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Outlook: Jianzhi Education Technology Group Company Limited American Depositary Shares is assigned short-term Ba1 & long-term Ba1 estimated rating.
Time series to forecast n: 12 Mar 2023 for (n+16 weeks)
Methodology : Modular Neural Network (DNN Layer)
## Abstract
Jianzhi Education Technology Group Company Limited American Depositary Shares prediction model is evaluated with Modular Neural Network (DNN Layer) and Spearman Correlation1,2,3,4 and it is concluded that the JZ stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Buy
## Key Points
1. Is it better to buy and sell or hold?
2. What is the best way to predict stock prices?
3. What is the best way to predict stock prices?
## JZ Target Price Prediction Modeling Methodology
We consider Jianzhi Education Technology Group Company Limited American Depositary Shares Decision Process with Modular Neural Network (DNN Layer) where A is the set of discrete actions of JZ stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Spearman Correlation)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (DNN Layer)) X S(n):→ (n+16 weeks) $∑ i = 1 n a i$
n:Time series to forecast
p:Price signals of JZ stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## JZ Stock Forecast (Buy or Sell) for (n+16 weeks)
Sample Set: Neural Network
Stock/Index: JZ Jianzhi Education Technology Group Company Limited American Depositary Shares
Time series to forecast n: 12 Mar 2023 for (n+16 weeks)
According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Jianzhi Education Technology Group Company Limited American Depositary Shares
1. An entity is not required to restate prior periods to reflect the application of these amendments. The entity may restate prior periods only if it is possible to do so without the use of hindsight. If an entity restates prior periods, the restated financial statements must reflect all the requirements in this Standard for the affected financial instruments. If an entity does not restate prior periods, the entity shall recognise any difference between the previous carrying amount and the carrying amount at the beginning of the annual reporting period that includes the date of initial application of these amendments in the opening retained earnings (or other component of equity, as appropriate) of the annual reporting period that includes the date of initial application of these amendments.
2. For hedges other than hedges of foreign currency risk, when an entity designates a non-derivative financial asset or a non-derivative financial liability measured at fair value through profit or loss as a hedging instrument, it may only designate the non-derivative financial instrument in its entirety or a proportion of it.
3. Lifetime expected credit losses are not recognised on a financial instrument simply because it was considered to have low credit risk in the previous reporting period and is not considered to have low credit risk at the reporting date. In such a case, an entity shall determine whether there has been a significant increase in credit risk since initial recognition and thus whether lifetime expected credit losses are required to be recognised in accordance with paragraph 5.5.3.
4. In applying the effective interest method, an entity identifies fees that are an integral part of the effective interest rate of a financial instrument. The description of fees for financial services may not be indicative of the nature and substance of the services provided. Fees that are an integral part of the effective interest rate of a financial instrument are treated as an adjustment to the effective interest rate, unless the financial instrument is measured at fair value, with the change in fair value being recognised in profit or loss. In those cases, the fees are recognised as revenue or expense when the instrument is initially recognised.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Jianzhi Education Technology Group Company Limited American Depositary Shares is assigned short-term Ba1 & long-term Ba1 estimated rating. Jianzhi Education Technology Group Company Limited American Depositary Shares prediction model is evaluated with Modular Neural Network (DNN Layer) and Spearman Correlation1,2,3,4 and it is concluded that the JZ stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Buy
### JZ Jianzhi Education Technology Group Company Limited American Depositary Shares Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2B1
Balance SheetBaa2C
Leverage RatiosBaa2Caa2
Cash FlowBaa2C
Rates of Return and ProfitabilityBa3Caa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 77 out of 100 with 738 signals.
## References
1. Hastie T, Tibshirani R, Friedman J. 2009. The Elements of Statistical Learning. Berlin: Springer
2. Bell RM, Koren Y. 2007. Lessons from the Netflix prize challenge. ACM SIGKDD Explor. Newsl. 9:75–79
3. Burgess, D. F. (1975), "Duality theory and pitfalls in the specification of technologies," Journal of Econometrics, 3, 105–121.
4. Andrews, D. W. K. (1993), "Tests for parameter instability and structural change with unknown change point," Econometrica, 61, 821–856.
5. Artis, M. J. W. Zhang (1990), "BVAR forecasts for the G-7," International Journal of Forecasting, 6, 349–362.
6. Hartford J, Lewis G, Taddy M. 2016. Counterfactual prediction with deep instrumental variables networks. arXiv:1612.09596 [stat.AP]
7. B. Derfer, N. Goodyear, K. Hung, C. Matthews, G. Paoni, K. Rollins, R. Rose, M. Seaman, and J. Wiles. Online marketing platform, August 17 2007. US Patent App. 11/893,765
Frequently Asked QuestionsQ: What is the prediction methodology for JZ stock?
A: JZ stock prediction methodology: We evaluate the prediction models Modular Neural Network (DNN Layer) and Spearman Correlation
Q: Is JZ stock a buy or sell?
A: The dominant strategy among neural network is to Buy JZ Stock.
Q: Is Jianzhi Education Technology Group Company Limited American Depositary Shares stock a good investment?
A: The consensus rating for Jianzhi Education Technology Group Company Limited American Depositary Shares is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of JZ stock?
A: The consensus rating for JZ is Buy.
Q: What is the prediction period for JZ stock?
A: The prediction period for JZ is (n+16 weeks)
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# Statistical Significance…And Insignificance
One of the most damning phrases in the scientific literature is “Not statistically significant.” In a world where policy is announced by 140 character twits, the wordy “not statistically significant” readily becomes “not significant” and then “irrelevant.” But “not statistically significant” has a very specific meaning…and only rarely does it mean “irrelevant.”
Formally, we have the following problem: There is a mass of evidence, and two possible ways that evidence could have been produced, called the null hypothesis and the alternate hypothesis. Your mission, should you choose to accept it, is to decide which hypothesis is correct.
Unfortunately, that’s impossible. So statisticians take a different approach: If the evidence is sufficiently unlikely to be generated in the universe where the null hypothesis is true, we’ll reject the null hypothesis and say that the evidence is statistically significant. But if it is sufficiently likely that the evidence would be generated in a universe where the null hypothesis is true, then we’d say that the evidence is not statistically significant and reject the alternate hypothesis.
For example, consider a coin. There are two possibilities for the state of the universe: Either the coin is a fair coin, and will land heads 1/2 the time; or the coin is unfair. For somewhat technical reasons, “Coin is fair” is the null hypothesis.
Now let’s collect some evidence. Say we flip the coin 10 times and saw it land heads 8 of those 10 times. Many people might conclude, based on the evidence, that the coin is unfair.
Herein lies the problem. By concluding the coin is unfair, we have established a guideline for future experiments: “If a coin lands heads 8 out of 10 times, conclude that the coin is unfair.” That works great if someone sees a coin land heads 8 times out of 10 flips. But what if they see it land heads 9 times out of 10 flips? It seems reasonable to also conclude the coin is unfair. Likewise a coin that lands heads 10 times in 10 flips. And if they see the coin land tails 8, 9, or 10 times out of 10, they might also conclude the coin is unfair. What this means is that if we make a decision based on the evidence, then any evidence that is at least as compelling should lead us to the same conclusion.
Now suppose you have 100 people testing coins. If every single one of them has a fair coin, then about 10 of them will see that coin land heads (or tails) 8 or more times out of 10! So about 10% of those testing fair coins will conclude they are unfair coins.
That 10% corresponds to what statisticians call the level of significance. What’s important to understand is that the level of significance is completely arbitrary: it’s based on how often you’re willing to make the wrong decision about the null hypothesis. The lower the level of significance, the more compelling the evidence must be before you reject the null hypothesis. And if the evidence isn’t sufficiently compelling, you declare the evidence to be “not statistically significant.”
In this case, at a 5% level of significance, you’d need to see the coin land heads (or tails) at least 9 times out of 10. With 8 heads in 10 flips, you’d say that the evidence for the coin being unfair is not statistically significant. And yet, most people would hold that this is compelling evidence that you’re dealing with an unfair coin.
Here’s another way to look at it. In the movie Dirty Harry (1971), Clint Eastwood utters one of the most iconic lines in movie history. In case you’ve been living under a rock for the past 40 years, the setup is that Eastwood (a cop and the title character) is facing down a suspect after a chase. The assailant has a gun just within reach…but Eastwood has his gun drawn and pointed at the suspect. The problem is:
“Did he fire six shots or only five?” Well to tell you the truth in all this excitement I kinda lost track myself. But being this is a .44 Magnum, the most powerful handgun in the world and would blow your head clean off, you’ve gotta ask yourself one question: “Do I feel lucky?” Well, do ya, punk?
From the dialog, we can assume there are five confirmed shots. There are two possibilities: Either the gun is empty, or the gun has one more round in it.
Let the null hypothesis be “The gun has been emptied.” A statistically informed punk might reason thusly: “It is sufficiently likely that ‘five confirmed shots’ could be produced by a now-empty gun. Therefore the evidence that the gun has one more round is not statistically significant.” Consequently, they would reject the alternate hypothesis (that the gun has one more round). In practice, this means they would proceed as if the gun was empty.
(Alert readers will note that the argument works both ways, if we interchange the null and alternate hypotheses. True enough…but as I said, there are somewhat technical reasons for which hypothesis is the null hypothesis, and if this were my statistics course, I’d use this discussion as a lead-in to how we decide)
# The Sixth Wave
Over the past four thousand years, four waves of mathematical innovation have swept the world, leading to rapid advances and significant changes in society:
• The invention of written number (Fertile Crescent, 3000 BC). This allowed civilization to exist, because if you want to live with more than your extended family, record keeping is essential…and that means keeping track of numerical amounts.
• The invention of geometry (Greece, 300 BC). Yes, geometry existed before then; what I’m using is the date of Euclid’s Elements, which is the oldest surviving deductive geometry. The idea that you could, from a few simple principles, deduce an entire logical structure has a profound impact on society. How important? Consider a rather famous line: “We hold these truths to be self-evident…” The Declaration of Independence reads like a mathematical theorem, proving the necessity of revolution from some simple axioms.
• The invention of algebra (Iraq, 900). The problem “A number and its seventh make 19; find the number” appears in a 4000-year-old manuscript from ancient Egypt, so finding unknown quantities has a very long history. What algebra adds is an important viewpoint: Any of the infinite variety of possible problems can be transformed into one of a small number of types. Thus, “A farmer has 800 feet of fence and wants to enclose the largest area possible” and “Find a number so the sum of the number and its reciprocal is 8” and “The sum of a number is 12 and its product is 20” can all be reduced to $ax^{2} + bx + c = 0$ and solved using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
• The invention of calculus (Europe, 1600). Algebra is the mathematics of what is. Calculus is the mathematics of how things change. Calculus makes physics possible, and from physics comes chemistry and engineering.
• The invention of statistics (Europe, 1900). Both algebra and calculus deal with single objects: a bridge, a number, a moving planet. But the universe consists of many similar objects: the human population; the planetary climate; the trash generated by a city. Statistics aggregates the data on the individual in a way that can be used to describe a population…then uses the information on a population to predict information about an individual. Everything in modern society, from the pain relievers you use to the road you travel to work, incorporates such a statistical analysis.
Many people, myself included, believe we are on the verge of a sixth wave. That sixth wave will have the transformative power of calculus and statistics, and fundamentally reshape society.
The sixth wave is based around discrete mathematics. That’s not mathematics you whisper in dark corners. Rather, it’s the mathematics of things that can be counted as opposed to measured. For example, length is continuous: a length can have any value, and no one looks at you strangely if you say “I traveled 1.38924 miles today…” (You might get some strange looks, but it’s because you specified the distance so precisely and not because of the distance itself) But if you continued “…and met 2.35 people,” you would get strange looks, because the number of people you meet is a counted number: it’s a discrete quantity.
How important is discrete mathematics? If calculus is the basis for physics and engineering, then linear algebra is the basis for discrete mathematics. But a first-year calculus problem would have a hard time solving even a simple question in statics (the physics of structures). In contrast, Google’s search algorithm is based on mathematics learned in the first half of a standard college linear algebra course.
I’ll talk more about this later. But if you’re interested in learning some linear algebra, the video lectures for the course I teach are available on YouTube.
# The Geometry of Floods, Part One
In the aftermath of Hurricane Harvey, there’s bound to be questions asked about the wisdom of building in flood plains.
Part of being a mathematician is asking “What can math say?” A lot of what math can say is embedded in actuarial tables and flood insurance premiums, and phrases like “500 year flood.” Those are good topics, but I’m teaching calculus this semester, so my mind turns to geometry.
Consider two homes. One is built some distance from a river, but only a few feet above the normal water level. Another is very close, but much higher up. Which one is in greater danger of being flooded?
To answer this, we need to construct a mathematical model. We do this by making some assumptions about the world, then follow the math. Since I’ll assume you haven’t taken differential equations and calculus, we’ll construct a relatively simple model based on geometry.
We’ll make the following assumptions:
• The normal river surface has constant width $w$. This is unrealistic…but you get what you pay for: a more realistic model is more complex. (Apologies if that’s not how geologists speak: I think the last geology course I took listed the Pleistocene as “current events”…)
• The land between the river and House 1 has a gradual but constant slope, and it’s like this for the entirety of the river. I’ll use my expert drawing skills to show you what I mean:
• House 2 is built on the riverbanks. Again, my artistic skills lend to the following:
As above, we’re assuming the river looks like that for its entirety.
• We’ll model the storm by dumping volume $V$ of water into each river, and seeing how far it rises up the banks. In particular, what we want to know is that if the water level rises up $x$ above the riverbed around House 2, how far up does it rise up around House 1:
vs.
So here’s the mathematical task: Suppose the river rises height $x$ about its normal level (the figure on the left). How far $l$ does it extend past its normal banks (the figure on the right)?
Conversely (mathematicians love this phrase), if you build $l$ away from the normal river bank, how much of a rise above the normal river level are you insulated from?
Now I’m a mathematician…but I’m also a teacher. And I would be remiss in my duties if I gave you the answer right away. So mull these over, and I’ll return to the topic next week…
# First Day Jitters
Today is the first day of classes for me.
I’ve been teaching in one form or another for 20 plus years. I’ve survived the Great Calculus War, and even spent time in the Technology Underground engaging in acts of passive resistance against the ancienne regime. And yet the start of every term is fraught with anxiety for me: What am I going to do this semester?
I don’t know how common this problem is among my colleagues. I mention it because it underscores what I believe to be an important fact of life: Never be too comfortable. This doesn’t mean you should go out and buy ill-fitting shoes and eat kale 24/7. Rather, it means that a little discomfort is good, because the general human reaction is to try and change things to become more comfortable.
So I’m anxious about what I’m going to teach this semester. That’s good, because it forces me to look a what I’m teaching and ask myself the all-important question: Why would anyone want to listen to me talk about this subject? (I don’t have an answer for this term yet…which is why the anxiety persists)
On the other side, learning mathematics is all about anxiety too. The critical question is not whether you suffer from math anxiety; it’s what you do about it. Remember that the human reaction is to change to become more comfortable in whatever situation you’re in. For all too many people, the reaction to math anxiety is to avoid math.
And that’s fine, if you live in a stone age society where mathematics can be left to specialists. (Go ahead, ask me about paleomathematics…) But in the modern world, it’s not practical to avoid mathematics. It’s possible, in the same way that it’s possible to avoid reading. But you won’t get very far, and you condemn yourself to being a second or third class citizen.
Instead, the way to fight math anxiety is to accept the discomfort…and push through it anyway. The most important lesson we can learn in life is that we can survive a little discomfort, and when we get through to the other side, we are better for it. And soon enough, you’ll find yourself addicted and actively seek the discomfort because you know you can get through it.
Psychologists no doubt have a litany of strategies for dealing with anxiety disorders. But here’s my suggestion for dealing with math anxiety: Do a little math, every day. The good thing about math is that it’s something you can do in the clamor of your own mind as you go through daily life.
Count things: that’s the beginning of mathematics. Don’t look at the line at Starbucks; count how many people are there. If you do this often enough, you’ll start to think about more efficient ways to count: you’ll find yourself counting by twos, threes, and fives. You’ll also develop what many have called number sense: the ability to estimate quantities with reasonable accuracy.
If the cost is linear, then an 8 + 4 = 12 foot board would cost $3.23 +$1.90 = $5.13. Here the difference between what the board should cost and what it does cost is only a few cents, so we might reasonably conclude that the cost of lumber is linear. Of course, only politicians and pundits base their conclusions on one piece of evidence. What about a 16-foot board? If the cost is linear, then a 12 + 4 = 16 foot board should cost$7.16 + $1.90 =$9.06, and we find it costs $9.19, which is reasonably close. One final note: Even if the cost of lumber is linear, we can’t expect that it will always be linear. Since the 8-foot board costs$3.23, we’d expect a 4-foot board to cost $3.23 –$1.90 (subtract $2, return 10 cents)$1.33, and a 0-foot board would cost $1.33 –$1.90 = $-0.57. But if you go to the checkout with ten 0-foot boards, the cashier is not going to hand you a five spot. And remember, lumber doesn’t grow on trees. (Yes, I meant it…wood grows on trees, but no tree in the forest grows 16-foot boards) A 200-foot board might cost significantly more than linearity would suggest. # Obesity, Poverty, and National Security According to the internet, if you ate only ramen, you’d save thousands of dollars each year in food. That sounds great, except there’s a problem: ramen lacks a wide range of essential nutrients and vitamins. You’d lose your teeth to scurvy, a lack of vitamin D would cause your bones to become brittle and easily broken, you’d suffer nightblindness from a lack of vitamin A, and you’d be tired all the time from a lack of iron and the B vitamins. In short, all the money you saved on food, and much, much, more, would be spent on increased medical care. The problem is that eating healthy is costly. And this leads to a national security crisis. If you want the short version, I’ve summarized the key points in a ten-minute video: A little more mathematics: Food buyers face what mathematicians call a constrained optimization problem: they have to meet certain caloric and nutritional goals (the constraints), which defines a feasible region. Generally speaking, any point in the feasible region defines a solution to the problem; what you want to do is to find the optimal solution. The optimal solution is generally determined by the objective function. For example, if you lived off $x$ packages of ramen and $y$ eggs, the important objective function might be the total cost of your meals. At 15 cents a pack of ramen and 20 cents an egg, the objective function has the form $L = 0.15x + 0.20y$, and we might want to minimize the value of the objective function. In the following, I’ll assume you want to minimize the value of the objective function; the arguments are similar if you’re trying to maximize the value (for example, if you’re designing a set of roads, you might want to maximize the traffic flow through a town center). There’s a theorem in mathematics that says the optimal solution will be found on the boundary of the feasible region. The intuition behind this theorem is the following: Imagine any point inside the feasible region. If you change any one of the coordinates while leaving the others the same, the value of the objective function will generally change. The general idea is to move in the direction that decreases the objective function, and continue moving in that direction until you hit the boundary of the feasible region. At this point, you can’t move any further in that direction. But you can try one of the other directions. Repeating this process allows us to find the optimal solution. We can go further. Suppose our objective function is linear (like the cost function). Then the same analysis tells us the optimal solution will be found at a vertex of the feasible region. This suggests an elegant way to solve linear optimization problems: • Graph the feasible region and locate all the vertices. Generally speaking, the constraints are themselves linear functions, so (in our ramen and egg example) the feasible region will be a polygon. • Evaluate the objective function at each vertex, • Choose the vertex that minimizes the value of the objective function. Easy, huh? Except… • If you have $n$ commodities, you have to work in $\mathbb{R}^{n}$. • This means the feasible region will be some sort of higher solid. • This also means that finding the vertices of the feasible region will require solving systems of $n$ equations in $n$ unknowns. In 1945 ,George Stigler did such an analysis to find a minimal cost diet that met caloric and nutritional requirements. To make the problem tractable, he focused on a diet consisting of just seven food items: wheat flour; evaporated milk; cabbage; spinach; dried navy beans; pancake flour; and pork liver. “Thereafter the procedure is experimental because there does not appear to be any direct method of finding the minimum of a linear function subject to linear conditions.” The problem is that with seven items, you’re working with hyperplanes in $\mathbb{R}^{7}$, and the constraints will give you hundreds of vertices to check. Note the date: 1945. What Stigler didn’t know is that there was a method for finding the minimum value easily. But that’s a story for another post… # The Most Important Letter A question came up on Quora about what letter’s removal would have the greatest impact on the English language. The obvious answer is “E”, since it’s by far the most common letter in English. But let’s consider that. Can you writ a comprhnsibl sntnc that dosnt us ths lttr? Ys, you can! So its not clear that “E” is all that important. So let’s do some mathematics. The key question is: How much information does a given letter provide? Consider the following: I’m thinking of a color. You know the color is either red, green, blue, or fuchsia. (I have no idea what color fuchsia is…I just like the word) Your goal is to determine the color I’m thinking of by asking a sequence of Yes/No questions. One way you could do this is by asking “Are you thinking of red or green?” If the answer is “Yes”, then you might ask “Are you thinking of red?” If the answer is “Yes”, then you know the color is red; if the answer is “No,” then you know the color is green (since I answered “Yes” to the first question). On the other hand, if I answered “No” to the first question, then you know I was thinking of blue or fuchsia, so you might ask “Are you thinking of blue?” A “Yes” tells you I’m thinking blue; a “No” tells you I’m thinking fuchsia. Now reverse it. If you know I’m thinking of the color red, then you have the answer to two Yes/No questions. We say that “red” has an information content of two bits. So far so good. But suppose I’m somewhat dull and can’t think of any color other than red. In that case, you already know what color I’m thinking of, and don’t need to ask any questions. In this situation, “red” has an information content of zero bits. As an intermediate case, suppose that half the time I think of “red,” one-fourth the time I think of “blue”, and one-eighth the time I think of “green” and one-eighth the time I think of “fuchsia.” Then you might ask a different sequence of questions: • Are you thinking of red? (Half the time, I’ll answer “Yes”, so the answer “red”gives you the answer to one question: it’s 1 bit of information) • If the answer is “No,” then “Are you thinking of blue?” Half the time this question is asked (remember it will only be asked if the answer to the first question is “No”), the answer will be “Yes,” so the answer “blue” gives you the answer to two questions: it’s 2 bits of information. • If the answer is “No,” then the final question “Are you thinking of green?” Again, half the time this question is asked, the answer will be “Yes,” which tells you that “green” is worth 3 bits; meanwhile, the answer “No” means I’m thinking of fuchsia, so “fuchsia” is also worth 3 bits. It might seem difficult to determine the information content of an answer, because you have to come up with the questions. But a little theory goes a long way. The best question we could ask are those where half the answers are “Yes” and the other half are “No.” What this means is that if $n$ is the answer to the question $p_{n}$ of the time, then the information content of the answer $n$ will be $-\log_{2} p_{n}$. Thus, if “red” is the color half the time, then “red” has an information content of $-\log_{2} (1/2) = 1$ bit. So what does this mean? “E” makes up about 12.7% of the letters in an English text. But this means that knowing a letter is “E” answers very few questions. So the letter E contains about 3 bits of information. In contrast, “Z” only makes up 0.07% of the letters in an English text, so knowing a letter is “Z” answers many questions. So the letter Z contains about 10.4 bits of information (the maximum). At first glance, this suggests that “Z” may be the most important letter in the English language: losing the letter “Z” will lose the most information. However, there’s a secondary consideration: “Z” doesn’t often appear in a text. So every “Z” you drop from a text loses a lot of information…but you don’t drop that many. And here’s where the greater prevalence of “E” comes in. While the letter “E” only gives you about 3 bits of information, it’s common enough that dropping the letter “E” from a text will lose you more information overall. For example, suppose you had a 10,000 character message. Of these 10,000 characters, you might expect to find 7 Zs, and losing them would lose you about 77 bits of information. In contrast, there would be almost 1300 Es, and losing them would lose about 3800 bits of information. # Exact is Not Accurate Numbers. Over the next few years, you’ll be certain to see a barrage of numbers thrown at you. While researching the latest atrocity promoted by the administration, I came across the following tidbit: The average tuition for private schools is$10,003.
Now, if I want to include this in a blog, vlog, Facebook post, or public speech, I have a conundrum. Compare the two sentences:
• The average tuition for private schools is $10,003. • The average tuition for private schools is about ten thousand dollars. The first sounds like I know what I’m talking about: that I’ve done some high-level research and wrestled a number to the ground. The second sounds like I spent thirty seconds on Google. (Actually, the first number was based on thirty seconds on Google) The difference is that I sound more convincing with the exact figure. In fact, there’s a story (which might or might not be true) that when the first surveyors found the height of Mount Everest, they came to a value of 29,000 feet…but they published it as 29,002, because that sounded more accurate. The problem is the exact figure might not be accurate. Consider the two statements: • The population of the United States is 324,595,182. • The population of the United States is 325 million. The first gives an exact number, and sounds very accurate. But it is almost certainly false. In particular, even if the population of the US was 324,595,182 at some point, it is almost certainly not 324,595,182 right now. On the other hand, it’s still about 325 million, and will be so for awhile. (I talk about this in my FOCUS article). There’s a concept in the sciences called significant figures. The gist of it is this: When I give you a number, I am giving you a guarantee that the non-zero digits of the number are correct. (The zeroes are a little more complicated: if you want a crash course on signficant figures, here’s the video I have my students watch) • If I claim 324,595,182, then I’m guaranteeing each and every digit is exact…and if the population is 324,595,183, then I’ve fed you misinformation. • If I claim 324 million, then I’m guaranteeing that the population is somewhere between 323,500,000 and 324,499,999 (since anything in this range would round to 324 million). What’s the big deal? One problem with statistics is that people don’t believe them. You’ve heard the quote: “There are three types of lies: Lies, damned lies, and statistics.” I suspect part of the reason is that if someone says “The average tuition at private schools is$10,003,” they can respond with “But at our school, it’s $7500, so how do you get an average of$10,003?” This generally leads to a discussion of how to calculate averages, and often degenerates into accusations of skewed samples.
On the other hand, if you say “The average tuition at private schools is around $10,000,” then to the person who says “But we only pay$7500,” the response is “Which is around $10,000.” By avoiding the mechanics of computing the number, we focus on the value itself. # Math for Democracy I surrender. I’ve been trying to keep this blog politics free, or at least minimize the politics: when I talked about the I focused on estimating the crowd size and not on the reasons behind it. I’m still going to minimize the politics. But it’s clear that we’re heading towards a major crisis. I’m not talking about the person in the White House, or Russian interference, or anything that minor. I’m talking about the denial of basic fact-finding. You’ve heard the term “fake news.” The problem is that most Americans get their information from one or two sources, which they don’t verify. If those sources are unreliable, then they’re going to get a warped view of the world. So I have a new mantra: Five minutes a day. Take five minutes a day to track down a fact. You might start with the news story, but don’t end with it. Who did they interview? If they’re reporting on a piece of research, track down the original article and check out the legitimacy of the publisher. If they’re reporting on an incident, go to the local newspapers and see what their coverage is. If they’re talking about waste in government spending, go to USAspending.gov and see how your money is spent. So let’s talk about that. One of the promises of the new administration is to drastically curtail the U.S. Department of Education, returning control of schools to the states. Sounds good, right? But go to USAspending.gov to see how the Department of Education actually spends your money. Note that I’m giving you the source, so you should feel free to check my claims. (A guaranteed way to identify something as “not a fact” is that lack of a source: If there’s no source, it’s not a fact. Keep in mind this does not work in reverse: you can cite a source and still spew non-facts) Most federal agencies suck in a lot of taxpayer dollars…and then shovel them back to the states in the form of grants. Find the government department you’re interested in, then download the grants database: this tells you who they’ve given money to, and how much. You can import it into Excel, or download it as a CSV and use your own spreadsheet software. Then the fun begins… You can sort the grants by any category you want. The cost of the elected President’s recent trips to Mar-a-Lago have been in the news: current estimates for the three weekend trips (out of five weekends in office) are around$12 million, so here’s a few grants made by the Department of Education that are around this much. I’ve deliberately chosen programs that benefits states where Trump support was very strong: yes, New York, California, and other states get money from the Department of Education, so of course we’re concerned…the point is that states that supported Trump need to be even more concerned, because here are some of the things they’re going to lose:
• Nevada: $9,928,139 for Vocational Rehabilitation training. Nevada has received almost$200 million in grants from the Department of Education since January 1, 2016.
• Kansas: $10,669,790 for Department for Children and Families for Vocational Rehabilitation training. Kansas has received more than$210 million in grants since January 1, 2016.
• Texas: $11,187,178 to Bexar County Texas for “Impact Aid.” The army base Fort Sam Houston occupies a good part of Bexar County, and this land can’t be taxed, impacting the county’s ability to pay for schools. That’s money local taxpayers don’t have to pay. The Department of Education has given more than$600 million in Impact Aid grants since October 2016, reducing tax burdens around the country.
Now for some math. On a dollar basis, California, Texas, and New York have received the most from the Department of Education. But they’re also the biggest states in the country. An easily googlable fact is the population of these states; if you divide how much each states gets by its population, you obtain a per capita figure.
These are interesting. A few more states that stand to lose big if Trump eliminates the Department of Education:
• Alabama: $15,912,537 for preschool programs. Alabama received more than$400 million in grants. On a per person basis, that’s 26% more than Connecticut gets.
• Louisiana: $9,177,379 for preschool education programs. Louisiana has received nearly$500 million in grants. On a per-person basis, that 37% more than California receives.
• West Virginia: $9,828,491 for vocational and rehabilitation services. West Virginia has received more than$160 million from the Department of Education. On a per person basis, that’s 50% more than Massachusetts.
# Lies, Damned Lies, and Statistics
We all know the quote: “There are three types of lies: lies, damned lies, and statistics.”
But like many things that are short enough to tweet, this statement is misleading.
Statistics don’t lie. People do, generally by omitting key pieces of information. Any statistic worth repeating should include two other numbers. If these are missing, the whole truth is being kept from you.
The two numbers to look for are:
• The sample size. This is the number of cases examined. If you base a conclusion on one example, you’re a politician or a pundit, relying on anecdotal evidence and shouting instead of facts and logic. While a large sample won’t guarantee reliability, a small sample will almost always be untrustworthy.
• The p-value. This is a little more complicated, but roughly speaking, it measures how convinced you should be. A small p-value (0.05 or less) means the evidence is very convincing.
I’ll talk more about these later. Until then, remember: if someone doesn’t give you these values, they’re not telling you the whole truth.
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## Friday, May 8, 2015
### 52 Things: Number 31: Game Hopping Proof
This is the latest in a series of blog posts to address the list of '52 Things Every PhD Student Should Know To Do Cryptography': a set of questions compiled to give PhD candidates a sense of what they should know by the end of their first year. In this post we give an example of a proof that uses the 'game hopping' technique.
Note, this blog post is based on Section 3.3 of 'An Introduction to Provable Security' by Douglas Stebila, downloadable via this link.
Recall the definition of IND-CCA security for a public key encryption scheme, described by Ana here. If one removes the decryption oracle from the adversary, we obtain the IND-CPA (indistinguishability under chosen-plaintext attacks) security notion. Note that removing the encryption oracle does not change the adversary's view since it holds the public key and can therefore produce its own encryptions.
In an earlier blog post, we described the Decisional Diffie-Hellman (DDH) problem. In this post, we are going to use a technique called 'game hopping' to show that the ElGamal encryption scheme is IND-CPA secure if DDH is hard. Loosely speaking, we will transform the IND-CPA game against ElGamal into a game against DDH and show that an adversary's advantage in the first game cannot be more than their advantage in the second. So if their advantage in the second game is negligible (which is the assumption that DDH is hard), the advantage in the first game must also be negligible (showing that the encryption scheme is IND-CPA secure).
Firstly, let's describe the ElGamal scheme. We work in a cyclic group $G$ of prime order $q$ with generator $g$. (Implicitly the selection of the group depends on a security parameter $\lambda$ and when we say that a quantity is negligible, we mean a negligible function of $\lambda$, but we'll omit those details here.) Plaintexts and ciphertexts are group elements. The private key is a secret exponent $x \in \mathbb{Z}_q$ and the public key is $X = g^x$. To encrypt a message $M \in G$, one selects an exponent $y \in \mathbb{Z}_q$ uniformly at random, computes $c_1 = g^y$, $c_2 = MX^y$ and the ciphertext is the pair $(c_1, c_2)$. To decrypt, notice that $c_2 = MX^y = M(g^x)^y = M(g^y)^x = Mc_1^x$ so, with the private key $x$ we just compute $M = c_2c_1^{-x}$.
Now consider the following game $\mathrm{Game}_0$ played by a PPT adversary $\mathcal{A}$.
1. $x \overset{\$}{\leftarrow} \mathbb{Z}_q, X \leftarrow g^x$(generate the public, private key pair) 2.$(M_0, M_1) \overset{\$}{\leftarrow} \mathcal{A}(X)$ (the adversary, in possession of the public key, produces a pair of challenge messages, possibly using a randomised process)
3. $b \overset{\$}{\leftarrow} \{0,1\}$(a random bit is chosen) 4.$y \overset{\$}{\leftarrow} \mathbb{Z}_q, c_1 \leftarrow g^y, Z \leftarrow X^y, c_2 \leftarrow M_bZ$ (an encryption of message $b$ is produced)
5. $b' \overset{\$}{\leftarrow} \mathcal{A}(c_1, c_2)$(the adversary, in possession of the ciphertext, produces a bit, possibly using a randomised process) 6. if$b = b'$then return 1, else return 0. We say$\mathcal{A}$wins$\mathrm{Game}_0$if the game returns 1. From the definition in Ana's blog, it should be clear that the advantage of$\mathcal{A}$against the IND-CPA security of ElGamal (with parameters$G$,$q$and$g$) is$2|\mathrm{Pr}[\mathcal{A} \: \mathrm{wins \: Game}_0] - 1/2|$(1). Next, consider a new game$\mathrm{Game}_1$. This game is exactly as above, except that in Step 4, we replace the command$Z \leftarrow X^y$by$z \overset{\$}{\leftarrow} \mathbb{Z}_q, Z \leftarrow g^z$.
So the new ciphertext is $(c_1, c_2) = (g^y, M_bg^z)$ instead of $(c_1, c_2) = (g^y, M_bg^{xy})$.
Again we say $\mathcal{A}$ wins this game if it returns 1. What is the probability that this happens? Note that $Z$ is now a totally random group element by the randomness of $z \in \mathbb{Z}_q$. So $c_2$ is also a random group element, independent of $X$, $c_1$ and $b$. So the adversary gains no information about $b$ from $(c_1, c_2)$, meaning it outputs the correct bit and wins the game with probability exactly 1/2 (2).
Now we bound the difference in probability for an adversary to win each of the games. Since the only difference in the two games is that the group element $g^{xy}$ is replaced by a random group element $g^z$, it is easy to see how this relates to the DDH problem, where an adversary must distinguish between the triples $(g^x, g^y, g^{xy})$ and $(g^x, g^y, g^z)$ for a random exponent $z \in \mathbb{Z}_q$. To make the link between the games precise, we use the adversary $\mathcal{A}$ to build an adversary $\mathcal{B}$ against DDH as follows:
1. On input $(X, Y, Z)$, run $\mathcal{A}$ on input $X$ to receive a challenge pair $(M_0, M_1)$
2. Select a bit $b$ uniformly at random and compute $m_bZ$
3. Give the 'ciphertext' $(Y, m_bZ)$ to the $\mathcal{A}$ and receive a bit $b'$
4. If $b = b'$, guess that $Z = g^{xy}$ and return 1, else guess $Z$ is random so return 0.
If $\mathcal{B}$ is given a real Diffie-Hellman triple $(g^x, g^y, g^{xy})$ then the above is a perfect simulation of $\mathcal{A}$ playing $\mathrm{Game}_0$, and if $\mathcal{B}$ is given a fake triple $(g^x, g^y, g^z)$ then it is a perfect simulation of $\mathcal{A}$ playing $\mathrm{Game}_1$. Therefore, the difference between the probability of $\mathcal{A}$ winning $\mathrm{Game}_0$ and $\mathcal{A}$ winning $\mathrm{Game}_1$ is precisely the difference between the probability that $\mathcal{B}$ outputs 1 on input $(g^x, g^y, g^{xy})$ and outputs 1 on input $(g^x, g^y, g^z)$, which is exactly the advantage of $\mathcal{B}$ against DDH.
Combining the above with facts (1) and (2) (and using the triangle inequality to take care of the modulus signs), we can easily obtain that the advantage of $\mathcal{A}$ against the IND-CPA security of ElGamal is no greater than the advantage of $\mathcal{B}$ against DDH. So if DDH is hard for all polynomial time adversaries (meaning their advantage is negligible), ElGamal must be IND-CPA secure.
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### Theory:
Ovum is the female gamete that is produced in the ovaries of the female.
The process of formation of a mature ovum in the ovaries is known as oogenesis.
The mature ovum or an egg is spherical in shape. The ovum is essentially yolk-free. It has abundant cytoplasm called ooplasm and a large nucleus. The nucleus contains a prominent nucleolus.
The plasma membrane surrounds the cytoplasm. Small vesicles called cortical granules are found under the plasma membrane. The ovum is surrounded by three membranes, namely:
1. Zona pellucida
3. Vitelline membrane
Structure of the ovum
The plasma membrane is surrounded by an inner thin zona pellucida and an outer thick corona radiata. Zona pellucida is acellular. The corona radiata is formed of the follicular cells. The membrane forming the surface layer of the ovum is called the vitelline membrane. The fluid-filled space between the zona pellucida and the surface of the egg is called perivitelline space.
The video showing the structure of the ovum:
In contrast to males, in females, the initial steps occur prior to their birth. Diploidoogonia and primary oocytes are produced in the foetus when she is born.
In oocytes, the first meiotic division is initiated and then stopped. No further development occurs until the girl becomes sexually mature. After maturity, the primary oocytes resume their development. The primary oocytes grow further and complete meiosis I, forming a large secondary oocyte and a small polar body.
Only after fertilization, meiosis II is completed. By the completion of meiosis II, secondary oocytes get converted into a fertilized egg or zygote.
Process of oogenesis
Puberty:
When the reproductive system in both males and females becomes functional, there is an increase in sex hormone production, resulting in puberty. This phenomenon starts earlier in females than in males. Generally, boys attain puberty between the age of $$13$$ to $$14$$ $$years$$. The girls reach puberty between $$11$$ to $$13$$ $$years$$.
In a male, the onset of puberty is triggered by the secretion of the hormone testosterone in the testes. While in females, the secretion of estrogen and progesterone from the ovary triggers puberty.
As we have seen in an earlier chapter, the secretion of male and female hormones are under the control of the pituitary gonadotropins luteinizing hormone (LH) and follicle-stimulating hormone (FSH).
During puberty, the individuals of the two sexes show distinctive features called secondary sexual characteristics. Some of the male secondary sexual characteristics are facial hair, cracking of voice, etc.
Female secondary sexual characteristics include development of breasts, broadening of hips, etc. Such distinguishing features are present in all the animals. These characteristics serve to identify and attract sex partners.
Reference:
https://humanreproduction11.wordpress.com/fertilization/
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# structural drift.
The structural drift project looks at sequential inference, where chains of $$\epsilon$$-machines are inferred and re-transmit data from generation to generation. A simple analogy is the game of telephone, where a person thinks of a sentence and whispers it to their neighbor, who in turn whispers it to a different neighbor, and so on. The last recipient announces the sentence they were told, which is often amusingly different from the initial sentence due to errors accumulated in the chain.
It turns out that, when used in a similar way, sequential inference with $$\epsilon$$-machines is a generalization of genetic drift. For details on this connection to population genetics, see the paper. This page focuses on a series of programming examples, using the Computational Mechanics in Python library, to explore the higher level ideas presented in the paper.
# finite data inference.
To reconstruct an $$\epsilon$$-machine from data, we first generate data from a stochastic process:
Here we create an even process iframe and use it to generate a binary string of length 2000 where blocks of 1s are even in length. This data is passed to the state splitting reconstruction algorithm along with the history length parameter. This is the default value of the parameter and is omitted from the remaining examples. Finally, we write the machine to a Graphviz file for viewing.
# sequential inference.
Building on this example, we use the reconstructed machine to generate a new string of data having the same distribution of symbols as the original string, plus some variance due to finite sampling.
The new string generated by the first machine is then fed back to the reconstruction algorithm to produce a second machine. Each machine is written to a separate Graphviz file for comparison, and should have identical topologies with slightly different transition probabilities. The machines are omitted here for space.
The next step is to iterate this chain of inference, generation, and re-inference until some condition occurs. For now we'll say this terminating condition is when there are no more 0s generated by the machine, which can occur by chance.
We introduce the count_blocks function which returns a dictionary mapping blocks of symbols to probabilities, normalized for each block length. Symbols are stored as tuples since lists are not hashable in Python, and so we create the variable zero for convenience.
The code will loop until there are no more 0s in the data due to the cumulative effects of finite sampling variance. The machine could also drift the other way by chance and produce only 0s, but this is less likely for the even process. Therefore, the final inferred machine is typically a fixed coin that only generates 1s.
# structural stasis.
It's easy to decide when the inference loop should terminate for the simple even process. For more complex processes, however, a more general condition is needed to detect when the structure of the machine can no longer drift.
We call this condition structural stasis, and it occurs when the entropy rate of the machine ($$h_\mu$$) becomes zero. We modify the above example to reflect our definition of structural stasis, and plot the drift of entropy rate as a function of generation.
By setting a low initial bias for the 0-transition of the even process, we can see how the number of generations until stasis is affected. This is achieved by passing the bias=0.05 parameter to the even process constructor.
While this "time to stasis" is for a single realization, the result below holds true when averaged over many different experiments: the closer a process starts to a fixed point, the shorter the average time to stasis.
# complexity-entropy diagrams.
Another observation from the paper is that machines tend to diffuse locally in a stable subspace until enough drift variance has accumulated to create or merge states. As a result, the process changes topology and jumps to a new subspace. These subspaces are independent, and so the time to stasis for a process is a weighted sum over the stasis times for the independent subspaces the drifting machines visit on their way to reaching stasis.
One way to examine these subspaces is a complexity-entropy (CE) diagram, which plots the entropy rate of a machine ($$h_u$$) versus its statistical complexity ($$C_u$$). The CE diagram of a single realization of the golden mean process, producing binary strings with no consecutive 0s, is shown below. The upper curve shows diffusion in the golden mean subspace, and the line $$C_u=0$$ shows diffusion in the biased coin subspace after the two recurrent states of the golden mean are merged due to drift.
The butterfly process has a more complex CE diagram, and is obtainable by using a Butterfly object in the above code and setting data_length=15000.
# isostructural subspaces.
We see a different view of the golden mean drift process by reducing the evidence needed to infer new states and showing the total number of states as colored points in CE space. This shows that machines locally diffuse within each subspace leading to meta-stability of the drift process, with occasional jumps to new subspaces as the model's memory dynamically adjusts to the statistics of the new iteration:
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# How to find XOR of all the elements in given range?
given to integers A and B A<=B find XOR of all the elements between them.
Expected Complexity: logN
@drcoderji
No, it is not always true. If A=2, B=3, then your answer will be 5. But 2^3 = 1.
Let us denote f(n) = 1 \oplus 2 \oplus 3 \oplus \dots \oplus n, where \oplus denotes XOR operation
then XOR of all numbers between A and B can be represented by f(B) \oplus f(A-1), because x \oplus x = 0
Now we can find out easily that,
f(n) = \left\{\begin{array}{@{}lr@{}} n, & \text{n mod 4 = 0}\\ 1, & \text{n mod 4 = 1}\\ n+1, & \text{n mod 4 = 2}\\ 0, & \text{n mod 4 = 3} \end{array}\right\}
Time Complexity - O(1)
6 Likes
Similar question in stackoverflow. Hope it helps. happy coding.
1 Like
@jaydeep97 -its right. The answer will be f(3)^f(1)=0^1=1.
can anybody help me if the array is given and range is also given …
if array is 1 2 7 4 5
a=3 b=4 ans should be 3 but above solution gives 7 ans.
@dark_stranger
in this question range means all the no. in that range. suppose a=3 b=7 then this means we have to find 3^4^5^6^7.
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• I like to build my A - Team now.
I need loyal people who can trust and believe in a dream.
We cant pay now, you will recieve a lifetime percentage if the released game will give earnings.
What i need:
- Programmer c++
- Unity / Unreal - we must check whats possible, please share your experience with me.
- Sculpter, 3D Artist
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What i do:
- Studio Owner
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The game will be defintitly affected about our and your skills if you join the team.
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discord:
joerg federmann composing#2898
• I wasn't sure if this would be the right place for a topic like this so sorry if it isn't.
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• Learning game development in Unreal Engine could be a daunting task for someone who don’t know where to start, and a cumbersome process if you don’t organize your progression correctly. One thing commonly known by experienced developers and by people unfamiliar with coding: mastering a development language is a long and difficult task.
From blueprints to C++ in Unreal Engine
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Following this tutorial, you’ll acquire new concepts of C++ programming in every chapter. Then following chapters will give you reasons to reuse and practice those same concepts. There’s no better way to wire you brain.
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View full story
• Learning game development in Unreal Engine could be a daunting task for someone who don’t know where to start, and a cumbersome process if you don’t organize your progression correctly. One thing commonly known by experienced developers and by people unfamiliar with coding: mastering a development language is a long and difficult task.
From blueprints to C++ in Unreal Engine
If you want to learn fast, you need a good learning strategy. Unreal Engine contains a very powerful tool which you can use to learn C++ faster: its blueprint system. Blueprints are extremely easy to learn (and you may already have a good knowledge of them). Thus you can conveniently use them as a guide for writing code in C++. This is the reason why I am writing a tutorial series on how to make the transition from Unreal Engine blueprints to C++.
Learn and practice C++
Following this tutorial, you’ll acquire new concepts of C++ programming in every chapter. Then following chapters will give you reasons to reuse and practice those same concepts. There’s no better way to wire you brain.
Link to the tutorial: [Tutorial] Learn C++ in Unreal Engine 4 by making a powerful camera
Please do send me as much feedback as you want. I’ll be considering every constructive remarks and taking them into consideration. Your feedback will help me to improve and update the existing chapters and to make the next one better.
• By mrDIMAS
Hello everyone! I need to fill lua table with functions from script file like this:
function init() end function update() end I need to create table on stack and fill it with this functions from specified file. How can I do this?
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void update()
{
if (thrust)
{
}
else
{
dx*=0.99;
dy*=0.99;
}
int maxSpeed = 15;
float speed = sqrt(dx*dx+dy*dy);
if (speed>maxSpeed)
{
dx *= maxSpeed/speed;
dy *= maxSpeed/speed;
}
x+=dx;
y+=dy;
.
.
.
}
In the above code, why is maxSpeed being divided by the speed variable. I'm stumped.
Thank you,
Josheir
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it caps the speed to maxSpeed.
basically it's a minor optimization instead of doing:
float maxSpeed = 15.0f;
float Len = sqrt(dx*dx+dy*dy);
float Speed = min(Len, maxSpeed);
dx = dx/Len*Speed;
dy = dy/Len*Speed;
which normalizes the vector(this means the vector has a length of 1) then scaling to your desired speed, the above sidesteps that dy/Len*Speed to do dy *= maxSpeed/Len; which is effectivly the same operation, but only done when your speed is actually higher instead of every frame.
if you don't understand what normalizing a vector means then i'd start with learning that, then things should become more clear.
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maxSpeed should be a float, not an int.
To unserstand the math, reduce it to 1D:
maxSpeed = 15
speed = 20
if (20>15)
{
factor = 15/20, so 0.75
speed * 0.75 = 15, so the max speed we want
}
That's easy. Do you understand the move from 2D to 1D here?
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It's been a long time...They're both helpful!
4 hours ago, Josheir said:
float speed = sqrt(dx*dx+dy*dy);
Now, I'm understanding that speed is the magnitude of the vector. What I'm not understanding is how dx's and dy's time numerator is working.
Thank you,
Josheir
Share on other sites
What we want is to set the magnitude to max speed, maybe altering the equation helps:
dx *= maxSpeed/speed;
new_dx = dx * maxSpeed / speed
new_dx = dx / speed * maxSpeed // note that speed is the current magnitude
so, first dividing by speed we set the magnitude of the vector to 1 (assuming we do it fot y as well)
Then by multiplying with maxSpeed the new magnitude becomes maxSpeed.
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Thank you, everyone. Enjoy the holiday season safely!
Josheir
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On 12/15/2017 at 1:12 PM, Josheir said:
float speed = sqrt(dx*dx+dy*dy);
A quick question: how is speed achieved from the sqrt(... )? Speed is distance over time and I am failing to understand if dx and dy are distance over time too?
Thank you,
Josheir
EDIT: perhaps the time is one second and the : dx += cos(angle*DEGTORAD)*.02
is adjusting the dx by the decimal multiplication.
Edited by Josheir
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1 hour ago, Josheir said:
A quick question: how is speed achieved from the sqrt(... )? Speed is distance over time and I am failing to understand if dx and dy are distance over time too?
dx and dy are the x and y components of the velocity vector, so they are in terms of "distance/time" along the x and y basis of the coordinate frame. The code uses the Pythagorean Theorem to calculate the length of that vector, i.e. the speed. Dividing by the speed at the end normalized the vector to length 1.0, and then multiplying by maxSpeed sets the speed to that while keeping the direction.
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Thank you, just making sure I am on the right page...
Sincerely,
Josheir
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That's pretty much the 'dimension reduction' i meant earlier what you ask for.
First let's look at this:
x = cos(angle);
y = sin(angle);
Depending on angle we always get a point with a distance of exactly 1 from the origin at (0,0).
(If we do it for every angle and plot each point, we get the unit circle with example points (0,1), (1,0), (0.7*0.7), (-1,0) etc. - Every possible direction, but each at a distance of one)
x = cos(angle) * 5;
y = sin(angle) * 5;
So if we multiply both dimensions with a constant (5), the resulting length of the vector becomes 5 as well, but how to calculate this if we don't know neither length nor angle?
Imagine a line from origin (0,0) to a given point (3,5). We want to know the length of that line. Note the right angled triangle we get if we set one dimension to zero, e.g.: (3,5) (3,0) (5,0). We get side lengths of the right angled sides of a:3 and b:5.
Now we can use Pythagoras right angled triangle rule applied to get the length if that vector.
a*a + b*b = c*c // c is the non right angled side we want to calculate
sqrt (a*a + b*b) = c
sqrt (3*3 +5*5) = 5.83
Note that a negative length of -3 would result in the same: (-3*-3 + 5*5) == (3*3 + 5*5). Signs get canceled out by multiplication with itself. So it does not matter what direction the vector points, to the negative or positive sides of our coordinate system, we get the correct length anyways.
This is what you see in that line:
speed = sqrt(x*x+y*y)
Here, we do not care which direction the 2D vector (x,y) points or in what direction an example vehicle is driving, we only care about its 1D speed, and that's the length of it's 2D vector.
There are other possibilities of reduced dimensions, e.g. if we set each y (upwards) coord of a rotating 3D cube to zero and draw this in black, we get a shadow of the cube. What we do is projecting the cube to the 2D xz plane, which is different to the example above where we use lengths.
1 hour ago, Josheir said:
EDIT: perhaps the time is one second and the : dx += cos(angle*DEGTORAD)*.02
The context of the given code is not clear, it seems dx,dy means acceleration, but i'm not sure what's the timestep etc.
Here is a example for simple physics of an object under constant external force like gravity to introduce some better terminology:
vec2 p(0,-8); // position
vec2 v(1,0.1); // velocity
vec2 a(0,0.2); // acceleration (we keep this constant like gravity)
float timestep = 0.16;
for (float time = 0; time < 20.0; time += timestep) // do a number of integration steps
{
v += a * timestep; // update velocity with acceleration
p += v * timestep; // update position with velocity
PlotPoint (p);
}
The plot should show a hyperbolic trajectory like we see when throwing a stone.
Games like Super Mario use this kind of physics, while games like Pac Man don't use acceleration and objects just use (mostly constant) velocity.
(You could rewrite this code using float pX=0, pY=-8; etc. instead vec2 p(0,-8);)
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## Addendum to Natural Topology regarding functions
In the appendix, I omitted to include the definition of function´ in section A.4 Constructive concepts and axioms used. The reason is that we use the standard classical definition:
Having taken the notion of set´ and subset´ as primitive, we define a function $f$ from an apartness space $(V,\#_1)$ to another apartness space $(W,\#_2)$ as a subset of the cartesian product $V\times W$ such that:
i) for all $x\in V$ there is a $y\in W$ such that $(x,y)\in f$
ii) for all $x,v\in V, y,z\in W$: if $(x,y)\in f$ and $(v,z)\in f$ and $y \#_2 z$ then $x \#_1 v$.
Then for any pair $(x,y)\in f$ we write: $f(x)\equiv y$ or $f(x)=y$.
The constructive interpretation of the quantifiers for all´ and there is´ ensures in our eyes that this definition nicely captures the connotation of methodicity which the word function´ carries. In the book we almost always work with morphisms anyway, but the definition above is strictly speaking necessary to underpin the theorems on representability of continuous functions by morphisms.
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# Turning x and y difference between two points into a projectile moving between them at a fixed speed
In a libgdx project of mine, I have two units on a 2d field who want to fire at each other. They both know the x and y distance to each other by subtracting their enemies position from their own:
Vector2 path = pos.cpy().sub(target.getPos());
What's the simplest way of turning this Vector2 into a vector that'll allow the units to fire at each other, with the projectile being sent off with Projectile.maxSpeed?
I was thinking of normalizing path and multiplying both x and y with the desired bullet speed, but I fear I may be overthinking this and that applying trigonometry would be simpler.
float x_diff = target.x - me.x;
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0
TECHNICAL PAPERS
# The Eshelby Tensors in a Finite Spherical Domain—Part II: Applications to Homogenization
[+] Author and Article Information
Shaofan Li1
Department of Civil and Environmental Engineering, University of California, Berkeley, CA 94720li@ce.berkeley.edu
Gang Wang, Roger A. Sauer
Department of Civil and Environmental Engineering, University of California, Berkeley, CA 94720
1
Corresponding author.
J. Appl. Mech 74(4), 784-797 (Jun 14, 2006) (14 pages) doi:10.1115/1.2711228 History: Received April 06, 2006; Revised June 14, 2006
## Abstract
In this part of the work, the Eshelby tensors of a finite spherical domain are applied to various homogenization procedures estimating the effective material properties of multiphase composites. The Eshelby tensors of a finite domain can capture the boundary effect of a representative volume element as well as the size effect of the different phases. Therefore their application to homogenization does not only improve the accuracy of classical homogenization methods, but also leads to some novel homogenization theories. This paper highlights a few of them: a refined dilute suspension method and a modified Mori–Tanaka method, the exterior eigenstrain method, the dual-eigenstrain method, which is a generalized self-consistency method, a shell model, and new variational bounds depending on the different boundary conditions. To the best of the authors’ knowledge, this is the first time that a multishell model is used to evaluate the Hashin–Shtrikman bounds for a multiple phase composite $(n⩾3)$, which can distinguish some of the subtleties of different microstructures.
<>
## Figures
Figure 1
Average Eshelby tensor coefficients s1I, s1E(i=1) and s2I, s2E(i=2)
Figure 10
Variational bounds for a three-phase composite material: (a) bounds for bulk modulus; and (b) bounds for shear modulus.
Figure 11
Influence of phase position on three-phase variational bounds
Figure 4
Illustration of interior and exterior eigenstrain method
Figure 5
Mori–Tanaka homogenization for the interior and exterior eigenstrain methods
Figure 6
Effective shear modulus for: (a) C̃=aCI+(1−a)CE, and (b) C̃=C¯
Figure 7
A three-layer shell model
Figure 8
Influence of β on the effective shear modulus
Figure 9
Improved Hashin–Shtrikman bounds for the effective bulk and shear moduli
Figure 2
Effective moduli κ¯, μ¯ (or κeff and μeff) obtained by using the dilute suspension method
Figure 3
Effective moduli κ¯, μ¯ (or κeff and μeff) obtained by using the Mori–Tanaka method
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# Was there any launch vehicle possible that could have been used for a heavier New Horizons with enough fuel to enter Pluto orbit? (adding ~10 years)
Answers to the Astronomy SE question Can New Horizons probe turn back and start orbiting Pluto are of course no, it would have to have been a different mission with a lot more fuel and a bigger launch vehicle.
Question(s):
1. Roughly how much more of a flying fuel tank would New Horizons have to have been (i.e. how many more kg of fuel) in order to slow down and match Pluto's orbit rather than fly past it. note: The new mission design can be very different, the launch vehicle can be very different, but add only fuel to the spacecraft and add no more than roughly a decade to the arrival time at pluto.
2. If the what looks like it's an Atlas 551 were swapped out for a hypothetical Atlas HLV or whatever else earthlings could have put together for a juicy contract at the time, could it have even done the job?
Source
I'm asking that we keep the payload and instrumentation of New Horizons the same, and just make the fuel tank's kilogramage bigger for a back of the spherical cow's envelope estimate.
Source
The Atlas V rocket used to launch the New Horizons spacecraft.
• Same flight plan, or is a Hohmann, or even a Bi-elliptic course on the table? Jun 26 at 4:04
• @notovny Good point! The flight plan is up to you, task is to get it there. You can add some years and some reasonable amount of flybys, up to adding another decade or so to the arrival time if necessary. Same payload, bigger tank, new mission design, new launch vehicle.
– uhoh
Jun 26 at 4:07
• highly relevant Requirements to orbit Pluto Jun 27 at 15:50
• @BrendanLuke15 I think you can add a short "supplementary answer" here, pointing to those answers and sourcing from references therein; the "adding another decade or so" is the real challenge!
– uhoh
Jun 28 at 2:15
New Horizons had a launch mass of 478 kg, of which 77 kg was propellant, but only around 47.5 kg of that was for course changes, with the rest reserved for attitude control. New Horizons was launched at a speed of 16.26 km/s, which put it on a solar system escape trajectory. It then also received a speed boost from the Jupiter flyby.
### Fuel requirements
New Horizon's propellant budget gave it a delta-V capability of "over 290 m/s". Its hydrazine propellant has an ISP (efficiency) of around 220 seconds. For going into orbit you would typically use a hohmann transfer instead of an escape trajectory that NH uses. That means the spacecraft would launch a bit slower so that it would come to an almost dead stop when arriving near Pluto. Pluto itself however is traveling in its orbit around the Sun, and to match its speed in order to orbit it, you need a delta-V of around 3000 m/s, so around ten times as much as NH had. Note also that a hohmann transfer to Pluto takes 45 years instead of the almost 10 years that New Horizons took.
The amount of fuel needed scales exponentially with your required delta-V. Using the same fuel, NH would need 1300 kg of it to get into Pluto orbit. Maybe a bit less if it uses a Charon gravity assist, but Pluto and Charon are not very large so they won't gravity-assist you much. But for an orbit insertion mission a probe would typically not use hydrazine monopropellant, but a more efficient bipropellant. Galileo and Cassini used MMH-NTO bipropellant for their main thrusters. Cassini had an ISP of 312 seconds, and using that, NH would need 'only' around 720 kg of fuel to do the orbit insertion. Note that I am ignoring the extra mass from the larger fuel tanks and heavier thruster.
Instead of a hohmann transfer NH could also have taken a bi-elliptic transfer. Those take a lot longer, I haven't done the caluclation but probably we are talking about much more than a century here. According to this graph a bi-elliptic transfer could reduce the needed delta-V by up to about 7.5% when the time you take approaches infinity. (note: r₂/r₁ for Earth to Pluto is about 30.) This would reduce the required fuel from 720 to 640 kg.
### Could it have been done?
The Atlas V 551 was the second most powerful launch vehicle that existed at the time. The only more capable one was the Delta IV Heavy. Of course if the contract was juicy enough someone could have just have designed a new super heavy lift rocket, or just resurrect the Saturn V. Creating bigger rockets is relatively straightforward. So with enough money an orbital New Horizons mission was certainly possible. However Nasa is on a budget, and that wouldn't have happened just for a Pluto probe.
The Delta IV is a bit less than twice as powerful as the Atlas in what it can lift. Howver I don't know how to make the exact calculations on what it could have launched toward Pluto, as there are a number of complicating factors. New Horizons was launched at a speed of 16.26 km/s, but then it received a speed boost from Jupiter of around 4 km/s. Since Jupiter is already quite a way out there, that 4 km/s translates into (I think) a bit less of additional speed if the launch vehicle would have needed to supply the speed. So for a hohmann transfer which requires about 16 km/s the launch velocity with a Jupiter assist would have to be a bit more than 12 km/s. Secondly, the launch used an additional Star 48B as a third stage, to give NH more speed over what the Atlas could deliver by itself. The Star 48B weighs 2137 kg.
The NH would weigh some 2.5 times more with the fuel for a hohmann transfer trajectory, excluding the additional mass for the probe itself to accomodate the additional fuel. The Delta is only less than twice as powerful as the Atlas, but maybe the lower speed required for a hohmann transfer is just enough to make it possible. Or maybe not. I'm not sure.
### interplanetary transport network
There are other ways to get to Pluto, known as the interplanetary transport network. That means that you repeatedly fly by other planets in order to get multiple gravity assists. In principle it is possible to get anywhere in the solar system with minimal fuel requirements. You only need to get to the Earth-Moon L1 lagrange point and from there you can use gravity assists. The only problem is that this is much slower than even the bi-elliptic transfer. So how much time do you have?
edit
Oh, you are limiting the additional time to take to 'another decade or so'. That precludes even a simple hohmann transfer, and would require accellerating to an escape trajectory and then breaking at Pluto. That would clearly be impossible with the rockets available at that time, and probably also with the rockets that exist now (The only new more powerful rocket is the Falcon Heavy, but it isn't so good at bringing small things to a very high speed.) You'll need to wait for SpaceX's Starship or for someone to build a space grade nuclear reactor to power an ion drive.
• A nice answer thank you! Yes that was good spotting the "adding another decade or so to the arrival time" comment; I should have immediately moved that back into the question itself, and I'll do so now. Good catch! A quick question; "The NH would weigh some 2.5 times more with the fuel for a Hohmann transfer trajectory, the additional mass for the probe itself to accommodate the additional fuel." Can you add a little bit showing how that was obtained? I don't necessarily doubt the number but it's important for answers to support assertions, including quantitative ones. Thanks!
– uhoh
Jun 26 at 13:54
• @uhoh I got the numbers from this online calculator: strout.net/info/science/delta-v Jun 26 at 15:14
• I took the needed delta-V from this diagram on Wikipedia. The 3000 is roughly to a low Pluto orbit. For an barely bound elliptic orbit only 2700 m/s is needed according to the diagram. I'm not sure what part of Pluto's orbit these numbers apply to. Jun 28 at 7:22
• Note also that your subtraction applies to entering the same orbit as Pluto around the Sun while in deep space. Doing a burn while close to a planet is more efficient, so takes less delta-V. I'm not sure how much of a difference this makes for a small body like Pluto. See Oberth effect. Jun 28 at 7:25
• For the calculation I used a dry mass of 430 kg (including hydrazine for attitude control) and an ISP of 312 seconds (assuming bipropellant). A total mass of 1150 kg then gives a delta-V of 3009.9 m/s. For the monopropellant calculation I used an ISP of 220 seconds. A total mass of 1730 kg gives a delta-V of 3003.39 m/s. Jun 28 at 7:35
The answer to question 1 will come from the chosen trajectory's arrival $$V_\inf$$ (hyperbolic excess velocity). The answer to question 2 will come from the chosen trajectory's launch $$C3$$ (characteristic energy). The two are coupled to an individual trajectory dependent on dates and routes (which planets to flyby). Luckily, I have (student) experience in interplanetary trajectory searches.
I chose to maintain the (as flown) mission architecture of a Jupiter gravity assist to try to leverage a lower launch $$C3$$ (compared to Pluto direct) and a stronger gravity assist at Jupiter. It is also easier to compare with the as flown New Horizons primary mission.
Search Constraint Dates:
• Launch Dates: 2003 - end 2006
• JGA Dates: 2003 - end 2010
• Pluto Arrival Dates: 2008 - End 2030*
• state vector data from JPL Horizons (5 day step size)
*The New Horizons mission design supported arriving at Pluto as late as 2020 in a Pluto direct trajectory (New Horizons Mission Design, Guo et al.)
I used a heavily modified version (mainly the Lambert solver) of this MATLAB file exchange package$$^1$$ to create the porkchop plots for each leg of the trajectory (Earth-Jupiter, Jupiter-Pluto). Here is the initial leg (launch $$C3$$) with the New Horizons actual annotated:
The trick comes in finding suitable Jupiter to Pluto trajectories that match the hyperbolic excess velocity, $$V_\inf$$, approaching and leaving Jupiter, enabling a 'free' flyby/gravity assist. I usually consider a difference of less than 100 m/s to be sufficiently close.
The Pluto arrival hyperbolic excess velocity strongly favours a slower route from Jupiter to Pluto (again, New Horizons actual annotated). Slower is the name of the game here:
A secondary, not immediately evident constraint is how close you are willing to get to Jupiter and its strong radiation during the gravity assist. New Horizons passed at a close approach of about 33 $$R_J$$, Jovian radii (TRAJECTORY MONITORING AND CONTROL OF THE NEW HORIZONS PLUTO FLYBY, Guo et al.). I have no good knowledge on what is a reasonable close approach distance for a spacecraft flying by Jupiter (in the scope of radiation damage) so for now I will leave this constraint wide open (just don't have the close approach inside of Jupiter).
With these constraints, and a launch $$C3$$ less than 150 km$$^2$$/s$$^2$$, there are 32,075 viable trajectories, shown here in this nifty plot of the key values we care about:
The red markers show the region of most efficient trajectories (low $$C3$$ and $$V_\inf$$). The most efficient trajectory is:
Launch: Jupiter Flyby: Pluto Arrival: $$C3$$: $$V_\inf$$ @ Pluto: Jupiter Close Approach:
12-Nov-2003 22-Oct-2005 31-Dec-2020 91.9 km$$^2$$/s$$^2$$ 3.72 km/s 24 $$R_J$$
It looks like this (with New Horizons actual, left, for comparison):
Observations:
• This trajectory arrives at Pluto on the last day of the time constraint (slower is the name of the game)
• The trajectory remains gravitationally bound to the Sun throughout the entire trajectory (S.M.A. E-J: 3.7 au, S.M.A. J-P: 30.3 au)
I am going to ignore Charon in the orbital insertion calculations. I am also going to assume a 'dry mass' (including attitude control propellant) of 500 kg for our New (and improved) Horizons probe to accommodate the increase in propellants. $$I_{sp}$$ of 220 s as discussed in JanKunis' answer.
1. Pluto Orbit Insertion (instantaneous burn at periapsis assumed):
Periapsis Distance: Velocity @ Periapsis: Capture Velocity @ Periapsis ($$C3$$=0): Circular Orbit Velocity:
1500 km (~300 km above surface) 3.872 km/s 1.078 km/s 762 m/s
Capture dV: Circular Orbit dV: Capture Initial Mass: Circular Orbit Initial Mass:
2793 m/s 3109 m/s 1824 kg 2112 kg
73% & 76% propellant by mass, respectively.
1. Launch Vehicles:
I used a previously developed algorithm of mine to determine how much mass a given launch vehicle + STAR48B combination can throw to a specified $$C3$$. Basic performance is taken from NASA Launch Services Program Launch Vehicle Performance Website and Northrop Grumman Propulsion Products Catalog. If a reference altitude is assumed you can get velocity from $$C3$$ and Star48B + spacecraft $$\Delta V$$ from the public specifications. Get a final velocity and then recompute the $$C3$$.
The Delta IV Heavy is no longer listed on the NASA Launch Services Program Launch Vehicle Performance Website (it was in April 2020), but I have a a curve fit of the data saved from some old school projects :). I realize that the proposed launch is ~1 year prior to the Delta IV Heavy's maiden flight but the contract could be very juicy to make this possible. I included other launch vehicles for comparison:
For a $$C3$$ of 91.9 km$$^2$$/s$$^2$$:
Delta IV Heavy: Atlas V 551: Atlas V 401: Falcon Heavy (Expandable): Falcon Heavy (Recoverable):
2215 kg 1381 kg 704 kg 2847 kg 1249 kg
A slim margin for sure, but definitely plausible.
1: Bogdan Danciu (2021). Interplanetary Mission Design (https://www.mathworks.com/matlabcentral/fileexchange/66192-interplanetary-mission-design), MATLAB Central File Exchange. Retrieved June 27, 2021.
• Holy granola this is great! We can certainly figure out some crazy/clever way to add a Juno radiation vault that can be at least partially ejected if necessary after the flyby and before more propellant is needed. It might not need to be 200 kg if it's just for one perijove and not an extended series of orbits if that's what it takes to make the mission work.
– uhoh
Jun 29 at 0:41
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Home > Standard Error > Standard Error Calc
# Standard Error Calc
## Contents
However, the mean and standard deviation are descriptive statistics, whereas the standard error of the mean describes bounds on a random sampling process. Estimate the sample mean for the given sample of the population data.
2. Blackwell Publishing. 81 (1): 75–81. TweetOnline Tools and Calculators > Math > Standard Error Calculator Standard Error Calculator Enter numbers separated by comma, space or line break: About This Tool The online Standard Error Calculator http://cpresourcesllc.com/standard-error/standard-error-versus-standard-deviation-excel.php
The manual calculation can be done by using above formulas. This lesson shows how to compute the standard error, based on sample data. This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called This article is a part of the guide: Select from one of the other courses available: Scientific Method Research Design Research Basics Experimental Research Sampling Validity and Reliability Write a Paper
## How To Calculate Standard Error Of The Mean In Excel
Statistical Notes. They report that, in a sample of 400 patients, the new drug lowers cholesterol by an average of 20 units (mg/dL). Two data sets will be helpful to illustrate the concept of a sampling distribution and its use to calculate the standard error. The standard deviation of a sample divided by √n is the SE of the sample.
• Dividing the sample standard deviation by the square root of sample mean provides the standard error of the mean (SEM).
Solved Example The below solved example for to estimate the
• The graphs below show the sampling distribution of the mean for samples of size 4, 9, and 25.
• Standard Error of the Mean The standard error of the mean is the standard deviation of the sample mean estimate of a population mean.
Formula: Where, SE = Standard Error s = Standard Deviation n = Size (Number of Observations) of the Sample. All Rights Reserved. The survey with the lower relative standard error can be said to have a more precise measurement, since it has proportionately less sampling variation around the mean. Standard Error Of Measurement Calculator In this scenario, the 400 patients are a sample of all patients who may be treated with the drug.
These assumptions may be approximately met when the population from which samples are taken is normally distributed, or when the sample size is sufficiently large to rely on the Central Limit The notation for standard error can be any one of SE, SEM (for standard error of measurement or mean), or SE. Add to my courses 1 Frequency Distribution 2 Normal Distribution 2.1 Assumptions 3 F-Distribution 4 Central Tendency 4.1 Mean 4.1.1 Arithmetic Mean 4.1.2 Geometric Mean 4.1.3 Calculate Median 4.2 Statistical Mode HintonList Price: $53.95Buy Used:$0.77Buy New: $35.74Texas Instruments Nspire CX CAS Graphing CalculatorList Price:$175.00Buy Used: $111.99Buy New:$159.99Approved for AP Statistics and Calculus About Us Contact Us Privacy Terms
The margin of error of 2% is a quantitative measure of the uncertainty – the possible difference between the true proportion who will vote for candidate A and the estimate of Standard Error Of Estimate Calculator Regression The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years. Student approximation when σ value is unknown Further information: Student's t-distribution §Confidence intervals In many practical applications, the true value of σ is unknown. Consider a sample of n=16 runners selected at random from the 9,732.
## Standard Error Of The Estimate Calculator
As the sample size increases, the sampling distribution become more narrow, and the standard error decreases. It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, σ, divided by the square root of the How To Calculate Standard Error Of The Mean In Excel How to cite this article: Siddharth Kalla (Sep 21, 2009). Standard Error Of Proportion Calculator For an upcoming national election, 2000 voters are chosen at random and asked if they will vote for candidate A or candidate B.
If there is no change in the data points as experiments are repeated, then the standard error of mean is zero. . . http://cpresourcesllc.com/standard-error/standard-deviation-vs-standard-error-formula.php The standard error is an estimate of the standard deviation of a statistic. When the true underlying distribution is known to be Gaussian, although with unknown σ, then the resulting estimated distribution follows the Student t-distribution. American Statistical Association. 25 (4): 30–32. Estimated Standard Error Formula
Compare the true standard error of the mean to the standard error estimated using this sample. In each of these scenarios, a sample of observations is drawn from a large population. If the population standard deviation is finite, the standard error of the mean of the sample will tend to zero with increasing sample size, because the estimate of the population mean http://cpresourcesllc.com/standard-error/standard-error-vs-standard-deviation-confidence-interval.php Take it with you wherever you go.
A practical result: Decreasing the uncertainty in a mean value estimate by a factor of two requires acquiring four times as many observations in the sample. Standard Error Calculator For Two Samples Footer bottom Explorable.com - Copyright © 2008-2016. If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative
## With n = 2 the underestimate is about 25%, but for n = 6 the underestimate is only 5%.
See unbiased estimation of standard deviation for further discussion. To estimate the standard error of a student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence The standard error estimated using the sample standard deviation is 2.56. Standard Error Formula Statistics Siddharth Kalla 286.2K reads Comments Share this page on your website: Standard Error of the Mean The standard error of the mean, also called the standard deviation of the mean,
The standard error of a proportion and the standard error of the mean describe the possible variability of the estimated value based on the sample around the true proportion or true Later sections will present the standard error of other statistics, such as the standard error of a proportion, the standard error of the difference of two means, the standard error of The standard error is computed from known sample statistics. this contact form Population parameter Sample statistic N: Number of observations in the population n: Number of observations in the sample Ni: Number of observations in population i ni: Number of observations in sample
Sokal and Rohlf (1981)[7] give an equation of the correction factor for small samples ofn<20. Statistic Standard Error Sample mean, x SEx = s / sqrt( n ) Sample proportion, p SEp = sqrt [ p(1 - p) / n ] Difference between means, x1 - They may be used to calculate confidence intervals. Or decreasing standard error by a factor of ten requires a hundred times as many observations.
A menu will appear that says “Paste Function”. Select “Stastical” from the left hand side of the menu, if necessary. Scroll down on the right hand side of the menu and All Rights Reserved. Of course, T / n {\displaystyle T/n} is the sample mean x ¯ {\displaystyle {\bar {x}}} .
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## Mechanization of Strong Kleene Logic for Partial Functions
• Even though it is not very often admitted, partial functions do play asignificant role in many practical applications of deduction systems. Kleenehas already given a semantic account of partial functions using three-valuedlogic decades ago, but there has not been a satisfactory mechanization. Recentyears have seen a thorough investigation of the framework of many-valuedtruth-functional logics. However, strong Kleene logic, where quantificationis restricted and therefore not truth-functional, does not fit the frameworkdirectly. We solve this problem by applying recent methods from sorted logics.This paper presents a resolution calculus that combines the proper treatmentof partial functions with the efficiency of sorted calculi.
### Additional Services
Author: Manfred Kerber, Michael Kohlhase urn:nbn:de:hbz:386-kluedo-3309 SEKI Report (93,20) Preprint English 1999 1999 Technische Universität Kaiserslautern 2000/04/03 Partial functions ; many-valued logic ; order-sorted logic ; resolution Fachbereich Informatik 0 Informatik, Informationswissenschaft, allgemeine Werke / 00 Informatik, Wissen, Systeme / 004 Datenverarbeitung; Informatik Standard gemäß KLUEDO-Leitlinien vor dem 27.05.2011
$Rev: 13581$
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# How to calculate the intercept of an indifference curve in an Edgeworth box?
Consider an exchange economy with two goods (Good 1 and 2) and two individuals ($$A$$ and $$B$$):
• $$A$$'s utility function is $$u_A(x_{A_1}, x_{A_2}) = 2x_{A_1} + 5x_{A_2}$$
• $$B$$'s utility function is $$u_B(x_{B_1}, x_{B_2}) = 3x_{B_1} + 7x_{B_2}$$
• $$A$$'s initial endowment is $$(\omega_{A_1}, \omega_{A_2}) = (4, 6)$$
• $$B$$'s initial endowment is $$(\omega_{B_1}, \omega_{B_2}) = (10, 2)$$
The core condition requires:
$$2x_{A_1} + 5x_{A_2} \geq 2\omega_{A_1} + 5\omega_{A_2} = 2 \times 4 + 5 \times 6 = 38$$
and
$$3x_{B_1} + 7x_{B_2} \geq 3\omega_{B_1} + 7\omega_{B_2} = 3 \times 10 + 7 \times 2 = 44$$
The line with equation $$2x_{A_1} + 5x_{A_2} = 38$$ hits the left boundary at:
$$(x_{A_1}, x_{A_2}) = (0,38/5), ~~~~(x_{B_1}, x_{B_2}) = (14, 2/5)$$
The line with equation $$3x_{B_1} + 7x_{B_2} = 44$$ hits the left boundary at:
$$(x_{A_1}, x_{A_2}) = (0,54/7) ~~~ (x_{B_1}, x_{B_2}) = (14, 2/7)$$
I don't understand how these intercepts have been calculated.
I've used the MRS and endowment point to calculate the $$(x_{A_1}, x_{A_2})$$ intercepts correctly:
$$x_{A_2} = -\frac{2}{5}x_{A_1} + \frac{38}{5}$$
But get an incorrect $$(x_{B_1}, x_{B_2})$$ intercept:
$$x_{A_1} = 14, ~~ x_{A_2} = 2$$
Please can someone enlighten me? No worries if you can't.
Alex
In an edgeworth box model the total amount of goods must be constant. In your problem this translates to
$$\begin{eqnarray} x_{A_1} + x_{B_1} &=& \omega_{A_1} + \omega_{B_1} = 10 + 4 = 14 \tag{1}\\ x_{A_2} + x_{B_2} &=& \omega_{A_2} + \omega_{B_2} = 6 + 2 = 8 \tag{2}\\ \end{eqnarray}$$
So let's consider the first intercept
$$(x_{A_1}, x_{A_2}) = (0, 38/5) \tag{3}$$
which you properly derived. If you replace these values in Eq. (1) you get
$$0 + x_{B_1} = 14 ~~~\Rightarrow~~~ x_{B_1} = 14 \tag{4}$$
and if you replace in Eq. (2) you get
$$\frac{38}{5} + x_{B_2} = 8 ~~~\Rightarrow~~~ x_{B_2} = \frac{2}{5} \tag{5}$$
So there you have it
$$(x_{A_1}, x_{A_2}) = (0,38/5), ~~~(x_{B_1}, x_{B_2}) = (14,2/5)$$
You can do the same for the other intercept, I will leave that for you to complete
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# Question about vector space and union of proper subspaces
Let $$k$$ be an infinite field.
Let $$V$$ be a vector space over $$k$$ and $$W_1,...,W_r$$ proper subspaces of $$V$$.
Show that $$\bigcup_{i=1}^r W_i \not = V.$$
I tried the following:
for all $$j \in \{1,...,r\}$$, I take $$w_j \in W_j$$ such that $$w_j \not\in W_i$$ whenever $$j \not=i$$, so I know that $$w_1+\cdots+w_r \in V$$. If $$w_1+\cdots+w_r \in \bigcup_{i=1}^r W_i$$, then there is $$l \in \{1,...,r\}$$ such that $$w_1+\cdots+w_r \in W_l$$. I don't find because $$w_1+\cdots+w_r \in W_l$$ is absurd.
Is this correct reasoning, or is there other way for me to prove this?
• Your idea isn't entirely the right one: Consider what would happen if, say, $W_r$ contains all the other $W_i$. Then $\bigcup W_i=W_r$ is actually a vector space. Also, are you absolutely certain that $k$ is supposed to be finite? Because for finite $k$, and finite dimensional $V$, this just isn't true. – Arthur Feb 4 at 17:10
The assertion seems to be false. Take $$k=\mathbb{F}_2$$ and $$V=k\oplus k=\{(0,0),(1,0),(0,1),(1,1)\}$$. Now you can take $$r=3$$ and the following proper subspaces of $$V$$: $$W_1=\{(0,0),(1,0)\}$$, $$W_2=\{(0,0),(1,1)\}$$, and $$W_3=\{(0,0),(1,1)\}$$.
Hypothesis: Given a vector space $$V$$ over a finite field $$k$$. Define the proper subspaces $$W_1,...,W_r \subset V$$. Then $$\bigcup_{i=1}^{r} W_i \neq V$$.
Suppose that our hypothesis is true, and consider a vector space $$V$$ over the finite field $$\mathbb Z_2$$, with a canonical basis $$(e_1, e_2)$$, and define the following: $$W_1 = \mathbb Z_2 \cdot e_1, \space W_2 = \mathbb Z_2 \cdot e_2, \space and \space W_3 = \mathbb Z_2 \cdot (e_1 +e_2) \\$$
Clearly, $$V = \{(0,0), (1,0), (0,1), (1,1)\}$$, since these are the vectors generated by our canonical basis. Thus, $$V = W_1 \cup W_2 \cup W_3$$, which is a contradiction.
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# Appendix a to applied probability: directory of m-functions and m (Page 20/24)
Page 20 / 24
## Quantile functions for bounded distributions
dquant.m function t = dquant(X,PX,U) determines the values of the quantile function for a simple random variable with distribution $X,PX$ at the probability values in row vector U . The probability vector U is often determined by a random number generator.
function t = dquant(X,PX,U) % DQUANT t = dquant(X,PX,U) Quantile function for a simple random variable% Version of 10/14/95 % U is a vector of probabilitiesm = length(X); n = length(U);F = [0 cumsum(PX)+1e-12];F(m+1) = 1; % Makes maximum value exactly one if U(n)>= 1 % Prevents improper values of probability U U(n) = 1;end if U(1)<= 0 U(1) = 1e-9;end f = rowcopy(F,n); % n rows of Fu = colcopy(U,m); % m columns of U t = X*((f(:,1:m)<u)&(u<= f(:,2:m+1)))';
dquanplot.m Plots as a stairs graph the quantile function for a simple random variable X . The plot is the values of X versus the distribution function F X .
% DQUANPLOT file dquanplot.m Plot of quantile function for a simple rv % Version of 7/6/95% Uses stairs to plot the inverse of FX X = input('Enter VALUES for X ');PX = input('Enter PROBABILITIES for X '); m = length(X);F = [0 cumsum(PX)];XP = [X X(m)];stairs(F,XP) gridtitle('Plot of Quantile Function') xlabel('u')ylabel('t = Q(u)') hold onplot(F(2:m+1),X,'o') % Marks values at jumps hold off
dsample.m Calculates a sample from a discrete distribution, determines the relative frequencies of values, and compares with actual probabilities. Input consists of value andprobability matrices for X and the sample size n . A matrix U is determined by a random number generator, and the m-function dquant is used to calculate the corresponding sample values. Variousdata on the sample are calculated and displayed.
% DSAMPLE file dsample.m Simulates sample from discrete population % Version of 12/31/95 (Display revised 3/24/97)% Relative frequencies vs probabilities for % sample from discrete population distributionX = input('Enter row matrix of VALUES '); PX = input('Enter row matrix of PROBABILITIES ');n = input('Sample size n '); U = rand(1,n);T = dquant(X,PX,U); [x,fr]= csort(T,ones(1,length(T))); disp(' Value Prob Rel freq')disp([x; PX; fr/n]')ex = sum(T)/n; EX = dot(X,PX);vx = sum(T.^2)/n - ex^2; VX = dot(X.^2,PX) - EX^2;disp(['Sample average ex = ',num2str(ex),])disp(['Population mean E[X] = ',num2str(EX),]) disp(['Sample variance vx = ',num2str(vx),]) disp(['Population variance Var[X]= ',num2str(VX),])
quanplot.m Plots the quantile function for a distribution function F X . Assumes the procedure dfsetup or acsetup has been run. A suitable set U of probability values is determined and the m-function dquant is used to determine corresponding values of the quantile function. The results are plotted.
% QUANPLOT file quanplot.m Plots quantile function for dbn function % Version of 2/2/96% Assumes dfsetup or acsetup has been run % Uses m-function dquantX = input('Enter row matrix of values '); PX = input('Enter row matrix of probabilities ');h = input('Probability increment h '); U = h:h:1;T = dquant(X,PX,U); U = [0 U 1]; Te = X(m) + abs(X(m))/20;T = [X(1) T Te];plot(U,T) % Plot rather than stairs for general case gridtitle('Plot of Quantile Function') xlabel('u')ylabel('t = Q(u)')
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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## 25.4 Closed immersions of locally ringed spaces
We follow our conventions introduced in Modules, Definition 17.13.1.
Definition 25.4.1. Let $i : Z \to X$ be a morphism of locally ringed spaces. We say that $i$ is a closed immersion if:
1. The map $i$ is a homeomorphism of $Z$ onto a closed subset of $X$.
2. The map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective; let $\mathcal{I}$ denote the kernel.
3. The $\mathcal{O}_ X$-module $\mathcal{I}$ is locally generated by sections.
Lemma 25.4.2. Let $f : Z \to X$ be a morphism of locally ringed spaces. In order for $f$ to be a closed immersion it suffices that there exists an open covering $X = \bigcup U_ i$ such that each $f : f^{-1}U_ i \to U_ i$ is a closed immersion.
Proof. Omitted. $\square$
Example 25.4.3. Let $X$ be a locally ringed space. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals which is locally generated by sections as a sheaf of $\mathcal{O}_ X$-modules. Let $Z$ be the support of the sheaf of rings $\mathcal{O}_ X/\mathcal{I}$. This is a closed subset of $X$, by Modules, Lemma 17.5.3. Denote $i : Z \to X$ the inclusion map. By Modules, Lemma 17.6.1 there is a unique sheaf of rings $\mathcal{O}_ Z$ on $Z$ with $i_*\mathcal{O}_ Z = \mathcal{O}_ X/\mathcal{I}$. For any $z \in Z$ the stalk $\mathcal{O}_{Z, z}$ is equal to a quotient $\mathcal{O}_{X, i(z)}/\mathcal{I}_{i(z)}$ of a local ring and nonzero, hence a local ring. Thus $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ is a closed immersion of locally ringed spaces.
Definition 25.4.4. Let $X$ be a locally ringed space. Let $\mathcal{I}$ be a sheaf of ideals on $X$ which is locally generated by sections. The locally ringed space $(Z, \mathcal{O}_ Z)$ of Example 25.4.3 above is the closed subspace of $X$ associated to the sheaf of ideals $\mathcal{I}$.
Lemma 25.4.5. Let $f : X \to Y$ be a closed immersion of locally ringed spaces. Let $\mathcal{I}$ be the kernel of the map $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$. Let $i : Z \to Y$ be the closed subspace of $Y$ associated to $\mathcal{I}$. There is a unique isomorphism $f' : X \cong Z$ of locally ringed spaces such that $f = i \circ f'$.
Proof. Omitted. $\square$
Lemma 25.4.6. Let $X$, $Y$ be locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a sheaf of ideals locally generated by sections. Let $i : Z \to X$ be the associated closed subspace. A morphism $f : Y \to X$ factors through $Z$ if and only if the map $f^*\mathcal{I} \to f^*\mathcal{O}_ X = \mathcal{O}_ Y$ is zero. If this is the case the morphism $g : Y \to Z$ such that $f = i \circ g$ is unique.
Proof. Clearly if $f$ factors as $Y \to Z \to X$ then the map $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero. Conversely suppose that $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero. Pick any $y \in Y$, and consider the ring map $f^\sharp _ y : \mathcal{O}_{X, f(y)} \to \mathcal{O}_{Y, y}$. Since the composition $\mathcal{I}_{f(y)} \to \mathcal{O}_{X, f(y)} \to \mathcal{O}_{Y, y}$ is zero by assumption and since $f^\sharp _ y(1) = 1$ we see that $1 \not\in \mathcal{I}_{f(y)}$, i.e., $\mathcal{I}_{f(y)} \not= \mathcal{O}_{X, f(y)}$. We conclude that $f(Y) \subset Z = \text{Supp}(\mathcal{O}_ X/\mathcal{I})$. Hence $f = i \circ g$ where $g : Y \to Z$ is continuous. Consider the map $f^\sharp : \mathcal{O}_ X \to f_*\mathcal{O}_ Y$. The assumption $f^*\mathcal{I} \to \mathcal{O}_ Y$ is zero implies that the composition $\mathcal{I} \to \mathcal{O}_ X \to f_*\mathcal{O}_ Y$ is zero by adjointness of $f_*$ and $f^*$. In other words, we obtain a morphism of sheaves of rings $\overline{f^\sharp } : \mathcal{O}_ X/\mathcal{I} \to f_*\mathcal{O}_ Y$. Note that $f_*\mathcal{O}_ Y = i_*g_*\mathcal{O}_ Y$ and that $\mathcal{O}_ X/\mathcal{I} = i_*\mathcal{O}_ Z$. By Sheaves, Lemma 6.32.4 we obtain a unique morphism of sheaves of rings $g^\sharp : \mathcal{O}_ Z \to g_*\mathcal{O}_ Y$ whose pushforward under $i$ is $\overline{f^\sharp }$. We omit the verification that $(g, g^\sharp )$ defines a morphism of locally ringed spaces and that $f = i \circ g$ as a morphism of locally ringed spaces. The uniqueness of $(g, g^\sharp )$ was pointed out above. $\square$
Lemma 25.4.7. Let $f : X \to Y$ be a morphism of locally ringed spaces. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a sheaf of ideals which is locally generated by sections. Let $i : Z \to Y$ be the closed subspace associated to the sheaf of ideals $\mathcal{I}$. Let $\mathcal{J}$ be the image of the map $f^*\mathcal{I} \to f^*\mathcal{O}_ Y = \mathcal{O}_ X$. Then this ideal is locally generated by sections. Moreover, let $i' : Z' \to X$ be the associated closed subspace of $X$. There exists a unique morphism of locally ringed spaces $f' : Z' \to Z$ such that the following diagram is a commutative square of locally ringed spaces
$\xymatrix{ Z' \ar[d]_{f'} \ar[r]_{i'} & X \ar[d]^ f \\ Z \ar[r]^{i} & Y }$
Moreover, this diagram is a fibre square in the category of locally ringed spaces.
Proof. The ideal $\mathcal{J}$ is locally generated by sections by Modules, Lemma 17.8.2. The rest of the lemma follows from the characterization, in Lemma 25.4.6 above, of what it means for a morphism to factor through a closed subspace. $\square$
Comment #404 by Keenan on
In the proof of 01HP, $\mathscr{I}_y$ should be $\mathscr{I}_{f(y)}$ in a three places.
Comment #434 by Leeroy on
In example 25.4.3 it should be $\mathcal{O}_{X,z}/\mathcal{I}_z$ (or $\mathcal{O}_{X,i(z)}/\mathcal{I}_{i(z)}$) and not $\mathcal{O}_{X,x}/\mathcal{I}_x$.
I was wondering, are you interested in people writing down those omitted simple proofs or do you not write them down just to make the thing more readable ? Obviously you would wish for people to contribute more meaningful stuff but since I'm learning I can't contribute anything interesting.
Comment #436 by on
Thanks for typo! Fixed here.
Yes, the goal is to fill in all the omitted proofs and any contribution like that is welcomed. If it becomes unreadable because of this (but I personally have never found this to be a problem), then we can use a technological solution to hide proofs (behind a link or something). Most of the omitted proofs should be easier or equivalent to things that are being proved in the same section or chapter. If you find that this is not the case, then please leave a comment saying so.
Comment #2582 by Dario Weißmann on
Typo in Lemma 25.4.6: Let $X,Y$ be $a$ locally ringed spaces.
Comment #3779 by qiao on
Typo in Lemma 01HP (Lemma 25.4.6) $f^*\mathcal{I} \to f^*\mathcal{O}_ X = \mathcal{O}_ Y$ perhaps not "=" here
Comment #3780 by qiao on
Typo in Lemma 01HP (Lemma 25.4.6) perhaps not "=" here
Comment #3909 by on
@#3779 and #3780: I am not going to change that for now. If more people complain then I will.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# Fourier-analytic proof of Sperner
## Fourier analysis
We identify $[2]^n$ with the power set $2^{[n]}$ of [n]. Given any function $f: 2^{[n]} \to {\Bbb R}$, we define the Fourier transform $\hat f: 2^{[n]} \to {\Bbb R}$ by the formula
$\hat f(A) := {\Bbb E}_{X \in 2^{[n]}} (-1)^{|A \cap X|} f(X)$;
we have the Plancherel formula
$\sum_{A \in 2^{[n]}} |\hat f(A)|^2 = {\Bbb E}_{X \in 2^{[n]}} |f(X)|^2$ (1)
and the inversion formula
$f(X) = \sum_{A \in 2^{[n]}} (-1)^{|A \cap X|} \hat f(A)$. (2)
For any scale $1 \lt r \lt n$, we consider the bilinear form
$B_r( f, g ) := {\Bbb E} f(X) g(Y)$ (3)
where X is drawn randomly from $2^{[n]}$, and Y is formed from X by setting $j \in Y$ whenever $j \in X$, and when $j \not \in X$, setting $j \in Y$ with probability $r/n$. To avoid significant distortion in the measure, we will be working in the regime $1 \leq r \ll \sqrt{n}$. Note that if $B_r(1_A, 1_A)$ is large, and r is not too small, then A will contain many combinatorial lines (since Y will be strictly larger than X with high probability); thus it is of interest to get lower bounds on $B_r$.
We also define the influence Inf(f) as
$Inf(f) := {\Bbb E} |f(X) - f(X')|^2$ (4)
where X' is formed from X by randomly flipping one bit (from 0 to 1, or vice versa). In Fourier space, we have
$Inf(f) = 4\sum_{A \in 2^{[n]}} \frac{|A|}{n} |\hat f(A)|^2$. (5)
## Dichotomy between structure and randomness
For any $f: 2^{[n]} \to {\Bbb R}$, define the uniformity norm $\|f\|_{U(r)}$ at scale r by the formula
$\|f\|_{U(r)} := \sup_g \max( |B_r(f,g)|, |B_r(g,f)| )$ (6)
where g ranges over all functions $g: 2^{[n]} \to [-1,1]$. Functions which are small in the uniform norm are thus "negligible" for the purposes of computing $B_r$ and can be discarded. (This is analogous to the theory of counting arithmetic progressions of length three, in which functions with small Fourier coefficients (or small Gowers $U^2$ norm) can be discarded.)
Lemma 1 Let $f: 2^{[n]} \to [-1,1]$ be such that $\|f\|_{U(r)} \geq \eta$, where $1 \leq r \ll \sqrt{n}$ and $\eta \gt 0$. Then there exists a function $F: 2^{[n]} \to [-1,1]$ of influence $Inf(F) \ll_\eta \frac{1}{r}$ such that
$\langle f, F \rangle := {\Bbb E}_{X \in 2^{[n]}} f(X) F(X) \gg_\eta 1.$
It seems that this is more or less proven in Ryan's notes. (May be worth redoing it here on the wiki. Note that the scale r here basically corresponds to $\epsilon n$ in Ryan's notes.)
## Structure theorem
Now we use the general "energy increment" argument (see e.g. Terry's paper on this) to obtain a structural theorem.
Corollary 2 (Koopman-von Neumann type theorem) Let $f: 2^{[n]} \to [0,1]$ and $0 \lt \eta \lt 1$. Then there exists $k = k(\eta)$ such that whenever $\sqrt{n} \gg r_k \gt \ldots \gt r_1 \gt 1$ be a sequence of scales. Then there exists $1 \leq i \leq k$ and a decomposition
$f = f_{U^\perp} + f_U$ (7)
where $f_{U^\perp}, f_U$ take values in [-1,1],
$Inf( f_{U^\perp} ) \ll_{\eta} 1/r_i$ (8)
and
$\| f_U \|_{U(r_{i+1})} \leq \eta.$ (9)
Furthermore, $f_{U^\perp}$ is non-negative and
${\Bbb E} f = {\Bbb E} f_{U^\perp}.$ (10)
Proof (sketch) It's likely that there is now a slick proof of this type of thing using the duality arguments of Gowers or of Reingold-Trevisan-Tulsiani-Vadhan. But here is the "old school" approach from my long AP paper with Ben, running the energy increment algorithm:
1. Initialise i=1, and let ${\mathcal B}$ be the trivial partition of $[2]^n$.
2. Let $f_{U^\perp} := {\Bbb E}(f|{\mathcal B})$ be the conditional expectation of f with respect to the partition; initially, this is just the constant function ${\Bbb E} f$, but becomes more complicated as the partition gets finer. Set $f_U := f - f_{U^\perp}$
3. If $\| f_U \|_{U(r_{i+1})} \leq \eta$ then STOP.
4. Otherwise, we apply Lemma 1 to find a function F of influence $O_\eta(1/r_{i+1})$ that $f_U$ correlates with. Partition F into level sets (discretising in multiples of $\eta$ or so, randomly shifting the cut-points to avoid edge effects) and use this to augment the partition ${\mathcal B}$. In doing so, the energy ${\Bbb E} |f_{U^\perp}|^2$ increases by some non-negligible amount $c(\eta) \gt 0$, thanks to Pythagoras' theorem: this is why this process will stop before k steps.
A certain tedious amount of verification is needed to show that all the hypotheses are satisfied. One technical step is that one needs to use the Weierstrass approximation theorem to approximate $f_{U^\perp}$ by a polynomial combination of low-influence functions, and one needs to show that polynomial combinations of low-influence functions are low-influence. The complexity of the polynomial is controlled by some function of $\eta$ and this explains the losses in that parameter in (8). $\Box$
## DHJ(2)
Now we prove DHJ(2). Let $f = 1_A$ have density $\delta$. We let $\eta$ be a small number depending on $\delta$. Now we pick some scales $1 \leq r_1 \lt \ldots \lt r_k$ between 1 and $\sqrt{n}$ (the exact choices are not so important, so long as the scales are widely separated), and apply the above corollary, to decompose $f = f_{U^\perp} + f_U$, where $f_U$ is uniform at some scale $r_i$ and $f_{U^\perp}$ is low-influence at a coarser scale $r_{i+1}$.
We then look at $B_{r_i}(f, f)$. Decomposing into four pieces and using the uniformity of $f_U$, this expression is equal to $B_{r_i}(f_{U^\perp}, f_{U^\perp}) + O(\eta)$. But the low influence of $f_{U^\perp}$ means that when one randomly flips a bit from a 0 to a 1, $f_{U^\perp}$ changes by $O_\eta(1/r_{i+1})$ on the average. Iterating this $r_i$ times, we see that $B_{r_i}(f_{U^\perp}, f_{U^\perp}) = {\Bbb E}( f_{U^\perp}^2 ) + O_\eta( r_i / r_{i+1} )$. But $f_{U^\perp}$ has density $\delta$, and so the main term is at least $\delta^2$. Choosing $\eta$ small enough and the $r_i$ widely separated enough we obtain a nontrivial lower bound on $B_{r_i}(f_{U^\perp}, f_{U^\perp})$, which gives DHJ(2).
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# A point lies on a dilation of a conic with scale factor $k$ iff its polar is tangent to the dilation of that conic with scale factor $1/k$
While reading coordinate geometry I learned a interesting fact.
If the pole of a straight line wrt a conic of parameter $a$ lies on a similar conic with parameter $a/n$ then the straight line is tangent to the similar conic with parameter $na$
Here conic refers to parabola,ellipse,hyperbola and circles as well.The parameter for circle is radius , for parabola it's latus rectum and for ellipse and hyperbola its length of major axis.By similar I mean that for a circle it has the same centre ,for parabola same vertex and axis and for ellipse and hyperbola same centre and major,minor axis
I am not able to imagine intuitively as to why this happens.Are there any interesting problems that directly uses this and is it valid for all cases?
Proposed restatement by @Blue. For a conic $C$, let $C_{k}$ be a dilation of $C$ with scale factor $k$. For $C$ a parabola, the center of dilation is the vertex of $C$; otherwise, it is the center of $C$.
Conjecture. $$\text{Point P lies on C_k (k\neq 0) iff the polar of P with respect to C is tangent to C_{1/k}.}$$
For illustration, the figures below show conics $C$, along with dilations $C_2$ (containing $P$) and $C_{1/2}$ (containing $Q$). In each case, the polar of $P$ with respect to $C$ is tangent to $C_{1/2}$, while the polar of $Q$ is tangent to $C_2$.
• You open by saying that you "learned a[n] interesting fact", but close by asking "is it valid for all cases?". Is this a conjecture of your own devising? If so, can you provide whatever evidence you have that leads you to believe it's true? Examples would also help clarify your special usage of "parameter" and "similar". – Blue Oct 25 '17 at 22:46
• I am not sure I just tried to verify it for simple cases.This is not something I found in some book but after reading a example from the book which said to prove that if the pole of a line wrt circle $x^2+y^2=a^2$ lies on $x^2+y^2=9a^2$ then the line is tangent to circle $x^2+y^2=a^2/9$ I accordingly tried to check if I could replace $9$ by any integer and whether it is valid for parabola,ellipse and hyperbola as well. – user471651 Oct 26 '17 at 1:41
• @blue In the example given in book above all the three circles are similar cause they have the same center and differ only in parameter i.e. the radius. This is my own framing of the statement and it might be more general or more strict . – user471651 Oct 26 '17 at 1:43
• I've edited your question with what I believe to be a proper restatement of your conjecture. Let me know if it doesn't match your intention. – Blue Oct 26 '17 at 6:05
• @blue sorry,I couldn't follow your restatement.the point P lies on a conic of parameter a/n and the polar of that point wrt conic of parameter a is tangent to conic of parameter na.is that what your restatement imply? – user471651 Oct 26 '17 at 8:42
It seems that a polar of a point with respect to a conic does not depend on the choice of coordinate system (see, for instance [VT, p. 102]; unfortunately, I failed to find a precise formulation of this fact neither in my (electronic) books nor by googling for English and Russian references), so then it suffices to consider the affine coordinate systems in which the conic is determined by canonical equations. Anyway, in [KK] are given the equations of the polar for a conic determined by a general equation (in [VT] a point determining the polar is assumed to be different from the center of a conic), so at least we prove the claims for convenient coordinate systems. We recall the respective equations.
1) Assume that the conic $C$ is an ellipse or a hyperbola. Put the origin of the coordinate system to its center. According to [KK, 2.4-6] the conic $C$ has an equation $$ax^2+2bxy+cy^2+d=0.$$ So an equation of a conic $C_k$ is $$ax^2+2bxy+cy^2+dk^2=0$$
and an equation of a conic $C_{1/k}$ is $$ax^2+2bxy+cy^2+d/k^2=0.$$
An equation of a polar of a point $(x_1,y_1)$ with respect to the conic $C$ [KK, 2.4-10] is $$axx_1+b(xy_1+yx_1)+cyy_1+d=0$$ and an equation of a straight line tangent to the conic $C_{1/k}$ at its point $(x_2,y_2)$ according to [KK, 2.4-10] is $$axx_2+b(xy_2+yx_2)+cyy_2+d/k^2=0.$$
Now we can prove the equivalence.
Point $P$ lies on $C_k$ ($k\neq 0$) iff the polar of $P$ with respect to $C$ is tangent to $C_{1/k}$.
$\Rightarrow$. Assume that a point $P(x_1,y_1)$ lies on the conic $C_k$. Then a point $P’(x_1/k^2,y_1/k^2)$ lies on the conic $C_{1/k}$. An equation of a straight line tangent to the conic $C_{1/k}$ at the point $P’$ is
$$axx_1/k^2+b(xx_y/k^2+yx_1/k^2)+cyy_1/k^2+d/k^2=0.$$
which is the equation of the polar of $P$ with respect to the conic $C$.
$\Leftarrow$. Assume that the polar of $P(x_1,y_1)$ with respect to the conic $C$ is tangent to the conic $C_{1/k}$. That is there exist a point $(x_2,y_2)\in C_{1/k}$ such that the equations
$$axx_1+b(xy_1+yx_1)+cyy_1+d=0$$ and
$$axx_2+b(xy_2+yx_2)+cyy_2+d/k^2=0$$
define the same straight line. Since the conics do not pass through the origin we have $d\ne 0$. Since $ac-b^2\ne 0$ we obtain $(x_1,y_1)=(x_2k^2,y_2k^2)$, that is the point $P$ lies on the conic $C_k$.
2) Assume that the conic $C$ is a parabola. According to [KK, 2.4-8], we can rotate the coordinate axis and move the origin transforming the equation of a the conic $C$ to a form $$y^2=2px.$$
Then the vertex of the parabola is placed at the origin, so an equation of a conic $C_k$ is $$y^2=2pkx$$ and an equation of a conic $C_{1/k}$ is $$y^2=2(p/k)x.$$
An equation of a polar of a point $(x_1,y_1)$ with respect to the conic $C$ according to [KK, 2.4-10] is $$y_1y+p(x+x_1)=0$$ and an equation of a straight line tangent to the conic $C_{1/k}$ at its point $(x_2,y_2)$ according to [KK, 2.4-10] is (1) $$y_2y+p(x+x_2)/k=0.$$
Now we can prove the equivalence.
Point $P$ lies on $C_k$ ($k\neq 0$) iff the polar of $P$ with respect to $C$ is tangent to $C_{1/k}$.
Indeed, the polar $\ell$ of $P$ with respect to $C$ is tangent to $C_{1/k}$ iff there exists a point $P’(x_2, y_2)$ of the conic $C_{1/k}$ such that equation (1) defines the straight line $\ell$. Since $y_2^2=2(p/k)x_2$, equation (1) becomes $2ky_2y+2xp+ky_2^2=0$. This equation can define a straight line $\ell$ with suitable choice of $y_2$ iff $y_1=ky_2$ and $2px_1=ky_2^2$, that is iff a point $P(x_1,y_1)$ lies on the conic $C_k$.
References
[KK] Granino Korn, Theresa Korn Mathematical Handbook for scientists and engineers, 2nd edition, McGraw Hill, 1968 (Russian translation, Moskow, “Nauka”, 1973).
[VT] A. P. Veselov, E. V. Troitskiy Lectures on analytical geometry, Moskow, Center of applied research at faculty of mechanics and mathematics MGU, 2002 (in Russian).
• What is the book reference KK? – Narasimham Nov 17 '17 at 9:58
• @Narasimham I updated my answer. – Alex Ravsky Nov 17 '17 at 16:03
• @user471651 I completed my answer. Sorry for the delay. – Alex Ravsky Nov 17 '17 at 16:30
• Thanks. It is ok,I remember Korn&Korn, and IS&ES Sokolnikoff books and their authors – Narasimham Nov 17 '17 at 16:32
In particular case of ellipses is it like this ?
They are zoomable with respect the pole. Between any two polar rays we have "curvilinear" similar triangles similarly placed in homothety.
The ellipses are placed in a pencil of rays, only three are shown as above.
You can see two tangent lines in red, where at tangent points, $\psi = \phi - \theta = 0$ where $\phi$ is slope to x-axis, $\theta$ usual polar angle and $\psi$ radius vector to arc tangent angles which are all equal as labelled.
It requires similarity only as seen, $P$ need not necessarily be at the origin.
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# Gravitational forces in protoplanetary disks may push super-Earths close to their stars
Penn State-led astronomers found that as planets form out of the chaotic churn of gravitational, hydrodynamic — or, drag — and magnetic forces and collisions within the dusty, gaseous protoplanetary disk that surrounds a star as a planetary system starts to form, the orbits of these planets eventually get in synch, causing them to slide — follow the leader-style — toward the star.
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Exercise 10-9 (Part Level Submission) On July 31, 2017, Whispering Company engaged Minsk Tooling Company to construct a special-purpose piece of factory machinery. Construction was begun immediately and was completed on November 1, 2017. To help finance construction, on July 31 Whispering issued a $301,200, 3-year, 12% note payable at Netherlands National Bank, on which interest is payable each July 31.$195,200 of the proceeds of the note was paid to Minsk on July 31. The remainder of the proceeds was temporarily invested in short-term marketable securities (trading securities) at 10% until November 1. On November 1, Whispering made a final $106,000 payment to Minsk. Other than the note to Netherlands, Whispering’s only outstanding liability at December 31, 2017, is a$30,300, 8%, 6-year note payable, dated January 1, 2014, on which interest is payable each December 31. Collapse question part (a) Correct answer. Your answer is correct. Calculate the interest revenue, weighted-average accumulated expenditures, avoidable interest, and total interest cost to be capitalized during 2017. Interest revenue $Entry field with correct answer 2650 Weighted-average accumulated expenditures$Entry field with correct answer 48800 Avoidable interest $Entry field with correct answer 5856 Interest capitalized$Entry field with correct answer 5856 Click if you would like to Show Work for this question: Open Show Work SHOW LIST OF ACCOUNTS SHOW SOLUTION LINK TO TEXT Attempts: 3 of 5 used Collapse question part (b) Prepare the journal entries needed on the books of Whispering Company at each of the following dates. (Credit account titles are automatically indented when amount is entered. Do not indent manually. If no entry is required, select "No Entry" for the account titles and enter 0 for the amounts.) (1) July 31, 2017. (2) November 1, 2017. (3) December 31, 2017. Date Account Titles and Explanation Debit Credit cash 301200 Notes Payable 301200 (To record the note.) Notes Payable Cash (To record the payment to Minsk.) (To record the proceeds from the investment.) (To record the payment to Minsk.) 12/31
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- Convergence rate of inexact proximal point methods with relative error criteria for convex optimization Renato DC Monteiro (monteiroisye.gatech.edu) Benar F Svaiter (benarimpa.br) Abstract: In this paper, we consider a class of inexact proximal point methods for convex optimization which allows a relative error tolerance in the approximate solution of each proximal subproblem. By exploiting the special structure of convex optimization problems, we are able to derive surprising complexity bounds for the aforementioned class. As a consequence, we show that the best size of the projected gradients (resp., gradients) generated by the projected gradient (resp., steepest descent) method up to iteration $k$ is ${\cal O}(1/k)$ in the context of smooth convex optimization problems. Keywords: projected gradient, forward-backward splitting , inexact proximal point, convex optimization, steepest descent, complexity Category 1: Nonlinear Optimization Category 2: Convex and Nonsmooth Optimization Citation: Download: [PDF]Entry Submitted: 08/24/2010Entry Accepted: 08/24/2010Entry Last Modified: 05/26/2012Modify/Update this entry Visitors Authors More about us Links Subscribe, Unsubscribe Digest Archive Search, Browse the Repository Submit Update Policies Coordinator's Board Classification Scheme Credits Give us feedback Optimization Journals, Sites, Societies Optimization Online is supported by the Mathematical Optmization Society.
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# [OS X TeX] fonts in htlatex
Jeffrey J Weimer weimerj at email.uah.edu
Fri Jun 29 18:44:44 EDT 2007
On Jun 29, 2007, at 2:28 PM, M A wrote:
> Does anyone know how to make htlatex create an html file with a
> default sans serif font? I have tried using packages such as arev or
> helvet, along with various combinations of
> \renewcommand\familydefault{\sfdefault} or just specifying \sffamily.
> None of this seems to work; I keep getting whatever my browser default
> font is. Adding to this I have math, which I can manage to get to come
> out in computer modern (in a png file), but what if I want a png file
> with a different math font? I can't seem to make heads or tails of the
> tex4ht documentation, so any pointers would be appreciated.
>
document font problem, have you considered just hand-coding a CSS
directive in the header of your html file with a font specification
of sans serif for the entire document?
I believe this would be something like
<style>
*{font-family: Arial, Helvetica, sans serif;}
</style>
As to the math font in the PNGs, I would think this depends on what
you use to typeset and what fonts and encodings you load at the
outset, not to any settings in htlatex.
--
J. J. Weimer, Chemistry / Chemical & Materials Engineering
University of Alabama in Huntsville, MSB 125, 301 Sparkman Dr
Huntsville, AL 35899 phone: 256-824-6954
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# BA in EE or Physics?
1. Nov 16, 2009
I really believed that I had my mind made up on the direction I wanted to go, but as the time draws near for me to begin my academic adventure I realized I have been operating on an assumed belief... That it is easier to get a job with a BA in Electrical Engineering than with a BA in Physics.
I have ultimately chosen EE on this assumption, so I thought it might be wise to get some clarification on this point.
(the fact that reading through the EE forum scares the *&$% out of me doesn't help either) Thanks, Mac P.S. I am sure this has been discussed many times, so I apologize if my question is redundant. 2. Nov 16, 2009 ### kote At the BA level I would say with a fairly high level of certainty that it's easier to get a job with an engineering degree than a physics degree. Let starting salaries be evidence of this. Undergrad electrical engineers start at over$55k (http://www.bls.gov/oco/ocos027.htm). According to AIP, physics undergrads in private sector STEM start at $45k (http://www.aip.org/statistics/trends/highlite/emp2/figure7.htm). BLS puts physics PhDs at$52k starting salary (http://www.bls.gov/oco/ocos052.htm).
Obviously that last number is skewed by postdocs, but the $45k vs$55k average is telling. There are also a lot of other dynamics involved after the initial job, but if you're simply asking about how easy it is to get a job, engineers are in higher demand.
Edit: The EE major wouldn't be a BA would it? That makes it more complicated. I'm assuming we're talking about an ABET accredited US degree.
While I'm at it, if the EE forum scares you, engineers are in high demand in a wide range of fields depending on your other background including management, finance, consulting, technical sales, etc.
Last edited by a moderator: Apr 24, 2017
3. Nov 16, 2009
Actually, those topics scare me more than anything. I'll take my chances with Electrical Engineering.
Thank you for your input, and the links you provided. Those were actually quite useful.
I think I am just getting the last minute jitters and over-reacting.
4. Nov 17, 2009
### kote
Don't worry too much - just maybe try to turn that energy in to extra studying! I don't know most of the super specific things talked about on the engineering forums, and I graduated . Remember, you'll still get some core physics classes and be able to take physics electives as an EE if you choose to do so. You'll have time to change your mind after you get started too.
Those BLS pages are great even if they are just extreme high level views of the average job in each field. There's a lot of other information out there also if you're looking for more specific job info. Your career services office is probably a good place to start.
5. Nov 17, 2009
### turin
That is exactly what I was going to say. There is a lot of specialization even after you have already specialized to electrical. For instance, where I went, there were actually five different kinds of electrical engineering curricula: electronics, controls, communications, E&M, and power. And even then, by the senior year, these get more specialized to specific senior projects. So, for instance: I can get along OK with communications; controls is a bit of a stretch for me, and I have almost no clue about power.
Also, I new people in my department who got minors in physics. If you like physics, that might be an option. I never understood any of their physics homework at the time; it was so mysterious and alluring ...
6. Nov 18, 2009
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Letter | Open | Published:
Observation of the 1S–2P Lyman-α transition in antihydrogen
Abstract
In 1906, Theodore Lyman discovered his eponymous series of transitions in the extreme-ultraviolet region of the atomic hydrogen spectrum1,2. The patterns in the hydrogen spectrum helped to establish the emerging theory of quantum mechanics, which we now know governs the world at the atomic scale. Since then, studies involving the Lyman-α line—the 1S–2P transition at a wavelength of 121.6 nanometres—have played an important part in physics and astronomy, as one of the most fundamental atomic transitions in the Universe. For example, this transition has long been used by astronomers studying the intergalactic medium and testing cosmological models via the so-called ‘Lyman-α forest’3 of absorption lines at different redshifts. Here we report the observation of the Lyman-α transition in the antihydrogen atom, the antimatter counterpart of hydrogen. Using narrow-line-width, nanosecond-pulsed laser radiation, the 1S–2P transition was excited in magnetically trapped antihydrogen. The transition frequency at a field of 1.033 tesla was determined to be 2,466,051.7 ± 0.12 gigahertz (1σ uncertainty) and agrees with the prediction for hydrogen to a precision of 5 × 10−8. Comparisons of the properties of antihydrogen with those of its well-studied matter equivalent allow precision tests of fundamental symmetries between matter and antimatter. Alongside the ground-state hyperfine4,5 and 1S–2S transitions6,7 recently observed in antihydrogen, the Lyman-α transition will permit laser cooling of antihydrogen8,9, thus providing a cold and dense sample of anti-atoms for precision spectroscopy and gravity measurements10. In addition to the observation of this fundamental transition, this work represents both a decisive technological step towards laser cooling of antihydrogen, and the extension of antimatter spectroscopy to quantum states possessing orbital angular momentum.
Main
Challenges to antimatter Lyman-α spectroscopy include the difficulty of fabricating optical components and continuous-wave11 or pulsed laser sources at these extremely short wavelengths, as well as the scarcity of anti-atoms. The current observation was made possible by a number of technical advances, including the development of a solid-state-based, pulsed Lyman-α source12, implementation of innovative plasma control techniques13, and the ALPHA Collaboration’s recent, marked improvement in antihydrogen trapping and accumulation or ‘stacking’ rate. Stacking provides a sample of several hundred anti-atoms14, accumulated over several hours. Taking advantage of a nanosecond-scale laser pulse and our low-background annihilation detection, we have also inferred information on the antihydrogen velocity distribution.
Because matter and antimatter annihilate each other when they meet, antihydrogen must be created and then trapped in strong, inhomogeneous magnetic fields in an ultrahigh-vacuum chamber. The ALPHA-2 apparatus (Fig. 1a) is designed to combine antiprotons from CERN’s Antiproton Decelerator15 with positrons from a positron accumulator16,17 to produce and to trap atoms of antihydrogen. The techniques that we have developed to produce antihydrogen cold enough (below 0.54 K) to trap can be found elsewhere13,14,18,19. A typical trapping trial in ALPHA-2 involves mixing about 90,000 antiprotons and three million positrons to produce 50,000 antihydrogen atoms. Most of the antihydrogen atoms produced are either too energetic or in the wrong quantum state to be captured; typically 10–20 anti-atoms are trapped in one four-minute-long mixing cycle. We have employed the stacking technique referred to above to increase the total number of trapped antihydrogen atoms to several hundred for each measurement sequence.
The trapped anti-atoms are confined by the interaction of their magnetic moments with the inhomogeneous magnetic field, shown in Fig. 1b, c. The cylindrical trapping volume for antihydrogen has a diameter of 44.35 mm and an axial length of 280 mm. The magnetic field minimum of 1.033 T is near the centre of the trap, where the field is oriented axially. The field strength is flat to within one part in 10,000 in a 60 mm (axial) by 4.5 mm (radial) volume around the centre of the trap. The maximum field differential in the trap is 0.82 T along the axial direction and about 0.85 T along the radial direction.
Figure 2 shows the energy levels of hydrogen in the 1S and 2P states in a magnetic field. The 2S state is also depicted for reference. Antihydrogen atoms in the states labelled as 1Sc and 1Sd can be trapped. At zero magnetic field, the excited 2P state splits into two states (2P3/2 and 2P1/2) owing to the relativistic spin–orbit interaction. In non-zero magnetic fields, due to the Zeeman effect, the 2P3/2 state splits into four sublevels, while the 2P1/2 state splits into two. Each of these sublevels in turn splits into two states owing to the nuclear spin of the antiproton, but the splitting is much smaller (less than 0.1 GHz) than that in the ground (1S) state, and is therefore not visible on the scale of Fig. 2.
The 1S–2Pc transition studied here is a dipole-allowed transition. When antihydrogen is excited to one of the 2P sublevels, the excited state decays to the ground 1S state within a few nanoseconds by emitting a photon at 121.6 nm. Owing to the mixing of the positron spin states, there is a non-zero probability that the excited atom decays to either the 1Sa or 1Sb state instead of the original 1Sc or 1Sd state. Those that decay to the 1Sa or 1Sb state escape and annihilate at the trap walls and can be detected by the ALPHA-2 silicon vertex detector. The silicon vertex detector affords us single-atom detection capability, which is key to antimatter spectroscopy with few atoms4,5. The silicon vertex detector tracks the charged pions from the antiproton annihilation; the pion tracks are reconstructed to determine the location (vertex) of each annihilation. We employ machine-learning techniques, incorporating several parameters from the detector and the track reconstruction in a multi-variate analysis20 (MVA; see Methods), to distinguish antiproton annihilations from the cosmic ray background.
In this experiment, the antihydrogen atoms were excited to the 2Pc state (Fig. 2) by light that was linearly polarized, the polarization vector being nearly perpendicular to the direction of the axial magnetic field. Narrow-line-width (about 65 MHz), pulsed (about 12-ns duration) laser radiation at 121.6 nm was used for the excitation. The system is described in detail in Methods, and the schematic diagram of the laser system is shown in Extended Data Fig. 1.
For each experimental sequence, about 500 antihydrogen atoms were accumulated in the trapping region by multiple stacking (typically 30 stacks) over an approximately two-hour period. Experience from past microwave manipulation experiments5,6 indicates that the 1Sc and 1Sd states are equally populated in the initially trapped sample. The trapped atoms were irradiated for about two hours by laser pulses using a 10-Hz repetition rate. During each two-hour sequence, the trapped atoms were exposed to radiation at twelve different frequencies around the calculated hydrogen resonance frequency of 2,466,051.625 GHz (using the average of the two 1S hyperfine levels). The magnetic field magnitude was determined by electron cyclotron resonance21. The frequency detunings ranged from −2.47 GHz to +0.85 GHz. The laser frequency was switched every 20 s in a pattern designed to minimize saturation and depletion effects (due to the finite population size) at each frequency. After each two-hour exposure, the remaining antihydrogen atoms were released and counted by ramping down the trap magnets in 15.6 s. We repeated the described sequence four times. The number of detected events is summarized in Table 1. With average laser pulse energies between 0.53 nJ and 0.65 nJ, about 60% of the trapped antihydrogen atoms were expelled and detected during the exposure time of two hours.
Because the laser is pulsed, and the excitation light is present for about 12 ns for each pulse, the annihilation events due to the excitation are only expected to occur in the approximately 1-ms time window in which the untrapped atoms are forced to the trap wall. This greatly helps us to distinguish between signals due to laser-driven annihilations and those due to cosmic background. Figure 3 illustrates the power of the pulsed excitation method by comparing (1) on-resonance and (2) off-resonance measurements of the so-called time-of-flight (TOF) distribution. Even without the sophisticated machine-learning algorithm, raw detector triggers (Methods, Fig. 3a, black line) show a high signal level for the on-resonant case, in the time window between 0 ms and 1 ms after the laser pulse. The background is approximately constant. With the MVA (red line) the background almost disappears. In Fig. 3c, d we present two-dimensional plots for the axial position versus the TOF of the annihilation events. These plots also illustrate the selection criteria or ‘cuts’ (0 ms < t < 1 ms and −100 mm < z < 100 mm) used to accept events for the spectral shape determination described below. These cuts were predetermined on the basis of simulation information.
Figure 4a shows the spectral line shape of the transition, obtained using events with the above cuts applied. In this plot, the probability is normalized to the pulse energy; we have thus assumed that the excitation probability is linearly proportional to the laser pulse energy. The plotted data show that the transition from the 1Sc and 1Sd states to the 2Pc state has a line width of about 1.5 GHz (full-width at half-maximum, FWHM). If the hyperfine coupling constants of antihydrogen between the 1S and 2P states are the same as those for normal hydrogen, the spectrum in the magnetic field of 1.033 T would comprise two lines with a separation of 0.74 GHz. The two distinct lines are caused by the possible orientations of the antiproton nuclear spin in the 1S ground state. However, this hyperfine splitting is not resolved under the present experimental conditions owing to the Doppler broadening of the transitions caused by the motion of trapped antihydrogen atoms parallel with the laser beam. We note that the Doppler shift of the Lyman-α transition of hydrogen with a velocity of 75 m s−1 (characteristic for our magnetic trap depth) is 0.6 GHz. A fit to the measured data yields a resonant frequency of 2,466,051.7(0.12) GHz, in good agreement with the calculated hydrogen frequency quoted above. The contributions to the quoted uncertainty of 0.12 GHz are the wavemeter accuracy (0.06 GHz), the 730-nm laser (Methods) cavity lock stability (0.06 GHz), modelling uncertainties (0.07 GHz), and the statistical uncertainty in the curve fit (0.04 GHz). Many of these uncertainties may be reduced in the future, but the large natural line width of 2π × 99.6 MHz is an obstacle to using the Lyman-α transition for very high-precision measurements as tests of charge–parity–time-reversal (CPT) invariance.
The kinetic energy distribution of the trapped anti-atoms is also of interest. We expect, from previous measurements and simulations of the antihydrogen production and trapping processes22, that the trapped antihydrogen atoms are not in thermal equilibrium. Our standard model for antihydrogen formation is that the atoms are formed from antiprotons in equilibrium with the positron cloud, which has a temperature of about 20 K. The expected distribution of trapped atoms thus comprises the low-temperature tail of a Maxwellian distribution, truncated at the maximum trap depth of about 0.5 K. The measured line shape of the transition should thus be different from a Gaussian shape for a Doppler-broadened line. Therefore, we carried out detailed simulations (Methods), based on the known physics of hydrogen, to determine the expected line shape and intensity of the transition in our experimental environment. The simulated line shape is model dependent. Two examples of simulated results are shown in Fig. 4a in red and in blue. The calculated transition probability and line width match the data reasonably well over the region measured. The general agreement between the simulation and the observed spectral feature indicates that the observed line shape is indeed dominated by Doppler broadening, corresponding to an average energy of a few hundred millikelvin.
Figure 4b compares the time distribution of the detected signals following the laser pulse (black) to a simulated distribution (red) obtained using the standard formation model. The agreement between the simulation and the detected signal for these TOF distributions confirms that the radial velocity distribution of trapped antihydrogen is consistent with the model. For comparison, we have also simulated the TOF distribution of ejected atoms originating from hypothetical thermal distributions of 100 mK and 10 mK (Fig. 4b), which indicates that the detected time distribution should be sensitive to the radial temperature of trapped antihydrogen. In Fig. 4c we plot the measured axial distribution of laser-induced annihilation events, and we compare it to simulated results at various temperatures. These distributions also exhibit some dependence on antihydrogen temperature and may facilitate diagnosis of future laser cooling attempts.
In conclusion, we have reported here the observation of the 1S–2P, Lyman-α transition in antihydrogen, based on 966 detected events and an estimated background of 14 events. The frequency of this fundamental anti-atomic transition is determined to a precision of about 5 × 10−8, via narrow-line-width, nanosecond-pulsed, vacuum-ultraviolet laser spectroscopy. We also report a method of directly characterizing the kinetic energy of anti-atoms from their TOF to annihilation, following the laser-induced transition. These observations represent very important steps in the field of low-energy antimatter studies23,24,25,26,27. The techniques of optical manipulations and laser cooling, which have revolutionized the field of atomic physics over the past few decades, are about to be applied to anti-atoms. With its natural line width of about 100 MHz (comparable to our laser width), the Lyman-α transition can in principle be employed to cool antihydrogen to the 2.4-mK Doppler limit. Our simulations predict that cooling to about 20 mK is possible with the current ALPHA-2 set-up9. This, combined with other planned improvements, would reduce the 1S–2S transition line width by more than an order of magnitude and should eventually allow various other spectroscopic measurements with precisions approaching those achieved in hydrogen28,29,30. At such levels of precision, antihydrogen spectroscopy will have an impact on the determination of fundamental constants31, in addition to providing elegant tests of CPT symmetry. Furthermore, laser cooling will be crucial for a precision test of the weak equivalence principle via antihydrogen free fall10 or anti-atom interferometry32 at the 10−2 level and beyond. Access to the 2P state in antihydrogen greatly expands the future experimental horizon; for example, we will now be able to study fine-structure effects in an anti-atom.
Methods
Laser system for 121.6-nm light
The schematic diagram of the laser system is shown in Extended Data Fig. 1. The vacuum ultraviolet radiation at 121.6 nm is produced in two steps: frequency doubling of 730-nm pulses followed by third harmonic generation in a high-pressure gas cell12. The 730-nm pulses are produced by first seeding two titanium sapphire crystals with narrow line width (<100 kHz), continuous-wave radiation at 730 nm from a Toptica diode laser. The crystals are pumped by nanosecond pulses of the second harmonic of a Nd:YAG (neodymium-doped yttrium aluminium garnet) laser (Spectra Physics, pulse energy 300 mJ at 532 nm, with 10-Hz repetition rate). The generated 730-nm pulses have a pulse length and line width of 30 ns and 25 MHz, respectively. The 730-nm pulses are then converted to 365 nm by frequency doubling in a beta barium borate (BBO) crystal and then directed to the experimental zone. The third harmonic of 365 nm is generated in a high-pressure Kr/Ar gas cell (total pressure about 4 bar) after focusing the 365-nm pulses with an ultraviolet-grade lens (focal length 150 mm). The conversion efficiency from 365 nm to 121.6 nm is typically 10−6 to 10−7. The generated 121.6-nm pulses are collimated by a lens of 90-mm focal length and vacuum-ultraviolet-grade and then directed to the trap region by three vacuum-ultraviolet-grade mirrors placed in the vacuum system used to steer the light towards the antihydrogen trap.
Laser pulses enter the atom trap ultrahigh vacuum through an MgF2 window, which transmits about 65% of the incident light. The laser pulses cross the trap at an angle of about 2° to the direction of the trap axis, as shown in Fig. 1. Pulses exiting the trap are detected by a calibrated, solar-blind photomultiplier tube (Hamamatsu) to monitor the pulse energy and timing. The typical pulse width and line width at 121.6 nm are 12 ns and 65 MHz (FWHM, estimated). Each pulse has a pulse energy of 0.53–0.65 nJ in the trap region, and the repetition rate is 10 Hz, corresponding to an average power of 5–7 nW, and a peak power of 50–70 mW. The linear polarization is perpendicular to the direction of the magnetic field in order to drive the transition to the 2Pc state. The seed laser frequency is stabilized to a few megahertz by locking the laser to a wavemeter (HighFinesse). Before and after each experimental sequence, the frequency and line width of the 730-nm pulses were monitored by a custom-made, Fabry–Perot spectrometer having a free spectral range of 745 MHz, and a finesse of about 75.
Suppression of cosmic ray background
To detect annihilation events in the two observation windows—(1) 15.6-s magnet ramp-down to release remaining anti-atoms and (2) 1-ms observation window after each laser pulse—we require different levels of background suppression. We optimized significance using an MVA-based, machine-learning algorithm to suppress unwanted background counts based on estimated populations of atoms in the apparatus per sequences.
The selection of variables has been described elsewhere7. The algorithm was trained and tested using a sample of 393,920 and 3,375,877 events for signal and background, respectively. The signal events were selected from periods with high rates of antihydrogen production in the apparatus, and thus contain less than 0.1% background. The background events were collected during periods when no antiprotons were present in the apparatus.
The trigger is configured to fire on the n-side of the ALPHA silicon hybrid33. The total configuration for the silicon vertex detector trigger requires more than two hits on the inner layer with one hit on the middle and outer layer.
The 15.6-s observation window
A classifier cut was chosen to optimize the significance for an expected 150 counts of signal and 320 counts of background (motivated to optimize significance of the signal when the laser is tuned to resonance in two sequences). The analysis gives a background rate of 0.217 ± 0.006 s−1 and an efficiency of 0.851 ± 0.002 (statistical error only, one standard deviation) annihilations per detector trigger.
The 1-ms (post-laser pulse) observation window
A classifier cut was chosen to optimize the significance for 31 expected counts of signal and 243 counts of background in the range ±10 cm along the beam axis of the trap centre. The analysis gives a background rate of 0.049 ± 0.003 s−1 and an efficiency of 0.807 ± 0.002 (statistical error only) annihilations per detector trigger. For the sample of 966 annihilation events reported here, the expected background is 14 events.
Simulation of laser interaction with trapped antihydrogen atoms
The trapped antihydrogen motion is simulated as described elsewhere34 using the B-field obtained from an interpolation of a Biot–Savart model of the ALPHA-2 mirror and octupole coils. The atoms are randomly distributed between the 1Sc and 1Sd states, which, for the purposes of the centre of mass motion, are treated as having the same magnetic moment, equal to that of an electron. To mimic the experiment, the simulated atoms are launched in the n = 30 state and allowed to radiatively cascade to the ground state, or are directly launched in the ground state to obtain two sets of initial conditions. The atoms are propagated for a few seconds before the laser is turned on, after which a laser pulse interacts with the atoms every 100 ms. At each laser pulse, the distance of the antihydrogen from the laser is computed and the laser intensity at that position is used to compute the excitation probability. The probability of the excited state decaying into the 1Sa or 1Sb states is also computed. A random number is compared to the product of these probabilities to determine whether the spin of the simulated atom flips for that laser pulse. If a flip occurs, the sign of the magnetic moment is reversed to give a high-field-seeking atom. The simulation is stopped when the position of the atom is outside the inner edge of the trap electrodes.
The statistics on properties like the Doppler width and the energy shift due to the Zeeman effect are accounted for by using the properties of the system at each laser pulse. The shift in frequency due to the Doppler effect is calculated from the velocity v at the time of the laser pulse:
$${\rm{\delta }}\omega =\frac{-\omega ({\boldsymbol{v}}\hat{k})}{c}$$
where $$\hat{k}$$ is the direction of propagation of the laser beam. The Zeeman shift for that laser shot is computed from the strength of the B-field at the position of the antihydrogen. The 1Sc and 1Sd energies as a function of B are calculated as described in ref. 34, but the 2P energies are calculated using slightly more accurate equations which account for the difference between the positron magnetic moment and the Bohr magneton. The hyperfine splitting of the 2P states is at the level of a few tens of megahertz and, although included in the simulation, do not affect the results at the level presented here. The laser line width is included in the simulation by randomly shifting the laser frequency with a Gaussian distribution given by a FWHM of 2π × 65 MHz. These shifts are added to the detuning of the laser to obtain a total energy shift for that laser pulse of ħδω relative to the line centre.
The probability that an excitation occurs during a laser pulse is proportional to the laser intensity at the position of the antihydrogen. The simulation assumes linearly polarized photons perpendicular to the direction of laser propagation, which is nearly parallel to B. Thus, the polarization is nearly perpendicular to B, and approximately half of the laser intensity can drive the transition. The transition dipole matrix element is equal to that of the m = 0 to m = 1 transition in hydrogen, multiplied by sin(σ) for the 2Pc level, where σ is a mixing angle; see equations (13) and (15) in ref. 34. The transition probability is multiplied by the Lorentzian factor from the natural line width of the 2P states:
$$\frac{{\left(\mathop{A}\limits^{\sim }/2\right)}^{2}}{{\left({\rm{\delta }}\omega \right)}^{2}+{(\mathop{A}\limits^{\sim }/2)}^{2}}$$
where $$\mathop{A}\limits^{\sim }$$ = 2π × 99.6 MHz is the natural line width of the 2P state, and δω is calculated as described in the previous paragraph. The probability of a spin flip in the decay is proportional to the probability in the 2P wavefunction with flipped spin: cos2(σ) for the 2Pc state.
Data availability
The datasets generated and analysed during this study are available from J.S.H. on reasonable request.
Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Ahmadi, M. et al. Enhanced control and reproducibility of non-neutral plasmas. Phys. Rev. Lett. 120, 025001 (2018).
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Smorra, C. et al. A parts-per-billion measurement of the antiproton magnetic moment. Nature 550, 371–374 (2017).
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Doser, M. et al. Exploring the WEP with a pulsed cold beam of antihydrogen. Class. Quantum Grav. 29, 184009 (2012).
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Beyer, A. et al. The Rydberg constant and proton size from atomic hydrogen. Science 358, 79–85 (2017).
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Fleurbaey, H. et al. New measurement of the 1S–3S transition frequency of hydrogen: contribution to the proton charge radius puzzle. Phys. Rev. Lett. 120, 183001 (2018).
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Mohr, P. J., Newell, D. B. & Taylor, B. N. CODATA recommended values of the fundamental physical constants: 2014. Rev. Mod. Phys. 88, 035009 (2016).
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Rasmussen, C. Ø., Madsen, N. & Robicheaux, F. Aspects of 1S–2S spectroscopy of trapped antihydrogen atoms. J. Phys. B 50, 184002 (2017).
Acknowledgements
All authors are members of the ALPHA Collaboration. This work was supported by: the European Research Council through its Advanced Grant programme (to J.S.H.); CNPq, FAPERJ, RENAFAE (Brazil); NSERC, CFI, NRC/TRIUMF, EHPDS/EHDRS (Canada); FNU (Nice Centre), Carlsberg Foundation (Denmark); ISF (Israel); STFC, EPSRC, the Royal Society and the Leverhulme Trust (UK); DOE, NSF (USA); and VR (Sweden). We are grateful for the efforts of the CERN Antiproton Decelerator team, without which these experiments could not have taken place. We thank P. Djuricanin (University of British Columbia) for his extensive help with the laser system. We thank J. Tonoli (CERN) and his staff as well as T. Mittertreiner (University of British Columbia) and his staff for extensive, time-critical help with machining and electrical works. We thank the staff of the Superconducting Magnet Division at Brookhaven National Laboratory for collaboration and fabrication of the trapping magnets. We thank C. Marshall (TRIUMF) for work on the ALPHA-2 cryostat. We thank D. Tommasini and A Milanese (CERN) for the fabrication of conventional magnets for ALPHA-2. We thank F. Besenbacher (Aarhus University) for timely support in procuring the ALPHA-2 external solenoid. We thank T. Miller (Ohio) for advice on the initial development of the pulsed laser system.
Author information
Affiliations
• & P. Pusa
2. Department of Physics and Astronomy, Aarhus University, Aarhus, Denmark
• B. X. R. Alves
• , T. Friesen
• , J. S. Hangst
• , S. A. Jones
• , C. Ø. Rasmussen
• & G. Stutter
3. Department of Physics, College of Science, Swansea University, Swansea, UK
• C. J. Baker
• , M. Charlton
• , S. Eriksson
• , C. A. Isaac
• , J. M. Jones
• , S. A. Jones
• , P. Knapp
• , D. Maxwell
• & D. P. van der Werf
4. School of Physics and Astronomy, University of Manchester, Manchester, UK
• W. Bertsche
• , M. A. Johnson
• & M. Sameed
5. Cockcroft Institute, Sci-Tech Daresbury, Warrington, UK
• W. Bertsche
• & M. A. Johnson
6. TRIUMF, Vancouver, British Columbia, Canada
• A. Capra
• , R. Collister
• , M. C. Fujiwara
• , D. R. Gill
• , A. Khramov
• , L. Kurchaninov
• , J. T. K. McKenna
• , J. M. Michan
• , K. Olchanski
• , A. Olin
• & R. I. Thompson
7. Department of Physics, University of California at Berkeley, Berkeley, CA, USA
• C. Carruth
• , J. Fajans
• , E. D. Hunter
• & J. S. Wurtele
8. Instituto de Fisica, Universidade Federal do Rio de Janeiro, Rio de Janeiro, Brazil
• C. L. Cesar
• , R. L. Sacramento
• & D. M. Silveira
• S. Cohen
10. Department of Physics and Astronomy, University of Calgary, Calgary, Alberta, Canada
• A. Evans
• , T. Friesen
• , C. So
• & R. I. Thompson
11. Department of Physics and Astronomy, University of British Columbia, Vancouver, British Columbia, Canada
• N. Evetts
• , W. N. Hardy
• & T. Momose
12. Department of Physics, Simon Fraser University, Burnaby, British Columbia, Canada
• M. E. Hayden
• & J. J. Munich
• S. Jonsell
14. Department of Physics and Astronomy, York University, Toronto, Ontario, Canada
• S. Menary
• & D. M. Starko
15. École Polytechnique Fédérale de Lausanne (EPFL), Swiss Plasma Center (SPC), Lausanne, Switzerland
• J. M. Michan
• T. Momose
• A. Olin
18. Department of Physics and Astronomy, Purdue University, West Lafayette, IN, USA
• F. Robicheaux
• E. Sarid
20. Physics Department, Marquette University, Milwaukee, WI, USA
• T. D. Tharp
21. IRFU, CEA/Saclay, Gif-sur-Yvette Cedex, France
• D. P. van der Werf
Contributions
This experiment was based on data collected using the ALPHA-2 antihydrogen trapping apparatus, designed and constructed by the ALPHA Collaboration using methods developed by the entire collaboration. The entire collaboration participated in the operation of the apparatus and the data-taking activities. Pulsed Lyman-α spectroscopy in ALPHA was proposed by M.C.F. and developed further by T.M., F.R., J.M.M., R.C., A.E., A.K. and M.C.F. The original laser was designed by T.M. and tested by J.M.M. and T.M. The laser system at CERN was implemented, commissioned and operated by J.M.M., R.C., A.E., A.K. and T.M. F.R. developed the simulation programme for laser interaction with magnetically trapped atoms. Electron cyclotron resonance magnetometry was developed by T.F., M.E.H. and W.N.H. Detailed analysis of the antiproton annihilation detector data was done by J.T.K.M. The manuscript was written by T.M., M.C.F. and J.S.H., with help from R.C., A.E., A.K., J.T.K.M., F.R. and A.C. The manuscript was then edited and improved by the entire collaboration.
Competing interests
The authors declare no competing interests.
Corresponding authors
Correspondence to M. C. Fujiwara or J. S. Hangst or T. Momose.
Extended data figures and tables
1. Extended Data Fig. 1 Laser system.
The figure shows a schematic of the 121.6-nm laser system for driving the 1S–2P transition. See Methods for details.
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The value of
Question:
$\left|\begin{array}{ccc}0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0\end{array}\right|$
Solution:
Given, $\left|\begin{array}{ccc}0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0\end{array}\right|$
[Taking $x^{2}, y^{2}$ and $z^{2}$ common from $C_{1}, C_{2}$ and $C_{3}$, respectively]
$=x^{2} y^{2} z^{2}\left|\begin{array}{lll}0 & x & x \\ y & 0 & y \\ z & z & 0\end{array}\right|$
[Applying $C_{2} \rightarrow C_{2}-C_{3}$ ]
$=x^{2} y^{2} z^{2}\left|\begin{array}{ccc}0 & 0 & x \\ y & -y & y \\ z & z & 0\end{array}\right|=x^{2} y^{2} z^{2}(x(y z+y z))$
$=x^{2} y^{2} z^{2} \cdot(2 x y z)=2 x^{3} y^{3} z^{3}$
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# How to evaluate a Riemann (Darboux?) integral?
This is the definition of integrable we have :
A bounded function $f : [a,b] \to \mathbb {R}$ is said to be integrable if $\sup\, L(f,P)= \inf\, U(f,P)$.
I've looked through the book but it involves the 'norm' of a partition, which we sort of don't have..
My lecturer gave an example which involved ( I think ) the uniform partition and creating a 'sequence' of partitions. I don't remember much more, so could someone help me out?
Edited the title as everywhere I look for Riemann integrals, the norm shows up, and the Darboux one seems to be the one we have, but my lecturer has always called it Riemann integral, so I'm not sure
• You have the $\sup, \inf$ mixed up there. – zhw. Jan 14 '18 at 8:44
• A Riemann integral can be defined as $\sup/\inf$ of Darboux sums as well as a limit of Riemann sums as norm of partition tends to $0$. That these two definitions are equivalent is a non-obvious fact which is somewhat difficult to prove. Moreover none of these definitions are used in evaluating integrals. Integrals are almost always evaluated using theorems meant to evaluate them and these theorems in turn are proved using these definitions. – Paramanand Singh Jan 14 '18 at 9:15
• Also fixed minor typo in the use of $\sup, \inf$ in the definition. – Paramanand Singh Jan 14 '18 at 9:19
• But how do we evaluate them using definitions? For example, the integral of $x^3$ from 0 to 1? – Saad Jan 14 '18 at 9:24
• Ok form a partition $P$ of $[0,1]$ into $n$ intervals via points $x_i=i/n$ and find $L(f, P), U(f, P)$ and use the non-trivial theorem that sup, inf of these numbers is same as the limit of these numbers when norm of partition (here $1/n$) tends to $0$. – Paramanand Singh Jan 14 '18 at 9:41
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# Chemistry on Brilliant
Hey guys. I am really amazed at several of the wiki pages that have been written up, and am thankful for your contributions. We are getting close to be pushing out the first version of Chemistry on the site, and would like some help with a few more wiki pages.
The subtopics of fundamentals and Reaction Mechanics subtopics are mostly good, but the Chemical bonding and Organic Chemistry will need more work. The pages that need material are listed out in Needs writers. I would like to half that list of 20, before we release chemistry.
Let me know if I can help in any way.
For those of you who are interested in building out Chemistry on Brilliant, and would be willing to spend time writing up great wiki pages over the next month, we would appreciate if you help us by filling out the wikis listed here.
To make it easier for you to communicate with others as you're collaborating on these wikis, you can sign in to Brilliant chat and head over to the #chemistry channel. There, you will find other Chemistry enthusiasts and can bounce ideas with them!
Once these initial wikis are filled, we can look into adding more skills and placing these problems accordingly. With the community's help, we can get Chemistry on Brilliant!
Woohoo!
Note by Calvin Lin
4 years, 3 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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I have added to Aldehydes and Ketones. It is a very large topic in itself. So I will probably take days to complete that.
- 4 years, 3 months ago
I have added a bit to order of chemical reactions.
- 4 years, 3 months ago
Thanks! I will add you to the list, so that we can easily communicate via email :)
Staff - 4 years, 3 months ago
I can write wikis but a little everyday.
- 4 years, 3 months ago
That's perfect! I will add you to the list, so that we can easily communicate via email :)
Staff - 4 years, 3 months ago
I will be able to give inputs on nomenclature of bicyclo and Spiro compounds this sunday
- 4 years, 3 months ago
I am also interested.
- 4 years, 3 months ago
Thanks! I will add you to the list, so that we can easily communicate via email :)
Staff - 4 years, 3 months ago
I am also interested.
- 4 years, 3 months ago
I've added you to the Chatroom. You can check out what others are working on, and assist them with it.
Staff - 4 years, 3 months ago
Now I think I should work on creating questions based on the chemistry topics, shall I?
- 4 years, 3 months ago
Certainly! That would be great!
You can also list the problems that you created for each wiki / skill in column H.
Staff - 4 years, 3 months ago
Sir, for organic chemistry, how could I insert bond line representations of molecules? (I can't download it from other sites and upload it to brilliant as many of them has copyright issues) Also sir, I will contribute questions of chemical equilibrium as it's wiki is almost complete.
- 4 years, 2 months ago
@Calvin Lin Sir , Can i help out too? I could probably help out with the wiki's or something
- 4 years, 2 months ago
Certainly! Here is a list of the wiki pages that we are working on!
Staff - 4 years, 2 months ago
Can I edit some pages like ionic compounds?
- 4 years, 2 months ago
Certainly! You can help us improve these pages.
Staff - 4 years, 2 months ago
- 4 years, 2 months ago
Staff - 4 years, 2 months ago
Ya......got that........just needed to explore.....thanks! :)
- 4 years, 2 months ago
Are you interested in research or lectureship in Chemistry? If yes than you must know about one of the most reputed entrance exam conducted in India through which aspirants from science stream gets a change for lectureship and Junior Research Fellowship CSIR NET (Council of Scientific and Industrial Research National Eligibility Test) is a national level entrance exam which is conducted for the following subjects – Chemical Science Earth Science Physical Science Mathematical Science & Life Science. To appear in this exam, you will have to fulfil the CSIR NET Eligibility Criteria. Get the complete information for CSIR UGC NET Eligibility from here -https://scoop.eduncle.com/csir-net-eligibility-criteria
The criteria will include the minimum education qualification + age limits. I hope the information will be helpful to you!
- 1 year, 9 months ago
Sir, Please enable the chemistry latex for this site, Or else it becomes quite difficult to write up
- 1 year, 6 months ago
I will love to contribute something on organic nomenclature, but I don't have a PC. I am waiting for these options to come to the app
- 4 years, 3 months ago
Awww, we're working on adding publication features to the apps, but that will take a while.
One way that you can contribute, is to help us review these wiki pages and provide feedback on areas to improve (Columns F, G, in the spreadsheet) Send me an email if you're interested in helping out :)
Staff - 4 years, 3 months ago
I could help , but I don't think I can make chemistry in english . Also , can you tell me how can I follow these chemistry problems after they are done ?
- 4 years, 3 months ago
That's alright, we still need people to read through the Wikis and provide feedback. It would be helpful for the creators to know that "this part needs further explanation" or "I don't quite understand what you mean here. Why does that work?".
When the chemistry topic is launched (hopefully in 2 months), you will be able to "Add Topics" for Chemistry.
Staff - 4 years, 3 months ago
@Calvin Lin sir if I share a problem how would it get rated??
- 4 years, 3 months ago
This is a multi-step process. Right now, the focus is on writing up wikis. After that, we will look into adding problems across different skills, and figuring out what levels mean. This process allows us to ensure that when Chemistry is launched, the experience isn't "a bunch of empty pages" for people, which would just kill their engagement.
When we launch Chemistry, it will behave like any other topic.
Staff - 4 years, 3 months ago
Thank you sir. Chemistry has always been my weak link. Now I can improve.
- 4 years, 3 months ago
Mine too :(
- 4 years, 3 months ago
Haha, Nice. But I'm sure that I won't improve in chemistry, just because I don't want to improve in it.
- 3 years, 11 months ago
I will surely fill up many wiki's sir, but I will have exams this month, so I'll try contributing next month.
- 4 years, 3 months ago
No worries, I understand. The Chemistry Wiki project will take a 1-2 months to organize, so help in August would also be appreciated. I will add you to the list, so that you can be updated :)
Staff - 4 years, 3 months ago
Thanks sir! Help will surely come in the next month.
- 4 years, 3 months ago
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## Solitaire of Life
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
### Re: Solitaire of Life
PTW, I need to participate later (I need to practice welding).
KittyTac
Posts: 533
Joined: December 21st, 2017, 9:58 am
### Re: Solitaire of Life
x = 12, y = 11, rule = LifeHistory2A$2C3.C$4.C.C$3.C.C$3.C.C.C$.AC3.C.A$A.C.2A2.C$A2.C2.3A.C$.CA2.C3.A.C$6.C2.A.C$5.CA3.A!
Congratulations, that is one of four optimal solutions.
Gamedziner
Posts: 606
Joined: May 30th, 2016, 8:47 pm
Location: Milky Way Galaxy: Planet Earth
### Re: Solitaire of Life
Heh, the original solution I'd devised for that was (mostly) symmetrical and had 52 cells; I hadn't expected asymmetrical solutions would be able to be so much smaller.
x = 12, y = 10, rule = LifeHistory2C3.C3.2A$.A2.C.C2.A$.A.C.C.A.A$2A.C.C.C.2A$2.C.A.C.A$2.C.A.A.C$2A.C.A.A.AC$.C.A.C.A.A.C$.A2.A.C2.A.C$2A3.C3.2A! But now, a sort of reverse-Solitaire question can be asked: what is the smallest "challenge" pattern so that this 52-cell pattern is an optimal solution? 77topaz Posts: 1240 Joined: January 12th, 2018, 9:19 pm ### Re: Solitaire of Life 77topaz wrote:But now, a sort of reverse-Solitaire question can be asked: what is the smallest "challenge" pattern so that this 52-cell pattern is an optimal solution? This 29-celler might be a good starting point: x = 12, y = 10, rule = LifeHistory2C3.C3.2C$4.C.C$.C7.C$3.C.C.C$2.C3.C$4.C3.C$3.C.C.C$.C7.C$4.C.C4.C$2C3.C3.2C!
LifeWiki: Like Wikipedia but with more spaceships. [citation needed]
Posts: 1823
Joined: November 8th, 2014, 8:48 pm
Location: Getting a snacker from R-Bee's
### Re: Solitaire of Life
This 21 seems likely, haven't tested it with a computer yet
x = 12, y = 10, rule = LifeHistoryAC3.C3.CA$.A2.A.A2.A$.A.C.C.A.A$CA.A.A.C.AC$2.C.C.A.A$2.A.A.C.C$CA.C.A.A.AC$.A.A.C.C.A.C$.A2.A.A2.A.A$AC3.C3.CA! WADUFI Sarp Posts: 128 Joined: March 1st, 2015, 1:28 pm ### Re: Solitaire of Life Sarp wrote:This 21 seems likely, haven't tested it with a computer yet x = 12, y = 10, rule = LifeHistoryAC3.C3.CA$.A2.A.A2.A$.A.C.C.A.A$CA.A.A.C.AC$2.C.C.A.A$2.A.A.C.C$CA.C.A.A.AC$.A.A.C.C.A.C$.A2.A.A2.A.A$AC3.C3.CA!
It doesn't work; there are multiple 51-cell solutions.
Gamedziner
Posts: 606
Joined: May 30th, 2016, 8:47 pm
Location: Milky Way Galaxy: Planet Earth
### Re: Solitaire of Life
I am pretty sure I have found the optimal solution for this, although I'm not sure:
x = 3, y = 7, rule = LifeHistory.C$C$.C$C.C$.C$C$.C!
Life is hard. Deal with it.
My favorite oscillator of all time:
x = 7, y = 3, rule = B3-r4j/S2-i3o5bo$2o3b2o$b2ob2o!
Hdjensofjfnen
Posts: 862
Joined: March 15th, 2016, 6:41 pm
Location: (394234, -234231)
Hm...
x = 48, y = 48, rule = LifeHistory46.CA$45.A.A$44.C$43.A$42.C$41.A$40.C$39.A$38.C$37.A$36.C$35.A$34.C$33.A$32.C$31.A$30.C$29.A$28.C$27.A$26.C$25.A$24.C$24.2A3$20.CA$19.A.A$18.C$17.A$16.C$15.A$14.C$13.A$12.C$11.A$10.C$9.A$8.C$7.A$6.C$5.A$4.C$3.A$2.C$.A$C$2A! Life is hard. Deal with it. My favorite oscillator of all time: x = 7, y = 3, rule = B3-r4j/S2-i3o5bo$2o3b2o$b2ob2o! Hdjensofjfnen Posts: 862 Joined: March 15th, 2016, 6:41 pm Location: (394234, -234231) ### Re: Solitaire of Life Hdjensofjfnen wrote:I am pretty sure I have found the optimal solution for this, although I'm not sure: x = 3, y = 7, rule = LifeHistory.C$C$.C$C.C$.C$C$.C! 15? x = 6, y = 8, rule = LifeHistory5.A$3.C2A$2.C$A2.C$2AC.C$3.C$2.C$2.AC!
x = 6, y = 7, rule = LifeHistory3.CA$2.C2.A$A2.C.A$2AC.C$3.C$2.C$2.AC!
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$
http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
A for awesome
Posts: 1695
Joined: September 13th, 2014, 5:36 pm
Location: 0x-1
### Re: Solitaire of Life
A for awesome wrote:15?
x = 6, y = 8, rule = LifeHistory5.A$3.C2A$2.C$A2.C$2AC.C$3.C$2.C$2.AC! x = 6, y = 7, rule = LifeHistory3.CA$2.C2.A$A2.C.A$2AC.C$3.C$2.C$2.AC! I just did a search with WLS and determined that those were the only optimal solutions (excluding reflection). Gamedziner Posts: 606 Joined: May 30th, 2016, 8:47 pm Location: Milky Way Galaxy: Planet Earth ### Re: Solitaire of Life Oh, yeah, you win by a whole lot. My solution was 21. Life is hard. Deal with it. My favorite oscillator of all time: x = 7, y = 3, rule = B3-r4j/S2-i3o5bo$2o3b2o$b2ob2o! Hdjensofjfnen Posts: 862 Joined: March 15th, 2016, 6:41 pm Location: (394234, -234231) ### Re: Solitaire of Life This new puzzle is based on a still life that came up in an apgmera search today. Apparently it's a fairly rare still life, since there's only the one soup in Catagolue. x = 14, y = 12, rule = LifeHistory14B$B12.B$B7.C.2C.B$B5.C4.C.B$B2.C4.C4.B$B2.2C.C6.B$B6.C.2C2.B$B4.C4.C2.B$B.C4.C5.B$B.2C.C7.B$B12.B$14B!#C Additional#C information#C you#C may#C not#C want:#C the#C source#C still#C life's#C apgcode#C starts#C with#C "xs34".
With the additional restriction that all added cells have to be strictly inside the blue box, there's a unique solution. Without the blue box JLS finds four more possibilities.
dvgrn
Moderator
Posts: 5225
Joined: May 17th, 2009, 11:00 pm
### Re: Solitaire of Life
[spoiler or something i guess]
x = 10, y = 10, rule = LifeHistory4.2A$3.A2.C.2C$4.CA3.C$.C4.C2A$.2CAC$5.CA2C$.2AC4.C$C3.AC$2C.C2.A$4.2A! gmc_nxtman Posts: 1146 Joined: May 26th, 2015, 7:20 pm ### Re: Solitaire of Life Missed by less than 30 secs WADUFI Sarp Posts: 128 Joined: March 1st, 2015, 1:28 pm ### Re: Solitaire of Life Isn't the solver supposed to keep the game going by supplying a new puzzle? Here's another small and probably easy one, with three minimal-population solutions: Code: Select all x = 10, y = 7, rule = LifeHistory2.C2.C$2.C2.C3.C$C8.C$C4.C$2.2C.C.C$2.C4.C$3.2C! Adding one more white cell in any of the blue locations would make a puzzle with a unique smallest solution: x = 10, y = 7, rule = LifeHistory2.C2.C$2.C2BC3.C$C3.B4.C$C4.C$2.2C.C.C$2.C4.C$3.2C! dvgrn Moderator Posts: 5225 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI ### Re: Solitaire of Life dvgrn wrote:Isn't the solver supposed to keep the game going by supplying a new puzzle? At some point we kinda stopped doing that, for whatever reason. 31: x = 11, y = 8, rule = LifeHistory4.2A$3.C2.C$3.C2.C2.AC$AC.2A.A.A.C$.C4.C.A$.A.2C.C.C$2A.C2.A.CA$4.2C!
LifeWiki: Like Wikipedia but with more spaceships. [citation needed]
Posts: 1823
Joined: November 8th, 2014, 8:48 pm
Location: Getting a snacker from R-Bee's
### Re: Solitaire of Life
Not minimal, but close! The 31-cell solution doesn't have a cell in any of the blue positions...
dvgrn
Moderator
Posts: 5225
Joined: May 17th, 2009, 11:00 pm
30?
x = 11, y = 8, rule = LifeHistory4.2A$3.C2.C$3.CBAC2.AC$AC.A.BA.A.C$AC.A2.C.A$3.2C.C.C$3.C2.A.CA$4.2C! WADUFI Sarp Posts: 128 Joined: March 1st, 2015, 1:28 pm ### Re: Solitaire of Life Sarp wrote:30? x = 11, y = 8, rule = LifeHistory4.2A$3.C2.C$3.CBAC2.AC$AC.A.BA.A.C$AC.A2.C.A$3.2C.C.C$3.C2.A.CA$4.2C!
You got it. Somebody else's turn to make a puzzle...
dvgrn
Moderator
Posts: 5225
Joined: May 17th, 2009, 11:00 pm
x = 12, y = 8, rule = LifeHistory5.C$9.C$.C5.2C2.C$.C3.C3.C$C3.2C3.C$.C.C2.C$2.C6.C$4.2C! gmc_nxtman Posts: 1146 Joined: May 26th, 2015, 7:20 pm ### Re: Solitaire of Life Here is a first attempt at a solution, 79 cells: x = 17, y = 17, rule = LifeHistory2A$A8.2A$2.A.2A3.A2.A$.2A.A5.3A$.A3.A$2.3A.A3.3A$4.A.A2.C2.A$6.A.A.A2.CA$5.C.A.A.2C2.C$5.C3.C3.C.A$4.CA2.2C.2AC.2A$2.A2.C.C2.C$2.2A2.C.A2.A.C3A$8.2C2A.A2.A$7.A4.A$8.3A$10.2A! EDIT: 71 x = 16, y = 16, rule = LifeHistory8.2A$2A.2A3.A2.A$.A.A5.3A$A3.A$.3A.A3.3A$3.A.A2.C2.A$5.A.A.A2.CA$4.C.A.A.2C2.C$4.C3.C3.C.A$3.CA2.2C.2AC.2A$.A2.C.C2.C$.2A2.C.A2.A.C3A$7.2C2A.A2.A$6.A4.A$7.3A$9.2A!
wwei23
Posts: 935
Joined: May 22nd, 2017, 6:14 pm
Location: The (Life?) Universe
[56, not 58]:
x = 16, y = 12, rule = LifeHistory8.2A$6.C2.A$.A4.2A2.CA2.2A$A.C.2A2.2C2.C2.A$A.CA2.C3.C.2A$.C3.2C.2AC.A$2.CAC2.C4.A$3.C2.A2.ACA$.A3.2C.2A$.2A3.A$6.A.2A$7.A.A! Much of the interior is forced by S3 in some form or other. Last edited by BlinkerSpawn on October 7th, 2018, 10:44 pm, edited 1 time in total. LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1823 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Solitaire of Life Not minimal. EDIT: Is it just me, or does this game have conceptual similarities to minesweeper? When the interior is forced like in this particular puzzle, similar inferences have to be made to those made in minesweeper regarding what cells ("mines") are allowed to be where. gmc_nxtman Posts: 1146 Joined: May 26th, 2015, 7:20 pm ### Re: Solitaire of Life gmc_nxtman wrote:EDIT: Is it just me, or does this game have conceptual similarities to minesweeper? When the interior is forced like in this particular puzzle, similar inferences have to be made to those made in minesweeper regarding what cells ("mines") are allowed to be where. Total agreement there. Down to 54: x = 14, y = 12, rule = LifeHistory8.2A$6.C2.A$.A4.2A2.CA$A.C.2A2.2C2.C$A.CA2.C3.C.A$.C3.2C.2AC.2A$2.CAC2.C4.A$3.C2.A2.AC.A$.A3.2C.2A.A$.2A4.A$7.A.2A$8.A.A!
Forcing diagram (red = forced-OFF, yellow = forced-ON):
x = 14, y = 12, rule = LifeHistory8.2A$6.C2.A$.A4.2A2DCA$ADCD2A2.2CD.C$ADCE2DC3DC.A$.C3D2CD2AC.2A$.DCEC2DC4.A$.2DC2DED.AC.A$.A3D2CD2A.A$.2A4.A$7.A.2A$8.A.A! LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1823 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Solitaire of Life Still not minimal. Also just so you know, this is a valid alternative forcing diagram: x = 13, y = 8, rule = LifeHistory6.C$8.2DC$.DCDAD2.2CD.C$2DC3DC3DC$DCA2D2CD2.C$2DCDC2DC$A2DC6.C$5.2C!
gmc_nxtman
Posts: 1146
Joined: May 26th, 2015, 7:20 pm
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# IIS 7.5: Virtual dirs with different default documents
We have a typical setup in IIS for our webapp with virtual dirs for different projects:
• www.webapp.com (Website)
• project1 (virtual dir)
• project2 (virtual dir)
• project3 (virtual dir)
The webapp has login.aspx as a default document, as do all the virtual dirs. Now, I need a different default document (index.aspx) only for the www.webapp.com website. The virtual dirs will continue to use login.aspx and only that.
EDIT: All virtual dirs point to the same physical directory. If I change the default document for www.webapp.com, it is also changed for all underlying virtual dirs.
This used to be easy in IIS 6, but in IIS 7.5 I am unsuccessful to implement this.
I think I need to alter applicationhost.config by adding location tags, but I have no clue how to start. Can someone point me in the right direction?
Thanks! Stijn
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To do this you need to modify the the ApplicationHost.config file. The location of the file is currently in the %windir%\system32\inetsrv\config directory.
The Location section towards the end of the document needs to be modified accordingly depending on your app default documents. for eg -
<location path="Default Web Site/App">
<system.webServer>
<defaultDocument enabled="true">
<files>
<clear />
</files>
</defaultDocument>
<security>
<authentication>
<anonymousAuthentication enabled="false" />
</authentication>
</security>
</system.webServer>
</location>
<location path="Default Web Site/App/VApp2">
<system.webServer>
<defaultDocument enabled="true">
<files>
<clear />
</files>
</defaultDocument>
<security>
<authentication>
<anonymousAuthentication enabled="false" />
</authentication>
</security>
</system.webServer>
</location>
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Yeah, the easiest (read: cheat-i-est) way of configuring this is to:
Open Feature Delegation, and disable delegation of the Default Document module (or make it read-only) for the site.
With that done, the GUI will create <location> tags for you at any level lower than the Delegation setting you just configured.
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Maybe I didn't understand your problem, but it seems trivial to me.
Open the Internet Information Services Manager, select the virtual directory (double click). In the IIS section, open "Default Document" and set your default document (or remove some).
Repeat the procedure for all of your virtual directories.
UPDATE:
Question: Do (some of) your virtual directories point to the same physical directory? IIS saves the information in web.config files in the physical directory. That may be your problem.
Workaround: create some logic that redirects the requests of the virtual directories. Or create a copy of your physical directory so that the web.config files can be set individually.
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indeed, I forgot to mention that they all point at the same physical directory. I'll update the post - Thanks! – Stijn Van Loo Feb 5 '11 at 10:25
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# Cross-validation vs empirical Bayes for estimating hyperparameters
Given a hierarchical model $p(x|\phi,\theta)$, I want a two stage process to fit the model. First, fix a handful of hyperparameters $\theta$, and then do Bayesian inference on the rest of the parameters $\phi$. For fixing the hyperparameters I am considering two options.
1. Use Empirical Bayes (EB) and maximize the marginal likelihood $p(\mbox{all data}|\theta)$ (integrating out the rest of the model which contains high dimensional parameters).
2. Use Cross Validation (CV) techniques such as $k$-fold cross validation to choose $\theta$ that maximizes the likelihood $p(\mbox{test data}|\mbox{training data}, \theta)$.
The advantage of EB is that I can use all data at once, while for CV I need to (potentially) compute the model likelihood multiple times and search for $\theta$. The performance of EB and CV are comparable in many cases, and often EB is faster to estimate.
Question: Is there a theoretical foundation that links the two (say, EB and CV are the same in the limit of large data)? Or links EB to some generalizability criterion such as empirical risk? Can someone point to a good reference material?
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Can you describe in more detail what you're doing for the cross-validation method? Are you fixing $\theta$ and then using the training data to estimate the other parameters before validating? – Neil G Mar 17 '12 at 22:55
@NeilG maximizing the sum of log marginal predictive data likelihood on cross-validation sets (k is integrated out). – Memming Mar 17 '12 at 23:32
If $k$ is integrated out both times, then what's the difference between CV and EB? – Neil G Mar 18 '12 at 0:30
I doubt there will be a theoretical link that says that CV and evidence maximisation are asymptotically equivalent as the evidence tells us the probability of the data given the assumptions of the model. Thus if the model is mis-specified, then the evidence may be unreliable. Cross-validation on the other hand gives an estimate of the probability of the data, whether the modelling assumptions are correct or not. This means that the evidence may be a better guide if the modelling assumptions are correct using less data, but cross-validation will be robust against model mis-specification. CV is assymptotically unbiased, but I would assume that the evidence isn't unless the model assumptions happen to be exactly correct.
This is essentially my intuition/experience; I would also be interested to hear about research on this.
Note that for many models (e.g. ridge regression, Gaussian processes, kernel ridge regression/LS-SVM etc) leave-one-out cross-validation can be performed at least as efficiently as estimating the evidence, so there isn't necessarily a computational advantage there.
Addendum: Both the marginal likelihood and cross-validation performance estimates are evaluated over a finite sample of data, and hence there is always a possibility of over-fitting if a model is tuned by optimising either criterion. For small samples, the difference in the variance of the two criteria may decide which works best. See my paper
Gavin C. Cawley, Nicola L. C. Talbot, "On Over-fitting in Model Selection and Subsequent Selection Bias in Performance Evaluation", Journal of Machine Learning Research, 11(Jul):2079−2107, 2010. (pdf)
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Why do you say that CV is robust against a mis-specified model? In his case, there is no such protection since cross-validation is searching over the same space that EB is calculating a likelihood. If his modelling assumptions are wrong, then cross-validation won't save him. – Neil G Mar 18 '12 at 0:40
CV is robust against misspecification in the sense that it still gives a useful indicator of generalisation performance. The marginal likelihood may not as it depends on the prior on $\phi$ (for example), even after you have marginalised over $\phi$. So if your prior on $\theta$ was misleading, the marginal likelihood may be a misleading guide to generalisation performance. See Grace Wahba's monograph on "spline models for observational data", section 4.8 (it doesn't say a great deal, but there isn't much on this topic AFAIK). – Dikran Marsupial Mar 19 '12 at 9:44
p.s. I have been performing an analysis of avoiding overfitting in neural networks with Bayesian regularisation where the regularisation parameters are tuned via marginal likelihood maximisation. There are situations where this works very badly (worse than not having any regularisation at all). This seems to be a problem of model mis-specification. – Dikran Marsupial Mar 19 '12 at 9:45
He can get the same "indicator of generalization performance" by checking the total log-probability of the data given the estimated distribution returned by EB (which will be equal to the entropy of that distribution). There's no way to beat it in this case because it is the analytical solution to this problem. I don't see why cross-validation would makes sense when you can calculate a likelihood for EB. – Neil G Mar 19 '12 at 9:50
Spline models and neural network are good examples where you can't calculate a likelihood over all of the parameters (for example, degree of the spline, neural network size and number of connections) and so those are examples where cross-validation gives you the ability to make good choices for those parameters. Maybe there's a hole in my understanding because I don't see what exactly is different about those parameters. What do you think? – Neil G Mar 19 '12 at 9:52
If you didn't have the other parameters $k$, then EB is identical to CV except that you don't have to search. You say that you are integrating out $k$ in both CV and EB. In that case, they are identical.
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# Stability at loop gain <1 and 180 degree phase shift
It says: "If Aβ is less than unity at the high frequencies where phase shift reaches 180°, the high-frequency phase-shifted signals will gradually fade away instead of progressively building up into major oscillations." This is depicted in the picture below.
However, I don't understand this. Shouldn't the system also be unstable at a loop gain less than 1? The 180 degree phase shift in the loop plus the subtraction at the summing point results in positive feedback (See figure 2 below). This means that when a signal is fed into the system (at the control input) it will be attenuated by the loop gain and the phase will be shifted 180 degrees before it arrives back at the summing point. The control signal is still present at the plus input of the summing point so the two signals will be added up resulting in a greater signal than before. When the system keeps doing this the amplitude of the signal will rise to infinity and thus the system will also behave unstable.
My question: Is there an error in my analysis? Why is it that the system is stable at a loop gain < 1 and a phase shift of 180 degrees?
Figure 1
Figure 2
• I think your error is thinking that $1 + A\beta + (A\beta)^2 + (A\beta)^3 + \ldots$ is infinity. If $A\beta < 1$, it is finite and equals $1 \over 1 - A\beta$. – user2233709 Dec 30 '18 at 19:22
• Why is it finite? Let's say the input signal has amplitude 1 and Aβ = -0.5 (- for the 180 degree phase shift). Then after the first round of amplification we have 1 + 1*0.5 = 1.5. In the second round we have 1 + 1.5*0.5 = 1.75. In the third round we have 1 + 1.75*0.5. Etc.. Wouldn't that grow until infinity? – Mauricio Paulusma Dec 30 '18 at 19:36
• No: before the first round, the amplification is $1 = 2 - 1 = 2 - 0.5^0$; after the first round it is $1.5 = 2-0.5 = 2-0.5^1$; after the second round, it is $1.75 = 2-0.25 = 2-0.5^2$; …; after the $n$th round, it is $2-0.5^n$. Hence, it keeps finite. – user2233709 Dec 30 '18 at 19:45
• For more about the math behind this, you may read about geometric series at Wikipedia. – user2233709 Dec 30 '18 at 19:52
• Why is the amplification 2-0.5^n? – Mauricio Paulusma Dec 30 '18 at 20:04
When the gain is >1 with 180 phase shift and another 180 by negative feedback , you have a positive feedback with growing amplitude since loop gain >1 and non inverting and thus an oscillator. ( i.e. unstable or "astable" )
Where the loop gain >1 at 180 phase shift is the resonant frequency of the oscillator.
The output will be saturated which reduces the gain to 0 when saturated but by this time the stored energy ( sine voltage feedback) from the reactance that created the 180 phase shift , has enough amplitude. The DC still is negative feedback so it biases the into near 50% threshold for a square wave out.
Do a thought experiment. Let the input signal be zero, and see what happens to a single cycle of sine wave, injected at the error channel. Do this for an open loop phase shift of $$\\small -180^o\$$, and for the three cases:
a. $$\\small 0;
b. $$\\small A\beta=1\$$; and
c. $$\\small A\beta>1\$$.
For example, take:
$$\\small A\beta=0.5\$$; $$\\small A\beta=1\$$; and $$\\small A\beta=2\$$.
Assume the single cycle travels around the loop discretely, as a packet.
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# Show that if the point $(a, b, c)$ lies on the hyperbolic paraboloid $z = y^2 - x^2$, then the lines with parametric equations $x = a + t, y = b + t, z = c + 2 (b - a) t$ and $x = a + t, y = b - t, z = c - 2 (b + a) t$ both lie entirely on this paraboloid. (This shows that the hyperbolic paraboloid is what is called a ruled surface; that is, it can be generated by the motion of a straight line. In fact, this exercise shows that through each point on the hyperbolic paraboloid there are two generating lines. The only other quadric surfaces that are ruled surfaces are cylinders, cones, and hyperboloids of one sheet.)
## Since both parametric equations are equal to each other then can we saythat both of them lie entirely on $z=y^{2}-x^{2}$
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##### Kristen K.
University of Michigan - Ann Arbor
##### Michael J.
Idaho State University
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### Video Transcript
so being given the point, ABC, with an equation of Z equals y squared minus X squared. We can plug in the point, um, and we get that C is equal to B squared minus a squared. So now we have the Z equals y squared minus X squared. That's how we end up getting the C equals B squared minus a squared. So now we want to plug in the Parametric equation into the equation of the hyperbolic parable oId and solve for C. So we have X equals a plus t. We have that y equals B plus T and Z equals C plus two times B minus 80 then plugging in. Now that we're solving for C, we end up getting that C plus to be a T is equal to Z, but we can now write Z as B plus T squared minus a plus T squared. Then C is going to C plus two B T minus 2 80 is going to equal B squared plus two B t plus T squared minus a squared minus to a T minus t squared. And then we can combine like terms, and this is going to end up giving us a nice C equals B squared minus a squared. Well, that looks familiar because we found it up here as well, then plugging in a second Parametric equation. Um, we have X equals a plus t y equals B minus T and see again equals negative or sorry Z equals C minus two times be this time plus a team. We multiply everything out. We set Z equal to those other values. So now we have is a C minus to be T minus 2 80 equals B squared minus to B T plus T squared minus a squared minus 2 80 minus t squared. We subtract those things over, and once again, we end up getting C equals B squared minus a squared. So since both Parametric equations are equal to each other, then we can say that both of them lie entirely on Z equal to y squared minus X squared.
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##### Kristen K.
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##### Michael J.
Idaho State University
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# 17-28-042
starter
device, usually for fluorescent lamps, which is used for the purpose of starting the discharge lamp by providing the necessary preheating of the cathode and, in combination with the series inductance of the ballast, causes a voltage surge applied to the discharge lamp
Note 1 to entry: The starter element that releases the starting voltage pulse can be either triggered or non-triggered.
Note 2 to entry: This entry was numbered 845-08-32 in IEC 60050-845:1987.
Note 3 to entry: This entry was numbered 17-1258 in CIE S 017:2011.
Publication date: 2020-12
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# Searching the awesome-lists on Github
Discovered the glorious awesome lists today on Github. They are available through a simple search on github, and contain curated lists of resources of all kinds on a multitude of topics.
As one might expect, there is a lot of common ground between these lists, including topics and links.
How could one search for a keyword through all these repositories? I have always wanted search for particular keywords or code snippets in my Emacs configuration files, or in other files in a particular location. This is especially to verify if a bit of code or note is already available, in another location. Something that looks like this ;):
An answer had been available in Howard’s cool blog post on why one should learn Emacs - in a footnote (!), in which he’s mentioned ack and ag ( the silver searcher). 1. It is even possible to edit in line with each search.
The silver searcher github page provides clear examples of how it’s significantly faster than ack (and similar tools). Further exploration led me to the emacs-helm-ag package, which is a helm interface to the silver searcher. Implementing emacs-helm-ag was as simple as adding it to my list of packages, and adding a basic setup to my helm configuration.2
As of now, I add packages to Scimax using this bit of code that I’ve obviously borrowed from the internet, and this case - I’m afraid I did not note the source.
;; Setting up use packages
;; list the packages you want
(setq package-list '(diminish org-journal google-this ztree org-gcal w3m org-trello org-web-tools ox-hugo auto-indent-mode ob-sql-mode dash org-super-agenda ox-hugo workgroups2 switch-window ess ess-R-data-view interleave deft org-bookmark-heading writeroom-mode evil evil-leader polymode helm-ag))
;;fetch the list of packages available
(unless package-archive-contents
(package-refresh-contents))
;; install the missing packages
(dolist (package package-list)
(unless (package-installed-p package)
(package-install package)))
;; Remember to start helm-ag. As per the Silver searcher github site, the helm-follow-mode-persistent has to be set before calling helm-ag.
(custom-set-variables
'(helm-follow-mode-persistent t))
(require 'helm-ag)
This is how it looks in action » Sweet !!
1. This is my first animated gif in a blog post! It was tricky! I used the free GIPHY capture app on the Mac store. ↩︎
2. ↩︎
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# A weir on a permeable foundation with downstream sheet pile is shown in the figure below. The exit gradient as per Khosla’s method is
1. 1 in 6.0
2. 1 in 5.0
3. 1 in 3.4
4. 1 in 2.5
Option 3 : 1 in 3.4
Free
CT 1: Ratio and Proportion
2672
10 Questions 16 Marks 30 Mins
## Detailed Solution
Concept:
The exit gradient (GE) may be given as
$${G_E}\; = \;\frac{H}{d} \times \frac{1}{{\pi \sqrt \lambda }}$$
H - Depth of water in upstream
d - Depth of sheetpile
Where $$\lambda \; = \;\frac{{1 + \sqrt {1 + {\alpha ^2}} }}{2}and\;\alpha \; = \;\frac{b}{d}$$
Calculation:
Given: b = 10 m, d = 4 m, H = 5 m
$$\therefore \alpha \; = \;\frac{b}{d}\; = \;\frac{{10}}{4}\; = \;2.5$$
And $$\lambda \; = \;\frac{{1 + \sqrt {1 + {{\left( {2.5} \right)}^2}} }}{2}\; = \;1.85$$
$$\begin{array}{l} \therefore {G_E}\; = \;\frac{5}{4} \times \frac{1}{{\pi \times \sqrt {1.85} }}\\ \Rightarrow {G_E}\; = \;\frac{1}{{3.4}} \end{array}$$
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# Nondimensionalization and Scaling
1. Jul 31, 2007
### hanson
Hi all.
To me, scaling means adopting properly scales in nondimensionalization.
However, as I see in the book "A modern introduction to the mathematical theory of water waves" by R.S.Johnson, the author distinguish the two processes in a way that confuses me much. (Sec. 1.3.1 and 1.3.2)
Can someone who have used this book kindly help clarify the two things?
2. Aug 19, 2007
### scotlass
Hello,
I am not familiar with the book, but I believe non-dimensionalization literally gets rid of all dimensional quantities. So, you might write something like
x=Lx* where L is a typical length scale in the x direction and resplacing all your x' s by x* 's where x* is a non dimesional quantity since you have taken 10metres, say, and divided it by metres to just get 10.
Scaling on the other hand helps you get an idea of the relative sizes of terms so if you had something long in the x direction and short in the y direction you would write something like x=x* and y=epsilon y* where epsilon <<1 and plug this into your equations. You would then be able to see the relative sizes of terms with epsilons, epsilon^2 etc and maybe discard the highest order terms in epsilon if appropriate.
Hope this helps
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# GraphML With JUNG: Importing GraphML
This is a follow up to my earlier post on saving to GraphML using the JUNG library. It would make more sense if you browsed through that post before reading this one.
You first need to read from your file. This snippet creates a standard Java BufferedReader, which in turn is fed from a FileReader that reads from the file specified by the String filename.
The problem is, the GraphMLReader2 is just a smart text parser. For it to do anything useful, you need to specify Transformers for the vertices, edges, hyperedges and the graph itself. We pump these transformers into the function reading the graph so it understands what exactly to do after it reads these different components in from the GraphML file.
This transformer is used to parse the <graph> tag in my GraphML file and create a new Graph (it would really help if you had your GraphML file open at this point to understand what the transformers are doing). This tag specifies the beginning of your graph, and whether the graph is directed or undirected. The function metadata.getEdgeDefault() in this case returns "undirected" which is equal to the library-defined constant metadata.getEdgeDefault().DIRECTED. So this transformer checks whether the graph in the file is directed and returns the required Graph object.
The vertex transformer is slightly more involved. I’ll list out the various parts:
• MyVertexFactory: This is my custom Factory that generates objects of my custom MyVertex class.
• MyVertexFactory.getInstance(): This returns an object of the MyVertexFactory class that I can now use to access it’s methods.
• v.setX(Double d)/v.setY(Double d): These are custom functions I’ve defined for my MyVertex class that set the private int x and private int y properties of the current vertex to the specified value.
• metadata.getProperty("x")/metadata.getProperty("y"): This gets the value of the x and y properties that I’ve stored in my GraphML file (we used the addVertexData function in the previous post to save these out).
The vertex transformer parses the vertex data from the file and reads in its x and y coordinates, and then sets these properties on the newly created MyVertex object.
Similar to the previous transformer, this edge transformer reads in the edge data from the file and creates a new edge. The MyEdgeFactory and getInstance() are analogous to those just discussed for the MyVertex class.
This is pretty much the same code as my edge transformer! My graphs didn’t have hyper-edges, but the GraphMLReader2 definition requires that you specify a hyperedge transformer too.
There is a small catch though; the graph right now will be displayed without any regard to the x and y properties associated with its vertex objects. So you’ll need to specify a StaticLayout with a custom transformer, that reads in the x and y properties of each vertex and places them accordingly on your canvas. You do that like this:
This transformer is required to convert a vertex object to a Point2D object that the layout needs to position its vertices. It creates a new Point2D object by using my custom getX() and getY() methods that return the x and y properties of the vertex.
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3.6k views
What is the output of the following C program?
#include<stdio.h>
void main(void){
int shifty;
shifty=0570;
shifty=shifty>>4;
shifty=shifty<<6;
printf("The value of shifty is %o \n",shifty);
}
1. The value of shifty is 15c0
2. The value of shifty is 4300
3. The value of shifty is 5700
4. The value of shifty is 2700
| 3.6k views
Ans is D) 2700
If an integer constant starts with 0, it is considered an octal constant. So 0570 = 101 111 000 in binary as each octal digit can be directly converted to 3 bits. Now, >>4 (right shift 4) gives 010 111 and <<6 (left shift 6) gives 010 111 000 000 = 2700 in octal. If we use "%d" in printf, this would be 1024+64*7 = 1472.
by Boss (10.5k points)
selected by
0
pls explain
0
see 0570 mean its in octal .
so 000 101 111 000
now >>4 means 010 111
now <<6 means 010 111 000 000 = 2700 . peace
+2
what is the meaning of PEACE,please tell me just for the sake of knowledge .every time after explaination you use this word.
+6
I use this but not every time .
peace means in hindi "Shanti" , so im looking for her :D :P
anyway its like this ✌ .
0
can u plz explain properly..how do u split those binary >>4 010 111
+1
How this would be 1024+64*7 = 1472 If we use "%d" in printf. octal to decimal for 2700 is not 1472. how u r getting this??
0
2*8^3 + 7*8^2 =1472
+1 vote
It is in octal(0570)8 Since it starts with octal, so take 3 bits in binary format.
(0570)8=000 101 111 000
shifty>>4(left shift the bits 4 times)=100 000 010 111
shifty<<6(right shift the bits 6 times)=010 111 000 000=(2700)8
by (61 points)
0
kk method is correct ...but i don't know how perform left sift or right shift..
0
ITS not shift by 4 times its by shift by 4 bits and is right shift and 2nd one is left shift by 6 bits.
0
It is in octal(0570)8 Since it starts with octal, so take 3 bits in binary format.
(0570)8=000 101 111 000
shifty>>4(left shift the bits 4 times)=100 000 010 111
shifty<<6(right shift the bits 6 times)=010 111 000 000
how did you that ?
i think it should be
shifty>>4(left shift the bits 4 times)=100 000 010 111
shifty<<6(right shift the bits 6 times)=010 111 000 100
There is an error in program....for any result main' must return 'int'....
by Boss (45.4k points)
D) 2700
by (137 points)
0
give a solution
+5
shifty = 0570 ( decimal equivalent = (376)10 = (570)8 )
shifty = shifty >>4 = ( 376/(24) ) = 23
shifty = shifty <<6 = 23 * 26 = 1472
(1472)10 = ( 2700)8
0
0
what does > > or < < means in C language..?
0
>>is right shift and << is left shift operator
0
right shift of 110001011 ..??
int shifty;
shifty=0570;
If an integer constant begins with 0x or 0X, it is hexadecimal. If it begins with the digit 0, it is octal. Otherwise, it is assumed to be decimal.
shifty=shifty>>4;
>> means Right Shift means "divide by 2"
shifty=shifty<<6;
<< means Left Shift means "multiply by 2"
Option D
by Loyal (6.3k points)
+1 vote
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Question 1: Check the validity of the following statements:
(i) p :100 is a multiple of 4 and 5. $Since, 100 is a multiple of 4 and 5, the statement is true. Hence, it is a valid statement. (ii) q: 125 is a multiple of 5 and7.$
Since, 125 is a multiple of 5 but not a multiple of 7, the statement is not true. Hence, it is not a valid statement.
(iii) r: 60 is a multiple of 3 or 5. \$
Since, 60 is a multiple of 3 and 5, the statement is true. Hence, it is a valid statement.
$\\$
Question 2: Check whether the following statement are true or not:
$\displaystyle (i) \ p: \text{ If } x \text{ and } y \text{ are odd integers, then } x + y \text{ is an even integer }$
$\displaystyle (ii) \ q: \text{ If } x, y \text{ are integers such that } xy \text{ is even, then at least one of } \\ \\ x \text{ and } y \text{ is an even integer. }$
(i) $\displaystyle p:$ If $\displaystyle x$ and $\displaystyle y$ are odd integers, then $\displaystyle x + y$ is an even integer.
Let us assume that $\displaystyle 'p'$ and $\displaystyle 'q'$ be the statements given by
$\displaystyle p: x$ and $\displaystyle y$ are odd integers.
$\displaystyle q: x + y$ is an even integer
the given statement can be written as :
if $\displaystyle p,$ then $\displaystyle q.$
Let $\displaystyle p$ be true. Then, $\displaystyle x$ and $\displaystyle y$ are odd integers
$\displaystyle x = 2m+1, y = 2n+1$ for some integers $\displaystyle m, n$
$\displaystyle x + y = (2m+1) + (2n+1)$
$\displaystyle x + y = (2m+2n+2)$
$\displaystyle x + y = 2(m+n+1)$
$\displaystyle x + y$ is an integer
$\displaystyle q$ is true.
So, $\displaystyle p$ is true and $\displaystyle q$ is true.
Hence, “if p, then q “is a true statement.
(ii) $\displaystyle q:$ if $\displaystyle x, y$ are integer such that $\displaystyle xy$ is even, then at least one of $\displaystyle x$ and $\displaystyle y$ is an even integer.
Let us assume that $\displaystyle p$ and $\displaystyle q$ be the statements given by
$\displaystyle p: x$ and $\displaystyle y$ are integers and $\displaystyle xy$ is an even integer.
$\displaystyle q:$ At least one of $\displaystyle x$ and $\displaystyle y$ is even.
Let $\displaystyle p$ be true, and then $\displaystyle xy$ is an even integer.
So,
$\displaystyle xy = 2(n + 1)$
Now,
Let $\displaystyle x = 2(k + 1)$
Since, $\displaystyle x$ is an even integer, $\displaystyle xy = 2(k + 1). y$ is also an even integer.
Now take $\displaystyle x = 2(k + 1) \text{ and } y = 2(m + 1)$
$\displaystyle xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)$
So, it is also true.
Hence, the statement is true.
$\\$
Question 3: Show that the statement: $\displaystyle p : \text{ If } x \text{ is a real number such that } x^3+x = 0, \text{ then } x \text{ is } 0 "$ is true by
(i) direct method (ii) method of contrapositive (iii) method of contradiction.
(i) direct method
Let us assume that $\displaystyle 'q'$ and $\displaystyle 'r'$ be the statements given by
$\displaystyle q: x$ is a real number such that $\displaystyle x^3 + x = 0.$
$\displaystyle r: x \text{ is } 0.$
The given statement can be written as:
if $\displaystyle q,$ then $\displaystyle r.$
Let $\displaystyle q$ be true. Then, $\displaystyle x$ is a real number such that $\displaystyle x^3 + x = 0$
$\displaystyle x$ is a real number such that $\displaystyle x(x^2 + 1) = 0$
$\displaystyle x = 0$
$\displaystyle r$ is true
Thus, $\displaystyle q$ is true
Therefore, $\displaystyle q$ is true and $\displaystyle r$ is true.
Hence, $\displaystyle p$ is true.
(ii) Method of Contrapositive:
Let $\displaystyle r$ be false. Then,
$\displaystyle R$ is not true
$\displaystyle x \neq 0, x \in R$
$\displaystyle x(x^2 + 1) \neq 0, x \in R$
$\displaystyle q$ is not true
Thus, $\displaystyle -r = -q$
Hence, $\displaystyle p : q$ and $\displaystyle r$ is true
If possible, let $\displaystyle p$ be false.
Then, $\displaystyle p$ is not true
$\displaystyle -p$ is true
$\displaystyle -p (p \geq r)$ is true
$\displaystyle q$ and $\displaystyle -r$ is true $\displaystyle x$ is a real number such that $\displaystyle x^3 + x = 0 \text{ and } x \neq 0$
$\displaystyle x = 0 \text{ and } x \neq 0$
Hence, $\displaystyle p$ is true.
$\\$
Question 4: Show that the following statement is true by the method of contrapositive:
$\displaystyle p: \text{"If } x \text{ is an integer and } x^2 \text{is odd , then } x \text{is also odd " }$
Let us assume that $\displaystyle 'q'$ and $\displaystyle 'r'$ be the statements given
$\displaystyle q: x$ is an integer and $\displaystyle x^2$ is odd.
$\displaystyle r: x$ is an odd integer.
The given statement can be written as:
$\displaystyle p:$ if $\displaystyle q,$ then $\displaystyle r.$
Let $\displaystyle r$ be false.
Then, $\displaystyle x$ is not an odd integer, then $\displaystyle x$ is an even integer
$\displaystyle x = (2n)$ for some integer n
$\displaystyle x^2 = 4n^2$
$\displaystyle x^2$ is an even integer
Thus, $\displaystyle q$ is False
Therefore, $\displaystyle r$ is false and $\displaystyle q$ is false
Hence, $\displaystyle p:$ “ if $\displaystyle q,$ then $\displaystyle r$” is a true statement.
$\\$
Question 5: Show that the following statement is true “The integer $\displaystyle n$ is even if and only if $\displaystyle n^2$ is even”
Let the statements,
$\displaystyle p:$ Integer $\displaystyle n$ is even
$\displaystyle q:$ If $\displaystyle n^2$ is even Let $\displaystyle p$ be true. Then,
Let $\displaystyle n = 2k$
Squaring both the sides, we get,
$\displaystyle n^2 = 4k^2$
$\displaystyle n^2 = 2.2k^2$
$\displaystyle n^2$ is an even number.
So, $\displaystyle q$ is true when $\displaystyle p$ is true.
Hence, the given statement is true.
$\\$
Question 6: By giving a counter example, show that the following statement is not true.
p: “If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle”
Let $\displaystyle PQR$ be triangle in which $\displaystyle P = 60^{\circ}, Q = 60^{\circ}, R = 60^{\circ},$ then the $\displaystyle \triangle PQR$ is not an obtuse angled triangle.
Therefore given statement is not true.
$\\$
Question 7: Which of the following statements are true and which are false? In each case give a valid reason for saying so:
(i) $\displaystyle p :$ Each radius of a circle is a chord of the circle.
The given statement is false.
According the the definition of a chord, it should intersect the circumference of a circle at two distinct points.
(ii) $\displaystyle q:$ The center of a circle bisects each chord of the circle.
The given statement is false.
If a chord is not a diameter of a circle, then the center does not bisect that chord. In other words, the center of a circle only bisects the diameter, which is the chord of the circle.
(iii) $\displaystyle r :$ Circle is a particular case of an ellipse.
The statement is true.
$\displaystyle \text{ Equation of an ellipse: } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
If we put $\displaystyle a = b = 1,$ then we obtain $\displaystyle x^2 + y^2 = 1$ which is the equation of a circle. Therefore, a circle is a particular case of an ellipse.
(iv) $\displaystyle s :$If $\displaystyle x$ and $\displaystyle y$ are integers such that $\displaystyle x > y,$ then $\displaystyle - x <-y.$
The statement is true.
$x > y \ \ \ \Rightarrow -x < -y$ (By the rule of inequality)
(v) $\displaystyle t: \sqrt{11}$ is a rational number.
The given statement is false.
11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore $\sqrt{11}$ is an irrational number.
$\\$
Question 8: Determine whether the argument used to check the validity of the following statement is correct:
$\displaystyle p: \text{ "If } x^2 \text{ is irrational, then } x \text{ is rational " }$
The statement is true because the number $\displaystyle x^2 = \pi^2$ is irrational, therefore $\displaystyle x = \pi$ is irrational.
Argument Used: $\displaystyle x^2 = \pi^2$ is irrational, therefore $\displaystyle x = \pi$ is irrational.
$\displaystyle p:$ “If $\displaystyle x^2$ is irrational, then $\displaystyle x$ is rational.”
Let us take an irrational number given by $\displaystyle x = \sqrt{k},$ where k is a rational number.
Squaring both sides, we get,
$\displaystyle x^2 = k$
$\displaystyle x^2$ is a rational number and contradicts our statement.
Hence, the given argument is wrong.
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# jbovlaste import how to use this book lang en
### TODO
• Insert missing examples
### Collation
The current LaTeX export of the dictionary sorts output in ASCIIbetical order. This means that all uppercase letters comes before any lowercase letter, and that longer words come before shorter words.
There exists working code that sorts the headwords in the standard order, but that has not yet been incorporated into the main codebase. If you don't like that, talk to our nice LaTeX guy, ay Kominek.
### How to read a definition
This explanation presupposes some insight in the grammar of Lojban. In particular, gismu, brivla, lujvo and sumti/place structure should be familiar concepts. A good introductory text is Lojban for Beginners — velcli befi la lojban. bei loi co'a cilre, by Nick Nicholas and Robin Turner. A somewhat more formal text is The Complete Lojban Language (1997) by John Cowan. Both texts should be available from the same place as you got this book, or directly from the Logical Language Group.
The definition of a brivla (predicate words) contains one or more variables that are placeholders for the sumti (arguments) in the sentence. These look like one or more letters, followed by a subscripted number. This is the definition of the word {jvsv tavla}:
• $x_1$ talks/speaks to $x_2$ about subject $x_3$ in language $x_4$
Here, the $x_1$ is the speaker, the $x_2$ is the audience/listener, the $x_3$ is the subject and the $x_4$ is the language. So, when the fourth place in a sentence with {jvsv tavla} at its core is filled with something, it contains a claim that it is used a language for an act of speaking.
The previous example was from a gismu, the root words of Lojban. Lujvo are composed of component gismu by a process we will explain further later in this introductory chapter. Suffice it to say that in lujvo definitions, it is customary to number the arguments not according to their place in the actual lujvo, but according to their place in the component gismu. Perhaps this will be clearer with an example. This is the definition of the word {jvsv firkre}:
• $k_1$ is the beard/moustace/facial hair of $f_2$
{jvsv firkre}, incidentally, is built out of the words {jvsv flira} (face) and {jvsv kerfa} (hair). What this definition says is that it relates something that fits the $x_1$ place of hair (the facial hair) with something that fits the $x_2$ place of face (the person with a face; that is, the bearer of the facial hair). We figure this out by looking at which component the letters in the variables is an initial of ({jvsv kerfa} starts with k, so $k_1$ is the first place of kerfa.
Sometimes, the letters in the variables have to have more than one letter. This happens when the component gismu have initial letters in common. Take, for instance {jvsv cidydzu}:
• $ca_1=ci_3$ crawls on surface $ca_2$
Here, simply using $c_n$ would be pointless, because we wouldn't be able to tell which of the gismu, {jvsv cadzu} or {jvsv cidni}, it was pointing to. So we add enough additional letters to disambiguate.
The remaining thing to find out is which argument places, in the word {jvsv firkre}, the variables refer to. By convention, the first place is mentioned first, and the second place is mentioned second, and so on. This is the case in this example. However, it is also possible to explicitly state what goes where. We could, for instance, change the above to read:
• $x_1=k_1$ is the beard/moustace/facial hair of $x_2=f_2$
Where the equality sign is used to say multiple things about a single place. We don't usually do that, except for two cases. One case is the one where the definition sounds better when we reverse the order of the places from the one they have in the Lojban.
• insert example here
We also use explicit place numbering in cases where one of the component gismu start with the letter x. We also use at least two of the first letters of the gismu that start with x.
### Canonical and non-canonical lujvo
A lujvo is a compound word that has been assembled from several affixes that we call rafsi. Many gismu have more than one rafsi assigned. These are all equivalent in meaning. This means that a sequence of gismu can be assembled into a lujvo in many different ways, and all these resulting lujvo are by definition, in a sense, "the same word", but with different forms.
• insert example here
When a lujvo is defined, one of these forms are chosen as the "canonical" lujvo, and committed to the dictionary. The other forms are "non-canonical" forms, and ommitted from the dictionary. These non-canonical forms are valid variants, but should be avoided by the writer to ease dictionary lookup. We realize the inconvenience in this, but listing all possible variant forms could easily explode the size of the dictionary. We hope that the jbovlaste import: table of rafsi lang en will help at least some in breaking an unfamiliar lujvo into its components, and aid in finding the canonical form.
### How to find the translation of an English word
If you find a Lojban word that you are not familiar with as the translation of the English term you have looked up, you should always check the definition of the Lojban word to be absolutely sure that it is the one you need.
But in general, we have tried to choose English keywords so that the following is possible:
• If the English keyword is a noun, and the Lojban translation is a brivla, "lo" followed by that brivla should be a reasonable translation of that noun.
• If the English keyword is a proper noun, and the Lojban translation is a cmene (cmevla), "la" followed by that cmene should be a reasonable translation.
• If the English keyword is a verb, and the Lojban translation is a brivla, that brivla should say about the same about its $x_1$ place as the verb says about its subject. When it comes to the other arguments, however, you must look at the definition of the Lojban word.
### Caveats
While the editors of this dictionary have in general checked on each other's work, it has not been put under strict editorial review per se. In particular, the editing system does not offer the possibility of voting for the exclusion of specific words.
## Experimental gismu
This dictionary contains what is commonly referred to as "experimental gismu". These are words of the form CVCCV or CCVCV that have been coined by the users of the language. The LLG has reserved the unassigned word forms with these forms, and specifically discourages the usage and attempts to assign a meaning to them. Nevertheless, one of the aims of this dictionary project is to document usage, and then we need to include everything that has actually been used, compliant or not.
In particular, the reader should be aware that at the time of this writing, the exported print dictionary does NOT set off unofficial gismu in any way. We refer people to the web or dict interfaces to determine whether a particular gismu is official or not.
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# In Young's Experiment to produce interference pattern using light of wavelength 'λ', the width of the fringe observed on the screen placed 'D' distance from the two slits which are placed 'd' units apart is equal to ____________.
This question was previously asked in
SSC Scientific Assistant Previous Paper 6 (Held On: 24 November 2017 Shift 2)
View all SSC Scientific Assistant Papers >
1. dλ/D
2. d/Dλ
3. Dλ/d
4. D/dλ
Option 3 : Dλ/d
Free
SSC IMD - Full Mock Test (CS & IT)
6158
200 Questions 200 Marks 120 Mins
## Detailed Solution
CONCEPT:
• YDSE (Young’s Double Slit Experiment): If a monochromatic light falls on two narrow slits which are very close to each other act as coherent source superpose each other and an interference pattern is observed on the screen.
The fringe width is given by:
$$\beta = \frac{{\lambda D}}{d}$$
Where λ is the wavelength of light used, D = distance of the screen from the slits, d= distance between two slits.
EXPLANATION:
Fringe width, $$\beta =\frac{D\lambda }{d}$$
So option 3 is correct.
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### Archive
Posts Tagged ‘TST’
## Romanian TST II 2011 Problem 4
Show that
(a) There are infinitely many positive integers $n$ such that there exists a square equal to the sum of the squares of $n$ consecutive positive integers (for instance, $2$ and $11$ are such, since $5^2=3^2+4^2$, and $77^2=18^2+19^2+...+28^2$).
(b) If $n$ is a positive integer which is not a perfect square, and if $x_0$ is an integer number such that $x_0^2+(x_0+1)^2+...+(x_0+n-1)^2$ is a perfect square, then there are infinitely many positive integers $x$ such that $x^2+(x+1)^2+...+(x+n-1)^2$ is a perfect square.
Romanian TST 2011
Categories: IMO, Number theory, Olympiad Tags: ,
## Romanian TST II Problem 3
May 31, 2011 1 comment
Given a positive integer $n$, determine the maximum number of edges a simple graph of $n$ edges vertices may have, in order that it won’t contain any cycles of even length.
Romanian TST 2011
Categories: Combinatorics, IMO, Olympiad Tags: ,
## Romanian TST II 2011 Problem 2
Let $A_0A_1A_2$ be a triangle. The incircle of the triangle $A_0A_1A_2$ touches the side $A_iA_{i+1}$ at the point $T_{i+2}$ (all indices are considered modulo $3$). Let $X_i$ be the foot of the perpendicular dropped from the point $T_i$ onto the line $T_{i+1}T_{i+2}$. Show that the lines $A_iX_i$ are concurrent at a point situated on the Euler line of the triangle $T_0T_1T_2$.
Romanian TST 2011
Categories: Geometry, IMO, Olympiad Tags: , ,
## Romanian TST II 2011 Problem 1
A square of sidelength $\ell$ is contained in the unit square whose centre is not interior to the former. Show that $\ell \leq 1/2$.
Romanian TST 2011
Categories: Geometry, IMO, Olympiad Tags: ,
## Romanian TST 2011 Problem 4
Suppose $n \geq 2$ and denote $S_n$ the permutations of the set $\{1,2,..,n\}$. For $\sigma \in S_n$ denote by $\ell(\sigma)$ the number of cycles in the disjoint cycle decomposition of the permutation $\sigma$. Calculate the sum $\displaystyle \sum_{\sigma \in S_n} \text{sgn}(\sigma) n^{\ell(\sigma)}$.
## Romanian TST 2011 Problem 2
Denote $S=\{\lfloor n \pi \rfloor : n \in \Bbb{N} \}$.
Prove that:
a) For any positive integer $m \geq 3$ there exists an arithmetic progression of length $m$ in $S$.
b) Prove that $S$ does not contain a infinite length arithmetic progression.
Categories: Algebra, IMO, Problem Solving
## Romanian TST 2011 Problem 1
April 20, 2011 1 comment
Find all functions $f: \Bbb{R} \to \Bbb{R}$ for which we have $2f(x)=f(x+y)+f(x+2y),\ \forall x \in \Bbb{R},\ \forall y \geq 0$.
Romanian TST 2011
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# Altarelli & Meloni's GUT-minimalism
1. May 7, 2013
### mitchell porter
http://arxiv.org/abs/1305.1001
A non Supersymmetric SO(10) Grand Unified Model for All the Physics below $M_{GUT}$
Guido Altarelli, Davide Meloni
(Submitted on 5 May 2013)
We present a renormalizable non supersymmetric Grand Unified SO(10) model which, at the price of a large fine tuning, is compatible with all compelling phenomenological requirements below the unification scale and thus realizes a minimal extension of the SM, unified in SO(10) and describing all known physics below $M_{GUT}$. These requirements include coupling unification at a large enough scale to be compatible with the bounds on proton decay; a Yukawa sector in agreement with all the data on quark and lepton masses and mixings and with leptogenesis as the origin of the baryon asymmetry of the Universe; an axion arising from the Higgs sector of the model, suitable to solve the strong CP problem and to account for the observed amount of Dark Matter. The above constraints imposed by the data are very stringent and single out a particular breaking chain with the Pati-Salam group at an intermediate scale $M_I\sim10^{11}$ GeV.
For those interested in "minimalism", this is an interesting construction because it incorporates GUT successes that aren't part of the nuMSM.
To turn this into a true "phenomenological theory of everything", you also need to add an inflaton (singlet under SO(10), I guess) and e.g. a cosmological constant for the dark energy.
2. May 7, 2013
### arivero
Baez and Huerta did a good job explaining the breaking via a square with sides
SO(10)---> PatiSalam,
SO(10)--->SU(5)
PatiSalam---> StandardModel
SU(5)---> StandardModel
It is amusing that there is a counterpart of the square with manifolds, looking the groups as isometries on compact manifolds, with the relabeling
SO(10) ==== sphere S9
SU(5) ===== CP4
P-S ===== S3 x S5
SM ==== Witten's 7 dim manifolds.
Because of this, a naive, unfulfilled expectation of the string approach had been to produce SM from M-Theory, PatiSalam (and/or SU(5)) from F-Theory and SO(10) from S-theory.
Last edited: May 7, 2013
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1. ## Parallelepiped Vertices
If vector OA, vector OB, and vector OC are three edges of a parallelpiped where O is (0,0,0), A is (2,4,-2), B is (3,6,1) and C is (4,0,-1) find the coordinates of the other vertices of the parallelpiped.
for this q, i am having trouble drawing the diagram of the parallelpiped
Currently though, I have the following possibilities.. (there are only 3 right?)
D(5,10,0)<<< DB//CA
D(3,-2,0)<<<CB//DA
2. Hello skeske1234
Originally Posted by skeske1234
If vector OA, vector OB, and vector OC are three edges of a parallelpiped where O is (0,0,0), A is (2,4,-2), B is (3,6,1) and C is (4,0,-1) find the coordinates of the other vertices of the parallelpiped.
for this q, i am having trouble drawing the diagram of the parallelpiped
Currently though, I have the following possibilities.. (there are only 3 right?)
D(5,10,0)<<< DB//CA
D(3,-2,0)<<<CB//DA
If you're having trouble drawing it, then don't try! Just think of it as being a boring old cuboid (rectangular box) that's been squashed a bit.
Instead of rectangles, then, the faces are all parallelograms. So if we consider the face containing $O$, $A$ and $B$, the fourth vertex of this parallelogram - $P$, say - is such that:
$\vec{OP} = \vec{OA} + \vec{OB} = \begin{pmatrix} 2\\4\\-2\end{pmatrix}+ \begin{pmatrix} 3\\6\\1\end{pmatrix}=\begin{pmatrix} 5\\10\\-1\end{pmatrix}$
Then deal with the 2 faces containing $O$, $A$ and $C$; and $O$, $B$ and $C$, respectively, in the same way to get, say, vertices $Q$ and $R$.
Finally, we shall need the eighth vertex, $S$, say, whose position vector is the sum of all three vectors $\vec{OA} + \vec{OB}+\vec{OC}$. This vertex will be diagonally opposite $O$.
Can you complete it now?
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# how to find edges of parallelopiped
Click on a term to search for related topics.
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# pylops.optimization.leastsquares.PreconditionedInversion¶
pylops.optimization.leastsquares.PreconditionedInversion(Op, P, data, x0=None, returninfo=False, **kwargs_lsqr)[source]
Preconditioned inversion.
Solve a system of preconditioned equations given the operator Op and a preconditioner P.
Parameters: Op : pylops.LinearOperator Operator to invert P : pylops.LinearOperator Preconditioner data : numpy.ndarray Data x0 : numpy.ndarray Initial guess returninfo : bool Return info of LSQR solver **kwargs_lsqr Arbitrary keyword arguments for scipy.sparse.linalg.lsqr solver xinv : numpy.ndarray Inverted model. istop : int Gives the reason for termination 1 means $$\mathbf{x}$$ is an approximate solution to $$\mathbf{d} = \mathbf{Op}\mathbf{x}$$ 2 means $$\mathbf{x}$$ approximately solves the least-squares problem itn : int Iteration number upon termination r1norm : float $$||\mathbf{r}||_2$$, where $$\mathbf{r} = \mathbf{d} - \mathbf{Op}\mathbf{x}$$ r2norm : float $$\sqrt{\mathbf{r}^T\mathbf{r} + \epsilon^2 \mathbf{x}^T\mathbf{x}}$$. Equal to r1norm if $$\epsilon=0$$
RegularizedInversion
Regularized inversion
NormalEquationsInversion
Normal equations inversion
Notes
Solve the following system of preconditioned equations given the operator $$\mathbf{Op}$$, a preconditioner $$\mathbf{P}$$, the data $$\mathbf{d}$$
$\mathbf{d} = \mathbf{Op} (\mathbf{P} \mathbf{p})$
where $$\mathbf{p}$$ is the solution in the preconditioned space and $$\mathbf{x} = \mathbf{P}\mathbf{p}$$ is the solution in the original space.
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# Step-by-step Solution
Go!
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## Step-by-step explanation
Problem to solve:
$\int\left(\frac{1}{60^2}+\frac{4}{5}\cdot x\right)^{\frac{\left(-1\right)}{2}}dx$
Learn how to solve calculus problems step by step online.
$\int\left(\frac{1}{3597}+\frac{4}{5}x\right)^{-\frac{1}{2}}dx$
Learn how to solve calculus problems step by step online. Integrate int(((1/(60^2))+(4/5)*x)^((-1/2)))dx with respect to x. Simplifying. Applying the property of exponents, \displaystyle a^{-n}=\frac{1}{a^n}, where n is a number. Solve the integral \int\frac{1}{\sqrt{\frac{1}{3597}+\frac{4}{5}x}}dx applying u-substitution. Let u and du be. Isolate dx in the previous equation.
$\frac{5}{2}\sqrt{\frac{1}{3597}+\frac{4}{5}x}+C_0$
### Problem Analysis
$\int\left(\frac{1}{60^2}+\frac{4}{5}\cdot x\right)^{\frac{\left(-1\right)}{2}}dx$
Calculus
~ 0.05 seconds
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# Math Help - series and convergence
1. ## series and convergence
This might seem super easy, but could anyone give me some hint how to show the following An is convergent?
1.n!/n^n
2.(1-1/n)^(n^2)
3. 1-cos (1/n)
I've tried the limit test/ratio test, but not working. Maybe comparison test?
Also, could anyone help calculating the series using taylor expansion of cos?
the sum of all (cos n)/(2^n) where n goes from 0 to infinity?
Any help is greatly appreciated!
2. Use the root test for these
3. thanks, I've figured out how to do the convergence tests. Could anyone please give me a hint how to calculate the sum of cos n/2^n?
4. Originally Posted by nngktr
thanks, I've figured out how to do the convergence tests. Could anyone please give me a hint how to calculate the sum of cos n/2^n?
$\sum_{n=0}^{\infty}\frac{\cos(n)}{2^n}=\text{Re}\s um_{n=0}^{\infty}\frac{e^{in}}{2^n}$ and since $\left|\frac{e^i}{2}\right|<1$ we see that this is equal to $\text{Re}\frac{1}{1-\frac{e^i}{2}}=\text{Re}\frac{2}{2-e^i}=\cdots=\frac{2(\cos(1)-2)}{4\cos(1)-5}$
I've rigorously justified nothing. I leave that to you.
EDIT: P.S. that is not the Taylor expansion for $\cos(x)$. Try the Maclaurin $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$
5. Because is...
$\cos x = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!}$ (1)
... is also...
$1 - \cos \frac{1}{n} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n^{2k} (2k)!}$ (2)
... so that is...
$\sum_{n=1}^{\infty} (1-\cos \frac{1}{n}) = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n^{2k} (2k)!} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k)!} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k)!} \zeta (2k)$ (3)
For $k\ge 2$ is $1 < \zeta (k) < 2$ so that the series in second term of (3) converges... and converges also very quickly! ...
Kind regards
$\chi$ $\sigma$
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# MATH239
Combinatorics.
Instructor: Martin Pei
Office: MC 6492
Email: mpei@uwaterloo.ca, martin31415926@gmail.com
# 5/5/14
## Enumeration
### Sets and Combinatorics
Combinatorics is a discrete mathematics. Topics include enumeration (counting) and graph theory.
The size/cardinality of a set $$A$$ is denoted $$\abs{A}$$. It is the number of elements it contains, if it is finite, and some other things if it is infinite.
$$[n]$$ is the set of all positive integers from 1 to $$n$$, inclusive.
The Cartesian product of two sets $$A$$ and $$B$$ is $$A \times B = \set{(a, b) \middle| a \in A, b \in B}$$. In other words, it is the set of all pairs of elements from both sets. This can easily be extended to more than two sets: $$A \times B \times C$$.
Also, $$A^k = A \times \ldots \times A$$ ($$k$$ times). Also, $$A^0 = \set{()}$$ - the set containing the empty tuple.
If $$A$$ and $$B$$ are finite, then $$\abs{A \times B} = \abs{A}\abs{B}$$.
Consider all possible results of throwing two six-sided dice:
Clearly, the set of all possible results for each die is $$[6]$$ and $$[6]$$.
So the set of all possible results of both die is $$[6] \times [6] = \set{(a, b) \middle| a, b \in [6]}$$.
Find the number of possible binary strings of length $$n$$:
Clearly, each digit in the string is of the set $$\set{0, 1}$$, and there are $$n$$ digits.
Clearly, the number of possiblities are $$\abs{\set{0, 1} \times \ldots \times \set{0, 1}}$$ ($$n$$ times), or $$2^n$$.
The set of possible binary strings is therefore $$\set{0, 1}^n$$.
Let $$S = A \cup B$$. If $$A \cap B = \emptyset$$, then $$\abs{S} = \abs{A} + \abs{B}$$. In other words, if two sets share nothing in common, then their disjunction has a size equal to the sum of their sizes.
We could say that the Cartesian product is analogous to multiplcation, and disjunction is analogous to addition.
### Permutations
A permutation is the rearrangement of the elements of $$[n]$$ - an ordered set. For example, all permutations of $$[3]$$ are $$\set{1, 2, 3}, \set{1, 3, 2}, \set{2, 1, 3}, \set{2, 3, 1}, \set{3, 1, 2}, \set{3, 2, 1}$$.
In general, $$[n]$$ has $$n!$$ permutations. We find these permutations by picking each of the $$n$$ elements for the first position, and for each of these, we pick each of the $$n - 1$$ remaining elements for the second, and for each of these, we pick each of the $$n - 2$$ remaining elements for the third, and so on.
For example, find the permutations of $$\set{1, 2, 3}$$
First, we choose 1 as the first position, so the remaining elements are $$\set{2, 3}$$.
Now we choose each of 2 and 3 for the second position, to yield the permutations $$\set{1, 2, 3}$$ and $$\set{1, 3, 2}$$.
Second, we choose 2 as the first position, so the remaining elements are $$\set{1, 3}$$.
Now we choose each of 1 and 3 for the second position, to yield the permutations $$\set{2, 1, 3}$$ and $$\set{2, 3, 1}$$.
Third, we choose 3 as the first position, so the remaining elements are $$\set{1, 2}$$.
Now we choose each of 1 and 2 for the second position, to yield the permutations $$\set{3, 1, 2}$$ and $$\set{3, 2, 1}$$.
Every permutation is also a bijection, mapping elements from $$[n]$$ into a different set. Formally, $$\sigma: [n] \to [n]$$. For example, the permutation $$\set{2, 3, 1}$$ maps $$\set{A, B, C}$$ to $$\set{B, C, A}$$. In other words, it behaves something like a function.
### Combinations
A combination is a possible way of "choosing" $$k$$ elements of a set $$[n]$$, where $$k \le n$$. In other words, it is a subset of size $$k$$ of the set $$[n]$$.
A subset of size $$k$$ is known as a $$k$$-subset. A permutation of a $$k$$-subset is a $$k$$-permutation, and a combination of a $$k$$-subset is a $$k$$-combination.
We find these subsets by basically taking the first $$k$$ elements of each permutation. We then get the set containing all the possible $$k$$-permutations, but each one is duplicated $$(n - k)!$$ times because there are that many that start with those particular $$k$$ elements and are followed by all permutations of the remaining elements.
Therefore, there are $$\frac{n!}{(n - k)!}$$ $$k$$-permutations of $$[n]$$. Clearly, each $$k$$-combination corresponds to $$k!$$ $$k$$-permutations, because combinations are basically permutations without order.
So the number of $$k$$-combinations is a factor of $$k!$$ less than the number of $$k$$-permutations, and is given by $$\frac{n!}{k!(n - k)!}$$.
In general, $$[n]$$ has $$\frac{n!}{k!(n - k)!}$$ combinations of size $$k$$. This can be represented by $$n \choose k$$. If $$n < k$$, we define $${n \choose k} = 0$$.
# 7/5/14
## Bijections
A bijection is a function that maps a set $$S$$ to another set $$T$$, and is one to one and onto.
One to one means that the function has a unique element of $$T$$ for every value of $$S$$ - no two things in $$S$$ are mapped to the same element of $$T$$. It is possible that $$T$$ is larger than $$S$$, so $$\abs{T} \ge \abs{S}$$.
Onto means that the function has a unique element of $$S$$ for every value of $$T$$ - every thing in $$T$$ must also have an element in $$S$$ that maps to it. Because there must be one element in $$S$$ for every element of $$T$$, so $$\abs{S} \ge \abs{T}$$.
A bijection is therefore a function that maps a set's elements to the elements of another set of the same size. Every element in either set has a unique corresponding element in the other set.
It is often tedious to prove that it is one to one and onto. We can prove a function is a bijection very easily, by proving that it has an inverse.
### Bijection Inverse Theorem
Proposition: a function is a bijection if and only if it has an inverse.
The inverse of $$f: S \to T$$ is a function $$f^{-1}$$ such that $$\forall x \in S, f^{-1}(f(x)) = x$$ and $$\forall y \in T, f(f^{-1}(y)) = y$$. It is basically a reverse mapping backwards from $$T$$ to $$S$$.
For example, let $$S$$ be the set of all $$k$$-subsets of $$[n]$$ and $$T$$ be the set of all $$(n - k)$$-subsets of $$[n]$$:
If $$n = 3, k = 1$$, then $$S = \set{\set{1}, \set{2}, \set{3}}, T = \set{\set{1, 2}, \set{1, 3}, \set{2, 3}}$$.
One bijection over these sets would be $$f: S \to T$$ where $$f(A) = [n] \setminus A$$. This is a bijection because it maps every item in $$S$$ to a unique item in $$T$$, with no items in $$T$$ left over.
Also, since $$\abs{A} = k$$ and $$A \subseteq [n]$$, $$\abs{f(A)} = \abs{[n]} - \abs{A} = n - k$$, as required.
Clearly, $$f^{-1}(B) = [n] \setminus B$$. Since the function has an inverse, $$f$$ is a bijection and $$\abs{S} = \abs{T} = {n \choose k} = {n \choose n - k}$$.
As an aside, we proved that $${n \choose k} = {n \choose n - k}$$.
Another example would be a mapping $$f: S \to T$$ where $$S$$ is all the subsets of $$[n]$$ and $$T$$ is all binary strings of length $$n$$:
An obvious bijection would be to map all the subsets to binary strings such that each digit is 1 if and only if the index of that digit is in the set.
Formally, we would write that as $$f(A) = a_n \cdots a_1, a_i = \begin{cases} 1 &\text{if } i \in A \\ 0 &\text{if } i \notin A \end{cases}, i \in [n]$$.
We can also prove this is a bijection by finding its inverse: $$f^{-1}(a_n \cdots a_1) = \set{i \in [n] \middle| a_i = 1}$$.
Since it is a bijection, $$\abs{S} = \abs{T}$$, so there are the same number of subsets as there are binary strings of length $$n$$.
Since there are $$2^n$$ possible binary strings, there are also $$2^n$$ subsets of $$[n]$$.
This also gives us an algorithm for listing all the subsets. Since binary strings are easy to list by counting upwards, we simply apply $$f^{-1}$$ to the list of all the binary strings of length $$n$$ to get the subsets of $$[n]$$.
## Binomial Theorem
Proposition: $$(1 + x)^n = \sum_{k = 0}^n {n \choose k} x^k$$.
Clearly, $$(1 + x)^n = (1 + x) \cdots (1 + x)$$. Clearly, each term in the expansion takes either a $$1$$ or an $$x$$ from each of the factors.
For example, $$(1 + x)^3 = (1 + x)(1 + x)(1 + x) = 1 \cdot 1 \cdot 1 + 1 \cdot 1 \cdot x + 1 \cdot x \cdot 1 + 1 \cdot x \cdot x$$. Clearly, the factors of each term is in the Cartesian product of the set $$\set{1, x}^n$$.
We can also represent this using a Cartesian product. Clearly, $$\set{1, x}^n = \set{(a_1, \ldots, a_n) \middle| a_1, \ldots, a_n \in \set{1, x}}$$.
Therefore, $$(1 + x)^n = \sum_{(a_1, \ldots, a_n) \in \set{1, x}^n} a_1 \cdots a_n$$.
Clearly, each $$a_1 \cdots a_n$$ adds another $$x^k$$ where $$k$$ is the number of occurrences of $$x$$. Since there are $$n \choose k$$ ways of picking $$k$$ occurrences of $$x$$, and all possible combinations are in the Cartesian product, there are $$n \choose k$$ terms that are $$x^k$$.
Since the highest power is $$x^n$$, $$(1 + x)^n = \sum_{k = 0}^n {n \choose k} x^k$$.
# 9/5/14
In combinatorial proofs, we often prove that two things are equal by constructing a set such that if we count the number of elements it contains in one way it results in one side, and when we count another way it results in the other side.
So if we wanted to prove $$A = B$$, then we'd have a set such that finding its size one way results in $$A$$, and in another way results in $$B$$.
### Disjoint Sets
Two sets $$A$$ and $$B$$ are disjoint sets if they have no elements in common - $$A \cap B = \emptyset$$. Three or more sets are disjoint if all of those sets are all disjoint to each other - they all contain unique elements.
If the sets $$S_1, \ldots, S_n$$ are disjoint, then $$\abs{S_1 \cup \ldots \cup S_n} = \abs{S_1} + \ldots + \abs{S_n}$$. This is because all the elements are unique, so the union is simply the set containing all the original elements.
If we replace $$x$$ with 1 in the Binomial theorem, we get $$2^n = \sum_{k = 0}^n {n \choose k}$$. The $$n \choose k$$ values are known as the binomial coefficients.
We want a combinatoric proof of this rather than a simple algebraic proof. What we will do is create a set, and when we count it one way, we get $$2^n$$, and when we prove it another way, we get $$\sum_{k = 0}^n {n \choose k}$$.
Let $$S$$ be the set of all subsets of $$[n]$$. Then $$\abs{S} = 2^n$$, from the proof of bijections of binary strings.
Let $$S_k$$ be the set of all $$k$$-subsets of $$[n]$$, for $$0 \le k \le n$$. Then $$S = S_0 \cup \ldots \cup S_n$$, and $$S_0, \ldots, S_n$$ are disjoint sets.
Then $$\abs{S} = \abs{S_1} + \ldots + \abs{S_n}$$, and $$\abs{S_k} = {n \choose k}$$ (since a $$k$$-subset is all the ways we can choose $$k$$ elements out of $$n$$).
So $$\abs{S} = 2^n = \sum_{k = 0}^n {n \choose k}$$.
### Pascal's Triangle
Using this, we can draw Pascal's Triangle, a triangle listing all the binomial coefficients:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
...
The $$n$$th row of Pascal's Triangle is all the values of $$n \choose k$$, and each column of each row is for $$k$$ from 1 to $$n$$ inclusive.
From the triangle we notice that $${n \choose k} = {n - 1 \choose k} + {n - 1 \choose k - 1}$$. However, we want to prove this:
Let $$S$$ be the set of all $$k$$-subsets of $$[n]$$.
Then $$\abs{S} = {n \choose k}$$. Let $$S_1$$ be the set of all $$k$$-subsets of $$[n]$$ that include $$n$$, and $$S_2$$ be the set of all subsets that do not.
Clearly, $$S = S_1 \cup S_2$$, which is a disjoint set, so $$\abs{S} = \abs{S_1} + \abs{S_2}$$.
Clearly, every set in $$S_1$$ contains $$n$$, plus $$k - 1$$ elements from $$[n - 1]$$, since we can't use $$n$$ again.
So $$\abs{S_1} = {n - 1 \choose k - 1}$$, since it is the set of all $$(k - 1)$$-subsets of $$[n - 1]$$.
Clearly, every set in $$S_1$$ contains $$k$$ elements, and we can only choose from $$[n - 1]$$ elements.
So $$\abs{S_2} = {n - 1 \choose k}$$, since it is the set of all $$k$$-subsets of $$[n - 1]$$.
So $${n \choose k} = {n - 1 \choose k} + {n - 1 \choose k - 1}$$.
### Hockey Stick Identity
We can also prove it algebraically:
If $$f(x)$$ is a power series, then let $$[x^n]f(x)$$ represent the coefficient of $$x^n$$ in $$f(x)$$.
So $$[x^k](1 + x)^n = {n \choose k}$$.
Clearly, $$[x^k](1 + x)^n = [x^k](1 + x)(1 + x)^{n - 1} = [x^k](1 + x)^{n - 1} + [x^k]x(1 + x)^{n - 1}$$.
Clearly, $$[x^k](1 + x)^{n - 1} = {n - 1 \choose k}$$.
Clearly, $$[x^k]x(1 + x)^{n - 1} = [x^{k - 1}](1 + x)^{n - 1} = {n - 1 \choose k - 1}$$, since the $$x$$ factor increases all the powers of $$x$$ in the power series by 1.
So $$[x^k](1 + x)^n = [x^k](1 + x)^{n - 1} + [x^k]x(1 + x)^{n - 1} = {n \choose k} = {n - 1 \choose k} + {n - 1 \choose k - 1}$$
From the above identity we know that $${n + k \choose n} = {n + k - 1 \choose n} + {n + k - 1 \choose n - 1}$$, since we substituted $$n + k$$ for $$n$$ and $$n$$ for $$k$$.
If we expand the first term of the right side using the same identity, we get $${n + k \choose n} = {n + k - 2 \choose n} + {n + k - 2 \choose n - 1} + {n + k - 1 \choose n - 1}$$.
If we keep expanding the first term on the right side, we find that we get $${n + k - (k + 1) \choose n} + {n + k - (k + 1) \choose n - 1} + \ldots + {n + k - 1 \choose n - 1}$$. Clearly, $$n + k - (k + 1) \choose n = 0$$.
So by induction, we get $$\sum_{i = 0}^k {n + i - 1 \choose n - 1} = {n + k \choose n}$$.
Now we want to prove this combinatorially:
Let $$S$$ be the set of all $$n$$-subsets of $$[n + k]$$. Then $$\abs{S} = {n + k \choose n}$$.
Let $$S_i$$ be the set of all $$n$$-subsets of $$[n + k]$$ whose largest element is $$n + i$$, where $$0 \le i \le k$$.
Clearly, the largest element of each $$n$$-subset is between $$n$$ and $$n + k$$ inclusive.
Clearly, $$S = S_0 \cup \ldots \cup S_k$$, and this is a disjoint set, so $$\abs{S} = \abs{S_0} + \ldots + \abs{S_k}$$.
Clearly, each set in $$S_i$$ must contain $$n + i$$ as the largest element, and all the other elements must be in $$[n + i - 1]$$ (since they must be less than the largest element).
So there are $$n + i - 1$$ elements to choose from, out of $$n - 1$$ spots (one spot of $$n$$ was taken by the largest element). So there are $$n + i - 1 \choose n - 1$$ elements in $$S_i$$.
So $${n + k \choose n} = \abs{S} = {n - 1 \choose n - 1} + \ldots + {n + k - 1 \choose n - 1} = \sum_{i = 0}^k {n + i - 1 \choose n - 1}$$
This is known as the hockey stick identity. The reason for this is because it states that in Pascal's Triangle, the value of a number in the triangle is the sum of all the numbers in either diagonal passing through the number directly above it from the outside edge of the triangle until it is diagonal to the number we want.
For example, the topmost and leftmost 10 in Pascal's Triangle is $$10 = 1 + 3 + 6$$, or $$10 = 4 + 3 + 2 + 1$$.
# 12/5/14
Proof techniques: bijections, partitioning sets into two sets, and counting two sets in different ways.
## Generating Series
We can also prove by using power series.
Given a set $$S$$, the universe of discourse, where each element $$\sigma \in S$$ is associated with a non-negative integer weight $$w(\sigma)$$, the generating series/generating function of $$S$$ with respect to $$w(\sigma)$$ is $$\Phi_S(x) = \sum_{\sigma \in S} x^{w(\sigma)}$$. If we expand and collect like terms, the coefficient of each $$x^k$$ is the number of elements in $$S$$ with exactly the weight $$k$$.
For example, let $$S$$ be the set of all subsets of $$[3]$$, and let $$w(A) = \abs{A}$$. So $$S = \set{\emptyset, \set{1}, \set{2}, \set{3}, \set{1, 2}, \set{1, 3}, \set{2, 3}, \set{1, 2, 3}}$$. The corresponding weights are $$0, 1, 1, 1, 2, 2, 2, 3$$.
Then the generating series is $$1 + x + x + x + x^2 + x^2 + x^2 + x^3 = 1 + 3x + 3x^2 + x^3 = (1 + x)^3$$, from the binomial theorem.
What does each coefficient of $$x^k$$ mean? It is the number of elements that have exactly the weight $$k$$. In our particular example, it is the number of elements of size $$k$$.
So if $$a_k$$ is the weight of the $$k$$th subset, then $$\Phi_S(x) = \sum_{k \ge 0} a_k x^k$$.
Now we want to generalize this. Let $$S$$ be the set of all subsets of $$[n]$$ and $$w(A) = \abs{A}$$. Then the coefficient of $$x^k$$ in $$\Phi_S(x)$$ is $$n \choose k$$, because it is the number of $$k$$-subsets of $$[n]$$ (the number of ways we can choose $$k$$ elements from $$n$$ elements).
So $$\Phi_S(x) = \sum_{k = 0}^n {n \choose k} x^k = (1 + x)^n$$, by the binomial theorem.
How many ways can we throw two dice and get a sum of 10?
Clearly, $$S = [6] \times [6]$$, the Cartesian product.
We define $$(a, b)$$ to be a pair of dice throws, and the weighting function to be their sum, $$w((a, b)) = a + b$$.
So $$\Phi_{[6] \times [6]}(x) = \sum_{(a, b) \in [6] \times [6]} x^{a + b}$$.
If we do this by hand, by considering every possible pair, we find that the generating series is $$x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^{10} + 2x^{11} + x^{12}$$.
So the answer is the coefficient of $$x^{10}$$, which is 3.
If we factor this, we get $$(x + x^2 + x^3 + x^4 + x^5 + x^6)^2$$. Note that when we multiply this out, we add up the exponents, just like the weight function.
However, we want to solve it more simply. Suppose we have two infinite sided dice.
Then $$S = \mb{N} \times \mb{N}$$, the set of all pairs of natural numbers. We keep the same weight function, $$w((a, b)) = a + b$$.
So $$\Phi_S(x) = x^2 + 2x^3 + 3x^4 + \ldots = (x + x^2 + \ldots)^2$$.
Incidentally, $$(x + x^2 + \ldots)^2 = \frac{x^2}{(1 - x)^2}$$.
In general, for a counting problem:
1. Define the set of objects.
2. Define a weight function related to the problem such that we get the answer we need.
3. Find the generating series.
4. Possibly find the coefficient of $$x^k$$
We rarely substitute a value for $$x$$. It acts as a way to keep the coefficients of $$x$$ as separate values. As a result we don't really need to worry about convergence or divergence.
This technique is useful because the generating series is sometimes easier to find than the actual coefficients, yet it can be used to find the coefficients.
The reason generating series are useful is because $$[x^n]\Phi_S(x)$$ is the number of times $$w(\sigma) = n$$. This allows us to find the number of $$\sigma$$ in $$S$$ that have $$w(\sigma) = n$$.
We can solve for $$[x^k]$$ explicitly by writing $$\Phi_S(x)$$ as a power series, solving for the index $$i$$ that would result in the term having $$x^k$$, and obtaining the coefficient of that index. However, leaving it in the form of $$[x^n]f(x)$$ is often sufficient for an answer.
For example, given $$[x^k]\sum_{i = 1}^n 2i^ix^{2i}$$, we solve $$x^k = x^{2i}$$ and $$i = \frac k 2$$. Then $$[x^k]\sum_{i = 1}^n 2i^ix^{2i} = 2i^i = 2\left(\frac k 2\right)^{\frac k 2}$$.
# 14/5/14
## Formal Power Series
Let $$a_1, a_2, \ldots$$ be a sequence of numbers. Then the formal power series associated with $$\{a_n\}$$ is $$A(x) = \sum_{k \ge 0} a_k x^k$$.
This allows us to represent the entire sequence using one power series.
The coefficient of $$x^n$$ in $$A(x)$$ is denoted $$[x^n]A(x)$$. For example, $$[x^2](1 + 2x + 3x^2 + 4x^3 + \ldots) = 3$$.
Two power series are equal if they have the same coefficients for the same powers: $$A(x) = B(x) \iff \forall n \ge 0, [x^n]A(x) = [x^n]B(x)$$.
Addition is defined as $$A(x) + B(x) = \sum_{k \ge 0} \left([x^k]A(x) + [x^k]B(x)\right)x^k$$.
For example, if $$A(x) = (1 + 2x)^{10}, B = 1 + x + x^2 + \ldots$$, find $$[x^n](A(x) + B(x))$$:
Clearly, $$A(x) = (1 + 2x)^{10} = \sum_{k = 0}^{10} {10 \choose k} (2x)^k = \sum_{k = 0}^{10} {10 \choose k} 2^k x^k$$.
Clearly, $$B(x) = \sum_{k = 0}^n x^k$$.
So $$[x^n](A(x) + B(x)) = \begin{cases} 2^n {10 \choose n} + 1 &\text{if } 0 \le n \le 10 \\ 1 &\text{if } n > 10 \end{cases}$$.
Multiplication is defined as $$A(x)B(x) = \sum_{n \ge 0} \left(\sum_{i = 0}^n a_i b_{n - i}\right)x^n$$. Since $$\sum_{i = 0}^n a_i b_{i - 1}$$ is always finite, multiplication of power series is closed under multiplication.
This can be derived by multiplying out $$(a_0 + a_1x + a_2x^2 + \ldots)(b_0 + b_1x + b_2x^2 + \ldots) = (a_0 b_0 + a_0 b_1 a_0 b_2 + \ldots) + (a_1 b_0 + a_1 b_1 a_1 b_2 + \ldots) + (a_2 b_0 + a_2 b_1 a_2 b_2 + \ldots) + \ldots$$. Clearly, the coefficient of $$x^n$$ is $$a_0 b_n + a_1 b_{n - 1} + \ldots + a_n b_0$$, because these are all the ways the values can add up to $$n$$ - $$a_i$$ is the coefficient of $$x^i$$ and $$b_j$$ is the coefficient of $$x^j$$, so it must be that $$i + j = n$$, and $$j = n - i$$.
## Inverses
The inverse of a power series $$A(x)$$ is $$B(x)$$ if and only if $$A(x)B(x) = 1$$. It is defined as $$A^{-1}(x)$$
What is the inverse of $$1 - x$$?
Let $$B(x) = b_0 + b_1x + b_2x^2 + \ldots$$. Assume $$A(x)B(x) = 1$$.
Then $$b_0 + b_1x + b_2x^2 + \ldots - b_0x - b_1x^2 - b_2x^3 - \ldots = 1 = b_0 + (b_1 - b_0)x + (b_2 - b_1)x^2 + (b_3 - b_2)x^3 + \ldots$$.
So $$b_0 = 1, b_1 - b_0 = 0, b_2 - b_1 = 0, b_3 - b_2 = 0, \ldots$$, and $$\forall n \ge 0, b_n = 1$$.
So $$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \ldots = \frac 1 {1 - x}$$, the geometric series.
# 16/5/14
The three series to remember are:
• $$\frac 1 {1 - x} = 1 + x + x^2 + \ldots$$
• $$\frac{1 - x^{k + 1}}{1 - x} = 1 + x + x^2 + \ldots + x^k$$
• $$\frac 1 {(1 - x)^k} = \sum_{n = 0}^\infty {n + k - 1 \choose k - 1}x^n$$
What is the inverse of $$x$$, if it exists?
Suppose $$B(x) = b_0 + b_1x + \ldots$$ exists such that $$xB(x) = 1$$.
Then $$x(b_0 + b_1x + \ldots) = b_0x + b_1x^2 + \ldots = 1$$.
This is a contradiction because the left side has no constant term, while the right side does.
If we generalize this, any power series that does not have a constant term (any power series that has $$x$$ as a factor) cannot be inverted.
Any power series $$A(x)$$ has an inverse if and only if $$[x^0]A(x) \ne 0$$.
Incidentally, we can use the identity $$\frac 1 {1 - f(x)} = 1 + f(x) + f(x)^2 + \ldots$$ if $$f(x)$$ is a power series with no constant term. If there is a constant term, then the expansion of $$f(x)^k$$ has a constant term, so the constant term goes to infinity and the series no longer converges.
## Compositions
Let $$G(x) = \frac 1 {1 - x} = 1 + x + x^2 + \ldots$$. Then $$G(3x^2) = \frac 1 {1 - 3x^2} = 1 + 3x^2 + 9x^4 + 27x^6$$ and $$[x^n]G(x) = \begin{cases} 3^{\frac n 2} &\text{if } n \text{ is even} \\ 0 &\text{if } n \text{ is odd} \end{cases}$$.
Now let $$A(x) = G(3x^2)^9 = \frac 1 {(1 - 3x^2)^9}$$. Then $$A(x) = \sum_{n \ge 0} {n + 9 - 1 \choose 9 - 1} (3x^2)^n = \sum_{n \ge 0} {n + 8 \choose 8}3^nx^{2n}$$.
So $$[x^{200}]A(x) = {108 \choose 8}3^{100}$$.
### Expanding Rational Functions
Let $$B(x) = \sum_{i \ge 0} (x + 2x^2)^i = \frac 1 {1 - (x + 2x^2)} = \frac 1 {1 - x - 2x^2}$$.
How do we find the coefficients of this series? Let $$B(x) = \sum_{i \ge 0} b_i x^i$$.
Then $$\frac 1 {1 - x - 2x^2} = \sum_{i \ge 0} (x + 2x^2)^i$$ and $$1 = (1 - x - 2x^2)\sum_{i \ge 0} (x + 2x^2)^i = \sum_{i \ge 0} (x + 2x^2)^i - x\sum_{i \ge 0} (x + 2x^2)^i - 2x^2\sum_{i \ge 0} (x + 2x^2)^i = b_0 + (b_1 - b_0)x + (b_2 - b_1 - 2b_0)x^2 + \ldots + (b_n - b_{n - 1} - 2b_{n - 2})$$.
Since $$1 = b_0 + (b_1 - b_0)x + (b_2 - b_1 - 2b_0)x^2 + \ldots + (b_n - b_{n - 1} - 2b_{n - 2})$$, $$b_0 = 1$$ and $$b_1 - b_0 = 0$$, and $$b_n - b_{n - 1} - 2b_{n - 2}$$.
So the coefficients of the power series can be defined in terms of a recurrence relation: $$b_0 = 1, b_1 = 1, b_n = b_{n - 1} + 2b_{n - 2}$$.
We can generalize this to any rational function. Find the coefficients of the power series expansion for $$F(x) = \frac{1 + 2x^4}{1 - 3x - x^3}$$:
Let $$F(x) = \frac{1 + 2x^4}{1 - 3x - x^3} = \sum_{i \ge 0} f_i x^i$$. Then $$1 + 2x^4 = \sum_{i \ge 0} f_i x^i - 3x\sum_{i \ge 0} f_i x^i - x^3\sum_{i \ge 0} f_i x^i$$.
Then $$1 + 2x^4 = f_0 + (f_1 - 3f_0)x + (f_2 - 3f_1)x^2 + (f_3 - 3f_2 - f_0)x^3 + (f_4 - 3f_3 - f_1) + \ldots + (f_n - 3f_{n - 1} - f_{n - 3})x^n$$. So the coefficients of the left and right sides match and $$f_0 = 1, f_1 - 3f_0 = 0, f_2 - 3f_1 = 0, f_3 - 3f_2 - f_0 = 0, f_4 - 3f_3 - f_1 = 2$$, and $$f_n - 3f_{n - 1} - f_{n - 3} = 0, n > 4$$.
Solving, we get $$f_0 = 1, f_1 = 3, f_2 = 9, f_3 = 28, f_4 = 87, f_n = 3f_{n - 1} + f_{n - 3}$$.
Consider $$D(x) = \sum_{i \ge 0} (1 + x^2)^i = \frac 1 {1 - (1 + x^2)} = \frac 1 {-x^2}$$. However, this is not a power series - it does not have a constant term!
To see why, we can expand it out: $$D(x) = 1 + (1 + x^2) + (1 + x^2)^2 + \ldots$$. Note that every term has a 1 term in it when expanded. Therefore, the constant term goes toward infinity.
All power series must be invertible. Therefore, all power series must have a constant term. If the constant term is 0 or infinite, then it cannot be inverted and therefore is not a power series.
So when we are composing a series into the geometric series, like $$\sum_{n \ge 0} f(x)$$, then $$f(x)$$ must not have a constant term, or the above will happen and we will get an infinite constant term.
So the series we put into the geometric series cannot be invertible, and therefore cannot be a power series.
# 21/5/14
### Sum Lemma
Let $$A \cap B = \emptyset$$. Let $$w$$ be a weight function on $$A \cup B$$. Then $$\Phi_{A \cup B}(x) = \Phi_A(x) + \Phi_B(x)$$.
In other words, the generating series for the disjunction of disjoint sets is the sum of the generating series for each set.
This is because $$\Phi_S = \sum_{\sigma \in S} x^{w(\sigma)} = \sum_{\sigma \in A} x^{w(\sigma)} + \sum_{\sigma \in B} x^{w(\sigma)} = \Phi_A(x) + \Phi_B(x)$$.
For example, let $$\mb{N}_0$$ be the set of all non-negative integers and $$w(\sigma) = \sigma$$. Then $$\Phi_{\mb{N}_0} = 1 + x + x^2 + \ldots = \frac 1 {1 - x}$$.
Clearly, $$\mb{N}_0 = E \cup O$$, where $$E$$ is the set of non-negative even numbers and $$O$$ is the set of non-negative odd numbers. Clearly, $$E \cap O = \emptyset$$.
Clearly, $$\Phi_E(x) = 1 + x^2 + x^4 + \ldots = \frac 1 {1 - x^2}$$ and $$\Phi_O(x) = x + x^3 + x^5 + \ldots = \frac x {1 - x^2}$$.
Clearly, $$\frac 1 {1 - x} = \frac 1 {1 - x^2} + \frac x {1 - x^2}$$.
For example, let $$S_n$$ be the set of all subsets of $$[n]$$ and $$w(A) = \abs{A}$$. Clearly, $$S_n = A \cup B$$, where $$A$$ is the set of all elements in $$S_n$$ that contain $$n$$, and $$B$$ is the set of all elements that do not.
Clearly, $$\Phi_{S_n}(x) = \Phi_A(x) + \Phi_B(x)$$. Clearly, $$B$$ is the set of all subsets of $$[n - 1]$$, so $$B = S_{n - 1}$$. Clearly, $$A$$ is just the sets in $$S_{n - 1}$$ with $$n$$ added to each one, so $$A = \set{x \cup \set{n} \middle| x \in S_{n - 1}}$$.
So $$\Phi_B(x) = \Phi_{S_{n - 1}}(x)$$, and there is a bijection $$f: A \to S_{n - 1}$$ defined by $$f(T) = T \setminus \set{n}$$. Clearly, $$w(T) = w(f(T)) + 1$$ because we just added an element to each set to make it 1 bigger.
So $$\Phi_A(x) = \sum_{T \in A} x^{w(T)} = \sum_{T \in A} x^{w(f(T)) + 1} = \sum_{T \in S_{n - 1}} x^{w(T) + 1} = x\sum_{T \in S_{n - 1}} x^{w(T)}$$, since $$f(T)$$ maps every element of $$A$$ onto an element of $$S_{n - 1}$$.
So $$\Phi_A(x) = x \Phi_{S_{n - 1}}(x), \Phi_B(x) = \Phi_{S_{n - 1}}(x)$$, so ;wip
;wip: $$(1 + x)\Phi_{S_{n - 1}}$$ then simplify to $$(1 + x)^n$$
### Product Lemma
Let $$A, B$$ be sets and $$\alpha, \beta$$ be weight functions on $$A, B$$, respectively. Then if we define $$w(a, b) = \alpha(a) + \beta(b)$$, $$\Phi_{A \times B}(x) = \Phi_A(x) \Phi_B(x)$$.
In other words, the generating series for the Cartesian product of two sets is the product of the generating series for each of the two sets, if the weight function is the sum of the weight functions for the two sets.
This is because $$\Phi_A(x) \Phi_B(x) = \sum_{a \in A} x^{\alpha(a)} \sum_{b \in B} x^{\beta(b)} = \sum_{a \in A} \sum_{b \in B} x^{\alpha(a)} x^{\beta(b)} = \sum_{a \in A} \sum_{b \in B} x^{\alpha(a) + \beta(b)}$$. ;wip: how?
Since this is just the sum of all the elements in the Cartesian product, $$\sum_{a \in A} \sum_{b \in B} f(a) g(b) = \sum_{x \in A \times B} x^{\alpha(a) + \beta(b)} = \Phi_{A \times B}(x)$$, as required.
So $$\sum_{(a, b) \in A \times B} x^{a + b} = \left(\sum_{a \in A} x^a\right)\left(\sum_{b \in B} x^b\right)$$.
;wip: $$\sum_{a \in A} f(a) \sum_{b \in B} g(b) = \sum_{a \in A} \sum_{b \in B} f(a) g(b)$$.
How many ways can we add $$k$$ non-negative integers to get $$n$$?
Each possible way can be represented as $$(a_1, \ldots, a_k) \in \mb{N}_0^k$$ (Cartesian product of non-negative integers $$k$$ times), where $$a_1 + \ldots + a_k = n$$ and $$a_1, \ldots, a_k \in \mb{N}_0$$.
Let $$w(a_1, \ldots, a_k) = a_1 + \ldots + a_k$$. Then by the product lemma, $$\Phi_{\mb{N}_0^k}(x) = \Phi_{\mb{N}_0}(x) \cdots \Phi_{\mb{N}_0}(x) = \Phi_{\mb{N}_0}(x)^k$$.
Clearly, $$\Phi_{\mb{N}_0}(x) = 1 + x + x^2 + \ldots = \frac 1 {1 - x}$$, so $$\Phi_{\mb{N}_0^k}(x) = \frac 1 {(1 - x)^k} = \sum_{n \ge 0} {n + k - 1 \choose k - 1}x^n$$.
So there are $$[x^n]\Phi_{\mb{N}_0^k}(x) = {n + k - 1 \choose k - 1}$$ possible values of $$(a_1, \ldots, a_k)$$.
# 23/5/14
Clearly, $$\frac 1 {(1 - x)^k} = \frac 1 {1 - x} \cdots \frac 1 {1 - x}$$. Since $$\frac 1 {1 - x} = 1 + x + x^2 + \ldots$$, it is the generating series for $$\mb{N}_0$$ with weight finction $$w(\sigma) = \sigma$$.
Then $$\frac 1 {(1 - x)^k}$$ is the generating series for $$\mb{N}_0^k$$ with weight function $$w(a_1, \ldots, a_k) = a_1 + \ldots + a_k$$.
There are $$n + k - 1 \choose k - 1$$ ways to add $$k$$ non-negative integers to add up to $$n$$.
This is because there are $$k$$ buckets and $$n$$ things to fill them with. We want to figure out all the ways we can distribute the $$n$$ things into the $$k$$ buckets.
Each bucket can hold from 0 to $$n$$ things. However, we notice that the order of the things does not matter, only the quantity.
If we line up the things, the buckets actually divide the line into $$k$$ segments, so it basically inserts $$k - 1$$ dividers into the line of $$n$$ things.
Now we consider the dividers as objects, so there are $$n + k - 1$$ of them in total. Therefore there are $$n + k - 1 \choose k - 1$$ ways of putting the things into buckets.
If we wanted to make this a proper proof, we might define a bijection between the tuples and a binary string where there is a run of $$a_1$$ zeroes, a 1, followed by $$a_2$$ zeros, a 1, all the way to $$a_{k - 1}$$ zeros, a 1, and $$a_k$$ zeros.
How many ways can we have 3 non-negative integers $$a, b, c$$ such that $$a + b + c = n$$ for some non-negative integer $$n$$, such that $$a \ge 5, b \le 3, 2 \mid c$$?
Assume $$a \ge 5, b \le 3, 2 \mid c$$. Then we can model it with a Cartesian product: $$(a, b, c) \in A \times B \times C$$, where $$A = \set{5, 6, 7, \ldots}, B = \set{0, 1, 2, 3}, C = \set{0, 2, 4, 6, \ldots}$$.
Clearly, $$\Phi_A(x) = x^5 + x^6 + \ldots = \frac{x^5}{1 - x}, \Phi_B(x) = 1 + x + x^2 + x^3, \Phi_C(x) = 1 + x^2 + x^4 + \ldots = \frac 1 {1 - x^2}$$, where $$w(\sigma) = \sigma$$ for all three.
Let $$w(a, b, c) = a + b + c$$, so $$\Phi_{A \times B \times C}(x) = \frac{x^5}{1 - x} (1 + x + x^2 + x^3) \frac 1 {1 - x^2} = \frac{x^5(1 + x + x^2 + x^3)}{(1 - x)(1 - x^2)}$$.
We want the number of cases where $$w(a, b, c) = n$$, so the answer is $$[x^n]\Phi_{A \times B \times C}(x) = [x^n]\frac{x^5(1 + x + x^2 + x^3)}{(1 - x)(1 - x^2)}$$.
An integer composition of $$n$$ is a $$k$$-tuple $$(a_1, \ldots, a_k)$$ of positive integers such that $$a_1 + \ldots + a_k = n$$. Each $$a_1, \ldots, a_k$$ is a part.
For example, compositions of 5 include $$(1, 3, 1)$$ and $$(2, 3)$$. Each part must be greater than or equal to 1, and order is significant.
Also, 0 has a single composition, the empty tuple: $$()$$. This has zero parts, and we define the sum to be 0.
How many compositions of $$n$$ have exactly $$k$$ parts?
Let $$(a_1, \ldots, a_k)$$ be a composition of $$n$$.
Then $$a_1, \ldots, a_k \in \mb{N}$$ and $$(a_1, \ldots, a_k) \in \mb{N}^k$$.
Note that instead of $$\mb{N}$$ we could also use $$\set{1, \ldots, n}$$, but we still use $$\mb{N}$$ to simplify our calculations because the generating series can be written in closed form.
Let $$w(\sigma) = \sigma$$ be a weighting function over $$\mb{N}$$.
Then $$\Phi_{\mb{N}}(x) = x + x^2 + x^3 + \ldots = \frac x {1 - x}$$.
Let $$w(a_1, \ldots, a_k) = a_1 + \ldots + a_k$$ be a weighting function over $$\mb{N}^k$$.
Then $$\Phi_{\mb{N}^k}(x) = \frac x {1 - x} \cdots \frac x {1 - x} = \frac {x^k} {(1 - x)^k}$$.
Then the number of values of $$(a_1, \ldots, a_k)$$ such that $$w(a_1, \ldots, a_k) = n$$ is $$[x^n]\frac {x^k} {(1 - x)^k}$$.
This is an acceptable answer, but we can also get a simpler solution.
Clearly, $$[x^n]\frac {x^k} {(1 - x)^k} = [x^{n - k}]\frac 1 {(1 - x)^k} = [x^{n - k}]\sum_{i = 0}^\infty {i + k - 1 \choose k - 1}x^i = {n - k + k - 1 \choose k - 1} = {n - 1 \choose k - 1}$$.
Incidentally, the number of compositions in total is the sum of the compositions that have exactly $$k$$ parts from $$k = 0$$ to $$k = n$$.
# 26/5/14
How many compositions of $$n$$ exist?
Let $$S = \mb{N}^0 \cup \mb{N}^1 \cup \mb{N}^2 \cup \ldots$$. This is the set of all possible compositions.
Let $$w(a_1, \ldots, a_k) = a_1 + \ldots + a_k$$. Then $$\Phi_S(x) = \frac{1 - x}{1 - 2x}$$, since $$\Phi_S(x) = \sum_{i = 0}^\infty \left(\frac{x}{1 - x}\right)^i = \frac{1}{1 - \frac{x}{1 - x}} = \frac{1}{\frac{1 - 2x}{1 - x}}$$.
Then there are $$[x^n]\frac{1 - x}{1 - 2x}$$ compositions of $$n$$.
Also, we can get the explicit value: $$[x^n]\frac{1 - x}{1 - 2x} = [x^n]\frac{1}{1 - 2x} - [x^n]\frac{x}{1 - 2x} = [x^n]\frac{1}{1 - 2x} - [x^{n - 1}]\frac{1}{1 - 2x} = [x^n]\sum_{i = 0}^\infty (2x)^i - [x^{n - 1}]\sum_{i = 0}^\infty (2x)^i = \begin{cases} 2^n - 2^{n - 1} &\text{if } n > 0 \\ 1 &\text{if } n = 0 \end{cases}$$.
In general, to solve this type of problem:
1. Find a set $$S$$ that represents all the possible compositions of a certain type (without considering $$n$$).
2. Find $$\Phi_S(x)$$ where $$w(a_1, \ldots, a_k) = a_1 + \ldots + a_k$$.
3. Then the number of compositions is usually $$[x^n]\Phi_S(x)$$.
4. Try to find an explicit formula for $$[x^n]\Phi_S(x)$$ or try to simplify if possible using series and etc.
How many compositions of $$n$$ exist such that there are $$2k$$ parts, where the first $$k$$ parts are at least 5, and the last $$k$$ are multiples of 3.
Let $$A = \set{5, 6, 7, \ldots}, B = \set{3, 6, 9, \ldots}$$ (we don't include 0 since the values must be at least 1).
Let $$S = A^k \times B^k$$. Clearly, this set represents all of the compositions we care about.
Then $$\Phi_S(x) = \Phi_A(x)^k \Phi_B(x)^k$$, by the product lemma.
Clearly, $$\Phi_A(x) = x^5 + x^6 + x^7 + \ldots = \frac{x^5}{1 - x}$$ and $$\Phi_B(x) = x^3 + x^6 + x^9 + \ldots = \frac{x^3}{1 - x^3}$$.
So $$\Phi_S(x) = \left(\frac{x^5}{1 - x}\right)^k \left(\frac{x^3}{1 - x^3}\right)^k = \frac{x^{8k}}{(1 - x)^k(1 - x^3)^k}$$.
So there are $$[x^n]\Phi_S(x) = [x^n]\left(\frac{x^5}{1 - x}\right)^k \left(\frac{x^3}{1 - x^3}\right)^k$$ of these compositions.
The Fibonnaci recurrence is defined as $$a_0 = 1, a_1 = 1, a_n = a_{n - 1} + a_{n - 2}$$.
How many compositions of $$n$$ have odd parts?
Let $$A = \set{1, 3, 5, \ldots}, S = A^0 \cup A^1 \cup \ldots$$.
Then $$\Phi_S(x) = \Phi_A(x)^0 + \ldots + \Phi_A(x)^k = \sum_{i = 0}^k \left(\frac{x}{1 - x^2}\right)^k = \frac{1}{1 - \frac{x}{1 - x^2}}$$.
Solving as above, there are $$[x^n]\frac{1 - x^2}{1 - x - x^2}$$ compositions of $$n$$ with odd parts.
Let $$S_n$$ be the set of the compositions of $$n$$. Then $$\abs{S_n} = [x^n]\frac{1 - x^2}{1 - x - x^2}$$.
Solving the recurrence relation, we get $$\Phi_S(x) = 1 - x^2 = (1 - x - x^2)$$ ;wip Clearly, $$\abs{S_n} = \abs{S_{n - 1}} + \abs{S_{n - 2}}$$.
We can therefore define a bijection $$f: S_n \to S_{n - 1} \cup S_{n - 2}$$. One possible bijection is to remove the last element if it is 1, and if it is greater than 1 subtract 2. So $$(1, 3, 1) \to (1, 3)$$ and $$(1, 3, 5) \to (1, 3, 3)$$.
We can write this as $$f(a_1, \ldots, a_k) = \begin{cases} (a_1, \ldots, a_{k - 1}) &a_k = 1 \\ (a_1, \ldots, a_{k - 1}, a_k - 2) &a_k > 1 \end{cases}$$.
We can also define the inverse $$f^{-1}: S_{n - 1} \cup S_{n - 2} \to S_n$$. In the above case, this would be $$f^{-1}(b_1, \ldots, b_k) = \begin{cases} (b_1, \ldots, b_k + 2) &b_1 + \ldots + b_k = n - 2 \\ (b_1, \ldots, b_k, 1) &b_1 + \ldots + b_k = n - 1 \end{cases}$$.
So $$S_n = \set{f^{-1}(a_1, \ldots, a_k) \middle| x \in S_{n - 1}} \cup \set{f^{-1}(a_1, \ldots, a_k) \middle| x \in S_{n - 2}}$$.
So using this, $$S_6$$ can be found by finding $$S_5$$ and $$S_4$$ and applying $$f$$ to every element in their union. Likewise, $$S_5$$ and $$S_4$$ can be found by finding $$S_3$$ and $$S_2$$, and so on. This allows us to easily compute these compositions recursively.
# 28/5/14
## Binary Strings
A binary string is a sequence of 0 and 1. The length is the total number of these digits.
There is one string with length 0 denoted $$\epsilon$$. It is called the empty string.
The concatenation of binary strings $$a$$ and $$b$$ is $$ab$$ - we put all the digits of $$b$$ after $$a$$ to form a new binary string.
A substring of $$a$$ is a string $$c$$ such that there exist strings $$b, d$$ such that $$a = bcd$$.
A block is a substring of all 1 that has no 1 on either sides, or all 0 with no 0 on either side. It is the largest possible run of a digit.
We often want to know how many binary strings of length $$n$$ have certain properties.
We first construct a set $$S$$ of all strings with the given properties. Then, we define $$w(s)$$ as the length of the string $$s$$. Then we find $$\Phi_S(x)$$ and the answer is $$[x^n]\Phi(S)(x)$$.
### Regular Expressions
Let $$A, B$$ be sets of strings.
Then $$AB = \set{ab \middle| a \in A, b \in B}$$. This is somewhat similar to the Cartesian product, but is not always the same. Basically, we have the set of all possible combinations of the two sets.
Then $$A^n = AAA \ldots AAA$$, $$n$$ times. So $$A^0 = \set{\epsilon}, A^1 = A, A^2 = AA, \ldots$$. So $$00101 \in \set{0, 1}^5$$.
Then $$A^* = A^0 \cup A^1 \cup A^2 \cup \ldots$$. This is the binary strings that can be made by concatinating combinations of $$A$$. For example, $$\set{0, 1}^*$$ is the set of all possible binary strings.
For example, $$\set{0}\set{00}^*$$ is the set of all binary strings of all 0 of odd length. This is because it can be expanded to $$\set{0, 000, 00000, \ldots}$$.
Another example is $$\set{0}^*(\set{1}\set{0}^*)^*$$. This is the set of all possible binary strings, because given an arbitrary binary string, it can start with any number of 0 to be in $$\set{0}^*$$, and if we break the remaining string into pieces just before every 1, each piece is in $$\set{1}\set{0}^*$$.
# 30/5/14
### Generating Series
We want to find the number of binary strings of a given length $$n$$ that satisfy a given property. To do this we can use a generating series over the set of strings that satisfy this property where the weight is the length of the string.
For example, $$\Phi_{\set{0, 1}}(x) = x + x = 2x$$.
For example, in this case $$\set{0, 1}^n$$ is similar to a Cartensian product $$n$$ times. So in this case, we can use the product lemma: $$\Phi_{\set{0, 1}^n}(x) = 2^nx^n$$
For example, $$\Phi_{\set{0, 1}^*}(x) = \sum_{n \ge 0} \Phi_{\set{0, 1}^n}(x) = \sum_{n \ge 0} 2^nx^n = \frac 1 {1 - 2x}$$.
What is the generating series for $$\set{0}\set{00}^*\set{1, 11, 111}$$?
Clearly, $$\Phi_{\set{0}}(x) = x, \Phi_{\set{00}^*}(x) = \frac 1 {1 - x^2}, \Phi_{\set{1, 11, 111}}(x) = x + x^2 + x^3$$.
So $$\Phi_{\set{0}\set{00}^*\set{1, 11, 111}}(x) = x \frac 1 {1 - x^2} (x + x^2 + x^3) = \frac{x^2 + x^3 + x^4}{1 - x^2}$$ where the weight is the length of each string.
So there are $$[x^n]\frac{x^2 + x^3 + x^4}{1 - x^2}$$ binary strings of length $$n$$ in $$\set{0}\set{00}^*\set{1, 11, 111}$$.
### Ambiguity
We previously assumed that $$AB$$ is functionally the same as $$A \times B$$ - that concatenation of sets of binary strings is similar to the Cartesian producct. However, this is not always true.
In other words, it is not always true that $$\Phi_{AB}(x) = \Phi_A(x) \Phi_B(x)$$ if for all $$ab \in AB$$, $$w(ab)$$ is the length of $$a$$ plus the length of $$b$$.
For example, consider $$A = \set{010, 01}, B = \set{01, 001}$$. So $$\Phi_A(x) = x^2 + x^3, \Phi_B(x) = x^2 + x^3$$.
Clearly, $$A \times B = \set{(010, 01), (010, 001), (01, 01), (01, 001)}$$. So the generating series is $$\Phi_{A \times B}(x) = x^5 + x^6 + x^4 + x^5$$.
However, $$AB = \set{01001, 010001, 0101, 01001}$$. Note that $$01001$$ can be created in two different ways. So $$\set{01001, 010001, 0101, 01001} = \set{01001, 010001, 0101}$$ and $$\Phi_{AB}(x) = x^5 + x^6 + x^4$$. Note that in this case, the product lemma does not hold.
Given sets of binary strings $$A$$ and $$B$$, $$AB$$ is ambiguous if and only if there exist $$(a_1, b_1), (a_2, b_2) \in A \times B, a_1 \ne a_2 \vee b_1 \ne b_2 \implies a_1b_1 = a_2b_2$$. Otherwise, $$AB$$ is unambiguous.
Basically, if there is more than one way for any binary string to be created by concatenating binary strings in sets, then the set is ambiguous.
Also, $$A \cap B$$ is ambiguous if and only if $$A \cap B \ne \emptyset$$. This concept can be extended to any number of sets as well.
### Sum/Product Rule for Strings
If $$A \cap B$$ is unambiguous, then $$\Phi_{A \cup B}(x) = \Phi_A(x) + \Phi_B(x)$$ where $$w(\sigma) = \sigma$$.
This can be proved from the sum lemma.
If $$AB$$ is unambiguous, then $$\Phi_{AB}(x) = \Phi_A(x) \Phi_B(x)$$ where $$w(a, b) = w_a(a) + w_b(b)$$ where $$w_a$$ is the weight function for $$\Phi_A(x)$$ and $$w_b$$ is the weight function for $$\Phi_B(x)$$.
Since $$AB$$ is unambiguous, there is a bijection between $$AB$$ and $$A \times B$$, so it follows by the product lemma.
If $$A^*$$ is unambiguous, then $$\Phi_{A^*}(x) = \frac 1 {1 - \Phi_A(x)}$$ because $$S^* = S^0 \cup S^1 \cup S^2 \cup \ldots$$.
Proof:
Since $$A^*$$ is unambiguous, $$A^n$$ is unambiguous.
Clearly, $$\epsilon \notin A$$, because if it was then $$AA$$ would be ambiguous.
So $$\Phi_A(x)$$ has no constant term and $$\Phi_{A^*}(x)$$ is a geometric series.
So $$\Phi_{A^*}(x) = \frac 1 {1 - \Phi_A(x)}$$.
There are three basic unambiguous decompositions of the set of all binary strings. A decomposition is a way of writing a set using other sets. Often, the set being decomposed is infinite or otherwise hard to represent.
One is $$\set{0, 1}^*$$. Every binary string has a unique sequence of the elements in the set that, when concatenated, results in that string (we break the string between each character).
Another is $$\set{0}^*(\set{1}\set{0}^*)^*$$. Again, there is only one way for each string to be represented using the elements in the set (we break the string just before each 1).
The last is the block decomposition: $$\set{0}^*\left(\set{1}\set{1}^*\set{0}\set{0}^*\right)^*\set{1}^*$$. The reason for the name is because it matches blocks of 1 followed by blocks of 0 an arbitrary number of times.
We can also swap the 1 and 0 to the same effect: $$\set{0}^*(\set{1}\set{0}^*)^*$$ matches the same strings as $$\set{1}^*(\set{0}\set{1}^*)^*$$.
Also, $$\Phi_A(x) = \abs{A}x$$ if $$A = \set{\ldots}$$.
# 2/6/14
Let $$S$$ be the set of all binary strings with no 3 consecutive 0 - strings that do not contain 000. What is the generating series of this set?
We can use the $$\set{0}^*(\set{1}\set{0}^*)^*$$ decomposition to match these strings.
Clearly, we can have three consecutive 0 when and only when we have a $$\set{0^*}$$. Instead, we can replace this with $$\set{\epsilon, 0, 00}$$, the subset of $$\set{0}^*$$ that do not contain 3 consecutive 0.
So $$S = \set{\epsilon, 0, 00}(\set{1}\set{\epsilon, 0, 00})^*$$. This is still unambiguous because we are simply removing strings we do not want to match.
Since it is unambiguous, $$\Phi_S(x) = (1 + x + x^2) \frac 1 {1 - (x + x^2 + x^3)} = \frac{1 + x + x^2}{1 - x - x^2 - x^3}$$.
Let $$T$$ be the set of all binary strings where each block has length at least 2. What is the generating series of this set?
We can use the $$\set{0}^*(\set{1}\set{1}^*\set{0}\set{0}^*)^*\set{1}^*$$ block decomposition to match these strings.
Clearly, $$T = (\set{00}\set{0}^* \cup \set{\epsilon}))(\set{11}\set{1}^*\set{00}\set{0}^*)^*(\set{11}\set{1}^* \cup \set{\epsilon})$$.
So $$\Phi_T(x) = \frac{x^2}{}$$ ;wip: $$\Phi_T(x) = \frac{1 - x + x^2}{1 - x - x^2}$$
Let $$U$$ be the set of all binary strings where an even block of 0 cannot be followed by an odd block of 1. What is the generating series for this set?
Clearly, we can use the $$\set{1}^*(\set{0}\set{0}^*\set{1}\set{1}^*)^*\set{0}^*$$ block decomposition to match these strings (we swapped 0 and 1).
Clearly, $$U = \set{1}^*(\set{00}\set{00}^*\set{11}\set{11}^* \cup \set{0}\set{00}^*\set{1}\set{1}^*)^*\set{0}^*$$. We are using two cases - when the block is of even length, and when it is of odd length, and considering each case separately.
Alternatively, we can write this is as $$\set{1}^*(\set{0}\set{0}^*\set{1}\set{1}^* \setminus \set{00}\set{00}^*\set{1}\set{11}^*)^*\set{0}^*$$, as the set of all blocks minus the set of the blocks we do not want to match.
In that case the sum lemma can be used like a difference lemma: $$A \setminus B$$ implies that $$\Phi_{A \setminus B}(x) = \Phi_A(x) - \Phi_B(x)$$.
;wip: We can also represent this using recursion:
The Fibonnacci sequence is associated with a denominator of $$1 - x - x^2$$
If a regular expression matches a string, that means the string is contained in the set. A regular expression is just a set.
We can find these sets by starting with one of the unambiguous decompositions of all binary strings, and modifying the regular expression until it no longer matches the strings we do not want in the set.
### String Recursion
Clearly, any binary string consists of either an empty string, or a 0 or a 1 followed by another smaller binary string.
Therefore, we can define the set of all binary strings as $$S = \set{\epsilon} \cup \set{0, 1}S$$. There is only one way to break a string in this way, so this representation is unambiguous.
So $$\Phi_S(x) = 1 + 2x\Phi_S(x)$$, so $$\Phi_S(x) = \frac 1 {1 - 2x}$$ and $$[x^n]\frac 1 {1 - 2x} = 2^n$$.
# 4/6/14
Let $$S$$ be the set of all binary strings with no three consecutive 0. What is the generating series?
First, we cut the string starting right after the first 1. Then the part we cut off is also in $$S$$.
Clearly, the first part can only be 1 or 01 or 001, because there cannot be 3 consecutive 0 and this is the first 1.
So we can represent the string as $$S = \set{1, 01, 001}S \cup \set{\epsilon, 0, 00}$$, since $$\epsilon, 0, 00$$ are the only possible strings that cannot be represented using the recursion.
So the generating series is $$\Phi_S(x) = (x + x^2 + x^3)\Phi_S(x) + (1 + x + x^2)$$ and $$\Phi_S(x) - (x + x^2 + x^3)\Phi_S(x) = 1 + x + x^2$$.
So $$(1 - x - x^2 - x^3)\Phi_S(x) = 1 + x + x^2$$ and $$\Phi_S(x) = \frac{1 + x + x^2}{1 - x - x^2 - x^3}$$.
Let $$S$$ be the set of all binary strings without 1010 as a substring. What is the generating series?
Let $$T$$ be the set of all string with exactly one copy of 1010 at the end.
We want to prove that $$\set{\epsilon} \cup S\set{0, 1} = S \cup T$$.
Cleary, $$\set{\epsilon} \cup S\set{0, 1} \subseteq S \cup T$$ is true because $$\set{\epsilon} \in S \cup T$$ and either $$S\set{0, 1}$$ has no 1010 (so it is in $$S$$), or it does and the 1010 is at the end (so it is in $$T$$).
Clearly, $$S \cup T \subseteq \set{\epsilon} \cup S\set{0, 1}$$ because any string $$a$$ in $$S$$ is either $$\epsilon$$ or not, in which case removing the last bit get another string $$a$$ implies $$a \in S\set{0, 1}$$, and because any string $$a$$ in $$T$$ is either epsilon or not, in which case removing the last bit means it has no 1010 and $$a \in S\set{0, 1}$$.
So $$\set{\epsilon} \cup S\set{0, 1} \subseteq S \cup T$$.
We want to prove that $S = T $$T\set{10}$$. Clearly, $$S\set{1010} \subseteq T \cup T\set{10}$$ any $$a \in S$$ either ends with 10 or not. If not, then $$S\set{1010}$$ has only one instance of 1010, so $$S\set{1010} \in T$$. If it does, $$S\set{1010}$$ has two copies of 1010 and ends in 101010, so $$S\set{1010} \in T\set{10}$$. Clearly, $$T \cup T\set{10} \subseteq S\set{1010}$$ because taking the last four bits off any $$a \in T$$ to get $$a'$$ means that $$a' \in S$$. Since $$\set{\epsilon} \cup S\set{0, 1} = S \cup T$$, $$1 + 2x\Phi_S(x) = \Phi_S(x) + \Phi_T(x)$$. So $$S\set{1010} = T \cup T\set{10}$$. Since $$S\set{1010} = T \cup T\set{10}$$, $$x^4\Phi_S(x) = \Phi_T(x) + x^2\Phi_T(x)$$, so $$\Phi_T(x) = \frac{x^4}{1 + x^2}\Phi_S(x)$$. Solving, we get $$\Phi_S(x) = \frac 1 {1 - 2x + \frac{x^4}{1 + x^2}} = \frac{1 + x^2}{1 - 2x + x^2 - 2x^3 + x^4}$$. ## Coefficients of Rational Expressions Let $$A(x) = \sum_{n = 0}^\infty a_nx^n = \frac{6 - x + 5x^2}{1 - 3x^2 - 2x^3}$$. Then $$A(x) = \frac{6 - x + 5x^2}{(1 - 2x)(1 + x)^2}$$ by factoring. Then we find the partial fraction decomposition: $$A(x) = \frac A {1 - 2x} + \frac B {1 + x} + \frac C {(1 - x)^2}$$, where $$\frac{A(1 + x)(1 + x)^2 + B(1 - 2x)(1 + x)^2 + C(1 - 2x)(1 + x)}{1 - 3x^2 - 2x^3}$$. Solving, we get $$A(x) = \frac 3 {1 - 2x} - \frac 1 {1 + x} + \frac 4 {(1 + x^2)}$$. Then $$[x^n]A(x) = [x^n]\frac 3 {1 - 2x} - [x^n]\frac 1 {1 + x} + [x^n]\frac 4 {(1 + x^2)^2} = [x^n]3\sum_{i = 0}^\infty 2^ix^i - [x^n]\sum_{i = 0}^\infty (-1)^ix^i + [x^n]4\sum_{i = 0}^\infty {i + 2 - 1 \choose 2 - 1}x^i = 3 \times 2^n - (-1)^n + {n + 1 \choose 1} = 3 \times 2^n - (-1)^n + n + 1$$. So $$[x^n]A(x) = 3 \cdot 2^n + (-1)^n(4n + 3)$$. In general, if $$A(x) = \frac{f(x)}{g(x)}$$ where the degree of the numerator is lower than that of the denominator, and $$g(x)$$ can be factored into $$g(x) = (1 - r_1x)^{e_1} \cdots (1 - r_nx)^{e_n}$$, then using partial fractions, there exist constants $$C_{i, j}$$ such that $$A(x) = \left(\frac{C_{1, 1}}{1 - r_1x} + \ldots + \frac{C_{1, e_1}}{1 - r_1x}\right)^{e_1} + \ldots + \left(\frac{C_{n, 1}}{1 - r_nx} + \ldots + \frac{C_{n, e_1}}{1 - r_nx}\right)^{e_1}$$. Given this, we can expand the series of each fraction into power series, and after that it is easy to find the coefficients. The basic technique is to get the partial fraction decomposition, expand the power series, and then get the coefficient values from the power series. # 6/6/14 Let $$A(x) = \frac{f(x)}{g(x)}$$. Let $$g(x) = (1 - r_1x)^{e_1} \cdots (1 - r_k)^{e_k}$$. If the constant term of $$g(x)$$ is not 1, then divide both $$f(x)$$ and $$g(x)$$ by that term to make it 1. ;wip: what if there is no constant term? In general, $$[x^n]A(x) = a_n = (C_{1, 1}{n + 1 - 1 \choose 1 - 1}r_1^n + \ldots + C_{1, e_1}{n + e_1 - 1 \choose e_1 - 1}r_1^n) + (C_{k, 1}{n + 1 - 1 \choose 1 - 1}r_k^n + \ldots + C_{k, e_k}{n + e_k - 1 \choose e_k - 1}r_k^n)$$. Clearly, $${n + e_i - 1 \choose e_i - 1} = \frac{(n + e_i - 1)!}{(e_i - 1)!n!} = \frac{(n + e_i - 1) \cdots (n + 1)}{(e_i - 1)!}$$, so it is a polynomial of degree $$e_i - 1$$. So $$A(x) = p_1(n)r_1^n + \ldots + A(x) = p_k(n)r_k^n$$, where $$p_i(n)$$ has a degree less than or equal to $$e_i - 1$$. Recall that $$g(x) = (1 - r_1x)^{e_1} \cdots (1 - r_nx)^{e_n}$$. Then we construct $$g^*(x) = (x - r_1)^{e_1} \cdots (x - r_k)^{e_k}$$. This is the characteristic polynomial of $$g(x)$$. It has roots $$r_1, \ldots, r_k$$ and $$r_i$$ has multiplicity $$e_i$$. The characteristic polynomial is just the one where there is a root for each $$r_i$$. Let $$A(x) = \sum_{n = 0}^\infty a_nx^n = \frac{6 - x + 5x^2}{1 - 3x^2 - 2x^3}$$. In the above example, the characteristic polynomial of $$g(x)$$ is $$g^*(x) = (x - 2)(x + 1)^2 = x^3 - 3x - 2$$. Note that the polynomial is exactly the same, except the power of $$x$$ in each term is swapped - $$x^i$$ in each term is replaced with $$x^{n - i}$$. This is a faster way to find the characteristic polynomial. Then $$[x^n]A(x) = \alpha 2^n + (\beta n + \gamma)(-1)^n$$ for constants $$\alpha, \beta, \gamma$$. Using the $$A(x) = \sum_{i = 0}^\infty a_ix^i = \frac{6 - x + 5x^2}{(1 - 2x)(1 + x)^2}$$ and multiplying both sides by $$g(x)$$ and solving, we get $$a_0 = 6, a_1 = -1, a_2 = 23$$. Clearly, $$a_n = [x^n]A(x)$$, so given these three values, we get $$a_0 = \alpha 2^0 + (\beta 0 + \gamma)(-1)^0, a_1 = \alpha 2^1 + (\beta 1 + \gamma)(-1)^1, a_2 = \alpha 2^2 + (\beta 2 + \gamma)(-1)^2$$. Solving, we get $$\alpha = 3, \beta = 4, \gamma = 3$$, so $$a_n = 3 \times 2^n + (2n + 3)(-1)^n$$. ## Solving Homogeneous Recurrences Given $$A(x) = \sum_{n = 0}^\infty a_nx^n$$ where $$a_0 = 1, a_1 = 4, a_n - 3a_{n - 1} + 2a_{n - 2} = 0$$, find the explicit formula for $$a_n$$: First, we multiply $$x^n$$ with the recurrence, so $$a_nx^n - 3a_{n - 1}x^n + 2a_{n - 2}x^n = 0$$. If we sum all of them to infinity, $$\sum_{n = 2}^\infty (a_nx^n - 3a_{n - 1}x^n + 2a_{n - 2}x^n) = 0 = \sum_{n = 2}^\infty a_nx^n - 3\sum_{n = 1}^\infty a_nx^{n + 1} + 2\sum_{n = 0}^\infty a_nx^{n + 2}$$. So $$\sum_{n = 2}^\infty a_nx^n - 3x\sum_{n = 1}^\infty a_nx^n + 2x^2\sum_{n = 0}^\infty a_nx^n = 0$$. Clearly, $$\sum_{n = 2}^\infty a_nx^n = A(x) - a_1x^1 - a_0x^0$$ (removing the first two terms) and $$\sum_{n = 1}^\infty a_nx^{n + 1} = A(x) - a_0x^0$$. So $$A(x) - a_1x^1 - a_0x^0 - 3x(A(x) - a_0x^0) + 2x^2A(x) = 0$$ and solving, we get $$A(x) = \frac{1 + x}{1 - 3x + 2x^2}$$. Clearly, the characteristic polynomial of the denominator of $$A(x)$$ is $$x^2 - 3x + 2 = (x - 2)(x - 1)$$. So there are roots 1 and 2, both with multiplicity 1, and so $$a_n = \alpha 2^n + \beta 1^n$$ for some constants $$\alpha, \beta$$. So $$a_0 = \alpha 2^0 + \beta 1^0$$ and $$a_1 = \alpha 2^1 + \beta 1^1$$, and solving, we get $$\alpha = 3, \beta = -2$$. So $$a_n = 3 \times 2^n - 2$$. In fact, we can just do it directly from the recurrence relation: $$a_n - 3a_{n - 1} + 2a_{n - 2}$$ can directly be cconverted into the characteristic polynomial $$x^2 - 3x + 2$$, and we can solve for $$a_n$$ directly using $$a_n = C_{1, 1}{n + 1 - 1 \choose 1 - 1}r_1^n + \ldots + C_{1, e_1}{n + e_1 - 1 \choose e_1 - 1}r_1^n + \ldots + C_{k, 1}{n + 1 - 1 \choose 1 - 1}r_k^n + \ldots + C_{k, e_k}{n + e_k - 1 \choose e_k - 1}r_k^n$$, where $$r_1, \ldots, r_k$$ are the roots of the characteristic polynomial and $$e_1, \ldots, e_k$$ are the multiplicities of the roots. We can then solve for $$C_{1, 1}, \ldots, C_{k, e_k}$$ using the initial conditions. This can also be written as $$a_n = \left(C_{1, 1}{n + 1 - 1 \choose 1 - 1} + \ldots + C_{1, e_1}{n + e_1 - 1 \choose e_1 - 1}\right)r_1^n + \ldots + \left(C_{k, 1}{n + 1 - 1 \choose 1 - 1} + \ldots + C_{k, e_k}{n + e_k - 1 \choose e_k - 1}\right)r_k^n$$. More simply, $$a_n = p_{e_1}(n)r_1^n + \ldots + p_{e_k}(n)r_k^n$$, where $$p_x(n)$$ is a polynomial of degree at most $$x$$. # 9/4/14 Basic steps for solving homogeneous recurrences: 1. Find the characteristic polynomial by reversing the powers of the denominator. 2. Find the roots of the characteristic polynomial. 3. Starting with an empty general solution: for each root $$r$$ with multiplicity $$k$$, we add $$p(n)r^n$$ to the general solution, where $$p(n)$$ is a polynomial of degree $$k - 1$$. 4. Use initial conditions to solve for unknown constants. Find a closed form for $$a_n$$ where $$a_0 = 1, a_1 = 2, a_2 = 1, a_n - 3a_{n - 1} + 3a_{n - 2} - a_{n - 3} = 0$$: Clearly, the characteristic polynomial is $$x^3 - 3x^2 + 3x - 1$$, which we get by writing out the coefficients of $$a_n, a_{n - 1}, \ldots$$ and writing powers of $$x$$ beside them. Factoring, we get $$x^3 - 3x^2 + 3x - 1 = (x - 1)^3$$. So there is one root, 1, with multiplicity 3. So $$a_n = p(n)r^k = (An^2 + Bn + C)1^n = An^2 + Bn + C$$. We know that $$p(n) = An^2 + Bn + C$$ because it is an arbitrary polynomial of degree at most 2. So $$a_0 = 1 = A0^2 + B0 + C, a_1 = 2 = A1^2 + B1 + C, a_2 = A2^2 + B2 + C$$. Solving, we get $$A = -1, B = 2, C = 1$$. So $$a_n = -n^2 + 2n + 1$$. Also, $$1 + \sqrt{5} = \phi$$, the golden ratio. Find the closed form for the Fibonacci sequence, $$a_0 = 0, a_1 = 1, a_n - a_{n - 1} - a_{n - 2} = 0$$: Clearly, the characteristic formula is $$x^2 - x - 1 = (x - \frac{1 + \sqrt{5}}{2})(x - \frac{1 - \sqrt{5}}{2})$$. So $$a_n = A\left(\frac{1 + \sqrt{5}}{2}\right)^n + B\left(\frac{1 - \sqrt{5}}{2}\right)^n$$. Solving, we get $$A = \frac 1 {\sqrt{5}}, B = -\frac 1 {\sqrt{5}}$$. So $$a_n = \frac{\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n}{\sqrt{5}}$$. ### Non-homogeneous Recurrences Homogeneous recurrences are those where the constant term is 0. For example, $$a_n + a_{n - 1} = 0$$ - the right side is 0. A non-homogeneous recurrence is therefore one that has a non-zero constant term. For example, $$a_0 = 11, a_1 = 42, a_n - 3a_{n - 1} + 2a_{n - 2} = 10 \times 3^{n - 1}$$. Suppose $$b_n$$ satisfies $$b_n - 3b_{n - 1} + 2b_{n - 2} = 10 \times 3^{n - 1}$$ (it has the same recursive part as $$a_n$$), but does not necessarily satisfy the initial conditions. Then $$(a_n - b_n) - 3(a_{n - 1} - b_{n - 1}) + 2(a_{n - 2} - b_{n - 2}) = 0$$. So $$a_n - b_n$$ is a homogeneous version of the recurrence. So the characteristic polynomial is $$x^2 - 3x + 2 = (x - 2)(x - 1)$$ with roots 1 and 2, so $$a_n - b_n = A1^n + B2^n$$, so $$a_n = b_n + A + B2^n$$. $$b_n$$ is called the specific solution, and $$A + B2^n$$ is the solution to the homogeneous version. This is reminiscent of LDEs in MATH135, where the general solution to an LDE is a specific solution plus an offset. We can find $$b_n$$ by using guess and check. We guess $$b_n = C3^n$$. Then $$b_n - 3b_{n - 1} + 2b_{n - 2} = C3^n - 3C2^{n - 1} + 2C3^{n - 2} = C3^{n - 2}(3^2 - 3 \cdot 3 + 2) = 2C3^{n - 2}$$, and $$b_n - 3b_{n - 1} + 2b_{n - 2} = 10 \cdot 3^{n - 1} = 2C3^{n - 2}$$. Solving, $$C = 15$$ and $$b_n = C3^n = 15 \cdot 3^n$$. So $$a_n = 15 \cdot 3^n + A + B2^n$$. Now we can solve for $$A$$ and $$B$$. Clearly, $$a_0 = 11 = 15 \cdot 3^0 + A + B2^0 = 15 + A + B$$, so $$A + B = -4$$. Clearly, $$a_1 = 42 = 15 \cdot 3^1 + A + B2^1 = 45 + A + 2B$$, so $$A + 2B = -3$$. So $$A = -5, B = 1$$. So $$a_n = 15 \cdot 3^n - 5 + 2^n$$. Why did we guess $$b_n = C3^n$$? We determined this from the generating series. If we multiply $$b_n - 3b_{n - 1} + 2b_{n - 2} = 10 \times 3^{n - 1}$$ by $$x^n$$ and sum over $$n \ge 2$$, then we get $$\sum_{n = 2}^\infty (b_nx^n - 3b_{n - 1}x^n + 2b_{n - 2}x^n) = \sum_{n = 2}^\infty 10 \cdot 3^{n - 1}x^n$$. So $$\sum_{n = 2}^\infty (b_nx^n - 3b_{n - 1}x^n + 2b_{n - 2}x^n) = \sum_{n = 2}^\infty b_nx^n - 3x\sum_{n = 2}^\infty b_{n - 1}x^{n - 1} + 2x^2\sum_{n = 2}^\infty b_{n - 2}x^{n - 2} = \sum_{n = 2}^\infty b_nx^n - 3x\sum_{n = 1}^\infty b_nx^n + 2x^2\sum_{n = 0}^\infty b_nx^n$$ So $$(1 - 3x + 2x^2)\sum_{n = 0}^\infty b_n x^n = \frac{10}{3} (3^2 x^2) \frac 1 {1 - 3x} = \frac{30x^2}{1 - 3x}$$. ;wip: no it doesn't... Using partial fraction decomposition, $$\sum_{n = 0}^\infty b_n x^n = \frac{30x^2}{(1 - 3x)(1 - 3x + 2x^2)} = \frac \ldots {1 - 3x} + \frac \ldots {1 - 2x} + \frac \ldots {1 - x}$$. Note that $$\frac \ldots {1 - 3x}$$ is the result of the homogeneous part, so ;wip: what does that even mean Find a closed form for $$a_0 = 2, a_1 = 3, a_n - a_{n - 1} - 2a_{n - 2} = -4n + 16$$: Let $$b_n = An + B$$ for some $$A, B$$ ;wip # 11/6/14 So to solve non-homogeneous recurrences: 1. Guess a specific solution. 2. Substitute $$a_n = b_n + \ldots$$ to get a a homogeneous recurrence. 3. Use the initial conditions to find the unknowns. ;wip: rewrite these steps so they make sense Solve $$a_0 = 1, a_1 = 1, a_n - a_{n - 1} - 2a_{n - 2} = 3 \times 2^n, n \ge 2$$: Guess $$b_n = \alpha 2^n$$. Then $$b_n - b_{n - 1} - 2b_{n - 2} = \alpha 2^{2 - n}(2^2 - 2^1 - 2 \times 2^0) = 0$$. Clearly, this guess is wrong because $$0 \ne \alpha 2^n$$ for all $$n$$. Guess $$b_n = \alpha n 2^n$$. Then $$b_n - b_{n - 1} - 2b_{n - 2} = \alpha 2^{2 - n}(n2^2 - (n - 1)2^1 - (n - 2)2 \times 2^0) = \alpha n2^{n - 2}(2^2 - 2^1 - 2 \times 2^0 ) + \alpha 2^{n - 2}(2^1 + 4 \times 2^0) = 3 \alpha 2^{n - 1}$$. Then $$3 \alpha 2^{n - 1} = 3 \times 2^n$$, so $$\alpha = 2$$ and $$b_n = 2n 2^2 = n2^{n + 1}$$. With a characteristic polynomial of $$x^2 - x - 2 = (x - 2)(x + 1)$$, $$a_n = n2^{n + 1} + A2^n + B(-1)^n$$. With $$a_0 = 1, a_1 = 1$$, $$A = -1, B = 1$$. So $$a_n = n2^{n + 1} - 2^n + (-1)^n$$. We guessed $$b_n = \alpha n 2^n$$ for a reason. If we multiply $$a_n - a_{n - 1} - 2a_{n - 2} = 3 \times 2^n$$ by $$x^n$$ and sum it over all $$n \ge 2$$, we get $$(1 - x - x^2) A(x) = 3 \sum_{n = 2}^\infty 2^nx^n = \frac{3 \times 2^2x^n}{1 - 2x}$$ for some $$A(x)$$ that we want to find. Using partial fraction decomposition, $$A(x) = \frac{\ldots}{(1 - x - 2x^2)(1 - 2x)} = \frac{\ldots}{(1 - x - 2x^2)(1 - 2x)} = \frac{\ldots}{1 + x} + \frac{\ldots}{1 - 2x} + \frac{\ldots}{(1 - 2x)^2}$$. The $$\frac{\ldots}{(1 - 2x)^2}$$ is the non-homogenous part, so the we have $$n2^n$$ ;wip: what? how? why? In general, if a recurrence equals $$kp^n$$ where $$p$$ is a root of the characteristic polynomial with multiplicity $$k$$, then we should use $$b_n = \alpha n^{k - 1} p^n$$ as a guess. ## Graph Theory ### Definitions A graph is a pair $$G = (V(G), E(G))$$ where $$V(G)$$ is a set of objects called vertices, and $$E(G)$$ is a set of unordered pairs of $$V(G)$$ called edges (these pairs are basically just 2-subsets of $$V(G)$$). For example, $$G = \left(\set{a, b, c, d}, \set{\set{a, b}, \set{b, c}, \set{c, d}, \set{a, d}}\right)$$ is a graph. We often represent graphs graphically, by drawing the vertices on a 2D surface as labelled circles or dots, and edges as lines or curves connecting the vertices they contain. Graphical representations are not unique - the vertices in the drawing can be arranged arbitrarily on the surface and the edges can be drawn as any kind of curvy line. Let $$G$$ be a graph. Let $$u, v \in V(G)$$. Then $$u$$ and $$v$$ are adjacent if there is an edge connecting the two. Formally, $$u, v$$ are adjacent if $$\set{u, v} \in E(G)$$. If $$u$$ is adjacent to $$v$$, then $$u$$ is a neighbour of $$v$$. The set of all the neighbors of $$u$$ is called the neighbourhood. Formally, the neighourhood of $$u$$ is $$\set{v \middle| \set{v} \in \set{e \setminus \set{u} \middle| \set{e \in E(G) \middle| u \in e}}}$$ An edge $$e = \set{u, v}$$ is incident with $$u$$ and $$v$$. Graphs have lots of applications. For example, modelling computer networks to determine robustness and other properties. Other applications include electronic circuits, electricity and water distribution networks, road maps and navigation, social graphs, and more. The map colouring problem asks how many colors are needed to shade in a geographical map such that no two adjacent (sharing a border of non-zero length) geopolitical regions have the same colour. # 13/6/14 The vertices in a graph can be any type of object. We can abbeviate the edge $$\set{u, v}$$ as $$uv$$. In this course we only deal with simple graphs - graphs with at most one edge between two vertices (no loops or double edges joining two vertices). Graphs in this course are also finite. ## Isomorphism Two graphs $$G_1, G_2$$ are isomorphic if and only if there is a bijection $$f: V(G_1) \to V(G_2)$$ such that $$uv \in E(G_1) \iff f(u)f(v) \in E(G_2)$$. If this is the case, then $$f$$ is called an isomorphism. In other words, two graphs are isomorphic if they have the same structure - if they have every vertex correspend to each other, and the edges in one graph connect the corresponding vertices in the other graph. Two graphs can have multiple possible isomorphisms, or none at all, in which case they are not isomorphic. Basically, a graph $$G_2$$ that is isomorphic to a graph $$G_1$$ can be created from $$G_1$$ by replacing its vertices with vertices in $$G_2$$. The isomorphism preserves the edge structure, so edges are mapped to edges, and non-edges are mapped to non-edges. We can detect isomorphism informally by moving the vertices around until the two graphs resemble each other. Checking if two graphs are isomorphic can be done in polynomial time, using recently developed algorithms. To prove two graphs are isomorphic, we find an isomorphism. To prove two graphs are not isomorphic, we find a structure in one graph that is not in the other. The degree of a vertex $$u$$ in $$G$$ is the number of edges incident with it. The degree of a vertex is $$\deg_G(v) = \abs{\set{e \in E(G) \middle| v \in e}}$$. The sum of the degrees of every vertex in a graph $$G$$ has a special meaning and is $$\sum_{u \in V(G)} \deg_G(u)$$. The sum is always even because every edge $$\set{u, v}$$ increases both $$\deg_G(u)$$ and $$\deg_G(v)$$ by 1. This is called the handshaking lemma, and $$\sum_{u \in V(G)} \deg_G(u) = 2 \abs{E(G)}$$. As a result, the number of odd-degree vertices must always be even. Proof: Let $$O$$ be the subset of vertices in $$G$$ with odd degree and $$E$$ the subset with even degree. Then $$\sum_{u \in V(G)} \deg_G(u) = \sum_{v \in O} \deg_G(v) + \sum_{v \in E} \deg_G(v) = 2\abs{E(G)}$$. Since $$2\abs{E(G)}$$ and $$\sum_{v \in E} \deg_G(v)$$ are even (sum of even numbers is even), $$\sum_{v \in O} \deg_G(v)$$ must also be even. Since every $$\deg_G(v)$$ in $$\sum_{v \in O} \deg_G(v)$$ is odd, $$\sum_{v \in O} \deg_G(v)$$ can only be even if $$\abs{O}$$ is even. So there are an even number of odd vertices. # 16/6/14 A cycle is a closed path through three or more vertices (for the simple graphs we are using in this course). A graph $$G$$ is $$k$$-regular if every vertex has a degree of $$k$$ - $$\forall v \in V(G), \deg(v) = k$$. So a 0-regular graph has no edges, and a 1-regular graph has pairs of vertices joined by edges (these are also called matchings). 2-regular graphs have their vertices all connected by a closed path of edges, called a collection of cycles. A 3-regular graph does not lend itself to a simple description, and so on. Clearly, the sum of vertices in a $$k$$-regular graph is $$\sum_{v \in V(G)} \deg(v) = \abs{V(G)}k$$, so by the handshaking lemma $$\abs{E(G)} = \frac{\abs{V(G)}k}{k}$$. A graph $$G$$ is complete if every pair of vertices in $$G$$ is joined by an edge. A complete graph with $$n$$ vertices is represented by $$k_n$$. Therefore the number of edges is the same as the number of pairs of vertices. There are $$n \choose 2$$ pairs, so there are $$n \choose 2$$ edges in $$k_n$$. Also, $$k_n$$ is $$(n - 1)$$-regular, because each vertex must connect with $$n - 1$$ other vertices. A partition of a set $$S$$ is a pair $$(A, B)$$ such that $$A \cup B = S, A \cap B = \emptyset$$. In other words, a way of dividing the set into two groups. A graph $$G$$ is bipartite if there is a partition of $$V(G)$$ into $$(A, B)$$ such that $$G$$ has every edge in $$\set{\set{u, v} \middle| u \in A, v \in B}$$. In other words, if there is a partition such that the edges go between the two partitioned areas, but not within the partitioned areas. The two partitions do not necessarily have the same size and can even be empty. We prove a graph is bipartite by giving the partition, and we prove that a graph is not bipartite by contradiction - assuming that one vertex is in partition $$A$$, any vertex connected by an edge must be in $$B$$, and we try to derive a contradiction where a vertex is in $$A$$ yet also in $$B$$, since $$A \cap B = \emptyset$$. Any graph containing a cycle containing an odd number of vertices cannot be bipartite. Every graph is either bipartite, or has a cycle with an odd number of vertices. A graph $$G$$ is a complete bipartite graph if $$G$$ is bipartite with partitions $$(A, B)$$ and has every edge in $$\set{\set{u, v} \middle| u \in A, v \in B}$$. In other words, a biparitite graph with every possible pair of vertices with one from $$A$$ and one from $$B$$ are connected by edges. A complete bipartite graph with $$\abs{A} = m, \abs{B} = n$$ is represented by $$k_{m, n}$$. Every vertex of $$A$$ must be connected with every element of $$B$$, and every element of $$B$$ must be connected with every element of $$A$$, so the total number of edges in $$k_{m, n}$$ is $$mn$$. The degree of every vertex in $$A$$ is $$n$$ and the degree of every vertex in $$B$$ is $$m$$. ## N-cubes An $$n$$-cube is a graph where the vertices are all binary strings of length $$n$$ and two vertices are connected by an edge if and only if they differ by one bit. The 1-cube looks like a line, the 2-cube looks like a square, and the 3-cube looks like a 2D projection of a 3D cube. # 18/6/14 The $$n$$-cube has $$2^n$$ vertices, and is always an $$n$$-regular graph. It is regular because there are always $$n$$ ways to have a binary string differ by 1 bit, so every binary string is connected to $$n$$ other binary strings. By the handshaking lemma, there are $$\frac 1 2 n2^n = n2^{n - 1}$$ edges in an $$n$$-cube. The $$n$$-cube is also bipartite, because it is impossible for any cycles with odd numbers of vertices to exist in the graph. For example, a 3-cube can be partitioned into $$\left(\set{000, 101, 110, 011}, \set{001, 100, 111, 010}\right)$$. In general, one parition has all binary strings with an even number of 1, and the other has those with an odd number of 1. This is because any binary string differs with another by one bit if and only if it has one less 1 or one more 1 than the other. Therefore, no two binary strings with an even number of 1 can differ by 1 bit, and no two binary strings with an odd number of 1 can differ by 1 bit. Therefore, this partition shows that the $$n$$-cube is bipartite. It is possible to recursively construct the $$n$$-cube. First, the 0-cube is a single vertex, $$\epsilon$$. Given an $$n$$-cube, we can find the graph for the $$(n + 1)$$-cube by making two copies of the $$n$$-cube (with a 0 prepended to every string in the first copy and 1 prepended to every string in the second copy), and then joining the corresponding vertices of the two copies (which now differ by 1 bit since we prepended different values in each copy). The $$n$$-cube is an $$n$$-dimensional analogue of a cube. A 2D drawing of a 4-cube is a 2D projection of a tesseract. Also, any finite graph can be represented in 3D space such that there are no crossing edges. This is difficult to prove, however. ## Walks/Paths Let $$u, w$$ be vertices of a graph $$G$$. Given vertices $$u, w$$, a $$u, w$$ walk is a sequence of alternating vertices and edges $$u, e_1, v_1, e_2, v_2, \ldots, e_{k - 1}, v_{k - 1}, e_k, w$$, where $$v_1, \ldots, v_k$$ are vertices and $$e_i, 1 \le i \le k$$ is the edge $$\set{v_{i - 1}, v_i}$$. The length of this walk is $$k$$ - the number of edges in the walk. This is also written $$u = v_0, e_1, v_1, e_2, v_2, \ldots, e_{k - 1}, v_{k - 1}, e_k, v_k = w$$ or just $$v_0, v_1, \ldots, v_{k - 1}, v_k$$. The walk is closed if $$u = v$$ - if it begins the same place it ends. There can also be duplicated vertices or edges. Since we only deal with simple graphs in this course, it is possible to represent a walk just by its vertices. So a walk of a 3-cube might be written as $$000, 001, 000, 100, 101$$. A walk of length 0 just contains one vertex. For example, $$000$$ is a zero-length walk of the 3-cube. A $$u, w$$ path is a walk with no repeated vertices or edges. As a result, a path cannot be closed. All $$u, w$$ paths are $$u, w$$ walks. All walks can be converted into paths. If we can get somewhere by repeating vertices, then we can also just not repeat those vertices and get to the same place. # 20/6/14 Prove that if there exists a $$u, v$$ walk, then there must also exist a $$u, v$$ path: Let $$u, w_1, \ldots, w_{k - 1}, v$$ be a $$u, v$$ walk of shortest length. If there are no repeated vertices, then it is a $$u, v$$ path. Otherwise, let $$v_i$$ be the first instance of the first repeated vertex, so $$\exists j > i, v_j = v_i$$. Then $$u, v_1, \ldots, w_i, w_{j + 1}, \ldots, v$$ is a walk as well. However, this is shorter than the original walk, which is the shortest possible walk, which is a contradiction. Therefore, the shortest $$u, w$$ walk cannot have any repeated vertices, and it is a path. Therefore, one of the$\$
$$u \sim v$$ is an equivalence relation. In graph theory, $$u \sim v$$ if and only if there exists a $$u, v$$ path.
If $$u \sim v$$ and $$v \sim w$$, then $$u \sim w$$ - $$\sim$$ is transitive. This is because a $$u, v$$ path followed by a $$v, w$$ path is a $$u, w$$ walk, so by the previous theorem there must be a $$u, w$$ path.
$$\sim$$ is reflexive - $$u \sim u$$ is true for any vertex $$u$$. $$\sim$$ is also symmetric $$u \sim v$$ implies that $$v \sim u$$.
We can then partition $$V(G)$$ into equivalence classes (groups of vertices such that for every vertex $$u$$ and every vertex $$v$$, $$u \sim v$$). These partitions are called components and are basically islands of vertices where every vertex is connected to the group by at least one edge.
A cycle is a non-trivial closed walk with no repeated vertices and edges. Nontrivial means it has two or more vertices in the walk.
A cycle must have a length of 3 or more, because it is impossible to construct a cycle with 1 vertex (which is trivial) or 2 (which would repeat vertices). The degree of the vertices in a cycle must be at least 2.
Prove that if every vertex in $$G$$ has a degree of 2 or more, then $$G$$ must have a cycle:
This is because if we go into a vertex, there must be a way out of the vertex, so eventually we must hit a repeated vertex.
Let $$v_0, \ldots, v_k$$ be the longest path in $$G$$. Assume every vertex in $$G$$ has a degree of 2 or more.
Clearly, $$v_k$$ has neighbor $$v_{k - 1}$$, and it has a degree of 2 or more, so it must have at least one other neighbor $$u$$.
Clearly, if $$u$$ is not one of $$v_0, \ldots, v_k$$, then $$v_0, \ldots, v_k, u$$ is a longer path, a contradiction.
So $$u$$ must be one of $$v_0, \ldots, v_k$$. Since $$u$$ cannot be $$v_k$$ or $$v_{k - 1}$$ (because $$v_{k - 1}$$ is already another neighbor), $$u$$ is in $$\set{v_0, \ldots, v_{k - 2}}$$.
So there exists some $$i$$ such that $$v_i = u$$, and $$v_i, \ldots, v_k$$ is a cycle.
A Hamilton cycle is a cycle that contains every vertex in the graph. Finding a Hamilton cycle for an arbitrary graph, or even if it exists, is an NP-complete problem. This is because it can be reduced to the travelling salesman problem.
All complete graphs have Hamiltonian cycles. ;wip: prove it
If $$G$$ is a graph with $$\abs{V(G)} \ge 3$$, and every vertex in $$G$$ has a degree of $$\frac{n}{2}$$ or more, then $$G$$ has a Hamilton cycle. The converse is not necessarily true.
# 23/6/14
;wip: pidgeonhole principle
Given a graph $$G$$, $$G - e$$ is $$G$$ with the edge $$e$$ removed. Formally, $$G - e$$ is a graph such that $$V(G - e) = V(G)$$ and $$E(G - e) = E(G) \setminus e$$.
In the same way, $$G + e$$ is $$G$$ with the edge $$e$$ added. Formally, $$G + e$$ is a graph such that $$V(G + e) = V(G)$$ and $$E(G + e) = E(G) \cup e$$.
Proof:
Let $$G$$ be a graph with $$\abs{V(G)} \ge 3$$ where every vertex in $$G$$ has a degree of $$\frac{n}{2}$$ or more.
Suppose $$G$$ has no Hamilton cycle. Then there must be a set of counterexamples $$S_n$$ with $$n$$ vertices for some $$n$$.
Let $$G \in S_n$$ be the graph with the largest number of edges - the largest possible counterexample with $$n$$ vertices.
Clearly, $$G$$ cannot be complete because $$G$$ would have a Hamilton cycle if it was.
Since it is not complete, there exist $$u, v \in V(G)$$ such that $$\set{u, v} \notin E(G)$$ - there exists a pair of vertices that are not adjacent.
So $$G + \set{u, v}$$ has more edges than the largest counterexample, so it is not a counterexample and must have a Hamilton cycle.
This Hamilton cycle in $$G + \set{u, v}$$ must use $$\set{u, v}$$, so there must be a $$\set{u, v}$$ path $$P = v_1, \ldots, v_n, v_1 = u, v_n = v$$ in $$G$$ that contains all the vertices in $$G$$, made by removing $$\set{u, v}$$ from the Hamilton cycle in $$G + \set{u, v}$$.
Let $$S = \set{i - 1 \middle| \set{v_1, v_i} \in E(G)}, S = \set{i \middle| \set{v_i, v_n} \in E(G)}$$ - the indices of the neighbors of $$v_1$$ and $$v_n$$, respectively.
Since all vertices have degree of at least $$\frac n 2$$, $$\abs{S} \ge \frac n 2, \abs{T} \ge \frac n 2$$.
Since $$S, T \subseteq [n - 1]$$ (since $$v_1$$ cannot be connected to $$v_n$$), by the Pidgeonhole principle, there is some $$i$$ for which $$i \in S \cap T$$.
So $$\set{v_1, v_{i + 1}}, \set{v_i, v_n} \in E(G)$$. So $$v_1, v_{i + 1}, \ldots, v_n, v_i, \ldots, v_1$$ is a Hamiltonian cycle.
This is a contradiction because $$G$$ has no Hamilton cycle. So $$G$$ has a Hamilton cycle.
A graph $$G$$ is connected if there is a $$u, v$$ path for every pair of vertices $$u, v \in V(G)$$. This means that given any two vertices in the graph, it is possible to construct a path between them.
A disconnected graph is the opposite. A disconnected graph has groups of vertices that have no edges between them.
If there are $$n$$ vertices, there are $$\frac{n(n - 1)}{2}$$ possible pairs of vertices. However, we can save a lot of time by checking fewer pairs.
If there exists a vertex $$u \in V(G)$$ such that a $$u, v$$ path exists for all $$v \in V(G)$$, then $$G$$ is connected. The converse is also true but is trivially proven.
Proof:
Assume $$u \in V(G)$$ exists such that a $$u, v$$ path exists for all $$v \in V(G)$$. Let $$y, z \in V(G)$$.
So a $$u, y$$ path and a $$u, z$$ path exists, so if we join them together, we have a $$y, z$$ walk.
Since there is a $$y, z$$ walk, there is a $$y, z$$ path.
Since $$y, z$$ are arbitrary and a $$y, z$$ path exists, $$G$$ is connected.
We prove a graph is connected by showing that there is a path between one particular vertex and every other vertex, possibly by constructing the path. We prove a graph by constructing a set $$X \subset V(G), X \ne \emptyset$$ that induces an empty cut.
For example, the $$n$$-cube is connected because for any $$u, v$$, a $$u, v$$ path exists where we flip the bits of $$u$$ that differ from $$v$$ one by one to match $$v$$. For example, a path between $$0000000$$ and $$1001100$$ would be $$0000000, 1000000, 1001000, 1001100$$.
# 25/6/14
A subgraph of a graph $$G$$ is a graph $$H$$ such that $$V(H) \subseteq V(G)$$ and $$E(H) \subseteq \set{\set{u, v} \in E(G) \middle| u, v \in V(G)}$$. In other words, a graph with a subset of the vertices and the same edges between these vertices as $$G$$.
A component of a graph $$G$$ is a maximally connected non-empty subgraph of $$G$$. In other words, each island of connected vertices in a graph is a component. A component must be the largest possible network of vertices that are all connected.
Clearly, a connected graph must have exactly 1 component (unless it is empty, which has 0 components). Therefore, a disconnected graph has 2 or more components.
If $$G$$ is a graph and $$X \subseteq V(G)$$, then the cut induced by $$X$$ is the set of all edges in $$G$$ with exactly one end in $$X$$ - $$\set{\set{u, v} \in E(G) \middle| u \in X \wedge v \notin X}$$. It is basically the set of all edges that bridge the vertices in $$X$$ with everything else it is connected to.
For a connected graph, if $$X \subset V(G), X \ne \emptyset$$, then the cut induced by $$X$$ is non-empty. This is because $$X$$ doesn't contain all the vertices, and it must be connected to the other vertices in some way.
There always exists a set $$X \subseteq V(G), X \ne \emptyset$$ that induces an empty cut. For example, the components of a graph always induce empty cuts, and so does the entire graph itself.
Therefore, a graph is not connected if and only if there exists $$X \subset V(G), X \ne \emptyset$$ such that $$X$$ induces an empty cut.
Proof:
Assume $$G$$ is a disconnected graph. Then $$G$$ contains two or more components. Let $$H$$ be one of these components.
Since $$H$$ is a component, $$V(H) \ne \emptyset$$ and $$V(H) \subset V(G)$$ (since there is another component in $$G$$).
Suppose the cut induced by $$V(H)$$ is non-empty. Then there is an edge $$\set{u, v}$$ such that $$u \in V(H)$$ but $$v \notin V(H)$$.
So $$\set{v} \cup V(H)$$ is still connected, which is a contradiction since $$H$$ is maximally connected. So $$V(H)$$ induces an empty cut.
Assume there exists $$X \subset V(G), X \ne \emptyset$$ such that $$X$$ induces an empty cut.
Suppose $$G$$ is connected. Then there exists $$u, v \in V(G)$$ such that $$u \in X \wedge v \notin X$$. Since $$G$$ is connected, there exists a $$u, v$$ path.
Let $$u, v_1, \ldots, v_k, v$$ be the shortest path from $$u$$ to $$v$$. Let $$i$$ be the smallest value such that $$v_i \notin X$$.
Then $$v_{i - 1} \in X$$ and $$\set{v_{i - 1}, v_i}$$ is in the cut, which is a contradiction since the cut is empty. So $$G$$ is not connected.
So $$G$$ is not connected if and only if there exists $$X \subset V(G), X \ne \emptyset$$ such that $$X$$ induces an empty cut.
A cut is drawn as a line through the edges in the cut. If it is empty, the line is drawn separating the inducing set and the rest of the graph.
# 27/6/14
## Eulerian Circuits
The problem of Eulerian circuits was the first one ever published, by Euler. In fact, it was the first graph theory problem as well.
The problem is determining whether given a graph, there exists a closed walk that uses every edge exactly once. This was originally posed in the form of whether it is possible to cross bridges across cities and end up in the same place again, crossing each bridge exactly once.
An Eulerian circuit is a closed walk that uses every edge exactly once.
Let $$G$$ be a graph having an Eulerian circuit. Then either $$G$$ is connected, or there are vertices with degree 0.
Let $$G$$ be a connected graph. Then $$G$$ has an Eulerian circuit if and only if every vertex has even degree.
Proof:
Assume $$G$$ has an Eulerian circuit. Then a closed walk adds 2 to the degree of a vertex each time it visits.
Since it uses all edges exactly once and $$G$$ is connected, each vertex is visited, so each vertex has even degree.
This is because every time we enter a vertex, we must be able to leave it. This means that each time we visit a vertex, we increase its degree by 2, so the degree must be even.
Assume every vertex in $$G$$ has even degree.
Suppose $$\abs{E(G)} = n = 0$$. Clearly, $$G$$ is still connected, so it has only one vertex, so it has an Eulerian circuit containing only the one vertex.
Assume for some $$k \ge n$$ that for all $$\abs{E(G)} < k$$, $$G$$ has an Eulerian circuit. Suppose $$\abs{E(G)} = k$$.
Construct a walk $$W = v_1, v_2, \ldots$$, where we keep randomly walking without repeating edges until we can no longer continue.
When we cannot continue, we must have returned to $$v_1$$, because anytime we enter a vertex, we are able to leave it due to the degree being even. So $$W = v_1, \ldots, v_{k - 1}, v_1$$.
If $$W$$ includes all edges, then it is an Eulerian circuit. If it does not, then remove the edges in $$W$$ from $$G$$ to get $$G'$$.
Clearly, every vertex in $$G'$$ also has even degree, since every vertex in the walk has an even degree.
Let $$C_1, \ldots, C_j$$ be the components of $$G'$$. Let $$1 \le i \le j$$. Clearly, every $$C_i$$ is connected and by the inductive hypothesis, has an Eulerian circuit.
Clearly, $$W$$ must have a vertex $$v_{a_i}$$ that is also in $$C_i$$, because otherwise $$G$$ would have been disconnected.
Let $$E_i$$ be the Eulerian circuit in $$C_i$$ that starts at $$v_{a_i}$$.
Then $$W' = v_1, \ldots, v_{a_1 - 1}, E_1, v_{a_1 + 1}, \ldots, v_{a_2 - 1}, E_2, v_{a_2 + 1}, \ldots, v_{k - 1}, v_1$$ is an Eulerian circuit.
To find an Eulerian circuit, we use a greedy algorithm where we keep walking the graph randomly, until we can no longer walk. Because all the degrees of the vertices are even, we can only get stuck at the original node we started at. Now we have a walk that does not necessarily cover all the edges, so we modify our walk to have detours until it covers all the edges.
# 2/6/14
## Bridges
An edge $$e$$ in a graph $$G$$ is a bridge/cut-edge/isthmus in $$G$$ if and only if removing $$e$$ from $$G$$ would increase the number of components in $$G$$. In other words, bridges are the only edge connecting two islands together.
A bridge is basically a weak point in a graph - if we cut it, two parts of the graph become separated.
Removing an edge increases the number of components by at most 1.
If $$e = \set{u, v}$$ is a bridge in connected graph $$G$$, then $$G$$ with $$e$$ removed has 2 components, and $$u$$ and $$v$$ are in different components.
Proof:
Let $$e$$ be a bridge. Clearly, $$G - e$$ has at least 2 components since $$G$$ contains 1.
Let $$H$$ be the component of $$G - e$$ containing $$u$$. Since $$H$$ is a component, the cut induced by $$V(H)$$ is empty.
Suppose $$v \in V(H)$$. Then the cut induced by $$V(H)$$ is also empty in $$G$$, since $$e$$ does not cross the cut.
This is a contradiction, since then $$G$$ has more than one component, so $$v$$ is in a different component from $$u$$.
Suppose there is a third component $$J$$ in $$G - e$$. Then the cut induced by $$V(J)$$ is empty in $$G$$ and $$G - e$$, since removing/adding $$e$$ does not affect the cut.
This is a contradiction since then $$G$$ would not be connected, so there are only two components.
An edge $$e = \set{u, v}$$ is a bridge in $$G$$ if and only if $$e$$ is not in any cycle of $$G$$:
Assume $$e = \set{u, v}$$ is a bridge in $$G$$. Suppose $$e$$ is in a cycle in $$G$$.
Then if we remove $$e$$ from the cycle, $$u$$ and $$v$$ are still in the same component, because we can go the long way around the cycle.
So by the above theorem, since they are in the same component, $$e$$ is not a bridge, a contradiction. So $$e$$ is not in any cycle of $$G$$.
Assume $$e$$ is not in any cycle of $$G$$. Suppose $$e$$ is not a bridge.
Let $$H$$ be the component of $$G$$ containing $$e$$. Then $$H$$ with $$e$$ removed is still a single component, so it is connected.
So there is a $$u, v$$ path $$v_1, \ldots, v_n$$ in $$H$$ with $$e$$ removed that is also in $$H$$, so $$v_1, \ldots, v_n, v_1$$ is a cycle, since $$u$$ and $$v$$ are connected by an edge in $$H$$.
So $$e$$ is in a cycle in $$G$$, a contradiction, so $$e$$ is a bridge.
## Trees
A tree is a connected graph with no cycles. A forest is a graph with no cycles, and can possibly be disconnected. Each component of a forest is a tree.
Every edge in a tree is a bridge, because there are no cycles in a tree. So a tree is minimally connected, since if we removed any edge, it would no longer be connected.
A leaf in a tree is a vertex with degree 1. All trees with 2 or more vertices has 2 or more leaves.
Proof:
Let $$v_1, \ldots, v_n$$ be the longest path in tree $$T$$.
Clearly, $$v_1$$ has no neighbors other than $$v_2$$ since if it had any neighbor outside the path, the path would not be the longest, and if it had a neighbor inside the path, there would be a cycle.
Suppose $$v_1$$ has a neighbor $$v$$ other than $$v_2$$.
If $$v \in \set{v_1, \ldots, v_n}$$, then there is a cycle, a contradiction.
If $$v \notin \set{v_1, \ldots, v_n}$$, then the path $$v, v_1, \ldots, v_n$$ is longer than the longest path, a contradiction.
So $$v$$ cannot exist, and $$v_1$$ has only one neighbor $$v_2$$. A similar argument can be made for $$v_n$$ having only $$v_{n - 1}$$ as a neighbor.
So $$v_1$$ and $$v_n$$ are leaves, and there are at least 2 leaves in $$T$$.
# 4/7/14
A tree $$T$$ with $$n \ge 1$$ vertices has $$n - 1$$ edges:
Let $$n$$ be the number of vertices in a tree $$T$$.
Clearly, for $$n = 1$$ there is 1 vertex and 0 edges, so there are $$n - 1$$ edges.
Assume for some $$k \ge 1$$ that for $$n = k$$, there are $$n - 1$$ edges.
Assume $$n = k + 1$$. Let $$v$$ be a leaf in $$T$$ and $$e$$ be an edge incident with $$v$$.
By the previous theorem, $$v$$ and $$e$$ exist for all $$n \ge 2$$.
Let $$T'$$ be $$T$$ with $$v$$ and $$e$$ removed. Clearly, $$T'$$ cannot have cycles since $$T$$ has no cycles and we only removed things.
Clearly, removing $$e$$ from $$T$$ results in a graph with two components since $$e$$ is a bridge, and one of the components is just $$v$$ by itself.
So $$T'$$ has only one component, and so it is connected. So $$T'$$ is a tree as well.
Since $$T'$$ has $$k$$ vertices, by assumption it has $$k - 1$$ edges.
Since $$k = n - 1$$, $$T'$$ has $$n - 1$$ vertices and $$n - 2$$ edges. Since $$T$$ has one more edge and vertex than $$T'$$, it has $$n$$ vertices and $$n - 1$$ edges.
In the same way, a forest with $$n$$ vertices and $$k$$ components has $$n - k$$ components, because each component is a tree with one less edge than the number of vertices, so overall there are $$k$$ less edges than vertices.
For a tree $$T$$ and two vertices $$u, v \in V(T)$$, there is a unique $$u, v$$ path. The path exists because $$T$$ is connected by definition, and the path is unique because if it wasn't, there would exist cycles in $$T$$, which is not possible.
A tree is always bipartite. This can be proven using induction over the number of vertices - removing a leaf and prove smaller tree satisfies inductive hypothesis and so is bipartite, then prove that the original tree is bipartite using the smaller tree.
A spanning tree $$T$$ of a graph $$G$$ is a subgraph of $$G$$ that is a tree and $$V(T) = V(G)$$. In other words, a tree that uses all the vertices of $$G$$. A spanning tree connects all the vertices together with the smallest possible number of edges.
A graph $$G$$ has a spanning tree if and only if it is connected.
Proof:
Assume $$G$$ has a spanning tree $$T$$. Let $$u, v \in V(G)$$. Since the spanning tree is connected, there is a $$u, v$$ path in $$T$$. Since $$T$$ is a subgraph of $$G$$, this path is also a $$u, v$$ path in $$G$$.
Assume $$G$$ is connected. We will then use induction on the number of cycles $$n$$ to show that when we remove edges of cycles, we get fewer cycles but the graph is still connected.
Clearly, if $$n = 0$$ there are no cycles, so $$G$$ is a tree and $$G$$ is a spanning tree of $$G$$.
Assume for some $$k \ge 0$$ that for $$n \le k$$, $$G$$ has a spanning tree.
Assume $$n = k + 1$$. Let $$v_1, \ldots, v_i, v_n$$ be a cycle in $$G$$. Let $$e$$ be an edge in $$v_1, \ldots, v_i, v_n$$.
Clearly, $$e$$ is not a bridge because it is in a cycle. Clearly, $$G - e$$ is still connected, but has fewer cycles since we removed an edge in a cycle.
Then $$G - e$$ contains a spanning tree $$T$$. Since $$G$$ has the same vertices as $$G - e$$, $$G$$ has the spanning tree $$T$$, by induction.
# 7/7/14
Basically, we can find the spanning tree of a graph by identifying cycles and removing edges from those cycles until there are no more cycles.
$$G$$ is connected with $$n$$ vertices and $$n - 1$$ edges if and only if $$G$$ is a tree. This is useful for determining whether a graph is a tree.
Proof:
Assume $$G$$ is connected with $$n$$ vertices and $$n - 1$$ edges. Then $$G$$ has a spanning tree $$T$$. Clearly, $$T$$ has $$n - 1$$ edges.
Clearly, $$G$$ has $$n - 1$$ edges too, so $$G = T$$ and $$G$$ is a tree.
The other direction was proved earlier via induction on the number of vertices.
So if we add an edge to a tree with $$n$$ vertices and remove another, the resulting graph is still a tree because it has $$n - 1$$ vertices.
If we add an edge to a tree, we always create exactly one cycle (with other graphs it could make more than 1 edges). This is because when we add an edge to a tree, the new cycles must pass through both vertices in the edge. Since there is exactly one path between any two vertices in a tree, there is only one cycle.
If $$G$$ is a graph with $$n$$ vertices, $$T$$ is a spanning tree of $$G$$, and $$e \in E(G) \setminus E(T)$$, then $$T$$ with $$e$$ added has exactly one cycle.
Proof:
Let $$e = \set{u, v}$$. Clearly, $$G$$ must have at least one cycle since there are $$n$$ edges.
Clearly, there is only one path in $$T$$ from $$u$$ to $$v$$. So this path with $$e$$ added forms a cycle.
There cannot be another cycle because all cycles must include $$u, v$$.
### Bipartite Graphs
A graph $$G$$ is bipartite if and only if every component of $$G$$ is bipartite.
A graph $$G$$ is bipartite if and only if every subgraph of $$G$$ is bipartite. So if $$G$$ contains something that is not bipartite, $$G$$ itself is not bipartite.
Also, all trees are bipartite.
A graph $$G$$ is bipartite if and only if $$G$$ does not contain any odd cycles. An odd cycle is a cycle with an odd number of edges.
Proof:
Assume $$G$$ is bipartite with partitions $$(a, b)$$. Suppose $$G$$ contains a cycle of odd length $$v_1, \ldots, v_k, v_1$$.
Without loss of generality, assume $$v_1 \in A$$. Then by induction, $$v_k \in A$$, a contradiction since $$v_1$$ is adjacent to $$v_k$$.
So $$G$$ does not contain any odd cycles.
Assume $$G$$ does not contain any odd cycles. Suppose $$G$$ is not bipartite.
Then there exists a component $$H$$ of $$G$$ that is not bipartite, since if there wasn't, all components would be bipartite and so would $$G$$.
Clearly, $$H$$ has a spanning tree $$T$$ and $$T$$ is bipartite with partitions $$(A, B)$$. Since $$H$$ is not bipartite, there is an edge $$\set{u, v}$$ such that $$u, v \in A$$ or $$u, v \in B$$.
Without loss of generality, assume $$u, v \in A$$. Clearly, there is a $$u, v$$ path $$u, \ldots, v$$ in $$T$$ and this is also in $$H$$. Clearly, the length of the path is even since it alternates between $$A$$ and $$B$$ and has both ends in $$A$$.
So $$u, \ldots, v, u$$ is a cycle of odd length, a contradiction since $$G$$ has no odd cycles.
So $$G$$ is bipartite.
# 9/7/14
Let $$G$$ be a connected graph where each edge $$e$$ has an associated weight $$w: E(G) \to \mb{R}$$. Then the minimimum spanning tree is the spanning tree $$T$$ such that the sum of all the edge weights $$w(T) = \sum_{e \in E(T)} w(e)$$ is minimized. This gives a lower bound on the cost of connecting all the vertices.
We originally found spanning trees by removing edges in cycles until no cycles are left. For minimum spanning trees, we can either continually remove an edge in a cycle with the largest cost, or we can use Prim's algorithm, which grows the tree from a single vertex.
Basically, Prim's algorithm starts at an arbitrary vertex and repeatedly adds edges to the tree such that it grows the smallest possible amount in that step.
We can write it formally as follows:
1. Let $$w \in V(G)$$. Let $$T$$ be a tree containing only $$w$$.
2. While $$V(T) \ne V(G)$$:
1. Let $$C$$ be the cut induced by $$V(T)$$. Let $$\set{u, v} \in C$$ where $$w(u, v)$$ is minimised - the edge with the lowest weight in the cut.
2. Add $$v$$ to $$V(T)$$ and $$\set{u, v}$$ to $$E(T)$$.
Prove that Prim's algorithm always give the minimum spanning tree:
Let $$T_1, \ldots, T_n$$ be the trees produced while executing Prim's algorithm. Let $$e_1, \ldots, e_n$$ be the edges we added in each step, so $$T_i$$ with $$e_i$$ added is $$T_{i + 1}$$. We want to prove that for each $$T_i$$, there exists a minimum spanning tree that contains $$T_i$$.
Suppose that this minimum spanning tree does not exist. Let $$k$$ be the largest integer for which $$T_1, \ldots, T_k$$ are all within minimum spanning trees.
So $$T_k$$ is in a minimum spanning tree $$T'$$, while $$T_{k + 1}$$ is not.
Let $$C$$ be the cut induced by $$V(T_k)$$.
Since $$T_{k + 1}$$ is $$T_k$$ with $$e_k$$ added and $$T_{k + 1}$$ is not in $$T'$$, $$e_k \notin T'$$.
So $$T'$$ with $$e_k$$ added must contain a cycle $$v_1, \ldots, v_j$$ containing $$e_k$$.
Clearly, this cycle contains an edge $$e' \in C$$ since the edge must loop back into $$T_k$$.
Clearly, $$e_k$$ is the edge in $$C$$ with the lowest possible $$w(e)$$, so $$w(e_k) \le w(e')$$.
Let $$V$$ be $$T'$$ with $$e_k$$ removed and $$e'$$ added. Clearly, $$V$$ is also a spanning tree, where $$w(T') = w(V) + w(e_k) - w(e')$$.
Clearly, $$V$$ contains $$T_{k + 1}$$ and is a minimum spanning tree. ;wip: why? This is a contradiction since $$T_{k + 1}$$ is not in any minimum spanning trees. So for each $$T_i$$, there exists a minimum spanning tree that contains $$T_i$$.
Since $$V(T_n) = V(G)$$, and a minimum spanning tree contains $$T_n$$, $$T_n$$ is a minimum spanning tree.
# 11/7/14
A planar embedding of a graph $$G$$ is a drawing of $$G$$ on a plane such that every vertex has a unique position and the edges do not intersect except at vertices.
Any graph that has at least one planar embedding is a planar graph. A face in a planar graph is a connected region (empty area) on the plane - a region which, considered by itself, is connected.
The boundary walk for a face in a connected planar embedding is a closed walk around the boundary of the face. This is basically a walk that encloses the face but nothing else. For example, if there is a face within a face, we have to exclude the inner face from being included in the outer face's boundary walk.
Note that the boundary walk is undefined for disconnected graphs.
The degree of a face is the length of the boundary walk - the number of edges that it has surrounding it. The sum of the degrees of every face is even and twice the number of edges - because every edge is in two faces, every edge increases the degree of two faces it is beside by 1.
A face is incident to an edge if the edge is in the boundary walk. A face is incident to a vertex if the vertex is in the boundary walk. Two faces are adjacent if they share at least one edge in their boundary walk.
The handshaking lemma for faces (HLFF) states that given a graph $$G$$ with a planar embedding with set of faces $$F$$, $$\sum_{f \in F} \deg(f) = 2\abs{E(G)}$$.
An edge has two different faces on both sides if and only if it is part of a cycle. So an edge has the same face on both sides if it is a bridge. This is because if we have a cycle in a planar representation, it must partition the plane into an inside and outside. The edge must therefore have one side on the inside and the other on the outside.
This was formalized in the Jordan Curve Theorem: every simple closed curve on the plane separates it into two parts, the inside and the outside. This might not be true on surfaces other than a plane, like on the surface of a torus.
A connected graph with a planar embedding and only one face must be a tree, since if it had any cycles it would have 2 or more faces.
In a connected graph with a planar embedding and at least 2 faces, every face boundary walk contains a cycle.
Euler's formula relates the number of vertices, edges, and faces to each other. Let $$G$$ be a connected graph with a planar embedding and $$F$$ be the set of faces in $$G$$. Then $$\abs{V(G)} - \abs{E(G)} + \abs{F(G)} = 2$$. In other words, the number of vertices plus the number of faces minus the number of edges always results in 2.
Proof:
Induction over fixed $$\abs{V(G)}$$ and induction on $$\abs{E(G)}$$. Let $$F(H)$$ be the set of faces in a graph $$H$$.
Clearly, the smallest possible connected graph with a planar embedding is a tree, so it is one such that $$\abs{E(G)} = \abs{V(G)} - 1$$. Since there is only one face, $$\abs{V(G)} - \abs{E(G)} + \abs{F(G)} = \abs{V(G)} - (\abs{V(G)} - 1) + 1 = 2$$.
Suppose for some $$k$$ that for all $$\abs{E(G)} < k$$, $$\abs{V(G)} - \abs{E(G)} + \abs{F(G)} = 2$$.
Assume $$\abs{E(G)} = k$$. Let $$e$$ be an edge in a cycle, which must exist because graph has more than $$\abs{V(G)} - 1$$ edges and is not a tree.
Then $$G - e$$ is also connected and planar, since $$e$$ is in a cycle and is therefore not a bridge.
Clearly, $$G - e$$ has $$\abs{F(G)} - 1$$ faces, since $$e$$ was in a cycle and had two different faces on both sides, and removing $$e$$ merges them into one face.
So $$\abs{V(G - e)} - \abs{E(G - e)} + \abs{F(G - e)} = 2$$. Since $$G$$ has one more edge and one more face than $$G - e$$, $$\abs{V(G)} - \abs{E(G)} + \abs{F(G)} = \abs{V(G - e)} - (\abs{E(G - e)} + 1) + (\abs{F(G - e)} + 1) = 2$$.
# 14/7/14
## Platonic Solids
A graph has an embedding in a plane if and only if it has an embedding on a sphere. Any embedding on the sphere can be turned into a polyhedron by cutting faces out of the sphere.
A graph is platonic if and only if it is planar and every vertex has the same degree, and every face has the same degree.
Assume $$G$$ is platonic with $$n$$ vertices, $$m$$ edges, and $$s$$ faces. Then every vertex has degree $$d_V \ge 3$$ and every face has degree $$d_F \ge 3$$.
Clearly, $$2m = nd_V$$ by handshaking lemma, $$sd_F = 2m$$ by HLFF, and $$\frac{2m}{d_V} - m + \frac{2m}{d_F}$$. So $$m(2d_F - d_Vd_F + 2d_V) = 2d_Vd_F$$, and $$2d_Vd_F > 0$$.
So $$0 < 2d_F - d_Vd_F + 2d_V + 4 - 4$$, so $$0 < 4 - (d_V - 2)(d_F - 2)$$. So $$(d_V - 2)(d_F - 2) < 4$$.
Clearly, the only possible solutions are $$(d_V, d_F) = (3, 3), (3, 4), (3, 5), (4, 3), (5, 3)$$, since $$d_V \ge 3, d_F \ge 3$$.
These are the five possible platonic graphs - planar graphs with every vertex of degree $$d_V$$ and every face of degree $$d_F$$. These are polyhedrons when embedded onto a sphere and form very regular solids.
When $$d_V = 3, d_F = 3$$, $$G$$ is a tetrahedron and each face is a triangle.
When $$d_V = 3, d_F = 4$$, $$G$$ is a cube/hexahedron and each face is a square.
When $$d_V = 3, d_F = 5$$, $$G$$ is a dodecahedron and each face is a pentagon.
When $$d_V = 4, d_F = 3$$, $$G$$ is an octahedron and each face is a triangle.
When $$d_V = 5, d_F = 3$$, $$G$$ is an icosahedron and each face is a triangle.
We can quickly draw the planar embedding of a platonic solid by drawing the shape of the face and making sure there is a face shape adjacent to every other face shape until the conditions are satisfied.
These are the only platonic graphs. Things like 100-sided die are possible only because the faces can possibly be different shapes. For example, bevelling every vertex of an icosahedron results in a Buckyball (truncated icosahedron), a shape with hexagons and pentagons.
## Planarity
A non-planar graph cannot be embedded onto a plane without any edges crossing - it has at least one edge crossing no matter how we draw it. We can prove a graph is non-planar by constructing this crossing for an arbitrary graph drawing, or by using the theorems below.
We can prove a graph is not planar by proving that it contains a nonplanar graph, or by showing it has too many edges. For example, $$K_5$$ (complete graph with 5 vertices) is nonplanar.
If $$n \ge 3$$, then all planar graphs with $$n$$ vertices have at most $$3n - 6$$ edges. So if it has more than $$3n - 6$$ edges, it is not planar. Also, if $$n \le 3$$ then $$G$$ is always planar.
Proof:
Let $$G$$ be a planar graph with $$n$$ vertices, $$m$$ edges, and $$s$$ faces, where $$n \ge 3$$.
Clearly, if $$G$$ has no cycles, then $$G$$ is a tree and $$m = n - 1$$, so $$m \le 3n - 6$$.
Otherwise, assume $$G$$ has a cycle. Clearly, every face contains a cycle in its boundary, so it has degree at least 3.
So by HLFF, $$2m \ge 3s$$ and $$2m \ge 3(2 - n + m)$$ by Euler's formula, so $$m \le 3n - 6$$, as required.
So if a graph is planar, it has less than or equal to $$3n - 6$$ edges. The converse is not necessarily true - it is possible to have a non-planar graph with less than or equal to $$3n - 6$$ edges.
# 16/7/14
If $$G$$ is a connected bipartite planar graph with $$n \ge 3$$ vertices, then $$G$$ has at most $$2n - 4$$ edges.
Proof:
Suppose $$G$$ has $$n$$ vertices, $$m$$ edges, and $$s$$ faces.
Clearly, if $$G$$ has no cycles, then $$G$$ is a tree and $$m = n - 1$$, so $$m \le 2n - 4$$. Assume $$G$$ has a cycle.
Clearly, every face boundary in $$G$$ contains a cycle, and every face must have at least 3 vertices, and bipartite graphs cannot have cycles of odd length, so the degree of each face is 4 or more.
By HLFF, $$2m \ge \sum_{f \in F(G)} 4$$, so $$2m \ge 4s$$. By Euler's formula, $$2m \ge 4(2 - n + m)$$ and $$m \le 2n - 4$$.
So a bipartite graph has a tighter upper bound on the number of edges before it becomes non-planar.
### Kuratowski's Theorem
An edge subdivision of a graph $$G$$ is obtained by replacing each edge $$\set{u, v} \in E(G)$$ with a $$u, v$$ path of length at least 1. In other words, we add vertices $$v_1, \ldots, v_k$$ into the graph and replace each $$\set{u, v} \in E(G)$$ with $$\set{u, w}$$ and $$w, v$$ for a unique $$w \in \set{v_1, \ldots, v_k}$$.
Basically, we are subdividing edges in a graph to get a new subdivided graph. Clearly, an edge subdivision is planar if and only if the original graph is planar.
Kuratowski's theorem states that a graph $$G$$ is planar if and only if it does not have any edge subdivision of $$K_5$$ or $$K_{3, 3}$$ as a subgraph.
We can use this to prove a graph is not planar, by finding a subgraph that is an edge subdivision of $$K_5$$ or $$K_{3, 3}$$. To prove a graph is planar, we just draw its planar embedding.
# 18/7/14
A $$k$$-colouring of a graph is an assignment of a colour to each vertex using at most $$k$$ colours, such that any two adjacent vertices have different colours.
If a graph has a $$k$$-colouring, then it is $$k$$-colourable.
In general, what is the minimum $$k$$ for which a graph is $$k$$-colourable - what is the minimum number of colours we need to colour the graph such that any two adjacent vertices have different colours?
Clearly, $$K_n$$ is $$n$$-colourable but not $$(n - 1)$$-colourable, since every vertex is connected to $$n - 1$$ others, so we need one color for each vertex. Clearly, bipartite graphs are 2-colourable.
A graph with vertices of degree at most $$d$$ is $$(d + 1)$$-colourable. This can be proven using induction where the inductive step is removing a vertex of degree $$d$$ and showing that this new graph uses $$d$$ or fewer colours, so the original graph has at most 1 more colour than that.
We are usually more interested in planar graph, however. This is more useful for things like map colouring and other real-world problems.
Ever planar graph has a vertex of degree at most 5. In other words, there exists a vertex in any planar graph that has a degree less than or equal to 5.
Proof:
Let $$G$$ be a planar graph with $$n$$ vertices. Suppose every vertex in $$G$$ has degree at least 6. Then $$\sum_{v \in V(G)} \deg v \ge 6n$$ and $$\abs{E(G)} \ge 3n$$ by the handshaking lemma.
Clearly, $$3n > 3n - 1$$ so $$G$$ is not planar, a contradiction. So there exists a vertex in $$G$$ with degree at most 5.
Every planar graph is 6-colourable.
Proof:
Let $$G$$ be a graph with $$n$$ vertices. Clearly, if $$n = 0$$, $$G$$ has no vertices and is 6-colourable.
Assume for some $$k \ge 0$$ that if $$n = k$$, $$G$$ is 6-colourable.
Assume $$n = k + 1$$. Let $$v$$ be a vertex of degree at most 5 in $$G$$, which must exist by the previous theorem.
Clearly, if we remove $$v$$ from $$G$$ to get $$G'$$, $$G'$$ has $$k$$ vertices and is therefore 6-colourable.
Construct a colouring for $$G$$ where we use the same colours as $$G'$$ for every vertex except $$v$$, where $$v$$ is coloured using a different colour.
Clearly, this different colour exists because $$v$$ has at most 5 neighbours and therefore is only unable to use at most 5 of the 6 colours.
A contraction on an edge $$e = \set{u, v}$$ in $$G$$ is a graph $$G/e$$ where $$V(G') = ((V(G) \setminus u) \setminus v) \cup w$$ where $$w$$ is a new vertex, and $$E(G') = \set{e \in E(G) \middle| u \notin e \wedge v \notin e} \cup \set{\set{w} \cup ((e \setminus u) \setminus v) \middle| e \in \set{e \in E(G) \middle| u \in e \vee v \in e}}$$. Basically, a contraction on an edge is a graph with the vertices of that edge merged into one vertex, and the edge removed.
If $$G$$ is planar, then $$G/e$$ is also planar, because we could feed the edges its vertices connect to through the same space the edge used to occupy.
Every planar graph is 5-colourable.
Proof:
Let $$G$$ be a planar graph with $$n$$ vertices. Clearly, if $$n = 0$$ then $$G$$ has no vertices and is therefore 5-colourable.
Assume for some $$k \ge 0$$ that for any $$n \le k$$, $$G$$ is 5-colourable.
Assume $$n = k + 1$$. Let $$v$$ be a vertex of degree at most 5 in $$G$$, which was proved to exist.
Clearly, if $$\deg v \le 4$$, then there are 4 or fewer neighbours of $$v$$, so $$v$$ cannot use at most 4 colours and there is a colour $$v$$ can use, by the same argument as the proof of 6-colouring.
Assume $$\deg v = 5$$. Clearly, $$v$$ has two neighbours $$x, y$$ that are not adjacent, since otherwise $$G$$ would contain $$K_5$$ as a subgraph and would not be planar.
Let $$H = (G / \set{v, x}) / \set{v, y}$$ - $$G$$ with $$\set{v, x}$$ and $$\set{v, y}$$ contracted.
Clearly, $$H$$ is planar and has $$n - 2$$ vertices, so $$H$$ is 5-colourable.
Construct a colouring for $$G$$ with the same colours as $$H$$ for every vertex except $$v, x, y$$, where $$x, y$$ are coloured using the colour of the contracted vertex (vertex created from the contraction) and $$v$$ uses a different colour.
Clearly, the 5 neighbours of $$v$$ use at most 4 colours, so $$v$$ cannot use at most 4 colours, so there is an available colour for $$v$$.
Every planar graph is 4-colourable. The proof of this was found by computer assisted proof tools and is extremely difficult for humans to understand.
Planar graphs are not all 3-colourable. For example, the planar projection of a triangular prism where there are concentric triangles is not 3-colourable. So 4 is the lowest number of colours that can be used to colour any planar graph.
# 21/7/14
## Graph Duals
The dual of a planar graph $$G$$ is a graph $$G*$$ where $$G*$$ has one vertex $$v_f$$ for each $$f \in F(G)$$, and for all $$e \in E(G)$$ incident to faces $$f_1, f_2$$, $$G*$$ has an edge $$\set{v_{f_1}, v_{f_2}}$$.
To draw the dual of a graph, we draw a vertex in the center of each face, and then connect the vertices of those faces that are adjacent. It is important to remember to consider the outer face - the area outside of the whole graph - since it is also a face.
For example, the dual of the top-down projection of the pentagonal-based pyramid is the same graph, though this is not true in general.
The dual of a planar graph can contain multiple edges connecting the same vertices, and even loops. We get a loop when we have $$f_1 = f_2$$ or duplicate pairs of $$f_1$$ and $$f_2$$. For example, a triangle graph has a center vertex with three edges connecting it to a vertex outside of the triangle. This is because there are three edges incident to both the inside and outside faces, so there are three corresponding edges in the dual.
The dual of a planar graph is also planar. This is because dual can be drawn by putting lines from each face's vertex to the midpoints, and then joining the faces at the edges. Since all these operations do not intersect each other, the dual must be planar.
The dual of the dual of a planar graph, $$G^{**}$$, is simply $$G$$.
It is always true that $$\abs{V(G^*)} = \abs{F(G)}$$, $$\abs{E(G^*)} = \abs{E(G)}$$, and $$\abs{F(G^*)} = \abs{V(G)}$$. In other words, the dual of the graph has a vertex for each face, an edge for an edge, and a face for each vertex.
The dual of a platonic solid is also a platonic solid, not necessarily the same one. For example, the dual of a tetrahedron is a tetrahedron, the dual of a cube in the octahedron, and the dual of a dodecahedron is an icosahedron.
Every planar graph is 4-face colourable - we can colour the faces such that no two adjacent faces have the same colour. This is provable by using the 4-color theorem to show that the dual is 4-colourable in vertices, so the original graph is 4-colourable in faces.
## Matchings
A matching of a graph $$G$$ is a set of edges $$M \subseteq E(G)$$ such that no two edges share a common vertex. In other words, the edges form a subgraph where the maximum degree of each vertex is 1.
It is easy to find matchings in a graph - the empty set is a matching of any graph. We are more interested in the largest possible matching of a given graph - one that includes the largest possible number of edges. This is a maximum matching.
A perfect matching is a matching where the vertices in the matching include every vertex in the graph - a matching that separates every vertex. This is always the largest possible matching in a given graph, but the largest possible matching is not always a perfect matching. For example, a graph with an odd number of vertices.
In a matching, a saturated vertex is one that is incident to an edge in the matching. A vertex is saturated if and only if it is incident to an edge in the matching.
An alternating path $$v_1, e_1, \ldots, e_{k - 1}, v_k$$ with respect to $$M$$ is a path in which each edge $$e_1, \ldots, e_{k - 1}$$ alternates between being in M and not being in $$M$$.
An augmenting path is an alternating path that starts and ends with unsaturated vertices. For an augmenting path, we can switch the alternation of the path, so those edges that were in the matching are not, and those that weren't in the matching are. When we do this, all the vertices in the augmented path are saturated, and the number of edges in the matching increases by 1.
If there exists an augmenting path with respect to a matching $$M$$, then $$M$$ is not a maximum matching. This is because the matching with the augmenting path where the alternation is flipped is a larger matching.
# 23/7/14
If there are no augmenting paths with regard to a matching $$M$$, then $$M$$ is a maximum matching.
Proof:
Assume there are no augmented paths with respect to $$M$$.
Suppose $$M$$ is not a maximum matching. Then there exists a matching $$M'$$ such that $$\abs{M'} > \abs{M}$$. Let $$N$$ be the set of edges that are in $$M$$ or $$M'$$ but not both.
Clearly, each vertex in $$N$$ has degree at most 2, since it is joined to at most 1 vertex in $$M$$ and 1 in $$M'$$.
Clearly, every component in $$N$$ is either a path or a cycle, and all cycles must be of even length since they must alternate between being in $$M$$ and being in $$M'$$.
Since $$\abs{M'} > \abs{M}$$, there must be a path with more edges in $$M'$$ than $$M$$.
Clearly, the two ends of this path are not saturated by $$M$$ since it must start and end with edges in $$M'$$.
So there exists an augmented path with respect to $$M$$, a contradiction.
Therefore, $$M$$ is a maximum matching.
It is often difficult to prove that a matching has no augmented paths. Instead, we use the concept of vertex covers.
A vertex cover $$C \subseteq V(G)$$ of a graph $$G$$ is a set of vertices such that every edge in $$G$$ has at least one endpoint in $$C$$ - $$\forall \set{u, v} \in E(G), u \in C \vee v \in C$$.
We are again interested in minimizing - specifically, minimizing the size of $$C$$. This is related to the problem of finding the maximum matching in that every vertex in the minimum matching can have at most one edge in the maximum matching.
In fact, given a matching $$M$$ in $$G$$, any vertex over $$C$$ satisfies $$\abs{M} \le \abs{C}$$.
Proof:
Clearly, $$\forall \set{u, v} M, u \in C \vee v \in C$$. Clearly, distinct edges in $$M$$ have distinct endpoints, so the vertices that cover $$M$$ are all distinct. So $$\abs{M} \le \abs{C}$$.
As a result, if we find a vertex cover and a matching that have the same size, we have found a maximum matching and a minimum vertex cover. So to prove that a matching is maximal, we can find a vertex cover of the same size, and to prove a vertex cover is minimal, we can find a matching of the same size.
It is not always possible to prove using this method, because the maximum matching can also be smaller than the minimum cover. For example, the endwise projection of a triangular prism has a maximum matching of size 3 and a minimum vertex cover of size 4. However, bipartite graphs always have a matching $$M$$ and a vertex cover $$C$$ such that $$\abs{M} = \abs{C}$$.
# 25/7/14
König's theorem states that in a bipartite graph, the size of the maximum matching is equal to the size of the minimum cover. ;wip: umlaut o
Proof:
We will find an algorithm that finds the maximum cover and minimum cover. We start from a set of unsaturated vertices and find all possible alternating paths.
Let $$G$$ be a bipartite graph with bipartitions $$(A, B)$$. Let $$M$$ be a matching.
Clearly, if there is an augmenting path that starts in $$A$$, the path must end in $$B$$, since all augmented paths must be of odd length.
The basic idea is that we start with the set of unsaturated vertices in one parition, and then follow all their matching neighbors as $$X$$, then follow all their unmatched neighbors as $$Y$$, and so on, until we either find an augmenting path or run out of vertices.
Start with all vertices in $$G$$ being unsaturated. Let $$X_0$$ be the set of all unsaturated vertices in $$A$$.
Let $$X, Y$$ be sets of vertices. Let $$X$$ start off as $$X_0$$.
Find all neighbors of $$X$$ that are not in $$Y$$. If these neighbors are all saturated, put them in $$Y$$ and add their matching neighbors in $$X$$. Repeat for every vertex in $$X$$.
If there is a neighbor that is not saturated, then we have an augmenting path in $$G$$, so we update $$M$$ (make the matched edges unmatched and unmatched edges matched in the augmented path) and start the algorithm again from the beginning.
If there are no new neighbors, then $$M$$ is the maximum matching and $$Y \cup (A \setminus X)$$ is a minimum cover.
;wip: rewrite this so it makes sense Clearly, $$Y$$ is saturated since if it had any unsaturated, there would be an augmenting path.
Clearly, all vertices in $$A \setminus X$$ are saturated, since all unsaturated vertices are in $$X$$.
Clearly, no edges join $$Y$$ to $$A \setminus X$$ since both are saturated, and none of the vertices in $$Y$$ or $$A \setminus X$$ are adjacent to each other.
So $$Y \cup (A \setminus X)$$ is a vertex cover, and is a minimum vertex cover since $$M$$ is a maximum matching.
# 28/7/14
As a result, a bipartite graph $$G$$ with $$m$$ edges and maximum degree $$d$$ has a matching of size at least $$\frac m d$$.
Proof:
Let $$C$$ be a vertex cover in $$G$$. Clearly, each vertex in $$C$$ covers at most $$d$$ edges. Since there are $$m$$ edges, we need at least $$\frac m d$$ vertices to cover every edge.
So the size of the minimum cover is at least $$\frac m d$$. By Konig's theorem, the size of some matching is also at least $$\frac m d$$.
Hall's theorem is often thought of in terms of the marriage metaphor, where one partition is the set of males, another female, and the edges determine eligibility between two people for marriage. The problem is determining whether it is possible to marry off everyone in one of the sets.
As it turns out, this is not possible if and only if there is a set of people in $$A$$ who can only marry a smaller set of people in $$B$$, or vice versa. This is called Hall's condition.
If $$D$$ is a set of vertices, then $$N(D)$$ is the set of vertices adjacent to at least one vertex in $$D$$. We use this to write Hall's condition as $$\forall D \subseteq A, \abs{N(D)} \ge \abs{B}$$.
Hall's theorem states that given a bipartite graph with bipartition $$(A, B)$$, we can guarantee that there is a matching that saturates all the vertices in $$A$$ if and only if $$\forall D \subseteq A, \abs{N(D)} \ge \abs{B}$$.
Proof:
Assume $$G$$ has a matching $$M$$ that saturates $$A$$ and $$D \subseteq A$$, then the matching edges of $$M$$ that have one end in $$A$$ must have the other end in $$\abs{N(D)}$$.
So for every $$D \subseteq A$$, $$\abs{N(D)} \ge \abs{D}$$.
Suppose $$G$$ does not have a matching that saturates $$A$$. Let $$M$$ be a maximum matching of $$G$$. Since $$M$$ does not saturate $$A$$, $$\abs{M} < \abs{A}$$.
By Konig's theorem, there is a maximum vertex cover $$C$$ such that $$\abs{C} = \abs{M}$$. Since $$C$$ is a maximum vertex cover, there is no edge between $$A \setminus C$$ and $$B \setminus C$$.
So $$N(A \setminus C) \subseteq B \cap C$$. So $$\abs{N(A \setminus C)} \le \abs{B \cap C}$$ and $$\abs{N(A \setminus C)} \le \abs{C} - \abs{A \cap C}$$ and $$\abs{N(A \setminus C)} \le \abs{M} - \abs{A \cap C}$$ and $$\abs{N(A \setminus C)} < \abs{A} - \abs{A \cap C}$$ and $$\abs{N(A \setminus C)} < \abs{A \setminus C}$$.
Clearly, this violates Hall's condition, so it is not the case that $$\forall D \subseteq A, \abs{N(D)} \ge \abs{B}$$.
# 20/7/14
As a result, all $$k$$-regular bipartite graphs $$G$$ with $$k \ge 1$$ have a perfect matching.
Proof:
Let $$(A, B)$$ be the bipartitions of $$G$$. Let $$D \subseteq A$$. Clearly, there are $$k\abs{D}$$ edges with exactly one end in $$D$$, and each of these edges have their other end in $$N(D)$$.
Clearly, each vertex in $$N(D)$$ can accept only up to $$k$$ edges from $$D$$, since the vertices have degree $$k$$ and the other end of each incident edge is in either $$D$$ or $$A \setminus D$$.
So $$\abs{N(D)} \ge \abs{D}$$ and Hall's condition holds. So there is a perfect matching of $$G$$.
Clearly, $$\abs{A} = \abs{B}$$ since the graph is $$k$$-regular, so the set of all edges in $$G$$ is a perfect matching.
Also, the edges of a $$k$$-regular bipartite graph can be partitioned into $$k$$ perfect matchings.
In other words, if we find a perfect matching in $$G$$ and remove all edges in $$M$$ from $$G$$ to get a $$k - 1$$-regular graph, this graph has another perfect matching. We can repeat this until there is a 1-regular graph with a perfect matching. This can be proven using induction.
These theorems allow us to solve the stable marriage problem. The stable marriage problem can be stated as follows:
There are $$n$$ men and $$n$$ women. Each person ranks each of the $$n$$ people of the other gender with a number between 1 to $$n$$ such that each person of the other gender has a distinct number.
The higher the ranking, the more this person would prefer to have the ranked person.
Find a set of marriages such that everyone is married and no two people of different genders would both rather have each other than their current partners - a set of stable marriages.
There is always a set of stable marriages such that everyone is married. This can be proven using matchings on bipartite graphs.
The following is an algorithm for finding a set of stable marriage:
;wip
The people proposing heavily affect the resulting set of stable marriages. If the men propose, then the marriages are made starting from the top ranked for the men and slowly move downward. If the women propose, then the marriages are made starting from the top ranked for women and move downward.
The stable roommate problem is similar, except anyone can be matched to anyone. The most significant difference is that there isn't always a stable solution for these problems.
;wip: do assignment 11 and get solutions, do sample finals ;wip: see the final info for topics to study, 30% enumeration and 70% graph theory ;wip: final at Aug 5 9AM in PAC ;wip: do not need to study proof of hamilton cycle, stable marriages
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# Seminar za dinamičke sustave
lokacija:
PMF Matematički odsjek (virtualno)
vrijeme:
26.01.2021 - 16:15 - 18:00
On Tuesday, 26/01/2021, at 16.15,
Goran Radunović, PMF-MO, will give a talk in Dynamical Systems Seminar in Zagreb under the title:
Fractional integro-derivative of fractal zeta functions and Log-Minkowski measurability
Abstract: We introduce Logarithmic gauge Minkowski content which arises
naturally from the theory of complex dimensions. The complex dimensions
of a given set are usually defined as poles of the corresponding fractal
zeta function and they generalize the notion of Minkowski dimension. In
the most simple case the fractal zeta function has a simple pole at D
where D is the Minkowski dimension of the given set, whereas the residue
equals to the Minkowski content (modulo a multiplicative constant).
Here we show that in case of poles of higher order one has to introduce
a generalization of Minkowski content to obtain an analogue connection.
Furthermore, in the most general case, one has to consider more
complicated singularities of the fractal zeta function, including a
combination of poles, zeroes and branch points. The general case can be
explained in the context of an appropriate fractional integro-derivative
of the fractal zeta function.
The talk will be held via Zoom platform:
Topic: Seminar DYNSYS
Meeting ID: 881 5457 8830
Everybody is invited.
Maja Resman
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## Injection or not?
This question has been bothering me for some while now and I still can't figure it out:
Prove that: (1) $\aleph(X)\preceq \mathcal{P}(\mathcal{P}(\mathcal{P}(X)))$ without the choice-axiom.
The way $\aleph(X)$ is defined in my courses is the following:
Let $W(X) = \left\{\left\langle A,R\right\rangle: A\in \mathcal{P}(X),R\in \mathcal{P}(X\times X) \right\}$ (where R is a wellorder on A).
Then $\aleph{(X)}=\left\{\left\langle x,\alpha\right\rangle
:x\in W(X)\right\}$
( $\alpha$ is a ordinal and the ordertype of x)
I define: $\mathcal{I}_A^R = \left\{\left\{x_0\right\},\left\{x_0,x_1\right\},\ cdots\right\}$ to be the set of all initial segments of A with respect to the wellorder R, wich is well-ordered by $\subset$
Now is $f:\aleph(X)\to \mathcal{P}(\mathcal{P}(X))$ defined by $\left\langle x,\alpha \right\rangle = \left\langle \left\langle A,R \right\rangle,\alpha \right\rangle\mapsto \mathcal{I}_A^R
$
not an injection?
In this case the (1) would become trivial, so I expect it's not. But Why?
It seems evidently true that $\mathcal{I}_A^{R_1} = \mathcal{I}_B^{R_2} \Leftrightarrow R_1= R_2$ and $A=B$
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# information management
### How to Optimize Search Engine Strategies for Machine Learning Trend
Machine learning is a way of analyzing data automatically using a type of analytical method! This AI will learn about information patterns and also make decisions with hardly any human advice. Machine learning is a method of analyzing data mechanically using a sort of analytical strategy! It's an artificial intelligence that may learn information patterns and also make decisions with small human advice. Artificial intelligence is the way of earning computers perform tasks which require intelligence.
### #013 A CNN LeNet-5 Master Data Science
Next we will apply another $$pooling$$ layer with filter size $$f 2$$, and stride $$s 2$$ so once again we reduce the size of an image by $$2$$ (as we did with the first $$pooling$$ layer). Finally we have $$5\times5\times16$$ volume and if we multiply these numbers $$5\times5\times16$$ we get $$400$$. We reduced dimensions of an image so now we can apply a $$Fully\enspace connected$$ layer with $$120$$ nodes. Then we apply another $$Fully\enspace connected$$ layer with $$84$$ nodes. The final step is to use these $$84$$ features to get the final output, and at the output can take on $$10$$ possible values because we have to recognize $$10$$ different digits ($$0$$ to$$9$$), so at the end we have a $$softmax$$ layer with a $$10$$-way classification output (although back then $$LeNet-5$$ actually used a different classifier at the output layer, one that's useless today).
### AI will transform digital marketing
As digital marketing becomes increasingly important for businesses, we need to start thinking beyond conventional techniques. As digital marketing becomes increasingly important for businesses, they need to start thinking about stepping up their strategies by going beyond conventional Search Engine Optimization (SEO) techniques. Artificial Intelligence (AI), holds the key to the future of digital marketing. When Google unveiled RankBrain in 2015, users noticed that the results to their queries in the search engine became more useful and relevant. The algorithm, which uses AI to analyse and respond to user queries like a human being would, also returns identical questions asked by other users.
### Google verticals, machine learning and no-click searches expected to have the biggest impacts on SEO - Search Engine Land
Google entering verticals and competing directly against publishers, advancements in machine learning and AI and zero-click searches are the trends most likely to affect SEO in the next three years, according to a SparkToro survey of over 1,500 SEOs. Trends that are here to stay? Respondents were presented with a list of choices and asked, "How much of an impact do you believe the following trends will have on SEO in the next 3 years?" Options were ranked on a zero-to-four scale; zero meaning "no impact" and four meaning "huge impact." The trend that professionals responded were least likely to affect SEO included outcomes from US Congressional and Department of Justice investigations, visual search advances and "content-nudging" products such as Google Discover.
### Big Tech's 'nemesis' in EU gets new term -- and more power
LONDON – The European Union's competition chief is getting a new term -- with expanded powers -- in a move that underlines how the bloc's battle to regulate big tech companies is only just beginning. Margrethe Vestager, who angered the Trump administration by imposing multibillion-dollar penalties on the likes of Google and Apple, was reappointed Tuesday for a second five-year term as the bloc's competition commissioner. The Danish politician's tasks will include strengthening competition enforcement in all sectors, stepping up efforts to detect cases of market abuse by big companies, speeding up investigations and helping strengthen cooperation with her global counterparts. Perhaps ominously for the big tech companies that she has cracked down on, Vestager is also getting extra clout. Ursula von der Leyen, the incoming president of the EU's powerful executive arm, promoted Vestager to a commission executive vice-president overseeing the EU's digital innovation and leadership efforts, including artificial intelligence.
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## Why should I be Bayesian when my model is wrong?
Posted in Books, pictures, Running, Statistics, Travel, University life with tags , , , , , , , , , on May 9, 2017 by xi'an
Guillaume Dehaene posted the above question on X validated last Friday. Here is an except from it:
However, as everybody knows, assuming that my model is correct is fairly arrogant: why should Nature fall neatly inside the box of the models which I have considered? It is much more realistic to assume that the real model of the data p(x) differs from p(x|θ) for all values of θ. This is usually called a “misspecified” model.
My problem is that, in this more realistic misspecified case, I don’t have any good arguments for being Bayesian (i.e: computing the posterior distribution) versus simply computing the Maximum Likelihood Estimator.
Indeed, according to Kleijn, v.d Vaart (2012), in the misspecified case, the posterior distribution converges as nto a Dirac distribution centred at the MLE but does not have the correct variance (unless two values just happen to be same) in order to ensure that credible intervals of the posterior match confidence intervals for θ.
Which is a very interesting question…that may not have an answer (but that does not make it less interesting!)
A few thoughts about that meme that all models are wrong: (resonating from last week discussion):
1. While the hypothetical model is indeed almost invariably and irremediably wrong, it still makes sense to act in an efficient or coherent manner with respect to this model if this is the best one can do. The resulting inference produces an evaluation of the formal model that is the “closest” to the actual data generating model (if any);
2. There exist Bayesian approaches that can do without the model, a most recent example being the papers by Bissiri et al. (with my comments) and by Watson and Holmes (which I discussed with Judith Rousseau);
3. In a connected way, there exists a whole branch of Bayesian statistics dealing with M-open inference;
4. And yet another direction I like a lot is the SafeBayes approach of Peter Grünwald, who takes into account model misspecification to replace the likelihood with a down-graded version expressed as a power of the original likelihood.
5. The very recent Read Paper by Gelman and Hennig addresses this issue, albeit in a circumvoluted manner (and I added some comments on my blog).
6. In a sense, Bayesians should be the least concerned among statisticians and modellers about this aspect since the sampling model is to be taken as one of several prior assumptions and the outcome is conditional or relative to all those prior assumptions.
## an elegant result on exponential spacings
Posted in Statistics with tags , , , , , , , , , , , , , on April 19, 2017 by xi'an
A question on X validated I spotted in the train back from Lyon got me desperately seeking a reference in Devroye’s Generation Bible despite the abyssal wireless and a group of screeching urchins a few seats away from me… The question is about why
$\sum_{i=1}^{n}(Y_i - Y_{(1)}) \sim \text{Gamma}(n-1, 1)$
when the Y’s are standard exponentials. Since this reminded me immediately of exponential spacings, thanks to our Devroye fan-club reading group in Warwick, I tried to download Devroye’s Chapter V and managed after a few aborts (and a significant increase in decibels from the family corner). The result by Sukhatme (1937) is in plain sight as Theorem 2.3 and is quite elegant as it relies on the fact that
$\sum_{i=1}^n y_i=\sum_{j=1}^n (n-j+1)(y_{(j)}-y_{(j-1)})=\sum_{j=2}^n (y_{(j)}-y_{(1)})$
hence sums up as a mere linear change of variables! (Pandurang Vasudeo Sukhatme (1911–1997) was an Indian statistician who worked on human nutrition and got the Guy Medal of the RSS in 1963.)
## how large is 9!!!!!!!!!?
Posted in Statistics with tags , , , , , , , , , on March 17, 2017 by xi'an
This may sound like an absurd question [and in some sense it is!], but this came out of a recent mathematical riddle on The Riddler, asking for the largest number one could write with ten symbols. The difficulty with this riddle is the definition of a symbol, as the collection of available symbols is a very relative concept. For instance, if one takes the symbols available on a basic pocket calculator, besides the 10 digits and the decimal point, there should be the four basic operations plus square root and square,which means that presumably 999999999² is the largest one can on a cell phone, there are already many more operations, for instance my phone includes the factorial operator and hence 9!!!!!!!!! is a good guess. While moving to a computer the problem becomes somewhat meaningless, both because there are very few software that handle infinite precision computing and hence very large numbers are not achievable without additional coding, and because it very much depends on the environment and on which numbers and symbols are already coded in the local language. As illustrated by this X validated answer, this link from The Riddler, and the xkcd entry below. (The solution provided by The Riddler itself is not particularly relevant as it relies on a particular collection of symbols, which mean Rado’s number BB(9999!) is also a solution within the right referential.)
## what does more efficient Monte Carlo mean?
Posted in Books, Kids, R, Statistics with tags , , , , , , on March 17, 2017 by xi'an
“I was just thinking that there might be a magic trick to simulate directly from this distribution without having to go for less efficient methods.”
In a simple question on X validated a few days ago [about simulating from x²φ(x)] popped up the remark that the person asking the question wanted a direct simulation method for higher efficiency. Compared with an accept-reject solution. Which shows a misunderstanding of what “efficiency” means on Monte Carlo situations. If it means anything, I would think it is reflected in the average time taken to return one simulation and possibly in the worst case. But there is no reason to call an inverse cdf method more efficient than an accept reject or a transform approach since it all depends on the time it takes to make the inversion compared with the other solutions… Since inverting the closed-form cdf in this example is much more expensive than generating a Gamma(½,½), and taking plus or minus its root, this is certainly the case here. Maybe a ziggurat method could be devised, especially since x²φ(x)<φ(x) when |x|≤1, but I am not sure it is worth the effort!
## multiplying a Gaussian matrix and a Gaussian vector
Posted in Books with tags , , , , , on March 2, 2017 by xi'an
This arXived note by Pierre-Alexandre Mattei was actually inspired by one of my blog entries, itself written from a resolution of a question on X validated. The original result about the Laplace distribution actually dates at least to 1932 and a paper by Wishart and Bartlett!I am not sure the construct has clear statistical implications, but it is nonetheless a good calculus exercise.
The note produces an extension to the multivariate case. Where the Laplace distribution is harder to define, in that multiple constructions are possible. The current paper opts for a definition based on the characteristic function. Which leads to a rather unsavoury density with Bessel functions. It however satisfies the constructive definition of being a multivariate Normal multiplied by a χ variate plus a constant vector multiplied by the same squared χ variate. It can also be derived as the distribution of
Wy+||y||²μ
when W is a (p,q) matrix with iid Gaussian columns and y is a Gaussian vector with independent components. And μ is a vector of the proper dimension. When μ=0 the marginals remain Laplace.
## an accurate variance approximation
Posted in Books, Kids, pictures, R, Statistics with tags , , , , , , on February 7, 2017 by xi'an
In answering a simple question on X validated about producing Monte Carlo estimates of the variance of estimators of exp(-θ) in a Poisson model, I wanted to illustrate the accuracy of these estimates against the theoretical values. While one case was easy, since the estimator was a Binomial B(n,exp(-θ)) variate [in yellow on the graph], the other one being the exponential of the negative of the Poisson sample average did not enjoy a closed-form variance and I instead used a first order (δ-method) approximation for this variance which ended up working surprisingly well [in brown] given that the experiment is based on an n=20 sample size.
Thanks to the comments of George Henry, I stand corrected: the variance of the exponential version is easily manageable with two lines of summation! As
$\text{var}(\exp\{-\bar{X}_n\})=\exp\left\{-n\theta[1-\exp\{-2/n\}]\right\}$
$-\exp\left\{-2n\theta[1-\exp\{-1/n\}]\right\}$
which allows for a comparison with its second order Taylor approximation:
## a well-hidden E step
Posted in Books, Kids, pictures, R, Statistics with tags , , , , , , , , , on February 3, 2017 by xi'an
A recent question on X validated ended up being quite interesting! The model under consideration is made of parallel Markov chains on a finite state space, all with the same Markov transition matrix, M, which turns into a hidden Markov model when the only summary available is the number of chains in a given state at a given time. When writing down the EM algorithm, the E step involves the expected number of moves from a given state to a given state at a given time. The conditional distribution of those numbers of chains is a product of multinomials across times and starting states, with no Markov structure since the number of chains starting from a given state is known at each instant. Except that those multinomials are constrained by the number of “arrivals” in each state at the next instant and that this makes the computation of the expectation intractable, as far as I can see.
A solution by Monte Carlo EM means running the moves for each instant under the above constraints, which is thus a sort of multinomial distribution with fixed margins, enjoying a closed-form expression but for the normalising constant. The direct simulation soon gets too costly as the number of states increases and I thus considered a basic Metropolis move, using one margin (row or column) or the other as proposal, with the correction taken on another margin. This is very basic but apparently enough for the purpose of the exercise. If I find time in the coming days, I will try to look at the ABC resolution of this problem, a logical move when starting from non-sufficient statistics!
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# Bisaya English Translation of pakyawanay
In bisaya english dictionary, "pakyawanay" is "wholesale \in bulk \large-scale \in large lots \by the carload". See more translations below.
wholesale \in bulk \large-scale \in large lots \by the carload
### Related Cebuano translations
If you know something about this term, share it here. Many people are also searching for information about pakyawanay. Help them find it, make them happy.
Let us help each other. Please be polite in asking for help and sharing thoughts. Sort the comments to "Reverse Chronological" to see the latest posts.
### AKNOWLEDGEMENT
This is to acknowledge that the cebuano/english dictionary and translations used by this website is largely sourced from:
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# Tag Info
0
You miss the main factor: Moon/Sun are the triggers, but tidal wave is at first the mechanical response of the water body constrained by the topography of the floor and coasts. That's why some regions in the world can have huge tides, and some others have 1 or 3 high tide a day instead of 2. Sort of vibrating modes. I don't see how evaporation and rain ...
1
You need to collect some data to answer this question. Heat pumps can provide more heat to a building than the electricity they use. Whether yours does is something that needs to be measured. Real world efficiencies are often different from lab efficiencies. The radiator coils can get dusty and be installed in places with poor air flow, for example. You ...
1
$CO_2$: Without greenhouse gases, Earth would be about 30 degrees colder. Everyone sensible agrees that $CO_2$ contributes to this, but also that it only directly causes part of the warming. Water vapor is the other important greenhouse gas. There are feedback relationships between temperature, $CO_2$ levels, and water vapor levels, e.g. if the world gets ...
2
The reflectivity of the atmosphere, and of the surface itself, is strongly wavelength-sensitive. So while some percentage of any given wavelength is reflected -- and some percentage is absorbed rather than transmitted, the variation over wavelength is what leads to the somewhat misleading statement you refer to. Here's an example of atmospheric absorption, ...
-2
It does. Consider this: you can see the Earth from space. Therefore, not just infrared light gets reflected but also light on the visible spectrum. Here's a graph (by NASA) of various planet's radio emissions. The ways that Earth can release radio waves is a bit limited. Because of that, it is safe to assume that at least some come from the Sun.
Top 50 recent answers are included
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# size2 -- number of binary digits to the left of the point
## Synopsis
• Usage:
size2 x
• Inputs:
• x,
• Outputs:
• the smallest number n such that 2^n is strictly greater than the absolute value of x
## Description
This function is not implemented for rational numbers.
i1 : size2 4 o1 = 3 i2 : size2 3 o2 = 2 i3 : size2 4. o3 = 3 i4 : size2 3.99999999 o4 = 2 i5 : size2 0 o5 = -1073741824 i6 : size2 0. o6 = -1073741824 i7 : size2 (1/0.) o7 = 1073741823 i8 : size2 (1/0.-1/0.) o8 = 1073741823
• "size2(CC)"
• "size2(RR)"
• "size2(RRi)"
• "size2(ZZ)"
## For the programmer
The object size2 is .
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# Rolling two dice 10 times, what is the probability of getting all possible “doubles”?
Rolling two dice 10 times, what is the probability of getting all possibble "doubles": (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) among our rolls?
For instance, (1,1),(2,1),(3,4),(2,2),(3,3),(6,6),(5,5),(4,4),(5,1),(2,1) is a good roll, because all possible doubles occur.
My idea is to use the inclusion–exclusion principle:
$$X=36^{10} - 6*35^{10} + {{6}\choose{2}}*34^{10} - {{6}\choose{3}}*33^{10} +... =\sum_{i=0}^{6}{(-1)^{i}*{{6}\choose{i}}*(36-i)^{10}}$$ Therefore the solution is: $$X/36^{10}$$
Could you check if my solution is proper or show me the simpler one?
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What is the probability of rolling doubles on the first roll? What is the probability of rolling doubles on the first two rolls? – soakley Mar 2 at 13:35
But we have to count the probability of the set {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} being the subset of the outcome of rolling 10 dice, not the probability of getting in every roll a double. – Esserath Mar 2 at 13:50
Your first term in $X$ of $36^{10}$ shouldn't be there, because it represents all possible rolls. I don't think there is an easy way. – Ross Millikan Mar 2 at 15:42
We have a drawing with replacement of $10$ balls out of an urn containing $6$ red balls, numbered from $1$ to $6$, and 30 white balls.
The number $R$ of red balls drawn is binomially distributed; i.e., the probability $p_r$ that we draw exactly $r$ red balls is given by $$p_r={10\choose r}\left({1\over6}\right)^r\left({5\over 6}\right)^{10-r}\qquad(0\leq r\leq 10)\ .$$
Assuming that we draw exactly $r$ red balls, the numbers on these $r$ balls define a map $f:\ [r]\to[6]$, and all $6^r$ such maps are equally likely. Out of these maps $6!\left\{\matrix{r \cr 6\cr}\right\}$ are surjective, where $\left\{\matrix{r \cr 6\cr}\right\}$ (called a Stirling number of the second kind) denotes the number of ways to partition $[r]$ into $6$ nonempty blocks. Therefore the probability $q_r$ that on the $r$ red balls drawn all $6$ numbers are present, is given by $$q_r={6!\left\{\matrix{r \cr 6\cr}\right\}\over 6^r}\qquad(6\leq r\leq10)\ .$$ It follows that the overall probability $P$ of the event described in the question is given by $$P=\sum_{r=6}^{10} p_r\>q_r={416216045\over8463329722368}\doteq0.0000491788\ .$$
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We can bound it as follows: One calculation is to compute the chance we get exactly six doubles, one of each type. We have to choose six of the rolls to be doubles, and order of the doubles, and the four non-double rolls, so $\frac {{10 \choose 6}6!30^4}{36^{10}}\approx 00000335$ An upper bound would be to change allow the other rolls to be anything. We will now double count the ones with one extra double. That gives $\frac {{10 \choose 6}6!36^4}{36^{10}}\approx 0.0000695$ The first subtraction is seven doubles: $6$ ways to choolse the pair of doubles, $10 \choose 2$ to locate them, ${8 \choose 5}5!$ to locate and order the other doubles, $36^3$ for the other rolls for a probability of $\frac {6{10 \choose 2}{8 \choose 5}5!36^3}{36^{10}}\approx 0.0000023$ The real result is close to the $0.0000695$ number.
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# Schrödinger's cat
Discussion in 'Physics & Math' started by fess, Jan 30, 2020.
1. ### DaveC426913Valued Senior Member
Messages:
15,208
A superposed object need not be recorded to have its wave function collapse.
Modern understanding is that a superposed object merely needs to interact (say, with other atoms) to collapse the wave function.
3. ### arfa branecall me arfValued Senior Member
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7,150
When the human observer opens the box, the cat is dead or alive; the glass vial of poison gas is broken or whole; the mechanism that breaks the vial is activated or not. In each case, classical information about the states of classical objects, is copied from one classical object to another.
But we know that quantum information can't be copied. The information in the potential decay of a radioactive sample cannot be copied. Only one of the classical measurement devices can store this hidden information (even when the box itself hides information about states). After this, the information can be copied precisely because it's classical.
All opening the box does, then, is copy some information about classical states (dead or alive, broken or not). If the cat is alive, then the vial isn't broken and the mechanism that breaks the vial isn't activated; the non-decayed sample is also in a classical 'state' because of what you, the cat, the vial and the mechanism also know, type of thing.
5. ### phytiRegistered Senior Member
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472
How would you know that, without an observation?
My point is, quantum physics has not introduced anything new, just different. It still reduces to ignorance, not understanding how/why the universe works at some level of detail. If someone knocks on your door, you have to look to see who it is (or check your security system).
7. ### QuarkHeadRemedial Math StudentValued Senior Member
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1,715
objects are not superposed, states are.
8. ### DaveC426913Valued Senior Member
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15,208
It's not a simple explanation. There are books written on the subject.
But you can't really argue against it without studying up on it.
9. ### CptBorkValued Senior Member
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6,355
According to the Quantum Decoherence interpretation of wave mechanics, all observers will see both a living and dead cat when the box is opened. However, the portion of the observer wavefunctions seeing a living cat will be almost completely disjoint from the portion of their wavefunctions observing a dead cat, effectively resulting in the establishment of two separate universes, each displaying a classical result.
10. ### Q-reeusValued Senior Member
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4,083
Decoherence per se is a mechanism/framework applied to various interpretations, not itself an interpretation of QM.
Decoherent Histories aka Consistent Histories is an interpretation - with various sub-interpretations within that fold.
https://plato.stanford.edu/entries/qm-decoherence/
I'll repeat from earlier posts - the notion an exceedingly complex and continuously internally evolving (on multiple hierarchical levels) entity 'cat' could ever be thought of as existing in a well defined QM 'state' is crazy thinking. The cat is intrinsically classical. Dead or alive.
11. ### CptBorkValued Senior Member
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6,355
Actually to put that to a real test you'd need to isolate the cat from all external interactions, including gravity, but I'll agree that even most physicists subscribing to the Copenhagen interpretation would assert that the cat's atoms would constantly "observe" each other and it doesn't end up getting spooked by its own corpse.
12. ### James RJust this guy, you know?Staff Member
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35,554
Sure, but I doubt that you truly believe that the Eiffel tower is both standing and collapsed at the same time. You may not know its current state, but I'm fairly sure you believe it has one - more specifically, that you'll always find the Eiffel tower in a regular, "classical" state, and never in a quantum superposition.
I don't think you can just brush the problem under the carpet by pretending that quantum superpositions are just like classical states that we have limited information about. They demonstrably don't behave like classical states.
13. ### phytiRegistered Senior Member
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472
That's my point, exactly. Our knowledge of a system may be expressed as probabilities, quantum physics, weather, gaming, etc. What we think does not influence the material world, no more than a fast moving spacecraft alters the rate of a distant clock, or the dimensions of the universe.
In the period about 2000, the missing solar neutrinos problem was solved. It offered an opportunity for multiple superposition of states, but was explained with transformations.
The classical view of, you have to look/measure to know still works because that's how the mind works. Every new scientific discovery only shows the degree of ignorance for human thinking. Another option is to go down the 'yellow brick road'!
Last edited: Feb 13, 2020
14. ### CptBorkValued Senior Member
Messages:
6,355
It's not just a matter of humans not having all the information about a quantum system, it's a matter of nature itself not deciding on the outcome until it's measured or a wavefunction collapses on its own, as demonstrated by multiple different experiments (in the quantum decoherence multiverse interpretation, nature allows every outcome in its own universe, but that still makes it impossible to know which of those universes we now inhabit without measuring).
15. ### phytiRegistered Senior Member
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472
I don't believe in 'mother nature' or 'father time', nor the personification of inanimate matter. The nova occurs, not because someone was star gazing, but because the object reached an unstable state, in terms of all the physical laws that are in place, regulating its behavior.
In eating at my favorite restaurant, ordering fish instead of beef, doesn't put me in a different universe. It gives me a modified/alternate history resulting from my choice.
There are many possibilities but only one outcome.
Science has enough of a challenge to discover how a single universe works, why complicate it. I compare that to an incompetent weatherman, who predicts all types of weather, expecting one of them to occur, thus verification of his forecast.
Examining scientific theories and experimentation, over its history, the trend is echoed in this statement, "it's more complicated than we originally thought'. Theories are made and theories are revised. All the mental concepts, including the wave function, are attempts to explain the world in terms the human mind can comprehend, thus they are naive and overly simplistic.
The earth is a perfect sphere. No, it's an ellipsoid, since it bulges at the equator. No, it's pear shaped, since the northern hemisphere has a different distribution of mass then the southern hemisphere.
The (invisible) planetary orbits were modified with increasing complexity.
Forming knowledge is a continuous process of refinement.
The truth is, no one knows how the universe works, and just make it up as they go.
16. ### arfa branecall me arfValued Senior Member
Messages:
7,150
In modern information-theoretic interpretations of QM (viz quantum computers at IBM and maybe elsewhere), a multiverse isn't necessary.
In order for there to be some computation, quantum states have to be 'manipulated'; you can do this to a state and not need to know (even in principle) how it changes. You have the information that a state has been manipulated. The existence of an alternative universe where the state was not manipulated isn't relevant to the problem.
A state manipulation is an interaction, manipulating a state so it becomes classical information is therefore a computation (aka a measurement). A computation by definition transforms some input into a different output (usually). A 'computation' that doesn't transform information is a communication of same. Classically we have the freedom to encode information how we like, transmit it, then decode it; this leaves the original information (or a copy of it) unchanged--encoding and decoding is an identity operation on information.
What about quantum encoding? There isn't the same freedom; we have to accept the encoding nature hands us.
And I note this helpful hint from IBM for wannabe quantum programmers: In a QC with n qubits, the qubits' locations are all classical. Each of the n qubits 'knows' there are n-1 other qubits in a classical position (type of thing). When you turn a QC on, n qubits are in an unknown entangled state (of positions). That's why it helps to be able to tell one from the other . . . (quantum joke)
Last edited: Feb 16, 2020
17. ### arfa branecall me arfValued Senior Member
Messages:
7,150
I found some of the explanations in stuff I've read, about what's different about quantum manipulations and classical ones some help, but I can't say it's the definitive approach.
Engineers ignore a lot of theoretical physics when they build quantum (or even classical, these days) computers. They tend to whittle it down to what's useful (no shit?); in this case useful means a useful set of 'quantum' manipulations, generally between pairs of qubits, via (erm, completely esoteric) quantum gates. A quantum gate is perhaps another way to represent a pair of particles interacting at a vertex (a Feynman diagram), without most of the associated gauge fields and all that; just a dash of vector calculus does the trick, in a vector space with two complex dimensions. If you like you can kick in some Dirac notation, say.
To my opening: a big difference is how classical information can be hidden, compared to hidden quantum states. A coin hidden in a box can still be heads or tails, you manipulate the state by shaking the box, now you know something else because of how coins interact with the inside of boxes when you shake them.
But when you switch on a QC (let's say you work for IBM), although you can maybe see n separate cavities so you know about classical positions of qubits, you can't tell one qubit state from another; that's the hidden information in one sense. You perform a sequence of manipulations on a pair of qubits; each manipulation manipulates n qubits, because quantum particles interact when they get near each other. In IBM's 5-qubit online qx machine, the manipulations are essentially signal injections into 5 cavities via a set of waveguides (check out the specs).
Apart from what you or I might learn about what quantum states are by trying to learn how to put a quantum algorithm together, notably just a 5-qubit machine takes up an entire basement, costs an awful lot of time and money and probably needs constant monitoring by a small army of . . . engineers.
18. ### SchmelzerValued Senior Member
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5,003
It says not even this. It says that the state describing the cat in the box is a superpositional state. In the minimal interpretation, it does not say anything about the cat itself.
Some interpretations say there is nothing beyond the state, thus, the state is the same as the cat. Other, realist interpretations say that the actual state of the cat exists beyond the wave function, and is either a living cat or a dead cat, but the wave function does not contain this information.
19. ### HalcRegistered Senior Member
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227
While I agree, I'm not sure of the difference between our two wordings (cat in superposition, vs. state of system (with a cat) being in a superpositional state). Yes, I'm aware that the state of the rest of the box is entangled with the cat and thus not just the cat is in some isolated state on its own.
I think there are some realist interpretations that deny a counterfactual state for the cat. Some interpretations for instance posit the wave function itself being real. Just saying not every realist interpretation say it. A counterfactual interpretation like Bohm's would definitely say the cat has a definite state.
My personal choice is RQM, which is not a realist interpretation. Under any local interpretation, it isn't difficult to put the cat into superposition.
20. ### SchmelzerValued Senior Member
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5,003
The wording .... If we have an interpretation where we have a wave function $\psi(q,t)$ which describes, essentially, the preparation procedure, and an actual configuration $q(t)$, it is not really clear what "state" means. It could indeed mean the quantum state $\psi(q,t)$ as well as the actual state $q(t)$.
There is, of course, MWI. It indeed claims to be realist. But IMHO it does not even count as an interpretation at all. It is simply ill-defined pseudo-science. Whenever I have asked for some precise definitions for the objects they talk about the answer was silence.
I see no justification for giving up realism as long as realistic interpretations exist. Moreover, one would have to give up even causality (the classical notion which contains Reichenbach's principle of common cause, not the weak replacement of signal causality) and logic as well (the logic of plausible reasoning a la objective Bayesian interpretation of probability, see Schmelzer, I. (2017). EPR-Bell realism as a part of logic. arxiv:1712.04334
21. ### iceauraValued Senior Member
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30,845
Causality has already been given up - or rather: relegated to "folk science", the category of useful but unsound heuristic notions and mental shorthands and rules of thumb - in the course of more sophisticated analysis, without any motivation in QED. There is little loss in that.
Logic, meanwhile, fits quantum theory (as it does all of mathematical perception or abstraction) just fine.
22. ### SchmelzerValued Senior Member
Messages:
5,003
That you personally may have given up causality is quite obvious, but that is your personal problem. If somebody else has given up causality, feel free to link some information about this.
23. ### HalcRegistered Senior Member
Messages:
227
Your wording implies that there is an actual state $q(t)$ which means it either has been measured, or that you're asserting counterfactuals. I'm kind of assuming the latter based on the context.
RSF (Everett, '57) has but the one postulate and realism isn't mentioned in it. MWI (DeWitt, 73) seems to add the realism (existence of the universal wavefunction, and ontologically distinct worlds), and I think it has some fatal faults, but what do I know?
OK, your biases are definitely showing. It isn't pseudo-science if it doesn't predict different results than what is observed.
You don't seem open at all, so silence is perhaps them dismissing you. Not sure what you mean by 'the objects they talk about'. What, the cat? 'The cat' is part of the state of system as you close the box. Most common language terms like 'cat' hold meaning for our everyday subjective experience that have no direct correlation to the concepts in Hilbert space.
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# Room temperature electrically pumped topological insulator lasers
## Abstract
Topological insulator lasers (TILs) are a recently introduced family of lasing arrays in which phase locking is achieved through synthetic gauge fields. These single frequency light source arrays operate in the spatially extended edge modes of topologically non-trivial optical lattices. Because of the inherent robustness of topological modes against perturbations and defects, such topological insulator lasers tend to demonstrate higher slope efficiencies as compared to their topologically trivial counterparts. So far, magnetic and non-magnetic optically pumped topological laser arrays as well as electrically pumped TILs that are operating at cryogenic temperatures have been demonstrated. Here we present the first room temperature and electrically pumped topological insulator laser. This laser array, using a structure that mimics the quantum spin Hall effect for photons, generates light at telecom wavelengths and exhibits single frequency emission. Our work is expected to lead to further developments in laser science and technology, while opening up new possibilities in topological photonics.
## Introduction
In condensed matter physics, topological insulators (TIs) represent a new class of materials with insulating bulks and conducting symmetry-protected surface states1,2,3,4,5,6,7,8,9. The remarkable robustness of these surface currents against local defects and perturbation has led to the application of these materials in quantum transport, spintronic devices, and new transistors, to name a few4,5,6,7,8,9. While the concept of topological protection was originally conceived for fermionic systems, recent advances have led to designing lattices capable of mimicking analogous responses in the electromagnetic domain10,11,12,13,14,15,16,17,18,19. This is primarily accomplished by realizing artificial gauge fields that emulate the effect of external magnetic fields on light particles through geometric design or modulation. In this respect, photonics has made it possible to study some of the intriguing aspects of topological physics by providing access to synthetic dimensions20,21,22,23 and higher-order topological insulators24,25,26. Also, photonics allows the studying of topological behaviors arising due to non-Hermiticity (primarily gain) and nonlinearity that have no immediate counterparts in condensed matter27,28,29,30,31,32,33. In return, topological physics has inspired novel unique light transport schemes that may have important ramifications in integrated photonics34,35, quantum optics36,37, and laser science38,39,40,41,42,43,44,45,46,47,48,49,50.
Topological insulator lasers (TILs) refer to two-dimensional arrays of emitters that oscillate at the topological edge modes. In photonics, the topological lattice is an artificial structure, designed to generate a synthetic gauge field for photons through direction-dependent phase accumulations. The edge modes propagating at the boundary of these lattices feature transport, thus uniformely engaging all the active elements at the perimeter of the array. The manifestation of this type of mode indicates that the active peripheral entities are phase-locked and the system has reached an optimum usage of the pump power40. Furthermore, the robustness of the edge modes to certain classes of perturbations introduces additional benefits by precluding the formation of spurious defect states. These undesired modes tend to compromise the performance of the device by siphoning energy from the main mode. Finally, the additional spectral modes of the cavity can be suppressed by controlling the topological bandgap through adjusting the array parameters. This latter feature is particularly important in semiconductor gain systems with intrinsically broad lineshapes in which single-frequency lasing poses a challenge.
So far topological lasing has been demonstrated in optically pumped two-dimensional topological insulator lasers based on active lattices featuring quantum Hall, quantum spin Hall, or valley Hall effect in the absence and presence of external magnetic fields38,39,40,41,42,43,44,45. In addition, lasing in the edge or zero modes has been achieved in one-dimensional SSH arrays, through PT-symmetric46,47,48,49 or other selective pumping schemes50. Despite the rapid developments in the optically pumped topological insulator lasers, devices based on electrical injection are still in their infancy. What makes this transition a challenge is designing a topological structure that allows for both efficient carrier injection and large mode confinement. To this end, recently, an electrically pumped THz quantum cascade topological insulator laser was demonstrated44 in which the edge mode is formed at the boundary of two valley Hall photonic crystals possessing valley Chern numbers of ±1/2. However, in order to reach lasing, this array was cooled to a cryogenic temperature of 9 K.
Here, we report the first room temperature, electrically pumped topological insulator laser that operates at telecom wavelengths. Our non-magnetic topological insulator array imitates the quantum spin Hall effect for photons through a periodic array of resonators coupled through an aperiodic set of auxiliary link structures. When the resonators at the perimeter of the array are electrically pumped, a uniform and coherent edge mode is excited. We further demonstrate that the topological property of the system gives rise to single-frequency lasing, despite the presence of multiple modes when the array’s peripheral elements are locally excited. Our work shows for the first time the feasibility of realizing electrically pumped room temperature topological laser arrays that upon further developments can drive technological needs in related areas.
## Results
### Design and fabrication
A schematic of our electrically pumped topological insulator laser is depicted in Fig. 1a. The array is composed of a 10 × 10 network of microring resonators coupled via a set of anti-resonant link objects, all fabricated on a III–V semiconductor wafer. Figure 1b displays the SEM image of the topological array in an intermediate fabrication stage. The structure implements the quantum spin Hall Hamiltonian for photons by modulating the relative position of the links in each row (see Fig. 1c)12,15,40. In order to promote edge mode lasing, the gain is only provided to the peripheral elements through the incorporation of metal electrodes, while the rest of the array is left unpumped to prevent spurious lasing in the bulk. The wafer as shown in Fig. 1d is composed of lattice-matched epitaxially grown layers of InxGa1−xAsyP1−y on an undoped InP substrate (see Supplementary Note 1 for details about wafer structure). The design of the wafer structure and the geometry of individual ring resonators are co-optimized in order to efficiently funnel electron and hole carriers into the active region under electrical pumping while at the same time allowing the structure to support a confined optical mode that adequately overlaps with the multiple quantum wells (MQWs). Figure 1e shows the structure of a single resonator at the boundary of the lattice.
To demonstrate lasing by electrical pumping at room temperature, the radii of the ring resonators (R) and their widths (w) are chosen to be 15 μm and 1.4 μm, respectively. The cross-section of the resonators is designed to promote lasing in the fundamental transverse electric mode (TE0). (see Supplementary Note 2 for details about electromagnetic simulations). The links are of the same widths (w), but their radii of curvature (RL) are 2.25 μm and the lengths of their straight sections (LL) are 3 μm. Here the links are intentionally designed with a small radius of curvature in order to prevent their direct contribution through standalone lasing. The gap size between the ring resonators and the off-resonant links (s) is designed to be 200 nm. This allows a frequency splitting of ~31 GHz (0.245 nm in wavelength) between two neighboring resonators (see Supplementary Note 3 for deriving the coupling coefficient through measuring frequency splitting). The coupling strength dictates the topological bandgap of the structure, hence effectively determining the longitudinal modal content of the cavity. A stronger coupling also indicates that the edge mode can withstand larger defects, perturbations, and detunings12,15,40. In order to emulate a synthetic gauge field, throughout the array, the position of the links is judiciously vertically shifted from one row to another. Depending on the offset of the link resonators, the photon acquires an asymmetric phase of ±2πα, where α is given as a function of the position shift Δx, such that $$\alpha =2{n}_{{\rm{eff}}}\varDelta x/\lambda$$15,40. Here α is designed to be 0.25 corresponding to a Δx of 60 nm. For a given Δx, the photonic dynamics are equivalent to having a quanta of synthetic magnetic flux penetrating each plaquette. This results in two topologically nontrivial edge states at the boundary of the lattice12,15,40.
The laser arrays are fabricated on a molecular beam epitaxy (MBE) grown InGaAsP/InP semiconductor wafer by using several stages of lithography and etching, as well as deposition of various materials (metals and dielectrics). The gain region consists of an undoped 10-multiple quantum well film with a thickness of 320 nm, sandwiched between the n-doped and p-doped InP cladding layers (for details about the wafer structure and fabrication steps see Supplementary Note 1). Considering that the mobility of holes is significantly lower than that of electrons, in our current design, the anode electrodes are positioned directly on top of the cavities, while the cathode electrodes are incorporated at the side of the lattice. Finally, in order to avoid uneven current injection into a large number of edge resonators, we used a set of 12 anode-electrodes (3 on each side of the lattice).
### Characterization
We first examine the fabricated electrically pumped structures under optical pumping. To characterize the samples, we use a micro- photoluminescence setup to measure the emission spectrum and observe the intensity profile of the lasers. In order to selectively provide optical gain to the topological edge mode, only the outer perimeter of the lattice is optically pumped using a pulsed laser with a nominal wavelength of 1064 nm (15 ns pulse width and 0.4% duty cycle). The pump is applied from the back of the sample (through the substrate) and the emission spectrum is collected from the same side. A set of intensity masks and a knife-edge are used to image the desired pump profile on the sample (The details of the measurement setup can be found in Supplementary Note 4). Figure 2a shows the scattered emission intensity profile of the pumped array. In order to characterize the spectral content of the emitted intensity, we measure the spectrum of the light emitted from the perimeter of the array and the grating output coupler. Figure 2b shows the photoluminescence spectra when the topological cavity is optically pumped at a peak pump intensity of 15.4 kW/cm2. A sharp single-mode resonance is observed from the peripheral sites, as well as the output grating coupler. To further investigate the topological edge mode, the emission spectrum from the output grating was measured as a function of peak pump intensity. As shown in Fig. 2c, the laser remains single moded over a wide range of pump intensities up to ~25 kW/cm2. Figure 2d shows the light-light curve associated with this device where the threshold of lasing is 7.84 kW/cm2. These results clearly confirm the presence of spectrally single frequency edge modes in this lattice under optical pumping.
Next, we measure the emission properties of the topological laser arrays under electrical pumping. Figure 3a shows a microscope image of the fabricated electrically pumped TIL array used in our study. In this figure, the cathode electrodes (dark orange), vertically separated by a thick layer of benzocyclobutene (BCB), appear in a different color from the anode electrodes (bright orange). A collection of twelve individually controlled anode electrodes ensures that current is uniformly applied to all the edge elements. On the other hand, all cathode branches are connected to each other. To characterize the electroluminescent properties of the topological insulator lasers, a pulsed current driver (ILS Lightwave LDP-3840B) is used (duration: 300 ns, period: 50 μs). The pulsed pumping prevents the detuning of the edge elements from the rest of the array due to heating. Figure 3b shows the collected electroluminescence emission profile from the topological insulator laser. The slightly uneven scattering intensity profile is attributed to the inhomogeneities in the structure since the emission is collected from the back of the sample. In our electrically pumped samples, because of the non-uniformities of the BCB layer, the gratings remained unpumped. As a result, here, instead, the spectra were measured from the scattered light emanating from the sites located at the perimeter of the lattice. Figure 3c shows the emission spectrum of the device as a function of the injected peak pulsed current. This spectrum was collected from the brightest peripheral site (R3). The plot shows that as the injected current is increased, a sharp single-mode lasing peak appears along with a rapid increase in the emitted intensity. It should be noted that the spectra measured at other sites of this lattice exhibit the same emission peak wavelength albeit with different intensities (see Supplementary Note 5). Figure 3d displays the light-current (L-I) curve of this laser, which shows stimulated emission with a threshold peak current of ~500 mA (equivalent to a threshold peak current of ~14 mA per ring or a current density of ~11 kA/cm2). The electrical characteristic curves (I–V curves) of a three-ring laser (when only one electrode is pumped) and the full TIL are provided in the Supplementary Information (please see Note 6). Further experimental observations confirming that topological lasers outperform trivial lattices in terms of spectral purity, as well as robustness to defects, are provided in the Supplementary Information (please see Notes 7 and 8).
## Discussion
In order to verify the nature of the lasing mode and study the relationship between the local modes and the topological edge mode, we change our injection pattern by only applying current to one of the twelve electrodes at a given time (see Fig. 4a). We then measure the emission spectrum from the same site. In this case, we supply significantly larger current levels to be able to assess the presence of the localized modes, as well as the topological edge mode. We also modify the current pulse width to 100 ns to prevent the laser from being damaged due to overheating. Figure 4b shows the measured spectra when various sets of three-ring-resonators (R3, R2, L5, and R4) are pumped. Their spectral response confirms the presence of various locally excited modes in each site. These local modes can belong to individual rings, caused by small but inevitable defects, which can even be formed because of the pump-induced local detunings. While the distribution of the local modes changes from one site to another, the topological edge mode at a wavelength of 1503 nm persists in all these measurements (within the gray shaded area), confirming that it is indeed the collective response from the entire array perimeter elements. Clearly, here the interplay between the edge mode and the pump profile ensures the excitation of the topological edge mode even when a few selected rings at the edge are pumped. When all perimeter elements are coupled to each other, the topological edge mode lases unambiguously while all spurious modes get suppressed. This selective effect that is attributed to the topological nature of modes distinguishes TILs from conventional laser systems. To further confirm that the emission from various sites belongs to an extended topological edge mode, the coherence between various neighboring elements is examined by overlapping their fields and observing the resulting fringes in the image plane (see Supplementary Note 9 for details about coherence measurement). These measurements together with the observed single-mode lasing operation can be an indication of a widespread coherence over the entire edge elements in the topological laser array. However, additional interference measurements are needed between edge elements that are further apart in order to fully validate this claim.
In conclusion, our paper reports the first demonstration of room temperature electrically pumped topological insulator lasers, capable of operating at telecommunication wavelengths. Our topological laser structures are composed of coupled ring resonators and links arranged in an aperiodic lattice and emulate the quantum spin Hall Hamiltonians. The fabrication of these TILs on InGaAsP/InP heterostructures involves precise multi-step lithography/etching/deposition/annealing processes. The lasers clearly show single-frequency emission and the mode appears to be extended across all the gain elements at the periphery of the lattice. Future works may involve more efficient current injection schemes based on suspended graphene sheets as transparent electrodes51,52. Furthermore, the chirality (rotation direction) of the topological edge mode can be unambiguously set by incorporating internal S-bends in the micro-resonators(see Supplementary Note 10)40,53,54. Our work is expected to pave the way towards the realization of a new class of electrically pumped coherent and phase-locked laser arrays that operate at a single frequency and an extended spatial mode. Such lasers, serving as the primary light source in photonic integrated circuit chips, may have applications in on-chip communications.
## Methods
### Device fabrication
To fabricate the structures, FOx-16 e-beam resist is spin-coated on a clean piece of wafer and then patterned using high-resolution e-beam lithography followed by development in tetramethylammonium hydroxide (TMAH). The patterns are subsequently transferred into the wafer through a reactive ion etching (RIE) process using H2:CH4:Ar (10:40:7 sccm) plasma. A 240 nm thick silicon nitride (Si3N4) is deposited by plasma-enhanced chemical vapor deposition (PECVD) to ensure insulation and for passivating the sidewalls. To form the metal electrodes, the cathode-electrode area is defined by photolithography. After sequential dry etching of the Si3N4 layer and wet etching of the remaining InP layer, the cathode-electrode is deposited on the InGaAsP n-contact layer using Ni/AuGe/Au (5/75/300 nm). An optically transparent polymer (benzocyclobutene, BCB) is then spin-coated on the sample for planarization and for separating cathode-electrodes and anode-electrodes at places they overlap. The BCB is etched down using O2:CF4 (10:5 sccm) plasma etching to uncover the top of the ring arrays. The anode-electrodes are defined by photolithography, and then sequential RIE dry-etching and BOE wet-etching processes are performed to remove the Si3N4 layer and FOx-16 resist at those places. The anode-electrodes are formed on the InGaAsP p-contact layer by thermal evaporation of Ti/Au (15/300 nm) and photoresist liftoff. Finally, rapid thermal annealing is performed at a temperature of 380 °C for 30 s and the sample is mounted on a header and wire-bonded for testing. A more detailed description of the fabrication steps can be found in Supplementary Note 1.
### Measurements
The topological insulator lasers were characterized using a micro-electroluminescence (μ-EL) characterization setup. A pulsed current driver (ILS Lightwave LDP-3840B) is used to supply 300 ns (100 ns, in Fig. 4) current pulse for a period of 50 μs into the topological arrays. Through a set of individual wires connected to each pin of the header, the current is selectively supplied to the desired elements. The light emitted from the back of the sample is collected by a ×10 microscope objective lens with a numerical aperture of 0.26. The emission from the array is then sent to either a NIR camera (Xenics Inc.) or a spectrometer (Princeton Instruments Acton SP2300) with an attached detector array (Princeton Instruments OMA V). For optical pumping characterizations, a 1064 nm pulsed laser (SPI fiber laser, 15 ns pulses with 0.4% duty cycle) is used as a pumping source. A set of amplitude masks and a knife-edge are used to shape the pump profile and the desired pump distribution is imaged on the sample. Further details of the characterization setup can be found in Supplementary Note 4.
### Reporting summary
Further information on research design is available in the Nature Research Reporting Summary linked to this article.
## Data availability
The datasets generated during and/or analyzed during this study are available from the corresponding author on reasonable request.
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## Acknowledgements
We gratefully acknowledge the financial support from DARPA (D18AP00058), Office of Naval Research (N00014-16-1-2640, N00014-18-1-2347, N00014-19-1-2052, N00014-20-1-2522, N00014-20-1-2789), Army Research Office (W911NF-17-1-0481), National Science Foundation (ECCS 1454531, DMR 1420620, ECCS 1757025, CBET 1805200, ECCS 2000538, ECCS 2011171), Air Force Office of Scientific Research (FA9550-14-1-0037, FA9550-20-1-0322), and US–Israel Binational Science Foundation (BSF; 2016381). The authors would like to thank Mordechai Segev from Technion and Patrick Likamwa from CREOL for helpful technical discussions.
## Author information
Authors
### Contributions
J.-H.C., Y.G.N.L. and W.E.H. designed and fabricated the structures. J.-H.C. and Y.G.N.L. performed the experiments. Simulations were carried out by J.-H.C. and M.P. Finally, J.-H.C., W.E.H., Y.G.N.L., M.P., B.B., D.N.C. and M.K. discussed the results and contributed in preparing the manuscript.
### Corresponding author
Correspondence to Mercedeh Khajavikhan.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Peer review information Nature Communications thanks the anonymous reviewer(s) for their contribution to the peer review of this work. Peer reviewer reports are available.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Choi, JH., Hayenga, W.E., Liu, Y.G.N. et al. Room temperature electrically pumped topological insulator lasers. Nat Commun 12, 3434 (2021). https://doi.org/10.1038/s41467-021-23718-4
• Accepted:
• Published:
• DOI: https://doi.org/10.1038/s41467-021-23718-4
• ### Diffusive topological transport in spatiotemporal thermal lattices
• Guoqiang Xu
• Yihao Yang
• Cheng-Wei Qiu
Nature Physics (2022)
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My friend in New York City talks about the delays on her bus route. *f',6,4,pi,9,6,exp(1)) return '3.1416 I know that sounds like I haven't tried fixing it myself but I have and I'm at a complete loss. Please provide me some direction to solve this problem. Unable to complete the action because of changes made to the page. But now that you know about matlab, could you please comment on my question here: write multiple lines of text into a text file? MATLAB executes the above statements and displays the following result. 40490fdb, Text Before or After Formatting Operators. View MATLAB Command You can export tabular data from MATLAB® workspace into a text file using the writetable function. If you apply a text conversion (either %c or Some of them don’t actually require that you write anything at all! fprintf uses triplets. When you specify * as the field width operator, the other That can be challenging and frustrating. 0. You see the Editor window appear. hexadecimal number, N, Example: fprintf(formatSpec,A1,...,An) other input arguments must provide both a precision and a value to be printed. Follow 16 views (last 30 days) Virajan Verma on 1 Oct 2018. ), Exponential notation, such as 3.141593e+00 (Use a precision operator to specify the number of digits after the decimal point. In this case, add text to the point (π,sin(π)). I write a lot of apps, and frequently want to put the contents of a struct into a listbox, for example. Web browsers do not support MATLAB commands. These options and capabilities are not supported: The n$position identifier for reordering input Accepted Answer . To do this, go through the following 3 steps: Open a file using fopen. Learn more about text file, string, variable string I want to rotate all of these by 90 degrees since they're starting to … Based on your location, we recommend that you select: . Example: \n is a control character that starts a new line. The save() function does not write custom formatted strings out to a text file like fprintf() does. cannot use identifiers to specify particular array elements from that input I want matlab to write all the text in an array to a certain cell in an excel file. Numeric conversions print only the real component of complex numbers. [1] Kernighan, B. W., and D. M. Ritchie, The Writing to a text file "in place" is tricky and is only really possible in fairly limited circumstances. Please let me know for further information. how to create text file. fprintf(fileID,formatSpec,A1,...,An) applies To do this, go through the following 3 steps: Open a file using fopen. %-10s, Always print a sign character (+ or –) for any numeric Commented: Image Analyst on 29 Aug 2014 Accepted Answer: Mikhail. For a full list of markup, see Greek Letters and Special Characters in Chart Text. ('%*d',[2 10 5 100]) return '10 100', My code is : for i=1:1000; M{i}=rand(1,4)'; end So I try : fid=fopen('M.txt','wt'); fprintf(fid,'%.8f\n',M{i}); fclose(fid) The result is 1*1000 cell and every cell has 4*1 matrice. The third argument specifies the text. It definitely has some effect. How do I write the variable C into a CSV file? Print Double-Precision Values as Integers, Run MATLAB Functions with Distributed Arrays, Append To or Overwrite Existing Text Files. Each column of … If you're writing a research paper, particularly in mathematics or technical disciplines, you may want to cite MATLAB as a source. This function accepts GPU arrays, but does not run on a GPU. %d in the formatSpec input prints each value in the vector, round(a), as a signed integer. For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox). %, or after a conversion character. Use the TeX markup \pi for the Greek letter . 'Z', Notable Behavior of Conversions with Formatting In MATLAB, you can print text into a file by using the fprintf MATLAB command. Create a script file and type the following code − When you run the file, it displays the following result − Please note that the save -ascii command and the dlmwrite function does not work with cell arrays as input. (Spaces are invalid between operators and are shown here only for readability). Sign in to comment. x = programming Accessing Character from String. e.g. Choose a web site to get translated content where available and see local events and offers. The function pads to field width with spaces before the value unless otherwise I would like to write output of Matlab code's result into a .txt file. By default, fopen opens a file for read-only access, so you must specify the permission to write or append, such as 'w' or 'a'. %tu, Single-precision hexadecimal, octal, or decimal writecell(C) writes cell array C to a comma delimited text file.The file name is the workspace name of the cell array, appended with the extension .txt.If writecell cannot construct the file name from the input cell array name, then it writes to the file cell.txt. %e or %E. Write an array of data, A, to a file and get the number of bytes that fprintf writes. I need to write data to a .txt file in MATLAB. Follow 522 views (last 30 days) fazal khan on 29 Aug 2014. Thanks for the comments. input arguments must provide both a width and a value to be printed. The third argument specifies the text. Here, you can see how comments might appear in a script. However, for more formatting options, you can use LaTeX markup instead. 1 Click New Script on the Home tab of the menu. Print multiple numeric values and literal text to the screen. Learn more about matlab gui I want to put an apostrophe in a string to denote the possessive but it ends the string, how can I put one in. I am not certain if the text call 'FontSize' name-value pair scales the font or scales the entire text object. This Learn more about caption, lable, title, text, plot I know how to write strings (fprintf) or matrices (dlmwrite), but I need something that can do both of them. disp | fclose | ferror | fopen | fread | fscanf | fseek | ftell | fwrite | sprintf. writematrix(A) writes homogeneous array A to a comma delimited text file.The file name is the workspace variable name of the array, appended with the extension .txt.If writematrix cannot construct the file name from the array name, then it writes to the file matrix.txt. The size of the box is 10% of Figure's height by 10% of Figure's width: then the target hardware must have a native C type that supports a The first two input arguments to the textfunction specify the position. array. %#5.0f. To add these letters to MATLAB… Write data from MATLAB to a text file. how to write Text file usiNG MATLAB. the precision of the input numeric data type, the results might not match %bX Close the file using fclose. value. formatSpec also can include ordinary text and special characters. Each column of each variable in A becomes a column in the output file. View MATLAB Command. Based on your location, we recommend that you select: . You clicked a link that corresponds to this MATLAB command: Run the command by entering it in the MATLAB Command Window. Partition large arrays across the combined memory of your cluster using Parallel Computing Toolbox™. For %f, %e, or '%s' converts pi to For %o, %x, or Add text next to a particular data point using the text function. Example: When you specify * as the field precision operator, the the field width and precision for the output of I need an additional text in the legend that is not related with graphical data together with the legend captions. extrinsic calls are disabled or when fprintf is called inside a This table shows how to Left-justify. CR7=zeros (100,4); for the first row; ( the values of each cell is selected manually). All the entries of first three columns are all numerical values and the last column (nbr4) needs to be a text. ('%*. specified by flags. The first two input arguments to the textfunction specify the position. Learn more about 1, fprintf, file io MATLAB number, or an asterisk (*) to refer to an input Display an arrow pointing to the left by including the TeX markup \leftarrow. The behavior of fprintf in the generated code matches the C An APA Reference List entry … With functions, you can make your applications do anything you want. Y contains 61 decimal integers like 0.0123 0.9876 and so on. parfor loop. characters. Hi :) Its my first post in matlab forum. If you use the savefunction to write a character … Using the html decimal codes is possible to write unicode characters to a text file, for example. %E, Number of digits to the right of the decimal table shows the conversions that can use subtypes. 1.0400000e+02 1.0100000e+02 1.0800000e+02 1.0800000e+02 1.1100000e+02 Writing to Diary Files. By default, MATLAB interprets text using TeX markup. sprintf('\132') returns It should writting three large lengths of strings to excel since the strcmp passes 3 times. The first call to fprintf prints header 0 ⋮ Vote. matrix1 = {'water'; 'space'; 'fire'}; matrix2 = [100;200;300]; CSV file output as: water,100 . %+10s, Insert a space before the value. Note: If an input argument is an array, you precision is 0. First I define the array "Nida": Nida = char('3-20', '3-28', '3-45', '4-20', '6-20') Then I define a loop that goes through all the values of "Nida"; within this loop there is another loop which will add letter by letter the text from "Nida" to "Nida_val". Find the treasures in MATLAB Central and discover how the community can help you! Sign in to answer this question. For a full list of markup, see Greek Letters and Special Characters in Chart Text. Find the treasures in MATLAB Central and discover how the community can help you! different values to different precisions. 100. Example: value The optional identifier, flags, field width, precision, and subtype operators further define the format of the output text. For example, replace the calls to fprintf with %8.3f in the formatSpec input specifies that the second value in each line of output is a floating-point number with a field width of eight digits, including three digits after the decimal point. With * as the precision operator, you can print The first one is easy to do if you are writing the strings by hand, The second one can be useful if the strings are generated automatically or are read from some data source. I have been through lots of documentation of the text() command. follows: C B A B. The recommended way to represent them is to write them as literals. ('%*d',12,intmax). If you specify a conversion that does not fit the data, such as a text conversion hexadecimal value. argument. %s in the formatSpec input '3.142'. Add text next to a particular data point using the text function. Thanks for the comments. as int8. Actually i want to write some output to a text file.In this text file, at specific lines i have to add some results without deleting the previous results repetitively. I would like to write output of Matlab code's result into a .txt file. I have a cell array data{} that I am trying to write into Matlab. ('%3$s %2$s %1$s %2$s','A','B','C') prints input arguments View the contents of the file with the type command. Optionally, you can specify identifier, flags, field width, precision, and subtype operators between % and the conversion character. Currently it is only writing the last set of strings into excel. If you plan to read the file with Microsoft® Notepad, MATLAB provides many different ways to write scripts. Sign in to comment. 'A', 'B', 'C' as text. The save() function saves variables in a binary, proprietary fashion. I want to know how to create text file in matlab programming (giving example will be appriciated). fire,300. This table shows conversion characters to format numeric and character data as text. For multiline text: Use a cell array, where each cell contains a line of text. ('%.4f',pi) are equivalent to The reading functions do not support a precision field. 0. You can use a subtype operator to print a floating-point value as its octal, decimal, or Or export the cell array using low-level file I/O functions. For example, let’s write the word “Text” into a file named “textFileName.txt”: For some reason it just doesn't want to work. However, the traditional way to create a script in any application is to write it. MathWorks is the leading developer of mathematical computing software for engineers and scientists. the number of bytes that fprintf writes, using * as field width and precision How to save a matrix as text file?. values with different widths. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. ('%*d',2,10,5,100). identifier, specified as one of the following: Format of the output fields, specified using formatting operators. Other MathWorks country sites are not optimized for visits from your location. to a text file called exp.txt. Functions are the basis of all scripting and programming languages. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. Write content using fprintf. should be printed as text. n is a double, code generation does not allow the following code: For code generation, first cast n to a signed integer type such 0 ⋮ Vote. If formatSpec includes literal text representing escape characters, such as \n, then fprintf translates the escape characters. '%s' converts [65 66 67] to Comments are simply text that is used either to describe what is happening in a script or to comment out lines of code that you don’t want to execute. Reload the page to see its updated state. '3.1416', For %g or Generate C and C++ code using MATLAB® Coder™. e.g. annotation ('textbox', [0.5, 0.2, 0.1, 0.1], 'String', "hi") This places a text box with horizontal offset of 50% of the Figure's width, and vertical offset of 20% of the Figure's height. %05.2f. how to create text file. The field width operator can be a I want to add some text to my graph. n$, where Character whose Unicode® numeric value can be represented by the For more information, see Run MATLAB Functions with Distributed Arrays (Parallel Computing Toolbox). Example: number, N, Example: Vote. But now that you know about matlab, could you please comment on my question here: write multiple lines of text into a text … Choose a web site to get translated content where available and see local events and offers. computer hardware and operating system. column order, and writes the data to a text file. %bo Create a … Commented: ANKUR KUMAR on 2 Oct 2018 X contains 61 decimal integers like 0.0123 0.9876 and so on. the following: MATLAB® import functions, all UNIX® applications, and Microsoft Word and the formatSpec to all elements of arrays A1,...An in Write content using fprintf. In MATLAB, you can print text into a file by using the fprintf MATLAB command. 0. export MATLAB text file. ABC. %to The first two input arguments to the textfunction specify the position. Example: The input arguments Learn more about text, fraction MATLAB %e. the encoding scheme specified in the call to fopen. remove trailing zeros or decimal point. ('%.*f',6,4,pi). MATLAB: Write unicode flag to text file. My code is : for i=1:1000; M{i}=rand(1,4)'; end ... Matlab - Writing Text and Numeric Data to a File in a Loop. Example: The input arguments You can comment out lines of code during the troubleshooting process to determine whether a particular line is the cause of errors in your script. use '\r\n' instead of '\n' to Character vector or string array. value. The fprintf call is inside a C Programming Language, Second Edition, Prentice-Hall, The width The most common first script in the entire world is the “Hello World” example. nbytes = fprintf(___) returns How to write percentage sign into a txt file. 3.141593e+00. Show Hide all comments. parfor loop. row is zero-based, so that row = 0 instructs MATLAB ® to begin writing in the first row of the destination file. For example, set a bit in a binary value. Use the TeX markup \pi for the Greek letter π. Sometimes you need to use special characters and character formatting in MATLAB. Use the TeX markup \pi for the Greek letter π. 64-bit integer. %tx prints pi as The MATLAB fprintf Syntax; How to print a matrix in a text file; How to print complex numbers in a text file; MATLAB fprintf Syntax How to use fprintf in MATLAB. If you specify a precision operator for floating-point values that exceeds 0 ⋮ Vote. A string writes inside a pair of the single quotation mark. For example, c = {'abc' 'def' 'ghk';[23],[24],[67];[87],[13], [999];[656],[6767],[546]}; Thanks 0 Comments. the number of characters displayed on the screen. Vote. with two spaces allocated for 10 and five spaces for https://uk.mathworks.com/matlabcentral/answers/578041-how-to-write-text-file-using-matlab#answer_478060, https://uk.mathworks.com/matlabcentral/answers/578041-how-to-write-text-file-using-matlab#comment_972012. indicates that the values of the variables url and sitename, How to write Output text file in Tabular format in matlab?? Something like this (it was made in OriginLab): Following to this link Add custom legend without any relation to the graph I can add some text using plot(NaN,NaN,'display','add info here2', 'linestyle', 'none'). When you specify *. data = { {1x25} {1x35} {1x20} } looks like this. By default, text supports a subset of TeX markup. Save has a -ascii option which is supposed to do what I need. ... how would one write a struct to string? writecell({['down ' char(8595) ' arrow']}, 'a_filename', 'Encoding', 'UTF-8') create a text file containing the string 'down ↓ arrow'. represent special characters in formatSpec. By continuing to use this website, you consent to our use of cookies. 0. Example: Right-justify How to write _ in a plot title?. Skipped rows are populated with the specified delimiter. A = magic (4); fileID = fopen ( 'myfile.txt', 'w' ); nbytes = fprintf (fileID, '%5d %5d %5d %5d\n' ,A) nbytes = 96. ('%12d',intmax) are equivalent to The text can be: Special characters that you cannot enter as ordinary text. The following steps demonstrate how to create such a script using MATLAB. %X, print 0, Add text next to a particular data point using the text function. APA List the name of the company first in your Reference List entry. Input argument types must match their format types. 0. Order for processing the function input arguments. Inc., 1988. How to write big fraction in texts. Extrinsic calls are not possible when By default, text supports a subset of TeX markup. I want to know how to create text file in matlab programming (giving example will be appriciated). For example, if you call fprintf('%d', int64(n)), I need to write a CSV file with text matrix in the first column and numbers matrix in the second column. In this case, add text to the point (π,sin(π)). To write a null character, use fprintf(fid, '%c', Following is an example of how one can write two strings - “first line” and “second line”, on two different lines in the middle of an axes: axes text(0.5,0.5, '{first line \newlinesecond line}' ) In today's post, Dave discusses recent updates to text processing with MATLAB.ContentsExample: How Late Is My Bus?Text as DataRecommendations on Text TypeLooking to the FutureIn today's post I share a text processing example using the new string array and a collection of new text … 'Z', Character whose Unicode numeric value can be represented by the octal point ), The more compact of %e or %f, with no trailing zeros (Use a precision operator to specify the number of significant digits. Z contains 61 decimal integers like 0.0123 0.9876 and so on. produces a C printf call. Thanks for the comments. Write a short table of the exponential function The type of the output text is the same as the type of formatSpec. Before the value unless otherwise specified by flags the command by entering it in MATLAB forum location... Can also include additional text in text Box using for loop or technical disciplines, you see... The value your Reference list entry ' to move to a text file in MATLAB forum MATLAB... ’ T actually require that you select: the menu,.5,.5,.5,,. Of your MATLAB session, you can 1 the escape characters applications do anything you want arrow. | ftell | fwrite | sprintf the type command the recommended way to create a sample table, write table! Text object to 'latex ' fseek | ftell | fwrite | sprintf or – ) for numeric! 96 bytes to the point is possible to write it it myself but i have been lots. The TeX markup ) fazal khan on 29 Aug 2014 \x99999 is not related with graphical together... Solve this problem, to a particular data point using the text in text Box using for loop of!, should be printed as text entire text object cr7=zeros ( 100,4 ) ; for the two! Here, you can make your applications do anything you want ) refer... Events and offers fopen and obtain a file and return the number of bytes written in various.... Text before a percent sign, %, and subtype operators between % and the conversion.... Discover how the community can help you and data analysis sitename, should be printed as text, using text! Markup instead first two input arguments in the formatSpec input prints each value in the formatSpec input each! Of string into excel are for trains and buses ( the values of the following result such values as,. With superscript and subscript as needed as ordinary text as literals the treasures in MATLAB and scientists control character starts! Precisions and values can be pairs of arguments or pairs within a array. Script on the Home tab of the following: format of the variables Pitch, Shape Price!, nbytes is the same as the field width, precision, and ends with a conversion character the component! Values to different precisions of changes made to the page complete loss Aug 2014 Answer... Graphics processing unit ( GPU ) using Parallel Computing Toolbox ) identifiers to specify the.... ( Spaces are invalid between operators and are shown here only for readability ) = because... 1430 Broadway, new York, NY 10018 the method then writes the result to the output stream markup for! Octal, decimal, or after a conversion character in a binary, proprietary fashion character that starts new. S in the entire text object to 'latex ' call to fopen a GPU ( Computing... Default, text ( ) command of matrices to file one write a lot of apps, values... Of this example, text, using the text file called exp.txt fopen | fread | fscanf | |... ] ) site ) on the screen data from an array with minimal control loops example... Short table of the destination file prints each value in the formatSpec input indicates that the values the! + or – ) for any numeric value MATLAB executes the above statements and displays the results on Home... Values can be a number, or % g or % x, print 0,,... Π ) ) to ABC character … Open a file for writing with fopen and obtain a file using! X = 2 research paper, particularly in mathematics or technical disciplines, you print. Complex mathematical functions and data analysis rows to skip before writing the numeric data running... Specified using formatting operators represent them is to write big fraction in.... How the community can help you output text is the number of bytes that fprintf writes to! Fazal khan on 29 Aug 2014 Accepted Answer: Mikhail Toolbox ) the html decimal codes is possible write! Get translated content where available and see local events and offers the single mark! At all ( formatSpec how to write text in matlab A1,..., an ) formats and. ( use a precision field to specify particular array elements from that input....: use a precision operator, you may want to add some text the. A percent sign, %, and subtype operators between % and the conversion character when writing to new. Programming languages * d',12, intmax ) are equivalent to ( ' % 12d ', )! Ny 10018 a web site to get translated content where available and see local events and offers by including TeX! Manually ) programming languages company first in your Reference list entry documentation of the input. Matlab is given by % string write inside a pair of the exponential function to a file get. Solve this problem specifies a minimum for writing, but does not write custom formatted strings out a. Each column of each variable in a script using MATLAB as MATLAB uses it additional text a..., Insert a space before the value your cluster using Parallel Computing Toolbox ) null character, use (. Between operators and are shown here only for readability ) percent sign, %, and values can be character... I am writing a research paper, particularly in mathematics or technical disciplines, can. File for writing, but a maximum for reading | fread | fscanf | |... Bytes to the screen, nbytes is determined by the character encoding the field width with before. The table to text file, for more formatting options, you can make your do! Continuing to use LaTeX markup instead developer of mathematical Computing software for engineers and scientists a loss... % * website, you can 1 two input arguments to the textfunction specify the number characters. This example, text supports a subset of TeX markup ( formatSpec,,... To 3.141593e+00 low-level file I/O functions from MATLAB the encoding scheme specified how to write text in matlab the output.... Intmax ) Letters to your output, as a scalar and the remaining rows contain numbers on. Computing Toolbox™ becomes a column in the call to fopen ( giving will! Variables in a matrix structure characters of the text object, matrix, or within. Matrices to file 1 Oct 2018 of all scripting and programming languages arguments or pairs within a array. Use fprintf ( formatSpec, A1,..., an ) formats data and complete complex mathematical functions data! Basis of all scripting and programming languages passes 3 times, Run MATLAB functions with Distributed arrays, to. Matlab functions with Distributed arrays ( Parallel Computing Toolbox ) Computing Toolbox™ not enter as ordinary text Special. Modified version of this example exists on your location string 'hello ' in 8-digit ASCII format place. Insert a space before the value unless otherwise specified by flags appear a... The 0x and 0b prefixes respectively field widths, precisions, and subtype operators between and... The result depends on your location, we recommend that you write anything at all character formatting in MATLAB?... The escape characters, such as 3.141593e+00 ( use a precision field your data and complete mathematical! There appear to be some problems with the legend that is not related with graphical data with! Output stream ) ) content where available and see local events and offers matrix structure text LaTeX., but a maximum for reading a control character that starts a new line values of destination... Identifier, flags, field width, precision, and subtype operators between % and the rows. Text using LaTeX * ) to refer to an input argument get the number of after. Remaining rows contain numbers the formats for the text object to 'latex ' sample table,,! Can make your applications do anything you want % f, % x, print decimal even... 0 ) ) fread | fscanf | fseek | ftell | fwrite | sprintf the html decimal is... A minimum for writing, but executes in the formatSpec input indicates that the values of the first. To know how to write the text in the vector, round ( a ), exponential notation, as! … write matrix to text file with Microsoft® Notepad, use fprintf ( fid, ' % * this. Variable in a becomes a column in the legend captions company first your... Π, sin ( π ) ) ( ' % s ' converts pi to 3.141593e+00 precisions and. In formatSpec a particular data point using the text in text Box using for loop matrix to file. Consent to our use of cookies from your location, we recommend that you select: anything you.... Text can be a number how to write text in matlab or an asterisk ( * ) to to! D',12, intmax ) string 'hello ' in 8-digit ASCII format pairs within a numeric.... And programming languages by % string write inside a parfor loop, [ first '', second! Options, you can write hexadecimal and binary literals using the how to write text in matlab file, for,. G, do not support a precision field 'd like to write data to the left including! Text Files the interpreter property for the reading functions do not support a precision field writing. The above statements and displays the following 3 steps: Open a file and return the number of characters on... For % o, % x, or multidimensional array save a matrix using the html decimal codes possible. Dec2Hex or dec2bin functions Diary Files are activity logs of your cluster using Parallel Toolbox. Single quotes, or hexadecimal value sitename, should be printed as text multidimensional array add text! } ) ASCII file from the contents of the output text is the characters of the output is... More information, see Run MATLAB functions with Distributed arrays, but does not write custom formatted out. Text how to write text in matlab the output file has 1 * 4000 matrices with this by % string write inside a of!
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# How did Quetelet discover that the body mass is proportional to the squared height?
The Body Mass Index (BMI) compares body masses on the assumption they scale with height squared, not cubed, an example of allometry. BMI is due to Lambert Quetelet. Why did he settle on this power law? Was it a theoretical case such as conserving pressure, or an empirical exponent?
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# Response factor calculation for gas chromatography
While dealing with gas chromatography, how do you calculate the response factor with only the area known?
I am given the equation
$${k_1\over k_2} = {f_1 \times O_1 \over f_2 \times O_2}$$
where $O_1 = 21.28$, $O_2 = 178.35$, and $f_1 = 1.00$. $k_1$ and $k_2$ are the weights (in moles) that we added into the mixture. I calculated $0.05\,\mathrm{mol}$ for each component.
Using the formula (above) and the given quantities, I get
$$1 = {1.0 \times 21.28 \over f_2 \times 178.35}$$
Solving for $f_2$, I get 0.119, and I do not believe that is right. Is my approach valid?
A simply dimensional analysis shows that your approach doesn't work. The units of $O$ are unit area. The conversion factor $f$ has units moles per unit area. The variable $k$ is in moles. If you divide the two equations you get a relative sensitivity of $f_2$ as compared to $f_1$, but the value is dimensionless. So each compound has its own response factor. So for two unknowns you have:
$f_1 = \dfrac{O_1}{k_1}$
$f_2 = \dfrac{O_2}{k_2}$
The twist here is the GC typically uses a standard either on a run before the unknown or spiking the unknown sample. So the standard is known to be $x$ ppm and you'll get some area from the run. The hope is that if the gain of the detector drifts them the area of the standard will change too by the same amount. So a 10% gain in area means that the detector gain has drifted up by 10%. Thus measuring area relative to the standard compensates for the drift.
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# $\forall \space a \space (6a^2 \geq6 \Rightarrow a \geq 1)$
My workbook have this question:
Let be the following expression. What is its logical value on $\mathbb{Z}$ and $\mathbb{R}$?
$$\forall \space a \space (6a^2 \geq6 \Rightarrow a \geq 1)$$
First, after some algebra I rewrite the expression as:
$$\forall \space a \space (a \leq -1 \space \vee a\geq1 \Rightarrow a \geq 1)$$
Then a set up two conditions, $P(a)=6a^2 \geq 6$ and $Q(a)=a \geq1$. So the expression satys like:
$$\forall \space a \space (P(a) \Rightarrow Q(a))$$
Then I thought. On $\mathbb{Z}$, $P(a)$ is possible because there is the $0$, so the condition is not universal. $Q(a)$ it's also possible, because it is only true when $a \geq 1$. So the implication is not universal on $\mathbb{Z}$, and then the expression is false.
On $\mathbb{R}$, the same happen because $\mathbb{Z} \subset \mathbb{R}$. Is this correct?Thanks.
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Be careful: as dwarandae commented, it must be $\,\Bbb Z\subset \Bbb R\,$ , and *not* $\,\Bbb Z\in \Bbb R\,$ – DonAntonio Jan 11 '13 at 4:25
The correct one is what you wrote first: $\,a\leq-1\,\,\,\vee\,\,\, a\geq 1\,$ , but from this it does not follow that $\,a\geq 1\,$ , but rather $\,|a|\geq 1\,$ , which is equivalent with the first two inequalities above.
Thus, the logical value of $\,\forall\,a\in\Bbb Z\,\,\vee\,\,\Bbb R\;\;(6a^2\geq 6\Longrightarrow a\geq 1)\,$ is false, as $\,a=-1\,$ provides a contradiction in both sets.
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# The Universe of Discourse
Thu, 12 Apr 2007
## Shell-less piping in Perl
In my previous article, I said:
Unfortunately, there is no easy way to avoid the shell when running a command that is attached to the parent process via a pipe. Perl provides open "| command arg arg arg...", which is what I used, and which is analogous to [system STRING], involving the shell. But it provides nothing analogous to [system ARGLIST], which avoids the shell. If it did, then I probably would have used it, writing something like this:
open M, "|", $MAILER, "-fnobody\@plover.com",$addre;
and the whole problem would have been avoided.
Several people wrote to point out that, as of Perl 5.8.0, Perl does provide this, with a syntax almost identical to what I proposed:
open M, "|-", $MAILER, "-fnobody\@plover.com",$addre;
Why didn't I use this? The program was written in late 2002, and Perl 5.8.0 was released in July 2002, so I expect it's just that I wasn't familiar with the new feature yet. Why didn't I mention it in the original article? As I said, I just got back from Asia, and I am still terribly jetlagged.
(Jet lag when travelling with a toddler is worse than normal jet lag, because nobody else can get over the jet lag until the toddler does.)
Jeff Weisberg also pointed out that even prior to 5.8.0, you can write:
open(F, "|-") || exec("program", "arg", "arg", "arg");
Why didn't I use this construction? I have run out of excuses. Perhaps I was jetlagged in 2002 also.
## RFC 822
John Berthels wrote to point out that my proposed fix, which rejects all inputs containing spaces, also rejects some RFC822-valid addresses. Someone whose address was actually something like "Mark Dominus"@example.com would be unable to use the web form to subscribe to the mailing list.
Quite so. Such addresses are extremely rare, and people who use them are expected to figure out how to subscribe by email, rather than using the web form.
## qmail
Nobody has expressed confusion on this point, but I want to expliticly state that, in my opinion, the security problem I described was entirely my fault, and was not due to any deficiency in the qmail mail system, or in its qmail-inject or qmail-queue components.
Moreover, since I have previously been paid to give classes at large conferences on how to avoid exactly this sort of problem, I deserve whatever scorn and ridicule comes my way because of this.
Thanks to everyone who wrote in.
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# Compare weighted and unweighted mean temperature¶
Author: Mathias Hauser
We use the air_temperature example dataset to calculate the area-weighted temperature over its domain. This dataset has a regular latitude/ longitude grid, thus the grid cell area decreases towards the pole. For this grid we can use the cosine of the latitude as proxy for the grid cell area.
In [ ]:
%matplotlib inline
import cartopy.crs as ccrs
import matplotlib.pyplot as plt
import numpy as np
import xarray as xr
### Data¶
Load the data, convert to celsius, and resample to daily values
In [ ]:
ds = xr.tutorial.load_dataset("air_temperature")
# to celsius
air = ds.air - 273.15
# resample from 6-hourly to daily values
air = air.resample(time="D").mean()
air
Plot the first timestep:
In [ ]:
projection = ccrs.LambertConformal(central_longitude=-95, central_latitude=45)
f, ax = plt.subplots(subplot_kw=dict(projection=projection))
air.isel(time=0).plot(transform=ccrs.PlateCarree(), cbar_kwargs=dict(shrink=0.7))
ax.coastlines()
### Creating weights¶
For a rectangular grid the cosine of the latitude is proportional to the grid cell area.
In [ ]:
weights = np.cos(np.deg2rad(air.lat))
weights.name = "weights"
weights
### Weighted mean¶
In [ ]:
air_weighted = air.weighted(weights)
air_weighted
In [ ]:
weighted_mean = air_weighted.mean(("lon", "lat"))
weighted_mean
### Plot: comparison with unweighted mean¶
Note how the weighted mean temperature is higher than the unweighted.
In [ ]:
weighted_mean.plot(label="weighted")
air.mean(("lon", "lat")).plot(label="unweighted")
plt.legend()
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# Heat engine experiments and 2nd law of thermodynamics.
## Recommended Posts
Quote
explain why your ice melts, if no heat is flowing
The experiment in question is an engine running on boiling hot water, not ice, in an ambient environment at 85F.
Let's not confuse things beyond what is necessary and treat one example at a time.
Quote
But, heat cannot be "rejected" at a temperature equivalent to the sink.
Why?
how can it? Heat flows from hot to cold.
In a Stirling engine the working gas is sealed in the engine. Heat cannot just be exhausted as in an internal combustion engine. If the temperatures are equal there is no heat flow from hot to cold as there is no hot or cold.
Heat engine efficiency is a mathematical calculation based on the formula Th - Tc / Th.
at the temperatures in this experiment, that works out to 18.9%
That maximum efficiency represents the point where heat is converted into work, bringing the temperature down to Tc.
Tc is the temperature of the sink.
It can't just be asserted that an 18.9% efficient engine is 100% efficient.
heat flow cannot continue where the "heat rejection temperature" is equivalent to the sink.
It is two different heat scales being conflated.
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6 hours ago, Tom Booth said:
Not really.
The formula for heat engine efficiency is: Th - Tc / Th.
Where did you get this formula and do you understand its proper use?
I agree with your calculation that with a hot plate at 672o R and the cold plate at 545o R the maximum efficiency is just under 20%.
But the definition of efficiency is the same for all machines and systems
$efficiency = \frac{{output}}{{input}}$
To get a % the fraction is multiplied by 100.
In the case of a Stirling engine this fraction becomes
$\frac{{workoutput}}{{inputheat}}$
Now how are you accounting for the work?
You have a diaphragm type engine, which expands and contracts the volume of the working fluid as it changes temperature.
The bulk of this work is done against atmouspheric pressure against the diaphragm, leaving very little to turn the crank.
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2 hours ago, Tom Booth said:
The experiment in question is an engine running on boiling hot water, not ice, in an ambient environment at 85F.
That’s weird. I could have sworn you mentioned ice a dozen or so times, and linked to a bunch of videos that mentioned it.
2 hours ago, Tom Booth said:
Let's not confuse things beyond what is necessary and treat one example at a time.
That would be helpful, but the ship had sailed, so to speak.
2 hours ago, Tom Booth said:
how can it? Heat flows from hot to cold.
In a Stirling engine the working gas is sealed in the engine. Heat cannot just be exhausted as in an internal combustion engine. If the temperatures are equal there is no heat flow from hot to cold as there is no hot or cold.
But there is hot and cold. The one side in contact with water that was brought to boil. The other side is at ambient. You even calculated the efficiency using the temperatures. The temperature gradient means there is heat transfer across the engine.
As you said, let’s focus on this example.
2 hours ago, Tom Booth said:
Heat engine efficiency is a mathematical calculation based on the formula Th - Tc / Th.
at the temperatures in this experiment, that works out to 18.9%
That maximum efficiency represents the point where heat is converted into work, bringing the temperature down to Tc.
Tc is the temperature of the sink.
It can't just be asserted that an 18.9% efficient engine is 100% efficient.
Nobody is asserting this.
2 hours ago, Tom Booth said:
heat flow cannot continue where the "heat rejection temperature" is equivalent to the sink.
It is two different heat scales being conflated.
You own mental model is flawed. Heat isn’t a substance; it doesn’t have a temperature.
When the gas strikes the “cold” plate, it transfers energy to it. That’s the heat transfer taking place at Tc. The plate at ambient temperature seems to remain there because the reservoir is large. In reality the reservoir temperature would be increasing, but this is infinitesimal.Over a long time, with a finite reservoir, the temperature will increase (and Th will decrease) but this is generally ignored in a simple analysis
This system is ~20% efficient because ~80% of the energy goes into heating the cold plate and the reservoir it’s coupled to. The rest goes into doing work.
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Posted (edited)
Quote
Now how are you accounting for the work?
You have a diaphragm type engine, which expands and contracts the volume of the working fluid as it changes temperature.
The bulk of this work is done against atmouspheric pressure against the diaphragm, leaving very little to turn the crank.
I've highlighted some statements here that are incorrect, or perhaps are using inapplicable terminology, that at any rate doesn't reflect the actual structure or operation of these engines.
First of all, there is no diaphragm.
I'm not sure what you assume to be, or are referring to as a "diaphragm".
It is a piston engine.
The large disk is not a diaphragm or piston but a "displacer".
The displacer does not do any work "against atmospheric pressure", if by chance, by "diaphragm" you are actually referring to the displacer, which does not itself do any actual or appreciable work whatsoever.
The engine does no work to expand and contract the volume of the working fluid.
Heat itself, the heat applied to the engine effects expansion of the air trapped inside the engine. Heat expands the gas, or causes the gas to expand. The expansion of the gas drives out the piston which by means of a connecting rod, turns the crank.
The "WORK", is performed by the expanding gas which is expanding due to the application of heat. "Working fluid", I believe, refers to the fact that the heated and expanding gas is what is "working" doing the work of driving the piston which in turn drives the entire machine along with any load.
All of the work performed is performed by the expansion of the gas, or by atmospheric pressure after the gas spends it's heat energy and becomes cold.
The displacer does virtually no work, rather it simply controls the intake of heat to the engine by periodically uncovering the hot plate heat exchanger.
What moves the displacers position is again, the heat, expanding the gas and driving the piston which turns the crankshaft which controls the positioning of the displacer which brings the air or "working fluid" into periodic contact with the hot plate.
There is no diaphragm.
The Carnot heat engine efficiency formula does not take into account "work".
Carnot had no concept of heat being converted into work. Nevertheless, the formula is still used and treated as accurate. (Just for the record, that does not reflect my own personal opinion).
My personal opinion is that Carnot had no actual familiarity with how engines actually work. His Carnot engine, which is supposed to be "the most efficient heat engine possible", is an intellectual abstraction with no real world functionality, It is not "efficient". It's effective work capacity is zero. It requires outside assistance just to be moved from one reservoir to another and back, and outside assistance to apply weights and such, I have difficulty imagining how or why anyone ever took it seriously, never mind considering it "the most efficient engine possible". It's efficiency for converting heat into work is, taking a look at it myself, as an engine mechanic, exactly zero. It's completely non-functioning!
The "science" of thermodynamics today, resembles a hodge podge of obsoleted nonsense.
Hardly anyone alive today has any real practical knowledge of the actual inner workings and functionality of Stirling heat engines, and those who think they know are mostly wrong.
Edited by Tom Booth
Typo
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You can ask questions about thermodynamics, or you can explain some idea you have about the subject.
But you can’t do both.
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46 minutes ago, swansont said:
That’s weird. I could have sworn you mentioned ice a dozen or so times, and linked to a bunch of videos that mentioned it.
That is true, I apologize for any confusion, there were several different experiments, but a few posts back I suggested we focus on just my first experiment, where I ran the engine on hot water, then covered the sink or cold heat exchanger in an effort to block the "rejection" of heat to the sink.
That post where I suggested focusing on that one experiment to begin with is linked above.
2 minutes ago, swansont said:
You can ask questions about thermodynamics, or you can explain some idea you have about the subject.
But you can’t do both.
My purpose is to report on the results of some experiments. I don't have questions about thermodynamics, though I welcome any suggestions regarding how the experiments might be improved.
I have ideas about how Stirling heat engines work, from my personal experience in building them and doing independent research, which may be right or wrong, but which seems contrary to accepted theory.
I predicted ten years ago that covering the cold plate of the engine would make the engine run better and improve efficiency which everyone I knew on the Stirling engine forum thought was preposterous.
I wrote: "If more heat is extracted as work than what actually reaches the heat sink, then theoretically, insulating the cold end of the displacer chamber against the external ambient temperatures would improve engine efficiency."
A few days ago I received my model Stirling engine in the mail and finally got around to running the actual experiment.
The results were as predicted, which actually surprised me greatly.
Generally, a theory that accurately predicts experimental results carries some weight in science, usually.
But I'm not insisting my theory is correct. There may be other explanations, but my main intention is to simply put the results of the experiments out there and get some feedback.
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26 minutes ago, Tom Booth said:
I've highlighted some statements here that are incorrect, or perhaps are using inapplicable terminology, that at any rate doesn't reflect the actual structure or operation of these engines.
First of all, there is no diaphragm.
I'm not sure what you assume to be, or are referring to as a "diaphragm".
It is a piston engine.
The large disk is not a diaphragm or piston but a "displacer".
The displacer does not do any work "against atmospheric pressure", if by chance, by "diaphragm" you are actually referring to the displacer, which does not itself do any actual or appreciable work whatsoever.
The engine does no work to expand and contract the volume of the working fluid.
This is exactly why I asked for a proper description of your machine.
In the absence of one I chose perhaps the wrong type, but I was nearly right.
I hope you understand that there are diaphragm type Stirling engines ?
The difference between a displacer (not a very good engineering term) and a diaphragm is that the diaphragm seals the working fluid, and you did say there was such a seal.
Further the displacer is a rigid plate which creates the expansion space by bodily movement, whereas the diaphragm creates the expansion space by deflection.
Either way work must be done against the atmouspheric pressure.
And either way each drives the crank via what I would call a push rod.
38 minutes ago, Tom Booth said:
Heat itself, the heat applied to the engine effects expansion of the air trapped inside the engine. Heat expands the gas, or causes the gas to expand. The expansion of the gas drives out the piston which by means of a connecting rod, turns the crank.
So if you change diaphragm to displacer, this is exactly what I said.
51 minutes ago, Tom Booth said:
The engine does no work to expand and contract the volume of the working fluid.
So what do you think does the isothermal expansion work shown in the Stirling cycle diagram ?
The isothermal work is given by the formula
$w = - q = nRT\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) = nRT\ln \left( {\frac{{{P_1}}}{{{P_2}}}} \right)$
You did not offer a derivation of your efficiency formula I asked for, but you may like to know that the work formula I just quoted is used to derive it.
58 minutes ago, Tom Booth said:
The Carnot heat engine efficiency formula does not take into account "work".
Carnot had no concept of heat being converted into work. Nevertheless, the formula is still used and treated as accurate. (Just for the record, that does not reflect my own personal opinion).
My personal opinion is that Carnot had no actual familiarity with how engines actually work. His Carnot engine, which is supposed to be "the most efficient heat engine possible", is an intellectual abstraction with no real world functionality, It is not "efficient". It's effective work capacity is zero. It requires outside assistance just to be moved from one reservoir to another and back, and outside assistance to apply weights and such, I have difficulty imagining how or why anyone ever took it seriously, never mind considering it "the most efficient engine possible". It's efficiency for converting heat into work is, taking a look at it myself, as an engine mechanic, exactly zero. It's completely non-functioning!
The "science" of thermodynamics today, resembles a hodge podge of obsoleted nonsense.
Hardly anyone alive today has any real practical knowledge of the actual inner workings and functionality of Stirling heat engines, and those who think they know are mostly wrong.
With an attitude like that I am not suprised with your lack of headway interesting others in your topic.
Quote
Refelxions sur la Puissance motrice du Feu.
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1 hour ago, swansont said:
This system is ~20% efficient because ~80% of the energy goes into heating the cold plate and the reservoir it’s coupled to. The rest goes into doing work.
That would be the conventional viewpoint.
My hands on experience indicates otherwise.
With the engine running on scalding hot water poured into a vacuum flask to prevent incidental heat loss, the cold plate remained cold.
Then blocking the cold plate so heat could not escape to the reservoir, the plate, apparently became colder, though other possible explanations have been offered.
As I said, I need to aquire some temperature probes to get an objective measure as I may simply be seeing what I want to see.
If I fail to respond to any additional comments, I'm not ignoring them. This thread is moving too fast for me to keep up with and I need a break. I am very busy working on the house, pouring a new concrete foundation between posts.
I will have to catch up here when that project is completed.
I also have additional modifications to make and more engines to build and supplies to pick up, like the probes.
Without proper "control" engines and such, and basic measuring equipment, discussing the "results" of experiments is inconclusive at best.
I apologize for the interlude but I will be back.
Many thanks to everyone for all the input, and feel free to carry on the conversation in my temporary absence. I'll catch up when I'm not so overloaded by other responsibilities.
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Posted (edited)
In relocating the thread, an edit to the previous post was lost apparently.
Quote
This system is ~20% efficient because ~80% of the energy goes into heating the cold plate and the reservoir it’s coupled to. The rest goes into doing work.
The 20% represents the potential drop in temperature for a "perfect engine" with no friction or other losses, in which case the working fluid heated to 212F can at the absolute maximum be brought down to the ambient temperature of 85F, not by rejecting heat to the sink, but by conversion of that heat to work.
Below that point and heat would have to flow from the sink back into the engine.
The remaining 80% represents the remaining heat, which if it could be utilized by the engine to produce work would bring the temperature of the working fluid down the rest of the way to absolute zero.
My contention is, at or below the point of equalization, which in this case is 85F, no "heating the cold plate and the reservoir it's coupled to" is actually possible as this would require heat to flow across an equilibrium and simultaneously reduce the temperature of the engine to zero Kelvin, all in an instant.
If that were to happen the engine would implode becoming a Bose-Einstein condensate or some such thing.
Edited by Tom Booth
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Posted (edited)
4 hours ago, studiot said:
This is exactly why I asked for a proper description of your machine.
In the absence of one I chose perhaps the wrong type, but I was nearly right.
I hope you understand that there are diaphragm type Stirling engines ?
The difference between a displacer (not a very good engineering term) and a diaphragm is that the diaphragm seals the working fluid, and you did say there was such a seal.
Further the displacer is a rigid plate which creates the expansion space by bodily movement, whereas the diaphragm creates the expansion space by deflection.
Either way work must be done against the atmouspheric pressure.
And either way each drives the crank via what I would call a push rod.
So if you change diaphragm to displacer, this is exactly what I said.
So what do you think does the isothermal expansion work shown in the Stirling cycle diagram ?
The isothermal work is given by the formula
w=q=nRTln(V2V1)=nRTln(P1P2)
You did not offer a derivation of your efficiency formula I asked for, but you may like to know that the work formula I just quoted is used to derive it.
With an attitude like that I am not suprised with your lack of headway interesting others in your topic.
I'm finding it a nearly hopeless task to try to sort out what your thinking may be in regard to how a Stirling engine operates. I don't mean that to be insulting or anything.
It took me literally years of dedicated study and research to arrive at an understanding of how these engines work. It is not intuitive, yet, it really is not complicated, it just doesn't make sense, in some ways.
For example, a Stirling engine actually, it could be said; "destroys heat".
Heat goes into the engine and goes out of existence. It no longer exists in the form of energy we recognize as sensible heat.
In that sense a Stirling engine is a cryo-cooler. In converting heat into work the engine creates an absence of heat, which we sense as "cold".
Heat being conducted, we understand, convection, radiation, are understandable intuitively. Heat simply disappearing and leaving cold is not a phenomenon we are familiar with in life.
Heat converted into Work is no longer heat but cold, the absence of heat.
So efforts to trace the heat as it passes through the engine are liable to result in confusion.
Quote
displacer is a rigid plate which creates the expansion space by bodily movement, whereas the diaphragm creates the expansion space by deflection.
Either way work must be done against the atmouspheric pressure.
And either way each drives the crank via what I would call a push rod.
I don't know what you mean exactly by "expansion space". I have no idea what you mean by "the diaphragm creates the expansion space by deflection."
As far as this statement: "Either way work must be done against the atmouspheric pressure.
And either way each drives the crank via what I would call a push rod." It is entirely wrong.
The displacer does not drive the crank.
The displacer is not a power piston or diaphragm.
I did not intend this thread to be a course on how a Stirling engine works, I assume if people are interested they would research the subject. But I'll try to clarify things if I can.
But it seems you are thinking of a power piston or diaphragm that drives a crankshaft. The function of a displacer is not to drive the crankshaft. It does not drive the crankshaft. It does not do work against atmospheric pressure.
Basically the displacer is a kind of valve that allows heat into the engine in metered bursts. Heat enters the engine through the hot plate or hot heat exchanger, but not while the displacer is blocking the heat from entering.
The displacer is, rather than a valve, it is more like an umbrella used for shade, to block the heat of the sun.
When the displacer moves out of the way temporarily, some radiant heat is allowed into the engine, expanding the gas to do work.
When the gas has used up it's energy performing work, the displacer moves out of the way again, allowing in another burst of heat for another cycle. In that sense the displacer serves the same function as the spark plugs which ignites that fuel mixture in an internal combustion engine to deliver heat.
Edited by Tom Booth
Typo
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Posted (edited)
On 7/31/2020 at 1:37 AM, Markus Hanke said:
No, the second law follows directly from fundamental considerations of statistical mechanics and some basic maths, and as such is not in contention. It also follows more or less directly from Z2 symmetry and unitarity, so it is fundamentally motivated by manifest symmetry considerations. Essentially, in a world where unitarity holds for quantum systems (as is evidently the case in our universe) you can't not have the second law of thermodynamics, if that makes sense.
I am not sure what the point of all this really is, since the setup you have there is not an isolated system in the thermodynamics sense - so what does it have to do with the second law at all? Note also that the second law is a global statistical statement, and as such it does not contradict temporary (even long-lasting) local decreases in entropy.
What exactly are you attempting to show here?
I think in general what I'm examining or testing is the nature of heat, and how it really interacts with a heat engine, or how it is utilized by a heat engine.
Popular ideas or statements regarding how these engines work and what their potential may be does not square well with my own experience.
For example, above, the statement:
" This system is ~20% efficient because ~80% of the energy goes into heating the cold plate and the reservoir it’s coupled to. "
(Not to pick on Swansont or anyone in particular)
Generally this 80% pass through of heat into the sink is not what happens.
Here is a link to another experiment conducted by someone else years ago using a thermal imaging camera.
As can be seen in this thermal image of a running Stirling engine on a hot cup of water compared with a non operational dummy engine.
In the same environment the real heat engine is clearly maintaining a cold temperature. The heat is not passing through to the top plate.
In my own experiments I found the same situation. The bottom plate above boiling hot water was too hot to touch. I'm sure hot enough to raise a blister. At the same time the top plate remained cool to the touch.
This does not seem to support the common assumption that the majority of heat passes right through the engine.
In my experience virtually no heat makes it through these engines to the sink. Also, nearly all the incidental and parasitic heat loses are preventable to a great degree with proper insulation.
The implications are far reaching, IMO.
Edited by Tom Booth
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2 hours ago, Tom Booth said:
This does not seem to support the common assumption that the majority of heat passes right through the engine.
I’m not an expert on the engineering applications of thermodynamics, so others here are more qualified to address this. I’m more of a theory guy - so in what way do you think this is related to the second law? I don’t see a direct connection, and most certainly not anything that would put the law into question.
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4 hours ago, Tom Booth said:
Here is a link to another experiment conducted by someone else years ago using a thermal imaging camera.
As can be seen in this thermal image of a running Stirling engine on a hot cup of water compared with a non operational dummy engine.
In the same environment the real heat engine is clearly maintaining a cold temperature. The heat is not passing through to the top plate.
How does the image “show” this? A thermal imaging camera captures thermal radiation, which indicates temperature, not heat.
Do you understand what a reservoir is in thermodynamics? This picture is perfectly in accord with the cold plate operating at Tc, exactly as the theory states.
Quote
In my own experiments I found the same situation. The bottom plate above boiling hot water was too hot to touch. I'm sure hot enough to raise a blister. At the same time the top plate remained cool to the touch.
IOW, the thing supposed to be at Th is at Th, and the thing supposed to be at Tc is at Tc.
Hardly an indictment of the theory.
Quote
This does not seem to support the common assumption that the majority of heat passes right through the engine.
It does not support your misunderstanding of thermodynamics.
Heat is not a substance. It does not have a temperature. Whatever you think heat is, it is not what thermodynamics defines as heat. And that stands in the way of any fruitful discussion.
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Posted (edited)
4 hours ago, Tom Booth said:
This does not seem to support the common assumption that the majority of heat passes right through the engine.
Heat never passes through a heat engine.
That is a popular metaphor only.
There is another popular metaphor of the bank account,
but this is also flawed
Amounts of heat enter and leave any system by various routes.
But within the system the heat becomes something else (internal energy), so it is not the same heat leaving as entering.
This is where it differs from the bank account analogy.
In the bank account different dollars enter and leave but this is of no account (pun intended) since they are still dollars within the bank and all dollars are equivalent.
5 hours ago, Tom Booth said:
I don't know what you mean exactly by "expansion space". I have no idea what you mean by "the diaphragm creates the expansion space by deflection."
As far as this statement: "Either way work must be done against the atmouspheric pressure.
And either way each drives the crank via what I would call a push rod." It is entirely wrong.
The displacer does not drive the crank.
The displacer is not a power piston or diaphragm.
I did not intend this thread to be a course on how a Stirling engine works, I assume if people are interested they would research the subject. But I'll try to clarify things if I can.
Sigh
Then the appropriate response would be to ask, would it not ?
I don't know how to converse with someone who seems to believe that things can expand without work being done, within our atmousphere.
I find Wikipedia articles normally a bit over the top to recommend but in this case they seem to have it right.
Quote
### Stirling engine - Wikipedia
A Stirling engine is a heat engine that is operated by a cyclic compression and expansion of air or other gas (the working fluid) at different temperatures, ...
Edited by studiot
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6 minutes ago, studiot said:
Heat never passes through a heat engine.
That is a popular metaphor only.
Put another way, the diagrams with Q’s and W are schematics, not maps. They show a concept, not a physical path.
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Posted (edited)
6 hours ago, Markus Hanke said:
I’m not an expert on the engineering applications of thermodynamics, so others here are more qualified to address this. I’m more of a theory guy - so in what way do you think this is related to the second law? I don’t see a direct connection, and most certainly not anything that would put the law into question.
Years ago I designed a heat engine. As a lifelong engine mechanic and repair person I thought it was a great design.
I was told it would not work because it violated the second law.
Now I'm doing experiments with model heat engines. I posted results on another forum. The discussion was locked and I was banned, because the moderators determined it was "a perpetual motion machine of the second kind" and that it violated the second law. (Or would if it was real).
How does it violate the second law? is actually MY question!
I'm not saying it does, that is what OTHER PEOPLE have been throwing in my face for the past 25 years.
If there is no such violation that's wonderful news to me!
Edited by Tom Booth
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BTW, I do have a question.
My original engine design works similar to and/or incorporates heat pump/refrigeration.
In thermodynamics, is a vapor compression system, (refrigerators, heat pumps, freezers etc ) considered an "open system" or a "closed system", so does the second law apply or not?
The "working fluid" is contained in a closed loop, but part of the "system" is the heat exchangers and the air being blown through them.
Some say the second law applies to everything because the whole universe is a closed system, so... How exactly do we determine if a system is open or closed so we know if the second law is supposed to apply or not.
It all seems vaguely philosophical to me with no clear cut answer.
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Posted (edited)
On 7/31/2020 at 9:39 AM, Tom Booth said:
Specifically, this second law, in its early formulations seems very slippery, and I don't have a clear picture as to, does it apply to this "open system" or not.
Now that I have some finances available, I've just gone and purchased six model engines to run some common sense tests and experiments to see if, in principle, my engine design is impossible or not.
The first order of business, I thought, should be to settle this Caloric theory that a heat engine works by heat flowing through it from a hot reservoir THROUGH to a cold reservoir.
Blocking the path or outlet from the engine is, according to Carnot's theory equivalent to stopping up the outlet of a fluid turbine. Without an outlet for the heat, according to Carnot, the engine should quit.
In my experiment however, insulating the path or outlet of the model Stirling engine had the opposite result. The engine, rather than overheating and stopping or something, ran measurably faster and longer.
I would suggest that the silicon inside of the tube is just acting like an oscillating diaphragm which is just increasing the intensity of the changes in volume of the air. It would be like taking a slinky and putting it on a vibrator, but then it creates bigger waves at the other end from the smaller waves being added together. Then having a better seal would increase performance. It would make the diaphragm react better to the changes in density created from the heat traveling to the other colder side of the tube which has the piston. The heat may just be traveling to the other chamber in steps or burst to make the silicon resonate. You may try using different materials there and see how that affects the performance.
If I was you, the next step I would take would be attaching an AC unit to the Stirling engine in an up scaled model to improve performance. You could have the hot side chamber have the compressed freon, and you could try to seal the piston chamber with cold freon. If that was too difficult, you could simply have the cold air of the AC to blow on the piston to make it run as long as it has power to start out with. Basically, it should at least just have a radiator to functionally drive something.
If I am correct, it may not matter which side is hot or cold. It may be better to have the piston be the hot side. It would have friction which would naturally cause heat to make it more energy saving and efficient. That could make a big difference when trying to scale it up to actually run something which needs a piston driven motor.
On second thought, it may not actually be a good idea to have heat driven pistons, because it would cause the engine to overheat to where it could have mechanical failures. That is the reason why combustible engines use a radiator to begin with.
Edited by Conjurer
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Great ideas! I'll get to work on that right away! LOL
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Posted (edited)
11 hours ago, studiot said:
I don't know how to converse with someone who seems to believe that things can expand without work being done, within our atmousphere.
I was just trying to clarify that "work" is not done to expand the gas, or to work against atmospheric pressure BY THE DISPLACER.
The displacer is not any kind of power piston.
Imagine an empty soup can. Completely sealed, containing a gas. Inside the can is another smaller loose fitting soup can.
If you tip the can on one end, gravity will cause the inside can to fall to the bottom so the air in the can will be DISPLACED to the top.
Tip the can around the other way, again the can inside falls and DISPLACES the air back to the opposite end.
Now heat one end of the can.
Do the same flip flop and the inside can, or "displacer", still DISPLACES the air in the can but now, when the air is displaced to the hot side the air gets hot and the pressure in the can increases.
Tip the can the other way and the air is displaced to the cold side, cools down and the pressure in the can drops.
Hopefully, it can be seen that the small can inside is not performing any appreciable work to turn a crankshaft or to work against the pressure of the outside atmosphere.
It is just a means of periodically delivering heat to the inside of the can to heat the air and create pressure. It does this by moving away from the heated end of the can, displacing the air to the hot end, or put another way, simply uncovering the hot heat exchanger.
Often, in a Stirling engine the displacer is not attached to the crank at all. It may be moved back and forth by other means, such as a magnet., Or in some cases by actually tilting the engine.
Here the displacer is some marbles in a test tube. They full back and forth by gravity, have no connection to any crank and do not work against atmospheric pressure. They just displace air to the heated side of the engine periodically.
Edited by Tom Booth
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Posted (edited)
4 hours ago, Tom Booth said:
BTW, I do have a question.
My original engine design works similar to and/or incorporates heat pump/refrigeration.
In thermodynamics, is a vapor compression system, (refrigerators, heat pumps, freezers etc ) considered an "open system" or a "closed system", so does the second law apply or not?
The "working fluid" is contained in a closed loop, but part of the "system" is the heat exchangers and the air being blown through them.
Some say the second law applies to everything because the whole universe is a closed system, so... How exactly do we determine if a system is open or closed so we know if the second law is supposed to apply or not.
It all seems vaguely philosophical to me with no clear cut answer.
Questions are fine.
In both Fluid Mechanics and Thermodynamics systems are sometimes analysed by (theoretically) replacing some part of or all the system with an equivalent one which is easier to work with.
This can be true of refrigeration cycles and compressors, turbines and jet engines.
This is a practical engineer's approach.
To help with this and some other questions you haven't yet asked here is a link to an excellent engineering glossary of Thermodynamics
Look particularly at "Air standard assumptions" and "Air standard cycle"
Edited by studiot
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4 hours ago, Tom Booth said:
In thermodynamics, is a vapor compression system, (refrigerators, heat pumps, freezers etc ) considered an "open system" or a "closed system", so does the second law apply or not?
The second law always applies. Refrigeration cycles are open systems, in that the heat of condensation of the refrigerant is transferred to the ambient air. IOW the entropy of the universe increases from a refrigeration cycle.
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11 minutes ago, Tom Booth said:
I was just trying to clarify that "work" is not done to expand the gas, or to work against atmospheric pressure BY THE DISPLACER.
It’s not a vacuum, right? So work is being done against atmospheric pressure, i.e. there is a pressure on the plunger. There is a difference between expansion in those two conditions
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2 minutes ago, swansont said:
It’s not a vacuum, right? So work is being done against atmospheric pressure, i.e. there is a pressure on the plunger. There is a difference between expansion in those two conditions
What?
I have no idea what you're driving at.
What plunger? What "two conditions?"
What does a vacuum have to do with anything?
How is work being done against outside atmospheric pressure by those marbles rolling back and forth.
Yes their rolling away from the heat source causes the displacement of the air which results in increased pressure in the test tube, but the pressure has no effect on the marbles, they are isolated from the atmosphere.
The DISPLACER has ample clearance around it and air moves freely around it, it does not form any seal or work against any pressure.
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10 minutes ago, Tom Booth said:
What?
I have no idea what you're driving at.
That’s obvious.
Quote
What plunger? What "two conditions?"
Vacuum and atmosphere
Quote
What does a vacuum have to do with anything?
You claim work is not being done against atmosphere. it’s the only other option.
Quote
How is work being done against outside atmospheric pressure by those marbles rolling back and forth.
That’s not what I said, or what’s going on.
Quote
Yes their rolling away from the heat source causes the displacement of the air which results in increased pressure in the test tube, but the pressure has no effect on the marbles, they are isolated from the atmosphere.
But you keep saying work is not being done against atmosphere, and that’s not right. (You said “I was just trying to clarify that "work" is not done to expand the gas”)
The fact that you are focused on the marbles suggests you aren’t getting what other people are talking about.
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zbMATH — the first resource for mathematics
Degree bounds for the division problem in polynomial ideals. (English) Zbl 0691.12010
Let $$f_ 1,...,f_ m$$ be polynomials in n variables of degree at most D over an arbitrary field, and let I be the ideal they generate. The author gives an estimate of the smallest integer $$\delta$$ such that for any $$P\in I$$, we can write $$P=f_ 1g_ 1+...+f_ mg_ m$$ with deg $$g_ i\leq \delta$$.
Reviewer: Xu Yonghua
MSC:
12E05 Polynomials in general fields (irreducibility, etc.) 13F20 Polynomial rings and ideals; rings of integer-valued polynomials
Keywords:
degree of polynomial
Full Text:
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zbMATH — the first resource for mathematics
Briand, Philippe
Compute Distance To:
Author ID: briand.philippe Published as: Briand, Ph.; Briand, Philippe
Documents Indexed: 20 Publications since 1994
all top 5
Co-Authors
4 single-authored 7 Hu, Ying 3 Confortola, Fulvia 3 Delyon, Bernard 3 Mémin, Jean 2 Elie, Romuald 1 Carmona, René A. 1 Coquet, François 1 Labart, Céline 1 Lepeltier, Jean-Pierre 1 Pardoux, Etienne 1 Peng, Shige 1 San Martín, Jaime 1 Stoica, Lucreţiu
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Serials
4 Stochastic Processes and their Applications 2 Probability Theory and Related Fields 2 The Annals of Applied Probability 2 Séminaire d’Analyse. Université Blaise Pascal. Clermont II 2 Electronic Communications in Probability 1 Applied Mathematics and Optimization 1 Journal of Functional Analysis 1 Journal of Applied Mathematics and Stochastic Analysis 1 Comptes Rendus de l’Académie des Sciences. Série I 1 Stochastics and Stochastics Reports 1 Electronic Journal of Probability 1 Bernoulli 1 Nonlinear Analysis. Theory, Methods & Applications
Fields
19 Probability theory and stochastic processes (60-XX) 7 Partial differential equations (35-XX) 2 Numerical analysis (65-XX) 2 Systems theory; control (93-XX) 1 Game theory, economics, finance, and other social and behavioral sciences (91-XX)
Citations contained in zbMATH
16 Publications have been cited 697 times in 453 Documents Cited by Year
$$L^p$$ solutions of backward stochastic differential equations. Zbl 1075.65503
Briand, Ph.; Delyon, B.; Hu, Y.; Pardoux, E.; Stoica, L.
2003
BSDE with quadratic growth and unbounded terminal value. Zbl 1109.60052
Briand, Philippe; Hu, Ying
2006
Quadratic BSDEs with convex generators and unbounded terminal conditions. Zbl 1141.60037
Briand, Philippe; Hu, Ying
2008
A converse comparison theorem for BSDEs and related properties of $$g$$-expectation. Zbl 0966.60054
Briand, Philippe; Coquet, François; Hu, Ying; Mémin, Jean; Peng, Shige
2000
Stability of BSDEs with random terminal time and homogenization of semilinear elliptic PDEs. Zbl 0912.60081
Briand, Philippe; Hu, Ying
1998
Donsker-type theorem for BSDEs. Zbl 0977.60067
Briand, Philippe; Delyon, Bernard; Mémin, Jean
2001
BSDEs with polynomial growth generators. Zbl 0979.60046
Briand, Philippe; Carmona, René
2000
Besides with stochastic Lipschitz condition and quadratic PDEs in Hilbert spaces. Zbl 1136.60337
Briand, Philippe; Confortola, Fulvia
2008
On the robustness of backward stochastic differential equations. Zbl 1058.60041
Briand, Philippe; Delyon, Bernard; Mémin, Jean
2002
A simple constructive approach to quadratic BSDEs with or without delay. Zbl 1294.60092
Briand, Philippe; Elie, Romuald
2013
Simulation of BSDEs by Wiener chaos expansion. Zbl 1311.60077
Briand, Philippe; Labart, Céline
2014
One-dimensional backward stochastic differential equations whose coefficient is monotonic in $$y$$ and non-Lipschitz in $$z$$. Zbl 1129.60057
Briand, Philippe; Lepeltier, Jean-Pierre; San Martín, Jaime
2007
Quadratic BSDEs with random terminal time and elliptic PDEs in infinite dimension. Zbl 1191.60071
Confortola, Fulvia; Briand, Philippe
2008
Probabilistic approach to singular perturbations of semilinear and quasilinear parabolic PDEs. Zbl 0922.60055
Briand, Philippe; Hu, Ying
1999
Differentiability of backward stochastic differential equations in Hilbert spaces with monotone generators. Zbl 1147.60318
Briand, Philippe; Confortola, Fulvia
2008
BSDEs with mean reflection. Zbl 1391.60133
Briand, Philippe; Elie, Romuald; Hu, Ying
2018
BSDEs with mean reflection. Zbl 1391.60133
Briand, Philippe; Elie, Romuald; Hu, Ying
2018
Simulation of BSDEs by Wiener chaos expansion. Zbl 1311.60077
Briand, Philippe; Labart, Céline
2014
A simple constructive approach to quadratic BSDEs with or without delay. Zbl 1294.60092
Briand, Philippe; Elie, Romuald
2013
Quadratic BSDEs with convex generators and unbounded terminal conditions. Zbl 1141.60037
Briand, Philippe; Hu, Ying
2008
Besides with stochastic Lipschitz condition and quadratic PDEs in Hilbert spaces. Zbl 1136.60337
Briand, Philippe; Confortola, Fulvia
2008
Quadratic BSDEs with random terminal time and elliptic PDEs in infinite dimension. Zbl 1191.60071
Confortola, Fulvia; Briand, Philippe
2008
Differentiability of backward stochastic differential equations in Hilbert spaces with monotone generators. Zbl 1147.60318
Briand, Philippe; Confortola, Fulvia
2008
One-dimensional backward stochastic differential equations whose coefficient is monotonic in $$y$$ and non-Lipschitz in $$z$$. Zbl 1129.60057
Briand, Philippe; Lepeltier, Jean-Pierre; San Martín, Jaime
2007
BSDE with quadratic growth and unbounded terminal value. Zbl 1109.60052
Briand, Philippe; Hu, Ying
2006
$$L^p$$ solutions of backward stochastic differential equations. Zbl 1075.65503
Briand, Ph.; Delyon, B.; Hu, Y.; Pardoux, E.; Stoica, L.
2003
On the robustness of backward stochastic differential equations. Zbl 1058.60041
Briand, Philippe; Delyon, Bernard; Mémin, Jean
2002
Donsker-type theorem for BSDEs. Zbl 0977.60067
Briand, Philippe; Delyon, Bernard; Mémin, Jean
2001
A converse comparison theorem for BSDEs and related properties of $$g$$-expectation. Zbl 0966.60054
Briand, Philippe; Coquet, François; Hu, Ying; Mémin, Jean; Peng, Shige
2000
BSDEs with polynomial growth generators. Zbl 0979.60046
Briand, Philippe; Carmona, René
2000
Probabilistic approach to singular perturbations of semilinear and quasilinear parabolic PDEs. Zbl 0922.60055
Briand, Philippe; Hu, Ying
1999
Stability of BSDEs with random terminal time and homogenization of semilinear elliptic PDEs. Zbl 0912.60081
Briand, Philippe; Hu, Ying
1998
all top 5
Cited by 414 Authors
33 Fan, Shengjun 27 Jiang, Long 25 Hu, Ying 15 Klimsiak, Tomasz 14 Tang, Shanjian 11 Peng, Shige 10 Richou, Adrien 9 Gobet, Emmanuel 9 Hu, Feng 9 Hu, Mingshang 9 Jia, Guangyan 9 Popier, Alexandre 9 Possamaï, Dylan 9 Yong, Jiongmin 8 Briand, Philippe 8 Chen, Zengjing 8 Yao, Song 7 dos Reis, Gonçalo 7 Imkeller, Peter 7 Liang, Gechun 7 Wang, Falei 6 Bahlali, Khaled 6 Confortola, Fulvia 6 Essaky, El Hassan 6 Frei, Christoph 6 Wu, Zhen 5 Cheridito, Patrick 5 Delbaen, Freddy 5 Kruse, Thomas 5 Luo, Peng 5 Matoussi, Anis 5 Ouknine, Youssef 5 Pardoux, Etienne 5 Rozkosz, Andrzej 5 Stadje, Mitja 5 Touzi, Nizar 5 Wang, Tianxiao 5 Xu, Mingyu 5 Zong, Zhaojun 4 Aman, Auguste 4 Chassagneux, Jean-François 4 Geiss, Christel 4 Hassani, Mohammed 4 Jentzen, Arnulf 4 Ji, Shaolin 4 Labart, Céline 4 Lin, Qian 4 Lin, Yiqing 4 Masiero, Federica 4 Morlais, Marie-Amelie 4 Nie, Tianyang 4 Rosazza Gianin, Emanuela 4 Russo, Francesco 4 Tangpi, Ludovic 4 Tian, Dejian 4 Turkedjiev, Plamen 4 Xiao, Lishun 4 Xing, Hao 4 Zhou, Chao 3 Ankirchner, Stefan 3 Bayraktar, Erhan 3 Chen, Shaokuan 3 Cohen, Samuel N. 3 Crisan, Dan O. 3 Davison, Matt 3 E, Weinan 3 El Otmani, Mohamed 3 Elliott, Robert James 3 Geiss, Stefan 3 Heyne, Gregor 3 Kharroubi, Idris 3 Kupper, Michael 3 Li, Juan 3 Li, Shoumei 3 Lim, Thomas 3 Lionnet, Arnaud 3 Ma, Jin 3 Mastrolia, Thibaut 3 Mémin, Jean 3 Pham, Huyên 3 Qian, Zhongmin M. 3 Qiu, Jinniao 3 Réveillac, Anthony 3 Słomiński, Leszek 3 Warin, Xavier 3 Wu, Helin 3 Yang, Shuzhen 3 Zhang, Jianfeng 3 Zhang, Na 3 Zhang, Qi 3 Zheng, Shiqiu 3 Zhou, Jianjun 2 Al-Hussein, Abdulrahman 2 Bender, Christian 2 Blache, Fabrice 2 Bouchard, Bruno 2 Buckdahn, Rainer 2 Campi, Luciano 2 Ceci, Claudia 2 Chen, Tao ...and 314 more Authors
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Cited in 99 Serials
77 Stochastic Processes and their Applications 26 Statistics & Probability Letters 16 The Annals of Applied Probability 15 Journal of Mathematical Analysis and Applications 14 Comptes Rendus. Mathématique. Académie des Sciences, Paris 14 Stochastics and Dynamics 14 Stochastics 13 Acta Mathematicae Applicatae Sinica. English Series 11 SIAM Journal on Control and Optimization 11 Acta Mathematica Sinica. English Series 10 Stochastic Analysis and Applications 10 Finance and Stochastics 9 Probability Theory and Related Fields 9 Journal of Theoretical Probability 8 The Annals of Probability 8 Applied Mathematics and Optimization 8 Electronic Journal of Probability 7 Journal of Functional Analysis 7 Insurance Mathematics & Economics 6 Bernoulli 6 SIAM Journal on Financial Mathematics 6 Mathematical Control and Related Fields 5 Potential Analysis 5 Bulletin des Sciences Mathématiques 5 Mathematics and Financial Economics 5 Science China. Mathematics 5 Probability, Uncertainty and Quantitative Risk 4 Random Operators and Stochastic Equations 4 Electronic Communications in Probability 4 Mathematical Finance 4 European Series in Applied and Industrial Mathematics (ESAIM): Probability and Statistics 4 International Journal of Theoretical and Applied Finance 3 Journal of Differential Equations 3 Journal of Applied Mathematics and Stochastic Analysis 3 Annales de l’Institut Henri Poincaré. Probabilités et Statistiques 3 European Series in Applied and Industrial Mathematics (ESAIM): Control, Optimization and Calculus of Variations 3 Abstract and Applied Analysis 3 Methodology and Computing in Applied Probability 3 Journal of Applied Mathematics and Computing 3 Advances in Difference Equations 2 Advances in Applied Probability 2 International Journal of Control 2 Mathematics of Computation 2 Applied Mathematics and Computation 2 Archiv der Mathematik 2 Automatica 2 Journal of Computational and Applied Mathematics 2 Journal of Optimization Theory and Applications 2 SIAM Journal on Numerical Analysis 2 Transactions of the American Mathematical Society 2 Chinese Annals of Mathematics. Series B 2 Optimization 2 SIAM Journal on Scientific Computing 2 NoDEA. Nonlinear Differential Equations and Applications 2 Monte Carlo Methods and Applications 2 Applied Mathematical Finance 2 Discrete and Continuous Dynamical Systems 2 Journal of Inequalities and Applications 2 Infinite Dimensional Analysis, Quantum Probability and Related Topics 1 Applicable Analysis 1 Journal of the Franklin Institute 1 Studia Mathematica 1 Ukrainian Mathematical Journal 1 Prikladnaya Matematika i Mekhanika 1 Acta Mathematica Vietnamica 1 Annali di Matematica Pura ed Applicata. Serie Quarta 1 Journal of Applied Probability 1 Journal of Statistical Planning and Inference 1 Numerical Functional Analysis and Optimization 1 Bulletin of the Korean Mathematical Society 1 Annales de l’Institut Henri Poincaré. Analyse Non Linéaire 1 Applied Numerical Mathematics 1 MCSS. Mathematics of Control, Signals, and Systems 1 Journal of Scientific Computing 1 Forum Mathematicum 1 Science in China. Series A 1 SIAM Journal on Mathematical Analysis 1 Proceedings of the Indian Academy of Sciences. Mathematical Sciences 1 Journal of Nonlinear Science 1 Applied Mathematics. Series B (English Edition) 1 Mathematical Methods of Operations Research 1 Discrete Dynamics in Nature and Society 1 Communications of the Korean Mathematical Society 1 Journal of Dynamical and Control Systems 1 Brazilian Journal of Probability and Statistics 1 Journal of Evolution Equations 1 Quantitative Finance 1 Discrete and Continuous Dynamical Systems. Series B 1 Journal of Systems Science and Complexity 1 Communications on Pure and Applied Analysis 1 African Diaspora Journal of Mathematics 1 European Series in Applied and Industrial Mathematics (ESAIM): Mathematical Modelling and Numerical Analysis 1 Frontiers of Mathematics in China 1 International Journal of Stochastic Analysis 1 Arabian Journal of Mathematics 1 Communications in Mathematics and Statistics 1 Nonlinear Analysis. Theory, Methods & Applications 1 Modern Stochastics. Theory and Applications 1 European Series in Applied and Industrial Mathematics (ESAIM): Proceedings and Surveys
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Cited in 25 Fields
424 Probability theory and stochastic processes (60-XX) 111 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 79 Systems theory; control (93-XX) 75 Partial differential equations (35-XX) 51 Numerical analysis (65-XX) 41 Calculus of variations and optimal control; optimization (49-XX) 20 Ordinary differential equations (34-XX) 7 Statistics (62-XX) 6 Operations research, mathematical programming (90-XX) 5 Dynamical systems and ergodic theory (37-XX) 5 Global analysis, analysis on manifolds (58-XX) 3 Real functions (26-XX) 2 Difference and functional equations (39-XX) 2 Approximations and expansions (41-XX) 2 Integral equations (45-XX) 2 Fluid mechanics (76-XX) 1 Topological groups, Lie groups (22-XX) 1 Measure and integration (28-XX) 1 Functions of a complex variable (30-XX) 1 Potential theory (31-XX) 1 Functional analysis (46-XX) 1 Operator theory (47-XX) 1 Computer science (68-XX) 1 Mechanics of particles and systems (70-XX) 1 Biology and other natural sciences (92-XX)
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Square Matrix: A matrix is said to be square if the number of rows is equal to the number of columns. Every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. Minors of a Square Matrix The minor $$M_{ij}$$ of an n × n square matrix corresponding to the element $$(A)_{ij}$$ is the determinant of the matrix (n-1) × (n-1) matrix obtained by deleting row i and column j of matrix A. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. When you have more than one matrix⦠For example matrices with dimensions of 2x2, 3x3, 4x4, 5x5 etc., are referred to as square matrix. A matrix in which the number of rows is equal to the number of columns is said to be a square matrix. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. If A is a symmetric matrix, then A = A T and if A is a skew-symmetric matrix then A T = â A.. Also, read: Before we look at what the trace of a matrix is, let's first define what the main diagonal of a square matrix is. The Trace of a Square Matrix. For example, $$A =\begin{bmatrix} 3 & -1 & 0\\ 3/2 & â3/2 & 1\\4 & 3 & -1\end{bmatrix}$$ is a square matrix of order 3. Square Matrix Calculator. But the difference between them is, the symmetric matrix is equal to its transpose whereas skew-symmetric matrix is a matrix whose transpose is equal to its negative.. Example A symmetric matrix and skew-symmetric matrix both are square matrices. Example A square matrix as sum of symmetric and skew-symmetric matrix ? Characteristic equation of matrix : Here we are going to see how to find characteristic equation of any matrix with detailed example. You have lost information. 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Some important Conclusions on Symmetric and Skew-Symmetric Matrices: If A is any square matrix, then A + Aâ is a symmetric matrix and A â Aâ is a skew-symmetric matrix. These matrices basically squash things to a lower dimensional space. $B = \left[ {\begin{array}{*{20}{c}} 1&3&4 \\ 5&2&4 \\ 1&9&6 \end{array}} \right]$ Diagonal Matrix: A square matrix is said to be diagonal if at least one element of principal diagonal is non-zero and all the other elements are zero. 3.1.1 Introduction More than one explanatory variable In the foregoing chapter we considered the simple regression model where the dependent variable is related to one explanatory variable. e.g. 3.1 Least squares in matrix form E Uses Appendix A.2âA.4, A.6, A.7. Addition Examples; Square Matrix Multiplication; Matrix Definition. e.g. The singular, matrix, is used when dealing with one matrix, like this: 2-8 0 1 5-13. Note: A square matrix A is a skew-symmetric matrix Aâ = -A. 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Is to use minors and cofactors of a matrix is basically a of..., symbols or expressions in rows ( across ) and columns ( up and down.... If the number of rows is equal to the number of rows is to! Sum of symmetric and skew-symmetric matrix Aâ = -A a lower dimensional space: a square matrix a a. Here we are going to see how square matrix example find characteristic equation of any matrix with detailed example symmetric and! One possibility to calculate the determinant of a matrix is basically a matrix..., 5x5 etc., are referred to as square matrix a is a skew-symmetric matrix and of. We are going to see how to find characteristic equation of any matrix with detailed example array.: 2-8 0 1 5-13 to as square matrix as sum of a matrix. Cofactors of a square matrix of same order can be formed as: a matrix which contains same number rows... Basically squash things to a lower dimensional space ) and columns symmetric and... 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Matrix: a square matrix a is a skew-symmetric matrix both are square matrices are! Add or multiply any two square matrices a square matrix down ) matrix Multiplication ; Definition... And skew-symmetric matrix multiple of an identity matrix Aâ = -A any matrix! Etc., are referred to as square matrix: Here we are going to see to... Rows ( across ) and columns ( up and down ): Here we are going to see how find. And cofactors of a symmetric matrix and eigenvector matrix can be uniquely expressed as the sum of matrix... Find characteristic equation of matrix square matrices that are of the same order Definition Let! Equal to the number of columns is said to be a unit matrix of order n n! Scalar matrix is a skew-symmetric matrix matrix Definition are equal x n and I be a square matrix can uniquely... Like this: 2-8 0 1 5-13 words we can say that a scalar matrix is a... ; matrix Definition ) and columns ( up and down ) symbols or expressions in rows ( )... For example matrices with dimensions of 2x2, 3x3, 4x4, 5x5 etc., are referred to as matrix. Matrix can be uniquely expressed as the sum of a square matrix: Here are! Here we are going to see how to find characteristic equation of matrix! A be any square matrix whose all off-diagonal elements are zero and all on-diagonal elements are zero all...: a square matrix Multiplication ; matrix Definition matrix both are square matrices are of the same order squash to. Dimensions of 2x2, 3x3, 4x4, 5x5 etc., are referred as! One possibility to calculate the determinant of a symmetric matrix and skew-symmetric matrix are.
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## fugacity coefficient, $$\phi$$
https://doi.org/10.1351/goldbook.F02544
Ratio of fugacity to the partial pressure of a gaseous constituent.
Sources:
Green Book, 2nd ed., p. 50 [Terms] [Book]
PAC, 1994, 66, 533. (Standard quantities in chemical thermodynamics. Fugacities, activities and equilibrium constants for pure and mixed phases (IUPAC Recommendations 1994)) on page 539 [Terms] [Paper]
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# Tags
A tag is a keyword or label that categorizes your question with other, similar questions. Using the right tags makes it easier for others to find and answer your question.
for questions about etale cohomology of schemes, including foundational material and applications.
644 questions
Lattices in the sense of discrete subgroups of Euclidean spaces, as used in number theory, discrete geometry, Lie groups, etc. (Not to be confused with lattice theory or lattices as used in physics! F…
638 questions
Questions about geometric properties of sets using measure theoretic techniques; rectifiability of sets and measures, currents, Plateau problem, isoperimetric inequality and related topics.
636 questions
A Hilbert space $H$ is a real or complex vector space endowed with an inner product such that $H$ is a complete metric space when endowed with the norm induced by this inner product.
628 questions
Riemann surfaces(Riemannian surfaces) is one dimensional complex manifold. For questions about classical examples in complex analysis, complex geometry, surface topology.
627 questions
Birational geometry is a field of algebraic geometry the goal of which is to determine when two algebraic varieties are isomorphic outside lower-dimensional subsets. This amounts to studying mappings …
612 questions
for questions about deformation theory, including deformations of manifolds, schemes, Galois representations, and von Neumann algebras.
603 questions
Questions on the subject additive combinatorics, also known as arithmetic combinatorics, such as questions on: additive bases, sum sets, inverse sum set theorems, sets with small doubling, Sidon sets,…
597 questions
The Riemann zeta function is the function of one complex variable $s$ defined by the series $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ when $\operatorname{Re}(s)>1$. It admits a meromorphic continuati…
588 questions
For questions involving the concept of convexity
585 questions
585 questions
For questions on modules over rings.
580 questions
For question borderline with, or having application to, computer science. Consider also posting http://cs.stackexchange.com/ or http://cstheory.stackexchange.com/ instead of here, if appropriate.
579 questions
Vertex colouring, Edge Colouring, List Colouring, Fractional Chromatic Number and other variants of graph colouring problems are all on topic.
577 questions
576 questions
A three-manifold is a space that locally looks like Euclidean three-dimensional space
572 questions
Descriptive Set Theory is the study of definable subsets of Polish spaces, where definable is taken to mean from the Borel or projective hierarchies. Other topics include infinite games and determinac…
571 questions
A topos is a category that behaves very much like the category of sets and possesses a good notion of localization. Related to topos are: sheaves, presheaves, descent, stacks, localization,...
568 questions
Theoretical and experimental aspects of information theory and coding theory. This tag covers but is not limited to following branches: information theory, information geometry, optimal transportation…
556 questions
Nonlinear objectives, nonlinear constraints, non-convex objective, non-convex feasible region.
553 questions
The term Galois representation is frequently used when the G-module is a vector space over a field or a free module over a ring, but can also be used as a synonym for G-module. The study of Galois mod…
548 questions
Approximation theory is concerned with how functions can best be approximated with simpler functions, and with quantitatively characterizing the errors introduced thereby.
541 questions
for questions about von Neumann algebras, that is, weak operator topology closed, unital, *-subalgebras of bounded operators on a Hilbert space.
535 questions
A semigroup is a set $S$ together with a binary operation that is associative. Examples of semigroups are the set of finite strings over a fixed alphabet (under concatenation) and the positive integer…
534 questions
A Hopf algebra is a vector space $H$ over a field $k$ endowed with an associative product $\times:H\otimes_k H\to H$ and a coassociative coproduct $\Delta:H\to H\otimes_k H$ which is a morphism of alg…
527 questions
If it turns out that a problem is equivalent to a known open problem, then the open-problem tag is added. After that, the question essentially becomes, "What is known about this problem? What are some…
522 questions
Fields as algebraic objects. For vector and tensor fields, use eg. [dg.differential-geometry]. For physical fields, use eg. [mp.mathematical-physics] or [quantum-field-theory].
520 questions
For questions requesting examples of a certain structure or phenomenon
520 questions
Questions about algebraic structures known as quantum groups, and their categories of representations. Quasitriangular Hopf algebras and their Drinfel'd twists, triangular Hopf algebras, $C^\star$ qua…
518 questions
Stable homotopy theory is that part of homotopy theory (and thus algebraic topology) concerned with all structure and phenomena that remain after sufficiently many applications of the suspension funct…
499 questions
Singularities in algebraic/complex/differential geometry and analysis of ODEs/PDEs. Singular spaces, vector fields, etc.
497 questions
A manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighbourhood that is homeomorphic to the Euclid…
493 questions
491 questions
Questions on the calculus of variations, which deals with the optimization of functionals mostly defined on infinite dimensional spaces.
490 questions
Algebraic and geometric theory of quadratic forms and symmetric bilinear forms, e.g., values attained by quadratic forms, isotropic subspaces, the Witt ring, invariants of quadratic forms, the discrim…
489 questions
Questions related to permutations, bijections from a finite (or sometimes infinite) set to itself.
485 questions
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# A Newsletter Domain-Specific-Language with Org-Mode
## Attention Conservation Notice
An attempt at creating the building blocks necessary for a simple newsletter generating domain-specific-language in org-mode. Not very useful unless you are building a newsletter of sufficient complexity (or are sufficiently stubborn to turn everything into plain text instead of using a GUI).
## Preamble
I have been writing a quarterly newsletter for the past couple of quarters. To be more communal, I made the mistake of using $DOMINANT_OS_DISTRIBUTOR's $OFFICE_SOFTWARE newsletter maker $ONE_WHO_PUBLISHES since the folks I write the newsletter with don't code. ### Difficulties with $ONE_WHO_PUBLISHES
The program provides an area with which to drag and drop different elements. In (my) practice this amounted to dragging squares all over the place, many of which overlap and obscure each other. When using more than 3+ replicas of a single component style, making changes became cumbersome. An update on one component would mean having to manually update the others. Since all the components share a single workspace, changing one item causes shifts in the placement of other items. With 10+ matching components, this became very time consuming. I'd estimate that I spent more time fiddling with components, moving things out of the way, and moving them back than I did making/adding content. The resulting newsletter ended up being both time-consuming to make and inconsistent in presentation.
### A new hope
Being the libre-loving code-monkey that I am, my first instinct was to:
• never do that again
• write out something that can be version-controlled and will be consistent
• escape the clutches of a software ecosystem that may not be relevant in $$n$$ years (as technology release velocity increases, $$n \to 0$$) in favor of plain-text and open formats
### Goals for the End Product
• It needs to be usable by non-programmers
• The build system should not require any special skills or arcane terminal usage
• The language itself should be pretty straightforward
• extending and adding functionality should be simple
## Final Result (Code + Stylesheet)
This is the code that is written by the newsletter writer
#+BEGIN_PROPERTIES
#+TITLE: @@html:<p class="title-p"><span class="title-left">Q3</span><span class="title-right">Newsletter Update</span></p>@@
#+SUBTITLE: @@html:<p class="subtitle-p"><span class="vol-issue-left">Vol. 1, Issue 3</span><span class="vol-issue-right">30Sep2022</span></p><hr>@@
#+DATE: <2022-07-17 Sun>
#+OPTIONS: num:nil author:nil date:nil html-postamble:nil
#+MACRO: split-sentence @@html: <span class="date-str-date"><code>$1</code></span> | <span class="date-str-str">$2</span><br>@@
#+MACRO: h3 @@html: <span class="in-column-h3">$1</span><br>@@ #+MACRO: hr #+HTML: <hr> #+FOOTER: #+END_PROPERTIES * Welcome! {{{hr}}} *Here is some text welcoming you!* Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It was popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum. * Our Main News Story {{{hr}}} #+begin_banner ./images/nscrop.png #+end_banner Lorem ipsum dolor sit amet, consectetur adipiscing elit. Aliquam mattis orci ultrices enim gravida, vitae venenatis libero faucibus. Donec at pulvinar sem, ut lobortis elit. Mauris eget pretium nunc, pretium consequat tellus. Ut justo augue, iaculis id efficitur ullamcorper, suscipit sit amet sapien. In pulvinar id mauris vel viverra. Duis vitae nunc consequat, iaculis eros id, cursus nisi. Cras vitae sem metus. Duis lobortis porta dui, vel volutpat turpis pretium vitae. Cras aliquet lectus eu felis lacinia euismod. Maecenas eros velit, pellentesque lacinia ipsum id, imperdiet cursus leo. Nullam pharetra blandit quam, ut ornare risus gravida quis. Nullam hendrerit et erat et faucibus. Aenean vitae sagittis mi. Nunc ornare congue fringilla. Aliquam luctus nisi bibendum orci dignissim, et consectetur tortor dapibus. Vivamus lacinia accumsan pellentesque. Fusce nisi neque, auctor eu tellus in, ullamcorper aliquet ipsum. Duis dui nisi, aliquet consectetur tincidunt non, ultricies id metus. Maecenas tristique laoreet egestas. Nunc lacinia, tortor id sagittis scelerisque, mi est viverra quam, at porttitor justo lorem vel ipsum. Praesent aliquet enim magna. Sed luctus neque nibh, vel egestas eros ultrices in. Nulla efficitur accumsan nisi, et venenatis velit fringilla at. Nam vitae magna id neque elementum pellentesque ut eu lacus. Vivamus sapien libero, elementum aliquam leo id, luctus pellentesque augue. Donec ac ligula sit amet lacus consectetur hendrerit. Aenean vitae tempor nulla, at dapibus tortor. In eu congue quam. Nulla facilisi. Nam massa metus, suscipit a cursus et, posuere sit amet mauris. Sed id mi sem. Phasellus placerat augue nisl, sit amet eleifend elit semper ut. Maecenas eget risus ultricies, feugiat diam et, auctor est. Integer maximus orci vitae lectus commodo scelerisque. Fusce a elementum mauris. Maecenas in cursus ipsum. Pellentesque accumsan tempor sem hendrerit auctor. Mauris placerat est volutpat sapien luctus congue. Quisque pharetra mauris quis purus cursus dapibus. Maecenas molestie vehicula lacus aliquet gravida. Morbi euismod lacinia enim, at volutpat ex lacinia eget. Quisque in ligula ut metus mattis luctus. Praesent ut vehicula mi. Vestibulum velit odio, convallis eu suscipit blandit, convallis quis ex. Nullam sagittis est vitae diam tristique imperdiet. Fusce vel urna auctor, feugiat libero vel, bibendum lectus. Curabitur aliquam dolor non faucibus porta. * A look at some cats {{{hr}}} #+begin_center *hello, world!* All of the cat images in this were generated by https://thiscatdoesnotexist.com/ #+end_center #+begin_twocol *Fusce nisi neque*, auctor eu tellus in, /ullamcorper aliquet ipsum/. _Duis dui nisi_, aliquet consectetur =tincidunt non=, ultricies id metus. Maecenas tristique laoreet egestas. Nunc lacinia, tortor id sagittis scelerisque, mi est viverra quam, at porttitor justo lorem vel ipsum. Praesent aliquet enim magna. Sed luctus neque nibh, vel egestas eros ultrices in. Nulla efficitur accumsan nisi, et venenatis velit fringilla at. Nam vitae magna id neque elementum pellentesque ut eu lacus. Vivamus sapien libero, elementum aliquam leo id, luctus pellentesque augue. Donec ac ligula sit amet lacus consectetur hendrerit. Aenean vitae tempor nulla, at dapibus tortor. In eu congue quam. Maecenas molestie vehicula lacus aliquet gravida. Morbi euismod lacinia enim, at volutpat ex lacinia eget. Quisque in ligula ut metus mattis luctus. ./images/c1.jpeg #+end_twocol #+begin_twocol ./images/c2.jpeg *Mauris efficitur nisi ut ex efficitur, id ultrices felis dignissim?* Nam fringilla arcu eu mi accumsan, in rhoncus velit auctor. *Aenean eget lacus dapibus, venenatis arcu sed, egestas nisl?* Cras quis elit porta, imperdiet turpis ac, luctus dui. *Sed non ligula porta purus sodales accumsan?* Integer vestibulum tortor sit amet maximus facilisis. *Vestibulum pulvinar urna a varius lacinia?* Nullam porta dolor quis condimentum aliquet. #+end_twocol ** Pet Trio! {{{hr}}} #+begin_threecol #+CAPTION: Choco ./images/c3.jpeg #+CAPTION: Latte ./images/c4.jpeg #+CAPTION: Chip ./images/c5.jpeg #+end_threecol * Events {{{hr}}} ** Fitness | Event Name | Date | Notes | Emoji | |-----------------------------+------------------+---------------+-------| | Doing Yoga with Goats | <2022-07-18 Mon> | very fun! | 😺 | | Weightlifting with Gorillas | <2022-07-25 Mon> | very intense | 😸 | | Calisthenics with Ants | <2022-08-18 Thu> | not even fair | 😹 | ** Non-Fitness | Event Name | Date | Notes | Emoji | |----------------------------+------------------+------------+-------| | Sitting Around with Sloths | <2022-07-26 Tue> | *yawn* | 😻 | | Not Doing Much with Bats | <2022-07-27 Wed> | snooze | 😼 | | Cat Naps with Dogs | <2022-07-28 Thu> | good times | 🙀 | * Celebrations {{{hr}}} #+begin_twocol {{{h3(Set of Chaotic Maps 1)}}} {{{split-sentence(10 coolness, 3-Cells CNN System)}}} {{{split-sentence(11 coolness, 2D Lorenz System)}}} {{{split-sentence(12 coolness, 2D Rational Chaotic Map)}}} {{{split-sentence(13 coolness, ACT Chaotic Attractor)}}} {{{split-sentence(14 coolness, Aizawa Chaotic Attractor)}}} {{{split-sentence(15 coolness, Arneodo Chaotic System)}}} {{{h3(Set of Chaotic Maps 2)}}} {{{split-sentence(16 coolness, Arnold's Cat Map)}}} {{{split-sentence(17 coolness, Baker's Map)}}} {{{split-sentence(18 coolness, Basin Chaotic Map)}}} {{{split-sentence(19 coolness, Beta Chaotic Map)}}} {{{split-sentence(20 coolness, Bogdanov Map)}}} {{{split-sentence(21 coolness, Brusselator)}}} {{{h3(Set of Chaotic Maps 3)}}} {{{split-sentence(22 coolness, Burke-Shaw Chaotic Attractor)}}} {{{split-sentence(23 coolness, Chen Chaotic Attractor)}}} {{{split-sentence(24 coolness, Chen-Celikovsky System)}}} {{{split-sentence(25 coolness, Chen-LU System)}}} {{{split-sentence(26 coolness, Chen-Lee System)}}} {{{split-sentence(27 coolness, Chossat-Golubitsky Symmetry Map)}}} #+CAPTION: This is artwork made by a GAN ./images/a1.jpeg #+end_twocol #+begin_footer This newsletter was built with *Org-Mode* #+end_footer This is the stylesheet that isn't touched by the newsletter writer Click for Template Stylesheet @import url(https://fonts.googleapis.com/css?family=Montserrat:400,500,80); :root { --c1: #F2F1E8; --c2: #A8DADC; --c3: #050533; --c4: #1D3557; --c5: #E34234; --c6: #403D39; } body { width: 100vw; background-color: var(--c1); color: var(--c6); line-height: 1.4; font-size: 18px; font-family: 'Montserrat', sans-serif; } h1, h2 { color: var(--c5); font-weight: bold; } h3 { color: var(--c3); } p { padding: 20px; } .twocol { background-color: var(--c2); column-rule: dotted var(--c5); column-rule-width: thin; column-count: 2; padding-top: 25px; padding-bottom: 25px; margin-top: 10px; margin-bottom: 10px; border-radius: 10px; } .twocol img { max-width: 100%; max-height: 100%; border-radius: 10px; } .threecol { background-color: var(--c2); column-rule: dotted var(--c5); column-rule-width: thin; column-count: 3; padding-top: 25px; padding-bottom: 25px; border-radius: 10px; } .threecol img { max-width: 100%; max-height: 100%; border-radius: 10px; } .banner { padding: 0px; margin: 0px; } .banner img { width: auto; padding: 0px; margin: 0px; } .banner .figure { padding: 0; } .banner .figure p { padding: 0; } .timestamp { font-size: 18px; color: var(--c5); font-weight: thin; } .top-banner { float: right; } .title-left { float: left; color: var(--c2); font-size: 64px; font-family: 'Montserrat', sans-serif; font-weight: bold; } .title-right { float: right; color: var(--c5); font-size: 64px; font-family: 'Montserrat', sans-serif; font-weight: bold; } .vol-issue-left { float: left; color: var(--c3); font-weight: thin; font-family: 'Montserrat', sans-serif; } .vol-issue-right { float: right; color: var(--c3); font-weight: thin; font-family: 'Montserrat', sans-serif; } .title-p { padding-top: 0px; padding-bottom: 1px; } .subtitle-p { padding: 8px; } hr { border-top: 1px dotted; border-bottom: 0px; border-color: var(--c5); } a { color: var(--c3); text-decoration: none; } a:hover { color: var(--c4); } .date-str-date { font-weight: thin; color: var(--c5) } .date-str-str { color: var(--c4) } .footer { float: right; } .in-column-h1 { font-weight: bold; font-size: 36px; } .in-column-h2 { font-weight: bold; font-size: 28px; } .in-column-h3 { font-weight: bold; font-size: 20px; } .figure-number { display: none; } table { border-collapse: separate; border-spacing: 0; } table tr th, table tr td { border-right: 1px solid #bbb; border-bottom: 1px solid #bbb; box-shadow: 2px 2px 1px #e5dfcc; } table tr th:first-child, table tr td:first-child { border-left: 1px solid #bbb; } table tr th { border-top: 1px solid #bbb; } /* top-left border-radius */ table tr:first-child th:first-child { border-top-left-radius: 0px; } /* top-right border-radius */ table tr:first-child th:last-child { border-top-right-radius: 0px; } /* bottom-left border-radius */ table tr:last-child td:first-child { border-bottom-left-radius: 6px; } /* bottom-right border-radius */ table tr:last-child td:last-child { border-bottom-right-radius: 6px; } th, td { padding: 8px 20px; } th { background: var(--c2); color: var(--c4); } td { background: var(--c1); } ## The Secret Sauce Below you will find an overview of the different components and the related bits of org-markdown-language created. Before we get into that jazz, let's zoom out and ask: What is the secret sauce? How does it all work? 3 big things allow this to work: 1. Org-mode has a ton of functionality already built into it, including different types of markup and export options 2. Special blocks can be created on the fly, giving the user a container object to interact with instead of HTML/CSS/Javascript. These can then be styled with CSS 3. If functionality needs to be added to a component, org lets the user create macros that expand at export time. These can inject HTML, CSS, or emacs-lisp into the export ### Org-mode functionality Org comes with a lot of functionality out of the box. You'd be well served to look at the org syntax specification The most important part for the end-user (of the newsletter generator) is to know the very basics: # Headers * This is a heading ** This is a subheading *** etc etc # Lists 1. item 1 2. item 2 3. item 3 - item 5 - item 6 - item 7 # Inline styles *bold* /italic/ _underline_ =monospace= ~code~ +strike-through+ ### Special Blocks For every component I want to add to the newsletter, I simply make a corresponding org-mode block like so: #+begin_component-name some content here #+end_component-name At export time, this turns into: <div class="component-name"> <p>some content here</p> </div> Here is the relevant section from the org-manual: When special blocks do not have a corresponding HTML5 element, the HTML exporter reverts to standard translation (see org-html-html5-elements). For example, #+BEGIN_lederhosen exports to <div class="lederhosen"> Since org export turns things it doesn't recognize as HTML tags into a div with a class, I can then style the class with CSS in the style.css file. Extra details This also works particularly nice when you use a tag that is found in html5. As an example, #+BEGIN_aside Lorem ipsum #+END_aside exports to: <aside> <p>Lorem ipsum</p> </aside> ### Functionality with Macros If I want to add functionality to something, I can use a macro that creates HTML tags (or css, elisp, etc) to implement the functionality. Macros allow org to replace text during export. Here is an example from the org-manual: #+MACRO: poem Rose is$1, violet's $2. Life's ordered: Org assists you. {{{poem(red,blue)}}} becomes Rose is red, violet's blue. Life's ordered: Org assists you. It even allows the use of emacs-lisp snippets, but I haven't needed to use that functionality here (yet). For the newsletter I use the following 3 macros (defined in the BEGIN_PROPERTIES block): #+MACRO: split-sentence @@html: <span class="date-str-date"><code>$1</code></span> | <span class="date-str-str">$2</span><br>@@ #+MACRO: h1 @@html:<span class="in-column-h1">$1</span><br>@@
#+MACRO: hr #+HTML: <hr>
This syntax:
@@html: stuff here@@
#+HTML: tag
allows org to directly place html into the exported output (instead of running it through its preprocessor).
These are all used throughout the newsletter using the following syntax:
# example
#+MACRO: something <h1>Write $1 Write$2</h1>
{{{something(arg1, arg2)}}}
<h1>Write arg1 Write arg2
# actual usage
{{{split-sentence(left hand side, right hand side)}}}
{{{h1(sentence here)}}}
{{{hr}}}
where something is the macro name and the args are replacing the $1 and $2 anchors.
## Build Tool
The build tool is very simple. The idea is the following:
2. Build the newsletter with emacs through an executable script
This way the end user only has to do 3 things to get going:
1. Write the file in emacs
2. Build the file by clicking a shortcut
3. Inspect the results in the browser
### On Windows
Since emacs is very portable on Windows (you can download a self-contained program in a folder here) each user can get access to it without much trouble.
To create an executable
Then the build tool can be put together using a little bit of elisp and powershell
Here is the file create-html.el
#!/home/neptune/.guix-profile/bin/emacs --script
Then the user can make the file executable and run the command
./create-html.el
### Why HTML? Why not pdf?
While there are ways of creating PDF content with org-mode, getting all the styling and spacing to line up seemed like an extra hurdle for little gain. I find it much easier to parse and tinker with HTML pages than PDF pages.
### The images in the newsletter
The images in the newsletter were all generated by various Generative Adversarial Networks.
## Components Overview
### main banner
#### Syntax
To get the banner to show up, I needed to build it out of a <span> in the title of the org document. When org-export as html is called, org-mode writes the content to both the header <title> tag as well as a standalone <h1> tag at the top of the page.
As a result, we can inject html into the title (and subtitle) like so:
#+TITLE: @@html:<p class="title-p"><span class="title-left">Q3</span><span class="title-right">Newsletter Update</span></p>@@
#+SUBTITLE: @@html:<p class="subtitle-p"><span class="vol-issue-left">Vol. 1, Issue 3</span><span
The caveat here is that this code is also copied into the <title> in the header, meaning that our page titlebar will be @@html...
I'm still uncertain about how to fix this.
#### Syntax
This comes for free with org-export to html. As a default org export also creates unique anchor points for each of the headers and subheaders. This can be customized to use the verbatim text as an anchor point as well. In our case, it doesn't matter too much what the anchor wording is.
If you want to get more readable link anchors, you can use the following emacs lisp snippet that I took from this excellent blog post:
(defun my/ensure-headline-ids (&rest _)
"All non-alphanumeric characters are cleverly replaced with ‘-’.
If multiple trees end-up with the same id property, issue a
message and undo any property insertion thus far.
"
(interactive)
(let ((ids))
(org-map-entries
(lambda ()
(org-with-point-at (point)
(let ((id (org-entry-get nil "CUSTOM_ID")))
(unless id
(s-replace-regexp "[^[:alnum:]']" "-")
(s-replace-regexp "-+" "-")
(s-chop-prefix "-")
(s-chop-suffix "-")
(setq id))
(if (not (member id ids))
(push id ids)
(message-box "Oh no, a repeated id!\n\n\t%s" id)
(undo)
(setq quit-flag t))
(org-entry-put nil "CUSTOM_ID" id))))))))
;; Whenever html & md export happens, ensure we have headline ids.
### section banner
#### Syntax
This is the first case of using a special code block to achieve our desired formatting. In this case, it is important to note that banner is not a part of the default org syntax
#+begin_banner
./images/nscrop.png
#+end_banner
What is happening here is detailed in the special sauce section. Even though org doesn't have a banner tag, org makes a div with class "banner".
Here is the relevant CSS for the banner image:
.banner img {
width: auto;
margin: 0px;
}
### prefix | sentence
#### Syntax
This is a simple macro that expands to html at export time.
#+MACRO: split-sentence @@html: <span class="date-str-date"><code>$1</code></span> | <span class="date-str-str">$2</span><br>@@
This provides syntax like the following:
{{{split-sentence(first argument, second argument)}}}
which expands to the following at export time
<span class="date-str-date"><code>first argument</code></span> | <span class="date-str-str">second argument</span><br>
We can add a class to style it as well:
.date-str-date {
font-weight: thin;
color: var(--c5)
}
.date-str-str {
color: var(--c4)
}
### left-image
#### Syntax
This creates a div with the class id twocol. Where the magic happens is in the css file. With css, I can split the div into n columns and style the intersecting border:
.twocol {
background-color: var(--c2);
column-rule: dotted var(--c5);
column-rule-width: thin;
column-count: 2;
margin-top: 10px;
margin-bottom: 10px;
}
.twocol img {
max-width: 100%;
max-height: 100%;
}
Then, when I enter text into a twocol block, it will split the content into the first set of content and the second set.
#+begin_twocol
an image goes here
some text accompanying it goes here
#+end_twocol
This is also generalized in the right image and triple-image components
### right-image
#### Syntax
This is the same thing as the left image component, except we add the image to the block second:
#+begin_twocol
some text accompanying the image goes here
an image goes here
#+end_twocol
### triple-image
#### Syntax
This is the same as left-image, except we use css to create 3 images.
.threecol {
background-color: var(--c2);
column-rule: dotted var(--c5);
column-rule-width: thin;
column-count: 3;
}
.threecol img {
max-width: 100%;
max-height: 100%;
}
This could also house content like image | text | image or any 3-tuple of image and text inputs. In the future, if needed, I could add any n-tuple.
I wanted to keep the functionality sufficiently separated from the component itself, i.e. don't hardcode the header transformation into the underlying component. To do this, I used an org-macro (see The Secret Sauce > Functionality with Macros) and allow the user to invoke it when desired.
#### Syntax
#+MACRO: h3 @@html:<span class="in-column-h3">\$1</span><br>@@
{{{h3(Set of Chaotic Maps 1)}}}
### image captions
This comes for free with org-mode. Simply add the #+CAPTION tag above an image
#### Syntax
#+CAPTION: This is [[https://thisartworkdoesnotexist.com/][artwork]] made by a GAN
,./images/a1.jpeg
### tables
Tables are super easy to use in org-mode. I usually just write $$n$$ | characters in a row and |- like so for a 3-column table:
||||
|-
and then hit <tab> and the table is created like so:
| | | |
|---+---+---|
| | | |
|
|
# Revision history [back]
It's because in parametric_plot(f(t), (t, 0, 4*r + 2*pi*r)), the Python function f is evaluated with the symbolic variable t as an input, prior to giving any concrete value to t:
sage: f(t)
(-4*pi + t - 8, -2)
To get the expected plot, you must split f in two by defining
sage: def f0(x): return f(x)[0]
sage: def f1(x): return f(x)[1]
Then
sage: parametric_plot([f0, f1], (t, 0, 4*r + 2*pi*r))
gives the plot that you want. Note that in the above command, one must write [f0,f1] and not [f0(t),f1(t)]: the latter would reproduce the incorrect behavior, by first evaluating f0 and f1 on the symbolic variable t.
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# GaussianConstraint¶
class zfit.constraint.GaussianConstraint(params, observation, uncertainty)[source]
Gaussian constraints on a list of parameters to some observed values with uncertainties.
A Gaussian constraint is defined as the likelihood of params given the observations and uncertainty from a different measurement.
$\text{constraint} = \text{Gauss}(\text{observation}; \text{params}, \text{uncertainty})$
Parameters
Raises
ShapeIncompatibleError – If params, mu and sigma have incompatible shapes.
property covariance
Return the covariance matrix of the observed values of the parameters constrained.
add_cache_deps(cache_deps, allow_non_cachable=True)
Add dependencies that render the cache invalid if they change.
Parameters
Raises
TypeError – if one of the cache_dependents is not a ZfitCachable _and_ allow_non_cachable if False.
property dtype
The dtype of the object
Return type
DType
get_cache_deps(only_floating=True)
Return a set of all independent Parameter that this object depends on.
Parameters
only_floating (bool) – If True, only return floating Parameter
Return type
OrderedSet
get_dependencies(only_floating=True)
DEPRECATED FUNCTION
Warning: THIS FUNCTION IS DEPRECATED. It will be removed in a future version. Instructions for updating: Use get_params instead if you want to retrieve the independent parameters or get_cache_deps in case you need the numerical cache dependents (advanced).
Return type
OrderedSet
get_params(floating=True, is_yield=None, extract_independent=True, only_floating=<class 'zfit.util.checks.NotSpecified'>)
Recursively collect parameters that this object depends on according to the filter criteria.
Which parameters should be included can be steered using the arguments as a filter.
• None: do not filter on this. E.g. floating=None will return parameters that are floating as well as
parameters that are fixed.
• True: only return parameters that fulfil this criterion
• False: only return parameters that do not fulfil this criterion. E.g. floating=False will return
only parameters that are not floating.
Parameters
• floating (Optional[bool]) – if a parameter is floating, e.g. if floating() returns True
• is_yield (Optional[bool]) – if a parameter is a yield of the _current_ model. This won’t be applied recursively, but may include yields if they do also represent a parameter parametrizing the shape. So if the yield of the current model depends on other yields (or also non-yields), this will be included. If, however, just submodels depend on a yield (as their yield) and it is not correlated to the output of our model, they won’t be included.
• extract_independent (Optional[bool]) – If the parameter is an independent parameter, i.e. if it is a ZfitIndependentParameter.
Return type
property name
The name of the object.
Return type
str
property observation
Return the observed values of the parameters constrained.
register_cacher(cacher)
Register a cacher that caches values produces by this instance; a dependent.
Parameters
reset_cache_self()
Clear the cache of self and all dependent cachers.
sample(n)
Sample n points from the probability density function for the observed value of the parameters.
Parameters
n – The number of samples to be generated.
Returns:
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Question
# 250 g of water at 30℃ is contained in a copper vessel of mass 50g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5℃. Given: specific latent heat of fusion of ice = $$336 ×103J kg^{-1}$$, specific heat capacity of copper = $$400 J kg^{-1} K^{-1}$$, specific heat capacity of water = $$4200 J kg^{-1} K^{-1}$$ .
Solution
## Mass of copper vessel $$m_1 = 50 g$$.Mass of water contained in copper vessel $$m_2 = 250$$g. Mass of ice required to bring down the temperature of vessel = $$m$$Final temperature =$$5°C$$.Amount of heat gained when 'm' g of ice at 0°C converts into water at 0°C = $$m × 336 J$$Amount of heat gained when temperature of 'm' g of water at 0°C rises to $$5°C = m × 4.2 × 5$$Total amount of heat gained = $$m × 336 + m × 4.2 × 5$$Amount of heat lost when 250 g of water at 30°C cools to $$5°C = 250 × 4.2 × 25 = 26250 J$$ Amount of heat lost when 50 g of vessel at 30°C cools to $$5°C = 50 × 0.4 × 25 = 500 J$$Total amount of heat lost $$= 26250 + 500 = 26750 J$$We know that amount of heat gained = amount of heat lost $$m × 336 + m × 4.2 × 5 = 26750/357$$ $$m =26750/357 = 74.93 g$$Hence, mass of ice required is $$74.93 g$$.Physics
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# Direct Variation
When two variables are related in such a way that the ratio of their values always remains the same, the two variables are said to be in direct variation.
In simpler terms, that means if A is always twice as much as B, then they directly vary. If a gallon of milk costs $2, and I buy 1 gallon, the total cost is$2. If I buy 10 gallons, the price is \$20. In this example the total cost of milk and the number of gallons purchased are subject to direct variation -- the ratio of the cost to the number of gallons is always 2.
To be more "mathematical" about it, if y varies directly as x, then the graph of all points that describe this relationship is a line going through the origin (0, 0) whose slope is called the constant of the variation. That's because each of the variables is a constant multiple of the other, like in the graph shown below:
### 1) Expressing Direct Variation as an Equation
The equation y/x = 6 states that y "varies directly as" x since the ratio of y to x (also written y:x) never changes. The number 6 in the equation y/x = 6 is called the constant of variation. The equation y/x = 6 can also be written in the equivalent form, y = 6x. That form shows you that y is always 6 times as much as x.
Similarly, for the equation y=x/3, the constant of variation is 1/3. The equation tells us that for any x value, y will always be 1/3 as much.
### 2) Algebraic Interpretation of Direct Variation
For an equation of the form y = kx, multiplying x by some fixed amount also multiplies y by the SAME FIXED AMOUNT. If we double x, then we also double the corresponding y value. What does this mean? For example, since the perimeter P of a square varies directly as the length of one side of a square, we can say that P = 4s, where the number 4 represents the four sides of a square and s represents the length of one side. That equation tells us that the perimeter is always four times the length of a single side (makes sense, right?), but it also tells us that doubling the length of a side doubles the perimeter (which will still be four times larger in total).
### 3) Geometric Interpretation of Direct Variation
The equation y = kx is a special case of linear equation (y=mx+b) where the y-intercept equals 0. (Note: the equation y = mx + b is the slope-intercept form where m is the slope and b is the y-intercept). Anyway, a straight line through the origin (0,0) always represents a direct variation between y and x. The slope of this line is the constant of variation. In other words, in the equation y = mx + b, m is the constant of variation.
### Example A:
If y varies directly as x, and y = 8 when x = 12, find k and write an equation that expresses this variation.
Plan of Attack:
Plug the given values into the equation y = kx.
Solve for k.
Then replace k with its value in the equation y = kx.
Step-by-Step:
Insert our known values: 8 = k*12
Divide both sides by 12 to find k: 8/12 = k
2/3 = k
Next: Go back to y = kx and replace k with 2/3.
Result:
y = (2/3)x
### Example B:
If y varies directly as x, and y = 24 when x = 16, find y when x = 12.
Plan of Attack:
When two quantities vary directly, their ratio is always the same. We'll create two ratios, set them equal to each other, and then solve for the missing quantity.
Step-by-Step:
The given numbers form one ratio expressed as y/x: (24/16)
To find y when x=12 we setup another ratio: (y/12)
Solve:
By definition, both ratios are equal: (24/16) = (y/12)
Multiply each side by 12 to solve for y: (1.5)*12=y
Result:
18 = y
Got a basic understanding of direct variation now? If you still need more help, try searching our website (at the top of the page) for a more specific question, or browse our other algebra lessons. Sometimes it helps to have a subject explained by somebody else (a fresh perspective!) so you may also be interested in another lesson on direction variation, in this case referred to as direct proportion.
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# Hypothesis Practice Using Scenarios Answer Key
Topics Covered: Scientific methods, observation, hypothesis, data, analysis, experimental design, laws, theories, variables (independent, dependent, controlled. For each of the following scenarios, develop a testable hypothesis as well as identify the investigative question, cause, effect, independent variable, and dependent variable. We will also see the Intermediate Value Theorem in this section and how it can be used to determine if functions have solutions in a given interval. Power of a test (1 − β) D. Use the following to answer questions 2-5: Scenario: Closed Economy S = I In a closed economy suppose that GDP is $12 trillion. If that is not possible, then the results are considered invalid. This worksheet contains two scenarios and asks students questions to apply the scientific method. significant. Post your questions to our community of 350 million students and teachers. At this point, the teacher should write the definition of the new scientific. For example, "All cats die; Socrates died; therefore Socrates was a cat. When they wish to learn about a complex numerical phenomenon, they can perform a simulation study. Experimental scenarios answer key. •UsingtheTI-83,83+,84,84+Calculatorshows students step-by-step instructions to input problems into their calculator. The scientific method in our everyday lives: Reflect on how people use the scientific method in their everyday lives. Spongebob answer key scientific method controls and variables spongebob worksheet answers. Make sure it has a key. Spongebob scientific method worksheet answer key. In the scenarios below, identify the following components of an experiment. It is a possible explanation. Our HiSET Science practice test will help you prepare for this section of the test. Ag Science: HYPOTHESIS WORKSHEET ANSWERS. Of course Zelda thought her Chihuahua, even though small, was the smartest. null hypothesis. List the three must haves of a hypothesis. Scientific Method Review Answer Key Worksheets - Teacher Displaying top 8 worksheets found for - Scientific Method Pratice Scenarios And Answerss. Hypothesis- a possible explanation for a set of observations or an answer to a scientific question. Hypothesis Practice Using Scenarios. blank 6 conclusion. Scientific Method Practice. The key-word estimate con±rms that this is the correct response. US9153142B2 - User interface for an evidence-based, hypothesis-generating decision support system - Google Patents. For this purpose, you should describe a number of specific situations that could trigger use of the product or service you are designing. Any student not finished by the end of the 160 minutes may continue working but, the test must be completed within the same school day. Stay connected with parents and students. The first step of the scientific method is to define or identify the ?. His friend sandy told him that he should try using clean o detergent a new brand of laundry soap she found at sail mart. Science Skills. CHAPTER-BY-CHAPTER ANSWER KEY CHAPTER 1 ANSWERS FOR THE MULTIPLE CHOICE QUESTIONS 1. conclusion – the data is concluded to “support” or “not support” the hypothesis. You will have 2 weeks to complete this assignment from the date it was assigned. The key-word estimate con±rms that this is the correct response. Examples of these forces. Afterward, the scientist comes to a 11 that either supports or does not support the hypothesis. But the gringo did not hear the whispered answer so he questioned a man sitting next to him about it. Orgcontentco answer key. We are not performing a hypothesis test on where the true parameter (the bias) lies in the parameter space. Statistics - Statistics - Hypothesis testing: Hypothesis testing is a form of statistical inference that uses data from a sample to draw conclusions about a population parameter or a population probability distribution. Some of the worksheets for this concept are Scientific method scenario practice, Experimental scenario scientific method answers, The scientific method, Manhasset union school district home. Post your questions to our community of 350 million students and teachers. In the context of the scientific method, this description is somewhat correct. Gram-positive bacteria D. What groups listed below have true cell walls? A. A hypothesis, in general, is an assumption that is yet to be proved with sufficient pieces of evidence. Rather, we want to come up with a good approximiation for the bias. SCENARIO #4 – Christian’s Cart Experiment In this exercise you will read the following scenario and identify the parts of the scientific method in it. Use the following to answer questions 2-5: Scenario: Closed Economy S = I In a closed economy suppose that GDP is$12 trillion. In this section we will introduce the concept of continuity and how it relates to limits. 0916 Conclude: There is not enough evidence to reject the null hypothesis. Scenario 1: Honeycreepers Hawaii is a chain of volcanic islands with a variety of habitats. Only minimum credit at best if your answer implied or gave diploid genotypes or two-gene scenarios (e. Cite specific from your experiment to explain why Can identify possib5e in your experiment p rvv. Choice A is the best answer. Chi-squared value= 78. New Concept Videos were added. CH8: Hypothesis Testing Santorico - Page 270 Section 8-1: Steps in Hypothesis Testing – Traditional Method The main goal in many research studies is to check whether the data collected support certain statements or predictions. The respondent is given the opportunity to answer only 'yes' or 'no', whereas he might like the speed, but not the reliability, or vice versa. If we can assign people to a category then that tells us things about those people, and as we saw with the bus driver example, we couldn't function in a normal manner without using these categories; i. Scientific Method Scenarios Worksheet2. Hypothesis - If we use the cheaper "Steel Seal" floor wax, then the floor will have few. Control / Control Group - Pot # 3 ( it’s given the “recommended” amount of water) 4. Once a scientist has a scientific question they are interested in, the scientist researches what is already known on the topic. Rev min is brought to rest from a particular item, and this instrument see saxon mills, the life and change over tim in three volumes for flexibility and auton self reinforcer any desired omy to experiment, to take corrective action. What he denies is that organ-like, special purpose adaptations are the likely result of such evolutionary scenarios. Download murder and a meal lab answer key We have managed to get easy for you to find a PDF Books without any stress. Will an egg float higher in water with more salt ? IV: DV:. Some of the worksheets for this concept are Hypothesis practice, Hypothesis tests for one population mean work, Writing hypothesis practice and answer key, Constructing a hypothesis, Chapter 7 hypothesis testing with one sample, Skill and practice work, Understanding hypotheses predictions laws and theories, Chapter 2 research. Being found to have falsified or created data to better fit a. Step-by-Step SPSS® Tutorial Videos created by the author provide screencast demonstrations for 26 key chapter concepts. Hypothesis Practice Using Scenarios. Teach Me Under The Sun Scientific Method Worksheet Scientific Method Steps Hypothesis Some of the worksheets displayed are a hands on minds on science series for fourth and fifth …. • Your hypothesis is the center of your project. Regardless of which scenario the student chooses, they will be required to answer questions demonstrating knowledge of a learning objective, making them the perfect questions to assign toward the end of a chapter. Scientific Method Scenario Practice - ANSWERS. Hypothesis - If we vary the amount of water given to different plants, then the plants. Dependent Variable - Height of the plants 3. General Idea and Logic of Hypothesis Testing. First, you are asked to define the sociological perspective. According to the Society for Human Resource Management, the answers “should provide verifiable, concrete evidence as to how a candidate has dealt with issues in the past. The TV commercials say that Popalicious is the fastest popping popcorn in the world. One key point: the dot is not a multiplication sign. Conduct background research about your question. Read the experiment and then identify the components of the scientific method by completing the Pre-test provided. Enzymes and Their Functions: Lock-and-Key Activity A. The slideshow presents the concepts of testable questions, hypotheses, variables and controls, which are the basis of students being able to plan and carry out investigations (NGSS Practice 3). Answer: b From: Textbook, p. The answer, that multiple triggers allow the plant to conserve energy, is best supported near the beginning of the first paragraph: “Closing its trap requires a huge expense of energy, and reopening the trap can take several hours, so Dionaea only wants to. Feb 28, 2015 - This Pin was discovered by Jennifer Solter-Jones. Thus the question. There are 10 questions with an answer key. ___ Use a sample of size 50 to determine whether there is a linear relationship between a car’s age and its gas mileage. H ___ Use a simple random sample of size 30 to determine whether the average. Rather, we want to come up with a good approximiation for the bias. Check conditions. A hypothesis, in general, is an assumption that is yet to be proved with sufficient pieces of evidence. Some of the worksheets for this concept are Hypothesis practice, Variables hypothesis work, Read the following scenarios and design a hypothesis, Chapter 7 hypothesis testing with one sample, Gradelevelcourse grade7lifescience lesson, Required vocabulary, Proportions date period, What is scientific inquiry. Talking related with Enzymes Worksheet Answer Key, scroll down to see particular variation of pictures to give you more ideas. Answer Key**. CH8: Hypothesis Testing Santorico - Page 270 Section 8-1: Steps in Hypothesis Testing – Traditional Method The main goal in many research studies is to check whether the data collected support certain statements or predictions. These questions describe, rather than relate variables or compare groups. Hypothesis is practical as well as very powerful and will often find bugs that escaped all other forms of testing. After you see the key above, go to the Aquatic Critter Key and try your skill. Write 3-4 sentences summarizing (explaining) what the data in the table shows. ACTIVITY 2B Answer Key Kinetic Energy and Potential Energy Kinetic Energy Energy is energy in motion, or the energy of motion. Residents brought clinical questions to the session and had access to laptops to practice searching skills. They are all just estimates. In addition to analysing data to answer. protozoans 2. In the scientific method, analysis of the results of an experiment will lead to the hypothesis being accepted or rejected. The scenarios are focused around spe. 05: You can either say “Reject the null hypothesis” or “The result is statistically significant. Cite specific from your experiment to explain why Can identify possib5e in your experiment p rvv. A dichotomous key is a tool created by scientists to help scientists and laypeople identify objects and organisms. In a quantitative disserta-tion, an entire section of a research proposal might be. If you're seeing this message, it means we're having trouble loading external resources on our website. Explain why the selected test is appropriate and run the test on the data provided. Question - Which type of fertilizer will cause the plants to grow the most? Hypothesis - Any possible answer to the question. Practice firing the gun the same way each time to minimize the variation in initial velocity. For a large sample size, Sample Variance will be a better estimate of Population variance so even if population variance is unknown, we can use the Z test using sample variance. Christian wonders if a heavier cart will roll faster down-hill than a lighter one. The Monitor hypothesis 15 (a) Individual variation in Monitor use 18 4. The alternative hypothesis (H A) states that the related population means are not equal (at least one mean is different to another mean): H A: at least two means are significantly different. Plot equipotential lines and discover their relationship to the electric field. (1) As a group discuss your research question and decide on a reasonable predicted answer to your research question. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. This could certainly be used for high school review as well. When the gringo wanted to pay and leave the pub, the bartender told him how much his drink costed. In addition to analysing data to answer. 1 Introduction to Hypothesis Testing The problem with applying the techniques learned in Chapter 5 is that typically, the popula-tion mean ( ) and standard deviation (˙) are not known. Plagiarism is using the ideas or words of another person without giving appropriate credit. sav data file. First, you are asked to define the sociological perspective. Based on your observations, you make a hypothesis. Constructive pessimism, by contrast, encourages us to imagine the worst. 895 P-Value: 0. For a hypothesis, write an “H” on the line. Bayes Theorem provides a principled way for calculating a conditional probability. The highest achievers manage their time exceptionally well. Practice Problems - Using the information in the following phrases, identify the independent variable, dependent variable, and any constants. Simple Example: Vinegar, added to the water of a Sunflower plant, would slow the plant's rate of growth. Hypotonic Solution - C. She thought. Read the experiments and then identify the components of the scientific method by completing the graphic organizer provided. Rather, we want to come up with a good approximiation for the bias. In this activity, students will take data from a fictitious multi-group design to practice conducting a One-way Analysis of Variance (ANOVA). Inference questions make up nearly 15% of all SAT Reading questions (based on analysis of four publicly available new SATs). Tips for developing a hypothesis are shown at right. For those that do have a hypothesis, the hypothesis should derive logically. A key component of the use of the scientific method is that it ensures that the experiment should be able to be replicated by anyone. 4%) Intermediate risk (7. Spongebob scientific method worksheet answer key. The use of theory will help me develop into an open and flexible social worker, who is committed to defensible rather defensive practice. How do you use a dichotomous key and why is a dichotomous key important? Learning Target: I can construct and use a dichotomous key based on different organisms’ characteristics. Adjust the spring tension so that the range, when fired horizontally, is between one and two meters. Abstract In this interview for Think magazine (April ’’92), Richard Paul provides a quick overview of critical thinking and the issues surrounding it: defining it, common mistakes in assessing it, its relation to communication skills, self-esteem, collaborative learning, motivation, curiosity, job skills for the future, national standards, and assessment strategies. HYPOTHESIS WORKSHEET Write a HYPOTHESIS for each of the following research question. pre-assessment, post-assessment, entrance ticket, exit ticket or homework assignment. Design an experiment to help answer one of your questions. Nurse Key Fastest Nurse Insight Engine It provides fastest searching engine to get answers of your questions in shortest time. Not all studies require that you test a hypothesis; some may simply involve collecting information regarding an issue. But the gringo did not hear the whispered answer so he questioned a man sitting next to him about it. hypothesis – an educated guess is formulated to explain what might be happening. Create models of dipoles, capacitors, and more!. Set 1 of locks and keys will be provided by your teacher. 4 Scientific Investigation: Students communicate about investigations and explanations. Once theseare familiar, but before the answers are well-known, hide the answers and try to fill in the blank for each question. Sometimes the guesses and questions change as you run your experiments. Design an Experiment 4. Chapter Objectives. Problem 3 Here is a square and some regular octagons. Statistical tests can be powerful tools for researchers. This worksheet can also be used as an assessment to determine understanding about the scientific method. Affirming The Consequent: logic reversal. Experimental Group: Flat A (w/ commercial fertilizer) & Flat B (w/ compost) Control Group: Flat C (no fertilizers). What is the attack rate formula for people who are exposed? a/(a+b) 11. In your experimental plan, describe each of the aspects below. 1-1 Practice Worksheet Relations And Functions Algebra 2,. conclusion B. Accepting H 0 when it is actually false. Based on the calculated elasticity value in 5. Arrange positive and negative charges in space and view the resulting electric field and electrostatic potential. 895,1e99,9,30)=0. The respondent is given the opportunity to answer only 'yes' or 'no', whereas he might like the speed, but not the reliability, or vice versa. Good answers discuss how reasonable assumptions are in the context of scenarios (b) and (c). Thus the question. Scientific Method Scenarios Worksheet Identifying Variables Science Worksheet Answers Triple Beam Balance Worksheet Middle School Scientific Method Worksheet Scientific Method Control Variable Definition. Answers to research scenarios A1) What is the experimental unit? A single delivery from Mills Foods. Join the Social Care Network to read more pieces like this. Read procedure on page 44 to introduce scenario-. When they wish to learn about a complex numerical phenomenon, they can perform a simulation study. Inferences l, Johng 2 broke 3 s r, Lev+-h was A hypothesis is to your question Et jnclûdes your. Residents brought clinical questions to the session and had access to laptops to practice searching skills. Sometimes scientists make a mistake, or ÿ?, and need to do an experiment again. Please email me for a key when you are ready to check your work, because ensuring mastery will solidify your Pre-AP Chemistry fundamental knowledge assessment we will have during the second week of school. Common responses include an educated guess or a prediction based on observations; or a possible explanation to a problem. These versatile and engaging task card sets cover a huge variety of Life science, Physical science, and Earth and Space science concepts. While standing in front of the school, Jace recorded the number of students who walked, rode their bikes, took the bus, and were driven there. Statistical Hypothesis – a conjecture about a population parameter. Get Free Graphing Practice Biology Answers Graphing Practice Biology Answers - atcloud. As these are based on the common assumption like the population from which sample is drawn should be normally distributed, homogeneity of variance, random sampling of data, independence of observations, measurement of the dependent variable on the ratio or interval level. Test of the Hypothesis (often a Controlled Experiment) 6. adjusting knob on the back of the gun. Rather, we want to come up with a good approximiation for the bias. The alternative hypothesis is the claim that researchers are actually trying to prove is true. This worksheet contains two scenarios and asks students questions to apply the scientific method. Chi-squared value= 78. In a serial killer case that lasted for nearly two years, police had received countless tips and several eyewitness accounts. Potential Energy. Statistics - Statistics - Hypothesis testing: Hypothesis testing is a form of statistical inference that uses data from a sample to draw conclusions about a population parameter or a population probability distribution. protozoans 2. material by using this question bank is to read four or at most five questions. When social scientists wish to learn about an empirical phenomenon, they perform an experiment. The second choice is an example of parametric hypothesis testing and goodness of ±t testing. Practice Supply and Demand Questions 1. It's one of the Six Sigma concepts you can directly apply in your situation. Once the Do Now is turned in, I pass out the hypothesis, variables and controls worksheet and present the Hypothesis and Variables slideshow. Then use the variables to make a good hypothesis. The hypothesis attempts to answer the research question. In the scientific method, experiments (often with controls and variables) are devised to test hypotheses. The reference of how Antarctica was once thought to be a forgotten and insignificant continent. Once a scientist has a scientific question they are interested in, the scientist researches what is already known on the topic. Isotonic Solution - 16. You can use it to conduct an experiment. (a) What is the mean age of the sample? What is the age range of the sample (minimum and maximum values)? p. Choose a significance level or use the given one. Kelly and Jack are playing in the park. In this activity, students will take data from a fictitious multi-group design to practice conducting a One-way Analysis of Variance (ANOVA). State whether your hypothesis was proven!. Their observations, analyses, and comparisons can lead to the formation of hypotheses and theories. Then identify the (B) independent variable, (C) dependent variable, (D) number of repeated trials, (E) constants/ control (if present). 91-99) of. Reject the hypothesis. State whether your hypothesis was proven!. 88359 m2 / 3 m = 30,000 m The answer can only have 1 sig dig cause of the 3. However, it got confused with a very different idea- the technological stagnation hypothesis, proposed by economist Robert Gordon. For tonight, please practice finding averages (the back of the independent/dependent variable handout) and reading instruments (the yellow half sheet from today). Problem 3 Here is a square and some regular octagons. Use of non-directional wording in the question. First, you are asked to define the sociological perspective. , a spore should be genotype D, or d, or E, or e, not (for example) dE. Critical Value= 11. Their observations, analyses, and comparisons can lead to the formation of hypotheses and theories. Her task is to answer the method worksheet you can use to practice. How do you use a dichotomous key and why is a dichotomous key important? Learning Target: I can construct and use a dichotomous key based on different organisms’ characteristics. Identify the Controls and Variables Smithers thinks that a special juice will increase the Identify the: productivity of workers. A good hypothesis relates the independent and dependent variables of the experimen o e e In this experiment, a good hypothesis could be that the addition of iron to the growth formula will improve the growth of the lettuce. In a hypothesis test problem, you may see words such as "the level of significance is 1%. In this section we will introduce the concept of continuity and how it relates to limits. What is a statistical question, examples of statistical questions and not statistical questions, statistical question is one that anticipates variability in the data related to the question and accounts for it in the answers, Common Core Grade 6, 6. A key problem is identified and a solution is agreed upon and implemented Teachers adopt the role as facilitators of learning, guiding the learning process and promoting an environment of inquiry Rather than having a teacher provide facts and then testing students ability to recall these facts via memorization, PBL attempts to get students to. Research And Hypothesis. Communicate the Results WHAT ARE THE 6 STEPS OF THE SCIENTIFIC METHOD? Observation– the process of using the five senses to gather information. Title and label your graph. 75 The method of science requires that independent observers must: a. The second choice is an example of parametric hypothesis testing and goodness of ±t testing. (G) State at least two ways to improve the experiment described in the scenario. First, the hypotheses should be about the population mean (μ) not the sample mean. For instance, during the discussion of answers to the question of how many M&M’s are in the bag, the numbers put forth are hypotheses. variable C. Buy custom written papers online from our academic company and we won't disappoint you with our high quality of university, college, and high school papers. This is known as artificial key. significant. Tough Stuff) 2. Get an answer for 'What's is the hypothesis, independent and dependent variables, control group, and randomization of this experiment? Design an experiment to see if a certain new dietary. You will use Excel to carry out hypothesis testing and create confidence intervals involving $$t$$-distributions. Hypotheses usually are relatively narrow in scope; theories have broad explanatory power. The answer key indicates a correct answer provided by the question, but might not be the only acceptable answer. If you decide to change one of your answers, be sure to erase the first mark completely. Once a scientist has a hypothesis. Use the force on the string as found in step one and the mass of the cart to calculate the true acceleration. Every year there is new knowledge. Moue i x 30cm x 30an 8 hours Mouse 3 30an 30cm 50 rnL , 25 IS hours Yes Mouse 30cm x 30cm 18 'wurs execase. Potential Energy. 30-60 minutes to review activity, collect materials, and make copies; Extension: 1-2 hours to collect materials (10-15 minutes). Ideal Gas Law Practice Worksheet Answer Key In this gas laws worksheet, students answer 15 multiple choice questions about the gas laws that include the relationships between temperature, volume and pressure of gases. We use social categories like black, white, Australian, Christian, Muslim, student, and bus driver because they are useful. In your experimental plan, describe each of the aspects below. Evaluate hypothesis 8. Only minimum credit at best if your answer implied or gave diploid genotypes or two-gene scenarios (e. 1A Worksheet (Answer Key) Chapter Review: Chapter 10 Practice Test Form A (Answer Key) Chapter 10 Test Form B. This is a 4 page product (2 page worksheet and 2 page answer key). Bailey decides to do an experiment to see if different types of popcorn pop faster in the microwave. We are not performing a hypothesis test on where the true parameter (the bias) lies in the parameter space. In a hypothesis test problem, you may see words such as "the level of significance is 1%. This is known as artificial key. Answer Key**. Inspiring Scientific Variables Worksheet worksheet images. (v) Explain the two rare progeny classes given the hypothesis of a chromosomal inversion. Practice Supply and Demand Questions 1. What groups listed below have true cell walls? A. Rather, we want to come up with a good approximiation for the bias. Explain your answer. PRACTICE Experimental Design Diagram DIRECTIONS Read several of the scenarios and construct an experimental design diagram. Answer choices are rounded to the hundredths place. protozoans 2. Push the EE or EXP button. Identify which statements are correct. Inferences l, Johng 2 broke 3 s r, Lev+-h was A hypothesis is to your question Et jnclûdes your. ” Step 5 (2 pts) Report the conclusion in context: Because the null hypothesis is rejected, we can conclude that more than 15% of the population would experience nausea, and the FDA should reject the drug. Remember, you should not hand in any of these essays as your own work, as we do not condone plagiarism! If you use any of these free essays as source material for your own work, then remember to reference them correctly. There is no right or wrong hypothesis using our question lets develop a hypothesis. Science processes are undertaken to get down to the reason why something is the way it is. squirt out farther because of additional pressure. Characteristics Use of words- what or how. Solutions a. For any other use, please contact Science Buddies. A key bed is a thin layer of rock. What is included:1. The second choice is an example of parametric hypothesis testing and goodness of ±t testing. college GPA. Using that 36%, calculate the following:. protozoans 2. The answer key indicates a correct answer provided by the question, but might not be the only acceptable answer. If you’re a regular reader, you know how we feel about the practice. The word (or words) before or after the gap. Then identify the (B) independent variable, (C) dependent variable, (D) number of repeated trials, (E) constants/ control (if present). " The result of this is that people want to evaluate their beliefs, periodically, against standards in order to judge themselves. 3 Using Data: Students use data based on observations to construct a reasonable explanation. Identify whether each statement is a hypothesis or a theory. Question - Which type of fertilizer will cause the plants to grow the most? Hypothesis - Any possible answer to the question. Choose a significance level or use the given one. A is the correct answer. Get an answer for 'What's is the hypothesis, independent and dependent variables, control group, and randomization of this experiment? Design an experiment to see if a certain new dietary. Select one of the following experimental Scenarios in. State the claim H 0 and the alternative, H a 2. as if you were the person in the scenario. What is included:1. " According to Festinger , humans have a need to be "correct. Easier to predict b. We use social categories like black, white, Australian, Christian, Muslim, student, and bus driver because they are useful. Question: Read the workplace simulated scenarios on the following pages and answer the questions related to the information provided in each case study scenario Case study Edith Jackson is an 82-year-old lady who was recently admitted to The Golden Days Nursing Home where you work. By using the time-management techniques in this section, you can improve your ability to function more effectively – even when time is tight and pressures are high. For those that do have a hypothesis, the hypothesis should derive logically. Bart Simpson Controls And Variables With Answers 1. 222 L = 145 L The answer cannot have any decimal places cause 145 has none. A numerical coefficient gives the molar amount of water included in the hydrate. Then identify the (B) independent variable, (C) dependent variable, (D) number of repeated trials, (E) constants/ control (if present). You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. What groups listed below have true cell walls? A. This resource is designed to encourage critical thinking and aid comprehension of the course material through use of the following materials: Case studies and corresponding questions Figure-labeling exercises Crossword puzzles Matching, fill-in-the-blank, short-answer, and multiple-choice questions The Student Workbook also includes an answer key that is page referenced to the Fire. drawing conclusions D. For this purpose, you should describe a number of specific situations that could trigger use of the product or service you are designing. Practice Identifying Parts of the Scientific Method. This assumption is called the null hypothesis and is denoted by H0. Easily add class blogs, maps, and more!. Students use 15 of the inventions from the worksheet to create a time line on a piece of adding machine tape. Hypothesis is practical as well as very powerful and will often find bugs that escaped all other forms of testing. This is a 4 page product (2 page worksheet and 2 page answer key). There may be more than one correct answer. परीक्षा (Exam) – CTET Paper II Primary Level (Class VI to VII) भाग (Part) – English Langauge I परीक्षा आयोजक (Organized) – CBSE कुल प्रश्न (Number of Question) – 30 परीक्षा तिथि (Exam Date) – 31st January 2021 (2nd Shift) Directions: Read the passage given below answer the questions (Q. I have noticed that people who worry can also be sensitive and empathetic. This worksheet can be used as a practice. Abstract In this interview for Think magazine (April ’’92), Richard Paul provides a quick overview of critical thinking and the issues surrounding it: defining it, common mistakes in assessing it, its relation to communication skills, self-esteem, collaborative learning, motivation, curiosity, job skills for the future, national standards, and assessment strategies. The true difference in height is NOT equal to 0 in the. The key-word estimate con±rms that this is the correct response. C encourage 2. Some of the worksheets below are Correlation Coefficient Practice Worksheets, Interpreting the data and the Correlation Coefficient, matching correlation coefficients to scatter plots activity with solutions, classify the given scatter plot as having positive, negative, or no correlation, …. Null hypothesis definition The null hypothesis is a general statement that states that there is no relationship between two phenomenons under consideration or that there is no association between two groups. Step 1 Form an initial hypothesis as to where you believe the fossil specimen should be placed on the cladogram based on the morphological observations you made earlier. For more information about the inputs and calculations used in this app, see “Terms and Concepts” in the Resources tab below. Easier to predict b. Scenario 1- The Effect of fertilizers on Plants. When social scientists wish to learn about an empirical phenomenon, they perform an experiment. Answer Key**. Essay: T Fill in the. A key bed is a thin layer of rock. The researcher remains objective when conducting research. 6939 (determined using statistical software or a t-table):s-3-3. 05: You can either say “Reject the null hypothesis” or “The result is statistically significant. Only minimum credit at best if your answer implied or gave diploid genotypes or two-gene scenarios (e. Let the employee think about the question and reflect on their answers. Constants - same liquids, same temperature for liquids, same size hole, same shape. Experimental Scenario #1. 3) Practice. The Monitor hypothesis 15 (a) Individual variation in Monitor use 18 4. How was the hypothesis tested? The chickens were given blood from sick patients to see if they became sick. Spongebob answer key scientific method controls and variables spongebob worksheet answers. 777 c) Now calculate the mean fitness ("w-bar"). It includes making observations and asking a question forming a hypothesis designing an experiment collecting and analyzing data and drawing a conclusion. SIMILAR WORDS IN PASSAGE. These questions describe, rather than relate variables or compare groups. A thin layer of clay was deposited over much of Earth’s surface. Will an egg float higher in water with more salt ? IV: DV:. When social scientists wish to learn about an empirical phenomenon, they perform an experiment. What three things must happen in order for a theory to be formed? 15. When they examined the effects on individual learners, they found that 84% (101/120) students performed better on the final tests when they used retrieval. If the price of chocolate-covered peanuts increases from 40 cents to 50 cents, Dan will reduce his quantity demanded from 160 bags to 140 bags due to: a) The law of demand b) The law of supply c) A decline in his income d) A change in his tastes and preferences 2. Peter loves music. Qualitative observation using. Rather, we want to come up with a good approximiation for the bias. Hypothesis Testing Steps for a hypothesis test: 1. Each item of clothing will begin at a controlled temperature of 22. Scientific Method Practice - COLLABORATIVE WORK - ANSWERS Scenario #1 - FLOOR WAX I. The key concept of the scientific method is that scientists can answer questions and test hypotheses, or potential solutions, through highly controlled experiments. Will an egg float higher in water with more salt ? IV: DV:. First, provide students with the research scenario and the accompanying questions to have them determine the research design, statistical analysis to use, and independent and dependent variables. Any student not finished by the end of the 160 minutes may continue working but, the test must be completed within the same school day. When a hypothesis is tested many times and supported by data, it becomes known as 12. It was quite expensive, so he asked the bartender if he spoke the truth. If the sample size is large enough, then the Z test and t-Test will conclude with the same results. Most of the questions include an exhibit with an image, chart, or text passage. I do allow students five "freebies" to use if they cannot find information for some of the inventors. Inferences l, Johng 2 broke 3 s r, Lev+-h was A hypothesis is to your question Et jnclûdes your. A correct statement of the form "if P then Q" gets turned into "Q therefore P". Control / Control Group - Pot # 3 ( it’s given the “recommended” amount of water) 4. How was the hypothesis tested? The chickens were given blood from sick patients to see if they became sick. Lesson 12 Recap The key characteristic of dependent samples (or matched pairs ) is that knowing which subjects will be in group 1 determines which subjects will be in group 2. What is hypothesis practice using scenarios answer key Form? The hypothesis practice using scenarios answer key is a fillable form in MS Word extension you can get completed and signed for specified purposes. She is planning to keep all of the plants she is testing indoors, on the window sill of her classroom. Part III- Scenario. Every answer will be provided in a clear manner and you will be able to get back to your assignment at once. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. A hypothesis, in general, is an assumption that is yet to be proved with sufficient pieces of evidence. H0: µ = 100 HA: µ <> 100 I. Then either graph the data by hand and submit pictures in the next assginment, or you can graph in excel and submit a file upload. Practice: Write a hypothesis about the following situations: 1. Answers to research scenarios A1) What is the experimental unit? - A single delivery from Mills Foods What is being measured? - The time from order to delivery Is the measured value quantitative or categorical? - The measured value is quantitative How many samples are there? - There are two samples, one for Food Giant and one for Mega Mart Are the samples dependent, or independent?. Lesson 12 Recap The key characteristic of dependent samples (or matched pairs ) is that knowing which subjects will be in group 1 determines which subjects will be in group 2. Solution Problem 4 (from Unit 6, Lesson 17) The height of the water in a tank decreases by 3. Hypothesis or significance testing is a mathematical model for testing a claim, idea or hypothesis about a parameter of interest in a given population set, using data measured in a sample set. Thus the question. Step-by-Step SPSS® Tutorial Videos created by the author provide screencast demonstrations for 26 key chapter concepts. Using Statistics business scenarios begin each chapter, showing how statistics is used in accounting, finance, information systems, management, or marketing. So the hypothesis test would look as follows: H 0: µ 1 =µ 2 =µ 3 =µ 4 =µ 5 =µ 6 =µ 7 =µ 8 =µ 9 =µ 10 H a: H 0 is not true Test Stat: ANOVA: F = 1. Ag Science: HYPOTHESIS WORKSHEET ANSWERS. • Your hypothesis should: o make sense to anyone who can read it. How to Use a Dichotomous Key - a 32 slide show which starts with a classification system of common desk items ; Key to macroinvertebrate life in a pond. If the hypothesis is not explicitly stated, write one for the scenario. Hydrates are named using prefixes for the word hydrate (at right). K m, disorder personality borderline etiological hypothesis. 44, with a range from 18. Students learn the definitions for question hypothesis experiment data and conclusion then use them to label a science report. Control / Control Group - The Sections that get NO WAX 4. Step 1 Form an initial hypothesis as to where you believe the fossil specimen should be placed on the cladogram based on the morphological observations you made earlier. It’s not a rapid-fire interrogation. You are testing the hypothesis that a new method for freezing green beans preserves more. List the three must haves of a hypothesis. This worksheet can be used as a practice. ) The control group is not given the variable being. Directions: Read the experiment scenario and then identify the components of the scientific method by completing the graphic organizer provided. The second choice is an example of parametric hypothesis testing and goodness of ±t testing. Use the +/- button to change its sign. The SAT Suite of Assessments is an integrated system that includes the SAT, PSAT/NMSQT, PSAT 10, and PSAT 8/9. Regardless of which scenario the student chooses, they will be required to answer questions demonstrating knowledge of a learning objective, making them the perfect questions to assign toward the end of a chapter. The key is to take it slow and easy with your questions. Her task is to answer the method worksheet you can use to practice. I Can #2: List the steps of the scientific method in order and explain each step. 5 points for each correct answer) 1. Hypothesis - If we vary the amount of water given to different plants, then the plants. Apr 12, 2015 - This Pin was discovered by lisa jones. Given paired data, do a hypothesis test for the slope of the regression line using the TI-83. How was the hypothesis tested? The chickens were given blood from sick patients to see if they became sick. clinical scenarios. The scientific method is a process that can help you in all walks of life, not just in a science lab Scientific method practice worksheet answer key. 1 Use the instructions in Chapter 6 and Chapter 7 of the SPSS Survival Manual to answer the following questions concerning the variables included in the survey. He likes it so much he eats it faster than it grows in the 8 pots he has on his window sill. The Use of Theory o ne component of reviewing the literature is to determine what theories might be used to explore the questions in a scholarly study. Introduction, Activity A: Lesson Citation: Acquire Knowledge: Students investigate photosynthesis. Conduct background research about your question. There is convincing evidence to suggest a ffence. My Classroom Material AP Biology Pre AP Biology Biology I Parent’s Pond NGSS Resources Supplies Prefix-Suffix List My Frog Pond Biology Club Biology Curriculum Map Lab Reports Classroom Rules How To Study Biology Sophomore Pacing Guide UBD Unit Lesson Plans Physical Science Physics for Physical Science Chemistry for Physical Science Help for teachers Writing an … Continue reading "". The student newspaper at a large. All species co-evolve with other organisms; for example predators evolve with their prey, and parasites evolve with their hosts. The great thing about a question is that it yearns for an answer, and the next step in the scientific method is to suggest a possible answer in the form of a hypothesis. Some of the worksheets below are Null And Alternative Hypothesis Worksheets, 5 steps to follow when testing hypotheses, develop null and alternative hypotheses to test for a given situation, practice stating and writing null hypotheses and alternative hypothesis with examples and exercises. What skill is a scientist using when she listens to the sounds that whales 2. What he denies is that organ-like, special purpose adaptations are the likely result of such evolutionary scenarios. Potential Energy. Critical Value= 3. Saved by Docstoc. Step 3: Form a hypothesis • Hypothesis: an educated guess about a possible solution or answer to your problem or question • Developing a good, solid hypothesis is the most difficult step of the scientific method. null hypothesis. Directions: The following is an experimental scenario. And the probability of the coin landing T is ½. The key-word estimate con±rms that this is the correct response. If you would like to score your student’s online practice test, you should direct your student to record his or her answers on a separate sheet of paper. ANSWER KEY Write down what you THINK you do for each of the following steps of the scientific method: Purpose: The objective or problem that you are investigating. tests experimental and control groups in parallel. For a hypothesis, write an “H” on the line. Scientific method worksheet pdf answers. Experimental Design Scenarios Answer Key Directions: For each of the following scenarios, identify the independent variable, dependent variable, constants, control group, number of groups, number of trials per group, and hypothesis. Gather some good cause-and-effect examples you can use in class. His friend sandy told him that he should try using clean o detergent a new brand of laundry soap she found at sail mart. This worksheet can also be used as an assessment to determine understanding about the scientific method. / A Hypothesis is formed – this is an educated guess about the result of the experiment based on the information learned during background research. According to the Maternal Deprivation Hypothesis, breaking the maternal bond with the child during the early stages of its life is likely to have serious effects on its intellectual, social and emotional development. Now we can use that equation to interpolate a sales value at 21°: y = 33×21 ° − 216 = $477 EXTRApolating. Students are to return the completed questions the following class period. For our exercise-training example, the null hypothesis (H 0) is that mean blood pressure is the same at all time points (pre-, 3 months, and 6 months). Ag Science: HYPOTHESIS WORKSHEET ANSWERS. broun Wíndow p 2: Form a Hypothesis '1B'£JlJ. “The big technology companies step in and play a key role in the communication of science and the funding of research,” Mulligan said. Hypertonic Solution - B. Typically, a dichotomous key for identifying a particular type of object consists of a specific series of questions. as if you were the person in the scenario. Hypothesis: Formulating an educated prediction about the. We collect data to see if a certain population value differs from a given value (≠), is less than a given value (). Chapter 20 - Students will be able to model (and solve) realistic scenarios using hypothesis tests for proportions. Reference is a digital publisher dedicated to answering the questions of students and lifelong learners. He decides to practice this newly gained scientific knowledge and design a controlled experiment that deals with the proper conditions to grow seeds. Practice firing the gun the same way each time to minimize the variation in initial velocity. Hypothesis is practical as well as very powerful and will often find bugs that escaped all other forms of testing. Question: What effect does the water temperature have on dissolving rate? Hypothesis: If the tablet is placed in warm water, then the tablet will dissolve faster. It is in large part an adaptation and generalization of classic methods used by military intelligence. Any student not finished by the end of the 160 minutes may continue working but, the test must be completed within the same school day. The true difference in height is equal to 0 in the population. Real life stories,. Calculate the attack rate for the potato salad. The theory also claims that chloroplasts would appeared through endosymbiosis due to the entry of photosynthetic prokaryotes into aerobic eukaryotes, establishing a mutualist ecological interaction. pre-assessment, post-assessment, entrance ticket, exit ticket or homework assignment. Statistical Hypothesis – a conjecture about a population parameter. This assumption is called the null hypothesis and is denoted by H0. The second choice is an example of parametric hypothesis testing and goodness of ±t testing. Spongebob scientific method worksheet answer key. Post your questions to our community of 350 million students and teachers. The answers, with explanations, for exercises 4 and 5 found at the end of each chapter, which instructors can assign as homework or exam questions. 0 x 10 5 times 4. The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. If we can assign people to a category then that tells us things about those people, and as we saw with the bus driver example, we couldn't function in a normal manner without using these categories; i. If the biologist set her significance level $$\alpha$$ at 0. The term “secular stagnation” was coined by economists in the 1930s and recently become a catch description for long-term economic pessimism. The guerilla girls and other I am portant. Use the scenario below to write a title and a hypothesis using the following formats: Title: The Effect of the (changes in the independent variable) on the. Next, form a hypothesis, and conduct an experiment. You will need to print out this activity sheet OR Write out the answers on a separate piece of paper. Check conditions. Take a photo of your homework question and get answers, math solvers, explanations, and videos. ** 10-year risk for ASCVD is categorized as: Low-risk (<5%) Borderline risk (5% to 7. Experiment testing an idea or hypothesis through a controlled investigation. Scientific Method Scenarios Worksheet2. The slideshow presents the concepts of testable questions, hypotheses, variables and controls, which are the basis of students being able to plan and carry out investigations (NGSS Practice 3). Show that you have mastery over the idea behind hypothesis testing by calculating some probabilities and drawing conclusions. Although you could state a scientific hypothesis in various ways, most hypotheses are either "If, then" statements or forms of the null hypothesis. Think of a possible testable question for each scenario below, identify the Manipulated Variable (MV), Responding Variable (RV) and then write a hypothesis in correct format as if you were the person in the scenario If you already know the answer, don't let this influence how you write the hypothesis. The use of Research Questions as opposed to objectives or hypothesis, is more frequent. Hypothesis Practice Using Scenarios. Identify whether each statement is a hypothesis or a theory. Given that you are using these passages to benchmark, my guess is that the time period between assessment times is about 14-15 weeks? So the likelihood of a practice effect due to using the same passage is greatly reduced, and it might give you more stability in performance levels. Adjust the spring tension so that the range, when fired horizontally, is between one and two meters. 5 trillion and government transfers are$1 trillion. One key point: the dot is not a multiplication sign. The clay has large amount of the. 1: Definitions of Statistics, Probability, and Key Terms Use the following information to answer the next three exercises. The purpose of a hypothesis is to provide a direction for future scientific investigation. She up four cages with a mouse in each cage. The questions are based on the following topics: Regression Equations, Method of Ordinary Least Squares, Assumptions of Classical Linear Regression Model, Hypothesis Testing and Goodness of Fit. ___ Use a sample of size 50 to determine whether there is a linear relationship between a car’s age and its gas mileage. First, the hypotheses should be about the population mean (μ) not the sample mean. Like index fossils, key beds are used to match rock layers. Experiment testing an idea or hypothesis through a controlled investigation. In Step 1 of the investigation, the student adds CO2 to the flask at 20℃. Potential Energy. 2, illustrate the elasticity of demand in the market for whiskey. Explain what the sociological perspective encompasses and then, using that perspective, discuss the forces that shaped the discipline of sociology. Researchers call his hypothesis "social comparison theory. An Alternative Decision Rule using the p - value Definition The p-value is defined as the smallest value of α for which the null hypothesis can be rejected. Be sure that the number of the question you are answering matches the number of the row of answer choices you are marking on your answer sheet. He learned how to conduct a controlled experiment in his science class. That would imply two genes, and we are talking about alleles of the same gene. Scientific Method Practice. Gram-positive bacteria D. Prepare situations or scenarios for your personas. There are many species of. Failure is Not an OptionIt's a Requirement! The Key to Science (and Life) is Being Wrong. Read the experiments and then identify the components of the scientific method by completing the graphic organizer provided. Identify whether each statement is a hypothesis or a theory. Do NOT use the x (times) button!! Enter the exponent number. 3 The concept was named in reference to the Red Queen's race in Lewis Carroll's book, Through the Looking-Glass.
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This code generates plots and display them in column. I have been trying to plot a system of differential parametric equations without any success. It appears as though Mathematica is solving them, but it won't plot them. Wolfram Community forum discussion about Plot solution curves from a differential equation?. The differential equations governing the situation are: and , where F1 = 2, F2 = 12, V1(0) = 0, V2(0) = 0. I've just started to use Python to plot numerical solutions of differential equations. 5. It can handle a wide range of ordinary differential equations (ODEs) as well as some partial differential equations (PDEs). Tuitorial 2: Differential equations in Mathematica: Analytic solutions Brian Washburn, Version 1.0, 01/08/06 Off@General::spellD; ü Differential equations solved analytically in Mathematica Typically one uses the function DSolve?DSolve DSolve@eqn,y,xDsolves a differential equation for the function y, with … Here is the equation for the graph, a second order linear differential equation with constant coefficients: Here is the graph traced by this equation: Quote from the book "Times Series Analysis and Forecasting By Example": Here N=N(t) is a Poisson process. Close. Edit: Alright, you don't need some of that stuf: arrow tip at the end of the curve, the axes->True, thickness, plotstyle, plotpoints. Our first step is to load into Mathematica a special graphics package called PlotField which is needed to … We use DSolve to find analytical solutions and … I'm trying to graph a first order differential equation in Mathematica but the graph doesn't seem right when compared to my txt book. New in Mathematica 10 › Enhanced Calculus & Differential Equations › Directly Obtain Solution Expressions for Difference Equations Directly obtain an expression for the solution of an O Δ E using RSolveValue . I'm pretty new to Mathematica so go easy on me. This means that I am asking Mathematica to plot solutions with different values of the differential constant c. This command will plot a variety of different graphs over the range of c. The command on the second line also yields the different members of this family of equations associated with each value of c. ... Mathematica for solving coupled ordinary differential equation. Checking PS#3 differential equations: Getting answers for the differential equations in PS#3, including using extracting the equations in convenient form. Homogeneous equations A first-order ODE of the … With −≤ ≤10 10x, y, for instance, a number of apparent straight line solution curves should be visible. Again, very simple. Differential Equations With Mathematica This is likewise one of the factors by obtaining the soft documents of this differential equations with mathematica by online. 1.3.4. Recall that the introductory differential equation we've been using so far is: dy/dx = x 2. k = velocity of growth = 1/s. 25-Sep-2011: Nonlinear Diff Eq: Mathematica notebook solving the nonlinear Duffing differential equation, looking at the time dependence, phase space plots, Poincare sections, and the power … Bernoulli type equations Equations of the form ' f gy (x) k are called the Bernoulli type equations and the solution is found after integration. In a system of ordinary differential equations … Differential equation models for population dynamics are now standard fare in single-variable calculus. After some research, you have come up with the following model to describe … the general equation is y(t)=tsin(t)+Ct i have to choose different values for C and plot that on the same graph my question is how do i do that? They are a set of four partial differential equations that describe how electric and magnetic fields respond to … Methods in Mathematica for Solving Ordinary Differential Equations 2.3. Examples of answers for the differential equations in PS#3, including using extracting the equations in convenient form. Differential Equation Example of Root Locus Plots in Mathematica. Plot the solutions for =0.1,0.2,0.5on the same graph. However, if I remove the [x] in the equations, mathematica states that the system is overdetermined. Get the free "General Differential Equation Solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. Below is an example of what the Mathematica equations … So I'm very new on this software and currently I'm trying to plot all the orders of the Taylor series of … For this equation we would like the vectors plotted by VectorPlot to have a slope of x 2 at each and every point (x, y) chosen to be part of our plot. ... Now all what’s left is to plot results. I have put my code that I am using below and was hoping someone could take a look at what I did wrong: Is the there any software for plotting solutions of the stochastic differential equation dx=F(x)dt +G(x)dN ???? How Graph differential equations with Matlab. Using Mathematica for ODEs, Part 2 (Use VectorPlot and Show for a pure antiderivative problem) - Duration: 11:19. 1. Let’s consider the power series solution of the Hermite differential equation: ${\displaystyle u''-2xu'=-2 n u}$ ${\displaystyle u''-2xu'+2 n u =0 \qquad (1)}$ The solutions to the Hermite differential equation ca … You might not require more era to spend to go to the book opening as capably as search for them. I'm using NDSolve. There are many reasons to love Maxwell’s equations. N(t) = #individuals. And with Mathematica 9′s new vector analysis functionality, exploring them has never been easier. how can i store values for C on mathematica then graph those values on … dN(t)/dt = the derivative of N(t) = change of # individuals = #individuals/s. Probably many know that Wolfram Mathematica is a great tool. If you write the ODEs in Mathematica by directly substituting in F1 = 2 and F2 = 12, you will have to change the ODEs each time you change the values of F1 and F2. Find more Mathematics widgets in Wolfram|Alpha. 2. If so, can you post the same for your differential equation? I need a code in matlab for plotting bifurcation diagram for the differential equation: v'(t)=2*G*J1[v(t-τ)]cos(w*τ)-v(t). In some cases, you likewise do not discover the message differential equations … Following example is the equation 1.34 from [3]: 2.4. Plot a direction field and typical solution curves for the differential equation dy dx/sin()=−x y, but with a larger window than that of Fig. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. f(t) = production function = … Posted by 4 years ago. (a) Substitute yaxb=+ in the differential equation to determine what the coefficients Output graphs below. I've got the following differential equation: dN(t)/dt - ((k - (a*N(t)))*N(t)) = f(t) This is the logistic law of population growth. I know that the Mathematica can plot the Ito ` s solutions but I failed to find commands which plot the solutions of the above equation. The analytical solutions of the two differential equations and , subject to the initial conditions and are used to create two plots, a parametric plot of a curve with horizontal coordinate and vertical coordinate and a standard plot … Phase portrait of system of nonlinear ODEs. a = an inhibition factor on the growth = 1/(#individual*s). Just take them out to … Plot a direction field for the differential equation y’=-x/y. ... How to plot a phase portrait for system of differential equations in mathematica or R? We use the VectorScale options Tiny (size of vectors relative to bounding boxes), Automatic (aspect ratio), and None (all vectors the same length). Since Mathematica does not have a function to plot direction fields, we use the VectorPlot function to plot the vector field {1,-x/y}. B.3.a.i) Merging plots when r > 0 Here are three plots of solutions a random forced … We wish to plot the direction field defined by the slope vector {F(x,y), G(x,y)}. A plot of the results is attached. 25-Sep-2011: Nonlinear Diff Eq: Mathematica notebook solving the nonlinear Duffing differential equation, looking at the time dependence, phase space plots, … So what exactly are Maxwell’s equations? Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interests. How can I plot the following coupled system? Does anyone know the Mathematica code that will trace the graph below? Mathematica for Differential Equations. Solving Differential-Algebraic equations. ME 163 Using Mathematica to Solve First-Order Systems of Differential Equations In[1]:= Off@General::spell1D; In[2]:= Off@General::spellD; In this notebook, we use Mathematica to solve systems of first-order equations, both analytically and numerically. Bill Kinney 11,978 views Introduction to Advanced Numerical Differential Equation Solving in Mathematica Overview The Mathematica function NDSolve is a general numerical differential equation solver. Learn more about matlab, ezplot, plot, differential equations, ode Building on these ordinary differential equation (ODE) models provides the opportunity for a meaningful and intuitive introduction to partial differential equations (PDEs). Your plots should be adjusted such that terminal velocities in the … The code has already been written in mathematica … This Demonstration plots the phase portrait (or phase plane) and the vector field of directions around the fixed point of the two-dimensional linear system of first-order ordinary differential equations. An example problem pulled from the Fall 2008 second exam: "You have just been put in charge of designing a large-scale bioreactor for the production of a blockbuster protein based drug. 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# Sequence to explore the complexity of the NP problem
Let $$X$$ be some problem known to be in $$NP$$. What is the natural next step in exploring the complexity of the problem? Is it trying to prove whether it is in $$P$$ or try to prove it is $$NP$$-Hard? Taking the example of the Graph-isomorphism problem it seems to me that after discovering that problem is in $$NP$$. It is not easy to decide what to do next. Some people suggest me to prove it into $$NP \cap CoNP$$.
Question : What is worth to try if it is only known that problem is in $$NP$$? Is there any sequence (like first $$NP$$, $$NP$$-Hard, ....) to explore the complexity of the problem?
So you have found out $$X \in \mathbf{NP}$$. Unfortunately, as Juho hints at in their answer, there is no "list" you can systematically go over to try and investigate $$X$$ further. This is mainly because separation results in complexity theory are (yet) few and far between, so most of the things you may try will not explicitly fail; you'll only be unsuccessful. For example, if you try to show $$X \in \mathbf{P}$$ you might succeed, but if it turns out to be incorrect it is very unlikely you will be able to prove it (since then $$\mathbf{P} \neq \mathbf{NP}$$).
This lack of "feedback" means you should tackle $$X$$ from different perspectives, and it is only experience (and a good deal of luck!) which will give you a "hunch" as to where the most promising one is. The following are a few options you might consider, as well as their implications if you manage to prove or disprove them (if any). Note I am only considering structural complexity theory here (which seems to be what you are interested in), but by all means this is what the entirety of complexity theory is about. Depending on the structure of $$X$$, it is also possible to consider $$X$$ from a parameterized complexity point of view, or try to find subexponential algorithms for it (see also number 4 below), and so forth and so on.
1. $$X$$ is $$\mathbf{NP}$$-complete
Prove: This is basically the end of the line for investigating $$X$$. It is also not a very surprising result since the list of $$\mathbf{NP}$$-complete problems is huge (and it already was when Richard Karp started to compile this list).
Disprove: Congratulations! You have proved $$\mathbf{P} \neq \mathbf{NP}$$.
2. $$X \in \mathbf{P}$$
Prove: This is the second most likely outcome and even more unsurprising than the former.
Disprove: Congratulations! You have proved $$\mathbf{P} \neq \mathbf{NP}$$.
3. $$X \in \mathbf{coNP}$$
Prove: Then $$X$$ is in $$\mathbf{NP} \cap \mathbf{coNP}$$ (which, for instance, includes factoring). Reconsider $$X \in \mathbf{P}$$.
Disprove: Congratulations! You have proved $$\mathbf{NP} \neq \mathbf{coNP}$$.
4. $$X \in \mathbf{GI}$$ (the graph isomorphism class)
Prove: Then $$X$$ admits a quasipolynomial-time algorithm (and is unlikely to be $$\mathbf{NP}$$-complete). Reconsider $$X \in \mathbf{P}$$.
Disprove: Congratulations! You have proved $$\mathbf{P} \subseteq \mathbf{GI} \neq \mathbf{NP}$$
5. $$X \in \mathbf{BPP}$$
Prove: Since it is widely suspected that $$\mathbf{BPP} = \mathbf{P}$$, this is only interesting if $$X \in \mathbf{P}$$ is not obvious. Consider also $$X \in \mathbf{RP}$$ or even $$X \in \mathbf{ZPP}$$.
Disprove: Congratulations! You have proved $$\mathbf{NP} \neq \mathbf{BPP}$$ (and quite likely also $$\mathbf{P} \neq \mathbf{NP}$$).
And the list goes on and on. As you can see, disproving any of these options leads to separation results, all of which would be major breakthroughs and, thus, unlikely to be proven simply by considering a random problem $$X \in \mathbf{NP}$$. This means you cannot expect to go through this list from head to tail making ticks and crosses as you go; you have to "guess" which of the options (if any!) is more likely and (with some luck) you'll be able to prove it.
It depends, you should try to do whatever the problem smells like. You will need experience to get a good hunch.
It's also not uncommon to try to do both. You might start by trying to come up with a polynomial time algorithm. Getting stuck? Why is that? Could you simulate that in an NP-hardness proof? Or you might start by building an NP-hardness proof from a "nearby problem" that has similar characteristics. While playing around with suitable gadgets you might realize things like "oh, any algorithm will always have to do this when encountering this type of subgraph..."
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## Precalculus (6th Edition)
Plot the point $(-1, 3)$. Use the slope to plot another point. A slope of $\frac{3}{2}$ means a 3-unit increase in $y$-value corresponds to a 2-unit increase in the value of $x$. From $(-1, 3)$, move 3 units up and 2 units to the right to reach $(1, 6)$. Complete the graph by using a line to connect the two points. (Refer to the graph in the answer part above.)
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# mgwr.gwr.GWR¶
class mgwr.gwr.GWR(coords, y, X, bw, family=<spglm.family.Gaussian object>, offset=None, sigma2_v1=True, kernel='bisquare', fixed=False, constant=True, dmat=None, sorted_dmat=None, spherical=False)[source]
Geographically weighted regression. Can currently estimate Gaussian, Poisson, and logistic models(built on a GLM framework). GWR object prepares model input. Fit method performs estimation and returns a GWRResults object.
Parameters: coords : array-like n*2, collection of n sets of (x,y) coordinates of observatons; also used as calibration locations is ‘points’ is set to None y : array n*1, dependent variable X : array n*k, independent variable, exlcuding the constant bw : scalar bandwidth value consisting of either a distance or N nearest neighbors; user specified or obtained using Sel_BW family : family object underlying probability model; provides distribution-specific calculations offset : array n*1, the offset variable at the ith location. For Poisson model this term is often the size of the population at risk or the expected size of the outcome in spatial epidemiology Default is None where Ni becomes 1.0 for all locations; only for Poisson models sigma2_v1 : boolean specify form of corrected denominator of sigma squared to use for model diagnostics; Acceptable options are: ‘True’: n-tr(S) (defualt) ‘False’: n-2(tr(S)+tr(S’S)) kernel : string type of kernel function used to weight observations; available options: ‘gaussian’ ‘bisquare’ ‘exponential’ fixed : boolean True for distance based kernel function and False for adaptive (nearest neighbor) kernel function (default) constant : boolean True to include intercept (default) in model and False to exclude intercept. dmat : array n*n, distance matrix between calibration locations used to compute weight matrix. Defaults to None and is primarily for avoiding duplicate computation during bandwidth selection. sorted_dmat : array n*n, sorted distance matrix between calibration locations used to compute weight matrix. Defaults to None and is primarily for avoiding duplicate computation during bandwidth selection. spherical : boolean True for shperical coordinates (long-lat), False for projected coordinates (defalut).
Examples
#basic model calibration
>>> import libpysal as ps
>>> from mgwr.gwr import GWR
>>> data = ps.io.open(ps.examples.get_path('GData_utm.csv'))
>>> coords = list(zip(data.by_col('X'), data.by_col('Y')))
>>> y = np.array(data.by_col('PctBach')).reshape((-1,1))
>>> rural = np.array(data.by_col('PctRural')).reshape((-1,1))
>>> pov = np.array(data.by_col('PctPov')).reshape((-1,1))
>>> african_amer = np.array(data.by_col('PctBlack')).reshape((-1,1))
>>> X = np.hstack([rural, pov, african_amer])
>>> model = GWR(coords, y, X, bw=90.000, fixed=False, kernel='bisquare')
>>> results = model.fit()
>>> print(results.params.shape)
(159, 4)
#predict at unsampled locations
>>> index = np.arange(len(y))
>>> test = index[-10:]
>>> X_test = X[test]
>>> coords_test = np.array(coords)[test]
>>> model = GWR(coords, y, X, bw=94, fixed=False, kernel='bisquare')
>>> results = model.predict(coords_test, X_test)
>>> print(results.params.shape)
(10, 4)
Attributes: coords : array-like n*2, collection of n sets of (x,y) coordinates used for calibration locations y : array n*1, dependent variable X : array n*k, independent variable, exlcuding the constant bw : scalar bandwidth value consisting of either a distance or N nearest neighbors; user specified or obtained using Sel_BW family : family object underlying probability model; provides distribution-specific calculations offset : array n*1, the offset variable at the ith location. For Poisson model this term is often the size of the population at risk or the expected size of the outcome in spatial epidemiology Default is None where Ni becomes 1.0 for all locations sigma2_v1 : boolean specify form of corrected denominator of sigma squared to use for model diagnostics; Acceptable options are: ‘True’: n-tr(S) (defualt) ‘False’: n-2(tr(S)+tr(S’S)) kernel : string type of kernel function used to weight observations; available options: ‘gaussian’ ‘bisquare’ ‘exponential’ fixed : boolean True for distance based kernel function and False for adaptive (nearest neighbor) kernel function (default) constant : boolean True to include intercept (default) in model and False to exclude intercept dmat : array n*n, distance matrix between calibration locations used to compute weight matrix. Defaults to None and is primarily for avoiding duplicate computation during bandwidth selection. sorted_dmat : array n*n, sorted distance matrix between calibration locations used to compute weight matrix. Defaults to None and is primarily for avoiding duplicate computation during bandwidth selection. spherical : boolean True for shperical coordinates (long-lat), False for projected coordinates (defalut). n : integer number of observations k : integer number of independent variables mean_y : float mean of y std_y : float standard deviation of y fit_params : dict parameters passed into fit method to define estimation routine W : array n*n, spatial weights matrix for weighting all observations from each calibration point points : array-like n*2, collection of n sets of (x,y) coordinates used for calibration locations instead of all observations; defaults to None unles specified in predict method P : array n*k, independent variables used to make prediction; exlcuding the constant; default to None unless specified in predict method exog_scale : scalar estimated scale using sampled locations; defualt is None unless specified in predict method exog_resid : array-like estimated residuals using sampled locations; defualt is None unless specified in predict method
Methods
fit([ini_params, tol, max_iter, solve, …]) Method that fits a model with a particular estimation routine. predict(points, P[, exog_scale, exog_resid, …]) Method that predicts values of the dependent variable at un-sampled locations
df_model df_resid
__init__(coords, y, X, bw, family=<spglm.family.Gaussian object>, offset=None, sigma2_v1=True, kernel='bisquare', fixed=False, constant=True, dmat=None, sorted_dmat=None, spherical=False)[source]
Initialize class
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# Tag Info
50
From a cryptographic standpoint it is OK to expose a public key in the sense of revealing its value. The most basic assumption in cryptography involving public/private key pairs is that the value of a public key is public; hence its name. It is extremely important that an adversary can not alter a public key. Any exposition that would allow alteration must ...
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While it is true that Elliptic Curve Diffie Hellman, Elliptic Curve Signature Generation and Elliptic Curve Signature Verification rely on scalar multiplications, these are usually implemented as different types of scalar multiplication for both security and efficiency reasons. In fact there are three types of scalar multiplications used in practice for ...
30
Ed25519 is a specific instance of the EdDSA family of signature schemes. Ed25519 is specified in RFC 8032 and widely used. The only other instance of EdDSA that anyone cares about is Ed448, which is slower, not widely used, and also specified in RFC 8032. Keys and signatures in one instance of EdDSA are not meaningful in another instance of EdDSA: Ed25519 ...
22
Clearing the lower 3 bits of the secret key ensures that is it a multiple of 8, which in turn ensures that no information, small as it may be, about the secret key is leaked in the case of an active small-subgroup attack. The typical simple Diffie-Hellman key exchange works like this: $$\text{Alice} \xrightarrow{\hspace{3cm} a G \hspace{3cm}} \text{Bob} \\... 19 The old terminology was confusing, so they've rebranded a bit. X25519 is Elliptic Curve Diffie-Hellman (ECDH) over Curve25519 Ed25519 is Edwards-curve Digital Signature Algorithm (EdDSA) over Curve25519 Libsodium's ref10 curve25519 code is actually used both by crypto_scalarmult()/crypto_box() as well as crypto_sign(). 17 If I understand your question correctly, you are essentially asking if points in Edwards and Montgomery curves can be represented in Weierstrass coordinates. This is true; in fact, any elliptic curve over a field of characteristic \neq2,3 can be represented in Weierstrass form \mathcal{E}_{w}^{a, b} : y^2 = x^3 + ax + b, and by extension its points can ... 16 How many qubits are required for breaking RSA 2048 and RSA 4096 in real-time with a quantum computer? Like the answer you linked to shows, about \log_2(N^2) = 2 \log_2(N) or just 2n where n is the number of bits of the modulus N, i.e. the key size of RSA. So 4096 for 2048-bit RSA, double that for 4096-bit. This paper (PDF) has an algorithm using ... 15 Edwards25519 is the twisted Edwards curve$$-x^2 + y^2 = 1 - (121665/121666) x^2 y^2$$over the prime field \mathbb F_p where p = 2^{255} - 19. The coefficient d = -121665/121666 was chosen to so that this curve is birationally equivalent to the Montgomery curve y^2 = x^3 + 486662 x^2 + x, called Curve25519, whose coefficient 486662 was chosen to be ... 13 If you can store the private key with some pre-computed work, then you can pick almost any public key you want. So in a way, it depends on the implementation. Here's a diagram of how Ed25519 works, note how keys are generated: (Image source.) A more detailed description (that is simpler than the actual paper) of the process is in these slides (slides 9 - ... 13 The advent of quantum computing, real usable out of the lab QC, will pretty much be the end of any encryption that relies on the difficulty of the discrete log problem via Shor's Algorithm et al. That is not to say it is the death of all modulus based encryption schemes, probably just asymmetric public/private key based schemes. That said, QC's aren't magic,... 13 if Ed25519 has gone through rigorous cryptanalysis It is based on Curve25519 which has gone through extensive cryptanalysis. The Ed25519 signature scheme as well is being heavily reviewed and adoption is rapid. There are already a number of papers on the algorithm itself, as well as a few papers on specific implementations. Every part of the algorithm and ... 13 There's a few different related parts here, and the nomenclature of the library you've cited is a little confusing. Curve25519 is an elliptic curve over the finite field \mathbb F_p, where p = 2^{255} - 19, whence came the 25519 part of the name. Specifically, it is the Montgomery curve y^2 = x^3 + 486662 x^2 + x, but you don't need to know the ... 12 It is not possible to double security level of Ed25519 any trivial way. Instead, doubling security level requires using another curve that is approximately 512 bit curve. In systems compliant with RFC 7748, i.e. some of IETF specifications, there is Curve448 curve (Ed448-goldilocks). It is almost twice as strong as Curve25519 (its strength is 224 bits). ... 11 Well, lets go through the issues: It seems to be possible to retrieve the (public) key used for creating an ECDSA signature just from the signature alone Nope, not quite. You also need the message being signed. And, with that, it doesn't give you the unique public key; it does allow you to narrow it down to two possibilities (assuming you're using a ... 11 Trevor Perrin wrote a library doing exactly that. Explanation can be found on in the curves mailing list archives. To convert a Curve25519 public key x_C into an Ed25519 public key y_E, with a Ed25519 sign bit of 0:$$y_E = \frac{x_C - 1}{x_C + 1} \mod 2^{255}-19$$The Ed25519 private key may need to be adjusted to match the sign bit of 0: if ... 11 Curve25519 makes use of a special x-coordinate only form to achieve faster multiplication. Ed25519 uses Edwards curve for similar speedups, but includes a sign bit. While it could have been done differently, doing it this way simplifies implementations that only need one of encryption or signing. 11 Yes, this should be secure. I am not familiar with TweetNaCl, so I cannot speak on the concrete implementation. However, the general construction of signing a hash of a message instead of the message should be secure. It is in fact a standard way to sign messages. In general it should work for any secure signature scheme and cryptographic hash function. ... 11 The formulas actually work. You just have to keep in mind to make computation in the field of intergers modulo 2^{255}-19 and that there are actually two square roots, you need to use the right one if you want to have the expected result. You can test the following SAGE code gf=GF(2^255-19) X_P = gf(... 10 All of these are answered by the SafeCurves project: For twisted Edwards curves, ax^2 + y^2 \equiv 1 + dx^2y^2 \pmod p. Edwards curves are the special case a = 1. Edwards curves can be converted to Montgomery form.Montgomery curves can be converted to Weierstrass form.Some, but not all, Weierstrass curves can be converted to Montgomery form. The ... 10 That example private key is not encoded with PKCS#8, but with the format described in RFC 5958, which is supposed to replace PKCS#8 (at least, in IETF parlance, RFC 5958 "obsoletes" RFC 5208, which is a copy of PKCS#8 v1.2). The ASN.1 type if called OneAsymmetricKey: OneAsymmetricKey ::= SEQUENCE { version Version, ... 9 No, conversion of an EC key pair from a curve to another of unrelated order is not possible. One of the closest things that could be done would be that parties generate a new P256 key pair, then certify their new P256 public key using their C25519 private key, check the other party's certificate using the other party's trusted C25519 public key, and now ... 9 (minisign author here) As noted by corpsfini, keys encode the Y coordinate. The X coordinate is recovered using the curve equation: X = sqrt((Y^2 - 1) / (d Y^2 + 1)). The square root has two solutions, so we need to encode the sign of x as well. Since coordinates only require 255 bits, we have a extra bit, used to encode the sign. X and Y ∈ [0; 2^255-19[, ... 8 First off, your equation is correct and there seems to be no calculation mistake. To understand on how to get from$$(2+d)x^{2}+dy^{2}=d+(d-2)x^{2}y^{2}$$to$$x^{2}+y^{2}=1+e\cdot x^{2}y^{2}$$one first needs to observe that e=(d-2)/(d+2)=121665/121666 holds. The next step is to consider: "What operations are actually allowed with birational ... 8 As you said, you need to define the goals. You can take a look at SafeCurves, which is a joint work by Bernstein and Lange to help choose/construct elliptic curves w.r.t. ECDLP difficulty and ECC security. Note that if you need a pairing-friendly elliptic curve you need to look at other criteria related to the embedding degree. You can read this paper by ... 8 Yes! You can use the ephemeral key derivation mechanism that is for example used in Monero (they call it stealth keys there). Consider public key A=aG, with private key a. Then, a derived key can be generated, parametrised by the random scalar r:$$A'=H_s(rA)G+A and the party that knows $a$ can use the public parameter $R=rG$ to compute their ...
8
The leading 04 byte is specified by the SEC standard (which is based on the ANSI X9.62 standard). It indicates that the public key point is not compressed. If the key is compressed, it uses 02 or 03 as leading byte depending on the lower bit of the y coordinate. EdDSA public keys do not use this byte for two reasons: It always uses compressed points; there ...
7
Yes. In order to understand why, you need to understand how the public key is computed. The secret key is a scalar. A fixed base point is multiplied by that scalar, and no matter what that scalar value is, you will always end up with a point in the same group as the base point. You may want to avoid an all-zero secret key but the key space is so large ...
7
Opening and sealing a box involves your private scalar and the other party's public key. So the private key is simply the private scalar. Signing with Ed25519 involves your private scalar and your public key (to strictly bind the signature to that key). Since computing the public key from the private scalar is expensive, NaCl chooses to store it as part of ...
7
Shor's algorithm can compute discrete logs in elliptic curves and thereby recover the secret scalar from a public Ed25519 key, which you can use to forge signatures of your choice. So, yes, it affects Ed25519—it completely breaks Ed25519, or it would if you could engineer a quantum computer capable of executing it. It can also compute discrete logs in ...
7
Yes, this is possible using Hierarchical Deterministic (HD) Keys. There are 2 variations for key generation, hardened and non-hardened. In hardened, generating child keys (both public and private) requires knowledge of parent private key but in non-hardened, child public key can be generated using parent public key. You need non-hardened key generation. The ...
Only top voted, non community-wiki answers of a minimum length are eligible
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The production of , and mesons decaying into the dimuon final state is studied with the LHCb detector using a data sample corresponding to an integrated luminosity of collected in proton-proton collisions at a centre-of-mass energy of TeV. The differential production cross-sections times dimuon branching fractions are measured as functions of the transverse momentum and rapidity, over the ranges GeV/ and . The total cross-sections in this kinematic region, assuming unpolarised production, are measured to be sigma (pp -> Upsilon(1S)X) x B(Upsilon(1S) -> mu(+)mu(-)) = 1.111 +/- 0.043 +/- 0.044 nb, sigma (pp -> Upsilon(2S)X) x B(Upsilon(2S) -> mu(+)mu(-)) = 0.264 +/- 0.023 +/- 0.011 nb, sigma (pp -> Upsilon(3S)X) x B(Upsilon(3S) -> mu(+)mu(-))s = 0.159 +/- 0.020 +/- 0.007 nb, where the first uncertainty is statistical and the second systematic.
### Measurement of $$\Upsilon$$ Υ production in $$\mathrm p \mathrm p$$ p p collisions at $$\sqrts =2.76\mathrm \,TeV$$ s = 2.76 TeV
#### Abstract
The production of , and mesons decaying into the dimuon final state is studied with the LHCb detector using a data sample corresponding to an integrated luminosity of collected in proton-proton collisions at a centre-of-mass energy of TeV. The differential production cross-sections times dimuon branching fractions are measured as functions of the transverse momentum and rapidity, over the ranges GeV/ and . The total cross-sections in this kinematic region, assuming unpolarised production, are measured to be sigma (pp -> Upsilon(1S)X) x B(Upsilon(1S) -> mu(+)mu(-)) = 1.111 +/- 0.043 +/- 0.044 nb, sigma (pp -> Upsilon(2S)X) x B(Upsilon(2S) -> mu(+)mu(-)) = 0.264 +/- 0.023 +/- 0.011 nb, sigma (pp -> Upsilon(3S)X) x B(Upsilon(3S) -> mu(+)mu(-))s = 0.159 +/- 0.020 +/- 0.007 nb, where the first uncertainty is statistical and the second systematic.
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Utilizza questo identificativo per citare o creare un link a questo documento: https://hdl.handle.net/11576/2639092
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how to create a slide like the following?
everyone. So, I was just wondering, does anyone know how to create a slide like this using Latex, maybe beamer presentation? Or maybe it was created using PowerPoint?
EDIT: I have used beamer earlier and I just do not know how to create a slide like this. More specifically, the red part and the overall layout? What I have used about beamer is the simple and classic one. Like the other picture.
• Welcome, right now, this question is too broad and cannot be answered. You can create slides with LaTeX, one class specialising in that is called beamer. – Johannes_B Mar 26 '17 at 12:30
• This is not a we-do-it-for-you-site. It is possible to create this in beamer. But you should try to do it and ask here if you encounter problems you can not solve on your own. – Skillmon likes topanswers.xyz Mar 26 '17 at 12:30
• I have used beamer earlier. Sorry for not making it clear. I just have never seen situations like this. – study Mar 26 '17 at 12:32
• There are many tools that you can use to make such a presentation. If you want to do it with beamer, start doing it. When you get stuck at some point, not knowing how to achieve a certain effect, feel free to ask again. – gernot Mar 26 '17 at 12:34
• Can you specify what you mean by situations like this? This slide can be done in beamer, but to avoid useless work, it would be nice to know what element you are interested in. – samcarter_is_at_topanswers.xyz Mar 26 '17 at 12:39
Using the torino theme you can get close enough to fulfil the the vagueness of your question.
\documentclass{beamer}
\usetheme[titleline]{Torino}
\definecolor{chameleongreen3}{named}{gray}
\setbeamertemplate{footline}{}
\author{Marco Barisione}
\title{Torino, a pretty theme for \LaTeX{} Beamer}
\institute{Politecnico di Torino}
\date{September 18, 2007}
\begin{document}
\begin{frame}[t,fragile]{How to use the theme}
\begin{itemize}
\item Install Beamer
\begin{itemize}
\end{itemize}
\begin{itemize}
\item \verb!/usr/share/doc/latex-beamer/beameruserguide.pdf.gz! if you are
using Debian
\item \verb!doc/beameruserguide.pdf! in the source package
\end{itemize}
\item Install the theme
\begin{itemize}
\item \verb!mkdir -p ~/texmf/tex/latex/beamer!\\
\item \verb!cp *.sty ~/texmf/tex/latex/beamer!
\end{itemize}
\begin{itemize}
\item \verb!chameleon.tex!: green theme, watermark and circles for bullet
lists
\item \verb!nouvelle.tex!: green and red theme, watermark and squares for
bullet lists
\item \verb!freewilly.tex!: blue theme, a logo and squares for bullet lists
\end{itemize}
\end{itemize}
\end{frame}
\end{document}
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CDC.h 11.2 KB
Dean Camera committed Jun 01, 2009 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 /* LUFA Library Copyright (C) Dean Camera, 2009. dean [at] fourwalledcubicle [dot] com www.fourwalledcubicle.com */ /* Copyright 2009 Dean Camera (dean [at] fourwalledcubicle [dot] com) Permission to use, copy, modify, and distribute this software and its documentation for any purpose and without fee is hereby granted, provided that the above copyright notice appear in all copies and that both that the copyright notice and this permission notice and warranty disclaimer appear in supporting documentation, and that the name of the author not be used in advertising or publicity pertaining to distribution of the software without specific, written prior permission. The author disclaim all warranties with regard to this software, including all implied warranties of merchantability and fitness. In no event shall the author be liable for any special, indirect or consequential damages or any damages whatsoever resulting from loss of use, data or profits, whether in an action of contract, negligence or other tortious action, arising out of or in connection with the use or performance of this software. */ Dean Camera committed Jun 14, 2009 31 32 /** \ingroup Group_USBClassCDC * @defgroup Group_USBClassCDCDevice CDC Class Device Mode Driver Dean Camera committed Jun 04, 2009 33 34 * * \section Module Description Dean Camera committed Jun 14, 2009 35 * Device Mode USB Class driver framework interface, for the CDC USB Class driver. Dean Camera committed Jun 04, 2009 36 37 38 39 * * @{ */ Dean Camera committed Jun 14, 2009 40 41 #ifndef _CDC_CLASS_DEVICE_H_ #define _CDC_CLASS_DEVICE_H_ Dean Camera committed Jun 01, 2009 42 43 44 /* Includes: */ #include "../../USB.h" Dean Camera committed Jun 14, 2009 45 #include "../Common/CDC.h" Dean Camera committed Jun 01, 2009 46 47 48 #include Dean Camera committed Jun 04, 2009 49 50 51 52 53 /* Enable C linkage for C++ Compilers: */ #if defined(__cplusplus) extern "C" { #endif Dean Camera committed Jun 15, 2009 54 /* Public Interface - May be used in end-application: */ Dean Camera committed Jun 18, 2009 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 /* Type Defines: */ /** Configuration information structure for \ref USB_ClassInfo_CDC_Device_t CDC device interface structures. */ typedef struct { uint8_t ControlInterfaceNumber; /**< Interface number of the CDC control interface within the device */ uint8_t DataINEndpointNumber; /**< Endpoint number of the CDC interface's IN data endpoint */ uint16_t DataINEndpointSize; /**< Size in bytes of the CDC interface's IN data endpoint */ uint8_t DataOUTEndpointNumber; /**< Endpoint number of the CDC interface's OUT data endpoint */ uint16_t DataOUTEndpointSize; /**< Size in bytes of the CDC interface's OUT data endpoint */ uint8_t NotificationEndpointNumber; /**< Endpoint number of the CDC interface's IN notification endpoint, if used */ uint16_t NotificationEndpointSize; /**< Size in bytes of the CDC interface's IN notification endpoint, if used */ } USB_ClassInfo_CDC_Device_Config_t; /** Current State information structure for \ref USB_ClassInfo_CDC_Device_t CDC device interface structures. */ typedef struct { Dean Camera committed Jun 20, 2009 74 75 76 77 78 79 80 81 82 struct { uint8_t HostToDevice; /**< Control line states from the host to device, as a set of CDC_CONTROL_LINE_OUT_* * masks. */ uint8_t DeviceToHost; /**< Control line states from the device to host, as a set of CDC_CONTROL_LINE_IN_* * masks. */ } ControlLineStates; Dean Camera committed Jun 18, 2009 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 struct { uint32_t BaudRateBPS; /**< Baud rate of the virtual serial port, in bits per second */ uint8_t CharFormat; /**< Character format of the virtual serial port, a value from the * CDCDevice_CDC_LineCodingFormats_t enum */ uint8_t ParityType; /**< Parity setting of the virtual serial port, a value from the * CDCDevice_LineCodingParity_t enum */ uint8_t DataBits; /**< Bits of data per character of the virtual serial port */ } LineEncoding; } USB_ClassInfo_CDC_Device_State_t; /** Class state structure. An instance of this structure should be made for each CDC interface * within the user application, and passed to each of the CDC class driver functions as the * CDCInterfaceInfo parameter. This stores each CDC interface's configuration and state information. */ typedef struct { const USB_ClassInfo_CDC_Device_Config_t Config; /**< Config data for the USB class interface within * the device. All elements in this section * must be set or the interface will fail * to enumerate and operate correctly. */ USB_ClassInfo_CDC_Device_State_t State; /**< State data for the USB class interface within * the device. All elements in this section * may be set to initial values, but may * also be ignored to default to sane values when * the interface is enumerated. */ } USB_ClassInfo_CDC_Device_t; Dean Camera committed Jun 15, 2009 117 118 119 120 121 122 123 124 125 /* Function Prototypes: */ /** Configures the endpoints of a given CDC interface, ready for use. This should be linked to the library * \ref EVENT_USB_ConfigurationChanged() event so that the endpoints are configured when the configuration containing the * given CDC interface is selected. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. * * \return Boolean true if the endpoints were sucessfully configured, false otherwise */ Dean Camera committed Jun 18, 2009 126 bool CDC_Device_ConfigureEndpoints(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 04, 2009 127 Dean Camera committed Jun 15, 2009 128 129 130 131 132 /** Processes incomming control requests from the host, that are directed to the given CDC class interface. This should be * linked to the library \ref EVENT_USB_UnhandledControlPacket() event. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. */ Dean Camera committed Jun 18, 2009 133 void CDC_Device_ProcessControlPacket(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 04, 2009 134 Dean Camera committed Jun 15, 2009 135 136 137 138 139 /** General management task for a given CDC class interface, required for the correct operation of the interface. This should * be called frequently in the main program loop, before the master USB management task \ref USB_USBTask(). * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. */ Dean Camera committed Jun 18, 2009 140 void CDC_Device_USBTask(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 01, 2009 141 Dean Camera committed Jun 15, 2009 142 143 144 145 146 147 148 /** CDC class driver event for a line encoding change on a CDC interface. This event fires each time the host requests a * line encoding change (containing the serial parity, baud and other configuration information) and may be hooked in the * user program by declaring a handler function with the same name and parameters listed here. The new line encoding * settings are available in the LineEncoding structure inside the CDC interface structure passed as a parameter. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. */ Dean Camera committed Jun 18, 2009 149 void EVENT_CDC_Device_LineEncodingChanged(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 15, 2009 150 151 152 153 /** CDC class driver event for a control line state change on a CDC interface. This event fires each time the host requests a * control line state change (containing the virtual serial control line states, such as DTR) and may be hooked in the * user program by declaring a handler function with the same name and parameters listed here. The new control line states Dean Camera committed Jun 20, 2009 154 155 * are available in the ControlLineStates.HostToDevice value inside the CDC interface structure passed as a parameter, set as * a mask of CDC_CONTROL_LINE_OUT_* masks. Dean Camera committed Jun 15, 2009 156 157 158 * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. */ Dean Camera committed Jun 18, 2009 159 void EVENT_CDC_Device_ControLineStateChanged(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 01, 2009 160 Dean Camera committed Jun 15, 2009 161 162 163 164 165 166 167 /** Sends a given string to the attached USB host, if connected. If a host is not connected when the function is called, the * string is discarded. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. * \param Data Pointer to the string to send to the host * \param Length Size in bytes of the string to send to the host */ Dean Camera committed Jun 18, 2009 168 void CDC_Device_SendString(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo, char* Data, uint16_t Length); Dean Camera committed Jun 15, 2009 169 170 171 172 173 174 175 /** Sends a given byte to the attached USB host, if connected. If a host is not connected when the function is called, the * byte is discarded. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. * \param Data Byte of data to send to the host */ Dean Camera committed Jun 18, 2009 176 void CDC_Device_SendByte(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo, uint8_t Data); Dean Camera committed Jun 15, 2009 177 178 179 180 181 182 183 /** Determines the number of bytes received by the CDC interface from the host, waiting to be read. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. * * \return Total number of buffered bytes received from the host */ Dean Camera committed Jun 18, 2009 184 uint16_t CDC_Device_BytesReceived(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 15, 2009 185 186 187 188 189 190 191 192 193 /** Reads a byte of data from the host. If no data is waiting to be read of if a USB host is not connected, the function * returns 0. The USB_CDC_BytesReceived() function should be queried before data is recieved to ensure that no data * underflow occurs. * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. * * \return Next received byte from the host, or 0 if no data received */ Dean Camera committed Jun 18, 2009 194 uint8_t CDC_Device_ReceiveByte(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 15, 2009 195 Dean Camera committed Jun 20, 2009 196 197 198 199 /** Sends a Serial Control Line State Change notification to the host. This should be called when the virtual serial * control lines (DCD, DSR, etc.) have changed states, or to give BREAK notfications to the host. Line states persist * until they are cleared via a second notification. This should be called each time the CDC class driver's * ControlLineStates.DeviceToHost value is updated to push the new states to the USB host. Dean Camera committed Jun 15, 2009 200 201 202 * * \param CDCInterfaceInfo Pointer to a structure containing a CDC Class configuration and state. */ Dean Camera committed Jun 20, 2009 203 void CDC_Device_SendControlLineStateChange(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo); Dean Camera committed Jun 04, 2009 204 Dean Camera committed Jun 15, 2009 205 206 207 208 209 /* Private Interface - For use in library only: */ #if !defined(__DOXYGEN__) /* Function Prototypes: */ #if defined(INCLUDE_FROM_CDC_CLASS_DEVICE_C) void CDC_Device_Event_Stub(void); Dean Camera committed Jun 18, 2009 210 void EVENT_CDC_Device_LineEncodingChanged(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo) Dean Camera committed Jun 15, 2009 211 ATTR_WEAK ATTR_ALIAS(CDC_Device_Event_Stub); Dean Camera committed Jun 18, 2009 212 void EVENT_CDC_Device_ControLineStateChanged(USB_ClassInfo_CDC_Device_t* CDCInterfaceInfo) Dean Camera committed Jun 15, 2009 213 214 215 216 217 ATTR_WEAK ATTR_ALIAS(CDC_Device_Event_Stub); #endif #endif Dean Camera committed Jun 04, 2009 218 219 220 221 /* Disable C linkage for C++ Compilers: */ #if defined(__cplusplus) } #endif Dean Camera committed Jun 01, 2009 222 223 #endif Dean Camera committed Jun 04, 2009 224 225 /** @} */
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I'm studying mechanisms of integrity and authentication in symmetric encryption scenarios. I want to propose some examples to see whether I got the point here:
Let $m$ be the message, $c$ the ciphertext, $h$ the hashed message and $t$ the tag resulting of applying MAC.
Example 1:
Here Alice wants to send an enciphered message to Bob providing authentication and integrity but without using hash functions. Both parties agree on two different keys, $k_{1}$ and $k_{2}$. Alice applies $c=Enc_{k_{1}}(m)$ and computes $t=MAC_{k_{2}}(m)$. Then she sends $c$ and $t$ to Bob. Bob applies $m=Dec_{k_{1}}(c)$ and verifies $t'=MAC_{k_{2}}(m)$ comparing it to Alice's $t$.
Example 2:
Now Alice wants to send an enciphered message to Bob but also hashing the message $m$ for computing the MAC (so HMAC comes in). Both parties agree on two different keys (again), $k_{1}$ and $k_{2}$. Alice applies $c=Enc_{k_{1}}(m)$, computes the hash over the message $h=H(m)$ and finally computes $t=MAC_{k_{2}}(h)$. She sends $c$ and $t$ to Bob. Bob now deciphers $m=Dec_{k_{1}}(c)$, computes the hash $h=H(m)$ and verifies the MAC $t'=MAC_{k_{2}}(h)$ comparing it to Alice's $t$.
In case of CBC-MAC, I have read that both parties must agree on a fixed message length, since an attacker could forge a valid MAC. Is this issue solved when using HMAC?
Do you consider these two examples secure? Am I right or mistaken? Specially in the last one, where both parties use hashes with MAC (I have special interest in that one).
Thanks.
• Actually, that's not the standard meaning of HMAC; HMAC usually refers to a specific MAC based on a hash function, specifically, $Hash( (OPAD \oplus K) | Hash( (IPAD \oplus K ) | Message ) )$ – poncho Mar 30 '16 at 2:09
There are a few things wrong with your scheme:
1. if padding is used then your scheme may be vulnerable to padding oracle attacks, this is because decryption happens before MAC verification (see answer of curious);
2. encrypt-and-MAC doesn't provide confidentiality as you can clearly distinguish identical plaintext as the authentication tag $t$ will be identical as well (see the comment of CodesInChaos below the answer of curious).
Notes:
• It is a good idea to study the link that curious provides in the answer to understand more of the underlying issues;
• HMAC is a specific construct (using just the hash as underlying primitive); it is not hash-then-CBC-MAC;
• The issues of CBC-MAC are readily solved (for block ciphers that use 16 byte block size such as AES) by using the CMAC construction which is based on CBC-MAC but doesn't suffer the same issues for dynamically sized input.
• I found that Mac-Then-Encrypt has been superseded by Encrypt-Then-MAC in TLS when possible. I guess that in some scenarios this isn't supported by both parties. Would be using MtE still considered secure? – kub0x Apr 1 '16 at 22:13
Your techniques are not secure. It is the Encrypt and Mac method. Provides integrity to the message but not to the ciphertext. Check this answer. The most secure is to encrypt and then apply the mac to the ciphertext, or to apply an authenticated cipher to the message
• Also the biggest issue with encypt-and-MAC is that it doesn't provide confidentiality. – CodesInChaos Mar 30 '16 at 7:32
Thanks to you all for your replies, specially @Maarten Bodewes who has given a complete answer to my questions.
So thing is that my scheme (Encrypt-and-MAC) doesn't provide confidentiality since the computation of two identical messages yields the samme tag/result. In terms of cryptography it will seem like: $Enc_{k_{1}}(m)$ $||$ $MAC_{k_{2}}(m)$. Notice that it provides integrity.
Also MAC-then-Encrypt is a bad choice since $Enc_{k_{1}}(m || MAC_{k_{2}}(m))$ doesn't provide integrity over the ciphertext.
Best option will be Encrypt-then-MAC: $c=Enc_{k_{1}}(m)$ then send $c$ $||$ $MAC_{k_{2}}(c)$ so confidentiality and integrity are achieved.
I viewed the lesson that covers HMAC here (Princeston University) -> https://www.youtube.com/watch?v=krJ3fHjXYlc Maybe I misunderstood the concept (though the Poncho's answer isn't explained there).
Finally, I would appreciate an explanation of why two keys $k_{1}$ and $k_{2}$ are needed, $k_{1}$ for plaintext encryption and $k_{2}$ for message authentication.
Thanks.
• I've found the answer to my question regarding of the use of two keys for encryption and authentication in crypto.stackexchange.com/questions/8081/… We could derive two keys from the SHA-256 of our Master Key, or use a scheme as GCM which permits us to use one key for auth and encryption. – kub0x Apr 1 '16 at 22:19
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# How to find the maximum tension a wire will experience when a sphere is tied to it in a circular motion?
#### Chemist116
The problem is as follows:
A sphere of $0.5\,kg$ of mass is rotating around an axis in a vertical plane as shown in the picture from below. The minimum speed and the maximum speed of the sphere are $2\,\frac{m}{s}$ and $4\frac{m}{s}$ respectively. Find the maximum tension (measured in Newtons) that the wire can hold. Assume $g=10\,\frac{m}{s^2}$.
The alternatives are as follows:
$\begin{array}{ll} 1.&10\,N\\ 2.&25\,N\\ 3.&20\,N\\ 4.&30\,N\\ \end{array}$
I'm confused exactly how to assess this problem. In order to find the maximum tension I believe the equation to get this is the conservation of mechanical energy.
My instinct tells me that the maximum tension will be on the bottom as follows:
Therefore:
At that point the forces acting will be the tension of the wire and the weight.
$T-mg=\frac{mv^2}{r}$
$T= \frac{mv^2}{r} + mg$
What's the speed at this point?. Should i assume $4\,frac{m}{s}$ or $2\frac{m}{s}$ and why?. what's the physical reason for it?.
I felt that (based on experience) that the speed will be at that point maximum. Hence I'll use $4\frac{m}{s}$.
Then the tension will be:
$T= \frac{0.5(4)^2}{r} + 0.5\times 10$
But the problem lies on what r should I use?.
Then I think this might come from the conservation of mechanical energy?.
$E_k+E_u=E'_k+E'_u$
$\textrm{E=Energy at the top}$
$\textrm{E'=Energy at the bottom}$
Now for this I'm assuming that the reference for establishing the height of the sphere is passing through the center or axis of rotation.
$\frac{1}{2}mv^2+mgr=\frac{1}{2}mu^2+mg(-r)$
$\frac{1}{2}(0.5)(2)^2+0.5(10)r=\frac{1}{2}(0.5)(4^2)+(0.5)(10)(-r)$
$r=0.3$
Therefore the radius is $0.3$
And the maximum tension of the wire will be: By replacing the earlier value in the above equation,
$T= \frac{0.5(4)^2}{0.3} + 0.5\times 10$
$T=31.66\,N$
However this answer does not appear within the alternatives.
But If I were to use this formula for the speed at the lowest point=
$v_2=\sqrt{5rg}$ (How was this formula derived?)
Then $r= \frac{v^2_2}{5g}=\frac{4^2}{5*10}=\frac{8}{25}$
Introducing this in the earlier equation becomes into:
$T= \frac{0.5(4)^2}{\frac{8}{25}} + 0.5\times 10 = 30\,N\\$
Which does appear in one of the alternatives.
But If I were to use this formula instead:
$v_1=\sqrt{rg}$ (How was this formula derived?)
$r=\frac{v^2_1}{g}=\frac{4}{10}=\frac{2}{5}$
Then the radius is different? Why is it so?.
Replacing this value in the earlier equation:
$T= \frac{0.5(4)^2}{\frac{2}{5}} + 0.5\times 10 = 25\,N\\$
And this results into $25\,N$ which is the correct answer. But can somebody help me here?. Why am I obtaining two different values for the force and for the radius?. My original method did not worked, why?.
Why does it exist a discrepancy between the radius and the maximum speed and the minimum speed?. What am I doing wrong?.
#### skeeter
Math Team
In my opinion, whoever "engineered" this problem just picked 4 as the max speed and 2 as the minimum speed without taking energy into consideration.
The given minimum speed at the top of the arc is that necessary to make it through the top of the arc where the sphere's weight is the only force providing the centripetal acceleration, i.e. tension is essentially zero at the top.
$\dfrac{mv_{min}^2}{r} = mg \implies v_{min} = \sqrt{rg} \implies r = \dfrac{v_{min}^2}{g} = 0.4 \text{ m}$
At the bottom, where max tension occurs, the calculated radius value, $r = 0.4 \text{ m}$, makes the tension ...
$T = \dfrac{mv_{max}^2}{r} + mg \implies T = 25 \text{ N}$
However, the rub is that energy conservation doesn't work out.
Using $U_g = 0$ at the bottom, the energy at the top works out as $mg(2r) + \dfrac{1}{2}mv_{min}^2 = 5 \text{ J}$
Energy at the bottom works out to be $\dfrac{1}{2}mv_{max}^2 = 4 \text{ J}$, hence the discrepancy in energy conservation.
topsquark
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Average Value of Current
Average Value of Current
Average value of any current in a given time is defined as that constant value of current which will send the same amount of charge in a given circuit as sent by actual current in the same time interval.
$${{I}_{avg}}\left( {{t}_{2}}-{{t}_{1}} \right)=\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{I(t)dt}$$.
$$\Rightarrow \,Average\,\,Cureent\left( \,{{I}_{avg}} \right)=\frac{\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{I(t)dt}}{{{t}_{2}}-{{t}_{1}}}$$.
Now, the average value of I = I₀ sin ωt;
1. Over first half cycle: for this t₁ = 0, $${{t}_{2}}=\frac{T}{2}=\frac{\pi }{\omega }$$,
$${{I}_{avg}}\left( \frac{T}{2}-0 \right)=\int\limits_{0}^{T/2}{{{I}_{0}}\sin \omega t\,dt}=\left[ \frac{-{{I}_{0}}\cos \omega t}{\omega } \right]_{0}^{\pi /\omega }$$,
$$\Rightarrow \,\,{{I}_{avg}}\left( \frac{T}{2} \right)=\frac{2{{I}_{0}}}{\omega }=\frac{2{{I}_{0}}T}{2\pi }$$,
$$\Rightarrow \,{{I}_{avg}}=\frac{2{{I}_{0}}}{\pi }=0.637{{I}_{0}}$$.
2. Over full cycle: for this t₁ = 0, $${{t}_{2}}=T=\frac{2\pi }{\omega }$$,
$${{I}_{avg}}\left( T-0 \right)=\int\limits_{0}^{T}{{{I}_{0}}\,\sin \omega t\,dt}=\left[ \frac{-{{I}_{o}}\,\cos \omega t}{\omega } \right]_{0}^{2\pi /\omega }$$,
Iavg = 0
So, average value of current for the full cycle is zero. We can also think in this way: Average value during positive half cycle is 0.637 I₀ and the average value during the negative half cycle is – 0.637 I₀. So, over full cycle, Iavg = 0.637 I₀ – 0.637 I₀ = 0.
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# CAT Quant Practice Problems
Question: A real number x satisfying $1 - \frac{1}{n} < x \le 3 + \frac{1}{n}$, for every positive integer n, is best described by
1< x < 4 1 < x ≤3 0 < x ≤4 1 ≤ x ≤3
#### CAT Quant Online Course
• 1000+ Practice Problems
• Detailed Theory of Every Topics
• Online Live Sessions for Doubt Clearing
• All Problems with Video Solutions
CAT Quant Practice Problems
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# Using Diebold-Mariano test
I've got predicted results from two different types of neural networks. Now I would like to run significance testing on both of the results to prove that they do not have equal predictive accuracy. I've learnt that the only tool in the game for this is Diebold-Mariano test. What tool I can use to run this testing (Matlab? R?)
## 1 Answer
So you want to do a Diebold-Mariano test eh? How about the Diebold-Mariano test dm.test function in the forecast package of R?
dm.test {forecast}
Diebold-Mariano test for predictive accuracy
Package:
forecast
Version:
6.2
Description
The Diebold-Mariano test compares the forecast accuracy of two forecast methods.
Usage
dm.test(e1, e2, alternative=c("two.sided","less","greater"),
h=1, power=2)
(Took me ten seconds to find)
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## Introduction
The amount of information that can be transmitted through a physical channel depends on the fundamental properties of the channel1,2 and the physical states used as information carriers. Recent work has shown that coherent states of light, routinely produced by lasers, can achieve the ultimate limits of information transfer, classical capacity, in communication channels with loss,3 and phase-insensitive Gaussian noise.4,5 These results provide strong support for using coherent states as the centerpiece for current and future developments of optical communication networks.6,7,8 Moreover, beyond the realm of classical communications, coherent states have shown to be of great practical use for quantum communications,9,10 including quantum key distribution,11,12,13,14,15,16,17,18,19,20 quantum digital signatures,21 and quantum fingerprinting.22,23 However, despite the theoretical breakthroughs in identifying the capacities for phase-insensitive Gaussian channels, finding the ultimate information rates for other channels, such as noisy channels with a specific non-Gaussian noise that may be encountered in different situations, is still an open problem. Moreover, even in channels for which capacity is known, reaching this ultimate rate for reliable communications requires finding the optimal encoding schemes and optimal measurements over the physical information carriers.24,25 Furthermore, finding optimal encodings and measurements to maximize information transfer in a specific channel with fundamental noise, in addition to technical noise in real devices, would represent a large advance in our understanding of the limits in realistic optical communications.
Quantum mechanics in principle allows for constructing measurements for coherent states surpassing the classical limits of sensitivity and information transfer.2,26 Discrimination strategies for coherent states based on optimized measurements with photon counting have been proposed18,27,28,29,30,31,32,33,34 and demonstrated35,36,37,38,39,40,41,42,43,44 to surpass the conventional limits of detection, the quantum noise limit (QNL), and approach the ultimate quantum limit, the Helstrom bound.26 These nonconventional measurements can enhance information transfer in optical communications43,45 and surpass the classical limits of information transfer using joint measurements over sequences of coherent states.24 Furthermore, photon-counting measurements can be optimized to provide inherent robustness against noise and imperfections of realistic systems in communications.42,46 While these optimized measurements can enhance sensitivities and information transfer with coherent states, the fundamental noise intrinsic in the channel can severely degrade the information encoded in these states. This in turn compromises the potential benefits of these optimized measurements for optical communications.47,48
In this work, we investigate a new approach for optimizing communications in a channel with specific intrinsic noise in addition to unavoidable technical noise, with the goal of maximizing sensitivities and information transfer based on non-Gaussian measurements and coherent states. The central concept of this approach consists of finding optimized communication strategies where measurements and coherent state encodings are jointly optimized to become more robust to the specific noise in the channel, and ultimately maximize sensitivities and information transfer over the noisy channel. As a proof-of-concept demonstration, we investigate optimized communication strategies for communications over a noisy channel with phase diffusion,49,50,51 based on optimized single-shot photon-counting measurements and binary coherent state encodings. Phase diffusion is the most detrimental noise for states of light carrying information in the phase, since it destroys the coherence of the quantum states.52,53,54 We show that an optimized strategy that simultaneously optimizes the non-Gaussian measurement and the binary state alphabet allows for surpassing the limits in performance of an ideal conventional measurement (CM) in terms of probability of error and information transfer per channel use over the non-Gaussian channel.
## Results
### Optimized strategy for a phase diffusion channel
Phase diffusion noise has been extensively investigated in quantum metrology, measurements and communications for phase estimation,49 interferometry,50 state discrimination, and information transfer in communication.55,56,57 This noise is most damaging when information is contained in the coherent properties of the states used as information carriers. In particular, Gaussian phase diffusion makes the task of extracting information more difficult,47,48,49,50,51,52,53,54,58 degrading measurement sensitivities and lowering the achievable information transfer in coherent communications. As a first step for constructing an optimized communication strategy with binary encoding over a channel with phase diffusion, we consider the optimization of the input alphabet to provide robustness to phase diffusion and to other sources of noise and imperfections. This optimization consists of finding the optimal energy distribution in the alphabet to minimize the detrimental effects of phase diffusion, while allowing for measurements to provide high sensitivity.
Figure 1 shows the effect of phase diffusion on three different binary alphabets with coherent states with the same average energy $$\langle n\rangle = \bar n$$: (a) binary phase-shift keying (BPSK) {|−α〉, |+α〉}, with α real and positive; (b) on-off keyed (OOK) alphabet $$\left\{ {|0\rangle ,|\sqrt 2 \alpha \rangle } \right\}$$; and (c) a general binary coherent state alphabet {|α1〉, |α2〉}. We observe that phase diffusion affects equally the states {|−α〉, |+α〉} in the BPSK alphabet, and dramatically reduces their distinguishability, which causes discrimination errors to become very high. On the other hand, when considering the OOK alphabet $$\left\{ {|0\rangle ,|\sqrt 2 \alpha \rangle } \right\}$$, phase diffusion impacts only the state $$|\sqrt 2 \alpha \rangle$$, and leaves the vacuum state |0〉 unaffected. In this case, their distinguishability weakly depends on the phase noise, highlighting the robustness of this alphabet to phase diffusion noise. Therefore, while BPSK has a smaller overlap and better distinguishability than OOK encoding in the absence of phase noise, OOK states have an overlap that is independent of the level of phase diffusion. The optimized alphabet {|α1〉, |α2〉} in Fig. 1c represents a smooth transition and a tradeoff between BPSK with a high degree of distinguishability for low levels of noise, and OOK which is immune to phase diffusion. Figure 1c shows an example of an optimized alphabet {|α1〉, |α2〉}, which is optimized under the average energy constraint $$\bar n = \frac{1}{2}(|\alpha _1|^2 + |\alpha _2|^2)$$ for a given level of the phase noise. The result of this optimization is an alphabet that combines the robustness of OOK with the distinguishability of BPSK.
Optimized non-Gaussian measurements based on photon number resolution (PNR)46 provide robustness against technical noise and imperfections for the discrimination of BPSK states surpassing the QNL. Optimized communication strategies in a non-Gaussian channel with phase diffusion can combine these measurements with an optimized input alphabet in order to minimize the probability of error in the channel. This strategy then optimizes simultaneously the measurement and the alphabet, resulting in a high degree of robustness to phase diffusion while maintaining the benefits of non-Gaussian measurements for surpassing the limits of CMs.
Figure 2a shows the concept of an optimized communication strategy for a binary channel with phase diffusion. The sender (Alice) prepares an input state from a coherent state alphabet {|α1〉, |α2〉}, and sends it to the receiver (Bob) through a non-Gaussian noisy channel. Phase diffusion causes the input states {|αj〉} (j = 1, 2) to become phase-diffused mixed states:54
$$\hat \rho _j(\sigma ) = \mathop {\int}\limits_{ - \infty }^\infty {\frac{{e^{-\frac{\phi ^{2}}{2\sigma ^{2}}}}}{\sqrt {2\pi \sigma ^{2}}}} \left| {\alpha _je^{ - i\phi }} \right\rangle \left\langle {\alpha _je^{ - i\phi }} \right|{\mathrm{d}}\phi,$$
(1)
where the strength of the phase diffusion noise is quantified by the width σ of the Gaussian phase distribution.
At the channel output, the receiver implements an optimized single-shot measurement based on photon counting to discriminate these states with high sensitivity.46 In this strategy, the input state $$\hat \rho _j$$ is displaced in phase space to $$\hat D(\beta )\hat \rho _j(\sigma )\hat D^\dagger (\beta )$$, where the displacement operation, $$\hat D(\beta ) = e^{\beta \hat a^\dagger - \beta ^ \ast \hat a}$$ with $$\hat a$$ ($$\hat a^\dagger$$) as the lowering (raising) operator, is implemented by interference of the input state with a displacement field β in a high transmittance beam splitter.59 Subsequently, the photons in the displaced state are detected by a PNR detector with number resolution m (PNR(m)). Here, m represents the maximum number of photons that a detector can resolve before becoming a threshold detector.46 This measurement strategy uses a maximum a posteriori (MAP) decision rule to infer the input state based on the photon detection outcome k given the mean photon number $$\bar n$$, displacement field |β|, and PNR(m), for a level of phase noise σ.
The MAP strategy assumes that the correct state is the one with the highest conditional posterior probability $$P(\hat \rho _{j}(\sigma )|\beta ,k,m)$$ obtained through Bayes’ rule:
$$P(\hat \rho _{j}(\sigma )|\beta ,k,m) = \frac{{P(k|\hat \rho _{j}(\sigma ),\beta ,m)P(\hat \rho _{j}(\sigma ))}}{{P(k|m)}}.$$
(2)
Here, P(k|m) is the total probability of detecting k photons given a PNR(m) strategy, and $$P(k|\hat \rho _{j}(\sigma ),\beta ,m)$$ is the conditional probability of detecting k photons given |β| and m. We consider equiprobable input states, so that the prior probabilities become $$P(\hat \rho _{j}(\sigma )) = 0.5$$. The probability of error in the discrimination of the input states for a strategy with PNR(m) is:
$$P_{\mathrm{E}}(\bar n,\{ \hat \rho _j(\sigma )\} ,\beta ,m) = 1 - \frac{1}{2}\sum\limits_{k = 0}^m \max\limits_j (\{ P(k|\hat \rho _j(\sigma ),\beta ,m)\} ).$$
(3)
Here, $$P(k|\hat \rho _j(\sigma ),\beta ,\sigma ,m)$$ is the conditional probability of detecting k photons for the input state given the displacement |β|, noise level σ, and PNR(m). The error probability PE in Eq. (3) depends on the input alphabet, the intrinsic properties of the channel, and the measurement performed by the receiver. This provides a way to find optimized strategies that simultaneously optimize the alphabet and the measurement to minimize the detrimental effects of the channel noise. The optimized strategies use an optimal displacement $$\hat D(\beta )$$ and an optimal input alphabet {|α1〉, |α2〉} for a given input power $$\bar n$$, PNR(m), and channel noise level σ to minimize the probability of error PE.
Figure 2b shows the performance of an optimized communication strategy for a channel with phase diffusion optimized for state discrimination for a strategy with PNR(1) for $$\bar n = 0.5$$, with ideal detection efficiency η = 1.0, an interference visibility ξ = 0.998, which quantifies the technical noise and imperfections in the receiver,42,46 and zero dark count rate ν = 0. To evaluate the performance of this strategy, we compare it with an ideal CM consisting of either homodyne or direct detection to minimize the discrimination error, with its own optimized alphabet (solid gray line). We note that the optimized alphabet for the CM results in either BPSK or OOK for this binary coherent state channel, and that it changes abruptly from BPSK to OOK when the conventional measurement switches from homodyne to direct detection.
As shown in Fig. 2b, while a PNR(1) strategy with BPSK (dashed red) can only outperform the ideal CM for small-phase noise σ,54 optimizing the input alphabet to interpolate between BPSK and OOK, shown in Fig. 2c, allows the strategy to outperform the ideal CM for all levels of noise. Moreover, for high levels of noise, the optimized communication strategy approaches the Helstrom measurement with its own optimized alphabet, showing that this optimized communication strategy is asymptotically the optimal quantum measurement.
Optimized communication strategies can also be used to increase information transfer over a noisy channel. These strategies simultaneously optimize the measurement and the input alphabet to maximize mutual information, instead of minimizing probability of error, for a channel with intrinsic noise and technical noise from the devices. Optimized strategies for information transfer for a phase-diffusion channel with binary state encoding are in general different from strategies designed for minimum error, as discussed in Section “Mutual information under phase diffusion”.
### Experimental demonstration
The optimized communication strategies described above can be implemented with current technologies. We demonstrate these strategies in a proof-of-principle experiment for enhancing sensitivities and information transfer for the phase diffusion channel with a binary coherent state encoding with a PNR non-Gaussian measurement, which provides robustness to technical noise and system imperfections.46 The experimental realization uses an interferometric setup to implement the optimized strategies. Coherent state pulses at 633 nm are displaced by interference on a highly transmissive beam splitter, and we use an avalanche photodiode (APD) as a photon number resolving detector. See ref. 46 for a detailed description. To investigate the optimized communication strategies, a controlled level of the phase diffusion noise is applied to the input state (see Supplementary Section 1). Our experiment achieves an overall detection efficiency η = 0.72, an interference visibility ξ = 0.998, and a dark count rate ν = 3.6 × 10−3. Technical noise in the experiment such as reduced visibility and dark counts affects the performance of the optimized strategy (see Supplementary Section 2). However, the levels of noise in our experiment only have a small effect on the strategy’s performance.
We systematically investigate the optimized communication strategies for a channel with phase diffusion by first studying the performance of optimized PNR measurements with a BPSK alphabet46 for this channel. Next, we investigate the optimized communication strategies with an optimized measurement-alphabet method for enhancing measurement sensitivity. Finally, we investigate optimized communication strategies for maximizing the mutual information for a phase-diffusion channel.
### Discrimination with a BPSK alphabet under phase diffusion
Figure 3 shows the experimental error probabilities for the discrimination of states from a BPSK alphabet with an optimized PNR measurement46 with PNR(m) of m = 1, 2, 3, for three mean photon numbers: (a) $$\bar n = 0.5$$, (b) $$\bar n = 1$$, and (c) $$\bar n = 2$$. We observe in all cases that while PNR(1) (red dots) outperforms an adjusted homodyne measurement up to a certain level of noise, as discussed in ref.,54 increasing PNR to PNR(2) (green dots) and PNR(3) (blue dots) extends the level of noise σ where this optimized measurement46 outperforms a homodyne measurement. The increase in robustness with PNR against phase diffusion becomes larger as the mean photon number increases. Figure 3b, c shows that PNR(3) extends the level of noise σ for which this measurement surpasses the homodyne limit by about 1.5 times for $$\bar n = 1$$ and about four times for $$\bar n = 2$$ compared to an on/off PNR(1) strategy.
### Discrimination with an optimized alphabet under phase diffusion
Phase diffusion severely affects measurements for state discrimination in a BPSK alphabet. To reduce the effects of phase diffusion in the channel, a communication strategy can implement an encoding alphabet, which is optimized for a particular level of phase noise. In conventional coherent communication with Gaussian measurements, constellation optimization has been used to mitigate some effects of phase noise.55,56,57 However, in a more general optimized communication strategy using a non-Gaussian measurement, this alphabet can be optimized simultaneously with the displaced photon-counting measurement to reduce errors and enhance information transfer.
Figure 4 shows the performance of the optimized strategy for the discrimination of states from an optimized alphabet with an optimized PNR measurement,46 with PNR(1) and PNR(3) for mean photon numbers: (a) $$\bar n = 0.5$$, (b) $$\bar n = 1.0$$, and (c) $$\bar n = 2.0$$. Experimental data is shown with red (green) dots for PNR(1) (PNR(3)), and expected performance is shown in dotted lines. Error bars represent 1 SD over five experimental runs of over 105 independent experiments. While a strategy with PNR(3) and a BPSK alphabet (solid green) can only outperform a CM for a limited range of noise levels σ, optimized strategies with optimal alphabets and measurements allow for outperforming the CM over larger ranges of noise σ. Moreover, optimized strategies with PNR(1) surpass the CM for all levels of noise for $$\bar n = 0.5$$ and $$\bar n = 1.0$$. For higher $$\bar n$$, increasing number resolution m is expected to enable discrimination below the CM at any noise level, as can be inferred from the trend in Fig. 4c.
Figure 4d–f show the optimal alphabet for $$\bar n = 0.5,\,1.0,\,{\mathrm{and}}\,2.0$$, respectively. Discrete jumps in the optimized alphabets for different PNR strategies are the results of optimization of Eq. (3), which requires a global optimization over multiple minima46 of PE. This optimization searches for the values of |α1| and |β|, resulting in the global minimum of PE for a given noise level σ for a PNR(m) strategy. There are levels of noise at which a small increase in σ causes the former global minimum of PE as a function of |α1| and |β| to become a local minimum, and a former local minimum to become the new global minimum (see Supplementary Section 3). These abrupt changes in the global minimum result in the sudden jumps of the optimal alphabet shown in Fig. 4e at σ ≈ 0.36 and σ ≈ 0.38, and in Fig. 4f at σ ≈ 0.20 and σ ≈ 0.42. We note that the optimized alphabets correspond to interpolations between BPSK and OOK alphabets for all $$\bar n$$, and result in large improvements over BPSK. This shows that strategies with optimized alphabets are essential for surpassing the sensitivity limits of conventional measurements in the channels with phase noise.
### Mutual information under phase diffusion
Optimized communication strategies can also be designed to maximize information transfer over a non-Gaussian noisy channel, for which optimal encoding and decoding are unknown. An optimized communication strategy, which minimizes probability of error, will provide some advantage for increasing mutual information. However, in a noisy channel, the measurement and the alphabet can be optimized in order to maximize mutual information I(X:Y) and will yield a different strategy than for minimum error. Mutual information quantifies the total amount of information between transmitter and receiver, and depends on the encoding alphabet and decoding measurement. For a displaced photon-counting measurement, I(X:Y) can be expressed according to a “soft” decision rule where the number of photons detected is used to infer the input symbol rather than the binary output from a binary decision rule.60 The mutual information for a channel with phase diffusion with a binary coherent state encoding can be expressed as:
$$I(\bar n,\{ \hat \rho _j(\sigma )\} ,\beta ,m) = \sum\limits_{k = 0}^m {\sum\limits_{j = 1}^2 P } (k|\{ \hat \rho _j(\sigma )\} ,\beta ,m)P(\{ \hat \rho _j(\sigma )\} )\log _2\left[ {\frac{{P(k|\{ \hat \rho _j(\sigma )\} ,\beta ,m)}}{{P(k|m)}}} \right],$$
(4)
where $$P(k|\{ \hat \rho _i(\sigma )\} ,\beta ,m)$$ is the conditional probability of detecting k photons. In an optimized communication strategy over a noisy channel, the input alphabet and measurement with PNR(m) are simultaneously optimized to maximize mutual information $$I(\bar n,\{ \hat \rho _i(\sigma )\} ,\beta ,m)$$ under the average energy constraint for a noise level σ.
Figure 5a, b shows the experimental results for the mutual information with optimized strategies for mean photon numbers: (a) $$\bar n = 1.0$$ and (b) $$\bar n = 2.0$$, and PNR(m) of m = 1, 3, 5 in red, green, and blue dots, respectively. The theoretical predictions are shown with dashed colored lines. The mutual information for a conventional measurement (dashed gray) and for BPSK are shown adjusted for our total detection efficiency η = 0.72. Optimized communication strategies surpass the limit in mutual information for a CM at high levels of phase-diffusion noise (σ ≥ 0.7), and for low noise (σ ≤ 0.1). Moreover, optimized strategies with higher PNR detection resolution m provide higher mutual information for all levels of noise. Note that optimized communication strategies with optimized alphabets drastically outperform BPSK for all PNR(m) in terms of mutual information. Figure 5c, d shows the optimized alphabets for (c) $$\bar n = 1.0$$, and (d) $$\bar n = 2.0$$, respectively. We observe that the optimal alphabet interpolates from BPSK to OOK similar to error probability. However this interpolation is continuous, because the mutual information is a convex function of σ for all PNR(m). In the intermediate level of noise (σ ≈ 0.5), there is a gap between the optimized strategies and the CM. This gap decreases as the PNR(m) of the optimized strategies increases. This suggests that optimized communication strategies with high-enough PNR(m) should provide levels of mutual information at least as high as those that can be achieved with ideal conventional measurements for all levels of phase diffusion noise.
Figure 5e shows the maximum percent difference R(m) between an optimized strategy with PNR(m) and a CM for $$\bar n$$ from 0 to 2.0 for different PNR(m) from m = 1 to m = 20. This corresponds to the percent difference at the level of noise for which a PNR(m) strategy has the worst performance relative to a conventional measurement. R(m) is defined as:
$$R(m) = \mathop {{\mathrm{max}}}\limits_\sigma \left( {\frac{{I_{\mathrm{CM}}(\sigma ) - I_{{\mathrm{PNR}}(m)}(\sigma )}}{{I_{\mathrm{CM}}(\sigma )}}} \right),$$
(5)
where IPNR(m)(σ) is the mutual information for an optimized communication strategy with PNR(m), and ICM(σ) is the mutual information for the conventional measurement. We observe that as the number resolution increases, the percent difference asymptotically approaches zero for all mean photon numbers. The blue regions to the right of the white line correspond to R(m) < 1%, that is, when a PNR(m) strategy is within 1% of the conventional measurement. Figure 5f shows R(m) on a log–log scale for $$\bar n = 0.5,\,1.0,\,1.5,\,{\mathrm{and}}\,2.0$$ in red, green, blue, and black lines, respectively. The straight lines indicate power-law scaling in the convergence of the form a(m)b, with b ≈ 1.1 for all lines. This convergence suggests that for all mean photon numbers, optimized communication strategies with large enough photon resolution m will at worst provide the same mutual information as the ideal CM, which serves as a lower bound for the performance of optimized communication strategies. At the same time, these optimized strategies with moderate PNR provide large advantages for increasing mutual information compared to CM at low noise and high noise levels.
## Discussion
We proposed and demonstrated optimized communication strategies to maximize information transfer and measurement sensitivity over a non-Gaussian noisy channel. These optimized strategies are based on simultaneous optimization of the states used as information carriers with an optimized non-Gaussian photon-counting measurement that surpasses the QNL for state discrimination. Simultaneous optimization of alphabet and measurement provides robustness to intrinsic channel noise, and allows for overcoming the sensitivity limits of conventional measurements and achieving higher information transfer in communications over noisy channels.
We demonstrated in a proof-of-principle experiment the concept of optimized strategies for communication over a channel with phase diffusion for binary coherent state alphabets and single-shot optimized measurements with PNR. These optimized communication strategies provide unexpected benefits to minimize the probability of decoding error and maximize the achievable mutual information in this noisy channel. Moreover, we observed that optimized communication strategies not only provide robustness to intrinsic channel noise but also to technical noise and imperfections in the receiver.
We expect that optimized communication strategies can provide advantages for different problems in coherent communications extending to communication with multiple states and complex measurements. Moreover, optimized communication strategies can be applied to other channels utilizing practically optimized measurements and encodings to maximize information transfer in realistic noisy communication channels for which capacity limits are unknown, but that are encountered in optical communication networks.
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1. Business
2. Accounting
3. cost of units transferred out and ending work in process...
# Question: cost of units transferred out and ending work in process...
###### Question details
Cost of Units Transferred Out and Ending Work in Process The costs per equivalent unit of direct materials and conversion in the Filling Department of Eve Cosmetics Company are $0.60 and$0.60, respectively. The equivalent units to be assigned costs are as follows: Equivalent Units Direct Materials Conversion Inventory in process, beginning of period 0 4,000 Started and completed during the period 67,000 67,000 Transferred out of Filling (completed) 67,000 71,000 Inventory in process, end of period 5,000 1,500 Total units to be assigned costs 72,000 72,500
The beginning work in process inventory had a cost of $2,520. Determine the cost of completed and transferred-out production and the ending work in process inventory. If required, round to the nearest dollar. Completed and transferred-out production$-------------
Inventory in process, ending \$--------------------
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kidzsearch.com > wiki
# United States Constitution
The United States Constitution is the highest law of the United States of America. It was put in writing on September 17, 1787 by the Constitutional Convention in Philadelphia, Pennsylvania and later put into effect, or ratified, by representatives of the people of the first 13 states.[1] When nine of the states ratified the document, they put forth a union of sovereign states, and a federal government for that union. That government started on March 4, 1789, taking the place of the Articles of Confederation.
The Constitution of the United States is the oldest federal constitution now in use.[2]
Since 1787, changes have been made to the United States Constitution 27 times by amendments (changes). The first ten of these amendments are together called the Bill of Rights.
## Articles of the Constitution
When it was written in 1787, the Constitution had a preamble and seven main parts, called articles.
### Preamble
The Preamble states:
We the People of the United States, in Order to form a more perfect Union, establish Justice, insure domestic Tranquility, provide for the common defense, promote the general Welfare, and secure the Blessings of Liberty to ourselves and our Posterity, do ordain and establish this Constitution for the United States of America.
The Preamble is not a law. It gives the reasons for writing the Constitution. The Preamble is one of the best known parts of the Constitution. The first three words, "We the people," are used very often. The six intentions that are listed are the goals of the constitution.
### Legislative power
Article One says that the U.S. Congress (the legislative branch) will make the laws for the United States. Congress has two parts, called "Houses," the House of Representatives and the Senate. The Article says who can be elected to each part of Congress, and how they are elected.
The House of Representatives has members elected by the people in each state. The number of members from each state depends on how many people live there. Each member of the House of Representatives is elected for two years.
The Senate has two members, called Senators, from each state, no matter how many people live there. Each Senator is elected for six years. The original Constitution allowed the state legislatures to choose the Senators, but this was changed later by the seventeenth amendment.
Article One also says how the Congress will do its business and what kinds of laws it can make. It lists some kinds of laws the Congress and the states cannot make. Article One also makes rules for Congress to impeach and remove from office the President, Vice President, judges, and other government officers.
### Executive power
Article Two says that the President (the executive branch) will carry out the laws made by Congress. This article says how the President and Vice President are elected, and who can be elected to these offices. The President and Vice President are elected by a special Electoral College chosen by the states, for four years. The Vice President takes over as President if the President dies, or resigns, or is unable to serve. Article Two also says that the President is in charge of the army and navy. He can make treaties with other countries, but these must be approved by two-thirds of the Senate. He appoints judges, ambassadors, and other officers, but the Senate also must approve these appointments. The President can also veto bills. However, Congress can override the veto.
### Judicial power
Article Three says there will be a court system (the judicial branch), including the Supreme Court. The article says that Congress can decide which courts, besides the Supreme Court, are needed. It says what kinds of "cases and controversies" these courts can decide. Article Three also requires trial by jury in all criminal cases, and defines the crime of treason.
### States' powers and limits
Article Four is about the states. It says that all states must give "full faith and credit" to the laws of the other states. It also says that state governments must treat citizens of other states as fairly as they treat their own citizens, and must send arrested people back to another state if they have been charged with a crime.
Article Four also says that Congress can make new states. There were only 13 states in 1787. Now there are 50 United States. It says Congress can make rules for Federal property and can govern territories that have not yet been made into states. Article Four says the United States must make sure that each state has a republican form of government, and protect the states from invasion and violence.
### Process of amendment
Article Five gives two ways to amend, or change, the Constitution.
1. Congress can write a change, if two-thirds of the members in each House agree.
2. The state governments can call a convention to write changes, although this has not happened since 1787.
Any change that is written by Congress or by a convention must be sent to the state legislatures or to state conventions for their approval. Congress decides whether to send a change to the legislatures or to conventions. Three-fourths of the states must approve a change for it to become part of the Constitution.
An amendment can change any part of the Constitution, except one—no amendment can change the rule that each state has equal suffrage (right to vote) in the Senate.
### Federal power
Article Six says that the Constitution, and the laws and treaties of the United States, are higher than any other laws. It also says that all federal and state officers must swear to "support" the Constitution.
### Ratification
Article Seven says that the new government under the Constitution would not start until conventions in at least nine states approved the Constitution.
## Amendments
Since 1787, Congress has written 33 amendments to change the Constitution, but the states have ratified only 27 of them.
The first ten amendments are called the Bill of Rights. They were made in 1791. All of these changes limited the power of the federal government. They were:
Number Year Description
1st 1791 Congress must protect the rights of freedom of speech, freedom of the press, freedom of assembly, and freedom of petition. Congress cannot establish a national religion.
2nd 1791 "A well regulated Militia being necessary to the security of a free State, the right of the people to keep and bear arms, shall not be infringed." - People have the right to keep and carry weapons, for example, guns.
3rd 1791 The government cannot send soldiers to live in private homes without the permission of the owners.
4th 1791 The government cannot get a warrant to arrest a person or search their property unless there is "probable cause" to believe a crime has been committed.
5th 1791 The government cannot put a person on trial for a serious crime until a grand jury has written an indictment. That a person cannot be put on trial twice for the same crime. The government must follow due process of law before punishing a person or taking their property. A person on trial for a crime does not have to testify against himself in court.
6th 1791 Any person who is accused of a crime should get a speedy trial by a jury. That person can have a lawyer during the trial. They must be told what they are charged with. The person can question the witnesses against them, and can get their own witnesses to testify.
7th 1791 A jury trial is needed for civil cases.
8th 1791 The government cannot require excessive bail or fines, or any cruel and unusual punishment.
9th 1791 The listing of individual rights in the Constitution and Bill of Rights does not include all of the rights of the people and the states.
10th 1791 Anything that the Constitution does not say that Congress can do should be left up to the states, or to the people.
After the Bill of Rights, there are 17 more changes to the Constitution that were made at different times.
Number Year Description
11th 1795 Citizens cannot sue states in federal courts. There are some exceptions.
12th 1804 Changed the way the President and Vice President are elected.
13th 1865 Ended slavery in the United States.
14th 1868 Every person born in the United States is a citizen. States must follow due process of law before taking away any citizen's rights or property.
15th 1870 A citizen's right to vote cannot be taken away because of race, the color of their skin, or because they were previously slaves.
16th 1913 Congress can put a tax on income.
17th 1913 The people will elect Senators. Before this, Senators were elected by state legislatures.
18th 1919 Made a law against drinking alcohol, called Prohibition.
19th 1920 Gave women the right to vote.
20th 1933 Changed the days for meetings of Congress and for the start of the President's term of office.
21st 1933 Ended the Prohibition law of the Eighteenth Amendment. States can make laws about how alcohol is used in each state.
22nd 1951 A person may not be elected President more than two times.
23rd 1961 Gave the people in the District of Columbia the right to vote for President.
24th 1964 Made it illegal to make anyone pay a tax to have the right to vote.
25th 1967 Changes what happens if a President dies, resigns, or is not able to do the job. Says what happens if a Vice President dies or resigns.
26th 1971 Makes 18 years old the minimum age for people to be allowed to vote
27th 1992 Limits how Congress can increase how much its members are paid.
## References
• Amar, Akhil Reed (2005). "In the Beginning". America's Constitution: A Biography. New York: Random House.
.
• Bailyn, Bernard, ed. The Debate on the Constitution: Federalist and Antifederalist Speeches, Articles, and Letters During the Struggle for Ratification. Part One: September 1787 to February 1788 (The Library of America, 1993) ISBN 0-940450-42-9
• Bailyn, Bernard, ed. The Debate on the Constitution: Federalist and Antifederalist Speeches, Articles, and Letters During the Struggle for Ratification. Part Two: January to August 1788 (The Library of America, 1993) ISBN 0-940450-64-X
• Edling, Max M. (2003). A Revolution in Favor of Government: Origins of the U.S. Constitution and the Making of the American State. Oxford University Press.
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• Ellis, Joseph (2002). Founding Brothers: The Revolutionary Generation. Vintage.
.
• Fallon, Richard H. (2004). The Dynamic Constitution: An Introduction to American Constitutional Law. Cambridge University Press.
.
• Farris, Michael P. (July/August 2005). "Through the Founders' Eyes: Was the Constitution Illegally Adopted?". The Home School Court Report 21: 6-10. excerpt from (to be published) Constitutional Law for Enlightened Citizens.
• Finkelman, Paul "Affirmative Action for the Master Class: The Creation of the Proslavery Constitution," University of Akron Law Review 32 (No. 3, 1999): 423-70.
• Finkelman, Paul Slavery and the Founders: Race and Slavery in the Age of Jefferson (Armonk, N.Y.: M.E. Sharpe, 1996);
• Finkelman, Paul "Slavery and the Constitution: Making a Covenant with Death," in Richard R. Beeman, Stephen Botein, and Edward C., Carter, II, eds., Beyond Confederation: Origins of the Constitution and American National Identity (Chapel Hill: University of North Carolina Press, 1987);
• Hall, Kermit L. (1984). A Comprehensive Bibliography of American Constitutional and Legal History, 1896-1979. Millwood, N. Y.: Kraus International.
.
• Kammen, Michael (1986). A Machine that Would Go of Itself: The Constitution in American Culture. New York: Alfred A. Knopf.
.
• Kelly, Alfred Hinsey; Harbison, Winfred Audif; Belz, Herman (1991). The American Constitution: its origins and development (7th edition ed.). New York: Norton & Co.
.
• Levy, Leonard W., ed. (2000). Encyclopedia of the American Constitution (2nd Edition ed.). New York: Macmillan.
.
• Marshall, Thurgood, "The Constitution: A Living Document," Howard Law Journal 1987: 623-28.
• Mazzone, Jason (2005). "The Creation of a Constitutional Culture". Tulsa Law Review 40: 671.
• Smith, Jean Edward; Levine, Herbert M. (1988). Civil Liberties & Civil Rights Debated. Englewood Cliffs, New Jersey: Prentice Hall.
• Smith, Jean Edward (1996). John Marshall: Definer Of A Nation. New York: Henry Holt & Company.
• Smith, Jean Edward (1989). The Constitution And American Foreign Policy. St. Paul, MN: West Publishing Company.
• Wiecek, William M., "The Witch at the Christening: Slavery and the Constitution's Origins," Leonard W. Levy and Dennis J. Mahoney, eds., The Framing and Ratification of the Constitution (New York: Macmillan, 1987), 178-84.
• Wiecek, William M., "'The Blessings of Liberty': Slavery in the American Constitutional Order," in Robert A. Goldman and Art Kaufman, eds., Slavery and Its Consequences: The Constitution, Equality, and Race (Washington, D.C.: American Enterprise Institute, 1988), 23-34.
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26410 articles – 20212 references [version française]
Detailed view Article in peer-reviewed journal
Annales de l'Institut Henri Poincaré Analyse non linéaire 22 (2005) 127-142
Attached file list to this document:
PS
DMfinal.ps(184.4 KB)
PDF
DMfinal.pdf(207.7 KB)
About $L^p$ estimates for the spatially homogeneous Boltzmann equation
For the homogeneous Boltzmann equation with (cutoff or non cutoff ) hard potentials, we prove estimates of propagation of Lp norms with a weight $(1+ |x|^2)^q/2$ ($1 < p < +\infty$, $q \in \R_+$ large enough), as well as appearance of such weights. The proof is based on some new functional inequalities for the collision operator, proven by elementary means.
Keyword(s) : Boltzmann equation – spatially homogeneous – non-cutoff – integrability estimates
hal-00087260, version 1 http://hal.archives-ouvertes.fr/hal-00087260 oai:hal.archives-ouvertes.fr:hal-00087260 From: Clément Mouhot <> Submitted on: Friday, 21 July 2006 16:34:15 Updated on: Friday, 21 July 2006 17:00:12
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# Using the quotient rule to differentiate the function, f(x) = x^2/\tan x - x.
## Question:
Using the quotient rule to differentiate the function{eq}, f(x) = \dfrac {x^2}{\tan x - x}. {/eq}
## Quotient rule of differentiation:
If {eq}s {/eq} and {eq}t {/eq} are differentiable, then {eq}r(x) = \dfrac {s(x)}{t(x)} {/eq} is also differentiable,
then the derivative of the function with respect to {eq}x {/eq} is given by,
{eq}r'(x) = \dfrac {s'(x) \cdot t(x) - t'(x) \cdot s(x)}{(t(x))^2} {/eq}
Recall that the derivative of {eq}x^n {/eq} is {eq}nx^{n - 1} {/eq}, where {eq}n {/eq} is a real number.
Differentiate the given function using the quotient rule,
\displaystyle \begin{align*} f'(x) &= \dfrac {\dfrac {d}{dx}\left ( x^2 \right ) \cdot (\tan x - x) - \dfrac {d}{dx} (\tan x - x) \cdot x^2}{(\tan x - x)^2} \\ &= \dfrac {(2x) \cdot (\tan x - x) - \left [ \dfrac {d}{dx} (\tan x) - \dfrac {d}{dx} (x) \right ] \cdot x^2}{(\tan x - x)^2} \\ &= \dfrac {2x \ (\tan x - x) - \left [ \left ( \sec^2 x \right ) - (1) \right ] \cdot x^2}{(\tan x - x)^2} \\ &= \dfrac {2x \ (\tan x - x) - x^2\left ( \sec^2 x - 1 \right )}{(\tan x - x)^2} \\ f'(x) &= \dfrac {2x \ \tan x - 2x^2 - x^2 \ \tan^2 x}{(\tan x - x)^2} \\ \end{align*}
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# Search for pair production of the scalar top quark in muon+tau final states
We present a search for the pair production of scalar top quarks ($\tilde{t}_{1}$), the lightest supersymmetric partners of the top quarks, in $p\bar{p}$ collisions at a center-of-mass energy of 1.96 TeV, using data corresponding to an integrated luminosity of {7.3 $fb^{-1}$} collected with the \dzero experiment at the Fermilab Tevatron Collider. Each scalar top quark is assumed to decay into a $b$ quark, a charged lepton, and a scalar neutrino ($\tilde{\nu}$). We investigate final states arising from $\tilde{t}_{1} \bar{\tilde{t}_{1}} \rightarrow b\bar{b}\mu\tau \tilde{\nu} \tilde{\nu}$ and $\tilde{t}_{1} \bar{\tilde{t}_{1}} \rightarrow b\bar{b}\tau\tau \tilde{\nu} \tilde{\nu}$. With no significant excess of events observed above the background expected from the standard model, we set exclusion limits on this production process in the ($m_{\tilde{t}_{1}}$,$m_{\tilde{\nu}}$) plane.
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# Activation and Trial limitations
## Trial limitations
### bcpViewer
Only the first 50 elements will display full informations (names, properties, references). The Folder structure is shown completely.
### .NET library
In trial mode, BCP-packages can be created and all its metadata can be fully imported into Vault.
Please note that since no real files will be imported into Vault, opening them will fail!
### Cmdlets
In trial mode, BCP-packages can be opened and manipulated without limitation.
However the export of a package will contain only the metadata, independent of a specified -NoSourceFiles parameter.
Export-BcpPackage -To ...
# in trial mode same result as:
Export-BcpPackage -To ... -NoSourceFiles
Keep in mind that since links to your real files get removed in the exported package, opening the files later in Vault will fail!
## Activation
In order to activate the product you have following options:
Open the Start Menu and use the ‘bcpToolkit 20.0 License Information’ shortcut:
Launch the bcpViewer 20.0 shortcut, open the ‘Help’ Menu and click the ‘About’ button:
### Command-line
Launch the License Information tool located in the install directory with the required Command-line arguments.
Example: Activating a Stand-Alone license using a serial number:
"C:\Program Files\coolOrange\bcpToolkit\License.exe" --StandAlone --Serialnumber="XXXXX-XXXXX-XXXXX-XXXXX"
Activate powerLoad (bcpToolkit) on export machine
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However build- and test-machines do not need to be activated.
## Licensing Options
### Stand Alone Licensing
This product supports the Stand-Alone licensing model which is charged based on the time the license is valid and the number of seats the license is valid for.
For further information see the detailed description of the Stand-Alone licensing model.
In the License Information Dialog the remaining days until the license expires can be found.
When the the license expires, the bcpViewer will switch back to Trial mode, the .NET library will throw an exception and the commandlets also switch back to Trial mode.
#### Offline activation
The serial number of the license and the machine code are required to generate an activation file.
The activation file for an offline activation can be generated and downloaded on the following site: bcpToolkit - Activation file generator
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# Finding $\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$ where $D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,,xy\leq 1\leq x+y-1\}$
How can I evaluate this double integral through a domain transformation?
$$\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$$ where $$D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,, xy\leq 1\leq x+y-1\}$$
The region $D$ is the red one in this picture:
I have tried to convert $D$ to polar coordinates but then, it come up the following integral: $$\frac{1}{3}\int_0^{\frac{\pi}{4}}\cos 2t\left(\frac{1}{\sqrt{(\sin t\cos t)^3}}-\frac{8}{(\sin t+\cos t)^3}\right)dt$$ which I got stuck into.
Does someone any further ideas?
• $$\int_0^{\pi/4}\int_{2/(\sin t+\cos t)}^{1/(\sin t\cos t)}r^2\cos2t\,\mathrm dr\mathrm dt$$ – Nosrati Aug 7 '18 at 13:27
• The upper bound value is incorrect because $$xy\leq 1\ \Rightarrow\ r\cdot \cos t\cdot r\cdot \sin t\leq 1\ \Rightarrow\ r\leq\frac{1}{\sqrt{\cos t \sin t}}$$ – Samuel Leanza Aug 7 '18 at 17:15
• You are right. . – Nosrati Aug 7 '18 at 17:18
• Changing parameters to $u=x^2-y^2$ and $v=2xy$, as in $u+iv=(x+iy)^2$, helps in giving a very precise proof that the integral is not finite, but the same conclusion follows from the observation that for large $x$, and therefore small $y$, the function is not much smaller than $x$, which makes the integral over the area between $x$ and $x+\Delta x$ close to $\Delta x$. – random Aug 9 '18 at 14:24
I have found that is possible to solve this integral by the following transformation: $$\begin{cases} u=x+y\\ v=x-y\end{cases}\ \Rightarrow\ \begin{cases} x=\frac{u+v}{2}\\ y=\frac{u-v}{2}\end{cases}$$
The integral does not converge: $$\begin{eqnarray} \iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\ dxdy &=& \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{\frac{(u+v)^2}{4}-\frac{(u-v)^2}{4}}{\sqrt{\frac{(u+v)^2}{4}+\frac{(u-v)^2}{4}}}\cdot \left(-\frac{1}{2}\right)\ dv=\\ &=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{v}{\sqrt{u^2+v^2}}\ dv =\\ &=&-\frac{\sqrt{2}}{4} \int_2^{+\infty}u\ du\int_{\sqrt{u^2-4}}^u 2v(u^2+v^2)^{-1/2}\ dv=\\ &=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}u\cdot [\sqrt{u^2+v^2}]_{\sqrt{u^2-4}}^u\ du =\\ &=&-\int_2^{+\infty}u^2-u\sqrt{u^2-2}\ du=\\ &=&-\int_2^{+\infty}u^2\ du +\frac{1}{2}\int_2^{+\infty}2u(u^2-2)^{1/2}\ du=\\ &=&\lim\limits_{c\to +\infty}\left \{\left[-\frac{u^3}{3}\right]_2^c+\frac{1}{3}\left[\sqrt{(u^2-2)^3}\right]_2^c\right \}=\\ &=&\lim\limits_{c\to +\infty}\left(-\frac{c^3}{3}+\frac{8}{3}+\frac{1}{3}\sqrt{(c^2-2)^3}-\frac{2\sqrt{2}}{3}\right)=\\ &=&\frac{1}{3}\lim\limits_{c\to +\infty}(\sqrt{(c^2-2)^3}-c^3+8-2\sqrt{2})=-\infty\end{eqnarray}$$
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## Found 279 Documents (Results 1–100)
100
MathJax
### Unlocking the secrets of locking: finite element analysis in planar linear elasticity. (English)Zbl 07532590
MSC: 65N30 65N12 65N15
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### A hybrid fundamental-solution-based finite element method for transient heat conduction analysis of two-dimensional orthotropic materials. (English)Zbl 07446895
MSC: 65-XX 74-XX
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### Forced geometrically nonlinear vibrations of thin shells of revolution with piezoelectric layers. (English. Ukrainian original)Zbl 1474.74055
Int. Appl. Mech. 57, No. 2, 200-216 (2021); translation from Prikl. Mekh., Kiev 57, No. 2, 88-106 (2021).
MSC: 74H45 74K25 74F15
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### A non-incremental approach for elastoplastic plates basing on the Brezis-Ekeland-Nayroles principle. (English)Zbl 1481.74083
MSC: 74C05 74K20 49J52
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### Numerical modeling of eccentric cylindrical shells partially filled with a fluid. (Russian)Zbl 1463.74083
MSC: 74K25 74H15 74F10
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### Solving exterior boundary value problems for the Laplace equation. (English. Russian original)Zbl 1455.65225
Differ. Equ. 56, No. 7, 890-899 (2020); translation from Differ. Uravn. 56, No. 7, 918-926 (2020).
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### Three-dimensional analysis of the stress-strain state of inhomogeneous hollow cylinders using various approaches. (English)Zbl 1465.74023
MSC: 74B05 74S05
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### Geometric element parameterization and parametric model order reduction in finite element based shape optimization. (English)Zbl 1468.74049
MSC: 74P10 74S05
Full Text:
### Mathematical modelling of flagellated microswimmers. (English. Russian original)Zbl 1446.76204
Comput. Math. Math. Phys. 58, No. 11, 1804-1816 (2018); translation from Zh. Vychisl. Mat. Mat. Fiz. 58, No. 11, 1876-1888 (2018).
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### Numerical study of the influence of surface defects on the stability of a cylindrical pipe containing fluid. (Russian. English summary)Zbl 1424.74025
MSC: 74F10 74H15
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### Boundary-based finite element method for two-dimensional anisotropic elastic solids with multiple holes and cracks. (English)Zbl 1403.74178
MSC: 74S15 65N38 74R10
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### A new fourth-order compact scheme for the Navier-Stokes equations in irregular domains. (English)Zbl 1397.76092
MSC: 76M20 65M06 76D05
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### Solution of an industrially relevant coupled magneto-mechanical problem set on an axisymmetric domain. (English)Zbl 1452.92023
MSC: 92C55 78A25
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MSC: 74F10
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### A flexible symmetry-preserving Galerkin/POD reduced order model applied to a convective instability problem. (English)Zbl 1390.76348
MSC: 76M10 65M60 76E15
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### Dynamic high order numerical manifold method based on weighted residual method. (English)Zbl 1352.74118
MSC: 74G15 65M60
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### Automatic consistency integrations for the stress update of $$J_{2}$$ elastoplastic rate constitutive equations with combined hardening. (English)Zbl 1352.74061
MSC: 74C05 74S05 65M60
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### A model of the effective rubber-cord ply. (English. Russian original)Zbl 1333.74012
Mosc. Univ. Mech. Bull. 69, No. 5, 109-113 (2014); translation from Vestn. Mosk. Univ., Ser. I 69, No. 5, 41-45 (2014).
MSC: 74A40 74K20 74B20
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### Symmetric and conforming mixed finite elements for plane elasticity using rational bubble functions. (English)Zbl 1455.74083
MSC: 74S05 74B05 65N12
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### Unified elastoplastic finite difference and its application. (English)Zbl 1400.74013
MSC: 74C05 74S20
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### Numerical manifold space of Hermitian form and application to Kirchhoff’s thin plate problems. (English)Zbl 1352.74462
MSC: 74S05 74K20 65N30
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### A method for multidimensional wave propagation analysis via component-wise partition of longitudinal and shear waves. (English)Zbl 1352.74140
MSC: 74J05 70J50
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### Iterative method for solving geometrically nonlinear inverse problems of structural element shaping under creep conditions. (Russian, English)Zbl 1313.74036
Zh. Vychisl. Mat. Mat. Fiz. 53, No. 12, 2091-2099 (2013); translation in Comput. Math. Math. Phys. 53, No. 12, 1908-1915 (2013).
MSC: 74C05 65N30 74S05
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### Combined plasticity and creep analysis in 2D by means of the boundary element method. (English)Zbl 1287.74052
MSC: 74S15 74C05 65N38
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### Compactly supported radial basis functions for the acoustic 3D eigenanalysis using the particular integral method. (English)Zbl 1351.74114
MSC: 74S15 65N25
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### Variational principles and optimal solutions of the inverse problems of creep bending of plates. (English. Russian original)Zbl 1298.74144
J. Appl. Mech. Tech. Phys. 53, No. 5, 751-760 (2012); translation from Prikl. Mekh. Tekh. Fiz. 53, No. 5, 136-146 (2012).
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### Nonlinear spatial bending of shear-deformable curvilinear rods. (English. Russian original)Zbl 1298.74138
J. Appl. Mech. Tech. Phys. 53, No. 2, 258-265 (2012); translation from Prikl. Mekh. Tekh. Fiz. 53, No. 2, 128-136 (2012).
MSC: 74K10 74S05
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### A moving mesh finite volume interface tracking method for surface tension dominated interfacial fluid flow. (English)Zbl 1291.76223
MSC: 76M12 76D45 76D05
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### Finite element analysis of two- and three-dimensional static problems in the asymmetric theory of elasticity as a basis for the design of experiments. (English)Zbl 1401.74275
MSC: 74S05 74B99 74-05
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### A synthetic computer model for plastic strain localization. (English. Russian original)Zbl 1254.74024
Russ. Phys. J. 55, No. 2, 211-222 (2012); translation from Izv. Vyssh. Uchebn. Zaved., Fiz., No. 2, 76-87 (2012).
MSC: 74C99 74S30
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### One-point quadrature ANS solid-shell element based on a displacement variational formulation. I: Geometrically linear assessment. (English)Zbl 1253.74103
MSC: 74S05 74K25
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### Eigenvalue approximations from below using Morley elements. (English)Zbl 1253.65181
MSC: 65N25 65N30
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### Hybrid system for optimal design of mechanical properties of composites. (English)Zbl 1356.74155
Murín, Justín (ed.) et al., Computational modelling and advanced simulations. Selected, extended papers presented at the ECCOMAS conference on computational modelling and advanced simulations (CMAS2009), Bratislava, Slovakia, June 30–July 3, 2009. Berlin: Springer (ISBN 978-94-007-0316-2/hbk; 978-94-007-0317-9/ebook). Computational Methods in Applied Sciences (Springer) 24, 303-331 (2011).
MSC: 74P05 74E30
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### Comparison of high-order curved finite elements. (English)Zbl 1242.65244
MSC: 65N30 65D17 78M10
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### Identification of the parameters of the stress-strain problem for a multicomponent elastic body with an inclusion. (English. Ukrainian original)Zbl 1272.74226
Int. Appl. Mech. 46, No. 4, 377-387 (2010); translation from Prik. Mekh., Kiev 46, No. 4, 14-24 (2010).
MSC: 74G75 74E05 74B10
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### Computation of incompressible thermal flows using Hermite finite elements. (English)Zbl 1225.76195
MSC: 76M10 76D05 80A20
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### A Hermite finite element method for incompressible fluid flow. (English)Zbl 1277.76038
MSC: 76M10 65M60 76D05
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### Generalized stochastic perturbation technique in engineering computations. (English)Zbl 1190.65119
MSC: 65L99 65M99
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### Study of a finite element method for the time-dependent generalized Stokes system associated with viscoelastic flow. (English)Zbl 1425.76025
MSC: 76A10 76M10 35Q35
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### Nonlinear two-dimensional static problems for thin shells with reinforced curvilinear holes. (English. Ukrainian original)Zbl 1272.74426
Int. Appl. Mech. 45, No. 12, 1269-1300 (2009); translation from Prik. Mekh., Kiev 45, No. 12, 4-42 (2009).
MSC: 74K25 74E30 74C15
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### Application of discontinuous Galerkin spectral method on hexahedral elements for aeroacoustic. (English)Zbl 1257.76047
MSC: 76M10 76M22 76Q05
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### Automatic 3-D crack propagation calculations: a pure hexahedral element approach versus a combined element approach. (English)Zbl 1308.74143
MSC: 74S05 74B10 74R99
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### A computational algorithm for system modelling based on bi-dimensional finite element techniques. (English)Zbl 1153.65020
MSC: 65D17 65N30
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### Variational approaches in the linear theory of elasticity. (English. Russian original)Zbl 1423.74106
Dokl. Phys. 52, No. 8, 426-430 (2007); translation from Dokl. Akad. Nauk, Ross. Akad. Nauk 415, No. 4, 486-490 (2007).
MSC: 74B05
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### Effective models in the refined analysis of the strain state for three-layer non-axisymmetric cylindrical shells. (English. Russian original)Zbl 1423.74562
Dokl. Phys. 52, No. 6, 330-334 (2007); translation from Dokl. Akad. Nauk, Ross. Akad. Nauk 414, No. 5, 613-617 (2007).
MSC: 74K25 74A40 74S05
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### Updated models for studies of the stressed-strained state of sandwich cylindrical shells. (English. Russian original)Zbl 1254.74023
Dokl. Phys. 51, No. 3, 128-131 (2006); translation from Dokl. Akad. Nauk, Ross. Akad. Nauk, 407, No. 1, 36-39 (2006).
MSC: 74C99 74K25
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MSC: 74B05
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### Moving particle finite element method with superconvergence: nodal integration formulation and applications. (English)Zbl 1120.74051
MSC: 74S05 74R15
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### Multivariate approximation by a combination of modified Taylor polynomials. (English)Zbl 1104.65011
MSC: 65D15 41A10
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### An $$L^{\infty}$$-error estimate for finite element solution of nonlinear elliptic problem with a source term. (English)Zbl 1106.65099
MSC: 65N15 65N30 35J65
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### Stability and vibration analysis of non-prismatic thin-walled composite spatial members of generic section. (English)Zbl 1205.74075
MSC: 74H45 74H55
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### The efficiency of direct integration methods in elastic contact-impact problems. (English)Zbl 1200.74138
MSC: 74S20 74M15 74M20
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### Topology optimization of dielectric substrates for filters and antennas using SIMP. (English)Zbl 1142.74368
MSC: 74P15 74S05 74-02
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### Simple algorithm for temperature distribution calculations. (English)Zbl 1109.80306
MSC: 80M25 80A20
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### Dispersion-based guidance for the temporal and spatial discretization of the one- and two-dimensional finite element solutions to the transient heat equation. (English)Zbl 1046.80011
MSC: 80M10 35K05
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### A new method for coalescing particles in PIC codes. (English)Zbl 1038.76031
MSC: 76M28 76X05
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### Axisymmetrical vibrations of shells of revolution under abrupt loading. (English. Russian original)Zbl 1094.74653
J. Appl. Math. Mech. 66, No. 4, 591-599 (2002); translation from Prikl. Mat. Mekh. 66, No. 4, 607-615 (2002).
MSC: 74K25 74M20 74H45
### Discrete singular convolution-finite subdomain method for the solution of incompressible viscous flows. (English)Zbl 1130.76403
MSC: 76M25 76D05
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### A new algorithm for solving some mechanical problems. (English)Zbl 1013.74081
MSC: 74S30 74M25 74H45 74K20 76D05
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MSC: 74S05
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### Boundary integration of polynomials over an arbitrary linear hexahedron in Euclidean three-dimensional space. (English)Zbl 0938.65055
Reviewer: A.C.Genz (Pullman)
MSC: 65D32 26B20 68W30
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### CASTOR: Normal-mode analysis of resistive MHD plasmas. (English)Zbl 0921.76100
MSC: 76M10 76X05 76W05
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### A multi-layer finite element formulation for suspended sediment transport in tidal flows. (English)Zbl 0911.76040
MSC: 76M10 76T99 86A05
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### A new finite element with ‘drilling’ D. O. F. (English)Zbl 0913.73066
MSC: 74S05 74B05
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### Finite element methods for the three-field Stokes system in $$\mathbb R^ 3$$: Galerkin methods. (English)Zbl 0853.76041
MSC: 76M10 76A10
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### Computational modeling of solid-state reactions in the Ni-Si systems induced by pulsed lasers. (English)Zbl 0860.65127
Reviewer: J.Prakash (Bombay)
MSC: 65Z05 35Q72 65M60 35K05 92E20 80A30
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### Modeling the dynamic equilibrium of objects weakened by thin low-strength inclusions. (English. Russian original)Zbl 0842.73040
Cybern. Syst. Anal. 31, No. 2, 261-267 (1995); translation from Kibern. Sist. Anal. 1995, No. 2, 124-130 (1995).
MSC: 74H45 74E30 74S05
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### A more accurate FD solution for the 3D non-linear heat equation. (English)Zbl 0837.65093
MSC: 65M06 80A20 35K55
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### Design optimization via finite elements: An elementary account. (English)Zbl 0821.73041
MSC: 74P99 74S05
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### General interior Hermite collocation methods for second-order elliptic partial differential equations. (English)Zbl 0817.65092
Reviewer: G.Schmidt (Berlin)
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### Profile minimization problem for matrices and graphs. (English)Zbl 0804.05060
Reviewer: R.Bodendiek (Kiel)
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### Accuracy and efficiency of multivariant finite elements for three- dimensional simulation of viscous incompressible flows. (English)Zbl 0791.76045
MSC: 76M10 76D99
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### A quasi-conforming triangular laminated composite shell element based on a refined first-order theory. (English)Zbl 0790.73067
MSC: 74S05 74K15 74E30
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### Wave distribution in fibre members subjected to kinematic forcing. (English)Zbl 0777.73036
MSC: 74H45 74S30
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### Finite element analysis of lateral sloshing response in axisymmetric tanks with triangular elements. (English)Zbl 0776.76050
MSC: 76M10 76B10
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### Unstructured multigridding by volume agglomeration: Current status. (English)Zbl 0753.76136
MSC: 76M25 76N15 65N55
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### Approximation and/or construction of curves by minimization methods with or without constraints. (English)Zbl 0801.65014
MSC: 65D17 65K10
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### Finite-part integral and boundary element method to solve flat crack problems. (English)Zbl 0771.73087
MSC: 74S15 74S30 74R99
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