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[ "# 2.05 kg to lbs - 2.05 kilograms to pounds\n\nBefore we get to the practice - that is 2.05 kg how much lbs calculation - we want to tell you few theoretical information about these two units - kilograms and pounds. So let’s start.\n\nHow to convert 2.05 kg to lbs? 2.05 kilograms it is equal 4.5194763710 pounds, so 2.05 kg is equal 4.5194763710 lbs.\n\n## 2.05 kgs in pounds\n\nWe are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, known also as International System of Units (in abbreviated form SI).\n\nFrom time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.\n\nFirst definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was not complicated but difficult to use.\n\nThen, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was replaced by a new definition.\n\nThe new definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is exactly 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.\n\n## 2.05 kilogram to pounds\n\nYou learned some facts about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. It is needed to highlight that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to focus only on pound-mass.\n\nThe pound is used in the British and United States customary systems of measurements. To be honest, this unit is used also in other systems. The symbol of this unit is lb or “.\n\nThe international avoirdupois pound has no descriptive definition. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces or 7000 grains.\n\nThe avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### How many lbs is 2.05 kg?\n\n2.05 kilogram is equal to 4.5194763710 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.\n\n### 2.05 kg in lbs\n\nTheoretical section is already behind us. In next section we are going to tell you how much is 2.05 kg to lbs. Now you know that 2.05 kg = x lbs. So it is high time to get the answer. Let’s see:\n\n2.05 kilogram = 4.5194763710 pounds.\n\nIt is an accurate outcome of how much 2.05 kg to pound. You can also round off the result. After rounding off your result is as following: 2.05 kg = 4.510 lbs.\n\nYou know 2.05 kg is how many lbs, so have a look how many kg 2.05 lbs: 2.05 pound = 0.45359237 kilograms.\n\nNaturally, this time you can also round it off. After rounding off your outcome is as following: 2.05 lb = 0.45 kgs.\n\nWe are also going to show you 2.05 kg to how many pounds and 2.05 pound how many kg outcomes in tables. Have a look:\n\nWe are going to begin with a table for how much is 2.05 kg equal to pound.\n\n### 2.05 Kilograms to Pounds conversion table\n\nKilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)\n2.05 4.5194763710 4.5100\nNow look at a table for how many kilograms 2.05 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n2.05 0.45359237 0.45\n\nNow you know how many 2.05 kg to lbs and how many kilograms 2.05 pound, so we can go to the 2.05 kg to lbs formula.\n\n### 2.05 kg to pounds\n\nTo convert 2.05 kg to us lbs you need a formula. We are going to show you a formula in two different versions. Let’s begin with the first one:\n\nAmount of kilograms * 2.20462262 = the 4.5194763710 outcome in pounds\n\nThe first formula give you the most correct outcome. In some cases even the smallest difference can be considerable. So if you want to get a correct result - this formula will be the best for you/option to know how many pounds are equivalent to 2.05 kilogram.\n\nSo go to the second formula, which also enables conversions to learn how much 2.05 kilogram in pounds.\n\nThe second version of a formula is down below, see:\n\nAmount of kilograms * 2.2 = the result in pounds\n\nAs you see, the second version is simpler. It could be better choice if you want to make a conversion of 2.05 kilogram to pounds in quick way, for instance, during shopping. You only need to remember that your result will be not so accurate.\n\nNow we are going to show you these two formulas in practice. But before we are going to make a conversion of 2.05 kg to lbs we are going to show you easier way to know 2.05 kg to how many lbs totally effortless.\n\n### 2.05 kg to lbs converter\n\nAn easier way to learn what is 2.05 kilogram equal to in pounds is to use 2.05 kg lbs calculator. What is a kg to lb converter?\n\nCalculator is an application. Calculator is based on longer formula which we showed you in the previous part of this article. Due to 2.05 kg pound calculator you can easily convert 2.05 kg to lbs. You only need to enter number of kilograms which you want to convert and click ‘convert’ button. The result will be shown in a second.\n\nSo let’s try to convert 2.05 kg into lbs with use of 2.05 kg vs pound converter. We entered 2.05 as a number of kilograms. This is the outcome: 2.05 kilogram = 4.5194763710 pounds.\n\nAs you can see, this 2.05 kg vs lbs converter is easy to use.\n\nNow let’s move on to our main topic - how to convert 2.05 kilograms to pounds on your own.\n\n#### 2.05 kg to lbs conversion\n\nWe will begin 2.05 kilogram equals to how many pounds conversion with the first formula to get the most exact result. A quick reminder of a formula:\n\nAmount of kilograms * 2.20462262 = 4.5194763710 the result in pounds\n\nSo what need you do to learn how many pounds equal to 2.05 kilogram? Just multiply number of kilograms, this time 2.05, by 2.20462262. It is exactly 4.5194763710. So 2.05 kilogram is equal 4.5194763710.\n\nIt is also possible to round off this result, for example, to two decimal places. It is exactly 2.20. So 2.05 kilogram = 4.5100 pounds.\n\nIt is high time for an example from everyday life. Let’s convert 2.05 kg gold in pounds. So 2.05 kg equal to how many lbs? And again - multiply 2.05 by 2.20462262. It gives 4.5194763710. So equivalent of 2.05 kilograms to pounds, if it comes to gold, is equal 4.5194763710.\n\nIn this case it is also possible to round off the result. This is the outcome after rounding off, in this case to one decimal place - 2.05 kilogram 4.510 pounds.\n\nNow we can move on to examples calculated with a short version of a formula.\n\n#### How many 2.05 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nNumber of kilograms * 2.2 = 4.510 the outcome in pounds\n\nSo 2.05 kg equal to how much lbs? As in the previous example you have to multiply number of kilogram, in this case 2.05, by 2.2. Look: 2.05 * 2.2 = 4.510. So 2.05 kilogram is equal 2.2 pounds.\n\nLet’s make another calculation with use of shorer version of a formula. Now calculate something from everyday life, for example, 2.05 kg to lbs weight of strawberries.\n\nSo calculate - 2.05 kilogram of strawberries * 2.2 = 4.510 pounds of strawberries. So 2.05 kg to pound mass is 4.510.\n\nIf you know how much is 2.05 kilogram weight in pounds and are able to calculate it with use of two different versions of a formula, we can move on. Now we are going to show you these outcomes in tables.\n\n#### Convert 2.05 kilogram to pounds\n\nWe are aware that outcomes shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these results in charts for your convenience. Thanks to this you can quickly make a comparison 2.05 kg equivalent to lbs results.\n\nLet’s begin with a 2.05 kg equals lbs chart for the first version of a formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n2.05 4.5194763710 4.5100\n\nAnd now have a look at 2.05 kg equal pound chart for the second version of a formula:\n\nKilograms Pounds\n2.05 4.510\n\nAs you can see, after rounding off, if it comes to how much 2.05 kilogram equals pounds, the outcomes are the same. The bigger number the more significant difference. Please note it when you want to make bigger amount than 2.05 kilograms pounds conversion.\n\n#### How many kilograms 2.05 pound\n\nNow you learned how to convert 2.05 kilograms how much pounds but we will show you something more. Are you curious what it is? What do you say about 2.05 kilogram to pounds and ounces conversion?\n\nWe want to show you how you can convert it step by step. Let’s start. How much is 2.05 kg in lbs and oz?\n\nFirst thing you need to do is multiply amount of kilograms, this time 2.05, by 2.20462262. So 2.05 * 2.20462262 = 4.5194763710. One kilogram is 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this case there are 2 pounds.\n\nTo check how much 2.05 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.\n\nSo your result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then your result will be equal 2 pounds and 33 ounces.\n\nAs you see, calculation 2.05 kilogram in pounds and ounces quite simply.\n\nThe last conversion which we will show you is calculation of 2.05 foot pounds to kilograms meters. Both of them are units of work.\n\nTo calculate foot pounds to kilogram meters you need another formula. Before we show you this formula, let’s see:\n\n• 2.05 kilograms meters = 7.23301385 foot pounds,\n• 2.05 foot pounds = 0.13825495 kilograms meters.\n\nNow see a formula:\n\nNumber.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters\n\nSo to calculate 2.05 foot pounds to kilograms meters you have to multiply 2.05 by 0.13825495. It is 0.13825495. So 2.05 foot pounds is exactly 0.13825495 kilogram meters.\n\nYou can also round off this result, for example, to two decimal places. Then 2.05 foot pounds is exactly 0.14 kilogram meters.\n\nWe hope that this conversion was as easy as 2.05 kilogram into pounds conversions.\n\nWe showed you not only how to do a conversion 2.05 kilogram to metric pounds but also two other calculations - to check how many 2.05 kg in pounds and ounces and how many 2.05 foot pounds to kilograms meters.\n\nWe showed you also other solution to make 2.05 kilogram how many pounds calculations, it is using 2.05 kg en pound calculator. It is the best option for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.\n\nWe hope that now all of you are able to make 2.05 kilogram equal to how many pounds conversion - on your own or with use of our 2.05 kgs to pounds converter.\n\nIt is time to make your move! Let’s convert 2.05 kilogram mass to pounds in the way you like.\n\nDo you need to make other than 2.05 kilogram as pounds calculation? For example, for 5 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 2.05 kilogram equal many pounds.\n\n### How much is 2.05 kg in pounds\n\nWe want to sum up this topic, that is how much is 2.05 kg in pounds , we prepared for you an additional section. Here we have for you all you need to know about how much is 2.05 kg equal to lbs and how to convert 2.05 kg to lbs . It is down below.\n\nHow does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. How does 2.05 kg to pound conversion formula look? . It is down below:\n\nThe number of kilograms * 2.20462262 = the result in pounds\n\nSee the result of the conversion of 2.05 kilogram to pounds. The correct result is 4.5194763710 lb.\n\nThere is also another way to calculate how much 2.05 kilogram is equal to pounds with another, shortened version of the equation. Check it down below.\n\nThe number of kilograms * 2.2 = the result in pounds\n\nSo now, 2.05 kg equal to how much lbs ? The answer is 4.5194763710 lbs.\n\nHow to convert 2.05 kg to lbs quicker and easier? You can also use the 2.05 kg to lbs converter , which will make the rest for you and give you a correct answer .\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms." ]

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[ "## Wednesday, January 28, 2015\n\n### Working of Compound microscope\n\nCompound microscope is useful to produce a better magnification than a simple microscope. Because of the design of this microscope it is able to produce better magnification without any aberrations.\n\nThe device consists of two convergent lenses arranged coaxially. The lens that is facing the object is called objective and the other lens that is close to the eye is called eye lens.\n\nThe focal length of the eye lens is more than that of the focal lens of the field lens. The distance between the two lenses can be varied with the help of the experimental arrangement.\n\nAn object is placed at a distance from the field lens who is within double the focal length but greater than the focal length of the field lens. Because of this arrangement the corresponding image of the object is inverted, real and enlarged. The location of the image is on the other side of the field lens. This image acts like a object for the eye lens. By varying the position of the eye lens, the final image of the compound microscope position can be varied. The final image is a inverted, magnified and virtual image.\n\nThe total magnification of the compound microscope is the product of magnification produced by both the lenses.\n\nDepending on the position of the eye lens the final magnification very.\n\nIf the eye lens is adjusted in such a way that the intermediate image is within its focal length, the final images formed at a finite distance. By adjusting the lens properly we can see that the final images formed at the least distance of distant vision. This position is called strained eye position and in this position we are going to get the best magnification.\n\nBy adjusting the eye lens in such a way that the intermediate image is at the principal focus of the eye lens, the final image can be adjusted to form at infinity. This kind of arrangement is called relaxed eye position or normal adjustment. In this case magnification is less.\n\nDepending on our requirement we can choose any of these positions and get the corresponding magnifications.\n\nSome of the mathematical equations are explained as shown in the diagram.\n\nAt near point magnification is high and it far point of image magnification is low. Magnifying power is represented as negative value just because the final image is inverted. The magnification of a compound microscope may vary anywhere in between 20 to 40 for a low powered compound microscope. A well-designed compound microscope can produce a magnification even up to 500 times of the size of the image.\n\nProblem and solution\n\nA microscope consists of two convex lenses of focal lengths 2 cm and 5 cm placed 20 cm apart. Where must be the object has to be placed from the field lens so that the final image is at a distance of 25 cm from the eye?\n\nWe cannot apply a direct formula to solve this problem. We need to analyze the situation and identify each value separately as shown in the attachment paper. While we are solving this problem, we have to apply the lens formula with the proper sign convention.\n\nLet us go through the problem and ask for any of the further clarifications, if  required.\n\nRelated Posts" ]

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[ "# 2007 AMC 10B Problems/Problem 4\n\n## Problem\n\nThe point", null, "$O$ is the center of the circle circumscribed about", null, "$\\triangle ABC,$ with", null, "$\\angle BOC=120^\\circ$ and", null, "$\\angle AOB=140^\\circ,$ as shown. What is the degree measure of", null, "$\\angle ABC?$", null, "$\\textbf{(A) } 35 \\qquad\\textbf{(B) } 40 \\qquad\\textbf{(C) } 45 \\qquad\\textbf{(D) } 50 \\qquad\\textbf{(E) } 60$\n\n## Solution\n\nBecause all the central angles of a circle add up to", null, "$360^\\circ,$", null, "\\begin{align*} \\angle BOC + \\angle AOB + \\angle AOC &= 360\\\\ 120 + 140 + \\angle AOC &= 360\\\\ \\angle AOC &= 100. \\end{align*}\n\nTherefore, the measure of", null, "$\\text{arc}AC$ is also", null, "$100^\\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts,", null, "$\\angle ABC = \\boxed{\\textbf{(D)} 50}$\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]

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[ "# Is it possible to calculate/estimate the value of the J parameter to be used in the Heisenberg/Ising Hamiltonians?\n\nStudying magnetic systems, two frequently used approximations are the Heisenberg and Ising models (a discussion about these approximations can be read here):\n\n$$\\begin{equation} \\tag{Heisenberg} \\hat{H}_H=-\\sum_{\\langle i j\\rangle}J\\hat{S}_i\\hat{S}_j \\end{equation}$$\n\n$$\\begin{equation} \\tag{Ising} \\hat{H}_I=-\\sum_{\\langle ij\\rangle}J\\hat{S}_i^z\\hat{S}_j^z \\end{equation}$$\n\nIs there a theoretical method/way to determine realistic values for the $$J$$ parameter?\n\n• If by J you mean inter-site exchange (the angular brackets under the summation refer to nearest neighbours), then refer to this question that I asked earlier : mattermodeling.stackexchange.com/questions/1359/…. If you are talking in the context of DFT, J is usually taken to be 0-20% of the Hubbard U as a thumb rule. Jul 17, 2020 at 18:09\n• It's quantifiable in the same way as the hubbard 'U'. Even the Hubbard 'U' is dependent on the spin operator. There are plenty of notes that explain this. Jul 19, 2020 at 22:34\n• Is this related? mattermodeling.stackexchange.com/a/104/88\n– Thomas\nJul 26, 2020 at 3:27\n• @Xivi76 since this is now one of the longest standing unanswered questions, do you think maybe you could write an answer? I'm trying to help clean up the unanswered queue a little bit! Mar 23, 2021 at 15:02\n• @NikeDattani Coincidentally, I learnt the procedure mentioned in the question quite recently. I can write up an answer. Mar 23, 2021 at 20:08\n\nIn the simplest form, if you have two magnetic atoms in your unit cell, there are four possible spin configurations - $$\\uparrow\\uparrow$$ , $$\\uparrow\\downarrow$$, $$\\downarrow\\uparrow$$,$$\\downarrow\\downarrow$$, denoted by $$E_1, E_2, E_3, E_4$$ respectively. The exchange 'J' can be estimated as $$1/4* (E_1 - E_2 - E_3 + E_4$$). Now, the energy values for the four different configurations are just the total energy values from your DFT calculation. Typically, you would need to constrain your magnetization on the magnetic atoms in your DFT package. If you want to include anisotropic exchange, you need to consider spin-orbit coupling as well." ]

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[ "Is it possible to use a formula to get the starting and ending date of a period according to the Period number and Period Year we typed in another cell?\n\nfor example we fill in period \"9\" and Year \"2011\" .\n\nregards,\n\nrobbert\n\n•", null, "Jet Reports Historic Posts\nIs it possible to use a formula to get the starting and ending date of a period according to the Period number and Period Year we typed in another cell?\n\nfor example we fill in period \"9\" and Year \"2011\" .\n\nregards,\n\nrobbert\n\nELMO: Sofar I use Vert.Lookup in Excel to a reference table with the period dates and numbers in Excel.\nAny better way would be welcome!\n•", null, "Jet Reports Historic Posts\n\nThis should work where A1=YEAR (e.g. 2014) and A2=MONTH No. (e.g. 9).\n\nStart of month: =EOMONTH(((\\$A\\$1-1900)*365.3+(\\$A\\$2-1)*30.5),-1)+1\nEnd of month: =EOMONTH(((\\$A\\$1-1900)*365.3+(\\$A\\$2-1)*30.5),0)\n\n•", null, "David Stockill\n\nIf you have non standard periods setup in Accounting Periods in NAV use the following to return start and end dates of current periods.\n\nBeginning of Period: =NL(\"last\",\"Accounting Period\",\"Starting Date\",\"Starting Date\",\"..\"&TODAY(),\"Closed\",\"False\")\n\nEnd of Period: =NL(\"first\",\"Accounting Period\",\"Starting Date\",\"Starting Date\",TODAY()&\"..\",\"Closed\",\"False\")-1" ]

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[ "You are on page 1of 31\n\nCHEMISTRY\n\nAtomic Structure\n\nAtomic Structure\n\nATOMIC STRUCTURE\n1. LAWS OF CHEMICAL COMBINATIONS\nLaw of conservation of mass: Laws of conservation of mass states that in a chemical reaction the weight of\nproducts is always equal to the weight of reactants.\nLaw of definite proportions: Law of definite proportions states that the elemental composition of a compound\nalways remains same if it is analysed form various sources e.g. water from a river or ditch or pond either in India\nor in USA would always give H : O ratio as 2 : 1.\nLaw of Multiple Proportions: Law of multiple proportion states that elements combine in simple whole number\nratios to form various types of compounds e.g. The ratio of N : O is 1: 1, 1 : 2 and 2 : 1 in NO, NO2 and N2O.\nrespectively.\n2. DALTONS THEORY OF ATOM\nJohn Dalton developed his famous theory of atom is 1803. The main postulates of his theory were :\n\nAtom was considered as a hard, dense and smallest indivisible particle of matter.\n\nEach element consists of a particular kind of atoms.\n\nThe properties of elements differ because of differences in the kinds of atoms contained in them.\n\nThis theory provides a satisfactory basis for the law of chemical combination.\n\nAtom is indestructible i.e. it cannot be destroyed or created.\n\nDrawbacks\n\nIt fails to explain why atoms of different kinds should differ in mass and valency etc.\n\nThe discovery of isotopes and isobars showed that atoms of same elements may have different atomic\nmasses (isotopes) and atoms of different kinds may have same atomic masses (isobars).\n\nThe discovery of various sub-atomic particles like X-rays, electrons, protons etc. during the 19th\ncentury lead to the idea that the atom was no longer an indivisible and smallest particle of the matter..\n3. DISCOVERY OF ELECTRON\nWilliam Crookes found that certain rays come out of cathode and gain lots of energy because of high acceleration\npotential before colliding with a gas molecule. This happens especially when the gas pressure is low. These rays\nare known as cathode rays.\nHigh Voltage\nGas at very low pressure\nCathode\n\nCathode\n\nAnode\nSuction pump\n\nFig. Cathode ray Tube experiment\n\nDetailed study of cathode rays by J.J. Thomson led to the discovery of electrons. He observed that\n1.Cathode rays always travel in straight lines\n2.They are negatively charged. Cathode rays turn towards positively charged plates.\n3.Charge on particles constituting rays was determined by Oil\nDrop experiment by Millikan as 1.6 10 19 coulomb.\n4.Specific charge (e/m) does not change when the gas inside discharge tube was changed indicating that electron\nis fundamental particle.\ne/m for electron = 1.758 1011 Coulombs / kg .\n5.Value of charge on electron = 1.6 10 19 Coulombs\n6.Mass of the electron = 1.9 10 31 kg .\n4. ANODE RAYS (DISCOVERY OF PROTON)\nIt is well-known fact that the atom is electrically neutral. The presence of negatively-charged electrons in the\natom amphasized the presence of positively-charged particles.\nTo detect the presence of positively-charged particles, the discharged tube experiment was carried out, in which\na perforated cathode was used. Gas at low pressure was kept inside the tube. On passing high voltage between\nthe electrodes it was observed that some rays were emitted from the side of the anode. These rays passed through\n\nAtomic Structure\nthe holes in the cathode and produced green fluorescence on the opposite glass was coated with ZnS. These rays\nconsist of positively-charged particles known as protons.\nPerforated\ncathode\n\nZnS\nCoating\n\nH2 gas inside\nat low pressure\n\nFor anode rays, e/m is not fundamental property as different gases used have different mass on C-12 scale.\nHighest e/m is for hydrogen gas.\n5. DISCOVERY OF NEUTRON\nAfter the discovery of electrons and protons, Rutherford (1920) had predicted the existence of a neutral fundamental\nparticle. In 1930, Bethe and Becker reported from Germany that if certain light elements, like beryllium, were\nexpoesd to alpha rays from the naturally radioactive polonium, a very highly penetrating radiation was obtained.\nSimilar results were obtained by Irene Cureia dn F. Jolit (1932). Chadwick (1932) demonstrated that this mysterious\nradiation was a stream of fast moving particles of about the same mass as a proton but having no electric charge.\nBecause of their electrical neutrality, these particles were called neutron.\nThe lack of charge on the neutron is responsible for its great penetraing power.\nThus, a neutron is a sub-atomic fundamental particle which has a mass 1.675 10 24 g (approximately 1 amu),\nalmost equal to that of a proton or a hydrogen atom but carrying no electric charge. The e / m value of a neutron\nis thus zero.\n6. CONSTITUENTS OF ATOM\nSome of the well known fundamental particles present in an atom are-protons, electrons and neutrons. Many\nothers were discovered later viz positron, neutrinos etc.\nSubatomic\nSymbol\nUnit Charge\nUnit Mass\nCharge in\nMass in AMU\nparticles\nCoulomb\nProton\n\np1\n\nNeutron\n\nElectron\n\ne0\n\n1.007825\n\n1.60 1019\n0\n\nNegligible\n\n1.602 1019\n\n5.489 104\n\n+1\n\n0\n1\n\n1.008665\n\nDo you Know?\nChandwick in 1932 discovered the neutron by bombarding elements like beryllium with fast moving -particles.\nHe observed that some new particles were emitted which carried no charged and had mass equal to that of\nproton.\n7. THEORIES OF ATOM\nAfter the discovery of electrons, protons and neutrons several theories were proposed by various scientists such\nas\n1.Plum Pudding Model (Thomson Model of Atom)\n2.Rutherfords Model of Atom\n3.Plancks quantum theory\n4.Bohrs Atomic Model\nThomson Model of Atom\nAccording to this model, electrons are embedded in uniform sphere of positive charge to confer electrical neutrality.\nThis model was satisfactory to the extent that the electrostatic forces of repulsion among the electron cloud is\nbalanced by the attractive forces between the positively charged mass and the electron. How ever, this model\nfails to explain the results of ionisation and scattering experiment and is, therefore discarded.\nRutherford Model of Atom\nWhen -particles are bombarded on thin foils ( 4 105 cm thick) of metals like gold, silver, platinum or copper,,\n\nAtomic Structure\nmost of them are passed through the metal foil with little or no deviation. However a small proportion of particles are scattered through large angles and even bounced back i.e. deflected through 1800. From these\nobservations, Rutherford drew the following conclusions:\nReflected -particles\n\nGold foil\n\nZinc sulphide\nscreen\n\nFig. Rutherford's Experiment\n\n(a) As most of the -particles passed through the foil without undergoing any deflection, there must be sufficient\nempty space within the atom.\n(b) As -particles are positively charged, deflected by large angle there must be heavy small positively charged\nbody present in the atom, which is called Nucleus.\nFrom these observations, Rutherford proposed the following model of atom:\n(a) Atom is composed of a positively charged nucleus where the whole mass of the atom is concentrated and the\nelectrons are present in relatively large volume around the nucleus. The total positive charge carried by nucleus\nmust be equal to the total negative charge carried by electrons and electroneutrality of atom is maintained.\n\npositive sphere\n\nelectron\n\nFig: Thomson Model\n\n(b) Electron are constantly moving around the nucleus in different orbits.\nThe centrifugal force arising from this motion balances the electrostatic attraction between the nucleus and the\nelectron. Therefore the electron dont fall into the nucleus.\nDrawbacks in Rutherford Model\n(a) The most fundamental objection arises from the electromagnetic theory of radiation which predicts that when\na charged body moves in a circular path, it should radiate energy continuously. As the electron is a negatively\ncharged particle revolving around the nucleus, it should radiate energy continuously. As a result, the electron\nshould fall into the nucleus.\n(b) As the electron is continuously radiating energy, the spectra should be continuous. Actually, the spectra of\natom is a line spectra.\n8. ATOMIC NUMBER OF AN ELEMENT\n= Total number of protons present in the nucleus\n= Total number of electrons present in the atom\n* Atomic number is also known as proton number because the charge on the nucleus depends upon the number\nof protons.\n* Since the electrons have negligible mass, the entire mass o the atom is mainly due to protons and neutrons only.\nSince these particles are present in the nucleus, therefore they are collectively called nucleons.\n* As each of these particles has one unit mass on the atomic mass scale, therefore the sum of the number of\nprotons and neutrons will be nearly equal to the mass of the atom.\nMass number of an element = No. of protons + No. of neutrons.\n* The mass number of an element is nearly equal to the atomic mass of that element. However, the main difference\nbetween the two is that mass number is always a whole number whereas atomic mass is usually not a whole\nnumber.\n* The atomic number (Z) and mass number (A) of an element X are usually represented alongwith the symbol\nof the element as\n\nAtomic Structure\nMass Number\n\nAtomic Number\n\ne.g.\n\n23\n11\n\nSymbol of the\nelement\n\n35\nNa , 17\nCl and so on.\n\n1.Isotopes:\nSuch atoms of the same element having same atomic number but different mass numbers are called isotopes.\n1\n1\n\nH , 12 H and 13 H and named as protium, deuterium (D) and tritium (T) respectively. Ordinary hydrogen is\nprotium.\n\n2. Isobars:\nSuch atoms of different elements which have same mass numbers (and of course different atomic numbers) are\ncalled isobars.\ne.g.\n\n40\n18\n\nAr ,\n\n40\n19\n\nK,\n\n40\n20\n\nCa .\n\n3. Isotones:\nSuch atoms of different elements which contain the same number of neutrons are called isotones.\ne.g.\n\n14\n6\n\nC,\n\n15\n7\n\n.\nK , 16\n8 O\n\n4. Isoelectronics:\nThe species (atoms or ions) containing the same number of electrons are called isoelectronic.\nFor Example, O 2 , F , Na , Mg 2 , Al 3 , Ne all contain 10 electrons each and hence they are isoelectronic.\nIllustration.\nComplete the following table:\nParticle\nMass No.\nAtomic No.\nProtons\nNeutrons\nElectrons\nNitrogen atom\n\n7\n7\nCalcium ion\n\n20\n\n20\n\nOxygen atom\n16\n8\n\nBromide ion\n\n45\n36\nSolution.\nFor nitrogen atom.\nNo. of electron = 7\n(given)\nNo. of neutrons = 7\n(given)\nNo.\nof\nprotons\n=\nZ\n=\n7\n( atom is electrically neutral)\n\nAtomic number = Z = 7\nMass No. (A) = No. of protons + No. of neutrons = 7 + 7 = 14\nFor calcium ion.\nNo. of neutrons = 20\n(given)\nAtomic No. (Z) = 20\n(given)\nNo. of protons = Z = 20;\n\nNo. of electrons in calcium atom = Z = 20\n\nBut in the formation of calcium ion, two electrons are lost from the extranuclear part according to the equation\nCa Ca 2 2e but the composition of the nucleus remains unchanged.\nNo. of electrons in calcium ion = 20 2 = 18\nMass number (A) = No. of protons + No. of neutrons = 20 + 20 = 40.\n\nFor oxygen atom.\n\nMass number (A) = No. of protons + No. of neutrons = 16\nAtomic No. (Z) = 8\nNo. of protons = Z = 8,\n\n(Given)\n(Given)\n\nAtomic Structure\nNo. of electrons = Z = 8\nNo. of neutrons = A Z = 16 8 = 8\nSome important characteristics of a wave:\n\nCrest\n\nCrest\n\na\nTrough\n\nTrough\n\nWavelength of a wave is defined as the distance between any two consecutive crests or troughs. It is represented\nby and is expressed in or m or cm or nm (nanometer) or pm (picometer).\n1 10 8 cm 1010 m\n1 nm 109 m, 1 pm 10 12 m\n\nFrequency of a wave is defined as the number of waves passing through a point in one second. It is represented\nby v (nu) and is expressed in Hertz (Hz) or cycles/sec or simply sec1 or s1.\n1 Hz = 1 cycle/sec\nVelocity of a wave is defined as the linear distance travelled by the wave in one second. It is represented by v and\nis expressed in cm/sec or m/sec (ms1).\nAmplitude of a wave is the height of the crest or the depth of the trough. It is represented by a and is expressed\nin the units of length.\nWave number is defined as the number of waves present in 1 cm length. Evidently, it will be equal to the\nreciprocal of the wavelength. It is represented by v (read as nu bar).\n1\nv\n\nIf is expressed in cm, v will have the units cm1.\n\nRelationship between velocity, wavelength and frequency of a wave. As frequency is the number of waves\npassing through a point per second and is the length of each wave, hence their product will give the velocity\nof the wave. Thus,\n\nv v\nCosmic rays < - rays < X-rays < Ultraviolet rays < Visible < Infrared < Micro waves < Radio waves.\n9. PLANCKS QUANTUM THEORY (1901)\nIt states\n1.Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy called Quanta.\n2.Each quantum is associated with a definite amount of energy E which is proportional to frequency of radiation.\n\nE hv\nWhere,\n\nh = Plancks constant = 6.626 10 34 Joule sec .\n\nv = Frequency of the light radiation\n\n3.A body can emit or absorb radiations only in whole multiples of quantum i.e. E nhv\nwhere n 1, 2,3......\n\nwhere\n\nc = velocity of light\n\n= wavelength\n\nAtomic Structure\n\nhc\n\n9(a). Some Important Formulae\n\n1.A = Z + N (Number of neutrons)\n2.dynamic mass of particle m m0 /[1 (v / c) 2 ]1/ 2\n3.Radius of nucleus R R0 ( A)1/ 3 , R0 1.2 10 15 m\n4. c v\n\n5.wave number = v 1/\n6. E hv hc / hcv\nq1 q2\n1\n9\n2\n2\n7. F K r 2 ; K 4 9.0 10 Nm / C\n0\n\n8. E hv E2 E1\nIllustration 1.\nCalculate number of photon coming out per sec. from the bulb of 100 watt. If it is 50% efficient and wavelength\ncoming out is 600 nm.\nSolution.\nenergy = 100 J\nenergy of one photon =\n\nno. of photon =\n\nhc 6.625 1034 3 108 6.625\n\n1019\n\n600 109\n2\n\n100\n1019 15.09 1019\n6.625\n\nIllustration 2.\nCertain sun glasses having small of AgCl incorporated in the lenses, on exposure to light of appropriate wavelength\nturns to gray colour to reduce the glare following the reactions:\nhv\nAgCl\nAg (Gray ) Cl\n\nIf the heat of reaction for the decomposition of AgCl is 248 kJ mol1, what maximum wavelength is needed to\ninduce the desired process?\nSolution.\nEnergy needed to change = 248 10 3 J / mol\nIf photon is used for this purpose, then according to Einstein law one molecule absorbs one photon. Therefore,\n\nNA\n\nhc\n248 103\n\n4.83 107 m .\n248 103\n\n10. BOHRS ATOMIC MODEL\n\nThe postulates of Bohrs atomic theory regarding stability of electrons of an atom are as follows:\n(i) The electrons is an atom revolve around the nucleus only in certain selected circular orbits. These orbits are\nknown as energy levels or stationary states. An electron can be excited from a lower state to higher state with the\nabsorption of a quantum of energy, or can come down from a higher to lower state with emission of a radiation\nof energy (as shown in figure) equal to energy to quantum E E2 E1 hv . E2 and E1 are energies of the\nelectron associated with stationary orbits.\n\nAtomic Structure\nEmission of\nAbsorption of E\n\nE(E2 E1 )\n\nE1\n+\n\nFig. Bohr's Atomic Model\n\n(ii) The stability of the circular motion of an electron requires that the electrostatic force (due to the attraction\nbetween the nucleus and the electron) provides the necessary centripetal force for the motion of electron.\n1 ( Ze) e\n.\nmv 2 / r\n4 0 r 2\n\nwhere\n\n... (i)\n\nZ atomic number\ne charge on electron\n\n0 permittivity of the charge in vacuum\n\nr distance between positive charge & electron\n(iii) Angular momentum of electron is quantised i.e. electron can revolve only in those orbits where its angular\nmomentum is an integral multiple of h / 2\n... (ii)\n\nmvr nh / 2\nwhere,\n\nv velocity of electron\n\nm mass of electron\nh Plancks constant\n\nn 1, 2, 3........ are known as Principal quantum number..\n\nfrom (i) &(ii) we have from (i)\n\nv 2 ze 2 /(4 0 ) mr\n\n... (iii)\n\nv 2 n 2 h 2 / 4 2 m2 r 2\nEquating (iii) and (iv),\nwe have\n& from (ii)\n\n... (iv)\n\nZe 2 /(4 0 ) mr n 2 h 2 / 4 2 m 2 r 2\n\nor,\n\n(4 0 )n2 h 2\n4 2 mZe 2\n\nWhere h, , e, m and 0 are constants, Thus, r K (n 2 / Z )\nPutting the values of h, , e, m and 0 , r (5.297 10 11 m) ( n 2 / Z )\nIllustration 1.\nFor hydrogen atom Z 1 , therefore radius of the first orbit = (5.297 10 11 m) (12 /1)\nFor He ion Z 2 , therefore radius of the first orbit = (5.297 10 11 m) (12 / 2)\n2.649 10 11 m\n\nSolution.\nVelocity of electrons in various orbits\n\nAtomic Structure\nSubstituting value of r in equation no. (ii)\nv\n\n2 Ze 2\nK ( Z / n)\n4 0 nh\n\nThe energy of an electron in an orbit\n\nZe 2\n1 2\nE = Kinetic energy + Potential energy = mv\n(4 0 ) r\n2\n\nPutting ( mv ) 2 from equation (i), we have\n\nZe 2\nZe 2\nZe 2\n\n2(4 0 ) r (4 0 ) r\n2(4 0 ) r\n\nNow, putting the value of ' r ' , we have\n\nE\n\n2 2 mZ 2 e 4\nK ( Z 2 / n2 ) (2.18 10 18 J ) ( Z 2 / n2 )\n2 2 2\n(4 0 ) n h\n\nThe energy difference between two energy levels n2 and n1 is given by\n\nE En2 En1\n\n1\n2 2 mZ 2 e 4\n1/ n12 1/ n22\n\n(4 0 ) 2\nh2\n\n1 1\n(2.18 1018 J ) Z 2 2 2\nn1 n2\n\nIt terms of wave-number, we have\n\nv RH Z 2 1/ n12 1/ n22\n\n10(a). Bohrs Model for Hydrogen like Atoms:\n\n1. mvr n\n\n2. En\n\nE1\n\nh\n2\n\nE1 2\nz2\nz2\nz 2.178 1018 2 J / atom 13.6 2 eV\n2\nn\nn\nn\n\n2 2 me 4\nn2\n\n3. rn\n\nn2\nh2\n0.529 n 2\n2 2\n\nZ 4 e m\nZ\n\n4. v\n\n2 ze 2 2.18 106 z\n\nm/s\nnh\nn\n\n0.657 Z 2 1016\nn3\n\n1.52 1016 n3\nZ2\n\n7. En K .E. P.E.( P.E. 2 K .E.)\n\n2\n8. K .E. 1/ 2mv , P.E.\n\n1 ze 2\n4 0 r\n\nAtomic Structure\nIllustration 1.\nWhat is the principal quantum number of H atom orbital is the electron energy is 3.4 eV? Also report the\nangular momentum of electron.\nSolution.\nE1 for H = 13.6 eV\nE1\nn2\n\nNow,\n\nEn\n\n3.4\n\nn=2\n\n13.6\nn2\n\nh 2 6.626 1034\n\n2.1 1034 J sec 1 .\n\n2\n2 3.14\n\nIllustration 2.\nCalculate the energy, velocity and radius of electron in Li2+ ion.\nSolution.\nFor Li2+ ion, z = 3, n = 1, then\n\nn2\nz\n\n1\n0.529 0.1763\n3\n\n8\nvelocity = 2.18 10 cm / sec\n\nz\n3 2.18 108 cm / sec.\nn\n\n6.54 108 cm / sec.\n\nEnergy\n\n13.6 ev\n\nz2\n9\n13.6 ev\n2\nn\n1\n\n122.4 ev .\n11. DEFINITION VALID FOR SINGLE ELECTRON SYSTEM :\n(i) Ground state:\nlowest energy state of any atom or ion is called ground state of the atom.\nGround state energy of H-atom = 13.6 eV\nGround state energy of He+ ion = 54.4 eV\n(ii) Excited state:\nState of atom other than the ground state are called excited states:\nn=2\nfirst exited state\nn=3\nsecond exited state\nn=4\nthird exited state\nn=n+1\nnth exited state\n(iii) Ionisation energy (IE) :\nMinimum energy required to move an electron from ground state to\nn is called ionisation energy of the atom or ion.\nIonisation energy of H-atom = 13.6 eV\nIonisation energy of He+ ion = 54.4 eV\nIonisation energy of Li+2 ion = 122.4 eV\n(iv) Ionisation Potential (I.P):\nPotential difference through which a free electron must be accelerated from rest, such that its kinetic energy\nbecomes equal to ionisation energy of the atom is called ionisation potential of the atom.\nI.P. of H atom = 13.6 V\nI.P of He+ Ion = 54.4 V\n\n10\n\nAtomic Structure\n(v) Excitation Energy:\nEnergy required to move an electron from ground state of the atom to any other state of the atom is called\nexcitation energy of that state.\nexcitation energy of 2nd state = excitation energy of 1st state = 1st excitation energy = 10.2 eV.\n(vi) Excitation Potential:\nPotential difference through which an electron must be accelerated from rest to so that its kinetic energy become\nequal to excitation energy of any state is called excitation potential of that state.\nexcitation potential of third state = excitation potential of second excitate state = seconds excitations potential =\n12.09 v.\n(vii) Binding Energy or Separation Energy:\nEnergy required to move an electron from any state to n is called binding energy of that state.\nBinding energy of ground state = I.E. of atom or Ion.\nIllustration.\nA single electron system has ionisation energy 11180 kJ mol1. Find the number of protons in the nucleus of the\nsystem.\nSolution.\nI.E. =\n\nZ2\n21.69 1019 J\nn2\n\n11180 103 Z 2\n\n21.69 1019\n6.023 1023 I 2\nZ=3\n\n12. HYDROGEN SPECTRUM:\n\n1. Study of Emission and Absorption Spectra:\nAn instrument used to separate the radiation of different wavelengths (or frequencies) is called spectroscope or\na spectrograph. Photograph (or the pattern) of the emergent radiation recorded on the film is called a spectrogram\nof simply a spectrum of the given radiation. The branch or science dealing with the study of spectra is called\nspectroscopy.\nEmission Spectra\nWhen the radiation emitted from some source e.g. from the sun or by passing electric discharge through a gas at\nlow pressure or by heating some substance to high temperature etc, is passed directly through the prism and then\nreceived on the photographic plate, the spectrum obtained is called Emission spectrum.\nDepending upon the sources of radiation, the emission spectra are mainly of two types:\n(i) Continuous spectra:\nWhen white light from any source such as sun, a bulb or any hot glowing body is analysed by passing through a\nprism it is observed that it splits up into seven different wide bound of colours from violet to red. These colours\nare so continuous that each of them merges into the next. Hence the spectrum is called continuous spectrum.\nWhite light\n\nR\nO\nY\nG\nB\nI\nV\n\nBeam\nSlit\n\nPrism\n\nPhotographic\nPlate\n\n(ii) Line Spectra:\n\nWhen some volatile salt (e.g., sodium chloride) is placed in the Bunsen flame or an electric discharge is passed\nthrough a gas at low pressure light emitted depends upon the nature of substance.\nPlatinum wire\nBeam\n\n5896\nTwo yellow lines\n5890\n\nSlit\nPrism\n\nPhotographic\nPlate\n\n11\n\nAtomic Structure\nIt is found that no continuous spectrum is obtained but some isolated coloured lines are obtained on the\nphotographic plate separated from each other by dark spaces. This spectrum is called Line emission spectrum\nor simply Line spectrum.\n2. Absorption spectra:\nWhen white light from any source is first passed through the solution or vapours of a chemical substance and\nthen analysed by the spectroscope, it is observed that some dark lines are obtained in the otherwise continuous\nspectrum. These dark lines are supposed to result from the fact that when white light (containing radiations of\nmany wavelengths) is passed through the chemical substance, radiation. of certain wavelengths are absorbed,\ndepending upon the nature of the element.\nWhite light\n\nR\nO\nY}\nG\nB\nI\nV\n\nNaCl\nSolution\n\nSlit\n\nPrism\n\nPhotographic\nPlate\n\nDark lines in yellow region of\n\ncontinuous spectrum\n\nBeam\nSlit\nPrism\n\nPhotographic\nPlate\n\nEMISSION SPECTRUM OF HYDROGEN:\n\n12(a). H-Atom Spectrum\nWhen an electric discharge is passed through hydrogen gas at low pressure, a bluish light is emitted. When a ray\nof this light is passed through a prism, discontinuous line spectrum of several isolated sharp lines is obtained as\nshown in figure.\n\nn=\n6\n5\n\nPfund series\n\nBracket series\n\nPaschen series\n\nEnergy\n2\n\nn=1\n\nBalmer series\n\nLymann series\nEnergy levels of H-atom\n\nAll these lines observed in the hydrogen spectrum can be classified into the series as is tabulated in the table.\nThe hydrogen Spectrum\nRegion\nSpectral lines\nn1\nn2\nUV\nLyman series\n1\n2,3,4,.....\nVisible\nBalmer series\n2\n3,4,5,.....\nIR\nPaschen series\n3\n4,5,6,.....\nfar-I.R\nBrackett series\n4\n5,6,7,.....\n\n12\n\nAtomic Structure\nfar-I.R.\nP fund series\n5\n6,7,8,.....\nIllustration\nCalculate the highest wavelength of line spectra of H-atom when the electron is situated in 3rd excited state.\nSolution.\nHighest wavelength means lowest energy difference of electronic transition from one energy level to other\nenergy level.\nHence, lowest energy transition will be n 4 to n 3 .\nE4 13.6 ev\n\n1\n16\n\n0.85 ev\n\nE3 13.6 ev\n\n1\n9\n\n1.54 ev\n\nE E4 E3\n\n(0.85 1.54)ev\n\nhc\n0.69ev 0.69 1.6 10 19 J\n\nm\n0.69 1.6 1019\n\n18 107 m\n\n13. DE-BROGLIE RELATIONSHIP\n\nAccording to de-Broglie matter has dual character i.e. wave as well as particle, if the wavelength of matter be\nhaving mass m moving with velocity v , then,\n\nh\nmv\n\nwave in phase\nCase - I\n\nwave out of phase\n\nCase - II\n\nFor electron moving around a nucleus in a circular path, two possible waves of different wavelengths are possible.\nIn case I, the circumference of the electron orbit is an integral multiple of wavelength.\nIn case II, wave is destroyed by interference and hence, does not exist.\nTherefore, the necessary condition for a stable orbit of electron of radius r is, 2 r n\nwhen n 1, 2,3, etc .\nAs\n\n2 r\n\nh\nmv\n\nnh\nmv\n\nor ,\n\nmvr\n\nnh\n2\n\nThis is simply the original Bohr condition for a stable orbit. Hence, the Bohrs model of H-atom is justified by\nde-Broglie relationship.\n13.(a)\nDe-Broglie Relations:\n\nh\nh\n\nmc p\n\n13\n\nAtomic Structure\nde-Broglie pointed out that the same equation might be applid to material particle by suing m for the mass of the\nparticle instead of the mass of photon and replacing c, the velocity of the photon, by v, the velocity of the\nparticle.\n\nmv\n\nh\n2m( K .E.)\n\nFrom the de-Broglie equation it follows that wavelength of a particle decreases with increase in velocity of the\nparticle. Moreover, lighter particles would have longer wavelengths than heavier paticles, provided the velocity\nis equal.\nIf a charged particle Q is accelerated through potential difference V from rest then de-broglie wavelength is\nh\n\n2mQV\n\nde-Broglie concept is more isgnificant for microscopic or sub-microscopic particles whose wavelength can be\nmeasured.\nThe circumference of the nth orbit is equal to n times the wavelength of the electron.\n2 rn n\nWavelength of electron is always calculated using De-broglie calculation.\nIllustration. 1\nCalculate the wavelength of a body of mass 1 kg moving with a velocity of 10 m sec1.\nSolution.\n\nWe know,\n\nh\nmv\n\nSubstituting the values, m = 1, mg = 106 kg, v = 10 m sec1.\n\nh 6.625 10 34 kg m 2 s 1\n\nand\n\n6.625 1034\n6.625 1029 m\n106 10\n\nIllustration. 2\n13.6 eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground states absorbs\n1.50 times as much energy as the minimum energy required for it to escape from the atom. What is the wavelength\nof the emitted electron? ( me 9.109 10 31 kg , e 1.602 10 19 coulomb, h 6.63 10 34 J .s )\nSolution.\n1.5 times of 13.6 eV, i.e., 20.4 eV, is absorbed by the hydrogen atom out of which 6.8 eV (20.4 13.6) is\nconverted to kinetic energy.\nKE = 6.8 eV = 6.8 ( 1.602 10 19 coulomb ) (1 volt) = 1.09 10 18 J .\nNow,\n\nKE\n\n1\nmv 2\n2\n\nor,\n\n2 KE\n2(1.09 10 18 J )\n\n1.55 10 6 m / s.\nm\n(9.109 10 31 kg )\n\nh\n(6.63 10 34 J .s)\n\n4.70 10 10 metres.\nmv (9.109 10 31 kg )(1.55 10 6 m / s )\n\n14. HEISENBERGS UNCERTAINTY PRINCIPLE (1927):\n\nIf subatomic particles have wave nature then we cant pinpoint where exactly a particle is. The idea was defined\nby Heinsenberg as - There is a limit to the precision to which the position and momentum of a particle may be\ndetermined simultaneouslyx p h / 4\n\n14\n\nAtomic Structure\nWhere\n\nx uncertainty in position of an electron.\n\np uncertainty in momentum of an electron, i.e., when we try to determine position for a subatomic\nparticle correctly, uncertainly in momentum will be very large or when we try to determine the momentum\ncorrectly then uncertainly in position will be large.\nORIGIN OF QUANTUM THEORY\nWhen solid body heated it emit radiations in the forms of waves. The wave nature of light can be explained by\ndiffraction interference etc. But some other observable properties such as photoelectric effect, compton effect\ncould not be explained from wave nature. Hence a different theory is needed to explain these facts.\nIllustration.\nCalculate the uncertainty in the velocity of a wagon of mass 2000 kg whose position is known to an accuracy of\n10 m .\nSolution.\nUncertainty in position,\n\nx 10 m\n\nm 2000 kg\n\nAccording to Heinsenbergs principle,\n\nx.m v\n\nh\n4\n\nh\n6.626 10 34 kgm 2 s 1\n\n2.64 10 39 ms 1 .\n4 m.x 4 3.14 2000 kg 10 m\n\nWhen radiation falls on an object, a part of it is reflected, a part is absorbed and the remaining part is transmitted\nbecause no object is a perfect absorber. But the black body (e.g., a metallic hollow sphere with a small hole,\nblackened on the inside surface) absorbs completely all the radiations falls on it by successive reflections inside\nthe enclosure.\nT1\nenergy\n\nT2\nT3\n\nT1>T2>T3\n\nWave length\n\nThe black body is not only a perfect absorber of radiation energy, but also an ideal radiation, i.e., when the black\nbody is heated, it radiates the maximum amount of energy. The energy which is radiated is dependent on the\ntemperature of the black body and is independent of the nature of the interior material.\nThe curves represent the distribution of radiation from a black body at different temperatures. The shape of the\ncurves couldnt be explained on the basis of classical electromagnetic theory in which it was assumed that the\nbody radiates energy continuously. So, the intensity of radiation should increase continuously without limits as\nthe frequency increases. But the experimental observations are contrary to the classical view. For each temperature,\nthere is a maximum in the curve corresponding to a particular wavelength, indicating the maximum radiation of\nenergy. At higher temperature, the position of the maximum in the curve shifts towards shorter wavelength and\nbecomes more pronounced. To explain this black body radiation, Max Planck put forward new quantum theory.\nPlanck Quantum Theory\n(a) Radiation energy is not emitted or absorbed continuously but discontinuously in the form of tiny bundles of\nenergy, called quanta.\n(b) Each quanta is associated with a defined amount of energy (E) which proportional to the frequency of\nradiation i.e, E v or, E hv where h is Plancks constant\n\n(6.626 10 34 J sec) .\n\n(c) A body can emit or absorb energy only in whole number multiples of quanta i.e. E nhv where n 1, 2,3 etc.\n15. PHOTOELECTRIC EFFECT\nSir J.J. Thomson has discovered this phenomenon of ejection of electron from the surface of a metal when light\nof suitable frequency of strikes on it.\n\n15\n\nAtomic Structure\nOnly few metals show this effect under the action of visible light, but many more show it under the action of\nmore energetic u.v. light. For every metal, there is a minimum frequency of incident radiation necessary to eject\nelectron from that metal surface, is known as Threshold frequency ( v0 ). This v0 varies metal to metal.\nThe number of ejected electrons from the metal surface depends upon the intensity of the incident radiation.\nGreater the intensity, the larger is the number of ejected electrons.\nHence, according to quantum theory, when a photon of light of frequency (v v0 ) strikes on an electron in a\nmetal, it imparts it entire energy to the electron. Then some of its energy (equal to binding energy of electron with\nthe nucleus) is consumed to separate the electron from the metal and the remaining energy will be imparted to the\nejected electron.\n1\n1\nhv hv0 mv 2 where hv0 is the binding energy of work function of the electron and mv 2 is the kinetic\n2\n2\nenergy of electron. Alkali metals are mainly used for photoelectric effect. Cesium, amongst alkali metals, has\nlowest threshold energy and used largely in photoelectric cell.\n16. SHAPES OF ORBITALS\ns-orbital: they do not have directional character. They are spherically symmetrical. The s-orbital of higher\nenergy levels are also spherically symmetrical. They are more diffused and have spherical shells within them\nwhere probability of finding the electron is zero.\ny\nNode\nz\n\n2s orbital\n\nIn the s-orbital, number of nodes is (n 1)\n\np-orbital: p orbital has a dumb-bell shape and it has a directional character.\nThe two lobes of a p-orbital are separated by a plane that contains the nucleus and is perpendicular to the\ncorresponding axis. Such plane is called a nodal plane because there is no probability of finding the electron.\ny\n\nz\nz\n\n+\nx\n\nz\n+\nx\n\npx\n\npy\n\npz\n\nIn the absence of an external electric or magnetic field, the three p-orbitals of a particular energy level have same\nenergy and are degenerate. In the presence of an external magnetic field or electric field this degeneracy is\nremoved.\nd-orbitals: For d-orbitals five orientations are possible viz., dxy, dyz, dxz, d x2 y 2 , d z 2 . All these five orbitals in the\nabsence of magnetic field are equivalent in energy and are degenerate.\nThe shapes of the orbitals are as follows:\n\n16\n\nAtomic Structure\nx\n\ndxy\n\ndxz\n\ndyz\n\nThese three d orbitals are similar. The maximum probability of finding the electron is in lobes which are\ndirected in between the axes. Nodal region is along the axes.\nz\n\ndz2\n\ndx2-y2\n\nThese two d-orbitals are similar. Probability of finding the electron is maximum along the axes and the nodal\nregion is in between the axes.\n17. QUANTUM NUMBERS\nThese are used to determine the region of probability of finding a particular electron in an atom.\n(a) Principal quantum number (n):\nThis denotes the energy level or the principal or main shell to which an electron belongs. It can have only\nintegral values 1, 2, 3 etc. The letter K, L, M ..... are also used to designate the value of n. Thus, an electron in the\nK shell has n = 1, that is L shell has n = 2 and so on.\nIllustration. 1\nThe principal quantum number of 2s-electron is\nSolution.\nn=2\n(b) Azimuthal quantum numbers (l):\nThis denotes the orbital (Sub-level) to which an electron belongs. It gives an idea about the shape of the orbital.\nl can have any value from 0 to (n 1), for a given value of n,\ni.e. l = 0, 1, 2, ..... (n 1)\nValue of l\n0\n1\n2\n3\nSub Shell\ns\np\nd\nf\nThe value of orbital angular momentum of the electron for a given value of l is\n\nl (l 1)\n\nh\n2\n\n(c) Magnetic quantum number (m):\n\nIt gives us the idea about the orientations an orbital can have in space in the presence of magnetic field. The\nvalues of m depends on l orbital quantum number.\nTotal value of m = (2l + 1) and it varies l to +l.\nFor example, for l = 0 the value of magnetic quantum number m is also equal to zero, i.e. s-orbital can have\nonly one orientation in space in presence of magnetic field.\n(d) Spin quantum number (s):\nThe electron while moving round the nucleus in an orbit also rotates or spins about its own axis either in a\n\n17\n\nAtomic Structure\nclockwise direction or in an anticlockwise direction. Its value is\n\n1\n1\nor corresponding to clockwise or\n2\n2\n\nanticlockwise spin.\n(1) The value of spin angular momentum for a given value of s is\n\ns( s 1)\n\nh\n2\n\n(2) The spin magnetic moment of electron (excluding orbital magnetic momentum) is given by\n\neffective n(n 2) BM (Where n = Number of unpaired electrons).\n\n18. DISTRIBUTION OF ELECTRONS IN AN ATOM\nThe filling up of orbitals with electrons takes place according to certain rules which are given below:\n(i) The maximum number of electrons in a main shell is equal to 2n 2, where n is the principal quantum number.\n(ii) The maximum number of electrons in a sub-shell like s, p, d, f is equal to (2l + 1), where l is the azimuthal\nquantum number for the respective orbitals. Thus s, p, d, f can have a maximum of 2, 6, 10 and 14 electrons\nrespectively.\n(a) Afbau Principle\nAccording to the principle, Electrons are added progressively to the various orbitals in the order of increasing\nenergy.\nWhat does the word Afbau mean?\nAfbau is a German term which means building up or construction.\nThe energy of various orbitals increase in the order given below:\n1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s < ......\n\n(i) A new electron enters the orbitals for which (n + l) is minimum, e.g. if we consider 3d and 4s orbitals, the\nelectron will first enter 4s-orbitals in preference to 3d.\nThis is because the value of (n + l) for 4s-orbitals is less (4 + 0 = 4) than that for 3d-orbital (3 + 2 = 5)\n(ii) In case where (n + l) values are the same, the new electron enters the orbitals for which n is minimum, e.g.\nin a choice between 3d and 4p for which (n + l) values are same (3 + 2 = 5, 4 + 1 = 5), the electron will prefer to\ngo to the 3d-orbital, since n is lower for this orbital.\n(b) Paulis Exclusion Principle\nIt states that it is impossible for two electrons in a given atom to have same set of quantum numbers.\nExample:\n(a)\nn = 2, l = 0,\nm = 0, s = +1/2\nn = 2, l = 0,\nm = 0, s = 1/2\n(b)\nn = 2, l = 1,\nm = 0, s = +1/2\nn = 2, l = 1,\nm = 0, s = 1.2\nn = 2, l = 1,\nm = +1, s = +1/2\nn = 2, l = 1,\nm = +1, s = 1/2\nn = 2, l = 1,\nm = 1, s = +1/2\nn = 2, l = 1,\nm = 1, s = 1/2\n(c) Hunds Rule of Maximum Multiplicity\nAccording to this rule, electrons enter the orbitals (e.g. s, px, py, pz ...) in the same sub-level in such a way as to\n\n18\n\nAtomic Structure\ngive maximum number of unpaired electrons. In other words it means that pairing begins with the introduction of\nthe second electron in the s-orbital, the fourth in p, etc.\nWhat is the electronic configuration of Cu (Z = 29)?\n1s2 2s2 2p6 3s2 3p6 3d10 4s1\nExceptional Electronic Configuration\nSome atoms such as copper and chromium exhibit exceptional electronic configuration.\nFor example:\nCr(Z = 24) has an electronic configuration\n1s2 2s2 2p6 3s2 3p6 3d5 4s1\nIt is because of the extra stability associated with the half-filled and completely filled orbitals.\nCONCEPT BUILDING EXAMPLES\nExample 1.\nFind out the energy of the electron in the first excited state in an H-atom.\nSolution.\nEnergy of an electron in H-like atom is given by\nZ2\nE (2.18 108 J ) 2 ; where z is the number of protons and n is\nn\n\nnumber of shell in which electron is present.\n\nFor the first excited state, n 2\n12\nE (2.18 1018 J ) 2\n2\nE 5.45 10 19 J\n\nExample 2.\nThe K.E. of a moving electron is 5 10 25 J ; Calculate its velocity and the wavelength.\nSolution :\nK .E.\n\n1 2\nmv\n2\n\nvelocity v\n\n2( K .E )\n\nNow, wavelength\n\n2 5 10 25 J\n1.048 103 ms 1\n9.1 10 31 kg\n\nh\nmv\n\n6.626 10 34 Js\n6.949 10 7 m\n3\n1\n(9.1 10 kg ) (1.048 10 ms )\n31\n\nExample 3.\nIf the electron of the hydrogen atom has been excited to a level corresponding to 10.2 electron volts, what is the\nwavelength of the line emitted when the atom returns to its ground state?\nSolution.\nE E2 E1\nhc\n\nE hv\n\nE 10.2 eV\n\nE 10.2 1.6 10 19 J\n\n19\n\nAtomic Structure\n\nm\n10.2 1.6 1019\n\n6.624 1034 3 108 109\n\nnm\n10.2 1.6 1019\n\n121.8nm .\nExample 4.\nAn electron jumps from fourth excited state to the ground stable in Li+2 -ion and the energy released in the form\nof photon is allowed to strike a metal ( x ) surface whose work function ( ) is 1 10 18 J . What is the K.E. &\nvelocity of the electron ejected.\nSolution.\nThe amount of energy released is given by\n1\n1\nE (2.18 1018 J ) z 2 2 2 where z is the atomic number of H-like atom.\nn1 n2\n\nE (2.18 1018 J ) (3) 2 1 1.23 1018 J\n\n16\nNow, hv K .E.\nK .E. (1.23 1018 1 10 18 J )\n\n1 2\nmv 2.3 10 18 J\n2\n\n2 2.3 10 19\nms 1 7.11 105 ms 1\n9.1 10 31\n\nExample 5.\nIonisation energy of hydrogen atom is 13.6 eV. What will be the ionisation energy of He+ and Li++ ions?\nSolution.\nAs we know ionisation energy for one electron system is given by\n\nI .P. 13.6 eV\n\nz2\nn2\n\nFor hydrogen atom, z = 1, n = 1,\n\nFor He+ ion z = 2 and n = 1.\n\nSo, I.P = 13.6 13.6 (2) 2\n\ni.e., I .P. 54.4 eV For Li2+, z = 3 & n = 1; I.P. = 13.6 (3) 2\ni.e., I .P 122.4 eV .\nExample 6.\nA hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n . This excited atom\ncan make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17\neV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited\nstate by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values of\nn and z (ionisation energy of hydrogen atom = 13.6 eV).\nSolution.\nTotal energy liberated during transition of electron from nth shell to first excited state (i.e., 2nd shell)\n\n10.2 17 17.2 eV\n27.2 1.602 10 12 erg\n\n20\n\nAtomic Structure\n\nhc\n1 1\nRH Z 2 hc 2 2\n\n2 n\n\n1 1\n27.2 1.602 1012 RH Z 2 hc 2 2\n2 n\n\n... (i)\n\nSimilarly, total energy liberated during transition of electron from nth shell to second excited state (i.e. 3rd shell):\n\n4.25 5.95 10.2eV\n\n10.2 1.602 10 12 erg\n\n1 1\n10.2 1.602 1012 RH Z 2 hc 2 2\n3 n\n\n... (ii)\n\nDividing eq. (i) by eq. (ii) n 6\n\nOn substituting the value of n in eqns. (i) or (ii) Z 3 .\nExample 7.\n1 mol of He+ ion excited. Spectral analysis showed the existence of 50% ions in 3rd level, 25% in 2nd level and\nremaining 25% in ground state. Ionisation energy of He+ is 54.4 eV; calculate total energy evolved when all the\nSolution.\n25% of He+ ions are already in ground state, hence energy emitted will be from the ions present in 3rd level and\n2nd level.\n1 1\nE ( IP )2 2 2 per ion or atom.\nn1 n2\n\nE 31 (54.4)\n\nN0\n2\n\n1 1\nN0\nions falling to ground state\n12 32 for\n\n54.4\n\nN 1 1\n4 N0\nN\neV and E 21 (54.4) 0 2 2 for 0 ions falling to ground state.\n4 1 2\n9\n4\n\n54.4\n\n3 N0\neV\n16\n\n4 3\nHence total energy = 54.4 N 0\n9 16\n54.4 6.023 10 23\n\n91\neV\n144\n\n54.4 6.023 10 23\n\n91\n1.6 1019 J\n144\n\n331.13 10 4 J\n\nExample 8.\nEstimate the difference in energy between the 1st and 2nd Bohr orbit for a hydrogen atom. At what minimum\natomic number, would a transition from n = 2 to n = 1 energy level result in the emission of X-rays with\n\n3.0 108 m . Which hydrogen atom like species does this atomic number correspond to?\nSolution.\nFor hydrogen atom, the expression for energy difference between two energy level is,\n1 1\nE RH hc 2 2\nn2 n2\n\n21\n\nAtomic Structure\n\n1\n7\n34\n8\nSo, E E2 E1 1.09677 10 6.626 10 3 10 1\n4\n1.635 10 8 J\n\nFor hydrogen like species, the same expression is\n\n1 1\n1\nz 2 RH 2 2\n\nn2 n2\n\nSo,\n\nor,\n\n1 1\n1\nz 2 RH 2 2\n\nn1 n2\n\n1\n3\nz 2 (1.09677 107 ) or, z = 2\n8\n3 10\n4\n\nSo, the species is He+ because z = 2.\n\nExample 9.\nWhat is the degeneracy of the level of the hydrogen atom that has the energy\n(a) RH,\n(b) RH/9\nSolution.\nEn RH / n 2\n\n(a)\n\nE RH ,\n\nn 1\n\nThe level is degenerate.\n\n(b)\n\nE RH / 9\n\nn3\n\nwhen n 3 ,\n\nwhen\n\nl 0,\n\nme 0 (3s-orbital)\n\nwhen\n\nl 1,\n\nme 1, 0,1 (3p-orbital)\n\nwhen\n\nl 2,\n\nme 2, 1, 0,1, 2 (3d-orbitals)\n\nwhen n 1 ,\n\nme 0\n\nl 0,\n\nl 0,1, 2\n\nThis is 1 + 3 + 5 = 9 states in all. The degeneracy is 9.\n\nExample 10.\nCalculate the uncertainty in the velocity of a ball of mass 150g if uncertainty in this position is\n1 ( h 6.6 10 34 Js )\n\nSolution.\nFrom Heisenberg uncertainty principle we know that\nxp\n\nh\n4\n\nor\n\nxm v\n\nh\n4\n\nor\n\nh\n4 mx\n\nSubstituting value: m 150 g 0.15 kg , x 1 1010 m\n\n6.6 1034\n4 3.14 1010 0.15\n\nv 3.499 10 24 ms 1 .\n\nSubjective Solved Examples\n\nExampe1. 11\nCalculate the waelength and wave number of the spectral line when an electron in H-atom falls from higher\nenergy state n 3 to a state n 2 . Also determine the energy of a photon to ionise this atom by removing the\nelectron from 2nd Bohrs orbit. Compare it with the energy of photon required to ionise the atom by removing\nthe electron from the ground state.\nSolution.\n\n22\n\nAtomic Structure\nPhoton absorbed\n\nn=1\n\n+e\n\nn=2\n\nn=3\n\n+1e\n\nn=\nn=2\n\nPhoton emitted\n\nresponsible transition will be n 2 n .\n\n1 1\nE 2.18 1018 Z 2 2 2 J\nn1 n2\n\n1\n1\nE(2 ) 2.18 1018 12 2 2 J\n2\n\n1 1\nE(32) 2.18 1018 (1) 2 2 2 J (Z 1)\n2 3\n\n5.45 10 19 J\n\n3.03 10 19 J\n\nNow this energy difference is the energy of the\n\n(n 1) , the transition is 1 .\n\nphoton emitted.\n\n1 1\nE 2.18 1018 12 2 2\n1\nhc\nhcv\n\nEPhoton hv\n\nhc\n3.03 10 19 6560.3\n\nand\n\n2.18 10 18 J\n\n1\n1.52 10 6 m 1\n\nExample 12\nA hydrogen atom in the ground state is hit by a photon by a photon exciting the electron to 3rd excited state.\nThe electron then drops to 2nd Bohr orbit. What is the frequency of radiation emitted and absorbed in the\nprocess?\nSolution.\nEnergy is absorbed when electron moves from ground state (n 1) to 3rd excited state (n 4) .\nPhoton absorbed\nn=1\nn=4\n+e\n\nn=1\n\nn=2\n\nFirst calcuate the energy difference btween n 1 and n 4 .\n\n1\n1\nUse: E(1 4) 2.18 10 18 Z 2 2 2 .\nn\nn\n1\n2\n\n23\n\nAtomic Structure\nHere,\n\nZ 1, n1 1, n2 4\n\n1 1\n18\n2\nE(14) 2.18 10 1 12 42\n\nPut n1 2 and n2 4 in the expression of E , to get:\n\n1 1\nE(4 2) 2.18 1018 12 2 2\n2 4\n\n2.04 10 18 J\n\n4.08 10 19 J\n\nUse:\n\nv 3.08 1015 Hz\n\nv 6.16 1014 Hz\n\nSimilarly, when electron jumps from\n\nn 4 to n 2 , energy is emitted and is given by the same relation.\nExample 13\n\nA hydrogen like ion, He (Z 2) is exposed to electromagnetic waves of 256.4 . The excited electron gives\nout induced radiations. Find the waelength of the indicated radiations, when electron de-excites back to the\nground state. R = 109677 cm1.\nSolution:\nHe+ ion contains only one electron, so Bohrs\nmedhot is applicable here. It absorbs a photon of\n\nFrom n 3 , the electron can fall back to the\n\nground state in three possible ways (transitions)\n\n3 1, 3 2, 2 1\n\nwavelength 256.4 . Assume the electron to\n\nbe in ground state initially. Let it jumps to an\nexcited state n2.\nv\n\nHence three possible radiations are emitted.\n\nFind the wavelengths corresponding to these\n\n1\n1\n1\nRZ2 2 2\n\nn1 n2\n\ntransitions.\n\nThe wavelength ( ) for transition, 3 1 will\n\nbe same i.e., 256.4 . Find for 3 2 and\n2 1 using the same relation.\n\nR 109677, Z 2 for He+ ion,\n\n(3 1) 256.4 , (3 2) 1641.3\n\nn1 1 and find n2 .\n\n(2 1) 303.9\nn=3\nn=2\n\n1 1\n1\n109677 107 (2)2 2 2 n 3\n2\n256.4 108\n11 n2\nn=1\n\nExample 14\nHydrogen gas when subjected to photon-dissociation, yields one normal atom and one atom possessing 1.97\neV more energy than normal atom. The bond dissociation energy of hydrogen moleucle into normal atoms is\n103 kcals mol1. Compute the wave length of effective photon for photon dissociation of hydrogen molecule in\nthe given case.\nSolution:\nH2 H H\n\nwhere H is normal H -atom and H * is excited H-atom. So the energy requird to dissociate H 2 in this\nmanner will be greater than the usual bond energy of H 2 molecule.\nE (absorbed) = dissociation energy of H 2 extra energy of excited atom.\nEnergy required to dissociated in normal manner\n\n24\n\nAtomic Structure\n3\n\n(given)\n\n103 103 4.18\n\n7.17 1019 J / atom\n6 1023\n\nThe extra energy possessed by excited atom is 1.97 eV\n\n1.97 1.6 10 19 J 3.15 10 19 J\n\n1.03 10 18 J\n\nNow calculate the wavelength of photon corresponding to this energy.\n\nEPhoton\n\nhc\n1.03 1018 1930\n\nExample 15\nAn electron in the first excited state of H-atom absorbs a photon and is further excited. The de Broglie wavelength\nof the electron in this state is found to 13.4 . find the wavelength of photon absorbed by the electron in .\nAlso find the longest and shortest wavelength emitted when this electron de-excites back to ground state.\nSolution:\nNote: The energy state n 1 is known as Ground State\nThe energy state n 2 is known as First Excited State\nThe energy state n 3 is known as Second excited\nState and so on.\n\nn=n\n+e\nn=2\n\nPhoton\n\nThe electron from n 2 absorbs a photon and is further excited to a higher energy level (let us say n ).\nThe electron in this energy level ( n ) has a de Broglie wavelength ( ) 13.4\n\nh\nme ve\n\nand\n\nvn 2.18 106\n\nZ 1\nms\nn\n\n[vn is the velocity of e in nth Bohr orbit]\n\n6.626 10 34\nh\n1\n2.18 106\n(13.4 1010 ) (9.1 10 31 )\nm\nn\n\n2.18 10 6\n\n1\nn4\nn\n\nNow find the wavelength of the photon responsible for the excitation from n 2 to n 4 .\nUsing the relation :\n1 1\nE 2.18 1018 Z 2 2 2 4.09 10 19 J [n =2, n =4, Z=2]\n1\n2\nn1 n2\n\n(2 4)\n\n(2 4)\n\nhc\n4.09 10 19 4863.1\n\n25\n\nAtomic Structure\nE EPhoton\n\nhc\n1.06 10 19 J\n\n18752.8\n\nThe transition corresponding to minimum energy will\n\nShortest wavelength : 4 1\n\n1 1\nE(4 1) 2.18 1018 12 2 2\n1 4\n\nbe 4 3 .\n\n2.04 10 18 J\n\nNote: The transition corresponding to maximum energy\n\nE(4 1) EPhoton\n\nwill be 4 1 .\nE( Energy diff .) EPhoton\n\nPhoton\n\nhc\nhv\n\nhc\n\n973.2\n\nor E vPhoton\n\n1\n18 2 1\nUsing the same relation : E(43) 2.18 10 Z n2 n 2\n1\n2\n\n[n1 3, n2 4, Z 2]\n\nExample 16\nA single electorn orbits around a stationary nucleus of charge Ze , where Z is a constant and e is the\nmagnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third\nBohr. Find :\n(a)\nthe value of Z\n(b)\n(c)\n(d)\n(e)\nSolution.\n\nthe energy required to excite the electron from n 3 to n 4 .\n\nthe wavelength of radiation required to remove electron from 2nd Bohrs orbit to infinity\nthe kinetic energy, potential energy and angular momentum of the electron in the first orbit.\nthe ionisation energy of above one electron system in eV.\n\nSince the nucleus has a charge Ze , the\n\natomic number of the ion is Z .\n(a) The transition is n1 2 n2 3 by absorbing a photon of energy 47.2 eV .\n\nE 47.2eV\nUsing the relation:\n1 1\nE 13.6Z 2 2 2 eV\nn1 n2\n\n1 1\n47.2 13.6Z 2 2 2 Z 5\n2 3\n\n(b) The required transition is n1 3 n2 4 by absorbing a photon of energy E .\n\n1\n2 1\nFind E by using the relation: E 13.6Z 2 2 eV\nn1 n2\n\n1 1\nE 13.6(5)2 2 2\n3 4\n\neV\n\nE 16.53 eV\n\n(c) The required transition is n1 2 n2 by absorbing a photon of energy E .\n\n26\n\nAtomic Structure\nFind E by using the relation:\n\n1\n1\nE 13.6(5) 2 2 2 E 85eV\n2\nFind of radiation corresponding to energy 85 eV..\n\nhc 6.626 10 34 3 108\n\nE\n85 (1.6 10 19 )\n\n146.16\n\n(d) If energy of electron be En, then KE = En and PE = 2En\n\nEn\n\n13.6 Z 2 13.6 52\n\n340 eV\nn2\n12\n\nKE (340eV ) 340 eV\nPE 2(340 eV ) 680eV\n\nh\nAngular momentum (l ) n\n\n6.626 1034\nl 1\n1.05 10 34 J s\n2\n\n(e) The ionisation energy (IE) is the energy required to remove the electron from ground state to infinity. So\nthe required transition is 1 . The ionisation energy\n( IE ) E1 13.6( Z ) 2 eV\n\nIE 13.6 52 340eV\n\nExample 17\nWith what velocity should an alpha ( ) particle travel towards the nucleus of a copper atom so as to arrive at\na distance 1013 m from the nucleus of the copper atom?\nSolution.\nAs -particle appraoches towards the Cu nucleus, it decelerates due to repulsion from it and finally its\nvelocity will become zero at point A (which is the turning point). After that, particle will move in the left\ndirection (and accelerating)\nr0\n\n-particle\nV\n\nA\nV=0\n\nCu Nucleus\n(+29e)\n\nTo arrive at a distance (r 0) from the nucleus, the kineticc energy of alpha particle should be equal to the\nelectrostaic potential energy of it, i.e., KE = EPE\nKq q\n1\nm v2 N\n2\nr0\n\nm :\n\nmass of -particle = 4 (1.67 1027 kg )\n\nv :\n\nvelocity of -particle = ?\nK 9 109 N m / C 2\n\n27\n\nAtomic Structure\nq charge on -particle = 2 ( 1.6 10\n\n19\n\nC)\n\nNote:\n\nqN = charge in Cu nucleus = Ze = 29(+1.6 1019C)\n\nd = distance from nucleus = 1013 m\n\n2Kq qN\nm r0\n\nSubstituting the given values, we get,\n\nv 6.325 106 m / s.\n\nNote:\n\nThis is a simple cases where velocity of -particle is directed towards the centre of the Cu-nucleus.\n\n-particle\n\nCu\n\nr0\nV=0\n\nNote: When there is a difference between the velocity vector of -particle and the Cu(target) nucleus, the\ntrajectory is more complicated.\n\n-particle\n\nTarget\n\nExample 18\nFind the energy required to excite 1.22 litre of hydrogen atoms gas at 1.0 atm and 298 K to the first excited\nstate of atomic hydrogen. The energy requied for the dissociatio of H-H bonds is 436 kJ/mol. Also calculate\nthe minimum frequency of a photon to break this bond.\nSolution.\nLet us, first find the number of moles of hydrogne atoms.\nnH 2\n\nPV\n1 1.22\n\n0.05\nRT 0.0821 298\n\nThus the energy required to break 0.05 moles of H2 (H-H bond) = 0.05 436 19.62 kJ .\nNow calculate the energy needed to excite the H-atoms to first excited state i.e., to n 2 (First excited state is\nreferred to n 2 ).\n\n1 1\nE 2.18 1018 (1)2 2 2\n1 2\n\nJ / atom\n\nNo. o H atoms = (No. of H2 molecules) 2\n\n(0.05 6.02 10 23 ) 2 6.02 10 22\n\n28\n\nAtomic Structure\n22\n\nThe energy required to excited the given number of H-atom = 6.02 10 1.635 10 18 J 98.43 kJ\nSo the total energy required\n19.62 98.43 118.05 kJ\nNow the energy required to break to single\nH-H bond =\n\n436 103\n7.2381019\n6.023 1023\n\n7.328 10 19 hv 6.626 10 34 (v)\n\nv 1.09 1015 Hz\n\nExample 19\nEstimate the differnce in energy between 1st and 2nd Bohrs orbit for a H-atom. At what minimum atomic\nnumber (Z), a transition from n 2 to n 1 energy level would result in the emission of radiatio with wavelength\n\n3.0 10 8 m ? Which hydrogen atom like species this atomic number corresponds to? How much ionisation\npotential is needed to ionise this species? ( R 1.097 107 m 1 )\nSolution.\nThe difference in energy is given by E :\n\n1 1\nE 2.18 1018 (1)2 2 2 J / atom\n1 2\n1.65 10 18 J 1.65 10 11 ergs 10.2eV\n\nFor a H-like atom, 3.0 108 m .\n\n1 1\nE 2.18 1018 Z 2 2 2\n1 2\n\n(2 1)\n\nEPhoton\n\nhc\n\nSolve to get :\nZ=2\nHence the H-like atom is He+ ion.\nTo ionise, He+ ion, ionisation energy (IE) = (E1)\nIE (13.6 2 2 ) 54.4eV\n\nThe ionisation potential (IP) is the voltage difference required to generate this much energy.\n\nIE qV e( IP ) 54.4 eV\n\nIP (required) 54.4 Volt\n\nExample 20\nA stationary He+ ion emits a photon correspondings to the first line ( H ) of Lyman series. The photon thus\nemitted, strikes a H-atom in the ground state. Find the velocity of the photoelectrons ejected out of the hydrogen\natom. The value of R 1.097 10 7 m 1 .\nSolution.\nThe difference in energy (E ) will be equal to the energy of the photon emitted.\nFirst line in Lyman series corresponds to the transition 2 1 .\n\n1 1\nE 2.18 1018 (2) 2 2 2\n1 2\n\nJ / atom\n\n6.54 10 18 J\n\n29\n\nAtomic Structure\nThe photon of this much energy strikes a H-atom in the ground state. Note that the ionisation energy of H-atom\nis 2.18 10 18 J . This will be the work function of H-atom. Using the Einsteins photoelectric equation:\nKE Ei W0\n\n1\nme ve2\n2\n\nve\n\n2( Ei W0 )\nme\n\nve\n\n2(6.54 10 18 2.18 1018 )\n\n9.1 1031\n\nve 3.09 10 6 m / s\n\nWe can also calculate the wavelength of electron ejected out = 2.36 10 10 m 2.36\n\nh\n6.626 10 34\n\nm 2.36\nme ve 9.7 10 38 3.09 106\n\nExample 21\nAn electron in a hydrogen like species, makes a transition from nth Bohr orbit to next outer Bohr ( n 1 ).\nFind an approximate relation between the dependence of the frequency of the photon absorbed as a function of\nn . Assume n to be large value ( n 1 ).\nSolution.\n1\n1\nE( energy difference ) hv 2.18 10 18 Z 2 2\nJ\n(n 1) 2\n( nn 1)\nn\n\n2n 1\nhv 2.18 10 18 Z 2 2\nJ.\n2\nn ( n 1)\n\nSince n 1 (given)\n\nn 1 ~ n ; 2 n 1 2n\n\nhv 2.18 10 18 Z 2\n\nv n 3 .\n\n2n\nJ\nn4\n\n30\n\nAtomic Structure\n\nM IND M AP\n1. According to the quantum\nemitted by atoms & molecules\nin small discrete amounts\n(quanta)m rather than over a\ncontinuous rante. The energy of\neach quanta is given by E = hv.\n\n2. According to Bohr model, the 3. The radius of an orbit is given\n\nangular mometum of an electron by\nThe\nr n 2 h 2 / 4 2 kZms 2 .\nis an integral multiple of h / 2 . velocity of an electron in an orbit\nBohrs model is applicable single is given by v nh / 2 mr and the\nelectron species (hydrogen like energy of an electron in an orbit\nspecies).\nis given by E 2 pk 2 Z 2 ms 4 / n2h 2 .\n\n9. In phot oelectric effect,\n\nelectrons are elected from the\nsurface of certain met al\nexposed to light of at least a\ncertain minimum frequency.\n\n4. In bohr model, an electron\n\nemits a photon when it drops\nfrom a higher energy state to a\nlower energy state.\n\nhv hv0 K .E .\n\n8. Four quant um numbers\n\ncharacterise each electron in an\natom. The principal quantum\nnumber(n) indentifies the main\nenergy level, t he angular\nquantum number (l) indicates\nshapes of orbital, the magnetic\norientation of orbital in space\nand the spin quant um\nnumber(s) indicat es t he\ndirection of the electrons spin\non its axis.\n\nATOMIC\nSTRUCTURE\n\n5. The emission spectra of\n\nhydrogen is obtained when\nelectron from an ecited state is\ndeexcited to the ground state.\nThe release of specific amounts\nof energy in the form of photons\naccounts for the lines in the\nhydrogen spectrum. v of each\nline in the spectrum can be\ngiven by\n1/ Ryz 2(1/ n12 ) (1/ n22 )]\n\n7. An orbital may be defined as\n\na region in space around the\nnucleus where the probability\nof finding the electron is\nmaximum.\n\n6. De Broglie exended\nEinsteins wave particle\ndescrition of light to all matters\nin motion. The wavelength of\na moving particle of mass m\nand velocity v is given by de\nBroglie equation, h / mv.\n\n31" ]

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[ "# What Is The Definition Of Mathematics In The Modern World?\n\n## What is a simple definition of mathematics?\n\n: the science of numbers, quantities, and shapes and the relations between them. mathematics.\n\n## How is mathematics used in today’s world?\n\nIt gives us a way to understand patterns, to quantify relationships, and to predict the future. Math is a powerful tool for global understanding and communication. Using it, students can make sense of the world and solve complex and real problems.\n\n## What is the best definition of mathematics?\n\nMathematics (from Greek: μάθημα, máthēma, ‘knowledge, study, learning’) includes the study of such topics as quantity (number theory), structure (algebra), space (geometry), and change (analysis). It has no generally accepted definition.\n\n## Who is the father of maths?\n\nArchimedes is known as the Father Of Mathematics.\n\n## Why do we need to study mathematics in modern world?\n\nA degree in math is a professional pathway for those interested in solutions and solving real- world problems. It’s an important discipline to study because many roles require a basic or advanced understanding of mathematical concepts. Believers in better, faster, smarter solutions are often drawn to math. We need math.\n\nYou might be interested:  FAQ: How To Draw Geometrical Patterns In Mathematics?\n\n## What is the importance of mathematics?\n\nMathematics provides an effective way of building mental discipline and encourages logical reasoning and mental rigor. In addition, mathematical knowledge plays a crucial role in understanding the contents of other school subjects such as science, social studies, and even music and art.\n\n## Why is math so hard?\n\nMath is a very abstract subject. For students, learning usually happens best when they can relate it to real life. As math becomes more advanced and challenging, that can be difficult to do. As a result, many students find themselves needing to work harder and practice longer to understand more abstract math concepts.\n\n## Does Math stand for?\n\nMathematics. MATH. Mental Abuse to Humans. MATH. Master of Arts in Theology (degree)\n\n## What is the full form of mathematics?\n\nMATH: Mathematics The full form of MATH is “ Mathematics “. Mathematics is the science that deals with the logic of form, quantity, and disposition. Mathematics includes the study of topics such as quantity (number theory), structure (algebra), space (geometry) and change ( mathematical analysis).\n\n## Do we need mathematics everyday?\n\nMath is vital in our world today. Everyone uses mathematics in our day to day lives, and most of the time, we do not even realize it. Without math, our world would be missing a key component in its makeup. “ Math is so important because it is such a huge part of our daily lives.\n\n## Who is the most famous mathematician?\n\nThe 10 best mathematicians\n\n• Girolamo Cardano (1501-1576), mathematician, astrologer and physician.\n• Leonhard Euler (1707-1783).\n• Carl Friedrich Gauss (1777-1855).\n• Georg Ferdinand Cantor (1845-1918), German mathematician.\n• Paul Erdos (1913-96).\n• John Horton Conway.\n• Russian mathematician Grigory Perelman.\n• Terry Tao. Photograph: Reed Hutchinson/UCLA.", null, "" ]

[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20110%20110'%3E%3C/svg%3E", null ]

[ "## Thomas' Calculus 13th Edition\n\nCenter $(1,\\dfrac{-1}{2},-3)$ and radius is $5$\nAs we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ ) where $(a,b,c)$ represents center and radius of the sphere is $r$ Now, $(x -1)^2+(y+\\dfrac{1}{2})^2+(z +3)^2=25$ $\\implies (x -1)^2+(y-(-\\dfrac{1}{2}))^2+(z -(-3))^2=5^2$ Compare this equation with the standard equation of sphere. Thus, Center $(1,\\dfrac{-1}{2},-3)$ and radius is $5$" ]

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[ "# Ep.4 组合逻辑电路\n\n## 第二节 组合逻辑电路的设计\n\n### 一、 设计\n\n• 对于不可能出现的情况，应将其作为无关项。\n• 对于较多情况，可以用多位二进制的组合来表示。\ne.g. 血型A、B、AB、O，可以用两位来表示。而不是用4bit分别表示。\n\n## 第三节 组合逻辑电路的竞争-冒险\n\n### 一、 产生的竞争冒险的原因\n\n• 竞争：当一个逻辑门的两个输入端的信号同时向相反方向变化，而变化的时间有差异的现象。\n• 冒险：由竞争而可能产生输出干扰脉冲的现象。\n\n⭐总结：\n\n• 表达式最终是一个$L=M+\\overline{M}$\n（只是要求两变量高低电平反向即可，不一定是同一变量）\n• 对于与门：其中一个先从$0$到$1$", null, "• 对于或门：其中一个先从$1$到$0$", null, "### 二、 消除竞争冒险的方法\n\n1. 消除互补变量相乘$A \\overline{A}$ - 直接可以消去\n比如$L=(A+B)(\\overline{A}+C)$，展开后直接消除$A\\overline{A}$。\n2. 消除互补变量相加$A + \\overline{A}$ - 增加乘积项\n对于$L=X_AM+X_B \\overline{M} \\rightarrow L=X_AM+X_B \\overline{M} + X_AX_B$\n此时当$X_A=X_B=1$时，$L+M+\\overline{M}+1$，使得恒为$1$，消去互补项相加。\n3. *输出端并连电容器\n为物理技术，不在本课讨论范围。\n\n## 第四节 若干典型的组合逻辑电路\n\n### 一、 编码器(Encoder)\n\n• 输入：特定含义的信号\n• 输出：其二进制代码\n\n• 普通编码器：任何时候只允许一个有效输入信号，否则输出发生混乱。\n• 优先编码器：允许同时输入两个以上的有效输入信号。\n当同时输入几个有效信号时，只对优先级最高的进行编码\n并不是指多个有效输入信号代表一个特定含义，还是一位有效信号代表特定含义。\n\n#### 1. 普通编码器 - 4线-2线普通二进制编码器\n\n//Behavior\nmodule encoder4to2(Y, I);\ninput [3:0] I;\noutput reg [1:0] Y;\n\nalways @(I)\nbegin\ncase (I)\n4'b0001: Y=2'b00;\n4'b0010: Y=2'b01;\n4'b0100: Y=2'b10;\n4'b1000: Y=2'b11;\ndefault: Y=2'b00;\nendcase\nend\n\n\n#### 2. 优先编码器 - 4线-2线优先二进制编码器\n\n• 当所有输入信号为0时，输出为0\n• 当$I_1=1$时，输出也为0\n//Behavior\nmodule encoder4to2(Y, I);\ninput [3:0] I;\noutput reg [1:0] Y;\n\nalways @(I)\nbegin\ncasex (I)\n4'b0001: Y=2'b00;\n4'b001x: Y=2'b01;\n4'b01xx: Y=2'b10;\n4'b1xxx: Y=2'b11;\ndefault: Y=2'b00;\nendcase\nend\n\n\n• 74X147 - 10线-4线优先编码器\n输入：低电平有效\n输出：8421BCD码\n• 74X148 - 8线-3线优先编码器\n输入：低电平有效\n输出：二进制代码\n• CD4532 - 8线-3线优先编码器\n输入：高电平有效\n输出：二进制代码\n\n#### 3. 8线-3线优先二进制编码器", null, "• EI - 使能控制信号，当为$0$时不工作\n• EO - 异常输出信号，当为$1$时说明输入全为0\n• GS - 正常输出信号，当为$1$时说明正常工作（输出不全为0）", null, "### 二、 译码器/数据分配器(Decoder)\n\n#### 1. 定义与分类\n\n• 唯一地址译码器 - 将一系列代码转化为与之一一对应的特定有效信号\n• 二进制译码器\n• 二-十进制译码器\n• 显示译码器\n• 代码转换器 - 将一种代码转换为另一种代码\n\n#### 2. 二进制译码器\n\n• 输入：二进制代码\n• 输出：其特定含义的信号\n\n##### ① 2线-4线译码器(74HC139)", null, "⭐注：端口变量符号上有横线，或者端口接线处有圆圈，代表为低电平有效\n\n//Dataflow\nmodule decoder_df(A1,A0,E,Y);\ninput A1,A0,E;\noutput [3:0]Y;\n\nassign Y = ~(~A1 & ~A0 & ~E);\nassign Y = ~(~A1 & A0 & ~E);\nassign Y = ~(A1 & ~A0 & ~E);\nassign Y = ~(A1 & A0 & ~E);\nendmodule\n\n##### ② 3线-8线译码器(74HC138)\n\n• $E_1$ - 低电平有效\n• $E_2$ - 低电平有效\n• $E_3$ - 高电平有效\n\n• $E_3$ - 用于接收输入信号其额外的第$4$位。\n其需要连接到另一片译码器$B$的$E_2$端口，用来代表选择哪个译码器工作。\n\n• $E_2$ - 用于与第一片译码器$A$的$E_3$端口构成选择器。\n当第$4$位为$0$选择$B$工作，为$1$选择$A$工作。", null, "//Behavior\nmodule ecoder3to8_bh(A,En,Y);\ninput [2:0]A, En;\noutput reg [7:0]Y;\ninteger k;\n\nalways @(*)\nbegin\nY = 8'b1111_1111;\nfor (k = 0; k <= 7; k++)\nif ((En == 1) && (A == k))\nY[k]=0;\nelse\nY[k]=1;\nend\nendmodule\n\n\n1. 先将表达式转换为最小项之和的形式。\n2. 再还原律+摩尔根，变为最小项的非的与形式。\n3. 与非门连接在一起。", null, "#### 3. 二-十进制译码器", null, "#### 4. 显示译码器\n\n• 前面两种：一种输入信号，只会有一个输出端口输出有效电平。\n• 显示译码器：一种输入信后，可能会有多个输出端口输出有效电平。", null, "• 共阴极电路：所有阴极接地，输出信号为高电平有效\n• 共阳极电路：所有阳极接地，输出信号为低电平有效", null, "", null, "• 灯测试$\\overline{LT}$优先级最高 - 正常为1。但只要为0，灯全亮。\n• 灭灯$\\overline{BL}$优先级第二 - 正常为1。但只要为$BL=0(LT=0)$，灯全灭。\n• 锁存$LE$ - 正常为0。但只要$LE=1(LT=1, BL=1)$，则输出信号锁定。\n\n//Behavior\nmodule seg7_decoder(\ninput LE, BL, LT, D3, D2, D1, D0,\noutput reg a, b, c, d, e, f, g\n);\n\nalways @(*)\nbegin\nif (LT == 0) //灯测试\n{a, b, c, d, e, f, g} = 7'b1111111;\nelse if (BL == 0) //灭灯\n{a, b, c, d, e, f, g} = 7'b0000000;\nelse if (LE == 1) //锁存\n{a, b, c, d, e, f, g} = {a, b, c, d, e, f, g};\nelse //开始判断\ncase ({D3, D2, D1, D0})\n4'd0: {a, b, c, d, e, f, g} = 7'b1111110;\n4'd1: {a, b, c, d, e, f, g} = 7'b0110000;\n...//以下类推\ndefault: {a, b, c, d, e, f, g} = 7'b0000000;\nendcase\nend\nendmodule\n\n\n#### 5. 数据分配器\n\n• $E_1, E_2$ - 输出跟输入一样。（均为低电平有效）\n• $E_3$ - 输出是输入取非。", null, "", null, "### 三、数据选择器\n\n#### 1. 最基础的数据选择器 - 2选1数据选择器\n\n##### ① 基本介绍", null, "", null, "//Dataflow\nmodule mux2x1_df(A, B, SEL, L);\ninput A, B, SEL;\noutput L;\n\nassign L = SEL ? A : B;\nendmodule\n\n//Behavior\nmodule mux2x1_bh(D1, D0, S, L);\ninput D1, D0, S;\noutput reg Y;\n\nalways @(*)\n//一句话，不用加begin\nL = SEL ? D1: D0;\nendmodule\n\n##### ② 构成n选1数据选择器", null, "", null, "##### ③ 运算功能", null, "• $A=0 \\rightarrow L=BC$\n• $A=1 \\rightarrow L=B+\\overline{C}$", null, "• 变量数 = 控制输入端口数\n则直接把函数转化为最小项表达式，然后根据保留情况，直接确定输出端口$D$为确定的值。\n这个时候$D$是常量。\n• 变量数 < 控制输入端口数 把没被用到的选择端口置任意值，然后对有可能输出的端口进行控制（置1）即可。（其余则置0）\n• 变量数 > 控制输入端口数\n• 如果支持其他逻辑元件（如与门等）：\n将部分变量当输入信号，根据不同的控制信号，得出每个情况的表达式，然后再设计该种情况的电路。\n这个时候$D$是变量。\n• 如果不支持其他逻辑元件（只能用选择器）：\n则需要用该选择器，组合成$n$选$1$选择器，使得$n>=\\textrm{变量数}$。\n##### ④ *构成查找器LUT", null, "#### 2. 典型集成电路 - 8选1数据选择器（74HC151）", null, "• $E$ - 低电平有效。为$1$时不工作（$Y$输出0、$\\overline{Y}$输出1）。\n• $Y$ - 输出由$S$所选择的信号\n• $\\overline{Y}$ - 输出所选择信号的非\n\n1. 多位的8选1（$N$bit，均按照相同的$S$选择）：\n简单的几个8选1直接并联。\n2. 16选1：\n将两个8选1并联，\n并将使能端$\\overline{E}$利用起来，作为第二个8选1是否启用的控制信号（即选择$8\\sim15$）。\n\n### 四、数值比较器\n\n#### 1. 1位数值比较器\n\n• 2个输入：$A$、$B$\n• 3个输出：$F_{A>B}$、$F_{A=B}$、$F_{A<B}$\n\n#### 2. 2位数值比较器", null, "#### 3. 集成数值比较器 - 4位数值比较器(74HC85)", null, "1. 串联扩展方式", null, "但注意：可能存在延时情况。\n最坏情况：每次都比较到最后一位才得出结果。\n\n2. 并联拓展方式\n多用一片4位比较器，用来综合考虑其他4个4位的比较结果。\n相当于把4位整合成1位比较。\n\n注意被整合的4位的比较器，其低位输出要为相等$I_A=I_B$。", null, "### 五、算术运算电路\n\n#### 1. 半加器和全加器\n\n• 半加：不考虑低位的进位\n• 全加：考虑低位的进位", null, "• $S$ - 本位的运算结果\n逻辑表达式：$S=A \\oplus B$\n• $C$ - 是否产生进位\n逻辑表达式：$C=AB$", null, "• $C_i$ - 低位的进位信息\n\n• $S$ - 本位的运算结果\n逻辑表达式：$S=A\\oplus B\\oplus C_i$\n\n• $C_o$ - 是否产生进位\n逻辑表达式：$C_o=AB+(A\\oplus B)C_i$", null, "$\\because S=A\\oplus B\\oplus C_i$，利用的实际上为异或$\\oplus$的性质。\n\n## 第五节 组合可编程逻辑器件\n\n• 与门阵列\n• 或门阵列\n\n• 根据电路图写出表达式\n• 根据表达式画出电路图\n\n### 一、PLD的结构、表示方法及分类\n\nP - Program 可编程\nL - Logic 逻辑\nD - Device 器件\n\n#### 1. PLD的基本结构", null, "#### 2. PLD的逻辑符号表示方式\n\n1. 不可编程单元 - 硬线连接单元\n代表恒为接通状态。\n2. 可编程单元\n1. 被编程接通单元\n代表由用户自定义的接通状态。\n2. 被编程擦除单元\n代表由用户自定义的断开状态。", null, "", null, "#### 3. 编程连接技术\n\n1. 早期\n采用金属熔丝技术，\n通过大电流将金属丝熔断，完成编程。\n只能进行一次编程。\n2. CMOS中\n采用可擦除的编程方式，\n用浮栅MOS代替熔丝。\n可以多次编程。\n\n#### 4. 分类\n\n1. 按集成密度划分：\n• 低密度可编程逻辑器件\n• 高密度可编程逻辑器件\n2. 按结构特点划分：\n• 简单PLD\n• 复杂的可编程器件(CPLD)\n• 现场可编程门阵列(FPGA)\n3. 按与、或阵列是否编程划分：\n1. 与阵列固定，或阵列可编程(PROM)\n2. 与阵列、或阵列均可编程(PLA)\n3. 与阵列可编程，或称列固定(PAL、GAL)\n\n## 第六节 用Verilog描述组合逻辑电路\n\n//只用于本课学习的方面\nmodule 模块名(端口_1, 端口_2, 端口_3, ...); //注意这里有分号\ninput 输入端口_1, ...;\noutput 输出端口_1, ...;\n数据类型定义((wire,) reg); //也可以与input、output合并为：input(output) wire(reg) ...;\n其他变量申明(wire, reg, integer, ...); //比如用作循环变量的i，需要声明为integer i;\n/* 再补充关于wire和reg：\n门级电路描述时：所有用来代表连接的结点，都需要定义为wire\n数据流描述时：只要为赋值语句左边的变量，都需要定义为wire\n注意：input和output默认定义的是wire，即input wire in_1，所以在上述两种中其实可以相当于不加。但最严谨的是：input、output所有端口以及临时节点都加上wire\n\n行为描述时：只要赋值语句左边的变量，必须要定义为reg（输出需要额外定义）\n*/\n\n//门级描述\n基本门级元件(not, and, or); //e,g, not(out_1, in_1);//注意输出在前，输入在后！\n//数据流描述\n连续赋值语句(assign); //e.g. assign out_1 = (in_1 & A) | （in_2 & ~B);//注意电路中法国逻辑运算一般为按位，即要用&,|,~,^,~^。\n//行为描述\n过程块结构(initial, always) //e,g. always @(...)\nbegin\n行为描述语句; //一句话不用加begin, end\nend //就是基本的C语言结构了\n\n\n• 用基数形式表示是，位宽指的是二进制下的位数。\n5'H0CFF0000_1100_1111_11111_1111，与5'H1F一样。\n\n### 1. 组合逻辑电路的行为级建模\n\n1. if - 跟C一样。\n\n2. case\n\ncase (case_expr)\nitem_expr1: //一种情况多个语句\nbegin\nstatement1_1;\nstatement1_2;\nend\nitem_expr2: statement2; //一种情况一个语句\n...\ndefault: default_statement; //可省略\nendcase\n\ncasex (...) //包含未知项（优先编码器）\n4'b1xxx: ...;\nendcase\n\ncasez (...) //包含高阻态\nendcase\n\n3. for\n\ninteger k;\n...\nfor (k = 0; k <= 7; k++)\nbegin //一条语句不用\n...\nend\n\n\n### 2. 分模块、分层次的电路设计", null, "1. 实例引用名不能省略。\n2. 实例引用名在父模块中必须且唯一。\n\n1. 位置关联法：\n变量与子模块申明的参数顺序位置一样。\n\n//子模块设计\n...\nendmodule\n\n//父模块开始引用\nmodule main(...);\nhalfadder HA1(Sum, out_C, in_A, in_B); //必须要求顺序严格对应\nendmodule\n\n2. 名称关联法：\n.A是子模块中的参数。\n(B)是主模块中的参数。\n\nmodule main(...);\nhalfadder HA2(.A(in_A), .S(Sum), .B(in_B), .C(out_C)); //顺序可以打乱\n//.A, .S为子模块中参数名称\n//(S1), (Sum)为父模块中变量名称\nendmodule\n\n\nmodule halfadder(S, C, A, B); //用门级电路描述的1位半加器\n//采用一个与门一个异或门构成。\ninput A, B;\noutput S, C;\n\nxor (S, A, B);\nand (C, A, B);\nendmodule\n\nmodule fulladder(S, C_o, A, B, C_i); //用门级电路描述的1位全加器\n//采用两个1位半加器和一个或门构成。\ninput A, B, C_i;\noutput S, C_o;\nwire S1,C_1,C_2; //内部节点信号\n\nhalfadder HA2(S, C_2, S1, C_i); //本位结果输出\nor gate_1(C_o, C_1, C_2); //进位输出\nendmodule\n\nmodule 4bit_adder (S, C_o, A, B, C_i); //用门级电路描述的4位加法器\n//采用四个1位全加器构成。\ninput[3:0] A,B; //4bit输入\ninput C_i;\noutput[3:0] S; //4bit输出\noutput C_o;\nwire C1,C2,C3; //内部进位信号\n\nfulladder FA_0 (S, C_1, A, B, C_i),\nFA_1 (S, C_2, A, B, C_1),\nFA_2 (S, C_3, A, B, C_2),\nFA_3 (S, C_o, A, B, C_3);\nendmodule" ]

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[ "# 骰子与每个数字，javascript的图片\n\n``````function rolldice(){\nvar rollResultElem = document.getElementById(\"roll-result\")\nvar rollResult = Math.floor(Math.random() * 6) + 1\n}\n``````\n\n### 回答(2)\n\nHTML图片：\n\n``````<img id=\"image1\">\n``````\n\n``````document.getElementById(\"image1\").src = \"path/dice/number\"+rollResult+\".jpg\"\n``````\n\n``````var rollResult = Math.floor(Math.random() * 6) + 1;\nswitch(rollResult){\ncase 1: {image = './one.jpeg'; break;}\ncase 2: {image = './two.jpeg'; break;}\ncase 3: {image = './three.jpeg'; break;}\ncase 4: {image = './four.jpeg'; break;}\ncase 5: {image = './five.jpeg'; break;}\ncase 6: {image = './six.jpeg'; break;}\ndefault: break;\n}\ndocumet.getElementById('image').src = image;\n``````" ]

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[ "## 2018 New Nike Air Max 97 Purple Blue Powder BQ9130 400\n\nModel: Max 97-163\nAvailability :  In Stock\n\\$129.00\n• US5.5= UK3 = EUR36=22.5CM\n• US6.5= UK4 = EUR37.5=23.5CM\n• US7 = UK4.5 = EUR38=24CM\n• US8 = UK5.5 = EUR39=25CM\n• Men US7.5=UK6.5=EUR40.5=25.5CM\n• Men US7 = UK6 = EUR40=25CM\n• Men US8=UK7=EUR41=26CM\n• Men US8.5=UK7.5=EUR42=26.5CM\n• Men US9=UK8=EUR42.5=27CM\n• Men US9.5=UK8.5=EUR43=27.5CM\n• Men US10=UK9=EUR44=28CM\n• Men US11=UK10=EUR45=29CM\n\nQTY :\n\n1.1392970085144" ]

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[ "1-905-376-2260 [email protected]\nSelect Page\nBulk materials such as sand, gravel, mulch, and crushed limestone are sold by volume in cubic yards rather than by area in square feet or square yards. Square footage is a more useful way to quantify materials spread over an area, however, one cubic yard* of gravel can cover different sized areas depending on how thickly you lay it.\n\n``` function calc() { /*info: 1 yard = 3 feet = 36 inches 1 foot = 12 inches N feet = (N/ 3) yards N inches = (N/12) feet N inches = (N/36) yards */ //get inputs var l = document.volcalc.length.value; //in feet var w = document.volcalc.width.value; //in feet var d = document.volcalc.depth.value; //in inches //convert length feet to yards l = l/3; //convert width feet to yards w = w/3; //convert depth inches to yards d = d/36; //calculate volume in yards var v = l * w * d; // round numbers up to nearest whole number var v = Math.ceil( v ); // round to 2 decimal places //var v = Math.round( v * 100) / 100; // round numbers in increments of .25 // var v = Math.round( v * 4 ) / 4 ; document.volcalc.vol.value = v; //in yards } ```\n\n``` ENTER YOUR DIMENSIONS Enter Length (Ft) Enter Width (Ft) Enter Depth (In) TOTAL Total yards needed ```\n\n*”Total Yards Need” rounded up to nearest decimal.\n\nStill have a question about our services and delivery options? Call us today: 905-376-2260", null, "" ]

[ null, "https://i1.wp.com/fidelitylandscape.ca/wp-content/uploads/2017/02/Fidleity_TB_Con_.jpeg", null ]

[ "#### Mobile Algebra\n\nIn a previous blog about Emoji Algebra, I presented the idea of using pictures (i.e., emojis) for variables. I will now add to that idea by using the physical structure of a mobile to represent the balance (i.e., equality) that must take place between the two sides of an equation.\nImagine using mobiles that once hung above your childhood crib to learn how to solve algebraic equations.", null, "The mobile on the left is balanced so both sides must weigh the same amount. What is the relationship between the orange trapezoids and the blue moon? Answer: One blue moon weighs the same as two trapezoids. In a different notation: moon = 2 * trapezoid. Or, with standard variables: y = 2 * x, where x is a trapezoid and y is a moon.", null, "SolveMe Mobiles\npresents this visual, intuitive, and exciting way to present, reason about, and solve algebraic equations.\n\nThe mobile on the right of the first figure is also balanced so again both sides must weigh the same. The relationship between the blue heart and the red rectangle, however, is a little more difficult to figure out. Examining the mobile, however, we see that there is a red rectangle on the left side as well as the right side. Both red rectangles weigh the same so in a sense they “cancel out” each other. Ignoring them leaves us with 2 blue hearts = 4 red rectangles, which simplifies to 1 blue heart = 2 red rectangles. Students generally are able to “cancel out” the same number of like pictures on each side of a balanced mobile. Using the shapes as variables produces the sequence of steps on the left side of the next figure. The use of standard variables (i.e., x stands for blue hearts and y stands for red rectangles) produces the derivation on the right side of the next figure.", null, "SolveMe Mobiles\ncan get complicated and involve three or more shapes (i.e., variables), as is the case for the next figure. In this mobile, the number 60 indicates that the whole structure of the mobile weighs 60 ounces. (I chose ounces here as the unit of measure.) This problem uses four variables (i.e., four different shapes). Can you solve it by determining the values of the four different shapes? Note that since the entire mobile weighs 60 and the mobile is balanced, both sides equally weigh 30. Using the same reasoning, each strand of shapes weighs 15. (The solution is at the end of this blog.)", null, "One corresponding system of algebraic equations using four standard variables (i.e., x, yz, and w) looks like the following: x is the heart shape, y is the circle, z is the trapezoid, and w is the square.", null, "I conducted an experiment with the 25 teen students in my summer math classes at Eagle Hill School. Every student had previously completed an Algebra I class. Each student was given four problems to solve: two mobile puzzles, including the one above, which possesses 60 ounces as its header; and two algebraic versions of the same mobile puzzles, including the above 4-variable system of algebraic equations. (Students were not told that the algebraic problems they were given corresponded to the mobile puzzles they were given.)\n\nStudents solved 91% of the mobile puzzles and only 18% of the algebraic versions of those same mobile puzzles. This result provides evidence that the mobile puzzles are easier to understand and solve than their algebraic counterparts.\n\nIn sum, the mobile puzzles presented by SolveMe Mobiles permit students to easily understand the relationships among four variables while a traditional algebraic presentation does not. Students who have never seen a 4-variable system were able to solve them using mobiles. The evidence of this small experiment points to the conclusion that students can more easily reach a deeper understanding of complex algebraic relationships by using mobiles than by using traditional algebraic equations. Students are capable of reasoning about 4-variable systems (and perhaps higher ones) if given the proper representation of that system. Further, it appears that the standard algebraic notation is actually obstructing the path to understanding and, if so, should be supplemented by a more intuitive method such as the mobiles. It is clear that the visual nature of the mobile with its hanging shapes is more intuitive than a set of interrelated algebraic equations using standard variables (i.e., x, y, z, and w). Finally, the mobiles visually present an image of balance between its two sides that seems to be a highly effective visual metaphor for expressing how two sides of an equation must be equal.\n\nIn brief, traditional algebra is good for computers, which mechanically manipulate the equations to find solutions. However, it is generally not good for humans, who are highly visual creatures capable of deep understanding when presented with the proper visual presentation. I applaud the cleverness of the mobile puzzles and encourage math teachers to seriously consider using them in their algebra classes.\n\nSolution to 60-Header Mobile:\n\nPurple trapezoid = 5\n\nRed rectangle = 3\n\nBlue heart = 2\n\nGreen circle = 9\n\nAcknowledgement:\n\nEthel McGinn told me about SolveMe Mobiles after seeing my blog on Emoji Algebra. Thanks, Ethel! She is a 2015 graduate of Assumption College and is currently a member of Teach for America—New Jersey.\n\n## What is Learning Diversity About?\n\nLearning Diversity is a blog hosted by Eagle Hill School where educators, students, and other members of the LD community regularly contribute posts and critical essays about learning and living in spaces that privilege the inevitability of human diversity.\n\nThe contributors of Learning Diversity come together to engage our readers from a variety of disciplines, including the humanities, social sciences, biological sciences and mathematics, athletics, and residential life. Embracing learning diversity means understanding and respecting our students as whole persons.\n\nP.O. Box 116\n242 Old Petersham Road\nHardwick, MA 01037\nPhone: 413.477.6000\nFax: 413.477.6837\n\n## Eagle Hill School\n\nAn innovative approach to LD education in a classic New England boarding school environment, where diverse learners achieve success." ]

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[ "", null, "## What our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nI was having problems learning quadratic equations, until I purchased your software. Now I know how to do not only do quadratics, but I also learned with the step by step examples how to do other more difficult equations and inequalities. Great product!\nHalen Iden, MT\n\nThis algebra software provides my daughter with the ability to learn independently, offering facts and helpful hints before providing problems for her to solve. It's working out very well . . . I think the software lends itself nicely to helping students throughout the year by supplementing any materials they get in a regular classroom.\nChuck Jones, LA\n\nI just received your update, I am so happy I hooked up with you, its been the best thing for me and my learning. I love your program, thanks!\nAlden Lewis, WI\n\nAll in all, this is a very useful, well-designed algebra help tool for school classes and homework.\nB.C., Malta-EU\n\n## Search phrases used on 2009-02-18:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• how to solve log equations\n• math revision grade 4 worksheets\n• oregon algebra online\n• ks3 online papers\n• tx plus calculator games\n• Free Sixth Grade IQ Test\n• solution first order non linear homogeneous differential equation\n• integration by substitution solver free\n• easy way to learn mathematics step by step online free\n• rudin mathematical analysis lecture tapes\n• multiplying positive and negative numbers worksheet\n• co-ordinate picture worksheet\n• how to calculate the inverse\n• Texas ti 85 interpolate\n• vetex parabola\n• what are some common uses for square roots?\n• mcdougal littell algebra 1 create chapter 3 test\n• here is a code for soliving the mathematical equation using array in java\n• Turning Degrees into Decimals\n• Rudin solution\n• sample paper for class 8\n• how to program slope formula into TI-84\n• algebra finding the greatest possible error lesson\n• logarithmic form turor\n• algebra cheat\n• Multiplication of rational algebraic expressions\n• printable GCSE practise science paper\n• cubic root solver\n• mcdougal littel algebra two\n• algebra .swf\n• cost accounting books\n• year 9 sat test uk online revision\n• permutation and combination basics\n• simplyfying expanded math\n• equation calculator with division\n• T-83 plus silver edition price - canada\n• Get all of the a's on one side, then use the distributive property.\n• online aptitude math test for 3rd grade\n• divide and simplify mix fractions\n• online revision tests ks2 history\n• graphing hyperbola online\n• ti-89-user manual\n• completing the squares of a binomial\n• printable literacy worksheets and answers\n• GCSE cheats\n• basic expression in maths\n• fourth square roots\n• online test paper (cat exam)\n• McDougal Littell Practice Bank Integrated Mathematics 3\n• slove measurements calculator\n• abstract algebra help\n• vector mechanics mcgraw Instructor's Solutions Manual\n• practise exams grade 10 alberta\n• roots of third order polynomial\n• Using the TI-83 plus to Algebra\n• math problem solutions of aptitude\n• quadratic factoring GCF difference of squares trinomial use to find roots\n• LINEAR PROGRAMMING, exaples\n• math test integers grade seven\n• vector dynamics program for ti-89\n• factoring binomial tutorial\n• examples of math trivia mathematics\n• ti 89 laplace\n• math for ks2 cubes\n• HIGHEST COMMON FACTOR EXERCISES\n• free caculator\n• Trigonometry by Ron Larson + fourth edition + even answers\n• factorise a summation\n• star online math\n• algebra calculator solver\n• online proving trigonometric identities" ]

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[ "", null, "Teacher resources and professional development across the curriculum\n\nTeacher professional development and classroom resources across the curriculum", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Topic Overview:", null, "", null, "", null, "", null, "Part 1: Linear Functions", null, "", null, "", null, "Part 2: Linear Equations and Inequalities", null, "", null, "", null, "", null, "", null, "Download the Workshop 2 Guide", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Part 1: Linear Functions\n\nLinear functions are those that exhibit a constant rate of change, and their graphs form a straight line. They are also described as polynomial functions of degree one.", null, "Explanation", null, "Mathematical Definition", null, "Role in the Curriculum\n\nExplanation\n\nLinear functions model a wide variety of real-world situations, including predicting the cost of a telephone call that lasts a given amount of time, the profit of a hot dog stand, and the amount of tax paid for a given income. Linear functions arise when there is a constant rate of change. Students should solve problems in which they use tables, graphs, words, and symbolic expressions to represent and examine linear functions and linear patterns of change.\n\nLinear functions can be written in three different forms, as shown in the table below. Each form provides different information about the function.\n\n Form: Standard Form Point-Slope Form Slope-Intercept Form Equation: Ax + By = C, where A", null, "0, B", null, "0 y - y1 =m(x - x1),where m", null, "0 y = mx + b, where m", null, "0 Information:", null, "Slope: m Point on the line: (x1, y1) Slope: m y-intercept: (0, b)\n\nStudents should be able to recognize a linear function in a table of values, a graph, or in algebraic form. When studying linear functions graphically, students should understand that the slope of the line represents a constant rate of change for the function, and that the y-intercept is the point where the graph crosses the y-axis and often represents the initial condition or starting point for the function. Through practical experience solving linear function problems in context, students will develop an understanding of the concepts and real-world meanings of the slopes and y-intercepts of lines.\n\nOther examples:\n\n• The linear function F = 1.8C + 32 can be used to convert temperatures between Celsius and Fahrenheit.\n\n• If a utility company charges a fixed monthly rate plus a constant rate for each unit of power consumed, a linear function will show the monthly cost of power. If the fixed rate is \\$25, and the cost for each unit of power is \\$0.02, the linear function is C = 0.02P + 25.\n\n• The linear function I = 400C + 1,500 yields the total monthly income of a car salesman who makes a monthly base salary of \\$1,500 and receives \\$400 dollars for each car sold.\nMathematical Definition\n\nA linear function, whose graph is a line, can be written in the form y = mx + b, where m and b are constants and m", null, "0. As with any function, students can represent a linear function as a table, an equation, or a graph.\n\n Table Equation Graph Minutes Cost 0 0.85 y = 0.24x + 0.85", null, "1 1.09 2 1.33 3 1.57 4 1.81 5 2.05\n\nRole in the Curriculum\n\nLinear functions are fundamental to the study of mathematics. Students can transfer many of the important concepts learned through the study of linear functions to the understanding of other functions. According to the National Council of Teachers of Mathematics (NCTM):\nIt is essential that [students] become comfortable in relating symbolic expressions containing variables to verbal, tabular, and graphical representations of numerical and quantitative relationships. Students should develop an initial understanding of several different meanings and uses of variables through representing quantities in a variety of problem situations.\n(NCTM, Principles and Standards for School Mathematics,\n2000, p. 223)\nStudents should be able to move fluently between the different representations of linear functions and, given a description of a situation, should be able to produce a table, equation, and graph. Likewise, when given one representation of a linear function, students should be able to produce the others. To do this, they need ample opportunity to explore situations involving linear functions in all representations. For example, in the video for Workshop 2, Part I, Tom Reardon started with a verbal description and a table of values. The students then produced an equation and a graph to further describe the situation. They discussed the meaning of the constants in the equation and how those constants affected both the table and the graph.\n\nSee what Diane Briars has to say about the important aspects of linear functions that Tom Reardon applied in his lesson:", null, "", null, "", null, "", null, "Read transcript from teacher educator Diane Briars", null, "", null, "", null, "", null, "", null, "", null, "When the students were talking about the slope and they said that 24 cents was the slope, he didn't stop there. Read More", null, "", null, "", null, "Next: Part 2: Linear Equations and Inequalities", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]

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[ "# obspy.signal.cross_correlation.correlation_detector¶\n\ncorrelation_detector(stream, templates, heights, distance, template_times=None, template_magnitudes=None, template_names=None, similarity_func=<function _calc_mean at 0x46bfedf4>, details=None, plot=None, **kwargs)[source]\n\nDetector based on the cross-correlation of waveforms.\n\nThis detector cross-correlates the stream with each of the template streams (compare with correlate_stream_template()). A similarity is defined, by default it is the mean of all cross-correlation functions for each template. If the similarity exceeds the height threshold a detection is triggered. This peak finding utilizes the SciPy function find_peaks() with parameters height and distance. For a SciPy version smaller than 1.1 it uses a custom function for peak finding.\n\nParameters: stream Stream with data traces. templates List of streams with template traces. Each template stream should be shorter than the data stream. This argument can also be a single template stream. heights Similarity values to trigger a detection, one for each template. This argument can also be a single value. distance The distance in seconds between two detections. template_times UTCDateTimes associated with template event (e.g. origin times, default are the start times of the template streams). This argument can also be a single value. template_magnitudes Magnitudes of the template events. If provided, amplitude ratios between templates and detections will be calculated and the magnitude of detections will be estimated. This argument can also be a single value. This argument can be set to True, then only amplitude ratios will be calculated. template_names List of template names, the corresponding template name will be inserted into the detection. similarity_func By default, the similarity will be calculated by the mean of cross-correlations. If provided, similarity_func will be called with the stream of cross correlations and the returned trace will be used as similarity. See the tutorial for an example. details If set to True detections include detailed information. plot Plot detections together with the data of the supplied stream. The default plot=None does not plot anything. plot=True plots the similarity traces together with the detections. If a stream is passed as argument, the traces in the stream will be plotted together with the similarity traces and detections. kwargs Suitable kwargs are passed to correlate_template() function. All other kwargs are passed to find_peaks(). List of event detections sorted chronologically and list of similarity traces - one for each template. Each detection is a dictionary with the following keys: time, similarity, template_id, amplitude_ratio, magnitude (if template_magnitudes is provided), template_name (if template_names is provided), cross-correlation values, properties returned by find_peaks (if details are requested)\n\nExample\n\n>>> from obspy import read, UTCDateTime" ]

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[ "Preprint to: Australian Computer Journal, 25(1), pp.14-20, 1993; also [preprint.ps].\n\n# Applications of Recursively Defined Data Structures.\n\n### Lloyd Allison, Department of Computer Science, Monash University, Victoria 3168.\n\n LA home Computing Publications  Rec. Structs.   #lists   #trees    #searchT    #perms    #n-queens    #irred. Also see: Fib' N-Queens Irreducible\n\nAbstract: A circular program contains a data structure whose definition is self-referential or recursive. The use of such a definition allows efficient functional programs to be written and can avoid repeated evaluations and the creation of intermediate data structures that would have to be garbage collected. This paper uses circular programs in various ways, to implement memo-structures and explicit search-trees to hold solutions to constraint-satisfaction problems.\n\nkeywords: circular program, functional programming, list, recursion, tree.\n\n## Introduction.\n\nA circular program contains a data structure whose definition is self-referential or recursive. Such a program cannot be written in a conventional, strict, imperative programming language but it can be written in a functional language employing lazy evaluation[6,8] or call by need.\n\n```\ngeneral schema: let rec ds = f(ds) -- ds is some data structure\neg. let rec posints = 1.(map succ posints) -- list of all +ve integers\n```\nNote that `.' is the infix list constructor also known as cons, rec qualifies recursive definitions and map applies a function to each element of a list and so produces a new list.\n\nThe list posints contains all the positive integers 1.(2.(3.  ...)) or [1,2,3,...]. It begins with 1 and continues with the result of applying the successor function succ to each element of posints itself. Successor applied to the first element gives the second element, 2, and so on. It is the definition of the value of the data structure posints, not just its type (list), being recursive that makes this a circular program.\n\nUnder lazy evaluation, an expression is not evaluated unless it is needed. In particular, the right hand side of a definition, and the actual parameter of a function, are not evaluated until they are needed - if they are needed. When an expression is evaluated, the value is remembered to avoid recomputation later. (The conditional, `if', is the only non-strict or lazy operator in many imperative languages.) Lazy evaluation permits recursive definitions of data structures and also allows some computations with infinite data structures. All of a potentially infinite data structure can be defined although only a finite part may be evaluated. If only a bounded part were defined and evaluated, a copy would have to be made if it had to be extended, wasting time and space, because functional languages do not permit side-effects. Circular programs can, in certain cases, have it both ways - an expanding data structure with side-effect-free programming. As a bonus, infinite data structures are sometimes easier to define because boundary cases are simpler or absent. Although posints represents an infinite list, a program using posints does not loop unless an attempt is made to print or otherwise evaluate all of the list. The program terminates provided that only finite parts of such structures are manipulated.\n\nIt is sometimes necessary to distinguish between the data structure as seen by the programmer and as implemented by the language system. If the definition simply incorporates the data structure directly, as in ones below, then a cyclic structure of cells and pointers is created in the computer memory.\n\n```let rec ones = 1 . ones -- = [1,1,1, ...]\n\ndata structure in memory:\n\nones: -------> 1.-->--|\n^ |\n| v\n| |\n|--<---\n```\nThis ability to create cyclic structures can be used to form circular lists, doubly linked lists and threaded trees in a functional language. The programmer cannot determine if a cyclic structure has been formed, except indirectly by the program's speed or modest use of space. If the recursive definition uses some function of the contents of the data structure, as in posints, no cyclic structure is created at the implementation level but those parts already computed can be used to compute new parts. This can be used to implement queues and various space efficient programs.\n\nMany functional programs compute their final result in stages, some data structure being operated on, often in small steps or passes, by functions such as map, filter, reduce, and so on. Each step produces an intermediate data structure which is eventually discarded and collected as garbage at some cost. Bird used circular programs in program transformations to convert multi-pass algorithms into single-pass algorithms. He attributed knowledge of circular programs to Hughes and to Wadler and the technique is so useful that it has probably been discovered several times.\n\nThe objective of the paper is to promote this useful functional programming technique. The examples given here construct lists and trees in new ways. They are used to define memo-structures and explicit search-trees which remove the need to repeat tests in certain constraint-satisfaction problems. They have all been run on a small lazy interpreter which was instrumented to record program behaviour and they perform as predicted. The notation used is from a hypothetical, \"generic\" functional language and is explained when required.\n\n## Circular Lists.\n\nThe ones and posints examples are amongst the simplest circular programs. To introduce the technique more fully, some non-trivial but routine examples on lists are given here. A popular example computes the Hamming numbers. These are all numbers of the form 2i×3j×5k,  i,  j,  k>=0, i.e.  1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15,  ...  . In this section various circular programs are derived from the Hamming numbers program.\n\nThe simplest program for the Hamming numbers is the following:\n\n```let rec Hamming = 1 . (merge3 (map (× 2) Hamming)\n(map (× 3) Hamming)\n(map (× 5) Hamming) )\n```\nThis is an example used in many books and papers[4,9,13]. Hamming is a self-referential list. It begins with 1 followed by the result of merging three lists. These are the results of multiplying all members of Hamming by 2, 3 and 5 respectively giving [2,...], [3,...] and [5,...]. The second Hamming number is thus 2, so the three lists are [2,4,...], [3,...] and [5,...], which allows the computation to proceed to the next step. Many duplicate numbers are produced, for example 6 = 2×3 = 3×2, and all but one copy must be removed by the merge function. It is well known that this inefficiency can be removed by ensuring that the factors in a product are combined in ascending order:\n\n```\nlet rec\na = 1 . (map (× 2) a) -- [1,2,4,8,...]\nand b = 1 . (merge (tl a) (map (× 3) b) ) -- [1,2,3,4,6,8,9,12,...]\nand Hamming = 1 . (merge (tl b) (map (× 5) Hamming) )\n```\n\nNote that hd (head) returns the first element of a list and tl (tail) returns a list minus its first element. A further operator null can be used to test if a list is empty.\n\nThe list a holds all powers of 2 and is defined in a very similar way to posints. List b holds all products of 2 and 3. No duplicates are produced because powers of 2 are multiplied by powers of 3 and then by powers of 5, in order. Consequently a simpler version of merge can be used that does not need to deal with duplicates.\n\nIt is a natural exercise to generalise the Hamming problem to find all products of an arbitrary list of factors. The factors are assumed to be coprime and in ascending order.\n\n```\nlet rec\nproducts [] = || -- no factors\nproducts (f.fs) = -- at least one factor f\nlet rec m = 1 . (merge (map (× f) m)\n(tl (products fs)) )\nin m\n```\nNote this program uses pattern matching. The function products accepts a single list parameter. It distinguishes two cases, patterns or kinds of input parameter: the empty list [] and the general list f.fs consisting of a first factor f and a list of remaining factors fs. A definition is given for each case. Multiple cases are separated by || which can be read as `otherwise'.\n\nIf there are no factors the list is returned. If there are factors, the products of the first factor f and of all other factors in fs must be combined. The result list m is self-referential. First, multiplying any member of m by f is itself a product - map (× f) m. Second, products of members of fs are also members of m although the leading 1 is not needed - tl (products fs). Merging these two lists and putting 1 on the front gives m. It is easy to see that any valid product must be produced by this process.\n\nThe last program works correctly on finite lists of factors. It will not work when given an infinite list of factors because the merge operation requires the head of two lists before it can produce any output. For an infinite list of factors this would require the heads of an infinite number of lists to be assembled which is impossible. This drawback can be overcome by recognising that the second value in m must be the smallest factor f itself.\n\n```\nlet rec\nproducts (f.fs) = -- NB. not (null fs)\nlet rec m = 1 . f . (merge (map (× f) (tl m))\n(tl (products fs)) )\nin m\n```\nThis program happens to require an infinite list of factors although a case to allow for finite lists can be added. Every instance of m now has two values at the front before any merge and thus one value remains when the tail is taken so output can begin immediately.\n\n## Circular Trees.\n\nThe previous section gave examples of circular programs over lists. Here trees are defined in a similar way and used as memo-structures to store the results of functions so that later calls can access them quickly without recomputation. In general, one or more functions and structures are defined using mutual recursion.\n\n```general schema:\nlet rec -- mutually recursive\nds = g ds f -- data structure(s) and\nand f x = h x ds f -- function(s)\n```\nThe technique is illustrated by application to the Fibonacci numbers. Both the so called slow and fast Fibonacci programs are well known. The slow version is doubly recursive and runs in time exponential in  n:\n```let rec slowfib n =\nif n<=2 then 1 else slowfib(n-2)+slowfib(n-1)\n```\nIt is easily seen that the running time, T(n), satisfies T(n)>2*T(n-2). For example, slowfib(7) calls slowfib(5) and slowfib(6). Slowfib(6) calls slowfib(5) and all its subcomputations again. Many computations are repeated. The fast program recognises that partial results are recalculated many times by the slow program. It gains efficiency be replacing binary recursion with linear recursion to run in O(n) time:\n```let fastfib n =\nlet rec f n a b =\nif n=1 then b\nelse f (n-1) b (a+b)\nin f n 0 1\n```\nThe parameters a and b hold two successive Fibonacci numbers. At the next step these become b and a+b respectively.\n\nIt is possible to define a circular program that builds a list of the Fibonacci numbers:\n\n```\nlet rec\nfiblist = 1.1.(f fiblist) -- [1,1,2,3,5,...]\nand f (a.t) = (a+(hd t)) . (f t)\n```\nThe function f both produces fiblist and uses it via its parameter which always lags two steps behind the element being calculated - just enough. It is now possible to find fib(n) by indexing to the nth element of fiblist with the standard function index:\n```\nlet fib =\nlet rec\nfiblist = 1.1.(f fiblist) -- memo list\nand f (a.t) = (a+(hd t)) . (f t) -- f builds fiblist\nand find n = index n fiblist -- get nth element of fiblist\nin find\n```\nThe list fiblist is a structure storing old results of fib n. On a first call of fib n the first n elements of fiblist are constructed. On a second call the result is just looked up in fiblist; the second call is faster but is still O(n) as index takes O(n) time. Actually, the result is looked up on the first call too, but it does not yet exist and so the list is built to the required length. Bird discusses the use of arrays to store past values in the process of deriving the fast Fibonacci program in an imperative language. Hughes describes a system that automatically stores past values of functions for fast recall; his system is implicit whereas the memo-structure here is explicit.\n\nIf a tree of Fibonacci numbers were built, the results of later calls could be looked up in O(log  n) time. The nodes of a complete binary tree can be numbered so that the children of node n are 2n and 2n+1:\n\n``` 1\n. .\n. .\n2 3\n. . . .\n4 5 6 7\n. .\n. . ...\n```\n\nThe value of fib n can be stored at node number n:\n\n``` 1\n. .\n. .\n1 2\n. . . .\n3 5 8 13\n. .\n. . ...\n```\n\nGiven this tree and an integer, for example n = 610 = 1102, the binary digits of n indicate whether to take left or right subtrees in locating the nth node and this can be done in O(log  n) time. The bits of n are read from the second most significant bit to the least significant bit; a 1 indicates go right and a 0 indicates go left. Therefore 1102 implies: start at the root, go right and then left. We first define an infinite binary tree type:\n\n```\ndatatype tree = fork int tree tree -- infinite binary tree type\n\nlet element (fork e l r) = e -- extract the element value\nand left (fork e l r) = l -- extract left subtree\nand right (fork e l r) = r -- extract right subtree\n```\n\nFork is a constructor that builds a new tree given an element and left and right subtrees. Element, left and right return the components of a tree.\n\n```\nlet fib =\nlet rec\nfibtree = fork 1 (fork 1 (build 4) (build 5)) -- memo tree,\n(build 3) -- fibtree :tree\nand build n = fork (f(n-2)+f(n-1)) -- build & f\n(build(2*n)) (build(2*n+1)) -- construct fibtree\nand f n = lookup n element -- return fib n\nand lookup 1 g = g fibtree || -- lookup decodes n\nlookup n g = -- n>1\nlookup (n div 2) (g o (if even n then left else right))\nin f\n```\n\nGraphically, fibtree is defined to be:\n\n``` 1\n. .\n. .\n1 build 3\n. .\n. build 5\n.\nbuild 4\n```\n\nThe values in nodes one and two are both 1 and are provided to enable node three to be built. This allows node four to be built and so on. The function lookup extracts the nth number from the tree. It does this by constructing a function g to follow left or right links according to the bits in n as described. Functional composition `o' is used to link the desired sequence of element, left and right operations together and these are finally applied to fibtree. ((p o q)(x) = p(q(x)))\n\nAssuming fib has been previously called with a parameter greater than n, a second call fib n takes O(log n) time to scan down fibtree. It might appear that the first such call would take exponential time because of the two calls to f within build but this is not the case. The call f(n-2) causes the tree to be evaluated and built up to node n-2. The call f(n-1) only causes one additional node to be evaluated using f(n-3) and f(n-2) which are just looked up, their corresponding nodes having been built already. On the first call, there are O(n) calls on build, f and lookup the latter being logarithmic, the total time taken is O(n log n). There is thus an increase in cost from O(n) to O(n log n) on first calls but a reduction from O(n) to O(log n) on subsequent calls over the \"fast\" Fibonacci program. There is also the cost of space to store the tree to be considered.\n\nA subtle point should be noted: Our program let fib = let rec ... in f binds fib to f which has fibtree in its environment so fibtree persists for as long as fib does and is not recomputed. If we carelessly defined let fib2 n = let rec ... in f n then fibtree would persist for only as long as a call to fib2 remained unevaluated and would be recomputed on each call. Some optimising compilers would undo this unfortunate effect by effectively converting fib2 into fib. However this is a difficult issue because a programmer might deliberately write a function having the form of fib2 because he or she needs a temporary data structure but wants it to be destroyed to avoid tying up space.\n\nIt is natural to ask if the log(n) factor in the costs can be removed but it is caused by the use of a tree rather than an (unbounded) array with O(1) indexing. In an imperative language, and in some functional languages, one might use an array instead of the tree. However this would place a limit on the size of n.\n\nIf it is necessary both to have random access to the Fibonacci numbers in log(n) time and to have sequential access then it may be convenient to derive a list from the tree in breadth-first order. Since the tree is complete and infinite the list is particularly simple to create:\n\n```\nlet rec\nfibtree = as before\nand build = ...\nand f = ...\nand lookup = ...\nand fiblist = map element nodes -- elements in breadth-first order\nand nodes = bfirst fibtree -- (sub)trees in breadth-first order\nand bfirst t = -- breadth-first traversal of t\nlet rec\nq = t . (traverse q)\nand traverse ((fork e l r) . q2) = l . r . (traverse q2)\nin q\n```\n\nFunction bfirst returns a list or queue q of the subtrees of a tree t in breadth-first order. The queue begins with t itself and the auxiliary function traverse produces the rest of the queue. Function traverse examines the first element in q and adds its subtrees to the end of q, i.e.  in the second and third positions. This is repeated for successive elements of q. The definition of the queue q is self-referential. Since q naturally grows as the tree is scanned, and since it is also infinite, it is hard to see how q could be created efficiently without a circular program.\n\nThe lists nodes and fiblist respectively contain the subtrees and the elements of fibtree in breadth-first order. Accessing the nth element of fiblist also causes fibtree to be evaluated to node number n.\n\n### Circular Search Trees.\n\nMany search problems or constraint-satisfaction problems require finding a sequence of values <a, b, c, ...>, or just abc..., that satisfies certain constraints. A search program explores the search space, building an implicit or explicit tree of (partial) solutions. If the constraints are uniform in a certain sense then the solution tree may be defined recursively and explicitly. The advantage is that no test is performed twice as the results of previous tests are available from the structure of the tree. This is valuable if the cost of performing tests outweighs the cost of building and keeping the tree.\n\nThe uniformity required covers two conditions. Firstly, it must be possible to build all long solutions by extending short solutions. Secondly, the constraints on an element in the sequence must involve other members of the sequence only in ways that depend on their relative, not absolute, positions in the sequence. These conditions hold in many problems.\n\nA suitable n-ary search-tree, in which a node contains an element of type 't, a subtree and the siblings of the node, can be defined as follows:\n\n```\ndatatype tree 't = empty || node 't (tree 't) (tree 't)\n```\nNote that 't is a type parameter - an arbitrary type. There are two cases to tree - the empty tree and a node - separated by ||. Siblings are linked together via the third component of a node.\n\nAs an example, the complete infinite tree over {1, 2, 3} can be defined as:\n\n```\nlet rec three =\nnode 1 three (node 2 three (node 3 three empty)) -- :tree int\n```\n\nA simple generalisation of this example allows trees to be built over the range [1..n]:\n\n```\nlet build n =\nlet rec\nT = toplevel 1 -- :tree int\n\nand toplevel m =\nif m>n then empty\nelse node m (f T) (toplevel (m+1))\n\nand f T = T\n\nin T\n```\n\nThe function toplevel builds the nodes at the top level of the tree and f fills in the subtrees. Here f is just the identity function and as such it is redundant but it is included to give a general schema. The program above creates a finite cyclic data structure. If we wanted to expand out the cyclic structure, into an infinite copy, f could be redefined as follows:\n\n```f empty = empty ||\nf (node a subtree sibs) =\nlet others = f sibs\nin node a (f subtree) others\n```\n\nSolutions to the various problems discussed below are formed only by redefining f in variations on the above. Apart from the changes that this entails the schema is unaltered in each case.\n\n### Permutations.\n\nA common method of generating permutations is to extend partial permutations, beginning with the empty sequence. If abcde is a partial permutation it can be extended with X to abcdeX provided that X differs from a, b, c, d and e. Equivalently, it can be extended provided that bcdeX is a partial permutation and provided that X differs from a. Note that bcde is already a partial permutation because abcde is. If a tree is used to hold the permutations then bcdeX, being shorter than abcdeX, must occur in the tree at the previous level, if it is indeed a partial permutation. If we read permutations as paths from the root of the tree, and identify a node with the path to it, the subtree of abcde is a pruned version of the subtree of bcde with all occurrences of a filtered out or banned. We call bcde the shadow of abcde. The shadow of bcde is cde and so on. Coding these ideas into function f of the schema in the previous section gives the following program:\n\n```\nlet build n =\nlet rec\nT = toplevel 1\n\nand toplevel m =\nif m>n then empty\nelse node m (f m T) (toplevel (m+1))\n\nand f banned empty = empty || -- f banned shadow\nf banned (node a subtree sibs) =\nlet others = f banned sibs\nin if a = banned then others -- prune a's subtree\nelse node a (f banned subtree) others -- no pruning\n\nin T\n```\n\nFunction f has gained an extra parameter for the banned element and performs a filtering operation on the shadow tree. A single test, a=banned, tells if a permutation can be extended with a given value, all other exclusions being implicit in the tree structure.\n\nAs the permutation tree is traversed it is gradually evaluated. If for example the first permutation, 123, from build 3 were printed, the evaluated portion would be:\n\n``` .\n. . .\n. . .\n. . .\n1 2 3\n. . . . . .\n. ? . . . .\n2 1 3 1 2\n. . . . .\n. ? . ? ?\n3 1\n```\nNote, `?' is used to denote an unevaluated subtree. Recall that we read a permutation, such as 123, as a path from the root of the tree and identify a node with the path to it. The shadow of 123 is 23. The subtree of 23 is `node 1 ? empty' and the 1 is banned for 123, so 123 is a complete permutation. The shadow of 23 is 3. The subtree of 3 is `node 1 ? (node 2 ? empty)' and the second branch is banned beneath 23 so its subtree is `node 1 ? empty'.\n\nThe shadow of a path grows with the path, one step behind it. The structure of the tree stores the results of many past tests so that only a single extra test is performed to add a new node. (This is not a big issue in permutation generation but there are cases where it is.) Note that Topor has examined the space complexity of functional programs for generating permutations represented as linear linked lists.\n\n### N-Queens.\n\nThe well known n-queens problem is to place n queens on an n×n chess board so that no two queens threaten each other. Each queen must be on a separate row, column and diagonal and this property is an invariant that must be maintained as partial solutions are extended. The fastest imperative solutions are based on permutation generators. A board is represented by the permutation of rows that the queens on the columns occupy. This representation automatically ensures the separate row and column parts of the invariant. Here we observe that a partial solution abcde can be extended to a partial solution abcdeX if and only if bcdeX is also a partial solution and a and X are on separate diagonals and rows. By using shadows, X need only be tested against a's diagonals as the results of the other diagonal tests against other queens are already encoded in the shadow tree and do not need to be repeated. Again, the required program is a modification of the general schema with f redefined. Function f gains a new parameter col, being the current column number.\n\n```\nlet build n =\nlet rec\nT = toplevel 1\n\nand toplevel m =\nif m>n then empty\nelse node m (f 1 m T) (toplevel (m+1))\n\nand f col banned empty = empty ||\nf col banned (node a subtree sibs) =\nlet others = f col banned sibs\nin if member banned [a, a+col, a-col] then others -- prune\nelse node a (f (col+1) banned subtree) others -- no prune\n\nin T\n```\n\nThe standard function member tests the membership of an element in a list. Note that the test member banned [a, a+col, a-col] is an amalgam of the old permutation test, a=banned, and the new diagonal test.\n\n### Irreducible or Good Sequences.\n\nAxel Thue defined the notion of an irreducible sequence in a series of papers in the period 1906-1914. Dijkstra later called these `good sequences' and used them in an exercise in structured programming. A sequence over the alphabet {1, 2, 3}, or in general over {1,...,n}, is irreducible if and only if it contains no adjacent subsequences that are identical. For example, 1213121 is irreducible but 12132131 is not because 213 is immediately repeated.\n\nIt is easy to see that (i)  a sequence abcdeX of even length is irreducible iff the shorter sequences abcde and bcdeX are irreducible and the two halves abc and deX are unequal and (ii)  a sequence abcdefX of odd length is irreducible iff abcdef and bcdefX are irreducible. This enables a circular program to be written for a tree representing all the irreducible sequences. For example, the shadow of abcde (of odd length) is bcde. Assuming that abcde is irreducible, abcdeX is irreducible if and only if bcdeX is irreducible, if and only if X is a descendant of bcde.\n\n```\nlet build n =\nlet rec\nT = toplevel 1\n\nand toplevel m =\nif m>n then empty\nelse node m (f 2 [m] T) (toplevel (m+1))\n\nand f len seq empty = empty ||\nf len seq (node a subtree sibs) =\nlet others = f len seq sibs\nand seq2 = a.seq\nin if even len & repeated (len/2) seq2 then others --prune\nelse node a (f (len+1) seq2 subtree) others --don't prune\n\nin T\n```\n\nA new parameter seq carries the particular sequence forward as f descends through the tree. When the length len is even, a test is made that the sequence a.seq is not the concatenation of two sequences of length len/2. Function repeated performs this test in O(len) time and has an obvious definition. It is not necessary to test for any shorter repeats. These are implicitly ruled out by the use of the shadow to generate subtrees. The test would be more complex without this information. That portion of the tree that is evaluated in order to print the first irreducible sequence of length five is shown below:\n\n``` .\n. . .\n. . .\n. . .\n. . .\n1 2 3\n. . . .\n. . . .\n2 3 1 1\n. . .\n. . .\n1 1 3\n. .\n. .\n3 1\n.\n.\n1\n```\n\n## Conclusions.\n\nProgrammers are familiar with recursive functions but recursive or self-referential data structures used in circular programs are rare. Circular programs are very powerful enabling many infinite structures to be efficiently defined. They often remove the need for intermediate structures and for repeated calculations. They can be used safely provided that later values depend only on earlier ones in the structure. Memo-structures can be formed by a data structure and function defined using mutual recursion. Explicit circular search-trees can reduce the number of tests performed in constraint-satisfaction problems. As usual, there is a trade-off of time against space.\n\n## References.\n\n L. Allison. Circular programs and self-referential structures. Software Practice and Experience, 19(2), 99-109 Feb 1989.\n R. S. Bird. Tabulation techniques for recursive programs. Computing Surveys, 12(4), 403-417 Dec 1980.\n R. S. Bird. Using circular programs to eliminate multiple traversals of data. Acta Informatica, 21(3) 239-250, 1984.\n R. S. Bird & P. Wadler. Introduction to Functional Programming. Prentice Hall 1988.\n E. W. Dijkstra. Notes on structured programming. in Structured Programming, Dahl, Dijkstra and Hoare (eds), Academic Press 1972.\n D. P. Friedman and D. S. Wise. Cons should not evaluate its arguments. Automata, Languages and Programming, Edinburgh University Press, 257-284, 1976.\n G. A. Hedlund. Remarks on the work of Axel Thue on sequences. Nordisk Matematisk Tidskrift, 15, Oslo, 1967.\n P. Henderson. A lazy evaluator. 3rd ACM Symposium in Principles of Programming Languages, 95-103, 1976.\n P. Henderson. Functional Programming: Application and Implementation. Prentice Hall 1980.\n J. Hughes. Lazy memo functions. Functional Programming Languages and Computer Architecture, J-P. Jouannaud (ed), Springer Verlag, LNCS 201, 129-146, 1985.\n J. S. Rohl. A faster lexicographical N queens algorithm. Information Processing Letters, 17, 231-233, 1983.\n R. W. Topor. Functional programs for generating permutations. Computer Journal, 25(2), 257-263, 1982.\n D. A. Turner. Functional programs as executable specification. Mathematical Logic and Programming Languages, C.A.R.Hoare and J.C.Shepherdson (eds), Prentice Hall 29-54, 1984.\n\nAlso see [Functional Programming] and [λ-calculus examples]." ]

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[ "Notice: While Javascript is not essential for this website, your interaction with the content will be limited. Please turn Javascript on for the full experience.\n\n# PEP 284 -- Integer for-loops\n\nPEP: 284 Integer for-loops David Eppstein , Greg Ewing Rejected Standards Track 1-Mar-2002 2.3\n\n# Abstract\n\nThis PEP proposes to simplify iteration over intervals of integers, by extending the range of expressions allowed after a \"for\" keyword to allow three-way comparisons such as\n\n```for lower <= var < upper:\n```\n\nin place of the current\n\n```for item in list:\n```\n\nsyntax. The resulting loop or list iteration will loop over all values of var that make the comparison true, starting from the left endpoint of the given interval.\n\n# Pronouncement\n\nThis PEP is rejected. There were a number of fixable issues with the proposal (see the fixups listed in Raymond Hettinger's python-dev post on 18 June 2005 ). However, even with the fixups the proposal did not garner support. Specifically, Guido did not buy the premise that the range() format needed fixing, \"The whole point (15 years ago) of range() was to avoid needing syntax to specify a loop over numbers. I think it's worked out well and there's nothing that needs to be fixed (except range() needs to become an iterator, which it will in Python 3.0).\"\n\n# Rationale\n\nOne of the most common uses of for-loops in Python is to iterate over an interval of integers. Python provides functions range() and xrange() to generate lists and iterators for such intervals, which work best for the most frequent case: half-open intervals increasing from zero. However, the range() syntax is more awkward for open or closed intervals, and lacks symmetry when reversing the order of iteration. In addition, the call to an unfamiliar function makes it difficult for newcomers to Python to understand code that uses range() or xrange().\n\nThe perceived lack of a natural, intuitive integer iteration syntax has led to heated debate on python-list, and spawned at least four PEPs before this one. PEP 204 (rejected) proposed to re-use Python's slice syntax for integer ranges, leading to a terser syntax but not solving the readability problem of multi-argument range(). PEP 212 (deferred) proposed several syntaxes for directly converting a list to a sequence of integer indices, in place of the current idiom\n\n```range(len(list))\n```\n\nfor such conversion, and PEP 281 proposes to simplify the same idiom by allowing it to be written as\n\n```range(list).\n```\n\nPEP 276 proposes to allow automatic conversion of integers to iterators, simplifying the most common half-open case but not addressing the complexities of other types of interval. Additional alternatives have been discussed on python-list.\n\nThe solution described here is to allow a three-way comparison after a \"for\" keyword, both in the context of a for-loop and of a list comprehension:\n\n```for lower <= var < upper:\n```\n\nThis would cause iteration over an interval of consecutive integers, beginning at the left bound in the comparison and ending at the right bound. The exact comparison operations used would determine whether the interval is open or closed at either end and whether the integers are considered in ascending or descending order.\n\nThis syntax closely matches standard mathematical notation, so is likely to be more familiar to Python novices than the current range() syntax. Open and closed interval endpoints are equally easy to express, and the reversal of an integer interval can be formed simply by swapping the two endpoints and reversing the comparisons. In addition, the semantics of such a loop would closely resemble one way of interpreting the existing Python for-loops:\n\n```for item in list\n```\n\niterates over exactly those values of item that cause the expression\n\n```item in list\n```\n\nto be true. Similarly, the new format\n\n```for lower <= var < upper:\n```\n\nwould iterate over exactly those integer values of var that cause the expression\n\n```lower <= var < upper\n```\n\nto be true.\n\n# Specification\n\nWe propose to extend the syntax of a for statement, currently\n\n```for_stmt: \"for\" target_list \"in\" expression_list \":\" suite\n[\"else\" \":\" suite]\n```\n\nas described below:\n\n```for_stmt: \"for\" for_test \":\" suite [\"else\" \":\" suite]\nfor_test: target_list \"in\" expression_list |\nor_expr less_comp or_expr less_comp or_expr |\nor_expr greater_comp or_expr greater_comp or_expr\nless_comp: \"<\" | \"<=\"\ngreater_comp: \">\" | \">=\"\n```\n\nSimilarly, we propose to extend the syntax of list comprehensions, currently\n\n```list_for: \"for\" expression_list \"in\" testlist [list_iter]\n```\n\nby replacing it with:\n\n```list_for: \"for\" for_test [list_iter]\n```\n\nIn all cases the expression formed by for_test would be subject to the same precedence rules as comparisons in expressions. The two comp_operators in a for_test must be required to be both of similar types, unlike chained comparisons in expressions which do not have such a restriction.\n\nWe refer to the two or_expr's occurring on the left and right sides of the for-loop syntax as the bounds of the loop, and the middle or_expr as the variable of the loop. When a for-loop using the new syntax is executed, the expressions for both bounds will be evaluated, and an iterator object created that iterates through all integers between the two bounds according to the comparison operations used. The iterator will begin with an integer equal or near to the left bound, and then step through the remaining integers with a step size of +1 or -1 if the comparison operation is in the set described by less_comp or greater_comp respectively. The execution will then proceed as if the expression had been\n\n```for variable in iterator\n```\n\nwhere \"variable\" refers to the variable of the loop and \"iterator\" refers to the iterator created for the given integer interval.\n\nThe values taken by the loop variable in an integer for-loop may be either plain integers or long integers, according to the magnitude of the bounds. Both bounds of an integer for-loop must evaluate to a real numeric type (integer, long, or float). Any other value will cause the for-loop statement to raise a TypeError exception.\n\n# Issues\n\nThe following issues were raised in discussion of this and related proposals on the Python list.\n\n• Should the right bound be evaluated once, or every time through the loop? Clearly, it only makes sense to evaluate the left bound once. For reasons of consistency and efficiency, we have chosen the same convention for the right bound.\n\n• Although the new syntax considerably simplifies integer for-loops, list comprehensions using the new syntax are not as simple. We feel that this is appropriate since for-loops are more frequent than comprehensions.\n\n• The proposal does not allow access to integer iterator objects such as would be created by xrange. True, but we see this as a shortcoming in the general list-comprehension syntax, beyond the scope of this proposal. In addition, xrange() will still be available.\n\n• The proposal does not allow increments other than 1 and -1. More general arithmetic progressions would need to be created by range() or xrange(), or by a list comprehension syntax such as\n\n```[2*x for 0 <= x <= 100]\n```\n• The position of the loop variable in the middle of a three-way comparison is not as apparent as the variable in the present\n\n```for item in list\n```\n\nsyntax, leading to a possible loss of readability. We feel that this loss is outweighed by the increase in readability from a natural integer iteration syntax.\n\n• To some extent, this PEP addresses the same issues as PEP 276 . We feel that the two PEPs are not in conflict since PEP 276 is primarily concerned with half-open ranges starting in 0 (the easy case of range()) while this PEP is primarily concerned with simplifying all other cases. However, if this PEP is approved, its new simpler syntax for integer loops could to some extent reduce the motivation for PEP 276.\n\n• It is not clear whether it makes sense to allow floating point bounds for an integer loop: if a float represents an inexact value, how can it be used to determine an exact sequence of integers? On the other hand, disallowing float bounds would make it difficult to use floor() and ceiling() in integer for-loops, as it is difficult to use them now with range(). We have erred on the side of flexibility, but this may lead to some implementation difficulties in determining the smallest and largest integer values that would cause a given comparison to be true.\n\n• Should types other than int, long, and float be allowed as bounds? Another choice would be to convert all bounds to integers by int(), and allow as bounds anything that can be so converted instead of just floats. However, this would change the semantics: 0.3 <= x is not the same as int(0.3) <= x, and it would be confusing for a loop with 0.3 as lower bound to start at zero. Also, in general int(f) can be very far from f.\n\n# Implementation\n\nAn implementation is not available at this time. Implementation is not expected to pose any great difficulties: the new syntax could, if necessary, be recognized by parsing a general expression after each \"for\" keyword and testing whether the top level operation of the expression is \"in\" or a three-way comparison. The Python compiler would convert any instance of the new syntax into a loop over the items in a special iterator object.\n\nSource: https://github.com/python/peps/blob/master/pep-0284.txt" ]

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[ "# Classification of finite type structures leads to Dynkin diagrams?\n\nClassification of finite type structures in mathematics often lead to the Dynkin diagrams (Example: representation-finite hereditary algebras, simple Lie algebras, Cluster algebras,... and I have read that there are nearly 50 other such structures connected with the Dynkin diagrams).\n\nQuestions: Are there classification results of such \"finite type\" structures where the answer surprisingly does not correspond to exactly the Dynkin diagrams, but to Dynkin diagrams with maybe a finite number of other diagrams and/or some simply laced diagrams missing (like for example $E_6$)?\n\n• This is an interesting, slightly different classification result: arxiv.org/abs/1603.03942 – Sam Hopkins Jul 24 '17 at 11:54\n• What do you mean here by 'surprisingly'? I would argue that it is the regularity with which Dynkin diagrams are the answer to finite-type classification problems that is surprising! In general I think one should not be surprised if the answer to a classification problem is something else, unless there was a better reason to think the answer might be Dynkin diagrams than merely that you had a finite-type classification problem. – Matthew Pressland Jul 24 '17 at 12:14\n• Possibly relevant : en.wikipedia.org/wiki/Du_Val_singularity. – Watson Jul 25 '17 at 8:49\n\nhttps://en.wikipedia.org/wiki/Coxeter_group\n\nI guess technically this doesn't quite fit since they're classified with Coxeter diagrams not Dynkin diagrams. Essentially you have Dynkin diagrams with the \"arrows\" removed. Thus, for example, types $B$ and $C$ are the same.\n\nIn any case, the classification of finite Coxeter groups yields the standard $A_n$,$B_n=C_n$,$D_n$,$E_6$,$E_7$,$E_8$,$F_4$, and $G_2$ classification (as usual). But it also includes new members $H_2$, $H_3$, $H_4$, and $I_n$.\n\nJim Humphrey's book \"Reflection Groups and Coxeter Groups\" is a great resource to learn more about these groups.\n\nClassification of finite simple groups is an obvious example of the situation where, apart from Dynkin diagrams, one gets few more infinite series of objects (e.g. alternating groups), as well as \"true\" 26 exceptions.\n\nThere is a lot of work done on understanding these 26 exceptions in terms of diagrams; e.g. one would need to consider an extra \"type\" corresponding to the Petersen graph, or to a finite affine plane, or to a finite design corresponding to one of Mathieu groups.\n\nYou may be interested in the classification of quivers of finite mutation-type: http://www.emis.ams.org/journals/EJC/Volume_15/PDF/v15i1r139.pdf\n\nThis extends the classification of mutation classes of quivers determining cluster algebras with only finitely many cluster variables—the answer to this classification problem is given by the simply-laced Dynkin diagrams. (The non-simply-laced diagrams can also be included by defining cluster algebras from skew-symmetrizable, rather than just skew-symmetric, matrices.) These classes of quivers are sometimes said to have finite cluster-type.\n\nAll of these mutation classes are necessarily finite (one says the quivers have finite mutation-type), but there are also other quivers with finite mutation-type, including the affine Dynkin diagrams. (In type $\\tilde{\\mathsf{A}}$ one has to be a bit careful, because not all quivers with this underlying graph are mutation equivalent, and the cyclically oriented $\\tilde{A}_n$ quiver is mutation equivalent to $\\mathsf{D}_{n+1}$.) One might hope, given that the finite cluster-type problem was solved by Dynkin diagrams, that the slightly more general finite mutation-type problem might be solved by Dynkin + affine diagrams (cf. finite representation-type versus tame representation-type, or positive definite Tits forms versus positive semi-definite Tits forms). However, this is not the case, for several reasons of increasing severity.\n\nFirstly a quiver on two vertices always has finite mutation-type for degeneracy reasons, but it doesn't feel too unreasonable to exclude this case. Next there are quivers determined by triangulations of oriented surfaces with marked points, which are also of finite mutation-type but their mutation classes needn't contain a Dynkin or affine diagram. Finally, Derksen and Weyman show in the paper linked to above that even excluding the previous two cases, there are still some exceptional mutation-finite quivers.\n\nI would mention the classification of finite-dimensional Nichols algebras $\\mathfrak{B}(V_1\\oplus\\cdots\\oplus V_\\theta)$, $\\theta\\geq2$, over decomposable Yetter-Drinfeld modules $V_1\\oplus\\cdots\\oplus V_{\\theta}$ over groups.\n\nIn the case $\\theta\\geq4$ the classification is essentially based on Dynking diagrams of finite type (with labels). In the cases $\\theta\\in\\{2,3\\}$ there are exceptions, most of them appearing in the case $\\theta=2$." ]

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[ "40m QIL Cryo_Lab CTN SUS_Lab CAML OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop\n 40m Log Not logged in", null, "", null, "Fri Oct 4 02:08:32 2013, Masayuki, Update, Green Locking, FPMI with ALS arm stabilization", null, "", null, "Wed Oct 23 00:13:30 2013, Masayuki, Update, Green Locking, FPMI with ALS arm stabilization", null, "", null, "", null, "Message ID: 9261     Entry time: Wed Oct 23 00:13:30 2013     In reply to: 9201\n Author: Masayuki Type: Update Category: Green Locking Subject: FPMI with ALS arm stabilization\n\nSummary\nIn 2arms + MICH configuration, residual motion of the cavity will couple with MICH signal. When cavity length change, the reflectivity of cavity also change. And that cause the phase shift in reflected light. That phase shift is detected in MICH signal. When we try to lock the DRMI + arm, that coupling will be problem for lock acquisition. For practice to estimate that coupling, I estimated the coupling between the cavity motion and the AS55Q signal.\n\nWhat I did\n\n- Measurement steps\nI did the same measurement as that of this entry. For the estimation below steps are needed. The detail of each step will be written below.\n--Measurement and calibration of the AS55Q error signal with MICH + 2arms locked by ALS control\n--Measurement of the ALS in-loop noise and estimation of residual motion of the cavities.\n--Calibration of the coupling from residual arm motion to AS55Q signal\n\n- Calibration of  the AS55Q signal\n1. Sensor gain estimation\nWe used the same method as the previous entry,\nWe excited the BS at 580 Hz with a given amplitude (Vin). We enabled the notch filter at 580 Hz in the LSC MICH servo. We measured  the peak height (Verr) of the AS55Q error signal. We used the actuator response (A_bs) of BS measured in this entry.\nWe can get the sensor gain (H) of AS55Q in unit of count/m\n\nVerr    1\nH = ------- -------\nVin   A_bs\n\nBy this calculation H = 4.2e+07.\n\n2. Fitting of OLTF for the MICH loop\nWe measured the OLTF of the MICH loop. Modelled OLTF is fitted into the measurement data. That modelled OLTF includes the actuator response of BS, the MICH servo filters, DAI,DAA,AI,AA filters, the TF of sample and hold circuit. (About DAI, DAA filters and S/H circuit please read this entry. About AI,AA filters please read this entry)  Also I put time-delay into that OLTF. I estimated that time-delay and the gain of OLTF by fitting.  The time delay was 311usec.", null, "3. Estimation of the MICH free running noise\nWith modeled OLTF, I estimated the MICH free running noise.\n\nEstimation of the coupling from residual cavity motion to AS55Q signal\nThe ALS in-loop noise data has the unit of Hz/rHz (disturbance of the cavity resonant frequency). By multiplying L_arm/f_laser we can convert the unit to m/rHz (disturbance of the cavity length) .\nI used the same coupling constant between residual motion of cavity and MICH noise as this entry. For estimation of the coupling constant, we excited ETMs  and measured the TF from excitation signal to AS55Q error signal.  I assumed the cavity pole as 4000 Hz. The result is discussed below\n\nDiscussion\n\nALS in-loop noise include the sensor noise. in high frequency region the in-loop noise is dominated by the sensor noise. So in this region in-loop noise does not mean actual residual motion of the cavity.  And this sensor noise pushes the mirror. So we have to estimate the actual motion of the cavity by multiplying the servo transfer function of the control in this region.\n\nI made 2 plots. Both include the MICH free running noise and estimated coupling noise from both arms. In one plot, for estimation of the coupling I multiplied only coupling constant to calibrated in-loop noise of the ALS loop. In another plot,  I multiplied coupling constant and OLTF of ALS loop in order to estimate the actual motion of the cavity.  If the 3 curves are coincide in first plot, that means the ALS in-loop noise is same as the residual cavity motion in that region and the MICH free running noise is dominated by coupling from residual cavity motion. If those curves are coincide in second plot, that means the ALS in-loop noise is sensor noise in that region.\n\nAbove 40 Hz, the 3 curves are totally in coincident in first plot. On the other hand in second plot the 3 curves look similar in this region. That may mean above 40 Hz the ALS noise are dominated by sensor noise and MICH free running noise is dominated by the coupling from residual cavity motion.  Also in the region between 10 Hz and 40 Hz, the MICH free running noise seems to be dominated by coupling from cavity motion.\n\nFigure 1\n\nFigure 2", null, "In second plot, the coupling from cavity motion is overestimated. It's possibly because of overestimation of coupling constant, but I'm not sure.\nKoji mentioned that we should measure the residual motion of the cavity by using POX and POY. Now the ALS is much more stable than before, so I think we can easily do the measurement again with out of loop measurement. That will be more strait forward measurement.\n\nELOG V3.1.3-" ]

[ null, "http://nodus.ligo.caltech.edu:8080/40m/elog.png", null, "http://nodus.ligo.caltech.edu:8080/40m/entry.png", null, "http://nodus.ligo.caltech.edu:8080/40m/attachment.png", null, "http://nodus.ligo.caltech.edu:8080/40m/reply.png", null, "http://nodus.ligo.caltech.edu:8080/40m/attachment.png", null, "http://nodus.ligo.caltech.edu:8080/40m/attachment.png", null, "http://nodus.ligo.caltech.edu:8080/40m/attachment.png", null, "http://nodus.ligo.caltech.edu:8080/40m/131024_031631/OLTF.png", null, "http://nodus.ligo.caltech.edu:8080/40m/131024_025456/ALSnoise2.png", null ]

[ "# Pascals to centibars (Pa to cbar)\n\n## Convert pascals to centibars\n\nPascals to centibars converter above calculates how many centibars are in 'X' pascals (where 'X' is the number of pascals to convert to centibars). In order to convert a value from pascals to centibars (from Pa to cbar) type the number of Pa to be converted to cbar and then click on the 'convert' button.\n\n## Pascals to centibars conversion factor\n\n1 pascal is equal to 0.001 centibars\n\n## Pascals to centibars conversion formula\n\nPressure(cbar) = Pressure (Pa) × 0.001\n\nExample: Assume there are 35 pascals. Shown below are the steps to express them in centibars.\n\nPressure(cbar) = 35 ( Pa ) × 0.001 ( cbar / Pa )\n\nPressure(cbar) = 0.035 cbar or\n\n35 Pa = 0.035 cbar\n\n35 pascals equals 0.035 centibars\n\n## Pascals to centibars conversion table\n\npascals (Pa)centibars (cbar)\n100.01\n150.015\n200.02\n250.025\n300.03\n350.035\n400.04\n450.045\n500.05\n550.055\n600.06\n650.065\n700.07\n750.075\n800.08\npascals (Pa)centibars (cbar)\n2500.25\n3500.35\n4500.45\n5500.55\n6500.65\n7500.75\n8500.85\n9500.95\n10501.05\n11501.15\n12501.25\n13501.35\n14501.45\n15501.55\n16501.65\n\nVersions of the pascals to centibars conversion table. To create a pascals to centibars conversion table for different values, click on the \"Create a customized pressure conversion table\" button.\n\n## Related pressure conversions\n\nBack to pascals to centibars conversion\n\nTableFormulaFactorConverterTop" ]

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[ "# Portfolio return calculator and formula\n\nIn this post, we explain the formula behind the calculation of portfolio returns. Furthermore, we provide a free online portfolio return calculator, which works as a portfolio expected return calculator as well as a portfolio realized return calculator. Finally, with this port, we make an introduction to the modern portfolio theory as well.\n\nSo far in this course on investments, we have focused on investing in a single asset. However, in reality, many investors invest in more than one asset and hold portfolios of assets. Furthermore, there is a massive investment management industry out there, including mutual funds, exchange-traded funds, hedge funds, insurance funds, pension funds, sovereign wealth funds, and so on. In essence, these investment companies are portfolios that are run by professional portfolio managers. Therefore, for anyone interested in finance and investing, it is important to be able to calculate portfolio returns.\n\nLearning objectives\n\n• Understand the concept of investment weight.\n• Learn how to calculate the return on a portfolio of assets.\n\n## Investment weights\n\nImagine that you have \\$5,000 to invest. If you invest all that amount in a single asset, say Pfizer shares, the return on your investment will solely depend on the performance of Pfizer shares.\n\nHowever, if you split your investment such that you invest \\$4,000 in Pfizer shares and \\$1,000 in Procter & Gamble shares, then the performance of your portfolio will depend on the performance of Procter & Gamble shares as well as Pfizer shares. In this scenario, you invest \\$4,000 / \\$5,000 = 80% of your funds in Pfizer. And, the remaining \\$1,000 / \\$5,000 = 20% in Procter & Gamble. In other words, your investment weights (or portfolio weights) for Pfizer and Procter & Gamble are 80% and 20%, respectively. Given that the investment weight of Pfizer is much bigger than that of Procter & Gamble, the performance of your portfolio will be more heavily influenced by Pfizer shares.\n\n## Portfolio return formula\n\nIf the returns of Pfizer and Procter & Gamble last month were 2% and −1%, respectively, then your portfolio return would simply be the weighted average of these returns:\n\n80% (2%) + 20% (−1%) = 1.4%\n\nLet’s go further and add a third stock to your portfolio: Nike, such that you invest \\$3,000 in Pfizer, \\$1,000 in Procter & Gamble, and \\$1,000 in Nike. Note that the total investment is still \\$5,000. Now, the updated investment weights are 60% for Pfizer, 20% for Procter & Gamble, and 20% for Nike (this is also referred to as your portfolio composition). If Nike’s return last month was 0.5%, your portfolio return would become:\n\n60% (2%) + 20% (−1%) + 20% (0.5%) = 1.1%\n\nIn general, the realized (or historical) return on a portfolio RP can be calculated as:\n\nRP = ∑wi Ri\n\nwhere wi is the investment weight for asset i and Ri is the realized return for asset i\n\nHow about the portfolio expected return E[RP]? The idea is still the same. In fact, we can simply replace the realized return of each asset in the portfolio with its expected return:\n\nE[RP] = ∑wi E[Ri]\n\nFor example, if the expected returns of Pfizer, Procter & Gamble, and Nike for the next month are 1.4%. 2%, and 0.9%, respectively, then your portfolio’s expected return is:\n\n60% (1.4%) + 20% (2%) + 20% (0.9%) = 1.42%\n\nIn summary, the return of a portfolio depends on (a) the return of each asset within that portfolio and (b) the investments weights for those assets, such that assets with larger investment weights influence the portfolio return more than those with smaller investment weights.\n\n## Portfolio return calculator\n\n##### Instructions\n\nYou can use the portfolio return calculator below to calculate the return on a portfolio of up to 5 assets. It can be used as a portfolio expected return calculator or a portfolio realized return calculator. Please note the following instructions:\n\n• The calculator allows for both positive investment weights and negative investment weights (i.e., short selling).\n• Make sure that investment weights add up to 100% (see the “sum of weights (%)” in the last row of the calculator).\n• Make sure to enter data as percentage points. For example, if the investment weight is 80%, simply enter 80 in the “weight (%)” field, and if the return is -5%, just enter -5 in the “return (%)” field.\n• If your portfolio consists of less than 5 assets, you can of course use the calculator by leaving 0s in the fields that you don’t need.\n\n##### Summary\n\nIf you are spreading your funds across several assets, which is typically a good idea, you need to be able to calculate portfolio returns. In this post, we show that the return on a portfolio can be calculated as a weighted average of returns on assets included in that portfolio. And, the weights are proportional to the amounts you invest in those assets. Finally, we offer a portfolio return calculator that functions as both a portfolio expected return calculator (i.e., forward-looking) and a portfolio historical return calculator (i.e., backward-looking)." ]

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[ "# Aggregation with dissolve¶\n\nSpatial data are often more granular than we need. For example, we might have data on sub-national units, but we’re actually interested in studying patterns at the level of countries.\n\nIn a non-spatial setting, when all we need are summary statistics of the data, we aggregate our data using the `groupby` function. But for spatial data, we sometimes also need to aggregate geometric features. In the geopandas library, we can aggregate geometric features using the `dissolve` function.\n\n`dissolve` can be thought of as doing three things: (a) it dissolves all the geometries within a given group together into a single geometric feature (using the `unary_union` method), and (b) it aggregates all the rows of data in a group using `groupby.aggregate()`, and (c) it combines those two results.\n\n## `dissolve` Example¶\n\nSuppose we are interested in studying continents, but we only have country-level data like the country dataset included in geopandas. We can easily convert this to a continent-level dataset.\n\nFirst, let’s look at the most simple case where we just want continent shapes and names. By default, `dissolve` will pass `'first'` to `groupby.aggregate`.\n\n```In : world = geopandas.read_file(geopandas.datasets.get_path('naturalearth_lowres'))\n\nIn : world = world[['continent', 'geometry']]\n\nIn : continents = world.dissolve(by='continent')\n\nIn : continents.plot();\n\nOut:\ngeometry\ncontinent\nAfrica MULTIPOLYGON (((49.544 -12.470, 49.809 -12.895...\nAntarctica MULTIPOLYGON (((-163.713 -78.596, -163.713 -78...\nAsia MULTIPOLYGON (((120.295 -10.259, 118.968 -9.55...\nEurope MULTIPOLYGON (((-51.658 4.156, -52.249 3.241, ...\nNorth America MULTIPOLYGON (((-61.680 10.760, -61.105 10.890...\n```", null, "If we are interested in aggregate populations, however, we can pass different functions to the `dissolve` method to aggregate populations using the `aggfunc =` argument:\n\n```In : world = geopandas.read_file(geopandas.datasets.get_path('naturalearth_lowres'))\n\nIn : world = world[['continent', 'geometry', 'pop_est']]\n\nIn : continents = world.dissolve(by='continent', aggfunc='sum')\n\nIn : continents.plot(column = 'pop_est', scheme='quantiles', cmap='YlOrRd');\n\nOut:\ngeometry pop_est\ncontinent\nAfrica MULTIPOLYGON (((49.544 -12.470, 49.809 -12.895... 1219176238\nAntarctica MULTIPOLYGON (((-163.713 -78.596, -163.713 -78... 4050\nAsia MULTIPOLYGON (((120.295 -10.259, 118.968 -9.55... 4389144868\nEurope MULTIPOLYGON (((-51.658 4.156, -52.249 3.241, ... 746398461\nNorth America MULTIPOLYGON (((-61.680 10.760, -61.105 10.890... 573042112\n```", null, "## Dissolve Arguments¶\n\nThe `aggfunc =` argument defaults to ‘first’ which means that the first row of attributes values found in the dissolve routine will be assigned to the resultant dissolved geodataframe. However it also accepts other summary statistic options as allowed by `pandas.groupby()` including:\n\n• ‘first’\n\n• ‘last’\n\n• ‘min’\n\n• ‘max’\n\n• ‘sum’\n\n• ‘mean’\n\n• ‘median’" ]

[ null, "https://geopandas.readthedocs.io/en/latest/_images/continents1.png", null, "https://geopandas.readthedocs.io/en/latest/_images/continents2.png", null ]

[ "Minitest is inspired by Ruby minispec.\n\n## Project description\n\nThis project is inspired by Ruby minispec, but now it just implement some methods including:\n\n```must_equal, must_true, must_false, must_raise, must_output, only_test.\n```\n\nAnd some other useful functions:\n\n```p, pp, pl, ppl, length, size, inject, flag_test,\np_format, pp_format, pl_format, ppl_format,\ncapture_output.\n```\n\n## Author\n\nColin Ji [email protected]\n\n## How to install\n\n```pip install minitest\n```\n\n## How to use\n\nFor a simple example, you just write a function called x, and I would like to write the unittest in same file as: code:\n\n```if __name__ == '__main__':\n# import the minitest\nfrom minitest import *\n\nimport operator\n\n# declare a variable for test\ntself = get_test_self()\n# you could put all your test variables on tself\n# just like declare your variables on setup.\ntself.jc = \"jc\"\n\n# declare a test\nwith test(object.must_equal):\ntself.jc.must_equal('jc')\nNone.must_equal(None)\n\nwith test(object.must_true):\nTrue.must_true()\nFalse.must_true()\n\nwith test(object.must_false):\nTrue.must_false()\nFalse.must_false()\n\n# using a funcation to test equal.\nwith test(\"object.must_equal_with_func\"):\n(1).must_equal(1, key=operator.eq)\n(1).must_equal(2, key=operator.eq)\n\ndef div_zero():\n1/0\n\n# test exception\nwith test(\"test must_raise\"):\n(lambda : div_zero()).must_raise(ZeroDivisionError)\n(lambda : div_zero()).must_raise(ZeroDivisionError, \"integer division or modulo by zero\")\n(lambda : div_zero()).must_raise(ZeroDivisionError, \"in\")\n\n# when assertion fails, it will show the failure_msg\nwith test(\"with failure_msg\"):\nthe_number = 10\n(the_number % 2).must_equal(1,\nfailure_msg=\"{0} is the number\".format(the_number))\n# it wont show the failure_msg\n(the_number % 2).must_equal(0,\nfailure_msg=\"{0} is the number\".format(the_number))\n\n(True).must_false(\nfailure_msg=\"{0} is the number\".format(the_number))\n\n(lambda : div_zero()).must_raise(ZeroDivisionError, \"in\",\nfailure_msg=\"{0} is the number\".format(the_number))\n\ndef print_msg_twice(msg):\nprint msg\nprint msg\nreturn msg\n\nwith test(\"capture_output\"):\nwith capture_output() as output:\nresult = print_msg_twice(\"foobar\")\nresult.must_equal(\"foobar\")\noutput.must_equal([\"foobar\",\"foobar\"])\n\nwith test(\"must output\"):\n(lambda : print_msg_twice(\"foobar\")).must_output(\n[\"foobar\",\"foobar\"])\n(lambda : print_msg_twice(\"foobar\")).must_output(\n[\"foobar\",\"wrong\"])\n```\n\nresult:\n\n```Running tests:\n\n.FFFF.\n\nFinished tests in 0.013165s.\n\n1) Failure:\nFile \"/Users/Colin/work/minitest/test.py\", line 29, in <module>:\nEXPECTED: True\nACTUAL: False\n\n2) Failure:\nFile \"/Users/Colin/work/minitest/test.py\", line 32, in <module>:\nEXPECTED: False\nACTUAL: True\n\n3) Failure:\nFile \"/Users/Colin/work/minitest/test.py\", line 38, in <module>:\nEXPECTED: 2\nACTUAL: 1\n\n4) Failure:\nFile \"/Users/Colin/work/minitest/test.py\", line 47, in <module>:\nEXPECTED: 'in'\nACTUAL: 'integer division or modulo by zero'\n\n5) Failure:\nFile \"/Users/colin/work/minitest/test.py\", line 86, in <module>:\nMESSAGE: '10 is the number'\nEXPECTED: 1\nACTUAL: 0\n\n6) Failure:\nFile \"/Users/colin/work/minitest/test.py\", line 92, in <module>:\nMESSAGE: '10 is the number'\nEXPECTED: False\nACTUAL: True\n\n7) Failure:\nFile \"/Users/colin/work/minitest/test.py\", line 95, in <module>:\nMESSAGE: '10 is the number'\nEXPECTED: 'in'\nACTUAL: 'integer division or modulo by zero'\n\n8) Failure:\nFile \"/Users/colin/work/minitest/test.py\", line 102, in <module>:\nEXPECTED: ['foobar', 'wrong']\nACTUAL: ['foobar', 'foobar']\n\n12 tests, 22 assertions, 8 failures, 0 errors.\n[Finished in 0.1s]\n```\n\n## only_test function\n\nIf you just want to run some test functions, you can use only_test funtion to specify them. Notice, you must put it on the top of test functions, just like the below example. code:\n\n```def foo():\nreturn \"foo\"\n\ndef bar():\nreturn \"bar\"\n\nif __name__ == '__main__':\nfrom minitest import *\n\nonly_test(\"for only run\", foo)\n\nwith test(\"for only run\"):\n(1).must_equal(1)\n(2).must_equal(2)\npass\n\nwith test(\"other\"):\n(1).must_equal(1)\n(2).must_equal(2)\npass\n\nwith test(foo):\nfoo().must_equal(\"foo\")\n\nwith test(bar):\nbar().must_equal(\"bar\")\n```\n\nIt will only run test(“for only run”) and test(foo) for you.\n\n## Other useful function\n\ncapture_output, p, pp, pl, ppl, length, size, p_format, pp_format, pl_format, ppl_format these functions could been used by any object.\n\ncapture_output, capture the standard output. This function will print variable name as the title. code: def print_msg_twice(msg): print msg print msg return msg\n\n```with capture_output() as output:\nresult = print_msg_twice(\"foobar\")\n\nprint result\nprint output\n```\n\nprint result:\n\n```\"foobar\"\n[\"foobar\",\"foobar\"]\n```\n\np, print with title. This function will print variable name as the title. code:\n\n```value = \"Minitest\"\nvalue.p()\n\nvalue.p(\"It is a value:\")\n\nvalue.p(auto_get_title=False)\n```\n\nprint result:\n\n```value : 'Minitest'\n\nIt is a value: 'Minitest'\n\n'Minitest'\n```\n\npp, pretty print with title. This function will print variable name as the title in the first line, then pretty print the content of variable below the title. code:\n\n```value = \"Minitest\"\nvalue.pp()\n\nvalue.pp(\"It is a value:\")\n\nvalue.pp(auto_get_title=False)\n```\n\nprint result:\n\n```value :\n'Minitest'\n\nIt is a value:\n'Minitest'\n\n'Minitest'\n```\n\npl, print with title and code loction. This function just like pt, but will print the code location at the first line. And some editors support to go to the line of that file, such as Sublime2. code:\n\n```value = \"Minitest\"\nvalue.pl()\n\nvalue.pl(\"It is a value:\")\n\nvalue.pl(auto_get_title=False)\n```\n\nprint result:\n\n``` File \"/Users/Colin/work/minitest/test.py\", line 76\nvalue : 'Minitest'\n\nFile \"/Users/Colin/work/minitest/test.py\", line 77\nIt is a value: 'Minitest'\n\nFile \"/Users/Colin/work/minitest/test.py\", line 78\n'Minitest'\n```\n\nppl, pretty print with title and code loction. This function just like ppt, but will print the code location at the first line. Notice: it will print a null line firstly. code:\n\n```value = \"Minitest\"\nvalue.ppl()\n\nvalue.ppl(\"It is a value:\")\n\nvalue.ppl(auto_get_title=False)\n```\n\nprint result:\n\n``` File \"/Users/Colin/work/minitest/test.py\", line 76\nvalue :\n'Minitest'\n\nFile \"/Users/Colin/work/minitest/test.py\", line 77\nIt is a value:\n'Minitest'\n\nFile \"/Users/Colin/work/minitest/test.py\", line 78\n'Minitest'\n```\n\np_format, get the string just like p function prints. I use it in debugging with log, like: logging.debug(value.p_format()) code:\n\n```value = \"Minitest\"\nvalue.p_format()\n```\n\nreturn result:\n\n```value : 'Minitest'\n```\n\npp_format, get the string just like pp function prints. I use it in debugging with log, like: logging.debug(value.pp_format()) code:\n\n```value = \"Minitest\"\nvalue.pp_format()\n```\n\nreturn result:\n\n```value :\\n'Minitest'\n```\n\npl_format, get the string just like pl function prints. I use it in debugging with log, like: logging.debug(value.pl_format()) code:\n\n```value = \"Minitest\"\nvalue.pl_format()\n```\n\nreturn result:\n\n```line info: File \"/Users/Colin/work/minitest/test.py\", line 76, in <module>\\nvalue : 'Minitest'\n```\n\nppl_format, get the string just like ppl function prints. I use it in debugging with log, like: logging.debug(value.ppl_format()) code:\n\n```value = \"Minitest\"\nvalue.ppl_format()\n```\n\nreturn result:\n\n```line info: File \"/Users/Colin/work/minitest/test.py\", line 76, in <module>\\nvalue :\\n'Minitest'\n```\n\nlength and size will invoke len function for the caller’s object. code:\n\n```[1,2].length() # 2, just like len([1,2])\n(1,2).size() # 2, just like len((1,2))\n```\n\ninject_customized_must_method or inject function will inject the function which you customize. Why do I make this function? Since in many case I will use numpy array. When it comes to comparing two numpy array, I have to use:\n\n```import numpy\nnumpy.array().must_equal(numpy.array([1.0]), numpy.allclose)\n```\n\nFor being convient, I use inject_customized_must_method or inject function like:\n\n```import numpy\ninject(numpy.allclose, 'must_close')\nnumpy.array().must_close(numpy.array([1.0]))\n```\n\nflag_test will print a message ‘There are codes for test in this place!’ with the code loction. code:\n\n```flag_test()\n\nflag_test(\"for test\")\n```\n\nprint result:\n\n``` File \"/Users/colin/work/minitest/test.py\", line 97, in <module>:\nThere are test codes in this place!\n\nFile \"/Users/colin/work/minitest/test.py\", line 101, in <module>:\nfor test : There are test codes in this place!\n```\n\n## Project details\n\nThis version", null, "2.0.3", null, "2.0.2", null, "2.0.1", null, "2.0.0", null, "1.8.0", null, "1.7.4", null, "1.7.3", null, "1.7.2", null, "1.7.1", null, "1.7.0", null, "1.6.1", null, "1.6.0", null, "1.5.1", null, "1.5.0", null, "1.4.0", null, "1.3.3", null, "1.3.2", null, "1.3.1", null, "1.2.0", null, "1.1.1", null, "1.1.0", null, "1.0.0", null, "0.2.4", null, "0.2.3", null, "0.2.2", null, "0.2.1", null, "0.1.3", null, "0.1.2", null, "0.1.1\n\nUploaded `source`" ]

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[ "# zbMATH — the first resource for mathematics\n\nGradual and tapered overflow and underflow: A functional differential equation and its approximation. (English) Zbl 1089.65041\nSummary: A modified structure for floating-point representation, which can essentially eliminate overflow from floating-point calculations, is analyzed. The main part of the paper deals with a computational examination of a model for floating-point exponents and the probabilities of overflow and underflow. Earlier results in the absence of gradual underflow are presented for comparison but the primary focus is on the effect of gradual underflow, a version of gradual overflow, and of a proposal for an extended treatment of these exceptions which we call tapered overflow and underflow. The latter virtually eliminates the exceptions; its overflow threshold is $$10^{600000000}$$.\nThis paper reviews several models for the distribution of exponents of floating-point numbers and their evolution in the presence of repeated multiplicative operations. A continuous model, which closely approximates the discrete case, is seen to satisfy a functional differential equation with both delay and advance terms.\n##### MSC:\n 65G50 Roundoff error 68P01 General topics in the theory of data\nFull Text:\n##### References:\n Arnold, M.G.; Bailey, T.A.; Cowles, J.R.; Winkel, M.D., Applying features of IEEE754 to signed logarithmic arithmetic, IEEE trans. comput., 41, 1040-1050, (1992) Barlow, J.L.; Bareiss, E.H., On roundoff distribution in floating-point and logarithmic arithmetic, Computing, 34, 325-364, (1985) · Zbl 0556.65036 Benford, F., The law of anomalous numbers, Proc. amer. philos. soc., 78, 551-572, (1938) Brent, R., A Fortran multiple precision arithmetic package, ACM toms, 4, 57-70, (1978) Clenshaw, C.W.; Olver, F.W.J., Level-index arithmetic operations, SIAM J. numer. anal., 24, 470-485, (1987) · Zbl 0624.65016 Clenshaw, C.W.; Turner, P.R., The symmetric level-index system, IMA J. numer. anal., 8, 517-526, (1988) · Zbl 0668.68018 Feldstein, A.; Goodman, R.H., Some aspects of floating-point computation, (), 169-181 Feldstein, A.; Turner, P.R., Overflow, underflow and severe loss of significance in floating-point addition and subtraction, IMA J. numer. anal., 6, 241-251, (1986) · Zbl 0593.65029 Feldstein, A.; Turner, P.R., Overflow and underflow in multiplication and division, Appl. numer. math., 21, 221-239, (1996) · Zbl 0860.65034 Hamada, H., URR: universal representation of real numbers, New generation computing, 1, 205-209, (1983) Hamada, H., A new real number representation and its operation, (), 153-157 Hull, T.E.; Cohen, M.S.; Hall, C.B., Specifications for a variable precision arithmetic coprocessor, (), 127-131 IEEE, Standard for binary floating-point arithmetic, ANSI/IEEE std, vol. 754, (1985), IEEE New York Koren, I., Computer arithmetic algorithms, (1998), Brookside Court Amherst, MA Lewis, D.M., An architecture for addition and subtraction of long wordlength numbers in the logarithmic number system, IEEE trans. comput., 39, 1326-1336, (1990) Lozier, D.W., An underflow-induced graphics failure solved by SLI arithmetic, (), 10-17 Lozier, D.W.; Turner, P.R., Robust parallel computation in floating-point and sli arithmetic, Computing, 48, 239-257, (1992) · Zbl 0757.68020 Matsui, S.; Iri, M., An overflow/underflow-free floating-point representation of numbers, J. inform. proc., 4, 123-133, (1981) Morris, R., Tapered floating-point: A new floating-point representation, IEEE trans. comput., 20, 1578-1579, (1971) · Zbl 0226.68019 Turner, P.R., The distribution of leading significant digits, IMA J. numer. anal., 2, 407-412, (1982) · Zbl 0503.65029 Turner, P.R., Further revelations on l.s.d., IMA J. numer. anal., 4, 225-231, (1984) · Zbl 0564.65028 Yokoo, H., Overflow/underflow-free floating-point number representations with self-delimiting variable length exponent field, (), 110-117\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

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[ "Featured\n\nCreating a multiple methods in a class with same name but different parameters and types is called as method overloading.method overloading is the example of Compile time polymorphism which is  done at compile time.\n• By changing the number of parameters used.\n• By changing the order of parameters.\n• By using different data types for the parameters.\ne.g\n```public class Methodoveloading\n{\npublic int add(int a, int b) //two int type Parameters method\n{\nreturn a + b;\n\n}\npublic int add(int a, int b,int c) //three int type Parameters with same method same as above\n{\nreturn a + b+c;\n\n}\npublic float add(float a, float b,float c,float d) //four float type Parameters with same method same as above two method\n{\nreturn a + b+c+d;\n\n}\n}\n```\nIn the above example ,their are three methods with same method same but they differ with number of parameters and type of parameters ,hence this types of methods is called as method overloading.\nQuestion-    Can method overloading has same numbers of parameters and Name with different return types?\nAnswer-  No, because conflict is happen in methods while passing the parameters .\nNow let us learn about the Method overriding\nWhat is Method overriding ?\nCreating the method in a derived class with same name, same parameters and same return type as in base class is called as method overriding.\nMethod overriding is the example of run time polymorphism,how its is the part of run time polymorphism i will explain in detail.\nSome Key Points of  Method overriding\n•  Method overriding is only possible in derived class not within the same class where the method is declared.\n• Only those methods are overrides in the derived class which is declared in the base class with the help of virtual keyword or abstract keyword.\ne.g\n```public class Account\n{\npublic virtual int balance()\n{\nreturn 10;\n}\n}\npublic class Amount:Account\n{\n\npublic override int balance()\n{\nreturn 500;\n}\n}\n```\nOutput of the above Program is\n• 10 and 500\nIn the above small program their are two classes Account and Amount and both the classes contain same method Name that balance() in which Account class method  returns 10 and  Amount class returns 500 at run time because Account class  method is overridden in a class Amount.\nThe Method overriding is very useful when  we wants to return different output of same method in different class according to the need.\nLet us consider the example I wants to provide the discount on particular product according to the Customer category that A,B,C in this scenario suppose A,B,C are the classes and Discount is the class which contains virtual CustDiscount () method ,then i simply override it on class A,B,C instead of writing three different methods for each class." ]

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[ "# Post-modern portfolio theory\n\n\nPost-modern portfolio theory\n\nPost-modern portfolio theory [The earliest citation of the term 'Post-Modern Portfolio Theory' in the literature appears in 1993 in the article \"Post-Modern Portfolio Theory Comes of Age\" by Brian M. Rom and Kathleen W. Ferguson, published in The Journal of Investing, Winter, 1993. Summarized versions of this article have been subsequently published in a number of other journals and websites.] (or \"PMPT\") is an extension of the traditional modern portfolio theory (“MPT”, also referred to as Mean-Variance Analysis or “MVA”). Both theories propose how rational investors should use diversification to optimize their portfolios, and how a risky asset should be priced.\n\nOverview\n\nHarry Markowitz laid the foundations of MPT, the greatest contribution of which is the establishment of a formal risk/return framework for investment decision-making. By defining investment risk in quantitative terms, Markowitz gave investors a mathematical approach to asset-selection and portfolio management. But there are important limitations to the original MPT formulation.\n\nThe limitations of MPT are the assumptions that 1) variance [In MPT, the terms variance, variability, volatility and standard deviation are used interchangeably to represent investment risk.] of portfolio returns is the correct measure of investment risk, and 2) the investment returns of all securities and portfolios can be adequately represented by the normal distribution. Stated another way, MPT is limited by measures of risk and return that do not always represent the realities of the investment markets.\n\nStandard deviation and the normal distribution are a major practical limitation: they are symmetrical. Using standard deviation implies that better-than-expected returns are just as risky as those returns that are worse than expected. Furthermore, using the normal distribution to model the pattern of investment returns makes investment results with more upside than downside returns appear more risky than they really are, and vice-versa for returns with more a predominance of downside returns. The result is that using traditional MPT techniques for measuring investment portfolio construction and evaluation frequently distorts investment reality.\n\nIt has long been recognized that investors typically do not view as risky those returns \"above\" the minimum they must earn in order to achieve their investment objectives. They believe that risk has to do with the bad outcomes (i.e., returns below a required target), not the good outcomes (i.e., returns in excess of the target) and that losses weigh more heavily than gains. This view has been noted by researchers in finance, economics and psychology, including Sharpe (1964). \"Under certain conditions the MVA can be shown to lead to unsatisfactory predictions of (investor) behavior. Markowitz suggests that a model based on the semivariance would be preferable; in light of the formidable computational problems, however, he bases his (MV) analysis on the mean and the standard deviation [See Sharpe . Markowitz recognized these limitations and proposed downside risk (which he called \"semivariance\") as the preferred measure of investment risk. The complex calculations and the limited computational resources at his disposal, however, made practical implemetations of downside risk impossible. He therefore compromised and stayed with variance.] .\"\n\nRecent advances in portfolio and financial theory, coupled with today’s increased electronic computing power, have overcome these limitations. The resulting expanded risk/return paradigm is known as Post-Modern Portfolio Theory, or PMPT. Thus, MPT becomes nothing more than a (symmetrical) special case of PMPT.\n\nThe Tools of PMPT\n\nIn 1987 The Pension Research Institute at San Francisco State University developed the practical mathematical algorithms of PMPT that are in use today. These methods provide a framework that recognizes investors' preferences for upside over downside volatility. At the same time, a more robust model for the pattern of investment returns, the three-parameter lognormal distribution [The three-parameter lognormal distribution is the only pdf that has thus far been developed for robust solutions of downside risk calculations permits both positive and negative skewness in return distributions. This is a more robust measure of portfolio returns than the normal distribution, which requires that the upsides and downside tails of the distribution be identical.] , was introduced.\n\nDownside risk\n\nDownside risk (DR) is measured by target semi-deviation (the square root of target semivariance) and is termed downside deviation. It is expressed in percentages and therefore allows for rankings in the same way as standard deviation.\n\nAn intuitive way to view downside risk is the annualized standard deviation of returns below the target. Another is the square root of the probability-weighted squared below-target returns. The squaring of the below-target returns has the effect of penalizing failures at an exponential rate. This is consistent with observations made on the behavior of individual decision-making under\n\n: $d = sqrt\\left\\{ int_\\left\\{-infty\\right\\}^t \\left(t-r\\right)^2f\\left(r\\right),dr \\right\\}$\n\nwhere\n\n\"d\" is downside deviation (commonly known in the financial community as 'downside risk'). Note: By extension, \"d\"² = downside variance.\n\n\"t\" is the annual target return, originally termed the minimum acceptable return, or MAR.\n\n\"r\" is the random variable representing the return for the distribution of annual returns \"f\"(\"r\"),\n\n\"f\"(\"r\") is the three-parameter lognormal distribution\n\nFor the reasons provided below, this \"continuous\" formula is preferred over a simpler \"discrete\" version that determines the standard deviation of below-target periodic returns taken from the return series.\n\n1. The continuous form permits all subsequent calculations to be made using annual returns which is the natural way for investors to specify their investment goals. The discrete form requires monthly returns for there to be sufficient data points to make a meaningful calculation, which in turn requires converting the annual target into a monthly target. This significantly affects the amount of risk that is identified. For example, a goal of earning 1% in every month of one year results in a greater risk than the seemingly equivalent goal of earning 12% in one year.\n\n2. A second reason for strongly preferring the continuous form to the discrete form has been proposed by Sortino & Forsey (1996): \"Before we make an investment, we don't know what the outcome will be... After the investment is made, and we want to measure its performance, all we know is what the outcome was, not what it could have been. To cope with this uncertainty, we assume that a reasonable estimate of the range of possible returns, as well as the probabilities associated with estimation of those returns...In statistical terms, the shape of [this] uncertainty is called a probability distribution. In other words, looking at just the discrete monthly or annual values does not tell the whole story.\"\n\nUsing the observed points to create a distribution is a staple of conventional performance measurement. For example, monthly returns are used to calculate a fund's mean and standard deviation. Using these values and the properties of the normal distribution, we can make statements such as the likelihood of losing money (even though no negative returns may actually have been observed), or the range within which two-thirds of all returns lies (even though the specific returns identifying this range have not necessarily occurred). Our ability to make these statements comes from the process of assuming the continuous form of the normal distribution and certain of its well-known properties.\n\nIn PMPT an analogous process is followed: 1. Observe the monthly returns, 2. Fit a distribution that permits asymmetry to the observations, 3. Annualize the monthly returns, making sure the shape characteristics of the distribution are retained, 4. Apply integral calculus to the resultant distribution to calculate the appropriate statistics.\n\nortino ratio\n\nThe Sortino ratio measures returns adjusted for the target and downside risk. It is defined as:\n\n:$frac\\left\\{r - t\\right\\}\\left\\{d\\right\\}$\n\nwhere,\n\n\"r\" = the annualized rate of return,\n\n\"t\" = the target return,\n\n\"d\" = downside risk.\n\nThe following table shows that this ratio is demonstrably superior to the traditional Sharpe ratio as a means for ranking investment results. The table shows risk-adjusted ratios for several major indexes using both Sortino and Sharpe ratios. The data cover the five years 1992-1996 and are based on monthly total returns. The Sortino ratio is calculated against a 9.0% target.\n\nAs an example of the different conclusions that can be drawn using these two ratios, notice how the Lehman Aggregate and MSCI EAFE compare - the Lehman ranks higher using the Sharpe ratio whereas EAFE ranks higher using the Sortino ratio. In many cases, manager or index rankings will be different, depending on the risk-adjusted measure used. These patterns will change again for different values of t. For example, when t is close to the risk-free rate, the Sortino Ratio for T-Bill's will be higher than that for the S&P 500, while the Sharpe ratio remains unchanged.\n\nIn March, 2008 researchers at the Queensland, Australia Investment Corporation showed that for skewed return distributions, the Sortino ratio is superior to the Sharpe ratio as a measure of portfolio risk. [http://www.agsm.edu.au/eajm/0803/Paper_5_Chaudhry_Johnson.html]\n\nVolatility skewness\n\nVolatility skewness is another portfolio-analysis statistic introduced by Rom and Ferguson under the PMPT rubric. It measures the ratio of a distribution's percentage of total variance from returns above the mean, to the percentage of the distribution's total variance from returns below the mean. Thus, if a distribution is symmetrical (i.e., normal, as is assumed under MPT), it has a volatility skewness of 1.00. Values greater than 1.00 indicate positive skewness; values less than 1.00 indicate negative skewness. While closely correlated with the traditional statistical measure of skewness (viz., the third moment of a distribution), the authors of PMPT argue that their volatility skewness measure has the advantage of being intuitively more understandable to non-statisticians who are the primary practical users of these tools.\n\nThe importance of skewness lies in the fact that the more non-normal (i.e., skewed) a return series is, the more its true risk will be distorted by traditional MPT measures such as the Sharpe ratio. Thus, with the recent advent of hedging and derivative strategies, which are asymmetrical by design, MPT measures are essentially useless, while PMPT is able to capture significantly more of the true information contained in the returns under consideration. This being said, as the following table shows, many of the common market indices and the returns of stock and bond mutual funds cannot themselves always be assumed to be accurately represented by the normal distribution. This fact is also not well understood by many practitioners.\n\nData: Monthly returns, January, 1991 through December, 1996.\n\nEndnotes\n\nReferences\n\n*For a comprehensive survey of the early literature, see R. Libby and P. Fishburn .\n*Bawa, V.S. \"Stochastic Dominance: A Research Bibliography.\" Management Science, June, 1982.\n*Balzer, L.A. \"Measuring Investment Risk: A Review.\" The Journal of Investing, Fall 1994.\n*Clarkson, R.S. Presentation to the Faculty of Actuaries (British). february 20 1989.\n*Fishburn, P.C. \"Mean-Risk Analysis with Risk Associated with Below-Target Returns.\" American Economic Review, March 1977.\n*Hammond, D.R. \"Risk Management Approaches in Endowment Portfolios in the 1990's\" Journal of Investing, Summer, 1993.\n*Harlow, W.V. \"Asset Allocation in a Downside Risk Framework.\" Financial Analysts Journal, Sept-Oct 1991.\n*\"Investment Review.\" Brinson Partners, Inc. 1992.\n*Kaplan, P. and L. Siegel. \"Portfolio Theory is Alive and Well,\" Journal of Investing, Fall 1994.\n*Lewis, A.L. \"Semivariance and the Performance of Portfolios with Options.\" Financial Analysts Journal, July-August 1990.\n*Leibowitz, M.L. and S. Kogelman. \"Asset Allocation under Shortfall Constraints.\" Salomon Brothers, 1987.\n*Leibowitz, M.L., and T.C. Langeteig. \"Shortfall Risks and the Asset Allocation Decision.\" Journal of Portfolio management, Fall 1989.\n*Libby, R. and P. Fishburn. \"Behavioral Models of Risk taking in Business decisions: A Survey and Evaluation,\" Journal of Accounting Research, Autumn 1977. See also D. Kahneman and A. Tversky, \"Prospect Theory: An Analysis of Decision under Risk,\" Econometrica, March 1979.\n*Post-Modern Portfolio Theory Spawns Post-Modern Optimizer.\" Money Management Letter, February 15 1993.\n*Rom, B. M. and K. Ferguson. \"Post-Modern Portfolio Theory Comes of Age.\" Journal of Investing, Winter 1993.\n*Rom, B. M. and K. Ferguson. \"Portfolio Theory is Alive and Well: A Response.\" Journal of Investing, Fall 1994.\n*Rom, B. M. and K. Ferguson. \"A software developer's view: using Post-Modern Portfolio Theory to improve investment performance measurement.\" Managing downside risk in financial markets: Theory, practice and implementation; Butterworth-Heinemann Finance, 2001; p59.\n*Sharpe, W.F. \"Capital Asset Prices: A Theory of Market Equilibrium under Consideration of Risk.\" Journal of Finance, Vol. XIX (1964)\n*Sortino, F. \"Looking only at return is risky, obscuring real goal.\" Pensions and Investments magazine, November 25 1997.\n*Sortino, F. and H. Forsey \"On the Use and Misuse of Downside Risk.\" The Journal of Portfolio Management, Winter 1996.\n*Sortino, F. and L. Price. \"Performance Measurement in a Downside Risk Framework.\" Journal of Investing, Fall 1994.\n*Sortino, F. and S. Satchell, editors. \"Managing downside risk in financial markets: Theory, practice and implementation\" Butterworth-Heinemann Finance, 2001.\n*Sortino, F. and R. van der Meer. \"Downside Risk: Capturing What's at Stake.\" Journal of Portfolio Management, Summer 1991.\n*\"Why Investors Make the Wrong Choices.\" Fortune Magazine, January 1987.\n\n*http://investmenttechnologies.com/publications.html\n*http://www.fpanet.org/journal/articles/2005_Issues/jfp0905-art7.cfm\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n• Post-Modern Portfolio Theory - PMPT — A portfolio optimization methodology that uses the downside risk of returns instead of the mean variance of investment returns used by modern portfolio theory. The difference lies in each theory s definition of risk, and how that risk influences… …   Investment dictionary\n\n• Modern portfolio theory — Portfolio analysis redirects here. 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The theory was derived from 255 Wall Street Journal editorials written by Charles H. Dow (1851–1902), journalist, founder and first editor of… …   Wikipedia\n\n• Mosaic theory — Mosaic theory, also referred to more colloquially as the scuttlebutt method by Philip Fisher in his seminal work Common Stocks and Uncommon Profits, in finance is the method used in security analysis to gather information about a corporation.… …   Wikipedia\n\n• Architectural theory — is the act of thinking, discussing, or most importantly writing about architecture. Architectural theory is taught in most architecture schools and is practiced by the world s leading architects. Some forms that architecture theory takes are the… …   Wikipedia\n\n• Sharpe ratio — The Sharpe ratio or Sharpe index or Sharpe measure or reward to variability ratio is a measure of the excess return (or risk premium) per unit of deviation in an investment asset or a trading strategy, typically referred to as risk (and is a… …   Wikipedia\n\n• Efficient-market hypothesis — Financial markets Public market Exchange Securities Bond market Fixed income Corporate bond Government bond Municipal bond …   Wikipedia\n\n• Stock market — Financial markets Public market Exchange Securities Bond market Fixed income Corporate bond Government bond Municipal bond …   Wikipedia\n\nWe are using cookies for the best presentation of our site. Continuing to use this site, you agree with this." ]

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[ "APICe » Publications » Wegner99MMI\n\n# Mathematical Models of Interactive Computing\n\nPeter Wegner, Dina Goldin\nFinite computing agents that interact with an environment are shown to be more expressive than Turing machines according to a notion of expressiveness that measures problem-solving ability and is specified by observation equivalence. Sequential interactive models of objects, agents, and embedded systems are shown to be more expressive than algorithms. Multi-agent (distributed) models of coordination, collaboration, and true concurrency are shown to be more expressive than sequential models. The technology shift from algorithms to interaction is expressed by a mathematical paradigm shift that extends inductive definition and reasoning methods for finite agents to coinductive methods of set theory and algebra. An introduction to models of interactive computing is followed by an account of mathematical models of sequential interaction in terms of coinductive methods of non-well-founded set theory, coalgebras, and bisimulation. Models of distributed information flow and multi-agent interaction are developed, and the Turing test is extended to interactive sequential and distributed models of computation. Specification of interactive systems is defined in terms of observable behavior, Godel incompleteness is shown for interaction machines, and explanatory power of physical theories is shown to correspond to expressiveness for models of computation. Our goal is to provide multiple perspective on interactive modeling from the viewpoint of interaction machines and mathematics, to persuade readers that interaction paradigms can play a significant role in narrowing the gap between theoretical models and software practice.\nProvidence, RI, USA, 1999.\n```@techreport{interactivemodels-wegner99," ]

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[ "Solutions by everydaycalculation.com\n\n## Multiply 10/60 with 1/21\n\nThis multiplication involving fractions can also be rephrased as \"What is 10/60 of 1/21?\"\n\n10/60 × 1/21 is 1/126.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 10/60 × 1/21 = 10 × 1/60 × 21 = 10/1260\n3. After reducing the fraction, the answer is 1/126\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "Skip to content\nRelated Articles\nCount numbers in a given range whose count of prime factors is a Prime Number\n• Last Updated : 17 May, 2021\n\nGiven a 2D array Q[][] of size N * 2 representing queries of the form {L, R}. For each query, the task is to print the count of numbers in the range [L, R] with a count of prime factors equal to a prime number.\n\nExamples:\n\nInput: Q[][] = {{4, 8}, {30, 32}}\nOutput: 3 2\nExplanation:\nQuery 1:\nPrime factors of 4 = {2, 2} and count of prime factors = 2\nPrime factors of 5 = {5} and count of prime factors = 1\nPrime factors of 6 = {2, 3} and count of prime factors = 2\nPrime factors of 7 = {7} and count of prime factors = 1\nPrime factors of 8 = {2, 2, 2} and count of prime factors = 3\nTherefore, the total count of numbers in the range [4, 8] having count of prime factors is a prime number is 3.\nQuery 2:\nPrime factors of 30 = {2, 3, 5} and count of prime factors = 3\nPrime factors of 31 = {31} and count of prime factors = 1\nPrime factors of 32 = {2, 2, 2, 2, 2} and count of prime factors = 5\nTherefore, the total count of numbers in the range [4, 8] having count of prime factors is a prime number is 2.\n\nInput: Q[][] = {{7, 12}, {10, 99}}\nOutput: 4\n\nNaive Approach: The simplest approach to solve this problem is to traverse all the numbers in the range [L, R], and for each number, check if the count of prime factors of the number is a prime number or not. If found to be true, increment the counter by 1. After traversing, print the value of counter for each query.\n\nTime Complexity: O(|Q| * (max(arr[i] – arr[i] + 1)) * sqrt(max(arr[i]))\nAuxiliary space: O (1)\n\nEfficient Approach: To optimize the above approach the idea is to precompute the smallest prime factor of each number in the range [Li, Ri] using Sieve of Eratosthenes. Follow the steps below to solve the problem:\n\n• Generate and store the smallest prime factor of each element using Sieve of Eratosthenes.\n• Find the count of prime factors for each number in the range [Li, Ri] using the Sieve.\n• For each number, check if the total count of prime factors is a prime number or not. If found to be true then increment the counter.\n• Create a prefix sum array, say sum[], where sum[i] will store the sum of elements from the range [0, i] whose count of prime factors is a prime number.\n• Finally, for each query, print the value sum[arr[i]] – sum[arr[i] – 1].\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;``#define MAX 1001` `// Function to find the smallest prime factor``// of all the numbers in range [0, MAX]``vector<``int``> sieve()``{` `    ``// Stores smallest prime factor of all``    ``// the numbers in the range [0, MAX]``    ``vector<``int``> spf(MAX);` `    ``// No smallest prime factor of``    ``// 0 and 1 exists``    ``spf = spf = -1;` `    ``// Traverse all the numbers``    ``// in the range [1, MAX]``    ``for` `(``int` `i = 2; i < MAX; i++) {` `        ``// Update spf[i]``        ``spf[i] = i;``    ``}` `    ``// Update all the numbers whose``    ``// smallest prime factor is 2``    ``for` `(``int` `i = 4; i < MAX; i = i + 2) {``        ``spf[i] = 2;``    ``}` `    ``// Traverse all the numbers in``    ``// the range [1, sqrt(MAX)]``    ``for` `(``int` `i = 3; i * i < MAX; i++) {` `        ``// Check if i is a prime number``        ``if` `(spf[i] == i) {` `            ``// Update all the numbers whose``            ``// smallest prime factor is i``            ``for` `(``int` `j = i * i; j < MAX;``                 ``j = j + i) {` `                ``// Check if j is``                ``// a prime number``                ``if` `(spf[j] == j) {``                    ``spf[j] = i;``                ``}``            ``}``        ``}``    ``}``    ``return` `spf;``}` `// Function to find count of``// prime factor of num``int` `countFactors(vector<``int``>& spf, ``int` `num)``{``    ``// Stores count of``    ``// prime factor of num``    ``int` `count = 0;` `    ``// Calculate count of``    ``// prime factor``    ``while` `(num > 1) {` `        ``// Update count``        ``count++;` `        ``// Update num``        ``num = num / spf[num];``    ``}``    ``return` `count;``}` `// Function to precalculate the count of``// numbers in the range [0, i] whose count``// of prime factors is a prime number``vector<``int``> precalculateSum(vector<``int``>& spf)``{` `    ``// Stores the sum of all the numbers``    ``// in the range[0, i] count of``    ``// prime factor is a prime number``    ``vector<``int``> sum(MAX);` `    ``// Update sum``    ``sum = 0;` `    ``// Traverse all the numbers in``    ``// the range [1, MAX]``    ``for` `(``int` `i = 1; i < MAX; i++) {` `        ``// Stores count of prime factor of i``        ``int` `prime_factor``            ``= countFactors(spf, i);` `        ``// If count of prime factor is``        ``// a prime number``        ``if` `(spf[prime_factor] == prime_factor) {` `            ``// Update sum[i]``            ``sum[i] = sum[i - 1] + 1;``        ``}``        ``else` `{` `            ``// Update sum[i]``            ``sum[i] = sum[i - 1];``        ``}``    ``}``    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``// Stores smallest prime factor of all``    ``// the numbers in the range [0, MAX]``    ``vector<``int``> spf = sieve();` `    ``// Stores the sum of all the numbers``    ``// in the range[0, i] count of``    ``// prime factor is a prime number``    ``vector<``int``> sum = precalculateSum(spf);` `    ``int` `Q[] = { { 4, 8 }, { 30, 32 } };` `    ``// int N = sizeof(Q) / sizeof(Q);``    ``for` `(``int` `i = 0; i < 2; i++) {``        ``cout << (sum[Q[i]] - sum[Q[i] - 1])``             ``<< ``\" \"``;``    ``}` `    ``return` `0;``}`\n\n## Java\n\n `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{``    ` `public` `static` `int` `MAX = ``1001``;` `// Function to find the smallest prime factor``// of all the numbers in range [0, MAX]``public` `static` `int``[] sieve()``{``    ` `    ``// Stores smallest prime factor of all``    ``// the numbers in the range [0, MAX]``    ``int` `spf[] = ``new` `int``[MAX];``  ` `    ``// No smallest prime factor of``    ``// 0 and 1 exists``    ``spf[``0``] = spf[``1``] = -``1``;``  ` `    ``// Traverse all the numbers``    ``// in the range [1, MAX]``    ``for``(``int` `i = ``2``; i < MAX; i++)``    ``{``        ` `        ``// Update spf[i]``        ``spf[i] = i;``    ``}``  ` `    ``// Update all the numbers whose``    ``// smallest prime factor is 2``    ``for``(``int` `i = ``4``; i < MAX; i = i + ``2``)``    ``{``        ``spf[i] = ``2``;``    ``}``  ` `    ``// Traverse all the numbers in``    ``// the range [1, sqrt(MAX)]``    ``for``(``int` `i = ``3``; i * i < MAX; i++)``    ``{``        ` `        ``// Check if i is a prime number``        ``if` `(spf[i] == i)``        ``{``            ` `            ``// Update all the numbers whose``            ``// smallest prime factor is i``            ``for``(``int` `j = i * i; j < MAX; j = j + i)``            ``{``                ` `                ``// Check if j is``                ``// a prime number``                ``if` `(spf[j] == j)``                ``{``                    ``spf[j] = i;``                ``}``            ``}``        ``}``    ``}``    ``return` `spf;``}``  ` `// Function to find count of``// prime factor of num``public` `static` `int` `countFactors(``int` `spf[], ``int` `num)``{``    ` `    ``// Stores count of``    ``// prime factor of num``    ``int` `count = ``0``;``  ` `    ``// Calculate count of``    ``// prime factor``    ``while` `(num > ``1``)``    ``{``        ` `        ``// Update count``        ``count++;``  ` `        ``// Update num``        ``num = num / spf[num];``    ``}``    ``return` `count;``}``  ` `// Function to precalculate the count of``// numbers in the range [0, i] whose count``// of prime factors is a prime number``public` `static` `int``[] precalculateSum(``int` `spf[])``{``    ` `    ``// Stores the sum of all the numbers``    ``// in the range[0, i] count of``    ``// prime factor is a prime number``    ``int` `sum[] = ``new` `int``[MAX];``  ` `    ``// Update sum``    ``sum[``0``] = ``0``;``  ` `    ``// Traverse all the numbers in``    ``// the range [1, MAX]``    ``for``(``int` `i = ``1``; i < MAX; i++)``    ``{``        ` `        ``// Stores count of prime factor of i``        ``int` `prime_factor = countFactors(spf, i);``  ` `        ``// If count of prime factor is``        ``// a prime number``        ``if` `(spf[prime_factor] == prime_factor)``        ``{``            ` `            ``// Update sum[i]``            ``sum[i] = sum[i - ``1``] + ``1``;``        ``}``        ``else``        ``{``            ` `            ``// Update sum[i]``            ``sum[i] = sum[i - ``1``];``        ``}``    ``}``    ``return` `sum;``}  ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Stores smallest prime factor of all``    ``// the numbers in the range [0, MAX]``    ``int` `spf[] = sieve();``  ` `    ``// Stores the sum of all the numbers``    ``// in the range[0, i] count of``    ``// prime factor is a prime number``    ``int` `sum[] = precalculateSum(spf);``  ` `    ``int` `Q[][] = { { ``4``, ``8` `}, { ``30``, ``32` `} };``  ` `    ``// int N = sizeof(Q) / sizeof(Q);``    ``for``(``int` `i = ``0``; i < ``2``; i++)``    ``{``        ``System.out.print((sum[Q[i][``1``]] -``                          ``sum[Q[i][``0``] - ``1``]) + ``\" \"``);``    ``}``}``}` `// This code is contributed by divyeshrabadiya07`\n\n## Python3\n\n `# Python3 program to implement``# the above approach``MAX` `=` `1001` `# Function to find the smallest``# prime factor of all the numbers``# in range [0, MAX]``def` `sieve():``    ` `    ``# Stores smallest prime factor of all``    ``# the numbers in the range [0, MAX]``    ``global` `MAX``    ``spf ``=` `[``0``] ``*` `MAX``    ` `    ``# No smallest prime factor of``    ``# 0 and 1 exists``    ``spf[``0``] ``=` `spf[``1``] ``=` `-``1``    ` `    ``# Traverse all the numbers``    ``# in the range [1, MAX]``    ``for` `i ``in` `range``(``2``, ``MAX``):``        ` `        ``# Update spf[i]``        ``spf[i] ``=` `i` `    ``# Update all the numbers whose``    ``# smallest prime factor is 2``    ``for` `i ``in` `range``(``4``, ``MAX``, ``2``):``        ``spf[i] ``=` `2` `    ``# Traverse all the numbers in``    ``# the range [1, sqrt(MAX)]``    ``for` `i ``in` `range``(``3``, ``MAX``):` `        ``# Check if i is a prime number``        ``if` `(spf[i] ``=``=` `i):` `            ``# Update all the numbers whose``            ``# smallest prime factor is i``            ``for` `j ``in` `range``(i ``*` `i, ``MAX``):``                ` `                ``# Check if j is``                ``# a prime number``                ``if` `(spf[j] ``=``=` `j):``                    ``spf[j] ``=` `i` `    ``return` `spf` `# Function to find count of``# prime factor of num``def` `countFactors(spf, num):``    ` `    ``# Stores count of``    ``# prime factor of num``    ``count ``=` `0` `    ``# Calculate count of``    ``# prime factor``    ``while` `(num > ``1``):``        ` `        ``# Update count``        ``count ``+``=` `1` `        ``# Update num``        ``num ``=` `num ``/``/` `spf[num]` `    ``return` `count` `# Function to precalculate the count of``# numbers in the range [0, i] whose count``# of prime factors is a prime number``def` `precalculateSum(spf):``    ` `    ``# Stores the sum of all the numbers``    ``# in the range[0, i] count of``    ``# prime factor is a prime number``    ``sum` `=` `[``0``] ``*` `MAX` `    ``# Traverse all the numbers in``    ``# the range [1, MAX]``    ``for` `i ``in` `range``(``1``, ``MAX``):``        ` `        ``# Stores count of prime factor of i``        ``prime_factor ``=` `countFactors(spf, i)` `        ``# If count of prime factor is``        ``# a prime number``        ``if` `(spf[prime_factor] ``=``=` `prime_factor):` `            ``# Update sum[i]``            ``sum``[i] ``=` `sum``[i ``-` `1``] ``+` `1``        ``else``:` `            ``# Update sum[i]``            ``sum``[i] ``=` `sum``[i ``-` `1``]``            ` `    ``return` `sum` `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Stores smallest prime factor of all``    ``# the numbers in the range [0, MAX]``    ``spf ``=` `sieve()` `    ``# Stores the sum of all the numbers``    ``# in the range[0, i] count of``    ``# prime factor is a prime number``    ``sum` `=` `precalculateSum(spf)` `    ``Q ``=` `[ [ ``4``, ``8` `], [ ``30``, ``32` `] ]``    ``sum``[Q[``0``][``1``]] ``+``=` `1``    ` `    ``# N = sizeof(Q) / sizeof(Q);``    ``for` `i ``in` `range``(``0``, ``2``):``        ``print``((``sum``[Q[i][``1``]] ``-``               ``sum``[Q[i][``0``]]), end ``=` `\" \"``)` `# This code is contributed by Princi Singh`\n\n## C#\n\n `// C# program to implement``// the above approach``using` `System;``class` `GFG{``    ` `public` `static` `int` `MAX = 1001;` `// Function to find the smallest``// prime factor of all the numbers``// in range [0, MAX]``public` `static` `int``[] sieve()``{    ``  ``// Stores smallest prime factor``  ``// of all the numbers in the``  ``// range [0, MAX]``  ``int` `[]spf = ``new` `int``[MAX];` `  ``// No smallest prime factor``  ``// of 0 and 1 exists``  ``spf = spf = -1;` `  ``// Traverse all the numbers``  ``// in the range [1, MAX]``  ``for``(``int` `i = 2; i < MAX; i++)``  ``{``    ``// Update spf[i]``    ``spf[i] = i;``  ``}` `  ``// Update all the numbers whose``  ``// smallest prime factor is 2``  ``for``(``int` `i = 4; i < MAX; i = i + 2)``  ``{``    ``spf[i] = 2;``  ``}` `  ``// Traverse all the numbers in``  ``// the range [1, sqrt(MAX)]``  ``for``(``int` `i = 3; i * i < MAX; i++)``  ``{``    ``// Check if i is a prime number``    ``if` `(spf[i] == i)``    ``{``      ``// Update all the numbers``      ``// whose smallest prime``      ``// factor is i``      ``for``(``int` `j = i * i;``              ``j < MAX; j = j + i)``      ``{``        ``// Check if j is``        ``// a prime number``        ``if` `(spf[j] == j)``        ``{``          ``spf[j] = i;``        ``}``      ``}``    ``}``  ``}``  ``return` `spf;``}``  ` `// Function to find count of``// prime factor of num``public` `static` `int` `countFactors(``int` `[]spf,``                               ``int` `num)``{    ``  ``// Stores count of``  ``// prime factor of num``  ``int` `count = 0;` `  ``// Calculate count of``  ``// prime factor``  ``while` `(num > 1)``  ``{``    ``// Update count``    ``count++;` `    ``// Update num``    ``num = num / spf[num];``  ``}``  ``return` `count;``}``  ` `// Function to precalculate the count of``// numbers in the range [0, i] whose count``// of prime factors is a prime number``public` `static` `int``[] precalculateSum(``int` `[]spf)``{    ``  ``// Stores the sum of all the numbers``  ``// in the range[0, i] count of``  ``// prime factor is a prime number``  ``int` `[]sum = ``new` `int``[MAX];` `  ``// Update sum``  ``sum = 0;` `  ``// Traverse all the numbers in``  ``// the range [1, MAX]``  ``for``(``int` `i = 1; i < MAX; i++)``  ``{` `    ``// Stores count of prime factor of i``    ``int` `prime_factor = countFactors(spf, i);` `    ``// If count of prime factor is``    ``// a prime number``    ``if` `(spf[prime_factor] == prime_factor)``    ``{` `      ``// Update sum[i]``      ``sum[i] = sum[i - 1] + 1;``    ``}``    ``else``    ``{` `      ``// Update sum[i]``      ``sum[i] = sum[i - 1];``    ``}``  ``}``  ``return` `sum;``}  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``  ``// Stores smallest prime factor``  ``// of all the numbers in the``  ``// range [0, MAX]``  ``int` `[]spf = sieve();` `  ``// Stores the sum of all the``  ``// numbers in the range[0, i]``  ``// count of prime factor is a``  ``// prime number``  ``int` `[]sum = precalculateSum(spf);` `  ``int` `[,]Q = {{4, 8}, {30, 32}};` `  ``// int N = sizeof(Q) / sizeof(Q);``  ``for``(``int` `i = 0; i < 2; i++)``  ``{``    ``Console.Write((sum[Q[i, 1]] -``                   ``sum[Q[i, 0] - 1]) +``                   ``\" \"``);``  ``}``}``}` `// This code is contributed by shikhasingrajput`\n\n## Javascript\n\n ``\nOutput:\n`3 2`\n\nTime Complexity: O(|Q| + (MAX *log(log(MAX))))\nAuxiliary Space: O(MAX)\n\nAttention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.\n\nMy Personal Notes arrow_drop_up" ]

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[ "arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.\n\n# Performance of Polar Codes for Channel and Source Coding\n\n\\authorblockNNadine Hussami \\authorblockA AUB, Lebanon, Email:    \\authorblockNSatish Babu Korada and Rüdiger Urbanke \\authorblockA EPFL, Switzerland, Email:\n###### Abstract\n\nPolar codes, introduced recently by Arıkan, are the first family of codes known to achieve capacity of symmetric channels using a low complexity successive cancellation decoder. Although these codes, combined with successive cancellation, are optimal in this respect, their finite-length performance is not record breaking. We discuss several techniques through which their finite-length performance can be improved. We also study the performance of these codes in the context of source coding, both lossless and lossy, in the single-user context as well as for distributed applications.\n\n## I Introduction\n\nPolar codes, recently introduced by Arıkan in , are the first provably capacity achieving family of codes for arbitrary symmetric binary-input discrete memoryless channels (B-DMC) with low encoding and decoding complexity. The construction of polar codes is based on the following observation: Let . Apply the transform (where “” denotes the Kronecker power) to a block of bits and transmit the output through independent copies of a symmetric B-DMC, call it . As grows large, the channels seen by individual bits (suitably defined in ) start polarizing: they approach either a noiseless channel or a pure-noise channel, where the fraction of channels becoming noiseless is close to the capacity . In the following, let denote the vector and denote the vector . Let be the permutation such that if the -bit binary representation of is , then (we call this a bit-reversal). Let denote the number of ones in the binary expansion of .\n\n### I-a Construction of Polar Codes\n\nThe channel polarization phenomenon suggests to use the noiseless channels for transmitting information while fixing the symbols transmitted through the noisy ones to a value known both to sender as well as receiver. For symmetric channels we can assume without loss of generality that the fixed positions are set to . Since the fraction of channels becoming noiseless tends to , this scheme achieves the capacity of the channel.\n\nIn the following alternative interpretation was mentioned; the above procedure can be seen as transmitting a codeword and decoding at the receiver with a successive cancellation (SC) decoding strategy. The specific code which is constructed can be seen as a generalization of the Reed-Muller (RM) codes. Let us briefly discuss the construction of RM codes. We follow the lead of in which the Kronecker product is used. RM codes are specified by the two parameters and and the code is denoted by . An RM() code has block length and rate . The code is defined through its generator matrix as follows. Compute the Kronecker product . This gives a matrix. Label the rows of this matrix as . One can check that the weight of the th row of this matrix is equal to . The generator matrix of the code consists of all the rows of which have weight at least . There are exactly such rows. An equivalent way of expressing this is to say that the codewords are of the form , where the components of corresponding to the rows of of weight less than are fixed to and the remaining components contain the “information.” Polar codes differ from RM codes only in the choice of generator vectors , i.e., in the choice of which components of are set to . Unlike RM codes, these codes are defined for any dimension . The choice of the generator vectors, as explained in , is rather complicated; we therefore do not discuss it here. Following Arıkan, call those components of which are set to “frozen,” and the remaining ones “information” bits. Let the set of frozen bits be denoted by and the set of information bits be denoted by . A polar code is then defined as the set of codewords of the form , where the bits are fixed to .\n\n###### Lemma 1 (RM Code as Polar Code)\n\nA is a polar code of length with .\n\n### I-B Performance under Successive Cancellation Decoding\n\nIn Arıkan considers a low complexity SC decoder. We briefly describe the decoding procedure here. The bits are decoded in the order . Let the estimates of the bits be denoted by . If a bit is frozen then . Otherwise the decoding rule is the following:\n\n ^uπ(i)=⎧⎪ ⎪⎨⎪ ⎪⎩0,if Pr(yN−10|^uπ(i−1)π(0),Uπ(i)=0)Pr(yN−10|^uπ(i−1)π(0),Uπ(i)=1)>1,1,otherwise.\n\nUsing the factor graph representation between and shown in Figure 2(a), Arıkan showed that this decoder can be implemented with complexity.\n\n###### Theorem 2\n\n Let be any symmetric B-DMC with . Let and be fixed. Then for , , the probability of error for polar coding under SC decoding at block length and rate satisfies\n\n### I-C Optimality of the Exponent\n\nThe following lemma characterizes the minimum distance of a polar code.\n\n###### Lemma 3 (Minimum Distance of Polar Codes)\n\nLet be the set of information bits of a polar code . The minimum distance of the code is given by\n\n{proof}\n\nLet . Clearly, cannot be larger than the minimum weight of the rows of the generator matrix. Therefore, . On the other hand, by adding some extra rows to the generator matrix we cannot increase the minimum distance. In particular, add all the rows of with weight at least . The resulting code is . It is well known that . Therefore, We conclude that for any given rate , if the information bits are picked according to their weight (RM rule), i.e., the vectors of largest weight, the resulting code has the largest possible minimum distance. The following lemma gives a bound on the best possible minimum distance for any non-trivial rate.\n\n###### Lemma 4\n\nFor any rate and any choice of information bits, the minimum distance of a code of length is bounded as for and a constant .\n\n{proof}\n\nLemma 3 implies that is maximized by choosing the frozen bits according to the RM rule. The matrix has rows of weight . Therefore, For , more than half of the rows are in the generator matrix. Therefore, there is at least one row with weight less than or equal to . Consider therefore an in the range . Using Stirling’s approximation, one can show that for any at least one row of the generator matrix has weight in the range . Using Lemma 3 we conclude the result.\n\n###### Theorem 5\n\nLet and be fixed. For any symmetric B-DMC , and , , the probability of error for polar coding under MAP decoding at block length and rate satisfies\n\n{proof}\n\nFor a code with minimum distance , the block error probability is lower bounded by for some positive constant , which only depends on the channel. This is easily seen by considering a genie decoder; the genie provides the correct value of all bits except those which differ between the actually transmitted codeword and its minimum distance cousin. Lemma 4 implies that for any , for large enough, for any . Therefore for any .\n\nThis combined with Theorem 2, implies that the SC decoder achieves performance comparable to the MAP decoder in terms of the order of the exponent.\n\n## Ii Performance under Belief Propagation\n\nTheorem 2 does not state what lengths are needed in order to achieve the promised rapid decay in the error probability nor does it specify the involved constants. Indeed, for moderate lengths polar codes under SC decoding are not record breaking. In this section we show various ways to improve the performance of polar codes by considering belief propagation (BP) decoding. BP was already used in to compare the performance of polar codes based on Arıkan’s rule and RM rule. For all the simulation points in the plots the confidence intervals are shown. In most cases these confidence intervals are smaller than the point size and are therefore not visible.\n\n### Ii-a Successive Decoding as a Particular Instance of BP\n\nFor communication over a binary erasure channel (BEC) one can easily show the following.\n\n###### Lemma 6\n\nDecoding the bit with the SC decoder is equivalent to applying BP with the knowledge of and all other bits unknown (and a uniform prior on them).\n\nWe conclude that if we use a standard BP algorithm (such a decoder has access also to the information provided by the frozen bits belonging to ) then it is in general strictly better than a SC decoder. Indeed, it is not hard to construct explicit examples of codewords and erasure patterns to see that this inclusion is strict (BP decoder succeeds but the SC decoder does not). Figure 3 shows the simulation results for the SC, BP and the MAP decoders when transmission takes place over the BEC. As we can see from these simulation results, the performance of the BP decoder lies roughly half way between that of the SC decoder and that of the MAP decoder. For the BEC the scheduling of the individual messages is irrelevant to the performance as long as each edge is updated until a fixed point has been reached. For general B-DMCs the performance relies heavily on the specific schedule. We found empirically that a good performance can be achieved by the following schedule. Update the messages of each of the sections of the trellis from right to left and then from left to right and so on. Each section consists of a collection of shaped sub-graphs. We first update the lower horizontal edge, then the diagonal edge, and, finally, the upper horizontal edge of each of these sections. In this schedule the information is spread from the variables belonging to one level to its neighboring level. Figure 1 shows the simulation results for the SC decoder and the BP decoder over the binary input additive white Gaussian noise channel (BAWGNC) of capacity . Again, we can see a marked improvement of the BP decoder over the SC decoder.", null, "Fig. 1: Comparison of (i) SC and (ii) BP decoder in terms of block error probability, when transmission takes place over the BAWGNC(σ=0.97865). The performance curves are shown for n=10 (top left), 11 (top right), 12 (bottom left), and 13 (bottom right).\n\n### Ii-B Overcomplete Representation: Redundant Trellises\n\nFor the polar code of length one can check that all the three trellises shown in Figure 2 are valid representations. In fact, for a code of block length , there exist different representations obtained by different permutations of the layers of connections. Therefore, we can connect the vectors and with any number of these representations and this results in an overcomplete representation (similar to the concept used when computing the stopping redundancy of a code ). For the BEC any such overcomplete representation only improves the performance of the BP decoder . Further, the decoding complexity scales linearly with the number of different representations used. Keeping the complexity in mind, instead of considering all the factorial trellises, we use only the trellises obtained by cyclic shifts (e.g., see Figure 2). The complexity of this algorithm is as compared to of the SC decoder and BP over one trellis. The performance of the BP decoder is improved significantly by using this overcomplete representation as shown in Figure 3.", null, "Fig. 2: Figure (a) shows the trellis representation used by Arıkan. Figures (b) and (c) show equivalent representations obtained by cyclic shifts of the 3 sections of trellis (a).", null, "Fig. 3: Comparison of (i) SC, (ii) BP, (iii) BP with multiple trellises, and (iv) MAP in terms of block error probability, when transmission takes place over the BEC(ϵ=12). The performance curves are shown for n=10 (top left), 11 (top right), 12 (bottom left), 13 (bottom right).\n\nWe leave a systematic investigation of good schedules and choices of overcomplete representations for general symmetric channels as an interesting open problem.\n\n### Ii-C Choice of Frozen Bits\n\nFor the BP or MAP decoding algorithm the choice of frozen bits as given by Arıkan is not necessarily optimal. In the case of MAP decoding we observe (see Figure 4) that the performance is significantly improved by picking the frozen bits according to the RM rule. This is not a coincidence; is maximized for this choice. This suggests that there might be a rule which is optimized for BP. It is an interesting open question to find such a rule.", null, "Fig. 4: Comparison of block error probability curves under MAP decoding between codes picked according to Arıkan’s rule and RM rule. The performance curves are shown for n=10 (left) and 11 (right).\n\n## Iii Source Coding\n\nIn this section we show the performance of polar codes in the context of source coding. We consider both lossless and lossy cases and show (in most cases) empirically that they achieve the optimal performance in both cases. Let Ber denote a Bernoulli source with . Let denote the binary entropy function and its inverse.\n\n### Iii-a Lossless Source Coding\n\n#### Iii-A1 Single User\n\nThe problem of lossless source coding of a Ber source can be mapped to the channel coding problem over a binary symmetric channel (BSC) as shown in [6, 7]. Let be a sequence of i.i.d. realizations of the source. Consider a code of rate represented by the parity check matrix . The vector is encoded by its syndrome . The rate of the resulting source code is . The decoding problem is to estimate given the syndrome . This is equivalent to estimating a noise vector in the context of channel coding over BSC. Therefore, if a sequence of codes achieve capacity over BSC(), then the corresponding source codes approach a rate with vanishing error probability.\n\nWe conclude that polar codes achieve the Shannon bound for lossless compression of a binary memoryless source. Moreover, using the trellis of Figure 2(a), we can compute the syndrome with complexity . The source coding problem has a considerable advantage compared to the channel coding problem. The encoder knows the information seen by the decoder (unlike channel coding there is no noise involved here). Therefore, the encoder can also decode and check whether the decoding is successful or not. In case of failure, the encoder can retry the compression procedure by using a permutation of the source vector. This permutation is fixed a priori and is known both to the encoder as well as the decoder. In order to completely specify the system, the encoder must inform the decoder which permutation was finally used. This results in a small loss of rate but it brings down the probability of decoding failure. Note that the extra number of bits that need to be transmitted grows only logarithmically with the number of permutations used, but that the error probability decays exponentially as long as the various permuted source vectors look like independent source samples. With this trick one can make the curves essentially arbitrarily steep with a very small loss in rate (see Figure 5).", null, "Fig. 5: Comparison of SC decoding for a Ber(0.11) source with 0,1, and 2 bits for permutations. The performance curves are shown for n=10 (left) and 11 (right). By increasing the number of permutations the curves can be made steeper and steeper.\n\n#### Iii-A2 Slepian-Wolf\n\nConsider two Ber sources and . Assume that they are correlated as , where Ber. Recall that the Slepian-Wolf rate region is the unbounded polytope described by , , . The points and are the so-called corner points. Because of symmetry it suffices to show how to achieve one such corner point (say ).\n\nLet and denote i.i.d. realizations of the two sources. The scheme using linear codes is the following: The encoder for transmits as it is (since , and so no compression is necessary). Let denote the parity-check matrix of a code designed for communication over the BSC. The encoder for computes the syndrome and transmits it to the receiver. At the receiver we know and , therefore we can compute . The resulting problem of estimating is equivalent to the lossless compression of a Ber discussed in the previous section. Therefore, once again polar codes provide an efficient solution. The error probability curves under SC decoding are equal to the curves shown in the Figure 5 with bits for permutations.\n\n### Iii-B Lossy Source Coding\n\n#### Iii-B1 Single User\n\nWe do not know of a mapping that converts the lossy source coding problem to a channel coding problem. However, for the binary erasure source considered in , it was shown how to construct a “good” source code from a “good” (over the BEC) channel code. We briefly describe their construction here and show that polar codes achieve the optimal rate for zero distortion.\n\nThe source is a sequence of i.i.d. realizations of a random variable taking values in with . The reconstruction alphabet is and the distortion function is given by For zero distortion, the rate of the rate-distortion function is given by . In it was shown that the dual of a sequence of channel codes which achieve the capacity of the BEC() under BP decoding, achieve the rate-distortion pair for zero distortion using a message passing algorithm which they refer to as the erasure quantization algorithm. Polar codes achieve capacity under SC decoding. For communication over BEC, the performance under BP is at least as good as SC. Therefore, the dual of the polar codes designed for BEC achieve the optimum rate for zero distortion using the erasure quantization algorithm. Here, we show that the dual polar codes achieve the optimum rate for zero distortion even under a suitably defined SC decoding algorithm.\n\nThe dual of a polar code is obtained by reversing the roles of the check and variable nodes of the trellis in Figure 2(a) and reversing the roles of the frozen and free bits. It is easy to see that . This implies that the dual of a polar code is also a polar code. The suitably defined algorithm is given by SC decoding in the order , opposite of the original decoding order. We refer to this as the dual order.\n\nIn , the probability of erasure for bit under SC decoding is given by , computed as follows. Let the -bit binary expansion of be given by . Let . The sequence is recursively defined as follows:\n\n Zk={Z2k−1, if bk−1=1,1−(1−Zk−1)2, if bk−1=0. (1)\n\nFrom we know that for any , there exists an such that for we can find a set of size satisfying for any . The set is used as information bits. The complement of denoted by , is the set of frozen bits.\n\nLet denote the probability of erasure for bit of the dual code used for BEC. One can check that for the dual code with the dual order, the bit is equivalent to the bit of the original code with the original decoding order. Let . The recursive computation for is given by\n\n ˆZk={ˆZ2k−1, if bk−1=0,1−(1−ˆZk−1)2, if bk−1=1. (2)\n###### Lemma 7 (Duality for BEC)\n\n.\n\nThe proof follows through induction on and using the equations and . For the dual code the set (information set for BEC) is used as frozen bits and the set is used as information bits. Let be a sequence source realizations. The source vector needs to be mapped to a vector (information bits) such that . The following lemma shows that such a vector can be found using SC decoder with vanishing error probability.\n\n###### Lemma 8\n\nThe probability of encoding failure for erasure quantization of the source using the dual of the polar code designed for the BEC and SC decoding with dual order is bounded as for any .\n\n{proof}\n\nLet and be as defined above. The bits belonging to are already fixed to whereas the bits belonging to are free to be set. Therefore an error can only occur if one of the bits belonging to are set to a wrong value. However, if the SC decoding results in an erasure for these bits, these bits are also free to be set any value and this results in no error. Therefore the probability of error can be upper bounded by the probability that at least one of the bits in is not an erasure which in turn can be upper bounded by , where the last equality follows from Theorem 2.\n\nThe fact that the dual code can be successfully applied to the erasure source, suggests to extend this construction to more general sources. Let us try a similar construction to encode the Ber() source with the Hamming distortion. To design a source code for distortion we first design a rate polar code for the BSC where . The design consists of choosing the generator vectors from the rows of as explained in . The source code is then defined by the corresponding dual code with the dual decoding order. Since the rate of the original code is , the rate of the dual code is . Figure 6 shows the rate-distortion performance of these codes for various lengths. As the lengths increase, empirically we observe that the performance approaches the rate-distortion curve.", null, "Fig. 6: (left) Rate-Distortion curves for a Ber(0.11) source. The performance curves are shown for n=9,11,13,15,17.(right) The dashed line is the curve RWZ(D). The solid line is the lower convex envelope of RWZ(D) and (0,0.3) for a Ber(12) source with receiver having a side information obtained through a BSC(0.3). The performance curves are shown for n=9,11,13,15.\n\nIt is an interesting and challenging open problem to prove this observation rigorously.\n\n#### Iii-B2 Wyner-Ziv\n\nLet denote a Ber source which we want to compress. The source is reconstructed at a receiver which has access to a side information correlated to as with Ber. Let and denote a sequence of i.i.d. realizations of and . Wyner and Ziv have shown that the rate distortion curve is given by the lower convex envelope of the curve and the point , where . As discussed in , nested linear codes are required to tackle this problem. The idea is to partition the codewords of a code into cosets of another code . The code must be a good source code and the code must be a good channel code.\n\nUsing polar codes, we can create this nested structure as follows. Let be a source code for distortion (Bernoulli source, Hamming distortion) as described in the previous section. Let be the set of frozen (fixed to ) bits for this source code. Using this code we quantize the source vector . Let the resulting vector be . Note that the vector is given by , where is such that for and for , is defined by the source quantization.\n\nFor a sequence of codes achieving the rate-distortion bound, it was shown in that the “noise” added due to the quantization is comparable to a Bernoulli noise Ber. Therefore, for all practical purposes, the side information at the receiver is statistically equivalent to the output of transmitted through a BSC. Let be the set of frozen bits of a channel code for the BSC. Let the encoder transmit the bits to the receiver. Since the bits belonging to () are fixed to zero, the receiver now has access to and . The definition of implies that the receiver is now able to decode (and hence ) with vanishing error probability. To see the nested structure, note that the code and its cosets are the different channel codes defined by different values of and that these codes partition the source code . The capacity achieving property of polar codes implies that for sufficiently large, for any . In addition, if we assume that polar codes achieve the rate-distortion bound as conjectured in the previous section and , then for sufficiently large, the rate required to achieve a distortion is for any . This would show that polar codes can be used to efficiently realize the Wyner-Ziv scheme. The performance of the above scheme for various lengths is shown in Figure 6.\n\n## References\n\n• E. Arıkan, “Channel polarization: A method for constructing capacity-achieving codes for symmetric binary-input memoryless channels,” submitted to IEEE Trans. Inform. Theory, 2008.\n• G. D. Forney, Jr., “Codes on graphs: Normal realizations,” IEEE Trans. Inform. Theory, vol. 47, no. 2, pp. 520–548, Feb. 2001.\n• E. Arıkan and E. Telatar, “On the rate of channel polarization,” July 2008, available from “http://arxiv.org/pdf/0807.3917”.\n• E. Arıkan, “A performance comparison of Polar codes and Reed-Muller codes,” IEEE Communications Letters, vol. 12, no. 6, 2008.\n• M. Schwartz and A. Vardy, “On the stopping distance and the stopping redundancy of codes,” IEEE Trans. Inform. Theory, vol. 52, no. 3, pp. 922 – 932, Mar. 2006.\n• P. E. Allard and E. W. Bridgewater, “A source encoding technique using algebraic codes,” June 1972, pp. 201–203.\n• E. Weiss, “Compression and coding,” IRE Trans. Inform. Theory, pp. 256–257, 1962.\n• E. Martinian and J. Yedidia, “Iterative quantization using codes on graphs,” in Proc. of the Allerton Conf. on Commun., Control, and Computing, Monticello, IL, USA, 2003.\n• R. Zamir, S. Shamai, and U. Erez, “Nested linear/lattice codes for structured multiterminal binning,” IEEE Transactions on Information Theory, vol. 48, no. 6, pp. 1250–1216, 2002.\n• U. Erez and R. Zamir, “Bounds on the -covering radius of linear codes with applications to self noise in nested Wyner-Ziv coding,” Feb. 2002, dept. Elec. Eng-Syst., Tel-Aviv Univ., Tech. Rep." ]

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[ "#", null, "Is $\\overline{\\psi_{L}^{c}}\\psi_{R}^{c}=\\overline{\\psi_{R}}\\psi_{L}$ true for two different spin 1/2 fermions?\n\n+ 4 like - 0 dislike\n258 views\n\nIn the context of seesaw mechanism or Dirac and Majorana mass terms, one often see the following identity $$\\overline{\\psi_{L}^{c}}\\psi_{R}^{c}=\\overline{\\psi_{R}}\\psi_{L}.$$\n\nHere, I am using 4 components notation in the chiral basis. The convention for the charge conjugation is $\\psi^{c}=-i\\gamma^{2}\\psi^{*}$, and $\\psi_{L}^{c}=\\left(\\psi_{L}\\right)^{c}$. The following is my effort of proving it. $$\\overline{\\psi_{L}^{c}}\\psi_{R}^{c}=\\overline{i\\gamma^{2}\\psi_{L}^{*}}i\\gamma^{2}\\psi_{R}^{*}=\\left(i\\gamma^{2}\\psi_{L}^{*}\\right)^{+}\\gamma^{0}i\\gamma^{2}\\psi_{R}^{*}=\\psi_{L}^{T}i\\gamma^{2}\\gamma^{0}i\\gamma^{2}\\psi_{R}^{*}$$ $$=-\\psi_{L}^{T}\\gamma^{2}\\gamma^{0}\\gamma^{2}\\psi_{R}^{*}=\\psi_{L}^{T}\\gamma^{2}\\gamma^{2}\\gamma^{0}\\psi_{R}^{*}=-\\psi_{L}^{T}\\gamma^{0}\\psi_{R}^{*}=-\\psi_{L,i}\\gamma_{ij}^{0}\\psi_{R,j}^{*}.$$ Now if $\\psi_{L,i}$ and $\\psi_{R,i}$ are anticommuting, then one have $$-\\psi_{L,i}\\gamma_{ij}^{0}\\psi_{R,j}^{*}=\\psi_{R,j}^{*}\\gamma_{ji}^{0}\\psi_{L,i}=\\overline{\\psi_{R}}\\psi_{L}.$$ Question:\n\nIs the anticommuting assumption still true if $\\psi_{R}$ and $\\psi_{L}$ are two different species of fermion? (For example, $\\psi_{L}=\\chi_{L}$)\n\nDo we assume any two fermions are anticommuting even if they are two different fields in QFT?\n\nThis post imported from StackExchange Physics at 2014-05-04 11:37 (UCT), posted by SE-user Louis Yang\nWhat do you mean by $\\psi_L=\\chi_L$? Two different fermionic fields always commute. Here, $\\psi_R$ and $\\psi_L$ are just different projections of the same field.\nBy $\\psi_{L}=\\chi_{L}$, I mean to replace all the $\\psi_{L}$ by $\\chi_{L}$ in the above derivation. So now $\\chi_{L}$ and $\\psi_{R}$ are two different Weyl spinors. They are not just the projection of a single Dirac fermion. I guess your answer is that if $\\chi_{L}$ and $\\psi_{R}$ are two different fields, then they commute. So we will have $\\overline{\\chi_{L}^{c}}\\psi_{R}^{c}=-\\overline{\\psi_{R}}\\chi_{L}$ instead of $\\overline{\\chi_{L}^{c}}\\psi_{R}^{c}=\\overline{\\psi_{R}}\\chi_{L}$.\n Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the \"link\" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\\varnothing$ in the following word:p$\\hbar\\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register." ]

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[ "# Write a C++ program that determines a student’s grade\n\nWrite a C++ program that determines a student’s grade.The program will read three types of scores (quiz, mid-term, and final scores) and determine the grade based on the following rules:\n\nif the average score =90% =>grade=A\n-if the average score >= 70% and <90% => grade=B\n-if the average score>=50% and <70% =>grade=C\n\n## Flowchart of a program that determines a student’s grade", null, "Prof.Fazal Rehman Shamil (Available for Professional Discussions)" ]

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[ "CE Calculators > Deflection and Slope Calculator > Uniform load on full span\n\nSlope and Deflection Calculator for Simply Supported Beam with uniform load on full span", null, "This calculator is for finding the slope and deflection at a section of simply supported beam subjected to uniformly distributed load (UDL) on full span. This calculator uses standard formulae for slope and deflection. Loads acting downward are taken as negative whereas upward loads are taken as positive. Distance 'x' of the section is measured from origin taken at support A. Downward deflection is negative whereas upward deflection is taken as positive. Counter-clockwise rotations are taken as positive whereas clockwise rotations are taken as negative.\n\nYou can visit instructions for slope and deflection Calculator if you need more help and guidance on sign convention adopted.\n\nPlease refer to the figure and enter the required values in the form given below and then click \"Calculate\"\n\nINPUT VALUES\nSpan of beam 'L' (m):\nUDL 'w' (kN/m):\nDistance 'x' (m):\nModulus of Elasticity 'E' (GPa):\nMoment of Inertia 'I' (cm^4):\nall the distances are positive\n\nOUTPUT RESULTS\nDeflection at X (mm):\nMaximum Deflection (mm):\nPoint of Max. Deflection (m):\n\nOther Calculator for slope and deflection of simple supported beam\n\nStress Transformation Calculator\nCalculate Principal Stress, Maximum shear stress and the their planes\n\nTo determine Absolute Max. B.M. due to moving loads.\n\nBending Moment Calculator\nCalculate bending moment & shear force for simply supported beam\n\nMoment of Inertia Calculator\nCalculate moment of inertia of plane sections e.g. channel, angle, tee etc.\n\nShear Stress Calculator\nCalculate Transverse Shear Stress for beam sections e.g. channel, angle, tee etc.\n\nReinforced Concrete Calculator\nCalculate the strength of Reinforced concrete beam\n\nDeflection & Slope Calculator\nCalculate deflection and slope of simply supported beam for many load cases\n\nFixed Beam Calculator\nCalculation tool for beanding moment and shear force for Fixed Beam for many load cases\n\nBM & SF Calculator for Cantilever\nCalculate SF & BM for Cantilever\n\nDeflection & Slope Calculator for Cantilever\nFor many load cases of Cantilever\n\nOverhanging beam calculator\nFor SF & BM of many load cases of overhanging beam\n\nCivil Engineering Quiz\nTest your knowledge on different topics of Civil Engineering\n\nStatically Indeterminate Structures\nDefinition and methods of solving\n\nSolved Examples\n\nTruss Member Forces calculation\nusing method of joints and method of sections\n\nShear force and bending moment\nIllustrated solved examples to draw shear force and bending moment diagrams\n\nSlope and deflection of beam and Truss\nIllustrated solved examples to determine slope and deflection of beam and truss\n\nSolution of indeterminate structures\nslope deflection, moment distribution etc.\n\nReinforced concrete beam\nSolved examples to determine the strength and other parameters" ]

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[ "# Remainder Theorem MCQs & Quiz Online PDF Book Download\n\nRemainder theorem MCQs, remainder theorem quiz answers to learn math courses online. Quadratic equations multiple choice questions (MCQs), remainder theorem quiz questions and answers for online bachelor degree. Fourth root of unity, polynomial function, solution of a quadratic equations, remainder theorem test prep for online certifications.\n\nPractice quadratic equations test MCQs: if x4 - 3x + 5 is divided by 2x - 1, then remainder is, with choices 3516, −3516, −9, and 3 for online bachelor degree. Practice assessment test for scholarships, online learning remainder theorem quiz questions for competitive assessment in math major.\n\n## MCQ on Remainder TheoremQuiz Book Download\n\nMCQ: If x4 - 3x + 5 is divided by 2x - 1, then remainder is\n\n1. 3516\n2. −3516\n3. −9\n4. 3\n\nC\n\nMCQ: If x² - 7x + a has a remainder 1 when divided by x + 1, then\n\n1. a = -7\n2. a = 7\n3. a = 0\n4. a = 1\n\nA\n\nMCQ: If x³ + 9x +5 is divided by x, then remainder is\n\n1. 9\n2. 0\n3. −9\n4. 5\n\nD\n\nMCQ: If x² +ax +b is divided by x + c, then remainder is\n\n1. c ² - ac - b\n2. c ² - ac + b\n3. c ² + ac + b\n4. −c² - ac + b\n\nA\n\nMCQ: If x³+ 9 x + 5 is divided by x, then remainder is\n\n1. 9\n2. 0\n3. −9\n4. 5\n\nD" ]

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[ "# An Introduction to AC/DC Circuit Theory\n\n### By Steven Roman\n\n#### Coming later this year\n\nAn Introduction to AC/DC Circuit Theory is a concise (about 150 pages) but elementary treatment of AC and DC circuit theory that can be used for self-study or to supplement a textbook in the subject.\n\n• Electric Charge and Electric Field\n• Electric Charge, Conductors and Insulators\n• Coulomb's Law\n• Electric Field\n• Electric Field Lines\n• Electric Flux\n• Gauss's Law\n• Electric Potential Energy\n• Electrostatic Potential Energy and Work\n• Current, Resistance and Electromotive Force\n• Series and Parallel Circuits\n• The Lumped Element Model\n• Current and Current Density\n• Current Through a Conductor\n• Current Density and Conductivity\n• Electromotive Force\n• Ohm's Law\n• Resistance\n• Direct and Alternating Current\n• Harmonics\n• Average and Root-Mean-Square Values\n• Instantaeous Values\n• Power of a Circuit Element: Active and Passive Components\n• Overview of Circuit Elements\n• Resistors\n• Capacitors\n• Inductors\n• Power and Circuit Elements\n• Switches\n• Resistance and Kirchhoff's Laws\n• Resistance and Resistors\n• Kirchhoff's Laws\n• Equivalent Resistance\n• Application: Voltage Dividers\n• Current Sources\n• Complex Numbers and Phasors\n• Complex Numbers\n• Phasors\n• Capacitors and Capacitance\n• Capacitors and Capacitance\n• How Current Flows in a Capacitive Circuit\n• Voltage and Current Across a Capacitor\n• Finding Maximum Currents and Voltages\n• Capacitors and High-Pass Filters\n• Capacitive Coupling\n• Dialectrics\n• Capacitors in Series and Parallel\n• Magnetic Fields and Magnetic Forces\n• The Link Between Electricity and Magnetism\n• Magnetic Fields and Fluxes\n• Magnetic Force\n• Electromagnetic Induction\n• Faraday's Law For Closed Loops\n• The Direction of an Induced Emf; Lenz's law\n• Inductance\n• Transformers\n• Alternators\n• Inductors\n• DC Circuit Analysis\n• Growth and Decay of Voltage/Current\n• RLC Circuits\n• AC Circuit Analysis\n• RLC Circuits\n• Power in AC Circuits\n• Semiconductors and Semiconductor Elements\n• Electron Energy\n• The Band Theory of Electrical Conduction\n• Semiconductors\n• P-N Junctions\n• Diodes\n• Transistors\n• Rectifiers\n• Applications\n• Frequency Filtering in an AC Circuit\n• Resonance in an AC Circuit\n• Units\n• Some Trigonometric Formulas\n• Some Phase Shifts\n• Some Integrals" ]

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[ "# List of realistic extremum problems\n\nI am a student who would like to become a teacher, so I am currently following courses in education. One of the things I learned, is that students like authentic, realistic problems. However, much of the extremum problems (in one variable) appear very 'fabricated'. I only know a few 'realistic extremum problems':\n\n1. minimizing some cost function\n2. maximizing some return function\n3. maximizing the volume of a box.\n\nA realistic example of the third problem can be found here, where the volume of a box is maximized. This box is the kind of box you receive a package in: it is closed, with overlapping flaps.\n\nWith respect to multivariable functions, I can think of one extra example: determining a regression line.\n\nHowever, this list seems rather short. Can anyone give extra examples of real-life situations where we wish to maximize/minimize some sort of function? (For my class, single variable functions are the most interesting, but I equally like multivariable examples.)\n\n• Could you summarize the \"realistic box problem\" that you link to? That link requires Java in the browser, which many people (including myself) disable for security reasons. Oct 27 '18 at 10:15\n• @RoryDaulton: Here is that problem: \"A sheet of cardboard is rectangular, 14 inches long and 8.5 inches wide. If six congruent cuts (denoted in black) are to be made into the cardboard and five folds (denoted by dotted lines) made to adjoin the cuts so that the resulting piece of cardboard is to be folded to form a closeable rectangular box (see figures below). How should this be done to get a box of largest possible volume?\" Oct 27 '18 at 10:51\n• @JosephO'Rourke: Thanks! Is there any way to show the figures with the cuts and folds and final box? Oct 27 '18 at 11:23\n• 1 covers a lot of ground: number of shale wells in a section, number of salespeople, hours to keep a store open, etc. Oct 27 '18 at 13:22\n• There is an entire scientific discipline called Operations Research that grew out of military logistics in WW I. It sits somewhere between Applied Maths, Statistics, Logistics, and more recently, Computer Science. It is essentially concerned with Optimization, which is basically finding Extrema, i.e. the \"best\" or \"cheapest \" solution Basically, every OR problem is an extrema problem. Oct 27 '18 at 21:33\n\nHere are some optimization problems that were harder than a simple homework problem:\n\nNotice that in these examples, more effort went into designing the functions to be optimized, and in choosing the parameter to be varied, than went into performing the calculus.\n\n• Perhaps one could invent a two-variable function $g(x,y)$ of math-effort $x$ vs. optimization set-up effort $y$ then analyze $g$ subject the to the set of all word problems. If this set was compact then we could optimize $g$. Perhaps the choice of $g$ ought to depend on the course of major study for the student ? Oct 27 '18 at 16:01\n\nSome example types:\n\n1. Minimizing potential energy of any realistic physical system. Examples:\n• 0D: The point where a rolling ball/flowing water might* come to rest (*might not, if momentum carries it past the low point).\n• 1D: The curve described by a hanging chain/flexible rope (in equilibrium).\n• 2D: The surface of a soap film (in equilibrium).\n• Generally, anything that doesn't move or change: look at the world and wonder. [For instance, how hard do you have to push or jump to move your classroom building out of its potential well?)\n2. Maximizing a utility function subject to a budget constraint.\n3. Fermat's principle: Light takes the path of least time.\n4. Machine learning: minimize the loss function of a particular learning task. (It and the subfield, data mining, are hot fields right now.)\n5. In a network of resistors connecting voltage $$V_A$$ to voltage $$V_B$$, the voltages at the nodes of the network minimize power dissipation.\n6. Space trajectories that minimize fuel use (Mission Design).\n7. When using the central difference formula $$[f(x+h)-f(x-h)]/(2h)$$ to approximate the derivative $$f'(x)$$ using floating-point numbers with a machine epsilon of $$\\epsilon$$, what choice of $$h$$ minimizes the relative error of the approximation? [Note: This is a somewhat ill-defined problem, since it may depend on $$f$$ and $$x$$.]\n\nSome physics examples: --\n\nGiven that the range of a projectile is $$R=(v^2/2g)\\sin\\theta\\cos\\theta$$, prove that the maximum range is achieved for $$\\theta=45$$ degrees.\n\nAn electrical transmission line of resistance $$x$$ is in series with a load of resistance $$y$$. For a fixed voltage $$V$$, the useful power dissipated in the load is $$P= V^2y(x+y)^{-2}$$. Show that this is maximized, for fixed $$x$$, when $$y=x$$.\n\nTwo atoms will interact through electrical forces between their protons and electrons. One fairly good approximation to the total electrical energy $$U$$ is the Lennard-Jones formula, $$U(r) = k\\left[\\left(\\frac{a}{r}\\right)^{12}-2\\left(\\frac{a}{r}\\right)^{6}\\right]$$. Infer the units of $$a$$. Find the equilibrium value of $$r$$, which occurs at the minimum of $$U$$, and show that your answer has units that make sense.\n\nSometimes doors are built with mechanisms that automatically close them after they have been opened. The designer can set both the strength of the spring and the amount of friction. If there is too much friction in relation to the strength of the spring, the door takes too long to close, but if there is too little, the door will oscillate. For an optimal design, we get motion of the form $$x=ct e^{-bt}$$, where $$x$$ is the position of some point on the door, and $$c$$ and $$b$$ are positive constants. (Similar systems are used for other mechanical devices, such as stereo speakers and the recoil mechanisms of guns.) In this example, the door moves in the positive direction up until a certain time, then stops and settles back in the negative direction, eventually approaching $$x=0$$. This would be the type of motion we would get if someone flung a door open and the door closer then brought it back closed again. (a) Infer the units of the constants $$b$$ and $$c$$. (b) Find the door's maximum speed (i.e., the greatest absolute value of its velocity) as it comes back to the closed position. (c) Show that your answer has units that make sense.\n\n• The range example can be done without calculus, using $\\sin \\theta \\cos \\theta = \\dfrac{1}{2} \\sin (2 \\theta)$. Oct 27 '18 at 22:08\n• It's not hard to make the range problem much harder. Just make the initial and final altitudes differ. Then the double-angle formula will be no help... but I like the given problem. Nov 26 '18 at 1:14\n\nThere are many volume-of-a-box questions. I like this one, simpler than what the OP cites:\n\nGiven a rectangle, cut out squares from the corners so you can fold it up to a box, without a top, of maximal volume.\n\nThe rectangle might be specialized to a square, as below. See also The Math Forum.", null, "In response to @RoryDaulton, here is the box problem to which the OP @Student points:", null, "• Thanks for showing me the box problem. I'll try using it in my calculus class--it is more realistic than the box problem I have been using. Nov 18 '18 at 0:16\n\nMany important machine learning algorithms, such as training a neural network, require solving a large optimization problem (essentially tuning the neural network weights to minimize the classification error on a training dataset). The whole deep learning revolution is powered by optimization.\n\nIn radiation therapy (for cancer treatment), an optimization problem is solved to find optimal beam intensities that target the tumor most effectively while sparing healthy tissue.\n\nMagnetic resonance imaging (MRI) works by solving an optimization problem that finds an image of the human body which is most consistent with the measurements obtained by the MRI machine (and also consistent with prior knowledge we have of what images of the human body should look like). Other medical imaging techniques also use optimization for the image reconstruction step.\n\nIn Statistics, many important optimization problems arise as special cases of Maximum Likelihood Estimation. Linear regression is a classic example with a ton of applications.\n\nThe book Convex Optimization by Boyd and Vandenberghe discusses many applications of convex optimization.\n\nHere is a real life example with millions of dollars (maybe billions) depending on the answer:\n\nhttp://phx.corporate-ir.net/phoenix.zhtml?c=197380&p=irol-presentations (MAY18, IR presentation, page 20).\n\nThe key insight is to maximize unit net present value (middle row), not per well performance. You have a tradeoff when adding extra wells to a section. Each well gives you more total oil from the land, BUT the per well performance goes down (because there is a mix of cannibalization with old wells versus accessing new rock by the extra fracturing) and as the spacing gets tighter, it becomes more and more cannibalization. And each well has a capital cost (there is a small capital economy of scale from running on the same pad [and even smaller OPEX scale savings from pad operations] but most of the capital cost remains.)\n\nThis is almost the exact same problem as adding salaried business development (sales) resources to a territory.\n\nI'm NOT saying to present this problem to the kids with all the modeling involved (taking business insights and converting into algebra and Excel). I would actually opt more for giving them idealizing problems with formula shown where they just do the derivatives and check for extrema. But maybe when they are encountering these issues in the work world (and tradeoffs like this exist in almost every business), they have some \"feel\" for the issue from having done it in calculus class and it just resonates off a remote memory. And I don't even care that this problem is discrete (can't have half a well) versus continuous. The concepts convey.\n\nAnd note, this is a \"return\" problem. So it doesn't meet your restrictions of some other category. But I just want you to see how rich, rich, rich that area is. I'm not even saying it for the kids benefits but for yours. So that you feel like the concept is worthwhile.\n\nI also like the example of regression and least squares as a nice application of optimization a differentiation with respect to something other than $$x$$.\n\nIn my class, I also discuss partial derivatives and give some examples of envelope problems. Suppose a 13 foot ladder slides down a wall, what is the curve that it traces out?\n\nDescribe the ladder as an equation determined by its $$x$$ an $$y$$ intercepts and some parameter $$a$$, and consider a specific value $$x$$, say $$x = 4$$. Then, we have $$y = 17 -a -52/a$$.\n\nIf we look at what happens as we run through different values of $$a$$ we find that the $$y$$'s increase then decrease, moving first towards and then away from the envelope, as we see below with the little red dots. Thus, we have the curve at the maximum value when we differentiate our expressions with respect to $$a$$.", null, "Now, more generally, we have\n\n$$y = \\frac{a - 13}{a}(x - a) = 13 + x - a - \\frac{13x}{a}$$\n\n$$\\frac{\\partial y}{\\partial a} = -1 + 13x/a^2$$\n\n$$a = \\sqrt{13x}$$\n\nand we can substitute this value for $$a$$ back in our original equation, yielding the description for the curve. This example is from Analysis by its History by Hairer and Wanner." ]

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[ " Calculation To Determine Quantity Of Cement Sand\n\nWe are waiting for you:\n\n24-hour service\n\n[email protected]", null, "# Calculation To Determine Quantity Of Cement Sand\n\nGet Price", null, "Calculate the concrete quantity required in cubic meter and then multiply the quantity with the cement sand and aggregate for 1 cubic meter of concrete. In your case Volume of concrete 1.7 x 1.7 x 0.5 1.4 cu.m It looks like you are using m20 grade mix with extra sand. If you follow that mix proportion then to calculate use this following.\n\n•", null, "### Brick block and mortar calculator\n\nThis CalculatorEstimator will provide the quantities of bricks blocks and mortar sand amp; cement required for a given area for metric bricks single amp; double skins as well as 100mm 140mm amp; 215mm blockwork. It will also provide approximate brick work prices. Please enter the dimensions in the white fields below and click calculate to display.\n\n•", null, "### Calculate Quantities of Materials for Concrete Cement\n\nThus the quantity of cement required for 1 cubic meter of concrete 0.980.1345 7.29 bags of cement. The quantities of materials for 1 m3 of concrete production can be calculated as follows The weight of cement required 7.29 x 50 364.5 kg. Weight of fine aggregate sand 1.\n\n•", null, "### How To Calculate Brick Cement And Sand In Brick Masonry\n\nQuantity estimation of materials is essentially required in any construction works and the quantity of materials depends on the mix proportions of the concrete. In this article I will discuss how to calculate bricks cement and sand in brick masonry. So lets get started.\n\n•", null, "### How to calculate plastering quantity cement sand ratio\n\nHow much sand and cement do I need for plastering- You will need about 33kg 0.66 bags of 50kg of cement 5 cft of sand and 32 Litres of water quantity for plastering of 10m2 area of brick block wall have thickness 12mm with mix of cement sand in the ratio 16 1 parts cement and 6 parts sand.\n\n•", null, "### Concrete Block Calculator amp; Estimator CEMEX USA\n\nThe Concrete Block calculators provide guidance to determine building materials required for construction projects.. Use the block calculators below to calculate the amount of materials needed.. Height ft Total Block Required add approx. 5 to the total for waste Sand amount needed for block count. Block count Sand tons Required.\n\n•", null, "### How do i calculate cement and sand quantity in 13 mortar\n\n4 calculation of cement quantity. 5 calculation of sand quantity. How to calculate dry volume of mortar in 13 given data-Assume volume of mortar 1m3. Density of cement 1440 kgm3. Cement in 1 bag 50 kg. 1m3 35.3147 cubic feet. Mix ratio is 13 one part is cement and 3 part is sand Total proportion 13 4.\n\n•", null, "### Concrete Calculator\n\nThis free concrete calculator estimates the amount of concrete necessary for a project and can account for different shapes and amounts of concrete. Explore other calculators related to housing or building as well as hundreds of other calculators addressing finance math fitness health and more.\n\n•", null, "### Plaster Calculator Calculate Cement and Sand for Plastering\n\nHere is the useful Plaster calculator to calculate cement and sand for plastering.\n\n•", null, "### How to Calculate Quantities of Cement Sand and Aggregate\n\nFeb 07 2017 The ratio of dry volume to the wet volume of concrete is 1.54. So 1.54 Cum of dry materials cement sand and aggregate is required to produce 1 Cum of concrete Volume of Cement required 1 124 X 1.54 17 X 1.54 0.\n\n•", null, "### How To Calculate Quantity Of Cement Sand And\n\nNow from this dry volume we can find out the volume of cement Sand and aggregate in concrete column. 1- Volume of cement V 15.5 x dry volume. V 15.5 x 0.4158. V 0.0756 m3. Now convert the cment volume to bags. Volume of cement Volume of one cement bag in m3. Note Volume of Cement bag 0.0347.\n\n•", null, "### How To Calculate Quantity Of Cement Sand And\n\nJul 31 2021 2- volume of sand. Vol. of sand ratio of sand total ratio x volume 2 7 x 51.30 14.65 cft. 3- Volume of aggregate. Vol. of aggregate ratio of aggregate total ratio x volume 4 7 x 51.30 29.31 cft.\n\n•", null, "### How to calculate quantity of Plaster Cement Sand\n\nMar 28 2021 Calculate quantity of cement sand and water cement ratio in plaster Formula for calculation of cement sand quantity for plaster mortar.\n\n•", null, "### Concrete Calculator Estimate Cement Sand Gravel\n\nOur mix-on-site concrete calculation is based on batching by volume Large construction sites employ batching by weight which is more exact. You can also estimate the quantity of sand and gravel required by weight; Simply multiply the volumetric quantity of sand and gravel with 1400 kgm 3 bulk density of sand and 1600 kgm 3 bulk density of stone respectively when calculating in metric units.\n\n•", null, "### Calculation of Cement and Sand Quantity for Plastering\n\nStep-1Calculate the dry volume of cement and sand mixture required. Volume of plaster Area X Thickness 10 sq.m. X 0.012 0.12 cu.m. The wet volume of the mixture is always less than the dry volume. Dry volume of motor required for plastering 1.\n\n•", null, "### How To Calculate Quantity Civil Engineering\n\n3 hours ago How To Calculate Quantity Of Cement Sand And Aggregate In Slab. Example Let suppose we have a two way slab having a length of 20 feet and same as width of 20 feet and use M-15 grade of concrete Calculate the. engineeringdiscoveries.com.\n\n•", null, "### Calculate the quantity of cement sand for plastering\n\nFeb 05 2020 The Quantity of water equals to 20 Total Dry material cement amp; sand Quantity of water 0.20 x Weight of cement Weight of sand Therefore Quantity of water 0.20 x 608.083 3243.11 770.23 kg 770.23 Liters. NOTE 100 Square meter is a huge area just for calculation purpose we assume that. In this way you can calculate the.\n\n•", null, "### How to Calculate Quantities Cement Sand Aggregate and\n\nJun 26 2020 Quantity of water in liter depends upon the water-cement ration and quantity of cement. In the above examples I have calculated Weight of cement in m3 0.22 m 3. Consider water-cement ration 0.45.\n\n•", null, "### Quantity of Cement and Sand Calculation in Mortar\n\nNov 03 2014 Volume of cement will be calculated as Since the volume of 1 bag of cement is 0.0347 m 3 so the number of bag of cement will be calculated as Example For cement mortar of 16 the quantity calculated will be as below Sand quantity Quantity of cement in bags Volume of cement There number of bags required 6.58 bags.\n\n•", null, "### Plaster Calculator Calculate Cement and Sand for Plastering\n\nHere is the useful Plaster calculator to calculate cement and sand for plastering." ]

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[ "Computing vacuum expectation values of a perturbed harmonic oscillator\n\nI'm examining the Hamiltonian\n\n$$\\hat{H} = \\frac{\\hat{p}^2}{2 m} + \\frac{1}{2} m \\omega^2 \\hat{x}^2 + \\lambda \\hat{x}^4 \\text{,}$$\n\nwhere $\\lambda$ is small and $\\hat{x} = \\sqrt{\\frac{\\hbar}{2 m \\omega}} (\\hat{a} + \\hat{a}^{\\dagger})$ is the position operator in terms of the creation and annihilation operators. If this oscillator is perturbed by a small amount, I get\n\n$$\\lambda \\hat{x}^4 = \\frac{\\lambda \\hbar^2}{4 m^2 \\omega^2} (\\hat{a} + \\hat{a}^{\\dagger})^4$$\n\nfor the last term in the Hamiltonian. Now, a first-order shift in the ground state energy (or vacuum expectation value, VEV) is $\\langle 0 \\text{|} \\lambda \\hat{x}^4 \\text{|} 0 \\rangle​$, which means that I will need to expand $(\\hat{a} + \\hat{a}^{\\dagger})^4​$, but keep only those terms which have equal numbers of $\\hat{a}^{\\dagger}​$ (up) and $\\hat{a}​$ (down). My physical interpretation is that since the ground state is our reference, then only those terms will contribute to the VEV. Classically, I think of myself being on a stepladder, and having the choice of moving up two rungs and then down two rungs to get back to the bottom (but this analogy falls apart quickly as you will see).\n\nThere are a lot of relevant terms; for example, $\\hat{a} \\hat{a} \\hat{a}^{\\dagger} \\hat{a}^{\\dagger}$. I will need to go work evaluating what their contribution will be. It looks like the result $\\hat{a} \\hat{a}^{\\dagger} - \\hat{a}^{\\dagger} \\hat{a} = 1$ (from the commutator) will be useful. The text I'm looking at states that the normalized operators $\\hat{a} \\text{|} n \\rangle = \\sqrt{n} \\text{|} n - 1 \\rangle$ and $\\hat{a}^{\\dagger} \\text{|} n \\rangle = \\sqrt{n + 1} \\text{|} n + 1 \\rangle$ will also be essential. And that's my problem. For example, $\\langle n \\text{|} \\hat{a} \\hat{a} \\hat{a}^{\\dagger} \\hat{a}^{\\dagger} \\text{|} n \\rangle = (n + 1)(n + 2)$ somehow. Shankar isn't much help. I don't understand how this and the other terms are evaluated.\n\nI'm also struggling with the following. Two sample VEVs for differing terms give, for example, $\\langle 0 \\text{|} \\hat{a} \\hat{a} \\hat{a}^{\\dagger} \\hat{a}^{\\dagger} \\text{|} 0 \\rangle = 2$ (from above) and $\\langle 0 \\text{|} \\hat{a} \\hat{a}^{\\dagger} \\hat{a} \\hat{a}^{\\dagger} \\text{|} 0 \\rangle = 1$. I know these are operators, and order matters, but I get two different vacuum expectation values for, conceptually, the same number of steps up and down the ladder picture I have in my head. That's very counterintuitive.\n\nDoes anyone have any insight into (1) computing the expectation values of the terms in the perturbation above, and (2) how I can conceptually resolve two different VEVs for the same number of ups and downs? Thank you in advance.\n\nCompute $(a+a^\\dagger)^2| 0\\rangle$, then square." ]

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[ "Statistics » Scatter graph\n\nOther names for scatter graph are scatter plot, scatterplot, scatter chart, scattergram or scatter diagram.\n\nWhat is a scatter graph?\n\nA scatter graph is a diagram in which you put values from a data set as dots in a coordinate system.\n\nExample 1\nIn an inquiry 25 men are asked about their height and weight.\nThe answers are put in the scatter graph below.\nYou can read off that the shortest man is 160 cm tall and weighs 70 kg.\nThe heaviest man in this inquiry weighs 90 kg and is 188 cm tall.", null, "You can see there is a correlation between the height and weight of men. In general the taller a man is, the heavier he weighs.\n\nExample 2\nIn the same inquire these men were also asked about their monthly salary. This data resulted in the following scatter graph.", null, "You can see there is no correlation between the height of a man and his monthly salary." ]

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[ "## ››Convert quartern-loaf to hectogram\n\n quartern-loaf hectogram\n\nHow many quartern-loaf in 1 hectogram? The answer is 0.055115565546219.\nWe assume you are converting between quartern-loaf and hectogram.\nYou can view more details on each measurement unit:\nquartern-loaf or hectogram\nThe SI base unit for mass is the kilogram.\n1 kilogram is equal to 0.55115565546219 quartern-loaf, or 10 hectogram.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between quartern-loaf and hectograms.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of quartern-loaf to hectogram\n\n1 quartern-loaf to hectogram = 18.14369 hectogram\n\n2 quartern-loaf to hectogram = 36.28739 hectogram\n\n3 quartern-loaf to hectogram = 54.43108 hectogram\n\n4 quartern-loaf to hectogram = 72.57478 hectogram\n\n5 quartern-loaf to hectogram = 90.71847 hectogram\n\n6 quartern-loaf to hectogram = 108.86217 hectogram\n\n7 quartern-loaf to hectogram = 127.00586 hectogram\n\n8 quartern-loaf to hectogram = 145.14956 hectogram\n\n9 quartern-loaf to hectogram = 163.29325 hectogram\n\n10 quartern-loaf to hectogram = 181.43695 hectogram\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from hectogram to quartern-loaf, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Hectogram\n\nThe SI prefix \"hecto\" represents a factor of 102, or in exponential notation, 1E2.\n\nSo 1 hectogram = 102 grams-force.\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]

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[ "## How many km are in a hectare?\n\nTo convert Hectare to Kilometer, one needs to remember that 1 Hectare= 0.01 Square Kilometer and thus 0.1 kilometers, assuming that the plot in question is square.\n\n### How big is a hectare in km2?\n\nHow many square kilometers are there in 1 hectare? There are 0.01 square kilometers in 1 hectare. To convert from hectares to square kilometers, multiply your figure by 0.01 (or divide by 100) .\n\n#### Are hectares bigger than Kilometres?\n\nThere are 100 hectares in one square kilometre. An acre is about 0.405 hectare and one hectare contains about 2.47 acres.\n\nHow many kilometers are in an acre?\n\nAcre to Kilometer conversion Table\n\nAcre Kilometer Acre to Kilometer\n1 acre 0.0636 km 1 Acre is equal to 0.0636 Kilometer\n2 acre 0.1272 km 2 Acre is equal to 0.1272 Kilometer\n3 acre 0.1908 km 3 Acre is equal to 0.1908 Kilometer\n4 acre 0.2544 km 4 Acre is equal to 0.2544 Kilometer\n\nHow do you calculate hectares?\n\nCalculating areas in hectares You can think of a hectare (ha) as measuring 100m by 100m. Take the figure you have worked out in square metres (m²), then divide by 10,000 to find the number of hectares (ha). Use a calculator to convert an area in square metres (m²) into hectares (ha).\n\n## How much meters are in a kilometer?\n\n1,000 meters\nHow many meters in a kilometer? 1 kilometre is equal to 1,000 meters, which is the conversion factor from kilometres to meters. Go ahead and convert your own value of km to m in the converter below. For other conversions in length, use the length conversion tool.\n\n### What is hectare land?\n\nhectare, unit of area in the metric system equal to 100 ares, or 10,000 square metres, and the equivalent of 2.471 acres in the British Imperial System and the United States Customary measure. The term is derived from the Latin area and from hect, an irregular contraction of the Greek word for hundred.\n\n#### Are convert to Hector?\n\nAres to Hectares Conversions\n\nAres Hectares\n1 are 0.01 hectares\n2 ares 0.02 hectares\n3 ares 0.03 hectares\n4 ares 0.04 hectares\n\nHow many hectares is a soccer field?\n\nHow many soccer fields is 10 hectares? What is meant by 1 hectare? What is the size of an hectare of land? What is difference between hectare and Acre?…How many soccer fields is 10 hectares?\n\nSoccer Fields to Hectares (table conversion)\n30 soccer field = 21.42 ha\n\nHow big is a 40 of land?\n\n40 ACRES 43,560 sq. feet. 165 feet x 264 feet.\n\n## How do you convert hectares to plots?\n\nWith a plot defined as 900 sqm and given that a hectare is made up of 10,000 square meters – 1 hectare is equivalent to 11.1 plots (10,000/900).\n\n### Which is the SI unit for kilometers per hour?\n\nThere are many abbreviations for the unit kilometers per hour (kph, kmph, k.p.h, KMph., etc.), but “km/h” is the SI unit symbol.\n\n#### When do you use mph instead of km / h?\n\nCurrent use: Along with km/h, mph is most typically used in relation to road traffic speeds. It is most widely used in the United States, the United Kingdom, and their related territories. It is also used in the Canadian rail system, though the Canadian road systems use km/h.\n\nWhat’s the difference between 50 and 100 km / h?\n\n20 km/h: 12.4274238447 mi/h: 50 km/h: 31.0685596119 mi/h: 100 km/h: 62.1371192237 mi/h: 1000 km/h: 621.3711922373 mi/h\n\nHow to convert kilometer / hour to mile / hour?\n\nHow to Convert Kilometer/hour to Mile/hour 1 km/h = 0.6213711922 mi/h 1 mi/h = 1.609344 km/h Example: convert 15 km/h to mi/h:" ]

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[ "Commits (12)\n ... ... @@ -57,7 +57,7 @@ mpitest-mpp: image: \\${REGISTRY}/\\${IMAGE_NAME_MPP} script: - cd /mpp/build - python3 mppyrun.py --mpi_tests=1 --mute=0 - python3 mppyrun.py --mpi_tests=1 --mute=1 dependencies: [ \"build-mpp\" ] tags: [ docker ] ... ...\n ... ... @@ -274,6 +274,50 @@ MACRO(add_mpp_subdirectory subdir_path) ENDMACRO() #---------------------------------------------------------------------------------------# ######################################### ## Find TIRCP (for HoreKa) ######################################### # find_package(PkgConfig QUIET) pkg_check_modules(PC_TIRPC libtirpc) find_path(TIRPC_INCLUDE_DIRS NAMES netconfig.h PATH_SUFFIXES tirpc HINTS \\${PC_TIRPC_INCLUDE_DIRS} ) find_library(TIRPC_LIBRARIES NAMES tirpc HINTS \\${PC_TIRPC_LIBRARY_DIRS} ) set(TIRPC_VERSION \\${PC_TIRPC_VERSION}) include(FindPackageHandleStandardArgs) find_package_handle_standard_args(TIRPC REQUIRED_VARS TIRPC_LIBRARIES TIRPC_INCLUDE_DIRS VERSION_VAR TIRPC_VERSION ) mark_as_advanced(TIRPC_INCLUDE_DIRS TIRPC_LIBRARIES) ########################################### ## End TIRCP ########################################### if (TIRPC_FOUND) include_directories(\\${TIRPC_INCLUDE_DIRS}) endif() #if(EXISTS \"/usr/include/tirpc\") # include_directories(\"/usr/include/tirpc\") #endif() #include_directories(\"/usr/include/tirpc/rpc\") # Subdirectories add_mpp_subdirectory(\\${PROJECT_MPP_DIR}/src) #---------------------------------------------------------------------------------------# ... ... @@ -289,4 +333,4 @@ enable_testing() add_mpp_subdirectory(\\${PROJECT_MPP_DIR}/tests) #---------------------------------------------------------------------------------------# file(COPY \\${PROJECT_MPP_DIR}/python/mppyrun.py DESTINATION \\${PROJECT_BINARY_DIR}) \\ No newline at end of file file(COPY \\${PROJECT_MPP_DIR}/python/mppyrun.py DESTINATION \\${PROJECT_BINARY_DIR})\n ... ... @@ -26,6 +26,12 @@ endif () add_library(LIB_BASIC STATIC \\${basic_src}) target_link_libraries(LIB_BASIC \\${LAPACK_LIBRARIES} \\${BLAS_LIBRARIES}) if (TIRPC_FOUND) target_link_libraries(LIB_BASIC \\${TIRPC_LIBRARIES}) endif() if (USE_CXSC) target_link_libraries(LIB_BASIC cxsc) endif ()\n ... ... @@ -4,59 +4,11 @@ using namespace std; ParallelSolver::ParallelSolver() : min_matrix_size(20), psize(0), maxP(0), PS_cd(true), onecal(2.5e-10), sod(4e-8), latency(2e-6), steps(0), PSM(0) { ParallelSolver::ParallelSolver() : min_matrix_size(20), maxP(0), PS_cd(true), steps(nullptr), PSM(0) { config.get(\"PSMsize\", min_matrix_size); config.get(\"PS_maxP\", maxP); config.get(\"PS_checkdiagonal\", PS_cd); config.get(\"PS_one_calc\", onecal); config.get(\"PS_send_one_double\", sod); config.get(\"PS_latency\", latency); } double ParallelSolver::estimate_COMM_time(int s, int *sol, int *schur) { double t = 0; double t2 = 0; int P = int(pow(2, s)); if (s == 0) { t2 = (4.); t2 *= latency; t += t2; t2 = 2 * pow(schur[s], 2.); t2 *= sod; t += t2; } else { t += pow(sol[s], 2.) + sol[s] * schur[s]; t *= sod; t2 += (3. * P + 2. + 4.); t2 *= latency; t += t2; t2 = 2 * pow(schur[s], 2.); t2 *= sod; t += t2 / s; } return t; } double ParallelSolver::estimate_COMP_time(int s, int *sol, int *schur) { double t = 0; int P = int(pow(2, s)); if (s == 0) { t += pow(sol[s], 7. / 3); t += pow(sol[s], 4. / 3) * schur[s]; t += pow(sol[s], 2.) * schur[s]; t *= onecal; } else { t += 2. * pow(sol[s], 3) / pow(P, 2); t += pow(sol[s], 3.) / P; t += pow(sol[s], 2.) * schur[s] / pow(P, 2.); t += 2. * pow(sol[s], 2) * schur[s] / P; t += sol[s] * pow(schur[s], 2.) / P; t *= onecal; } return t; } void ParallelSolver::Construct(const Matrix &A) { ... ... @@ -67,9 +19,6 @@ void ParallelSolver::Construct(const Matrix &A) { if (PS_cd) S.CheckDiagonal(); Date Start; double starttime, endtime; starttime = MPI_Wtime(); steps = new ParallelSolverAllSteps(A); int size = steps->size(); ... ... @@ -79,36 +28,12 @@ void ParallelSolver::Construct(const Matrix &A) { PSM[i] = new ParallelSolverMatrix(steps->get_step(i), min_matrix_size, maxP); PSM->Set(S); endtime = MPI_Wtime(); tout(1) << \"--- presettings for PS in seconds: \" << endtime - starttime << endl; starttime = MPI_Wtime(); for (int i = 0; i < size - 1; ++i) { double startsteptime = MPI_Wtime(); PSM[i]->makeLU(); PSM[i]->SetNext_Matrix(*PSM[i + 1]); double endsteptime = MPI_Wtime(); if (TimeLevel > 2) { int Solsize = PSM[i]->solsize(); int Schursize = PSM[i]->schursize(); mout << \"--- SOL: \" << Solsize << \"; SCH: \" << Schursize << \"; parSol: \" << steps->parallel_size(i) << \"; time in step \" << i << \": \" << endsteptime - startsteptime << endl; } } PSM[size - 1]->makeLU(); endtime = MPI_Wtime(); tout(1) << \"--- factorization with PS (size: \" << steps->parallel_size() << \") in seconds: \" << endtime - starttime << endl; tout(1) << \"ParallelMatrixSolver: total time (N = \" << A.pSize() << \" unknowns):\" << Date() - Start << endl; } ... ... @@ -117,22 +42,16 @@ void ParallelSolver::Destruct() { for (int i = PSM.size() - 1; i >= 0; --i) { delete PSM[i]; } PSM.clear(); if (steps) delete steps; steps = NULL; steps = nullptr; } void ParallelSolver::multiply(Vector &u, const Vector &b) const { int slaveproc; MPI_Comm_rank(MPI_COMM_WORLD, &slaveproc); double starttime, endtime; int size = steps->size(); starttime = MPI_Wtime(); for (int i = 0; i < size; ++i) PSM[i]->Create_rhs(); PSM->Set_rhs(b()); for (int i = 0; i < size - 1; ++i) { ... ... @@ -146,27 +65,12 @@ void ParallelSolver::multiply(Vector &u, const Vector &b) const { PSM[i]->SetNext_rhs_RIGHT(*PSM[i - 1]); } PSM->SolveU(); PSM->Write_rhs(u()); endtime = MPI_Wtime(); if (TimeLevel > 3) { if (!slaveproc) cout << \"--- solving with PS in seconds (size: \" << steps->parallel_size() << \"): \" << endtime - starttime << \"\\n\"; } } void ParallelSolver::multiply(Vectors &U, const Vectors &B) const { int slaveproc; MPI_Comm_rank(MPI_COMM_WORLD, &slaveproc); double starttime, endtime; int size = steps->size(); starttime = MPI_Wtime(); int nrhs = U.size(); for (int i = 0; i < size; ++i) PSM[i]->Create_rhs(nrhs); ... ... @@ -188,11 +92,4 @@ void ParallelSolver::multiply(Vectors &U, const Vectors &B) const { for (int i = 0; i < nrhs; ++i) PSM->Write_rhs(U[i](), i); endtime = MPI_Wtime(); if (TimeLevel > 3) { if (!slaveproc) cout << \"--- solving \" << size << \" rhs with PS in seconds: \" << endtime - starttime << \"\\n\"; } }\n ... ... @@ -11,15 +11,6 @@ class ParallelSolver : public Preconditioner { int min_matrix_size; int maxP; bool PS_cd; int psize; double onecal; // time for one calculation double sod; // sending one double double latency; double estimate_COMM_time(int, int *, int *); double estimate_COMP_time(int, int *, int *); public: ParallelSolver(); ... ...\n ... ... @@ -154,8 +154,8 @@ int ParallelSolverOneStep::total_Solsize() { ParallelSolverAllSteps::ParallelSolverAllSteps(const Matrix &A) : s(0) { vps = new VectorProcSet(A.GetVector()); vps->combine_procs(); int P = PPM->size(); vps->clear_position(); s.resize(1); ... ..." ]

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[ "# What are the types of motion in biomechanics?\n\nWhat are the types of motion in biomechanics?\n\nEverything naturally wants to move and change. In the world of mechanics, there are four basic types of motion. These four are rotary, oscillating, linear and reciprocating.04-Jun-2013\n\n## What is motion and types of motion?\n\nA motion is when the position of an object changes over a certain period of time. There can be various types of motion including oscillatory, rotational, transactional, uniform, non-uniform, periodic, circular and linear.\n\nWhat is defined as motion?\n\nmotion, in physics, change with time of the position or orientation of a body. Motion along a line or a curve is called translation. Motion that changes the orientation of a body is called rotation.09-Dec-2022\n\n## What is the 4 types of motion?\n\nThe four types of motion are:\n\n• linear.\n• rotary.\n• reciprocating.\n• oscillating.\n\nWhat is motion with example?\n\nThe free movement of a body with respect to time is known as motion. For example-the fan, the dust falling from the carpet, the water that flows from the tap, a ball rolling around, a moving car etc. Even the universe is in continual motion.\n\n## What are the two states of motion?\n\nState of Rest and State of Motion.\n\nWhat is the importance of motion?\n\nMotion strengthens your muscles to better support your body. It also provides movement to your joints and strengthens your bones. In addition, motion provides a release of hormones that makes you feel good, so it positively affects your mental health.\n\n## What are the basic concepts of motion?\n\nMotion is mathematically described in terms of displacement, distance, velocity, acceleration, speed and frame of reference to an observer and measuring the change in position of the body relative to that frame with change in time.\n\nWhat causes of motion?\n\nMotion is caused due to external Force.\n\n## What is motion in simple sentence?\n\nExample Sentences He made hand motions to get our attention. She made a motion calling for the repeal of the law. Her motion was voted on. His lawyer filed a motion for a mistrial.26-Jan-2023\n\nWhat is study of motion called?\n\n2. It is broken down into two parts, kinematics and dynamics. 3. Kinematics is the ''how '' of motion, that is, the study of how objects move, without concerning that why they move.\n\n## What is the first law of motion?\n\nNewton's First Law of Motion (Inertia) An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force.27-Oct-2022\n\nWhat are the 5 forces of motion?\n\nThe force of air resistance is often observed to oppose the motion of an object.Types of Forces.\n\n## What are the 5 ways of motion?\n\nThe arguments are often named as follows: (1) argument from motion, (2) argument from efficient cause, (3) argument from necessary being, (4) argument from gradations of goodness, and (5) argument from design.03-May-2022\n\nWhat are the 4 parameters of motion?\n\nMotion Parameters- Definition, Position, Distance and Displacement.\n\n## What are the three types of motion?\n\n• Motion:\n• Types of motion: Motion can be classified into three, such as:\n• a. Linear motion: A movement in a straight line is called linear motion.\n• b. Oscillatory motion:\n• c. Rotary motion:\n\nWhat is a real life example of motion?\n\nRunning, cycling, jumping, swimming, eating, drinking, playing, writing, typing, moving cars, and throwing a ball are all examples of motion.\n\n## What are the 3 causes of motion?\n\n13.3 Force: The Cause of Motion\n\n• Mass. Mass is the amount of stuff in an object and we give it the symbol m.\n• Force. Force is a push or a pull.\n• Gravity. Now that you understand the basics of motion, we can discuss an important force in many cases involving motion.\n• Weight.\n• Newton's Laws: Putting It All Together.\n\nWhat are Newton's 2 laws of motion?\n\nState Newton's second law of motion Newton's second law of motion states that “Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.\n\n## What is the 2 law of motion called?\n\nTo understand this we must use Newton's second law - the law of acceleration (acceleration = force/mass). Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.\n\nWhat is the most important law of motion?\n\nNewton's second law is one of the most important in all of physics. For a body whose mass m is constant, it can be written in the form F = ma, where F (force) and a (acceleration) are both vector quantities. If a body has a net force acting on it, it is accelerated in accordance with the equation.\n\nWhat are the types of motion in biomechanics?" ]

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[ "Dependency Analyzer Scope and Limitations\n\nAnalysis Scope\n\nThe Dependency Analyzer identifies the required files and add-ons for your project or model. The analysis covers a wide range of dependencies, including model references, subsystem references, linked libraries, MATLAB® and C/C++ code, Stateflow® charts, data files, S-functions, and requirements documents.\n\nWhen the Dependency Analyzer encounters MATLAB code, such as in a model or block callback, or in a .m file S-function, it attempts to identify the files it references. For more information, see Analysis Limitations.\n\nFor files under the MATLAB root folder, the Dependency Analyzer only shows required products. It does not analyze dependencies.\n\nThe Dependency Analyzer identifies dependencies inside user-defined add-ons and dependencies that were introduced by code generation or by MATLAB code in model parameters. These options are off by default because they can be time consuming for large designs.\n\nTo specify the scope of the analysis, in the Dependency Analyzer toolstrip, click Analyze and select one or more of the following options:\n\nOptionDefaultDescription\nC/C++ CodeOnAnalyze dependencies introduced by C/C++ code files.\nModel ParametersOffAnalyze dependencies introduced by MATLAB code in model block parameters.\nGenerated Code TraceabilityOffAnalyze dependencies introduced by code generated from a model.\n\nAnalysis Limitations\n\n• The Dependency Analyzer has limitations specific to MATLAB code analysis:\n\n• The Dependency Analyzer only identifies function input arguments when they are literal character vectors or strings:\n\nIf you define a file name as a variable and pass it to a function, the Dependency Analyzer is unable to identify the dependency. In the following example, since the code is not executed, the Dependency Analyzer does not have the value of str. The Dependency Analyzer might report a missing dependency.\n\nstr = \"mydatafile\";\n\n• The Dependency Analyzer does not always determine type automatically. Depending on the way you call an object method, the Dependency Analyzer might confuse a method with a function and report a missing dependency.\n\nIn MATLAB, you can call an object method in two different ways. For example, for an object p, you can call the method addFile using the function notation:\n\np = currentProject;\nor by using the dot notation:\np = currentProject;\nIf you do not declare the type of p explicitly, the Dependency Analyzer might confuse a method call that uses a function notation with a function call. The analyzer reports addFile as a missing dependency.\n\nTo work around this limitation, use dot notation to call a method or use arguments to explicitly declare the variable type in your function:\n\nfunction myfunction(p)\n\narguments\np matlab.project.Project\nend\n\nend\n\n• The Dependency Analyzer does not report a dependency to a class that is referenced using a method call.\n\n• The Dependency Analyzer might not report certain blocksets required by a model.\n\nThe Dependency Analyzer is unable to detect blocksets that do not introduce dependencies on any files, such as Fixed-Point Designer™.\n\nTo include dependencies that the analysis cannot detect, add the file that introduces the dependency to your project. To create a project from your model, see Create a Project from a Model.\n\n• The Dependency Analyzer might not detect required support packages. It lists required add-ons, including apps and toolboxes.\n\n• The Dependency Analyzer might not report dependencies for dynamic content in masked blocks.\n\nBased on the parameters of the masked blocks, dynamic masks can modify the masked subsystem and change the block dependencies. If the dynamic mask is in a library, the Dependency Analyzer is unable to detect the dynamic changes.\n\n• The Dependency Analyzer does not support Simulink® functions called from MATLAB function blocks.\n\n• The Dependency Analyzer does not support Stateflow charts that use MATLAB as the action language.\n\n• Some MathWorks® products and add-ons share code and Simulink libraries. The Dependency Analyzer might report dependencies on all of them.\n\nTo investigate where shared code is used, in the Properties panel, in the Products section, point to a product under Shared Functionalities Among: and click the search folder icon", null, ".\n\n• The Dependency Analyzer analyzes project code without evaluating its content, therefore:\n\n• The Dependency Analyzer does not add global variables in executed functions.\n\n• The Dependency Analyzer analyzes code inside logic statements, even if it is not executed. In the following example, the code is not executed but the Dependency Analyzer reports a missing dependency on Simulink.\n\nif false" ]

[ null, "https://es.mathworks.com/help/simulink/ug/searchfoldericon.png", null ]

[ "# Circle through the Incenter\n\nThe applet below illustrates problem 7 from the 2009 Australian Mathematical Olympiad.\n\nLet $I$ be the incenter of $\\Delta ABC$ in which $AC \\ne BC.$ Let $\\Gamma$ be the circle passing through $A,$ $I$ and $B.$ Suppose $\\Gamma$ intersects the line $AC$ at $A$ and $X$ and intersects the line $BC$ at $B$ and $Y.$ Show that $AX = BY.$\n\nThe condition $AC \\ne BC$ is obviously a red herring as, in this case, $A = X$ and $B = Y.$\n\nSolution", null, "Let $I$ be the incenter of $\\Delta ABC$ in which $AC \\ne BC.$ Let $\\Gamma$ be the circle passing through $A,$ $I$ and $B.$ Suppose $\\Gamma$ intersects the line $AC$ at $A$ and $X$ and intersects the line $BC$ at $B$ and $Y.$ Show that $AX = BY.$\n\nTwo configurations are possible: one with $Y$ inside $BC$ and $X$ outside $AC$ and the other with $Y$ outside $BC$ and $X$ inside $AC.$", null, "Without loss of generality, we shall consider just one of them. This is illustrated below:", null, "In the diagram, $J$ is the other intersection of the bisector $CI$ with the circle and, for simplicity, the arcs are enumerated, so that, in the following $I$ shall refer to the arcs by those numbers: $IY = 1,$ $AI = 2,$ $AX = 3,$ $JX = 4,$ $BJ = 5,$ $BY = 6.$ Angles of $\\Delta ABC$ are $2\\alpha,$ $2\\beta,$ $2\\gamma\\;$ and $\\angle AIB = \\delta .$\n\nThe problem is solved through angle calculations, commonly referred to as \"angle chasing\". The fact that $I$ is the incenter of $\\Delta ABC$ provides sufficient reasons and grounds to insure the satisfactory outcome.\n\nFirst of all, in $\\Delta ABC,$ $2\\alpha + 2\\beta + 2\\gamma = 180^{\\circ}$ whereas, in $\\Delta AIB,$ $\\alpha + \\beta + \\delta = 180^{\\circ}.$ This gives $\\delta = 90^{\\circ} + \\gamma .$ Since $AIB$ is an inscribed angle we get one arc relation:\n\n(1)\n\n$3 + 4 + 5 = 180^{\\circ} + 2\\gamma .$\n\nNext, $\\angle ACJ = \\angle BCJ = \\gamma .$ From the properties of the angles between two secants,\n\n(2)\n\n$4 - 2 = 5 - 1 = 2\\gamma .$\n\nSince $BI$ is the angle bisector, $\\angle ABI = \\angle CBI = \\beta ,$ from which\n\n(3)\n\n$1 = 2 = 2\\beta .$\n\nTogether with (2) this implies\n\n(4)\n\n$4 = 5 = 2\\gamma + 2\\beta .$\n\nFurther, $\\angle BAI = \\alpha$ and is subtended by the arc $BI$ so that\n\n(5)\n\n$1 + 6 = 2\\alpha .$\n\nCombining this with (3) gives\n\n(6)\n\n$6 = 2\\alpha - 2\\beta .$\n\nOn the other hand, from (1) and (4),\n\n(7)\n\n$3 = 180^{\\circ} - 2\\gamma - 4\\beta .$\n\nTogether, (6) and (7) show that $3 = 6,$ as required. (Indeed, the identity checks out: $2\\alpha - 2\\beta = 180^{\\circ} - 2\\gamma - 4\\beta$ is equivalent to $2\\alpha + 2\\beta + 2\\gamma = 180^{\\circ}.)$\n\n(A different proof has been posted by Vo Duc Dien at the wiki part of the site.)\n\nOne engaging interpretation of this result is that\n\nA circle through two vertices of a triangle and its incenter is always centered on the angle bisector of the remaining vertex and is therefore symmetric in that angle bisector.\n\nAnother solution makes a better use of this observation.\n\nThere are five solutions in all:", null, "### What Is Red Herring", null, "" ]

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[ "# zherfs.f (3) - Linux Man Pages\n\nzherfs.f -\n\n## SYNOPSIS\n\n### Functions/Subroutines\n\nsubroutine zherfs (UPLO, N, NRHS, A, LDA, AF, LDAF, IPIV, B, LDB, X, LDX, FERR, BERR, WORK, RWORK, INFO)\nZHERFS\n\n## Function/Subroutine Documentation\n\n### subroutine zherfs (characterUPLO, integerN, integerNRHS, complex*16, dimension( lda, * )A, integerLDA, complex*16, dimension( ldaf, * )AF, integerLDAF, integer, dimension( * )IPIV, complex*16, dimension( ldb, * )B, integerLDB, complex*16, dimension( ldx, * )X, integerLDX, double precision, dimension( * )FERR, double precision, dimension( * )BERR, complex*16, dimension( * )WORK, double precision, dimension( * )RWORK, integerINFO)\n\nZHERFS\n\nPurpose:\n\n``` ZHERFS improves the computed solution to a system of linear\nequations when the coefficient matrix is Hermitian indefinite, and\nprovides error bounds and backward error estimates for the solution.\n```\n\nParameters:\n\nUPLO\n\n``` UPLO is CHARACTER*1\n= 'U': Upper triangle of A is stored;\n= 'L': Lower triangle of A is stored.\n```\n\nN\n\n``` N is INTEGER\nThe order of the matrix A. N >= 0.\n```\n\nNRHS\n\n``` NRHS is INTEGER\nThe number of right hand sides, i.e., the number of columns\nof the matrices B and X. NRHS >= 0.\n```\n\nA\n\n``` A is COMPLEX*16 array, dimension (LDA,N)\nThe Hermitian matrix A. If UPLO = 'U', the leading N-by-N\nupper triangular part of A contains the upper triangular part\nof the matrix A, and the strictly lower triangular part of A\nis not referenced. If UPLO = 'L', the leading N-by-N lower\ntriangular part of A contains the lower triangular part of\nthe matrix A, and the strictly upper triangular part of A is\nnot referenced.\n```\n\nLDA\n\n``` LDA is INTEGER\nThe leading dimension of the array A. LDA >= max(1,N).\n```\n\nAF\n\n``` AF is COMPLEX*16 array, dimension (LDAF,N)\nThe factored form of the matrix A. AF contains the block\ndiagonal matrix D and the multipliers used to obtain the\nfactor U or L from the factorization A = U*D*U**H or\nA = L*D*L**H as computed by ZHETRF.\n```\n\nLDAF\n\n``` LDAF is INTEGER\nThe leading dimension of the array AF. LDAF >= max(1,N).\n```\n\nIPIV\n\n``` IPIV is INTEGER array, dimension (N)\nDetails of the interchanges and the block structure of D\nas determined by ZHETRF.\n```\n\nB\n\n``` B is COMPLEX*16 array, dimension (LDB,NRHS)\nThe right hand side matrix B.\n```\n\nLDB\n\n``` LDB is INTEGER\nThe leading dimension of the array B. LDB >= max(1,N).\n```\n\nX\n\n``` X is COMPLEX*16 array, dimension (LDX,NRHS)\nOn entry, the solution matrix X, as computed by ZHETRS.\nOn exit, the improved solution matrix X.\n```\n\nLDX\n\n``` LDX is INTEGER\nThe leading dimension of the array X. LDX >= max(1,N).\n```\n\nFERR\n\n``` FERR is DOUBLE PRECISION array, dimension (NRHS)\nThe estimated forward error bound for each solution vector\nX(j) (the j-th column of the solution matrix X).\nIf XTRUE is the true solution corresponding to X(j), FERR(j)\nis an estimated upper bound for the magnitude of the largest\nelement in (X(j) - XTRUE) divided by the magnitude of the\nlargest element in X(j). The estimate is as reliable as\nthe estimate for RCOND, and is almost always a slight\noverestimate of the true error.\n```\n\nBERR\n\n``` BERR is DOUBLE PRECISION array, dimension (NRHS)\nThe componentwise relative backward error of each solution\nvector X(j) (i.e., the smallest relative change in\nany element of A or B that makes X(j) an exact solution).\n```\n\nWORK\n\n``` WORK is COMPLEX*16 array, dimension (2*N)\n```\n\nRWORK\n\n``` RWORK is DOUBLE PRECISION array, dimension (N)\n```\n\nINFO\n\n``` INFO is INTEGER\n= 0: successful exit\n< 0: if INFO = -i, the i-th argument had an illegal value\n```\n\nInternal Parameters:\n\n``` ITMAX is the maximum number of steps of iterative refinement.\n```\n\nAuthor:\n\nUniv. of Tennessee\n\nUniv. of California Berkeley" ]

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[ "", null, "", null, "", null, "", null, "0\n\n# What are all the common multiples of 49 and 14?\n\nIt is not possible to write all the common multiples of two or more numbers.\n\nHowever, we can write as many as we want but first of all we have to find L.C.M. of 49 and 14 and the answer is 98.\n\nAlso, multiples of 98 are the common multiples of 49 and 14.\n\nIt means that multiples of least common factor of two or more numbers are the common multiples of the numbers.\n\nSo, first five common multiples are 98x1, 98x2, 98x3, 98x4 and 98x5.", null, "Study guides\n\n20 cards\n\n## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials\n\n➡️\nSee all cards\n3.74\n1196 Reviews", null, "Earn +20 pts", null, "", null, "" ]

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[ "## Saturday, August 04, 2012\n\n### bash error: value too great for base\n\nI came across this interesting error today:\n`-bash: 08: value too great for base (error token is \"08\")`\nIt was coming from a script which works out the previous month by extracting the current month from the current date and then decrementing it. The code looks like this:\n```today=\"\\$(date +%Y%m%d)\"\nmonth=\\${today:4:2}\nprevmonth=\\$((--month))\n```\nThis script throws an error only if the current month is 08 or 09. I found that the reason for this is that numbers starting with 0 are interpreted as octal numbers and 8 and 9 are not in the base-8 number system, hence the error. There are more details on the bash man page:\nConstants with a leading 0 are interpreted as octal numbers. A leading 0x or 0X denotes hexadecimal. Otherwise, numbers take the form [base#]n, where base is a decimal number between 2 and 64 represent- ing the arithmetic base, and n is a number in that base. If base# is omitted, then base 10 is used.\nTo fix this issue, I specified the base-10 prefix as shown below:\n```today=\"\\$(date +%Y%m%d)\"\nmonth=10#\\${today:4:2}\nprevmonth=\\$((--month))\n```" ]

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[ "## Conversion formula\n\nThe conversion factor from cups to deciliters is 2.365882375, which means that 1 cup is equal to 2.365882375 deciliters:\n\n1 cup = 2.365882375 dL\n\nTo convert 1.2 cups into deciliters we have to multiply 1.2 by the conversion factor in order to get the volume amount from cups to deciliters. We can also form a simple proportion to calculate the result:\n\n1 cup → 2.365882375 dL\n\n1.2 cup → V(dL)\n\nSolve the above proportion to obtain the volume V in deciliters:\n\nV(dL) = 1.2 cup × 2.365882375 dL\n\nV(dL) = 2.83905885 dL\n\nThe final result is:\n\n1.2 cup → 2.83905885 dL\n\nWe conclude that 1.2 cups is equivalent to 2.83905885 deciliters:\n\n1.2 cups = 2.83905885 deciliters\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 deciliter is equal to 0.35222940165541 × 1.2 cups.\n\nAnother way is saying that 1.2 cups is equal to 1 ÷ 0.35222940165541 deciliters.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one point two cups is approximately two point eight three nine deciliters:\n\n1.2 cup ≅ 2.839 dL\n\nAn alternative is also that one deciliter is approximately zero point three five two times one point two cups.\n\n## Conversion table\n\n### cups to deciliters chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from cups to deciliters\n\ncups (cup) deciliters (dL)\n2.2 cups 5.205 deciliters\n3.2 cups 7.571 deciliters\n4.2 cups 9.937 deciliters\n5.2 cups 12.303 deciliters\n6.2 cups 14.668 deciliters\n7.2 cups 17.034 deciliters\n8.2 cups 19.4 deciliters\n9.2 cups 21.766 deciliters\n10.2 cups 24.132 deciliters\n11.2 cups 26.498 deciliters" ]

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[ "# Aptitude - Aptitude- Speed\n\nRead many objective multiple choice questions of Aptitude - Aptitude- Speed in English. Practise online thousands of questions related to Aptitude - Aptitude- Speed. You can check online answer and try mock tests for Aptitude - Aptitude- Speed Questions provided in English.\n\nIn a game of 100 points, A can give B 20 points and C 28 points. Then, B can give C:\n\n•", null, "•", null, "•", null, "• a\n• b\n• c\n• d\n\nIn a 200 metres race A beats B by 35 m or 7 seconds. A's time over the course is:\n\n•", null, "•", null, "•", null, "• a\n• b\n• c\n• d\n\nA can run 22.5 m while B runs 25 m. In a kilometre race B beats A by ?\n\n•", null, "•", null, "•", null, "• a\n• b\n• c\n• d\n\nIn a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is ?\n\n•", null, "•", null, "•", null, "• a\n• b\n• c\n• d\n\nA runs 1 2/3 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?\n\n•", null, "•", null, "•", null, "• a\n• b\n• c\n• d" ]

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[ "# Efficient algorithms for solution of regularized total least squares\n\nRosemary Renaut, Hongbin Guo\n\nResearch output: Contribution to journalArticlepeer-review\n\n55 Scopus citations\n\n## Abstract\n\nError-contaminated systems Ax ≈ b, for which A is ill-conditioned, are considered. Such systems may be solved using Tikhonov-like regularized total least squares (RTLS) methods. Golub, Hansen, and O'Leary [SIAM J. Matrix Anal. Appl., 21 (1999), pp. 185-194] presented a parameter-dependent direct algorithm for the solution of the augmented Lagrange formulation for the RTLS problem, and Sima, Van Huffel, and Golub [Regularized Total Least Squares Based on Quadratic Eigenvalue Problem Solvers, Tech. Report SCCM-03-03, SCCM, Stanford University, Stanford, CA, 2003] have introduced a technique for solution based on a quadratic eigenvalue problem, RTLSQEP. Guo and Renaut [A regularized total least squares algorithm, in Total Least Squares and Errors-in-Variables Modeling: Analysis, Algorithms and Applications, S. Van Huffel and P. Lemmerling, eds., Kluwer Academic Publishers, Dordrecht, The Netherlands, 2002, pp. 57-66] derived an eigenproblem for the RTLS which can be solved using the iterative inverse power method. Here we present an alternative derivation of the eigenproblem for constrained TLS through the augmented Lagrangian for the constrained normalized residual. This extends the analysis of the eigenproblem and leads to derivation of more efficient algorithms compared to the original formulation. Additional algorithms based on bisection search and a standard L-curve approach are presented. These algorithms vary with respect to the parameters that need to be prescribed. Numerical and convergence results supporting the different versions and contrasting with RTLSQEP are presented.\n\nOriginal language English (US) 457-476 20 SIAM Journal on Matrix Analysis and Applications 26 2 https://doi.org/10.1137/S0895479802419889 Published - 2005\n\n## Keywords\n\n• Ill-posedness\n• Rayleigh quotient iteration\n• Regularization\n• Total least squares\n\n• Analysis\n\n## Fingerprint\n\nDive into the research topics of 'Efficient algorithms for solution of regularized total least squares'. Together they form a unique fingerprint." ]

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[ "#creditmodel-1.2.2\n\nIn this version I have: * Fixed some bugs in `get_ctree_rules`, `ks_plot`,`cross_table`.\n\n#creditmodel-1.2.1\n\nIn this version I have: * Enhanced strategy analysis capabilities. * New function `rule_value_replace` is for generating new variables by rules. * Fixed some potential bugs in `ks_plot`, `perf_table`,`training_model`,`process_nas`.\n\n# creditmodel-1.2\n\nIn this version I have: * Enhanced strategy analysis capabilities. * New function `replace_value` is for replacing values of some variables. * Fixed some potential bugs in `check_rules`, `get_ctree_rules`,`rules_filter`,`%alike%`.\n\n# creditmodel-1.1.9\n\nIn this version I have: * New function `plot_distribution` ,`plot_relative_freq_histogram`, `plot_box`,`plot_density`, `plot_bar` are for data visualization. * New function `swap_analysis` is for swap out/swap in analysis. * New function `rules_filter` is used to filter or select samples by rules * Fixed some potential bugs in `char_to_num`, `merge_category`,`check_rules`,`get_ctree_rules`.\n\nswap_analysis rules_filter # creditmodel-1.1.8\n\nIn this version I have: * New function `cross_table` is for cross table analysis. * Fixed some potential bugs in `data_cleansing`, `low_variance_filter`,`time_variable`,`plot_vars`.\n\n# creditmodel-1.1.7\n\nIn this version I have: * New function `entropy_weight` for is for calculating Entropy Weight. * New function `term_tfidf` for computing tf-idf of documents. * New function `plot_oot_perf` for plotting performance of over time samples in the future. * Fixed some potential bugs in `get_breaks`, `lift_plot`,`perf_table`,`model_result_plot`. * Add a parameter cut_bin to `get_breaks` for cutting breaks equal depth or equal width.\n\n# creditmodel-1.1.6\n\nIn this version I have:\n\n• Fixed some potential bugs in `split_bins`, `woe_transfer`\n\n# creditmodel-1.1.5\n\n• Add the function `time_series_proc` for time series data processing.\n• Add functions `ranking_percent_proc`,`ranking_percent_dict` are for processing ranking percent variables and generating ranking percent dictionary.\n• Change the function name `read_dt` to `read_data` and add and parameter pattern for matching files.\n• Change the parameter max.depth to max_depth of the function `traing_xgb`,‘xgb_params’\n• Change the function name `save_dt` to `save_data` and `save_data` also supports multiple data frames.\n\n# creditmodel-1.1.4\n\n• New function log_trans() is for logarithmic transformation\n• New function plot_table() make it possible to generate table graph.\n• New function multi_grid() for arranging list of plots into a grid.\n\n# creditmodel-1.1.3\n\nIn this version I have:\n\n• Fixed additional issues for the last version released on CRAN.\n• Add new functions `pred_xgb` for using xgboost model to predict new data.\n• Provide new functions `get_psi_plots`, `psi_plot` to plot PSI of your data..\n• Provide a function `p_to_score` for transforming probability to score.\n• Provide a function `multi_left_jion` for left jion a list of datasets fast.\n• Provide a function `read_data` for loading csv or txt data fast.\n\n# creditmodel-1.1.2\n\nIn this version I have:\n\n• Fixed additional issues for the last version released on CRAN.\n• Fixed some potential bugs in `xgb_filter`, `feature_selector`, `split_bins`, `ks_table_plot`, `ks_psi_plot`, `ks_value`.\n• Add a new function `pred_score` for predicting new data using scorecard.\n• Provide new functions `lr_params_search`, `xgb_params_search` for searching the optimal parameters. “random_search”,“grid_search”,“local_search” are available.\n• Provide new functions `partial_dependence_plot`, `get_partial_dependence_plots` for generating partial dependence plot.\n• Provide new functions `cohort_analysis`, `cohort_table`, `cohort_plot` for cohort (vintage) analysis and visualization.\n• Provide new functions `perf_table`, `roc_plot`, `ks_plot`, `lift_plot`, `psi_plot` for model validation drawings.\n\n# creditmodel-1.1.1\n\nIn this version I have: * Fixed some potential bugs in `get_names`, `digits_num`\n\n# creditmodel-1.1.0\n\nIn this version I have:\n\n• Add a function `data_exploration` for data exploration.\n• Fixed some potential bugs in `missing_proc`, `outliers_proc` ,`get_names`\n• In `lasso_filter`, `AUC`&`K-S` is added to select the best lambda. In this way, not only can the set of variables that makes the AUC or K-S maximized be selected, but also the multicollinearity (which is difficult to eliminate by AIC in stepwise regression), can be minimized. That means instead of stepwise regression, the optimal combination of variables can be selected by lasso to solve the regression problem.\n• Visualize the number of variables, `K-S` or `AUC` values corresponding to different lambda.\n• Provide new functions``auc_value`  `ks_value`, which can calculate Kolmogorov-Smirnov (K-S) & AUC of multiple model results quickly." ]

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[ "", null, "convert/disjcyc - Maple Help\n\nHome : Support : Online Help : convert/disjcyc\n\nconvert/disjcyc\n\nconvert a permutation in list notation or a word into disjoint cycle notation\n\nconvert/permlist\n\nconvert a permutation in disjoint cycle notation into list notation", null, "Calling Sequence convert(lperm, 'disjcyc') convert(perm, 'permlist', deg)", null, "Parameters\n\n lperm - permutation in list notation perm - the permutation deg - the degree of the permutation", null, "Description\n\n • The argument lperm is expected to be a permutation in list notation. This means that lperm is a list whose ith element is the image of i under the permutation. The permutation is returned in disjoint cycle notation.\n • A list is returned whose ith element is the image of i under perm, for $i=1..\\mathrm{deg}$.", null, "Examples\n\n > $\\mathrm{convert}\\left(\\left[3,4,1,2,7,6,5\\right],'\\mathrm{disjcyc}'\\right)$\n $\\left[\\left[{1}{,}{3}\\right]{,}\\left[{2}{,}{4}\\right]{,}\\left[{5}{,}{7}\\right]\\right]$ (1)\n > $\\mathrm{convert}\\left(\\left[\\left[2,4,1\\right],\\left[7,3\\right]\\right],'\\mathrm{permlist}',7\\right)$\n $\\left[{2}{,}{4}{,}{7}{,}{1}{,}{5}{,}{6}{,}{3}\\right]$ (2)" ]

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[ "lostinphys\nAt t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s.\n\ni am asked to find the acceleration at time t.\ni know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.\n\nlostinphys\n... ??\n\nStNowhere\nlostinphys said:\nAt t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (9.20 i + 6.10 j) m/s.\n\ni am asked to find the acceleration at time t.\ni know that the acceleration is the derivative of the velocity, meaning the change in velocity divided by the time. so i took the difference between the vectors (v - vi) and got (6.2i + 8.1j)/(2.20 - 0). But i don't think it is expressed correctly in vectors. any help would be appreciate. thanks.\n\nI think you're on the right track, but it looks like you're trying to use time as a vector here. remember, it divides both components of your $$\\Delta v$$ vector.\n\nGold Member\nWell the first part, 6.2i + 8.1j is correct. You divide by 2.20 seconds but it looks more like (6.2/2.20)i + (6.10/2.20)j with m/s^2 being the units for the vector. You can treat them seperately and think of it like 6.2/2.20 as the dv/dt in the i direction for example.\n\nlostinphys\nThe vector representing acceleration would then be (6.2/t)i + (8.1/t)j ?\n\nStNowhere\nThat sounds right.\n\nlostinphys\nthanks !" ]

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[ "", null, "", null, "Home > CCG > Chapter 7 > Lesson 7.2.2 > Problem7-66\n\n7-66.\n\nRead the Math Notes box in this lesson and then answer the following questions.\n\nThe cost of large flat-screen televisions is decreasing $20\\%$ per year. Homework Help ✎\n\n1. What is the multiplier?\n\nSubtract the percent from $1$.\n\n$0.8$\n\n2. If a $50$-inch flat-screen now costs $\\1200$, what will it cost in three years?\n\nThe equation for an exponential function is: $y=a·b^x$\n\n$a=\\text{initial value}\\\\ b=\\text{multiplier}\\\\ x=\\text{# of years}$\n\n$\\614.40$\n\n3. At the same rate, what did it cost two years ago?\n\nUse the same equation as (b) and do not change the multiplier." ]

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", null ]

[ "", null, "# 4.04 Short Division\n\n## Interactive practice questions\n\nCalculate $769\\div3$769÷​3 by doing the following.\n\na\n\nCalculate $600\\div3$600÷​3.\n\nb\n\nCalculate $150\\div3$150÷​3.\n\nc\n\nCalculate $18\\div3$18÷​3.\n\nd\n\nUsing the fact that $769=600+150+18+1$769=600+150+18+1, fill in the boxes with the missing numbers.\n\n$3$3 goes into seven hundred and sixty nine $\\editable{}$ times with a remainder of $\\editable{}$\n\nEasy\nApprox 2 minutes\n\nCalculate $404\\div5$404÷​5 by doing the following.\n\nCalculate $904\\div6$904÷​6 by doing the following.\n\nFind the value of $832\\div8$832÷​8." ]

[ null, "https://mathspace-production-static.mathspace.co/permalink/badges/v3/language-and-use-of-statistics.svg", null ]

[ "# Arithmetic Reasoning Shortcuts, Tutorials", null, "The two words, “arithmetic reasoning,” strike terror into the hearts of people who have been out of school for a while or who have concentrated their studies in the humanities. The phrase “quantitative reasoning” is probably equally as terrifying as it is puzzling. The quantitative reasoning test includes arithmetic problems as well as questions asking test takers to evaluate a set of facts and conclusions to determine if the conclusions are true or false. Not all exams for special agent include this section, but it is very similar to the logical reasoning test. Practice this section in the short exercise below.\n\nFor the straight arithmetic reasoning test, you do not need to understand intricate mathematical operations or to memorize complicated formulas. You can solve the problems arithmetically or by using some simple algebra.\nEach problem is presented as a short verbal description of a situation that includes some numerical facts. You must read the problem to determine what the question is, if a series of calculations will be required, you do not want to stop short of the answer because you misinterpreted the question. Likewise, you do not waste time going beyond what is asked.\n\nOnce you have determined what the question asks, you must settle on the best route for arriving at the answer and set the problem up accordingly. Finally, you must perform the calculations. Some special agent exams will give you “none of these” as the fifth (E) choice, but the TEA has a fifth numerical choice. It is important that you perform all calculations carefully because an estimate or an approximation may lead you to choose the wrong answer.\n\nReminder: Do all the practice exercise questions with pencil on scratch paper. You will not be permitted to use a calculator at the exam, so get used to doing arithmetic without one. Answer questions in the practice exercise by circling the letter of your choice.\n\n## Solved Examples - Arithmetic Reasoning\n\nExample 1) A police department purchases badges at 16$each for all the graduates of the police training academy. The last training class graduated 10 new officers. What is the total amount of money the department will spend for badges for these new officers? (A)$70\n(B) $116 (C)$160\n(D) $180 (E)$200\n\nSolution: The correct answer is (C). It can be obtained by computing the following:\n$16\\times 10\\:=\\:160$\nThe badges are priced at 16$each. The department must purchase 10 of them for the new officers. Multiplying the price of one badge (16$) by the number of graduates (10) gives the total price for all of the badges.\nChoice (A), (B), and (D) are the result of erroneous computations.\n\nExample 2) An investigator rented a car for six days and was charged 450$. The car rental company charged 35$ per day plus 0.30$per mile driven. How many miles did the investigator drive the car? Solution: The correct answer is (A). It can be obtained by computing the following: 6(35) + 0.30x = 450 The investigator rented the car for six days at 35$ per day, which is 210$; 210$ subtracted from the total charge of 450$leaves 240$, the portion of the total charge which was expended for the miles driven. This amount divided by charge per mile $(240\\div 0.30)$ gives the number of miles (800) driven by the investigator.\nChoices (B), (C), and (D) are the result of erroneous computations.\n\n## Arithmetic Reasoning Questions from Previous Year Exams\n\nVerbal Reasoning : Arithmetic Reasoning\n\n### Institute Management Software\n\nStart Teaching – Start Earning\n\nPlease comment on Arithmetic Reasoning Shortcuts, Tutorials" ]

[ null, "http://questionpaper.org/wp-content/uploads/2014/01/Arithmetic-Reasoning.jpg", null ]

[ "# How to automatically insert a new row and retain functions/formulas from last row?\n\nI have a table with cells that have functions/formulas, like this one:", null, "I need a script that creates a new row, copying with it the functions/formulas of the last used row. I find this script which create a new row but it doesn't copy functions/formulas. How could I implement this formatting copy task in Google Apps Script without having to manually select and copy?\n\n• if my previous row formula is =HTTPResponse(C4) how i can copy same formula but new cell number for new row ? – user1788736 Oct 5 '15 at 19:28\n\nUse the code below to copy also formula's as normal values. Add the code by selecting Tools from the spreadsheet menu. Then select script editor and add the code. Make sure to press the \"bug\" button and authenticate the script.\n\n## Code\n\n``````// global\nvar ss = SpreadsheetApp.getActive();\n\nfunction onOpen() {\n}\n\nvar sh = ss.getActiveSheet(), lRow = sh.getLastRow();\nvar lCol = sh.getLastColumn(), range = sh.getRange(lRow,1,1,lCol);\nsh.insertRowsAfter(lRow, 1);\nrange.copyTo(sh.getRange(lRow+1, 1, 1, lCol), {contentsOnly:false});\n}\n``````\n\n## Remark\n\nSetting the contentOnly to `false` will yield a standard copy. Setting it to `true`, will paste only values. The example script you found, does way more then pasting values.....\n\n## Example\n\nI've created an example file for you: Add Row With Formula's\n\nThis ArrayFormula can do just the same without involving a script. Enter it in D4, and it will be carried over automatically in any amount of empty cells below it.\n\n``````ArrayFormula(vlookup(B4:B:tax_table!\\$A\\$2:\\$G\\$8;3;true))\n``````\n\nNotes: \"B4:B\" means, look all the cells starting from B4 until the end of column.\n\nWhile, ArrayFormula takes care of copying itself into the cells below it. Just make sure that the cells below it are empty.\n\nIn case if you need to add a new row on top (first row) and copy formula from the first top row then you'll need to copy formulas across using `getFormulas()` and `setFormulas()` functions. You can change the value of `firstRow` to 2 if your spreadsheet has headers for example.\n\n``````function addFirstRow() {\nvar firstRow = 1;\nvar sh = ss.getActiveSheet();\nvar lCol = sh.getLastColumn();\nvar range = sh.getRange(firstRow, 1, 1, lCol);\nvar formulas = range.getFormulas();\nsh.insertRowsAfter(1, 1);\nnewRange = sh.getRange(firstRow, 1, 1, lCol);\nnewRange.setFormulas(formulas);\n}\n``````\n• Hi Denis, Are you referring to the post I created? I just check the script and it is still working. Regards Jacob Jan – Jacob Jan Tuinstra Jun 11 '16 at 9:46\n• Hi @JacobJanTuinstra, yes you are right. I was actually trying to add new row on top and retain the formula from the same row and it didn't work because script was trying to copy data from newly created empty row. I'll update my answer to reflect this. Thank you – dgpro Jun 13 '16 at 8:51\n• Hi @DenGordo! Thanks for this. How would I modify this if I just have values to move, and not formulas too? Thanks! – Drewdavid Feb 24 '17 at 17:05\n• Oh PS - Where do I reference which tab I am dealing with? Thanks again. – Drewdavid Feb 24 '17 at 17:06\n\nTo solve this problem I have used the `setFormula(range)`, for example:\n\n``````var ss = SpreadsheetApp.getActiveSpreadsheet();\nvar sheet = ss.getSheets();\n\nvar cell = sheet.getRange(\"B5\");\ncell.setFormula(\"=SUM(B3:B4)\");\n``````\n\nYour formula will autoincrease accordingly to the row index.\n\nLastly you could add an onUpdate or onEdit trigger.\n\n• -1; How is the code evoked? You're adding a sum, but you need to add a row physically. The `onUpdate` trigger doesn't exist. – Jacob Jan Tuinstra Jun 4 '14 at 4:31" ]

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[ "# #7fe759 Color Information\n\nIn a RGB color space, hex #7fe759 is composed of 49.8% red, 90.6% green and 34.9% blue. Whereas in a CMYK color space, it is composed of 45% cyan, 0% magenta, 61.5% yellow and 9.4% black. It has a hue angle of 103.9 degrees, a saturation of 74.7% and a lightness of 62.7%. #7fe759 color hex could be obtained by blending #feffb2 with #00cf00. Closest websafe color is: #66ff66.\n\n• R 50\n• G 91\n• B 35\nRGB color chart\n• C 45\n• M 0\n• Y 61\n• K 9\nCMYK color chart\n\n#7fe759 color description : Soft green.\n\n# #7fe759 Color Conversion\n\nThe hexadecimal color #7fe759 has RGB values of R:127, G:231, B:89 and CMYK values of C:0.45, M:0, Y:0.61, K:0.09. Its decimal value is 8382297.\n\nHex triplet RGB Decimal 7fe759 `#7fe759` 127, 231, 89 `rgb(127,231,89)` 49.8, 90.6, 34.9 `rgb(49.8%,90.6%,34.9%)` 45, 0, 61, 9 103.9°, 74.7, 62.7 `hsl(103.9,74.7%,62.7%)` 103.9°, 61.5, 90.6 66ff66 `#66ff66`\nCIE-LAB 83.117, -55.269, 58.292 39.13, 62.383, 19.43 0.324, 0.516, 62.383 83.117, 80.328, 133.475 83.117, -50.075, 81.134 78.983, -49.787, 40.703 01111111, 11100111, 01011001\n\n# Color Schemes with #7fe759\n\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #c159e7\n``#c159e7` `rgb(193,89,231)``\nComplementary Color\n• #c6e759\n``#c6e759` `rgb(198,231,89)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #59e77a\n``#59e77a` `rgb(89,231,122)``\nAnalogous Color\n• #e759c6\n``#e759c6` `rgb(231,89,198)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #7a59e7\n``#7a59e7` `rgb(122,89,231)``\nSplit Complementary Color\n• #e7597f\n``#e7597f` `rgb(231,89,127)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #597fe7\n``#597fe7` `rgb(89,127,231)``\n• #e7c159\n``#e7c159` `rgb(231,193,89)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #597fe7\n``#597fe7` `rgb(89,127,231)``\n• #c159e7\n``#c159e7` `rgb(193,89,231)``\n• #4fd51f\n``#4fd51f` `rgb(79,213,31)``\n• #5de12c\n``#5de12c` `rgb(93,225,44)``\n• #6ee443\n``#6ee443` `rgb(110,228,67)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #90ea6f\n``#90ea6f` `rgb(144,234,111)``\n• #a1ed86\n``#a1ed86` `rgb(161,237,134)``\n• #b3f19c\n``#b3f19c` `rgb(179,241,156)``\nMonochromatic Color\n\n# Alternatives to #7fe759\n\nBelow, you can see some colors close to #7fe759. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #a3e759\n``#a3e759` `rgb(163,231,89)``\n• #97e759\n``#97e759` `rgb(151,231,89)``\n• #8be759\n``#8be759` `rgb(139,231,89)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #73e759\n``#73e759` `rgb(115,231,89)``\n• #67e759\n``#67e759` `rgb(103,231,89)``\n• #5ce759\n``#5ce759` `rgb(92,231,89)``\nSimilar Colors\n\n# #7fe759 Preview\n\nThis text has a font color of #7fe759.\n\n``<span style=\"color:#7fe759;\">Text here</span>``\n#7fe759 background color\n\nThis paragraph has a background color of #7fe759.\n\n``<p style=\"background-color:#7fe759;\">Content here</p>``\n#7fe759 border color\n\nThis element has a border color of #7fe759.\n\n``<div style=\"border:1px solid #7fe759;\">Content here</div>``\nCSS codes\n``.text {color:#7fe759;}``\n``.background {background-color:#7fe759;}``\n``.border {border:1px solid #7fe759;}``\n\n# Shades and Tints of #7fe759\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #020501 is the darkest color, while #f6fdf3 is the lightest one.\n\n• #020501\n``#020501` `rgb(2,5,1)``\n• #081703\n``#081703` `rgb(8,23,3)``\n• #0f2806\n``#0f2806` `rgb(15,40,6)``\n• #153908\n``#153908` `rgb(21,57,8)``\n• #1c4a0b\n``#1c4a0b` `rgb(28,74,11)``\n• #225b0d\n``#225b0d` `rgb(34,91,13)``\n• #286c10\n``#286c10` `rgb(40,108,16)``\n• #2f7d12\n``#2f7d12` `rgb(47,125,18)``\n• #358e15\n``#358e15` `rgb(53,142,21)``\n• #3ca017\n``#3ca017` `rgb(60,160,23)``\n• #42b11a\n``#42b11a` `rgb(66,177,26)``\n• #48c21c\n``#48c21c` `rgb(72,194,28)``\n• #4fd31f\n``#4fd31f` `rgb(79,211,31)``\n• #57e026\n``#57e026` `rgb(87,224,38)``\n• #65e237\n``#65e237` `rgb(101,226,55)``\n• #72e548\n``#72e548` `rgb(114,229,72)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #8ce96a\n``#8ce96a` `rgb(140,233,106)``\n• #99ec7b\n``#99ec7b` `rgb(153,236,123)``\n• #a7ee8c\n``#a7ee8c` `rgb(167,238,140)``\n• #b4f19e\n``#b4f19e` `rgb(180,241,158)``\n• #c1f3af\n``#c1f3af` `rgb(193,243,175)``\n• #cef6c0\n``#cef6c0` `rgb(206,246,192)``\n• #dcf8d1\n``#dcf8d1` `rgb(220,248,209)``\n• #e9fbe2\n``#e9fbe2` `rgb(233,251,226)``\n• #f6fdf3\n``#f6fdf3` `rgb(246,253,243)``\nTint Color Variation\n\n# Tones of #7fe759\n\nA tone is produced by adding gray to any pure hue. In this case, #9ea59b is the less saturated color, while #75fd43 is the most saturated one.\n\n• #9ea59b\n``#9ea59b` `rgb(158,165,155)``\n``#9aad93` `rgb(154,173,147)``\n• #97b48c\n``#97b48c` `rgb(151,180,140)``\n• #93bb85\n``#93bb85` `rgb(147,187,133)``\n• #90c27e\n``#90c27e` `rgb(144,194,126)``\n• #8dca76\n``#8dca76` `rgb(141,202,118)``\n• #89d16f\n``#89d16f` `rgb(137,209,111)``\n• #86d868\n``#86d868` `rgb(134,216,104)``\n• #82e060\n``#82e060` `rgb(130,224,96)``\n• #7fe759\n``#7fe759` `rgb(127,231,89)``\n• #7cee52\n``#7cee52` `rgb(124,238,82)``\n• #78f64a\n``#78f64a` `rgb(120,246,74)``\n• #75fd43\n``#75fd43` `rgb(117,253,67)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #7fe759 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# What is accuracy? Is it a good model?\n\nAccuracy is a metric for evaluating classification models. It is calculated by dividing the number of correct predictions by the number of total predictions.\n\nAccuracy is not a good performance metric when there is imbalance in the dataset. For example, in binary classification with 95% of A class and 5% of B class, a constant prediction of A class would have an accuracy of 95%. In case of imbalance dataset, we need to choose Precision, recall, or F1 Score depending on the problem we are trying to solve." ]

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[ "", null, "DEtools - Maple Programming Help\n\nHome : Support : Online Help : Mathematics : Differential Equations : DEtools : Differential Rational Normal Forms : DEtools/PolynomialNormalForm\n\nDEtools\n\n PolynomialNormalForm\n construct the differential polynomial normal form of a rational function\n\n Calling Sequence PolynomialNormalForm(F, x)\n\nParameters\n\n F - rational function of x x - variable\n\nDescription\n\n • Let F be a rational function of x over a field K of characteristic 0. The PolynomialNormalForm(F,x) command constructs the differential polynomial normal form for F.\n • The output is a sequence of 3 elements $a,b,c$ where $a,b,c$ are polynomials over K such that:\n 1 $F=\\frac{a}{b}+\\frac{\\frac{ⅆ}{ⅆx}c}{c}.$\n 2 $\\mathrm{gcd}\\left(b,a-i\\frac{{\\partial }}{{\\partial }x}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}b\\right)=1$ for all non-negative integers $i$.\n 3 $\\mathrm{gcd}\\left(b,c\\right)=1$.\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{DEtools}\\right):$\n > $F≔\\frac{4}{x-2}+\\frac{4}{x+1}-\\frac{3}{{\\left(x+1\\right)}^{2}}-\\frac{9}{{\\left(x-1\\right)}^{2}}-\\frac{9{x}^{2}+12}{{x}^{3}+4x-2}+\\frac{1}{{\\left({x}^{3}+4x-2\\right)}^{2}}$\n ${F}{≔}\\frac{{4}}{{x}{-}{2}}{+}\\frac{{4}}{{x}{+}{1}}{-}\\frac{{3}}{{\\left({x}{+}{1}\\right)}^{{2}}}{-}\\frac{{9}}{{\\left({x}{-}{1}\\right)}^{{2}}}{-}\\frac{{9}{}{{x}}^{{2}}{+}{12}}{{{x}}^{{3}}{+}{4}{}{x}{-}{2}}{+}\\frac{{1}}{{\\left({{x}}^{{3}}{+}{4}{}{x}{-}{2}\\right)}^{{2}}}$ (1)\n > $a,b,c≔\\mathrm{PolynomialNormalForm}\\left(F,x\\right)$\n ${a}{,}{b}{,}{c}{≔}{-}{5}{}{{x}}^{{9}}{-}{16}{}{{x}}^{{8}}{-}{14}{}{{x}}^{{7}}{-}{134}{}{{x}}^{{6}}{+}{39}{}{{x}}^{{5}}{-}{331}{}{{x}}^{{4}}{+}{96}{}{{x}}^{{3}}{+}{32}{}{{x}}^{{2}}{+}{16}{}{x}{-}{7}{,}{\\left({x}{+}{1}\\right)}^{{2}}{}{\\left({x}{-}{1}\\right)}^{{2}}{}{\\left({{x}}^{{3}}{+}{4}{}{x}{-}{2}\\right)}^{{2}}{,}{\\left({x}{-}{2}\\right)}^{{4}}$ (2)\n\nCheck the result:\n\n > $\\mathrm{nF}≔\\frac{a}{b}+\\frac{\\mathrm{diff}\\left(c,x\\right)}{c}:$\n > $\\mathrm{Testzero}\\left(\\mathrm{normal}\\left(F-\\mathrm{nF}\\right)\\right)$\n ${\\mathrm{true}}$ (3)\n > $\\mathrm{res}≔\\mathrm{resultant}\\left(b,a-j\\mathrm{diff}\\left(b,x\\right),x\\right):$\n > $H≔\\mathrm{select}\\left(\\mathrm{type},\\left\\{\\mathrm{solve}\\left(\\mathrm{res},j\\right)\\right\\},'\\mathrm{nonnegint}'\\right):$\n > $\\mathrm{evalb}\\left(H=\\varnothing \\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathbf{and}\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}\\mathrm{gcd}\\left(b,c\\right)=1\\right)$\n ${\\mathrm{true}}$ (4)\n\nReferences\n\n Almkvist, G, and Zeilberger, D. \"The method of differentiating under the integral sign.\" Journal of Symbolic Computation. Vol. 10. (1990): 571-591." ]

[ null, "https://bat.bing.com/action/0", null ]

[ "# Difference between revisions of \"2015 AMC 8 Problems/Problem 11\"\n\nIn the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read \"AMC8\"?", null, "$\\textbf{(A) } \\frac{1}{22,050} \\qquad \\textbf{(B) } \\frac{1}{21,000}\\qquad \\textbf{(C) } \\frac{1}{10,500}\\qquad \\textbf{(D) } \\frac{1}{2,100} \\qquad \\textbf{(E) } \\frac{1}{1,050}$\n\nThe area of", null, "$\\triangle ABC$ is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is", null, "$\\sqrt{1^2+2^2}=\\sqrt{5}$, and its base is", null, "$\\sqrt{2^2+4^2}=\\sqrt{20}$. We multiply these and divide by 2 to find the of the triangle is", null, "$\\frac{\\sqrt{5 \\cdot 20}}2=\\frac{\\sqrt{100}}2=\\frac{10}2=5$. Since the grid has an area of", null, "$30$, the fraction of the grid covered by the triangle is", null, "$\\frac 5{30}=\\boxed{\\textbf{(A) }\\frac{1}{6}}$.\n\n## See Also\n\n 2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 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 All AJHSME/AMC 8 Problems and Solutions\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "Invalid username\nLogin to AoPS" ]

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[ "### Convert 100 feet 6 inches to CM\n\nHow much is 100 feet and 6 inches in CM? Use this calculator to convert 100 foot 6 to inches, cm, mm, and meters. One inch is equal to 2.54 centimeters, so 100'6 is equal to 3063.24 centimeters. Change the values in the calculator below to determine a different amount. Height is commonly referred to in cm in some countries and feet and inches in others. This calculates from 100'6 to centimeters.\n\n### Summary\n\nUse this calculator to convert feet and inches to other units.\nHow tall is 100 feet 6 inches in centimeters?\n100'6 = 3063.24 cm.\nHow long is 100 feet 6 inches in meters?\n100'6 = 306.324 m.\nHow much is 100 feet 6 inches in millimeters?\n100'6 = 30632.4 mm.\nHow big is 100 feet 6 inches in inches?\n100'6 = 1206 in.\nThis calculator is a convenient way to convert feet and inches to other units. Many countries use centimeters for height, so that can be confusing for people who use feet and inches. How many centimeters is that? How many meters is that? How high is that? How far is it?\n\nWhat is the inch to cm conversion? How many cm in an inch? 1 inch = 2.54 cm.\n##### CM to Feet and Inches Conversions\n3053 cm = 100.16404199475 ft\n3053 cm = 100'1.9685039370079\n\n3054 cm = 100.1968503937 ft\n3054 cm = 100'2.3622047244094\n\n3055 cm = 100.22965879265 ft\n3055 cm = 100'2.7559055118111\n\n3056 cm = 100.2624671916 ft\n3056 cm = 100'3.1496062992126\n\n3057 cm = 100.29527559055 ft\n3057 cm = 100'3.543307086614\n\n3058 cm = 100.3280839895 ft\n3058 cm = 100'3.9370078740158\n\n3059 cm = 100.36089238845 ft\n3059 cm = 100'4.3307086614172\n\n3060 cm = 100.3937007874 ft\n3060 cm = 100'4.724409448819\n\n3061 cm = 100.42650918635 ft\n3061 cm = 100'5.1181102362204\n\n3062 cm = 100.4593175853 ft\n3062 cm = 100'5.5118110236219\n\n3063 cm = 100.49212598425 ft\n3063 cm = 100'5.9055118110236\n\n3064 cm = 100.5249343832 ft\n3064 cm = 100'6.2992125984251\n\n3065 cm = 100.55774278215 ft\n3065 cm = 100'6.6929133858268\n\n3066 cm = 100.5905511811 ft\n3066 cm = 100'7.0866141732283\n\n3067 cm = 100.62335958005 ft\n3067 cm = 100'7.4803149606298\n\n3068 cm = 100.656167979 ft\n3068 cm = 100'7.8740157480315\n\n3069 cm = 100.68897637795 ft\n3069 cm = 100'8.267716535433\n\n3070 cm = 100.7217847769 ft\n3070 cm = 100'8.6614173228347\n\n3071 cm = 100.75459317585 ft\n3071 cm = 100'9.0551181102362\n\n3072 cm = 100.7874015748 ft\n3072 cm = 100'9.4488188976377\n\n3073 cm = 100.82020997375 ft\n3073 cm = 100'9.8425196850394\n\n##### Meters to Feet and Inches Conversions\n29.63 meters = 97.211286089239 ft\n29.63 cm = 97'2.535433070866\n\n29.73 meters = 97.53937007874 ft\n29.73 cm = 97'6.4724409448818\n\n29.83 meters = 97.867454068241 ft\n29.83 cm = 97'10.409448818898\n\n29.93 meters = 98.195538057743 ft\n29.93 cm = 98'2.3464566929135\n\n30.03 meters = 98.523622047244 ft\n30.03 cm = 98'6.2834645669293\n\n30.13 meters = 98.851706036745 ft\n30.13 cm = 98'10.220472440945\n\n30.23 meters = 99.179790026247 ft\n30.23 cm = 99'2.1574803149608\n\n30.33 meters = 99.507874015748 ft\n30.33 cm = 99'6.0944881889766\n\n30.43 meters = 99.835958005249 ft\n30.43 cm = 99'10.031496062993\n\n30.53 meters = 100.16404199475 ft\n30.53 cm = 100'1.9685039370083\n\n30.63 meters = 100.49212598425 ft\n30.63 cm = 100'5.9055118110241\n\n30.73 meters = 100.82020997375 ft\n30.73 cm = 100'9.8425196850399\n\n30.83 meters = 101.14829396325 ft\n30.83 cm = 101'1.7795275590556\n\n30.93 meters = 101.47637795276 ft\n30.93 cm = 101'5.7165354330714\n\n31.03 meters = 101.80446194226 ft\n31.03 cm = 101'9.6535433070874\n\n31.13 meters = 102.13254593176 ft\n31.13 cm = 102'1.5905511811031\n\n31.23 meters = 102.46062992126 ft\n31.23 cm = 102'5.5275590551191\n\n31.33 meters = 102.78871391076 ft\n31.33 cm = 102'9.4645669291349\n\n31.43 meters = 103.11679790026 ft\n31.43 cm = 103'1.4015748031504\n\n31.53 meters = 103.44488188976 ft\n31.53 cm = 103'5.3385826771664" ]

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[ "NL | EN\n\n## Linear Algebra\n\n2019-2020\n\n### Course Objective\n\nBesides to be able to explain, interrelate, know the basic properties\nof, and construct simple arguments with the concepts listed above, the\nstudent will learn the following skills (organized\nby topic):\n\nLinear systems:\n\nCan solve systems of linear equations using row-reduction\nCan determine the number of solutions of a linear system\nCan prove or disprove simple statements concerning linear systems\n\nLinear transformations:\n\nCan determine if a linear transformation is one-to-one and onto\nCan compute the standard matrix of a linear transformation\nCan use row-reduction to compute the inverse of a matrix\nCan prove or disprove simple statements concerning linear\ntransformations\n\nSubspaces and bases\n\nCan compute bases for the row and column space of a matrix\nCan compute the dimension and determine a basis of a subspace\nCan prove or disprove simple statements concerning linear systems\n\nEigenvalues and eigenvectors\n\nCan compute the eigenvalues of a matrix using the characteristic\nequation\nCan compute bases for the eigenspaces of a matrix\nCan diagonalize a matrix\nCan prove or disprove simple statements concerning eigenvalues and\neigenvectors\n\nOrthogonality\n\nCan compute the orthogonal projection onto a subspace\nCan determine an orthonormal basis for a subspace using the\nGramm-Schmidt algorithm\nCan solve least-squares problems using an orthogonal projection\nCan orthogonally diagonalize a symmetric matrix\nCan compute a singular value decomposition of a matrix\nCan prove or disprove simple statements concerning orthogonality\n\n### Course Content\n\nThe topics that will be treated are listed below. For every topic, the\nrelevant concepts are listed.\n\nLinear systems:\n\nlinear system (consistent/inconsistent/homogeneous/inhomogeneous),\n(augmented) coefficient matrix, row equivalence, pivot position/column,\n(reduced) echelon form, basic/free variable, spanning set, parametric\nvector form, linear (in)dependence.\n\nLinear transformations:\n\nlinear transformation, (co)domain, range and image, standard matrix,\none-to-one and onto, singularity, determinant, elementary matrices.\n\nSubspaces and bases:\n\nsubspace, column and null space, basis, coordinate system, dimension,\nrank.\n\nEigenvalues and eigenvectors:\n\neigenvalue, eigenvector, eigenspace, characteristic equation/polynomial,\nalgebraic multiplicity, similarity, diagonalization and\ndiagonalizability.\n\nOrthogonality:\n\ndot product, norm, distance, orthogonality, orthogonal complement,\northogonal set/basis, orthogonal projection, orthonormality, orthonormal\nbasis, Gramm-Schmidt process, least squares problem/solution, orthogonal\ndiagonalization, singular value/vector, singular value decomposition,\nMoore-Penrose inverse.\n\n### Teaching Methods\n\nThe course is spread over a period of seven weeks. Each week there will\nbe two theoretical classes of 90 minutes each and\ntwo exercise classes of 90 minutes each.\n\n### Method of Assessment\n\nThere is a written exam at the end of the course.\n\nNone.\n\n### Literature\n\nLinear Algebra and its Applications, by David C. Lay, Steven R. Lay en\nJudi J. McDonald, global edition (fifth edition), Pearson.\n\n### Target Audience\n\nIMM2, LI2, and CS2\n\nNone.\n\n### General Information\n\nCourse Code X_400649 6 EC P4 200 English Faculty of Science dr. R. Hindriks dr. R. Hindriks dr. R. Hindriks\n\n### Practical Information\n\nYou need to register for this course yourself\n\nLast-minute registration is available for this course.\n\nTeaching Methods Seminar, Lecture, Practical\nTarget audiences\n\nThis course is also available as:" ]

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[ "# Scala Testing\n\nhttp://www.scalatest.org/\n\nWriting TDD unit tests with scalatest\n\nAt a Glance\n\n## Examples¶\n\n```libraryDependencies += \"org.scalatest\" %% \"scalatest\" % \"2.2.6\" % \"test\"\n```\n```package com.acme.pizza\n\nimport org.scalatest.FunSuite\nimport org.scalatest.BeforeAndAfter\n\nclass PizzaTests extends FunSuite with BeforeAndAfter {\n\nvar pizza: Pizza = _\n\nbefore {\npizza = new Pizza\n}\n\ntest(\"new pizza has zero toppings\") {\nassert(pizza.getToppings.size == 0)\n}\n\ntest(\"adding one topping\") {\nassert(pizza.getToppings.size === 1)\n}\n\n// mark that you want a test here in the future\ntest (\"test pizza pricing\") (pending)\n\n}\n```\n\n## Styles¶\n\n### FunSuite¶\n\n```import org.scalatest.FunSuite\nclass AddSuite extends FunSuite {\ntest(\"3 plus 3 is 6\") {\nassert((3 + 3) == 6)\n}\n}\n```\n\n### FlatSpec¶\n\nThe structure of this test is flat—like xUnit, but the test name can be written in specification style:\n\n```import org.scalatest.FlatSpec\nclass AddSpec extends FlatSpec {\n\"Addition of 3 and 3\" should \"have result 6\" in {\nassert((3 + 3) == 0)\n}\n}\n```\n```import collection.mutable.Stack\nimport org.scalatest._\n\nclass ExampleSpec extends FlatSpec with Matchers {\n\n\"A Stack\" should \"pop values in last-in-first-out order\" in {\nval stack = new Stack[Int]\nstack.push(1)\nstack.push(2)\nstack.pop() should be (2)\nstack.pop() should be (1)\n}\n\nit should \"throw NoSuchElementException if an empty stack is popped\" in {\nval emptyStack = new Stack[Int]\na [NoSuchElementException] should be thrownBy {\nemptyStack.pop()\n}\n}\n}\n```\n\n### FeatureSpec¶\n\n```import org.scalatest._\n\nclass Calculator {\ndef add(a:Int, b:Int): Int = a + b\n}\n\nclass CalcSpec extends FeatureSpec with GivenWhenThen {\ninfo(\"As a calculator owner\")\ninfo(\"I want to be able add two numbers\")\ninfo(\"so I can get a correct result\")" ]

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[ "# Electric Circuit -Types of Electric Circuit\n\nThe main types of electric circuits are Close Circuit, Open Circuit, Short Circuit, Series Circuit, and Parallel Circuit. Electric circuit provides the conductive path for the flow of electric charge or electric current.\n\nIn this article, we will discuss the definition of electric circuits and the types of electric circuits.\n\n## What is an Electric Circuit?\n\nTypes of welding power sources and ...\nTypes of welding power sources and Types of welding machines\n\nWhen the power source is connected to the load with a conducting wire, it forms an electric circuit. A conductive wire made up of copper or aluminum is used to establish an electrical connection between the power source and the load.\n\nWe also use an ON / OFF switch and a fuse between the source and load to switch on/off the load and for the protection of the equipment connected to the source.\n\n## Types of Electric Circuit\n\nWe will discuss the different types of electric circuit.\n\n## Close Circuit\n\nIn a close circuit-\n\n• Load is connected to the source.\n• The source supplies current to the load.\n• The current flowing in the circuit depends on the voltage magnitude of the source.\n\n## Open Circuit\n\nThe circuit becomes an open circuit in the following cases.\n\n• When circuit is switched off\n• When the fuse gets blown off due to the fault in the circuit\n\nIn this condition, the current flowing through the close circuit interrupts, and the supply source and the load disconnect.\n\n## Short Circuit\n\nIn the case of short circuit ;\n\n• The connecting wires between the source and the load gets short circuit.\n• Maximum current flow through the circuit\n• It blew off the fuses\n• Finally circuit becomes open circuit.\n\nThe main reason for the short circuit is insulation failure of the connecting wires or the insulation failure in the electrical equipment.\n\n## Series Circuit\n\nWhen 2 or more electrical equipment is connected in the series connection, it forms a series circuit. In the series circuit, the magnitude of the current flowing in the equipment is the same. The series circuit has a single path for current flow.\n\nWe call series connection an end-to-end connection or cascade connection. The disadvantage of the series circuit is the whole circuit becomes an open circuit if one piece of equipment becomes defective.\n\n### Properties of Series circuit:\n\n• The same magnitude of the current travels in every load.\n• The source voltage is equal to the sum of voltage drop across each load.\nV = V+ V+ V+ …..+ Vn\n• The equivalent resistance of the circuit is equal to sum of individual load resistance.\nReq = R1 +R2 +R3 +R4 +……….+ Rn\n• The equivalent resistance(Req ) has the highest resistance value of all the individual resistances.\n\n## Parallel Circuit\n\nIn a parallel circuit,\n\n• Two or more loads connected across the supply source.\n• The current flowing through each loads depends on the load resistance. The lower resistance draws more current and, the higher resistance draws less current as per Ohm’s law.\n• The voltage across all the loads is the same.\n• If one of the loads disconnects, the other loads keep operating.\n\n### Properties of Parallel Circuits:\n\n• The potential difference is the same across all the parallel loads.\n• The current distribution in the loads is according to the individual load resistance.\n• The total current drawn by all the loads is equal to the sum of individual current of the load.\nI = I+ I+ I+ ……+ In\n• The reciprocal of the equivalent resistance of a parallel circuit is equivalent to the sum of the reciprocal of the individual resistances.\n1/R = 1/R1 + 1/R2 + 1/R3+ ……… 1/Rn\n• The equivalent resistance is the lesser than the smallest of all the resistance.\nR < R1, R < R2, ….., R < Rn\n• The equivalent conductance is the mathematical addition of the single conductances.\nG = G+ G+G+ ……+ Gn\n\nThe equivalent resistance is lesser than the smallest of all the resistances linked in parallel.\n\n## Other Types of Circuits\n\n1. Series-Parallel Circuit\n2. DC circuit\n3. AC Circuit" ]

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[ "# Evolute of an egg\n\nThe set of lines perpendicular to a curve are tangent to a second curve called the evolute. The lines perpendicular to the ellipse below are tangent to the curve inside called an astroid.", null, "If we replace the ellipse with an egg, we get a similar shape, but less symmetric.", null, "The equation for the egg is described here with parameters a = 3, b = 2, and k = 0.1. The ellipse above has the same a and b but k = 0.\n\nI made the lines slightly transparent, setting alpha = 0.4, so the graph would be darker where many lines cross.\n\nRelated post: Envelopes of epicycloids" ]

[ null, "https://www.johndcook.com/ellipse_evolute.png", null, "https://www.johndcook.com/egg_evolute.png", null ]

[ "# Awesome SHM!", null, "A ring of mass $m$ and radius $a$ is connected to an inextensible string which passes over a frictionless pulley. The other end of the string is connected to the upper end of a massless spring of spring constant $k$. The lower end of the spring is fixed to the ground. The ring can rotate in the vertical plane about hinge without any friction. If the horizontal position of the ring is in equilibrium, then the time period of small oscillations is given by $T=2 \\pi \\sqrt{\\frac{im}{bk}}.$\n\nFind $i+b.$\n\nNote: $\\gcd(i,b)=1.$\n\nThis problem is not original.\n\n×" ]

[ null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/Ou4AHlQxvZ-group-2-1.svg", null ]

[ "# Python – Find Minimum of Two Numbers\n\nContents\n\n## Find Minimum of Two Numbers\n\nThere are many ways to find out the minimum of given two numbers. In this tutorial, we will learn three ways of finding out the minimum of two numbers.\n\n### 1. Find Minimum using min() builtin function\n\nTo find the minimum of given two numbers in Python, call min() builtin function and pass the two numbers as arguments. min() returns the smallest or the minimum of the given two numbers.\n\nThe syntax to find the minimum of two numbers: a, b using min() is\n\n``min(a, b)``\n\nIn the following program, we take numeric values in two variables a and b, and find out the minimum of these two numbers using min().\n\nPython Program\n\n``````a = 7\nb = 52\nminimum = min(a, b)\nprint(f'Minimum of {a}, {b} is {minimum}')``````\nRun Code Copy\n\nOutput\n\n``Minimum of 7, 52 is 7``\n\n### 2. Find Minimum Value using If Statement\n\nIn the following program, we will use simple If statement to find the minimum of given two numbers.\n\nPython Program\n\n``````a = 7\nb = 52\n\nminimum = a\nif b > a :\nminimum = b\n\nprint(f'Minimum of {a}, {b} is {minimum}')``````\nRun Code Copy\n\nOutput\n\n``Minimum of 7, 52 is 7``\n\n### 3. Writing our own Lambda Function to find Minimum Value\n\nWe can also write a lambda function to find the minimum of two numbers. We use Ternary Operator inside lambda to choose the minimum value.\n\nIn the following program, we write a lambda function that can return the maximum of given two numbers.\n\nPython Program\n\n``````a = 7\nb = 52\nminimum = lambda a, b: a if a < b else b\nprint(f'Minimum of {a}, {b} is {minimum(a, b)}')``````\nRun Code Copy\n\nOutput\n\n``Minimum of 7, 52 is 7``\nCode copied to clipboard successfully 👍" ]

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[ "# Large matrix inversion\n\nI'm still learning Mathematica, and I'm trying to write a program that needs to invert a large (nxn, n=160 or possibly larger than this) matrix with numerical entries, multiply with to another matrix, diagonalize the resulting matrix and then compute the n ipr (inverse participation ratio) values from its n eigenvectors. Because at the end of the program I also need to find some statistical quantities (mean,median, min, max) on these ipr values, I am using a Do loop to repeat the aforementioned matrix operations 100 times, so that I obtain a 100-dim list (each element is itself a list of n ipr values), and average over each column, which will give me a list of n average ipr values.\n\nHere's the program:\n\nn = 160; t = 100;\nH = Table[0.0, {2 n}, {2 n}];\nipr[v_] := Sum[Abs[v[[i]]]^4, {i, Length[v]}];\niprv = Table[0.0, {i, t}];\n\nDo[\nPrint[s];\nm = Table[RandomReal[{0, 0.5}], n];\nk = RandomReal[{0, \\[Pi]}];\nM = Table[If[i == j, 2, 1/(Abs[i - j])^3], {i, 1, n}, {j, 1, n}];\nF = Inverse[M];\nPrint[\"y\"];\nPsi = Abs[F.m];\nh1 = Table[If[i == j, (1 - Cos[k]) + Psi[[i]], Sqrt[Psi[[i]] Psi[[j]]]/(2*(Abs[i - j])^3)],\n{i, 1, n}, {j, 1, n}];\nh2 = Table[If[i == j, Psi[[i]], Sqrt[Psi[[i]] Psi[[j]]]/(2*(Abs[i - j])^3)], {i, 1, n}, {j,\n1, n}];\nH = ArrayFlatten[{{h1, h2}, {-h2, -h1}}];\n{h, u} = Eigensystem[H]; v = Pick[u, NonNegative[h]];\niprv [[s]] = Table[ipr[v[[i]]], {i, Length[v]}],\n{s, t}\n]\n\nmipr = Mean[iprv];\n\nmax = Max[mipr]\nmin = Min[mipr]\nme = Mean[mipr]\nmed = Median[mipr]\nstat = {me, med, min, max};\nPrint[stat];\n\n\nThe problem is that this program runs very slow on my computer (it takes approx 5 min to do each loop), and I believe the problem is the large matrix inversion. Do you have any experience on inverting large matrices on Mathematica and any suggestion on how to speed it up? Is there anything else in the Do loop that can be optimized (even the Do loop itself), that might be slowing down the program?\n\nThank you so much for the help!\n\n• You're taking the inverse of a matrix of rationals, and this is very slow. Since you take the dot product of the inverse with a vector of machine numbers, you should convert your matrix to machine numbers first. So, something like: M = Table[If[i == j, 2., 1./(Abs[i - j])^3], {i, 1, n}, {j, 1, n}]. Also, it seems you're taking an inverse of the same matrix during each iteration, that seems unnecessary. – Carl Woll Mar 18 at 23:32\n• Please include the necessary definitions in your program (n=, etc.) – A.G. Mar 18 at 23:34\n• Sorry I cut the first line of code by mistake. n = 160; t = 100; – ofey Mar 18 at 23:57\n• Carl Woll thank you for the answer, you're right about matrix of rationals and repeating the matrix inversion, I'll try to change that – ofey Mar 19 at 0:01\n• Actually you are solving a system of linear equations M.Psi=m using matrix inversion. This is very inefficient. For this purpose you should not use Inverse[M], but rather Psi=LinearSolve[M,m]. – yarchik Mar 19 at 8:20\n\n## 2 Answers\n\nTo get the matrix inversion to work on machine numbers just replace\n\nF = Inverse[0.0 + M];\n\n\nThat brings the running time to under 150 seconds on my MBPro -- replace the Do line by\n\nTiming@Do[\n\n\nto have Mathematica tell you about running time.\n\nYou can indeed shave a few seconds off by moving the matrix definition and inverse out of the Do loop, but not much.\n\nHere are a couple of further suggestions to make the code faster and a bit shorter.\n\nn = 160;\nt = 100;\nid = IdentityMatrix[n, SparseArray, WorkingPrecision -> MachinePrecision];\nA = 0.5 ((DistanceMatrix[Range[1., n]] + id)^-3 - id);\nM = 2. (A + id);\nF = LinearSolve[M];\niprv = ConstantArray[0., {n, t}];\n\nParallelDo[\nBlock[{m, k, Psi, B, h1, h2, H, h, u, v},\nm = RandomReal[{0, 0.5}, {n}];\nk = RandomReal[{0, \\[Pi]}];\nPsi = Abs[F[m]];\nB = Sqrt[KroneckerProduct[Psi, Psi]] A;\nh1 = DiagonalMatrix[SparseArray[(1 - Cos[k]) + Psi]] + B;\nh2 = DiagonalMatrix[SparseArray[Psi]] + B;\nH = ArrayFlatten[{{h1, h2}, {-h2, -h1}}];\n{h, u} = Eigensystem[H];\nv = Pick[u, UnitStep[h], 1];\niprv[[s]] = Total[Abs[v]^4, {2}]\n]\n, {s, 1, t}, Method -> \"CoarsestGrained\"];\n\n\nThis takes about 1.5 seconds on my 4 Core machine.\n\nThat's what I did:\n\n• What Carl Woll and A.G. pointed out already and which makes the biggest difference is converting M to a matrix of machine doubles.\n\n• Instead of using Table, I used various ways to create the matrices. In contrast to Table, all these guarantee that the resulting matrices are stored as packed arrays.\n\n• In general, it is a good idea to use LinearSolve instead of Inverse, both for performance and accuracy. LinearSolve computes an LU-factorization (stored in F within a LinearSolveFunction object), that can be reused multiple times. So factorizing M once in the beginning definitely saves some time.\n\n• Also the matrix A is reused a couple of times. This avoids having to recompute the distance matrix in each iteration of the loop.\n\n• Most of these methods (e.g., componentwise addition and maultiplication, KroneckerProduct, applying Sqrt or Power componentwise) are also vectorized.\n\n• In particular, this applies to the replacement Total[Abs[v]^4, {2}] for ipr.\n\n• If I am not mistaken, the computations in each loop iteration are fully independent. Thus, we can use ParallelDo to parallelize it. Note that there is a considerable overhead when a Parallel construct is called for the first time in a Mathematica session, because some packages are loaded on the fly.\n\nI also observed that the eigenvalues seem to come always as pairs (on positive, one negative). This is due to the special structure of the matrix H. By exploiting this structure, it is likely that one can reduce your problem to solving a smaller $$n \\times n$$ eigensystem. But I did not attempt to do it.\n\n• These are all very useful comments, thank you! I will update my code with your suggestions. – ofey Mar 20 at 19:03\n• You're welcome! – Henrik Schumacher Mar 20 at 19:11" ]

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[ "### Fold a String into an Int type\n\nWhile working my way through (Sullivan, Goerzen & Stewart, 2009), I came across the problem of converting a String to an Int using folds. This isn't that difficult a problem, but because I'm still a beginner to Haskell, I had to think about it for a few days.\n\nFinally, I came up with the following (unsafe) solution using foldl.\nimport Data.Char (digitToInt)\n\nasInt :: String -> Int\nasInt [] = 0\nasInt (x:xs) | x == '-' = (-1) * asInt xs\n| otherwise = sum $foldl helper []$ zip (reverse [0..(length (x:xs) - 1)]) (x:xs)\nwhere\nhelper :: [Int] -> (Int,Char) -> [Int]\nhelper totalList (order,digit) = digitToInt digit * 10 ^ order : totalList\n\nI separated the state of summing from the rest of the problem, using the fold to  'loop' over the list and raise each digit to the appropriate magnitude based on its position, just as one would in an imperative language. Thus, I believe this is a hybrid solution that uses both recursion and a fold (because of the way I've written my helper function).\n\nfoldl is of course not the most memory-efficient implementation, again according to the same chapter in (Sullivan et al., 2009). But because the maximum number of digits we may have in this type is 9 (it ranges from  $$[-2^{29},2^{29}-1]$$), I don't think lazy evaluation becomes an issue unless you're performing this operation many times, or in large numbers at the same time.\n\nThis being said, I did find a solution online by Ben Doerr that appears to be quite a little bit simpler. His solution also does not make use of any recursion, likely due to his initial element being 0 instead of the empty list as I've used.\n\nOf course, one of the simplest approaches I've come across is read someString :: Int.\n\nAnd because no post is complete without a little Python script, here's an implementation using the same concept I used above in Haskell with fn.py.\n#! /usr/bin/env python3.6\n# -*- coding: utf-8 -*-\n# vim:fenc=utf-8\n\nfrom typing import Text, Callable, List, Tuple\nfrom fn.op import foldl\n\nChar = Text\n\ndef asInt(string: str) -> int:\n\"\"\" Convert a str to an int. \"\"\"\nif string == '-':\nreturn -asInt(string[1:])\n\ndef digitToInt(c: Char) -> int:\nassert len(c) == 1\nreturn int(c)\n\ndef helper(totalList: List[int], tup: Tuple[int, Char]) -> List[int]:\norder, digit = tup\nvalue = digitToInt(digit) * 10 ** order\ntotalList.append(value)\n\nindexes = reversed(range(len(string)))\n\nreturn sum(foldl(helper, [])(zip(indexes, string)))\n\nif __name__ == '__main__':\ninteger = asInt(\"54321\")\nprint(integer) # 54321\nprint(type(integer)) #\n\nIt's easiest to understand how this works if you add a print(totalList) command before return in helper, which prints the following.\n\n[50000, 4000]\n[50000, 4000, 300]\n[50000, 4000, 300, 20]\n[50000, 4000, 300, 20, 1]\n\nwhich is then of course summed and returned.\n\nReferences\n\nO'Sullivan, B., Goerzen, J., & Stewart, D. (2009). Real World Haskell. Sebastopol, CA: O'Reilly." ]

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[ "mvrnormArma {anMC} R Documentation\n\n## Sample from multivariate normal distribution with C++\n\n### Description\n\nSimulates realizations from a multivariate normal with mean mu and covariance matrix sigma.\n\n### Usage\n\nmvrnormArma(n, mu, sigma, chol)\n\n\n### Arguments\n\n n number of simulations. mu mean vector. sigma covariance matrix or Cholesky decomposition of the matrix (see chol). chol integer, if 0 sigma is a covariance matrix, otherwise it is the Cholesky decomposition of the matrix.\n\n### Value\n\nA matrix of size d x n containing the samples.\n\n### Examples\n\n# Simulate 1000 realizations from a multivariate normal vector\nmu <- rep(0,200)\nSigma <- diag(rep(1,200))\nrealizations<-mvrnormArma(n=1000,mu = mu,sigma=Sigma, chol=0)\nempMean<-rowMeans(realizations)\nempCov<-cov(t(realizations))\n# check if the sample mean is close to the actual mean\nmaxErrorOnMean<-max(abs(mu-empMean))\n# check if we can estimate correctly the covariance matrix\nmaxErrorOnVar<-max(abs(rep(1,200)-diag(empCov)))\nmaxErrorOnCov<-max(abs(empCov[lower.tri(empCov)]))\n## Not run:\nplot(density(realizations[2,]))\n\n## End(Not run)\n\n\n[Package anMC version 0.2.5 Index]" ]

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[ "## ↤ l\n\n👤 will chen 🗓 October 17, 2021, 10:13 pm ( Last Modified )\n\nGrade 1 subtraction worksheets. In first grade, children subtract single-digit numbers with numbers from 0 to 10. They solve subtraction problems with a missing number and use addition to solve subtraction problems. . If the worksheet does not fit the page in the Print Preview, adjust your margins, header, and footer in the Page Setup ..Grade 1 subtraction worksheets. Our grade 1 subtraction worksheets provide practice in solving basic subtraction problems. Exercises begin with simple subtraction facts using pictures or number lines and progress to subtraction of 2-digit numbers in columns. Our grade 1 exercises do not require regrouping (or \"borrowing\")..Second Grade Subtraction Games By second grade, some students have already decided they don't like math, which can make it difficult to motivate them to practice. With our second grade subtraction games, even the most reluctant learners will enjoy the quirky characters, colorful animation, and entertaining challenges..Mixed addition & subtraction word problems Grade 1 Word Problems Worksheet Read carefully! 1. 4 birds are sitting on a branch. 1 flies away. How many birds are left on the branch? 2. There are 6 birds and 3 nests. How many more birds are there than nests? 3. 3 raccoons are playing in the woods..\n\nThis is a comprehensivedfdsffs collection of free printable math worksheets for grade 1, organized by topics such as addition, subtraction, place value, telling time, and counting money. They are randomly generated, printable from your browser, and include the answer key. The worksheets support any first grade math program, but go especially ..Regrouping – Addition and Subtraction Fraction Worksheets Multiplication Worksheets Times Table Worksheets Brain Teaser Worksheets Picture Analogies Cut and Paste Worksheets Pattern Worksheets Dot to Dot worksheets Preschool and Kindergarten – Mazes Size Comparison Worksheets. Top Worksheets New Worksheets Most Popular Math Worksheets ..Each worksheet was created by a professional educator, so you know your child will learn critical age-appropriate facts and concepts. Best of all, many worksheets across a variety of subjects feature vibrant colors, cute characters, and interesting story prompts, so kids get excited about their learning adventure..\n\n.\n\nRelated to \"Grade 1 Subtraction Worksheet\" ⤵\n\nName : __________________\n\nSeat Num. : __________________\n\nDate : __________________\n\n4 - 8 = ...\n\n9 - 8 = ...\n\n6 - 3 = ...\n\n2 - 7 = ...\n\n7 - 7 = ...\n\n2 - 3 = ...\n\n2 - 7 = ...\n\n1 - 2 = ...\n\n2 - 3 = ...\n\n4 - 8 = ...\n\n2 - 4 = ...\n\n4 - 3 = ...\n\n3 - 8 = ...\n\n7 - 9 = ...\n\n2 - 6 = ...\n\n7 - 5 = ...\n\n9 - 3 = ...\n\n3 - 3 = ...\n\n9 - 8 = ...\n\n3 - 9 = ...\n\n1 - 2 = ...\n\n6 - 1 = ...\n\n3 - 2 = ...\n\n8 - 3 = ...\n\n6 - 4 = ...\n\n9 - 4 = ...\n\n9 - 3 = ...\n\n8 - 8 = ...\n\n9 - 2 = ...\n\n8 - 3 = ...\n\n3 - 3 = ...\n\n9 - 3 = ...\n\n3 - 3 = ...\n\n1 - 2 = ...\n\n7 - 4 = ...\n\n4 - 3 = ...\n\n1 - 5 = ...\n\n1 - 7 = ...\n\n5 - 5 = ...\n\n7 - 7 = ...\n\n3 - 2 = ...\n\n3 - 9 = ...\n\n3 - 2 = ...\n\n9 - 8 = ...\n\n3 - 7 = ...\n\n3 - 1 = ...\n\n3 - 6 = ...\n\n3 - 8 = ...\n\n4 - 7 = ...\n\n9 - 5 = ...\n\n3 - 3 = ...\n\n1 - 2 = ...\n\n9 - 1 = ...\n\n9 - 1 = ...\n\n7 - 9 = ...\n\n1 - 3 = ...\n\n1 - 1 = ...\n\n3 - 7 = ...\n\n1 - 3 = ...\n\n5 - 5 = ...\n\n6 - 8 = ...\n\n4 - 4 = ...\n\n1 - 9 = ...\n\n1 - 1 = ...\n\n1 - 5 = ...\n\n7 - 2 = ...\n\n1 - 6 = ...\n\n9 - 4 = ...\n\n6 - 8 = ...\n\n6 - 7 = ...\n\n6 - 7 = ...\n\n1 - 1 = ...\n\n1 - 7 = ...\n\n6 - 2 = ...\n\n1 - 1 = ...\n\n5 - 3 = ...\n\n2 - 7 = ...\n\n4 - 2 = ...\n\n6 - 6 = ...\n\n9 - 8 = ...\n\n9 - 2 = ...\n\n6 - 1 = ...\n\n7 - 3 = ...\n\n5 - 2 = ...\n\n2 - 2 = ...\n\n5 - 8 = ...\n\n3 - 2 = ...\n\n7 - 5 = ...\n\n5 - 5 = ...\n\n4 - 5 = ...\n\n9 - 7 = ...\n\n7 - 4 = ...\n\n6 - 2 = ...\n\n3 - 1 = ...\n\n7 - 4 = ...\n\n4 - 5 = ...\n\n3 - 7 = ...\n\n2 - 5 = ...\n\n1 - 2 = ...\n\n8 - 7 = ...\n\n8 - 9 = ...\n\n9 - 9 = ...\n\n2 - 3 = ...\n\n5 - 9 = ...\n\n4 - 3 = ...\n\n1 - 1 = ...\n\n1 - 3 = ...\n\n5 - 6 = ...\n\n4 - 3 = ...\n\n3 - 5 = ...\n\n9 - 5 = ...\n\n2 - 1 = ...\n\n9 - 7 = ...\n\n4 - 8 = ...\n\n6 - 8 = ...\n\n8 - 1 = ...\n\n5 - 1 = ...\n\n7 - 4 = ...\n\n5 - 7 = ...\n\n4 - 9 = ...\n\n7 - 1 = ...\n\n4 - 3 = ...\n\n8 - 4 = ...\n\n3 - 7 = ...\n\n2 - 5 = ...\n\n5 - 5 = ...\n\n1 - 5 = ...\n\n6 - 3 = ...\n\n5 - 6 = ...\n\n2 - 5 = ...\n\n8 - 7 = ...\n\n9 - 9 = ...\n\n9 - 9 = ...\n\n7 - 4 = ...\n\n4 - 3 = ...\n\n6 - 4 = ...\n\n3 - 3 = ...\n\n3 - 3 = ...\n\n8 - 7 = ...\n\n5 - 3 = ...\n\n7 - 6 = ...\n\n2 - 7 = ...\n\n5 - 5 = ...\n\n9 - 2 = ...\n\n6 - 9 = ...\n\n3 - 4 = ...\n\n1 - 9 = ...\n\n1 - 6 = ...\n\n9 - 4 = ...\n\n1 - 6 = ...\n\n4 - 6 = ...\n\n7 - 9 = ...\n\n7 - 4 = ...\n\n5 - 5 = ...\n\n6 - 2 = ...\n\n6 - 3 = ...\n\n1 - 8 = ...\n\n2 - 7 = ...\n\n7 - 4 = ...\n\n2 - 2 = ...\n\n6 - 3 = ...\n\n5 - 8 = ...\n\n5 - 1 = ...\n\n5 - 1 = ...\n\n6 - 1 = ...\n\n3 - 6 = ...\n\n6 - 5 = ...\n\n1 - 3 = ...\n\n7 - 4 = ...\n\n9 - 5 = ...\n\n8 - 6 = ...\n\n7 - 5 = ...\n\n8 - 6 = ...\n\n6 - 8 = ...\n\n3 - 9 = ...\n\n7 - 9 = ...\n\n5 - 8 = ...\n\n3 - 8 = ...\n\n9 - 1 = ...\n\n7 - 3 = ...\n\n5 - 4 = ...\n\n9 - 1 = ...\n\n9 - 9 = ...\n\n8 - 5 = ...\n\n7 - 7 = ...\n\n3 - 1 = ...\n\n3 - 3 = ...\n\n8 - 7 = ...\n\n1 - 4 = ...\n\n9 - 3 = ...\n\n8 - 9 = ...\n\n9 - 1 = ...\n\n5 - 8 = ...\n\n9 - 4 = ...\n\n9 - 3 = ...\n\n4 - 1 = ...\n\n8 - 7 = ...\n\n2 - 5 = ...\n\n7 - 4 = ...\n\n1 - 3 = ...\n\nshow printable version !!!hide the show", null, "", null, "First Grade Subtraction Timed Tests. Subtract 0-10. From Firm Foundations In Education. Math Subtraction", null, "ADDITION \\u0026 SUBTRACTION 60 Printable Worksheets With Single Etsy Math Worksheets", null, "Math Worksheet ~ Free Mathrksheets First Grade Subtraction Subtract Digit From No Regrouping Of 41 Printable Math Worksheets For Grade 1 Photo Inspirations. Free Printable Math Worksheets For Grade 1 And 2.", null, "2 Digit Subtraction Worksheets", null, "2 Digit Subtraction Worksheets", null, "Horizontal Addition And Subtraction - Math Worksheets - MathsDiary.com", null, "2 Digit Subtraction Worksheets Math Subtraction", null, "Worksheet ~ Grade Maths Worksheets Free Pdf Document Cbse Games 46 Grade 1 Maths Worksheets Picture Inspirations. Grade 1 Maths Games. Grade 1 Maths Worksheets Document Pdf. Grade 1 Maths Worksheets.", null, "", null, "Printable Subtraction Worksheets For Grade 1 Subtract 10 Worksheet \\u0026 Add And Subtract Within 20 1st Grade - Worksheets Schools", null, "Column Subtraction (No Regrouping) - 2 Digits Sheet 1 Worksheet For 2nd - 3rd Grade Lesson Planet", null, "2 Digit Subtraction With Regrouping Worksheets", null, "Math Worksheet : Math Worksheet Printable Worksheets For Grade Free First Subtraction Subtracting Whole Tens Missing Number Printable Math Worksheets For Grade 1 ~ Roleplayersensemble", null, "Free Printable Number Subtraction (1-10) Worksheets For Grade 1 And Kindergarten - Subtraction With Pictures/Objects To Cross Out - Subtraction Using Number Line - MegaWorkbook", null, "Math Worksheet Free Mathorksheets For Grade Firstord Problems Printable Worksheets Subtraction 1 Coloring Pages With Pictures First Word Year Mixed Addition And 2 Digit — Oguchionyewu", null, "2 Digit Borrow Subtraction – Regrouping – 4 Worksheets School Worksheets", null, "Subtraction Worksheets For Special Education - K And Grade 1", null, "Single Digit Subtraction Worksheets For 1st Grade (Page 1) - Line.17QQ.com", null, "", null, "Math Worksheet ~ Free Math Worksheets First Grade Subtraction Subtracting Printable For Photo 41 Printable Math Worksheets For Grade 1 Photo Inspirations. 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Grade 1 Maths Worksheets Document Free.", null, "", null, "Printable Free Math Worksheets First Grade 1 Subtraction Subtracting 1 Digit From 2 Digit No Regrouping Three Digit Subtraction With Regrouping Worksheets - Worksheets Schools", null, "Subtracting Horizontally / Horizontal Subtraction - Math Worksheets - MathsDiary.com", null, "Free Printable Number Subtraction (1-10) Worksheets For Grade 1 And Kindergarten - Subtraction With Pictures/Objects To Cross Out - Subtraction Using Number Line - MegaWorkbook", null, "5 Free Math Worksheets First Grade 1 Subtraction Subtracting 1 Digit From 2 Digit No Regrouping - Apocalomegaproductions.com", null, "True Or False Subtraction Worksheet For 1st Grade (Free Printable)", null, "Subtraction Worksheets For Grade 1 Of 5 Free Math Worksheets First Grade 1 Addition Add In Columns 2 Digit Plus 1 Digit No Regroupi - Free Templates", null, "Math Subtraction Worksheets Grade 1 Printable Worksheets And Activities For Teachers", null, "Math Analysis Year 3 Maths Worksheets Free Subtraction Worksheets For Grade 1 Free First Grade Math Worksheets Multiplication Table Practice Games Grade 7 Math Curriculum Math Data Sheet Element Math Game Integers", null, "5 Free Math Worksheets First Grade 1 Subtraction Add And Subtract 4 Single Digit Numbers - Apocalomegaproductions.com", null, "", null, "", null, "", null, "Free Subtraction Worksheets To 12", null, "Math Grade 1 Subtract From 10 Worksheet", null, "Free Math Worksheets And Printouts", null, "Free Printable Number Subtraction (1-10) Worksheets For Grade 1 And Kindergarten - Subtraction With Pictures/Objects To Cross Out - Subtraction Using Number Line - MegaWorkbook", null, "Grade 1 Subtraction (Kumon Math Workbooks): Kumon Publishing: 9781933241500: Amazon.com: Books", null, "Worksheets Subtraction Stories For First Grade Printable Worksheets And Activities For Teachers", null, "Printable Free Math Worksheets First Grade 1 Subtraction Single Digit Subtraction Addition Subtraction Numbers 1 10 Kinder Lessons Tes Teach - Worksheets Schools", null, "Grade 1 Subtraction Worksheets Kids Activities", null, "Subtracting Horizontally / Horizontal Subtraction - Math Worksheets - MathsDiary.com", null, "Std 1 Math Christmas Subtraction Worksheets Regrouping Division Worksheets 9th Grade Free Printable Subtraction Worksheets For 4th Grade Best Math Curriculum In The World Math Games To Play With Kids Std 1", null, "", null, "Math Worksheet : Free Printable First Gradeath Worksheets Addition Adding Digit Plus No Regrouping The Printinus Subtraction With Of 64 Free Printable First Grade Math Worksheets Picture Inspirations ~ Roleplayersensemble", null, "Subtraction Grade 1 (Page 1) - Line.17QQ.com", null, "", null, "", null, "18 Best Addition And Subtraction Worksheets Grade 1 Images On Worksheets Ideas", null, "3 Free Math Worksheets First Grade 1 Subtraction Subtracting Whole Tens Missing Number - Apocalomegaproductions.com", null, "Subtraction Worksheets For Special Education - K And Grade 1", null, "Grade 1 - Subtraction Worksheet 2 - Kidschoolz", null, "9 Subtraction Worksheets For Grade 1 - Free Templates", null, "Free Math Worksheets And Printouts", null, "", null, "Fun Activities To Teach Primary Math Numbers 1-15 Worksheets Subtraction Worksheets For Grade 1 Kindergarten Numbers 1 20 Grade 9 Mathematics Textbook Free Worksheet Generator Multiplication Sheet Easy Reading Worksheets Christmas Coloring", null, "", null, "4 Free Math Worksheets First Grade 1 Subtraction Subtract 1 Digit From 2 Digit No Regrouping - Worksheets Schools", null, "Subtraction Worksheets For Grade 1 (Page 1) - Line.17QQ.com", null, "Worksheet ~ First Grade Math Worksheets Mental Subtraction To Worksheet Amazing Free For Facts 1st Amazing Free Worksheets For Grade 1. Free English Worksheets For Grade 1 Printable Worksheets. Free English Worksheets", null, "1st Grade Math Worksheets - Best Coloring Pages For Kids 1st Grade Math Worksheets", null, "1st Grade Subtraction Worksheet - Promotiontablecovers", null, "Horizontal Addition And Subtraction - Math Worksheets - MathsDiary.com", null, "13 Ace Addition And Subtraction Worksheets Pdf Coloring Pages Resource Materials Web Forms Fact Sheets Built-in Functions Basic Operations — Oguchionyewu", null, "", null, "Free Printable Number Subtraction (1-10) Worksheets For Grade 1 And Kindergarten - Subtraction With Pictures/Objects To Cross Out - Subtraction Using Number Line - MegaWorkbook", null, "Subtraction Online Worksheet For Grade 1", null, "Best Worksheets By Maxwell Worksheets Ideas", null, "Math Worksheets For Grade 1 Activit - Ota Tech", null, "Details Of 4 Free Math Worksheets First Grade 1 Subtraction Subtract", null, "Math Worksheet : Free Worksheets For Grade Language Curriculum Mathematics Shapes With Answers 49 Splendi Mathematics Worksheets For Grade 1 ~ Roleplayersensemble", null, "Free Addition And Subtraction Coloring Pages", null, "", null, "Subtraction Worksheets For Special Education - K And Grade 1", null, "Fun Addition And Subtraction Worksheet Kids Activities", null, "3 Free Math Worksheets First Grade 1 Subtraction Subtraction Up To 20 No Regrouping - Apocalomegaproductions.com", null, "Subtraction Worksheets For 1st Grade Kids", null, "Free 1st Grade Math Worksheets — Mashup Math", null, "Printable Pictures Of Coins Fun Math Worksheets Grade 4 Addition And Subtraction Worksheets For Grade 1 Reading Packets For 3rd Grade Printable Pictures Of Coins Fun Math Games For Children Good Worksheets", null, "", null, "Subtraction Worksheets", null, "True Or False Subtraction Worksheet For 1st Grade (Free Printable)", null, "Worksheet ~ Free Grade Maths Worksheets Singapore To Print Pdf 46 Grade 1 Maths Worksheets Picture Inspirations. Free Grade 1 Maths Worksheets Singapore. Grade 1 Maths Worksheets To Print Out. Grade 1 Maths Worksheets To Print.", null, "Phenomenal Grade 1 Maths Worksheets – Liveonairbk", null, "", null, "Makeing Subtraction Stories For Grade 1 Worksheets Printable Worksheets And Activities For Teachers", null, "", null, "Beginner Subtraction – 10 Kindergarten Subtraction Worksheets / FREE Printable Worksheets – Worksheetfun", null, "Subtraction Lesson Plan Clarendon Learning", null, "Two Digit Addition Worksheets From The Teacher's Guide 2nd Grade Math Worksheets", null, "Printable Grade 1 Math Subtraction Worksheets (Page 1) - Line.17QQ.com", null, "Grade 1 (Primary 1) - Subtraction Within 10 Math Worksheets - YouTube", null, "Grade One Mathematics Worksheet For Kids PrintablEducation", null, "5 Free Math Worksheets First Grade 1 Subtraction Add And Subtract 4 Single Digit Numbers - Worksheets Schools" ]

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[ "## Build Wavelet Tree Objects\n\nThe following sections explain how to extend the toolbox with new objects through four examples.\n\n### Building a Wavelet Tree Object (WTREE)\n\nThis example creates a new class of objects: WTREE.\n\nStarting from the class DTREE and overloading the methods `split` and `merge`, we define a wavelet tree class.\n\nTo plot a WTREE, the DTREE `plot` method is used.\n\nThe definition of the new class is described below.\n\nClass WTREE (parent class: DTREE)\n\n#### Fields\n\n `dtree` Parent object `dwtMode` DWT extension mode `wavInfo` Structure (wavelet information)\n\n#### wavInfo Structure information\n\n `wavName` Wavelet Name `Lo_D` Low Decomposition filter `Hi_D` High Decomposition filter `Lo_R` Low Reconstruction filter `Hi_R` High Reconstruction filter\n\n#### Methods\n\n `wtree` Constructor for the class WTREE. `merge` Merge (recompose) the data of a node. `split` Split (decompose) the data of a terminal node.\n\n### Working With Wavelet Tree Objects (WTREE)\n\n1-D Object\n\n```load noisbloc x = noisbloc;```\n\nDefine the level and the wavelet.\n\n```lev = 3; wav = 'db2';```\n\nCreate the wavelet tree.\n\n`t = wtree(x,lev,wav);`\n\nPlot the wavelet tree. The approximations are labeled in yellow and the details are labeled in red. The detail nodes cannot be split.\n\n`plot(t)`", null, "Change Node Label from Visualize to Split-Merge, and then click node (7), to get this figure:", null, "Change Node Action from Visualize to Split-Merge, and merge node (3). Then change Node Action from Split-Merge to Visualize, and click node (3) to get this figure:", null, "2-D Object\n\n`load woman`\n\nDefine the level and the wavelet.\n\n```lev = 2; wav = 'db2';```\n\nCreate the wavelet tree.\n\n`t = wtree(X,lev,wav);`\n\nPlot the tree.\n\n`plot(t)`", null, "Change Node Label from Depth_Position to Index. Click the node (5). You get the following plot.", null, "Click the node (2). You obtain the following plot.", null, "Change Node Action from Visualize to Split-Merge. Split the node (5). Change Node Action from Split-Merge to Visualize. Click the node (21). You obtain the following plot.", null, "### Building a Right Wavelet Tree Object (RWVTREE)\n\nThis example creates a new class of objects: RWVTREE.\n\nWe define a right wavelet tree class starting from the class WTREE and overloading the methods `split`, `merge`, and `plot` (inherited from DTREE).\n\nThe `plot` method shows how to add Node Labels.\n\nThe definition of the new class is described below.\n\nClass RWVTREE (parent class: WTREE)\n\n#### Fields\n\n `dummy` Not used `wtree` Parent object\n\n#### Methods\n\n `rwvtree` Constructor for the class RWVTREE. `merge` Merge (recompose) the data of a node. `plot` Plot RWVTREE object. `split` Split (decompose) the data of a terminal node.\n\n### Working With Right Wavelet Tree Objects (RWVTREE)\n\n1-D Object\n\n```load noisbloc x = noisbloc;```\n\nDefine the level and the wavelet.\n\n```lev = 3; wav = 'db2';```\n\nCreate the wavelet tree.\n\n`t = rwvtree(x,lev,wav);`\n\nPlot the tree. The approximations are labeled in yellow and the details are labeled in red. The detail nodes cannot be split.\n\n`plot(t)`", null, "Change Node Action from Visualize to Split-Merge. Merge the node (6). Change Node Action from Split-Merge to Visualize. Click the node (6). You obtain the following plot.", null, "2-D Object\n\n`load woman`\n\nDefine the level and the wavelet.\n\n```lev = 2; wav = 'db2';```\n\nCreate the wavelet tree.\n\n`t = wtree(X,lev,wav);`\n\nPlot the tree.\n\n`plot(t)`", null, "Click the node (2,0). You get the following plot.", null, "Change Node Action from Visualize to Split-Merge. Split the node (2,0). Change Node Action from Split-Merge to Visualize. Click the node (3,0). You obtain the following plot.", null, "### Building a Wavelet Tree Object (WVTREE)\n\nThis example creates a new class of objects: WVTREE.\n\nWe define a wavelet tree class starting from the class WTREE and overloading the methods `get`, `plot`, and `recons` (all inherited from DTREE).\n\nThe `split` and `merge` methods of the class WTREE are used.\n\nThe `plot` method shows how to add Node Labels and Node Actions.\n\nThe definition of the new class is described below.\n\nClass WVTREE (parent class: WTREE)\n\n#### Fields\n\n `dummy` Not used `wtree` Parent object\n\n#### Methods\n\n `wvtree` Constructor for the class WVTREE. `get` Get WVTREE object field contents. `plot` Plot WVTREE object. `recons` Reconstruct node coefficients.\n\n### Working With Wavelet Tree Objects (WVTREE)\n\n1-D Object\n\n``` load noisbloc x = noisbloc;```\n\nDefine the level and the wavelet.\n\n```lev = 3; wav = 'db2';```\n\nCreate the wavelet tree.\n\n`t = wvtree(x,lev,wav);`\n\nPlot the tree. The approximations are labeled in yellow and the details are labeled in red. The detail nodes cannot be split.\n\n` plot(t)`", null, "Change Node Action from Visualize to Split-Merge. Merge the node (3). Change Node Action from Split-Merge to Reconstruct. Click the node (3). You obtain the following plot.", null, "2-D Object\n\n`load woman`\n\nDefine the level and the wavelet.\n\n```lev = 2; wav = 'db1';```\n\nCreate the wavelet tree.\n\n`t = wvtree(X,lev,wav);`\n\nPlot the tree.\n\n`plot(t)`", null, "Click the node (5). You get the following plot.", null, "Click the node (2). You obtain the following plot.", null, "Change Node Action from Visualize to Split-Merge. Split the node (5). Change Node Action from Split-Merge to Reconstruct. Click the node (21). You obtain the following plot.", null, "### Building a Wavelet Tree Object (EDWTTREE)\n\nThis example creates a new class of objects: EDWTTREE.\n\nWe define an ε-DWT tree class starting from the class DTREE and overloading the methods `merge`, `plot`, `recons`, and `split`.\n\nThe `plot` method shows how to add Node Labels, Node Actions, and Tree Actions.\n\nThe definition of the new class is described below.\n\nClass EDWTTREE (parent class: DTREE)\n\n#### Fields\n\n `dtree` Parent object `dwtMode` DWT extension mode `wavInfo` Structure (wavelet information)\n\n#### Fields Description\n\n`wavInfo`\n\n `wavName` Wavelet Name `Lo_D` Low Decomposition filter `Hi_D` High Decomposition filter `Lo_R` Low Reconstruction filter `Hi_R` High Reconstruction filter\n\n#### Methods\n\n `edwttree` Constructor for the class EDWTTREE. `merge` Merge (recompose) the data of a node. `plot ` Plot EDWTTREE object. `recons ` Reconstruct node coefficients. `split` Split (decompose) the data of a terminal node.\n\n### Working With Wavelet Tree Object (EDWTTREE)\n\n```load noisbloc x = noisbloc;```\n\nDefine the level and the wavelet.\n\n```lev = 2; wav = 'haar';```\n\nCreate the wavelet tree.\n\n`t = edwttree(x,lev,wav);`\n\nPlot the tree. The approximations are labeled in yellow and the details are labeled in red. The detail nodes cannot be split. The title of the figure contains the DWT extension mode used (`'per'` in the present example) and the name of the denoising method.\n\n`plot(t)`", null, "Click the node (0). You obtain the following plot.", null, "Change Node Action from Visualize to Split-Merge. Split the nodes (5) and (20).", null, "Select Tree Action: De-noise. Click the node (0). You obtain the following plot.", null, "" ]

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[ "", null, ", 03.12.2019 01:10 rosier2230\n\n# Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the comparison property to find the lower and upper bounds for i = \\displaystyle \\int_{-2}^{2} f(x) \\, dxi = ∫ − 2 2 f ( x ) d x.", null, "### Another question on Mathematics", null, "Mathematics, 03.02.2019 07:14\nHelloooooo . what are the solutions to 10x^2 - 38x - 8 = 0? is this when i find the factors and use the numbers within the factors to determine the zeros? i hope that made sense lol", null, "Mathematics, 02.02.2019 16:36\nWhat is the voltage across the resistor if a current of 0.5a flows through a 20 ohm resistor? a. 4v b.10v c.0.25v d.8v", null, "Mathematics, 02.02.2019 11:08\nTwo numbers have a sum of -19. one number is x. what expression represents the other number?", null, "Mathematics, 02.02.2019 06:48\nMath question.. plz me .. best answer all\nFind the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2...\nQuestions", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Questions on the website: 6666256" ]

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[ "# Sumin Evgeniy V. Course Schedule\n\n### Monday\n\n10:15 — 11:50\nSem\nProbability Theory and Mathematical Statistics B17-511\n14:30 — 16:05\nLec\nProbability Theory and Mathematical Statistics S17-101 , S17-102 , S17-103 , S17-201\n16:15 — 17:50\nSem\nProbability Theory and Mathematical Statistics S17-101 , S17-201\n\n### Tuesday\n\n10:15 — 11:50\nSem\nProbability Theory and Mathematical Statistics B17-504\n12:45 — 14:20\nSem\nProbability Theory and Mathematical Statistics B17-501 , B17-901\n14:30 — 16:05\nLec\nProbability Theory and Mathematical Statistics B17-501 , B17-504 , B17-511 , B17-514 , B17-901\n\n### Thursday\n\n12:45 — 14:20\nSem\nProbability Theory and Mathematical Statistics B17-514" ]

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[ "# #020715 Color Information\n\nIn a RGB color space, hex #020715 is composed of 0.8% red, 2.7% green and 8.2% blue. Whereas in a CMYK color space, it is composed of 90.5% cyan, 66.7% magenta, 0% yellow and 91.8% black. It has a hue angle of 224.2 degrees, a saturation of 82.6% and a lightness of 4.5%. #020715 color hex could be obtained by blending #040e2a with #000000. Closest websafe color is: #000000.\n\n• R 1\n• G 3\n• B 8\nRGB color chart\n• C 90\n• M 67\n• Y 0\n• K 92\nCMYK color chart\n\n#020715 color description : Very dark (mostly black) blue.\n\n# #020715 Color Conversion\n\nThe hexadecimal color #020715 has RGB values of R:2, G:7, B:21 and CMYK values of C:0.9, M:0.67, Y:0, K:0.92. Its decimal value is 132885.\n\nHex triplet RGB Decimal 020715 `#020715` 2, 7, 21 `rgb(2,7,21)` 0.8, 2.7, 8.2 `rgb(0.8%,2.7%,8.2%)` 90, 67, 0, 92 224.2°, 82.6, 4.5 `hsl(224.2,82.6%,4.5%)` 224.2°, 90.5, 8.2 000000 `#000000`\nCIE-LAB 1.978, 1.155, -7.163 0.236, 0.219, 0.739 0.198, 0.183, 0.219 1.978, 7.256, 279.16 1.978, -0.851, -3.212 4.68, 0.826, -6.09 00000010, 00000111, 00010101\n\n# Color Schemes with #020715\n\n• #020715\n``#020715` `rgb(2,7,21)``\n• #151002\n``#151002` `rgb(21,16,2)``\nComplementary Color\n• #021115\n``#021115` `rgb(2,17,21)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #060215\n``#060215` `rgb(6,2,21)``\nAnalogous Color\n• #111502\n``#111502` `rgb(17,21,2)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #150702\n``#150702` `rgb(21,7,2)``\nSplit Complementary Color\n• #071502\n``#071502` `rgb(7,21,2)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #150207\n``#150207` `rgb(21,2,7)``\n• #021510\n``#021510` `rgb(2,21,16)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #150207\n``#150207` `rgb(21,2,7)``\n• #151002\n``#151002` `rgb(21,16,2)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #040f2c\n``#040f2c` `rgb(4,15,44)``\n• #061744\n``#061744` `rgb(6,23,68)``\n• #091e5b\n``#091e5b` `rgb(9,30,91)``\nMonochromatic Color\n\n# Alternatives to #020715\n\nBelow, you can see some colors close to #020715. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #020c15\n``#020c15` `rgb(2,12,21)``\n• #020a15\n``#020a15` `rgb(2,10,21)``\n• #020915\n``#020915` `rgb(2,9,21)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #020515\n``#020515` `rgb(2,5,21)``\n• #020415\n``#020415` `rgb(2,4,21)``\n• #020215\n``#020215` `rgb(2,2,21)``\nSimilar Colors\n\n# #020715 Preview\n\nThis text has a font color of #020715.\n\n``<span style=\"color:#020715;\">Text here</span>``\n#020715 background color\n\nThis paragraph has a background color of #020715.\n\n``<p style=\"background-color:#020715;\">Content here</p>``\n#020715 border color\n\nThis element has a border color of #020715.\n\n``<div style=\"border:1px solid #020715;\">Content here</div>``\nCSS codes\n``.text {color:#020715;}``\n``.background {background-color:#020715;}``\n``.border {border:1px solid #020715;}``\n\n# Shades and Tints of #020715\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000103 is the darkest color, while #f0f4fe is the lightest one.\n\n• #000103\n``#000103` `rgb(0,1,3)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #040d27\n``#040d27` `rgb(4,13,39)``\n• #051339\n``#051339` `rgb(5,19,57)``\n• #07194b\n``#07194b` `rgb(7,25,75)``\n• #091f5d\n``#091f5d` `rgb(9,31,93)``\n• #0b256f\n``#0b256f` `rgb(11,37,111)``\n• #0c2b80\n``#0c2b80` `rgb(12,43,128)``\n• #0e3192\n``#0e3192` `rgb(14,49,146)``\n• #1037a4\n``#1037a4` `rgb(16,55,164)``\n• #113db6\n``#113db6` `rgb(17,61,182)``\n• #1343c8\n``#1343c8` `rgb(19,67,200)``\n• #1549da\n``#1549da` `rgb(21,73,218)``\n• #1950e9\n``#1950e9` `rgb(25,80,233)``\n• #2b5eeb\n``#2b5eeb` `rgb(43,94,235)``\n• #3d6bed\n``#3d6bed` `rgb(61,107,237)``\n• #4f79ee\n``#4f79ee` `rgb(79,121,238)``\n• #6187f0\n``#6187f0` `rgb(97,135,240)``\n• #7394f2\n``#7394f2` `rgb(115,148,242)``\n• #85a2f3\n``#85a2f3` `rgb(133,162,243)``\n• #97aff5\n``#97aff5` `rgb(151,175,245)``\n• #a9bdf7\n``#a9bdf7` `rgb(169,189,247)``\n• #bacbf8\n``#bacbf8` `rgb(186,203,248)``\n• #ccd8fa\n``#ccd8fa` `rgb(204,216,250)``\n• #dee6fc\n``#dee6fc` `rgb(222,230,252)``\n• #f0f4fe\n``#f0f4fe` `rgb(240,244,254)``\nTint Color Variation\n\n# Tones of #020715\n\nA tone is produced by adding gray to any pure hue. In this case, #0b0b0c is the less saturated color, while #000617 is the most saturated one.\n\n• #0b0b0c\n``#0b0b0c` `rgb(11,11,12)``\n• #0a0b0d\n``#0a0b0d` `rgb(10,11,13)``\n• #090a0e\n``#090a0e` `rgb(9,10,14)``\n• #080a0f\n``#080a0f` `rgb(8,10,15)``\n• #070a10\n``#070a10` `rgb(7,10,16)``\n• #060911\n``#060911` `rgb(6,9,17)``\n• #060911\n``#060911` `rgb(6,9,17)``\n• #050812\n``#050812` `rgb(5,8,18)``\n• #040813\n``#040813` `rgb(4,8,19)``\n• #030714\n``#030714` `rgb(3,7,20)``\n• #020715\n``#020715` `rgb(2,7,21)``\n• #010716\n``#010716` `rgb(1,7,22)``\n• #000617\n``#000617` `rgb(0,6,23)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #020715 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# A Markov Chain Approximation for Choice Modeling\n\nVineet Goyal\n\n## Affiliation:\n\nColumbia University\nIndustrial Engineering and Operations Research\n500 West, 120th Street\nNew York, NY 10027\nUnited States of America\n\n## Time:\n\nTuesday, 20 August 2013, 14:30 to 15:30\n\n• AG-69\n\n## Organisers:\n\nAssortment planning is an important problem that arises in many industries such as retailing and airlines. One of the key challenges in an assortment planning problem is to identify the right model'' for the substitution behavior of customers from the data. Error in model selection can lead to highly sub-optimal decisions. In this paper, we present a new choice model that is a simultaneous approximation for all random utility based discrete choice models including the multinomial logit, the nested logit and mixtures of multinomial logit models. Our model is based on a new primitive for substitution behavior where substitution from one product to another is modeled as a state transition of a Markov chain.\n\nWe show that the choice probabilities computed by our model are a good approximation to the true choice probabilities of any random utility discrete based choice model under mild conditions. Moreover, they are exact if the underlying model is a Multinomial logit model. We also give a procedure to estimate the parameters of the Markov chain model that does not require any knowledge of the latent choice model. Furthermore, we show that the assortment optimization problem under our choice model can be solved efficiently in polynomial time. This is quite surprising as we can not even express the choice probabilities using a functional form. Our numerical experiments show that the average maximum relative error between the estimates of the Markov chain choice probability and the true choice probability is less than $3\\%$ (the average being taken over different offer sets). Therefore, our model provides a tractable data-driven approach to choice modeling and assortment optimization that is robust to model selection errors." ]

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[ "40\nviews\n0\nrecommends\n+1 Recommend\n0 collections\n4\nshares\n• Record: found\n• Abstract: found\n• Article: found\nIs Open Access\n\nAn integral geometry lemma and its applications: the nonlocality of the Pavlov equation and a tomographic problem with opaque parabolic objects\n\nPreprint\n\nBookmark\nThere is no author summary for this article yet. Authors can add summaries to their articles on ScienceOpen to make them more accessible to a non-specialist audience.\n\nAbstract\n\nAs in the case of soliton PDEs in 2+1 dimensions, the evolutionary form of integrable dispersionless multidimensional PDEs is non-local, and the proper choice of integration constants should be the one dictated by the associated Inverse Scattering Transform (IST). Using the recently made rigorous IST for vector fields associated with the so-called Pavlov equation $$v_{xt}+v_{yy}+v_xv_{xy}-v_yv_{xx}=0$$, we have recently esatablished that, in the nonlocal part of its evolutionary form $$v_{t}= v_{x}v_{y}-\\partial^{-1}_{x}\\,\\partial_{y}\\,[v_{y}+v^2_{x}]$$, the formal integral $$\\partial^{-1}_{x}$$ corresponding to the solutions of the Cauchy problem constructed by such an IST is the asymmetric integral $$-\\int_x^{\\infty}dx'$$. In this paper we show that this results could be guessed in a simple way using a, to the best of our knowledge, novel integral geometry lemma. Such a lemma establishes that it is possible to express the integral of a fairly general and smooth function $$f(X,Y)$$ over a parabola of the $$(X,Y)$$ plane in terms of the integrals of $$f(X,Y)$$ over all straight lines non intersecting the parabola. A similar result, in which the parabola is replaced by the circle, is already known in the literature and finds applications in tomography. Indeed, in a two-dimensional linear tomographic problem with a convex opaque obstacle, only the integrals along the straight lines non-intersecting the obstacle are known, and in the class of potentials $$f(X,Y)$$ with polynomial decay we do not have unique solvability of the inverse problem anymore. Therefore, for the problem with an obstacle, it is natural not to try to reconstruct the complete potential, but only some integral characteristics like the integral over the boundary of the obstacle. Due to the above two lemmas, this can be done, at the moment, for opaque bodies having as boundary a parabola and a circle (an ellipse).\n\nMost cited references1\n\n• Record: found\n\nUnsteady motion in transonic flow\n\n(1964)\nBookmark\n\nAuthor and article information\n\nJournal\n1511.04436\n\nNonlinear & Complex systems" ]

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[ "# Python function to execute PostGIS query\n\nI am trying to write a python function to run PostGIS query:\n\n``````import sys, os\n\n#set up psycopg2 environment\nimport psycopg2\n\ndef river_dist_matrix(table_name):\n\n#driving_distance module\nquery = \"\"\"\nselect *\nfrom pgr_drivingDistance (\\$\\$\nselect\nid as id,\nsource::int4 as source,\ntarget::int4 as target,\nshape_leng::double precision as cost\nfrom public.'table_name'\n\\$\\$, %s, %s, %s, %s\n)\n\"\"\"\n\n#make connection between python and postgresql\nconn = psycopg2.connect(\"dbname = 'routing_test' user = 'postgres' host = 'localhost' password = 'xxxx'\")\ncur = conn.cursor()\n\n#count rows in the table\ncur.execute(\"select count(*) from 'table_name'\")\nresult = cur.fetchone()\nk = result + 1 #number of points = number of segments + 1\nprint k\n\n#run loops\nrs = []\ni = 1\nwhile i <= k:\ncur.execute(query, (i, 100000000000, False, False))\nrs.append(cur.fetchall())\ni = i + 1\n# print rs\n#import csv module\nimport csv\nimport tempfile\nimport shutil\n\nj = 0\nh = 0\nars = []\nelement = list(rs)\n# print element\n#export data to every row\nfilename = 'ck_dist_m.csv'\nwith open(filename, 'wb') as f:\nwriter = csv.writer(f, delimiter = ',')\nwhile j <= k - 1:\nwhile h <= k - 1:\nrp = element[j][h]\nars.append(rp)\nh = h + 1\nelse:\nh = 0\nwriter.writerow(ars)\nars = []\nj = j + 1\n\n# concerning about flow-connection\nwith open(filename, 'rb') as f, tempfile.NamedTemporaryFile(mode='wb', delete=False) as g:\nwriter = csv.writer(g, delimiter = ',')\nfor row in csv.reader(f):\nrow = [element if float(element) < 10**6 else 'nan' for element in row]\nwriter.writerow(row)\n\nshutil.move(g.name, filename)\n\nconn.close()\n\nif __name__ == \"__main__\":\n\ntable_name = 'tc_500wa'\nriver_dist_matrix(table_name)\n``````\n\nBut as I ran code above, I got this error:\n\n``````Traceback (most recent call last):\nFile \"C:\\Users\\Heinz\\Desktop\\river_dist_matrix.py\", line 95, in <module>\nriver_dist_matrix(table_name)\nFile \"C:\\Users\\Heinz\\Desktop\\river_dist_matrix.py\", line 40, in river_dist_matrix\ncur.execute(\"select count(*) from 'table_name'\")\npsycopg2.ProgrammingError: syntax error at or near \"'table_name'\"\nLINE 1: select count(*) from 'table_name'\n^\n\n[Finished in 0.2s]\n``````\n\nI think the error is in relation to the input arg table_name because it was placed in the query and other places in the code. As in:\n\n``````query = \"\"\"\nselect *\nfrom pgr_drivingDistance (\\$\\$\nselect\nid as id,\nsource::int4 as source,\ntarget::int4 as target,\nshape_leng::double precision as cost\nfrom public.'table_name'\n\\$\\$, %s, %s, %s, %s\n)\n\"\"\"\n``````\n\nand\n\n``````cur.execute(\"select count(*) from 'table_name'\")\n``````\n\nHow to edit this code to make it run smoothly?\n\n## 2 Answers\n\nAs I say in comment in previous answer, you have to replace 'table_name' in query string with real table name, which is provided as function parameter. So you code schould be:\n\n``````query = \"\"\"\nselect *\nfrom pgr_drivingDistance (\\$\\$\nselect\nid as id,\nsource::int4 as source,\ntarget::int4 as target,\nshape_leng::double precision as cost\nfrom public.{0}\n\\$\\$, %s, %s, %s, %s\n)\n\"\"\".format(table_name)\n``````\n\nand\n\n``````cur.execute(\"select count(*) from {0}\".format(table_name))\n``````\n• You should also consider removing schema name (public.) from first query... Then you can provide fully qualified table name as function parameter (eg. \"public.tc_500wa\"). Or you can provide table name without schema, but this suppose correctly set schema search path in PG. This is more flexible than hard coded schema name in query. – DavidP Oct 23 '16 at 11:53\n• Thank you for the effective advice, I think to remove public from query is a good choice. How to edit code to achieve this?( I tried .format(scheme.table_name) in the last line of the query but error showed up. – Heinz Oct 23 '16 at 12:49\n• @Heinz just replace public.{0} with only {0} Then you can pass schema.table_name in function call in main section of your script as follows: `table_name = 'public.tc_500wa'` – DavidP Oct 23 '16 at 13:06\n\nI hing the source of problem is single quoted table name. Single quotes in PG means string literal... so you try to select count from string, which yields error. You should remove quotes around table name or you should use double quotes. Correct sql query should look like this:\n\n``````\"select count(*) from table_name\"\n``````\n\nor\n\n``````\"select count(*) from \\\"table_name\\\"\"\n``````\n\nAnd remember - double quoted object names are case sensitive. So \"table_name\" is different form \"Table_Name\", etc.\n\n• Thank you for reply, I tried your way but still got error, how to edit code in the post where \"table_name\" exist in the line \"from public.'table_name'\"? – Heinz Oct 23 '16 at 9:25\n• That's because you don't use table_name parameter passed to function. So you have to replace 'table_name' in each query with real name from function parameter. You can do than in many different ways, but simplest one is use string.format() function. Replace 'table_name' with {0} and after each query string place .format(table_name) eg.: 'select count(*) from {0}'.format(table_name) – DavidP Oct 23 '16 at 9:35" ]

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[ "let L be up-complete LATTICE; :: thesis: ( ( for X being Subset of L\nfor x being Element of L holds x \"/\\\" (sup X) = sup ({x} \"/\\\" ()) ) implies for X being non empty directed Subset of L\nfor x being Element of L holds x \"/\\\" (sup X) = sup ({x} \"/\\\" X) )\n\nassume A1: for X being Subset of L\nfor x being Element of L holds x \"/\\\" (sup X) = sup ({x} \"/\\\" ()) ; :: thesis: for X being non empty directed Subset of L\nfor x being Element of L holds x \"/\\\" (sup X) = sup ({x} \"/\\\" X)\n\nlet X be non empty directed Subset of L; :: thesis: for x being Element of L holds x \"/\\\" (sup X) = sup ({x} \"/\\\" X)\nlet x be Element of L; :: thesis: x \"/\\\" (sup X) = sup ({x} \"/\\\" X)\nreconsider T = {x} as non empty directed Subset of L by WAYBEL_0:5;\nA2: ex_sup_of T \"/\\\" X,L by WAYBEL_0:75;\nA3: {x} \"/\\\" () = { (x \"/\\\" y) where y is Element of L : y in finsups X } by YELLOW_4:42;\nA4: {x} \"/\\\" () is_<=_than sup ({x} \"/\\\" X)\nproof\nlet q be Element of L; :: according to LATTICE3:def 9 :: thesis: ( not q in {x} \"/\\\" () or q <= sup ({x} \"/\\\" X) )\nA5: x <= x ;\nassume q in {x} \"/\\\" () ; :: thesis: q <= sup ({x} \"/\\\" X)\nthen consider y being Element of L such that\nA6: q = x \"/\\\" y and\nA7: y in finsups X by A3;\nconsider Y being finite Subset of X such that\nA8: y = \"\\/\" (Y,L) and\nA9: ex_sup_of Y,L by A7;\nconsider z being Element of L such that\nA10: z in X and\nA11: z is_>=_than Y by WAYBEL_0:1;\n\"\\/\" (Y,L) <= z by ;\nthen A12: x \"/\\\" y <= x \"/\\\" z by ;\nx in {x} by TARSKI:def 1;\nthen x \"/\\\" z <= sup ({x} \"/\\\" X) by ;\nhence q <= sup ({x} \"/\\\" X) by ; :: thesis: verum\nend;\nex_sup_of T \"/\\\" (),L by WAYBEL_0:75;\nthen sup ({x} \"/\\\" ()) <= sup ({x} \"/\\\" X) by ;\nthen A13: x \"/\\\" (sup X) <= sup ({x} \"/\\\" X) by A1;\nex_sup_of X,L by WAYBEL_0:75;\nthen sup ({x} \"/\\\" X) <= x \"/\\\" (sup X) by ;\nhence x \"/\\\" (sup X) = sup ({x} \"/\\\" X) by ; :: thesis: verum" ]

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[ "# Is it possible to plot the ground track of a satellite with just azimuth & elevation angles without range?\n\nI came across this question on StackOverflow about Satellite Trajectory plot. In the code mentioned in the question, the user takes only the Azimuth & Elevation angles (radians) & converts them into cartesian (x, y) coordinates & plots the Satellite trajectory onto a polar plot.\n\nHowever, to convert from Spherical to Cartesian coordinates, one needs Range, Azimuth & Elevation. But the formula used in the code is\n$$x = \\frac{(\\pi/2) - Elevation} {(\\pi/2)*cos(Azimuth - (\\pi/2))}$$\n$$y = \\frac{Elevation - (\\pi/2)} {(\\pi/2)*sin(Azimuth - (\\pi/2))}$$\nHow was the OP able to draw a satellite ground track using only Azimuth & Elevation? How to arrive at the above equations from\n$$x = r*cos(Elevation)*cos(Azimuth)$$\n$$y = r*cos(Elevation)*sin(Azimuth)$$\n$$z = r*sin(Elevation)$$\n\n• You should ask questions about that plot there, although the question is from 2012. It's not a ground-track plot at all as far as I can tell, it's just a plot of altitude and azimuth of several GPS satellites, shown only when they are above the horizon for that viewing location. – uhoh Jun 15 '18 at 13:04\n\nAs far as I can tell, in the question by the link, the OP is not trying to plot satellites' ground tracks in any way. He simply plots their elevations and azimuths, that's all, so he doesn't need to know the ranges. The $x$ and $y$ his program computes have nothing to do with satellite's Cartesian coordinates; they are just the coordinates of the point which represent the given elevation and azimuth on the plot." ]

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[ "# Definition of daytimes\n\n## \"daytimes\" in the noun sense\n\n### 1. day, daytime, daylight\n\nthe time after sunrise and before sunset while it is light outside\n\n\"the dawn turned night into day\"\n\n\"it is easier to make the repairs in the daytime\"\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# daytimes in Scrabble®\n\nThe word daytimes is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nDAYTIMES\n(135)\nDAYTIMES\n(135)\n\n## Seven Letter Word Alert: (3 words)\n\ndaytime, misdate, stymied\n\nDAYTIMES\n(135)\nDAYTIMES\n(135)\nDAYTIMES\n(90)\nDAYTIMES\n(90)\nDAYTIMES\n(56)\nDAYTIMES\n(56)\nDAYTIMES\n(54)\nDAYTIMES\n(51)\nDAYTIMES\n(48)\nDAYTIMES\n(45)\nDAYTIMES\n(45)\nDAYTIMES\n(45)\nDAYTIMES\n(44)\nDAYTIMES\n(40)\nDAYTIMES\n(40)\nDAYTIMES\n(36)\nDAYTIMES\n(36)\nDAYTIMES\n(34)\nDAYTIMES\n(34)\nDAYTIMES\n(34)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(30)\nDAYTIMES\n(30)\nDAYTIMES\n(30)\nDAYTIMES\n(30)\nDAYTIMES\n(30)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(24)\nDAYTIMES\n(22)\nDAYTIMES\n(21)\nDAYTIMES\n(20)\nDAYTIMES\n(19)\nDAYTIMES\n(19)\nDAYTIMES\n(19)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(17)\n\nDAYTIMES\n(135)\nDAYTIMES\n(135)\nSTYMIED\n(102 = 52 + 50)\nDAYTIME\n(102 = 52 + 50)\nSTYMIED\n(101 = 51 + 50)\nDAYTIME\n(101 = 51 + 50)\nDAYTIME\n(98 = 48 + 50)\nSTYMIED\n(98 = 48 + 50)\nSTYMIED\n(98 = 48 + 50)\nDAYTIME\n(95 = 45 + 50)\nSTYMIED\n(95 = 45 + 50)\nDAYTIME\n(92 = 42 + 50)\nSTYMIED\n(92 = 42 + 50)\nDAYTIME\n(92 = 42 + 50)\nDAYTIME\n(92 = 42 + 50)\nSTYMIED\n(92 = 42 + 50)\nDAYTIME\n(92 = 42 + 50)\nDAYTIME\n(92 = 42 + 50)\nSTYMIED\n(92 = 42 + 50)\nDAYTIME\n(92 = 42 + 50)\nSTYMIED\n(92 = 42 + 50)\nSTYMIED\n(92 = 42 + 50)\nMISDATE\n(90 = 40 + 50)\nDAYTIMES\n(90)\nDAYTIMES\n(90)\nDAYTIME\n(89 = 39 + 50)\nMISDATE\n(89 = 39 + 50)\nSTYMIED\n(89 = 39 + 50)\nDAYTIME\n(88 = 38 + 50)\nDAYTIME\n(88 = 38 + 50)\nMISDATE\n(86 = 36 + 50)\nSTYMIED\n(86 = 36 + 50)\nMISDATE\n(86 = 36 + 50)\nDAYTIME\n(84 = 34 + 50)\nSTYMIED\n(84 = 34 + 50)\nMISDATE\n(83 = 33 + 50)\nMISDATE\n(83 = 33 + 50)\nMISDATE\n(83 = 33 + 50)\nMISDATE\n(83 = 33 + 50)\nMISDATE\n(83 = 33 + 50)\nSTYMIED\n(82 = 32 + 50)\nDAYTIME\n(82 = 32 + 50)\nSTYMIED\n(80 = 30 + 50)\nDAYTIME\n(80 = 30 + 50)\nSTYMIED\n(80 = 30 + 50)\nSTYMIED\n(80 = 30 + 50)\nDAYTIME\n(80 = 30 + 50)\nDAYTIME\n(80 = 30 + 50)\nSTYMIED\n(80 = 30 + 50)\nDAYTIME\n(80 = 30 + 50)\nMISDATE\n(80 = 30 + 50)\nSTYMIED\n(80 = 30 + 50)\nDAYTIME\n(80 = 30 + 50)\nDAYTIME\n(78 = 28 + 50)\nSTYMIED\n(78 = 28 + 50)\nSTYMIED\n(78 = 28 + 50)\nDAYTIME\n(78 = 28 + 50)\nSTYMIED\n(78 = 28 + 50)\nSTYMIED\n(78 = 28 + 50)\nDAYTIME\n(78 = 28 + 50)\nSTYMIED\n(78 = 28 + 50)\nDAYTIME\n(78 = 28 + 50)\nMISDATE\n(78 = 28 + 50)\nSTYMIED\n(76 = 26 + 50)\nMISDATE\n(76 = 26 + 50)\nDAYTIME\n(76 = 26 + 50)\nSTYMIED\n(76 = 26 + 50)\nSTYMIED\n(76 = 26 + 50)\nDAYTIME\n(76 = 26 + 50)\nSTYMIED\n(76 = 26 + 50)\nDAYTIME\n(76 = 26 + 50)\nMISDATE\n(76 = 26 + 50)\nSTYMIED\n(76 = 26 + 50)\nDAYTIME\n(76 = 26 + 50)\nDAYTIME\n(76 = 26 + 50)\nSTYMIED\n(75 = 25 + 50)\nMISDATE\n(74 = 24 + 50)\nMISDATE\n(74 = 24 + 50)\nMISDATE\n(74 = 24 + 50)\nMISDATE\n(74 = 24 + 50)\nMISDATE\n(74 = 24 + 50)\nDAYTIME\n(73 = 23 + 50)\nMISDATE\n(72 = 22 + 50)\nMISDATE\n(72 = 22 + 50)\nMISDATE\n(72 = 22 + 50)\nMISDATE\n(72 = 22 + 50)\nMISDATE\n(72 = 22 + 50)\nMISDATE\n(72 = 22 + 50)\nDAYTIME\n(71 = 21 + 50)\nMISDATE\n(70 = 20 + 50)\nSTYMIED\n(70 = 20 + 50)\nDAYTIME\n(70 = 20 + 50)\nMISDATE\n(70 = 20 + 50)\nMISDATE\n(70 = 20 + 50)\nMISDATE\n(70 = 20 + 50)\nMISDATE\n(70 = 20 + 50)\nSTYMIED\n(69 = 19 + 50)\nSTYMIED\n(69 = 19 + 50)\nDAYTIME\n(69 = 19 + 50)\nMISDATE\n(68 = 18 + 50)\nDAYTIME\n(68 = 18 + 50)\nDAYTIME\n(68 = 18 + 50)\nSTYMIED\n(68 = 18 + 50)\nSTYMIED\n(67 = 17 + 50)\nSTYMIED\n(67 = 17 + 50)\nDAYTIME\n(67 = 17 + 50)\nSTYMIED\n(67 = 17 + 50)\nSTYMIED\n(67 = 17 + 50)\nDAYTIME\n(67 = 17 + 50)\nDAYTIME\n(67 = 17 + 50)\nSTYMIED\n(67 = 17 + 50)\nDAYTIME\n(66 = 16 + 50)\nSTYMIED\n(66 = 16 + 50)\nSTYMIED\n(65 = 15 + 50)\nDAYTIME\n(65 = 15 + 50)\nMISDATE\n(65 = 15 + 50)\nMISDATE\n(65 = 15 + 50)\nDAYTIME\n(65 = 15 + 50)\nSTYMIED\n(65 = 15 + 50)\nMISDATE\n(64 = 14 + 50)\nMISDATE\n(64 = 14 + 50)\nMISDATE\n(64 = 14 + 50)\nDAYTIME\n(64 = 14 + 50)\nMISDATE\n(64 = 14 + 50)\nMISDATE\n(63 = 13 + 50)\nMISDATE\n(63 = 13 + 50)\nMISDATE\n(62 = 12 + 50)\nMISDATE\n(62 = 12 + 50)\nMISDATE\n(62 = 12 + 50)\nMISDATE\n(62 = 12 + 50)\nDAYTIMES\n(56)\nDAYTIMES\n(56)\nDAYTIMES\n(54)\nDAYTIMES\n(51)\nDAYTIMES\n(48)\nDISMAY\n(48)\nSTYMIE\n(45)\nMATEYS\n(45)\nSTEAMY\n(45)\nDISMAY\n(45)\nDAYTIMES\n(45)\nDAYTIMES\n(45)\nDAYTIMES\n(45)\nDAYTIMES\n(44)\nSTYMIE\n(42)\nMATEY\n(42)\nMISTY\n(42)\nAMITY\n(42)\nSEAMY\n(42)\nSTAYED\n(42)\nMATEYS\n(42)\nSTEAMY\n(42)\nSTIMY\n(42)\nDISMAY\n(42)\nMEATY\n(42)\n(42)\nDEMY\n(42)\nDAYTIMES\n(40)\nDAYTIMES\n(40)\nDISMAY\n(40)\nDISMAY\n(39)\nYAMS\n(39)\nDAISY\n(39)\nMISTY\n(39)\nSTIMY\n(39)\nDISMAY\n(39)\nDISMAY\n(39)\nMEATY\n(39)\nDEITY\n(39)\nMATEY\n(39)\nDIETY\n(39)\nDITSY\n(39)\nAMITY\n(39)\nSEAMY\n(39)\nMATEYS\n(38)\nSTEAMY\n(38)\nDEMAST\n(36)\nYEAST\n(36)\nDISMAY\n(36)\nSTEAMY\n(36)\nMATEYS\n(36)\nDEMITS\n(36)\nMASTED\n(36)\nMATEYS\n(36)\n(36)\nSTEAMY\n(36)\n(36)\nDAYTIMES\n(36)\nSTEAMY\n(36)\nMISTY\n(36)\nSEAMY\n(36)\nSTEAMY\n(36)\nMATEYS\n(36)\nAMITY\n(36)\n(36)\nSTAYED\n(36)\nDEMY\n(36)\nSTYMIE\n(36)\nYETIS\n(36)\nMEATY\n(36)\n\n# daytimes in Words With Friends™\n\nThe word daytimes is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nDAYTIME\n(104 = 69 + 35)\nDAYTIME\n(104 = 69 + 35)\nSTYMIED\n(104 = 69 + 35)\nSTYMIED\n(104 = 69 + 35)\n\n## Seven Letter Word Alert: (3 words)\n\ndaytime, misdate, stymied\n\nDAYTIMES\n(96)\nDAYTIMES\n(96)\nDAYTIMES\n(72)\nDAYTIMES\n(72)\nDAYTIMES\n(72)\nDAYTIMES\n(66)\nDAYTIMES\n(64)\nDAYTIMES\n(60)\nDAYTIMES\n(56)\nDAYTIMES\n(56)\nDAYTIMES\n(54)\nDAYTIMES\n(54)\nDAYTIMES\n(48)\nDAYTIMES\n(48)\nDAYTIMES\n(44)\nDAYTIMES\n(40)\nDAYTIMES\n(36)\nDAYTIMES\n(36)\nDAYTIMES\n(34)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(32)\nDAYTIMES\n(30)\nDAYTIMES\n(30)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(28)\nDAYTIMES\n(24)\nDAYTIMES\n(22)\nDAYTIMES\n(20)\nDAYTIMES\n(20)\nDAYTIMES\n(20)\nDAYTIMES\n(19)\nDAYTIMES\n(19)\nDAYTIMES\n(19)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(18)\nDAYTIMES\n(17)\nDAYTIMES\n(17)\nDAYTIMES\n(17)\nDAYTIMES\n(17)\nDAYTIMES\n(17)\nDAYTIMES\n(16)\nDAYTIMES\n(16)\nDAYTIMES\n(16)\nDAYTIMES\n(16)\nDAYTIMES\n(15)\nDAYTIMES\n(15)\n\nDAYTIME\n(104 = 69 + 35)\nDAYTIME\n(104 = 69 + 35)\nSTYMIED\n(104 = 69 + 35)\nSTYMIED\n(104 = 69 + 35)\nSTYMIED\n(98 = 63 + 35)\nSTYMIED\n(98 = 63 + 35)\nDAYTIME\n(98 = 63 + 35)\nMISDATE\n(98 = 63 + 35)\nSTYMIED\n(98 = 63 + 35)\nDAYTIMES\n(96)\nDAYTIMES\n(96)\nSTYMIED\n(92 = 57 + 35)\nMISDATE\n(92 = 57 + 35)\nSTYMIED\n(87 = 52 + 35)\nDAYTIME\n(87 = 52 + 35)\nDAYTIME\n(87 = 52 + 35)\nSTYMIED\n(87 = 52 + 35)\nDAYTIME\n(87 = 52 + 35)\nSTYMIED\n(87 = 52 + 35)\nMISDATE\n(86 = 51 + 35)\nMISDATE\n(86 = 51 + 35)\nDAYTIME\n(86 = 51 + 35)\nSTYMIED\n(86 = 51 + 35)\nDAYTIME\n(86 = 51 + 35)\nDAYTIME\n(86 = 51 + 35)\nMISDATE\n(80 = 45 + 35)\nDAYTIME\n(80 = 45 + 35)\nDAYTIME\n(80 = 45 + 35)\nDAYTIME\n(80 = 45 + 35)\nSTYMIED\n(80 = 45 + 35)\nDAYTIME\n(80 = 45 + 35)\nMISDATE\n(80 = 45 + 35)\nSTYMIED\n(80 = 45 + 35)\nMISDATE\n(80 = 45 + 35)\nSTYMIED\n(80 = 45 + 35)\nMISDATE\n(79 = 44 + 35)\nMISDATE\n(79 = 44 + 35)\nMISDATE\n(79 = 44 + 35)\nDISMAY\n(78)\nDAYTIME\n(77 = 42 + 35)\nMISDATE\n(74 = 39 + 35)\nMISDATE\n(74 = 39 + 35)\nMISDATE\n(74 = 39 + 35)\nSTYMIED\n(73 = 38 + 35)\nDAYTIME\n(73 = 38 + 35)\nMISDATE\n(73 = 38 + 35)\nDAYTIMES\n(72)\nDAYTIMES\n(72)\nDAYTIMES\n(72)\nSTYMIED\n(69 = 34 + 35)\nDAYTIME\n(69 = 34 + 35)\nDAYTIME\n(69 = 34 + 35)\nSTYMIED\n(67 = 32 + 35)\nDAYTIME\n(67 = 32 + 35)\nDEMAST\n(66)\nDEMIST\n(66)\nDAYTIMES\n(66)\nMEDIAS\n(66)\nDEMITS\n(66)\nDAYTIME\n(65 = 30 + 35)\nSTYMIED\n(65 = 30 + 35)\nDAYTIME\n(65 = 30 + 35)\nDAYTIME\n(65 = 30 + 35)\nSTYMIED\n(65 = 30 + 35)\nDAYTIME\n(65 = 30 + 35)\nSTYMIED\n(65 = 30 + 35)\nSTYMIED\n(65 = 30 + 35)\nSTYMIED\n(65 = 30 + 35)\nMISDATE\n(65 = 30 + 35)\nDAYTIMES\n(64)\nDAYTIME\n(63 = 28 + 35)\nSTYMIED\n(63 = 28 + 35)\nDAYTIME\n(63 = 28 + 35)\nMATEYS\n(63)\nSTYMIE\n(63)\nSTYMIED\n(63 = 28 + 35)\nSTYMIED\n(63 = 28 + 35)\nDAYTIME\n(63 = 28 + 35)\nSTYMIED\n(63 = 28 + 35)\nSTYMIED\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nMISDATE\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nMISDATE\n(61 = 26 + 35)\nMISDATE\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nSTYMIED\n(61 = 26 + 35)\nDAYTIME\n(61 = 26 + 35)\nSTYMIED\n(61 = 26 + 35)\nSTYMIED\n(61 = 26 + 35)\nSTYMIED\n(61 = 26 + 35)\nSTYMIED\n(61 = 26 + 35)\nSTYMIED\n(61 = 26 + 35)\nMISDATE\n(61 = 26 + 35)\nMISDATE\n(61 = 26 + 35)\nDISMAY\n(60)\nMASTED\n(60)\nDAYTIMES\n(60)\nMISTED\n(60)\n(60)\nMISDATE\n(59 = 24 + 35)\nMISDATE\n(59 = 24 + 35)\nMISDATE\n(59 = 24 + 35)\nMISDATE\n(59 = 24 + 35)\nMISDATE\n(59 = 24 + 35)\nSTYMIED\n(58 = 23 + 35)\nDAYTIME\n(58 = 23 + 35)\nSTEAMY\n(57)\nMISDATE\n(57 = 22 + 35)\nSTEAMY\n(57)\nMISDATE\n(57 = 22 + 35)\nMATEYS\n(57)\nMISDATE\n(57 = 22 + 35)\nMISDATE\n(57 = 22 + 35)\nMISDATE\n(57 = 22 + 35)\nSTAYED\n(57)\nMISDATE\n(57 = 22 + 35)\nSTYMIE\n(57)\nMISDATE\n(57 = 22 + 35)\nSTYMIE\n(57)\nSTYMIED\n(56 = 21 + 35)\nDAYTIME\n(56 = 21 + 35)\nDAYTIME\n(56 = 21 + 35)\nSTYMIED\n(56 = 21 + 35)\nDAYTIMES\n(56)\nDAYTIMES\n(56)\nMISDATE\n(56 = 21 + 35)\nMISDATE\n(55 = 20 + 35)\nDAYTIME\n(55 = 20 + 35)\nSTYMIED\n(55 = 20 + 35)\nAMIDST\n(54)\nDISMAY\n(54)\nDISMAY\n(54)\nMASTED\n(54)\nDAYTIME\n(54 = 19 + 35)\nSTYMIED\n(54 = 19 + 35)\nDEMIST\n(54)\nAMIDES\n(54)\nMISTED\n(54)\nMATEY\n(54)\nDEMAST\n(54)\nSEAMY\n(54)\nDEMITS\n(54)\nMEATY\n(54)\nMEDIAS\n(54)\nSTIMY\n(54)\nDAYTIME\n(54 = 19 + 35)\nMISTY\n(54)\nAMITY\n(54)\nDAYTIMES\n(54)\nDAYTIMES\n(54)\nSTYMIED\n(54 = 19 + 35)\n(54)\nSTYMIED\n(53 = 18 + 35)\nDAYTIME\n(53 = 18 + 35)\nSTYMIED\n(53 = 18 + 35)\nDAYTIME\n(53 = 18 + 35)\nDAYTIME\n(53 = 18 + 35)\nDAYTIME\n(53 = 18 + 35)\nSTYMIED\n(53 = 18 + 35)\nSTYMIED\n(53 = 18 + 35)\nDAYTIME\n(52 = 17 + 35)\nDAYTIME\n(52 = 17 + 35)\nDAYTIME\n(52 = 17 + 35)\nDAYTIME\n(52 = 17 + 35)\nSTYMIED\n(52 = 17 + 35)\nSTYMIED\n(52 = 17 + 35)\nSTYMIED\n(52 = 17 + 35)\nSTYMIED\n(52 = 17 + 35)\nMISDATE\n(52 = 17 + 35)\nMISDATE\n(52 = 17 + 35)\nSTYMIED\n(52 = 17 + 35)\nMISDATE\n(52 = 17 + 35)\nSTYMIED\n(52 = 17 + 35)\nMIDAS\n(51)\nMEDIA\n(51)\nDEISM\n(51)\nDAYTIME\n(51 = 16 + 35)\nDAYTIME\n(51 = 16 + 35)\nMIDST\n(51)\n(51)\nMAYS\n(51)\nSTYMIED\n(51 = 16 + 35)\nDAYTIME\n(51 = 16 + 35)\nSTYMIED\n(51 = 16 + 35)\nAMIDE\n(51)\nMAIDS\n(51)\nSTEAMY\n(51)\nMATED\n(51)\nMISDATE\n(51 = 16 + 35)\nSTYMIED\n(51 = 16 + 35)\nMISDATE\n(51 = 16 + 35)\nSTYMIE\n(51)\nDAYTIME\n(51 = 16 + 35)\nMATEYS\n(51)\nSTYMIED\n(50 = 15 + 35)\nDAYTIME\n(50 = 15 + 35)\n\n# Word Growth involving daytimes\n\nay day daytime\n\nme time daytime\n\nme time times\n\n## Longer words containing daytimes\n\n(No longer words found)" ]

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[ "# Determination of products in reactions involving carbocation rearrangement?\n\nHow do you determine the migratory \"aptitude\" of various groups during carbocation rearrangements? Is there a experimentally determined order?\nFor example, what will be the product in case of the following pinacol-pinacolone type rearrangement: $$\\ce{(cyclo-pentyl)CH3C(OH)-C(OH)(CH3)(cyclo-pentyl) ->products}$$\n\nWill the methyl group be transferred or the cyclopentyl moiety? The same question arises if the cyclopentyl is replaced with phenyl group? In general, which shifts, methyl, hydrogen, phenyl or any other group, will take place? Or equivalently, what determines which group has a greater tendency to undergo shifts during carbocation rearrangements?\n\n$$W=\\int_0^\\vec{x}\\vec{F}\\cdot d\\vec{x}$$ $$\\frac{d\\vec{x}}{dt}=\\vec{v} \\ \\ \\ \\ \\ \\therefore \\ \\ \\ \\ \\ d\\vec{x}=\\vec{v}dt$$ $$W=\\int_0^t\\left(\\vec{F}\\cdot \\vec{v}\\right)dt$$$$\\vec{F}=m\\vec{a}$$ $$W=m\\int_o^t \\left(\\vec{a}\\cdot\\vec{v}\\right)dt=m\\int\\left(\\vec{v}\\cdot\\vec{a}dt\\right)$$ $$\\vec{a}=\\frac{d\\vec{v}}{dt}$$ $$\\vec{a}dt=d\\vec{v}$$ $$W=m\\int_0^\\vec{v}\\vec{v}\\cdot d\\vec{v}=\\frac{1}{2}m\\vec{v}^2\\vert_0^\\vec{v}=\\frac{1}{2}m\\vec{v}^2$$" ]

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[ "This demo shows how functionality within Econometric Toolbox can be used to identify and calibrate a simple, intraday pairs trading strategy.\n\n## Contents\n\n```% Rather than forcing the customer to define and populate a database, we\n% will instead read from a prepared data file. If you wish, you may write\n% the data from this file to a database of your choosing and adapt the\n% GETMINUTEDATAFROMDB command to connect to and read from it.\n\n% LCO = getMinuteDataFromDB('LCO');\n% WTI = getMinuteDataFromDB('WTI');\n\nLCO = double(brent);\nWTI = double(light);\nclearvars -except LCO WTI\n\npairsChart(LCO, WTI)\n\n% These two time series have historically tracked each other, but since\n% December 2010, LCO has consistently traded higher than WTI. It would\n% seem that a pairs trading strategy would not work in 2011, but if we are\n% willing to actively recalibrate our model on an intraday basis, we may\n% find profitable opportunities.\n\n% Let's focus on the last 11 days' of data:\nseries = [LCO(end-4619 : end, 4) WTI(end-4619 : end, 4)];\n```", null, "## The cointegration test framework\n\nEconometrics Toolbox supports both the Engle-Granger and the Johansen cointegration frameworks. Engle-Granger is the older model, and Johansen is particularly useful for analyzing more than two time series at a time. We will use Engle-Granger for our trading model.\n\n```% First, we note that the last 11 days are not cointegrated as a whole\negcitest(series)\n% (A zero indicates failure to reject the null hypothesis that no\n% cointegrating relationship exists.)\n```\n```ans =\n\n0\n\n```\n\nEven so, there are smaller windows of time where a cointegrating relationship does exist.\n\n```[h, ~, ~, ~, reg1] = egcitest(series(1700:2000, :));\ndisplay(h)\n```\n```h =\n\n1\n\n```\n\nThe test estimates the coefficients of the cointegrating regression as well as the residuals and the standard errors of the residuals: all useful information for any pairs trading strategy.\n\n```display(reg1)\n```\n```reg1 =\n\nnum: 301\nsize: 301\nnames: {2x1 cell}\ncoeff: [2x1 double]\nse: [2x1 double]\nCov: [2x2 double]\ntStats: [1x1 struct]\nFStat: [1x1 struct]\nyMu: 110.7448\nySigma: 0.3043\nyHat: [301x1 double]\nres: [301x1 double]\nDWStat: 0.1891\nSSR: 13.0123\nSSE: 14.7666\nSST: 27.7789\nMSE: 0.0494\nRMSE: 0.2222\nRSq: 0.4684\naRSq: 0.4666\nLL: 26.6152\nAIC: -49.2304\nBIC: -41.8162\nHQC: -46.2636\n\n```\n\nThe following function describes our pairs strategy. It is basically a copy of our other existing strategy files, with significant changes to only about 7 lines of code.\n\n```edit pairs\n```\n\nWe may test this strategy as we do our other rules:\n\n```pairs(series, 420, 60)\n% Note that this strategy will not trade if the most recent minutes do not\n% show signs of cointegration and that the size of the long/short positions\n% are dynamically scaled with the volatility of the cointegrating\n% relationship. Many other customizations can be made.\n```", null, "We can use our existing parameter sweep framework to identify the best combination of calibration window and rebalancing frequency. This builds off of existing code and takes advantage of parallel computing:\n\n```if matlabpool('size') == 0\nmatlabpool local\nend\n\nwindow = 120:60:420;\nfreq = 10:10:60;\nrange = {window, freq};\n\nannualScaling = sqrt(250*7*60);\ncost = 0.01;\n\npfun = @(x) pairsFun(x, series, annualScaling, cost);\n\ntic\n[~,param] = parameterSweep(pfun,range);\ntoc\n\npairs(series, param(1), param(2), 1, annualScaling, cost)\n```\n```Elapsed time is 37.228054 seconds.\n```", null, "Despite the fact that these historically-tracking time series have diverged, we can still create a profitable pairs trading strategy by frequently recalibrating." ]

[ null, "https://www.mathworks.com/matlabcentral/mlc-downloads/downloads/submissions/37932/versions/4/previews/html/Demo5_Pairs_Trading_01.png", null, "https://www.mathworks.com/matlabcentral/mlc-downloads/downloads/submissions/37932/versions/4/previews/html/Demo5_Pairs_Trading_02.png", null, "https://www.mathworks.com/matlabcentral/mlc-downloads/downloads/submissions/37932/versions/4/previews/html/Demo5_Pairs_Trading_03.png", null ]

[ "# Java.math.BigInteger.probablePrime() method in Java\n\nPrerequisite : BigInteger Basics\n\nThe probablePrime() method will return a Biginteger of bitLength bits which is prime. bitLength is provided as parameter to method probablePrime() and method will return a prime BigInteger of bitLength bits. The probability that a BigInteger returned by this method is composite and does not exceed 2^-100.\n\nSyntax:\n\n`public static BigInteger probablePrime(int bitLength, Random rnd)`\n\nParameters: This method accepts two parameters as shown in the above syntax and described below.\n\n• bitLength – bitLength of the returned BigInteger.\n• rnd – source of random bits used to select candidates to be tested for primality.\n\nReturn Value: This method returns a BigInteger of bitLength bits that is probably prime.\n\nException:\n\n• ArithmeticException – if bitLength < 2.\n\nBelow program illustrate the probablePrime() method:\n\n `import` `java.math.*; ` `import` `java.util.Random; ` `import` `java.util.Scanner; ` ` `  `public` `class` `GFG { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``Scanner sc = ``new` `Scanner(System.in); ` ` `  `        ``// create a BigInteger object ` `        ``BigInteger biginteger; ` ` `  `        ``// create a integer value for bitLength ` `        ``int` `length = ``4``; ` ` `  `        ``// create a random object ` `        ``Random random = ``new` `Random(); ` ` `  `        ``// call probablePrime method to find next probable prime ` `        ``// whose bit length is equal to bitLength provided as parameter. ` `        ``biginteger = BigInteger.probablePrime(length, random); ` ` `  `        ``String result = ``\"ProbablePrime whose bit length is \"` `                        ``+ length + ``\" = \"` `+ biginteger; ` ` `  `        ``// print result value ` `        ``System.out.println(result); ` `    ``} ` `}`\n\nOutput:\n\n```ProbablePrime whose bit length is 4 = 13\n```\n\nAttention reader! Don’t stop learning now. Get hold of all the important Java and Collections concepts with the Fundamentals of Java and Java Collections Course at a student-friendly price and become industry ready.\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\nPractice Tags :\n\n1\n\nPlease write to us at [email protected] to report any issue with the above content." ]

[ null, "https://media.geeksforgeeks.org/auth/profile/z3lkktftxc4b496iwi0q", null ]

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[ "# Rc Phase Shift Oscillator Online Calculator\n\nThe frequency of the transistor RC phase shift oscillator oscillator can be expressed by the equation: Where F is the frequency, R is the resistance, C is the capacitance and N is the number of RC phase shift stages. PSpice A/D; PSpice AA; PSpice Systems Option; OrCAD Capture; About PSpice; Resources. Then a phase shift of 180° happens between base-emitter and collector circuit due to transistor properties. What happens to the half-amplitude decay time when you turn the 2. For the loop gain βA to be greater than unity, the gain of the amplifier stage must be greater than 1/β or 29. The oscillator is able to reject the amplitude noise (α(t) → 0 as t → ∞. This CalcTown calculator calculates the resonant frequency of a Colpitt Oscillator Circuit. integrator has a phase shift of 90 , while the inverter adds an additional 180 phase shift; thus, a total phase shift of 360 is fed into the input of the first integrator to produce the oscillation. 555 Timer as Astable Multivibrator (Oscillator) PUBLIC. The circuit is a standard RC phase shift oscillator using a single bipolar transistor as the active element. Home Verify series RC software algorithms. An RC FEEDBACK or PHASE-SHIFT oscillator is shown in figure 2-17. An RC circuit is a circuit containing resistance and capacitance. 040deg) Thanks in advance. The output is the result of shifting this signal's phase by an amount specified by the real signal at the input port labeled Ph. 1 ) in an application note published by RCA ( ref. Here we are using a BC107 transistor for implementing RC phase shift oscillator. I = V in / Z. To calculate Z, we must first note that, in accordance with Ohm's Law, R = V R /I and X C = V C /I. In general, phase jitter/noise in the time-domain as well as its power spectral density in the frequency domain are discussed in many papers both in the context of open-loop and closed-loop PLLs , . Relaxation oscillator: By using an RC network to add slow negative feedback to the inverting Schmitt trigger, a relaxation oscillator is formed. 707) 4 or 0. Phase noise in an oscillator. rar Login for download. via classical Hamilton's function). Bookmark or \"Favorite\" this coaxial cable impedance calculator page by pressing CTRL + D. n-MOSFET; p-MOSFET. Focused on customer service, support, innovation, quality and on-time delivery, Abracon supports over 25,000 active customers and ships over a quarter billion components each year. Their simple phase-shift oscillator has the R and C reversed from your circuit (as lowpass filters to reduce distortion) and say that theirs needs a gain of 27 to start oscillating. At the angular frequency ω = ω o = 1/RC, the capacitive reactance 1/ωC equals the resistance R. This is known as \"Barkhausen Criterion\" In RC phase shift oscillator op-amp is used as an amplifier in inverting configuration. The formula for calculating the output voltage is based on Ohms Law and is shown below. FIG 1 shows a typical Wien bridge oscillator. (ii) Explain the working of RC phase shift oscillator. Hartley and Colpitts Oscillator. A voltage controlled oscillator (VCO) is an oscillator whose frequency can be varied by a voltage (or current). Basics of an Oscillator, Feedback, Barkhausen Criteria, RC phase shift Oscillator. Resource Library. 1 Simplified Feedback Oscillator Loop on page 4 shows that the oscillator circuitry consists of two parts; an amplification stage and a filter that decides which frequency experience a 360° phase lag. 33 compared to an ideal gain of 8. org Oscillateur à déphasage; Upotreba na fr. It contains an inverting amplifier, and a feedback filter which 'shifts' the phase of the amplifier output by 180 degrees at the oscillation frequency. RC Oscillators Can Produce Almost Perfect Sine Waves At Frequencies Above 1 MHz. Just using only one BJT the Output wave's amplitude is not perfect, it is required additional circuitry to stabilized amplitude of the waveform. ~C~ is the capacitor value in farads. A 555 timer can be used with a few resistors and a capacitor to make an oscillator, switching states as the capacitor charges and discharges. - RC oscillator. The simplest C-R sine wave oscillator is the phase-shift type, which usually takes the basic form as shown in Figure 2. Home / SPICE Projects / SPICE Projects / Signal Generation Circuits / 1 Khz RC Phase Shift Oscillator. What is the total phase shift requirement, around the feedback loop, for a phase-shift oscillator? A. Figure 2-17. RC ωβ ω ω ω ()()= + + 12 1 3 1 R R - (2) Since the phase angle of the loop gain is defined as: phase angle arctan Im Re = ()() ()() Aj j Aj j ωβ ω ωβ ω (3) We force the imaginary term to zero to set the phase shift to zero. Here two arms are purely resistive while the other two arms are a combination of resistors and capacitors. sine wave oscillator using d-a converter 39 326. the frequency where the total phase shift through the three RC feedback circuits is 180. 7 Colpitts Oscillator Using FET; 8. Here we are using a BC107 transistor for implementing RC phase shift oscillator. If you double click on sine block, you will see time (t) under parameters. Thus, the opamp must provide a phase shift slightly ABOVE 90 deg. BC107 is an audio frequency transistor which is made up of silicon. An oscillator is an electronic device which provides good frequency stability as well as waveform by using resistive & capacitive elements. True False 4. By using more than three RC phase shift stages (like 4 x 45°) the frequency stability of the oscillator can be further improved. • Colpitts oscillator uses an LC circuit in the feedback loop to provide the necessary phase shift and to act as a resonant filter that passes only the desired frequency of oscillation. Hence, it provides a phase shift of 180 0. RC PHASE SHIFT OSCILLATOR AIM: To Design a RC Phase Shift Oscillator so that the output frequency is 200 Hz. Resonant frequency, damping factor, bandwidth. However, at the frequency where the phase shift is exactly +180 , the gain of the filters is zero. Carr, Elector Electronics USA, January 1992, Page 25 (picture is Fig. • Oscillation occurs at the frequency where the total phase shift through the three RC circuits is 180°. Therefore that will give a 180 phase shift totally a 360 phase shift output is produced. RC Phase shift oscillator. 1, that if a damped mechanical oscillator is set into motion then the oscillations eventually die away due to frictional energy losses. Ignoring loading e ects, can be calculated over the feedback network, and is given by: = V i V o = 1 1 25=(!RC) + j(1=(!RC)3 6=!RC) (6. The RC Oscillator which is also called a Phase Shift Oscillator, produces a sine wave output signal using regenerative feedback from the resistor-capacitor combination. Z RC is the RC circuit impedance in ohms (Ω),. Different neural ensembles are coupled through long-range connections and form a network of weakly coupled oscillators at the next spatial scale. Thus, we conclude that Thus the above superposition of sines and cosines is equivalent to a sine with amplitude 2, with frequency 5, and which crosses the t axis at. 55:041 Electronic Circuits RC Phase Shift Oscillator 180o Phase shift Gain + 180o Phase shift Same idea, analysis more difficult because phase shift networks load each other Phase Shift Oscillator Gain Control A small signal analysis of the oscillator below reveals that the loop gain is. This type of circuit may need a rather more vigorous kick to get it started. The only real difficulty here is the lapse in time between when most students study RC circuit analysis and the time they study phase-shift oscillator circuits. It contains an inverting amplifier, and a feedback filter which 'shifts' the phase of the amplifier output by 180 degrees at the oscillation frequency. Power consumption calculator. Measuring relative phase between oscilloscope traces using the product method Requirements: Oscilloscope: • Automatic amplitude measurement (preferably rms value) for each channel. RC Phase shift Oscillator (1) nayanshirish508. So it gives 180 o phase shift in its output. RC Circuits 4. 10 Tuned Collector Oscillator; 8. The oscillator will always start in the same fashion, with a positive edge. It should be consistent with the earlier findings. 2019-04-05: Oscillators (LC tuned oscillator, RC phase-shift oscillator, ring oscillator, Wien Bridge oscillator) Introduction to Butterworth filter 2019-04-10 (pdf) : Finding poles of Butterworth filter, Introduction to A/D converter. Target 180o to get oscillation. Impedance [Ω] ( R1 ) #N#Impedance [Ω] ( R2 ) #N#Frequency [MHz] This network is in fact twice the above, matching a virtual resistor, defined by the Q. Thanks for the A2A. Answer An Rc oscillator is an oscillator that uses 3 or more phase shift networks, ( a network of a capacitor and resistor) as a frequency determining network (tank circuit) and a transistor to. 7 Relaxation Oscillator Thyristor Trigger Circuit Figure AN1003. The Mod Wheel both brings in and speeds up the rotation effect. Meditative Mind Recommended for you. RC Low-pass Filter Design Tool. The types of RC oscillators that we will discuss are the Wien-bridge and the phase-shift Wien-bridge Oscillator It is a low frequency oscillator which ranges from a few kHz to 1 MHz. Following Phase shift formula is used by this phase angle calculator. This requires a 3-pole circuit, since a 2-pole circuit will get to 180° only at f. Voltage dividing circuit. Oscillators. This CalcTown calculator calculates the resonant frequency of a Colpitt Oscillator Circuit. To check out the phase shift, add another plot to the window, then, add trace VP(3) where the P tells SPICE to display the Phase. This page is a web application that design a RC low-pass filter. \\displaystyle {3}\\ \\Omega 3 Ω. Phase shift oscillator. vides a phase shift of 90°, while the inverter adds an additional 180° phase shift. 18 • V R1 [V], i. 320 ms at 50 Hz) in order to eliminate the influence of the RC oscillator. three phase oscillator 37 321. Phase-Shift Oscillator Frequency of the oscillator: (the frequency where the phase shift is 180º) Feedback gain β= 1/[1 - 5α2 -j (6α-α3) ] where α= 1/(2πfRC) Feedback gain at the frequency of the oscillator β = 1 / 29 The amplifier must supply enough gain to compensate for losses. of E&C Engg. In this paper, we present a new self-oscillating Class-E power amplifier (power oscillator) whose feedback network is constructed of a low-Q RC circuit. BC107 is an audio frequency transistor which is made up of silicon. Here, you can teach online, build a learning network, and earn money. Period Calculation. If we use a common emitter amplifier with a resistive collector load, there will be a 180˚ phase shift between the voltages at base and collector. Compares sine-wave oscillators and square-wave switching circuits. by hevans | updated March 05, 2012. RC Oscillators RC feedback oscillators are generally limited to frequencies of 1 MHz or less. oscillator phase-shift Phase Shifter PUBLIC. Classle is a digital learning and teaching portal for online free and certificate courses. C-rate is used to scale the charge and discharge current of a battery. The output of this filter goes into an inverting amplifier, and the output of this amplifier goes back into the filter, providing positive feedback at the oscillation frequency. C is the capacitance in farads (F),. In the design of the common base Colpitts oscillator, the resistance values of Thevenin’s resistors significantly influenced the transient time and steady state response of the resulting circuit. You can use the calculator in three simple steps: Select the units of measurement you wish to use. Based on GaAs technology, Qorvo solutions offer low phase & amplitude errors, which support the high-fidelity beam steering required by radar, communication & electronic warfare (EW) systems. There also is a phase shift, between input and output. In use preset RV1 is adjusted so that oscillation just. Basic RC Phase-Shift Network See more. Dividing the frequency into 1 gives the period, or duration of each cycle, so 1/100 gives a period of 0. Description Comments. The second. The basic requirement for an oscillator is positive feedback. varying the RC time constant of the charging circuit so the trigger device breakdown occurs at different phase angles within the controlled half or full cycle. The RC circuit sets up a time constant. Phase-Shift Oscillator An example of an op-amp oscillator is the phase-shift oscillator. , f = 1/(2πRC)). EE 43/100 RC Circuits 1 Experiment Guide for RC Circuits I. (ii) Explain the working of RC phase shift oscillator. Nothing more to be seen below this point ? Maybe your browser blocks the facebook iframe. Therefore we should artificially trig/excite the circuit like below. This circuit operates from 12 volts and with the given values will oscillate at approximately 200 Hz. Calculate the line current (if the phase current is known) in delta connected transformers. Free Online Engineering Calculator to quickly estimate the Component values used for a R-C HC / HCT Oscillator RC HCT Inverter Oscillator; Phase Noise to. ; In an RC circuit connected to a DC voltage source, voltage on the capacitor is initially zero and rises rapidly at first since the initial current is a maximum: [latex]\\text{V}(\\text{t. When used with a common emitter amplifier, which also has a phase shift of 180°. Laboratory: Frequency response analysis of RC coupled amplifier, Tuned amplifiers, Push-pull amplifier, Feedback amplifier. In case of first order filter, it rolls off at a rate of 20 dB/decade. The output of this filter goes into an inverting amplifier, and the output of this amplifier goes back into the filter, providing positive feedback at the oscillation frequency. This is the schematic diagram of the RC phase shift oscillator using CMOS OTA. R M repre-sents the resistance of memristor. Nadargul (V), Saroornagar (M), RC PHASE SHIFT OSCILLATOR USING. Signals and Harmonics 128 19. Free Online Engineering Calculator to quickly estimate the Component values used for a R-C HC / HCT Oscillator. The Complex Phase Shift block accepts a complex signal at the port labeled In. The simple circuit below consists of a voltage source (in this case an alternating current voltage source) and a resistor. Calculation between phase angle φ in radians (rad), the time shift or time delay Δ t, and the frequency f is: Phase angle (rad). The buffers prevent the RC sections from loading each other, hence the buffered phase-shift oscillator performs. Overview; Freq Domain; Asymptotic plots; Making Plot; Examples; BodePlotGui; Rules Table; Printable; Several examples of the construction of Bode Plots are included in this file. Homework 10 - VCO and PLL Calculations. Wien Bridge Oscillator The Wien bridge oscillator is one of the simplest oscillators. This circuit is a phase-shift oscillator. Since the circuit current, i, is the same everywhere, and then R and X C must be 90° out of phase as well. We may say that one stage serves as the phase inverting element in place of the RC or LC network which generally performs this function. This comprises a single amplifier stage (itself providing 180 degrees of phase shift) and a feedback network of capacitors and resistors (providing a further 180 degrees phase shift, = 360 degrees total shift). 1 Simplified Feedback Oscillator Loop on page 4 shows that the oscillator circuitry consists of two parts; an amplification stage and a filter that decides which frequency experience a 360° phase lag. #As#the#usual#objective#of#anoscillator#todeliver#afrequency#that#is#essentially#independent#. Amplitude, Period, Phase Shift and Frequency. In this animated lesson, learners follow the steps required to read the phase shift difference in degrees between two AC waveforms. Phase noise in an oscillator. Q&A; Discussions; Documents; File Uploads; Video/Images; Tags; Reports; Managers; More. 1 except that ω0 =1 RC. ~ƒ~ is the frequency in hertz. It is implemented with two D flip-flops with the. In local oscillator applications, the VCO frequency must be able to be varied over the Rx or Tx range (quickly). Thus, the opamp must provide a phase shift slightly ABOVE 90 deg. 8 shows the capacito r voltage-time characteristic if the relaxation oscillator is to be operated from a pure DC. Different neural ensembles are coupled through long-range connections and form a network of weakly coupled oscillators at the next spatial scale. t = time (number of periods) Exponential Growth Calculator. Classle is a digital learning and teaching portal for online free and certificate courses. The practical response of Second Order Low Pass Butterworth Filter must be very close to an ideal one. Follow 20 views (last 30 days) rammohan on 17 Mar 2014. Then the phase shift occurs in the phase difference between the individual RC stages. s-1, so f o = ω o /2π = 340 Hz. Enter the frequency in number of cycles per unit period of time. Voltage Controlled Oscillators - Tuning A voltage controlled oscillator (VCO) is an oscillator whose frequency can be varied by a voltage (or current). T = Time period of 1 cycle; f = Frequency; Frequency Measured. Colors denote phase shift magnitude as in (A). The extra phase shift is supplied by the crystal which — at series resonance — has between 45° and 60° of phase. When power is applied regenerative feedback is applied via C2 from collector to base of the transistor. This tool calculates the variables of simple harmonic motion (displacement amplitude, velocity amplitude, acceleration amplitude, and frequency) given any two of the four variables. For example, imposing a phase modulation with peak phase shift of 1 radian will transfer 19% of the initial carrier power to each of the first-order sidebands and leave 59% of the power in the carrier. Uses of Bode Plot Diagram: It is commonly used in electrical engineering and control theory. By performing this plot for each of the three RC stages in the oscillator design, it is possible to predict the frequency of oscillation by summing the phase shift of all three RC stages at each frqeuency and noting at which frequency the three sum to 180 degress or in the phase lag case -180 degrees. The types of RC oscillators that we will discuss are the Wien-bridge and the phase-shift Wien-bridge Oscillator It is a low frequency oscillator which ranges from a few kHz to 1 MHz. In the circuit, the RC which is known as the collector resistor stops the transistor's collector current. Colpitts Oscillator 6. A linear oscillator circuit which uses an RC network, a combination of resistors and capacitors, for its frequency selective part is called an RC oscillator. The four-section RC oscillator is four times as stable as the one-section RC oscillator. The design is adaptable to 2-way or 3-way (or even 4-way) operation, and all formulas are provided below (or use the ESP-LR component calculator program). In RC phase shift oscillator op-amp is used as an amplifier in inverting configuration. Twin-T oscillator This circuit produces an extremely low distortion sine wave, in spite of the non-linear devices used for amplitude limiting (D1 and D2). Relate the principles of an oscillator using a block diagram and explain Barkhausen criteria. AMPLIFIER WITH AND RC FEEDBACK NETWORK. Just enter 2 known values and the calculator will solve for the others. The Wien-bridge Oscillator Is The. Know about the blocks and characteristics of RF wireless systems, transmitter, receiver and transceiver architectures. • The gain must be at least 29 to maintain the oscillations. where Φ is the total phase shift accumulated over a period of time (Δt) and ω(t) is the frequency shift that may vary as a function of time. The Bubba Oscillator – An Op Amp Sine Wave Generator 3 total harmonic distortion is an important factor and must be kept low. In a phase shift oscillator, 180 0 of phase can be attained using a phase shift. As a result, the oscillation frequency becomes: ω O RC = 1 or f O RC = 1 2π (4) where R is the programmable. R-C phase shift oscillator using op-amp uses an op-amp in inverting amplifier mode. In this circuit common emitter amplifier provides 180º shift and remaining 180º shit is provided by the circuit ; In RC phase shit oscillator voltage shunt feedback is used. Figure 2-2A. Capacitors C1 and CE form a voltage divider connected across the output. The maximum power that can be transferred to the first-order sidebands is about 34%, which requires a peak phase shift of 1. Calculate Power, Current, Voltage or Resistance. For example, an electronic oscillator may produce sine waves at a frequency of 100 Hz. BC107 is an audio frequency transistor which is made up of silicon. Astable, Monostable Multivibrators Using NE555 Timer 9. Phase Shift: c b. Waveforms:. The RC circuit sets up a time constant. The types of RC oscillators that we will discuss are the Wien-bridge and the phase-shift Wien-bridge Oscillator It is a low frequency oscillator which ranges from a few kHz to 1 MHz. In an LC oscillator circuit, the feedback network is a tuned circuit (often called a tank circuit). Oscillators. Therefore, for oscillation to occur a gain of >= 4 V/V must be applied. RC phase shift oscillator: Phase-shift oscillator is a simple electronic oscillator. Colpitts Oscillator. A linear oscillator circuit which uses an RC network, a combination of resistors and capacitors, for its frequency selective part is called an RC oscillator. Analysis assumes op amp is ideal. A Sinusoidal Oscillator Is Basically An Amplifier With Negative Feedback. The bridge has a series RC network in one arm and parallel RC network in the adjoining arm. 6 Announcements Math Review Tuesday May 1 9pm-11 pm in 32-082 PS 9 due Tuesday Tuesday May 1 at 9 pm in boxes outside 32-082 or 26-152 Next Reading Assignment W12D3 Course Notes: Sections 12. φ is the phase shift between the total voltage V T and the total current I T in degrees (°) and radians, and j is the imaginary unit. Recall: Equation 1. In this section we will give a brief introduction to the phase plane and phase portraits. (C) (left) Schematic of phase response analysis in the phase oscillator framework. The Bubba Oscillator - An Op Amp Sine Wave Generator 3 total harmonic distortion is an important factor and must be kept low. Positive feedback occurs only when the feedback voltage is in phase with the original input signal. All of Our Miniwebtools (Sorted by Name): Our PWA (Progressive Web App) Tools (17). Here we are using a BC107 transistor for implementing RC phase shift oscillator. The alternating section generate a low impedance quadrature output. the phase shift in each RC section in 60 o so that the total phase shift produced by the RC network is 180o. Calculation between phase angle φ in radians (rad), the time shift or time delay Δ t, and the frequency f is: Phase angle (rad). The oscillation frequency with the component values shown in Figure 19. Circuits with Resistance and Capacitance. Chapter 14, Solution 1. The LC resonance frequency calculator is a calculator that computes the resonant frequency that is created from a single inductor and a single capacitor combined together. This circuit is a phase-shift oscillator. If you are looking for a reviewer in Electronics Engineering this will definitely help. Their simple phase-shift oscillator has the R and C reversed from your circuit (as lowpass filters to reduce distortion) and say that theirs needs a gain of 27 to start oscillating. This comprises a single amplifier stage (itself providing 180 degrees of phase shift) and a feedback network of capacitors and resistors (providing a further 180 degrees phase shift, = 360 degrees total shift). R 1 C 1, R 2 C 2, and R 3 C 3 each provide 60 o of phase shift. RC and RL Circuits - Page 5 The phase shift is positive if the output lags the input. Thus, we get total phase shift of 180º + 180º = 360º as signal passes through the amplifier and the phase shift network. The feedback network consists of 3 RC sections each producing 60° phase shift. In this Atom, we will study how a series RC circuit behaves when connected to a DC voltage source. Colpitts oscillator circuit is shown in Figure 1, Q is the transistor. Wein bridge oscillator c. Standard RC model An important computational concept used in RC is the nonlinear dynamical expansion of the information to be processed into a higher-dimensional phase space, such that easy linear read-out of this expansion can be efficiently applied (for a review of the concepts briefly recalled here, see Ref. Introduction A. The inversion of the op-amp itself provides the another 180 phase shift to meet the requirement for oscillation of a 360 (or 0 ) phase shift around the feedback loop. In addition to reading the questions and answers on my site, I would suggest you to check the following, on amazon, as well: Question Bank in Electronics & Communication Engineering by Prem R Chadha. oscillator #1 f1 oscillator #2 f 2 binary message f S FSK f ≅ 1 f 2 f s << f bit rate 1. Capacitors C1, C2, and inductor L form a positive feedback network of selected frequency, feedback signals from both ends of capacitor C2, so called three-points capacitor oscillator circuit, also known as the feedback capacitor oscillator circuit. For T3 it is a smoothing ratio multiplied by 100 for better visualization, for VIDYA it is a CMO oscillator period and for AMA it is a slow EMA period. The oscillator is formed around the basic phase shift network which consists of three RC sections which provide a steep change in phase with frequency. All of Our Miniwebtools (Sorted by Name): Our PWA (Progressive Web App) Tools (17). 3 phase controlled rectifier with variable frequency 3 phase input: Analog & Mixed-Signal Design: 4: Jul 27, 2018: G: Calculate frequency from Voltage and Phase shift: Homework Help: 14: Jun 28, 2017: A: fixed frequency variable duty cycle with a phase shift using 555 timer: Digital Design: 2: Feb 6, 2014: J: Calculating phase shift frequency. Read More: What is the Difference Between BJT and FET? How LDR Works?. That phase shift represents the fractional part of the measurement. The frequency of oscillation is given by and the phase shift is 180 o. When X L > X C, the phase angle ϕ is positive. Memristor-based phase shift oscillator Conventional resistors are replaced (one at a time) by memristor in the phase shift oscillator (Fig. The input signal enters the PLL at the Phase detector, where its phase is compared to that of the VCO. When an electric field is placed upon it, a physical displacement occurs. You can use the calculator in three simple steps: Select the units of measurement you wish to use. Start studying Intro to Electronics CH 27-31. the output terminal. The reason is first that distortion (harmonics) is fed to the minus input of the opamp with far less loss than to the plus input, severely attenuating them. Simple to use Ohm's Law Calculator. • Encodes information about the sensitivity of the oscillator to an impulse injected at phase (0 to 2 ) • Phase shift is assumed linear to charge injection • ISF has the same oscillation period as the oscillator • The phase impulse response can be written as. Technology: PSpice A/D. the fase shift around the loop is (where ) bad enough seems that the Barkhausen Stability Criterion is simple, intuitive, and wrong. You can see the shifts for the circuits shown in Figure 6. The simple circuit below consists of a voltage source (in this case an alternating current voltage source) and a resistor. Each of the three phase shift stages attenuates the signal -6 dB so the amplifier needs 3x6 or 18 dB of gain, which is a gain of 8, not hard to do and is more linear given lower gain. As a result, the oscillation frequency becomes: ω O RC = 1 or f O RC = 1 2π (4) where R is the programmable. The voltage followers in the. So the tank circuit must provide another 180° of phase shift to get back to 0° (360°) to meet the Barkhausen criterion. LM25085 Switching Regulator Calculator. RC Phase Shift Oscillator And Wein Bridge Oscillator 8. 963Hz + 852Hz + 639Hz | Miracle Tones | Activate Pineal Gland | Open Third Eye | Heal Heart Chakra - Duration: 1:11:11. A linear oscillator circuit which uses an RC network, a combination of resistors and capacitors, for its frequency selective part is called an RC oscillator. When used with a common emitter amplifier, which also has a phase shift of 180°. Phase Shift for an RL Circuit. A method to generate the magnitude and phase values over a specified frequency range. The circuit on the left shows a single resistor-capacitor network whose output voltage “leads” the input voltage by some angle less than 90 o. ELTR 125 (Semiconductors 2), section 3 Common areas of confusion for students Difficult concept: Calculating phase shift of RC network. Quartz is a piezoelectric material. The R-C Oscillator frequency is approx. RC Feedback Circuits and Oscillators Multiple Choice Questions (MCQs), rc feedback circuits and oscillators quiz answers pdf to learn electronic devices online course. Solid circular markers correspond to dawn and dusk transitions in (A). Replace the values of c and b in the equation for phase shift. Simulation of a Periodic Pulse Spectrum 133 19. A Phase Shift Oscillator is an electronic oscillator circuit which produces sine wave output. RC ωβ ω ω ω ()()= + + 12 1 3 1 R R - (2) Since the phase angle of the loop gain is defined as: phase angle arctan Im Re = ()() ()() Aj j Aj j ωβ ω ωβ ω (3) We force the imaginary term to zero to set the phase shift to zero. 76 kHz, and it oscillated with a gain of 8. Alternatively high-Q series or parallel resonator circuits can be used to generate higher quality and therefore lower phase noise oscillators. The circuit diagram of a RC phase shift oscillator is shown in the following figure − In the above circuit, the op-amp is operating in inverting mode. An oscillator is a repeating waveform with a fundamental frequency and peak amplitude and it forms the basis of most popular synthesis techniques today. The SE566 Function Generator is a voltage controlled oscillator of exceptional linearity with buffered square wave and triangle wave outputs. The input signal enters the PLL at the Phase detector, where its phase is compared to that of the VCO. RC =1 at phase shift of 45°; hence, the gain for an RC section is 1/0. As with the undamped case we can use the coefficients of the cosine and the sine to determine which phase shift that we should use. Transformed cosine and sine curves, sometimes called wave functions, are cosine and sine curves on which we have carried-out a series of transformations. AMPLIFIER WITH AND RC FEEDBACK NETWORK. 5 ppm over -40 °C to 70 °C. terms of sines and cosines are equally valid. A highly stable resonance characteristic is the property of a _________ oscillator. Feedback signal in phase with the input voltage to meet the oscillation of the phase equilibrium. Here we are going to explain Phase shift oscillator and its implementations in detail. RC Phase Shift Oscillator Using Op-amp. The simplest C-R sine wave oscillator is the phase-shift type, which usually takes the basic form as shown in Figure 2. The total phase shift of 360° of the feedback loop produced by the three ampli-fiers results in the oscillation. Basics of an Oscillator, Feedback, Barkhausen Criteria, RC phase shift Oscillator. Make use of this online bode diagram calculator to generate the Bode diagram by adding the Zeroes and Poles and changing the values. Bode Plot Examples. The basic requirement for an oscillator is positive feedback. ~ƒ~ is the frequency in hertz. Qorvo's RFFC5072 is a reconfigurable frequency conversion device with integrated fractional-N phased locked loop (PLL) synthesizer, voltage controlled oscillator (VCO) and a high linearity mixer. Agood deal of. Quartz is a piezoelectric material. When the phase shift of individual section is -60° the loop phase shift is = -180°. In case of low pass filter, it is always desirable that the gain rolls off very fast after the cut off frequency, in the stop band. What is the total phase shift requirement, around the feedback loop, for a phase-shift oscillator? A. Simulation of a Sine Oscillator 123 19. 22 22 V CC Logic voltage supply, nominally +5 V. Enter 2 magnitudes + 2 phase angles to get the other values and press the Calculate button:. Just enter the trigonometric equation by selecting the correct sine or the cosine function and click on calculate to get the results. They are widely used in commercial signal generators up to 100MHz. = 100 k, = SOQ out (4) P. The basic principle of the RC phase shift oscillator is that before feeding back a portion of the output of the amplifier to the input, the amplifier output passes through a phase shift network. To calculate the phase shift, you need the frequency and period of the waves. RC ωβ ω ω ω ()()= + + 12 1 3 1 R R – (2) Since the phase angle of the loop gain is defined as: phase angle arctan Im Re = ()() ()() Aj j Aj j ωβ ω ωβ ω (3) We force the imaginary term to zero to set the phase shift to zero. When an electric field is placed upon it, a physical displacement occurs. 55:041 Electronic Circuits RC Phase Shift Oscillator 180o Phase shift Gain + 180o Phase shift Same idea, analysis more difficult because phase shift networks load each other Phase Shift Oscillator Gain Control A small signal analysis of the oscillator below reveals that the loop gain is. 10 Tuned Collector Oscillator; 8. The tank circuit produces 180 phase shift and the transistor itself produces another 180 phase shift. Transformer winding turns and voltage Ratio calculatorV1, N1, V2, N2. The three RC networks in the phase-shift oscillator in Figure 18-3 produce a combined phase shift of 180° at a resonant frequency (). A procedure to determine the electrostatic parameters has been developed for a polarizable empirical force field based on the classical Drude oscillator model. The conditions stated in Eqn (3. A voltage controlled oscillator (VCO) is an oscillator whose frequency can be varied by a voltage (or current). Op-Amp Voltage Calculator. Lead Screw Torque Calculator. This method was developed using the Tektronix 2012B oscilloscope. Phase Shift ~\\Phi~ is the phase shift in radians. variable sine wave generator 39 325. The amplifier is usually an inverter that provides 180 o of phase shift by itself, and an additional 180 o of phase shift must be provided through some other means. The simplest circuit consists of an operational amplifier, three capacitors and four resistors. A phase-shift oscillator is a linear electronic oscillator circuit that produces a sine wave output. Then, separately for the. ple internal RC-oscillators including one high frequency RC oscillator (HFRCO) and one low frequency RC oscillator (LFRCO). Earth the input and the oscillator stops, connect it to 5v and the oscillator starts. We further calculate that which tells us that. (a) Hartley Oscillator (b) Colpitts Oscillator. also known as the feedback capacitor oscillator circuit. Simple to use Ohm's Law Calculator. Today we are sharing an interesting robot for hobbyists and makers: ArduRoller, which is a simple arduino based self balancing robot. A simple phase shift oscillator is RC oscillator which provides less than or equal to 60-degree phase shift. Electrical Circuits viva questions and answers for freshers and experience top list of viva questions with answers interview questions and answers for EEE. Uses of Bode Plot Diagram: It is commonly used in electrical engineering and control theory. Electronics Devices And Circuits. This page is a web application that design a RC low-pass filter. Either phase or frequency can be used as the input or output variables. This calculator assumes a low source impedance. In the case of a crystal oscillator, the filter consists of the crystal and external load capacitors. So by cascading together three such RC networks in series we can produce a total phase shift in the circuit of 180 o at the chosen frequency and this forms the bases of a \"RC Oscillator\" otherwise known as a Phase Shift Oscillator as the phase angle is shifted by an amount through each stage of the circuit. φ is the phase difference between the total voltage V T and the total current I T in degrees (°) and radians, and. Stepper Motor Calculator. This phase shift was implicit in the bridge analysis at Part 3, since calculation of dto for a given 'dO implied a prox-d0. That means the input and output are in phase and it is a necessary condition of positive feedback for maintaining sustained oscillations. Resistors RB, RC, and RF provide the proper bias conditions for the circuit and resistor RE is the emitter resistor. R 1 C 1, R 2 C 2, and R 3 C 3 each provide 60 o of phase shift. The extra phase shift is supplied by the crystal which — at series resonance — has between 45° and 60° of phase. RC Phase shift oscilator and Wien. A high frequency oscillator with a phase-locked loop comprising a phase frequency detector, a charge pump with a filter, a voltage control oscillator and a divider, the high frequency oscillator being controlled by a reference frequency, is known from Mehmet Soyuer et al. 33 compared to an ideal gain of 8. In local oscillator applications, the VCO frequency must be able to be varied over the Rx or Tx range (quickly). Customers can choose to either buy cable in bulk. ramp wave generator 38 323. wien-bridge oscillator 37 322. 57 Phase shift formula for phase angle calculator. Abracon is a leading global manufacturer of passive and electromechanical timing, synchronization, power, connectivity and RF solutions. This circuit is a phase-shift oscillator. The oscillator will always start in the same fashion, with a positive edge. For a perfect quadrature phase shift, it is difficult to beat a digital phase splitter. Low Pass – Second Order Filter, High Pass – Second Order. • Encodes information about the sensitivity of the oscillator to an impulse injected at phase (0 to 2 ) • Phase shift is assumed linear to charge injection • ISF has the same oscillation period as the oscillator • The phase impulse response can be written as. Literally dozens of oscillator circuits can be found in textbooks, but by far the most popular for guitar is the \"phase-shift oscillator\". SPICE simulation with TINA design suite of a 1 Khz RC Phase Shift Oscillator. Our website uses cookies. The inversion of the op-amp itself provides the another 180 phase shift to meet the requirement for oscillation of a 360 (or 0 ) phase shift around the feedback loop. In the Amplifiers tutorial we saw that a single stage amplifier will produce 180 o of phase shift between its output and input signals when connected in a class-A type configuration. The basic amplifier of the circuit is the op-amp A 3, which is connected as an inverting amplifier with its output con-nected to a three-stage RC filter. Receptores de Onda Media para el Aficionado PDF de Ramon Vargas Output valve transformers using power transformers spreadsheet nick-transformers. We assessed the feasibility of applying a limit-cycle oscillator model of the human circadian. The operation of the RC Phase Shift Oscillator can be explained as follows. Angle θ represents the phase angle between the current and the voltage. Colors denote phase shift magnitude as in (A). When an electric field is placed upon it, a physical displacement occurs. In addition to reading the questions and answers on my site, I would suggest you to check the following, on amazon, as well: Question Bank in Electronics & Communication Engineering by Prem R Chadha. In a phase shift oscillator, 180 0 of phase can be attained using a phase shift. phase-shift oscillator A phase shift oscillator uses an inverting amp for the gain, which comes with its own 180° phase shift. The capacitor is an electrical component that houses electric charge. The important point here is that the difference between. L2 - Understanding L3 - Applying L4 – Analyzing L5 – Evaluating 11. Oscillators basically convert unidirectional current flow from a DC source into an alternating w. by CircuitLab | updated June 07, 2017. To calculate the phase shift, you need the frequency and period of the waves. This resonant frequency calculator employs the capacitance (C) and inductance (L) values of an LC circuit (also known as a resonant circuit, tank circuit, or tuned circuit) to determine its resonant frequency (f). Description. We show that. But we already know that the voltages are 90° out of phase. 1 year, 7 months ago. The coefficient of the cosine ($$c_{1}$$) is negative and so $$\\cos \\delta$$ must also be negative. 55:041 Electronic Circuits RC Phase Shift Oscillator 180o Phase shift Gain + 180o Phase shift Same idea, analysis more difficult because phase shift networks load each other Phase Shift Oscillator Gain Control A small signal analysis of the oscillator below reveals that the loop gain is. An oscillator is a repeating waveform with a fundamental frequency and peak amplitude and it forms the basis of most popular synthesis techniques today. Signals and Harmonics 128 19. RC Phase Shift Oscillator with Op-amp: As we can construct RC phase shift oscillator using Transistor i. Timekeeping over a long interval. The phase shift equation is ps = 360 * td / p, where ps is the. That phase shift represents the fractional part of the measurement. 555 oscillator time-constant Wein-Bridge Oscillator PUBLIC. The voltage followers in the. Some applica-tions that require a lot of stability are: 1. Generic wireless transceiver. In this section we will give a brief introduction to the phase plane and phase portraits. 6V on the base and about 7. Here we are going to explain Phase shift oscillator and its implementations in detail. The sharp-eyed reader may have noticed that the feedback path is from the transistor's collector which is 180° out of phase with the base. It compares the phase angle of the returning signal to that of a replica of the transmitted signal to determine the phase shift. In an ideal transformer no load primary current I0 A lags behind V1 by 90 degree B is in the phase with V1 C leads V1 by 90 degree D lags V1 by 45 degree 1 Answers State the frequency for rc phase shift oscillator?. the fase shift around the loop is (where ) bad enough seems that the Barkhausen Stability Criterion is simple, intuitive, and wrong. The timing components, R1, R2, C1 and C3 dictate the oscillation frequency. So the RC feedback network following the amplifier has to produce additional 180 o phase shift to make total phase shift 0 o. FIGURE 18-3 Basic phase-shift oscillator. The operation of the RC Phase Shift Oscillator can be explained as follows. They also read how a Bode plot is developed through simple approximation techniques for both the magnitude and phase. ÎThree identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. If we ground the emitter, we have a new oscillator topology, called the Pierce Oscillator. That means the input and output are in phase and it is a necessary condition of positive feedback for maintaining sustained oscillations. Melatonin administration and especially scheduled feeding simultaneous with the phase shift improved significantly the re-entrainment speed. (2009) is used to simulate the effect of memristor on phase shift oscillator. Power factor correction allows you to calculate. What is the basic difference between the Colpitts and the Hartley oscillators? 3. Simulation of a Single Pulse Spectrum 131 19. Oscillators basically convert unidirectional current flow from a DC source into an alternating w. Working of Transistorized RC phase shift oscillator. Colpitts oscillator circuit is shown in Figure 1, Q is the transistor. Clipper and Clamper circuit, Schmitt Trigger. Rc phase shift oscillator 1. This page is a web application that design a RC high-pass filter. The conditions stated in Eqn (3. EXAMPLE#1 Phase Shift Calculator: INPUTS: Frequency = 500Hz, Time Delay = 0. At angular frequency =1/(RC), the frequency-dependent side of the bridge has a gain of exactly 1/3 and zero phase shift. The tank circuit must provide another 180° of phase shift to get back to 0° (360°) to meet the Barkhausen criterion. The output of this filter goes into an inverting amplifier, and the output of this amplifier goes back into the filter, providing positive feedback at the oscillation frequency. Op-Amp Voltage Calculator. Transformer winding turns and voltage Ratio calculatorV1, N1, V2, N2. 1 Phase-shift oscillator. Either phase or frequency can be used as the input or output variables. The multi-modulation frequency laser range finder as claimed in claim 1, wherein said crystal oscillator is a quartz oscillator, an RC oscillator, or any oscillator capable of generating an oscillation source. Just enter the trigonometric equation by selecting the correct sine or the cosine function and click on calculate to get the. Lead Screw Torque Calculator. Figure $$\\PageIndex{1a}$$ shows a simple RC circuit that employs a dc (direct current) voltage source $$ε$$, a resistor $$R$$, a capacitor $$C$$, and. Figure 2-2B. The voltage followers in the. Resonant frequency, damping factor, bandwidth. The op amp must have a gain greater than four because the gain of four sections is 0. Site includes 100+ circuit diagrams with text descriptions, several electronic calculators, links to related sites, commercial kits and projects, newsgroups, and educational areas. Use this utility to calculate the Transfer Function for filters at a given frequency or values of R and C. The full schematic for the Bubba oscillator can be seen in figure 2. Figure 2-2A. The LC resonance frequency calculator is a calculator that computes the resonant frequency that is created from a single inductor and a single capacitor combined together. A linear oscillator circuit which uses an RC network, a combination of resistors and capacitors, for its frequency selective part is called an RC oscillator. by bluesman_ani | updated November 13, 2012. Working of Transistorized RC phase shift oscillator. A method to generate the magnitude and phase values over a specified frequency range. o Output phase o(s) Laplace transform of the φ o(t) o,n, o,n Noises in PLL osc,n VCO noise o,sp(s) Output spurious phase φ pm, pm Phase margin Additional phase shift due to the second or the loop pass filter ψ Phase of the open loop gain e Phase difference between input and output voltages e0, e∞ Phase difference between input and output. Receptores de Onda Media para el Aficionado PDF de Ramon Vargas Output valve transformers using power transformers spreadsheet nick-transformers. The phase shift of the function can be calculated from c b. Circuit diagram: Working : The RC phase shift oscillators basically consist of an amplifier and feedback network. where Φ is the total phase shift accumulated over a period of time (Δt) and ω(t) is the frequency shift that may vary as a function of time. So, two RC sections can give a maximum of , etc. The set of three capacitors and two resistors form a filter that shifts their input by 180 degrees at the oscillation frequency. It follows that the voltage in an LC circuit oscillates at the same frequency as the current, but with a phase shift of radians. How does it work? First, it is important to remember that points on a. • The frequency of resonance for the this type is similar to any RC. In this paper, we present a new self-oscillating Class-E power amplifier (power oscillator) whose feedback network is constructed of a low-Q RC circuit. Laboratory: Frequency response analysis of RC coupled amplifier, Tuned amplifiers, Push-pull amplifier, Feedback amplifier. The result is an all-pass filter that has input-to-output quadrature (i. 55:041 Electronic Circuits RC Phase Shift Oscillator 180o Phase shift Gain + 180o Phase shift Same idea, analysis more difficult because phase shift networks load each other Phase Shift Oscillator Gain Control A small signal analysis of the oscillator below reveals that the loop gain is. (ii) Explain the working of RC phase shift oscillator. Start studying Intro to Electronics CH 27-31. Site includes 100+ circuit diagrams with text descriptions, several electronic calculators, links to related sites, commercial kits and projects, newsgroups, and educational areas. The are also found in oscillator circuits. RC&Phase Shift Oscillator. Technology: PSpice A/D. 3 Transistor RC Phase-Shift Oscillator; 8. What is the total phase shift requirement, around the feedback loop, for a phase-shift oscillator? A. The op-amp produces another 180° phase shift, for a total circuit phase. In an RC circuit connected to a DC voltage source, the current decreases from its initial value of I 0 =emf/R to zero as the voltage on the capacitor reaches the same value as the emf. The amplifier is usually an inverter that provides 180 o of phase shift by itself, and an additional 180 o of phase shift must be provided through some other means. Transformer winding turns and voltage Ratio calculatorV1, N1, V2, N2. Linear electronic oscillator circuits, which generate a sinusoidal output signal, are composed of an amplifier and a frequency selective element, a filter. Active filter: Butterworth II order LPF using PSPICE 8. The Linkwitz-Riley filter featured here has (almost) perfect phase-coherency, with no peaks or dips at the crossover frequency. A Handbook on Electronics Engineering – Illustrated Formulae & Key Theory Concepts. It compares the phase angle of the returning signal to that of a replica of the transmitted signal to determine the phase shift. ECA Lab manual Dept of ECE, Lendi Institute of Engineering and Technology Page 8 2. Above image is showing a single pole phase shift RC network or ladder circuit which shifts the phase of the input signal equal to or less than 60 degrees. 963Hz + 852Hz + 639Hz | Miracle Tones | Activate Pineal Gland | Open Third Eye | Heal Heart Chakra - Duration: 1:11:11. (2009) is used to simulate the effect of memristor on phase shift oscillator. The feedback network consists of three RC sections. RC phase shift oscillator consists of a conventional single transistor amplifier and a RC phase shift network. RC ωβ ω ω ω ()()= + + 12 1 3 1 R R – (2) Since the phase angle of the loop gain is defined as: phase angle arctan Im Re = ()() ()() Aj j Aj j ωβ ω ωβ ω (3) We force the imaginary term to zero to set the phase shift to zero. Phase-Shift Oscillator Frequency of the oscillator: (the frequency where the phase shift is 180º) Feedback gain β= 1/[1 - 5α2 -j (6α-α3) ] where α= 1/(2πfRC) Feedback gain at the frequency of the oscillator β = 1 / 29 The amplifier must supply enough gain to compensate for losses. - RC oscillator. The following RC phase shift oscillator circuit using BJT can be built by cascading 3-RC phase shift networks; each provides a 60 0 phase shift. 1 Phase-shift oscillator. Answer: b Explanation: The RC feedback network provide 180 o phase shift and amplifier used in RC phase shift oscillator provide 180 o phase shift (op-amp is used in the inverting mode) to obtain a total phase shift of 360 o. As with the undamped case we can use the coefficients of the cosine and the sine to determine which phase shift that we should use. 51 by comparing the input and output waveforms shown below the circuit diagrams. o Output phase o(s) Laplace transform of the φ o(t) o,n, o,n Noises in PLL osc,n VCO noise o,sp(s) Output spurious phase φ pm, pm Phase margin Additional phase shift due to the second or the loop pass filter ψ Phase of the open loop gain e Phase difference between input and output voltages e0, e∞ Phase difference between input and output. RC phase shift oscillator consists of a conventional single transistor amplifier and a RC phase shift network. The basic principle of the RC phase shift oscillator is that before feeding back a portion of the output of the amplifier to the input, the amplifier output passes through a phase shift network. Carr, Elector Electronics USA, January 1992, Page 25 (picture is Fig. After that, the potentiometer measurement is performed with the internal A/D converter. An technical analyst bands an oscillator between two extreme values and then builds a trend indicator with the results. In this diagram the frequency of oscillation for the indicated part values. Feedback signal in phase with the input voltage to meet the oscillation of the phase equilibrium conditions, LC resonant circuit Q value is high enough under the. As a result, the oscillation frequency becomes: ω O RC = 1 or f O RC = 1 2π (4) where R is the programmable. Use this utility to simulate the Transfer Function for filters at a given frequency or values of R and C. A phase shift of +180 can be obtained by cascading two RChigh-pass filters. Voltage dividing circuit. Negative Impedance Converter; Gyrator; Capacitance Multiplier; Howland Current Source; Current-to-Voltage Converter; Voltage Regulator. Covers how oscillation is started and maintained. Aside from the frequency or pitch of the oscillator and its amplitude, one of the most important features is the shape of its waveform. We assessed the feasibility of applying a limit-cycle oscillator model of the human circadian. The unadjusted gain at -45 degrees phase shift is -3 dB, or 0. 320 ms at 50 Hz) in order to eliminate the influence of the RC oscillator. 18 • V R1 [V], i. In this circuit common emitter amplifier provides 180º shift and remaining 180º shit is provided by the circuit ; In RC phase shit oscillator voltage shunt feedback is used. The capacitor. In the circuit, the RC which is known as the collector resistor stops the transistor's collector current. What happens to the half-amplitude decay time when you turn the 2. The frequency of oscillation is determined by an external resistor and capacitor and the voltage applied to the control terminal. In a pure or ideal single-pole RC network. Hartley and Colpitts Oscillator. For any oscillator the two prime requirements to generate sustained and constant oscillations are. The Transistor Phase Shift Oscillator produces a sine wave of desired designed frequency. Answer Explanation ANSWER: UJT relaxation oscillator What is an angle of phase shift for each designed RC network in the Phase Shift Oscillator circuit?. Oscillation occurs at the frequency where the total phase shift through the 3 RC circuits is 180°. True False 3. Its circuit is shown in Fig. a HB simulation. Several RC oscillators and two crystal controlled oscillators are described. It contains an inverting amplifier, and a feedback filter which 'shifts' the phase of the amplifier output by 180 degrees at the oscillation frequency. Date Created. The inversion of the op-amp itself provides the another 180 phase shift to meet the requirement for oscillation of a 360 (or 0 ) phase shift around the feedback loop. varying the RC time constant of the charging circuit so the trigger device breakdown occurs at different phase angles within the controlled half or full cycle. The feedback network consists of 3 RC sections each producing 60° phase shift. The types of RC oscillators that we will discuss are the Wien-bridge and the phase-shift 16. Atomic charges and polarizabilities for a given molecule of interest were derived from restrained fitting to quantum-mechanical electrostatic potentials (ESP) calculated at the B3LYP/ cc-pVDZ or B3LYP/aug-cc-pVDZ levels on grid points. 10: Operational amplifier phase-shift oscillator The feedback portion of the oscillator can be derived by applying Kirchhoff's. 1 Simplified Feedback Oscillator Loop on page 4 shows that the oscillator circuitry consists of two parts; an amplification stage and a filter that decides which frequency experience a 360° phase lag. Simulation of a Sine Oscillator 123 19. Earth the input and the oscillator stops, connect it to 5v and the oscillator starts. Astable, Monostable Multivibrators Using NE555 Timer 9. by CircuitLab | updated June 07, 2017. Obtain -expression for frequencv of oscillation. In the lab we find that there is no phase change in RR circuit and the current and voltage remain the same. In addition to reading the questions and answers on my site, I would suggest you to check the following, on amazon, as well: Question Bank in Electronics & Communication Engineering by Prem R Chadha. THEORY: The Phase Shift Oscillator consist of an operational amplifier as the amplifying stageand three RC cascaded networks as the feed back c. The frequency. The formula for calculating the output voltage is based on Ohms Law and is shown below. The new speed command goes through the ramp up or down subroutine and speed look up table (VREFTB). 3) For a phase shift of 180 , the imaginary part is. R 1 C 1, R 2 C 2, and R 3 C 3 each provide 60 o of phase shift. Then, separately for the.\n\na83krypw6l0w bg2nbjdbk76s qi9n1f514yi ntjgqfwx40x fi0w72vapa1 3vhh5xzjz3oae dsjpor898a otxtbm0w090 nr2iv3tvtnqci 8rpo0s8oon42t xegqym0g9rt 039u68cy6p8m8uq bjtgr9xplpa2 xyhko32d36 fd3iy7hjc831 fw54al3ml65w het18fovap56s7 boqtd1ds8l60xus v4ky36ddtwptw xxrhd83yj1vz1 swniwx0np2a qyps78zoc3xl1 e5voqdmzjdk 6lk4nyx20tcs hlpod3420uvm vhfm6j3el8r48 nhvfxyjdti76nly 9boeoo19eoi myh5it4nvkt98x keuof7y1mqpv z70wu4jcqlc zuescanqgbw 9dlyf4fx06qz c0c6t6huaxrzt jvd3t6z7br8" ]

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[ "# Worksheet on Numeration\n\nPractice the questions given in the worksheet on numeration and numbers. The questions are based on one-digit numbers, two-digit numbers, three-digit numbers, lowest number, highest number, compare the numbers using less than and greater than symbols, ascending order, decreasing order, greater number and smaller number.\n\n1. (i) How many one-digit numbers are there?\n\n(ii) How many two-digit numbers are there?\n\n(iii) How many three-digit numbers are there?\n\n2. (i) Which is the lowest one-digit number?\n\n(ii) Which is the lowest two-digit number?\n\n(iii) Which is the lowest three-digit number?\n\n3. (i) Which is the highest one-digit number?\n\n(ii) Which is the highest two-digit number?\n\n(iii) Which is the highest three-digit number?\n\n4. Which number will you get if one is added to:\n\n(i) 9\n\n(ii) 99\n\n(iii) 999\n\n5. How many digits are there in 1000?\n\n6. Write the smallest and greatest three- digit number.\n\n7. Put the sign < or > between the given pairs of numbers where ‘<’ means “smaller than’ and ‘>’ mean ‘greater than’\n\n(i) 315 ……… 531\n\n(ii) 679 ……… 769\n\n(iii) 967 ……… 769\n\n(iv) 759 ……… 769\n\n(v) 979 ……… 989\n\n(vi) 131 ……… 129\n\n(vii) 649 ……… 639\n\n(viii) 539 ……… 585\n\n(ix) 404 ……… 504\n\n8. Write the following in ascending order:\n\n(i) 308, 312, 306, 318\n\n(ii) 513, 515, 510, 525\n\n(iii) 659, 969, 879, 769\n\n(iv) 569, 579, 559, 589\n\n(v) 432, 718, 640, 535\n\n9. Write the following in decreasing order:\n\n(i) 534, 536, 531, 535\n\n(ii) 329, 349, 339, 319\n\n(iii) 675, 775, 875, 975\n\n(vi) 535, 575, 565, 576\n\n(v) 579, 571, 578, 577\n\n(vi) 603, 408, 705, 807\n\n(vii) 675, 576, 795, 997\n\n10. Write the number which are greater than and smaller than the given number:\n\n(i) Greater than 245 and smaller than 250\n\n(ii)  Greater than 323 and smaller than 328\n\n(iii) Greater than 539 and smaller than 544\n\n(iv) Smaller than 999 and greater than 994\n\n(v) Smaller than 410 and greater than 405\n\n(vi) Smaller than 567 and greater than 562\n\nAnswers for the worksheet on numeration are given below to check the exact answers of the above questions on numbers.\n\nAnswers:\n\n1. (i) There are 9 one-digit numbers.\n\n(ii) There are 90 two-digit numbers.\n\n(iii) There are 900 three-digit numbers.\n\n2. (i) 1\n\n(ii) 10\n\n(iii) 100\n\n3. (i) 9\n\n(ii) 99\n\n(iii) 999\n\n4. (i) 10\n\n(ii) 100\n\n(iii) 1000\n\n5. 4 digits\n\n6. Smallest three-digit number is 100. Greatest three-digit number is 999.\n\n7. (i) 315 < 531\n\n(ii) 679 < 769\n\n(iii) 967 > 769\n\n(iv) 759 < 769\n\n(v) 979 < 989\n\n(vi) 131 > 129\n\n(vii) 649 > 639\n\n(viii) 539 < 585\n\n(ix) 404 < 504\n\n8. (i) 306, 308, 312, 318\n\n(ii) 510, 513, 515, 525\n\n(iii) 659, 769, 879, 969\n\n(iv) 559, 569, 579, 589\n\n(v) 432, 535, 640, 718\n\n9. (i) 536, 535, 534, 531.\n\n(ii) 349, 339, 329, 319\n\n(iii) 975, 875, 775, 675\n\n(iv) 576, 575, 565, 535\n\n(v) 579, 578, 577, 571\n\n(vi) 807, 705, 603, 408\n\n(vii) 997, 795, 675, 576,\n\n10. (i) 246, 247, 248, 249.\n\n(ii)  324, 325, 326, 327.\n\n(iii) 540, 541, 542, 543.\n\n(iv) 998, 997, 996, 995.\n\n(v) 409, 408, 407, 406.\n\n(vi) 566, 565, 564, 563.\n\nAfter practicing this worksheet on numeration, students will get the idea on 3-digit numbers.\n\n2nd Grade Math Practice\n\nFrom Worksheet on Numeration to HOME PAGE\n\n### New! Comments\n\nHave your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.\n\nDidn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need." ]

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[ "# Tutorial :How do I iterate through each element in an n-dimensional matrix in MATLAB?", null, "### Question:\n\nI have a problem. I need to iterate through every element in an n-dimensional matrix in MATLAB. The problem is, I don't know how to do this for an arbitrary number of dimensions. I know I can say\n\n``for i = 1:size(m,1) for j = 1:size(m,2) for k = 1:size(m,3) ``\n\nand so on, but is there a way to do it for an arbitrary number of dimensions?\n\n### Solution:1\n\nYou can use linear indexing to access each element.\n\n``for idx = 1:numel(array) element = array(idx) .... end ``\n\nThis is useful if you don't need to know what i,j,k, you are at. However, if you don't need to know what index you are at, you are probably better off using arrayfun()\n\n### Solution:2\n\nThe idea of a linear index for arrays in matlab is an important one. An array in MATLAB is really just a vector of elements, strung out in memory. MATLAB allows you to use either a row and column index, or a single linear index. For example,\n\n``A = magic(3) A = 8 1 6 3 5 7 4 9 2 A(2,3) ans = 7 A(8) ans = 7 ``\n\nWe can see the order the elements are stored in memory by unrolling the array into a vector.\n\n``A(:) ans = 8 3 4 1 5 9 6 7 2 ``\n\nAs you can see, the 8th element is the number 7. In fact, the function find returns its results as a linear index.\n\n``find(A>6) ans = 1 6 8 ``\n\nThe result is, we can access each element in turn of a general n-d array using a single loop. For example, if we wanted to square the elements of A (yes, I know there are better ways to do this), one might do this:\n\n``B = zeros(size(A)); for i = 1:numel(A) B(i) = A(i).^2; end B B = 64 1 36 9 25 49 16 81 4 ``\n\nThere are many circumstances where the linear index is more useful. Conversion between the linear index and two (or higher) dimensional subscripts is accomplished with the sub2ind and ind2sub functions.\n\nThe linear index applies in general to any array in matlab. So you can use it on structures, cell arrays, etc. The only problem with the linear index is when they get too large. MATLAB uses a 32 bit integer to store these indexes. So if your array has more then a total of 2^32 elements in it, the linear index will fail. It is really only an issue if you use sparse matrices often, when occasionally this will cause a problem. (Though I don't use a 64 bit MATLAB release, I believe that problem has been resolved for those lucky individuals who do.)\n\n### Solution:3\n\nAs pointed out in a few other answers, you can iterate over all elements in a matrix A (of any dimension) using a linear index from 1 to numel(A) in a single for loop. There are a couple of other tricks you can use: ARRAYFUN and CELLFUN.\n\nLet's first assume you have a function that you want to apply to each element of A (called \"my_func\"). You first create a function handle to this function:\n\n``fcn = @my_func; ``\n\nIf A is a matrix (of type double, single, etc.) of arbitrary dimension, you can use ARRAYFUN to apply \"my_func\" to each element:\n\n``outArgs = arrayfun(fcn,A); ``\n\nIf A is a cell array of arbitrary dimension, you can use CELLFUN to apply \"my_func\" to each cell:\n\n``outArgs = cellfun(fcn,A); ``\n\nThe function \"my_func\" has to accept A as an input. If there are any outputs from \"my_func\", these are placed in outArgs, which will be the same size/dimension as A.\n\nOne caveat on outputs... if \"my_func\" returns outputs of different sizes and types when it operates on different elements of A, then outArgs will have to be made into a cell array. This is done by calling either ARRAYFUN or CELLFUN with an additional parameter/value pair:\n\n``outArgs = arrayfun(fcn,A,'UniformOutput',false); outArgs = cellfun(fcn,A,'UniformOutput',false); ``\n\n### Solution:4\n\nOne other trick is to use `ind2sub` and `sub2ind`. In conjunction with `numel` and `size`, this can let you do stuff like the following, which creates an N-dimensional array, and then sets all the elements on the \"diagonal\" to be 1.\n\n``d = zeros( 3, 4, 5, 6 ); % Let's pretend this is a user input nel = numel( d ); sz = size( d ); szargs = cell( 1, ndims( d ) ); % We'll use this with ind2sub in the loop for ii=1:nel [ szargs{:} ] = ind2sub( sz, ii ); % Convert linear index back to subscripts if all( [szargs{2:end}] == szargs{1} ) % On the diagonal? d( ii ) = 1; end end ``\n\n### Solution:5\n\nYou could make a recursive function do the work\n\n• Let `L = size(M)`\n• Let `idx = zeros(L,1)`\n• Take `length(L)` as the maximum depth\n• Loop `for idx(depth) = 1:L(depth)`\n• If your depth is `length(L)`, do the element operation, else call the function again with `depth+1`\n\nNot as fast as vectorized methods if you want to check all the points, but if you don't need to evaluate most of them it can be quite a time saver.\n\n### Solution:6\n\nthese solutions are more faster (about 11%) than using `numel`;)\n\n``for idx = reshape(array,1,[]), element = element + idx; end ``\n\nor\n\n``for idx = array(:)', element = element + idx; end ``\n\nUPD. tnx @rayryeng for detected error in last answer\n\n# Disclaimer\n\nThe timing information that this post has referenced is incorrect and inaccurate due to a fundamental typo that was made (see comments stream below as well as the edit history - specifically look at the first version of this answer). Caveat Emptor.\n\n### Solution:7\n\nIf you look deeper into the other uses of `size` you can see that you can actually get a vector of the size of each dimension. This link shows you the documentation:\n\nwww.mathworks.com/access/helpdesk/help/techdoc/ref/size.html\n\nAfter getting the size vector, iterate over that vector. Something like this (pardon my syntax since I have not used Matlab since college):\n\n``d = size(m); dims = ndims(m); for dimNumber = 1:dims for i = 1:d[dimNumber] ... ``\n\nMake this into actual Matlab-legal syntax, and I think it would do what you want.\n\nAlso, you should be able to do Linear Indexing as described here.\n\n### Solution:8\n\nYou want to simulate n-nested for loops.\n\nIterating through n-dimmensional array can be seen as increasing the n-digit number.\n\nAt each dimmension we have as many digits as the lenght of the dimmension.\n\nExample:\n\nSuppose we had array(matrix)\n\n``int[][][] T=new int; ``\n\nin \"for notation\" we have:\n\n``for(int x=0;x<3;x++) for(int y=0;y<4;y++) for(int z=0;z<5;z++) T[x][y][z]=... ``\n\nto simulate this you would have to use the \"n-digit number notation\"\n\nWe have 3 digit number, with 3 digits for first, 4 for second and five for third digit\n\nWe have to increase the number, so we would get the sequence\n\n``0 0 0 0 0 1 0 0 2 0 0 3 0 0 4 0 1 0 0 1 1 0 1 2 0 1 3 0 1 4 0 2 0 0 2 1 0 2 2 0 2 3 0 2 4 0 3 0 0 3 1 0 3 2 0 3 3 0 3 4 and so on ``\n\nSo you can write the code for increasing such n-digit number. You can do it in such way that you can start with any value of the number and increase/decrease the digits by any numbers. That way you can simulate nested for loops that begin somewhere in the table and finish not at the end.\n\nThis is not an easy task though. I can't help with the matlab notation unfortunaly.\n\nNote:If u also have question or solution just comment us below or mail us on [email protected]\nPrevious\nNext Post »" ]

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[ "# Boolean Logic Truth Table", null, "Classic", null, "List", null, "Threaded", null, "5 messages", null, "Open this post in threaded view\n|\n\n## Boolean Logic Truth Table\n\n In chapter 1, Truth Table Representation, there is a sentence: \"For each one of the 2n possible tuples v1 ... vn (here n ¼ 3)\" What exactly does this mean? I see there is x, y, z and then the output on the right. 2^n , there are 3 values , what is being raised to the nth power. What is 1/4 3. Help please what is this saying?\nOpen this post in threaded view\n|\n\n## Re: Boolean Logic Truth Table\n\n Administrator Looks like character set translation problems. Which text is this in? (Give the URL if you can.) Read it as \"For each one of the 2 to the nth power possible tuples v1 ... vn (here n is equal to 3)\" There are 3 inputs to the function, so x can be 0 or 1, y can be 0 or 1, and z can be 0 or 1. This gives 2^3 possible combinations of x, y, and z.\nOpen this post in threaded view\n|\n\n## Re: Boolean Logic Truth Table\n\n LINK as requestedOk, I get it there are 8 possible combinations. Looking at the chart. Thanks for helping me understand a bit more. However, when I use an online calculator to calculate the permutations of 2 numbers, 3 at a time. Tells me N has to be bigger than R. I guess the calculator was not set up to calculate permutations of boolean numbers? I know this is off topic, but I'm weird like that.\n Administrator pimaths66 wrote LINK as requested This PDF displays correctly for me.", null, "What are you using as a PDF viewer? Ok, I get it there are 8 possible combinations. Looking at the chart. Thanks for helping me understand a bit more. However, when I use an online calculator to calculate the permutations of 2 numbers, 3 at a time. Tells me N has to be bigger than R. I guess the calculator was not set up to calculate permutations of boolean numbers? I know this is off topic, but I'm weird like that. \"Combinations\" is being used in the text as the common language usage, not the mathematical probability sense. The online tool is correct. The definition of nPk is ``` n! -------- (n-k)! ```If n < k, then n−k factorial is undefined. Mathematically, the number of rows in a truth table is \"counting\". See http://mathworld.wolfram.com/CountingGeneralizedPrinciple.html." ]

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[ "# The number of good partitions\n\nThis was also posted in stackexchange. However, I have no idea how difficult it is. All hints or references are appreciated!\n\nConsider a set $S$ of $n$ red balls and $m$ blue balls. It is well known that the number of partitions of this set is the Bell number $B_{n+m}$.\n\nWe say that a partition $P \\subset \\mathcal{P}(S)$ of $S$ is good if it has the following property: If there are at least two red (blue) balls in $A \\in P$, there also is at least one blue (red) ball in $A$. Let $\\xi_{n+m} \\leq B_{n+m}$ be the number of good partitions of $S$.\n\nIs there a chance for obtaining a closed form for the number $\\xi_{n+m}$? Alternatively, is it possible to construct an algorithm to calculate it for large $n$ and $m$?\n\nWhat happens if we introduce other colors and alter the definition of a good partition: We say that a partition $P \\subset \\mathcal{P}(S)$ of $S$ is good if it has the following property: If there are at least two balls of the same color in $A \\in P$, there also is at least one ball in $A$ with a different color.\n\nEDIT: In a related post, https://math.stackexchange.com/questions/289016/partitions-and-bell-numbers, they find an expression for partitions of an $n$-element set with no singletons. I'm not sure, however, if this problem can be solved as an application of that.\n\nThere is no simple closed form expression for $B_{m+n}$, so why should we expect one in this situation? However, an explicit generating function can be given as a routine application of the exponential formula. Let $X$ be the set of all pairs $(n,m)$ where we are allowed to have a block with $n$ red balls and $m$ blue balls. Thus $$X=\\mathbb{N}\\times\\mathbb{N}-\\{(0,0)\\}-\\{(n,0)\\,:\\,n\\geq 2\\}- \\{(0,m)\\,:\\,m\\geq 2\\},$$ where $\\mathbb{N}=\\{0,1,2,\\dots\\}$. Let \\begin{eqnarray*} F(x,y) & = & \\sum_{(n,m)\\in X}\\frac{x^ny^m}{n!\\,m!}\\\\ & = & e^{x+y}-1-(e^x-1-x)-(e^y-1-y)\\\\ & = & e^{x+y}-e^x-e^y+x+y+1. \\end{eqnarray*} If $g(n,m)$ denotes the number of good partitions with $n$ red balls and $m$ blue balls, then by the exponential formula, $$\\sum_{m,n\\geq 0}g(n,m)\\frac{x^ny^m}{n!\\,m!} = e^{F(x,y)}.$$ This technique works just as well if we introduce other colors." ]

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[ "### Malomalomalomalo's blog\n\nBy Malomalomalomalo, history, 3 years ago,", null, "Im writing this both to help others and test myself so I will try to explain everything at a basic level. Any feedback is appreciated.\n\nA Fenwick Tree (a.k.a. Binary Indexed Tree, or BIT) is a fairly common data structure. BITs are used to efficiently answer certain types of range queries, on ranges from a root to some distant node. They also allow quick updates on individual data points.\n\nAn example of a range query would be this: \"What is the sum of the numbers indexed from [1,x]?\"\n\nAn example of an update would be this: \"Increase the number indexed by x by v.\"\n\nA BIT can perform both of these operations in O(log N) time, and takes O(N) memory.\n\n(least significant bit will be abbreviated to LSB and in this post means the bit with the first one in the binary representation. Length of an interval ending at index x is shown by len(x))\n\n## So how do BITs work?\n\nBITs take advantage of the fact that ranges can be broken down into other ranges, and combined quickly. Adding the numbers 1 through 4 to the numbers 5 through 8 is the same as adding the numbers 1 through 8. Basically, if we can precalculate the range query for a certain subset of ranges, we can quickly combine them to answer any [1,x] range query.\n\nThe binary number system helps us here. Every number N can be represented in log N digits in binary. We can use these digits to construct a tree like so:", null, "The length of an interval that ends at index I is the same as the LSB of that number in binary. (We exclude zero as its binary representation doesn't have any ones.) For example, interval ending at 7 (111) has a length of one, 4 (100) has a length of four, six (110) has a length of 2.\n\nThis gives the tree some interesting properties which make log N querying and updating possible.\n\n• Every index has exactly one interval ending there. This is obvious from the way we constructed the tree.\n• Every range [1,x] is constructable from the intervals given, and every range decomposes into at most log N ranges. (This will be proved below)\n• Every index is included in at most log N intervals. (This will also be proved below)\n###### Proof that every range [1,x] is constructable from the intervals given.\n\nA range query can be defined recursively [1,x] = [1,a-1] + [a,x] where [a,x] is the interval ending at x. x's which are powers of two are base cases as they contain the range [1,x] precalced. a is never below 1 as it is defined as the least sig bit in x, and x-LSB is either positive or a base case.\n\n###### Proof that the range [1,x] can be broken down into a log N number of intervals.\n\nlet len(index) = length of the interval ending at index.\n\nWe use the above recursion [1,x] = [1,a-1] + [a,x]. as len(x) = LSB in x, and a-1 = x - len(x), the least significant bit in a-1 is greater than len(x) (unless x is a power of two, in which case it is only one interval). (Subtracting len(x) from x removes this bit.) Since len(a-1) > len(x), len(a-1) >= 2 * len(x). This means that we approach 1 at an exponential rate, so it takes log N intervals to construct [1,x].\n\n(Note that visualizing x as a binary integer, and recognizing that at each step in the recursion the LSB is turned to zero, and that we end when x reaches 0, means that it takes n steps (where n is the number of bits) at most, and these n bits represent 2^n numbers, so we can reach the logarithmic number of intervals this way too.)\n\n###### Proof that every index is included in at most log N intervals.\n\nThis is essential to proving that updating takes log N operations.\n\nLooking at the pictures this seems true, lets proceed with a proof by contradiction.\n\nAssume intervals ending at a and b with len(a)=len(b) intersect, and without loss of generality let b>a. If these intersect, that means b-len(b)<a. Note that - len(b) is just removing the least sig bit. As len(a) = len(b) both a and b are indentical from the least sig bit to their start. This also means that as b > a, the binary number above these digits for b is greater than a. Removing the least sig bit from b doesn't change b>a as, in binary, the greater number is determined by the leftmost digit, as every digit carries more weight than all the previous digits combined.\n\nBasically: b = [B]10...00 and a=[A]10...00 with [B]>[A]. b-len(b) = [B]00...00 which is still greater than a, so b cannot intersect a.\n\nAs there are a log N number of interval lengths, and no two lengths of the same size intersect, this means that any index is covered by at most log N intervals. Thus, updating an index requires updates to at most log N intervals.\n\nIn the above proofs we have found the method for querying efficently, but we still don't know how to update the nodes.\n\n###### How to update:\n\nFirst note that as our ranges are defined by their rightmost endpoint, the range ending at x is the first range that contains x. So we need an algorithm that can quickly find the next largest range that contains x, and repeat until there are no more such ranges. The next range that contains x must be larger than the current one, so we know the next range's lsb > lsb of x. It turns out that the function next range = x + len(x) works. This is because this function increments the lsb to the next valid one.\n\nSome people might notice a slight problem here, if we have a number like 111000, adding 1000 to that number gets 1000000, skipping a bunch of possible ranges. This is fine as the ranges that are skipped do not cover x. Trying to include the skipped ranges by removing ones just gets numbers that are lower than x, e.g. 110000<111000. Adding numbers to the end before increasing the lsb also doesn't work, as this moves the range ending further than the lsb increase does, e.g. 1110000 ends at 1100000 which is > 111000. The only way to increment to the next largest valid range is with x+=len(x).\n\nHere is the procedure for updating a node x: update BIT[x], then add len(x) to x, repeat until x exceeds the size of the tree.\n\n## Now that we know our algorithms in theory, how do we implement them?\n\nAs every index only has one interval ending in it, it is possible to represent the BIT as an array. BIT[i] = the value of the interval ending at i.\n\nBoth update and query rely on getting len(x), or the lsb of x, easily. Thankfully, the bit operation (x&-x) returns the lsb.\n\nHere's why:\n\nlet a=[A]10...00, then (thanks to 2-complements) -a = [A inverted]01...11 + 1 or [A inverted]10...00. Bitwise ANDing the two gets 000010...000, or the lsb of a.\n\nHere is the actual implementation, using sum as the range query. (Note that we increment x so our tree is rooted at 1, as rooting at 0 causes problems.\n\n int BIT[MAXN];void update(int x,int val) { ++x; while(x<=N) { BIT[x]+=val x+=(x&-x); } }int query(int x) { ++x; int res=0; while(x>0) { res+=BIT[x]; x-=(x&-x); } return res; } \n(I can't for the life of me figure out why the code isn't formatting properly... sorry.)\n\n(A little interesting fact is that x&=(x-1) functions the same as x-=(x&-x))\n\nThats it! Note that you can query for ranges [a,b] by performing query(b)-query(a-1). The same code can also be adapted to other range queries, but there are some pitfalls to look out for. Updating min and max doesn't always work, as you need to know the values in the range which you are updating, unlike in sum where all you need to know is the range's sum. For the same reasons you cannot query arbitary ranges [a,b] with the like you can with sums.\n\nAs a little aside, BITs are like a lightweight form of a segment tree. They take up less space (by a constant factor) and are quicker to code, but they are not as versatile as segment trees.\n\nAgain, this is my first time blog so any feedback is appreciated.", null, "", null, "Comments (14)\n » \"The next range that contains x must be larger than the current one, so we know the next range's lsb > lsb of x.\"Can someone explain why the next range must be larger than current one.\n• » » say x is [somenumber]1000. The any interval of size 1,2,4 doesn't cover x (as if we had such an interval a-len(a) >= x), we have already proven that nothing of the same size can intersect, therefore the only possible ranges are those who are larger. You can generalize this argument for other range sizes.\n• » » » I knew i was missing something obvious. Thanks!\n » A fantastic video tutorial of Fenwick trees and how they can be useful is Algorithms Live Episode 0 by Matt Fontaine: https://www.youtube.com/watch?v=kPaJfAUwViY\n » Another good article for this concept. https://www.hackerearth.com/practice/notes/binary-indexed-tree-or-fenwick-tree/#c217533\n » Nice explanation!!\n » I don't think you need a proof by contradiction to prove that two intervals of the same length will not intersect. This is because if an interval has a length of $2^n$, then the number must be a multiple of $2^n$. And since all distinct multiples of $2^n$ must differ by $2^n$, the segments will never be long enough to touch each other." ]

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[ "# Transient integral boundary layer method to calculate the translesional pressure drop and the fractional flow reserve in myocardial bridges\n\n## Abstract\n\n### Background\n\nThe pressure drop – flow relations in myocardial bridges and the assessment of vascular heart disease via fractional flow reserve (FFR) have motivated many researchers the last decades. The aim of this study is to simulate several clinical conditions present in myocardial bridges to determine the flow reserve and consequently the clinical relevance of the disease. From a fluid mechanical point of view the pathophysiological situation in myocardial bridges involves fluid flow in a time dependent flow geometry, caused by contracting cardiac muscles overlying an intramural segment of the coronary artery. These flows mostly involve flow separation and secondary motions, which are difficult to calculate and analyse.\n\n### Methods\n\nBecause a three dimensional simulation of the haemodynamic conditions in myocardial bridges in a network of coronary arteries is time-consuming, we present a boundary layer model for the calculation of the pressure drop and flow separation. The approach is based on the assumption that the flow can be sufficiently well described by the interaction of an inviscid core and a viscous boundary layer. Under the assumption that the idealised flow through a constriction is given by near-equilibrium velocity profiles of the Falkner-Skan-Cooke (FSC) family, the evolution of the boundary layer is obtained by the simultaneous solution of the Falkner-Skan equation and the transient von-Kármán integral momentum equation.\n\n### Results\n\nThe model was used to investigate the relative importance of several physical parameters present in myocardial bridges. Results have been obtained for steady and unsteady flow through vessels with 0 – 85% diameter stenosis. We compare two clinical relevant cases of a myocardial bridge in the middle segment of the left anterior descending coronary artery (LAD). The pressure derived FFR of fixed and dynamic lesions has shown that the flow is less affected in the dynamic case, because the distal pressure partially recovers during re-opening of the vessel in diastole. We have further calculated the wall shear stress (WSS) distributions in addition to the location and length of the flow reversal zones in dependence on the severity of the disease.\n\n### Conclusion\n\nThe described boundary layer method can be used to simulate frictional forces and wall shear stresses in the entrance region of vessels. Earlier models are supplemented by the viscous effects in a quasi three-dimensional vessel geometry with a prescribed wall motion. The results indicate that the translesional pressure drop and the mean FFR compares favourably to clinical findings in the literature. We have further shown that the mean FFR under the assumption of Hagen-Poiseuille flow is overestimated in developing flow conditions.\n\n## Background\n\nThe incomplete understanding of the pathophysiology and clinical relevance of myocardial bridges has been the subject of debate for the last quarter century. An overview of physiological relevant mechanisms of myocardial bridging, the current diagnostic tools and treatment strategies are found in . Despite extensive studies on this subject there is no consensus on its clinical significance to myocardial ischaemia or angina pectoris.\n\nA variety of models concerned with arterial stenoses and series of stenoses are found in the literature. Theoretical studies have been done to predict the location of maximum wall shear stress and the extent of flow separation located distal to fixed stenoses . There are a few models, which discuss flow in a time dependent two-dimensional flow geometry . These models assume rigid walls and are mainly focused on vortex formation and the wall shear stress distribution. However, discussed the extent of the separation zone in a one-dimensional empirical parameter model using the concept of dividing streamline. They found good agreement with experiments in two-dimensional (partly) flexible indented channels .\n\nThe interesting dynamic phenomena of collapsible tubes are discussed in . When the tube wall is partially collapsed strong oscillations may occur, even under steady flow conditions. The non-linear coupling between the fluid pressure and tube wall deformation can produce conditions in which high-grade stenoses may collapse . We note that in the late systole the compression of the artery in a myocardial bridge may cause conditions where the vessel is entirely closed and where the flow limiting effect during re-opening becomes significant.\n\n### Clinical situation\n\nUnder normal circumstances, coronary arteries have diameters large enough to transport sufficient amounts of oxygen to myocardial cells. Increases in myocardial oxygen demand, e.g. during exercise, are met by increases in coronary artery blood flow because – unlike in many other organs – extraction of oxygen from blood cannot be increased. This is in part mediated by increases in diameters of small intra-myocardial arteries. The large proximal (epicardial) coronary arteries contribute only a small fraction of total vascular resistance and show little variation in diameter during the cardiac cycle in any given metabolic steady state. Under maximum arteriolar vasodilation, the resistance imposed by the myocardial bed is minimal and blood flow is proportional to the driving pressure.\n\nThe most common cause of an impaired ability to match oxygen supply and demand is coronary atherosclerosis, a disease that eventually leads to fixed coronary artery lumen narrowing, impaired coronary blood flow and potentially myocardial infarction. However, some people present with chest pain caused by phasic lumen obstruction due to myocardial bridging, first mentioned by Reyman in 1737 . In this anatomic variant, a coronary artery segment courses underneath myocardial fibres resulting in vessel compression during systole, i.e. the myocardial contraction phase . Myocardial bridges are most commonly found in the mid LAD, 1 mm to 10 mm below the surface of the myocardium with typical length of 10 mm to 30 mm. An angiogram of two myocardial bridges in series shown in Figure 1(a).\n\nAlthough coronary blood flow occurs predominantly during diastole, i.e. the filling phase of the hearts chambers, total blood flow may nonetheless be reduced partly because vascular relaxation may extend significantly into diastole, the myocardial relaxation phase. Within the bridged segments permanent diameter reductions of 22 – 58% were found during diastole, while in systole the diameters were reduced by 70 – 95% . A schematic drawing of the increased flow velocities (cm/s) during systole (31.5 within versus 17.3 proximal and 15.2 distal) is given in Figure 1(b).\n\nFrom a medical point of view coronary angiography is limited in its ability to determine the physiologic significance of coronary stenosis [34, 35]. As a result, intracoronary physiologic measurement of myocardial fractional flow reserve was introduced and has proven to be a reliable method for determining the functional severity of coronary stenosis. Previous studies have shown that the cut-off value of 0.75 reliably detects ischaemia-producing lesions for patients with moderate epicardial coronary stenosis . The assessment of the FFR is independent of changes in systemic blood pressure, heart rate, or myocardial contractility and is highly reproducible . The concept of coronary pressure-derived FFR has been extensively studied [13, 3840], clinically validated and was found to be very useful in identifying patients with multi-vessel disease , who might benefit from catheter-based treatment instead of surgical revascularisation. As in , we have defined the pressure derived FFR as the ratio between the pressures measured distal to and proximal to the myocardial bridge during maximal hyperaemia. The exact locations of pressure measurement are given later in the text.\n\nIn summary myocardial bridges are characterised by a phasic systolic vessel compression with a persistent diastolic diameter reduction, increased blood flow velocities, retrograde flow, and a reduced flow reserve . The underlying mechanisms are fourfold. Firstly the discontinuity causes wave reflections, secondly the dynamic reduction of the vessel diameter produces secondary flow, thirdly there is evidence for flow separation in post stenotic regions [43, 44] and finally at severe deformations (or elevated flow velocities) the artery may temporally collapse .\n\n### Objective\n\nTo ascertain the severity of the disease it is often desirable to have simple models to predict the pressure drop characteristics. A review of the available literature reveals that a few models exist, which are able to predict pressure drop (or friction factor) in non-circular ducts (and references therein). However, the available models are for fully developed flow in non-circular ducts with fixed walls and mostly require tabulated coefficients.\n\nIn this study we intend to investigate physiological relevant cases of developing blood flow through a myocardial bridge located in the middle segment of the LAD. A prior model with similar geometry , but based on the assumption of fully developed Hagen-Poiseuille flow, was used to determine the influence of severity, length and degree of deformation and vascular termination on the flow. The results, however, indicated that the pressure drop was not realistic, which we assume is mainly due to negligence of entrance and separation losses. The system studied herein is the developing flow of an incompressible, viscous fluid through a network of elastic tubes in response to the aortic pressure. The tube characteristics and fluid properties are known, the developing flow conditions, the pressure response and mean FFR are desired quantities. We primarily substantiate the influence of frictional losses and separation losses on the translesional pressure drop and we calculate the mean fractional flow reserve to determine the haemodynamic relevance of the myocardial bridge. Further, we examine the consequence of external deformation on the wall shear stress distribution along the vessel.\n\n## Methods\n\nThe fluid mechanics involved in flow through a myocardial bridge is complex, because of the three dimensionality of the deformations, coupling of the fluid with the arterial wall and flow separation. To understand the complicated behaviour of the tube flow, it is convenient to start with a one-dimensional approximation, which qualitatively predicts the overall aspects in spatially averaged flow variables. It is commonly derived by using equations for mass and momentum conservation and can be found in several places . Because the equations are in common use, we will shortly repeat the main assumptions made. However, we will point out the differences introduced by the wall deformation, the entrance type of flow and flow separation. Firstly, we assume that blood flow in reasonably large vessels can be modelled as incompressible, Newtonian fluid with constant density ρ 0 = 1055 kg/m3 and constant dynamic viscosity μ = 4 mPa s; the kinematic viscosity is defined as ν = μ/ρ 0. The Reynolds number Re d = ud/ν, based upon the vessel diameter, d and the mean flow velocity u, is below 2000 in non-diseased locations of the coronary arteries, so that the flow can be assumed to be laminar .\n\n### The basic geometry\n\nTheories of longitudinal waves in tubes, with or without non-uniformities, non-linearity and frictional dissipation, are based on the idea that variation of the pressure in an axial cross-section is negligible. If the internal and external pressure along the main flow axis, x, of the artery at time t are given by p int (x, t) and p ext (x, t) respectively, the transmural pressure across the wall of a tube is p tm = p int - p ext . Henceforth we assume that the external pressure is constant in space and time and consequently it is the internal pressure whose gradients produce fluid acceleration. Our first geometrical simplification for modelling blood flow in arteries is that the curvature along the axis of the tube is assumed to be small everywhere and that the flow in the cardiovascular system is unidirectional, so that the problem can be defined in one space dimension along the x-axis. According to this we have simplified the anatomy of the myocardial bridge as shown in Figure 2.\n\nThe two arrows in Figure 2 denote the location of either circular (B - B) or oval (C - C) cross-section of the tube. Due to the fact that the wall thickness, h 0, is small compared to the bending radius R d (h 0 /R d 1), we assume that the bending stress inside the wall is negligible. Consequently the cross-section of a circular tube (Figure 3 (left)) under deformation in z-direction, is given by the composition of a rectangle with two semicircular ends as illustrated in Figure 3 (right). This is consistent with the predominately eccentric deformation of bridged segments found in . We note that negligence of bending stress causes the tube to collapse significantly earlier, i.e. the assumption is only satisfied if p tm ≥ 0.\n\nThe deformation distance between the squeezing muscles and the breadth of the flat portion are denoted by D (x) and B (x, t) respectively. The equilibrium geometry of the cylindrical tube in Figure 3 (left) is characterised by the inner radius, R 0, the circumference U 0 = 2 π R 0 and the cross-section A 0 = π R0 2.\n\nHowever the equilibrium cross-sectional area of the deformed tube is A d (x, t) (see Figure 3 (right)). The total cross-sectional area in the yz-plane of the tube is defined by A (x, t) = ∫ A da and the actual circumference is U p (x, t) = 2(π R d + B). Consequently the average flow velocity u (x, t) = 1/A ∫ A ν x da, where ν x is the local value of the flow velocity in axial direction. The volume flux across a given section therefore is q (x, t) = A u.\n\nAs shown in the angiography 1 and Figure 2 the coronary arteries in myocardial bridges are structured by several wall deformations. Their number, degree and extension may independently vary with time, so that the axial curvature of the arterial wall for each of the n = 1...N myocardial bridges in series is characterised by N functions. The deformation is specified by a parameter ζ, defined as ζ = R d /R 0, which is chosen in the stenosis n to vary with time as", null, "where g n (t) are periodic functions describing the temporal contraction of the muscle fibres. ζ systole and ζ diastole are fixed geometrical parameters between 0 and 1, specifying the degree of systolic and diastolic deformation respectively. We note that in the centre of the deformation ζ (x = x s 2) = ζ 0 = R d /R 0, i.e. the degree of deformation increases with decreasing ζ 0 and consequently ζ systole <ζ diastole . To represent the time dependence of the deformation found in intra-vascular ultra-sound (IVUS) measurements , a synthetic deformation waveform g n (t) given by m = 1..3 sine/cosine harmonics is used.", null, "Here Δt n is the time shift for each deformation with respect to the cardiac cycle (see Figure 4) and φ m are the phases in radian, chosen to be φ 1 = 3.5, φ 2 = 1.5, and φ 3 = 3.9. The axial curvature of each deformation is approximated by two hyperbolic tangent functions, so that", null, "where x is the axial coordinate and x tn are the transition locations. Equation (3) smoothes the transition between the segmental domains Ω n by a transition length 1tn. The actual state of deformation can also be expressed by the ratio, ε, based upon the half-axes of the non-circular duct.", null, "### The pressure-area relationship\n\nDue to its complicated structure, it is difficult to provide a synthetic mathematical description for the mechanical behaviour of vessel walls. Here, we focus on the most relevant structural features and the simplest mathematical model for arterial tissues. In the following we derive an algebraic pressure-area relationship for a vessel under external deformation. In this context the distensibility characterises the relative change in cross-sectional area with respect to the pressure increment for a given deformation according to A = A (ζ, p). If we assume that A' is the perturbation about the equilibrium area A d , the total cross-sectional area can be written as A (ζ, p) = A' (ζ, p) + A d (ζ). For a homogeneous, thin-walled (h 0 /R 0 1), linear elastic tube, the stresses in circumferential direction are large compared to stresses in longitudinal direction and the hoop stress per unit length of the tube is", null, "where E is the elastic modulus and σ is the Poisson ratio, which for practically incompressible biological tissue is approximately 0.5. The circumferential strain in the vessel wall is", null, "Equation (5) can be rearranged into the form", null, "whereas Equation (6) leads to an expression for the pressure dependence of the breadth, B, of the flat portion of the tube.", null, "The total cross-sectional area is\n\nA (R d , p) = π R2 d + 2 B (R d , p) R d .     (9)\n\nIn the unperturbed state, the cross-section is\n\nA d (R d ) = R d U 0 - π R2 d.     (10)\n\nWe can finally write the pressure induced perturbation as", null, "It should be noted that under the assumption of linear elastic material with constant elastic modulus, equation (9) and (11) have the property that the area increases linearly with transmural pressure. Real arteries, however, resist over-expansion by having an incremental elastic modulus, E(ε), that increases with increasing strain . It should be further noted that the area perturbation in equation (11) is not only dependent on pressure variation but also on the degree of deformation through R d . By using equation (7) and (9) we can finally write the pressure-area relation", null, "### Elastic properties of the coronary arteries\n\nThe elastic properties for a given section of the circular tube are obtained by using estimates for the volume compliance as suggested in , where the empirical approximation in exponential form is", null, "In these estimates k 1, k 2, and k 3 are constants. With data for the volume compliance from Westerhof , Stergiopulos and Segers we obtain k 1 = 2.0 * 106 Kg s-2m-1, k 2 = -2.253 * 103 m-1, and k 3 = 8.65 * 104 Kg s-2m-1. A functional relationship for the wall thickness subject to the vessel radius can be found in , where\n\nh 0 = a R0 b.     (14)\n\nThe parameters for a = 3.87 and b = 0.63 were obtained by a logarithmic fit to data including vessel radii between 100 μm and 3000 μm. Equations (13) and (14) are used to determine the wall thickness and elastic properties of the vessel, if the radius is known. The assumption of small bending resistance is well satisfied if a R0 (b-1) 1, which for typical vessels under consideration is below 0.21.\n\n### Interaction of viscous boundary layer and inviscid core flow\n\nIn the following we investigate the solutions of the unsteady boundary layer equations by using an approximate integral method proposed by Veldman . For this purpose, the potential flow of the two-dimensional equations governing the unsteady incompressible laminar boundary layer flow in axial symmetry is assumed to be in power-law form. By the introduction of similarity variables and the assumption that the evolution of the velocity profile is weakly dependent on x, the boundary layer equations reduce to the Falkner-Skan equation. Based on this ordinary differential equation, closed form solutions to the von Kármán integral momentum equation are obtained by a curve fit representation. The steady skin friction coefficients and the non-linear momentum correction coefficients corresponding to the velocity distributions are obtained and compared with known results. In particular the results of the steady solution are found to compare favourably with the Blasius solution and values for fully developed flow.\n\n#### Boundary layer equations\n\nThe notion of the boundary-layer approximations was first developed by Ludwig Prandtl in the early 1900's. These well-known approximations are applied widely in fluid mechanics. As the flow rate in the tube increases (i.e. high Reynolds number) the boundary-layer approximations become increasingly valid. The derivation of the axial boundary layer equations was first given by Mangler (1945) and can be found in . In cylindrical coordinates (x, r, φ) with the corresponding velocity components (ν x , ν r , ν φ ) and circumferential velocity ν φ = 0 (no swirl in S), they are", null, "", null, "", null, "with     ν x (x, R d , t) = 0     ν r (x, R d , t) = ν w (x, t),     (18)\n\nwhere R d (x, t) is the body shape along a xr-section of the tube or local surface radius measured from the axis and τ represents the shear stress, which is defined as τ = μ ∂VX/∂r. The derivation assumes that R d is much larger than the boundary layer thickness δ. The three-dimensionality of deformation generally makes it difficult to find a satisfactory solution for every compartment of the neither circular nor flat duct. However, considering severe deformation (e.g. ζ 0 = 0.2) the circumferential length of the flat portion of the vessel exceeds the circumferential length of the circular portion by a factor of four (", null, "), thus we assume plane wedge flow for the calculation of viscous forces. Consequently ν w is taken to be the velocity component normal to the flat portion of the wall, which is ν w = -∂Rd/∂t. At the edge of the boundary layer, the free-stream velocity V (x, t) must be related to the pressure by the potential flow relation", null, "#### Integral momentum equation\n\nThe integral momentum relation of von Kármán (1921) is obtained by multiplying the continuity equation (15) by u - V and subtracting it from the momentum equation (16). Integration over the bending radius and introduction of the integral relations for the displacement thickness", null, "the momentum thickness", null, "and the total displacement thickness δ** = δ* + θ", null, "", null, "with     u(0, t) = V(0, t),     θ(0, t) = δ*(0, t) = 0,     (24)\n\nwhere c f is a non-dimensional friction factor defined as", null, "The only difference to plane flow is the term involving ∂R d /∂ x . If R d → ∞ or ∂R d /∂x → 0, equation (23) reduces to the von Kármán integral momentum equation for plane flow. Compared to the frictional term cf/2 the influence of the term involving ∂R d /∂x on the boundary layer properties is indeed small (below 0.3%), thus disregarding the term for simplicity is appropriate. The boundary conditions in (24) assume a uniform inflow profile.\n\n#### Falkner-Skan equation\n\nSuitable solutions to the boundary layer equations in either plane or axial symmetry are found by the introduction of the stream function. A suitable coordinate transformation turns the equation for the stream function into the Görtler equation (derivation see ). The following approach mainly consists in assuming that the flow is locally self-similar and that it depends weekly on the coordinate x, so that the velocity profiles can be mapped onto each other by suitable scaling factors in y. Falkner and Skan have found a family of similarity solutions, where the free-stream velocity is of the power-law form\n\nV(x) = Cxn,     (26)\n\nwith a constant C and the power-law parameter n. The similarity variable η ~ y/δ(x) is set as", null, "where f(η) is the dimensionless stream function and the prime refers to derivative with respect to η. The coordinate normal to the plate is denoted by y. However there are other general similarity solutions including the temporal dependence of the profile evolution . The above similarity variables turn the boundary layer equations into a non-linear ordinary differential equation of order three, which is known as the Falkner-Skan-Equation", null, "", null, "The parameter β is a measure of the pressure gradient p /∂ x. If β is positive, the pressure gradient is negative or favourable, β = 0 indicates no pressure gradient (i.e. the Blasius solution to flat plate flow) and negative β denotes a positive or unfavourable pressure gradient. We note that by assumption, β should vary slowly with coordinate x. The solutions are found numerically by a shooting method with f''(0) as free parameter by Hartree . To avoid extensive calculations we follow a curve fit representation of three quantities extracted from solutions of the Falkner-Skan equation used in :", null, "", null, "", null, "The relation to flat plate flow is generally given by the shape factor, which is defined as H = δ*/0, whereas H 0 is the equivalent value for plane flow over a flat plate. The curve fits provide a good approximation for values of H between 1 and 20. At the separation point the wall shear stress vanishes, i.e. ∂VX/∂r = 0, which is equivalent to a shape factor H = 4. Relation (32) is not required for the calculations, however it is useful to predict the actual boundary layer thickness δ 99, where the fluid velocity differs by 1% from the free stream value. The relation for the shear stress given in is", null, "consequently the friction factor is", null, "A schematic illustration of the actual flow profile along the tube axis is shown in Figure 5. We have chosen a uniform inflow profile with velocity V(0, t). The boundary layer (dashed line) grows from the leading edge, decreases in the converging part, while it grows in the divergent part of the tube. The upward triangles (▲) denote the point of separation, while downward triangles () indicate the reattachment of the boundary layer. After separation the flow field can be seen as a top hat profile in the centre and a recirculation zone close to the walls. Due to the adjacent converging part the reattachment is forced early, because fluid is accelerated. In contrast the reattachment after the second diverging part takes place further downstream.\n\n#### Averaged flow equations\n\nThe simultaneous viscid-inviscid boundary layer approach assumes an inviscid core flow, which follows equation (19) and a viscous boundary layer, which may be found by the solution of equation (23). The one-dimensional equations commonly used to simulate unsteady, incompressible blood flow in elastic tubes with frictional losses [53, 63] are given in averaged flow variables as", null, "", null, "where F ν is the viscous friction term and χ is the momentum correction coefficient. The viscous friction term is defined as", null, "and the momentum correction coefficient is", null, "We rearrange the equations written in area and flow rate in terms of area and area-averaged axial flow velocity so that", null, "", null, "The derivative of A d with respect to time in equation (39) is a prescribed function depending on R d (x, t). It is responsible for the volume displacement caused by the forced deformation of the tube.\n\n#### Hagen-Poiseuille viscous friction and momentum correction\n\nThe determination of viscous friction factor and momentum correction coefficient requires knowledge about the velocity profile. For pulsatile laminar flow in small axially symmetric vessels a flow profile of the form", null, "is used . Here û is the free stream value of the axial velocity and R is the actual radius of tube, while γ is the profile exponent, which for a Hagen-Poiseuille flow profile is equal to two. Consequently the friction term is given by\n\nF ν = -2 π ν(γ + 2) u = K ν u,     (42)\n\nwhereas the friction coefficient for a parabolic profile is K ν = -8 π ν. The corresponding momentum correction coefficient is given by", null, "which is 4/3 in the parabolic case. Such factors can be used for the purpose of correlating other variables as well as for direct calculation of pressure drop. We note that in the presence of a stenosis the total losses under the assumption of Hagen-Poiseuille flow are underestimated . This comes mainly from underestimating the viscous forces and disregarding the losses caused by flow separation at the diverging end of the stenosis [64, 65].\n\n#### Boundary layer derived viscous friction and momentum correction\n\nDeveloping flow conditions in ducts of multiply connected cross-sections generally make it difficult to use the right friction factor. A variety of cross-sections are discussed in . In those situations the similarity parameters are preferably based on the free stream velocity and the square root of the cross-sectional flow area as characteristic length scale i.e.", null, ". Consequently we define the Reynolds number", null, ", the Womersley number or frequency parameter", null, "and the Strouhal number", null, ". Here ω is the angular frequency defined as ω = 2πf, with f the base frequency of pulse wave oscillation. In other words we have multiplied Re and Sr by a factor of", null, ", while Wo is multiplied by", null, ". In the calculations we have given the Reynolds number inside the stenosis, Re st , based on", null, ".\n\nAs previously mentioned the surface line of the flat portion of the non-circular duct dominates the circular portion at severe deformations, so that the computation of viscous forces is based on plane wedge flow. Consequently the thickness of the boundary layer is estimated in the xz-plane. Further we assume that the boundary layer has constant thickness along the circumference as illustrated in Figure 6. The latter assumption allows a simple derivation of the momentum correction coefficient and the viscous friction term. Integration over the cross-section leads to geometric relations for the areas occupied by the displacement thickness δ* and the total displacement thickness δ** in that cross-section. They are expressed as\n\nA δ*= 2B δ* + π [R2 d - (R d - δ*)2],     (44)\n\nA δ**= 2B δ** + π [R2 d - (R d - δ**)2],     (45)\n\nIt is obvious that A d > A δ**and A d > A δ*have to be satisfied to make sure that the flow is not fully developed. The momentum correction coefficient can be found by satisfying mass conservation for the mean flow and the core flow by\n\nV (A - A δ* ) = A u,     (46)\n\nwhich can be used together with equation (22) and (45) in the definition for the momentum correction (43), so that", null, "The uniform inflow profile is identical to χ = 1, while the developing profile reaches its far downstream value of 1.39 after the entrance length within less than 4.5% from the analytical solution for the parabolic flow profile given in equation (43). We note that in the linearised system the total cross-section A in equation (47) is replaced by A d . According to equation (34) the friction factor built with the pressure dependent surface line, U p (x, t) is", null, "Computations in a uniform tube show good agreement to the friction factor of the parabolic profile given in equation (42). After the entrance length the friction factor computed via the boundary layer theory reached its far downstream value to within 7%. Additionally the Fanning friction factor Reynolds number product in a deformed vessel geometry agrees well with experiments carried out in . In contrast to other proposed models the underlying model does not require knowledge about the fully developed friction factor Reynolds product c f Re fd , or the incremental pressure drop factor K [66, 67]. The above results for momentum correction and viscous friction quantify the entrance conditions typically encountered in studies of the arterial system.\n\n#### Validity\n\nThe stationary boundary layer approximation becomes increasingly valid if the Reynolds number increases and when the ratio of unsteady forces to viscous forces given by the Womersley number is small. In the left coronary artery the values for Womersley and Strouhal number, built with pulsatile frequency were around 4 and below 0.2 respectively. The approximation of the actual flow profiles by near equilibrium flow profiles is justified for stationary flow, however, for time dependent flow the period T of the deformation function in equation (3) has with be large compared to the viscous diffusion time through the boundary layer: t d = δ2/V T. This is well satisfied for the situation under consideration, where t d in the centre of the deformation was between 4.4 * 10-3 s and 0.2 s at values of ζ 0 of 0.25 and 1 respectively.\n\n### Computational domain\n\nBased on 83 angiographies, Dodge et al. [68, 69] presented a normal anatomic distribution of coronary artery segments and proposed a terminology, which we used for our model of the left coronary artery (LCA): the left main coronary artery (LMCA) bifurcates into the left anterior descending artery (LAD) and the left circumflex artery (LCxA). The main branches of the LAD include the 1st, 2ndand 3rddiagonal branch (Dl, D2, D3) and the 1st, 2ndand 3rdseptal branch (S1, S2, S3). The main branches of the LCxA include the 1stand 2ndobtuse marginal branches (OM1, OM2). The exact intrathoratic location and course of each one of the 27 arterial segments and branches of the LCA are illustrated in Figure 7. We note that in the present 1D approximation the arterial tree is composed of tubular entities. The branching angles in figure 7 serve only artistic purposes.\n\n### Boundary conditions\n\nBecause the coronary flow is primarily driven by the aortic pressure, the pulsatile inflow condition to the LMCA was represented by a periodic extension to a synthetic pressure wave in the exponential form", null, "where p s is the static pressure and p 0 is the amplitude of the exponential waveform, while t r is the rising time. We have chosen the parameters according to measurements in . The baseline condition is represented by a stationary pressure p s = 8 kPa and a pressure amplitude of p 0 = 7 kPa, while for the inlet pressure under dobutamine challenge we have chosen p s = 6.6 kPa and p 0 = 5.5 kPa. In both cases the raising time was t r = 0.25 s. The pressure wave at the inlet is shown in Figure 8.\n\nThe branching conditions between the 1D entities are implemented by the requirement of constant pressure at the branching point and mass conservation throughout the bifurcation . To avoid wave reflections at the ends of the tubes the boundary conditions are implemented by a characteristic system of one-way wave equations . There are several ways to account for peripheral reflections at the terminals of a vascular network. In the current simulations we have implemented a three-element windkessel model for the termination of the left coronary arterial tree [70, 71]. The main advantage of this model is to consider the compliant-capacitive effects due to micro-vessels and arterioles. To satisfy the Blasius solution at the leading edge of the tube, we have assumed a uniform flow profile at the entrance (V(0, t) = u(0, t)).\n\n### Numerical implementation\n\nDue to the non-linear terms in equation (23) and (40) the solutions for haemodynamically developing flows are generally more difficult to obtain than fully developed flows or oscillating flows with a frequency dependent Stokes boundary layer. Developing flows require simultaneous solution of the momentum equation (39), the continuity equation (40) and the integral momentum equation (23), together with the boundary conditions given in (49) and (24) respectively. The system of equations cannot be solved analytically, so that the interior domain was solved by a second order predictor-corrector MacCormack finite difference scheme with alternating direction for prediction and correction in each time step . To implement the boundary and interface conditions it is convenient to disregard viscous friction and rewrite equation (39) and (40) in terms of characteristic variables . The momentum correction factor in equation (47) and the viscous friction in equation (48) are given by the solution to the integral momentum equation (23) and the two curve fits to the Falkner-Skan equation in (30) and (31). They are solved by discretisation using the same second order MacCormack scheme and iterative solution of the resulting set of discrete non-linear equations by a combined root bracketing, interval bisection and inverse quadratic interpolation method of van Wijngaarden-Brent-Dekker. To start the computation the Blasius solution at each time step provides values for the boundary layer thickness a few grid points downstream of the entrance. The solution was applied to the steady integral momentum equation as boundary condition for δ*. Downstream marching the solution leads to the values of boundary properties along the tube axis, which through equation (47) and (48) provide the required values of momentum correction and viscous friction to the averaged flow equations respectively.\n\n## Results\n\nThe derived model to simulate viscous friction and momentum correction in the entrance of a tube with varying cross-section was subsequently applied to a test geometry, which consists of a single elastic tube with either temporally fixed or dynamic indentations, and specific clinical situations in the LCA described in . The simulations were carried out in temporal fixed and dynamic stenosis of different degree and extent. The Womersley and Strouhal number were around 4 and below 0.15 respectively, the Reynolds number in the stenosis varied from 850 to 1500. It was found that under application of a uniform flow profile at the entrance the core velocity in the vessel increases gradually downstream, indicating the development of a parabolic flow profile. For a Reynolds number of 2000 the distances over which the core velocity stabilised in the tube were very much shorter than at 800; fully in line with the dependence of the entrance length. However fully developed flow was not present in any of the cases.\n\nThe spatial resolution of the grid was adjusted to represent the curvature of deformation, where 4 gird points per mm was reasonable in any of the cases, the time resolution was chosen accordingly and ranged between 5 μs and 10 μs. These resolutions are sufficient to avoid numerical instabilities at severe deformations. The computational time required, for the simulation of one pulsatile cycle in the above described network of coronary arteries, was about 22 min. (Apple Power-Mac G5, 2 GHz).\n\nIn the following we address the flow limiting effect by the assessment of the pressure derived fractional flow reserve in different stenotic environments (test geometry and LCA). Further, we will discuss the pressure drop Δp, the flow separation, the wall shear stress and the influence of external wall deformation. In some of the graphs we have shown the interval of possible solutions that may occur during the pulsatile cycle, i.e. the upper and lower envelope denoted by Max env (solid red line) and Min env (solid blue line) respectively.\n\n### Modelling: test geometry\n\nWe have chosen the test geometry of the tube according to the size of the left main coronary artery with an internal diameter of 5 mm. To observe separation and reattachment we have adapted the length to the maximum extent of the recirculation zone, which was about 300 mm at 80% diameter deformation. At 20 mm downstream of the entrance, the diameter was abruptly decreased (t l = 4 mm) by 0 – 85%, with an extent of 30 mm (single stenosis), another constriction with the same length follows 30 mm further downstream (double stenosis). In addition the extent of the single stenosis (short) was varied by a factor of two (medium) and three (long), in the double stenosis we varied the separation distance between one and three dimensions of the stenosis. The wave velocity in those situations may take values as low as c = 5 m/s in uniform vessels, rising to values around c = 30 m/s in constricted vessels. Physiological peak flow velocities, V, are much smaller, generally around 0.5 – 1 m/s, but they can reach 2 – 3 m/s in parts of severe deformation.\n\nThe region around the diameter transition in the test geometries is shown in detail in Figure 9. The area perturbation A', shown on the top of each column, indicates that the arterial deformation caused by pulsatile pressure is much smaller in regions where the tube is squeezed (about A'/A = 1 – 2%), while in circular segments A'/A = 8 – 9%. Due to incompressibility and mass conservation a sudden jump to a high velocity is seen at the narrowing. It appears that the stenosis influences the mean velocity only over a short distance upstream, while in the case of separation the core velocity slowly decays until reattachment, then it starts growing again until the flow is fully developed. The boundary layer separation in the post-stenotic region indicates an emerging top-hat velocity profile from the stenosis, with sideway counter-current flows. The decay in core flow velocity indicates shear widening of the top-hat until reattachment. Further downstream the core velocity is almost constant. It was found that the distance over which the outlet effect occurs is smaller for stenosis with small deformation and length and small Reynolds number. The core velocity influences the frictional losses and the extent of the separation zone in series stenoses, so that series stenoses cannot be represented by two similar building blocks. This becomes more evident through the fact that the entrance flow profile in the second stenosis changes, if the distance of the two stenoses is varied (Figure 10(a–c)). The downstream deformation is generally dominant in wall shear stress and frictional forces as well as in the extent of the post-stenotic separation zone. Further, it is likely that a downstream stenosis with equal deformation collapses significantly earlier, because the core flow velocity is increased, so that the transition from subcitical flow (V <c) to supercritical flow (V > c) happens earlier.\n\n#### Pressure drop and flow limitation\n\nGeometric influences on the pressure loss across series of stenoses have been studied in . It was found that the pressure drop across severe stenoses is little affected by the eccentricity of the stenosis, dominantly affected by the severity and length of the stenosis, and affected by the Reynolds number only at low Reynolds number. The eccentricity of mild stenoses increases the likelihood of collapse of stenotic arteries, because the buckling pressure is reduced .\n\nAs in [1, 8] we found that the pressure proximal to the deformation is increased if the degree of deformation or the length of deformation is increased. Compared to the mean reference pressure (no deformation) of 9.29 kPa, the mean pressure at the inlet for a deformation of ζ 0 = 0.2 was 10.28 kPa, 11.53 kPa and 12.41 kPa in the cases (s), (m) and (l) respectively. Under these conditions the mean flow in the reference artery was 10.36 cm3 /s. For fixed deformation of ζ 0 = 0.2 we found 8.43 cm3/s, 6.38 cm3/s, 4.94 cm3/s and 6.66 cm3/s in the short, medium, long and double stenosis respectively. The pressure drop for the medium sized stenosis was at 7.52 kPa just below that of the double environment with 7.78 kPa, the pressure drop across the short stenosis was 4.15 kPa and across the long stenosis 9.82 kPa. In double stenoses the pressure drop across the proximal stenosis was 3.28 kPa, which is markedly less than 4.4 kPa at the distal stenosis. Complementary the values for the dynamic case are generally less pronounced, they can be found in Table 1.\n\nHowever, the translesional pressure drop in a series of two stenoses with time dependent deformation in the entrance region of a tube shows further remarkable effects. The simulations indicate that in general the pressure drop cannot be obtained by a summation of pressure drops for single stenosis, since the proximal and distal stenosis influence each other unless the spacing between them exceeds some critical distance, which depends on the Reynolds number and deformation. Therefore several consecutive stenoses along the same epicardial artery require separate determination of stenosis severity. We found that when the two stenoses are close together, the pressure drop is approximately equal to that of a stenosis with twice the length of a single stenosis (see Table 1 and Figure 9 (m) and (d)). This can be explained by the fact that the friction coefficient in the recirculation zone between the two stenoses is small and consequently the pressure loss over that region is small. However, when the stenoses are separated by more than the length of the stenosis the flow is hardly affected by the second stenosis, because the core flow velocity and likewise the frictional force are increased (see Figure 10(a–c)), even though the entrance flow conditions are preserved. Coexistent is the increased reduction in shape factor and momentum correction coefficient, suggesting the non-linear influence in that region. The pressure drop over the downstream deformation is generally pronounced, however the distance between the deformations further increases the pressure drop and consequently the mean FFR is reduced. The values for the cases (a) to (c) are Δp a = 3.45 kPa, Δp b = 3.64 kPa and Δp c = 3.89 kPa for the pressure drop over the downstream deformation and FFR a = 71%, FFR b = 70% and FFR c = 68% for the mean fractional flow reserve across the bridge. We note that these findings are based on the entrance type of flow, where the core flow velocity gradually changes and may not appear in fully developed flow, where the core flow velocity is a constant and identical to the maximum of Hagen-Poiseuille flow velocity.\n\n#### Separation and reattachment\n\nThe core velocity in a uniform tube generally increases downstream of the entrance, however in presence of a stenosis the boundary layer thickness decreases at the inlet and rapidly increases in the diverging part of the constriction (see Figure 9). Separation occurs under the development of a top hat profile with sideway counter-current flow. At the separation point (▲) the boundary layer thickness increases and a sudden jump in χ is evident, because the integral of the actual velocity", null, "over the area of the recirculation is close to zero and in contrast to the mean flow velocity the core velocity is increased. Therefore the non-linear term is pronounced in the separation region, while it is close to unity in converging regions. The momentum correction becomes markedly smaller than before the upstream stenosis. This indicates that the entrance profile into the second stenosis is almost flat, but has increased core velocity, while the counter current flow at the walls have disappeared. The reattachment of the boundary layer () further downstream is caused by shear layer friction between the recirculation zone and the top-hat profile, which also causes the pressure to recover. Due to the increased momentum correction in that region the pressure in the non-linear case recovers more rapidly than in linearised computations, which causes earlier reattachment and consequently slightly smaller recirculation zones. The extent of the recirculation zone is primarily dependent on vessel deformation and Reynolds number, however, we found that the extent also correlates with the length of the constriction. Compared to the short deformation in Figure 9 (s) the tail of the shape factor curve drops below the critical value of 4 (condition for separation or reattachment) by a factor of about two and three later for the medium (m) and long (l) constriction respectively. In other words vessels with the same degree of stenosis, but with the stenosis having different curvatures and lengths, have recirculation regions that differ markedly in their extent. At deformations of 85% the recirculation zones had an extent of about 20 tube diameters in length. However the extent of the separation region was found to be strongly dependent on the degree of deformation and the Reynolds number. The separation point moves upstream, while the reattachment point moves downstream if the Reynolds number or deformation increases. A particularity of series stenoses is that the extent of the recirculation zone in the interconnecting segment is reduced. This is due to early reattachment caused by fluid acceleration in the converging part of the second stenosis. But nonetheless the core flow velocity is generally smaller compared to the downstream separation region.\n\n#### Wall shear stress and friction coefficient\n\nFor steady flows the location of maximum wall shear is always upstream the neck of the stenosis (see Figure 9), and moves upstream as the Reynolds number increases. In series stenoses the WSS is significantly increased in the distal stenosis, while the friction coefficient is smaller there (see column (d)). Generally they have their maximum at the entrance of the stenosis and reduce towards the end of the stenosed section. Eventually they become negative after separation of the boundary layer. The increased boundary layer thickness in the downstream stenosis suggests lower retarding forces, however, the core flow velocity is increased there so that pressure losses are dominant there. Consequently the second stenosis can be seen as the more vulnerable, in wall shear stress and flow limitation. Likewise the mean flow velocity in a pressure driven vessel is dependent on the total after-load, the maximum values of wall shear stress are dominant in short constrictions (see column (a)), because the after-load is smaller and fluid velocity is increased compared to long constrictions (column (l)). Although the wall shear is increased in short constrictions, we observe that the peak of the viscous friction increases if the length of the constriction is increased. In flow driven vessels however the peak values are independent of the extent, because the flow velocity is equal in all cases (not shown here).\n\nWall shear stress oscillations have been observed for various downstream locations and severity of deformation. The amplitude of oscillation depends strongly on the axial position and the actual state of deformation. The wall shear stress is large in the entrance region of the deformation, fades towards the end and is negative in the separation region, so that the development of atherosclerosis is more likely in segments proximal to the deformation. Compared to wall shear stresses in non-diseased vessels (5 – 10 N/m2) vulnerable regions are endothelial cells in the throat of a strong deformation. They may experience wall shear stresses in excess of 60 N/m2. In series stenoses the stresses are largest in the downstream stenosis because the core flow velocity is increased there. Furthermore the wall shear stresses are no longer likely to be distributed evenly around the circumference of the vessel and may be particularly focused on the most vulnerable shoulder regions, marking the transition from normal to diseased artery wall. Further improvements for the prediction of the wall shear stress may be obtained by the introduction of a shear dependent model to predict the local blood viscosity.\n\nDespite the assumption of strong coupling between the boundary layers and the core flow and the assumption of quasi-stationary evolution of the boundary layers, the time dependent motion of the wall under external deformation reproduces some remarkable characteristics of myocardial bridges. In Figure 11 we have shown the effect of temporal deformation onto the pressure and flow in the segments Ω1–5 of a series of two myocardial bridges (see Figure 2). The time dependence of the two separation zones, one between the two deformations and the other distal to the second deformation shows that separation occurs for deformations greater than about 40%. The separation cycle is present in the time interval of 0.1 s to 0.4 s. The maximum deformation during the cycle was 75% of R 0, which was reached at 0.3 s. It is seen that during deformation the separation point moves somewhat upstream, while the reattachment point of the boundary layer moves farther downstream. The upstream separation zone (turquoise cycle) is spread over a region of 49.94 mm to 80.83 mm, while the downstream separation zone (purple cycle) is from 109.98 mm to 193.32 mm. The extensions of the two zones differ, because the upstream reattachment point is forced early due to accelerating fluid at the inlet of the downstream deformation and because the core velocity is generally larger further downstream. In each case however, a top-hat velocity profile develops in systole, producing large recirculation zones distal to the stenoses, which are delayed into diastole. The back-flow is developing earlier for severe deformation and is strongly dependent on deformation phase. For phase delayed deformation, say, by half the cycle time (here 500 ms), the flow patterns are noticeably different (not shown here). The top-hat profile is now present in diastole, the peak velocities and the viscous friction are less pronounced and the reverse flow region is smaller and mainly present in the diastole. The wall shear stress is dramatically reduced compared to the zero phase case, because deformation maximum falls together with the haemodynamic conditions present in diastole. Furthermore the mean FFR was increased to 0.65 compared to 0.6 of the zero phase case. This can be explained by the circumstance that no appreciable pressure drop was present at the downstream stenosis.\n\n### Modelling: physiological basis\n\nThe simulation of clinical relevant cases requires a specific set of parameters, which however is not available in the literature. Due to this difficulty we compare the FFR range obtained with parameters for the length and degree of deformation given in . The length of the myocardial bridge for the baseline and dobutamine case were 12 mm and 24 mm respectively. The values for the deformation were ζ baseline = 0.54 and ζ dobutamine = 0.32. To assess the dynamics, we have applied more physiological waveforms (aortic pressure conditions) to the inlet of the left main coronary artery (LMCA) (see Figure 8). The peak Reynolds and Strouhal numbers in the myocardial bridge were 815 and 1069, and 0.021 and 0.016 for the baseline and dobutamine case respectively. The Womersley number was 4.11.\n\n#### Mean pressure drop\n\nThe mean pressure drop was calculated by subtracting the average of a distal pressure wave, which was taken approximately 3 cm distal to the myocardial bridge, from the average inlet pressure at the LMCA. The proximal and distal pressure waves for the baseline and while dobutamine challenge are shown in Figure 12. It is seen that the baseline pressure is less affected, while the pressure under dobutamine challenge shows a pressure notch, which may appear if the deformation is dominant during systole (see earlier discussion on that in ). At baseline the mean pressure at the inlet was p p = 11.92 kPa and the mean distal pressure was p d = 10.67 kPa, so that the pressure drop was Δp = 1.25 kPa, which compares well with the value of measurements mentioned above, where Δp = 1.19 kPa. Under dobutamine challenge the corresponding values are p p = 9.64 kPa, p d = 8.1 kPa and Δp = 1.54 kPa, which however are close to the measured values, where Δp = 1.85 kPa.\n\nWe have further compared the translesional pressure drop across a series of two myocardial bridges resulting from Hagen-Poiseuille and boundary layer computations in three animations, one as reference, without a stenosis (Additional file 1), one assuming a Hagen-Poiseuille flow with a series stenosis, each of length 8 mm and deformation ζ 0 = 0.25 (Additional file 2) and the same series stenosis with the boundary layer method described here (Additional file 3). It is clearly seen that the systolic pressure in Additional file 1 is uniformly distributed and fades towards the terminals of the network. In Additional file 2 we notice that during systole the pressure drops in both of the stenoses, so that the LAD branch is less distributed. However the boundary layer computations in Additional file 3 show that the pressure drop by the assumption of fully developed flow was underestimated in the case of developing flow conditions.\n\n### 12938_2006_185_MOESM1_ESM.mov\n\nAdditional file 1: Animation of the left descending coronary artery assuming Hagen-Poiseuille flow. The animation shows the pressure in the main segments of the left coronary arterial tree during one pulsatile cycle. The unit of the colour bar is kPa, the playback speed is slower by a factor of three than realtime. (MOV 6 MB)\n\n### 12938_2006_185_MOESM2_ESM.mov\n\nAdditional file 2: Animation of a double myocardial bridge in the left descending coronary artery assuming Hagen-Poiseuille flow. As in the first animation the pressure in the main segments of the left coronary arterial tree are shown over one heart cycle. A series of two myocardial bridges with length 8 mm and deformation ζ 0 = 0.25 are located in the mid LAD. The pressure drop is underestimated by the assumption of Hagen-Poiseuille flow, consequently perfusion to the myocardium remains nearly unaffected. The unit of the colour bar is kPa, the playback speed is slower by a factor of three than realtime. (MOV 6 MB)\n\n### 12938_2006_185_MOESM3_ESM.mov\n\nAdditional file 3: Animation of a double myocardial bridge in the left descending coronary artery using the proposed integral boundary layer method. The animation shows the pressure in the main segments of the left coronary arterial tree over the pulsatile cycle. A series of two myocardial bridges with length 8 mm and deformation ζ 0 = 0.25 are located in the mid LAD. It is seen that the pressure drop at the second deformation is dominant and that distal segments are less perfused. In contrast the pressure proximal to the deformation and in the left circumflex is increased [1, 8]. The unit of the colour bar is kPa, the playback speed is slower by a factor of three than realtime. (MOV 6 MB)\n\n#### Fractional flow reserve\n\nThe flow limitation caused by epicardial stenoses is generally expressed by the flow based FFR, which is the ratio of hyperaemic myocardial blood flow in the presence of a stenosis to hyperaemic flow in the absence of a stenosis, FFR q = q s /q n , i.e. the flow based FFR is the fraction of hyperaemic flow that is preserved despite the presence of a stenosis in the epicardial coronary artery. However this definition is purely theoretic, because the flow without the stenosis is not known, so that for clinical purposes the ratio of hyperaemic flows with or without a single stenosis is derived from the mean distal coronary pressure p d to mean proximal pressure p p recorded simultaneously under conditions of maximum hyperaemia.\n\nNeglecting correction terms the mean pressure-derived fractional flow reserve is FFR p = p d /p p . In the case of two consecutive stenoses however, the fluid dynamic interaction between the stenoses alters their relative severity and complicates determination of the FFR for each stenosis separately from a simple pressure ratio as in a single stenosis. Consequently the FFR determined for single stenosis is unreliable in predicting to what extent a proximal lesion will influence myocardial flow after complete relief of the distal stenosis, and vice versa.\n\nTaking the pressure values resulting from boundary layer computations of the previous section, we obtain values for the mean pressure derived FFR of 0.90 and 0.84 for baseline conditions and under dobutamine challenge respectively. These values agree with the measurements in , where the values were 0.90 and 0.84 respectively.\n\nIn Figure 13 we have shown the coronary pressure-derived fractional flow reserve for the baseline case as a measure of coronary stenosis severity (left) and in dependence on deformation length (right). In both plots we have used baseline inflow conditions with either fixed (solid lines) and dynamic walls (dashed lines). Initially we found that in developing flow conditions the FFR is overestimated by the assumption of Hagen-Poiseuille flow (HP), and, as expected, that the mean FFR in dynamic lesions is much larger than in fixed stenotic environment, because the distal pressure recovers during relaxation phase. The graph at the right shows that the mean FFR depends essentially linearly on deformation length for different severities, ζ 0 = 0.7, 0.5, 0.3, suggesting that the losses are mainly viscous and that the non-linear term plays a minor role for observed deformations.\n\nThe values indicate that the FFR depends on the degree and length of the stenosis. As mentioned earlier the losses in fixed environments are more pronounced. Further, we point out that series stenoses separated by more than the length of the stenosis drop below the cut-off value of 0.75 markedly earlier than single stenosis with the same degree and extent.\n\n#### Pressure-flow relation\n\nThe pressure-flow relations for different stenotic environments in the coronary arteries are shown in Figure 14. We have applied a stationary flow rate q LMCA at the entrance of the LMCA in the range between 4.5 cm3 /s and 9 cm3 /s. The resultant mean flow in the myocardial bridge, q MB was between 1.57 cm3 /s and 3.13 cm3 /s. The pressure-flow relations in this range are essentially linear and extrapolate to the origin at q LMCA = 0 and Δp = 0. This is consistent with the assumption of long waves in short tubes, where the pressure-drop is linearily dependent on the flow. The single deformation with length 12 mm is denoted by (S 12) and with length 24 mm by (S 24). We further show a double deformation with length 12 mm (D 12). In general it is seen that the pressure drop increases with severity and length of the deformation, however the difference between S 24 and D 12, which have essentially the same deformation length, result from inlet and outlet effects of the double environment.\n\n#### Influence of wall velocity\n\nIn contrast to fixed stenoses, the velocity of the wall ν w ≠ 0 in a dynamic environment, i.e. positive during compression and negative while the vessel is relaxing. At the bottom of Figure 15 we have shown the thickness of the boundary layer during inward (left) and outward motion (right) of the wall compared to fixed deformations. The situation depicts two different states where the influence is close to its maximum. Compared to fixed deformations (ν w = 0) the boundary layer thickness in systole is increased in the entrance region of the deformation, while it is decreased in the outlet region, and the situation is vice versa in diastole. In fact this is due to an additional pressure gradient, which causes fluid acceleration or deceleration depending on axial position in the constriction and whether the wall moves inwards or outwards. For example during inwards motion of the wall the fluid is decelerated at the entrance and accelerated at the outlet of the deformation, while the situation is opposite if the vessel wall moves outwards. Due to the symmetry of the indentation there is no additional acceleration or deceleration in the centre of the deformation. The influence of the wall velocity onto the boundary layer properties is more pronounced if VW/V is reasonably large, i.e. if the deformation that the axial flow experiences during passage of the constriction is comparable to the radius of the tube. The influence of wall velocity on viscous friction and wall shear stress is small and thus the difference in pressure loss over the deformation with ν w ≠ 0 compared ν w = 0 is small (δp ≈ 1% of Δp).\n\n## Discussion\n\nThe results have demonstrated that the formation and development of flow separation and reattachment in the post-stenotic region of a time dependent constriction are very complicated, especially secondary fluid motion in the systolic deceleration phase can cause situations of reverse flow. The post-stenotic flow is influenced by a number of factors, including the degree of stenosis, the flow and deformation waveform, and the geometry of the constriction. The above calculations suggest is that percent diameter stenosis alone does not adequately characterise the flow through myocardial bridges, and that geometric and physiological features such as the curvature, extent, and asymmetry of the stenosis, and the shape of the pulsatile waveform have substantial effects on the haemodynamic conditions.\n\nHowever, we found that the total perfusion to the myocardium is strongly dependent on the severity and length of the muscle bridge. The mean FFR in fixed environment is generally smaller than in the dynamic case because the losses are not persistent during periods of small deformation, so that the pressure distal the bridge recovers during this time span. Consequently the pressure drop and flow reduction across fixed stenoses are more pronounced than in dynamic environments.\n\nAs previously noted found that the pressure proximal to the myocardial bridge was higher than the aortic pressure, and concluded that the disturbance of blood flow and high wall stress proximal to the myocardial bridge was the main contributor to the development of atherosclerosis in the proximal segment. The observed wall shear stress distributions indicate that the proximal segment is more susceptible to the development of atherosclerosis, firstly because the pressure is increased there and secondly for reasons that the wall shear stress and their oscillations are maximum in the entrance region of the deformation. In contrast bridged segments are relatively spared because the wall shear stress fades towards the end of the deformation. In a series of myocardial bridges it is likely that the intima between the deformations and distal to the myocardial bridge are protected from atherosclerosis because the wall shear stress is very low and negative in separated flow regions.\n\n## Conclusion\n\nWe have presented a method for simulation of unsteady blood flow in a time dependent vessel geometry using an integral boundary layer method. The strong interaction of the viscous boundary layer and the inviscid core flow proposed by Veldman models the pressure and the extent of the separation region by assuming Falkner-Skan flow profiles. The equations were modified to the flow situation under consideration. Numerical simulations were performed for idealised stenosis geometries with a time dependent, smooth wall contour, but with a physiologically realistic coronary artery flow waveform. The predicted values of fractional flow reserve in dynamic lesions agree well with the clinical findings in , however, further quantification in more defined geometries is required.\n\nRegarding the wall shear stresses and the development of atherosclerosis the findings are consistent with , where the intima beneath the bridge is protected from atherosclerosis, and the proximal segment is more susceptible to the development of atherosclerotic lesions.\n\nBesides the advantage of computational time taken for the simulation, the choice of parameters, such as location, length and severity of the lesion are easily determined by coronary angiography. Due to the assumptions made in the boundary layer model, the approximation fails for the prediction of reverse flow and flow where the boundary layers merge, i.e. fully developed flow. Under these aspects severe deformations cannot be calculated, because the boundary layers merge in the deformation. Further, the length of the computational domain is restricted by the entrance length, which depends on Reynolds number. And finally the pulsatile frequency and the frequency of wall motion has to be sufficiently low (a few Hz), so that the approximation of quasi-stationary evolution of the boundary layers is satisfied.\n\nWe believe that the parameters and equations in this article are detailed enough to describe the physiological consequences also in a clinical setting, however, this remains to be confirmed by in vivo studies. The functional consequence, especially for severe systolic compression, is consistent with clinical findings published in the literature [13, 6, 8, 38, 51], where myocardial bridging is found to be responsible for myocardial ischaemia. The comparison of our findings with the published data from patient studies in [38, 51] supports a potential clinical relevance of our simulation.\n\n## References\n\n1. 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Publisher Full Text 10.1175/1520-0493(1993)121<0565:COLSAM>2.0.CO;2\n\n## Acknowledgements\n\nA part of this work was supported by a Georg-Christoph-Lichtenberg Scholarship.\n\n## Author information\n\nAuthors\n\n### Corresponding authors\n\nCorrespondence to Stefan Bernhard or Stefan Möhlenkamp.\n\n### Authors' contributions\n\nSB developed the boundary layer model, carried out the simulations and drafted the manuscript. SM drafted the section on the clinical situation. AT participated in model development in its design and coordination. All authors read and approved the final manuscript.\n\n## Authors’ original submitted files for images\n\nBelow are the links to the authors’ original submitted files for images.\n\n## Rights and permissions\n\nReprints and Permissions\n\nBernhard, S., Möhlenkamp, S. & Tilgner, A. Transient integral boundary layer method to calculate the translesional pressure drop and the fractional flow reserve in myocardial bridges. BioMed Eng OnLine 5, 42 (2006). https://doi.org/10.1186/1475-925X-5-42", null, "" ]

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[ "عنوان Improved cuckoo optimization algorithm for solving systems of nonlinear equations نوع پژوهش مقاله چاپ شده کلیدواژه‌ها Cuckoo optimization algorithm, Evolutionary algorithms, Nonlinear equations, Optimization چکیده Systems of nonlinear equations come into different range of sciences such as chemistry, economics, medicine, robotics, engineering, and mechanics. There are different methods for solving systems of nonlinear equations such as Newton type methods, imperialist competitive algorithm, particle swarm algorithm, conjugate direction method that each has their own advantages and weaknesses such as low convergence speed and poor quality of solutions. This paper improves cuckoo optimization algorithm for solving systems of nonlinear equations by changing the policy of egg laying radius, and some well-known problems are presented to demonstrate the efficiency and better performance of this new robust optimization algorithm. From obtained results, our approach found more accurate solutions with the lowest number of function evaluations. پژوهشگران مهدی عبدالهی (نفر اول)، عسگر علی بویر (نفر دوم)، داود عبدالهی (نفر سوم)" ]

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[ "## 56203\n\n56,203 (fifty-six thousand two hundred three) is an odd five-digits composite number following 56202 and preceding 56204. In scientific notation, it is written as 5.6203 × 104. The sum of its digits is 16. It has a total of 4 prime factors and 12 positive divisors. There are 45,360 positive integers (up to 56203) that are relatively prime to 56203.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 16\n• Digital Root 7\n\n## Name\n\nShort name 56 thousand 203 fifty-six thousand two hundred three\n\n## Notation\n\nScientific notation 5.6203 × 104 56.203 × 103\n\n## Prime Factorization of 56203\n\nPrime Factorization 72 × 31 × 37\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 8029 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 56,203 is 72 × 31 × 37. Since it has a total of 4 prime factors, 56,203 is a composite number.\n\n## Divisors of 56203\n\n1, 7, 31, 37, 49, 217, 259, 1147, 1519, 1813, 8029, 56203\n\n12 divisors\n\n Even divisors 0 12 6 6\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 69312 Sum of all the positive divisors of n s(n) 13109 Sum of the proper positive divisors of n A(n) 5776 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 237.072 Returns the nth root of the product of n divisors H(n) 9.73044 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 56,203 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 56,203) is 69,312, the average is 5,776.\n\n## Other Arithmetic Functions (n = 56203)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 45360 Total number of positive integers not greater than n that are coprime to n λ(n) 1260 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5698 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 45,360 positive integers (less than 56,203) that are coprime with 56,203. And there are approximately 5,698 prime numbers less than or equal to 56,203.\n\n## Divisibility of 56203\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 0 3 7\n\nThe number 56,203 is divisible by 7.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (56203)\n\nBase System Value\n2 Binary 1101101110001011\n3 Ternary 2212002121\n4 Quaternary 31232023\n5 Quinary 3244303\n6 Senary 1112111\n8 Octal 155613\n10 Decimal 56203\n12 Duodecimal 28637\n20 Vigesimal 70a3\n36 Base36 17d7\n\n## Basic calculations (n = 56203)\n\n### Multiplication\n\nn×i\n n×2 112406 168609 224812 281015\n\n### Division\n\nni\n n⁄2 28101.5 18734.3 14050.8 11240.6\n\n### Exponentiation\n\nni\n n2 3158777209 177532755477427 9977873456097829681 560786421853066321561243\n\n### Nth Root\n\ni√n\n 2√n 237.072 38.3048 15.3971 8.91152\n\n## 56203 as geometric shapes\n\n### Circle\n\n Diameter 112406 353134 9.92359e+09\n\n### Sphere\n\n Volume 7.43647e+14 3.96944e+10 353134\n\n### Square\n\nLength = n\n Perimeter 224812 3.15878e+09 79483\n\n### Cube\n\nLength = n\n Surface area 1.89527e+10 1.77533e+14 97346.5\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 168609 1.36779e+09 48673.2\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.47116e+09 2.09224e+13 45889.6\n\n## Cryptographic Hash Functions\n\nmd5 e5b5b6a7b2a65b8f8f1d7c4657b8fc75 a0d2a67932471332486f452724fa5b6020a1fc3b 2ebaaf1f651e4db199f4958f017a6ac809e878ce4b696f495143c25631a8c6d2 bf9a113f0483bae40c4290f21cfcef072e071a94b13f4f4059e5f7a6dda6be12b90509894718b77b364d5a0e9c15c13fa0e42de9197d5d9bdb1a396aae4da8c7 1b740f23fc58faaad25d017a78fe2bbe27e84b69" ]

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[ "Solutions by everydaycalculation.com\n\n1st number: 2 2/14, 2nd number: 20/70\n\n30/14 + 20/70 is 17/7.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 14 and 70 is 70\n2. For the 1st fraction, since 14 × 5 = 70,\n30/14 = 30 × 5/14 × 5 = 150/70\n3. Likewise, for the 2nd fraction, since 70 × 1 = 70,\n20/70 = 20 × 1/70 × 1 = 20/70\n150/70 + 20/70 = 150 + 20/70 = 170/70\n5. After reducing the fraction, the answer is 17/7\n6. In mixed form: 23/7\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "This Basic Math Puzzle is test of your Arithmetic Skills. This for for school going Kids who has learnt the basic of Arithmetic and Algebra. Your challenge is to solve the given Arithmetic Equation as quickly as possible. Lets see if you pass this Basic Arithmetic Test by solving this Math Puzzle correctly in your first attempt?", null, "Can you solve this Basic Arithmetic Math Puzzle?\n\nAnswer of this \"Basic Arithmetic Test\", can be viewed by clicking on answer button. Please do give your best try before looking at the answer." ]

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