# Second derivatives when pouring juice into a cup
1. Oct 31, 2017
### mindy
Its a question about volume increase (in units cm^3) and height increase (cm) when pouring juice into a cup. Its stated that the volume of the juice in the cup increases at a constant rate, so I know the volume derivatives are zero. But the shape of the cup is inconsistent and there is alot of variance in it, so the height increase is not constant. I know the first derivative of the height increase is positive... my professor hinted that the second derivative of height increase would be negative... how is that so?
Last edited by a moderator: Oct 31, 2017
2. Oct 31, 2017
The derivatives of interest are $\frac{dh}{dt}$ and $\frac{d^2 h }{dt^2}$. You are given $\frac{dV}{dt}=constant$. Suggestion is to write $V=V(h)$, and compare derivatives $\frac{dV}{dt}$ and $\frac{d^2 V}{dt^2}$ using the chain rule. Also $\frac{d^2 V}{dt^2}=0$. $\\$ Additional hint: What can you say about $\frac{d^2 V}{dh^2}$ if the volume is written as $V(h)=\int\limits_{0}^{h} A(h') \, dh'$, and $\frac{d A(h)}{dh} >0$ ?
Last edited: Oct 31, 2017
3. Oct 31, 2017
### Staff: Mentor
The first derivative measures the rate of change, so it's not zero for the volume. It's a positive constant. The second derivative of the volume is zero, because it measures the rate to which the rate of change changes. The same is true for the height. As it doesn't change constantly, it's first derivative actually changes with time. And as this change isn't constant, the first derivative of this change isn't zero, which is the second derivative of the height function. Whether this is positive or negative depends on whether the rate of change is sublinear or hyperlinear.
4. Oct 31, 2017
### mindy
What does the A stand for in your last equation?
5. Oct 31, 2017
### Staff: Mentor
The surface area at height $h$.
6. Oct 31, 2017
I had a slight error. I corrected it. (I originally wrote $V(h)=A(h) h$. The correct expression is $V(h)=\int\limits_{0}^{h} A(h') \, dh' \,$).
7. Oct 31, 2017
### mindy
i know that the height increase and volume increase first derivatives are positive. thats helpful
the units of the first volume derivative is cm^3/s (seconds)
second derivative is cm^3/s^2
height first derivative is cm/s
second derivative of height is cm/s^2
8. Oct 31, 2017
### mindy
not sure why the photo of the problem didn't go through last time. here it is
#### Attached Files:
• ###### calculus.jpg
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Last edited by a moderator: Oct 31, 2017
9. Oct 31, 2017
### mindy
I guess I was confused because I thought the derivative of a constant was always zero
10. Oct 31, 2017
The trickiest part of this for someone just learning calculus is applying the chain rule a couple of times. Suggestion: $dV/dt =(dV/dh)(dh/dt)=constant$. Finding $d^2V/dt^2 =0=...$ using the chain rule takes a little more work, but it should give you the result you need. $\\$ Additional hint: By the chain rule, $d^2 V/dt^2=(d(dV/dh)/dt)(dh/dt)+(dV/dh)(d^2 h/dt^2)=0$. The first term there needs to be evaluated with the chain rule one more time..
Last edited: Oct 31, 2017
11. Oct 31, 2017
### FactChecker
You will get answers that are more specific to your questions if you show us the Figure 7 of the cup and the rest of the problem statement.
12. Oct 31, 2017
### mindy
thanks everyone
thanks charles =)
13. Oct 31, 2017
See my edited post of post 10. That is very close to the result you need.
14. Oct 31, 2017
### mindy
here it is, for those curious
#### Attached Files:
• ###### Inkedcalc2_LI.jpg
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34.6 KB
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15. Oct 31, 2017
### mindy
nevermind. my phones too blurry. i give up
16. Oct 31, 2017
I was able to read it. They want to know the sign of the terms $(dh/dt)$ and $(d^2 h/dt^2)$, etc. If you look at my hint in post 10, you need to compute $\frac{d (\frac{dV}{dh})}{dt }$ using the chain rule. Once you have that, it will allow you to compare $d^2 h/dt^2$ with $d^2 V/dh^2$. (See also my hint for $d^2V/dh^2$ in the last part of post 2).
Last edited: Oct 31, 2017
17. Oct 31, 2017
### FactChecker
First you should consider how the shape of the cup effects the rate of height increase and the derivative of that. I think that your statement in the original post about the first derivative of height increase being positive is not always true as the cup fills up. [CORRECTION: Of course it is always positive. The second dirivative is not always positive.]
Consider the derivatives where the cup diameter is expanding, where it is contracting, the sudden change in the middle, and the constant curvature of the top and bottom halves.
Last edited: Oct 31, 2017
18. Oct 31, 2017
@FactChecker input can be quantified by computing $d^2 V/dh^2 =dA(h)/dh$ as my hint in the last part of post 2 is suggesting. Using that, along with the equation in post 10, and the problem is really very straightforward.
19. Oct 31, 2017
### FactChecker
Yes, I know. I can do it easily. But I think that the original poster is less familiar with the application and meaning of derivatives in real problems and I'm not sure that the problem was asking for actual formulas as opposed to a more qualitative characterization.
20. Oct 31, 2017
### mindy
you're right factchecker, I am new to all this. I took a 6 year break from pre calc to this class (calc 1) and prior to that, it'd been like 6 years since I took algebra or trig. so I am very behind. I own calculus for dummies, and a complete idiots guide to calculus, and I watch alot of kahn academy and youtube videos. but i am struggling and am very behind academically
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## Props¶
color : Color
Bootstrap color schema (see Color)
dismissible : boolean | "animated"
Whether the the alert has an button to allow the user dismissing it.
Note
You have to handle the dismiss action yourself through onDismiss.
onDismiss : () => void
Function called when the user clicks the dismiss button.
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## Structure of the moduli space in a neighborhood of a cusp-curve and meromorphic hulls.(English)Zbl 0930.32017
Let us call a Kähler surface $$(X,\omega)$$ strongly minimal if for any almost-complex structure $$J$$ on $$X$$ tamed by $$\omega$$, any $$J$$-holomorphic sphere $$C\subset X$$ satisfies $$[C]^2\geq 0$$. For example $$CP^2$$ and $$\mathbb{C}\mathbb{P}^1 \times\mathbb{C} \mathbb{P}^1$$ are strongly minimal in this sense.
Consider an $$\omega$$-symplectic immersion $$u:S^1\to X$$ of a standard two-dimensional sphere into $$X$$, and suppose that $$M:= u(S^2)$$ has only double and positive self-intersections. Denote by $$\delta$$ the number of double points of our $$M$$. $$c_1(X)$$ will denote the first Chern class of $$X$$ and $$c_1(X)[M]$$ its value on $$M$$.
Theorem 1. Under the conditions as above suppose that $$c_1(X) [M]- \delta= p\geq 1$$. Then the envelope of meromorphy $$(\widehat U,\pi)$$ of any neighborhood $$U$$ of $$M$$ is either unbounded (i.e. $$\pi (\widehat U)$$ is non-compact in $$X)$$, or $$\widehat U$$ contains a one-dimensional analytic set $$C$$ such that
(1) $$\pi(C)= \cup^N_{k=1} C_k$$ is a union of rational curves;
(2) $$\sum^N_{k=1} (c_1(\widehat U) [C_k]- \delta_k- \kappa_k)) \geq p$$.
Here $$c_1 (\widehat U)= \pi^*c_1(X)$$ is the first Chern class of $$\widehat U$$, $$\delta_k$$ the sum of local intersection numbers of $$C_k$$, and $$\kappa_k$$ – the sum of Milnor numbers of the cusps of $$C_k$$.
Let $$(X,J)$$ be an almost complex manifold of complex dimension $$n$$. An almost complex structure is always supposed to have smoothness of class $$C^1$$. In what follows $$(S,J_S)$$ denotes a Riemann surface with complex structure $$J_S$$.
It can be shown that for a pseudo-holomorphic curve $$u:(S,J_S) \to(X, J_0)$$ the pull-back bundle $$E:= u^*TX$$ possesses a natural holomorphic structure (the corresponding sheaf of holomorphic sections will be denoted as $${\mathcal O}(E))$$, such that the differential $$du:TS\to E$$ is a holomorphic homomorphism. This allows us to define the order of vanishing of the differential $$du$$ at point $$s\in S$$. We denote this number by $$\mathbf{ord}_sdu$$.
This also gives the following short exact sequence: $0\to {\mathcal O}(TS) @>du>> {\mathcal O}(E) @>\text{pr}>> {\mathcal O}(N_0) \oplus {\mathcal N}_1 \to 0,$ where $${\mathcal O}(N_0)$$ denotes a free part of the quotient $${\mathcal O}(E)/du ({\mathcal O}(TS))$$ and $${\mathcal N}_1$$ is supported on a finite set of cusps of $$u$$ (i.e. points of vanishing of $$du)$$.
On the Sobolev space $$L^{1,p} (S,N_0)$$ of $$L^{1,p}$$-smooth sections of the bundle $$N_0$$ the natural Gromov operator $$D_N:L^{1,p} (S,N_0)\to L^p(S, \Lambda^{0,1} S \otimes N_0)$$ is defined. Put $$H^0_D(S,N_0): =\text{Ker} D_N$$ and $$H^1_D (S, N_0): =\text{Coker} D_N$$.
The following Theorem is the second main result of this paper.
Theorem 2. Let $$u:(S,J_S) \to(X,J_0)$$ be a non-constant pseudo-holomorphic map, such that $$H^1_D (S,N_0)=0$$. Then:
i) in a neighbourhood of $$M:=u(S)$$ the moduli space of nonparametrized $$J_0$$-holomorphic curves $${\mathcal M}_{[\gamma], g,J_0}$$ is a manifold the tangent space of which is $$T_M {\mathcal M}_{[\gamma], g,J_0}= H^0_D (S,N_0) \oplus H^0(S, {\mathcal N}_1)$$;
ii) further, there is a neighborhood $$V\ni J$$ in the Banach manifold $${\mathcal I}$$ of $$C^1$$-smooth almost-complex structures on $$X$$ and a neighborhood $$W$$ of $$M$$ in $${\mathcal M}_{[\gamma],g,V}: =\cup_{J\in V} {\mathcal M}_{[\gamma], g,J}$$ such, that the natural projection $$\text{pr}_J: W\to V$$ is regular and with smooth fibers, in particular, for any $$M'\in W$$ the conclusion i) is valid;
iii) if $$\dim_\mathbb{R} X=4$$ and $$c_1(E)[M]>\sum_{s\in S}\text{ord}_sdu$$, then $$H^1_D (S, N_0)=0$$ and so the conclusions i) and ii) hold.
Remark. When $$X$$ is a complex surface and $$N_0$$ is a line bundle the sufficient condition for surjectivity of $$D_N$$ is $$c_1(N_0) >2g-2$$, where $$g$$ is the genus of $$S$$. So, if $$c_1(E)> \sum_{s\in S} \text{ord}_sdu$$, one has the surjectivity of projection $$\text{pr}_{\mathcal J}$$.
### MSC:
32Q55 Topological aspects of complex manifolds 32Q65 Pseudoholomorphic curves
### Keywords:
envelope of meromorphy; pseudo-holomorphic curve
Full Text:
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Let $$x_{1}=97$$ ,and for n>1, let $$x_{n}=\frac{n}{x_{n-1}}$$. Calculate the product $$x_{1}x_{2}x_{3}…x_{8}$$.
(第三届AIME1985 第1题)
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Messages - Yifei Gu
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1
MAT244--Lectures & Home Assignments / Question 8, Sec 3.6
« on: November 29, 2018, 09:22:21 PM »
Having a bit trouble finding solutions, can anyone take a look.
Question: $y'' + 4y = 3\csc 2t, 0 < t <\frac{\pi}{2}$
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Quiz-1 / Re: Q1: TUT0301
« on: September 29, 2018, 03:13:34 PM »
$$y' - 2y = e^{2t}\\\mu (x) = e^{\int-2\ dt} = e^{-2t}\\\frac{d}{dt} (e^{-2t}y) = e^{-2t}y' -2e^{-2t} y = e^{-2t}e^{2t}= 1\\ \text{integral on both side gives:}\\ e^{-2t}y = t + C\\y = te^{2t} + C{e^{2t}}\\y(0) = 0 + C = 2 \implies C = 2\\\text{thus} \ \ y =e^{2t}(t+2)\\\text{and} \ \ t \to \infty \implies y \to \infty$$
3
Quiz-1 / Re: Q1: TUT 0501
« on: September 28, 2018, 07:50:46 PM »
here is the solution to the question.
Pages: [1]
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1. ## clarification
i was just wondering which is correct:
a function f is defined on an interval I is contonuous if lim as x->c for f(x) is f(c) for all x in I or for all c in I.
i thought it should be the former but the book that im reading states that its the latter.
thanks
2. It is the latter.
Fernando Revilla
3. thanks for the clarification..but may i know why, if for this case, it is the latter then for this case:
a function f is cont at c if every seq of points x_n in I such that lim as n->infinity for x_n is c if the lim n->inifinity f(x_n) is f(c).
in this case, we are talking about x_n in I and not for all c in I.
i hope im making sense
4. Originally Posted by alexandrabel90
thanks for the clarification..but may i know why, if for this case, it is the latter then for this case:
a function f is cont at c if every seq of points x_n in I such that lim as n->infinity for x_n is c if the lim n->inifinity f(x_n) is f(c).
in this case, we are talking about x_n in I and not for all c in I.
i hope im making sense
Think about what you're saying in a simpler sense. This $x$ is a "dummy" variable, in the same sense as the $x$ in $\displaystyle \int_a^b f(x)\text{ }dx$. What's really fixed is the $c\in I$. Think about your definition you gave in terms of sequences. A function is continuous on $I$ if at every $\mathbf{c}\in I$ that's true. No?
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# How to prove that the series is decreasing
$a_n=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n!}{2^n}$, I am trying to prove it using Leibniz theorem, Condition (i) - all $u_n$ are positive is satisfied, Condition (ii) $u_n\geq u_{n+1}$, here how do we prove that $\frac{n!}{2^n}\geq\frac{(n+1)!}{2^{n+1}}$. To prove that a function is decreasing we take derivative and prove it that for some $x=x_0$ to $x\to\infty$ the function deceases and we have condition (ii) satisfied. Taking derivative is simple but I think is not possible always as in this case where I don't know how to find the slope of $f(x)=\frac{x!}{2^n}$ with basic calculus, whereas knowing how to prove $u_n\geq u_{n+1}$ will help a lot.
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$\frac{(n+1)!}{2^{n+1}}=\frac{n!}{2^n}\cdot\frac{n+1}2>\frac{n!}{2^n}$ whenever $n>1$, so you certainly can’t prove that $\frac{(n+1)!}{2^{n+1}}\le\frac{n!}{2^n}$ in general. – Brian M. Scott Apr 8 '12 at 5:08
The general term is not decreasing! The series is non convergent. Maybe you meant the reciprocal. – Pedro Tamaroff Apr 8 '12 at 22:23
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# Schrödinger equation
Arranging a Wikipedia selection for schools in the developing world without internet was an initiative by SOS Children. To compare sponsorship charities this is the best sponsorship link.
In physics, especially quantum mechanics, the Schrödinger equation is an equation that describes how the quantum state of a physical system changes in time. It is as central to quantum mechanics as Newton's laws are to classical mechanics.
In the standard interpretation of quantum mechanics, the quantum state, also called a wavefunction or state vector, is the most complete description that can be given to a physical system. Solutions to Schrödinger's equation describe atomic and subatomic systems, electrons and atoms, but also macroscopic systems, possibly even the whole universe. The equation is named after Erwin Schrödinger, who discovered it in 1926.
Schrödinger's equation can be mathematically transformed into Heisenberg's matrix mechanics, and into the Feynman's path integral formulation. The Schrödinger equation describes time in a way that is inconvenient for relativistic theories, a problem which is less severe in Heisenberg's formulation and completely absent in the path integral.
## The Schrödinger equation
There are several different Schrödinger equations.
### General quantum system
For a general quantum system:
$i\hbar {d\Psi \over dt} = \hat H \Psi$
where
• $\psi$ is the wavefunction, which is the probability amplitude for different configurations.
• $\scriptstyle \hbar$ is Planck's constant over $2\pi$, and it can be set to a value of 1 when using natural units.
• $\scriptstyle \hat H$ is the Hamiltonian operator.
### Single particle in three dimensions
For a single particle in three dimensions:
$i\hbar\frac{\partial}{\partial t} \psi = -\frac{\hbar^2}{2m}\nabla^2\psi + V(x,y,z)\psi$
where
• $\psi$ is the wavefunction, which is the amplitude for the particle to have a given position at any given time.
• $m$ is the mass of the particle.
• $V(x,y,z)$ is the potential energy the particle has at each position.
## Historical background and development
Einstein interpreted Planck's quanta as photons, particles of light, and proposed that the energy of a photon is proportional to its frequency, a mysterious wave-particle duality. Since energy and momentum are related in the same way as frequency and wavenumber in relativity, it followed that the momentum of a photon is proportional to its wavenumber.
DeBroglie hypothesized that this is true for all particles, for electrons as well as photons, that the energy and momentum of an electron are the frequency and wavenumber of a wave. Assuming that the waves travel roughly along classical paths, he showed that they form standing waves only for certain discrete frequencies, discrete energy levels which reproduced the old quantum condition.
Following up on these ideas, Schrödinger decided to find a proper wave equation for the electron. He was guided by Hamilton's analogy between mechanics and optics, encoded in the observation that the zero-wavelength limit of optics resembles a mechanical system--- the trajectories of light rays become sharp tracks which obey a principle of least action. Hamilton believed that mechanics was the zero-wavelength limit of wave propagation, but did not formulate an equation for those waves. This is what Schrödinger did, and a modern version of his reasoning is reproduced in the next section. The equation he found is (in natural units):
$i \frac{\partial}{\partial t}\psi=-\frac{1}{2m}\nabla^2\psi + V(x)\psi$
Using this equation, Schrödinger computed the spectral lines for hydrogen by treating a hydrogen atom's single negatively charged electron as a wave, $\psi\;$, moving in a potential well, V, created by the positively charged proton. This computation reproduced the energy levels of the Bohr model.
But this was not enough, since Sommerfeld had already seemingly correctly reproduced relativistic corrections. Schrödinger used the relativistic energy momentum relation to find what is now known as the Klein-Gordon equation in a Coulomb potential:
$(E + {e^2\over r} )^2 \psi = - \nabla^2\psi + m^2 \psi$
He found the standing-waves of this relativistic equation, but the relativistic corrections disagreed with Sommerfeld's formula. Discouraged, he put away his calculations and secluded himself in an isolated mountain cabin with a lover.
While there, Schrödinger decided that the earlier nonrelativistic calculations were novel enough to publish, and decided to leave off the problem of relativistic corrections for the future. He put together his wave equation and the spectral analysis of hydrogen in a paper in 1926.. The paper was enthusiastically endorsed by Einstein, who saw the matter-waves as the visualizable antidote to what he considered to be the overly formal matrix mechanics.
The Schrödinger equation tells you the behaviour of $\psi$, but does not say what $\psi$ is. Schrödinger tried unsuccessfully, in his fourth paper, to interpret it as a charge density. In 1926 Max Born, just a few days after Schrödinger's fourth and final paper was published, successfully interpreted $\psi$ as a probability amplitude. Schrödinger, though, always opposed a statistical or probabilistic approach, with its associated discontinuities; like Einstein, who believed that quantum mechanics was a statistical approximation to an underlying deterministic theory, Schrödinger was never reconciled to the Copenhagen interpretation.
## Derivation
### Short Heuristic Derivation
#### Assumptions
(1) The total energy E of a particle is
$E= T + V = \frac{p^2}{2m}+V$
This is the classical expression for a particle with mass m where the total energy E is the sum of the kinetic energy, $\frac{p^2}{2m}$, and the potential energy V. The momentum of the particle is p, or mass times velocity. The potential energy is assumed to vary with position, and possibly time as well.
Note that the energy E and momentum p appear in the following two relations:
(2) Einstein's light quanta hypothesis of 1905, which asserts that the energy E of a photon is proportional to the frequency f of the corresponding electromagnetic wave:
$E = h f = {h \over 2\pi} (2\pi f) = \hbar \omega \;$
where the frequency f of the quanta of radiation (photons) are related by Planck's constant h,
and $\omega = 2\pi f\;$ is the angular frequency of the wave.
(3) The de Broglie hypothesis of 1924, which states that any particle can be associated with a wave, represented mathematically by a wavefunction Ψ, and that the momentum p of the particle is related to the wavelength λ of the associated wave by:
$p = { h \over \lambda } = { h \over 2\pi } {2\pi \over \lambda} = \hbar k\;$
where $\lambda\,$ is the wavelength and $k = 2\pi / \lambda\;$ is the wavenumber of the wave.
Expressing p and k as vectors, we have
$\mathbf{p} =\hbar \mathbf{k}\;$
#### Expressing the wave function as a complex plane wave
Schrödinger's great insight, late in 1925, was to express the phase of a plane wave as a complex phase factor:
$\Psi(\mathbf{x},t) = Ae^{i(\mathbf{k}\cdot\mathbf{x}- \omega t)} = Ae^{i \mathbf{k} \cdot \mathbf{x}}e^{-i\omega t} = \psi(\mathbf{x}) \phi(t)$
where
$\psi(\mathbf{x}) = Ae^{i \mathbf{k} \cdot \mathbf{x}}$
and
$\phi(t) = e^{-i\omega t} \,$
and to realize that since
$\frac{\partial}{\partial t} \Psi = -i\omega \Psi$
then
$E \Psi = \hbar \omega \Psi = i\hbar\frac{\partial}{\partial t} \Psi$
and similarly since:
$\frac{\partial}{\partial x} \Psi = i k_x \Psi$
then
$p_x \Psi = \hbar k_x \Psi = -i\hbar\frac{\partial}{\partial x} \Psi$
and hence:
$p_x^2 \Psi = -\hbar^2\frac{\partial^2}{\partial x^2} \Psi$
so that, again for a plane wave, he obtained:
$p^2 \Psi = (p_x^2 + p_y^2 + p_z^2) \Psi = -\hbar^2\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) \Psi = -\hbar^2\nabla^2 \Psi$
And by inserting these expressions for the energy and momentum into the classical formula we started with we get Schrödinger's famed equation for a single particle in the 3-dimensional case in the presence of a potential V:
$i\hbar\frac{\partial}{\partial t}\Psi=-\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi$
### Longer Discussion
The particle is described by a wave, and in natural units, the frequency is the energy E of the particle, while the momentum p is the wavenumber k. These are not two separate assumptions, because of special relativity.
$E= \omega \;\;\;\;P=k \,$
The total energy is the same function of momentum and position as in classical mechanics:
$E = T(p) + V(x) = {p^2\over 2m} + V(x)$
where the first term T(p) is the kinetic energy and the second term V(x) is the potential energy.
Schrödinger required that a Wave packet at position x with wavenumber k will move along the trajectory determined by Newton's laws in the limit that the wavelength is small.
Consider first the case without a potential, V=0.
$E = {1\over 2m} (p_x^2+p_y^2 + p_z^2)$
$\omega = {1\over 2m} (k_x^2 + k_y^2 + k_z^2)$
So that a plane wave with the right energy/frequency relationship obeys the free Schrodinger equation:
$i {\partial \over \partial t} \psi = -{1 \over 2m} ( {\partial^2 \psi \over \partial x^2} + {\partial^2 \psi \over \partial y^2} + {\partial^2 \psi \over \partial z^2} )$
and by adding together plane waves, you can make an arbitrary wave.
When there is no potential, a wavepacket should travel in a straight line at the classical velocity. The velocity v of a wavepacket is:
$v = {\partial \omega \over \partial k } = {\partial \over \partial k} { k^2\over 2m} = { k\over m}$
which is the momentum over the mass as it should be. This is one of Hamilton's equations from mechanics:
${dx \over dt} = {\partial H \over \partial p}$
after identifying the energy and momentum of a wavepacket as the frequency and wavenumber.
To include a potential energy, consider that as a particle moves the energy is conserved, so that for a wavepacket with approximate wavenumber k at approximate position x the quantity
${ k^2\over 2m } + V(x)$
must be constant. The frequency doesn't change as a wave moves, but the wavenumber does. So where there is a potential energy, it must add in the same way:
$i{\partial \over \partial t}\psi=-{1\over 2m}\nabla^2\psi + V(x)\psi$
This is the time dependent schrodinger equation. It is the equation for the energy in classical mechanics, turned into a differential equation by substituting:
$E\rightarrow i{\partial\over \partial t} \;\;\;\;\;\; p\rightarrow -i{\partial\over \partial x}$
Schrödinger studied the standing wave solutions, since these were the energy levels. Standing waves have a complicated dependence on space, but vary in time in a simple way:
$\psi(x,t) = \psi(x) e^{- iEt } \,$
substituting, the time-dependent equation becomes the standing wave equation:
${E}\psi(x) = - {1\over 2m} \nabla^2 \psi(x) + V(x) \psi(x)$
Which is the original time-independent Schrodinger equation.
In a potential gradient, the k-vector of a short-wavelength wave must vary from point to point, to keep the total energy constant. Sheets perpendicular to the k-vector are the wavefronts, and they gradually change direction, because the wavelength is not everywhere the same. A wavepacket follows the shifting wavefronts with the classical velocity, with the acceleration equal to the force divided by the mass.
an easy modern way to verify that Newton's second law holds for wavepackets is to take the Fourier transform of the time dependent Schrodinger equation. For an arbitrary polynomial potential this is called the Schrodinger equation in the momentum representation:
$i{\partial \psi(p) \over \partial t} = {p^2\over 2m} \psi(p) + V(i{\partial\over \partial x}) \psi(p)$
The group-velocity relation for the fourier trasformed wave-packet gives the second of Hamilton's equations.
${dp \over dt} = -{\partial H \over \partial x}$
## Versions
There are several equations which go by Schrodinger's name:
### Time Dependent Equation
This is the equation of motion for the quantum state. In the most general form, it is written:
$i{d\over dt} \psi = \hat H \psi$
Where $\hat H$ is a linear operator acting on the wavefunction $\psi$. $\hat H$ takes as input one $\psi$ and produces another in a linear way, a function-space version of a matrix multiplying a vector. For the specific case of a single particle in one dimension moving under the influence of a potential V:
$i{d\over dt} \psi = -{1\over 2m} {\partial^2 \over \partial x^2} \psi(x) + V(x) \psi(x) \,$
and the operator H can be read off:
$\hat H = -{1\over 2m} {\partial^2 \over \partial x^2} + V(x) \,$
it is a combination of the operator which takes the second derivative, and the operator which pointwise multiplies $\psi$ by V(x). When acting on $\psi$ it reproduces the right hand side.
For a particle in three dimensions, the only difference is more derivatives:
$i{d\over dt} \psi = -{1\over 2m} \nabla^2 \psi + V(x) \psi \,$
and for N particles, the difference is that the wavefunction is in 3N-dimensional configuration space, the space of all possible particle positions.
$i{d\over dt} \psi(x_1,...,x_n) = (-{\nabla_1^2\over 2m_1} - {\nabla_2^2 \over 2m_2} ... - {\nabla_N^2\over 2m_N} ) \psi + V(x_1,..,x_N)\psi \,$
This last equation is in a very high dimension, so that the solutions are not easy to visualize.
### Time Independent Equation
This is the equation for the standing waves, the eigenvalue equation for H. In abstract form, for a general quantum system, it is written:
$E \psi = \hat H \psi \,$
For a particle in one dimension, with the mass absorbed into rescaling either time or space:
$E \psi = -{\partial^2 \psi \over \partial x^2} + V(x)\psi \,$
But there is a further restriction--- the solution must not grow at infinity, so that it has a finite norm:
$|| \psi ||^2 = \int_x \psi^*(x) \psi(x) \,$
For example, when there is no potential, the equation reads:
$- E \psi = {\partial^2 \psi \over \partial x^2} \,$
which has oscillatory solutions for E>0 (the C's are arbitrary constants):
$\psi_E(x) = C_1 e^{i\sqrt{E} x} + C_2 e^{-i\sqrt{E}x} \,$
and exponential solutions for E<0
$\psi_{-|E|}(x) = C_1 e^{\sqrt{|E|} x} + C_2 e^{-\sqrt{|E|} x} \,$
The exponentially growing solutions have an infinite norm, and are not physical. They are not allowed in a finite volume with periodic or fixed boundary conditions.
For a constant potential V the solution is oscillatory for E>V and exponential for E quantum tunneling. If the potential V grows at infinity, the motion is classically confined to a finite region, which means that in quantum mechanics every solution becomes an exponential far enough away. The condition that the exponential is decreasing restricts the energy levels to a discrete set, called the allowed energies.
### Energy Eigenstates
A solution $\scriptstyle \psi_E(x)$ of the time independent equation is called an energy eigenstate with energy E:
$H\psi_E = E \psi_E \,$
To find the time dependence of the state, consider starting the time-dependent equation with an initial condition $\psi_E(x)$. The time derivative at t=0 is everywhere proportional to the value:
$i {d\over dt} \psi(x) = H \psi(x) = E \psi(x) \,$
So that at first the whole function just gets rescaled, and it maintains the property that its time derivative is proportional to itself. So for all times,
$\psi(x,t) = A(t) \psi_E(x) \,$
substituting,
$i {dA \over dt } = - E A \,$
So that the solution of the time-dependent equation with this initial condition is:
$\psi(x,t) = \psi_E(x) e^{-iEt} \,$
or, with explicit $\scriptstyle\hbar$s.
$\psi(x,t) = \psi_E(x) e^{-i{E t\over\hbar}} \,$
This is a restatement of the fact that solutions of the time-independent equation are the standing wave solutions of the time dependent equation. They only get multiplied by a phase as time goes by, and otherwise are unchanged.
Superpositions of energy eigenstates change their properties according to the relative phases between the energy levels.
## Properties
### First Order in Time
The Schrodinger equation describes the time evolution of a quantum state, and must determine the future value from the present value. A classical field equation can be second order in time derivatives, the classical state can include the time derivative of the field. But a quantum state is a full description of a system, so that the Schrodinger equation is always first order in time.
### Linear
The Schrödinger equation is linear in the wavefunction: if $\psi(x, t)$ and $\phi(x,t)$ are solutions to the time dependent equation, then so is $a \psi + b \phi$, where a and b are any complex numbers.
In quantum mechanics, the time evolution of a quantum state is always linear, for fundamental reasons. Although there are nonlinear versions of the Schrodinger equation, these are not equations which describe the evolution of a quantum state, but classical field equations like Maxwell's equations or the Klein-Gordon equation.
The Schrodinger equation itself can be thought of as the equation of motion for a classical field not for a wavefunction, and taking this point of view, it describes a coherent wave of nonrelativistic matter, a wave of a Bose condensate or a superfluid with a large indefinite number of particles and a definite phase and amplitude.
### Real Eigenstates
The time-independent equation is also linear, but in this case linearity has a slightly different meaning. If two wavefunctions $\psi_1$ and $\psi_2$ are solutions to the time-independent equation with the same energy E, then any linear combination of the two is a solution with energy E. Two different solutions with the same energy are called degenerate.
$\hat H (a\psi_1 + b \psi_2 ) = ( a \hat H \psi_1 + b \hat H \psi_2) = E (a \psi_1 + b\psi_2) \,$
In an arbitrary potential, there is one obvious degeneracy: if a wavefunction $\psi$ solves the time-independent equation, so does $\psi^*$. By taking linear combinations, the real and imaginary part of $\psi$ are each solutions. So that restricting attention to real valued wavefunctions does not affect the time-independent eigenvalue problem.
In the time-dependent equation, complex conjugate waves move in opposite directions. Given a solution to the time dependent equation $\psi(x,t)$, the replacement:
$\psi(x,t) \rightarrow \psi^*(x,-t) \,$
produces another solution, and is the extension of the complex conjugation symmetry to the time-dependent case. The symmetry of complex conjugation is called time-reversal.
### Unitary Time Evolution
The Schrodinger equation is Unitary, which means that the total norm of the wavefunction, the sum of the squares of the value at all points:
$\int_x \psi^*(x) \psi(x) = \langle \psi | \psi \rangle \,$
has zero time derivative.
The derivative of $\psi^*$ is according to the complex conjugate equations
$-i \hbar {d\over dt} \psi^* = \hat H^\dagger \psi^* \,$
where the operator $H^\dagger$ is defined as the continuous analog of the Hermitian conjugate,
$\langle \eta H^{\dagger} | \psi \rangle = \langle \eta | H \psi \rangle \,$
For a discrete basis, this just means that the matrix elements of the linear operator H obey:
$\hat H_{ij} = \hat H^*_{ji} \,$
The derivative of the inner product is:
${d\over dt} \langle \psi | \psi \rangle = i \langle \psi \hat H^\dagger | \psi \rangle - i \langle \psi| H \psi \rangle \,$
and is proportional to the imaginary part of H. If H has no imaginary part, if it is self-adjoint, then the probability is conserved. This is true not just for the Schrodinger equation as written, but for the Schrodinger equation with nonlocal hopping:
$i{d \over dt} \psi(x) = \int_y H(x,y) \psi(y) \,$
so long as:
$H(x,y) = H(y,x)^* \,$
the particular choice:
$H(x,y) = - {1\over 2m} \nabla_x^2 \delta(x-y) + V(x) \delta(x-y) \,$
reproduces the local hopping in the ordinary Schrodinger equation. On a discrete lattice approximation to a continuous space, H(x,y) has a simple form:
$H(x,y) = -{1\over 2m} \,$
whenever x and y are nearest neighbors. On the diagonal
$H(x,x) = +{n\over 2m} + V(x) \,$
where n is the number of nearest neighbors.
### Positive Energies
The solutions of the Schrodinger equation in a potential which is bounded below have a frequency which is bounded below. For any linear operator $\scriptstyle \hat A$, the smallest eigenvector minimizes the quantity:
$\langle \psi |A|\psi \rangle$
over all $\psi$ which are normalized:
$\|\psi\|^2 = \int_x |\psi(x)|^2 =\langle \psi | \psi \rangle = 1 \,$
by the variational principle.
The value of the energy for the Schrodinger Hamiltonian is then the minimum value of:
$\langle \psi|H|\psi\rangle = \int_x \psi^*(x) (- \nabla^2 \psi + V(x)\psi) = \int_x |\nabla \psi|^2 + V(x) |\psi|^2 dx$
after an integration by parts, and the right hand side is positive definite when V is positive.
### Positive Definite Nondegenerate Ground State
For potentials which are bounded below and are not infinite over a region, there is a ground state which minimizes the integral above. This lowest energy wavefunction is real and positive definite.
Suppose for contradiction that $\psi$ is a lowest energy state and has a sign change, then $\eta(x)=|\psi(x)|$, the absolute value of $\psi$ obeys
$V(x)|\eta|^2 = V(x)|\psi|^2\,$
everywhere, and
$|\nabla \eta|^2 = |\nabla \psi|^2$
except for a set of measure zero. So $\eta$ is also a minimimum of the integral, and it has the same value as $\psi$. But by smoothing out the bends at the sign change, the gradient contribution to the integral is reduced while the potential energy is hardly altered, so the energy of $\eta$ can be reduced, which is a contradiction.
The lack of sign changes also proves that the ground state is nondegenerate, since if there were two ground states with energy E which were not proportional to each other, a linear combination of the two would also be a ground state with a zero.
These properties allow the analytic continuation of the Schrodinger equation to be identified as a stochastic process, which can be represented by a path integral.
### Completeness
The energy eigenstates form a basis--- any wavefunction may be written as a sum over the discrete energy states or an integral over continuous energy states, or more generally as an integral over a measure. This is the spectral theorem in mathematics, and in a finite state space it is just a statement of the completeness of the eigenvectors of a Hermitian matrix.
### Local Conservation of Probability
The probability density of a particle is $\psi^*(x)\psi(x)$. The probability flux is defined as:
$\mathbf{j} = {\hbar \over m} \cdot {1 \over {2 \mathrm{i}}} \left( \psi ^{*} \nabla \psi - \psi \nabla \psi^{*} \right) = {\hbar \over m} \operatorname{Im} \left( \psi ^{*} \nabla \psi \right)$
in units of (probability)/(area × time).
The probability flux satisfies the continuity equation:
${ \partial \over \partial t} P\left(x,t\right) + \nabla \cdot \mathbf{j} = 0$
where $P\left(x, t\right)$ is the probability density and measured in units of (probability)/(volume) = r−3. This equation is the mathematical equivalent of the probability conservation law.
For a plane wave:
$\psi (x,t) = \, A e^{ \mathrm{i} (k x - \omega t)}$
$j\left(x,t\right) = \left|A\right|^2 {\hbar k \over m}$
So that not only is the probability of finding the particle the same everywhere, but the probability flux is as expected from an object moving at the classical velocity $p/m$.
The reason that the Schrodinger equation admits a probability flux is because all the hopping is local and forward in time.
### Heisenberg Observables
There are many linear operators which act on the wavefunction, each one defines a Heisenberg matrix when the energy eigenstates are discrete. For a single particle, the operator which takes the derivative of the wavefunction in a certain direction:
$\hat p = {i\hbar {\partial \over \partial x}}$
Is called the momentum operator. Multiplying operators is just like multiplying matrices, the product of A and B acting on $\psi$ is A acting on the output of B acting on $\psi$.
An eigenstate of p obeys the equation:
$\hat p \psi = k \psi \,$
for a number k, and for a normalizable wavefunction this restricts k to be real, and the momentum eigenstate is a wave with frequency k.
$\psi(x) = e^{i kx} \,$
The position operator x multiplies each value of the wavefunction at the position x by x:
$\hat x(\psi) = x\psi$
So that in order to be an eigenstate of x, a wavefunction must be entirely concentrated at one point:
$\hat x \delta(x-x_0) = x_0 \delta(x-x_0)$
In terms of p, the Hamiltonian is:
$\hat H = {\hat p^2\over 2m} + V(x)$
It is easy to verify that p acting on x acting on psi:
$\hat p (\hat x( \psi)) = -i \hbar {\partial \over \partial x}( x \psi) = -i x {\partial \over \partial x}\psi -i\hbar \psi$
while x acting on p acting on psi reproduces only the first term:
$\hat x(\hat p (\psi)) = -i \hbar {\partial \over \partial x} \psi$
so that the difference of the two is not zero:
$( x p - p x ) \psi = i \hbar \psi \,$
or in terms of operators:
$[x,p] = xp - px = i \hbar \,$
since the time derivative of a state is:
$i{d \over dt} \psi = \hat H \psi \,$
while the complex conjugate is
$- i{d\over dt} \psi^* = \hat H \psi^* \,$
The time derivative of a matrix element
${d\over dt} \langle \eta | A |\psi \rangle = - \eta \hat H A \psi + \eta AH \psi = [H,A] \,$
obeys the Heisenberg equation of motion. This establishes the equivalence of the Schrodinger and Heisenberg formalisms, ignoring the mathematical fine points of the limiting procedure for continuous space.
### Correspondence principle
The Schrödinger equation satisfies the correspondence principle. In the limit of small wavelength wavepackets, it reproduces Newton's laws. This is easy to see from the equivalence to matrix mechanics.
All operators in Heisenberg's formalism obey the quantum analog of Hamilton's equations:
${dA \over dt} = -i\hbar (AH - HA)$
So that in particular, the equations of motion for the X and P operators are:
${dX \over dt} = {P\over m}$
${dP \over dt} = - {\partial V \over \partial x}$
in the Schrodinger picture, the interpretation of this equation is that it gives the time rate of change of the matrix element between two states when the states change with time. Taking the expectation value in any state shows that Newton's laws hold not only on average, but exactly, for the quantities:
$\langle X\rangle = \int_x \psi^*(x)\psi(x) X = \langle \psi|X|\psi \rangle \,$
$\langle P\rangle = \int_x \psi^*(x) i\hbar {\partial \psi \over \partial x}(x) = \langle \psi |P|\psi\rangle \,$
## Relativity
The Schrödinger equation does not take into account relativistic effects, as a wave equation, it is invariant under a Galilean transformation, but not under a Lorentz transformation. But in order to include relativity, the physical picture must be altered in a radical way.
A naive generalization of the Schrodinger equation uses the relativistic mass-energy relation (in natural units):
$E^2 = P^2 + m^2 \,$
to produce the differential equation:
$- {\partial^2 \over \partial t^2}\psi = - \nabla^2 \psi + m^2 \psi \,$
which is relativistically invariant, but second order in $\psi$, and so cannot be an equation for the quantum state. This equation also has the property that there are solutions with both positive and negative frequency, a plane wave solution obeys:
$\omega^2 - k^2 = m^2 \,$
which has two solutions, one with positive frequency the other with negative frequency. This is a disaster for quantum mechanics, because it means that the energy is unbounded below.
A more sophisticated attempt to solve this problem uses a first order wave equation, the Dirac equation, but again there are negative energy solutions. In order to solve this problem, it is essential to go to a multiparticle picture, and to consider the wave equations as equations of motion for a quantum field, not for a wavefunction.
The reason is that relativity is incompatible with a single particle picture. A relativistic particle cannot be localized to a small region without the particle number becoming indefinite. When a particle is localized in a box of length L, the momentum is uncertain by an amount roughly proportional to h/L by the uncertainty principle. This leads to an energy uncertainty of hc/L, when |p| is large enough so that the mass of the particle can be neglected. This uncertainty in energy is equal to the mass-energy of the particle when
$L = {\hbar \over mc} \,$
and this is called the Compton wavelength. Below this length, it is impossible to localize a particle and be sure that it stays a single particle, since the energy uncertainty is large enough to produce more particles from the vacuum by the same mechanism that localizes the original particle.
But there is another approach to relativistic quantum mechanics which does allow you to follow single particle paths, and it was discovered within the path-integral formulation. If the integration paths in the path integral include paths which move both backwards and forwards in time as a function of their own proper time, it is possible to construct a purely positive frequency wavefunction for a relativistic particle. This construction is appealing, because the equation of motion for the wavefunction is exactly the relativistic wave equation, but with a nonlocal constraint that separates the positive and negative frequency solutions. The positive frequency solutions travel forward in time, the negative frequency solutions travel backwards in time. In this way, they both analytically continue to a statistical field correlation function, which is also represented by a sum over paths. But in real space, they are the probability amplitudes for a particle to travel between two points, and can be used to generate the interaction of particles in a point-splitting and joining framework. The relativistic particle point of view is due to Richard Feynman.
Feynman's method also constructs the theory of quantized fields, but from a particle point of view. In this theory, the equations of motion for the field can be interpreted as the equations of motion for a wavefunction only with caution--- the wavefunction is only defined globally, and in some way related to the particle's proper time. The notion of a localized particle is also delicate--- a localized particle in the relativistic particle path integral corresponds to the state produced when a local field operator acts on the vacuum, and exacly which state is produced depends on the choice of field variables.
## Solutions
Some general techniques are:
• Perturbation theory
• The variational principle
• Quantum Monte Carlo methods
• Density functional theory
• The WKB approximation and semi-classical expansion
In some special cases, special methods can be used:
• List of quantum mechanical systems with analytical solutions
• Hartree-Fock method and post Hartree-Fock methods
• Discrete delta-potential method
## Free Schrödinger equation
When the potential is zero, the Schrödinger equation is linear with constant coefficients:
$i \frac{\partial \psi}{\partial t}=-{1\over 2m}\nabla^2\psi$
where $\scriptstyle \hbar$ has been set to 1. The solution $\psi_t(x)$ for any initial condition $\psi_0(x)$ can be found by Fourier transforms. Because the coefficients are constant, an initial plane wave:
$\psi_0(x) = A e^{i k x} \,$
stays a plane wave. Only the coefficient changes. Substituting:
${dA \over dt} = -{i k^2 \over 2m} A \,$
So that A is also oscillating in time:
$A(t) = A e^{- i {k^2 \over 2m} t} \,$
and the solution is:
$\psi_t(x) = A e^{i k x - i \omega t} \,$
Where $\omega=k^2/2m$, a restatement of DeBroglie's relations.
To find the general solution, write the initial condition as a sum of plane waves by taking its Fourier transform:
$\psi_0(x) = \int_k \psi(k) e^{ikx} \,$
The equation is linear, so each plane waves evolves independently:
$\psi_t(x) = \int_k \psi(k)e^{-i\omega t} e^{ikx} \,$
Which is the general solution. When complemented by an effective method for taking Fourier transforms, it becomes an efficient algorithm for finding the wavefunction at any future time--- Fourier transform the initial conditions, multiply by a phase, and transform back.
### Gaussian Wavepacket
An easy and instructive example is the Gaussian wavepacket:
$\psi_0(x) = e^{-x^2 / 2a} \,$
where a is a positive real number, the square of the width of the wavepacket. The total normalization of this wavefunction is:
$\langle \psi|\psi\rangle = \int_x \psi^* \psi = \sqrt{\pi a}$
The Fourier transform is a Gaussian again in terms of the wavenumber k:
$\psi_0(k) = (2\pi a)^{d/2} e^{- a k^2/2} \,$
With the physics convention which puts the factors of $2\pi$ in Fourier transforms in the k-measure.
$\psi_0(x) = \int_k \psi_0(k) e^{-ikx} = \int {d^dk \over (2\pi)^d} \psi_0(k) e^{-ikx}$
Each separate wave only phase-rotates in time, so that the time dependent Fourier-transformed solution is:
$\psi_t(k) = (2\pi a)^{d/2} e^{- a { k^2\over 2} - it {k^2\over 2m}} = (2\pi a)^{d/2} e^{-(a+it/m){k^2\over 2}} \,$
The inverse Fourier transform is still a Gaussian, but the parameter a has become complex, and there is an overall normalization factor.
$\psi_t(x) = \left({a \over a + i t/m}\right)^{d/2} e^{- {x^2\over 2(a + i t/m)} } \,$
The branch of the square root is determined by continuity in time--- it is the value which is nearest to the positive square root of a. It is convenient to rescale time to absorb m, replacing t/m by t.
The integral of $\psi$ over all space is invariant, because it is the inner product of $\psi$ with the state of zero energy, which is a wave with infinite wavelength, a constant function of space. For any energy state, with wavefunction $\eta(x)$, the inner product:
$\langle \eta | \psi \rangle = \int_x \eta(x) \psi_t(x)$,
only changes in time in a simple way: its phase rotates with a frequency determined by the energy of $\eta$. When $\eta$ has zero energy, like the infinite wavelength wave, it doesn't change at all.
The sum of the absolute square of $\psi$ is also invariant, which is a statement of the conservation of probability. Explicitly in one dimension:
$|\psi|^2 = \psi\psi^* = {a \over \sqrt{a^2+t^2} } e^{-{x^2 a \over a^2 + t^2}}$
Which gives the norm:
$\int |\psi|^2 = \sqrt{\pi a}$
which has preserved its value, as it must.
The width of the Gaussian is the interesting quantity, and it can be read off from the form of $|\psi^2|$:
$\sqrt{a^2 + t^2 \over a} \,$.
The width eventually grows linearly in time, as $\scriptstyle t/\sqrt{a}$. This is wave-packet spreading--- no matter how narrow the initial wavefunction, a Schrodinger wave eventually fills all of space. The linear growth is a reflection of the momentum uncertainty--- the wavepacket is confined to a narrow width $\scriptstyle \sqrt{a}$ and so has a momentum which is uncertain by the reciprocal amount $\scriptstyle 1/\sqrt{a}$, a spread in velocity of $\scriptstyle 1/m\sqrt{a}$, and therefore in the future position by $\scriptstyle t/m\sqrt{a}$, where the factor of m has been restored by undoing the earlier rescaling of time.
### Galilean Invariance
Galilean boosts are transformations which look at the system from the point of view of an observer moving with a steady velocity -v. A boost must change the physical properties of a wavepacket in the same way as in classical mechanics:
$p'= p + mv \,$
$x'= x + vt \,$
So that the phase factor of a free Schrodinger plane wave:
$p x - E t = (p' - mv)(x' - vt) - {(p'-mv)^2\over 2} t = p' x' + E' t + m v x - {mv^2\over 2}t \,$
is only different in the boosted coordinates by a phase which depends on x and t, but not on p.
An arbitrary superposition of plane wave solutions with different values of p is the same superposition of boosted plane waves, up to an overall x,t dependent phase factor. So any solution to the free Schrodinger equation, $\psi_t(x)$, can be boosted into other solutions:
$\psi'_t(x) = \psi_t(x + vt) e^{ i mv x - i {mv^2\over 2}t} \,$
Boosting a constant wavefunction produces a plane-wave. More generally, boosting a plane-wave:
$\psi_t(x) = e^{ipx - i {p^2\over 2m} t} \,$
produces a boosted wave:
$\psi'_t(x) = e^{ i p(x + vt) - i{p^2\over 2m}t + imv x - i {mv^2\over 2}t} = e^{i(p+mv)x + i {(p+mv)^2\over 2m}t } \,$
$\psi_t(x) = {1\over \sqrt{a+it/m}} e^{ - {x^2\over 2a} } \,$
produces the moving Gaussian:
$\psi_t(x) = {1\over \sqrt{a + it/m}} e^{ - {(x + vt)^2 \over 2a} + i m v x - i {mv^2\over 2} t } \,$
Which spreads in the same way.
### Free Propagator
The narrow-width limit of the Gaussian wavepacket solution is the propagator K. For other differential equations, this is sometimes called the Green's function, but in quantum mechanics it is traditional to reserve the name Green's function for the time Fourier transform of K. When a is the infinitesimal quantity $\epsilon$, the Gaussian initial condition, rescaled so that its integral is one:
$\psi_0(x) = {1\over \sqrt{2\pi \epsilon} } e^{-{x^2\over 2\epsilon}} \,$
becomes a delta function, so that its time evolution:
$K_t(x) = {1\over \sqrt{2\pi (i t + \epsilon)}} e^{ - x^2 \over 2it+\epsilon } \,$
gives the propagator.
Note that a very narrow initial wavepacket instantly becomes infinitely wide, with a phase which is more rapidly oscillatory at large values of x. This might seem strange--- the solution goes from being concentrated at one point to being everywhere at all later times, but it is a reflection of the momentum uncertainty of a localized particle. Also note that the norm of the wavefunction is infinite, but this is also correct since the square of a delta function is divergent in the same way.
The factor of $\epsilon$ is an infinitesimal quantity which is there to make sure that integrals over K are well defined. In the limit that $\epsilon$ becomes zero, K becomes purely oscillatory and integrals of K are not absolutely convergent. In the remainder of this section, it will be set to zero, but in order for all the integrations over intermediate states to be well defined, the limit $\scriptstyle \epsilon\rightarrow 0$ is to only to be taken after the final state is calculated.
The propagator is the amplitude for reaching point x at time t, when starting at the origin, x=0. By translation invariance, the amplitude for reaching a point x when starting at point y is the same function, only translated:
$K_t(x,y) = K_t(x-y) = {1\over \sqrt{2\pi it}} e^{-i(x-y)^2 \over 2t} \,$
In the limit when t is small, the propagator converges to a delta function:
$\lim_{t\rightarrow 0} K_t(x-y) = \delta(x-y)$
but only in the sense of distributions. The integral of this quantity multiplied by an arbitrary differentiable test function gives the value of the test function at zero. To see this, note that the integral over all space of K is equal to 1 at all times:
$\int_x K_t(x) = 1 \,$
since this integral is the inner-product of K with the uniform wavefunction. But the phase factor in the exponent has a nonzero spatial derivative everywhere except at the origin, and so when the time is small there are fast phase cancellations at all but one point. This is rigorously true when the limit $\epsilon\rightarrow zero$ is taken after everything else.
So the propagation kernel is the future time evolution of a delta function, and it is continuous in a sense, it converges to the initial delta function at small times. If the initial wavefunction is an infinitely narrow spike at position $x_0$:
$\psi_0(x) = \delta(x - x_0) \,$
it becomes the oscillatory wave:
$\psi_t(x) = {1\over \sqrt{2\pi i t}} e^{ -i (x-x_0) ^2 /2t} \,$
Since every function can be written as a sum of narrow spikes:
$\psi_0(x) = \int_y \psi_0(y) \delta(x-y) \,$
the time evolution of every function is determined by the propagation kernel:
$\psi_t(x) = \int_y \psi_0(x) {1\over \sqrt{2\pi it}} e^{-i (x-x_0)^2 / 2t} \,$
And this is an alternate way to express the general solution. The interpretation of this expression is that the amplitude for a particle to be found at point x at time t is the amplitude that it started at $x_0$ times the amplitude that it went from $x_0$ to x, summed over all the possible starting points. In other words, it is a convolution of the kernel K with the initial condition.
$\psi_t = K * \psi_0 \,$
Since the amplitude to travel from x to y after a time $t+t'$ can be considered in two steps, the propagator obeys the identity:
$\int_y K(x-y;t)K(y-z;t') = K(x-z;t+t') \,$
Which can be interpreted as follows: the amplitude to travel from x to z in time t+t' is the sum of the amplitude to travel from x to y in time t multiplied by the amplitude to travel from y to z in time t', summed over all possible intermediate states y. This is a property of an arbitrary quantum system, and by subdividing the time into many segments, it allows the time evolution to be expressed as a path integral.
### Analytic Continuation to Diffusion
The spreading of wavepackets in quantum mechanics is directly related to the spreading of probability densities in diffusion. For a particle which is random walking, the probability density function at any point satisfies the diffusion equation:
${\partial \over \partial t} \rho = {1\over 2} {\partial \over \partial x^2 } \rho$
where the factor of 2, which can be removed by a rescaling either time or space, is only for convenience.
A solution of this equation is the spreading gaussian:
$\rho_t(x) = {1\over \sqrt{2\pi t}} e^{-x^2 \over 2t}$
and since the integral of $\rho_t$, is constant, while the width is becoming narrow at small times, this function approaches a delta function at t=0:
$\lim_{t\rightarrow 0} \rho_t(x) = \delta(x) \,$
again, only in the sense of distributions, so that
$\lim_{t\rightarrow 0} \int_x f(x) \rho_t(x) = f(0) \,$
for any smooth test function f.
The spreading Gaussian is the propagation kernel for the diffusion equation and it obeys the convolution identity:
$K_{t+t'} = K_{t}*K_{t'} \,$
Which allows diffusion to be expressed as a path integral. The propagator is the exponential of an operator H:
$K_t(x) = e^{-tH} \,$
which is the infinitesimal diffusion operator.
$H= -{\nabla^2\over 2} \,$
A matrix has two indices, which in continuous space makes it a function of x and x'. In this case, because of translation invariance, the matrix element K only depend on the difference of the position, and a convenient abuse of notation is to refer to the operator, the matrix elements, and the function of the difference by the same name:
$K_t(x,x') = K_t(x-x') \,$
Translation invariance means that continuous matrix multiplication:
$C(x,x'') = \int_{x'} A(x,x')B(x',x'') \,$
is really convolution:
$C(\Delta) = C(x-x'') = \int_{x'} A(x-x') B(x'-x'') = \int_{y} A(\Delta-y)B(y) \,$
The exponential can be defined over a range of t's which include complex values, so long as integrals over the propagation kernel stay convergent.
$K_z(x) = e^{-zH} \,$
As long as the real part of z is positive, for large values of x K is exponentially decreasing and integrals over K are absolutely convergent.
The limit of this expression for z coming close to the pure imaginary axis is the Schrodinger propagator:
$K_t^{\rm Schr} = K_{it+\epsilon} = e^{-(it+\epsilon)H} \,$
and this gives a more conceptual explanation for the time evolution of Gaussians. From the fundamental identity of exponentiation, or path integration:
$K_z * K_{z'} = K_{z+z'} \,$
holds for all complex z values where the integrals are absolutely convergent so that the operators are well defined.
So that quantum evolution starting from a Gaussian, which is the diffusion kernel K:
$\psi_0(x) = K_a(x) = K_a * \delta(x) \,$
gives the time evolved state:
$\psi_t = K_{it} * K_a = K_{a+it} \,$
This explains the diffusive form of the Gaussian solutions:
$\psi_t(x) = {1\over \sqrt{2\pi (a+it)} } e^{- {x^2\over 2(a+it)} } \,$
## Variational Principle
The variational principle asserts that for any any Hermitian matrix A, the lowest eigenvalue minimizes the quantity:
$\langle v,Av \rangle = \sum_{ij} A_{ij} v^*_i v_j \,$
on the unit sphere $=1$. This follows by the method of Lagrange multipliers, at the minimum the gradient of the function is parallel to the gradient of the constraint:
${\partial\over \partial v_i} \langle v,Av\rangle = \lambda {\partial \over \partial v_i} \langle v,v\rangle \,$
which is the eigenvalue condition
$\sum_{j} A_{ij} v_j = \lambda v_i \,$
so that the extreme values of a quadratic form A are the eigenvalues of A, and the value of the function at the extreme values is just the corresponding eigenvalue:
$\langle v,Av\rangle = \lambda\langle v,v\rangle = \lambda \,$
When the hermitian matrix is the Hamiltonian, the minimum value is the lowest energy level.
In the space of all wavefunctions, the unit sphere is the space of all normalized wavefunctions $\psi$, the ground state minimizes
$\langle \psi | H |\psi \rangle = \int \psi^* H \psi = \int \psi^* (-\nabla^2 + V(x)) \psi \,$
or, after an integration by parts,
$\langle \psi | H |\psi \rangle = \int |\nabla \psi|^2 + V(x) |\psi|^2 \,$
All the stationary points come in complex conjugate pairs since the integrand is real. Since the stationary points are eigenvalues, any linear combination is a stationary point, and the real and imaginary part are both stationary points.
The lowest energy state has a positive definite wavefunction, because given a $\psi$ which minimizes the integral, $|\psi|$, the absolute value, is also a minimizer. But this minimizer has sharp corners at places where $\psi$ changes sign, and these sharp corners can be rounded out to reduce the gradient contribution.
## Potential and Ground State
For a particle in a positive definite potential, the ground state wavefunction is real and positive, and has a dual interpretation as the probability density for a diffusion process. The analogy between diffusion and nonrelativistic quantum motion, originally discovered and exploited by Schrodinger, has led to many exact solutions.
A positive definite wavefunction:
$\psi = e^{-W(x)} \,$
is a solution to the time-independent Schrodinger equation with m=1 and potential:
$V(x) = {1\over 2} |\nabla W|^2 - {1\over 2} \nabla^2 W \,$
with zero total energy. W, the logarithm of the ground state wavefunction. The second derivative term is higher order in $\scriptstyle \hbar$, and ignoring it gives the semi-classical approximation.
The form of the ground state wavefunction is motivated by the observation that the ground state wavefunction is the Boltzmann probability for a different problem, the probability for finding a particle diffusing in space with the free-energy at different points given by W. If the diffusion obeys detailed balance and the diffusion constant is everywhere the same, the Fokker Planck equation for this diffusion is the Schrodinger equation when the time parameter is allowed to be imaginary. This analytic continuation gives the eigenstates a dual interpretation--- either as the energy levels of a quantum system, or the relaxation times for a stochastic equation.
### Harmonic Oscillator
Main article: Quantum harmonic oscillator.
W should grow at infinity, so that the wavefunction has a finite integral. The simplest analytic form is:
$W(x) = \omega x^2 \,$
with an arbitrary constant $\omega$, which gives the potential:
$V(x) = {1\over 2} \omega^2 x^2 - {\omega \over 2} \,$
This potential describes a Harmonic oscillator, with the ground state wavefunction:
$\psi(x) = e^{-\omega x^2 } \,$
The total energy is zero, but the potential is shifted by a constant. The ground state energy of the usual unshifted Harmonic oscillator potential:
$V(x) = {\omega x^2 \over 2} \,$
$E_0 = {a\over 2} \,$
which is the zero point energy of the oscillator.
### Coulomb Potential
Another simple but useful form is
$W(x) = 2a|x| \,$
where W is proportional to the radial coordinate. This is the ground state for two different potentials, depending on the dimension. In one dimension, the corresponding potential is singular at the origin, where it has some nonzero density:
$V(x) = 2a^2 + a \delta(x) \,$
and, up to some rescaling of variables, this is the lowest energy state for a delta function potential, with the bound state energy added on.
$V(x) = a \delta(x) \,$
with the ground state energy:
$E_0 = - 2a^2 \,$
and the ground state wavefunction:
$\psi = e^{-2a|x|} \,$
In higher dimensions, the same form gives the potential:
$V(x) = 2a^2+ { 2a (d-1) \over r}; \,$
which can be identified as the attractive Coulomb law, up to an additive constant which is the ground state energy. This is the superpotential that describes the lowest energy level of the Hydrogen atom, once the mass is restored by dimensional analysis:
$\psi_0 = e^{-r/r_0} \,$
where $r_0$ is the Bohr radius, with energy
$E_0 = - {2a\over d-1} \,$
The ansatz
$W(x) = a r + b \log(r) \,$
modifies the Coulomb potential to include a quadratic term proportional to $1/r^2$, which is useful for nonzero angular momentum.
## Operator Formalism
### Bra-ket Notation
In the mathematical formulation of quantum mechanics, a physical system is fully described by a vector in a complex Hilbert space, the collection of all possible normalizable wavefunctions. The wavefunction is just an alternate name for the vector of complex amplitudes, and only in the case of a single particle in the position representation is it a wave in the usual sense, a wave in space time. For more complex systems, it is a wave in an enormous space of all possible worlds. Two nonzero vectors which are multiples of each other, two wavefunctions which are the same up to rescaling, represent the same physical state.
The wavefunction vector can be written in several ways:
1. as an abstract ket vector:
$|\psi\rangle$
2. As a list of complex numbers, the components relative to a discrete list of normalizable basis vectors $|\eta_i\rangle$:
$c_i = \langle \eta_i |\psi \rangle$
3. As a continuous superposition of non-normalizable basis vectors, like position states $|x\rangle$:
$|\psi\rangle = \int_x \psi(x) |x\rangle$
The divide between the continuous basis and the discrete basis can be bridged by limiting arguments. The two can be formally unified by thinking of each as a measure on the real number line.
In the most abstract notation, the Schrödinger equation is written:
$i\hbar {d\over dt} |\psi\rangle = H |\psi\rangle$
which only says that the wavefunction evolves linearly in time, and names the linear operator which gives the time derivative the Hamiltonian H. In terms of the discrete list of coefficients:
$i\hbar {d\over dt} C_i = \sum_j H_{ij} C_j$
which just reaffirms that time evolution is linear, since the Hamiltonian acts by matrix multiplication.
In a continuous representation, the Hamiltonian is a linear operator, which acts by the continuous version of matrix multiplication:
$\langle x| i\hbar {d\over dt} |\psi\rangle = \langle x|H|\psi\rangle = \hat{H} \psi (x)$
Taking the complex conjugate:
$-i\hbar {d\over dt} \langle \psi | = \langle \psi | H^\dagger$
In order for the time-evolution to be unitary, to preserve the inner products, the time derivative of the inner product must be zero:
$i\hbar {d\over dt} \langle \psi | \psi \rangle = \langle\psi | H - H^\dagger |\psi\rangle = 0$
for an arbitrary state $|\psi\rangle$, which requires that H is Hermitian. In a discrete representation this means that $\scriptstyle H_{ij}= H_{ji}$. When H is continuous, it should be self-adjoint, which adds some technical requirement that H does not mix up normalizable states with states which violate boundary conditions or which are grossly unnormalizable.
The formal solution of the equation is the matrix exponential ( natural units):
$|\psi(t)\rangle = e^{-i H t} |\psi(0)\rangle = U(t) |\psi(0)\rangle$
For every time-independent Hamiltonian operator, $\hat H$, there exists a set of quantum states, $\left|\psi_n\right\rang$, known as energy eigenstates, and corresponding real numbers $E_n$ satisfying the eigenvalue equation.
$H |\psi_n \rangle = E_n |\psi_n \rangle \,$
This is the time-independent Schrodinger equation.
For the case of a single particle, the Hamiltonian is the following linear operator ( natural units):
$H = -{\nabla^2 \over 2m} + V(x) = {p^2\over 2m} + V(x)$
which is a Self-adjoint operators when V is not too singular and does not grow too fast. Self-adjoint operators have the property that their eigenvalues are real in any basis, and their eigenvectors form a complete set, either discrete or continuous.
Expressed in a basis of Eigenvectors of H, the Schrodinger equation becomes trivial:
$\mathrm{i} \hbar \frac{\partial}{\partial t} \left| \psi_n \left(t\right) \right\rangle = E_n \left|\psi_n\left(t\right)\right\rang.$
Which means that each energy eigenstate is only multiplied by a complex phase:
$\left| \psi \left(t\right) \right\rangle = \mathrm{e}^{-\mathrm{i} Et / \hbar} \left|\psi\left(0\right)\right\rang.$
Which is what matrix exponentiation means--- the time evolution acts to rotate the eigenfunctions of H.
When H is expressed as a matrix for wavefunctions in a discrete energy basis:
$i\hbar {d\over dt} C_i = E_i C_i \,$
so that:
$C_n(t) = e^{-iE_n t} C_n(t) \,$
The physical properties of the C's are extracted by acting by operators, matrices. By redefining the basis so that it rotates with time, the matrices become time dependent, which is the Heisenberg picture.
### Galilean Invariance
Galilean symmetry requires that H(p) is quadratic in p in both the classical and quantum Hamiltonian formalism. In order for Galilean boosts to produce a p-independent phase factor, px - Ht must have a very special form--- translations in p need to be compensated by a shift in H. This is only true when H is quadratic.
The infinitesimal generator of Boosts in both the classical and quantum case is:
$B = \sum_i m_i x_i(t) - t \sum_i p_i \,$
where the sum is over the different particles, and B,x,p are vectors.
The poisson bracket/commutator of $\scriptstyle B\cdot v$ with x and p generate infinitesimal boosts, with v the infinitesimal boost velocity vector:
$[B\cdot v ,x_i] = vt \,$
$[B\cdot v ,p_i] = v m_i \,$
Iterating these relations is simple, since they add a constant amount at each step. By iterating, the dv's incrementally sum up to the finite quantity V:
$x \rightarrow x_i + Vt \,$
$p \rightarrow p_i + m_i V \,$
B divided by the total mass is the current center of mass position minus the time times the centre of mass velocity:
$B = M X_\mathrm{cm} - t P_\mathrm{cm} \,$
In other words, B/M is the current guess for the position that the centre of mass had at time zero.
The statement that B doesn't change with time is the centre of mass theorem. For a Galilean invariant system, the centre of mass moves with a constant velocity, and the total kinetic energy is the sum of the center of mass kinetic energy and the kinetic energy measured relative to the centre of mass.
Since B is explicitly time dependent, H does not commute with B, rather:
${dB\over dt} = [H,B] + {\partial B \over \partial t} = 0 \,$
this gives the transformation law for H under infinitesimal boosts:
$[B\cdot v,H] = - P_\mathrm{cm} v \,$
the interpretation of this formula is that the change in H under an infinitesimal boost is entirely given by the change of the centre of mass kinetic energy, which is the dot product of the total momentum with the infinitesimal boost velocity.
The two quantities (H,P) form a representation of the Galilean group with central charge M, where only H and P are classical functions on phase-space or quantum mechanical operators, while M is a parameter. The transformation law for infinitesimal v:
$P' = P + M v \,$
$H' = H - P\dot v \,$
can be iterated as before--- P goes from P to P+MV in infinitesimal increments of v, while H changes at each step by an amount proportional to P, which changes linearly. The final value of H is then changed by the value of P halfway between the starting value and the ending value:
$H' = H - (P+{MV\over 2})\cdot V = H - P\cdot V - {MV^2\over 2}. \,$
The factors proportional to the central charge M are the extra wavefunction phases.
Boosts give too much information in the single-particle case, since Galilean symmetry completely determines the motion of a single particle. Given a multi-particle time dependent solution:
$\psi_t(x_1,x_2...,x_n) \,$
with a potential that depends only on the relative positions of the particles, it can be used to generate the boosted solution:
$\psi'_t = \psi_t(x_1 + v t, ..., x_2 + vt) e^{i P_\mathrm{cm}\cdot X_\mathrm{cm} - {Mv_\mathrm{cm}^2\over 2}t}. \,$
For the standing wave problem, the motion of the center of mass just adds an overall phase. When solving for the energy levels of multiparticle systems, Galilean invariance allows the centre of mass motion to be ignored.
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# What Does It Mean For A Number To Be Transcendental?
## Are there more algebraic numbers are transcendental numbers?
The non-transcendental numbers (otherwise known as the algebraic numbers – Wikipedia link) comprise a countably infinite set, whereas the transcendental numbers are uncountably infinite.
Therefore, there are certainly more transcendental numbers than there are algebraic numbers that also are not rational..
## What is meant by transcendental equation?
A transcendental equation is an equation containing a transcendental function of the variable(s) being solved for. Such equations often do not have closed-form solutions.
## Who found pi?
Archimedes of SyracuseThe Egyptians calculated the area of a circle by a formula that gave the approximate value of 3.1605 for π. The first calculation of π was done by Archimedes of Syracuse (287–212 BC), one of the greatest mathematicians of the ancient world.
## What is the most mysterious number?
Therefore the number 6174 is the only number unchanged by Kaprekar’s operation — our mysterious number is unique. The number 495 is the unique kernel for the operation on three digit numbers, and all three digit numbers reach 495 using the operation. Why don’t you check it yourself?
## How do you prove a number is transcendental?
The Lindemann–Weierstrass theorem is the primary tool utilized for this purpose. It states that if are non-zero algebraic numbers, and are distinct algebraic numbers, then: As an example let us show that and are transcendental numbers. The Lindemann–Weierstrass theorem is the primary tool utilized for this purpose.
## How many numbers are there between 0 and 1?
So the number of numbers between 0 and 1 is INFINITE. In fact, the number of numbers between any two numbers (eg between 3 and 8) is also called INFINITE.
## What is a true number?
The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as √2 (1.41421356…, the square root of 2, an irrational algebraic number). Included within the irrationals are the transcendental numbers, such as π (3.14159265…).
## Is zero a number Yes or no?
Zero is an even number because it is divisible by 2 with no remainder. 0 is neither positive nor negative. Many definitions include 0 as a natural number, in which case it is the only natural number that is not positive. Zero is a number which quantifies a count or an amount of null size.
Zero (0) is used as a number and also as the numerical digit. Zero gives the additive identity of the integers, real numbers, and many algebraic structures. It is used as a placeholder for writing numbers. Natural numbers start from 1, then 2 and so on.
## Is there a 666 in pi?
Pi is the 16th letter of the Greek alphabet just as p is the 16th of our alphabet. … The first 144 digits of pi add up to 666, the Number of the Beast in the Book of Revelation. 6. Albert Einstein was born on Pi Day: March 14, 1879.
## Is G irrational?
Numbers are dimensionsless abstract mathematical entities. The acceleration due to gravity is a physical measurement with units of length divided by time squared. The acceleration due to gravity is not a number at all, let alone an irrational number!
## What is the difference between an irrational and transcendental number?
A number x is irrational if it is not the solution of any algebraic equation of the first degree with integer coefficients, such as ax+b=0. A number z is transcendental if it is not the solution of any polynomial equation with integer coefficients of any degree n, such as a z^n+b z^(n-1)+ c z^(n-2)… +u z + v =0.
## Is Pi an infinite?
Value of pi Pi is an irrational number, which means that it is a real number that cannot be expressed by a simple fraction. That’s because pi is what mathematicians call an “infinite decimal” — after the decimal point, the digits go on forever and ever.
## Is 0 an algebraic number?
Zero is algebraic, being a root of the polynomial (for instance). Every real or complex number is either algebraic or transcendental because the definition of a transcendental number is a number that is not algebraic. … It’s therefore ratio of two integers (like 0/AnyInteger), so it’s rational.
## Who proved that pi is transcendental?
Ferdinand von LindemannThe theorem is named for Ferdinand von Lindemann and Karl Weierstrass. Lindemann proved in 1882 that eα is transcendental for every non-zero algebraic number α, thereby establishing that π is transcendental (see below).
## Will Pi ever end?
Because while these other national holidays come to an end, Pi Day actually doesn’t come to an end, because though Pi technically isn’t infinite, it does, in a sense, never fully end. Pi, formally known as π in the world of mathematics, is the ratio of the circumference of a circle and the diameter of a circle.
## Why are transcendental numbers important?
Transcendental numbers are useful in the study of straightedge-and-compass constructions, particularly in proving the impossibility of squaring the circle (i.e. it proves that it is impossible to construct a square with area equal to the area of any given circle, including 1 π 1\pi 1π, using only a straightedge and a …
## What number is between 1 and 2?
Between any two whole numbers there is a fraction. Between 0 and 1 there is 12, between 1 and 2 there is 112=3/2, and so on. In fact, there are infinitely many fractions between any two whole numbers.
## Are transcendental numbers constructible?
Computable Numbers. Crucially, transcendental numbers are not constructible geometrically nor algebraically…
## What is the value of E?
approximately 2.718The exponential constant is an important mathematical constant and is given the symbol e. Its value is approximately 2.718. It has been found that this value occurs so frequently when mathematics is used to model physical and economic phenomena that it is convenient to write simply e.
## What are algebraic and transcendental numbers?
Definition: A number is said to be Algebraic if there exists a nonzero polynomial such that . If no such polynomial exists, then is said to be Transcendental. It is easy to check that every is algebraic since is a polynomial satisfying .
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The RegExp Library
## Overview
The RegExp Library …
## Contents
Using the RegExp Library
A tutorial guide to using this library.
`signature REGEXP_PARSER`
Defines the interface to a parser for a concrete regular-expression syntax.
`signature REGEXP_ENGINE`
Defines the interface to a regular-expression search engine.
`structure RegExpSyntax`
Defines the abstract syntax of regular expressions.
`structure MatchTree`
Provides a tree-structured representation of the result of a successful regular expression match.
`structure AwkSyntax`
A parser for the regular-expression syntax defined by the AWK language.
`structure BackTrackEngine`
something
`structure DfaEngine`
something
`structure ThompsonEngine`
something
`functor RegExpFn`
Provides a functor for combining a regular-expression parser with an engine to form a complete implementation of an regular-expression search module.
## Usage
For SML/NJ, include `$/regexp-lib.cm` in your CM file. For use in MLton, include `$(SML_LIB)/smlnj-lib/RegExp/regexp-lib.mlb` in your MLB file.
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2 added 2462 characters in body
The question is: what about the stack of coherent sheaves, without flatness hypothesis? First of all, one should interpret "coherent" as meaning "quasi-coherent of finite presentation". The notion of coherent sheaf, as defined in EGA, is not functorial, that is, pullbacks of coherent sheaves are not necessarily coherent. Hartshorne's book defines "coherent" as "quasi-coherent and finitely generated", but this is a useless notion when working with non-noetherian schemes.
The stack of quasi-coherent finitely presented sheaves is not algebraic either. For example, let $k$ be an algebraically closed field, $k[\epsilon] = k[t]/(t^2)$ the ring of dual numbers. If $S = \mathop{\rm Spec} k[\epsilon]$, consider the coherent sheaf $F$ corresponding to the $k[\epsilon]$-module $k = k[\epsilon]/(\epsilon)$. Then I claim that the functor of automorphisms of $F$ over $S$ is not represented by an algebraic space.
Suppose it is represented by an algebraic space $G \to S$. Denote by $p$ the unique rational point of $S$ over $k$; the tangent space of $S$ at $p$ has a canonical generator $v_0$. Furthermore, if $X$ is a $k$-scheme with a rational point $x_0$ and $v$ is a tangent vector of $X$ at $x_0$, then there exists a unique $k$-morphism $S \to X$ sending $p$ to $x_0$ and $v_0$ to $v$. The inverse image of $S_{\rm red} = \mathop{\rm Spec} k$ in $G$ is isomorphic to $\mathbb G_{\mathrm m, k}$; so $G_{\rm red}$ is an affine scheme, hence $G$ is an affine scheme. The differential of the projection $G \to S$ at the origin of $G$ has a $1$-dimensional kernel, the tangent space of $\mathbb G_{\mathrm m, k}$ at the origin. On the other hand there is a unique section $S \to G$ sending $p$ to the origin, corresponding to $1 \in k^* = \mathrm{Aut}_{k[\epsilon]}k$; this means that there is a unique tangent vector of $G$ at the origin mapping to $v_0$. These two facts give a contradiction.
Here is another way to look at this. Given a contravariant functor $F$ on $k$-schemes and an element $p$ of $F(\mathop{\rm Spec}k)$, one can define the tangent space of $F$ at $p$ as the set of element of $F(\mathop{\rm Spec}k[\epsilon])$ that restrict to $p$. However, in order for this tangent space to be a $k$-vector space, one needs a Schlessinger-like gluing condition on $F$ (this is standard in deformation theory). The analysis above shows that this condition is not satisfied for the functor of automorphisms of $k$ over $k[\epsilon]$.
1
Artin's axioms do not apply in this case, because the stack is not limit-preserving. They only work with stacks that are locally finitely presented.
In any case, it is easy to give examples of quasi-coherent sheaves whose functor of automorphisms is not representable (for example, an infinite dimensional vector space), and this of course prevents the stack from being algebraic.
As Matthieu says, one should consider coherent sheaves.
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Insights
July / August 2017
# Epigenetics in Diseases of Aging
Scott Fotheringham, PhD
A vibrant avenue of research and drug development has opened up with the discovery that epigenetic changes affect gene expression, alter cellular development, and can lead to disease. Epigenetics are chemical modifications that affect gene expression, such as cytosine methylation of DNA, and acetylation and methylation of the histone proteins that package DNA. These processes are important to the development of cell types and, while epigenetic modifications do not alter the cell’s underlying nucleotide sequence, the changes can persist for generations.
According to Jean-Pierre Issa, MD, an expert in the field, most diseases of the elderly probably have an epigenetic component. Issa studies myelodysplastic syndrome (MDS), a group of blood diseases that predominantly affect people over 65 in which bone marrow can become almost completely filled with cancerous cells. The cause has not been determined, but exposure to agents that damage DNA—radiation, benzene, long-term exposure to pesticides, and some chemotherapies—are known risk factors.
There are four FDA-approved epigenetic therapies. One class inhibits enzymes that modify histones and includes romidepsin and vorinostat, which are used in the treatment of cutaneous T-cell lymphoma. The other class interferes with the function of DNA methyltransferases (DNMT) and includes decitabine, a hypomethylation tumor therapy approved in 2006 for the treatment of MDS, and azacitidine, a treatment for MDS and other blood cancers.
Transparency Market Research pegs the market value of epigenetic drugs and diagnostic technologies at $5.7 billion by 2018.1 Among companies keen on moving into epigenetic therapies, Merck will invest as much as$515 million on the development and commercialization of these drugs.2
Issa, director of the Fels Institute for Cancer Research and Molecular Biology at Temple University in Philadelphia, outlines the challenges for the development of these novel drugs: “There is no one magic bullet,” he said. “Some epigenetic drugs have broad activity, affecting a thousand or ten thousand genes, while others have a narrow effect on only a hundred. Each drug may not work in the same disease or in the same way as another one.”
Designing clinical trials to test these agents is also a challenge since they don’t work the same way as chemotherapy, which uses toxins to kill as many cancerous cells as possible, and the effects of epigenetic drugs may take a long time to show up.
Cell type and function rely on epigenetics, and changing modifications on DNA or histones can alter cellular identity. While most of the genes affected by DNMT inhibitors are abnormally methylated—and those are the ones you want to target—off-target effects are a concern. While normal cells usually return to their previous epigenetic pattern after they’ve been exposed to epigenetic drugs, there is evidence from animal models that changing epigenetic modifications can lead to new cancers.
These off-target effects might also have benefits, however. “Epigenetic drugs can sensitize patients to immunotherapy or chemotherapy,” Issa said. One thought is that the drugs trigger an inflammatory response by activating endogenous retroviruses that are normally kept inactive by DNA methylation. “Activation of these genes may be one of the ways these drugs work to complement cancer therapies. While this is promising, it might not be a universal property of all epigenetic drugs, which means we have to find out which drugs sensitize cells to a particular therapy. We also have to watch for autoimmunity, which is one of the concerns with chronic hypomethylation.”
There are likely more drugs in use that have epigenetic properties than the four approved by the FDA. At least one traditional Chinese medicine might work because it’s an epigenetic drug. “Arsenic trioxide was discovered serendipitously centuries ago and works well in some forms of leukemia,” Issa said. It is currently available as the FDA-approved Trisenox, manufactured by Teva. “It’s possible that we’ve been using epigenetic therapies for hundreds of years without knowing it.”
Issa’s group has found that cardiac glycosides—sodium-potassium pump inhibitors—have prominent epigenetic activity, and valproic acid, an anti-seizure medication that has been used in children for decades, has weak histone deacetylase inhibitor activity that may be responsible for its efficacy.
As people age, accumulated epigenetic changes accompany the onset of cancers, Alzheimer’s disease, and respiratory conditions. Issa’s team is particularly interested in type 2 diabetes, another age-related disorder. AstraZeneca and MRC Technology are collaborating to find epigenetic candidates for chronic obstructive pulmonary disease and asthma,3 while Oryzon has five drug candidates in clinical trials, including ORY-2001, a potent and highly selective dual LSD1-MAOB inhibitor for the treatment of Alzheimer’s disease, Huntington’s chorea, multiple sclerosis, and some forms of cancer.4,5
“There’s likely some degree of epigenetic deregulation in all aging diseases,” Issa said. “There are still many challenges ahead of us, but epigenetic therapies are a rich and promising area of drug development.”
• 1. MedGadget. “Epigenetics Drugs and Diagnostic Technologies Market to Account for $5.7 Billion by 2018.” 14 July 2016. http://www.medgadget.com/2016/07/epigenetics-drugs-and-diagnostic-technologies-marketto-account-for-us5-7-bn-by-2018.html • 2. Garde, D. “Merck Bets Up to$515M on an Epigenetic Project for Cancer and Blood Disease.” FierceBiotech. 28 January 2016. http://www.fiercebiotech.com/partnering/merck-bets-up-to-515m-onan-epigenetic-project-for-cancer-and-blooddisease?mkt_tok=3RkMMJWWfF9wsRoku63Ac%2B/hmjTEU5z16u0vW6Oxh5l41El3fuXBP2XqjvpVQcFqMr/LRw8FHZNpywVWM8TILtkUt9hsLQziDW0%3D&utm_medium=nl&utm_source=internal
• 3. MRC website. Epigenetic Drugs http://science.mrctechnology.org/drugdiscovery/call-for-targetsrespiratory/
• 4. Oryzon. “Therapeutic Programs: Our Pipeline.” https://www.oryzon.com/en/therapeutic-programs/ourpipeline
• 5. BioNap, Inc. “Oryzon’s Pipeline Taking Shape.” 20 March 2017. https://www.oryzon.com/sites/default/files/Oryzon’s%20Pipeline%20Taking%20Shape%20-%20March%202017_BioNap.pdf
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One thing I’ve wanted to visualize from the hbg-crime.org dataset is what times of day have the most and least crime, in which parts of the city. Using the Gadfly plotting package with Julia makes that easy.
First, pull down the current dataset:
\$ wget http://hbg-crime.org/reports.csv
Then launch Julia, and import all the libraries we’ll be using.
We’ll read the reports into a DataFrame:
Then we need to convert the time of the report into an hour of the day, from 0 (midnight to 1:00 am) to 23 (11:00 pm to midnight):
We’re just creating a quick function that takes a String timestamp, converts it to a DateTime, then extracts the hour; after that, we just vectorize that function and apply it to the “End” column from the data.
The final data we need is just to group those results by Neighborhood and Hour:
The complete_cases! function just strips all of the non-classified data out, as it tends to give Gadfly some problems. Speaking of which, all that’s left is to create the plot and draw it to an SVG file:
The color= attribute tells Gadfly to use the “Neighborhood” column to group different columns.
Crime spikes everywhere after dark and decreases during the day, but unsurprisingly Downtown sees a disproportionate spike around 1:00-2:00 am when the bars let out.
Full source is available on Github.
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fill multicolumn cell with dotfill
I am trying to fill a multicolumn cell with \dotfill. For some reason I do not understand, the dots stop too early. (In normal cells it works fine, but multicolumn cells are reacting differently) This is what I get:
I had a look at the solution mentioned in this thread but with no success. (the @{} didn't make any difference).
Here is my MWE:
\documentclass[11 pt, a4paper, oneside, openany]{book}
\usepackage[table, svgnames]{xcolor}
\usepackage{tabularx}
\usepackage{array}
\begin{document}
\begin{table}[H]
\centering
\begin{tabularx}{\textwidth}{|>{\hsize=1\hsize}X|>{\hsize=1\hsize\arraybackslash}X|}
\hline
$\bigcirc$ Aanvraag offerte & $\bigcirc$ Bestelling\\
\hline
Offertebon nr.: \dotfill & Bestelbon nr.: \dotfill \\
\hline
\multicolumn{2}{|l|}{Firmanaam: \dotfill}\\
\multicolumn{2}{|l|}{Straat + nr.: \dotfill}\\
\multicolumn{2}{|l|}{Postcode + Plaats: \dotfill}\\
\hline
Tel.: \dotfill & Fax.: \dotfill \\
\hline
\multicolumn{2}{|@{}l|}{Naam aanvrager: \dotfill}\\
\hline
\end{tabularx}
\end{table}
\begin{table}
\centering
\begin{tabularx}{\textwidth}{|>{\centering\hsize=0.4\hsize}X|>{\centering\hsize=1\hsize}X|>{\centering\hsize=2\hsize}X|>{\centering\hsize=0.6\hsize\arraybackslash}X|}
\hline
\rowcolor{SteelBlue} Aantal & Referentiecode & Beschrijving wisselstuk & Levertijd\\
\hline
& & &\\
\hline
& & &\\
\hline
& & &\\
\hline
& & &\\
\hline
& & &\\
\hline
& & &\\
\hline
\end{tabularx}
\end{table}
\end{document}
• If I remove [H] from your example document to make it compilable and then compile, I get the following output that elready seems to be the desired result: i.stack.imgur.com/PMqGw.png Oct 12 '19 at 20:09
• No problem either with this code. Unrelated: if both X have the same width, >{\hsize=1\hsize} is totally unnecessary. Oct 12 '19 at 20:19
• Oh, strange. I got that output with that MWE yesterday. But it seems the compiling files were not up to date and still got older code. I did the update of the MWE and retried. Now I do get the unwanted behaviour... It turns out that the error comes from the simultanously loading of colortbl (via xcolor) and tabularx. Any idea how to fix this? Oct 13 '19 at 8:48
• @Philippe Yes, colortbl is the culprit. Oct 13 '19 at 9:02
In order to get the desired output you could either go without colortbl or, if you need this package for a different table in your document, try th following workaround:
\documentclass[11 pt, a4paper, oneside, openany]{book}
\usepackage[table, svgnames]{xcolor}
\usepackage{tabularx}
\usepackage{array}
\usepackage{calc}
\newlength{\mylength}
\setlength{\mylength}{\hsize-2\tabcolsep-4\arrayrulewidth}
\begin{document}
\begin{table}
\centering
\begin{tabularx}{\textwidth}{|X|X|}
\hline
$\bigcirc$ Aanvraag offerte & $\bigcirc$ Bestelling\\
\hline
Offertebon nr.: \dotfill & Bestelbon nr.: \dotfill \\
\hline
\multicolumn{2}{|>{\hsize=\mylength}X|}{Firmanaam: \dotfill}\\
\multicolumn{2}{|>{\hsize=\mylength}X|}{Straat + nr.: \dotfill}\\
\multicolumn{2}{|>{\hsize=\mylength}X|}{Postcode + Plaats: \dotfill}\\
\hline
Tel.: \dotfill & Fax.: \dotfill \\
\hline
\multicolumn{2}{|>{\hsize=\mylength}X|}{Naam aanvrager: \dotfill}\\
\hline
\end{tabularx}
\end{table}
\end{document}
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User marting - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:27:21Z http://mathoverflow.net/feeds/user/4716 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition/107492#107492 Answer by MartinG for Families of ideal sheaves: What's the correct definition? MartinG 2012-09-18T17:04:12Z 2012-09-19T07:49:13Z <p>[This is now an answer to the edited question(s), with some details added. My answer to the original question is kept at the very end.]</p> <p>Firstly: The question is a good one, and it is not easy to find references on this. I had spent too much time pondering about the failure of the double dual argument (see below) before I finally heard the arguement given in the last section below, indirectly from Fantechi, via Faber.</p> <p>Assume $X$ is smooth projective.</p> <p><strong>Definition</strong>: An $S$-valued point in $M_I(X)$ is an $S$-flat coherent sheaf on $S\times X$, with stable fibres of rank one, and with determinant line bundle isomorphic to $\mathcal{O}_{S\times X}$, modulo isomorphism.</p> <p>(I do not know if this is what Bridgeland meant, but to me this is resonably standard.)</p> <p><strong>Comment</strong>: Stability for rank one means torsion free.</p> <p><strong>Existence</strong>: Let $M(X)$ be the (Simpson) moduli space for stable rank one sheaves. Then $M_I(X)$ is the fibre over $\mathcal{O}_X$ for the determinant map $M(X) \to \mathrm{Pic}(X)$. This map sends a sheaf $I$ (stable rank one fibres) on $S\times X$ to the determinant line bundle $\det(I)$, and it is trivial as a point in $\mathrm{Pic}(X)$ if it is of the form $p^*L$ with $L\in\mathrm{Pic}(S)$. Then $I\otimes p^*L^{-1}$ is equivalent to $I$ in $M(X)(S)$, and it has trivial determinant. This shows that $M_I(X)$ indeed is a fibre of the determinant map.</p> <p>Of course the determinant of an ideal $I_Y\subset \mathcal{O}_X$ is nontrivial if $Y$ is a non principal divisor, so you cannot map such ideals to $M_I(X)$. In any case, the ideal of a divisor, without the embedding, would only remember the linear equivalence class.</p> <p>For brevity, let $\mathrm{Hilb}(X)$ be the part of the Hilbert scheme parametrizing subschemes $Y\subset X$ of codimension at least $2$. Then there is a natural map $F: \mathrm{Hilb}(X) \to M(X)$ that sends an ideal $I_Y\subset\mathcal{O}_{S\times X}$ to $I_Y$, forgetting the embedding. Since $Y$ is flat, so is $I_Y$, and its fibres are torsion free (by flatness again) of rank one. By the codimension assumption, the determinant of $I$ is trivial.</p> <p><strong>Theorem</strong>: $F$ is an isomorphism.</p> <p><strong>Comment</strong>: In the literature one sometimes finds the argument that if $I$ is a rank one torsion free sheaf with trivial determinant, then $I$ embeds into its double dual, which coincides with its determinant $\mathcal{O}_X$. This establishes bijectivity on points. (For Hilbert schemes of points on surfaces this is enough to conclude, since you can check independently that both $\mathrm{Hilb}(X)$ and $M_I(X)$ are smooth, and that the induced map on tangent spaces is an isomorphism.) I do not know how to make sense of this argument in families.</p> <p><strong>Sketch proof of theorem</strong>: The essential point is to show that every $I$ in $M_I(X)(S)$ has a canonical embedding into $\mathcal{O}_{S\times X}$ such that the quotient is $S$-flat.</p> <p>Let $U\subset S\times X$ be the open subset where $I$ is locally free. Its complement has codimension at least $2$ in all fibres. By the trivial determinant assumption, the restriction of $I$ to $U$ is trivial. By codimension $2$, the trivialization extends to a map $I\to \mathcal{O}_{S\times X}$. This map is injective, in fact injective in all fibres: The restriction to each fibre ${s}\times X$ is nonzero (as $U$ intersects all fibres) and hence an embedding ($I$ is torsion free in fibres). It follows that the quotient is flat. There are some details to check, but this is the main point, I think.</p> <p>[End of new answer, here is the original one:]</p> <p>If we attempt to define $M_I(X)(S)$ as the set of $S$-flat ideals $I_Z$ in $\mathcal{O}_{S\times X}$, then that would not be functorial in $S$, as the inclusion $I_Z \subset \mathcal{O}_{S\times X}$ may not continue to be injective after base change (in the counter example in the other answer, restriction to the problematic fibre gives the zero map). We could impose "universal injectivity", but that is just another way of requiring the quotient $\mathcal{O}_Z$ to be $S$-flat, so then we have (re)defined the Hilbert scheme. </p> <p>Another common way of defining moduli of ideals is as the moduli space for rank one stable sheaves (i.e. torsion free) with trivial determinant line bundle. The resulting moduli space is isomorphic to the Hilbert scheme of subschemes of codimension at least 2.</p> http://mathoverflow.net/questions/60757/extending-a-polynomial-function-from-an-open-subset/60773#60773 Answer by MartinG for Extending a polynomial function from an open subset MartinG 2011-04-06T06:30:31Z 2011-04-06T06:30:31Z <p>I was in fact pondering the same question just yesterday, and ended up with this: there is a thickening $Z'\subset X$ of $\pi^{-1}Z$ such that every regular function vanishing on $Z'$ comes from $Y$.</p> <p>Proof is straight forward, but anyway: in terms of rings, let $\phi\colon A \subset B$ be a finite extension of rings, with $B$ generated by finitely many $x_i$ as an $A$-module, let $I\subset A$ be an ideal and suppose $A_f \cong B_f$ for all $f\in I$. Let $f\in I$. For each $i$ we can then write $\phi(a_i) = x_i \phi(f)^{n_i}$ for some $a_i\in A$. Choose $n > n_i$ for all $i$ and let $I'\subset A$ be generated by all $f^n$ for $f$ running through a generator set of $I$ (with $n$ varying with $f$). Now $I'B$ defines $Z'$: its elements are $A$-linear combinations of $\phi(f^n)x_i =\phi(f^{n-n_i})\phi(a_i)$, which is in $I$.</p> <p>(Still interested if there is a better statement. Also I was still looking for an example where the thickening was needed, so thanks, Tom!)</p> http://mathoverflow.net/questions/57325/when-do-stability-and-semistability-coincide/57357#57357 Answer by MartinG for When do stability and semistability coincide? MartinG 2011-03-04T14:13:40Z 2011-03-04T14:13:40Z <p>If the rank and degree (wrt $H$) are coprime, then $\mu$-stability and $\mu$-semistability coincide. The argument is the same as for Gieseker-stability, but simpler.</p> http://mathoverflow.net/questions/35029/endomorphisms-of-bundles-associated-to-codimension-2-subvarieties/35100#35100 Answer by MartinG for Endomorphisms of bundles associated to codimension 2 subvarieties MartinG 2010-08-10T11:23:19Z 2010-08-10T11:23:19Z <p>This might be obvious, but one can sometimes argue via stability:</p> <p>Notation as in Sasha's answer, with $L=\mathcal{O}(-n)$ and $n>0$. Then if you know that no curve of degree $\le n/2$ contains $Z$, then $E$ is stable (a destabilizing line bundle $\mathcal{O}(-m)\subset E$ has $-m \ge -n/2$).</p> <p>And then of course $\Gamma(\mathbb{P}^2, \mathrm{End}(E))=k$.</p> http://mathoverflow.net/questions/34621/what-in-the-meaning-of-schematic-support/34639#34639 Answer by MartinG for What in the meaning of "schematic support"? MartinG 2010-08-05T15:06:13Z 2010-08-05T15:06:13Z <p>Assume $\mathcal{F}$ is coherent on $X$. To complement the answers you have already got, there are two useful definitions of its schematic support: either use the annihilator ideal or the Fitting ideal. They have the same underlying reduced scheme, but in general different scheme structures. The annihilator is usually (always, I think) meant if nothing else is said.</p> <p>The annihilator support $Z = V(\mathrm{Ann}(\mathcal{F}))$ can be viewed as the minimal closed subscheme $i\colon Z \subset X$ such that the natural map <code>$\mathcal{F} \to i_*i^*\mathcal{F}$</code> is an isomorphism (more or less a tautology).</p> <p>The Fitting ideal (locally the maximal minors of a free presentation) gives a subscheme that contains $V(\mathrm{Ann}(\mathcal{F}))$, but with a possibly thicker scheme structure. A feature is that it is compatible with pullback; the annihilator construction is not.</p> <p>Example: a rank $r$ vector bundle $i_*E$ on a divisor $i\colon D\subset X$ has $D$ as annihilator support, and $rD$ as Fitting support.</p> http://mathoverflow.net/questions/25365/characterisation-of-coherent-sheaves-on-an-algebraic-variety/25401#25401 Answer by MartinG for Characterisation of coherent sheaves on an algebraic variety MartinG 2010-05-20T18:19:05Z 2010-05-20T18:19:05Z <p>This is false. Every sheaf in that class would have zero first Chern class, since $c_1$ is additive over short exact sequences.</p> http://mathoverflow.net/questions/23627/group-and-hopf-algebra-structures-for-projective-varieties/23689#23689 Answer by MartinG for Group and Hopf Algebra Structures for Projective Varieties MartinG 2010-05-06T11:17:27Z 2010-05-06T11:17:27Z <p>In the projective case, i.e. abelian varieties, this question is addressed in Mumford's <em>On the equations defining abelian varieties. I</em>, Invent. math. 1, 287--354, 1966. In §3, "the addition formula", the aim is to describe the group law, given a projective embedding, explicitly.</p> <p>The starting point is exactly that the homogeneous coordinate ring is not a Hopf algebra, because a line bundle $L$ (defining the embedding) on $X$ does not pull back to $p_1^*L\otimes p_2^*L$ under the group law $X\times X\to X$. (This is what one would need to have induced comultiplication maps $H^0(X,L^n) \to H^0(X\times X, p_1^*L^n\otimes p_2^*L^n)\cong H^0(X, L^n)\otimes H^0(X, L^n)$.)</p> <p>Instead Mumford finds that one <em>can</em> describe $X\times X\to X\times X$, $(x,y)\mapsto (x+y,x-y)$ explicitly in terms of the homogeneous coordinate ring.</p> http://mathoverflow.net/questions/22039/are-any-two-k3-surfaces-over-c-diffeomorphic/22045#22045 Answer by MartinG for Are any two K3 surfaces over C diffeomorphic? MartinG 2010-04-21T12:06:30Z 2010-04-21T12:06:30Z <p>I find this reference quite readable:</p> <p>Le Potier: "Simple connexité des surfaces K3", in Asterisque 126, 1985.</p> <p>I haven't read Kodaira's paper (in Joel's answer), so I don't know whether it is the same argument, but Le Potier also deforms to quartic surfaces.</p> http://mathoverflow.net/questions/20288/reference-request-is-the-punctual-hilbert-scheme-irreducible/20707#20707 Answer by MartinG for Reference request: is the punctual Hilbert scheme irreducible? MartinG 2010-04-08T09:56:19Z 2010-04-08T09:56:19Z <p>(I cannot post comments, regard this as one.)</p> <p>It was correctly suggested in the comments to damiano's answer that the case of dimension d=2 was settled in the 70s, here are references:</p> <ul> <li>I believe the first proof is that of Briancon [<em>Description de <code>$Hilb^n(C\{x,y\})$</code></em>, Invent. Math. 41 (1977)]</li> <li>Ellingsrud and Strømme [<em>On the homology of the Hilbert scheme of points in the plane</em>, Invent. Math. 87 (1987)] have an argument involving a cell decomposition, where there is only one cell of the expected dimension (by a separate argument, any extra component cannot have smaller dimension)</li> <li>Perhaps the state of the art is the inductive argument by Ellingsrud and Lehn [<em>Irreducibility of the punctual quotient scheme of a surface</em>, Ark. Mat. 37 (1999)], building on previous work by Ellingsrud and Strømme.</li> </ul> http://mathoverflow.net/questions/18928/union-of-closed-subschemes-with-the-structure-sheaf-over-it/19009#19009 Answer by MartinG for Union of closed subschemes with the structure sheaf over it MartinG 2010-03-22T13:18:53Z 2010-03-22T13:18:53Z <p>Emerton explained well that $V(I\cap J)$ is the natural scheme structure; I'd just like to add that if $V(I)$ and $V(J)$ are divisors, it is also reasonable to use the scheme structure $V(IJ)$ on their union, i.e. their sum as divisors. So both versions have their merits.</p> http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition Comment by MartinG MartinG 2012-09-19T12:34:57Z 2012-09-19T12:34:57Z I just had a look at the paper: Maybe this is also stated more explicitly elsewhere, but I think at least the proof of Lemma 6.3 reveals that an ideal sheaf means a torsion free sheaf for which there exists a nonzero map to $\mathcal{O}_X$, that for families, this condition is just imposed fibrewise (as Sasha suggested) and moreover that this is a closed condition. Anyway, I am leaving my answer as it is, covering the smooth case. (And I still have no idea whether the natural map could fail to be an iso in the singular case.) http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition/107492#107492 Comment by MartinG MartinG 2012-09-19T10:08:19Z 2012-09-19T10:08:19Z @Sasha: You are right. I did point out that I didn't know about the singular case (the OP seemed interested in the smooth case also), but it is worthwhile to note that the very definition of $M_I(X)$ via determinants applies only in the smooth case. Thanks. http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition/107529#107529 Comment by MartinG MartinG 2012-09-19T08:34:44Z 2012-09-19T08:34:44Z @Sasha: This viewpoint makes sense, except that for the embedding of $I$ into $\mathcal{O}_X$ to be unique, you need to assume codimension at least 2. This is what 36min complained about with maximal ideals in $\mathbb{Z}$. So we have to restrict to codimension at least 2 to get a bijection. (What I wrote in my answer, then, is basically that this $M_I(X)$ can be defined without reference to ideals, and that the morphism from the Hilbert scheme is in fact an isomorphism, although that is not entirely obvious and probably independent from what Bridgeland is doing, as you say.) http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition/107492#107492 Comment by MartinG MartinG 2012-09-19T06:42:32Z 2012-09-19T06:42:32Z @36min: And to answer your new question 2: With this definition, you can realize M_I as the fibre over $\mathcal{O}_X$ for the determinant map $M\to \mathrm{Pic}(X)$, where $M$ is the Simpson moduli space for stable/torsion free rank one sheaves. http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition/107492#107492 Comment by MartinG MartinG 2012-09-19T06:38:26Z 2012-09-19T06:38:26Z @36min: Yes, I meant to say that moduli of rank one sheaves with trivial determinant is more or less the standard definition. But, on autopilot, I assumed $X$ to be smooth, I guess Bridgeland does not. If singularitites are allowed, I do not know whether the natural map from the Hilbert scheme is an isomorphism. (And even for $X$ smooth it is not entirely obvious.) http://mathoverflow.net/questions/107442/families-of-ideal-sheaves-whats-the-correct-definition/107445#107445 Comment by MartinG MartinG 2012-09-18T18:42:09Z 2012-09-18T18:42:09Z @temp: No, $M_I(X)$ does not exist: it is not a functor. http://mathoverflow.net/questions/97902/has-the-cotangent-complex-been-used-in-context-other-than-morphism-of-schemes/97930#97930 Comment by MartinG MartinG 2012-05-25T10:57:03Z 2012-05-25T10:57:03Z The cotangent complex has nothing to with the cotangent function or complex numbers.. http://mathoverflow.net/questions/63324/singular-locus-of-the-punctual-hilbert-scheme/63713#63713 Comment by MartinG MartinG 2011-05-02T20:42:08Z 2011-05-02T20:42:08Z Indeed I have.. I deleted the answer referred to. http://mathoverflow.net/questions/35811/square-of-an-elliptic-curve-and-projective-plane/35818#35818 Comment by MartinG MartinG 2010-08-17T08:09:31Z 2010-08-17T08:09:31Z No genericity condition is needed: any line intersects the cubic in an effective degree 3 divisor summing to zero, and conversely. So the result is isomorphic to the projective plane (it is also the linear system |0+0+0|, where 0 means the group identity). http://mathoverflow.net/questions/33658/does-open-imply-smooth Comment by MartinG MartinG 2010-07-29T09:22:10Z 2010-07-29T09:22:10Z A class of counter examples: flat implies open, but smoothness is stronger than flatness. In particular one can take any ramified, flat and finite morphism, such as the projection of a plane curve to a line (project away from a point not on the curve). http://mathoverflow.net/questions/25365/characterisation-of-coherent-sheaves-on-an-algebraic-variety Comment by MartinG MartinG 2010-05-20T18:21:43Z 2010-05-20T18:21:43Z It isn't silly, after all it was claimed at Wikipedia, and you were skeptical.. http://mathoverflow.net/questions/25365/characterisation-of-coherent-sheaves-on-an-algebraic-variety Comment by MartinG MartinG 2010-05-20T14:25:59Z 2010-05-20T14:25:59Z Wouldn't the first Chern class of every sheaf in that class be zero, since $c_1$ is additive over short exact sequences?
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# Question c7f67
Mar 12, 2015
You have approximately $6 \cdot {10}^{5} \text{kg}$ of mercury in the lake.
Now, before getting started, notice that you have a wide array of units that you must work with, so I suggest deciding what units would be most useful to convert to.
Since density was given to you in micrograms per mL, I'll calculate the volume of the lake in mL by going from cubic miles to cubic meters, and finally to mL.
The volume of the lake can be caulculated by multiplying the surface with the average depth. Go from feet to miles first
$\text{20 ft" * "0.0001893939 miles"/"1 ft" = "0.0037878 miles}$
The volume in cubic miles will be
$V = \text{area" * "average depth" = "100 mi"^(2) * "0.0037878 mi}$
$V = {\text{0.37878 mi}}^{3}$
Now use two conversion factors to get to mL
$\text{0.37878 mi"^(3) * (4.16818183 * 10^(9)"m"^(3))/"1 mi"^(3) * (10^(6)"mL")/("1 m"^(3)) = 1.5788 * 10^(15)"mL}$
The mass of mercury in micrograms will be
rho = m/V => m = rho * V = (0.4mu"g")/"mL" * 1.5788 * 10^(15)"mL"
$m = 6.3152 \cdot {10}^{14} \mu \text{g}$
Now covert to kilograms to get the final answer
$6.3152 \cdot {10}^{14} \mu \text{g" * "1 kg"/(10^(9)mu"g") = 6.3152 * 10^(5)"kg}$
Rounded to one sig fig, the number of sig figs in 100 square miles and in 20 ft, the answer is
m_("mercury") = 6 * 10^(5) "kg"#
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Harry is now in his 6th year at Hogwarts. He has been made the captain of the Gryffindor Quidditch team by Headmaster Dumbledore. Its time for Quidditch practice. So all players must be divided into two teams for practice sessions. Now the problem that he faces before selecting the team is that some players don't feel comfortable playing on the same team as some other players. Harry knows about these restrictions that he has while forming the teams. He wonders if he can make two teams taking into account these restrictions and further how many different teams he can make. But since Harry has detention from Professor Snape, he leaves the task to you.
Calculate the total number of different possible team distributions for Harry. Output the answer modulo (10^9)+7.
Note
The two teams must not necessarily have equal number of players i.e. there can be 3 players on one team and only 1 player on other team. But there has to be atleast one player in each team.
Input
The first line of each test file contains a single integer T. Then T test cases follow. Each line contains two integers N - the number of players and M - the number of different restrictions that Harry must take care of. The next M line each contain two integers A and B, which indicate that player A would not like to play with player B and vice versa.
Output
For each test case output the number of different possible team formations modulo (10^9)+7. If no such formation is possible, output "NOT POSSIBLE" (quotes for clarity only). Each output must be followed by a newline.
Constraints
1<= T <=100
2<= N <= 10^4
0<= M <= 10^5
0<= A,B <=N-1
Problem Setter: Jayam Modi
SAMPLE INPUT
2
10 5
0 4
1 2
1 3
4 8
2 5
10 5
1 2
1 3
2 3
4 7
6 8
SAMPLE OUTPUT
16
NOT POSSIBLE
Explanation
In the second test case, 1 and 2 dont like to play together. Similarly 2 and 3 dont like to play together. So now 1 and 2+3 should be in different teams. But now if 2 and 3 also dont like to play together then there is no possilble team formation.
Time Limit: 1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, C++14, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Julia, Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Swift, Visual Basic, Kotlin
## CODE EDITOR
Initializing Code Editor...
|
## Chemistry 12th Edition
We know that entropy is a measure of the energy of a system, and its SI unit is $\frac{J}{K\space mol}.$
|
Math Help - What program can you use LaTeX in?
1. What program can you use LaTeX in?
What programs do you use to write mathematical papers in LaTeX? I can't seem to find anything useful.
2. Re: What program can you use LaTeX in?
Try TeXworks.
3. Re: What program can you use LaTeX in?
I use Emacs with AUCTeX package. The package shows typeset formulas right in the editor.
4. Re: What program can you use LaTeX in?
Texworks worked wonderfully for my purposes.
5. Re: What program can you use LaTeX in?
You could also try TeXeR on artofproblemsolving.com
|
Next: Applications to LJ and Up: PROPERTIES OF REARRANGEMENT PATHWAYS Previous: Localisation Contents
# Cooperativity
atoms can participate in a rearrangement according to a continuous range of cooperativity. At one end of the scale there are rearrangements where atoms all move simultaneously [see Figure 3.2 (a)]. Although these paths exhibit the highest degree of correlated atomic motion they do not usually pose a problem for double-ended transition state search algorithms [7,30]. Linear interpolation between the two minima tends to generate initial guesses that lie close to the true pathway, particularly if . At the opposite extreme, atoms can move almost one at a time, following a domino' pattern [see Figure 3.2 (b)]. Locating a transition state for such rearrangements may require a better initial guess, since linear interpolation effectively assumes that all the coordinates change at the same rate.
The degree of correlation in the atomic displacements can be quantified by considering the displacement overlap'
(5.11)
where the index indicates that was calculated for atoms numbered , ,...,. is a -dimensional vector that represents a particular choice of atoms from , and hence there are possible values of . The index can be thought of as a measure of how the displacements of the atoms , , etc. overlap along the pathway. For example, if two atoms move at different times then is small for this pair because the minimum displacement in Equation 3.11 is always small. However, if both atoms move in the same region of the path then is larger.
We now explain how the statistics of the overlaps, , can be used to extract a measure of cooperativity (Figure 3.3). Suppose that atoms move simultaneously in a hypothetical rearrangement. Then all the overlaps for will be relatively small, because one or more atoms are included in the calculation whose motion is uncorrelated with the others. For overlaps with the set of for all possible choices of atoms from will exhibit some large values and some small. The large values occur when all the chosen atoms are members of the set that move cooperatively, while other choices give small values of . Hence the kurtosis of the set , , calculated from moments taken about the origin, will be large for , and small for .
To obtain a measure of how many atoms move cooperatively we could therefore calculate , , etc. and look for the value of where falls in magnitude. However, to avoid an arbitrary cut-off, it is better to calculate the kurtosis of the set , , or for short. There are members of this set, and by analogy with the definition of , we could define a cooperativity index . Then, if is large, and all the other are small, we obtain and , correctly reflecting the number of atoms that move together.
In practice, there are several problems with the above definition of . Calculating in this way quickly becomes costly as the number of atoms and/or number of frames in the pathway increases, because the number of elements in the set varies combinatorially with . Secondly, as approaches the distribution of all the possible values for becomes more and more uniform. Under these circumstances deviations from the mean that are negligible in comparison with the overall displacement can produce large kurtosises. Instead, we suggest a modified (and simpler) definition of , which better satisfies our objectives.
We first define the overlap of atomic displacements in a different manner. It can be seen from Equation 3.11 that the simultaneous displacement of atoms is included in each set of overlaps with . For example, if three atoms move cooperatively then both the and sets will include large elements corresponding to these contributions. Another redundancy is present within , since values in this set are calculated for all possible subsets of atoms and the displacement of each atom is therefore considered more than once. However, we can avoid this redundancy by defining a single overlap, rather than dealing with different values.
Recall that is the displacement of atom between frames and . The ordering of the atoms is arbitrary but remains the same for each frame number . We now define as the displacement of atom in frame , where index numbers the atoms in frame in descending order, according to the magnitude of , e.g. atom in frame is now the atom with the maximum displacement between frames and , atom has the second largest displacement etc. As the ordering may vary from frame to frame, the atoms labelled in different frames can now be different. This relabelling greatly simplifies the notation we are about to introduce. Consider the -overlap defined as
(5.12)
where ranges from to , and is defined to be zero for all . A schematic illustration of this construct is presented in Figure 3.4. For example, if only two atoms move in the course of the rearrangement, and both are displaced by the same amount (which may vary from frame to frame), the only non-zero overlap will be .
We can now define an index to quantify the number of atoms that move cooperatively as
(5.13)
If only one atom moves during the rearrangement then , while if atoms displace cooperatively during the rearrangement then . This definition is independent of the total displacement, the integrated path length, and the number of atoms, which makes it possible to compare indices calculated for different systems.
Next: Applications to LJ and Up: PROPERTIES OF REARRANGEMENT PATHWAYS Previous: Localisation Contents
Semen A Trygubenko 2006-04-10
|
Calculate tree similarity and distance measures based on the amount of phylogenetic or clustering information that two trees hold in common, as proposed in Smith (2020).
TreeDistance(tree1, tree2 = tree1)
SharedPhylogeneticInfo(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE,
diag = TRUE
)
DifferentPhylogeneticInfo(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE
)
PhylogeneticInfoDistance(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE
)
ClusteringInfoDistance(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE
)
ExpectedVariation(tree1, tree2, samples = 10000)
MutualClusteringInfo(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE,
diag = TRUE
)
SharedPhylogeneticInfoSplits(
splits1,
splits2,
nTip = attr(splits1, "nTip"),
reportMatching = FALSE
)
MutualClusteringInfoSplits(
splits1,
splits2,
nTip = attr(splits1, "nTip"),
reportMatching = FALSE
)
MatchingSplitInfo(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE,
diag = TRUE
)
MatchingSplitInfoDistance(
tree1,
tree2 = NULL,
normalize = FALSE,
reportMatching = FALSE
)
MatchingSplitInfoSplits(
splits1,
splits2,
nTip = attr(splits1, "nTip"),
reportMatching = FALSE
)
## Arguments
tree1, tree2
Trees of class phylo, with leaves labelled identically, or lists of such trees to undergo pairwise comparison. Where implemented, tree2 = NULL will compute distances between each pair of trees in the list tree1 using a fast algorithm based on Day (1985).
normalize
If a numeric value is provided, this will be used as a maximum value against which to rescale results. If TRUE, results will be rescaled against a maximum value calculated from the specified tree sizes and topology, as specified in the 'Normalization' section below. If FALSE, results will not be rescaled.
reportMatching
Logical specifying whether to return the clade matchings as an attribute of the score.
diag
Logical specifying whether to return similarities along the diagonal, i.e. of each tree with itself. Applies only if tree2 is a list identical to tree1, or NULL.
samples
Integer specifying how many samplings to obtain; accuracy of estimate increases with sqrt(samples).
splits1, splits2
Logical matrices where each row corresponds to a leaf, either listed in the same order or bearing identical names (in any sequence), and each column corresponds to a split, such that each leaf is identified as a member of the ingroup (TRUE) or outgroup (FALSE) of the respective split.
nTip
(Optional) Integer specifying the number of leaves in each split.
## Value
If reportMatching = FALSE, the functions return a numeric vector specifying the requested similarities or differences.
If reportMatching = TRUE, the functions additionally return an integer vector listing the index of the split in tree2 that is matched with each split in tree1 in the optimal matching. Unmatched splits are denoted NA. Use VisualizeMatching() to plot the optimal matching.
## Details
Generalized Robinson–Foulds distances calculate tree similarity by finding an optimal matching that the similarity between a split on one tree and its pair on a second, considering all possible ways to pair splits between trees (including leaving a split unpaired).
The methods implemented here use the concepts of entropy and information (MacKay 2003) to assign a similarity score between each pair of splits.
The returned tree similarity measures state the amount of information, in bits, that the splits in two trees hold in common when they are optimally matched, following Smith (2020) . The complementary tree distance measures state how much information is different in the splits of two trees, under an optimal matching.
## Concepts of information
The phylogenetic (Shannon) information content and entropy of a split are defined in a separate vignette.
Using the mutual (clustering) information (Meila 2007; Vinh et al. 2010) of two splits to quantify their similarity gives rise to the Mutual Clustering Information measure (MutualClusteringInfo(), MutualClusteringInfoSplits()); the entropy distance gives the Clustering Information Distance (ClusteringInfoDistance()). This approach is optimal in many regards, and is implemented with normalization in the convenience function TreeDistance().
Using the amount of phylogenetic information common to two splits to measure their similarity gives rise to the Shared Phylogenetic Information similarity measure (SharedPhylogeneticInfo(), SharedPhylogeneticInfoSplits()). The amount of information distinct to each of a pair of splits provides the complementary Different Phylogenetic Information distance metric (DifferentPhylogeneticInfo()).
The Matching Split Information measure (MatchingSplitInfo(), MatchingSplitInfoSplits()) defines the similarity between a pair of splits as the phylogenetic information content of the most informative split that is consistent with both input splits; MatchingSplitInfoDistance() is the corresponding measure of tree difference. (More information here.)
### Conversion to distances
To convert similarity measures to distances, it is necessary to subtract the similarity score from a maximum value. In order to generate distance metrics, these functions subtract the similarity twice from the total information content (SPI, MSI) or entropy (MCI) of all the splits in both trees (Smith 2020) .
### Normalization
If normalize = TRUE, then results will be rescaled such that distance ranges from zero to (in principle) one. The maximum distance is the sum of the information content or entropy of each split in each tree; the maximum similarity is half this value. (See Vinh et al. (2010, table 3) and Smith (2020) for alternative normalization possibilities.)
Note that a distance value of one (= similarity of zero) will seldom be achieved, as even the most different trees exhibit some similarity. It may thus be helpful to rescale the normalized value such that the expected distance between a random pair of trees equals one. This can be calculated with ExpectedVariation(); or see package 'TreeDistData' for a compilation of expected values under different metrics for trees with up to 200 leaves.
Alternatively, to scale against the information content or entropy of all splits in the most or least informative tree, use normalize = pmax or pmin respectively. To calculate the relative similarity against a reference tree that is known to be 'correct', use normalize = SplitwiseInfo(trueTree) (SPI, MSI) or ClusteringEntropy(trueTree) (MCI).
## Troubleshooting
Trees being compared must have identical tips. (If you have a use case for comparing trees with non-identical tips, do file a GitHub issue or drop the package maintainer an e-mail.)
To determine which tips do not occur in both trees, try:
library('TreeTools')
setdiff(TipLabels(tree1), TipLabels(tree2)) # In tree1 but not tree2
setdiff(TipLabels(tree2), TipLabels(tree1)) # In tree2 but not tree1
## References
MacKay DJC (2003). Information Theory, Inference, and Learning Algorithms. Cambridge University Press, Cambridge. https://www.inference.org.uk/itprnn/book.pdf.
Meila M (2007). “Comparing clusterings---an information based distance.” Journal of Multivariate Analysis, 98(5), 873--895. doi:10.1016/j.jmva.2006.11.013 .
Smith MR (2020). “Information theoretic Generalized Robinson-Foulds metrics for comparing phylogenetic trees.” Bioinformatics, 36(20), 5007--5013. doi:10.1093/bioinformatics/btaa614 .
Vinh NX, Epps J, Bailey J (2010). “Information theoretic measures for clusterings comparison: variants, properties, normalization and correction for chance.” Journal of Machine Learning Research, 11, 2837--2854. doi:10.1145/1553374.1553511 .
Other tree distances: JaccardRobinsonFoulds(), KendallColijn(), MASTSize(), MatchingSplitDistance(), NNIDist(), NyeSimilarity(), PathDist(), Robinson-Foulds, SPRDist()
## Examples
tree1 <- ape::read.tree(text='((((a, b), c), d), (e, (f, (g, h))));')
tree2 <- ape::read.tree(text='(((a, b), (c, d)), ((e, f), (g, h)));')
tree3 <- ape::read.tree(text='((((h, b), c), d), (e, (f, (g, a))));')
# Best possible score is obtained by matching a tree with itself
DifferentPhylogeneticInfo(tree1, tree1) # 0, by definition
#> [1] 0
SharedPhylogeneticInfo(tree1, tree1)
#> [1] 22.53747
SplitwiseInfo(tree1) # Maximum shared phylogenetic information
#> [1] 22.53747
# Best possible score is a function of tree shape; the splits within
# balanced trees are more independent and thus contain less information
SplitwiseInfo(tree2)
#> [1] 19.36755
# How similar are two trees?
SharedPhylogeneticInfo(tree1, tree2) # Amount of phylogenetic information in common
#> [1] 13.75284
attr(SharedPhylogeneticInfo(tree1, tree2, reportMatching = TRUE), 'matching')
#> [1] 1 4 2 3 5
VisualizeMatching(SharedPhylogeneticInfo, tree1, tree2) # Which clades are matched?
DifferentPhylogeneticInfo(tree1, tree2) # Distance measure
#> [1] 14.39934
DifferentPhylogeneticInfo(tree2, tree1) # The metric is symmetric
#> [1] 14.39934
# Are they more similar than two trees of this shape would be by chance?
ExpectedVariation(tree1, tree2, sample=12)['DifferentPhylogeneticInfo', 'Estimate']
#> [1] 31.7014
# Every split in tree1 conflicts with every split in tree3
# Pairs of conflicting splits contain clustering, but not phylogenetic,
# information
SharedPhylogeneticInfo(tree1, tree3) # = 0
#> [1] 0
MutualClusteringInfo(tree1, tree3) # > 0
#> [1] 0.6539805
# Converting trees to Splits objects can speed up multiple comparisons
splits1 <- TreeTools::as.Splits(tree1)
splits2 <- TreeTools::as.Splits(tree2)
SharedPhylogeneticInfoSplits(splits1, splits2)
#> [1] 13.75284
MatchingSplitInfoSplits(splits1, splits2)
#> [1] 17.09254
MutualClusteringInfoSplits(splits1, splits2)
#> [1] 3.031424
|
# Quark Matter 2012
12-18 August 2012
US/Eastern timezone
## Baryonic resonances at the LHC energies with the ALICE experiment
16 Aug 2012, 16:00
2h
### Speaker
Enrico Fragiacomo (Universita e INFN (IT))
### Description
The study of resonances production in p-p collisions provides constraints on QCD-inspired particle production models. In Pb-Pb collisions, resonances are good probes to estimate the collective properties of the fireball and may add constraints to the estimate of its lifetime. $p_T$ spectra have been measured for the baryonic resonances $\Lambdastar$, $\Sigmastar$ and $\Xistar$ using data collected by the ALICE experiment in p-p collisions at $\sqrt{s} = 7$ TeV. The spectra will be compared to QCD-inspired models such as PYTHIA and PHOJET, which in general underpredict the experimental results on the yields of strange resonances. The ratios of yields of baryonic resonances to stable particles, namely $\Sigmastar / \Lambda$, $\Lambdastar / \Lambda$ and $\Xistar / \Xi$ will be compared with both thermal models and corresponding values from previous experiments at different colliding energies. These results will serve as baseline for the forthcoming heavy-ion results. The status and prospects of the measurements of baryonic resonances in Pb-Pb collisions at $\sqrt{s_{NN}} = 2.76$ TeV will be discussed.
### Primary author
Collaboration ALICE (CERN, Geneva, Switzerland)
### Co-author
Enrico Fragiacomo (Universita e INFN (IT))
Slides
|
# How to have Run-in Section in a separate command?
In the question asked here, How to have the same solution but in a different command without affecting the \section command? I mean I would like to create another command e.g. \runinsection that would make the text continue beside the section header while maintaining the \section command with its normal behavior.
• Although this is your second question, nobody has welcomed you here yet... So, Welcome to TeX.SX! – karlkoeller Dec 7 '14 at 19:54
• I'm not sure to understand what can be the use for this. – egreg Dec 7 '14 at 22:21
• In my case it was a one line section that caused me to go into a new page having also only one line, it's a 3 page document, the document did not look nice – Mystic Odin Dec 10 '14 at 0:34
You can simply define a new command \runinsection
\makeatletter
\newcommand\runinsection{\@startsection {section}{1}{\z@}%
{3.5ex \@plus 1ex \@minus .2ex}%
{-1em}%
{\normalfont\Large\bfseries}}
\makeatother
and use it as in the following MWE:
\documentclass{article}
\makeatletter
\newcommand\runinsection{\@startsection {section}{1}{\z@}%
{3.5ex \@plus 1ex \@minus .2ex}%
{-1em}%
{\normalfont\Large\bfseries}}
\makeatother
\usepackage{lipsum} %just for the example
\begin{document}
\section{This is a normal section}
\lipsum[1]
\runinsection{This is a run-in section}
\lipsum[1]
\end{document}
Output
• @MysticOdin You're welcome. – karlkoeller Dec 7 '14 at 19:59
|
Algebra 1
To find the average speed of the car, train, and boat, you divide the total distance traveled by the time it took to travel that distance. This will give you the distance traveled per unit of time. In this case, km per hr (km/hr) Car: $\frac{\text{360 km}}{\text{6 hr}}=60$ km/hr Train: $\frac{\text{400 km}}{\text{8 hr}}=50$ km/hr Boat: $\frac{\text{375 km}}{\text{5 hr}}=75$ km/hr The fastest average speed was 75 km/hr, which was the boat.
|
# All Questions
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### Why does CTR mode XOR the plaintext into the output of the block cipher rather than XORing the plaintext into the input of the block cipher?
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### Questions about OAEP for RSA
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### OpenSSL PBKDF2 documentation
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### If $G'(s)=G(s0^{|s|})$ and $G$ is a PRNG, is $G'$ necessarily a PRNG? [closed]
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### Which data to share in CTR mode?
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### Need Help Reversing my Encryption Algorithm [closed]
So I wrote a simple encryption algorithm (Not very secure, just a starting point. I don't want criticism on it.) and I have been having trouble finding a way to decrypt the text. The algorithm is here ...
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### File Encryption (EFS method)
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### Derive both MAC and AES keys from same PBKDF2?
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### Where i can find an AES implementation in python?
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### How to calculate the time it'll take to crack RSA or DH?
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### Why Elliptic curve cryptography are not popular in practice
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### Z*p Generator (Cyclic Group) in java
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### ElGamal message signatures retrieving the secret value x
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### Why can't Diffie-Hellman be used for signing?
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### Homomorphic encryption for vector addition
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### Retrieve cryptographic key knowing cyphertext and plaintext [duplicate]
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### How to generate initial vectors during DPA attack on stream cipher in real life?
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|
2.5 Comparing Expectations to Data
Now that you have data in hand (the check at the restaurant), the next step is to compare your expectations to the data. There are two possible outcomes: either your expectations of the cost match the amount on the check, or they do not. If your expectations and the data match, terrific, you can move onto the next activity. If, on the other hand, your expectations were a cost of 30 dollars, but the check was 40 dollars, your expectations and the data do not match. There are two possible explanations for the discordance: first, your expectations were wrong and need to be revised, or second, the check was wrong and contains an error. You review the check and find that you were charged for two desserts instead of the one that you had, and conclude that there is an error in the data, so ask for the check to be corrected.
One key indicator of how well your data analysis is going is how easy or difficult it is to match the data you collected to your original expectations. You want to setup your expectations and your data so that matching the two up is easy. In the restaurant example, your expectation was $30 and the data said the meal cost$40, so it’s easy to see that (a) your expectation was off by $10 and that (b) the meal was more expensive than you thought. When you come back to this place, you might bring an extra$10. If our original expectation was that the meal would be between $0 and$1,000, then it’s true that our data fall into that range, but it’s not clear how much more we’ve learned. For example, would you change your behavior the next time you came back? The expectation of a \$30 meal is sometimes referred to as a sharp hypothesis because it states something very specific that can be verified with the data.
|
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