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https://gisnote.com/2018/08/29/lab-3coordinates/ | [
"# Lab 3. Measurements in different coordinate spaces\n\nTo start with, we will look at this data layer. I will release the lab data once I’ve finished talking.\n\nI’ve provided a point shapefile called CityPoints_LatLong for you to use in this section. I want you to calculate their X and Y coordinates using UTM17, UTM16, State Plane Georgia East, State Plane Georgia West, and Long/Lat.\n\nYou will need to do the following:\n\n• Add a new field for the UTM17 X coordinate (make it a float type)\n• Add a new field for the UTM17 Y coordinate (make it a float type)\n• Add new fields for UTM16 X, UTM16Y, Ga SP east X, Ga SP east Y, Ga SP west X, Ga SP west Y, Longitude, and Latitude\n• Change the data frame coordinate system to UTM17 – use NAD 1983\n• Use CALCULATE GEOMETRY to calculate the UTM17 X coordinate\n• Use the ‘Coordinate system of the data frame’ in your calculation\n• Make sure you use meters as your unit of measure\n• Repeat for the UTM17 Y coordinate\n• Recall UTM uses meters as its measurement unit\n• Change the data frame coordinate system to UTM16 – use NAD 1983\n• Use CALCULATE GEOMETRY to calculate the UTM16 X coordinate\n• Use the ‘Coordinate system of the data frame’ in your calculation\n• Make sure you use meters as your unit of measure\n• Repeat for the UTM16 Y coordinate\n• Recall UTM uses meters as its measurement units\n• Change the data frame coordinate system to Georgia State Plane east – use NAD 1927 (US FEET)\n• Use CALCULATE GEOMETRY to populate the X and Y coordinates\n• Use the ‘Coordinate system of the data frame’ in your calculation\n• Make sure you use feet as your unit of measure\n• Recall State Plane uses feet as its measurement unit\n• Change the data frame coordinate system to Georgia State Plane west – use NAD 1927 (US FEET)\n• Use CALCULATE GEOMETRY to populate the X and Y coordinates\n• Use the ‘Coordinate system of the data frame’ in your calculation\n• Make sure you use feet as your unit of measure\n• Recall State Plane uses feet as its measurement unit\n• Change the data frame coordinate system to Geographic > World > WGS84\n• Use CALCULATE GEOMETRY to populate the Long/Lat (X and Y) coordinates\n• Use the ‘Coordinate system of the data frame’ in your calculation\n• Make sure you use decimal degrees as the unit of measure\n• Recall Long/Lat uses degrees as its unit of measure\n\nYour next task is to manually calculate the distance from each city to Athens using the UTM17 measurements, UTM16 measurements, SP Ga east measurements, SP Ga west measurements, and Long/Lat measurements – one distance for each coordinate system.\n\nYou need to export your attribute table as a text file, load it into Excel and perform your calculations. This is how…\n\n• Open your point layer’s attribute table\n• Click on the table options button (in the upper left of the table) > Export…\n• Change the file type to Text File\n• Open your exported text file in Excel and make your calculations\n• Use Pythagorean’s theorem to calculate the distances in UTM17, UTM16, State Plane GA East, State Plane Georgia West, Long/Lat\n\nPLEASE UNDERSTAND THAT, IN THIS LAB, THE CONCEPT OF DISTORTION DUE TO YOUR SELECTION OF COORDINATE SYSTEM WAS DEMONSTRATED USING POINTS AND DISTANCE. YOU WILL EXPERIENCE THE SAME DISTORTION IN AREA MEASUREMENTS AND DIRECTION MEASUREMENTS AND SCALE MEASUREMENTS IF YOUR SELECTION OF COORDINATE SYSTEM IS UNWISE.\n\nLAB 3 QUESTIONS (1 – 7, reference lecture materials) (8 – 10, reference your Excel table)\n\n1. What UTM zone should I use if I am creating GIS data for the state of Alabama?\n2. What UTM zone should I use if I am creating GIS data for a site in western Idaho?\n3. What UTM zone should I use if I am creating GIS data for a site in eastern Idaho?\n4. Assume you are working along the Mississippi/Alabama border.\n1. What UTM zone should you be working in?\n2. At this site, would your easting value (the X coordinate) be larger than 500,000m or smaller than 500,000m?\n5. What does a UTM northing value (the Y coordinate) of 3,000,000m mean?\n6. What State Plane zone should you use if your site is in eastern Georgia?\n7. The State Plane zones for some states are oriented in a north-south direction and others are oriented in an east-west direction. Why are the state plane zones established in this manner?\n8. Consider the four Athens-to-Pittsburgh distances (UTM17, UTM16, SPGaEast, SPGaWest). Which measurement would you present as the most accurate? Please tell me why you made this decision?\n9. Why are the Lat/Long distance measurements nonsense?\n10. Create a table that presents the results from your Athens-to-city distances. The table should contain the to-city, utm17-distance, utm16-distance, SPGaE-distance, SPGaW-distance, and the decimal degree distance. Round your values to the nearest tenth. Format your table like this one",
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"11. Flipgrid response (make sure to respond to the ‘Lab 3 Response’ topic): In today’s lab, you measured the straight-line distance from a handful of cities to Athens, GA. You made these measurements based on XY coordinates measured in the UTM 17 , UTM16, State Plane Georgia East, State Plane Georgia West, and Latitude/Longitude mapping planes. Your distances varied wildly. I would like to know:\n1. Which distance measure do you have most confidence in?"
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"https://142.93.66.153/wp-content/uploads/2018/08/image-result-for-properly-format-a-table.jpeg",
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https://www.owlgen.in/central-tendency-measures-meaning-functions-characteristics-types/ | [
"# Concept of Central Tendency Measures: Meaning, Functions, Characteristics & Types\n\n## MEANING OF MEASURES OF CENTRAL TENDENCY\n\nA measure of central tendency describes a summary measure That tries to spell out an entire set of information using one value that reflects the center or center of its supply.\n\nChaplin (1975) defines central tendency as the representative value of the distribution of scores.\n\nEnglish & English (1958) define measure of central tendency as a statistic calculated from a set of distinct and independent observations and measurements of certain items or entity and intend to typify those observations.\n\nFor example, when we talk about the achievement scores of the students of a class, we find some students with very high or very low score. However, the score of the most students live somewhere between the highest and the lowest scores of the whole class. Here we see a score around which the data converge around and this will be used as a measure of central tendency.\n\n## FUNCTIONS OF MEASURES OF CENTRAL TENDENCY\n\nMeasure of central tendency provides a figure that describes the whole data. It makes it easy for the researcher and the reader to comprehend the data.\n\nIt helps in minimizing the large data into a single value. For example, it may not be easy to know a family’s need for electricity, but if we know the average use, the government will plan for generation and procurement of electricity accordingly.\n\nWith the help of a sample, it provides us the idea about the mean of the whole population.\n\nIt helps in decision making. For example, a telecom company wants to operate in a city and before that it wants to know about average number of calls by a person. Measures of central tendency will be helpful in getting this figure.\n\nMeasure of central tendency is used to understand mental functioning. For example, an investigator can use mean score to conclude whether eight-year-old girls are better in linguistic ability in comparison to boys of the same age.\n\nAlso read | What are the Requisites of a Good Average?\n\n### Definitions of Mean, Mode and Median\n\nStatistical distribution states how a group of data is distributed in a population. For example, if you want to know about the number of children, adolescents and adults in Indian population, we can get the information through the data in a statistical distribution in terms of actual numbers or in terms of percentage. Statistical distribution describes the properties of the distribution in terms of mean, median, mode and range.\n\nThere are two types of statistical distributions:\n\n1. The discrete random variable distribution, and\n2. The continuous random variable distribution.\n\nThe discrete random variable distribution means variables are usually counts. If a random variable can take only a finite number of distinct values, then it must be discrete. Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor’s surgery, the number of defective light bulbs in a box of ten.\n\nIn continuous random variable distribution, values are recorded within an interval or span. For example, scores of five students are 63, 72, 76, 84, and 39. These are discrete scores. If the same scores are given as 35-55, 55-75 and 75-95.\n\nThis is called continuous random distribution. In the continuous distribution, the value will be within an unbroken interval or span. This is also called a probability density function, generally used in the forecasts of weather.\n\n#### Mean:\n\nMean is the average of all values given in both discrete and continuous distribution. It is calculated differently in both discrete distribution and the continuous distribution. In the discrete data, all the scores are added and divided by the total number. In the continuous distribution, there are different methods to calculate the mean.\n\n#### Median:\n\nThe Median is the middle value in a series of data. In the discrete data, when the totals of the list are odd, the median is the middle entry in the list after sorting the list into increasing order.\n\nWhen the totals of the list are even, the median is equal to the sum of the two middle (after sorting the list into increasing order) numbers divided by two. In the continuous distribution, some different formula is applied.\n\nAlso read | Testing and Measurement Concepts of Assessment.\n\n#### Mode:\n\nThe mode in a list of numbers refers to the list of numbers that occur most frequently. For example, in the following data —7, 2, 2, 43, 11, 11, 44, 18, 18, 18, 27, 39, 6 -18 occurs the most at 3 times. There can be more than one mode for a distribution with a discrete random variable.\n\nA distribution with two modes is called bimodal and a distribution with three modes is called trimodal. The mode of a distribution with a continuous random variable is calculated differently.\n\n#### Range:\n\nThe range of a set of data is the difference between the largest and smallest values. However, in descriptive statistics, this concept of range has a more complex meaning.\n\nIt is the same in discrete random variable series and continuous random variable series.\n\n## CHARACTERISTICS OF GOOD MEASURES OF CENTRAL TENDENCY\n\nRightly and rigidly defined: It implies that the definition of the measure should be so clear and same to everyone. It should be interpreted in the same manner.\n\nSimple to calculate: It should lead to one interpretation whoever may be calculating it.\n\nEasy to understand: It should be simple to calculate. Too much complexity and high calculations do not make the measure a good one.\n\nBased on all the observations: It means whatever may be the measure of central tendency, it must be easily understood what it conveys.\n\nAlso read | Meaning of Variance in Psychology\n\nLeast affected by fluctuation in sampling: It should be based on all observations means it should take into all the scores. For example, if five students have scored 30, 45, 20, 64, 75 out of 100 marks in a test. All these marks should be considered as it is in the case of Mean. If any mark is left out say the extreme that is 20 and 75, then we cannot calculate the correct measure.\n\nA sample is a smaller group of members of a population selected to represent the population. The most commonly used sample is a simple random sample. It requires that every possible sample of the selected size has an equal chance of being used. For example, we take only 50 in a class of 200 students that is 1/4th of the total class. They represent the whole class.\n\nIf we take the sample randomly then this sample will be called random sample. Sample not correctly selected or defective in some way may show fluctuations. For example, if we select only the best student or only the worst, many students will be totally left out who are good or bad? These would affect the measure of central tendency.\n\nIf we take the two samples randomly from the same universe or population the value of average for both of them should be near each other. Fluctuation of sampling happens if there difference in the averages of two samples drawn from the same population.\n\n## TYPES OF MEASURES OF CENTRAL TENDENCY\n\nThere are different measures of central tendency. The three most commonly used are: the mean, median and mode.\n\nAlso read | What is Survey Research?\n\n### The Mean\n\nMean usually refers to average. There are different types of mean such as arithmetic mean, geometric mean and harmonic mean.\n\nThe arithmetic mean is the sum of all the scores in a data divided by the total number of scores. M is use for the Mean in psychology, and also recommended by the American Psychological Association. The Greek letter mue is use to denote the mean of the population. X or M is use to denote the mean of a sample.\n\n#### Properties of the Mean:\n\nMean is responsive to the changes in any score. Its value will change with increase or decrease in the value of any score.\n\nIt is sensitive to the presence or absence of extreme scores. For example, if the data series is 10, 20, 30, 40, the mean is 25 and if we change one score to 20, 30, 40, 50, the mean will 35. One extreme change also changes the mean values drastically.\n\nAmong the measures of central tendency, the mean is the best choice. Since it is the only measure which is based on the total scores. For example: if we want to see the impact of training or a group of students. We will compare the mean scores obtained before and after the training.\n\nThe difference will give an idea of the effect of training on the students. If the difference is greater than one may be able to state that the training had a good effect on the students.\n\nFor an insurance company, on the basis of life expectancy as a Mean the company knows how much it gets from policy-holders and pays to survivors.\n\nThe mean is also the most useful when we have to do the further statistical computations.\n\nThe mean goes along with other statistical formulas and procedures; it is based on arithmetic and algebraic manipulation. Mean is also includes implicitly or explicitly when we have to do further calculations.\n\nAlso read | Importance of Tabulation of Data.\n\n#### Limitations of the Mean:\n\nThe mean is too responsive to the extreme scores. It means if any one score is extreme and all other scores are near each other, it may give a wrong idea about the average. For example, 5 students scored: 25, 35, 45, 55, and 90. The score of 90 will affect adversely the Mean. Here the mean is 50. If 90 is remove, the mean will become 40.\n\nEvery value given equal importance in the calculation of the mean. However, an extreme value becomes misleading.\n\nThe mean is often misleading. For example, the average score obtained by group A in first, second and third term is 45, 55 and 50; another group scored 75, 30, 45. The mean in both the cases is 50. We cannot compare the two groups with the mean.\n\n### The Median\n\nThe median is the numerical value separating the higher half of a data sample, a population, or a probability distribution, from the lower half.\n\nMinimum, King & Bear (2001) defines median as the value that divides the distribution into two halves. Garrett (1981) says the median by definition is the 50% point in the distribution when scores in a continuous series are grouped into frequency distribution.\n\nFor example, the score of seven students in ascending order is 4, 6, 7, 8, 10, 12; 15, then the marks obtained by the fourth student – 8 – is the median of the scores of the group.\n\nAlso read | Quartile Deviation (QD) and Standard Deviation (SD)\n\n#### Properties of the Median:\n\nUnlike the mean, median is less responsive to extreme values in the distribution. For example, the median is 8 for 4, 7, 8, 10, 14 and 2, 7, 8, 10, 39.\n\nSometimes median is also better than the mean as a representative value for a group of scores. For example the mean of the first scores is 12 (4 + 7 + 9 + 14 + 26 5) and the mean of the second scores is 18 (4+7+9+14+54/ 5), but the median is 9 in both the cases, which is much more representative of most of the scores.\n\nAn extreme score like this (54) is called an outlier. Outlier may be much higher or much lower than the other scores. Thus, when outliers are present in a data, the Median should used as the central tendency measure.\n\n#### Limitations of the Median:\n\nThe median is misleading as it does not say how many scores lie below or above it. For example, in 10, 25, 30, 49, 50, the median is 30 but the difference between 25 and 30 is 5 and difference between 30 and 49 is 19. Thus, median can be misleading when the data has a very wide range of scores with minimum at one extreme and maximum extreme.\n\nThe median does not represent the complete data since it is not based on each and every item of the distribution.\n\nWe leave out most of the scores in the median taking only the midpoint value. Thus it cannot be the representative of the sample.\n\nAlso read | Spearman’s Two Factor Theory.\n\n### The Mode\n\nThe mode is the value that appears most often in a set of data. Like the mean and median, the mode is a way of expressing, in a single number, important information, about a random variable or a population.\n\nThe word mode originates from the French word La Mode. The literal meaning of the word mode is frequent and fashionable.\n\nMinimum et al (1997) defines mode as the score that occurs with the greater frequency. Guildford (1965) says the mode is strictly defined as the point on the scale of measurement with maximum frequency in a distribution.\n\nFor example, in a factory maximum number of laborers (out of 1000, 700 laborers) earn Rs. 200 per day and those earning more than 200 or less than 200, is less than 7. Thus the mode wage of the factory is Rs 200.\n\n#### Properties of the Mode:\n\nWe can easily get the mode in a series of data. The mode can used for nominal level variables. For example, there are more Hindi speaking people in India than people of any other language. Here, Hindi language is referred to the mode.\n\nNo other measure of central tendency is appropriate for distribution of language in India, as one can use mode to describe the most common scores in any distribution.\n\nAlso read | Types of Social Distance.\n\n#### Limitations of the Mode:\n\nThe mode is not stable. It is responsive to sampling fluctuations. In a data series, there may be more than one mode.",
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https://www.gla.ac.uk/schools/mathematicsstatistics/events/details/?id=9980 | [
"# Symmetric phylogenetic group-based models using numerical algebraic geometry.\n\n### Dimitra Kosta (University of Glasgow)\n\nThursday 16th November, 2017 14:00-15:00 311B Mathematics and Statistics Building\n\n#### Abstract\n\n(Lunch with the speaker will be at One A The Square, leaving from the school front foyer at 12.45.)\n\nPhylogenetic models have polynomial parametrization maps. For symmetric group-based models, Matsen studied the polynomial inequalities that characterize the joint probabilities in the image of these parametrizations. We employ this description for maximum likelihood estimation via numerical algebraic geometry. In particular, we explore an example where the maximum likelihood estimate does not exist, which would be difficult to discover without using algebraic methods. We also study the embedding problem for symmetric group-based models, i.e. we identify which mutation matrices are matrix exponentials of rate matrices that are invariant under a group action."
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https://developer.aliyun.com/article/979941 | [
"# JavaScript 中 Math.random() 生成随机数据\n\n+关注继续查看\n\nMath.random() 将生成一个介于0(包括)和 1(不包括)之间的伪随机浮点数(带有小数的数),随机数并不意味着总是得到一个唯一的数字,它会在一段时间后产生相同的数字。这里的间隔很长,所以可能不会得到两次相同的数。\n\n### 1. 随机布尔值\n\nconst randomBoolean = () => Math.random() >= 0.5;\nconsole.log(randomBoolean()); // true\nconsole.log(randomBoolean()); // false\nconsole.log(randomBoolean()); // false\n\n### 2. 随机数\n\nconst randomNumber = (max) => Math.round(Math.random() * max);\nconsole.log(randomNumber(100));\n\nconst randomNumber = (min, max) =>\nMath.round(Math.random() * (max - min) + min);\nconsole.log(randomNumber(51, 100));\n\n### 3. 随机 ID\n\nconst randomID = () => Math.random().toString(36).substring(2);\nconsole.log(randomID()); // 961b4gd0t3\n\nnum.toString(radix)\n\nBase36是一个二进制到文本编码表示方案的二进制数据以ASCII通过将其转化为一个字符串格式基数 -36 表示。选择 36 十分方便,因为可以使用阿拉伯数字 0–9 和拉丁字母 A–Z (ISO基本拉丁字母)表示数字。\n\nMath.random 结果不包括 1 但包括 0。这意味着 randomID 结果是空 \"\" 的可能性很小,那是因为依赖序列的开头为 0. ,在这种情况下,可以简单地返回0或任何其他的值作为默认字符串id :\n\nconst randomID = () => Math.random().toString(36).substring(2) || \"0\";\n\n### 4. 随机十六进制数\n\nconst randomHex = () =>\n#${Math.random().toString(16).slice(2, 9).padEnd(6, \"0\")}; console.log(randomHex()); // #2b988b5 如果不想使用 ES7,可以改下以兼容其他版本: const randomHex = () => #${(0x1000000 + Math.random() * 0xffffff).toString(16).slice(1, 7)};\nconsole.log(randomHex()); // #4efabd",
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"《JavaScript数据可视化编程》——1.3 使用饼图强调部分数据\n2144 0",
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"《JavaScript数据可视化编程》——1.4 用离散图表绘制x/y值\n1565 0",
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"《JavaScript数据可视化编程》——1.2 用折线图来绘制连续数据\n1422 0",
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"《JavaScript数据可视化编程》——1.7 小结\n970 0",
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"《JavaScript数据可视化编程》——导读\n\n2547 0",
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"《你不知道的JavaScript》整理(三)——对象\n901 0",
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"javascript语言扩展:可迭代对象(3)\n1119 0",
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"javascript语言扩展:可迭代对象(5)\n1204 0",
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"javascript语言扩展:可迭代对象(4)\n1037 0",
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"javascript语言扩展:可迭代对象(1)\n1282 0\n\n255\n\n0"
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https://www.polk.k12.ga.us/olc/41/page/376339 | [
"Unit 1: Rounding and Place Value\n\nThis unit is all about rounding numbers, understanding the midpoint of a number, adding and subtracting whole numbers, and solving word problems while completing these operations! Throughout the year our standards specify that there are certain ways I need to teach your child. Some of these various methods are to ensure deeper understanding of the concept or of numbers in general. Please don't fall into the idea that this is \"common core math\". I am simply going to teach your child various methods of understanding material! Feel free to work with them \"the old way\" or how you were taught in school. I'm sure we will cover that method as well. :)\n\nStandards for Unit 1\n\nMGSE4.NBT.1 Recognize that in a multi- digit whole number, a digit in any one place represents ten times what it represents in the place to its right. For example, recognize that 700 ÷ 70 = 10 by applying concepts of place value and division.\n\nMGSE4.NBT.2 Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.\n\nMGSE4.NBT.3 Use place value understanding to round multi-digit whole numbers to any place.\n\nMGSE4.NBT.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm.\n\nMGSE4.OA.3 Solve multistep word problems with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a symbol or letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.\n\nMGSE4.MD.2 Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale.\n\nVocabulary for Unit 1\n\n1) Value: the amount a digit is worth\n\n2) Word Form: how a number is written with words or said aloud\n\n3) Standard Form: how a number is written with numerals\n\n4) Expanded Form: how a number is written to show the place value of each digit\n\n5) Period: digits in groups of three in a large number\n\n6) Compare: to decide if one number is greater than, less than, or equal to another number\n\n7) Greater Than: a comparison that says one number has greater value than another number (>)\n\n8) Less Than: a comparison that says one number has less value than another number (<)\n\nPractice Websites\n\nRounding Kahoot"
]
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https://www.solipsys.co.uk/new/ArchimedesHatBoxTheorem.html?RememberingConway | [
"# Archimedes Hat Box Theorem",
null,
"Subscribe!",
null,
"My latest posts can be found here:\nPrevious blog posts:\n\n# Archimedes' Hat Box Theorem - 2018/05/30",
null,
"",
null,
"",
null,
"Take a sphere, and encase it in a cylinder. Take a horizontal slice between two planes. You end up with a narrow cylinder, and a smaller, \"slopey\" cylinder.\n\nThe Archimedes Hat-Box Theorem says these two cylinders have the same surface area. In fact the theorem says that the surface areas are the same no matter how thick the slice is, so as a consequence we can take a \"slice\" that captures the entire cylinder and sphere, and that means that the surface area of the curved surface of the cylinder is exactly the same as the surface area of the sphere.\n\nThe cylinder has height $2r$ and circumference $2{\\pi}r$, so the area is $4{\\pi}r^2.$\n\nBut how do we prove this?\n\nSuppose the radius of the sphere is $R$, and take a really thin slice. Suppose the base of the slice is at height $h$, and it's of thickness, $\\delta{h}$. We can see that the surface area of the slice of cylinder is $2{\\pi}R\\cdot\\delta{h}$.\n\nThe slice of the sphere is a smaller cylinder with sloping sides. We can see that the radius is smaller, but the slope \"adds height\" to the cylinder, and we need to allow for that. So if the smaller radius is $r$ and the sloped \"height\" is $\\delta{l}$ then the surface area of the cylinder from the sphere is $2{\\pi}r\\cdot\\delta{l}$\n\nWe can see from the diagram that we have similar triangles, marked here in red. Note that the hypotenuse of the little triangle, the one marked $\\delta{l}$, is at right angles to the main radius, $R$.",
null,
"The large triangle has hypotenuse $R$, and one side $r$. The smaller triangle has hypotenuse $\\delta{l}$, and the equivalent side is $\\delta{h}$. So $\\frac{r}{R}=\\frac{\\delta{h}}{\\delta{l}}$, or, rearranging, we have $r\\cdot\\delta{l}=R\\cdot\\delta{h}$.\n\nSo the surface area of the slopey cylinder is $2{\\pi}r\\cdot\\delta{l}$, which we can bracket as $2{\\pi}(r\\cdot\\delta{l})$. Replacing what's in the brackets we get $2{\\pi}(R\\cdot\\delta{h})$. That is, of course, $2{\\pi}R\\cdot\\delta{h}$, which is the area of the slice through the enclosing cylinder, and we are done.\n\nFor larger slices we can simply subdivide as finely as we need and get the sum to be as close as necessary to the area of the larger slice. Linearisation is powerful when it works, and it works here.\n\nAnd thus we have Archimedes' Hat Box Theorem.\n\nYou might also want to read: Considering A Sphere\n\n <<<< Prev <<<< Considering A Sphere : >>>> Next >>>> Unexpected Interaction Of Features ...",
null,
"You can follow me on Mathstodon.\n\n Of course, you can alsofollow me on twitter:",
null,
"",
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"## Send us a comment ...\n\n You can send us a message here. It doesn't get published, it just sends us an email, and is an easy way to ask any questions, or make any comments, without having to send a separate email. So just fill in the boxes and then\n\n Your name : Email : Message :\n\n# Contents",
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"Suggest a change ( <-- What does this mean?) / Send me email"
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https://www.nagwa.com/en/videos/547198541763/ | [
"",
null,
"Question Video: Subtraction of Rational Numbers into Simplest Form | Nagwa Question Video: Subtraction of Rational Numbers into Simplest Form | Nagwa\n\n# Question Video: Subtraction of Rational Numbers into Simplest Form Mathematics • 7th Grade\n\nEvaluate (2/5) − (4/5) giving the number in its simplest form.\n\n00:44\n\n### Video Transcript\n\nEvaluate two-fifths minus four-fifths giving the number in its simplest form.\n\nWhen we look at this expression two-fifths minus four-fifths, both of our values are fractions. And they have a common denominator. Since both fractions are in the same form and have a common denominator, to subtract, we simply subtract their numerators and the denominator stays the same. This means we’ll have two minus four over five. We know that two minus four equals negative two. And then, the denominator is five. And so, we say two-fifths minus four-fifths is equal to negative two-fifths."
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http://ebook2.worldlibrary.net/AlsoRead.aspx?BookId=2824743 | [
"",
null,
"#jsDisabledContent { display:none; } My Account | Register | Help\n\n### Other People Who Read Permutations and combinations : Combinations Also Read",
null,
"•",
null,
"### Basic probability : Example: Picking a non-blue marble\n\n##### By: Sal Khan\n\nExample of figuring out the probability of picking a non-blue marble from a bag.\n\nBasics of probability and combinatorics.\n\n•",
null,
"### Permutations and combinations : Example: Ways to arrange colors\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nThinking about how many ways you can pick four colors from a group of 6\n\nBasics of probability and combinatorics. If want to display your Chuck Norris dolls on your desk at school and there is only room for five of them. Unfortunately, you own 50. How many ways can you pick the dolls and arrange them on your desk? What if you\n\n•",
null,
"### Probability using combinatorics : Exactly Three Heads in Five Flips\n\n##### By: Sal Khan\n\nProbability of exactly 3 heads in 5 flips using combinations\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Birthday Probability Problem\n\n##### By: Sal Khan\n\nThe probability that at least 2 people in a room of 30 share the same birthday.\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Permutations and combinations : Permutations\n\n##### By: Sal Khan\n\nIntroduction to permutations\n\nBasics of probability and combinatorics. If want to display your Chuck Norris dolls on your desk at school and there is only room for five of them. Unfortunately, you own 50. How many ways can you pick the dolls and arrange them on your desk? What if you\n\n•",
null,
"### Dependent events : Example: Is an event independent or dependent?\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nIndependent Events 1\n\nBasics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in the first (if you\n\n•",
null,
"### Dependent events : Example: Dependent probability\n\n##### By: Sal Khan\n\nThinking about how the probability of an event can be dependent on another event occuring.\n\nBasics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in the first (if you\n\n•",
null,
"### Permutations and combinations : Example: 9 card hands\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nThinking about how many ways we can construct a hand of 9 cards\n\nBasics of probability and combinatorics. If want to display your Chuck Norris dolls on your desk at school and there is only room for five of them. Unfortunately, you own 50. How many ways can you pick the dolls and arrange them on your desk? What if you\n\n•",
null,
"### Dependent events : Introduction to dependent probability\n\n##### By: Sal Khan\n\nDeciding whether you want to play a game at a strange casino.\n\nBasics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in the first (if you\n\n•",
null,
"### Dependent events : Monty Hall Problem\n\n##### By: Sal Khan\n\nPresentation and analysis of the famous Monty Hall Problem\n\nBasics of probability and combinatorics. What's the probability of picking two e from the bag in scrabble (assuming that I don't replace the tiles). Well, the probability of picking an 'e' on your second try depends on what happened in the first (if you\n\n•",
null,
"### Permutations and combinations : Example: Ways to pick officers\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nHow many ways can we pick officers for our organization?\n\nBasics of probability and combinatorics. If want to display your Chuck Norris dolls on your desk at school and there is only room for five of them. Unfortunately, you own 50. How many ways can you pick the dolls and arrange them on your desk? What if you\n\n•",
null,
"### Probability using combinatorics : Getting Exactly Two Heads (Combi...\n\n##### By: Sal Khan\n\nA different way to think about the probability of getting 2 heads in 4 flips\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Generalizing with Binomial Coeff...\n\n##### By: Sal Khan\n\nConceptual understanding of where the formula for binomial coefficients come from\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Example: Different ways to pick ...\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nThinking about the different ways we can pick officers in order to find the probability of one situation in particular.\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Conditional Probability and Comb...\n\n##### By: Sal Khan\n\nProbability that I picked a fair coin given that I flipped 4 out of 6 heads.\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Example: All the ways you can fl...\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nManually going through the combinatorics to determine the probability of an event occuring\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Venn diagrams and the addition rule : Probability with Playing Car...\n\n##### By: Sal Khan\n\nProbability of compound events. The Addition Rule. Common Core Standard 457 S-CP. 7\n\nBasics of probability and combinatorics. What is the probability of getting a diamond or an ace from a deck of cards? Well I could get a diamond that is not an ace, an ace that is not a diamond, or the ace of diamonds. This tutorial helps us think these t\n\n•",
null,
"### Probability using combinatorics : Example: Probability through cou...\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Mega Millions Jackpot Probability\n\n##### By: Sal Khan\n\nProbability of winning the Mega Millions jackpot\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int\n\n•",
null,
"### Probability using combinatorics : Example: Lottery probability\n\n##### By: Sal Khan, Monterey Institute for Technology and Education\n\nWhat is the probability of winning a 4-number lottery?\n\nBasics of probability and combinatorics. This tutorial will apply the permutation and combination tools you learned in the last tutorial to problems of probability. You'll finally learn that there may be better investments than poring all your money int",
null,
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92956877,"math_prob":0.6559211,"size":2503,"snap":"2019-26-2019-30","text_gpt3_token_len":530,"char_repetition_ratio":0.17687075,"word_repetition_ratio":0.025700934,"special_character_ratio":0.19616461,"punctuation_ratio":0.0625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99057025,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46],"im_url_duplicate_count":[null,null,null,10,null,null,null,9,null,null,null,null,null,8,null,7,null,10,null,9,null,9,null,8,null,10,null,6,null,7,null,null,null,null,null,10,null,8,null,null,null,null,null,null,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-23T04:58:41Z\",\"WARC-Record-ID\":\"<urn:uuid:627b4409-c691-421a-aa19-220310733efa>\",\"Content-Length\":\"77648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:287b1492-c221-4c15-82bd-571fb76cc1cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:d88f7750-512a-4621-a8b6-2c46c1a1c11d>\",\"WARC-IP-Address\":\"66.27.42.21\",\"WARC-Target-URI\":\"http://ebook2.worldlibrary.net/AlsoRead.aspx?BookId=2824743\",\"WARC-Payload-Digest\":\"sha1:GYIFAABSPTQMDT2JEUJU5SC6BOTJN4FX\",\"WARC-Block-Digest\":\"sha1:DMKZXFAH7VSJWBRTHL6ERUBANVWRYVKF\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195528869.90_warc_CC-MAIN-20190723043719-20190723065719-00247.warc.gz\"}"} |
https://theses.lib.polyu.edu.hk/handle/200/9111 | [
"Author: Wei, Zikai Title: Volatility forecasting with fuzzy methods Degree: M.Phil. Year: 2017 Subject: Hong Kong Polytechnic University -- DissertationsEconomic forecastingOptions (Finance) -- Mathematical modelsSecurities -- Prices -- Mathematical models Department: Department of Applied Mathematics Pages: xxii, 131 pages : color illustrations Language: English Abstract: The thesis will present the volatility forecasting using some fuzzy methods. Three topics are considered: 1. The proposed volatility modeling technique based on fuzzy method is used to replace the model averaging technique in multivariate volatility forecasting. 2. Use the Hidden Markov Model (HMM) for the volatility forecasting of the univariate and multivariate time series. 3. Application of this proposed volatility forecasting technique with fuzzy methods. For topic 1, a fuzzy-method-based multivariate volatility model is intended to improve the model averaging technique for multivariate volatility forecasting. Volatility modeling is crucial for risk management and asset allocation. This is an influential area in financial econometrics. The central requirement of volatility modeling is to be able to forecast volatility accurately. The literature review of volatility modeling shows that the approaches of model averaging estimation are commonly used to reduce model uncertainty to achieve a satisfactory forecasting reliability. However, those methods attempt to produce a more reliable forecast by confirming all forecasting outcomes equally from several volatility models. Forecasting patterns generated by each model may be similar. Using all forecasting results may cause redundant computations without improving prediction reliability. The proposed multivariate volatility modeling method which is called the Fuzzy-method-based Multivariate Volatility Model (abbreviated as FMVM) classifies the individual models into smaller scale clusters and selects the most representative model in each group. Hence, repetitive but unnecessary computational burden can be reduced, and forecasting patterns from representative models can be integrated. The proposed FMVM is benchmarked against existing multivariate volatility models on forecasting volatilities of Hong Kong Hang Seng Index (HSI) constituent stocks. Numerical results show that it can obtain relatively smaller forecasting errors with less model complexity.For topic 2, HMM-based Multivariate Volatility Model is proposed to present another orientation for multivariate volatility forecasting. The foundation of this method is retrieved from technical analysis, which believes historical data will influence the future performance. Recognition of distinct patterns and search similar pattern is crucial for this approach. The proposed HMM-based volatility forecasting algorithm can obtain a volatility matrix from the past distinct patterns, which is found to generate accurate volatility forecast. We use the K-means clustering algorithm to classify the latest attributes into predetermined clusters and label each attributes vector with the index of its cluster. Then, each attributes vector labeled with a distinct index, and those indexes can form a new pattern with a window of a predetermined size. Subsequently, compute the likelihood values of each pattern using trained HMM and define the similarities by using these likelihood values. The closest value of past likelihood value to the likelihood value of current pattern means the corresponding pattern with this closest value share the same similarity with the current pattern. As the foundation of this method implies, we believe the behavior of the most similar pattern will reoccur. The HMM-based multivariate volatility model is also compared to the FMVM and the volatility averaging model with the same data used in topic 1. Numerical results show that HMM-based multivariate model can obtain a better forecasting accuracy than that of the multivariate volatility model with model averaging technique. For topic 3, A case of portfolio management of a hedge fund is taken as one application of the proposed volatility forecasting algorithms. A framework of a quantitative trading hedge fund is designed, and a changing allocation problem is considered. The application of both our multivariate volatility model and classical portfolio management models is considered at the same time. Regarding the application of these proposed models, a full flow line of product development with quantitative research is shown in this topic. Access: open access\n\nFiles in This Item:\nFile Description SizeFormat\n\nPlease use this identifier to cite or link to this item: `https://theses.lib.polyu.edu.hk/handle/200/9111`"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89255315,"math_prob":0.85881764,"size":5489,"snap":"2020-24-2020-29","text_gpt3_token_len":1004,"char_repetition_ratio":0.16463082,"word_repetition_ratio":0.0,"special_character_ratio":0.17799234,"punctuation_ratio":0.095022626,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97567946,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-30T21:46:43Z\",\"WARC-Record-ID\":\"<urn:uuid:31f1acfa-721d-44f3-93c1-30d9a81e90c1>\",\"Content-Length\":\"21404\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c83f94db-1f18-4cd7-8ad1-5149012f1862>\",\"WARC-Concurrent-To\":\"<urn:uuid:475550ed-987b-4af7-8e65-4ba97c58bad7>\",\"WARC-IP-Address\":\"158.132.160.72\",\"WARC-Target-URI\":\"https://theses.lib.polyu.edu.hk/handle/200/9111\",\"WARC-Payload-Digest\":\"sha1:TA6JJVBSOVMDMZBNWULCZTQXWRTVG7JP\",\"WARC-Block-Digest\":\"sha1:2US6RLWIZCH43UZKCRKQHOAYKWJCQX23\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347410352.47_warc_CC-MAIN-20200530200643-20200530230643-00588.warc.gz\"}"} |
https://answers.everydaycalculation.com/add-fractions/50-36-plus-5-70 | [
"Solutions by everydaycalculation.com\n\n1st number: 1 14/36, 2nd number: 5/70\n\n50/36 + 5/70 is 92/63.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 36 and 70 is 1260\n2. For the 1st fraction, since 36 × 35 = 1260,\n50/36 = 50 × 35/36 × 35 = 1750/1260\n3. Likewise, for the 2nd fraction, since 70 × 18 = 1260,\n5/70 = 5 × 18/70 × 18 = 90/1260\n1750/1260 + 90/1260 = 1750 + 90/1260 = 1840/1260\n5. 1840/1260 simplified gives 92/63\n6. So, 50/36 + 5/70 = 92/63\nIn mixed form: 129/63\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
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.79692847,"math_prob":0.99912447,"size":746,"snap":"2020-34-2020-40","text_gpt3_token_len":313,"char_repetition_ratio":0.1495957,"word_repetition_ratio":0.0,"special_character_ratio":0.54959786,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99686176,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-20T21:35:17Z\",\"WARC-Record-ID\":\"<urn:uuid:f160182a-cb14-4cb1-93db-b406f8d1e66c>\",\"Content-Length\":\"7589\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:814511bd-232b-4c7b-abdc-8dace6c45082>\",\"WARC-Concurrent-To\":\"<urn:uuid:f110225c-bbf6-4432-a0ea-32bc1f8dee9a>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/50-36-plus-5-70\",\"WARC-Payload-Digest\":\"sha1:TFLAIZNMH43HKBMFR5WYOSULDHW5QAQH\",\"WARC-Block-Digest\":\"sha1:ICFEDKZUM4VFEZXVFAI6BSB6UHQDTTLT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198652.6_warc_CC-MAIN-20200920192131-20200920222131-00453.warc.gz\"}"} |
https://www.datasciencemadesimple.com/power-function-in-excel/ | [
"# POWER Function in Excel – find the power of a number in Excel\n\nPOWER function in Excel returns a number raised to a given power\n\n#### Syntax of POWER Function in Excel\n\nPOWER (number, power)\n\nnumber – Number to which the power to be raised.\n\npower – Exponent to raise power to.\n\n#### Example of POWER Function in Excel\n\n##### Formula",
null,
"• Row number 2 calculates 2 ^4 which is 16\n• Row number 3 calculates 25 ^0.5e. square root of 25 which will be 5\n• Row number 4 calculates -2^2 which is 4\n• Row number 5 calculates 2^4 which is 16\n• Row number 6 calculates 5 ^ -2 i.e. (1/0.5)^2 which will be 4\n\nSo the result will be\n\n##### Result",
null,
"Note: #VALUE! Error occurs if the supplied number argument is non-numeric."
]
| [
null,
"https://www.datasciencemadesimple.com/wp-content/uploads/2017/06/POWER-Function-in-excel.png",
null,
"https://www.datasciencemadesimple.com/wp-content/uploads/2017/06/POWER-Function-in-excel-1.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7676851,"math_prob":0.99860674,"size":613,"snap":"2022-40-2023-06","text_gpt3_token_len":178,"char_repetition_ratio":0.19376026,"word_repetition_ratio":0.052173913,"special_character_ratio":0.29526916,"punctuation_ratio":0.08461539,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99966526,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T10:47:01Z\",\"WARC-Record-ID\":\"<urn:uuid:7484d259-a53e-4bcd-a64f-4904dd9fce0e>\",\"Content-Length\":\"88691\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95a39572-8915-4b02-b71b-a4c38fa5d4cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:89739a6b-47a9-4977-889c-1a6edc89f911>\",\"WARC-IP-Address\":\"104.21.73.127\",\"WARC-Target-URI\":\"https://www.datasciencemadesimple.com/power-function-in-excel/\",\"WARC-Payload-Digest\":\"sha1:KDOCNXNC5JUQJGOKZ7WY3FCSEOSFC2Y2\",\"WARC-Block-Digest\":\"sha1:MF5UARGCCE26VXYDHJTVDSMKS2YVDXIF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337490.6_warc_CC-MAIN-20221004085909-20221004115909-00213.warc.gz\"}"} |
http://thermalfluidscentral.org/encyclopedia/index.php/Condensation_Removal_by_Capillary_Action | [
"# Condensation Removal by Capillary Action\n\nIn a zero gravity environment, capillary action is one mechanism of condensate removal. Heat pipes fall under the category of capillary driven devices. Gas-loaded heat pipes have been applied in many diverse fields, and are useful when the temperature of a device must be held constant while a variable heat load is dissipated. In this section, a noncondensable gas-loaded heat pipe modeled by Harley and Faghri (1994) is presented below where the effect of capillary and noncondensable action is applied simultaneously.\n\nThe physical configuration and coordinate system of the gas-loaded heat pipe studied is shown in the figure on the right. Gas-loaded heat pipes offer isothermal operation for varying heat loads by changing the overall thermal resistance of the heat pipe. As the heat load increases, the vapor temperature and total pressure increase in the heat pipe. This increase in total pressure compresses the noncondensable gas in the condenser, increasing the surface area available for heat transfer, which maintains a nearly constant heat flux and temperature.\n\nVapor Space\n\nThe conservation equations for transient, compressible, two-species flow for mass, momentum, energy, and species in vapor space are as follows:",
null,
"$\\frac{\\partial \\bar{\\rho }}{\\partial t}+\\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( \\bar{\\rho }rv \\right)+\\frac{\\partial }{\\partial z}\\left( \\bar{\\rho }w \\right)=0$ (1)",
null,
"$\\bar{\\rho }\\frac{DV}{Dt}=-\\nabla \\bar{p}+\\frac{1}{3}\\bar{\\mu }\\nabla \\left( \\nabla \\cdot V \\right)+\\bar{\\mu }{{\\nabla }^{2}}V$ (2)",
null,
"$\\rho {{c}_{p}}\\frac{DT}{Dt}-\\nabla \\cdot \\bar{k}\\nabla T-\\nabla \\cdot \\left( \\sum\\limits_{j=1}^{2}{{{D}_{d}}{{c}_{pj}}T\\nabla {{\\rho }_{j}}} \\right)-\\frac{D\\bar{p}}{Dt}-\\bar{\\mu }\\Phi =0$ (3)\n\nwhere the subscript j denotes either vapor (v) or gas (g).",
null,
"$\\frac{D{{\\rho }_{g}}}{Dt}-\\nabla \\cdot {{D}_{gv}}\\nabla {{\\rho }_{g}}=0$ (4)\n\nand",
null,
"$\\Phi =2\\left[ {{\\left( \\frac{\\partial v}{\\partial r} \\right)}^{2}}+{{\\left( \\frac{v}{r} \\right)}^{2}}+{{\\left( \\frac{\\partial w}{\\partial z} \\right)}^{2}} \\right]+{{\\left( \\frac{\\partial v}{\\partial z}+\\frac{\\partial w}{\\partial r} \\right)}^{2}}-\\frac{2}{3}{{\\left[ \\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( rv \\right)+\\frac{\\partial w}{\\partial z} \\right]}^{2}}$ (5)\n\nFurthermore, v and w are the radial and axial vapor velocities,",
null,
"$\\bar{p}$ is the total mixture pressure,",
null,
"$\\bar{\\mu }$ is the mass-fraction-weighted mixture viscosity,",
null,
"${{\\bar{c}}_{p}}$ is the specific heat of the mixture,",
null,
"$\\bar{k}$ is the thermal conductivity of the mixture, Dd is the self-diffusion coefficient for both vapor and gas species, Dgv is the mass diffusion coefficient of the vapor-gas pair, ρg is the density of the noncondensable gas, and the mixture density is",
null,
"$\\bar{\\rho }={{\\rho }_{g}}+{{\\rho }_{v}}$. The partial gas density is determined from the species equation, and the vapor density is found from the ideal gas relation using the partial vapor pressure.\n\nThe two choices in species conservation formulation are mass and molar fraction. Molar fraction offers the possibility of a direct simplification in the formulation by the assumption of constant molar density. The assumption is valid over a wider range of temperature and pressure than the corresponding assumption of constant mass density. However, when molar fractions are used, the momentum equation must be written in terms of molar-weighted velocities. The resulting equation cannot be written in terms of the total material derivative, and is significantly more difficult to solve. A benefit of the general equation formulation is its allowance for variable properties. Typical of compressible gas applications, the density is related to the temperature and pressure through the equation of state",
null,
"$\\bar{p}=\\frac{\\bar{\\rho }{{\\text{R}}_{\\text{u}}}T}{{\\bar{M}}}$ (6)\n\nwhere",
null,
"$\\bar{M}$ is the molecular weight of the vapor-gas mixture and Ru is the universal gas constant.\n\nIn the species equation, Dgvis a function of pressure and temperature. For a vapor-gas mixture of sodium-argon, Harley and Faghri (1994) used the following relationship for Dgv",
null,
"${{D}_{gv}}=1.3265\\times {{10}^{-3}}{{T}^{3/2}}{{\\left( {\\bar{p}} \\right)}^{-1}}$ (7)\n\nwhere T is in degrees Kelvin,",
null,
"$\\bar{p}$ is in N/m2, and Dgv is in m2/s. Following a similar procedure, the variable diffusion coefficient formulation for the sodium-helium pair is",
null,
"${{D}_{gv}}=1.2795\\times {{10}^{-3}}{{T}^{3/2}}{{\\left( {\\bar{p}} \\right)}^{-1}}$ (8)\n\nThe interspecies heat transfer that occurs through the vapor-gas mass diffusion was modeled with a self-diffusion model. In the present model, however, the self-diffusion coefficient, Dd, was assumed constant at the initial temperature of the heat pipe.\n\nWick\n\nThe solid structure in the wick is saturated with the working fluid. The condensate is pumped to the evaporator through capillary action. The liquid velocity is taken to be the intrinsic phase-averaged velocity",
null,
"${{\\left\\langle {{v}_{\\ell }} \\right\\rangle }^{\\ell }}$ through the porous wick, which is assumed to be isotropic and homogeneous. For simplicity,",
null,
"${{\\left\\langle {} \\right\\rangle }^{\\ell }}$ is dropped for velocity. Furthermore, the working fluid and wick structure are assumed to be in local thermal equilibrium. The continuity, momentum, and energy equations for the liquid saturated wick are",
null,
"$\\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( r{{v}_{\\ell }} \\right)+\\frac{\\partial {{w}_{\\ell }}}{\\partial z}=0$ (9)",
null,
"\\begin{align} & \\frac{1}{\\varepsilon }\\frac{\\partial {{v}_{\\ell }}}{\\partial t}+\\frac{1}{{{\\varepsilon }^{2}}}\\left( {{v}_{\\ell }}\\frac{\\partial {{v}_{\\ell }}}{\\partial r}+{{w}_{\\ell }}\\frac{\\partial {{v}_{\\ell }}}{\\partial z} \\right) \\\\ & =-\\frac{1}{{{\\rho }_{\\ell }}}\\frac{\\partial {{p}_{\\ell }}}{\\partial r}-\\frac{{{v}_{\\ell }}{{v}_{\\ell }}}{K}+\\frac{{{\\nu }_{\\ell }}}{\\varepsilon }\\left[ \\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( r\\frac{\\partial {{v}_{\\ell }}}{\\partial r} \\right)-\\frac{{{v}_{\\ell }}}{{{r}^{2}}}+\\frac{{{\\partial }^{2}}{{v}_{\\ell }}}{\\partial {{z}^{2}}} \\right] \\\\ \\end{align} (10)",
null,
"\\begin{align} & \\frac{1}{\\varepsilon }\\frac{\\partial {{w}_{\\ell }}}{\\partial t}+\\frac{1}{{{\\varepsilon }^{2}}}\\left( {{v}_{\\ell }}\\frac{\\partial {{w}_{\\ell }}}{\\partial r}+{{w}_{\\ell }}\\frac{\\partial {{w}_{\\ell }}}{\\partial z} \\right) \\\\ & =-\\frac{1}{{{\\rho }_{\\ell }}}\\frac{\\partial {{p}_{\\ell }}}{\\partial r}-\\frac{{{v}_{\\ell }}{{w}_{\\ell }}}{K}+\\frac{{{\\nu }_{\\ell }}}{\\varepsilon }\\left[ \\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( r\\frac{\\partial {{w}_{\\ell }}}{\\partial r} \\right)-\\frac{{{w}_{\\ell }}}{{{r}^{2}}}+\\frac{{{\\partial }^{2}}{{v}_{\\ell }}}{\\partial {{z}^{2}}} \\right] \\\\ \\end{align} (11)",
null,
"${{\\left( \\rho {{c}_{p}} \\right)}_{\\text{eff}}}\\frac{\\partial {{T}_{\\ell }}}{\\partial t}+{{v}_{\\ell }}\\frac{\\partial {{T}_{l}}}{\\partial r}+{{w}_{\\ell }}\\frac{\\partial {{T}_{\\ell }}}{\\partial z}=\\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( r{{k}_{\\text{eff}}}\\frac{\\partial {{T}_{\\ell }}}{\\partial r} \\right)+\\frac{\\partial }{\\partial z}\\left( {{k}_{\\text{eff}}}\\frac{\\partial {{T}_{\\ell }}}{\\partial z} \\right)$ (12)\n\nwhere ε is the porosity of the wick and K is the permeability.\n\nWall\n\nIn the heat pipe wall, heat transfer is described by the transient two-dimensional conduction equation",
null,
"${{\\rho }_{w}}{{c}_{pw}}\\frac{\\partial T}{\\partial t}={{k}_{w}}\\left[ \\frac{1}{r}\\frac{\\partial }{\\partial r}\\left( r\\frac{\\partial T}{\\partial r} \\right)+\\frac{{{\\partial }^{2}}T}{\\partial {{z}^{2}}} \\right]$ (13)\n\nwhere the subscript w denotes the heat pipe wall material.\n\nBoundary Conditions\n\nAt the end caps of the heat pipe, the no-slip condition for velocity, the adiabatic conduction for temperature, and the overall gas conservation conditions are imposed",
null,
"$v=w=\\frac{\\partial T}{\\partial z}=\\frac{\\partial {{\\rho }_{g}}}{\\partial z}=0,\\text{ }z=0$ (14)",
null,
"$v=w=\\frac{\\partial T}{\\partial z}=0,{{\\rho }_{g}}={{\\rho }_{g,BC}},\\text{ }z=L$ (15)\n\nwhere ρg,BC is iteratively adjusted to satisfy overall conservation of noncondensable gas. This boundary condition is implemented through the calculation of the total mass of noncondensable gas. This boundary condition is implemented through the calculation of the total mass of noncondensable gas in the heat pipe. If the total mass is found to be less than the mass initially present in the pipe, the boundary value is increased by 10% of the previous value. Conversely, if the calculated mass is larger than that initially present, the boundary value is decreased. This ensures the conservation of the overall mass to within a preset tolerance, which is 1% in the present formulation. The symmetry of the cylindrical heat pipe requires that the radial vapor velocity and the gradients of the axial vapor velocity, temperature, and gas density be zero at the centerline:",
null,
"$v=\\frac{\\partial w}{\\partial r}=\\frac{\\partial T}{\\partial r}=\\frac{\\partial {{\\rho }_{g}}}{\\partial r}=0,\\text{ }r=0$ (16)\n\nThe liquid-vapor interface",
null,
"$\\left( r={{R}_{v}} \\right)$ is impermeable to the noncondensable gas",
null,
"${{\\dot{m}}_{g}}={{S}_{\\delta }}A{{D}_{gv}}\\nabla {{\\rho }_{g}}+{{\\rho }_{g}}{{S}_{\\delta }}V=0,\\text{ }r={{R}_{v}}$ (17)\n\nwhere",
null,
"$\\dot{m}_g$ is the mass flow rate of gas, A is the cross-sectional area of the heat pipe and is the surface area of the liquid-vapor interface. This formulation of",
null,
"${{\\dot{m}}_{g}}$ accounts for both the convective and diffusive noncondensable gas mass fluxes at the liquid-vapor interface. To ensure saturation conditions in the evaporator section (and part of the adiabatic section since the exact transition point is determined iteratively), the Clausius-Clapeyron equation is used to determine the interface temperature as a function of pressure. The interface radial velocity is then found through the evaporation rate required to satisfy heat transfer requirements. The no-slip condition is still in effect for the axial velocity component.\n\nAt r = Rv for",
null,
"$z\\le {{L}_{e}}+{{L}_{a}}$:",
null,
"${{T}_{\\text{sat}}}={{\\left( \\frac{1}{{{T}_{0}}}-\\frac{{{\\text{R}}_{\\text{u}}}}{{{M}_{v}}{{h}_{\\ell v}}}\\ln \\frac{{{p}_{v}}}{{{p}_{0}}} \\right)}^{-1}}$ (18)",
null,
"${{v}_{\\delta }}=\\frac{\\left( {{k}_{\\text{eff}}}\\frac{\\partial {{T}_{\\ell }}}{\\partial r}-{{{\\bar{k}}}_{\\delta }}\\frac{\\partial {{T}_{v}}}{\\partial r} \\right)}{\\left( {{h}_{\\ell v}}+{{{\\bar{c}}}_{p\\delta }}{{T}_{\\text{sat}}} \\right){{{\\bar{\\rho }}}_{\\delta }}}$ (19)\n w = 0 (20)\n\nwhere",
null,
"${{\\bar{k}}_{\\delta }}$,",
null,
"${{\\bar{c}}_{p\\delta }}$ and",
null,
"${{\\bar{\\rho }}_{\\delta }}$ are the vapor-gas mixture properties at the liquid-vapor interface. In eq. (18), the saturation temperature of the vapor is found from the partial vapor pressure. A solution of the momentum equation gives the total mixture pressure, but the partial vapor pressure can be found using the local gas density:",
null,
"${{p}_{v}}=\\frac{{\\bar{M}}}{{{M}_{v}}}\\bar{p}\\left( 1-\\frac{{{\\rho }_{g}}}{{\\bar{\\rho }}} \\right)$ (21)\n\nwhich was derived assuming a mixture of ideal gases following Dalton’s model for mixtures.\n\nAt the liquid-vapor interface in the active portions of the condenser section, vapor condenses and releases its latent heat energy. This process is simulated by applying a heat source at the interface grids in the condenser section. The interface velocity can be obtained through a mass balance between the evaporator and condenser section, allowing for inactive sections of the condenser.\n\nAt r = Rv for z > Le + La:",
null,
"${{q}_{so}}=-\\left( {{h}_{\\ell v}}+{{{\\bar{c}}}_{p\\delta }}{{T}_{\\delta }} \\right)\\left( \\bar{\\rho }-{{\\rho }_{g}} \\right){{v}_{g}}$ (22)\n\nDue to the conjugate nature of the solution procedure, the boundary condition between the wick and the wall is automatically satisfied. In addition to the equality of temperature, this condition requires the equality of the heat fluxes into and out of the wick-wall interface:",
null,
"${{k}_{w}}\\frac{\\partial T}{\\partial r}={{k}_{\\text{eff}}}\\frac{\\partial T}{\\partial r},\\text{ }r={{R}_{w}}$ (23)\n\nAt the outer pipe wall surface, the boundary conditions depend on both the axial position and the mechanism of heat transfer being studied. In the evaporator, a constant heat flux is specified. In the condenser, a radiative boundary condition is imposed.",
null,
"${{k}_{w}}{{\\left. \\frac{\\partial T}{\\partial r} \\right|}_{r={{R}_{O}}}}=\\left\\{ \\begin{matrix} {{{{q}''}}_{e}}\\text{ evaporator} \\\\ 0\\text{ adiabatic} \\\\ \\sigma {{\\varepsilon }_{r}}\\left( T_{w}^{4}-T_{\\infty }^{4} \\right)\\text{ condenser} \\\\ \\end{matrix} \\right.$ (24)\n\nwhere σ is the Stefan-Boltzman constant and εr is the emissivity.\n\nInitial Conditions\n\nThere is no motion of either the gas or vapor, and the noncondensable gas is evenly distributed throughout the vapor space by diffusion. The initial temperature of the heat pipe is above the free-molecular/continuum-flow transition temperature for the specific heat pipe vapor diameter. The gas-loaded heat pipe experimentally studied by Ponnappan (1989) was simulated using the above analysis. The results show that the wall and vapor temperatures decreased significantly in the condenser section due to the presence of the noncondensable gas.\n\n## References\n\nFaghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA\n\nFaghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.\n\nHarley, C., and Faghri, A., 1994, “Transient Two-Dimensional Gas-Loaded Heat Pipe Analysis,” ASME Journal of Heat Transfer, Vol. 116, pp. 716-723.\n\nPonnappan, R., 1989, “Studies on the Startup Transients and Performance of a Gas Loaded Sodium Heat Pipe,” WRDC-TR-89-2046, Wright-Patterson AFB, OH."
]
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https://gatewayschool-bucks.co.uk/school-life/latest-news/year-3-mastering-column-method | [
"",
null,
"Year 3 - mastering… | Independent Primary School Buckinghamshire",
null,
"# Year 3 - mastering Column Method\n\nOver the past few weeks we have been learning to use the Column Method to solve addition and subtraction problems. We started off adding and subtracting 2 digit numbers and ended up subtracting 3 digit numbers from 3 digit numbers.\n\nThe first thing we learned was that we need to make sure that each digit in each number is lined up correctly; the ones underneath the ones, the tens under the tens and the hundreds under the hundreds. When subtracting, the bigger number must always be on top. Next we start adding or subtracting from the ones column and it is always from top to bottom. When we add sometimes the numbers will add up to 10 or more which means we need to carry over either a ten or a hundred. As some of us learned it's easy to forget to add your extra ten or hundred so we made sure we wrote it below the answer lines. When it comes to subtracting sometimes the top number is smaller than the bottom number for example 3 - 8. If you have 3 sweets you cannot give someone 8. When this happens you need to go next door to your neighbour the tens or the hundreds and exchange one of them for a ten or a hundred. Don’t forget to cross one of them off and then add it to the column where the number was not enough. Now you should have enough to perform your subtraction.\n\nOnce we had mastered the technique we were able to add and subtract 3 digit numbers to and from each other confidently crossing the tens and the hundreds. Wow! How cool is maths?",
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"https://gatewayschool-bucks.co.uk/images/site/logo-gateway-school-bucks.svg",
null,
"https://gatewayschool-bucks.co.uk/images/site/home-updates-overlay.png",
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https://imparquet.ru/calculate-4-5-of-113 | [
"# 4.5% of 113\n\nWhat is 4.5 percent of 113? What is 113 plus 4.5% and 113 minus 4.5%?\n\n 4.5% of 113 = 5.085 113 + 4.5% = 118.085 113 − 4.5% = 107.915\n\nCalculate online:\n % of\n\nMore calculations with 113:\n 13% of 113 = 14.69 113 + 13% = 127.69 113 − 13% = 98.31\n 40% of 113 = 45.2 113 + 40% = 158.2 113 − 40% = 67.8\n 82% of 113 = 92.66 113 + 82% = 205.66 113 − 82% = 20.34\n 268% of 113 = 302.84 113 + 268% = 415.84 113 − 268% = -189.84\nMore calculations with 4.5%:\n 4.5% of 97 = 4.365 97 + 4.5% = 101.365 97 − 4.5% = 92.635\n 4.5% of 130 = 5.85 130 + 4.5% = 135.85 130 − 4.5% = 124.15\n 4.5% of 622 = 27.99 622 + 4.5% = 649.99 622 − 4.5% = 594.01\n 4.5% of 97328 = 4379.76 97328 + 4.5% = 101708 97328 − 4.5% = 92948.2"
]
| [
null
]
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https://www.isprimenumber.com/prime/45667 | [
"# Is 45667 a Prime Number?\n\nYes, 45667 is a prime number. 45667 is divisible by 1 and itself.\n\nYes, 45667 is a prime number.\n\n### Why 45667 is a prime number?\n\nBecause 45667 has no positive divisors rather than 1 and itself.\n\nThe next prime number of 45667 is 45673\nThe previous prime number of 45667 is 45659.\n\n### Related Prime Numbers\n\nBiggest 10 prime numbers smaller than 45667\n\nSmallest 10 prime numbers bigger than 45667\n\n### Related Non-Prime Numbers\n\nBiggest 10 composite numbers smaller than 45667\n\nSmallest 10 composite numbers bigger than 45667"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8585717,"math_prob":0.9450023,"size":531,"snap":"2019-13-2019-22","text_gpt3_token_len":147,"char_repetition_ratio":0.2523719,"word_repetition_ratio":0.11494253,"special_character_ratio":0.33898306,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95886827,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T02:13:32Z\",\"WARC-Record-ID\":\"<urn:uuid:196f3803-2157-42d2-80fa-3992e26a8539>\",\"Content-Length\":\"12629\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18eb620c-6aab-4ece-aefc-427c8fd3abd3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c76f0e3-0326-47ea-b35f-84db37f69579>\",\"WARC-IP-Address\":\"104.197.145.71\",\"WARC-Target-URI\":\"https://www.isprimenumber.com/prime/45667\",\"WARC-Payload-Digest\":\"sha1:RMQF6YAFT3URXCUQJD7ZDIFUTTAPYNAA\",\"WARC-Block-Digest\":\"sha1:OWMPIPPVALLV3QNHYWA2OKGCLOIHJX2N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202711.3_warc_CC-MAIN-20190323020538-20190323042538-00209.warc.gz\"}"} |
https://yunpeng.site/index.php/2021/04/16/dmsum/ | [
"#### Welcome!\n\n0 %\n##### Yun Peng\nPh.D. Candidate in CUHK\n彭昀\n• ###### Major\nComputer Science\nHong Kong\n22\n• ###### Mail\[email protected]\n\n# 数据挖掘总结\n\nApril 16, 2021\nData Mining Cheat Sheet\n\n# Data Mining Cheat Sheet\n\n## Introduction\n\nDefinition 1: The nontrivial extraction of implicit, previous unknown, and potentially useful information from data.\n\nDefinition 2: Data mining is the process of automatically discovering useful information in large data repositories.\n\nDirected Knowledge Discovery: Goal-oriented. Example: classification\n\nUndirected Knowledge Discovery: Bottom-up approach, no target field. Example: clustering\n\nClassifcation, Regression, Clustering, Association Rule Discovery, Anomaly Detection\n\n## Data Preprocessing\n\n4 types of attributes:\n\nNominal, Ordinal, Interval and Ratio\n\nData quality examples:\n\nMissing values, duplication and inconsistent data, noise and outliers\n\nMethods for improving data quality:\n\nSampling, aggregation, dimensionality reduction, feature subset selection, feature creation, discretization and binarization, attribute transformation\n\n## Clustering\n\nSimilarity: often falls in the range [0,1] or [-1,1]\n\nDissimilarity: upper bound can be 1 or $\\infty$\n\n### Different types of similarities\n\nEuclidean Distance: $L_2$ Norm\n\nCity Block: $dist(x,y) = \\sum ^p _{i=1} |x_m - y_m|$\n\nMinkowski Distance: $dist(x,y) = (\\sum ^p _{i = 1}|x_i - y_i|^ \\lambda ) ^{\\frac{1}{\\lambda}}$\n\nSupremum Distance: $dist(x,y) = max_i |x_i - y_i|$\n\nCosine Similarity: $dist(x,y) = \\frac{\\sum _{i=1} ^p x_iy_i }{[\\sum _{i=1} ^p x_i ^2 \\sum ^p _{i=1} x_i^2]^{\\frac{1}{2}}}$\n\nPearson's Correlation: $corr(x,y) = \\frac{\\sum ^p _{i=1} (x_i - \\bar{x})(y_i - \\bar{y})}{[\\sum ^p _{i=1} (x_i - \\bar{x})^2 \\sum^p _{i=1} (y_i - \\bar{y})^2]^{\\frac{1}{2}}}$\n\nMahalanobis Distance: $dist(x,y) = (x - y)^TS ^{-1}(x - y)$ $S_{jk} = \\frac{1}{n-1}\\sum ^n _{i=1} (X_{ij} - \\bar{X}_j)(X_{ik} - \\bar{X}_k)$\n\n### Sequential K-Means\n\n1. make the initial guess for the means $m_1,...,m_k$\n\n2. set the counts $n_1,...,n_k$ to 0\n\n3. until Interrupted\n\n1. acquire the next example $x$\n2. if $x$ is closest to $m_i$, then increase $n_i$ and update $m_i$ by $m_i + \\frac{1}{n_i}(x - m_i)$\n4. end\n\n### Forgetful Sequence K-Means\n\nReplace the counts $n_1,...,n_k$ with a constant $a \\in [0,1]$, update $m_i$ by $m_i + a (x - m_i)$\n\n### Expection-Maximization Algorithm (Gaussian Mixture Methods)\n\nInitialize the model parameters\n\nUntil there is no change in any parameters\n\nExpectation Step : calculate the probability that each object belongs to each cluster\n\nMaximization Step : find the new estimates of the parameters that maximize the expected likelihood\n\nend until\n\n### Kernel K-Means\n\nTransform datapoint $x_i$ into a feature space $\\phi(x_i)$ and then apply the basic K-means\n\n$K(x_i,x_j) = \\phi(x_i)^T \\phi(x_j)$\n\n$||\\phi(x_i) -\\phi(x_j)||^2 = K(x_i,x_i) + K(x_j,x_j) - 2K(x_i,x_j)$\n\nHomogeneous polynomial kernel: $K_q(x,y) = (x^Ty)^q$\n\nInhomogeneous polynomial kernel (c >= 0): $K_q(x,y) = (c + x^Ty)^q$\n\nGaussian Kernel: $K(x,y) = e^{-\\frac{||x-y||^2}{2\\sigma ^2}}$\n\n### Self-Organizing Feature Maps\n\nUnlike sequential K-Means, the centres (called units or neurons) have a pre-determined topographic ordering relationship\n\n1. make initial guesses for centroids $m_1,...,m_k$\n\n2. repeat\n\n1. acquire the next example $x$\n2. if $m_i$ is closest to $x$, then update $m_i$ by $m_i + a(x - m_i)$; update $m_j$ (neighbor of $m_i$) by $m_j + a_n (x - m_j)$\n3. end until centroids do not change or a threshold is exceeded\n\n### Agglomerative Hierarchical Clustering\n\nStart out with each data point forming its own cluster and gradually merge clusters until all points have been gathered together in one big cluster\n\n1. compute the proximity matrix\n\n2. repeat\n\n1. merge the two closest clusters\n2. update the proximity matrix to reflect the proximity between the new cluster and the original clusters\n3. until only one cluster remains\n\nDifferent methods to define the distances between clusters:\n\nSingle Linkage Clustering: distance between groups is defined as that of the closest pair of data\n\nPros: can handle non-elliptical shapes Cons: sensitive to noise and outliers\n\nComplete Linkage Clustering: distance between two clusters is given by the distance between their most distant members\n\nPros: less susceptible to noise and outliers Cons: tend to break large clusters; biased towards globular clusters\n\nGroup Average Clustering: distance between two clusters is defined as the average of the distances between all pairs of records\n\nPros: less susceptible to noise and outliers Cons: biased towards globular clusters\n\nCentroid Clustering: distance between two clusters is defined as the distance between the mean vectors of the two clusters\n\nMedian Clustering: distance between two clusters is defined as the distance between the prototypes of the two clusters\n\nthe prototype is the data point when there is only one point in the clusterl; When we merge two clusters, the new prototype is the mid point of the two prototypes\n\nWard's Method: Similarity of two clusters is based on the increase in squared error when two clusters are merged\n\n$\\Delta SSE(C_i,C_j) = \\frac{n_in_j}{n_i+n_j}||\\mu _i - \\mu _j||^2$\n\n### DBSCAN\n\nCore point: has more than a specified number of points (MinPts) within Eps\n\nBorder point: has fewer than MinPts within Eps, but is in the neighborhood of a core point\n\nNoise point: neither is core point nor border point\n\n1. Label all points as core, border, or noise points\n2. Eliminate noise points\n3. Put an edge between all core points that are within Eps of each other\n4. Make each group of connected core points into a separated cluster\n5. Assign each border point to one of the clusters containing a core point from which the border point is directly density- reachable\n\nPros: resistant to noise, can handle clusters of different shapes and sizes\n\nCons: sensitive to denses, can not handle high dimensional data well\n\n### Cluster Evaluation\n\n$SSE = \\sum _i \\sum _{x \\in C_i}(x - c_i)^2$\n\n$SSB = \\sum_i |C_i| (c - c_i)^2$\n\n$TSS = \\sum _{i=1} ^K \\sum _{x \\in C_i} (x-c)^2$\n\nSilhouette Coefficient:\n\n$a_i:$ average distance of $i$ to the points in its cluster\n\n$b_i$: min (average distance of $i$ to points in another cluster)\n\n$s = \\frac{b_i - a_i}{max(a_i,b_i)}$\n\n## Classification\n\n### Minimum Distance Classifier\n\n$m_j = \\frac{1}{N_j} \\sum _{x \\in C_j} x$\n\nThe goal is to minimize $D_j(x) = ||x - m_j||$\n\nWhich is equivalent with maximizing $d_j(x) = m_j^Tx - \\frac{1}{2} m_j^Tm_j$ the decision function for class j\n\n### Linear Discriminant Classifier\n\n$g(x) = W^Tx + W_o$\n\n$x$ is assigned to one class if $g(x) \\gt 0$ and another class if $g(x) \\lt 0$\n\n### k Nearest Neighbor\n\nlazy learners, does not build model explicitly\n\ncan have arbitary decision boundries\n\n### Bayes Classifier\n\n$P(A|C) = P(A)P(C|A) / P(C)$\n\nA point $x$ belongs to class $i$ if $P(m_i | x)$ is the largest\n\nNormally we are given $P(x|m_i)$ and $P(m_i)$ to compute $P(m_i|x)$\n\nthe decision function is $d_j(x) = p(w_j)p(x|w_j)$\n\nHow to estimate $p(x|w_j)$: $p(x|w_j) = P(x_1|w_j)...P(x_n|w_j)$\n\nPosted in TechnologyTags:"
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http://sage-doc.sis.uta.fi/reference/schemes/sage/schemes/toric/ideal.html | [
"# Toric ideals¶\n\nA toric ideal (associated to an integer matrix $$A$$) is an ideal of the form\n\n$I_A = \\left< x^u - x^v : u,v \\in \\ZZ_\\geq^n , u-v \\in \\ker(A) \\right>$\n\nIn other words, it is an ideal generated by irreducible “binomials”, that is, differences of monomials without a common factor. Since the Buchberger algorithm preserves this property, any Groebner basis is then also generated by binomials.\n\nEXAMPLES:\n\nsage: A = matrix([[1,1,1],[0,1,2]])\nsage: IA = ToricIdeal(A)\nsage: IA.ker()\nFree module of degree 3 and rank 1 over Integer Ring\nUser basis matrix:\n[-1 2 -1]\nsage: IA\nIdeal (-z1^2 + z0*z2) of Multivariate Polynomial\nRing in z0, z1, z2 over Rational Field\n\n\nHere, the “naive” ideal generated by $$z_0 z_2 - z_1^2$$ does already equal the toric ideal. But that is not true in general! For example, this toric ideal ([Stu1997], Example 1.2) is the twisted cubic and cannot be generated by $$2=\\dim \\ker(A)$$ polynomials:\n\nsage: A = matrix([[3,2,1,0],[0,1,2,3]])\nsage: IA = ToricIdeal(A)\nsage: IA.ker()\nFree module of degree 4 and rank 2 over Integer Ring\nUser basis matrix:\n[-1 1 1 -1]\n[-1 2 -1 0]\nsage: IA\nIdeal (-z1*z2 + z0*z3, -z1^2 + z0*z2, z2^2 - z1*z3) of\nMultivariate Polynomial Ring in z0, z1, z2, z3 over Rational Field\n\n\nThe following family of toric ideals is from Example 4.4 of [Stu1997]. One can show that $$I_d$$ is generated by one quadric and $$d$$ binomials of degree $$d$$:\n\nsage: I = lambda d: ToricIdeal(matrix([[1,1,1,1,1],[0,1,1,0,0],[0,0,1,1,d]]))\nsage: I(2)\nIdeal (-z3^2 + z0*z4,\nz0*z2 - z1*z3,\nz2*z3 - z1*z4) of\nMultivariate Polynomial Ring in z0, z1, z2, z3, z4 over Rational Field\nsage: I(3)\nIdeal (-z3^3 + z0^2*z4,\nz0*z2 - z1*z3,\nz2*z3^2 - z0*z1*z4,\nz2^2*z3 - z1^2*z4) of\nMultivariate Polynomial Ring in z0, z1, z2, z3, z4 over Rational Field\nsage: I(4)\nIdeal (-z3^4 + z0^3*z4,\nz0*z2 - z1*z3,\nz2*z3^3 - z0^2*z1*z4,\nz2^2*z3^2 - z0*z1^2*z4,\nz2^3*z3 - z1^3*z4) of\nMultivariate Polynomial Ring in z0, z1, z2, z3, z4 over Rational Field\n\n\nFinally, the example in [SH1995b]\n\nsage: A = matrix(ZZ, [ [15, 4, 14, 19, 2, 1, 10, 17],\n....: [18, 11, 13, 5, 16, 16, 8, 19],\n....: [11, 7, 8, 19, 15, 18, 14, 6],\n....: [17, 10, 13, 17, 16, 14, 15, 18] ])\nsage: IA = ToricIdeal(A) # long time\nsage: IA.ngens() # long time\n213\n\n\nAUTHORS:\n\n• Volker Braun (2011-01-03): Initial version\nclass sage.schemes.toric.ideal.ToricIdeal(A, names='z', base_ring=Rational Field, polynomial_ring=None, algorithm='HostenSturmfels')\n\nThis class represents a toric ideal defined by an integral matrix.\n\nINPUT:\n\n• A – integer matrix. The defining matrix of the toric ideal.\n\n• names – string (optional). Names for the variables. By default, this is 'z' and the variables will be named z0, z1, …\n\n• base_ring – a ring (optional). Default: $$\\QQ$$. The base ring of the ideal. A toric ideal uses only coefficients $$\\pm 1$$.\n\n• polynomial_ring – a polynomial ring (optional). The polynomial ring to construct the ideal in.\n\nYou may specify the ambient polynomial ring via the polynomial_ring parameter or via the names and base_ring parameter. A ValueError is raised if you specify both.\n\n• algorithm – string (optional). The algorithm to use. For now, must be 'HostenSturmfels' which is the algorithm proposed by Hosten and Sturmfels in [SH1995b].\n\nEXAMPLES:\n\nsage: A = matrix([[1,1,1],[0,1,2]])\nsage: ToricIdeal(A)\nIdeal (-z1^2 + z0*z2) of Multivariate Polynomial Ring\nin z0, z1, z2 over Rational Field\n\n\nFirst way of specifying the polynomial ring:\n\nsage: ToricIdeal(A, names='x,y,z', base_ring=ZZ)\nIdeal (-y^2 + x*z) of Multivariate Polynomial Ring\nin x, y, z over Integer Ring\n\n\nSecond way of specifying the polynomial ring:\n\nsage: R.<x,y,z> = ZZ[]\nsage: ToricIdeal(A, polynomial_ring=R)\nIdeal (-y^2 + x*z) of Multivariate Polynomial Ring\nin x, y, z over Integer Ring\n\n\nIt is an error to specify both:\n\nsage: ToricIdeal(A, names='x,y,z', polynomial_ring=R)\nTraceback (most recent call last):\n...\nValueError: You must not specify both variable names and a polynomial ring.\n\nA()\n\nReturn the defining matrix.\n\nOUTPUT:\n\nAn integer matrix.\n\nEXAMPLES:\n\nsage: A = matrix([[1,1,1],[0,1,2]])\nsage: IA = ToricIdeal(A)\nsage: IA.A()\n[1 1 1]\n[0 1 2]\n\nker()\n\nReturn the kernel of the defining matrix.\n\nOUTPUT:\n\nThe kernel of self.A().\n\nEXAMPLES:\n\nsage: A = matrix([[1,1,1],[0,1,2]])\nsage: IA = ToricIdeal(A)\nsage: IA.ker()\nFree module of degree 3 and rank 1 over Integer Ring\nUser basis matrix:\n[-1 2 -1]\n\nnvariables()\n\nReturn the number of variables of the ambient polynomial ring.\n\nOUTPUT:\n\nInteger. The number of columns of the defining matrix A().\n\nEXAMPLES:\n\nsage: A = matrix([[1,1,1],[0,1,2]])\nsage: IA = ToricIdeal(A)\nsage: IA.nvariables()\n3"
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https://www.jiskha.com/questions/16739/the-shorter-diagonal-of-a-rhombus-is-10-inches-long-and-the-angle-the-diagonal-forms-with | [
"# math\n\nThe shorter diagonal of a rhombus is 10 inches long, and the angle the\ndiagonal forms with the side measures 60º. Find the area of the\nrhombus.\n\nI really don't know where to start. I know that the area formula for a\nrhombus is\nA=1/2*d1*d2\nSo if I know that one of the D is 10, how do i get the area? I don't know\nwhere the angle measurement is very helpful either.\n\nHelp?\n\nDraw the picture.\n\nThe angle helps a lot. You know the diagonal bisects that corner angle, so now you know the corner angles. If you know the corner angles, you know the central angles. With these, you can determine the lengths of all parts of the figure.\n\nRepost if you need more help.\n\nIsn't the central angle 90 because the diagonals intersect? How do I find the 2nd diagonal?\n\n1. 👍 0\n2. 👎 0\n3. 👁 246\n1. Surfing random posts from a while back, hello there xD\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### math\n\nEach side of a rhombus is 5 inches long and the shorter diagonal is 4 inches. Find the angle of the rhombus and the longer diagonal.\n\nasked by jessa on February 13, 2011\n2. ### geometry\n\nThe longer diagonal of a rhombus is 2 times as long as the shorter diagonal. Find the length of the shorter diagonal if the area is 24in squared.\n\nasked by Mandy on March 24, 2010\n3. ### geometry\n\nthe perimeter of a rhombus is 36 inches. if the longer diagonal is 14 inches what is the length of the shorter diagonal of the rhombus?\n\nasked by mc on March 23, 2010\n\nProving quadrilaterals PLEASE PLEASE PLEASE HELP *the diagonals of a square bisect the vertex angle * the diagonal of a rhombus are perpendicular *each diagonal of a rhombus bisects the opposite angles of the rhombus help me prove\n\nasked by Julia on September 30, 2013\n5. ### Math\n\nThe larger angles of a rhombus are twice the smaller angle of the rhombus. If the shorter diagonal is 20, find the perimeter of the rhombus.\n\nasked by Xinyi panggg on December 13, 2013\n6. ### Math\n\nthe larger angles of a rhombus are twice the smaller angle of the rhombus. if the shorter diagonal is 20, find the perimeter of the rhombus\n\nasked by Simon on December 10, 2013\n7. ### maths\n\nA triangle has sides 11cm long. The shorter diagonal of rhombus is 8cm long. Find the size of one of the smallest angle\n\nasked by bashiru on May 21, 2017\n8. ### Geometry need help pls\n\nA. A parallelogram has a 7-inch side and a 9-inch side, and the longer diagonal is 14 inches long. Find the length of the other diagonal. Do you need your calculator to do it? B. (Continuation) Evaluate 72 + 92 + 72 + 92 − 142\n\nasked by Ella on April 21, 2010\n9. ### Geometry help\n\nA parallelogram has a 7-inch side and a 9-inch side, and the longer diagonal is 14 inches long. Find the length of the other diagonal. Do you need your calculator to do it? then, (Continuation) Evaluate 72 + 92 + 72 + 92 − 142\n\nasked by Ella on April 21, 2010\n10. ### MATHS\n\nA diagonal of a rectangle forms an angle of 60 degree with each of the two shorter sides of the rectangle.If the length of a shorter side of the rectangle is 3,what is the length of the diagonal?\n\nasked by HEJUMACLA on October 15, 2015\n\nMore Similar Questions"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8909097,"math_prob":0.9194263,"size":2238,"snap":"2019-51-2020-05","text_gpt3_token_len":645,"char_repetition_ratio":0.1996419,"word_repetition_ratio":0.29672897,"special_character_ratio":0.27301162,"punctuation_ratio":0.082039915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99900526,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T14:59:35Z\",\"WARC-Record-ID\":\"<urn:uuid:0161679b-e93a-4c14-affd-80fd43a645c7>\",\"Content-Length\":\"21352\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4cd54c7f-2875-4410-a32e-8594d2f3fb5e>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e029a3e-0180-40ee-96cc-938a02c8b556>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/16739/the-shorter-diagonal-of-a-rhombus-is-10-inches-long-and-the-angle-the-diagonal-forms-with\",\"WARC-Payload-Digest\":\"sha1:2IE7SIMDJWFSK7H3AJ54HTL7KAQZR6WT\",\"WARC-Block-Digest\":\"sha1:MTRPBBEUWDUTOF6B4AZVNQXEEXHMEVWS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251778272.69_warc_CC-MAIN-20200128122813-20200128152813-00372.warc.gz\"}"} |
https://tools.carboncollective.co/compound-interest/76170-at-35-percent-in-29-years/ | [
"# What is the compound interest on $76170 at 35% over 29 years? If you want to invest$76,170 over 29 years, and you expect it will earn 35.00% in annual interest, your investment will have grown to become $458,630,826.31. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$76,170 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 29 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 35.00% into a decimal, which would be 0.35.\n\n$$A = 76170(1 + \\dfrac{ 0.35 }{1})^{ 29}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 29 because our example is for annual compounding, or one period per year, so 29 × 1 = 29.\n\n## How the compound interest on $76,170 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 29 years it is compounding: Start Balance Interest End Balance 1$76,170.00 $26,659.50$102,829.50\n2 $102,829.50$35,990.33 $138,819.83 3$138,819.83 $48,586.94$187,406.76\n4 $187,406.76$65,592.37 $252,999.13 5$252,999.13 $88,549.70$341,548.83\n6 $341,548.83$119,542.09 $461,090.92 7$461,090.92 $161,381.82$622,472.74\n8 $622,472.74$217,865.46 $840,338.20 9$840,338.20 $294,118.37$1,134,456.56\n10 $1,134,456.56$397,059.80 $1,531,516.36 11$1,531,516.36 $536,030.73$2,067,547.09\n12 $2,067,547.09$723,641.48 $2,791,188.57 13$2,791,188.57 $976,916.00$3,768,104.57\n14 $3,768,104.57$1,318,836.60 $5,086,941.16 15$5,086,941.16 $1,780,429.41$6,867,370.57\n16 $6,867,370.57$2,403,579.70 $9,270,950.27 17$9,270,950.27 $3,244,832.59$12,515,782.86\n18 $12,515,782.86$4,380,524.00 $16,896,306.87 19$16,896,306.87 $5,913,707.40$22,810,014.27\n20 $22,810,014.27$7,983,504.99 $30,793,519.27 21$30,793,519.27 $10,777,731.74$41,571,251.01\n22 $41,571,251.01$14,549,937.85 $56,121,188.86 23$56,121,188.86 $19,642,416.10$75,763,604.96\n24 $75,763,604.96$26,517,261.74 $102,280,866.70 25$102,280,866.70 $35,798,303.35$138,079,170.05\n26 $138,079,170.05$48,327,709.52 $186,406,879.57 27$186,406,879.57 $65,242,407.85$251,649,287.41\n28 $251,649,287.41$88,077,250.59 $339,726,538.01 29$339,726,538.01 $118,904,288.30$458,630,826.31\n\nWe can also display this data on a chart to show you how the compounding increases with each compounding period.\n\nAs you can see if you view the compounding chart for $76,170 at 35.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure. ## How long would it take to double$76,170 at 35% interest?\n\nAnother commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $76,170 assuming an interest rate of 35.00%. We can calculate this very approximately using the Rule of 72. The formula for this is very simple: $$Years = \\dfrac{72}{Interest\\: Rate}$$ By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this: $$Years = \\dfrac{72}{ 35 } = 2.06$$ Using this, we know that any amount we invest at 35.00% would double itself in approximately 2.06 years. So$76,170 would be worth $152,340 in ~2.06 years. We can also calculate the exact length of time it will take to double an amount at 35.00% using a slightly more complex formula: $$Years = \\dfrac{log(2)}{log(1 + 0.35)} = 2.31\\; years$$ Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value. As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember. Hopefully, this article has helped you to understand the compound interest you might achieve from investing$76,170 at 35.00% over a 29 year investment period."
]
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https://inches.guru/28-inches-in-feet/ | [
"# Convert 28 inches in feet\n\n28 inches = 2.3333 feet\n\n## How many feet is 28 inches?\n\nWhat is 28 inches to feet? The answer is that 28 inches equals 2.3333 feet.\n\nHow to convert 28 inches to feet? We know that one foot is equal to 12 inches. Therefore, to convert 28 inches to feet, we divide the inch value by 12. This gives us 28 inches ÷ 12 = 2.3333 feet. So, 28 inches is equal to 2.3333 feet."
]
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http://www.padmad.org/2015/11/explaining-percentages-concepts-through.html | [
"## Saturday, November 07, 2015\n\n### Explaining Percentages Concepts through Question Solving Video (16 questions)\n\nIn this video I have tried explaining and revising concepts of Percentages by solving some questions that are frequently asked in competitive examinations. I have related them with the concepts that I have explained in my previous percentages video. In order to properly understand them I would advise all of you to watch the video again and again and solve more and more questions. You can ask doubts related to any question in the comment section below.\n\nQues 1. Ram spends 20 % of his monthly income on his household expenditure, 15 % of the rest on books, 30 % of the rest on clothes and saves the rest. On counting, he comes to know that he finally saves Rs. 9520. Find his monthly income.\n\nQues 2. In an examination, 80 % students passed in Physics, 70 % in Chemistry while 15 % failed in both the subjects. If 325 students passed in both the subjects. Find the total number of students who appeared in the examination.\n\nQues 3. Ram sells his goods 20 % cheaper than Bobby and 20 % dearer than Krishna. How much percentage is Krishna's goods cheaper or dearer than Bobby's ??\n\nQues 4. The entrance ticket of a circus in New Delhi is of Rs. 250. When the price of the ticket was lowered, the sale of tickets increased by 50 % while collections recorded a decrease of 17.5 %. Find the deduction in the ticket price in rupees ??\n\nQues 5. Sneha went to a furniture shop to buy a sofa set costing Rs. 13,080 . The rate of sales tax is 9 % . She tells the shopkeeper to reduce the price of the sofa set to such an extent that she has to pay Rs. 13,080 inclusive of sales tax. Find the percentage reduction needed in the price of the sofa set to satisfy her requirement.\n\nQues 6. A and B have between them Rs. 1200. A spends 12 % of his money while B spends 20 % of his money. They are then left with a sum that constitutes 85 % of the whole sum. Find what amount is left with A ?\n\nQues 7. A fraction is such that if the double of the numerator and triple of the denominator is changed by +10 % and -30 % respectively then we get 11 % of (16/21). Find the value of the fraction.\n\nQues 8. The minimum quantity of milk in litres that should be mixed in a mixture of 60 litres in which the initial ratio of milk to water is 1:4 so that the resultant mixture has 15 % milk is\n\nQues 9. After three successive equal percentage rise in the salary, the sum of Rs. 100 turned into Rs 140 and 49 paise. Find the percentage rise in the salary.\n\n1. 12 %\n2. 22 %\n3. 66 %\n4. 82 %\n\nQues 10. To pass an examination 40 % marks are essential. A obtains 10 percent marks less than the pass marks and B obtains 11.11 % marks less than A. What percentage less than the sum of A's and B's marks should C obtain to pass the exam ??\n\nQues 11. India scored a total of x runs in 50 overs. Australia tied the scores in 20 percent less overs. If Australia's average run rate had been 33.33 % higher the scores would have been tied 10 overs earlier. How many runs Australia scored ??\n\nQues 12. A salesman is appointed on the basic salary of Rs. 1200 per month and the condition that for every sales of Rs. 10000 above Rs. 10000, he will get 50 % of the basic salary and 10 % of the sales as a reward. This incentive scheme does not operate for the first Rs. 10000 of sales. What should be the value of sales if he wants to earn Rs. 7600 in a particular month.\n\nQues 13. In the previous question, which of the following amount of money can be achieved by the salesman in a month\n\n1. 6000\n2. 10000\n3. 11000\n4. All of the above\n\nQues 14. Price of a commodity is first increased by x % and then decreased by x % . If the new price is K/100 then find the original price of the commodity.\n\n1. (x-100)100/k\n2. (x2-1002)100/k\n3. (100-x)100/k\n4. 100k/(1002-x2)\n\nQues 15. The prices of raw materials has gone up by 15 %, labour cost has also increased from 25 % of the cost of the raw material to 30 % of the cost of raw material. By how much percentage should there be a reduction in the usage of raw materials so as to keep the cost same.\n\nQues 16. In an election of 3 candidates A,B and C, A gets 50 % more votes than B. A also beats C by 18000 votes. If it is known that A gets 5 percentage point more votes than C, find the number of voters on the voting list given that 90 % voters on the voting list voted and no votes polled were invalid.\n\n1.",
null,
"2.",
null,
"Please explain 6. Que??? How we get .88x+4/5(1200-x)=1020.????\n\n1.",
null,
"80/100 is cancelled out and become 4/5\n\n3.",
null,
"Very good...very useful..\nThanks sir\n\n4.",
null,
"thx a lot sir\n\nAdd a Comment or Query"
]
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null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
null,
"http://lh3.googleusercontent.com/zFdxGE77vvD2w5xHy6jkVuElKv-U9_9qLkRYK8OnbDeJPtjSZ82UPq5w6hJ-SA=s35",
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https://hal-cea.archives-ouvertes.fr/cea-01472977 | [
"# Many-Body-Localization Transition : sensitivity to twisted boundary conditions\n\nAbstract : For disordered interacting quantum systems, the sensitivity of the spectrum to twisted boundary conditions depending on an infinitesimal angle $\\phi$ can be used to analyze the Many-Body-Localization Transition. The sensitivity of the energy levels $E_n(\\phi)$ is measured by the level curvature $K_n=E_n\"(0)$, or more precisely by the Thouless dimensionless curvature $k_n=K_n/\\Delta_n$, where $\\Delta_n$ is the level spacing that decays exponentially with the size $L$ of the system. For instance $\\Delta_n \\propto 2^{-L}$ in the middle of the spectrum of quantum spin chains of $L$ spins, while the Drude weight $D_n=L K_n$ studied recently by M. Filippone, P.W. Brouwer, J. Eisert and F. von Oppen [arxiv:1606.07291v1] involves a different rescaling. The sensitivity of the eigenstates $\\vert \\psi_n(\\phi) >$ is characterized by the susceptibility $\\chi_n=-F_n\"(0)$ of the fidelity $F_n =\\vert < \\psi_n(0) \\vert \\psi_n(\\phi) >\\vert$. Both observables are distributed with probability distributions displaying power-law tails $P_{\\beta}(k) \\simeq A_{\\beta} \\vert k \\vert^{-(2+\\beta)}$ and $Q(\\chi) \\simeq B_{\\beta} \\chi^{-\\frac{3+\\beta}{2}}$, where $\\beta$ is the level repulsion index taking the values $\\beta^{GOE}=1$ in the ergodic phase and $\\beta^{loc}=0$ in the localized phase. The amplitudes $A_{\\beta}$ and $B_{\\beta}$ of these two heavy tails are given by some moments of the off-diagonal matrix element of the local current operator between two nearby energy levels, whose probability distribution has been proposed as a criterion for the Many-Body-Localization transition by M. Serbyn, Z. Papic and D.A. Abanin [Phys. Rev. X 5, 041047 (2015)].\nDocument type :\nJournal articles\nDomain :\n\nhttps://hal-cea.archives-ouvertes.fr/cea-01472977\nContributor : Emmanuelle de Laborderie <>\nSubmitted on : Tuesday, February 21, 2017 - 3:28:45 PM\nLast modification on : Monday, February 10, 2020 - 6:13:40 PM\nLong-term archiving on: : Monday, May 22, 2017 - 3:46:44 PM\n\n### File\n\n1607.00750.pdf\nFiles produced by the author(s)\n\n### Citation\n\nCécile Monthus. Many-Body-Localization Transition : sensitivity to twisted boundary conditions. Journal of Physics A: Mathematical and Theoretical, IOP Publishing, 2017, 50 (9), pp.95002. ⟨10.1088/1751-8121/aa583f⟩. ⟨cea-01472977⟩\n\nRecord views"
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https://stat.ethz.ch/pipermail/r-help/2020-September/468876.html | [
"# [R] Problem with contour()\n\nHelmut Schütz he|mut@@chuetz @end|ng |rom beb@c@@t\nMon Sep 28 21:05:31 CEST 2020\n\n```Dear all,\n\nI can't get my head around how contour lines are drawn.\nWorking example(x and y are parameters of a certain test and z the\nresulting power):\n\nlibrary(PowerTOST)\nx <- 0.90\ny <- 0.35\nres <- as.numeric(sampleN.TOST(theta0 = x, CV = y, design = \"2x2x4\",\nmethod = \"central\", details = FALSE,\nprint = FALSE)[7:8])\nmesh <- 28\nys <- unique(sort(c(y, seq(y*.8, y*1.2, length.out = mesh))))\nxs <- unique(sort(c(x, seq(x*0.95, 1, length.out = mesh))))\nz <- matrix(nrow = length(ys), ncol = length(xs),\ndimnames = list(paste0(\"y.\", signif(ys, 5)),\npaste0(\"x.\", signif(xs, 5))))\nfor (i in seq_along(ys)) {\nfor (j in seq_along(xs)) {\nz[i, j] <- suppressMessages(\npower.TOST(CV = ys[i], theta0 = xs[j], design = \"2x2x4\",\nmethod = \"central\", n = res))\n}\n}\nz <- z[nrow(z):1, ncol(z):1] # reverse rows & columns\nz[paste0(\"y.\", y), paste0(\"x.\", x)] == res # should be TRUE\ncontour(xs, ys, z, nlevels = 20, las = 1, labcex = 1,\nxlab = \"x\", ylab = \"y\", main = paste(\"n =\", n))\nabline(h = y, v = x, lty = 2)\npoints(x, y, col = \"red\", cex = 1.5)\ntext(x, y, labels = signif(z[paste0(\"y.\", y), paste0(\"x.\", x)], 6),\ncol = \"red\", adj = c(-0.1, 1.5))\n\nAt x = 0.9 and y = 0.35 z = 0.8130092. Obviously this does not agree\nwith the contour-lines.\nI'm sure that I screwed up - but where?\n\nAll the best,\nHelmut\n\n--\nIng. Helmut Schütz\nBEBAC – Consultancy Services for\nBioequivalence and Bioavailability Studies\nNeubaugasse 36/11\n1070 Vienna, Austria\nE helmut.schuetz using bebac.at\nW https://bebac.at/\nF https://forum.bebac.at/\n\n```"
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"CHAPTER 9 FUNDAMENTALS OF ALGEBRA The numbers and operating rules of arithmetic form a part of a very important branch of mathematics called ALGEBRA. Algebra extends the concepts of arithmetic so that it is possible to generalize the rules for operating with numbers and use these rules in manipulating symbols other than numbers. It does not involve an abrupt change into a distinctly new field, but rather provides a smooth transition into many branches of mathematics with a continuation of knowledge already gained in basic arithmetic. The idea of expressing quantities in a general way, rather than in the specific terms of arithmetic, is fairly common. A typical example is the formula for the perimeter of a rectangle, P = 2L + 2W, in which the letter P represents perimeter, L represents length, and W represents width. It should be understood that 2L = 2(L) and 2W = 2(W). If the L and the W were numbers, parentheses or some other multiplication sign would be necessary, but the meaning of a term such as 2L is clear without additional signs or symbols. All formulas are algebraic expressions, although they are not always identified as such. The letters used in algebraic expressions are often referred to as LITERAL NUMBERS (literal implies \"letteral\"). Another typical use of literal numbers is in the statement of mathematical laws of operation. For example, the commutative, associative, and distributive laws, introduced in chapter 3 with respect to arithmetic, may be restated in general terms by the use of algebraic symbols. COMMUTATIVE LAWS The word \"commutative\" is defined in chapter 3. Remember that the commutative laws refer to those situations in which the factors and terms of an expression are rearranged in a different order. ADDITION The algebraic form of the commutative law for addition is as follows: a+b=b+a From this law, it follows that a + (b + c) = a + (c + b) = (c + b) + a In words, this law states that the sum of two or more addends is the same regardless of the order in which the addends are arranged. The arithmetic example in chapter 3 shows only one specific numerical combination in which the law holds true. In the algebraic example, a, b, and c represent any numbers we choose, thus giving a broad inclusive example of the rule. (Note that once a value is selected for a literal number, that value remains the same wherever the letter appears in that particular example or problem. Thus, if we give a the value of 12, in the example just given, as value is 12 wherever it appears.) MULTIPLICATION The algebraic form of the Commutative law for multiplication is as follows: ab = ba In words, this law states that the product of two or more factors is the same regardless of the order in which the factors are arranged. ASSOCIATIVE LAWS The associative laws of addition and multiplication refer to the grouping (association) of terms and factors in a mathematical expression. ADDITION The algebraic form of the associative law for addition is as follows: a+b+c=(a+b)+c=a+(b+c) In words, this law states that the sum of three or more addends is the same regardless of the manner in which the addends are grouped. MULTIPLICATION The algebraic form of the associative law for multiplication is as follows: a � b � c = (a � b) � c = a � (b � c) In words, this law states that the product of three or more factors is the same regardless of the manner in which the factors are grouped. DISTRIBUTIVE LAW The distributive law refers to the distribution of factors among the terms of an additive expression. The algebraic form of this law is as follows: a(b + c) = ab + ac From this law, it follows that: If the sum of two or more quantities is multiplied by a third quantity, tine product is found by applying the multiplier to each of the original quantities separately and summing the resulting expressions. ALGEBRAIC SUMS The word \"sum\" has been used several times in this discussion, and it is important to realize the full implication where algebra is concerned. Since a literal number may represent either a positive or a negative quantity, a sum of several literal numbers is always understood to be an ALGEBRAIC SUM. That is, it is the sum that results when the algebraic signs of all the addends are taken into consideration. The following problems illustrate the procedure for finding an algebraic sum: Let a = 3, b = -2, and c = 4. Then a + b + c = (3) + (-2) + (4) = 5 Also, a - b - c = a + (-b) + (-c) = 3 + (+2) + (-4) =1 The second problem shows that every expression containing two or more terms to be combined by addition and subtraction may be rewritten as an algebraic sum, all negative signs being considered as belonging to specific terms and all operational signs being positive. It should be noted, in relation to this subject, that the laws of signs for algebra are the same as those for arithmetic.",
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"See more ideas about math geometry, math classroom, teaching math. ). SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. 15. Adjacent angles | theorems on adjacent angles | [email protected] What are examples of vertical angles in real life? Examples of acute angles in real life are all around you. What are some real life examples of acute angles? Dividing the right angle will give us two or more acute angles since each newly formed angle will be less than 90˚. Here is an example of an angle in this image: Ask students if they can identify or name this type of angle. For instance, when we are observing the angles of a triangle, it makes more sense for us to consider the angles inside the triangle. Isosceles triangle wikipedia. Trigonometry word problems mathbitsnotebook(geo ccss math). My real life example is the bottom of a coffee table. Acute angles in the real world | sciencing. If you were to lower the ramp until it was exactly parallel to the plane of the trailer floor, you would have a straight angle. Arrows contained in road signs such as \"One Way\" and \"No Right Turn\" display an acute angle at its point. The two rays are called the arms of the angle, and the point is called the vertex. Obtuse | definition of obtuse by merriam-webster. Vertical Angles 6. The letter \"K\" and a diamond-shaped kite contain two acute angles, and each tip of a football is an acute triangle. Real life examples of line and angles … Each slice of the pizza makes an acute angle. . | reference. Real life examples of geometric angles. Angles of elevation and depression (5 powerful examples! The Law of Sines can be used to compute the remaining sides of a triangle when two angles and a side are known (AAS or ASA) or when we are given two sides and a non-enclosed angle (SSA). arrow blades, harpoons, bullets, bolts, airplane nose, anything that needs to travel fast and needs the air resistance to be minimal will be born out of an acute angle initial design, Geometry is all around, if you take a moment to look. And a acute triangle is an example of a umbrella. 10,699 acute angle stock photos, vectors, and illustrations are available royalty-free. Here are some real-life examples of acute angles. Some real-life examples of acute angles are: If we slice a pizza into 5 or more slices, each slice of pizza will make an acute angle. 4 unit: geometric measurement – understand concepts of angle. Bisecting Angle 8. Examples of acute angles found in your home: high school math. EXAMPLE 1 triangle GOAL 1 Classify triangles by their sides and angles, as applied in Example 2. Four Angles Acute Right Obtuse Straight 2. You can find real-world examples of acute angles in many different arenas of everyday life. Examples Of Acute Angles 14. Anywhere, where right triangles occur, you can see complementary acute angles of this triangle. In geometry, angles can be classified according to the size of the angle. A yield sign contains three acute angles and an exit ramp creates an acute angle as it swerves from the highway. 7. What is angle? Acute Angle Examples. Therefore, angles that measure \\(145^\\circ \\),\\(150^\\circ \\), \\(178^\\circ \\), \\(149^\\circ \\), \\(91^\\circ \\) are considered as obtuse angle examples. 3. Students may also be able to measure the angles within the activity, whether that be on the storyboard or going out into the playground and measuring the angle on the slide in real life. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. Go Geometry: Acute Angles in the Real World -- Slideshow. Using embodiment, manipulatives and real-life examples: teaching. An acute angle is less than 90 degrees and obtuse angle is more than 90 degrees. Feb 7, 2020 - In the CCSS angles are mentioned in the 2nd grade standards and then a big emphasis is put on them in 4th and 5th grade. Types of angle 45 and 46. What is a real life example of a reflex angle. We can observe reflex angles in our everyday environment. A sharpened wood pencil has an acute end, and so does a pair of scissors: Take a piece of paper and drawn a dot anywhere on it. A way to remember Sometimes we can confuse acute and obtuse angles. Examples of total internal reflection and critical angle calculator. Conic's Fulfills Aides. Reflex. A slice of watermelon See acute angle stock video clips. Drawing acute, right and obtuse angles (video) | khan academy. Types of Angles - right angle, acute angle, obtuse angle, straight angle, reflex angle, full angle, how to classify angles, in video lessons with examples and step-by-step solutions. Inside the car, the dashboard’s turn signal indicator and the speedometer also create acute angles. 13 best triangles in real life images | practical life, real life, triangle. Desks, books, classroom materials, and floor tiles are all things students can measure or label as they start to recognize Real life examples of geometric angles 1. Examples of different angles inspired from real life Bringing all types of angles together! This may be because we can always use the acute or obtuse angle that goes along with any given reflex angle. 5 use facts about supplementary, complementary. An acute angle is anything angle that is less than 90 degrees. We know that when two rays meet at a point, they form an angle. A triangle formed by all angles measuring less than 90˚ is also known as an acute triangle. The architectural pitch of an A-frame house is an acute angle as is the play, rewind and fast-forward buttons on the DVD remote control. Also the lines on a basketball court are parallel ... Real world examples of acute angles? Other real world examples of an obtuse angle include the angle between the screen and the base of an opened laptop, a hockey stick, an accordion hand fan and between the wings of a boomerang. Straight Angle 7. A pair of tweezers, the tip of a Chihuahua ear, salad tongs, a miter box, some houseplant leaves and a pair of open scissors can create an acute angle within your home. 10,699 acute angle stock photos, vectors, and illustrations are available royalty-free. Have you ever thought why we say sun rays and not sun li… 90 degrees. Drawing acute, right and obtuse angles (video) | khan academy. Examples include a folding easel, a pencil tip, number 7, and the top of letter A. What are the real life applications of linear pair of angles? Types of Angles - right angle, acute angle, obtuse angle, straight angle, reflex angle, full angle, how to classify angles, in video lessons with examples and step-by-step solutions. Each slice of the pizza makes an acute angle. Definition, facts & example. It is valid for all types of triangles: right, acute or obtuse triangles. The angle … A real-life example of a straight angle is when a clock reads 6:00pm. Another example is the wall clock. Straight Angle A straight angle an angle that is 180 degrees. For example, 2 o’ Clock. Quora. Acute angles are the smallest, being between (but not including) zero and 90° Note also that acute triangles are those where all the interior angles are acute. What did you get? We know that the measure of an acute angle is less than right angle, i.e. Real life example of parallel lines are railroad tracks and rows in a garden. Examples we may notice are the outside angle of the letter 'V' or the varied outside angles of roofs. Acute angle things are sharp; they come to a sharp point. Source(s): https://shrinke.im/a71Nl. Right angle An angle formed by the perpendicular intersection of two straight lines; an angle of 90°. The Law of Sines states that The following figure shows the Law of Sines for the triangle ABC The law of sines states that We can also write the law of sines or sine rule as: The Law of Sines is also known as the sine rule, sine law, or sine formula. of 107. maths angles british garden july mathematics angles angles geometry angles in math angle illustration angle geometry measuring angle outdoor installation art regent’s park. We know that angles measuring greater than 0° and less than 90° are called acute angles in geometry. Maurizio G. 1 decade ago. A way to remember is that small things tend to be cute. Using embodiment, manipulatives and real-life examples: teaching. Angles in real life … This intersecting line is known as a transversal. Com. Squares in real life. This dot O is called point. Here are some real-life examples of obtuse angles. Commonly, elementary students in grades three through five learn in math class that an acute angle is made of two rays or line segments that intersect at one end point and is smaller than 90 degrees when measured with a protractor. Sorry, we could not process your request. https://www.pinterest.com/jacobgross000/triangles-in-real-life ∠ABC measures 30 ̊and hence it is an acute angle. Observe the range of each type of angle and comprehend the classification instantly. Quora. For example, 2 o’ Clock. of 107. maths angles british garden july mathematics angles angles geometry angles in math angle illustration angle geometry measuring angle outdoor installation art regent’s park. Obtuse angles are seen on most house rooftops, as the two roof surfaces slope down from it. Acute Angle 3. Let us not forget coasters, the chessboard, the keys of the laptop you are working on! The hangover part 2 utorrent. The size of an angle is measured in degrees; and the symbol used to represent degree is º. The two red arrows are showing the parallel lines and the yellow Acute and Obtuse vs. Quora. Angle wikipedia. Therefore, angles that measure \\(145^\\circ \\),\\(150^\\circ \\), \\(178^\\circ \\), \\(149^\\circ \\), \\(91^\\circ \\) are considered as obtuse angle examples. Mafs. Example Of Acute Angle In Real Life is a totally free PNG image with transparent background and its resolution is 1200x798. Commonly, elementary students in grades three through five learn in math class that an acute angle is made of two rays or line segments that intersect at one end point and is smaller than 90 degrees when measured with a protractor. In the early years, the sailing ships … Vertical angles. How to Construct an Angle Bisector The Angle Bisector Theorem helps you find unknown lengths of sides of triangles, because an angle bisector divides the side opposite that angle into two segments that are proportional to the triangle's other two sides. Mar 27, 2013 - This Pin was discovered by Jacob Gross. Sailing Boat. Acute Angle In Real Life. 0 0. Can you think of more objects in real life that include obtuse angles? Real life examples of geometric angles. A real life example of a right-angled triangle would be a ladder leaning against a wall Real life example of an acute angle. It is an acute angle. The arms of a wall clock make acute angles at several hours of a day. Other real world examples of an obtuse angle include the angle. Name this dot O. Drawing acute, right and obtuse angles (video) | … What are some examples of obtuse angles around the house? Is 0 degrees acute angle? Drawing an obtuse angle. What are some real life examples of acute angles? Four equal straight sides with four right angles are the most common to observe after round/ circular shape! Jan 10, 2017 - Obtuse angles are seen on most house rooftops, as the two roof surfaces slope down from it. Acute angle is the smallest type. Examples we may notice are the outside angle of the letter 'V' or the varied outside angles of roofs. Angle wikipedia. Trigonometric functions of any angle. I chose this picture because it is a acute angle and it is 35 degrees. Before studying angle, let us do one interesting thing. Two acute angles formed by dividing the right angle. An acute angle is an angle that measures less than 90 degrees. Can you observe the obtuse angles in all these images? The hour hand and the minute hand forming an acute angle at 2 o’ Clock. Acute Angles The measure of an angle with a measure between 0° and 90° or with less than 90° radians. Real life examples of geometric angles. Using embodiment, manipulatives and real-life examples: teaching. Can you think of more objects in real life that include obtuse angles? Anywhere, where right triangles occur, you can see complementary acute angles of this triangle. Angles in real life: terminologies, types, acute angle, videos. Can you observe the obtuse angles in all these images? If you were to lower the ramp until it was exactly parallel to the plane of the trailer floor, you would have a straight angle. A right angle is any angle … Six different types of angles based on degrees. Examples of acute angles found in your home: high school math. The amount of turn from one arm of the angle to the other is said to be the size of an angle. This is an example in non-parallel lines cut by a transversal because it has two lines that will eventually intersect and has one line that intersects both of them. Almost every boat nowadays have a triangular sail. For example, when a rectangular piece of bread in divided in two pieces by cutting along the diagonal, we get to right triangles, each with a pair of complementary angles. Acute angle … Modern architectural structures contain an acute angles that add interest and varied shapes. Angle Bisector Examples; Angle Bisector of a Triangle; Angle Bisector Definition. Draw the marker lines over the angles in the image and identify each one, such as shown below. From this point draw as many straight linesas you can. Real life examples of obtuse angles: It is important to have a clear idea about the angle formation to understand the obtuse angle. Even though each of these angles has a ref… Positive and Negative angles at BYJU’S. the adjacent angles are two angles that have common vertex and common side and don't overlap example Wrong examples of adjacent angles These images were taken from: https: ... How are adjacent angles used in real life? Some flagstone pieces used to create a walkway or driveway contain acute angles, as well. Linear Pair of Angles. An acute angle is an angle that is less than 90 degrees. A line has many line segments in it. ... Why can two acute angles be supplementary? Example Of Acute Angle In Real Life This Example Of Acute Angle In Real Life is high quality PNG picture material, which can be used for your creative projects or simply as a decoration for your design & website content. Give two examples of acute angle and reflex angle based on real life. Discover (and save!) The common endpoint is called the vertex, and the two rays are called the arms of the angle. A compass, used to by architects and construction workers to draw home plans, can be narrowed to an acute angle. Comparing angles in real life — classroom activity by david. On-ramp/off-ramp to an elevated road An escalator or travelator Any kind of ramp that you can walk on or drive. Therefore, 45∘ 45 ∘, 5∘ 5 ∘, 18∘ 18 ∘, 49∘ 49 ∘, 89∘ 89 ∘ are all examples of acute angles. Examples of acute angles in real life are all around you. For example, the surface of a table might be a trapezoid, but its legs and supports are not.A trapezoid is a two-dimensional shape with four straight sides and two parallel sides. A line segment OP with endpoints O and P. A line segment is a part of a line. Parents, we need your age to give you an age-appropriate experience. your own Pins on Pinterest The sharp edge of a knife is an acute angle. . You can find real-world examples of acute angles in many different arenas of everyday life. Quora. A clock is made up of two different hands. For example, when a rectangular piece of bread in divided in two pieces by cutting along the diagonal, we get to right triangles, each with a pair of complementary angles. Definition of acute triangle. SHAPES THE THEME IS FURNITURE Acute Angle An acute angle is an angle that is less than 90 Same example applies to a birthday card. StudyPad®, Splash Math®, SplashLearn™ & Springboard™ are Trademarks of StudyPad, Inc. In geometry, two rays sharing a common endpoint form an angle. There are 360º in a full turn (or circle). As students find angles, take a marker and highlight at least one angle of each type (acute, right, obtuse, and straight). The arms of a wall clock make acute angles at several hours of a day. Desks, books, classroom materials, and floor tiles are all things students can measure or label as they start to recognize angles around us. 13. Complementary & supplementary angles (video) | khan academy. What are examples of an obtuse angle in real life? Therefore, the examples of acute angles are 25°, 36°, 47°, 64°, 55°, 72°, 80° and so on. Copyright © 2020 Studypad Inc. All Rights Reserved. Most noteworthy, there are real-world examples of acute angles appearing all around us. Other real world examples of an obtuse angle include the angle between the screen and the base of an opened laptop, a hockey stick, an accordion hand fan and between the wings of a boomerang. Other real world examples of an obtuse angle include the angle. Geometry: Types Of Angles. In a 180˚angle, if one angle is obtuse (more than 90˚), the other will always be an acute angle (less than 90˚). Acute, right, & obtuse angles (video) | khan academy. Why you should learn it GOAL 2 GOAL1 What you should learn 4.1 Classification by Sides Definition of Right Angle explained with real life illustrated examples. Math expression: reflex angles. Supplementary angles definition, properties, theorem & examples. Angles in real life: terminologies, types, acute angle, videos. Angles in Life at MICDS Hannah Akre & Nealey Wallis 2. Obtuse angles are seen on most house rooftops, as the two roof surfaces slope down from it. The hour hand and the minute hand forming an acute angle at 2 o’ Clock. When company owners sign contractual agreements, the pen is held at an acute angle to the paper. Topics in trigonometry. At least two angles of any triangle are acute angles. Quora. Real life examples of a straight angle: Some trailers come with ramps that can be lowered to allow the contents to be removed more easily. The interior angles of all triangles must sum to 180 °, so in an equilateral triangle, where all three angles have the same measure, we know that each interior angle is an acute 60 ° angle. Real life examples of geometric angles. Gr. Complementary Angles 10. Angles in real life: terminologies, types, acute angle, videos. Examples of how to use “acute angle” in a sentence from the Cambridge Dictionary Labs \"acute angle in real life\" \"vertex in real life\" \"[geometric shape] in real life\" A couple more just came in as I was starting this post! Linear Pair 9. G. 2. My real life example is a counter top. Draw another dot and name it P. Join OP. There are three basic types of angles; right angle, acute angle, obtuse angle. Acute angles in the real … A degreeis defined such that the angle of one full turn (or circle) is 360 degrees. Drawing acute, right and obtuse angles (video) | khan academy. See acute angle stock video clips. There are many examples of acute angles found within the classroom, including the side of a folding easel, a pencil tip, the top of the letter \"A\" and the number \"7.\" What would be a real life example of an obtuse angle. To solve real-life problems, such as finding the measures of angles in a wing deflector in Exs. We can use the L… - 32887555 Comparing angles in real life — classroom activity by david. Obtuse Angle 5. Find angle measures in triangles. Below are the various arenas of everyday life where acute angles appear: In the Classroom- There are plenty of examples of acute angles in the classroom. Real life example of acute angle. Here are some real-life examples of obtuse angles. The Complete K-5 Math Learning Program Built for Your Child. Do you know a line has no endpoints like line segment? A doctor's stethoscope that doctors use to listen to your heartbeat contains acute angles, and landscape professionals often use hedge shears and tree-trimming tools that open to an acute angle. Right Angle 4. Some real-life examples of acute angles are: If we slice a pizza into 5 or more slices, each slice of pizza will make an acute angle. We can observe reflex angles in our everyday environment. Examples Of Right Angle 16. Square rubber stamps, tiles on the floor are all squares that you see around as examples of squares in real life. Full's. Angles in Real Life 1. For example, in an equilateral triangle, all three angles measure 60˚, making it an acute triangle. What is a real life example of a obtuse angle. Supplementary Angles In real life examples, the angle between hour and minute hands at 1 O’clock will be an acute angle. Real-life examples of trapezoids include certain table tops, bridge supports, handbag sides and architectural elements. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. Some types of an umbrella are divided into a few sections . Angles in real life: terminologies, types, acute angle, videos. How many did you get? This is a good example because it is a straight angle and it is exactly 180 degrees. Some examples of student-made art may contain acute triangles with acute angles. I'm never sure exactly what folks are looking for when they use those search terms, but I am always hopeful that once they get here they find what they're looking for. Another example is the wall clock. However, there are few real world applications involving reflex angles. Classroom activity by david hand forming an acute angles in geometry from the highway all three angles 60˚! 80° and so on reads 6:00pm of parallel lines are railroad tracks and rows a. & Supplementary angles ( video ) | khan academy two acute angles appearing all you. Roof surfaces slope down from it than 90° are called the vertex degreeis defined such that measure... Measures less than 90 degrees draw the marker lines over the angles in real life examples of acute the!, videos include obtuse angles goes along with any given reflex angle yield sign three... … angle Bisector examples ; angle Bisector Supplementary angles definition, properties, theorem & examples a garden pair! | theorems on adjacent angles | [ email protected ] what are the real life include. Critical angle calculator folding easel, a pencil tip, number 7, and are. Rays meet at a point, they form an angle that goes along with any given reflex angle arrows in... Minute hand forming an acute triangle is an award winning math learning program used by more than 90.. ’ clock will be an acute angle … Other real world -- Slideshow geo ccss math ) another dot name... Angles appearing all around you angles around the house different hands name this type of angle triangles with angles... Dashboard ’ s turn signal indicator and the minute hand forming an acute at. 64°, 55°, 72°, 80° and so on our everyday environment the vertex, and point! Letter `` K '' and a acute triangle an obtuse angle properties, &. Free PNG image with transparent background and its resolution is 1200x798 and rows a. Of roofs or more acute angles of this triangle line has no endpoints like line segment image: students. Geometric angles 1 math ) to be cute right triangles occur, you can find examples., there are few real world examples of acute angles, as the two roof surfaces slope down from.! Angle include the angle 35 degrees most house rooftops, as applied example!, math classroom, teaching math and each tip of a wall clock make acute angles found your... And minute hands at 1 o ’ clock will be an acute triangle is an example of a clock. Up of two different hands they come to a sharp point that goes along with any given reflex angle on! Angle, videos each one, such as `` one way '' and `` no right turn '' display acute! Tracks and rows in a wing deflector in Exs triangle ; angle Bisector of a coffee table so.... Way '' and a acute triangle is an example of acute angles the measure of angle... Such as finding the measures of angles ; right angle an angle is acute! Road an escalator or travelator any kind of ramp that you see around as examples of acute angles several. Your home: high school math a degreeis defined such that the angle, us. Of an obtuse angle is less than 90 degrees at SplashLearn can see complementary acute angles are seen on house! Come to a sharp point facts to easily understand math glossary with fun math worksheet online at SplashLearn in... Letter `` K '' and `` no right turn '' display an acute triangle have a idea! Are parallel... real world examples of acute angle 1 Classify triangles by their sides and angles, as two! Ltd. / Leaf Group Media, all Rights Reserved angle Bisector examples ; angle Bisector Supplementary angles ( )... Math worksheet online at SplashLearn some types of triangles: right, & angles! A sharp point clock is made up of two straight lines ; an that., SplashLearn™ & Springboard™ are Trademarks of StudyPad, Inc the paper MICDS Hannah Akre & Nealey Wallis 2 geometry. S turn signal indicator and the minute hand forming an acute angle at 2 o ’ clock be... Depression ( 5 powerful examples us not forget coasters, the dashboard ’ s turn signal indicator and symbol. Finding the measures of angles varied shapes things are sharp ; they come to a sharp point the. Round/ circular shape are real-world examples of acute angles in many different arenas of everyday life by all angles less. Problems mathbitsnotebook ( geo ccss math ) 90 degrees newly formed angle will give us two or more acute,! The angle slope down from it because we can observe reflex angles measurement – understand concepts of angle and is... Of more objects in real life — classroom activity by david also lines. The most common to observe after round/ circular shape by the perpendicular of... And obtuse angles ( video ) | khan academy most common to observe after circular... Dot anywhere on it jan 10, 2017 - obtuse angles around the house my real?... Geo ccss math ) than 90° are called the arms of a line has no endpoints acute angle examples in real life line segment with... Three acute angles in all these images life — classroom activity by david embodiment, and... A walkway or driveway contain acute triangles with acute angles and an exit ramp creates an acute triangle mar,... And less than 90 degrees up of two straight lines ; an angle that is than... And it is 35 degrees its resolution is 1200x798 this Pin was discovered Jacob... ∠Abc measures 30 ̊and hence it is exactly 180 acute angle examples in real life triangle GOAL 1 Classify triangles their... Add interest and varied shapes: Ask students if they can identify or name this of... A slice of the letter `` K '' and `` no right turn '' display an acute triangle to... That angles measuring greater than 0° and 90° or with less than 90° are called the.. Remember is that small things tend to be cute to observe after round/ shape! A pencil tip, number 7, and the two rays sharing a common endpoint form an angle a... Way to remember Sometimes we can observe reflex angles in real life example of an obtuse in. Another dot and name it P. Join OP making it an acute angle at 2 o ’ clock StudyPad Inc. Stock photos, vectors, and illustrations are available royalty-free angles definition, properties, theorem &.! Angles measuring less than 90° are called the vertex, and illustrations are available.! Name it P. Join OP worksheet online at SplashLearn award winning math learning program used by more acute angle examples in real life degrees. Math learning program used by more than 90 degrees when company owners sign contractual,!, there are real-world examples of acute angles since each newly formed angle will us... Clock will be less than 90 degrees word problems mathbitsnotebook ( geo math... Tip of a day 10, 2017 - obtuse angles i chose this picture because it is an angle is! Is held at an acute triangle 64°, 55°, 72°, 80° and so on with acute,. Owners sign contractual agreements, the dashboard ’ s turn signal indicator and the minute forming... Good example because it is 35 degrees basic types of triangles:,! Life examples of acute angles the measure of an umbrella are divided into a few sections two lines! Is 35 degrees example 2 us do one interesting thing each of these angles has a ref… 10,699 angle. Part of a reflex angle based acute angle examples in real life real life examples of acute angles reflection critical! By Jacob Gross trigonometry word problems mathbitsnotebook ( geo ccss math ) life images | life! In degrees ; and the minute hand forming an acute angle of the angle pieces used to create walkway... 180 degrees varied outside angles of elevation and depression ( 5 powerful examples you take a moment to.... With a measure between 0° and 90° or with less than right angle, videos construction to... Than right angle an angle that is less than right angle an angle with a measure between and! 5 powerful examples point is called the arms of a straight angle an angle Bisector definition have clear! Interesting thing learning program used by more than 90 degrees and obtuse angles have a clear idea about angle... Will give us two or more acute angles in our everyday environment rubber stamps, on. Folding easel, a pencil tip, number 7, and the two roof surfaces slope down from.... In example 2 contain acute angles, as the two rays are called acute angles in real —. Basic types of an angle of one full turn ( or circle ) is 360 degrees range of each of. Though each of these angles has a ref… 10,699 acute angle things are sharp they. What are some real life illustrated examples geo ccss math ) concepts of angle 90° or with less than degrees. Reflex angle outside angles of any triangle are acute angles found in your home: high math! Splashlearn™ & Springboard™ are Trademarks of StudyPad, Inc angle explained with real life example of acute angles solve problems!... real world examples of an angle that is less than 90˚ is also known as an angle... Learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn V ' or the outside. A point, they form an angle Bisector examples ; angle Bisector examples ; angle examples. Are seen on most house rooftops, as well of triangles: right, acute stock. This triangle are acute angles since each newly formed angle will give two. Group Media, all Rights Reserved give us two or more acute angles formed the! Formed by dividing the right angle, 55°, 72°, 80° and so on and construction workers to home... May contain acute triangles with acute angles in the real world examples of acute angles angle and comprehend classification... Football is an acute angle, two rays are called acute angles found in acute angle examples in real life... Stock photos, vectors, and illustrations are available royalty-free a clear idea about the …... Perpendicular intersection of two different hands chose this picture because it is a part of a coffee table triangles!\n\nRogue Playthrough Skyrim, Lyon County, Mn, Raine Motel Valentine, Ne, 17 Euro To Naira, Hanger 2 - Unblocked Games World, 3/8 Forstner Bit Extension, Merseyrail Christmas Timetable 2020, Window Tinting Near Me, Tuesday Specials 2020, Breville Smart Grinder Pro Manual, 8 Roles Of The President Examples, That Boy Sus, Get The Pump, Lab Rats: Elite Force Songs,"
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http://blog.yannis-lionis.gr/2008/04/16/python-exercise-1/ | [
"As part of a Python focus group started at work, we had the following exercise - as the first of many - set to us. The problem is the wll known fizz buzz one:\n\nWrite a program that processes a list of numbers from 1 to 100. For each number, if the number is a multiple of 3, print “FIZZ”; if the number is a multiple of 5, print “BANG”; otherwise, print the number. You are *NOT* allowed to use any *IF/ELSE* statements in your code. You can use the list-accessing ternary operator hack, but whilst I’ll accept your homework if you do, you’ll miss out on the prize (alcoholic), which goes to the most concise code (not including whitespace).\n\nKerry and Nigel have already posted their solutions, which I’m afraid are much more concise than mine - I was mostly happy to find a way to do it without any conditional logic (including an or) and too lazy to look further. So here’s mine:\n\n```def fizzbang(): # Numbers from 1 to 100 numbers = range(1,101)``````# 3,6,9,...,99 multiples_of_3 = range(3,101,3) # 5,10,15,...,95,100 multiples_of_5 = range(5,101,5) # 15,30,45,60,75,90 multiples_of_3_and_5 = range(15,101,15)``````# Replace all muliples of 3 with FIZZ for i in multiples_of_3: numbers[i-1] = \"FIZZ\"```\n\n``` # Replace all muliples of 5 with BANG for j in multiples_of_5: numbers[j-1] = \"BANG\" # Replace all muliples of both 3 and 5 with FIZZ BANG for k in multiples_of_3_and_5: numbers[k-1] = \"FIZZ BANG\" for x in numbers: print x ```\n\nGo on, poke some holes in it."
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https://cs.stackexchange.com/questions/92525/how-to-prove-a-side-effect-in-a-function | [
"# How to prove a side effect in a function\n\nI asked a question earlier about Saving to the Database, which was very general and about the requirements for a proof when you go through many layers of non-verified systems such as the network and databases.\n\nIn this question I am wondering about a more middle-level proof, this time about transforming an object $f : A \\to B$ with side effect $C$. Say I have as input a string $A$, and as output an Abstract Syntax Tree (AST) $B$. All of this happens in memory with a small string of say a few KB. Right now, ignoring all the details of the hardware implementation and all the details of any particular language.\n\nI am wondering at a high level what it takes to prove something like this. Specifically I wanted to focus on side effects in this question. Say during the parse process, we create a global symbol table to store classes. Then as we are parsing through the code and we encounter the class, we add to the symbol table. So instead of $f : A \\to B$, we really have:\n\n\\begin{align} f : A &\\to B\\\\ &\\downarrow\\\\ &C \\end{align}\n\nThat is for the symbol table $C$ and AST output $B$. Somewhere in the function $f$ implementation there is another function $g : \\{C,c\\} \\to C'$, which adds the new symbol $c$ to $C$.\n\nWhat I would like to prove (in this question, just at a high level, some key points) is that the function generates the symbol table $C$, even though the output of the function is the AST $B$. In type theory, the proof for AST $B$ could possibly just be the sequence of type definitions and transformations, similar to Hoare Logic. But to prove that the function $f$ has side effect $C$ seems much harder/trickier.\n\nIt seems that you have to go and step through the algorithm one step at a time, and (assuming everything is strongly typed), figure out what the \"current state\" would look like at that point (of the whole program). Then you would compare your pattern (the assertion of the post-condition if it were Hoare Logic) with the current state of the program at that point, and see if it was a match. And see if that stayed true until the end of the function/algorithm. But this sort of seems like it's becoming Model Checking, which I only know the basics about, not sure if that is correct to assume though. Also, this one-step-at-a-time stepping through the algorithm seems like program simulation, so wondering if that is true or not or if simulation has a role here.\n\nSo I'm wondering, at a high level, what is required to prove that function $f$ generates a side effect $C$. As a specification, I would write \"$f$ generates the symbol table $C$\".\n\nIf the function has two outputs, a standard way to represent this is as a function $f: A \\to (B \\times C)$, i.e., $f$ outputs a pair of an AST and a symbol table.\n\nIf the function updates an existing symbol table, you could represent this as a function $f : (A \\times C) \\to (B \\times C)$. It takes as an input a pair of a string and a symbol table, and outputs a pair of an AST and an updated symbol table.\n\nYou could also read about monads, which are a way to structure and reason about side effects in a functional programming language. They are notoriously challenging to understand.\n\nYou don't prove a side effect. You can prove a fact about the possible side effects, but you need to state what you are claiming; a side effect isn't something that can be proven or not. Similarly, you don't prove a function. You can prove the correctness of a function, if you specify what correctness means for that particular function, but you don't prove the function.\n\nIt depends on how you model the system and what proof approach you're using. For early versions of Hoare Logic, there isn't really any notion of \"scope\", so there's absolutely nothing special you need to do. You simply have a pre-conditions and post-conditions on the global state. For Hoare Logic, types have nothing to do with it.\n\nIn a bit more detail, in a Hoare triple like $\\{P\\}\\ y\\,\\mathtt{:=}\\,f(x)\\ \\{Q\\}$, $P$ and $Q$ are predicates on the entire program state. If $f(x)$ updates some (global) variable, you simply include the appropriate condition in the post-condition $Q$. It's no different than $y$ or any of the other, presumably \"local\", variables $f(x)$ may mutate.\n\nMost type theories have no notion of mutable state. If you want to talk about mutable state in such systems, you have to model it, e.g. via state passing (i.e. a state monad). In this case, a function $f:A\\to B$ which may \"mutate\" some \"global\" state may be modeled as a function of type $S\\times A\\to S\\times B$ where $S$ is a type representing \"global\" state. You can consider more refined versions of this type to capture the actual behavior."
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https://125fps.com/casino/what-is-the-probability-that-all-3-dice-show-the-same-number.html | [
"What is the probability that all 3 dice show the same number?\n\nContents\n\nWhat is the probability that both dice show the same number?\n\nThe probability of one dice being a particular number is 1/6. The probability of two dice being the same particular number is 1/6 x 1/6 = 1/36. This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be 6/36 or 1/6.\n\nWhat is the probability that all 3 dice show a different number?\n\nIf three dice are tossed, then the probability that the numbers shown will all be different is 5/9.\n\nWhat is the probability of rolling a pair of ones?\n\nSince there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.\n\nWhat is the probability of getting a prime number from 1 to 100?\n\nThe prime numbers from 1 to 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Hence, the probability of the event that a number chosen from 1 to 100 is a prime number .\n\nTHIS IS INTERESTING: Your question: Can you bet on sports using crypto?\n\nWhat is the probability of rolling a sum of 3?\n\nWe divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4%\n\nWhen two dices are thrown what is the probability of not appearing the same number on the dices?\n\nThe probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). The probability of rolling any number twice in a row is 1/6, because there are six ways to roll a specific number twice in a row (6 x 1/36).\n\n= 4 / 36 = 1/9."
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https://numbermatics.com/n/38880/ | [
"38880\n\n38,880 is an even composite number composed of three prime numbers multiplied together.\n\nWhat does the number 38880 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 72 divisors.\n\n38880 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of seventy-two divisors.\n\nPrime factorization of 38880:\n\n25 × 35 × 5\n\n(2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5)\n\nSee below for interesting mathematical facts about the number 38880 from the Numbermatics database.\n\nNames of 38880\n\n• Cardinal: 38880 can be written as Thirty-eight thousand, eight hundred eighty.\n\nScientific notation\n\n• Scientific notation: 3.888 × 104\n\nFactors of 38880\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 11\n• Sum of prime factors: 10\n\nDivisors of 38880\n\n• Number of divisors d(n): 72\n• Complete list of divisors:\n• Sum of all divisors σ(n): 137592\n• Sum of proper divisors (its aliquot sum) s(n): 98712\n• 38880 is an abundant number, because the sum of its proper divisors (98712) is greater than itself. Its abundance is 59832\n\nBases of 38880\n\n• Binary: 10010111111000002\n• Base-36: U00\n\nSquares and roots of 38880\n\n• 38880 squared (388802) is 1511654400\n• 38880 cubed (388803) is 58773123072000\n• The square root of 38880 is 197.1801207019\n• The cube root of 38880 is 33.8772970397\n\nScales and comparisons\n\nHow big is 38880?\n• 38,880 seconds is equal to 10 hours, 48 minutes.\n• To count from 1 to 38,880 would take you about ten hours.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 38880 cubic inches would be around 2.8 feet tall.\n\nRecreational maths with 38880\n\n• 38880 backwards is 08883\n• 38880 is a Harshad number.\n• The number of decimal digits it has is: 5\n• The sum of 38880's digits is 27\n• More coming soon!\n\nHTML: To link to this page, just copy and paste the link below into your blog, web page or email.\n\nBBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:"
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http://www.teos-10.org/pubs/gsw/html/gsw_alpha_wrt_CT_t_exact.html | [
"# gsw_alpha_wrt_CT_t_exact\n\n```thermal expansion coefficient with respect to\nConservative Temperature```\n\n## USAGE:\n\n`alpha_wrt_CT_t_exact = gsw_alpha_wrt_CT_t_exact(SA,t,p)`\n\n## DESCRIPTION:\n\n```Calculates the thermal expansion coefficient of seawater with respect to\nConservative Temperature.```",
null,
"```Click for a more detailed description of alpha with respect to Conservative Temperature.```\n\n## INPUT:\n\n```SA = Absolute Salinity [ g/kg ]\nt = in-situ temperature (ITS-90) [ deg C ]\np = sea pressure [ dbar ]\n(ie. absolute pressure - 10.1325 dbar)```\n```SA & t need to have the same dimensions.\np may have dimensions 1x1 or Mx1 or 1xN or MxN, where SA & t are MxN.```\n\n## OUTPUT:\n\n```alpha_wrt_CT_t_exact = thermal expansion coefficient [ 1/K ]\nwith respect to Conservative Temperature```\n\n## EXAMPLE:\n\n```SA = [34.7118; 34.8915; 35.0256; 34.8472; 34.7366; 34.7324;]\nt = [28.7856; 28.4329; 22.8103; 10.2600; 6.8863; 4.4036;]\np = [ 10; 50; 125; 250; 600; 1000;]```\n`alpha_wrt_CT_t_exact = gsw_alpha_wrt_CT_t_exact(SA,t,p)`\n`alpha_wrt_CT_t_exact =`\n`1.0e-003 *`\n``` 0.324707942226035\n0.322724175251768\n0.281178901333399\n0.173138222139544\n0.146269210200123\n0.129426849037398```\n\n## AUTHOR:\n\n`David Jackett, Trevor McDougall and Paul Barker [ [email protected] ]`\n\n## VERSION NUMBER:\n\n`3.05 (16th February, 2015)`\n\n## REFERENCES:\n\n```IOC, SCOR and IAPSO, 2010: The international thermodynamic equation of\nseawater - 2010: Calculation and use of thermodynamic properties.\nIntergovernmental Oceanographic Commission, Manuals and Guides No. 56,\nUNESCO (English), 196 pp. Available from the TEOS-10 web site.\nSee Eqn. (2.18.3) of this TEOS-10 manual.```\n`The software is available from http://www.TEOS-10.org`"
]
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"http://www.teos-10.org/pubs/gsw/html/TEOS-10_front_cover.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6277569,"math_prob":0.7949837,"size":1508,"snap":"2019-13-2019-22","text_gpt3_token_len":503,"char_repetition_ratio":0.119015954,"word_repetition_ratio":0.01,"special_character_ratio":0.3998674,"punctuation_ratio":0.25838926,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95903665,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-20T15:00:57Z\",\"WARC-Record-ID\":\"<urn:uuid:9a776f65-ff43-4456-8d3d-22f87ca023e2>\",\"Content-Length\":\"4447\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d036694a-b9c3-4f1b-9970-f64a47bef30f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed547e10-fea7-4e26-a8a2-493a53728e50>\",\"WARC-IP-Address\":\"202.124.241.200\",\"WARC-Target-URI\":\"http://www.teos-10.org/pubs/gsw/html/gsw_alpha_wrt_CT_t_exact.html\",\"WARC-Payload-Digest\":\"sha1:JC2GDJBA2MJET3HM5NLQJLRIOMB4NHGI\",\"WARC-Block-Digest\":\"sha1:AB6LFXB2A2UWVKPR5GFYLLX72QHHEZTH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256040.41_warc_CC-MAIN-20190520142005-20190520164005-00119.warc.gz\"}"} |
https://www.zbmath.org/?q=an%3A1261.35079 | [
"# zbMATH — the first resource for mathematics\n\nCurve shortening flow in heterogeneous media. (English) Zbl 1261.35079\nThis paper concerns the curvature shortening flow of planar curves in a heterogeneous medium, which is modeled by a spatially-dependent additive forcing term: $v = (\\kappa +g)\\nu,$ where $$\\nu$$ is the inward normal vector to the curve, $$\\kappa$$ is the curvature, $$v$$ is the normal velocity vector, and $$g \\in L^\\infty(\\mathbb R^2)$$ represents the forcing term. This comes from a homogenization problem related to the averaged behavior of an interface moving by curvature plus a rapidly oscillating forcing term $$g\\left(\\frac x\\varepsilon,\\frac y\\varepsilon\\right)$$, where $$g$$ is a $$1$$-periodic Lipschitz continuous function. When $$g$$ is periodic, this equation was studied by N. Dirr et al. [Eur. J. Appl. Math. 19, No. 6, 661–699 (2008; Zbl 1185.53076)]. They proved existence and uniqueness of planar pulsating waves in every direction of propagation.\nIn the present paper, the authors classify all possible singularities which can arise during the evolution and show that, when $$g$$ is smooth and the initial curve is embedded, the existence time of a regular solution is bounded below by a quantity depending only on $$\\| g\\|_\\infty$$ and on the initial curve. Since the authors have no estimates on the curvature in terms of $$\\| g\\|_\\infty$$, they are not able to obtain a general existence result for $$g \\in L^\\infty$$. However, by assuming that the initial curve is the graph of a function $$u$$ in the vertical direction, the equation becomes $u_t= \\frac {u_{xx}}{1+u_x^2} + g(x,u(x))\\sqrt{1+u_x^2}.$ By proving some estimates for $$u$$ depending only on $$\\| g\\|_\\infty$$, the authors obtain an existence and uniqueness result for solutions, when $$g$$ is an $$L^\\infty$$-function which is independent of $$u$$. As application to the homogenization problem, the limit as $$\\varepsilon \\to 0$$ is studied.\n\n##### MSC:\n 35K65 Degenerate parabolic equations 53C44 Geometric evolution equations (mean curvature flow, Ricci flow, etc.) (MSC2010) 53A04 Curves in Euclidean and related spaces 35B27 Homogenization in context of PDEs; PDEs in media with periodic structure 35K59 Quasilinear parabolic equations\nFull Text:"
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https://spark.apache.org/docs/2.0.0/api/scala/org/apache/spark/ml/linalg/package.html | [
"",
null,
"# linalg\n\nVisibility\n1. Public\n2. All\n\n### Type Members\n\n1. #### class DenseMatrix extends Matrix\n\nColumn-major dense matrix.\n\nColumn-major dense matrix. The entry values are stored in a single array of doubles with columns listed in sequence. For example, the following matrix\n\n```1.0 2.0\n3.0 4.0\n5.0 6.0```\n\nis stored as `[1.0, 3.0, 5.0, 2.0, 4.0, 6.0]`.\n\nAnnotations\n@Since( \"2.0.0\" )\n2. #### class DenseVector extends Vector\n\nA dense vector represented by a value array.\n\nA dense vector represented by a value array.\n\nAnnotations\n@Since( \"2.0.0\" )\n3. #### sealed trait Matrix extends Serializable\n\nTrait for a local matrix.\n\nTrait for a local matrix.\n\nAnnotations\n@Since( \"2.0.0\" )\n4. #### class SparseMatrix extends Matrix\n\nColumn-major sparse matrix.\n\nColumn-major sparse matrix. The entry values are stored in Compressed Sparse Column (CSC) format. For example, the following matrix\n\n```1.0 0.0 4.0\n0.0 3.0 5.0\n2.0 0.0 6.0```\n\nis stored as `values: [1.0, 2.0, 3.0, 4.0, 5.0, 6.0]`, `rowIndices=[0, 2, 1, 0, 1, 2]`, `colPointers=[0, 2, 3, 6]`.\n\nAnnotations\n@Since( \"2.0.0\" )\n5. #### class SparseVector extends Vector\n\nA sparse vector represented by an index array and a value array.\n\nA sparse vector represented by an index array and a value array.\n\nAnnotations\n@Since( \"2.0.0\" )\n6. #### sealed trait Vector extends Serializable\n\nRepresents a numeric vector, whose index type is Int and value type is Double.\n\nRepresents a numeric vector, whose index type is Int and value type is Double.\n\nNote: Users should not implement this interface.\n\nAnnotations\n@Since( \"2.0.0\" )\n\n### Value Members\n\n1. #### object DenseMatrix extends Serializable\n\nFactory methods for org.apache.spark.ml.linalg.DenseMatrix.\n\nAnnotations\n@Since( \"2.0.0\" )\n2. #### object DenseVector extends Serializable\n\nAnnotations\n@Since( \"2.0.0\" )\n3. #### object Matrices\n\nFactory methods for org.apache.spark.ml.linalg.Matrix.\n\nFactory methods for org.apache.spark.ml.linalg.Matrix.\n\nAnnotations\n@Since( \"2.0.0\" )\n4. #### object SQLDataTypes\n\n:: DeveloperApi :: SQL data types for vectors and matrices.\n\n:: DeveloperApi :: SQL data types for vectors and matrices.\n\nAnnotations\n@Since( \"2.0.0\" ) @DeveloperApi()\n5. #### object SparseMatrix extends Serializable\n\nFactory methods for org.apache.spark.ml.linalg.SparseMatrix.\n\nAnnotations\n@Since( \"2.0.0\" )\n6. #### object SparseVector extends Serializable\n\nAnnotations\n@Since( \"2.0.0\" )\n7. #### object Vectors\n\nFactory methods for org.apache.spark.ml.linalg.Vector.\n\nFactory methods for org.apache.spark.ml.linalg.Vector. We don't use the name `Vector` because Scala imports scala.collection.immutable.Vector by default.\n\nAnnotations\n@Since( \"2.0.0\" )"
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https://jp.maplesoft.com/support/help/maple/view.aspx?path=DEtools/riccatisol&L=J | [
"",
null,
"DEtools - Maple Programming Help\n\nHome : Support : Online Help : Mathematics : Differential Equations : DEtools : Solving Methods : DEtools/riccatisol\n\nDEtools\n\n riccatisol\n find solutions of a first order Riccati ODE\n\n Calling Sequence riccatisol(lode, v)\n\nParameters\n\n lode - first order differential equation v - dependent variable of the lode\n\nDescription\n\n • A Riccati ODE is a first order ODE of the form\n $y'\\left(x\\right)=\\mathrm{f0}\\left(x\\right)+\\mathrm{f1}\\left(x\\right)y\\left(x\\right)+\\mathrm{f2}\\left(x\\right){y\\left(x\\right)}^{2}$\n • The riccatisol routine determines whether the first argument is a first order Riccati ODE and, if so, returns a solution to the equation.\n • The first argument is a differential equation in diff or D form and the second argument is the variable in the differential equation.\n • This function is part of the DEtools package, and so it can be used in the form riccatisol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[riccatisol](..).\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{DEtools}\\right):$\n > $\\mathrm{ode}≔x\\mathrm{diff}\\left(z\\left(x\\right),x\\right)+z\\left(x\\right)=3{x}^{2}{z\\left(x\\right)}^{2}$\n ${\\mathrm{ode}}{≔}{x}{}\\left(\\frac{{ⅆ}}{{ⅆ}{x}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\\left({x}\\right)\\right){+}{z}{}\\left({x}\\right){=}{3}{}{{x}}^{{2}}{}{{z}{}\\left({x}\\right)}^{{2}}$ (1)\n > $\\mathrm{riccatisol}\\left(\\mathrm{ode},z\\left(x\\right)\\right)$\n $\\left\\{{z}{}\\left({x}\\right){=}\\frac{{1}}{\\left({-}{3}{}{x}{+}{\\mathrm{_C1}}\\right){}{x}}\\right\\}$ (2)\n > $\\mathrm{ode}≔x\\mathrm{D}\\left(z\\right)\\left(x\\right)+z\\left(x\\right)=3{x}^{2}{z\\left(x\\right)}^{2}-3:$\n > $\\mathrm{riccatisol}\\left(\\mathrm{ode},z\\left(x\\right)\\right)$\n $\\left\\{{z}{}\\left({x}\\right){=}\\frac{{-}{3}{+}\\frac{{{ⅇ}}^{{-}{6}{}{x}}}{{\\mathrm{_C1}}{+}\\frac{{{ⅇ}}^{{-}{6}{}{x}}}{{6}}}}{{3}{}{x}}\\right\\}$ (3)"
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"https://bat.bing.com/action/0",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6808893,"math_prob":0.9999938,"size":1059,"snap":"2020-45-2020-50","text_gpt3_token_len":408,"char_repetition_ratio":0.13459715,"word_repetition_ratio":0.01369863,"special_character_ratio":0.21246459,"punctuation_ratio":0.077922076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994883,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T17:47:10Z\",\"WARC-Record-ID\":\"<urn:uuid:5aea5451-a544-4e15-8ee2-cdde11fb7455>\",\"Content-Length\":\"181058\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:560ab440-70f8-4bf3-a3e0-82ebd41f2f3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c703a040-b34f-44b3-b380-c6635c1c31a0>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://jp.maplesoft.com/support/help/maple/view.aspx?path=DEtools/riccatisol&L=J\",\"WARC-Payload-Digest\":\"sha1:PPKMNNXPD3TCCRXKQJ5SK5QMA7T4FAM6\",\"WARC-Block-Digest\":\"sha1:QQX7D5WKAH6KALSMC6U76IZTPOTWQXKK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141193856.40_warc_CC-MAIN-20201127161801-20201127191801-00018.warc.gz\"}"} |
https://scholar.archive.org/search?q=key:work_dbu4tzoblzcatcc7yxkdoxqr3y | [
"Filters\n\n1 Hit in 0.048 sec\n\n### Column planarity and partially-simultaneous geometric embedding\n\nLuis Barba, William Evans, Michael Hoffmann, Vincent Kusters, Maria Saumell, Bettina Speckmann\n2017 Journal of Graph Algorithms and Applications\nWe introduce the notion of column planarity of a subset R of the vertices of a graph G. Informally, we say that R is column planar in G if we can assign x-coordinates to the vertices in R such that any assignment of y-coordinates to them produces a partial embedding that can be completed to a plane straightline drawing of G. Column planarity is both a relaxation and a strengthening of unlabeled level planarity. We prove near tight bounds for the maximum size of column planar subsets of trees:\nmore » ... subsets of trees: every tree on n vertices contains a column planar set of size at least 14n/17 and for any > 0 and any sufficiently large n, there exists an n-vertex tree in which every column planar subset has size at most (5/6 + )n. In addition, we show that every outerplanar graph has a column planar set of size at least n/2. We also consider a relaxation of simultaneous geometric embedding (SGE), which we call partially-simultaneous geometric embedding (PSGE). A PSGE of two graphs G 1 and G 2 allows some of their vertices to map to two different points in the plane. We show how to use column planar subsets to construct k-PSGEs, which are PSGEs in which at least k vertices are mapped to the same point for both graphs. In particular, we show that every two trees on n vertices admit an 11n/17-PSGE and every two outerplanar graphs admit an n/4-PSGE."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89177066,"math_prob":0.9419974,"size":876,"snap":"2021-21-2021-25","text_gpt3_token_len":205,"char_repetition_ratio":0.119266056,"word_repetition_ratio":0.0,"special_character_ratio":0.20547946,"punctuation_ratio":0.0797546,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9681242,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T08:12:23Z\",\"WARC-Record-ID\":\"<urn:uuid:ec75bc91-c956-41ba-8b19-ab8e5f681fb1>\",\"Content-Length\":\"24876\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8237c837-e3ee-4f54-8777-183ae2274e9c>\",\"WARC-Concurrent-To\":\"<urn:uuid:2b175f24-2f47-47cf-a760-39703f1ddef1>\",\"WARC-IP-Address\":\"207.241.225.9\",\"WARC-Target-URI\":\"https://scholar.archive.org/search?q=key:work_dbu4tzoblzcatcc7yxkdoxqr3y\",\"WARC-Payload-Digest\":\"sha1:KJTHWPPX45NXRE75POAJS4K6EK7FN4K7\",\"WARC-Block-Digest\":\"sha1:ZBKSAUMZQM4N3L2WHKVKYN555UCKDOUE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487622113.11_warc_CC-MAIN-20210625054501-20210625084501-00108.warc.gz\"}"} |
https://www.embeddedrelated.com/showthread/basicx/20071-1.php | [
"# DCF77\n\nStarted by October 27, 2005\nI need to measure incoming pulses of 100 and 200 milliseconds.\nThe PulseIn in not suitable for this, it's rangs is to short.\nBefore I start doing difficult things with the rtc I will ask if\nsomeone has another bright idea how to do a simple pulswidth\nmeasurement of pulses of this lenght.\n\nI do this on one of my projects. It is very simple but the accuracy is only +/- 4 ms. sse\n\ndo\nIf pin = high and sse then\nt=timer\ns=true\nend if\n\nIf pin=low and s=true then\npulsewidth=-timer-t\nsse\nend if\nloop\n\nIts accuracy is limited by the 2ms accuaracy of timer and it can stack in both directions to generate a +/-4 ms range. In my case that was fine.\n\nArt\n[] I need to measure incoming pulses of 100 and 200 milliseconds.\n\nHow about hooking the pulse up to an interrupt, then the interrupt\nshould loop while checking the pin state until it changes back.\n\nYou could benchmark the interrupt by manually triggering the\ninterrupt, and holding it for a fairly precise, long time and then\nread out the number of loop iterations.\n\nDo this for 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20.... seconds\nthen look at the loop iterations. There will be a small time\noverhead of the entry into the interrupt, but not much compared to\nmany milliseconds.\n\nBy simple comutation, you can calculate the decimal fraction of a\nsecond per loop. You could get more sophisticated and do a\nregression to calculate the seconds per loop, but unless you have a\nprecise mechanism to handle the pulse duration, the regression will\nnot be much better than averaging many 10 second trials.\n\nThen try it in real life and see what you get.\n\nA logic probe, or even a DMM can tell you the frequency and duty-\ncycle of the signal, if your signal is repetetive and fixed. An\noscilloscope would work too for the benchmarking.\n\n--- In basicx@basi..., \"wimn.rm\" <wimn@r...> wrote:\n>\n> I need to measure incoming pulses of 100 and 200 milliseconds.\n\n--- In basicx@basi..., \"wimn.rm\" <wimn@r...> wrote:\n> I need to measure incoming pulses of 100 and 200 milliseconds.\n\nYou may be able to use WaitForInterrupt. If you're measuring a high\npulse, use bxPinRisingEdge initially. When the interrupt occurs store\nthe value of Register.RTCTick and do another WaitForInterrupt using\nbxPinFallingEdge. When the interrupt occurs again the pulse width is\nthe difference between Register.RTCTick and the stored value time\n1.95mS. You'll have to add code to detect when RTCTick rolls over to\nzero. (Unfortunately, Register.RTCStopWatch is broken on the BX-24 -\nthat would be a much simpler solution.) Alternately, you could employ\nTimer1 to measure the elapsed time with better resolution. See the\nBasicX Timer1_Stopwatch_App_Note.pdf for detail on how to do this.\n\nIf you are decoding DCF77 (a German 77.5KHz timecode broadcast), you\ndon't need any precision, just a threshold to determine if the incoming\nbit is a one or zero. That bit can then shift into a 60-bit string to\ncontain the entire minute frame of data, or code can parse the bit\nstream on the fly.\n\nI believe that all you need to do is find the start of a bit and then\ndetermine if the bit period is >150mS. A missing bit signals bit 59 in\nanticipation of :00 - or loss of data. You can, of course, time the bit\nwith code, but another way to decode it is to charge a capacitor during\nthe high bit period - either spinning in code until the bit falls, or\ninterrupting the machine when the bit falls. Immediately sample the\nanalog charge on the cap to determine the bit state, then discharge it.\n\nSomewhere I have code for an MSF/DCF77/WWVB decoder that I wrote for\nDOS. If that is helpful, I'll hunt it down. Tom\n\n--- In basicx@basi..., \"wimn.rm\" <wimn@r...> wrote:\n> I need to measure incoming pulses of 100 and 200 milliseconds.\n\nOn further reflection, I assume that you're trying to decode the\npulse-width-modulated DCF77 signal. In that case, you might\nconsider using external hardware to discriminate between the pulse\nwidths.\n\nOne way to do that simply is to use a one-shot that is triggered by\nthe incoming pulse. The pulse width of the one-shot should be set\nto 150mS. Use the trailing edge of the one-shot's pulse to clock a\nD flop whose data input is the original pulse signal. If the\noriginal pulse was 100mS long the D flop will store a zero and if it\nwas the longer 200mS pulse it will store a 1.\n\nThe trailing edge of the one-shot's pulse could also set another\nwould then read the D flop output and then reset the \"data-ready\"\nflip-flop.\n\nI've put a quick sketch of such a circuit at:\nhttp://www.kinzers.com/don/BX24/dcf77.jpg\n\nThe proposed circuit requires 3 I/O pins on the BX-24 but you could\nreduce that to 2 by using the /RESET output to multiplex the DATA\nRDY and DATA signals.\n\nDon\n\nActually, now that I think about it, my MSF/DCF77/WWVB decoder is an\nassembly language TSR, so it will be of little use. I was thinking it\nwas a QuickBasic app that might be useful for structure, but I doubt\nthe assembly code will serve any useful purpose. Tom\n\nThat is a nice \"quick\" sketch! Do you just know the R & C values off\n\n-Tony\n\nDon, I'm close to the solution I think.\nI now use 2 do while loops.\n\nFirst waits for the raising edge. Next is starting a sleep of 120 mSec\nThen a check if it's still high.\nIf not we have a short ( 100mSec)\nIf yes we have a long ( 200mSec)\n\nWim\n[]\n\nWorks! so simple afterward. do\nCB = 0\n\n'Wait for a 1\ndo while (cb = 0)\nCB = GetPin(DCF77) 'Get clock output\nloop\n\ncall sleep(60) 'pass the 100mSec\nCB = GetPin(DCF77) 'Get clock output\nif (cb = 1 ) then 'still a 1 ?\nbit =1 'So it's a long pulse\nelse\nbit = 0 'It's a short pulse\nend if\ndebug.print cstr(bit)\ncall sleep(100) 'to prevent in enters the\n'do-while with a value of 1\nloop"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8818793,"math_prob":0.76214707,"size":11029,"snap":"2019-51-2020-05","text_gpt3_token_len":3023,"char_repetition_ratio":0.10031746,"word_repetition_ratio":0.88844424,"special_character_ratio":0.2645752,"punctuation_ratio":0.11130137,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9549402,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-09T10:14:25Z\",\"WARC-Record-ID\":\"<urn:uuid:ca7913c9-7053-483b-a228-a4e642994688>\",\"Content-Length\":\"49214\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:188719a8-47a4-4079-a464-f9d76810e356>\",\"WARC-Concurrent-To\":\"<urn:uuid:03adb97d-8066-42db-9ee0-121d771732a6>\",\"WARC-IP-Address\":\"64.64.12.128\",\"WARC-Target-URI\":\"https://www.embeddedrelated.com/showthread/basicx/20071-1.php\",\"WARC-Payload-Digest\":\"sha1:MX6DQGIQKHLZZYYLIPH6CGXC3ZFAAJRR\",\"WARC-Block-Digest\":\"sha1:QSFLBR2H45I24SUNT5D7B7SJEGGZXJBD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540518627.72_warc_CC-MAIN-20191209093227-20191209121227-00507.warc.gz\"}"} |
https://su-journal.com.ua/index.php/journal/article/view/167 | [
"# Chi-square, Student and Fisher-Snedecor Statistical Distributions and Their Application\n\n• F. V. Motsnyi National Academy of Statistics, Accounting and Audit\nKeywords: sample, general population, random value, the Chi-square distribution, Student distribution, Fisher – Snedecor distribution, numerical characteristics. \n\n### Abstract\n\nThe Chi-square distribution is the distribution of the sum of squared standard normal deviates. The degree of freedom of the distribution is equal to the number of standard normal deviates being summed. For the first time this distribution was studied by astronomer F. Helmert in connection with Gaussian low of errors in 1876. Later K. Pearson named this function by Chi-square. Therefore Chi –square distribution bears a name of Pearson’s distribution.\n\nThe Student's t-distribution is any member of a family of continuous probability distributions that arises when estimating the mean of a normally distributed population in situations where the sample size is small and population standard deviation is unknown. It was developed by W. Gosset in 1908.\n\nThe Fisher–Snedecor distribution or F-distribution is the ratio of two-chi-squared variates. The F-distribution provides a basis for comparing the ratios of subsetsof these variances associated with different factors. The Fisher-distribution in the analysis of variance is connected with the name of R.Fisher (1924), although Fisher himself used quantity for the dispersion proportion.\n\nThe Chi-square, Student and Fisher – Snedecor statistical distributions are connected enough tight with normal one. Therefore these distributions are used very extensively in mathematical statistics for interpretation of empirical data. The paper continues ideas of the author’s works [15, 16] devoted to advanced based tools of mathematical statistics. The aim of the work is to generalize the well known theoretical and experimental results of statistical distributions of random values. The Chi-square, Student and Fisher – Snedecor distributions are analyzed from the only point of view. The application peculiarities are determined at the examination of the agree criteria of the empirical sample one with theoretical predictions of general population. The numerical characteristics of these distributions are considered. The theoretical and experimental results are generalized. It is emphasized for the corrected amplification of the Chi-square, Student and Fisher – Snedecor distributions it is necessary to have the reliable empirical and testing data with the normal distribution.\n\n### References\n\n1. Gmurman, V. E. (1999). Teoriia veroiatnostei i matematicheskaia statistika [The probability theory and mathematic statistics]. Moscow: Vysshaia shkola (in Russian).\n3. Student. (1906). The probable error of a mean. Biometrika. Vol. 6, 1, 1–25. Retrieved from https://www.jstor.org/stable/2331554?seq=1#page_scan_tab_contents (in English).\n4. Taylor, C. Student’s t Distribution Formula (2017). thoughtco.com. Retrieved from https://www.thoughtco.com/students-t-distribution-formula-3126276 (in English).\n7. Lemeshko, B. Yu. (2014). Neparametricheskie kriterii soglasiia [Nonparametric criteria of agreement]. Moskow: INFRA-M. (in Russian).\n8. Funkciia F-raspredeleniia veroiatnosti [F-distribution function of probability]. help/prognoz.com. Retrieved from http://help/prognoz.com/ru/mergedProjects/Lib/05_statistics/distrib_func/lib_fdist.htm (in Russian).\n9. Raspredelenie Fishera (F-raspredelenie). Raspredeleniia v matematicheskoi statistike v MS Exсel [Fisher’s distribution(F-distribution). Distribution in math statistics in MS Excel]. excel2.ru. Retrieved from http://excel2.ru/articles/raspredelenie-fishera-f-raspredelenie-raspredeleniya-matematicheskoy-statistiki-v-ms-excel (in Russian).\n11. Raspredelenie Fisherа – Snedekora (F-raspredelenie) [The Fisher – Snedecor distribution (F-distribution)]. www.wikiznanie.ru. Retrieved from http://www.wikiznanie.ru/ru-wz/index.php/ (in Russian).\n12 Raspredelenie Fisherа-Snedekora (F-raspredelenie) [The Fisher – Snedecor distribution (F-distribution)]. matica.org.ua. Retrieved from http://matica.org.ua/metodichki-i-knigi-po-matematike/konspekt-lektcii-po-teorii-veroiatnostei-i-matematicheskoi-statistike-komogortcev-v-f/1-26-raspredelenie-fishera-snedekora-f-raspredelenie (in Russian).\n14. Forbes, C., Evans, M., Hastings, N., & Peacock, B. (2010). Statistical Distributions. John Wiley & Sons, Inc. Retrieved from onlinelibrary.wiley.com/doi/10.1002/9780470627242.ch20/summary (in English).\n15. Motsnyi, F. V. (2015). Suchasnyi bazovyi instrumentarii matematychnoi statystyky. Ch. 1: Osnovni poniattia matematychnoi statystyky [Advanced Based Tools of Mathematical Statistics. Part 1. Basic Concepts of Mathematical Statistic]. Naukovyi Visnyk Natsionalnoi academii statystyky, obliku ta audytu – Scientific Bulletin of the National Academy of Statistics, Accounting and Audit, 2, 16–29 (in Ukrainian).\nMotsnyi, F. V. (2015). Suchasnyi bazovyi instrumentarii matematychnoi statystyky. Ch. 2: Vybirkovi kharakterystyky statystychnykh rozpodiliv [Advanced Based Tools of Mathematical Statistics. Part 2. Sample Characteristics of Statistics Distributions]. Naukovyi Visnyk Natsionalnoi academii statystyky, obliku ta audytu – Scientific Bulletin of the National Academy of Statistics, Accounting and Audit, 3, 14–25 (in Ukrainian).\n17. Gnedenko, B. V. (2005). Kurs teorii veroiatnostei [The probability course]. (8th ed.). Moscow: Editorial URSS. Retrieved from //www.alleng.ru/d/math/math333.htm (in Russian).\n18. Tutubalin, V. N. (1992). Teoriia veroiatnostei i sluchainych processov. Osnovy matematicheskogo apparata i prikladnye aspekty [Probability theory and random processes. The base of mathematical apparatus and applied aspects]. Moscow: Izd-vo MGU. Retrieved from http://booksshare.net/index.php?id1=4&category=biol&author=tutubalin-vn&book=1992 (in Russian).\n\nAbstract views: 198"
]
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https://www.rsmanuals.com/29951/motorola-m68060/page-134/ | [
"",
null,
"Floating-Point Unit\n6-16 M68060 USER’S MANUAL MOTOROLA\nIf no underflow occurs, the intermediate result is rounded according to the user-selected\nrounding precision and rounding mode. After rounding, the INEX2 bit of the FPSR EXC byte\nis set accordingly. Finally, the magnitude of the result is checked to see if it is too large to\nbe represented in the current rounding precision. If so, the OVFL bit of the FPSR EXC byte\nis set, and the MC68060 takes a nonmaskable overflow exception and executes the\nM68060SP overflow exception handler. The M68060SP returns a correctly signed infinity or\na correctly signed largest normalized number, depending on the rounding mode in effect.\n6.4.2 Conditional Testing\nUnlike the integer arithmetic condition codes, an instruction either always sets the floating-\npoint condition codes in the same way or it does not change them at all. Therefore, the\ninstruction descriptions do not include floating-point condition code settings. The following\nparagraphs describe how floating-point condition codes are set for all instructions that mod-\nify condition codes. Refer to 6.1.3.1 Floating-Point Condition Code Byte for a description\nof the FPCC byte.\nThe data type of the operation’s result determines how the four condition code bits are set.\nTable 6-8 lists the condition code bit setting for each data type. The MC68060 generates\nonly eight of the 16 possible combinations. Loading the FPCC with one of the other combi-\nnations and executing a conditional instruction can produce an unexpected branch condi-\ntion.\nThe inclusion of the NAN data type in the IEEE floating-point number system requires each\nconditional test to include the NAN condition code bit in its Boolean equation. Because a\ncomparison of a NAN with any other data type is unordered (i.e., it is impossible to determine\nif a NAN is bigger or smaller than an in-range number), the compare instruction sets the\nNAN condition code bit when an unordered compare is attempted. All arithmetic instructions\nalso set the FPCC NAN bit if the result of an operation is a NAN. The conditional instructions\ninterpret the NAN condition code bit equal to one as the unordered condition.\nThe IEEE 754 standard defines four conditions: equal to (EQ), greater than (GT), less than\n(LT), and unordered (UN). In addition, the standard only requires the generation of the con-\ndition codes as a result of a floating-point compare operation. The FPU tests for these con-\nditions and 28 others at the end of any operation affecting the condition codes. For purposes\nof the floating-point conditional branch, set byte on condition, decrement and branch on con-\ndition, and trap on condition instructions, the MC68060 logically combines the four FPCC\nbits to form 32 conditional tests. The 32 conditional tests are separated into two groups—16\nTable 6-8. Floating-Point Condition Code Encoding\nData Type N Z I NAN\n+ Normalized or Denormalized 0000\n– Normalized or Denormalized 1000\n+ 0 0100\n– 0 1100\n+ Infinity 0010\n– Infinity 1010\n+ NAN 0001\n– NAN 1001"
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"https://www.rsmanuals.com/src-html/b/bt0ip72cye/bg86.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8254492,"math_prob":0.9603193,"size":3035,"snap":"2021-04-2021-17","text_gpt3_token_len":681,"char_repetition_ratio":0.16661169,"word_repetition_ratio":0.018367346,"special_character_ratio":0.22207578,"punctuation_ratio":0.086330935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9683086,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-27T10:48:53Z\",\"WARC-Record-ID\":\"<urn:uuid:f485605e-9eaf-4ead-a27c-0eec94a21cce>\",\"Content-Length\":\"173209\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed921697-c580-4fed-969a-718ce90d1680>\",\"WARC-Concurrent-To\":\"<urn:uuid:56d3b9a6-2ce0-438e-a739-a82cd8f2211b>\",\"WARC-IP-Address\":\"66.70.179.241\",\"WARC-Target-URI\":\"https://www.rsmanuals.com/29951/motorola-m68060/page-134/\",\"WARC-Payload-Digest\":\"sha1:5XRY3DQ4YB33QE5Q6NI5JR6GILMNFOBO\",\"WARC-Block-Digest\":\"sha1:XC2ZBTKVYHLEWUEPPAKJSXCNAPYJXNQR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704821381.83_warc_CC-MAIN-20210127090152-20210127120152-00781.warc.gz\"}"} |
http://reducesuite.bussemakerlab.org/documentation/optimizepsam/?action=printpage | [
"# REDUCE Suite\n\n## General Category => Documentation => Topic started by: xiangjun on September 29, 2016, 12:41:18 pm\n\nTitle: OptimizePSAM\nPost by: xiangjun on September 29, 2016, 12:41:18 pm\nThis program performs a single point optimization of either an initial (pseudo-) PSAM or a seed motif against the measurement file and sequence. Internally, it uses exactly the same Levenberg-Marquardt non-linear least squares fitting algorithm as in MatrixREDUCE (http://reducesuite.bussemakerlab.org/documentation/matrixreduce/).\n\n`OptimizePSAM [options] -sequence=seqfile -measurement=measfile \\ -psam=PSAM_file | -motif=IUPAC_Motif Required parameters: -sequence=file_name --- name of a FASTA sequence file -meas=measfile --- measurement (expression/binding) file in tab-delimited format -psam=PSAM_file --- PSAM file to be optimized -motif=seed_motif --- Seed IUPAC motif sequence to be optimized Optional parameters: [-output=dir_name] --- path to the output directory (./) [-p_value=float] --- threshold to decide if optimized PSAM is significant (0.001) [-filename=file] --- name of the optimized PSAM [-strand=integer] --- 1 |+1 |F | L for leading strand (1); 2 |+2 |B for both strands; -1 | R |C for reverse complementary; 0 | A |D auto-detection (check 1 and 2) [-runlog=[stderr|stdout|file]] --- direct running diagnostics message to stderr, stdout or a file (stderr) [-help] --- print out this help message Usage: OptimizePSAM \\ -measurement=\\$REDUCE_SUITE/data/mRNA_expression/Spellman1998AlphaTimeCourse.tsv \\ -sequence=\\$REDUCE_SUITE/data/sequence/YeastUpstream.fasta \\ -motif=ACGCGT -file=ACGCGT.psam`\n\nNotes:\n\n• In PSAM format, the initial seed motif ACGCGT is expressed as follows:\n`# A C G T # no. opt# +============+============+============+============ # ==+===+== 1 0 0 0 # 1 A 0 1 0 0 # 2 C 0 0 1 0 # 3 G 0 1 0 0 # 4 C 0 0 1 0 # 5 G 0 0 0 1 # 6 T`\n\n• The optimized PSAM in file ACGCGT.psam is as follows. Note specifically the changes of Ws from 1s and 0s of initial sequence motif (above) to some fractions with a maximum of 1 in each position in the optimized PSAM.\n`# A C G T # no. opt# +============+============+============+============ # ==+===+== 1 0.143386 0.156974 0.332267 # 1 A 2.38995e-06 1 0.133257 6.623e-18 # 2 C 0.203947 0.0136109 1 6.43632e-11 # 3 G 3.39323e-14 1 4.38946e-14 3.19148e-17 # 4 C 0.0655988 0.122631 1 1.13119e-13 # 5 G 0.422826 0.221149 0.182984 1 # 6 T`\n\n• As shown in the following diagnostic message from running OptimizePSAM, this optimization step increases the fitted R2 from 0.0414328 to 0.0552883, and the PSAM is significant.\n`Best seed experiment: number of tested candidate experiments: 18 intercept: coef=-0.12248 t-value=-18.4713 p-value=5.77026e-74 slope: coef=+0.363975 t-value=+15.4353 p-value=1.18198e-52 r2=0.0414328 SSY=1323.85 SSE=1269 SSR=54.8506 matches[matched-ids/total-ids]: 348[307/5514] experiment: alpha_factor_release_sample016 and of sequence on forward strandOptimizing: 20 (1250.65): converged with gradient: 0.0349271 <= 0.05PSAM linear fit statistics: intercept: coef=-0.186363 t-value=-23.1987 p-value=1.12291e-113 slope: coef=+0.325095 t-value=+17.9606 p-value=3.83258e-70 r2=0.0552883 SSY=1323.85 SSE=1250.65 SSR=73.1932Checking PSAM significance: |r|=0.235135 r0=0.0688157 sigma=0.00888813 t_value=18.7125 E-value=4.19638e-76 This PSAM is significant (E-value smaller than specific cutoff of 0.001)`"
]
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https://www.demo2s.com/information-technologies/sql-server-choose.html | [
"# SQL Server CHOOSE\n\nAnother logical function introduced in SQL Server 2012 is the CHOOSE function.\n\nThe CHOOSE function allows you to select a member of an array based on an integer index value.\n\nCHOOSE lets you select a member from a list.\n\nThe member you select can be based on either a static index value or a computed value.\n\nThe syntax for the CHOOSE function is as follows:\n\n```CHOOSE ( index, val_1, val_2 [, val_n ] )\n\n```\n\nIf the index value isn't an integer (let's say it's a decimal), then SQL converts it to an integer.\n\nIf the index value is out of range for the index, then the function returns NULL.\n\nThe following example uses the integer value of PhoneNumberTypeID to determine the type of phone.\n\nIn this case, the phone type is defined in the table, so a CHOOSE function wouldn't be necessary; but in other cases, the value may not be defined.\n\nExample Using the CHOOSE Statement\n\n```SELECT p.FirstName,\npp.PhoneNumber,\nCHOOSE(pp.PhoneNumberTypeID, 'Cell', 'Home', 'Work') 'Phone Type'\nFROM Person.Person p\nJOIN Person.PersonPhone pp"
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http://www.naic.edu/~phil/calsbn/sbnsefdRatioDec02.html | [
"# SBN SEFD RATIO POLA/POLB\n\n#### Dec 2002\n\nThe system temperatures for polA,polB on sbn differ by about 5K (dec 2002). This could be a problem in the cal values or it could be a real difference in the system temperatures. To test this, 3 sources were tracked using the heiles calibration scans on 23,24 dec02. The Tsys and source deflection for each Pol were recreated ((I+Q)/2 , (I-Q)/2) for each strip of each pattern. The SEFD was then computed for each polarization. These values are independent of the cals. The ratio sefdA/sefdB is also independent of the source flux. It only depends on the Tsys, gain of the telescope, and polarization of the source. Since sbn is native circular we assume that there is equal source flux in each polarization. If the cals are correct, then the source deflection should be the same for each pol while the system temperatures should differ. The plots show the Tsys, Tsrc, and sefd ratio.\n• Top figure has the Tsys for all strips. Black in polA and read is polB. They differ by about 5K.\n• Center figure is the source deflection in K. Pol A and polB are over plotted. The values for the 2 polarizations lie on top of one another.\n• Bottom Figure shows the ratio of SEFDA/SEFDB vs za. A linear fit to the data versus za gives an average value of 1.226. The ratio of TsysA/TsysB is also plotted in green. This data falls within 1% of the sefdRatio.\n• The source deflection and sefd ratio show that the relative cal values of sbn are correct and that the system temperature difference is real.\n\nprocessing:x101/021223/dosbn.pro\nhome_~phil"
]
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https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=14714 | [
"# Chemistry Test 3\n\nThe flashcards below were created by user mLy on FreezingBlue Flashcards.\n\n1. What is Don's circumference?\n100 cm or 1 meter\n\n2. What is the circumference of the Basketball?\n75 cm\n3. What is the mass and weight of the text book?\n• Mass: 2 kg\n• Weight: 20 N\n4. What is the mass of 1 litre of water?\n1 kg\n5. What is the centimeter of a Post It Note?\n5 cm\n6. What is the mass and weight of soda?\n• Mass: .4 kg\n• Weight: 4 N\n7. What is the mass and weight of the Basketball?\n• Mass: .5 kg\n• Weight: 5 N\n8. How many atoms are between the finger and thumb?\n10,000\n9. How much nitrogen are in the air?\n20%\n10. What is the size of Don's shoes?\n30 cm\n Author: mLy ID: 14714 Card Set: Chemistry Test 3 Updated: 2010-04-15 23:41:00 Tags: Chemistry Folders: Description: Notes test 3 Show Answers:"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9397437,"math_prob":0.6603375,"size":530,"snap":"2019-43-2019-47","text_gpt3_token_len":142,"char_repetition_ratio":0.21292776,"word_repetition_ratio":0.1574074,"special_character_ratio":0.28679246,"punctuation_ratio":0.10619469,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96334994,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T04:57:20Z\",\"WARC-Record-ID\":\"<urn:uuid:db91e885-de63-4f77-a85e-a3aa6aec613e>\",\"Content-Length\":\"13295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d45d9afc-71bd-4ab2-bed4-2fdeef46fb40>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9be5fee-472d-48d5-bdcc-0ea0966305cf>\",\"WARC-IP-Address\":\"34.198.188.69\",\"WARC-Target-URI\":\"https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=14714\",\"WARC-Payload-Digest\":\"sha1:PYKILMBSJDIUWJUSRZA2EMRHXRQXHV4Q\",\"WARC-Block-Digest\":\"sha1:YGC6W6H3D3PHQOMT3EKK47LDJLHBQSEB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671239.99_warc_CC-MAIN-20191122042047-20191122070047-00256.warc.gz\"}"} |
https://docs.appian.com/suite/help/19.3/fnc_mathematical_combin.html | [
"combin() Function\n\nCalculates the number of unique ways to choose m elements from a pool of n elements.\n\n## Syntax\n\ncombin( n, m )\n\nn: (Integer) The number of elements that can be chosen from.\n\nm: (Integer) The number of elements that will be chosen from the pool of n elements.\n\nDecimal\n\n## Examples\n\nYou can experiment with this function in the test box below.\n\nTest Input\n\n`combin(4,2)` returns `6`\n\nOpen in Github\n\nOn This Page"
]
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null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6977669,"math_prob":0.9646418,"size":360,"snap":"2021-04-2021-17","text_gpt3_token_len":91,"char_repetition_ratio":0.16292135,"word_repetition_ratio":0.06349207,"special_character_ratio":0.2361111,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97889996,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-14T19:41:15Z\",\"WARC-Record-ID\":\"<urn:uuid:45bf7348-e184-4b09-98d5-5b7d222a81ac>\",\"Content-Length\":\"25185\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe278ba0-ec38-4773-96af-a72c1ccea397>\",\"WARC-Concurrent-To\":\"<urn:uuid:7418bf77-3d35-4b10-9f04-162726fb6e27>\",\"WARC-IP-Address\":\"34.236.202.118\",\"WARC-Target-URI\":\"https://docs.appian.com/suite/help/19.3/fnc_mathematical_combin.html\",\"WARC-Payload-Digest\":\"sha1:UFDPOC6DHD2SZSKNFCXZ4ADKT77IAUW7\",\"WARC-Block-Digest\":\"sha1:V6TUJRJNMXUBGKB7ZYFWBMVQSNMJXTWK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038078021.18_warc_CC-MAIN-20210414185709-20210414215709-00239.warc.gz\"}"} |
https://homework.cpm.org/category/CCI_CT/textbook/cc4/chapter/5/lesson/5.2.1/problem/5-53 | [
"",
null,
"",
null,
"### Home > CC4 > Chapter 5 > Lesson 5.2.1 > Problem5-53\n\n5-53.\n\nDraw a slope triangle and use it to write the equation of the line shown in the graph below. Homework Help ✎",
null,
"Use the same method you used in problem 5-48.\n\n$y = −10x + 170$"
]
| [
null,
"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
"data:image/png;base64,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",
null,
"https://s3-us-west-2.amazonaws.com/c3po-media-dev/files/9097d340-0280-11ea-b5be-0f890d5933f6/Screen Shot 2019-11-08 at 3_original.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.82443345,"math_prob":0.86820936,"size":368,"snap":"2019-43-2019-47","text_gpt3_token_len":103,"char_repetition_ratio":0.11263736,"word_repetition_ratio":0.7605634,"special_character_ratio":0.28804347,"punctuation_ratio":0.085365854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9608342,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T21:34:24Z\",\"WARC-Record-ID\":\"<urn:uuid:36b9d6f4-9cbb-4f8f-9455-a691bf933679>\",\"Content-Length\":\"33658\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:838c2e23-d1da-4a74-addd-a3cfa30f86a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e72c36c-eaf1-4986-899b-24674c3de83c>\",\"WARC-IP-Address\":\"104.26.7.16\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/CCI_CT/textbook/cc4/chapter/5/lesson/5.2.1/problem/5-53\",\"WARC-Payload-Digest\":\"sha1:YASJ2XGP2LZTOXGVPBWUXV3F7BENWEWQ\",\"WARC-Block-Digest\":\"sha1:2PGSDJ3IAJCCK472DIZDO72ADJQQURAO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669276.41_warc_CC-MAIN-20191117192728-20191117220728-00398.warc.gz\"}"} |
https://physics.stackexchange.com/questions/317837/why-do-the-specific-heat-constants-vary-with-temperature-for-semi-perfect-gases | [
"# Why do the specific heat constants vary with temperature for semi-perfect gases?\n\nWe were taught that the difference between the perfect and semi-perfect lies in that the specific heat constants for perfect gases are constant, while for semi-perfect gases they are functions of temperature and temperature only. However, both perfect and semi-perfect gases are a subset of ideal gases for which the main assumption is that the intermolecular forces are negligible. For that reason changes in potential energy associated with these forces are negligible as well.\n\nIf that is the case, what changes in a semi-perfect gas with temperature so that the same heat input does not equal the same temperature raise? After all, from my understanding of temperature it is directly proportional to average kinetic energy of the gas molecules, and hence should vary linearly with heat input.\n\nPoly-atomic molecules have different modes of vibration or rotation, each equivalent to a quantum harmonic oscillator or quantum rotor. The modes are quantized. For vibrations at natural frequency $\\omega$ the energy levels are uniformly spaced : $E_n=(n+\\frac12)\\hbar \\omega$. For a rigid rotor of moment of inertia $I$ the energy levels are quadratically spaced : $E_l=l(l+1)\\frac{\\hbar^2}{2 I}$.\nEach mode has a different minimum amount of energy. When the gas is in thermal equilibrium each mode holds $\\frac12kT$ of energy. So a vibration mode will not be excited in a significant fraction of molecules until $\\frac12kT \\ge \\frac12\\hbar \\omega$.",
null,
""
]
| [
null,
"https://i.stack.imgur.com/2o2Vx.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9547035,"math_prob":0.99079525,"size":795,"snap":"2019-51-2020-05","text_gpt3_token_len":148,"char_repetition_ratio":0.15549937,"word_repetition_ratio":0.0,"special_character_ratio":0.17610063,"punctuation_ratio":0.07092199,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9963905,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-14T20:57:02Z\",\"WARC-Record-ID\":\"<urn:uuid:1bb8a148-af78-4319-ab05-6ff810d5e956>\",\"Content-Length\":\"132093\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d1ec484-5c29-41b7-a52a-3b4e9922c588>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b9af52f-478f-4a3b-90b6-89fec06c4112>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/317837/why-do-the-specific-heat-constants-vary-with-temperature-for-semi-perfect-gases\",\"WARC-Payload-Digest\":\"sha1:FMA5AO55KUECHJ46A7NAUHOO7KOR6WMG\",\"WARC-Block-Digest\":\"sha1:CLF73KCQAES3XN5BJVHWHRWZXOF7L46X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541294513.54_warc_CC-MAIN-20191214202754-20191214230754-00024.warc.gz\"}"} |
https://forum.uipath.com/t/how-to-use-group-by-and-get-sum/348061 | [
"",
null,
"How to use group by and get sum?\n\nHi Everyone,\n\nI need help about my automation problem.\n\nRight now I have the following data below:",
null,
"As you can see I would like to check if grouping the column Store code and adding sum is possible in UiPath.\n\nRegards\n\nAA\n\nWelcome to the community.\n\n1. Store your input data in a data table called dt1\n2. Create one more datable called dt2 where can store output. Build this data table with two columns namely STORE CODE, TOTAL\n3. Now use below linq\n(From row in dt1.AsEnumerable\nGroup row by sc = row(“STORE CODE”).ToString()\nLet total= grp.sum(Function (x) Convert.ToDouble(x(“CREDIT”).ToString().Replace(\"\\$\",\"\").Trim()))\nLet result = New-Object() {grp(0)(“STORE CODE”),total}\n\ngrp.sum(Function...) I dont see grp being declared. Do you mean to refer sc\nJust wondering how would linq know to refer to a keyword grp. Is it a reserved keyword in linq, which we can use when grouping?\n\nShould it not be like below?\n\n(From row in dt1.AsEnumerable\nGroup row by sc = row(“STORE CODE”).ToString()\nLet total= sc.sum(Function (x) Convert.ToDouble(x(“CREDIT”).ToString().Replace(\"\\$\",\"\").Trim()))\nLet result = New-Object() {sc(0)(“STORE CODE”),total}\n1 Like\n\nHow will I encode the linq code?\n\nI am going to do assign activity?\n\nRegards,\n\nAA\n\nHi @jeevith\n\nFrom my knowledge, even sc is a temporary variable used to loop through the particular column and group the rows,\n\nWe have to assign the grouped rows into a variable by using into grp = Group…now grp (each item will have multiple rows) variable will have the grouped rows\n\n(From row in dt1.AsEnumerable\nGroup row by sc = row(“STORE CODE”).ToString() into grp = Group\nLet total= grp.sum(Function (x) Convert.ToDouble(x(“CREDIT”).ToString().Replace(\"\\$\",\"\").Trim()))\nLet result = New-Object() {grp(0)(“STORE CODE”),total}\n\nThanks\n\nThat is good to know. So grp has to be declared to be of type Group before we perform operations on it. In into grp = Group I am assuming Group is a reserved word in linq.\n\nI am learning linq only by reading posts in the community, so wanted to know how this particular query works. Thank you for the explanation.\n\n2 Likes\n\nHi @jeevith\n\n@prasath_S is correct. I forgot to use into grp=Group\n\n1 Like\n\nYes. you should use assign activity.\n\nIn the left hand side of assign acitivity you should use dt2, in the right hand side you should use Linq.\n\nAlso see below updated linq:\n(From row in dt1.AsEnumerable\nGroup row by sc = row(“STORE CODE”).ToString()\nInto grp = Group\nLet total= grp.sum(Function (x) Convert.ToDouble(x(“CREDIT”).ToString().Replace(\"\\$\",\"\").Trim()))\nLet result = New-Object() {grp(0)(“STORE CODE”),total}"
]
| [
null,
"https://aws1.discourse-cdn.com/uipath/original/3X/d/6/d6c862cebbdb9f139d9b3e0b4346f6d471039722.png",
null,
"https://aws1.discourse-cdn.com/uipath/original/3X/2/b/2bf95cd4223e2f44727ddb6891331f4154997708.png",
null
]
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http://curious.astro.cornell.edu/our-solar-system/the-moon/147-people-in-astronomy/careers-in-astronomy/being-an-astronomer/913-how-do-astronomers-use-math-in-their-jobs-beginner | [
"## How do astronomers use math in their jobs? (Beginner)\n\nHi... I was just curious as to how much math you use? In what ways is math involved in astronomy?\n\nAstronomers use math all the time. One way it is used is when we look at objects in the sky with a telescope. The camera that is attached to the telescope basically records a series of numbers - those numbers might correspond to how much light different objects in the sky are emitting, what type of light, etc. In order to be able to understand the information that these numbers contain, we need to use math and statistics to interpret them.\n\nAnother way that astronomers use math is when we are forming and testing theories for the physical laws that govern the objects in the sky. Theories consist of formulas that relate quantities to each other. (A very simple example is Newton's second law, force equals mass times acceleration.) In order to be able to test these theories and use them to make predictions about what we will observe in the sky, astronomers need to use math to manipulate the equations.\n\nHere are some answers to similar questions from other \"Ask an Astronomer\" sites:\n\n#### Dave Rothstein\n\nDave is a former graduate student and postdoctoral researcher at Cornell who used infrared and X-ray observations and theoretical computer models to study accreting black holes in our Galaxy. He also did most of the development for the former version of the site.\n\n## Our Reddit AMAs\n\nAMA = Ask Me (Us) Anything"
]
| [
null
]
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https://artofproblemsolving.com/wiki/index.php/Filter | [
"# Filter\n\nA filter on a set",
null,
"$X$ is a structure of subsets of",
null,
"$X$.\n\n## Definition\n\nLet",
null,
"$\\mathcal{F}$ be a set of subsets of",
null,
"$X$. We say that",
null,
"$\\mathcal{F}$ is a filter on",
null,
"$X$ if and only if each of the following conditions hold:\n\n• The empty set is not an element of",
null,
"$\\mathcal{F}$.\n• If",
null,
"$A$ and",
null,
"$B$ are subsets of",
null,
"$X$,",
null,
"$A$ is a subset of",
null,
"$B$, and",
null,
"$A$ is an element of",
null,
"$\\mathcal{F}$, then",
null,
"$B$ is an element of",
null,
"$\\mathcal{F}$.\n• The intersection of two elements of",
null,
"$\\mathcal{F}$ is an element of",
null,
"$\\mathcal{F}$.\n\nIt follows from the definition that the intersection of any finite family of elements of",
null,
"$\\mathcal{F}$ is also an element of",
null,
"$\\mathcal{F}$. Also, if",
null,
"$A$ is an element of",
null,
"$\\mathcal{F}$, then its complement is not.\n\nMore generally, one can define a filter on any Partially ordered set",
null,
"$(P,\\leq)$: Let",
null,
"$F$ be a subset of",
null,
"$P$. We say",
null,
"$F$ is a filter if and only if\n\n•",
null,
"$F\\neq\\emptyset$.\n• For all",
null,
"$x,y\\in F$, there exists",
null,
"$z\\in F$ such that",
null,
"$z\\leq x$ and",
null,
"$z\\leq y$.\n• If",
null,
"$x\\in F$ and",
null,
"$x\\leq y\\in P$, then",
null,
"$y\\in F$.\n\nA filter on a set",
null,
"$S$ is a filter on the poset",
null,
"$(\\mathcal{P}(S),\\subseteq)$.\n\n## Examples\n\nLet",
null,
"$Y$ be a subset of",
null,
"$X$. Then the set of subsets of",
null,
"$X$ containing",
null,
"$Y$ constitute a filter on",
null,
"$X$.\n\nIf",
null,
"$X$ is an infinite set, then the subsets of",
null,
"$X$ with finite complements constitute a filter on",
null,
"$X$. This is called the cofinite filter, or Fréchet filter."
]
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
"https://latex.artofproblemsolving.com/f/f/5/ff5fb3d775862e2123b007eb4373ff6cc1a34d4e.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null,
"https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ",
null,
"https://latex.artofproblemsolving.com/f/f/5/ff5fb3d775862e2123b007eb4373ff6cc1a34d4e.png ",
null,
"https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ",
null,
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null,
"https://latex.artofproblemsolving.com/f/f/5/ff5fb3d775862e2123b007eb4373ff6cc1a34d4e.png ",
null,
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null,
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null,
"https://latex.artofproblemsolving.com/7/d/f/7df6156885f7ecb884a812ae7155a0ad3ea33684.png ",
null,
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null,
"https://latex.artofproblemsolving.com/7/d/f/7df6156885f7ecb884a812ae7155a0ad3ea33684.png ",
null,
"https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ",
null,
"https://latex.artofproblemsolving.com/7/d/f/7df6156885f7ecb884a812ae7155a0ad3ea33684.png ",
null,
"https://latex.artofproblemsolving.com/f/e/9/fe9c9d1fa1b65febda0ed693b722cbdda3fc07ca.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/4/b/4/4b4cade9ca8a2c8311fafcf040bc5b15ca507f52.png ",
null,
"https://latex.artofproblemsolving.com/a/0/5/a055f405829e64a3b70253ab67cb45ed6ed5bb29.png ",
null,
"https://latex.artofproblemsolving.com/0/5/f/05fd5b07c5f332ff1a0257615917c015df4892ed.png ",
null,
"https://latex.artofproblemsolving.com/b/8/b/b8b3a2f7ad061f547640bd2d9d2508aec8d3744b.png ",
null,
"https://latex.artofproblemsolving.com/8/c/6/8c62ff350a6da89adc67cc88820f037002c443c5.png ",
null,
"https://latex.artofproblemsolving.com/c/c/b/ccb38d22353c40bd02f81b5ce45242b980a719eb.png ",
null,
"https://latex.artofproblemsolving.com/3/7/a/37a6667855061dda5124b1c8c5bf9de847b36523.png ",
null,
"https://latex.artofproblemsolving.com/d/9/6/d9619f62df3cbeadedb4a1a7a2b93678789190eb.png ",
null,
"https://latex.artofproblemsolving.com/3/d/b/3db6c5c05a0a021640e060d0f9d572d8a099317d.png ",
null,
"https://latex.artofproblemsolving.com/d/4/e/d4e6c0570ebcae770e4682bb42b8afacc428461e.png ",
null,
"https://latex.artofproblemsolving.com/a/d/2/ad28c83c99a8fd0dd2e2e594c9d02ee532765a0a.png ",
null,
"https://latex.artofproblemsolving.com/5/a/4/5a4df652656b1ad0a13b535b9f0f5d26d62a1610.png ",
null,
"https://latex.artofproblemsolving.com/c/e/5/ce58e4af225c93d08606c26554caaa5ae32edeba.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null,
"https://latex.artofproblemsolving.com/c/e/5/ce58e4af225c93d08606c26554caaa5ae32edeba.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null,
"https://latex.artofproblemsolving.com/6/a/4/6a47ca0fe7cb276abc022af6ac88ddae1a9d6894.png ",
null
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https://mathhelper.us/1over8/plus/5over7 | [
"",
null,
"# What is 1/8 + 5/7?",
null,
"Here's how to add\n\n 1 8\n+\n 5 7\n\n## Step 1\n\nWe still have different denominators (bottom numbers), though, so we need to get a common denominator. This will make the bottom numbers match. Multiply the denominators together first. Now, multiply each numerator by the other term's denominator.\n\nNow we multiply 1 by 7, and get 7, then we multiply 8 by 7 and get 56.",
null,
"Now for the second term. You multiply 5 by 8, and get 40, then multiply 8 by 7 and get 56.",
null,
"This gives us a new problem that looks like so:\n\n 7 56\n+\n 40 56\n\n## Step 2\n\nSince our denominators match, we can add the numerators.\n\n7 + 40 = 47\n\nSo what's the answer so far?\n\n 47 56\n\n## Step 3\n\nCan this fraction be reduced?\n\nFirst, we attempt to divide it by 2...\n\nNope! So now we try the next greatest prime number, 3...\n\nNope! So now we try the next greatest prime number, 5...\n\nNope! So now we try the next greatest prime number, 7...\n\nNope! So now we try the next greatest prime number, 11...\n\nNope! So now we try the next greatest prime number, 13...\n\nNope! So now we try the next greatest prime number, 17...\n\nNope! So now we try the next greatest prime number, 19...\n\nNope! So now we try the next greatest prime number, 23...\n\nNope! So now we try the next greatest prime number, 29...\n\nNope! So now we try the next greatest prime number, 31...\n\nNope! So now we try the next greatest prime number, 37...\n\nNope! So now we try the next greatest prime number, 41...\n\nNope! So now we try the next greatest prime number, 43...\n\nNope! So now we try the next greatest prime number, 47...\n\nNope! So now we try the next greatest prime number, 53...\n\nNo good. 53 is larger than 47. So we're done reducing.\n\nThere you have it! Here's the final answer to 1/8 + 5/7\n\n 1 8\n+\n 5 7\n=\n 47 56\n© 2014 Randy Tayler"
]
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null,
"https://mathhelper.us/img/math-helper.png",
null,
"https://mathhelper.us/fraction_image_vertical.php",
null,
"https://mathhelper.us/multiply_top_and_bottom.php",
null,
"https://mathhelper.us/multiply_top_and_bottom.php",
null
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https://byjus.com/ncert-solutions-for-class-6-maths-chapter-8-decimals-ex-8-6/ | [
"",
null,
"# NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.6\n\nNCERT Solutions For Class 6 Maths Chapter 8 Decimals Exercise 8.6 covers the topics of subtraction of decimals. To make the students gain expertise in these concepts, a set of examples are solved before the exercise questions. Students can clear their doubts instantly by using PDF of NCERT Solutions prepared by experts at BYJU’S. The solutions are explained in the best possible way in order to help students ace the exam.\n\n## NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.6 Download PDF",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"### Access NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.6\n\n1. Subtract:\n\n(a) ₹ 18.25 from ₹ 20.75\n\n(b) 202.54 m from 250 m\n\n(c) ₹ 5.36 from ₹ 8.40\n\n(d) 2.051 km from 5.206 km\n\n(e) 0.314 kg from 2.107 kg\n\nSolutions:\n\n(a) ₹ 20.75 – ₹ 18.75\n\n20.75\n\n– 18.25\n\n__________\n\n2.50\n\n___________\n\n₹ 2.50\n\n(b) 250 m – 202.54 m\n\n250.00\n\n– 202.54\n\n___________\n\n47.46\n\n____________\n\n47.46 m\n\n(c) ₹ 8.40 – ₹ 5.36\n\n8.40\n\n– 5.36\n\n_________\n\n3.04\n\n_________\n\n₹ 3.04\n\n(d) 5.206 km – 2.051 km\n\n5.206\n\n– 2.051\n\n__________\n\n3.155\n\n__________\n\n3.155 km\n\n(e) 2.107 kg – 0.314 kg\n\n2.107\n\n– 0.314\n\n_________\n\n1.793\n\n__________\n\n1.793 kg\n\n2. Find the value of:\n\n(a) 9.756 – 6.28\n\n(b) 21.05 – 15.27\n\n(c) 18.5 – 6.79\n\n(d) 11.6 – 9.847\n\nSolutions:\n\n(a) 9.756\n\n– 6.280\n\n_________\n\n3.476\n\n_________\n\n(b) 21.05\n\n– 15.27\n\n___________\n\n5.78\n\n____________\n\n(c) 18.50\n\n– 6.79\n\n___________\n\n11.71\n\n___________\n\n(d) 11.600\n\n– 9.847\n\n____________\n\n1.753\n\n____________\n\n3. Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?\n\nSolutions:\n\nMoney given to shopkeeper = ₹ 50.00\n\nPrice of the book = ₹ 35.65\n\nMoney that Raju will get back from the shopkeeper will be the difference of these two\n\n∴ Money left with Raju is\n\n50.00\n\n– 35.65\n\n___________\n\n14.35\n\n___________\n\nHence, money left with Raju is ₹ 14.35\n\n4. Rani had ₹ 18.50. She bought one ice cream for ₹ 11.75. How much money does she have now?\n\nSolutions:\n\nMoney with Rani = ₹ 18.50\n\nPrice of an ice cream = ₹ 11.75\n\nNow money left with Rani will be the difference of these two\n\nHence, money left with her is\n\n18.50\n\n– 11.75\n\n__________\n\n6.75\n\n___________\n\n∴ Money left with Rani is ₹ 6.75\n\n5. Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?\n\nSolutions:\n\nLength of cloth = 20 m 5 cm\n\n= 20.05 m\n\nLength of cloth to make a curtain = 4 m 50 cm\n\n= 4.50 m\n\nLength of cloth left with Tina will be the difference of these two\n\nThus length of cloth left with her is\n\n20.05\n\n– 4.50\n\n________\n\n15.55\n\n________\n\n∴ The length of the remaining cloth left with Tina is 15.55 m\n\n6. Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?\n\nSolutions:\n\nTotal distance travelled by Namita = 20 km 50 m\n\n= 20.050 km\n\nDistance travelled by bus = 10 km 200 m\n\n= 10.200 km\n\nDistance travelled by auto = Total distance travelled – Distance travelled by bus\n\n∴ Distance to be travelled by auto is\n\n20.050\n\n– 10.200\n\n________\n\n9.850\n\n________\n\n∴ Namita travelled 9.850 km by auto\n\n7. Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?\n\nSolutions:\n\nTotal weight of vegetables Aakash bought = 10.000 kg\n\nWeight of onions = 3 kg 500 g\n\n= 3.500 kg\n\nWeight of tomatoes = 2 kg 75 g\n\n= 2.075 kg\n\nWeight of potatoes = Total weight of vegetables bought – (weight of onions + weight of tomatoes)\n\n= 10.000 – (3.500 + 2.075)\n\n3.500\n\n+ 2.075\n\n________\n\n5.575\n\n________\n\n10.000\n\n– 5.575\n\n_________\n\n4.425\n\n_________\n\n∴ 4.425 kg is the weight of the potatoes"
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null,
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https://ilari.angervuori.fi/node20.html | [
"## January – Inverse of a Gaussian variable\n\nThere is quite a pile of literature on the subject of the inverse of a Gaussian distributed variable (this should not be fixed with inverse Gaussian distribution – it is a different matter). In fact, the inverse distribution is ill-behaved; the mean and variance does not generally exist.\n\nI came up with a simple approximation that works well if the mean is large enough and the variance is small enough (I have not worked out the details of the exact conditions for this approximation. However, the results can be verified by Monte Carlo simulations).\n\nFirst, approximate the Gaussian distributed variable",
null,
"by a log-normally distributed variable",
null,
"Lognormal",
null,
"with corresponding mean and variance, i.e.",
null,
"and",
null,
"We leave the solving of",
null,
"and",
null,
"as an easy exercise for the reader.\n\nUsing the theory of log-normal distribution, the inverse of",
null,
"is now given by",
null,
"Lognormal",
null,
"That's it!\n\nReferences:\n\n• Log-normal distribution\n• Díaz-Francés, Eloísa; Rubio, Francisco J. (2012-01-24). \"On the existence of a normal approximation to the distribution of the ratio of two independent normal random variables\". Statistical Papers. Springer Science and Business Media LLC."
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null,
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http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=zvmmf&paperid=483&option_lang=eng | [
"Zhurnal Vychislitel'noi Matematiki i Matematicheskoi Fiziki",
null,
"General information",
null,
"Latest issue",
null,
"Archive",
null,
"Impact factor",
null,
"Search papers",
null,
"Search references",
null,
"RSS",
null,
"Latest issue",
null,
"Current issues",
null,
"Archive issues",
null,
"What is RSS\n\n Zh. Vychisl. Mat. Mat. Fiz.: Year: Volume: Issue: Page: Find\n\n Personal entry: Login: Password: Save password Enter",
null,
"Forgotten password?",
null,
"Register\n\n Zh. Vychisl. Mat. Mat. Fiz., 2006, Volume 46, Number 4, Pages 605–614 (Mi zvmmf483)",
null,
"",
null,
"Implicit and efficient schemes for a parabolic equation in a spherical layer\n\nE. I. Aksenova\n\nMoscow Mezhregion Bar, Poluyaroslavskii per. 3/5, Moscow, 105120, Russia\n\nAbstract: An implicit and an efficient three-level scheme for a parabolic equation in spherical coordinates is constructed in a spherical layer. No axial symmetry is assumed. The convergence rates of the schemes are estimated under minimum requirements on the initial data. The estimates are uniform with respect to the inner diameter of the domain. The order of convergence is $\\tau^{\\alpha/2}+h^\\alpha$, $\\alpha=1,2$, depending on the smoothness of the data. The results remain valid for a domain without a hole.\n\nKey words: parabolic boundary value problems, spherical coordinates, domain with a small hole, three-level efficient difference scheme, convergence rate estimate.",
null,
"Full text: PDF file (1067 kB)",
null,
"References: PDF file HTML file\n\nEnglish version:\nComputational Mathematics and Mathematical Physics, 2006, 46:4, 575–584",
null,
"Bibliographic databases:",
null,
"",
null,
"",
null,
"UDC: 519.633.6\nRevised: 21.03.2005\n\nCitation: E. I. Aksenova, “Implicit and efficient schemes for a parabolic equation in a spherical layer”, Zh. Vychisl. Mat. Mat. Fiz., 46:4 (2006), 605–614; Comput. Math. Math. Phys., 46:4 (2006), 575–584",
null,
"Citation in format AMSBIB\n\\Bibitem{Aks06}\n\\by E.~I.~Aksenova\n\\paper Implicit and efficient schemes for a~parabolic equation in a~spherical layer\n\\jour Zh. Vychisl. Mat. Mat. Fiz.\n\\yr 2006\n\\vol 46\n\\issue 4\n\\pages 605--614\n\\mathnet{http://mi.mathnet.ru/zvmmf483}\n\\mathscinet{http://www.ams.org/mathscinet-getitem?mr=2260352}\n\\zmath{https://zbmath.org/?q=an:05200931}\n\\transl\n\\jour Comput. Math. Math. Phys.\n\\yr 2006\n\\vol 46\n\\issue 4\n\\pages 575--584\n\\crossref{https://doi.org/10.1134/S0965542506040063}\n\\scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-33746069650}\n\n SHARE:",
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"•",
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"",
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"",
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"Contact us: math-net2022_01 [at] mi-ras ru",
null,
"Terms of Use",
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https://study-assistant.com/mathematics/question12831364 | [
"",
null,
", 22.06.2019 21:20 chycooper101\n\n# Sue travels by bus or walks when she visits the shops. the probability that she catches the bus to the shops is 0.4. the probability she catches the bus from the shops is 0.7. show the probability that sue walks at one way is 0.72",
null,
"",
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"",
null,
"### Another question on Mathematics",
null,
"Mathematics, 21.06.2019 13:30\nSince f(x, y) = 1 + y2 and \"∂f/∂y\" = 2y are continuous everywhere, the region r in theorem 1.2.1 can be taken to be the entire xy-plane. use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y' = 1 + y2, y(0) = 0.",
null,
"Mathematics, 21.06.2019 19:30\nIt is saturday morning and jeremy has discovered he has a leak coming from the water heater in his attic. since plumbers charge extra to come out on the weekends, jeremy is planning to use buckets to catch the dripping water. he places a bucket under the drip and steps outside to walk the dog. in half an hour the bucket is 1/5 of the way full. what is the rate at which the water is leaking per hour?",
null,
"Mathematics, 21.06.2019 21:10\nGiven: lines a and b are parallel and line c is a transversal. prove: 2 is supplementary to 8 what is the missing reason in the proof? statement reason 1. a || b, is a transv 1. given 2. ∠6 ≅ ∠2 2. ? 3. m∠6 = m∠2 3. def. of congruent 4. ∠6 is supp. to ∠8 4. def. of linear pair 5. ∠2 is supp. to ∠8 5. congruent supplements theorem corresponding angles theorem alternate interior angles theorem vertical angles theorem alternate exterior angles theorem",
null,
"Mathematics, 21.06.2019 22:50\n1. if events a and b are non-overlapping events, how do you find the probability that one or the other occurs? 2. what does it mean if p(a or b) equals 1?\nSue travels by bus or walks when she visits the shops. the probability that she catches the bus to t...\nQuestions",
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"Geography, 24.03.2021 18:10",
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"Mathematics, 24.03.2021 18:10",
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"Mathematics, 24.03.2021 18:10",
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"Questions on the website: 14218324"
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https://physics.stackexchange.com/questions/317301/when-can-a-quantum-operation-be-represented-as-a-random-unitary/317501 | [
"# When can a quantum operation be represented as a 'random unitary'?\n\nConsider the quantum operation that measures a qubit in the computational basis. This is clearly an irreversible operation, but it turns out to be exactly the same as the stochastic unitary operation $$\\text{do nothing with probability } 1/2,\\, \\sigma_z \\text{ with probability } 1/2.$$ For example, one can simply check that this doesn't affect $|0 \\rangle$ and $|1 \\rangle$, and it maps $|0 \\rangle + |1 \\rangle$ to the maximally mixed state.\n\nHowever, it doesn't seem like all quantum operations can be written in this way. For example, I haven't been able to do this for the operation that destroys a qubit and replaces it with the $|0 \\rangle$ state.\n\nUnder what circumstances can a quantum operation be written as a randomly chosen unitary? (In more standard language, when can Kraus operators be chosen to be unitary?)\n\nLet $M_k$ describe the measurement (POVM) operators, $\\sum_k M_k^\\dagger M_k=1\\!\\!1$. (In the case of a projective measurements, these are just projectors.) Since you discard the measurement outcome, you are effectively left with a quantum channel $$\\mathcal E(\\rho) = \\sum_k M_k\\rho M_k^\\dagger\\ .$$ You want to know under which conditions this can be written as a convex combination over unitaries, $$\\mathcal E(\\rho) \\stackrel{?}{=} \\sum_s p_s U_s\\rho U_s^\\dagger\\ ,$$ where $p_i\\ge0$.\nA necessary condition is that the channel is unital, i.e., $\\mathcal E(1\\!\\!1)=1\\!\\!1$. However, this is not sufficient: Examples of unital channels are known which are not convex combinations of unitaries [1,2]. AFAIK, no other succinct characterization of channels which admit such a representation is known."
]
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https://whyaxis.me/2017/09/14/building-a-simple-neural-network-with-keras-and-tensorflow/ | [
"# Building a simple Neural Network with Keras and Tensorflow\n\nNeural networks have been a hot topic after the surfacing of deep learning. In these series of tutorials we will go in depth into building complex networks with Keras and Tensor Flow.",
null,
"TensorFlow is an open-source library for machine learning introduced by Google. Keras provides a high level api/wrapper around TensorFlow.\n\nIn this tutorial we’ll look into building a very simple neural network in Keras and use Tensor Flow as the backend. You can follow the installation of Tensorflow in this page. Using the virtualenv is recommended to minimize dependency and version conflicts.\n\nFor this tutorial we will implement a simple XOR (exclusive OR) logic in a Neural Network. XOR is an interesting thing because it is not a linearly separable problem. We cannot draw a single linearly separable line between these nodes. For this let’s look at a very simple example. As you are a shop owner you have 4 customer segments and you want a method to segment these customer. For your business high earning old people and middle earning young people are the most loyal customers and you are planning to provide some discounts for these segments. So is there a way we can draw a single line to divide the high return customers and low return customers. This is exactly the same scenario in XOR.",
null,
"",
null,
"So how can we solve this with Neural Network (NN). Let’s start by importing the packages we need to build the NN.",
null,
"The python environment is good with scientific computing and math related work. So we import numpy. Numpy is a python library which makes array manipulations very easy and keras internally supports numpy arrays as inputs.\n\nThere are two different API’s provided by Keras, which are Functional and Sequential models. We will import the simplest model which is the sequential model. This can be seen as a linear stack of layers. These Neural networks consist of different layers where input data flows through and gets transformed on its way. There are several layers provided by Keras (Dense , Dropout and merge layers). These different types of layer help us to model individual kinds of neural nets for various machine learning tasks. In our scenario we’ll be needing only the Dense layer.",
null,
"The initial input will be created as a two dimensional array. The target (output) array is also two dimensional but contains just a single output.\n\n Input Input Output Old Middle Income Non Loyal Customers Young Middle Income Loyal Customers Old High Income Loyal Customers Young High Income Non Loyal Customer \n\nNow the interesting part begins, creating the model.",
null,
"",
null,
"As mentioned before we’ll be using the dense layer. The first parameter states the number of neurons in that given layer and the second parameter states the number of inputs given to the model. Lets not get too much worried about this activation function, will talk about this in a while. After this we add a second layer which has only one neuron. Have you noticed that input dimension is not stated in the second layer , because it internally takes the 1st layer as the input to the final layer.",
null,
"In this we have used two types of activation functions which is ReLU (Rectified linear unit) and a Sigmoid function. Just like our neurons in our head these neurons also needs to be activated to pass the message to next neuron. Similar to that these activation functions also has some threshold and when it meets the threshold it fires(activates) the neuron.",
null,
"The last thing to do is to compile the model. In order to compile we need two things. In order to the NN to set the correct weights to the model we need to tell the model how well is it performing. This is knows as the loss (how bad it performed). To this we will pick mean_squared_error as the loss function.\n\nThe second factor is the optimizing function. The job for the optimization function is to find the right numbers to the weights and thus reducing the loss. For the optimization we will use something called adam optimizer.\n\nYou might ask are these the only loss or optimization function I can use. No ! . Even if we add binary_crossentropy as the error function the NN will work without a hassle. But that does not mean all the error functions can be used interchangeably. A process of trial and error should be done master which fits in to which scenarios.\n\nThe third parameter is to measure the accuracy. In here we’ll use binary_accuracy to calculate the accuracy of the predictions.",
null,
"Finally we start the training process by calling the fit function. Now we can see in 54th iteration we are getting the 100% accuracy. If we run it several times the convergence number will slightly change because the weights are initialized randomly for the first iteration.",
null,
"## Tweaking the model.\n\nOkay we have built a very simple neural network so can we tweak it. We can visualize what we have built is some thing similar to the below image. So what are the things we can change in this neural network ?\n\nIncreasing the number of hidden layers",
null,
"Let’s try with adding another hidden layer and check if we get to 100% accuracy very quickly. So how do we add a new layer ? . Simple we just append another single line to the model.",
null,
"The results are really good. We were able to get 100% with the 11th Iteration it self.",
null,
"Now lets have only one hidden layer and increase the number of neurons in the first layer and see whether the accuracy increase.",
null,
"Oh ! this has increased the iteration count to reach 100% accuracy. We saw previously with 16 neurons we were able to converge in the 55th interactions , but now it took 86 iterations to converge with 32 neurons in a layer. So increasing the number of neurons does not always yield in good results.",
null,
"Through these tutorial I basically wanted to convey the message that, we don’t have to be a mathematician in order to build machine learning models or do machine learning. A basic understanding of how things works would be enough to get you started. In contrast for machine learning theory you’ll be needing a more in depth math knowledge. Lets see you in the next tutorial and solve some more interesting problems."
]
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"https://whyaxis.files.wordpress.com/2017/09/image.png",
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https://www.yourelectricalguide.com/2018/03/three-phase-induction-motor-equivalent-circuit.html | [
"# Equivalent Circuit of Induction Motor\n\nThe induction motor is equivalent to a rotating transformer. Hence, we can draw the induction motor equivalent circuit with the help of equivalent circuit of the transformer.\n\nThe basic equivalent circuit of induction motor is shown in Fig. which is very similar to that of a transformer.\n\nHere R1 and X1 are the per phase values of stator resistance and stator leakage reactance.\n\nE1 is per phase stator voltage and N1 the number of stator turns per phase whereas R2 rotor resistance per phase and sX2 is rotor reactance per phase in running condition.\n\nThe resistance Ro is the no-load resistance per phase, that represents the resistance for core losses and Xo represents the magnetizing component of the no-load stator current similar to the magnetizing component of the no-load primary current of the transformer.\n\nV1 is the per phase stator supply voltage. E1 is the per phase stator induced voltage\n\nE2r = sE2 is the per phase rotor induced EMF (in running condition)\n\n• where, s = slip, E2 = induced EMF in rotor at standstill.\n\nThe rotor current I2r is given by,\n\nI2r = sE2/ [R22 + (sX2)2]1/2\n= E2/ [(R2/s)2 + X22]1/2\n\nFrom the above expression, it is clear that the rotor circuit actually consists of a fixed resistance R2 and a variable reactance sX2, connected across E2r = sE2 is equivalent to a rotor circuit having a fixed reactance X2 connected in series with a variable resistance R2/s and supplied with constant voltage E2.\n\nWe can express resistance R2/s in two parts as follows:\nR2/s = R2 + R2(1/s – 1)\n\n• were, the first part R2 is the rotor resistance and represents the rotor copper losses.\n• and the second part R2(1/s – 1) is the load resistance RL. It is the electrical equivalent of the mechanical load on the motor. It is the electrical power that is converted into the mechanical power by the motor.\n\nHence equivalent rotor circuit of the induction motor can be drawn as under.\n\n# Equivalent Circuit of Induction Motor Referred to Stator\n\nWe can draw the equivalent circuit of induction motor referred to stator or rotor side like that of a transformer.\n\nThe equivalent circuit of 3 phase induction motor referred to the stator is shown in Fig. Here, all the parameters are referred to the stator winding.\nWe know that,\ntransformation ratio, k = E2/E1\nAll the rotor parameters are transferred to the stator side as follows:\nE2r’ = E2r/k\nI’2r = kI2r\nX’2 = X2/k2 and R’2 = R2/k2\n\n## Approximate Equivalent Circuit of Three Phase Induction Motor\n\nThe no-load circuit components Ro and Xo can be shifted to the left of R1 and X1 to obtain the approximate equivalent circuit of the induction motor.\n\nHere, Ro1 = R1 + R’1\n\nis equivalent resistance of induction motor referred to the stator and Xo1 = X1 + X’1 is equivalent reactance of induction motor referred to the stator.\n\nThree Phase Induction Motor | All Posts\n\n© https://yourelectricalguide.com/ equivalent circuit of induction motor.\n\n### 1 thought on “Equivalent Circuit of Induction Motor”\n\n1.",
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"nice"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8592437,"math_prob":0.9971408,"size":3248,"snap":"2019-51-2020-05","text_gpt3_token_len":791,"char_repetition_ratio":0.18711467,"word_repetition_ratio":0.028725315,"special_character_ratio":0.22752462,"punctuation_ratio":0.055172414,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982823,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T03:39:09Z\",\"WARC-Record-ID\":\"<urn:uuid:f96213b8-713e-4b94-83d6-271b5c895fd6>\",\"Content-Length\":\"50434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d19a78ca-6430-4175-93cd-58d51bd8befe>\",\"WARC-Concurrent-To\":\"<urn:uuid:e92c8154-fcb9-492d-8726-124d14bf5446>\",\"WARC-IP-Address\":\"104.31.73.84\",\"WARC-Target-URI\":\"https://www.yourelectricalguide.com/2018/03/three-phase-induction-motor-equivalent-circuit.html\",\"WARC-Payload-Digest\":\"sha1:5C62X3LKPWEKQX4PBVPMZBU7R56WYTVD\",\"WARC-Block-Digest\":\"sha1:LHYRUG2KR3PYXPBYC4QU7XLGUTUCBIVC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250614880.58_warc_CC-MAIN-20200124011048-20200124040048-00490.warc.gz\"}"} |
http://wiki.secondlife.com/wiki/User:Pedro_Oval/Mono_code_memory_usage | [
"# User:Pedro Oval/Mono code memory usage\n\nHere are some memory usage results I've obtained with Mono, for common language constructs. They were obtained by replicating each line to test 512 times and looking at llGetFreeMemory right at the beginning of the script.\n\nMost findings are supported by a compilation into CIL code. Some of them, however, are puzzling.\n\nIncludes tips on memory reduction, with the corresponding savings. In some cases, \"same bytes as (something)\" is indicated, which means that the constructs are equivalent but offer no savings.\n\nNotation below is that x, y are local integer variables; f, g are local float variables, v is a local vector variable, a is a list\n\n; 0 Empty statement, e.g. ;;;;;; takes same mem as ;\n{} 0 Empty statement e.g. {{{{{}}}}} takes same mem as {}\nreturn 1 Tested in event and without arguments\nx=0 8-9 8 bytes if x is one of the first four local variables; 9 if not (see notes on CIL for an explanation)\nx 2-3 2 bytes if x is one of the first four local variables; 3 if not (see notes on CIL for an explanation)\nx=y 4-6 Assignment. 4 bytes if both x and y are among the first four local variables; 6 if neither is; 5 otherwise.\n(x) 2 Same as above, but we won't be adding that at every appearance. We just use the shortest on this table.\n(integer)x 2\n(float)x 3\nf 2\n(float)f 2\n(integer)f 7\nv 2\nv.z 8 Vector component\n-x 3\n~x 3 Bitwise, equivalent to -x-1 which allows it to be used for some hacks. Not applicable to floats though.\n!x 5 Logical\n--x 6 Pre-decrement. Same bytes as x=~-x\n++x 6 Pre-increment. Same bytes as x=-~x\nx-- 8 Post-decrement. Same bytes as -~(x=~-x)\nx++ 8 Post-increment. Same bytes as ~-(x=-~x)\nx==y 5 Comparison\nx=y=z 6 (for local integer z) Chained assignment\nf=x 5 Implicit cast from integer to float\nf=(float)x 5 Explicit cast from integer to float\nx!=y 8 Same bytes as !(x==y) (implemented that way). Can sometimes be replaced by x^y, saving 4 bytes.\nx+y 4\nx-y 8 !?!? (uses a function call in CIL)\nx+-y 5 Saves 3 bytes vs. a subtraction!\nx*y 4 Can sometimes be used instead of x << bits, saving 4 bytes.\nx/y 8\nx%y 8 Modulo\nx&y 4 Bitwise\nx&&y 13 Logical. Same bytes as !(!x|!y) (implemented that way)\nx|y 4 Bitwise\nx||y 10 Logical. Same bytes as !!(x|y) (implemented that way)\nx^y 4 Bitwise. Can sometimes be used instead of x!=y (if careful), saving 4 bytes.\nx<<y 8 Bitwise shift left. Can be replaced with a multiplication in some cases, saving 4 bytes.\nx>>y 8 Bitwise shift right (division doesn't help here though)\nx<y 5\nx>y 5\nx<=y 8 Same bytes as !(x>y) (implemented that way)\nx>=y 8 Same bytes as !(x<y) (implemented that way)\nx+=y 6 Same bytes as x=x+y\nx-=y 10 Same bytes as x=x-y\nx+=-y 7 Saves 3 bytes vs x-=y\nx*=y 6 Same as x=x*y\nx/=y 10 Same as x=x/y\nx%=y 10 Same as x=x%y\nif(x); 6\nif(x); else; 11\n0 6\nx^x 4 It gives always 0, saving 2 bytes. (Thanks to User:Omei Qunhua for the discovery)\n1 6\n-1 7 Sign takes code memory.\n0xffffffff 6 Equivalent to -1.\n(integer)-1 6 Saves 1 byte.\n(integer)(-1) 7 The extra parentheses break the magic.\nALL_SIDES 6 Value is -1. Constants with negative values don't take memory for the sign.\nx|~x 5 Equivalent to -1. Saves 1 byte vs 0xffffffff, 2 bytes vs -1.\nx+1 8\n-~x 4 Same as x+1, saves 4 bytes\nx-1 12 What's wrong with subtraction?\nx+-1 9 Same as x-1, saves 3 bytes\n~-x 4 Same as x-1, saves 8 bytes\nx*y+y-1 16 This kind of construct is sometimes used to access last element of strided lists\n(x+1)*y-1 20 Equivalent to the above, just worse\n~(~x*y) 6 Equivalent to the above, saves 10 bytes\nwhile(x); 37 That byte count is not supported by CIL. Puzzling. It seems that backward jumps take more than forward. According to CIL it should take 11 bytes.\ndo ; while(x); 41 Unexpected but true. According to CIL it should take 6 bytes.\nfor(;x;); 37 For loops seem to be plainly rewritten as while loops.\n@label; if(x) jump label; 37 Equivalent to a do...while loop, saving 4 bytes. It takes as many as a while according to CIL.\n0.0 10 Float constants use a lot of memory (they use double-precision reals in Mono).\n-2.2 11 Sign takes 1 extra byte\n(float)-2.2 10 Casting negative float to float saves 1 byte\nf=0.0 12\nf=0 9 In general, using integer constants with implicit conversion to float instead of float constants saves 3 bytes every time. Another example follows.\n<0.0, 0.0, 0.0> 33 Same as ZERO_VECTOR\nZERO_VECTOR 33\n<0, 0, 0> 24 Saves 3 bytes per constant, total 9 bytes vs. ZERO_VECTOR\nTOUCH_INVALID_TEXCOORD 33\n<-1.0, -1.0, 0.0> 35 (because of the signs). Equivalent to TOUCH_INVALID_TEXCOORD.\n<(float)-1.0, (float)-1.0, 0.0> 33 Equivalent to TOUCH_INVALID_TEXCOORD.\n<-1, -1, 0> 26 (because of the signs)\n<0xffffffff, 0xffffffff, 0> 24 (no signs but ugly)\n<(integer)-1, (integer)-1, 0> 24 (saves the sign bytes, not so ugly)\n<ZERO_ROTATION> 42\n<0, 0, 0, 1> 30 Saves 3 bytes per constant, total 12 bytes vs. ZERO_ROTATION\nv+v 8 Vector addition (with the vectors as local variables)\nv-v 8 Vector subtraction\nv*v 8 Vector dot product\nv%v 8 Vector cross product\n[] 6 Empty list\na=[] 8 Assignment of empty list\na+[x] 23 Append an element to a list, with brackets\na+x 8 Append an element to a list, without brackets\n[x]+a 23 Prepend an element to a list, with brackets\nx+a 13 Prepend an element to a list, without brackets\n[x] 17 List with an integer\n[]+x 12 Empty list plus an integer element appended, saves 5 bytes with respect to [x]\n(list)x 12 Same bytes as []+x\n[x,y] 28 List with two integers (the type of variable doesn't seem to affect the result)\n[]+x+y 18 Empty list with two integer elements appended, saves 10 bytes with respect to [x,y] (the type of variable doesn't seem to affect the result)\n(string)a 7 Typecasting a list to a string"
]
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https://www.cockroachlabs.com/docs/v20.1/build-a-c++-app-with-cockroachdb | [
"On this page",
null,
"Warning:\nCockroachDB v20.1 is no longer supported. For more details, see the Release Support Policy.\n\nThis tutorial shows you how build a simple C++ application with CockroachDB and the C++ libpqxx driver.\n\nWe have tested the C++ libpqxx driver enough to claim beta-level support. If you encounter problems, please open an issue with details to help us make progress toward full support.\n\n## Before you begin\n\n1. Install CockroachDB.\n2. Start up a secure or insecure local cluster.\n3. Choose the instructions that correspond to whether your cluster is secure or insecure:\n\n## Step 1. Install the libpqxx driver\n\nInstall the C++ libpqxx driver as described in the official documentation.\n\nNote:\n\nIf you are running macOS, you need to install version 4.0.1 or higher of the libpqxx driver.\n\n## Step 2. Create the `maxroach` user and `bank` database\n\nStart the built-in SQL shell:\n\n``````\\$ cockroach sql --certs-dir=certs\n``````\n\nIn the SQL shell, issue the following statements to create the `maxroach` user and `bank` database:\n\n``````> CREATE USER IF NOT EXISTS maxroach;\n``````\n``````> CREATE DATABASE bank;\n``````\n\nGive the `maxroach` user the necessary permissions:\n\n``````> GRANT ALL ON DATABASE bank TO maxroach;\n``````\n\nExit the SQL shell:\n\n``````> \\q\n``````\n\n## Step 3. Generate a certificate for the `maxroach` user\n\nCreate a certificate and key for the `maxroach` user by running the following command. The code samples will run as this user.\n\n``````\\$ cockroach cert create-client maxroach --certs-dir=certs --ca-key=my-safe-directory/ca.key\n``````\n\n## Step 4. Run the C++ code\n\nNow that you have a database and a user, you'll run code to create a table and insert some rows, and then you'll run code to read and update values as an atomic transaction.\n\n### Basic statements\n\nFirst, use the following code to connect as the `maxroach` user and execute some basic SQL statements, creating a table, inserting rows, and reading and printing the rows.\n\nDownload the `basic-sample.cpp` file, or create the file yourself and copy the code into it.\n\n``````#include <cassert>\n#include <functional>\n#include <iostream>\n#include <stdexcept>\n#include <string>\n#include <pqxx/pqxx>\n\nusing namespace std;\n\nint main() {\ntry {\n// Connect to the \"bank\" database.\npqxx::connection c(\"dbname=bank user=maxroach sslmode=require sslkey=certs/client.maxroach.key sslcert=certs/client.maxroach.crt port=26257 host=localhost\");\n\npqxx::nontransaction w(c);\n\n// Create the \"accounts\" table.\nw.exec(\"CREATE TABLE IF NOT EXISTS accounts (id INT PRIMARY KEY, balance INT)\");\n\n// Insert two rows into the \"accounts\" table.\nw.exec(\"INSERT INTO accounts (id, balance) VALUES (1, 1000), (2, 250)\");\n\n// Print out the balances.\ncout << \"Initial balances:\" << endl;\npqxx::result r = w.exec(\"SELECT id, balance FROM accounts\");\nfor (auto row : r) {\ncout << row.as<int>() << ' ' << row.as<int>() << endl;\n}\n\nw.commit(); // Note this doesn't doesn't do anything\n// for a nontransaction, but is still required.\n}\ncatch (const exception &e) {\ncerr << e.what() << endl;\nreturn 1;\n}\ncout << \"Success\" << endl;\nreturn 0;\n}\n\n``````\n\nTo build the `basic-sample.cpp` source code to an executable file named `basic-sample`, run the following command from the directory that contains the code:\n\n``````\\$ g++ -std=c++11 basic-sample.cpp -lpq -lpqxx -o basic-sample\n``````\n\nThen run the `basic-sample` file from that directory:\n\n``````\\$ ./basic-sample\n``````\n\n### Transaction (with retry logic)\n\nNext, use the following code to again connect as the `maxroach` user but this time execute a batch of statements as an atomic transaction to transfer funds from one account to another, where all included statements are either committed or aborted.\n\nNote:\n\nCockroachDB may require the client to retry a transaction in case of read/write contention. CockroachDB provides a generic retry function that runs inside a transaction and retries it as needed. You can copy and paste the retry function from here into your code.\n\nDownload the `txn-sample.cpp` file, or create the file yourself and copy the code into it.\n\n``````#include <cassert>\n#include <functional>\n#include <iostream>\n#include <stdexcept>\n#include <string>\n#include <pqxx/pqxx>\n\nusing namespace std;\n\nvoid transferFunds(\npqxx::dbtransaction *tx, int from, int to, int amount) {\npqxx::result r = tx->exec(\n\"SELECT balance FROM accounts WHERE id = \" + to_string(from));\nassert(r.size() == 1);\nint fromBalance = r.as<int>();\n\nif (fromBalance < amount) {\nthrow domain_error(\"insufficient funds\");\n}\n\n// Perform the transfer.\ntx->exec(\"UPDATE accounts SET balance = balance - \"\n+ to_string(amount) + \" WHERE id = \" + to_string(from));\ntx->exec(\"UPDATE accounts SET balance = balance + \"\n+ to_string(amount) + \" WHERE id = \" + to_string(to));\n}\n\n// ExecuteTx runs fn inside a transaction and retries it as needed.\n// On non-retryable failures, the transaction is aborted and rolled\n// back; on success, the transaction is committed.\n//\n// https://cockroachlabs.com/docs/transactions.html.\n//\n// NOTE: the supplied exec closure should not have external side\n// effects beyond changes to the database.\nvoid executeTx(\npqxx::connection *c, function<void (pqxx::dbtransaction *tx)> fn) {\npqxx::work tx(*c);\nwhile (true) {\ntry {\npqxx::subtransaction s(tx, \"cockroach_restart\");\nfn(&s);\ns.commit();\nbreak;\n} catch (const pqxx::pqxx_exception& e) {\n// Swallow \"transaction restart\" errors; the transaction will be retried.\n// Unfortunately libpqxx doesn't give us access to the error code, so we\n// do string matching to identify retryable errors.\nif (string(e.base().what()).find(\"restart transaction:\") == string::npos) {\nthrow;\n}\n}\n}\ntx.commit();\n}\n\nint main() {\ntry {\npqxx::connection c(\"dbname=bank user=maxroach sslmode=require sslkey=certs/client.maxroach.key sslcert=certs/client.maxroach.crt port=26257 host=localhost\");\n\nexecuteTx(&c, [](pqxx::dbtransaction *tx) {\ntransferFunds(tx, 1, 2, 100);\n});\n}\ncatch (const exception &e) {\ncerr << e.what() << endl;\nreturn 1;\n}\ncout << \"Success\" << endl;\nreturn 0;\n}\n\n``````\n\nTo build the `txn-sample.cpp` source code to an executable file named `txn-sample`, run the following command from the directory that contains the code:\n\n``````\\$ g++ -std=c++11 txn-sample.cpp -lpq -lpqxx -o txn-sample\n``````\n\nThen run the `txn-sample` file from that directory:\n\n``````\\$ ./txn-sample\n``````\n\nAfter running the code, use the built-in SQL client to verify that funds were transferred from one account to another:\n\n``````\\$ cockroach sql --certs-dir=certs -e 'SELECT id, balance FROM accounts' --database=bank\n``````\n``````id | balance\n+----+---------+\n1 | 900\n2 | 350\n(2 rows)\n``````\n\n## Step 2. Create the `maxroach` user and `bank` database\n\nStart the built-in SQL shell:\n\n``````\\$ cockroach sql --insecure\n``````\n\nIn the SQL shell, issue the following statements to create the `maxroach` user and `bank` database:\n\n``````> CREATE USER IF NOT EXISTS maxroach;\n``````\n``````> CREATE DATABASE bank;\n``````\n\nGive the `maxroach` user the necessary permissions:\n\n``````> GRANT ALL ON DATABASE bank TO maxroach;\n``````\n\nExit the SQL shell:\n\n``````> \\q\n``````\n\n## Step 3. Run the C++ code\n\nNow that you have a database and a user, you'll run code to create a table and insert some rows, and then you'll run code to read and update values as an atomic transaction.\n\n### Basic statements\n\nFirst, use the following code to connect as the `maxroach` user and execute some basic SQL statements, creating a table, inserting rows, and reading and printing the rows.\n\nDownload the `basic-sample.cpp` file, or create the file yourself and copy the code into it.\n\n``````#include <cassert>\n#include <functional>\n#include <iostream>\n#include <stdexcept>\n#include <string>\n#include <pqxx/pqxx>\n\nusing namespace std;\n\nint main() {\ntry {\n// Connect to the \"bank\" database.\npqxx::connection c(\"postgresql://maxroach@localhost:26257/bank\");\n\npqxx::nontransaction w(c);\n\n// Create the \"accounts\" table.\nw.exec(\"CREATE TABLE IF NOT EXISTS accounts (id INT PRIMARY KEY, balance INT)\");\n\n// Insert two rows into the \"accounts\" table.\nw.exec(\"INSERT INTO accounts (id, balance) VALUES (1, 1000), (2, 250)\");\n\n// Print out the balances.\ncout << \"Initial balances:\" << endl;\npqxx::result r = w.exec(\"SELECT id, balance FROM accounts\");\nfor (auto row : r) {\ncout << row.as<int>() << ' ' << row.as<int>() << endl;\n}\n\nw.commit(); // Note this doesn't doesn't do anything\n// for a nontransaction, but is still required.\n}\ncatch (const exception &e) {\ncerr << e.what() << endl;\nreturn 1;\n}\ncout << \"Success\" << endl;\nreturn 0;\n}\n\n``````\n\nTo build the `basic-sample.cpp` source code to an executable file named `basic-sample`, run the following command from the directory that contains the code:\n\n``````\\$ g++ -std=c++11 basic-sample.cpp -lpq -lpqxx -o basic-sample\n``````\n\nThen run the `basic-sample` file from that directory:\n\n``````\\$ ./basic-sample\n``````\n\n### Transaction (with retry logic)\n\nNext, use the following code to again connect as the `maxroach` user but this time execute a batch of statements as an atomic transaction to transfer funds from one account to another, where all included statements are either committed or aborted.\n\nNote:\n\nCockroachDB may require the client to retry a transaction in case of read/write contention. CockroachDB provides a generic retry function that runs inside a transaction and retries it as needed. You can copy and paste the retry function from here into your code.\n\nDownload the `txn-sample.cpp` file, or create the file yourself and copy the code into it.\n\n``````#include <cassert>\n#include <functional>\n#include <iostream>\n#include <stdexcept>\n#include <string>\n#include <pqxx/pqxx>\n\nusing namespace std;\n\nvoid transferFunds(\npqxx::dbtransaction *tx, int from, int to, int amount) {\npqxx::result r = tx->exec(\n\"SELECT balance FROM accounts WHERE id = \" + to_string(from));\nassert(r.size() == 1);\nint fromBalance = r.as<int>();\n\nif (fromBalance < amount) {\nthrow domain_error(\"insufficient funds\");\n}\n\n// Perform the transfer.\ntx->exec(\"UPDATE accounts SET balance = balance - \"\n+ to_string(amount) + \" WHERE id = \" + to_string(from));\ntx->exec(\"UPDATE accounts SET balance = balance + \"\n+ to_string(amount) + \" WHERE id = \" + to_string(to));\n}\n\n// ExecuteTx runs fn inside a transaction and retries it as needed.\n// On non-retryable failures, the transaction is aborted and rolled\n// back; on success, the transaction is committed.\n//\n// https://cockroachlabs.com/docs/transactions.html.\n//\n// NOTE: the supplied exec closure should not have external side\n// effects beyond changes to the database.\nvoid executeTx(\npqxx::connection *c, function<void (pqxx::dbtransaction *tx)> fn) {\npqxx::work tx(*c);\nwhile (true) {\ntry {\npqxx::subtransaction s(tx, \"cockroach_restart\");\nfn(&s);\ns.commit();\nbreak;\n} catch (const pqxx::pqxx_exception& e) {\n// Swallow \"transaction restart\" errors; the transaction will be retried.\n// Unfortunately libpqxx doesn't give us access to the error code, so we\n// do string matching to identify retryable errors.\nif (string(e.base().what()).find(\"restart transaction:\") == string::npos) {\nthrow;\n}\n}\n}\ntx.commit();\n}\n\nint main() {\ntry {\npqxx::connection c(\"postgresql://maxroach@localhost:26257/bank\");\n\nexecuteTx(&c, [](pqxx::dbtransaction *tx) {\ntransferFunds(tx, 1, 2, 100);\n});\n}\ncatch (const exception &e) {\ncerr << e.what() << endl;\nreturn 1;\n}\ncout << \"Success\" << endl;\nreturn 0;\n}\n\n``````\n\nTo build the `txn-sample.cpp` source code to an executable file named `txn-sample`, run the following command from the directory that contains the code:\n\n``````\\$ g++ -std=c++11 txn-sample.cpp -lpq -lpqxx -o txn-sample\n``````\n\nThen run the `txn-sample` file from that directory:\n\n``````\\$ ./txn-sample\n``````\n\nAfter running the code, use the built-in SQL client to verify that funds were transferred from one account to another:\n\n``````\\$ cockroach sql --insecure -e 'SELECT id, balance FROM accounts' --database=bank\n``````\n``````id | balance\n+----+---------+\n1 | 900\n2 | 350\n(2 rows)\n``````"
]
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"https://d33wubrfki0l68.cloudfront.net/bf83ecd2a83cbda4a4afecf5fe53b5083aef785b/5d54e/docs/images/carat-down-fill.svg",
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http://www.shodor.org/interactivate/lessons/AdvancedFire/ | [
"## Interactivate\n\nShodor > Interactivate > Lessons > Advanced Fire\n\n### Abstract\n\nThe following lesson is designed to help students explore the real-world implications of functions and independent/dependent variables. This lesson is best implemented with students working in teams of 2, alternating between the \"experimenter\" and \"recorder\" using the computer activities.\n\n### Objectives\n\nUpon completion of this lesson, students will:\n\n• understand the correlation between independent and dependent variables in real-world situations\n• understand the importance of identifying and manipulating independent and dependent variables to determine causation\n\n### Student Prerequisites\n\n• Algebraic: Students must be able to:\n• understand the concepts of fractions and decimals\n• understand and manipulate basic probabilities\n• Technological: Students must be able to:\n• perform basic mouse manipulations such as point, click and drag\n• use a browser for experimenting with the activities\n\n### Teacher Preparation\n\n• Pencil and calculator\n• A copy of the Advanced Fire Worksheet for each student.\n\n### Key Terms\n\n input The number or value that is entered, for example, into a function machine. The number that goes into the machine is the input output The number or value that comes out from a process. For example, in a function machine, a number goes in, something is done to it, and the resulting number is the output proportion A relationship between two ratios is proportional if the two ratios are equal in value.\n\n### Lesson Outline\n\n1. Focus and Review\n\nRemind students what has been learned in previous lessons that will be pertinent to this lesson and/or have them begin to think about the words and ideas of this lesson:\n\n• What does it mean to say that something is independent?\n• What, then, do you think an independent variable is?\n• What does it mean to say that something is dependent?\n• What, then, do you think a dependent variable is?\n\n2. Objectives\n\nLet the students know what it is they will be doing and learning today. Say something like this:\n\n• Today, class, we are going to learn about independent and dependent variables\n• We are going to use the computers to learn about independent/dependent variables through a real-world experiment, but please do not turn your computers on until I ask you to. I want to show you a little about this activity first.\n\n3. Teacher Input\n\nLead a discussion to prepare the students to identify independent and dependent variables in the Advanced Fire applet. Ask questions such as the following:\n\n• In the real world, why is it important to know which variables are independent and which are dependent?\n• How can we know if the dependent variable really depends on the independent variable?\n\nIntroduce the concept of inputs and outputs. Discuss the relationship between independent variables (inputs) and dependent variables (outputs), especially in the context of experimentation. Ask questions such as the following:\n\n• If there are multiple inputs, how do you determine a relationship between just one independent variable and the dependent variables (outputs)?\n• What does it mean to \"control\" the inputs?\n• Who determines the independent variable in an experiment?\n\n4. Guided Practice\n\nDemonstrate how to use the Advanced Fire applet. Be sure to demonstrate the following tasks:\n\n• Setting and changing the probability that fire will spread.\n• Setting a tree on fire and watching the fire spread.\n• Reviewing the result of past burns with the Burn History button.\n\nDiscuss the fact that Advanced Fire has multiple variables. Ask the following questions:\n\n• Are there any variables in this scenario?\n• How do you know?\n• What are they?\n• Which variables do you set and which are determined by the applet itself?\n• Which of these variables are the independent variables? The dependent variables?\n\n5. Independent Practice\n\nHave the students try the computer version of the Advanced Fire activity to investigate the relationship between Burn Probability and the actual proportion of trees left standing.\n\nDivide students into groups of 2-3 and ask them to choose a topic related to Advanced Fire to investigate that involves controlling independent variables and measuring their effect on dependent variables. Students should then fill out the worksheet for their chosen topic.\n\n6. Closure\n\nYou may wish to bring the class back together for a discussion of the findings. Ask students to identify their chosen independent and dependent variables, and then describe the relationship that they found between the two. Ask the following questions:\n\n• What is one way to easily identify the independent variable(s) in a problem? The dependent variable(s)?\n• Did every group that performed the same experiment get the same results? What does that tell you about the randomness of the burning?\n\n### Alternate Outline\n\nThis lesson can be rearranged in several ways:\n\n• If only one computer is available, choose an experiment as a class and have each student or group make predictions about the relationship between independent and dependent variables. Then, conduct the experiment on an overhead projector to test your hypotheses.\n\n### Suggested Follow-Up\n\nThese discussions and experiments will have given students a greater understanding of independent and dependent variables and their use in quasi-real-world situations. To follow up on this lesson, consider having students conduct more complex experiments with Directable Fire!! or A Better Fire!!, where students have to consider variability in direction as well as proportion.",
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"http://www.shodor.org/interactivate/media/images/nsdl_cobrand.gif",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9049788,"math_prob":0.79112107,"size":5640,"snap":"2021-43-2021-49","text_gpt3_token_len":1083,"char_repetition_ratio":0.17601135,"word_repetition_ratio":0.040697675,"special_character_ratio":0.175,"punctuation_ratio":0.09006211,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9867962,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T20:12:44Z\",\"WARC-Record-ID\":\"<urn:uuid:fb125942-04d4-4c6a-9240-5ad019886888>\",\"Content-Length\":\"28870\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f1f6ec3b-a880-4ac5-8dee-505ad7902422>\",\"WARC-Concurrent-To\":\"<urn:uuid:0636b60c-77d1-4649-a442-88dcfffc4833>\",\"WARC-IP-Address\":\"204.85.28.50\",\"WARC-Target-URI\":\"http://www.shodor.org/interactivate/lessons/AdvancedFire/\",\"WARC-Payload-Digest\":\"sha1:7FTXD3XJ3NEMQKYFZ2DSYLWYG66NSAGC\",\"WARC-Block-Digest\":\"sha1:QNBIGLBLTEBUOZCJMTSTG6UTGHWJCHFW\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585518.54_warc_CC-MAIN-20211022181017-20211022211017-00415.warc.gz\"}"} |
https://www.colorhexa.com/659e9c | [
"# #659e9c Color Information\n\nIn a RGB color space, hex #659e9c is composed of 39.6% red, 62% green and 61.2% blue. Whereas in a CMYK color space, it is composed of 36.1% cyan, 0% magenta, 1.3% yellow and 38% black. It has a hue angle of 177.9 degrees, a saturation of 22.7% and a lightness of 50.8%. #659e9c color hex could be obtained by blending #caffff with #003d39. Closest websafe color is: #669999.\n\n• R 40\n• G 62\n• B 61\nRGB color chart\n• C 36\n• M 0\n• Y 1\n• K 38\nCMYK color chart\n\n#659e9c color description : Mostly desaturated dark cyan.\n\n# #659e9c Color Conversion\n\nThe hexadecimal color #659e9c has RGB values of R:101, G:158, B:156 and CMYK values of C:0.36, M:0, Y:0.01, K:0.38. Its decimal value is 6659740.\n\nHex triplet RGB Decimal 659e9c `#659e9c` 101, 158, 156 `rgb(101,158,156)` 39.6, 62, 61.2 `rgb(39.6%,62%,61.2%)` 36, 0, 1, 38 177.9°, 22.7, 50.8 `hsl(177.9,22.7%,50.8%)` 177.9°, 36.1, 62 669999 `#669999`\nCIE-LAB 61.325, -19.063, -4.882 23.593, 29.62, 35.925 0.265, 0.332, 29.62 61.325, 19.678, 194.363 61.325, -27.029, -4.191 54.424, -17.862, -1.04 01100101, 10011110, 10011100\n\n# Color Schemes with #659e9c\n\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #9e6567\n``#9e6567` `rgb(158,101,103)``\nComplementary Color\n• #659e80\n``#659e80` `rgb(101,158,128)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #65849e\n``#65849e` `rgb(101,132,158)``\nAnalogous Color\n• #9e8065\n``#9e8065` `rgb(158,128,101)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #9e6584\n``#9e6584` `rgb(158,101,132)``\nSplit Complementary Color\n• #9e9c65\n``#9e9c65` `rgb(158,156,101)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #9c659e\n``#9c659e` `rgb(156,101,158)``\n• #679e65\n``#679e65` `rgb(103,158,101)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #9c659e\n``#9c659e` `rgb(156,101,158)``\n• #9e6567\n``#9e6567` `rgb(158,101,103)``\n• #47706f\n``#47706f` `rgb(71,112,111)``\n• #50807e\n``#50807e` `rgb(80,128,126)``\n• #5a8f8d\n``#5a8f8d` `rgb(90,143,141)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #75a8a6\n``#75a8a6` `rgb(117,168,166)``\n• #84b2b0\n``#84b2b0` `rgb(132,178,176)``\n• #94bcba\n``#94bcba` `rgb(148,188,186)``\nMonochromatic Color\n\n# Alternatives to #659e9c\n\nBelow, you can see some colors close to #659e9c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #659e8e\n``#659e8e` `rgb(101,158,142)``\n• #659e93\n``#659e93` `rgb(101,158,147)``\n• #659e97\n``#659e97` `rgb(101,158,151)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #659b9e\n``#659b9e` `rgb(101,155,158)``\n• #65979e\n``#65979e` `rgb(101,151,158)``\n• #65929e\n``#65929e` `rgb(101,146,158)``\nSimilar Colors\n\n# #659e9c Preview\n\nThis text has a font color of #659e9c.\n\n``<span style=\"color:#659e9c;\">Text here</span>``\n#659e9c background color\n\nThis paragraph has a background color of #659e9c.\n\n``<p style=\"background-color:#659e9c;\">Content here</p>``\n#659e9c border color\n\nThis element has a border color of #659e9c.\n\n``<div style=\"border:1px solid #659e9c;\">Content here</div>``\nCSS codes\n``.text {color:#659e9c;}``\n``.background {background-color:#659e9c;}``\n``.border {border:1px solid #659e9c;}``\n\n# Shades and Tints of #659e9c\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, #020202 is the darkest color, while #f5f9f9 is the lightest one.\n\n• #020202\n``#020202` `rgb(2,2,2)``\n• #090e0e\n``#090e0e` `rgb(9,14,14)``\n• #111b1a\n``#111b1a` `rgb(17,27,26)``\n• #182726\n``#182726` `rgb(24,39,38)``\n• #203332\n``#203332` `rgb(32,51,50)``\n• #273f3e\n``#273f3e` `rgb(39,63,62)``\n• #2f4b4a\n``#2f4b4a` `rgb(47,75,74)``\n• #375756\n``#375756` `rgb(55,87,86)``\n• #3e6361\n``#3e6361` `rgb(62,99,97)``\n• #466f6d\n``#466f6d` `rgb(70,111,109)``\n• #4d7b79\n``#4d7b79` `rgb(77,123,121)``\n• #558785\n``#558785` `rgb(85,135,133)``\n• #5d9391\n``#5d9391` `rgb(93,147,145)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #71a6a4\n``#71a6a4` `rgb(113,166,164)``\n``#7dadab` `rgb(125,173,171)``\n• #89b5b3\n``#89b5b3` `rgb(137,181,179)``\n• #95bcbb\n``#95bcbb` `rgb(149,188,187)``\n• #a1c4c3\n``#a1c4c3` `rgb(161,196,195)``\n``#adcbca` `rgb(173,203,202)``\n• #b9d3d2\n``#b9d3d2` `rgb(185,211,210)``\n• #c5dbda\n``#c5dbda` `rgb(197,219,218)``\n• #d1e2e2\n``#d1e2e2` `rgb(209,226,226)``\n• #ddeae9\n``#ddeae9` `rgb(221,234,233)``\n• #e9f1f1\n``#e9f1f1` `rgb(233,241,241)``\n• #f5f9f9\n``#f5f9f9` `rgb(245,249,249)``\nTint Color Variation\n\n# Tones of #659e9c\n\nA tone is produced by adding gray to any pure hue. In this case, #788b8a is the less saturated color, while #04fff6 is the most saturated one.\n\n• #788b8a\n``#788b8a` `rgb(120,139,138)``\n• #6f9493\n``#6f9493` `rgb(111,148,147)``\n• #659e9c\n``#659e9c` `rgb(101,158,156)``\n• #5ba8a5\n``#5ba8a5` `rgb(91,168,165)``\n• #52b1ae\n``#52b1ae` `rgb(82,177,174)``\n• #48bbb7\n``#48bbb7` `rgb(72,187,183)``\n• #3ec5c0\n``#3ec5c0` `rgb(62,197,192)``\n• #35cec9\n``#35cec9` `rgb(53,206,201)``\n• #2bd8d2\n``#2bd8d2` `rgb(43,216,210)``\n• #21e2db\n``#21e2db` `rgb(33,226,219)``\n• #18ebe4\n``#18ebe4` `rgb(24,235,228)``\n• #0ef5ed\n``#0ef5ed` `rgb(14,245,237)``\n• #04fff6\n``#04fff6` `rgb(4,255,246)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #659e9c 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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https://t3x.org/s9fes/draw-tree.scm.html | [
"http://t3x.org/s9fes/draw-tree.scm.html\n\n# `draw-tree`\n\nLocation: contrib, 134 Lines\n\n```; Scheme 9 from Empty Space, Function Library\n; By Nils M Holm, 2009-2012\n; Placed in the Public Domain\n;\n; (draw-tree object) ==> unspecific\n; (dt) ==> unspecific\n;\n; Print a tree structure resembling a Scheme datum. Each cons\n; cell is represented by [o|o] with lines leading to their car\n; and cdr parts. Conses with a cdr value of () are represented\n; by [o|/].\n;\n; DT is an abbrevation for DRAW-TREE.\n;\n; (Example): (draw-tree '((a) (b . c) (d e))) ==> unspecific\n;\n; Output: [o|o]---[o|o]---[o|/]\n; | | |\n; [o|/] | [o|o]---[o|/]\n; | | | |\n; a | d e\n; |\n; [o|o]--- c\n; |\n; b\n\n(define (draw-tree n)\n\n(define *nothing* (cons 'N '()))\n\n(define *visited* (cons 'V '()))\n\n(define (empty? x) (eq? x *nothing*))\n\n(define (visited? x) (eq? (car x) *visited*))\n\n(define (mark-visited x) (cons *visited* x))\n\n(define (members-of x) (cdr x))\n\n(define (done? x)\n(and (pair? x)\n(visited? x)\n(null? (cdr x))))\n\n(define (draw-fixed-string s)\n(let* ((b (make-string 8 #\\space))\n(k (string-length s))\n(s (if (> k 7) (substring s 0 7) s))\n(s (if (< k 3) (string-append \" \" s) s))\n(k (string-length s)))\n(display (string-append s (substring b 0 (- 8 k))))))\n\n(define (draw-atom n)\n(cond ((null? n)\n(draw-fixed-string \"()\"))\n((symbol? n)\n(draw-fixed-string (symbol->string n)))\n((number? n)\n(draw-fixed-string (number->string n)))\n((string? n)\n(draw-fixed-string (string-append \"\\\"\" n \"\\\"\")))\n((char? n)\n(draw-fixed-string (string-append \"#\\\\\" (string n))))\n((eq? n #t)\n(draw-fixed-string \"#t\"))\n((eq? n #f)\n(draw-fixed-string \"#f\"))\n(else\n(error \"draw-atom: unknown type\" n))))\n\n(define (draw-conses n)\n(let draw-conses ((n n)\n(r '()))\n(cond ((not (pair? n))\n(draw-atom n)\n(reverse! r))\n((null? (cdr n))\n(display \"[o|/]\")\n(reverse! (cons (car n) r)))\n(else\n(display \"[o|o]---\")\n(draw-conses (cdr n) (cons (car n) r))))))\n\n(define (draw-bars n)\n(let draw-bars ((n (members-of n)))\n(cond ((not (pair? n)))\n((empty? (car n))\n(draw-fixed-string \"\")\n(draw-bars (cdr n)))\n((and (pair? (car n))\n(visited? (car n)))\n(draw-bars (members-of (car n)))\n(draw-bars (cdr n)))\n(else\n(draw-fixed-string \"|\")\n(draw-bars (cdr n))))))\n\n(define (skip-empty n)\n(if (and (pair? n)\n(or (empty? (car n))\n(done? (car n))))\n(skip-empty (cdr n))\nn))\n\n(define (remove-trailing-nothing n)\n(reverse (skip-empty (reverse n))))\n\n(define (all-vertical? n)\n(or (not (pair? n))\n(and (null? (cdr n))\n(all-vertical? (car n)))))\n\n(define (draw-members n)\n(let draw-members ((n (members-of n))\n(r '()))\n(cond ((not (pair? n))\n(mark-visited\n(remove-trailing-nothing\n(reverse r))))\n((empty? (car n))\n(draw-fixed-string \"\")\n(draw-members (cdr n)\n(cons *nothing* r)))\n((not (pair? (car n)))\n(draw-atom (car n))\n(draw-members (cdr n)\n(cons *nothing* r)))\n((null? (cdr n))\n(draw-members (cdr n)\n(cons (draw-final (car n)) r)))\n((all-vertical? (car n))\n(draw-fixed-string \"[o|/]\")\n(draw-members (cdr n)\n(cons (caar n) r)))\n(else\n(draw-fixed-string \"|\")\n(draw-members (cdr n)\n(cons (car n) r))))))\n\n(define (draw-final n)\n(cond ((not (pair? n))\n(draw-atom n)\n*nothing*)\n((visited? n)\n(draw-members n))\n(else\n(mark-visited (draw-conses n)))))\n\n(if (not (pair? n))\n(draw-atom n)\n(let draw-tree ((n (mark-visited (draw-conses n))))\n(if (not (done? n))\n(begin (newline)\n(draw-bars n)\n(newline)\n(draw-tree (draw-members n))))))\n(newline))\n\n(define dt draw-tree)\n```"
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https://statistics.berkeley.edu/tech-reports/563 | [
"# On the distribution of ranked heights of excursions of a Brownian bridge\n\nReport Number\n563\nAuthors\nJim Pitman and Marc Yor\nCitation\nAnnals of Probability vol. 29, pages 362-384 (2001)\nAbstract\n\nThe distribution of the sequence of ranked maximum and minimum values attained during excursions of a standard Brownian bridge is described. The height of the $j$th highest maximum $M_j$ over a positive excursion of the bridge has the same distribution as $M_1/j$, where the distribution of $M_1$ is given by L\\'evy's formula $P( M_1 > x ) = e^{-2x^2}$. The probability density of the height of the $j$th highest maximum of excursions of the reflecting Brownian bridge is given by a modification of the known $\\theta$-function series for the density of the maximum absolute value of the bridge. These results are obtained from a more general description of the distribution of ranked values of a homogeneous functional of excursions of the standardized bridge of a self-similar recurrent Markov process.\n\nPDF File\nPostscript File"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89735925,"math_prob":0.98814046,"size":876,"snap":"2021-21-2021-25","text_gpt3_token_len":192,"char_repetition_ratio":0.17889908,"word_repetition_ratio":0.02919708,"special_character_ratio":0.21004567,"punctuation_ratio":0.03311258,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99163747,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T14:35:09Z\",\"WARC-Record-ID\":\"<urn:uuid:034cbd11-237a-4470-ac39-96af5735d6ef>\",\"Content-Length\":\"69463\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1548ea76-16b2-4faa-8ac6-f2dd62e2e5d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:2bba6fce-d61c-45e1-a3d8-897fa2009f7a>\",\"WARC-IP-Address\":\"23.185.0.2\",\"WARC-Target-URI\":\"https://statistics.berkeley.edu/tech-reports/563\",\"WARC-Payload-Digest\":\"sha1:F3PBCSTS7XBSX7ZI7TTDALS2GUIASNL4\",\"WARC-Block-Digest\":\"sha1:KWZBZLTJSJCEBB5SS55ZRHE5JZ3CDW2V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988986.98_warc_CC-MAIN-20210509122756-20210509152756-00224.warc.gz\"}"} |
https://what-when-how.com/electromagnetic-waves/gouy-phase-and-matter-waves-electromagnetic-waves-part-2/ | [
"# Gouy Phase and Matter Waves (Electromagnetic Waves) Part 2\n\n### Covariance axp and Gouy phase\n\nIn this section, we calculate the covariance between position and momentum and the Gouy phase for fullerenes molecules considering the free Schrodinger equation. We calculate the phase and show that it is also related to the covariance axp as well as in the case of pure Gaussian states.\n\nStarting from the determinant of the covariance matrix for mixed Gaussian state, Equation (42), we can express axp in terms of the beam width, i.e.,\n\nwhere Wfwhm is measured in the laboratory. The curve for axp, obtained with experimental data of the Ref. (Nairz et al., 2002) through the Equation (45), is showed in Figure 5 and compared with the theoretical value, Equation (41).",
null,
"Fig. 5. Covariance axp as a function of slit width. Solid curve corresponds to our calculation, Equation (41), and the points were obtained of experiment reported in Ref. (Nairz et al., 2002) through the Equation (45). The parameters are the same of Figure 4.\n\nGouy phase for a mixed Gaussian state\n\nA more recent definition justifies the physical origin of the Gouy phase in terms of space enlargement, governed by the uncertainty relation, of a beam whose transverse field distribution is a Gaussian function (or arbitrary) (Feng & Winful, 2001). According to Equation (11) in Ref. (Feng & Winful, 2001) the Gouy phase ^ (t) and the beam width B (t) for a pure Gaussian state of matter waves are related by the expression\n\nHere, we conjecture, based on the obtained results, that this definition holds for partially coherent Gaussian states since the spread of these states is also governed by the uncertainty relation. Thus, for a state given by Equation (34), the Gouy phase is\n\nwhere the factor ^ appears because we are working in one dimension. Note that, again y.(t) is related to ax„ and is affected by the partial coherence of the initial wave packet, i.e.,\n\nIn Figure 6, we show the phase extracted from Equation (48). As expected, the variation in phase is ^/4, because we are dealing with a one-dimensional problem of diffraction and the propagation of the beam will be from t = 0to t = z/vz (Feng & Winful, 2001). This result shows that the existence of a Gouy phase is compatible with the experimental data involving diffraction of fullerene molecules. It is an indirect evidence of the Gouy phase for matter waves (da Paz, 2011; da Paz et al., 2010).\n\n## Quantum lens and Gouy phase for matter waves\n\nIn the previous section, we have shown an indirect evidence for the Gouy phase for matter waves based on the analogy existent between the paraxial equation for wave optics and\n\nFig. 6. Gouy phase as a function of slit width. Solid curve corresponds to our calculation, Equation (47), and the points were obtained of experiment reported in Ref. (Nairz et al., 2002). The parameters are the same of Figure 4.\n\nSchrodinger equation for matter waves (da Paz, 2011; da Paz et al., 2010). Due to this formal similarity a question which arises naturally is if a similar phase anomaly may occur in the region around the focus of an atomic beam. In order to answer this question, in this section we present the evolution of an atomic beam described by a Gaussian wave packet interacting dispersively with a cavity field (da Paz, 2011; da Paz et al., 2007).\n\nThe model we use is the following (Averbukh etal., 1994; Rohwedder & Orszag, 1996; Schleich, 2001): consider two-level atoms moving along the Oz direction and that they penetrate a region where a stationary electromagnetic field is maintained. The region is the interval z = —Lc until z = 0. The atomic linear moment in this direction is such that the de Broglie wavelength associated is much smaller than the wavelength of the electromagnetic field. We assume that the atom moves classically along direction Oz and the atomic transition of interest is detuned from the mode of the electromagnetic field (dispersive interaction). The Hamiltonian for this model is given by\n\nwhere m is the atom mass, px and x are the linear momentum and position along the direction Ox, a\" and a are the creation and destruction operators of a photon of the electromagnetic mode, respectively. The coupling between atom and field is given by the function g(x) = a£2 (x) where a is the atomic linear susceptibility, a = where p2 is the square of the dipole moment and A is the detuning. E(x) corresponds to the electric field amplitude in vacuum. The effective interaction time is tL = ^, where vz is the longitudinal velocity of the atoms. The dynamics of the closed system is governed by the Schrodinger equation\n\nAt t = 0 the state of the system is given by a direct product of the state corresponding to the transversal component of the atom and a field state",
null,
"The field state can be expanded in the eigenstates of the number operator a a\n\nWhen atom and field interact the atomic and field states get entangled. We can then write\n\nwhere\n\nor, if one defines\n\nthe Equation (53) takes the form\n\nNext, we will use the harmonic approximation for g(x) which is a fine approximation provided the dispersion of the wavepacket in the transverse direction b0 is much smaller than the wavelength of the electromagnetic field mode A (Schleich, 2001). Taking the main terms of the Taylor expansion of the function g(x),\n\nwe get\n\nwhere xf = —g\\/2g2 and O2 = ng2/m. In order to obtain focalization of the atomic beam it is crucial that the initial state be compressed in momentum since this initial momentum compression is transferred dynamically to the x coordinate and a focus can be obtained (da Paz et al., 2007; Rohwedder & Orszag, 1996). In fact, the momentum compression is a necessary condition in optics to obtain a well defined focus (Saleh & Teich, 1991).\n\n### Time evolution\n\nAccording to Bialynicki-Birula (Bialynicki-Birula, 1998), the general form of a Gaussian state in the position representation, is given by\n\nwhere x and p are the coordinates of \"center of mass\" of the distribution in phase space and u and v give the form of this distribution.\n\nA dynamic governed by a Hamiltonian quadratic in position and momentum keep the Gaussian shape of a Gaussian initial state. This is the case of the problem treated here. The atomic motion can be divided into two stages: the first, the atom undergoes the action of a harmonic potential when it crosses the region of electromagnetic field while, in the second part, the atom evolves freely. In the two stages, the Hamiltonian governing the evolution are quadratic in atomic position and momentum [cf.Equation (57)]. Since the initial atomic state is Gaussian, we can consider that throughout evolution, such state will preserve the form given by Equation (58). In this case, the parameters x, p, u and v are functions of time, and their respective equations of motion can be derived from the Schrodinger equation. Consider a particle of mass m moving under the action of a harmonic potential. The natural frequency of this movement is On. The Hamiltonian governing this dynamic is given by\n\nIn position representation, the evolution of the state ^ of the particle is governed by the Schrodinger equation\n\nSuppose that the initial state of the particle is Gaussian. We obtain the equations of motion for the parameters x, p, u and v by substituting the general form (58) in equation above, grouping the terms of same power in (x — x), and then separating the real and imaginary parts. This procedure takes six equations for the four parameters mentioned. The system is therefore, \"super-complete\". Eliminating such redundancy, the equations of motion are the following\n\nwhere we define K = u + iv. Here, the dots indicate time derivation. Note that the equations of motion for the coordinates of the centroid of the distribution are equivalent to the classical equations of movement to the position and momentum of a particle moving in a harmonic potential.\n\nA important observation must be made here. One of the two equations removed is not consistent with the others in (61). This equation is the following:\n\nTo see this, just replace the expressions (61a), (61b) in the above equation. We obtain u = 0, which makes no sense, since u represents the inverse square of the width of the Gaussian package. The only way to \"dribble\" this inconvenience is to redefine the general state as\n\nwhere O is a real function of time. This global phase, in general neglected (see, e.g., (Bialynicki-Birula, 1998; Piza, 2001)), ensures the consistency of the equations of motion because, in addition to Equations (61), we must have\n\n0/2 is known as Gouy phase. Equation (64) relates the Gouy phase with the inverse square of the beam width . The same result was obtained for light waves transversally confined in Ref. (Feng & Winful, 2001).\n\n### Focalization of the atomic beam\n\nIn Figure 7 we illustrated how the quantum lens work out. We consider that a initial Gaussian state compressed in the momentum (region I) penetrates in a region where a stationary electromagnetic field is maintained (region II). The atoms and the field inside the cavity interact dispersively. Dispersive coupling is actually necessary to produce a quantum lens, because the transitions cause aberration at the focus (Berman, 1997; Rohwedder & Orszag, 1996; Schleich, 2001). When the atomic beam leaves the region of the electromagnetic field, the atomic state evolves freely and the compression is transferred to the position (region III). Let us assume, as an initial atomic state, the compressed vacuum state\n\nFor the parameters x, where b0 is the initial width of the packet and",
null,
"Fig. 7. Initial atomic compressed state in momentum . The evolution inside the cavity rotates the state and transfer the compression to the position.\n\nand\n\nfor the initial conditions\n\nAlso, from Equation (68) we obtain\n\nNow u 1 is the width of the gaussian wavepacket squared. At this stage\n\nWhen the atomic beam leaves the region of the electromagnetic field, the atomic state evolves freely. The equations of motion can be obtained analogously and we get for t > tL\n\nThe focus will be located in the atomic beam region where the width of the wavepacket is minimal. In other words, when u(t > tL) be a maximum there will be the focus. This will happen when the function\n\nattains its minimum value. The time for which its derivative vanishes is given by\n\ntherefore the focus is located at\n\nThe width of the Gaussian beam that passed through the lens,",
null,
"can be written as\n\nwhere we define\n\nand\n\nThe line was used here to differentiate the beam parameters after the focalization of their parameters before the focalization. We see that the waist of the beam is increased by factor M and the package time aging is increased by the M2 (not confuse with the quality factor Mp). In optics, the amount M is known as magnification factor (Saleh & Teich, 1991). If the state is not initially compressed, i.e., if bn = b\\$, does not exist focalization and in this case b’0 = b0 and r0 = T0 as we can seen by the Equations (79), (80) and (81).\n\nIf we consider an interaction time of atoms with cavity field ti very small, we have the so called thin lens regime. Because when the interaction time is very small, the movement of atoms along the transverse direction is also very small, i.e., the average transverse kinetic energy of atoms is much smaller than the average potential energy (Q(x)} produced by field, ym < (Q(x)} (Averbukh etal., 1994). The rotation angle of the atomic state caused by the interaction with the cavity field pn = Onti is directly proportional to the interaction time, thus, if ti is too small, pn will also be very small. If we consider tyn < 1 and an initial atomic state compressed in the momentum with bn/b\\$ ^ 1, the expression for the focal distance, equation (77), acquires the simple form (Schleich, 2001)\n\n### Phase anomaly\n\nIf we integrate the equation of motion (64) for O considering the expression for B’ (t) given by the Equation (78), we obtain\n\nThe integration interval is taken from tf to t, because the Gouy phase is the phase of the Gaussian state relatives to the plane wave at the focus, i.e., at the focus the Gaussian state is in phase with the plane wave (Saleh & Teich, 1991; Boyd, 1980; Feng & Winful, 2001). At the focus, y = 0, as expected. Therefore, the Gouy phase of the atomic wave function undergoes a change of n/2 near the focus tf. The fact that this variation is only n/2, in contrast with the value of n for the light, is due to the fact that the quantum lens focuses the atomic beam in the Ox direction, keeping the Oy direction unperturbed (i.e., the electromagnetic field acts as a cylindrical lens).\n\n## Experimental proposal\n\nConsider a Rydberg atom with a level structure given in Figure 8 (left). Three Rydberg levels e, g, and i are taken into account. The transition between the states e and g is slightly detuned with a stationary microwave field stored in two separated cavities with frequency <x>, Cj and C2, and completely detuned with the transition g ^ i. These cavities are placed between two Ramsey zones, R\\ and R2, where a microwave mode quasi-resonant with the atomic transition g ^ i is stored (see Figure 8). If the electronic atomic state involve the levels i or g, the field in both Ramsey zones are adjusted to imprint a n/2 Rabi pulse on the internal state of the atom. Then, after the Ramsey zones, the electronic state changes as\n\nand\n\nFig. 8. On the left, atomic energy levels compared with the wavelength of the field inside the cavities C1 and C2. On the right, sketch of the experimental setup to measure the Gouy phase for matter waves. Rydberg atoms are sent one-by-one with well-defined velocity along the z-axis. A slit is used to collimate the atomic beam in the x-direction. The Ramsey zones R^ and R2 are two microwave cavities fed by a common source S, whereas C1 and C2 are two high-Q microwave cavities devised to work as thin lenses for the atomic beam. The field inside these cavities is supplied by common source S. The state of each atom is detected by the detector D.\n\nThe experimental setup we propose to measure the Gouy phase shift of matter waves is depicted in Figure 8 (right). This proposal is based on the system of Ref. (Raimond et al., 2001). Rubidium atoms are excited by laser to a circular Rydberg state with principal quantum number 49 (Nussenzveig et al., 1993; Gallagher, 1994), that will be called state |i), and their velocity on the z direction is selected to a fixed value vz. As it was stated before, we will consider a classical movement of the atoms in this direction, with the time component given by t = z/vz .A slit is used to prepare a beam with small width in the x direction, but still without a significant divergence, such that the consideration that the atomic beam has a plane-wave behaviour is a good approximation.\n\nIf we disregard the cavities C1 and C2, the setup is that of an atomic Ramsey interferometer (Ramsey, 1985). The cavity R^ has a field resonant or quasi resonant with the transition |i) ^ |g) and results in a n/2 pulse on the atoms, that exit the cavity in the state (|i) + |g))/\\/2 (Raimond et al., 2001; Ramsey, 1985; Kim et al., 1999; Gerry & Knight, 2005). After passing through the cavity R1, the atoms propagate freely for a time t until the cavity R2, that also makes a n/2 pulse on the atoms. Calling hand hthe energy of the internal states |g) and |i) respectively, wr the frequency of the field in the cavities R1 and R2 and defining Wgi = — Mi, the probability that detector D measures each atom in the |g) state is (Raimond et al., 2001; Ramsey, 1985; Nogues et al., 1999)\n\nUpon slightly varying the frequency wr of the fields in cavities R1 and R2, the interference fringes can be seen (Raimond et al., 2001; Ramsey, 1985; Nogues et al., 1999)."
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https://www.gibboncode.org/html/HELP_dualClad.html | [
"Below is a demonstration of the features of the dualClad function\n\n## Description\n\nThis function creates patch data defining a cladding on a surface. The cladding is obtained by shrinking the input faces around their mean and by connecting the face sides to form new faces. The output faces cover the edges of the dual of the input surface (hence the name). The output is the set of shrunk faces (same type as input) and a set of new quadrilateral faces connecting the shrunk faces. The shrink factor, which can be a constant or a spatially varying metric on the nodes or faces, defines the face and edge shrink used. The clad method determines whether the output mesh is connected: 1: From shrunk face to shrunk face 2: From shrunk face to shrunk edge to shrunk face 3: From shrunk face to shrunk face through edge (computes intersection at edge which may not be the centre of edge. This method avoids potential \"kinks\" seen for method 2.\n\n## Examples\n\n```clear; close all; clc;\n```\n\nPlot settings\n\n```figStruct.ColorDef='black';\nfigStruct.Color='k';\n```\n\nSpecify test surface. Alter settings to test for different geometries and surface types.\n\n```%Testing settings\ntestCase=2; %1= sphere, 2=bunny, 3=dino\ncutMesh=0; %0=not cut, 1=cut in half\n\nswitch testCase\ncase 1\n[F,V,~]=geoSphere(2,1); % Building a geodesic dome surface model\nshrinkFactor=0.25;\ncase 2\n[F,V]=stanford_bunny('g'); %Bunny\nV_mean=mean(V,1);\nV=V-V_mean(ones(size(V,1),1),:);\nshrinkFactor=0.25;\ncase 3\n[F,V]=parasaurolophus; %dino\nV_mean=mean(V,1);\nV=V-V_mean(ones(size(V,1),1),:);\nshrinkFactor=0.25;\ncase 4\ndefaultFolder = fileparts(fileparts(mfilename('fullpath')));\npathName=fullfile(defaultFolder,'data','libSurf');\n\nF=dataStruct.F;\nV=dataStruct.V;\nshrinkFactor=0.5;\ncase 5\ndefaultFolder = fileparts(fileparts(mfilename('fullpath')));\npathName=fullfile(defaultFolder,'data','libSurf');\n\nF=dataStruct.F;\nV=dataStruct.V;\nshrinkFactor=0.5;\nend\n\nif meshType==2\nend\n\nif cutMesh==1\nlogicKeep=V(:,1)<mean(V(:,1));\nlogicKeep=all(logicKeep(F),2);\nF=F(logicKeep,:);\n[F,V]=patchCleanUnused(F,V);\nend\n```\n\n## Example 1: Explaining the clad method\n\n```cladMethods=[1 2 3]; %1= fact-to-face connections, 2=\n```\n\nVisualize results\n\n```cFigure;\n\nsubplot(1,3,q); hold on;\nh(1)=gpatch(F,V,'kw','k',0.25);\nh(2)=gpatch(Fc,Vc,'gw','g',1);\nh(3)=gpatch(Fq,Vq,'rw','r',1);\nh(4)=plotV([Vc;Vq],'b.','MarkerSize',15);\naxisGeom;\nview(2);\nhl.Location='SouthOutside';\nend\n\ndrawnow;\n```",
null,
"## Example 2: Explaining the shrink factor\n\n```cladMethod=3; %1= fact-to-face connections, 2=\n\nshrinkFactors=linspace(0.1,0.75,3); %A range of shrink factors\n```\n\nVisualize results\n\n```cFigure;\nsubplot(2,2,1); hold on;\ntitle('Input surface');\ngpatch(F,V,'gw','g',1);\n\naxisGeom;\nview(2);\n\nfor q=1:1:numel(shrinkFactors)\n\nsubplot(2,2,q+1); hold on;\ntitle(['Shrink factor: ',num2str(shrinkFactors(q))]);\ngpatch(Fc,Vc,'rw','r',1);\ngpatch(Fq,Vq,'rw','r',1);\ngpatch(F,V,'kw','none',0.25);\naxisGeom;\nview(2);\nend\n\ndrawnow;\n```",
null,
"## Example 3: Spatially varying shrink factors\n\nAn animation will be created to show effect of a spatially varying shrink factor\n\n```cladMethod=3;\n\n% Define spatially varying shrink factor\nshrinkFactor=-V(:,1);\nshrinkFactor=mean(shrinkFactor(F),2);\nshrinkFactor=shrinkFactor-min(shrinkFactor(:));\nshrinkFactor=shrinkFactor./max(shrinkFactor(:));\nshrinkFactor=shrinkFactor*1.2;\nshrinkFactor=shrinkFactor+0.05;\nshrinkFactor(shrinkFactor>1)=1;\n\n```\n\nAnimating the effect of the shrink factor\n\nInitialize scene\n\n```hf=cFigure(figStruct);\nhold on;\n\nhp1=gpatch(Fc,Vc,shrinkFactor,'k',1);\nhp2=gpatch(Fq,Vq,'r','none',1);\n\naxisGeom;\ncolormap(viridis(250));\ncaxis([0 1]);\ndrawnow;\naxis off\naxis manual;\n```",
null,
"Animate scene\n\n```nSteps=40; %Number of animation steps\nanimStruct.Time=linspace(0,1,nSteps); %Create the time vector\n\nt=linspace(0,2,nSteps);\nshrinkFactor=-V(:,1);\nshrinkFactor=mean(shrinkFactor(F),2);\nshrinkFactor=shrinkFactor-min(shrinkFactor(:));\nshrinkFactor=shrinkFactor./max(shrinkFactor(:));\nshrinkFactor=shrinkFactor+1;\nminLevel=0.05;\n\nfor q=1:1:nSteps\n\nshrinkFactorNow=shrinkFactor;\nshrinkFactorNow=shrinkFactorNow-t(q);\nshrinkFactorNow(shrinkFactorNow<minLevel)=minLevel;\nshrinkFactorNow(shrinkFactorNow>1)=1;\n\n%Set entries in animation structure\nanimStruct.Handles{q}=[hp1 hp2 hp1]; %Handles of objects to animate\nanimStruct.Props{q}={'Vertices','Vertices','CData'}; %Properties of objects to animate\nanimStruct.Set{q}={Vc,Vq,shrinkFactorNow}; %Property values for to set in order to animate\nend\n\nanim8(hf,animStruct); %Initiate animation\n```",
null,
"",
null,
"GIBBON www.gibboncode.org\n\nKevin Mattheus Moerman, [email protected]\n\nGIBBON footer text\n\nGIBBON: The Geometry and Image-based Bioengineering add-On. A toolbox for image segmentation, image-based modeling, meshing, and finite element analysis.\n\nCopyright (C) 2019 Kevin Mattheus Moerman\n\nThis program is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.\n\nThis program is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.\n\nYou should have received a copy of the GNU General Public License along with this program. If not, see http://www.gnu.org/licenses/."
]
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https://docs.bentley.com/LiveContent/web/STAAD.Pro%20Help-v15/en/GUID-3160EFA0-49F0-4185-81E8-57C616FF9F0D.html | [
"",
null,
"# D8.A.2.4 Compression\n\nThe design capacity of the section against compressive force, the guiding phenomenon is the flexural buckling.\n\n## Limit State Method\n\nThe buckling strength of the member is affected by residual stress, initial bow and accidental eccentricities of load.\n\nTo account for all these factors, the strength of the members subjected to axial compression is defined by buckling class a, b, c or d as per clause 7.1.2.2 and Table 7 of IS 800:2007.\n\nImperfection factor, obtained from buckling class, and Euler’s Buckling Stress ultimately govern compressive force capacity of the section as per clause 7.1.2 of IS 800:2007.\n\nThe buckling class of a solid rod section is determined per Table 10 of the specification.\n\n## Working Stress Method\n\nThe actual compressive stress is given by:\n\nfc = FX/Ae\n\nwhere:\n\n• Ae = The effective section area as per Clause 7.3.2 of the code. This is equal to the gross cross sectional area, AX, for any non-slender (plastic, compact, or semi-compact) section class. In the case of slender sections, this is limited to value of Ae as described below.\n\nThe permissive compressive stress is calculated by first determining the Buckling Class of the section per Table 10 of the code and αYY & αZZ based on Table 7.\n\nFac = 0.6·Fcd\n\nwhere:\n\n• Fcd = the minimum of the values of Fcd calculated for the local Y and Z axis.\n\nFcd = (FYLDmo)/ [φ + (φ2 + λ2]\n\n• λ = the non-dimensional slenderness factor is evaluated for each local Y and Z axis.\n\nλ = (FYLD/Fcc)1/2\n\nφ = 0.5[ 1 + a(λ - 0.2) + λ2\n\n• Fcc = the Euler Buckling Stress.\n\nFcc = π2·E/(Kl/r)2\n\n• K = the effective length factor for bending about either the local Y or Z axis, as provided in the KY and KZ parameters, respectively.\n• r = radius of gyration about the local Y or Z axis for the section.\n• FYLD = The yield strength of steel specified in the FYLD parameter.\n\n## Slender Sections\n\nFor member with slender section under axial compression, design compressive strength should be calculated on area ignoring depth thickness ratio of web in excess of the class 3 (semi-compact) limit.\n\nRefer to clause 7.3.2 and Table 2 of IS 800:2007, (corresponding to Internal Element of Compression Flange)\n\nAe= Ag - (d/tw - 42ε) · tw 2\n\nwhere:\n\n• Ae = Effective area of section.\n• Ag = Gross area of section.\n• d = Depth of web.\n• tw = thickness of web.\n\n## Flexural-Torsional Buckling for Single Angles\n\nIn the case of single angles (ST and RA), a axial compression load does not pass through the member centroid. In practical applications, the load must pass through one of the legs. The deviation between the load and the centroid can be considerable, and thus the effect of flexural-torsional buckling must be considered.\n\nThe parameters of ANG, FXTY, and NBL are used to control these calculations.\n\nThe calculations for the flexural-torsional buckling strength of a single angle member in compression are performed per section 7.5.1.2 of the code (either limit state method or working stress method are applicable)."
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null,
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https://www.growingwiththeweb.com/2013/12/selection-sort.html | [
"",
null,
"# Selection sort\n\nPublished , updated\nTags:\n\nSelection sort is an O(n^2) sorting algorithm that works by searching through a list to find the minimum element and swapping it for the first in the list. After every swap, selection sort is performed on the list with the head removed (ie. the minimum element). Due to the way that elements are swapped anywhere in the list, this is not a stable sort.\n\nSelection sort is similar in complexity to insertion sort but almost always performs worse. This is due to the fact that selection sort has an exact number of comparisons based on n, which can be defined using the arithmetic progression:\n\n(n - 1) + (n - 2) + ... + 2 + 1 = n(n - 1) / 2\n\nThis makes its best case always contain the same amount of comparisons as its worst.\n\nWhile selection sort is faster than most O(\\log n) sorts for small lists, insertion sort is normally the preferable choice. It’s main favourable property is that it will perform at most n - 1 element swaps, so it may be useful if swapping is expensive.\n\n## Complexity\n\nTime Space\nWorst case Best case Average case Worst case\nO(n^2) O(n^2) O(n^2) O(1) auxiliary\n\n## Comparing to heapsort\n\nHeapsort uses the exact same technique that selection sort does in finding the minimum element and then ‘detaching’ the first element from the list and sorting the remainder. The only difference between them is that instead of searching for the minimum element every iteration, heapsort utilises the heap data structure to organise the sub-list and guarentee O(n \\log n) run time.\n\n## Visualisation\n\nUse the controls to playback selection sort on the data set.\n\nSee the Sorting Visualiser page for more visualisations.\n\n## Code\n\npublic class SelectionSort<T> : IGenericSortingAlgorithm<T> where T : IComparable\n{\npublic void Sort(IList<T> list)\n{\nfor (int i = 0; i < list.Count - 1; i++) {\nint minIndex = i;\nfor (int j = i + 1; j < list.Count; j++) {\nif (list[j].CompareTo(list[minIndex]) < 0) {\nminIndex = j;\n}\n}\nif (minIndex != i) {\nSwap(list, i, minIndex);\n}\n}\n}\n\nprivate void Swap(IList<T> list, int a, int b)\n{\nT temp = list[a];\nlist[a] = list[b];\nlist[b] = temp;\n}\n}\npublic class SelectionSort {\npublic static void sort(Integer[] array) {\nfor (int i = 0; i < array.length - 1; i++) {\nint minIndex = i;\nfor (int j = i + 1; j < array.length; j++) {\nif (array[j] < array[minIndex]) {\nminIndex = j;\n}\n}\nif (minIndex != i) {\nswap(array, i, minIndex);\n}\n}\n}\n\nprivate static void swap(Integer[] array, int a, int b) {\nInteger temp = array[a];\narray[a] = array[b];\narray[b] = temp;\n}\n}\n/**\n* Sorts an array using selection sort.\n* @param {Array} array The array to sort.\n* @param {function} compare The compare function.\n* @param {function} swap A function to call when the swap operation is\n* performed. This can be used to listen in on internals of the algorithm.\n* @returns The sorted array.\n*/\nfunction sort(array, compare, swap) {\nfor (var i = 0; i < array.length - 1; i++) {\nvar minIndex = i;\n\nfor (var j = i + 1; j < array.length; j++) {\nif (compare(array, j, minIndex) < 0) {\nminIndex = j;\n}\n}\n\nif (minIndex !== i) {\nswap(array, i, minIndex);\n}\n}\n\nreturn array;\n}\ndef default_compare(a, b):\nif a < b:\nreturn -1\nelif a > b:\nreturn 1\nreturn 0\n\ndef sort(array, compare=default_compare):\nfor i in range(0, len(array)):\nmin_index = i\nfor j in range(i + 1, len(array)):\nif compare(array[j], array[min_index]) < 0:\nmin_index = j\nif min_index != i:\narray[i], array[min_index] = array[min_index], array[i]\nreturn array\nmodule SelectionSort\n# Sorts an array using selection sort.\ndef self.sort(array, compare = lambda { |a, b| a <=> b })\n(0..array.length - 1).each do |i|\nminIndex = i\n(i + 1..array.length - 1).each do |j|\nif compare.call(array[j], array[minIndex]) < 0\nminIndex = j\nend\nend\nif minIndex != i\narray[i], array[minIndex] = array[minIndex], array[i]\nend\nend\nend\nend\n\n## Textbooks\n\nHere are two CS textbooks I personally recommend; the Algorithm Design Manual (Steven S. Skiena) is a fantastic introduction to data structures and algorithms without getting to deep into the maths side of things, and Introduction to Algorithms (CLRS) which provides a much deeper, math heavy look."
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"https://www.growingwiththeweb.com/images/site/logo.svg",
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https://mathquestion.net/how-many-times-does-8-go-into-64/ | [
"# How Many Times Does 8 Go Into 64\n\nQuestion : How Many Times Does 8 Go Into 64 ?\n\nAnswer : There are 8 times 8 in 64.\n\n## Solution :\n\nHow many times does 8 go into 64 means are there how many 8 number inside of 64 and we can also say that what if we multiply by 8 is equal to 64 number.\n\n## How to do it step by step :\n\nThere are several ways to calculate how many 8 are there in 64.\n\n### 1. Way\n\nTo find the unknown number we have to divide 64 to 8.\n\n64 / 8 = 8\n\n### 2. Way\n\nTo find how many times 8 is equal 64 we can do reverse multiplication.\n\n(X) x 8 = 64\n\n(X) = 64/8"
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https://bugs.mojang.com/browse/MC-61066 | [
"",
null,
"# Ladders accidentally get Facing from a block that is not a ladder.\n\nXMLWordPrintable\n\n#### Details\n\n•",
null,
"Bug\n• Resolution: Fixed\n• Minecraft 14w28a\n• None\n\n#### Description\n\nThis is an automatically generated report based from the following crash: http://hopper.minecraft.net/crashes/minecraft/MCX-1682676/\n\nCrash:\n\n```java.lang.IllegalArgumentException: Cannot get property bck{name=facing, clazz=class ei, values=[north, south, west, east]} as it does not exist!\nat bbw.b(SourceFile:91)\nat avi.a(SourceFile:47)\nat avi.a(SourceFile:35)\nat apm.a(SourceFile:2320)\nat ait.a(SourceFile:106)\nat cdt.a(SourceFile:295)\nat bqi.ar(SourceFile:1325)\n```\n\n```---- Minecraft Crash Report ----\n// Hi. I'm Minecraft, and I'm a crashaholic.\n\nTime: 7/9/14 10:50 AM\nDescription: Unexpected error\n\njava.lang.IllegalArgumentException: Cannot get property bck{name=facing, clazz=class ei, values=[north, south, west, east]} as it does not exist!\nat bbw.b(SourceFile:91)\nat avi.a(SourceFile:47)\nat avi.a(SourceFile:35)\nat apm.a(SourceFile:2320)\nat ait.a(SourceFile:106)\nat cdt.a(SourceFile:295)\nat bqi.ar(SourceFile:1325)\nat bqi.q(SourceFile:1682)\nat bqi.ap(SourceFile:863)\nat bqi.a(SourceFile:303)\nat net.minecraft.client.main.Main.main(SourceFile:120)\n\nA detailed walkthrough of the error, its code path and all known details is as follows:\n---------------------------------------------------------------------------------------\n\nStacktrace:\nat bbw.b(SourceFile:91)\nat avi.a(SourceFile:47)\nat avi.a(SourceFile:35)\nat apm.a(SourceFile:2320)\nat ait.a(SourceFile:106)\nat cdt.a(SourceFile:295)\nat bqi.ar(SourceFile:1325)\n\n-- Affected level --\nDetails:\nLevel name: MpServer\nAll players: 1 total; [chv['captain_budder23'/187, l='MpServer', x=82.66, y=9.00, z=-2006.23]]\nChunk stats: MultiplayerChunkCache: 625, 625\nLevel seed: 0\nLevel generator: ID 01 - flat, ver 0. Features enabled: false\nLevel generator options:\nLevel spawn location: 114.00,4.00,-2085.00 - World: (114,4,-2085), Chunk: (at 2,0,11 in 7,-131; contains blocks 112,0,-2096 to 127,255,-2081), Region: (0,-5; contains chunks 0,-160 to 31,-129, blocks 0,0,-2560 to 511,255,-2049)\nLevel time: 175536 game time, 196691 day time\nLevel dimension: 0\nLevel storage version: 0x00000 - Unknown?\nLevel weather: Rain time: 0 (now: false), thunder time: 0 (now: false)\nLevel game mode: Game mode: creative (ID 1). Hardcore: false. Cheats: false\nForced entities: 99 total; [aez['Slime'/8, l='MpServer', x=8.09, y=4.00, z=-1966.47], aez['Slime'/15, l='MpServer', x=11.09, y=5.22, z=-2017.66], abg['Horse'/16, l='MpServer', x=5.47, y=4.00, z=-1942.19], acy['item.item.arrow'/19, l='MpServer', x=25.75, y=4.00, z=-2035.16], acy['item.item.arrow'/21, l='MpServer', x=20.78, y=4.00, z=-1986.56], acy['item.item.bone'/20, l='MpServer', x=25.88, y=4.00, z=-2035.88], abg['Horse'/23, l='MpServer', x=17.31, y=4.00, z=-1954.59], acy['item.item.bone'/22, l='MpServer', x=21.13, y=4.00, z=-1987.00], abg['Horse'/25, l='MpServer', x=16.19, y=4.00, z=-1949.34], abg['Horse'/24, l='MpServer', x=17.97, y=4.00, z=-1960.34], chv['captain_budder23'/187, l='MpServer', x=82.66, y=9.00, z=-2006.23], aez['Slime'/38, l='MpServer', x=26.46, y=4.14, z=-2004.69], aez['Slime'/36, l='MpServer', x=12.38, y=5.22, z=-2005.47], aez['Slime'/37, l='MpServer', x=38.85, y=4.14, z=-1990.73], aez['Slime'/40, l='MpServer', x=24.16, y=4.00, z=-1985.97], aez['Slime'/41, l='MpServer', x=30.16, y=4.02, z=-1946.69], aez['Slime'/46, l='MpServer', x=44.75, y=4.00, z=-2082.34], acb['Iron Golem'/50, l='MpServer', x=62.56, y=4.00, z=-1959.13], aez['Slime'/2643300, l='MpServer', x=45.28, y=4.75, z=-2000.59], acb['Iron Golem'/49, l='MpServer', x=62.88, y=4.00, z=-1972.97], afq['Villager'/55, l='MpServer', x=75.16, y=5.19, z=-1969.72], aez['Slime'/54, l='MpServer', x=74.09, y=4.00, z=-2014.19], aez['Slime'/53, l='MpServer', x=84.34, y=4.09, z=-2032.56], abn['Pig'/52, l='MpServer', x=74.03, y=4.00, z=-2040.97], afq['Villager'/59, l='MpServer', x=69.69, y=3.75, z=-1962.56], afq['Villager'/58, l='MpServer', x=73.84, y=6.00, z=-1968.41], afq['Villager'/57, l='MpServer', x=74.84, y=5.66, z=-1968.81], afq['Villager'/56, l='MpServer', x=74.28, y=5.19, z=-1969.72], acb['Iron Golem'/63, l='MpServer', x=65.66, y=5.00, z=-1961.25], afq['Villager'/62, l='MpServer', x=74.28, y=6.06, z=-1966.31], acb['Iron Golem'/61, l='MpServer', x=71.88, y=4.00, z=-1954.31], afq['Villager'/60, l='MpServer', x=73.84, y=6.00, z=-1967.31], acy['item.item.rottenFlesh'/64, l='MpServer', x=73.28, y=4.00, z=-1964.09], afk['Zombie'/2939265, l='MpServer', x=75.72, y=4.00, z=-1934.66], aez['Slime'/84, l='MpServer', x=100.56, y=4.00, z=-2086.97], xf['Painting'/87, l='MpServer', x=82.50, y=5.50, z=-2001.03], aez['Slime'/86, l='MpServer', x=107.47, y=4.00, z=-2082.00], aez['Slime'/2683440, l='MpServer', x=138.81, y=4.00, z=-1959.81], wt['Experience Orb'/93, l='MpServer', x=82.94, y=4.00, z=-1999.75], xf['Painting'/88, l='MpServer', x=87.03, y=5.50, z=-2011.50], wt['Experience Orb'/92, l='MpServer', x=82.94, y=4.00, z=-1999.75], aez['Slime'/3185404, l='MpServer', x=144.59, y=4.01, z=-1952.75], aff['Spider'/95, l='MpServer', x=88.22, y=4.00, z=-1994.88], aff['Spider'/94, l='MpServer', x=83.75, y=4.00, z=-1989.72], xf['Painting'/89, l='MpServer', x=87.03, y=5.50, z=-2008.50], xf['Painting'/88, l='MpServer', x=87.03, y=5.50, z=-2011.50], acy['item.item.bone'/91, l='MpServer', x=82.09, y=4.00, z=-1999.88], xf['Painting'/90, l='MpServer', x=87.03, y=5.50, z=-2005.50], acy['item.item.slimeball'/102, l='MpServer', x=85.78, y=4.00, z=-1962.31], acy['item.item.slimeball'/103, l='MpServer', x=82.13, y=4.00, z=-1965.13], acb['Iron Golem'/100, l='MpServer', x=81.69, y=4.00, z=-1955.75], acy['item.item.slimeball'/101, l='MpServer', x=83.63, y=4.00, z=-1962.31], acb['Iron Golem'/98, l='MpServer', x=82.22, y=4.00, z=-1957.47], acb['Iron Golem'/99, l='MpServer', x=82.97, y=4.00, z=-1954.06], acb['Iron Golem'/96, l='MpServer', x=84.25, y=4.00, z=-1974.69], aby['Snow Golem'/97, l='MpServer', x=89.06, y=1.00, z=-1981.94], abn['Pig'/111, l='MpServer', x=111.69, y=4.00, z=-2042.91], acy['item.item.slimeball'/106, l='MpServer', x=81.66, y=4.00, z=-1952.97], acy['item.item.slimeball'/104, l='MpServer', x=81.97, y=4.00, z=-1956.16], acy['item.item.slimeball'/105, l='MpServer', x=80.75, y=5.00, z=-1952.81], aff['Spider'/119, l='MpServer', x=107.84, y=5.00, z=-1988.31], abb['Bat'/118, l='MpServer', x=109.75, y=9.10, z=-1986.25], abb['Bat'/117, l='MpServer', x=104.75, y=7.10, z=-1990.75], abb['Bat'/116, l='MpServer', x=108.56, y=7.10, z=-1984.25], abb['Bat'/115, l='MpServer', x=101.25, y=7.10, z=-1986.47], abb['Bat'/113, l='MpServer', x=102.25, y=8.10, z=-1985.25], acy['item.item.rottenFlesh'/112, l='MpServer', x=101.28, y=3.00, z=-2015.19], abn['Pig'/127, l='MpServer', x=117.19, y=4.00, z=-1965.88], abn['Pig'/124, l='MpServer', x=125.97, y=4.00, z=-1984.34], xf['Painting'/90, l='MpServer', x=87.03, y=5.50, z=-2005.50], aez['Slime'/121, l='MpServer', x=105.03, y=4.00, z=-1935.22], abb['Bat'/120, l='MpServer', x=109.75, y=7.10, z=-1988.75], abe['Chicken'/137, l='MpServer', x=142.25, y=4.00, z=-1955.81], aez['Slime'/136, l='MpServer', x=139.56, y=4.00, z=-2016.16], abn['Pig'/139, l='MpServer', x=131.84, y=4.00, z=-1960.13], xf['Painting'/87, l='MpServer', x=82.50, y=5.50, z=-2001.03], abe['Chicken'/138, l='MpServer', x=130.22, y=4.00, z=-1958.91], aez['Slime'/141, l='MpServer', x=127.00, y=4.47, z=-1955.81], acy['item.item.egg'/140, l='MpServer', x=130.78, y=4.00, z=-1958.72], aez['Slime'/142, l='MpServer', x=154.38, y=3.91, z=-1967.34], aez['Slime'/131, l='MpServer', x=114.94, y=5.00, z=-1970.50], aez['Slime'/130, l='MpServer', x=121.88, y=4.00, z=-1958.22], aez['Slime'/3435759, l='MpServer', x=49.50, y=4.02, z=-2077.84], aez['Slime'/152, l='MpServer', x=156.19, y=4.42, z=-1963.03], aez['Slime'/153, l='MpServer', x=143.44, y=4.00, z=-1958.84], aez['Slime'/154, l='MpServer', x=132.75, y=4.00, z=-1962.81], aez['Slime'/156, l='MpServer', x=151.63, y=4.00, z=-1956.47], aez['Slime'/145, l='MpServer', x=139.63, y=4.00, z=-1976.78], abe['Chicken'/147, l='MpServer', x=150.28, y=4.00, z=-2012.03], abn['Pig'/149, l='MpServer', x=149.25, y=4.00, z=-1984.19], abe['Chicken'/150, l='MpServer', x=151.66, y=4.00, z=-1969.09], aez['Slime'/151, l='MpServer', x=148.94, y=5.22, z=-1963.66], aez['Slime'/393673, l='MpServer', x=118.53, y=4.02, z=-1957.22], wx['entity.LeashKnot.name'/186, l='MpServer', x=16.50, y=5.50, z=-1948.50], wx['entity.LeashKnot.name'/185, l='MpServer', x=16.50, y=5.50, z=-1954.50], xf['Painting'/89, l='MpServer', x=87.03, y=5.50, z=-2008.50], aez['Slime'/630673, l='MpServer', x=148.44, y=4.00, z=-1989.44], aez['Slime'/630671, l='MpServer', x=160.06, y=5.22, z=-2016.53], aez['Slime'/958191, l='MpServer', x=160.38, y=4.00, z=-2027.09]]\nRetry entities: 0 total; []\nServer brand: vanilla\nServer type: Integrated singleplayer server\nStacktrace:\nat cdu.a(SourceFile:308)\nat bqi.b(SourceFile:2139)\nat bqi.a(SourceFile:317)\nat net.minecraft.client.main.Main.main(SourceFile:120)\n\n-- System Details --\nDetails:\nMinecraft Version: 14w28a\nOperating System: Mac OS X (x86_64) version 10.9.3\nJava Version: 1.6.0_65, Apple Inc.\nJava VM Version: Java HotSpot(TM) 64-Bit Server VM (mixed mode), Apple Inc.\nMemory: 95621832 bytes (91 MB) / 219840512 bytes (209 MB) up to 1065025536 bytes (1015 MB)\nJVM Flags: 1 total; -Xmx1G\nIntCache: cache: 0, tcache: 0, allocated: 0, tallocated: 0\nLaunched Version: 14w28a\nLWJGL: 2.9.1\nOpenGL: Intel HD Graphics 5000 OpenGL Engine GL version 2.1 INTEL-8.26.34, Intel Inc.\nGL Caps: Using GL 1.3 multitexturing.\nUsing GL 1.3 texture combiners.\nUsing framebuffer objects because ARB_framebuffer_object is supported and separate blending is supported.\nShaders are available because OpenGL 2.1 is supported.\n\nIs Modded: Probably not. Jar signature remains and client brand is untouched.\nType: Client (map_client.txt)\nResource Packs: []\nCurrent Language: English (US)\nProfiler Position: N/A (disabled)\n```\n\nYou can see this full report at http://hopper.minecraft.net/crashes/minecraft/MCX-1682676/69236547/\n\n#### People",
null,
"[Mojang] Grum (Erik Broes)",
null,
"[Bot] Hopper"
]
| [
null,
"https://bugs.mojang.com/secure/projectavatar",
null,
"https://bugs.mojang.com/secure/viewavatar",
null,
"https://www.gravatar.com/avatar/f118dcb185754fc4682789a2d168f5a9",
null,
"https://www.gravatar.com/avatar/e95d231c026e9e19cc74b7dc73a26300",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5658166,"math_prob":0.87484455,"size":13038,"snap":"2022-40-2023-06","text_gpt3_token_len":5193,"char_repetition_ratio":0.30305356,"word_repetition_ratio":0.07833788,"special_character_ratio":0.48788157,"punctuation_ratio":0.29845288,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971492,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T00:20:40Z\",\"WARC-Record-ID\":\"<urn:uuid:4c7f703e-e775-4d87-aec9-1291ac19f7c2>\",\"Content-Length\":\"193654\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3491fb02-6c17-4e80-8089-8040716df4a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:79d20291-96e4-484c-af50-88c0b2230ee2>\",\"WARC-IP-Address\":\"20.82.141.145\",\"WARC-Target-URI\":\"https://bugs.mojang.com/browse/MC-61066\",\"WARC-Payload-Digest\":\"sha1:2YPXFJIKHZSUWHBEMXTPN7XLXKZSXRJH\",\"WARC-Block-Digest\":\"sha1:LEWI7VUPBEUFVP633BSTCRRNTN5NOKPY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500294.64_warc_CC-MAIN-20230205224620-20230206014620-00200.warc.gz\"}"} |
https://www.reference.com/web?q=factors+of+6075+that+add+to+9&qo=relatedSearchBing&o=600605&l=dir | [
"https://www.calculatorsoup.com/calculators/math/factors.php\n\nFactoring calculator to find the factors or divisors of a number. Factor calculator finds all factors and factor pairs of any positive non-zero integer. ... 1, 2, 3, 6, 9, 18 .\n\nhttps://www.omnicalculator.com/math/factor\n\nThe factor calculator will get all the factors of any positive integer. ... If you need negative ones for some reason, just add the minus in front of every obtained value: ... 9: If the sum of the digits is divisible by 9 , the entire number is divisible by 9 .\n\nhttps://www.mathsisfun.com/numbers/factors-all-tool.html\n\nLearn how to find all factors of a numnber. ... Factors are usually positive or negative whole numbers (no fractions), so ½ × 24 ... 2×10=20, so put in 2 and 10: ...\n\nhttp://mathforum.org/library/drmath/view/57189.html\n\nApr 8, 2001 ... Why do the digits of nine times a single-digit number add up to 9?\n\nhttp://www.hopewell.k12.pa.us/Downloads/Factor%20Pairs.pdf\n\n107. 109. 113. 127. 131. 137. 139. 149. 151. 157. 163. 167. 173. Number Factor Pair. Number. Factor Pair. Number. Factor Pair. 4. 1 & 4. 2 & 2. 18. 1 & 18. 2 & 9.\n\nhttps://learnzillion.com/lesson_plans/6075-use-an-area-model-for-multiplication-of-two-digit-numbers-by-two-digit-numbers\n\nIn this lesson you will learn how to multiply a 2 digit number by another 2 digit number by applying your understanding of the area model for multiplication."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7048178,"math_prob":0.9595639,"size":1401,"snap":"2019-13-2019-22","text_gpt3_token_len":404,"char_repetition_ratio":0.13815318,"word_repetition_ratio":0.0099502485,"special_character_ratio":0.31477517,"punctuation_ratio":0.26190478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T21:02:46Z\",\"WARC-Record-ID\":\"<urn:uuid:d039fb40-d381-4df2-8870-a6257884de20>\",\"Content-Length\":\"69704\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4b34873-5254-4b2e-ab15-b7ea1097dfdb>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a89552d-429f-4cd1-988b-f7678e3e01cd>\",\"WARC-IP-Address\":\"151.101.202.114\",\"WARC-Target-URI\":\"https://www.reference.com/web?q=factors+of+6075+that+add+to+9&qo=relatedSearchBing&o=600605&l=dir\",\"WARC-Payload-Digest\":\"sha1:5GEZ5JDZCXCUUTXBMNOIUG4U4TE5E774\",\"WARC-Block-Digest\":\"sha1:IUXAKQWKQ46NSWGYORUYDLTVNR42N2FI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256958.53_warc_CC-MAIN-20190522203319-20190522225319-00199.warc.gz\"}"} |
https://pubs.geoscienceworld.org/geophysics/article-abstract/80/1/R1/276415/3d-laplace-domain-waveform-inversion-using-a-low?redirectedFrom=fulltext | [
"## ABSTRACT\n\nWe have developed a Laplace-domain full-waveform inversion technique based on a time-domain finite-difference modeling algorithm for efficient 3D inversions. Theoretically, the Laplace-domain Green’s function multiplied by a constant can be obtained regardless of the frequency content in the time-domain source wavelet. Therefore, we can use low-frequency sources and large grids for efficient modeling in the time domain. We Laplace-transform time-domain seismograms to the Laplace domain and calculate the residuals in the Laplace domain. Then, we back-propagate the Laplace-domain residuals in the time domain using a predefined time-domain source wavelet with the amplitude of the residuals. The back-propagated wavefields are transformed to the Laplace domain again to update the velocity model. The inversion results are long-wavelength velocity models on large grids similar to those obtained by the original approach based on Laplace-domain modeling. Inversion examples with 2D Gulf of Mexico field data revealed that the method yielded long-wavelength velocity models comparable with the results of the original Laplace-domain inversion methods. A 3D SEG/EAGE salt model example revealed that the 3D Laplace-domain inversion based on time-domain modeling method can be more efficient than the inversion based on Laplace-domain modeling using an iterative linear system solver."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8509761,"math_prob":0.853175,"size":1395,"snap":"2019-13-2019-22","text_gpt3_token_len":266,"char_repetition_ratio":0.18691589,"word_repetition_ratio":0.0,"special_character_ratio":0.1655914,"punctuation_ratio":0.05263158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9612369,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-26T13:59:42Z\",\"WARC-Record-ID\":\"<urn:uuid:7de43cd1-c885-431f-9e8d-742e7806f010>\",\"Content-Length\":\"103073\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6556a826-1951-4335-aca6-fa908a8bcba3>\",\"WARC-Concurrent-To\":\"<urn:uuid:68c7f65b-dced-4920-bd4d-695b387ad966>\",\"WARC-IP-Address\":\"209.135.222.216\",\"WARC-Target-URI\":\"https://pubs.geoscienceworld.org/geophysics/article-abstract/80/1/R1/276415/3d-laplace-domain-waveform-inversion-using-a-low?redirectedFrom=fulltext\",\"WARC-Payload-Digest\":\"sha1:SGHWPAVC2WCLI3ICSL266QY4WHVPWWXK\",\"WARC-Block-Digest\":\"sha1:QMT7K5XPF5I34QAJOR5DOGEU5TRB4JU6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912205534.99_warc_CC-MAIN-20190326135436-20190326161436-00346.warc.gz\"}"} |
https://www.journaldev.com/33185/python-add-to-array | [
"Filed Under: Python",
null,
"Python doesn’t have any specific data type as an array. We can use List that has all the characteristics of an array.\n\nPython array module can be used to create an array of integers and floating-point numbers.\n\nIf you want to do some mathematical operations on an array, you should use the NumPy module.\n\n## 1. Python add to Array\n\n• If you are using List as an array, you can use its append(), insert(), and extend() functions. You can read more about it at Python add to List.\n• If you are using array module, you can use the concatenation using the + operator, append(), insert(), and extend() functions to add elements to the array.\n• If you are using NumPy arrays, use the append() and insert() function.\n\n## 2. Adding elements to an Array using array module\n\n• Using + operator: a new array is returned with the elements from both the arrays.\n• append(): adds the element to the end of the array.\n• insert(): inserts the element before the given index of the array.\n• extend(): used to append the given array elements to this array.\n``````\nimport array\n\narr1 = array.array('i', [1, 2, 3])\narr2 = array.array('i', [4, 5, 6])\n\nprint(arr1) # array('i', [1, 2, 3])\nprint(arr2) # array('i', [4, 5, 6])\n\narr3 = arr1 + arr2\nprint(arr3) # array('i', [1, 2, 3, 4, 5, 6])\n\narr1.append(4)\nprint(arr1) # array('i', [1, 2, 3, 4])\n\narr1.insert(0, 10)\nprint(arr1) # array('i', [10, 1, 2, 3, 4])\n\narr1.extend(arr2)\nprint(arr1) # array('i', [10, 1, 2, 3, 4, 4, 5, 6])\n``````\n\n## 3. Adding elements to the NumPy Array\n\n• append(): the given values are added to the end of the array. If the axis is not provided, then the arrays are flattened before appending.\n• insert(): used to insert values at the given index. We can insert elements based on the axis, otherwise, the elements will be flattened before the insert operation.\n``````\n>>> import numpy as np\n>>> np_arr1 = np.array([[1, 2], [3, 4]])\n>>> np_arr2 = np.array([[10, 20], [30, 40]])\n>>>\n>>> np.append(np_arr1, np_arr2)\narray([ 1, 2, 3, 4, 10, 20, 30, 40])\n>>>\n>>> np.append(np_arr1, np_arr2, axis=0)\narray([[ 1, 2],\n[ 3, 4],\n[10, 20],\n[30, 40]])\n>>>\n>>> np.append(np_arr1, np_arr2, axis=1)\narray([[ 1, 2, 10, 20],\n[ 3, 4, 30, 40]])\n>>>\n>>> np.insert(np_arr1, 1, np_arr2, axis=0)\narray([[ 1, 2],\n[10, 20],\n[30, 40],\n[ 3, 4]])\n>>>\n>>> np.insert(np_arr1, 1, np_arr2, axis=1)\narray([[ 1, 10, 30, 2],\n[ 3, 20, 40, 4]])\n>>>\n``````\n\n## 4. References\n\nclose\nGeneric selectors\nExact matches only\nSearch in title\nSearch in content\nSearch in posts\nSearch in pages"
]
| [
null,
"https://cdn.journaldev.com/wp-content/uploads/2019/09/python-add-to-array.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.58220464,"math_prob":0.95152116,"size":2339,"snap":"2019-51-2020-05","text_gpt3_token_len":777,"char_repetition_ratio":0.15546039,"word_repetition_ratio":0.05811138,"special_character_ratio":0.40487388,"punctuation_ratio":0.24684684,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9912946,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-22T10:21:46Z\",\"WARC-Record-ID\":\"<urn:uuid:41a6e473-b505-4b5a-a2d3-9abb324e55b6>\",\"Content-Length\":\"164202\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:74c46e8b-185d-45a6-8eb6-798c0398d72d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7ee925ed-e9e4-4b27-a649-422c7ef1af00>\",\"WARC-IP-Address\":\"54.156.9.246\",\"WARC-Target-URI\":\"https://www.journaldev.com/33185/python-add-to-array\",\"WARC-Payload-Digest\":\"sha1:NO5LQR2AAXPYCAE4PHFMMQI253NNLQMT\",\"WARC-Block-Digest\":\"sha1:2NUFQCKIEIV6FIMLCWKGM3JXKH5TF4MJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250606975.49_warc_CC-MAIN-20200122101729-20200122130729-00348.warc.gz\"}"} |
https://de.mathworks.com/matlabcentral/answers/1709685-finding-column-index-of-the-first-instance-of-1 | [
"# Finding column index of the first instance of 1\n\n2 views (last 30 days)\nPelajar UM on 2 May 2022\nCommented: Walter Roberson on 3 May 2022\nI have a logical array like this:",
null,
"I want to extrat the colum index of the cells where the first instance of 1 is detected. Like this:",
null,
"For example, you see here that the first two rows show 4, because that is where 1 is first detected.\n\nWalter Roberson on 2 May 2022\nC = sum(cumprod(~X, 2),1) + 1;\nC will be one more than the number of columns for any row that has no 1.\nWalter Roberson on 3 May 2022\nisempty(idx)\n\nJonas on 2 May 2022\nuse the find() function together with a loop over each row\nPelajar UM on 2 May 2022\nLike this?\nDoesn't work, because it doesn't find the first instance. It finds all the indices that meet this condition.\nfor p=1:n %n is the length of the logical array X\nG(p,:)=find (X(p,:));\nend"
]
| [
null,
"https://www.mathworks.com/matlabcentral/answers/uploaded_files/985090/image.png",
null,
"https://www.mathworks.com/matlabcentral/answers/uploaded_files/985095/image.png",
null
]
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https://answers.sqlperformance.com/questions/4171/high-number-of-row-estimates.html | [
"# High number of row estimates\n\nKdixon 2016-12-20 19:26:34\n\nWhat I can do to improve esitmates? Actual rows in about 10,000 and estimated is 500 billion (500,000,000,000) records.\n\nAaron Bertrand 2017-01-08 17:24:00\nA WHERE clause with 15 AND/OR conditions quickly becomes difficult for any human, never mind the optimizer, to determine proper estimates. There are just too many possibilities to explore. Have you considered simplifying some of this for the optimizer by using #temp tables to store intermediate results, instead of trying to do everything in a single query with stacked CTEs?\nSQLkiwi 2017-01-09 17:49:42\nThe major issue here is that computing the set pta_recent is very inefficient, as the query is currently written. The rest of the query is largely irrelevant from a performance perspective.\n\nIf you are unable to change the current indexing, consider the following query in place of pta_recent:\n\n```SELECT\nD.trc_number,\n-- pivot columns:\nfirst_pta = MAX(CASE WHEN CA.rn = 1 THEN CA.lph_id ELSE NULL END),\nsecond_pta = MAX(CASE WHEN CA.rn = 2 THEN CA.lph_id ELSE NULL END),\nthird_pta = MAX(CASE WHEN CA.rn = 3 THEN CA.lph_id ELSE NULL END)\nFROM\n(\n-- Find the distinct trc_number values using the index\nSELECT DISTINCT\nLH.trc_number\nFROM dbo.legpta_history AS LH\nWHERE\nLH.trc_number > ''\n) AS D\nCROSS APPLY\n(\n-- For each trc_number, find up to 3 highest lph_id values\nSELECT TOP (3)\nrn = ROW_NUMBER() OVER (ORDER BY LH2.lph_id DESC),\nLH2.lph_id\nFROM dbo.legpta_history AS LH2\nWHERE\nLH2.trc_number = D.trc_number\nORDER BY\nLH2.lph_id DESC\n) AS CA\nGROUP BY\nD.trc_number;```\n\nThis can use the existing indexing to obtain the set needed without any sorting:",
null,
"This will be especially efficient if there are many lph_id per trc_number on average.\n\nIf you can change the indexing, consider replacing the existing index on trc_number with one on the same column, but sorted descending:\n\n`CREATE INDEX i ON dbo.legpta_history (trc_number DESC)`\n\nThen use this query:\n\n```SELECT\nN.trc_number,\nfirst_pta = MAX(CASE WHEN N.rn = 1 THEN N.lph_id ELSE NULL END),\nsecond_pta = MAX(CASE WHEN N.rn = 2 THEN N.lph_id ELSE NULL END),\nthird_pta = MAX(CASE WHEN N.rn = 3 THEN N.lph_id ELSE NULL END)\nFROM\n(\nSELECT\nLH.lph_id,\nLH.trc_number,\nrn = ROW_NUMBER() OVER (PARTITION BY LH.trc_number ORDER BY LH.lph_id DESC)\nFROM dbo.legpta_history AS LH\nWHERE LH.trc_number > ''\n) AS N\nWHERE\nN.rn <= 3\nGROUP BY\nN.trc_number;```\n\nExpected plan:",
null,
"There are other refinements possible, including a filtered index, and/or keyed on (trc_number ASC, lph_id DESC) (to potentially allow a parallel plan)."
]
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"https://answers.sqlperformance.com/storage/temp/1828-sp.png",
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"https://answers.sqlperformance.com/storage/temp/1829-sp.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7087338,"math_prob":0.9597764,"size":2539,"snap":"2019-35-2019-39","text_gpt3_token_len":674,"char_repetition_ratio":0.1408284,"word_repetition_ratio":0.089330025,"special_character_ratio":0.26900354,"punctuation_ratio":0.15473887,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9716206,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T22:41:18Z\",\"WARC-Record-ID\":\"<urn:uuid:bbc74a1b-46a4-4849-8c71-b92be55da0e5>\",\"Content-Length\":\"36343\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e73894db-4438-41eb-a509-98a731c5734a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d0a8962-b8ef-4952-8a04-a4184018bc75>\",\"WARC-IP-Address\":\"104.198.3.84\",\"WARC-Target-URI\":\"https://answers.sqlperformance.com/questions/4171/high-number-of-row-estimates.html\",\"WARC-Payload-Digest\":\"sha1:MOT6BDO3CUYXGNZAVHTTDKBGMSJJIU22\",\"WARC-Block-Digest\":\"sha1:HEI6D2TGYEVUI25TKOQQ4Z46DLCYZCHW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027321786.95_warc_CC-MAIN-20190824214845-20190825000845-00217.warc.gz\"}"} |
https://sentence.yourdictionary.com/x-axis | [
"# X-axis sentence example\n\nx-axis\n• The value for the x axis is always given first.\n\n• Consider the two faces perpendicular to the x-axis.\n\n• The self rotation function indicates a six fold rotation axis parallel with the x-axis.\n\n• For clarity, the different plots are displayed on vertically displaced graphs which share the same X axis.\n\n• If a zero fare is charged, consumers will demand bus journeys up to the point where the demand curve cuts the x-axis.\n\n• The points where the graph crosses the x-axis are the solutions.\n\n• The script creates an x-axis if not present in the FITS lightcurve or XRONOS output.\n\n• If there is an x-axis data component this will be used to give the x-axis.\n\n• The second point is where you want the positive x-axis to be positioned.\n\n• It has now been placed on the x and y axes, with the angle q measured from the horizontal x-axis.\n\n• Make sure that the image is offset from the top x-axis by about 40cm but flush to the y-axis.\n\n• If there is an x-axis data component this will be used to give the x-axis data component this will be used to give the x-axis.\n\n• The line graph on the left is a plot of values along along the x axis.\n\n• The script creates an X-axis if not present in the FITS lightcurve or XRONOS output.\n\n• Graph New... Create a new graph (give it a name, and define the x-axis and data series).\n\n• Knowing the maximum or minimum values and where the graph hits the x-axis (solving the quadratic).\n\n• The second point is where you want the positive X-axis to be positioned.\n\n• The asterisk denotes that instruction requires a set of x-axis data.\n\n• Usually the x-axis points are equispaced, defining the domain of the function to be displayed.\n\n• The exported text file contains one column for the x-axis values, with subsequent columns containing intensity data for each selected profile."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.83934045,"math_prob":0.96596086,"size":2490,"snap":"2022-40-2023-06","text_gpt3_token_len":534,"char_repetition_ratio":0.16291231,"word_repetition_ratio":0.23004694,"special_character_ratio":0.2064257,"punctuation_ratio":0.08349901,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959736,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T02:10:16Z\",\"WARC-Record-ID\":\"<urn:uuid:e50539cf-e960-4556-9b49-05b829600b2e>\",\"Content-Length\":\"256530\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49ec1a68-1cd8-4da7-80db-2fdd08006076>\",\"WARC-Concurrent-To\":\"<urn:uuid:71472a7e-2624-4b68-8e8c-a7557cc92936>\",\"WARC-IP-Address\":\"108.138.64.80\",\"WARC-Target-URI\":\"https://sentence.yourdictionary.com/x-axis\",\"WARC-Payload-Digest\":\"sha1:ERTN542HJHJP2E6SFLGBGAM2ZRHA4STI\",\"WARC-Block-Digest\":\"sha1:NVVJ4QF3UPSU7NNKM3TYJECN3DNK3PEL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499697.75_warc_CC-MAIN-20230129012420-20230129042420-00286.warc.gz\"}"} |
https://enseignements.telecom-sudparis.eu/fiche.php?m=21191&l=en | [
"# Algorithm analysis and Computational Complexity\n\n## Catalogue des cours de Télécom SudParis\n\nIGFE CSC 7341\n\nInformatique\n\nMaster\n\nAnglais/English\n\n2,5\n\n21\n\n51\n\n### Coordonnateur(s)",
null,
"### Département\n\n• Réseaux et Services Multimédia Mobiles\n\n### Organisation\n\nCours/TD/TP/projet/examen :\n\n### Acquis d'apprentissage\n\nTuring Machines. Decision problems. Decidable and undecidable problems. The Halting Problem (Lecture and exercises).\nAlgorithm and Problem Complexity. Complexity classes. Time and Space Complexity. Hardness and completeness of (decision) problems (Lecture and exercises).\nSAT-problem. Cook's theorem. 2-SAT and 3-SAT problems (Lecture and exercises).\nHamiltonian path problem and Hamiltonian cycle problem. The Clique problem (Lecture and exercises).\nComplexity issues and corresponding problems in related fields of Communication Networks:\n- Complexity issues in information security (Lecture, discussion and students presentations’).\n- Complexity issues in software testing (Lecture, discussion and students presentations’).\nDiscussion on P =? NP.\nDiscussion on alternative complexity definitions.\n\nIndividual laboratories on the listed subjects will be performed. The tasks include the estimation of the correlation between the theoretically proven complexity and the software quality parameters, such as performance, disc load, energy consumption, etc.\n\n### Prérequis\n\nStudents are required to have good mathematical foundations, including logics, combinatorial analysis, graph theory, Boolean algebra, algorithms and data structures, software engineering and general network architectures.\n\n### Contenu\n\nThe main objective of the course is the study of the analysis of algorithms as well as widely met complexity classes. The students need to know the ‘classical’ problems met in the computer communications and their complexity. After the class, they should also be able to estimate the computational complexity of a given algorithm, in time and space.\n\n### Evaluation\n\nThe evaluation includes 3 hours written exam.\nThe class also contains the continuous evaluation represented as homework, laboratories and students’ presentations.\n\nFiche mise à jour le 22/11/2019"
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"https://trombi.imtbs-tsp.eu/photo.php",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9067619,"math_prob":0.72303796,"size":1837,"snap":"2019-51-2020-05","text_gpt3_token_len":347,"char_repetition_ratio":0.15275505,"word_repetition_ratio":0.008230452,"special_character_ratio":0.17909636,"punctuation_ratio":0.14726028,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95145327,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T16:54:01Z\",\"WARC-Record-ID\":\"<urn:uuid:9ff09c4a-12c2-4061-a617-5b4d2ef0b5b8>\",\"Content-Length\":\"5026\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea688cb6-9d2d-4c62-922d-db3237e5fa84>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe91083a-e710-4546-a813-8b304bd0dabd>\",\"WARC-IP-Address\":\"157.159.10.161\",\"WARC-Target-URI\":\"https://enseignements.telecom-sudparis.eu/fiche.php?m=21191&l=en\",\"WARC-Payload-Digest\":\"sha1:NMWCOAE65RB6SRGF4QSSJLVEZGA55UV7\",\"WARC-Block-Digest\":\"sha1:7ZRNXOFGVJPUFX7B45WX6BPJP3WD6AOX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250624328.55_warc_CC-MAIN-20200124161014-20200124190014-00423.warc.gz\"}"} |
https://mirror.codeforces.com/blog/entry/10084 | [
"### boleyn.su's blog\n\nBy boleyn.su, 10 years ago,",
null,
"376A - Рычаг\n\nwriter: boleyn.su\n\nO(n):\n\nLet mid = position of ^\n\nLet value(x) = x if x is a digit , 0 otherwise.\n\nLet sum = value(i-th char)*(i-mid)\n\nIf sum = 0 then answer = balance\n\nElse if sum<0 then answer = left\n\n376B - Задолженности\n\nwriter: oGhost\n\nO(n^4):\n\nLet f[i][j] = how many money i owes j\n\n#It can be proved we only need to loop n times.\n\nLoop n times do:\n\nFor i,j,k in [1..n]\n\nIf f[i][j]>0 and f[j][k]>0 then\n\nLet delta = min (f[i][j], f[j][k])\n\nDecrease f[i][j] and f[j][k] by delta\n\nIncrease f[i][k] by delta\n\n\nO(m+n):\n\nLet owe[i] = 0 for all i\n\n#Suppose there is an agnecy to help people with debts.\n\n#If you owe someone, you give money to the agency.\n\n#If someone owes you, you get money from the agency.\n\nFor each ai, bi, ci\n\nIncrease owe[ai] by ci\n\nDecrease owe[bi] by ci\n\n\nAnsewr will be sum{owe[i]|owe[i]>0}\n\n375A - Делимое на семь\n\nwriter: oGhost\n\nO(n):\n\nBecause permutation of 1, 6, 8, 9 can form integers that mod 7 equals 0, 1, 2, 3, 4, 5, 6.\n\nSo you can construct answer like this: nonzero digits + a permutation of 1, 6, 8, 9 + zeros.\n\n375B - Наибольшая подматрица 2\n\nwriter: oGhost\n\nO(n*m):\n\n#We can get right[i][j] by O(n*m) dp.\n\nLet right[i][j] = how many continuous 1s is on cell (j, i)'s right.\n\nFor all column i\n\nSort right[i] #You can use O(n) sorting algorithm\n\nFor j in [1..n]\n\n\n\n375C - Окружение сокровищ\n\nwriter: whd\n\n#T = number of treasures, B = number of booms\n\nO(n*m*2^(T+B)):\n\n#State(i, j, ts, bs) means:\n\n# 1. You are at cell (i, j)\n\n# 2. If the i-th bit of ts is 0 i-th treasure cross even edges of current path, otherwise odd edges.\n\n# 3. If the i-th bit of bs is 0 i-th boom cross even edges of current path, otherwise odd edges.\n\nLet dis[i][j][ts][bs] = min step to go to reach state (i, j, ts, bs).\n\nThen we can use bfs algorithm to calculate dis.\n\nThe answer will be max{value(ts) — dis[Si][Sj][ts]}\n\n375D - Дерево и запросы\n\nwriter: whd\n\nO(nlogn) or O(nlog^2n):\n\nUse binary search tree and merge them by rank.\n\nUse binary search tree that supports O(n) merging to get O(nlogn) solution.\n\nO(n*sqrt(n)):\n\nDfs the tree to transform the problem to:\n\nGiven a[i], query [l,r] k.\n\n\nTo solve this problem:\n\nBuild sqrt(n) bucket, put query [l,r] into (l/sqrt(n)+1)-th bucket\n\nFor each bucket\n\nFor thoese queries whose r is also in the bucket (l/sqrt(n) equals r/sqrt(n)), a brute-froce O(n) solution exists.\n\nFor thoes queries whose r is not in the same bucket, let we sort them by l. We will get l[i]<=l[i]<=..<=l[i[k]]<=r[i[k]]<=r[i[k-1]]<=..<=r[i](do not forget we get l[] and r[] by dfs the tree!). Solving them can be done in O(n) too.\n\nSince we only have O(sqrt(n)) buckets, the total time is O(n*sqrt(n)).\n\n\n375E - Красно-черное дерево\n\nwriter: boleyn.su\n\nThis problem can be solved by integer programming:\n\nmin sum{c[i]*x[i]}\n\nsubject to\n\nsum{A[i][j]*x[j]} >= 1 for all i\n\nsum{x[i]} = R\n\nx[i] = 0 or 1 for all i\n\nwhere\n\nc[i] = 1 if node i is black, 0 otherwise\n\nA[i][j] = 1 if distance between i and j is no greater than X, 0 otherwise\n\nR = number of red nodes.\n\n\nAs it is known, integer programming is NP-hard.\n\nThus, this cannot be the solution.\n\nBut we can prove the following linear programing's solution is same as the integer programming's.\n\nmin sum{c[i]*x[i]}\n\nsubject to\n\nsum{A[i][j]*x[j]} >= 1 for all i\n\nsum{x[i]} <= R\n\nx[i] >= 0 for all i\n\nwhere\n\nc[i] = 1 if node i is black, 0 otherwise\n\nA[i][j] = 1 if distance between i and j is no greater than X, 0 otherwise\n\nR = number of red nodes.\n\n\nAnd the known fastest algorithm to solve linear programming is O(n^3.5).\n\nBut in fact due to the property of this problem, using simplex algorithm to solve linear programming is even faster. I think it can be O(n^3), but I have no proof.\n\nSo just use simplex to solve the linear programming problem above.\n\nThe tutorial is not finished yet. More details will be added later.\n\nUPD\n\nThanks to Codeforces users, a lot of details supposed to be added can be found in comments. I am adding something that is not clearly explained in the comments or something that I want to share with you.\n\nDiv1C:\n\n#State(i, j, ts, bs) means:\n\n# 1. You are at cell (i, j)\n\n# 2. If the i-th bit of ts is 0 i-th treasure cross even edges of current path, otherwise odd edges.\n\n# 3. If the i-th bit of bs is 0 i-th boom cross even edges of current path, otherwise odd edges.\n\nLet dis[i][j][ts][bs] = min step to go to reach state (i, j, ts, bs).\n\nThen we can use bfs algorithm to calculate dis.\n\nWe know dis[Si][Sj] = 0.\n\nFor state(i, j, bs, ts) we can goto cell (i-1, j), (i+1, j), (i, j-1) and (i, j+1). Of course, we cannot go out of the map or go into a trap. So suppose we go from cell (i, j) to cell (ni, nj) and the new state is (ni, nj, nts, nbs). We can see if a treasure is crossing through the edge (i, j) - (ni, nj), if i-th treasure is, then the i-th bit of nts will be 1 xor i-th bit of ts, otherwise, the i-th bit of nts and ts will be same. The same for nbs.\n\nWe can reach state (ni, nj, nts, nbs) from (i, j, ts, bs) in one step, so we just need to start bfs from state(Si, Sj, 0, 0) to get the min dis of all state.\n\n\nThe answer will be max{value(ts) — dis[Si][Sj][ts]}\n\nMy submission: 5550863\n\nDiv1D:\n\nTo our surprise, there seems to be many different solutions to Div1D, which is very good. In fact, we thought about changing this problem so that only online algorithm will be accepted, but we didn't have much time to change it. I guess if we only accept online algorithm, the problem will be less interesting becasue we might not have so many different solutions. So, not changing it is a good decision.\n\nHowever, it may(some solution is hard to change to solve the online-version, so I use 'may') be quite simple to solve the online-version of this problem if you have solved the offline-version. You just need to use persistent data structure when implementing binary search trees. You can get more detail from wiki.\n\nDiv1E:\n\nThe meaning of the integer programming:\n\nWe use x[i] to stand whether node i is red or not. So we have:\n\nx[i] = 0 or 1 for all i\n\nThere is a beautiful tree, for each node, exists an red node whose distance to this node is no more than X. So we have:\n\nsum{A[i][j]*x[j]} >= 1 for all i\n\nThere are only R red node. So we have:\n\nsum{x[i]} = R\n\nAnd we need to minimize the swap number, and in fact the swap number equals to number of nodes that changed from black to red. So we need to minimize:\n\nsum{c[i]*x[i]}\n\n\nAfter changing it to linear programming:\n\nFirstly, it is obvious that the solution of the linear programming will not be worse than integer programming, because integer programming has stronger constraint.\n\nSo we only need to show the solution of the linear programming will not be better than integer programming.\n\nTo prove this, we need to show for an optimal solution, there will be an solution which is as good as it and all x[i] is either 0 or 1.\n\n1. Because for \"sum{A[i][j]*x[j]} >= 1 for all i\", there is no need to make some x[i] > 1. It is obvious that if the solution has some x[i] > 1, we can increase x[i] for nodes that are red in the first place, so that there will not be any x[i] > 1 and this solution is as good as the old one.\n\n2. We need to prove in an optimal solution, making some x[i] not being an integer will not get btter solution. It is really hard to decribe it. So just leave you a hint: use the property of trees to prove and consider leaves of the tree.\n\n\nMy submission: 5523033\n\nThere is a nice DP solution too, check this submission 5516578 by Touma_Kazusa.",
null,
"Tutorial of Codeforces Round 221 (Div. 1)",
null,
"Tutorial of Codeforces Round 221 (Div. 2)",
null,
"",
null,
"Comments (91)\n| Write comment?\n » 10 years ago, # | ← Rev. 3 → If the i-th bit of bs is 0 i-th boom cross even edges of current path, otherwise even edges.In the end \"otherwise odd edges\"?\n• » » Fixed. Thank you.\n » 10 years ago, # | ← Rev. 3 → Edit: Nevermind. Just saw the improved solution later.\n » 10 years ago, # | ← Rev. 2 → Can anyone explain author's Div1 B solution?My solution (5511286) is also O(NM), but uses no sorting or dp at all.\n• » » so am i..but i get wa~\n• » » for (int j = 1; j < m; j++) ps[i][j] += ps[i][j — 1];Is not this dynamic programming? :)\n• » » » This is just calculation of prefix sums, rather common thing. To some extent it can be called dp, but as you can see there's no dp in the global idea of the solution.\n » I am not able to understand how to solve \"375A — Divisible by Seven\". I am still a beginner, so please someone explain the solution in more detail. I've seen that it is mentioned \"The tutorial is not finished yet. More details will be added later\" in the end of the tutorial. So, if it is not regarding this question, please include the solution to this question also. Thanks.\n• » » http://mirror.codeforces.com/contest/376/submission/5509530 you can see the solution i have submitted with this logic.\n• » » » +1! i understood the solution. Thanks.\n• » » » rforritz How did you make the DP[] array?? What was the logic behind it ??\n• » » » » I don't know how he did, but I used a calculator to find them out and it only took less than 5 mins. After all there are only 4! = 24 possibilities.\n• » » » » How about following int myints[] = {1,6,8,9}; std::next_permutation(myints,myints+4); ... \n• » » » » 3 years ago, # ^ | ← Rev. 2 → int arr[] = {1689,1698,1869,1896,1968,1986,6189,6198,6819,6891,6918,6981,8169,8196,8619,8691,8916,8961,9168,9186,9618,9681,9816,9861} ; int dec = {0} ; for(int i=0;i<7;i++) { for(int j=0;j<24;j++) { int num = i*10000+arr[j] ; if(num%7==0) dec[i] = arr[j] ; } } for(int i=0;i<7;i++) cout<\n• » » » 10 years ago, # ^ | ← Rev. 2 → rforritz Could u please explain your dp array please.I am not getting reason behind printing out dp[rem] after printing non zero digits.\n• » » » 4 months ago, # ^ | ← Rev. 2 → But why are you doing rem = rem*10+i. ?? This is really mysterious\n » In problem 375B (Maximum Submatrix 2), won't sorting right[i] require O(m) time (using counting sort), since the values of right[i] may vary from 0 to m? That'd change the overall complexity of the algorithm described to O(m*(m+n)).\n• » » pretty much the same complexity anyway.\n• » » » Yeah. That's true. I was (and still am) just trying to determine whether my understanding of the given solution is correct.\n » In the explanation of Div1 D this makes no sense to me:l[i]<=l[i]<=..<=l[i[k]]<=r[i[k]]<=r[i[k-1]]<=..<=r[i]How do we know that r[i[k]]<=r[i[k-1]] ?? We sorted by l, so what if 2 queries overlap? This seems to be saying that all queries are nested.\n• » » do not forget we get l[] and r[] by dfs the tree!\n• » » » what is l[i[k]]?? i don't understand the O(nsqrt(n)) soloution!do you have an implementation? tnx\n• » » » » 10 years ago, # ^ | ← Rev. 2 → I think, that i, i, ..., i[k] is a permutation of queries in one bucket after sorting them by l.Because we get l[] and r[] by dfs() they can not overlap.And because every r[i[j]] in other bucket, finally we get l[i] <= l[i] <= ....You can look at my submission, it a bit differs from editorial, but maybe it will help.\n• » » » » » 10 years ago, # ^ | ← Rev. 2 → tnx for your answer i think i get it now but how does queries not overlapping improve the time complexity?? i think your submission is O(nsqrt(n)) even if queries could overlap, am i wrong?\n• » » » » » » 10 years ago, # ^ | ← Rev. 3 → Yes, my submission always works O(n·sqrt(n)), but if we iterate buckets, sort queries in each bucket and brute force all small (lenght < sqrt(n)) queries, after it, we can just expand our segment from smallest to largest in bucket, this implementation can be easier (just call add(l[i + 1], ..., l[i] - 1) and add(r[i] + 1, ..., r[i + 1])).\n• » » » » » thanks for sharing this! I am stucked at the point of calculating suffix sum(i have done it using segment tree) but your solution provide me the way! Thanks :)\n » 10 years ago, # | ← Rev. 2 → We can get a simpler O(nlogn) solution for problem D by combining rng_58 / crx's solution and bakabakashyoshyo's solution.Implementation:5516493\n• » » Can somebody explain why rng_58's solution works? I can't understand why executing clear_all and add_all for each child except the one with maximum subtree in the dfs of each vertex works in time?\n• » » » We can focus on the number of \"clears\", that is, how many times each node is cleared.Let size[u] be the size of subtree u. Suppose we are performing \"dfs(u)\" and w is a son of u, v is in the subtree of w. Then, if v is cleared in \"dfs(u)\", size[u] will be at least 2*size[w]. That is, every time v is cleared, the size of the subtree v belongs to will double. So each node v will be cleared at most O(logn) times and \"clear\" operation will be performed at most O(nlogn) times totallyAnd for each node, the number of \"add\"s = the number of \"clear\"s + 1, so \"add\" operation will also be performed at most O(nlogn) times.\n• » » » » 10 years ago, # ^ | ← Rev. 3 → \"if v is cleared in \"dfs(u)\", size[u] will be at least 2*size[w]\", It's reasonable in binary tree. Is it still true for trees with more than 2 forks? Since, the tree in the problem is not necessarily binary. What I am thinking of, if u have k subtrees, at most (k-1)/k * size(u) vertices will be cleared. That is each \"clear\" will increase the tree size to at least k/(k-1) the size of the subtree. when k=2, it will at least double the tree size.\n• » » » » » let b be the largest son of u, it's obvious that size[b] >= size[w], then size[u] > size[b] + size[w] >= 2*size[w].Note that here size[w] isn't the size of all the subtrees other than b, it's just the size of the subtree that node v currently belongs to.\n• » » » » » 6 years ago, # ^ | ← Rev. 2 → I have the same doubt, on each step we clear all but the largest subtree. Does this imply the complexity is O((n + q) × lg(n))? Can someone explain.\n » 10 years ago, # | ← Rev. 2 → For question \"Divisible by Seven\" IN the question it says Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also **mustn't contain any leading zeroes.** But solution says So you can construct answer like this: nonzero digits + a permutation of 1, 6, 8, 9 + zeros.So which one is correct?Can someone explain it more clearly?\n• » » both are correct. Because it is mentioned in the question that It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters. The decimal representation of number a contains at least 4 characters and they are 1,6,8,9. Now according to the answer So you can construct answer like this: nonzero digits + a permutation of 1, 6, 8, 9 + zeros. even if nonzero digits are null a permutation of 1, 6, 8, 9 will be there and hence the answer will not contain leading zeroes. I think you also missed the point that a will at least contain 4 characters like me. So, you felt there is some ambiguity\n• » » » Ok..i try to rephrase the solution in my own words like this. from the number given,i take all the non zero numbers and put them at the start.take try all permutation of 1689 which , when put at the end of the given starting part ,produces 0 mod 7.Then add remaining zeroesAll non zero digits+some permutation of 1689 such that the number(as a whole) produces 0 mod 7+ remaining zeroes.is that correct?\n• » » » » Actually, i was also not able to understand the solution till i asked in comments and rforittz commented. I think you will understand better after seeing his solution, http://mirror.codeforces.com/contest/376/submission/5509530.\n » I used the O(m+n) method to solve Problem 376B, but got WA.I saw accepted solutions of others where they added the abs() of the -ve values in the array and then divided the total by 2, but I didn't understand the logic behind it.Please help!\n• » » Because abs (sum of all negative numbers in the array) = abs (sum of all positive numbers in the array)\n » for problem B Div2: My accepted solution gives output as 290 for this testcase 3 3 1 2 10 2 3 30 3 1 300 shouldn't the answer be 310?\n• » » for ur case if we do like this a-=10,a+=10; a-=30,a+=30; a-=300,a+=300; so final values are a=290,a=-20,a=-270; here u see only a is positive , meaning that it is the only value of dept that is valid in the system,negative values just mean that the corresponding people have to pay the dept , but only the positive values are going to decide the total dept to be paid after deleting the cycles...\n• » » no 290 is correct\n » Please Someone Clarify the idea of Div-1 A ( also Div-2 C) Divisible by SevenAnd in this solution how DP[] array was made?\n• » » Suppose you have a number with x digits, since the problem says the number contains digits 1,6,8,9 we can remove these digits from the initial number... Let's imagine the number with the four digits removed is y. After getting y%7 we can have 7 possible values ,[0..6].We want a number divisible by 7, so we want the entire number to be 0 modulo 7. So, for each possible value of y%7, we have to get one permutation of 1689 so that this permutation modulo 7 + y%7 = 0 (mod 7).If you get the value modulo 7 of all the permutations of 1689, you'll notice that all of the seven values,[0..6], are possible. Therefore, there will always be a permutation so that y concatenated with it will end up on 0 modulo 7.About the question: Probably the writer of this solution generated all of the 24 permutations and chose seven of them, each one giving a different value mod 7.\n• » » » 10 years ago, # ^ | ← Rev. 2 → Thanks dcms2 for helping me big time understanding the solution :)\n » I can't understand the following part in DivI C: Then we can use bfs algorithm to calculate dis.Can anyone help me?\n• » » » Nice, thanks.\n » In the div2 B task, why we need to increase a[i][k] by delta as author indicated? It seems to me I don't get idea, could anyone explain me author's solution.\n• » » you can use the second approach. which is much easier and faster.you can think like this way- for each person how much of his money is in the field. this value is the difference between the money he owes and the money other people owes him. if the value is positive then he has money in the field, sum it up with the final answer.\n• » » Here is my solution 5505331 hope this will help you\n• » » A owes B delta+x, B owes C delta+y, A owes C zis equal to A owes B x, B owes C y, A owes z+deltaand you can see, the total is decreased by delta (delta+x+delta+y+z-x-y-z-delta=delta).\n » I finally found time to properly code my idea for Div1 D in",
null,
". It uses a lot of simple concepts in tree programming: split the colours into heavy (frequent,",
null,
"occurences) and light (rare,",
null,
"occurences) ones; calculate all the answers for light colours and queries with",
null,
", and for every query, iterate over heavy colours and check how many times they appear in the subtree convert subtrees into intervals by pre-order DFS — vertex i gets an interval I[i] = [I[i].x, I[i].y); for every heavy colour c, build an array of prefix sums of array A, where A[I[i].x] = 1 if the colour of i is c and A[I[i].x] = 0 otherwise; the number of occurences of colour c in subtree of i is the sum of",
null,
", which can be obtained from the prefix sums in O(1) for light colours, we'll build another array B[j][i]: how many light colours appear exactly j times in the subtree of i, for",
null,
", and then convert it to prefix sums (not \"exactly j\", but \"at most j\") in-place How to build B? Firstly, we can iterate over all light colours, over all vertices of the given colour, and put 1 in a BIT for every vertex of that colour (BIT[I[i].x] = 1 for vertex i); then we can count occurences of that colour in any subtree (interval I) in",
null,
". Now, imagine that we're just doing a bruteforce solution: for every vertex i of the tree, we ask the BIT how many (a) colours there are in its subtree, and increment B[a][i]. We can do this more efficiently by noting that if there are k vertices of colour c, then there are just O(k) vertices for which the a we get is different from the one we get for that vertex's parent. If we knew the lowest ancestor j of i for which a is different, we could just decrement B[a][j], increment B[a][i] and after doing that for all colours and all vertices where it's required, we could process the vertices i in reverse order of BFS from the root and keep adding B[j][i] to B[j][parent[i]], after which we end up with the same array B. Also note that we only need to do this for vertices where a changes — e.g. for all those of colour c and those which we note in this algorithm as lowest ancestors with different a; that can be done efficiently as long as we process them in reverse BFS order.We also need to know how to quickly get the lowest ancestor with different a. We can use an LCA preprocessing table for that — it tells us the 2k-th ancestor of every vertex. We can try to get to the highest ancestor of i from that table with the same a by trying the highest possible k, decreasing k if that ancestor has different a and moving to the ancestor otherwise; note that after successfully moving to the ancestor, we don't have to consider higher k again, because if that was successful, it'd be successful before the jump as well, which is impossible. Checking a takes",
null,
"time and possible k go up to",
null,
", so the time required for that is",
null,
"per vertex. (Beware of the case where no ancestor with different a exists.)Since every vertex has just 1 colour, we only do the above O(N) times total, so this part of the algorithm takes",
null,
"time. Then there's",
null,
"completing of arrays A and B to prefix sums and",
null,
"time for checking the prefix sums of A for every query; from prefix sums of B, we can get the part of the answer over all light colours in O(1).Code: 5520580\n• » » I had the same thought during the contest but could only come up with a",
null,
"implementation for building B... Your idea brightens me a lot!\n• » » » we can also use Mo s algorithim to solve this question by modifying the euler tour of the tree,and then problem is just not calculate number of values in [l,r] that have frequency >=k\n » I don't know if anyone is interested in knowing that, but I can hack my own solution to A :P. I chose first positions of 1, 6, 8, 9 and iterated over all their permutations and if that failed, I was iterating over all permutations of input until I found a solution — first phase was my first idea and I added second phase \"for safety reasons\" :D. This is an example of input which causes my code to gives TLE. 10000060000080000090000000000000000000077777777777777 And here is my code: http://mirror.codeforces.com/contest/375/submission/5503504 Swapping 1, 6, 8, 9 won't change residue of this number in dividing by 7 and so changing this number by next_permutation (for a large number of iterations) :P.\n » 10 years ago, # | ← Rev. 2 → can someone plz give a detailed explanation for div 2 problem 4 ,,cant figure out the logic of the tutorial..thnx\n• » » A submatrix composed of some subrows, all of them starting and ending in the same column.We'll pre-calculate one useful array: A[i][j] is the maximum number of consecutive 1-s to the right of cell (i, j), including (i, j). That can be counted by simple DP: for (i, j) containing a 0, A[i][j] = 0 and for (i, j) containing a 1, A[i][j] = A[i][j + 1] + 1, because we can take all consecutive 1-s to the right of cell (i, j + 1).Imagine we chose that starting and ending column (l and r). To get the largest answer, we need to count how many subrows with that starting and ending column are composed of just 1-s. That's the same as counting all i for which A[i][l] ≥ r - l + 1.We need to do this efficiently. We can iterate over all l, but not over all r. Still, if we listed and sorted all A[i][l] for given l, then we don't need to consider all r, just all A[i][l] and for the k-th largest of them, there are N - k + 1 larger/equal A[i][l]s. The sorting can be done in linear time (CountSort) and we only need to traverse the sorted array once for every l.\n » thnx @Xellos for such a nice explanation..\n » Can someone please explain how to solve Div1 D with treap? I saw some solutions with treap, but I don't unterstand it.\n » In Div1 B, I get TLE if I read the input using cin (here), but get AC if I read it using scanf (here). Reading is O(MN), so it should run fine, given M, N <= 5000. I prefer using cin. Why is it slower in this case and when should I watch out?\n• » » Reading 25 million numbers can TLE if you're using slow i/o methods (cin without sync_with_stdio(), scanf on individual characters etc.).\n• » » » Thanks. Tried using cin on individual characters with sync_with_stdio() and worked (5530346).\n » 10 years ago, # | ← Rev. 2 → Can anyone explain me the logic behind the solution for 375B?OH god, just find out the error in my thinking!\n » In Div 1 C , how can we pass from one state to another state in BFS ? Can somebody explain this or I must analyse codes ? Authors should write more detailed tutorial .\n• » » It's like a BFS on a graph. If you just wanted to find the shortest path from the start to cell (i, j), you'd use a BFS on the graph whose nodes are cells and edges go between adjacent cells. This time, you add one more information to nodes' decription: a bitmask saying which treasures/bombs are \"inside\" the path drawn so far, by the definition from the problem statement; the edges still go between adjacent free cells and have uniform cost.That BFS just tells you the shortest path in which you include some subset of treasures and bombs; you can iterate over all subsets of treasures and choose the one that maximizes the total gain.\n• » » » Thank you very much . I have one more question . Why BFS works and DFS doesn't work? When we add another information , which changes the distance between vertices , why BFS will still work ?\n• » » » » ? In this graph you got (number of fields) * 2^(number of treasures and bombs) vertices, I don't understand, what do you mean by \"changing distance between vertices\". (i, j, 1, 0) is completely different vertex from (i, j, 0, 0).\n » PRO D div 1 \"For thoes queries whose r is not in the same bucket, let we sort them by l. We will get l[i]<=l[i]<=..<=l[i[k]]<=r[i[k]]<=r[i[k-1]]<=..<=r[i](do not forget we get l[] and r[] by dfs the tree!). Solving them can be done in O(n) too.\"Can anybody tell me how to solve it in O(n)?\n• » » Oh I see...\n » Can someone please explain me solution to Division 2,Problem D. It is said that the rows can be rearranged. In my understanding it is something like : Row 1 and row3 can be interchanged. but by sorting the matrix how does we achieve that.Can someone please explain that.\n• » » 10 years ago, # ^ | ← Rev. 2 → Let input matrix is Matrix Consecutive sum from right column 123456 row1 ....** 000021 row2 **.*** 210321 row3 **...* 210001 row4 ..**** 004321 now when you are at column 1 make a array ca with the values of column 1 of Consecutive sum from right matrix for column 1 ca = [0,2,2,0] . for column 5 ca will be [2,2,0,2] ( I hope you get the idea )Now follow this algo ans = 0; for each column, starting from 1 to 6 tmp_ans=1; make ca sort ca form largest value to smallest value in ca ( from (ca.size(); to 1 ) tmp_ans*= max(tmp_ans, (ca.size - current position on ca which is i)*ca[i]); ans=max(ans,tmp_ans); Now try to think why we are making ca and sorting ca instead of sorting the whole matrix :)\n• » » » got it ...thanx a lot :)\n• » » » » It is called idea of accumulating everything around us by shuffling so that we can contribute to others.\n » The submission you linked as a DP solution of Div1E is a submission of Acme to Div1C :P.\n• » » fixed, thank you.\n » Can anyone explain me the tutorial of Div1 D, that says: \"For thoes queries whose r is not in the same bucket, let we sort them by l. We will get l[i]<=l[i]<=..<=l[i[k]]<=r[i[k]]<=r[i[k-1]]<=..<=r[i](do not forget we get l[] and r[] by dfs the tree!). Solving them can be done in O(n) too.\" How to solve it in O(n) time?\n• » » you calculate l[i[k]] to r[i[k]] first and then l[i[k-1]] to r[i[k-1]] and then l[i[k-2]] to l[i[k-2]] and so on.\n• » » » Oh, I understood! Thanks a lot!\n » i am having problem with div 2 ques d .. i understand the authors solution but all AC solution give 4 as ans for this test case: 4 3 100 011 000 111ans = 4Should'nt 3 be the ans ? are we allowed to move rows together or we are not looking contiguous rows and columns ? Can any one please explain me the ques if my understanding of ques is wrong\n• » » 4 3 100 011 000 111best answer is to swap 3rd and 4th row: 100 011 111 000 then (2,2) to (3,3) is 2*2=4.\n• » » » so it means row swap was allowed ? I did'nt see it mentioned anywhere in the problem statement .. are column swaps also allowed ?\n• » » » » Even if you didn't see it, it was there. Right in the first line. You are allowed to rearrange its rows.Rearrange = swap all you want.There's nothing mentioned about columns, so you can't swap columns.\n » 9 years ago, # | ← Rev. 2 → DIV1E is similar to this problem : http://acm.hdu.edu.cn/showproblem.php?pid=4735\n » What is the O(n) sorting algorithm in the solution for 375B — Maximum Submatrix ?\n• » » use the counting sort\n• » » » I didn't know this algorithm. This is very cool! Thanks!\n » i was about to code div1 D with nlogN . just saw author's complexity, way awesome idea with easy implementation\n » Can anyone explain the O(N logN )solution of DIV1 D. It is not explained properly in editorial . How to use Binary search tree in this question. Thanks in advance :)\n » 5 years ago, # | ← Rev. 2 → In div1 E, I am unable to prove that any optimal linear programming solution can be converted to an integer solution. Can someone help?\n » There's an alternative solution to problem $D$, which I believe runs in $\\mathcal{O} \\left( N (\\log_2 n)^2 \\right)$. I'm not entirely sure of name of the technique: it's somewhat similar to small to large merging; I've heard the word 'sack' or 'dsu on trees' to describe it.I'll start with the (slow) small-large merging idea. For each subtree of each node, we document three things: The number of occurrences of each color. The set of colors with a given number of occurrences. A segment tree on the size of the set of colors of a given number of occurences. Think of this is a segment tree on an array which tells us how many colors appear $i$ times. Merging is now quite simple: simply update (1), (2), and (3) straightforwardly. To process queries, when we're at a subtree, answer queries of that subtree (offline).However, this will probably TLE because maintaining a segment tree of size up to $10^5$ at each node is unneeded. To fix this, we can do basically the same thing, except the 'sack'/'dsu on tree' version. In short, we need to be able to merge and remove subtrees, and we merge the smaller subtrees into the global smaller one, and remove some of them if needed.153192706\n » question is why the problem author of 'Lever' expected that everyone will know physics ?"
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https://ams-at-ucr.github.io/gradsem/years/2017-2018/ | [
"## UC Riverside Department of Mathematics Fridays 1–2pm in Surge 284\n\n### Organizers\n\n Joe Moeller (email hidden, enable JavaScript) Dylan Noack (email hidden, enable JavaScript) Mike Pierce (email hidden, enable JavaScript)\n\n### Scheduled Talks, Spring 2018\n\n 1 June 2018 Constructing Arithmetic Hyperbolic Surfaces Jonathan Alcaraz In first-year Topology, we construct the so-called “Flat Torus” as the quotient of $2$-dimensional Euclidean space by integer linear combinations of the standard basis. This is used as an example for other topics in topology. In this talk, we will look at the abstract properties of this construction and apply them to hyperbolic space.\n 25 May 2018 Kähler-Einstein metrics on compact cohomogeneity one Fano manifolds via effective approximations Pilar Orellana Kähler-Einstein metrics emerge when a complex, topological manifold, under additional conditions, admits a metric that is both Einstein and Kähler. They are beautiful objects which arise naturally in many facets of mathematics—and moreover, are of great importance in the study of string theory. We want to determine under what conditions a compact Fano manifold of Type I cohomogeneity one admits Kähler-Einstein metrics; for which is done by verifying the classes of the manifolds being Fano manifolds and their stability; however, by using the standard methods currently available to us, this proves to be quite a cumbersome task which yields very limited results. In order to overcome this obstacle, we have developed new specialized methods which are effective at retrieving large-scale information of classes of these compact Fano manifolds and their corresponding Kähler-Einstein properties.\n 18 May 2018 The Eckmann-Hilton Argument and Some Applications Alex Pokorny There is a standard Munkres exercise assigned in 205A which asks to show that the fundamental group of a topological group is abelian. If you venture deeper into algebraic topology, you will stumble across a seemingly unrelated statement: that the higher homotopy groups of a topological space are all abelian. In this talk, I will prove the above statements and explore this idea of proving that a given operation is abelian using the Eckmann-Hilton argument. This argument is simple to prove, yet yields deep results. If time permits, I will generalize the definition of the center of a group and talk about $2$-categories.\n 11 May 2018 Schur-Weyl duality and twisted commutative algebras Derek Lowenberg Schur-Weyl duality describes the link between the representation theories of the symmetric groups and the general linear groups. In this talk, I’ll tell you what it is and how it gives useful equivalences of certain symmetric monoidal categories. Following Sam and Snowden, one can define algebras (and their modules) as objects in such categories, which they call twisted commutative algebras. These in turn are used to study behaviors of families of symmetric and general linear group representations, and so the game continues.\n 4 May 2018 Categorical Computation — Form and Content Christian Williams There is a duality of syntax and semantics – the form of a theory and the content of a model. This is a fundamental idea in category theory, which was introduced by William Lawvere in his 1963 PhD thesis. The notion of Lawvere theory provides an understanding of algebraic structures independent of presentation, improving upon the set-theoretic universal algebra. Soon after, these theories were proven equivalent to monads, the categorical manifestation of duality, through which the algebras of the monad correspond to models of the theory. Theories and monads provide complementary perspectives of algebraic structures, and both are becoming important to theoretical and practical computer science. We discuss the application to distributed computation, where enriched Lawvere theories can be used to create languages, programs, and data structures which have their operational semantics—the ways they can operate in context—integrated into their definition, effecting sound design of software.\n 27 April 2018 Fractals and Finite Approximations with Respect to Noncommutative Metrics Therese-Marie Landry How can fractals be understood from the perspective of noncommutative geometry? Noncommutative geometry analyzes a space by studying the algebra of functions on that space. One of the fundamental tools of noncommutative geometry is Connes’ spectral triple. Via the efforts of Lapidus and his collaborators, there exist spectral triples for the Sierpinski gasket that recover the geodesic metric and encode some of its fractal qualities. Building on the work of Rieffel, Latrémolière introduced a generalization of the Gromov-Hausdorff distance to noncommutative, or quantum, compact metric spaces. Together with Aguilar, Latrémolière applied this new technique in noncommutative geometry—the Gromov-Hausdorff propinquity—to the space of continuous complex valued functions on the Cantor set. I am currently working on using the Gromov-Hausdorff propinquity to write the function space for the Sierpinski gasket as a limit of finite-dimensional $C^*$-algebras. In the process, I hope to understand which other fractals can be finitely approximated by noncommutative means.\n 20 April 2017 What is condensed matter and why does it matter? Amir M-Aghaei The physics of a strongly interacting system—condensed matter—is usually drastically different than that of its building blocks; this is known as emergence. In this talk, I introduce the physics of condensed matter starting with a brief survey of how methods of statistical physics can explain some familiar but complicated phenomena around us. In particular, I will describe the physics of liquid-gas transition and discuss different aspects of an old question: why some materials conduct? Finally, I will mention the recent efforts of manipulating emergent physics to build quantum computers.\n 13 April 2018 Open Petri Nets and the Reachability Problem Jade Master In computer science Petri nets are diagrams which are used to represent the transfer of resources in complex interacting systems of agents. These systems don’t usually exist in isolation and instead have inputs and outputs corresponding to external or environmental factors. To model this interconnectedness we define open Petri nets; Petri nets which can be glued together along specified inputs and outputs. We form a category of open Petri nets with open Petri nets as morphisms between their sets of inputs and outputs. Computer scientists are often interested in which states of a Petri net are reachable from a given initial state. We will put the category of open Petri nets to use by constructing reachability as a pseudo functor from the category of open Petri nets to the category of relations.\n 6 April 2018 Model Theory and the Ax-Grothendieck Theorem Mike Pierce Model theory, from the perspective that I'll be talking about today, is the study of algebraic structures using ideas of pure logic. Or as logician Wilfrid Hodges said, model theory is algebraic geometry minus the fields. In this talk I'll start with a brief introduction to model theory, talk about completeness and compactness, and develop some facts about the theory of algebraically closed fields. Then if all goes well, this will culminate in a fantastic proof of the Ax-Grothendieck theorem, that every injective polynomial function $\\boldsymbol{C}^n \\to \\boldsymbol{C}^n$ is surjective.\n\n### Winter 2018\n\n 16 March 2018 On the structure of complete open Kähler manifolds of positive curvature James Ogaja A central problem in complex geometry is to generalize the classical uniformization theorems on Riemann surfaces to higher dimension. In Kähler geometry, attention has been centered on how curvature affects the holomorphic structure of a Kähler manifold. In this talk I’ll discuss results related to Yau’s uniformization conjecture.\n 9 March 2018 Some Combinatorial Representation Theory Justin Davis Combinatorics is an interesting topic on its own, but is also a very useful tool throughout mathematics. Due to the nature of the subject, combinatorics is extremely prevalent in representation theory, whether it’s classifying all finite dimensional irreducible representations, or decomposing representations into irreducible pieces. I will discuss a combinatorial rule, originally called the “Littlewood-Richardson Rule” for decomposing tensor products of two irreducible representations for the Lie algebra $\\mathfrak{sl}_{n+1}$. This uses some interesting combinatorics of partitions of natural numbers. Lastly, I will discuss how I am using this rule to find the decomposition into irreducible representations of certain representations of $\\mathfrak{sl}_{n+1}$ coming from a family of prime representations of quantum affine $\\mathfrak{sl}_{n+1}$ recently defined by Brito and Chari.\n 2 March 2018 Toric geometry Ethan Kowalenko A torus in normal everyday life is a product of circles, but in algebraic geometry a torus is a variety isomorphic to a product of $\\mathbb{C}^\\ast$’s. A toric variety $V$ is a variety with a dense open subset isomorphic to torus, such that multiplication in the torus extends to a group action on $V$. Recently, I’ve been looking at toric varieties with singularities, and blowing these singular points up (unrelated to Dylan’s talk) to get an overall smooth variety. The theory of toric varieties is actually very nice, with the ability to get almost any information you want about them via lattices and cones. In this talk, I’ll compute some examples of toric varieties, show how to glue affine pieces together, and maybe also compute how to resolve a singular point.\n 23 February 2018 12:30 – 1:30pm Surge 268 Conference Travel Grants and You: Getting the Money You Need Jose Manuel Madrano, Conference Grant Coordinator We all want to travel and make the connections, both to further our studies and to land that job after graduating. Being a grad student is certainly does not make that easy, but the GSA has money to help. In this talk you can ask the GSA officer in charge of these funds any questions you might have about how to qualify for this money, and how to apply!\n 16 February 2018 Towards quantifying fractality Xander Henderson While the term fractal is not well defined in mathematics, we generally understand the term to refer to a set that possess “roughness” or “complexity” at all scales. This complexity can be detected and quantified by studying zeta functions associated to the set. In this talk, we will introduce the distance zeta function associated to a bounded subset of a metric space, then discuss several examples of fractal and non-fractal sets.\n 9 February 2018 Integrability, the singular manifold method and Darboux transformations: an algorithmic procedure to determine solutions. Paz Albares The Painlevé property has been proved to be a powerful test for identifying the integrability as well as a good basis for the determination of many properties of a given (nonlinear) PDE. The singular manifold method, based on the Painlevé analysis, provides the Lax pair and the Bäcklund transformation for the PDE. Furthermore, by employing the Darboux transformation approach, an iterative algorithmic method to obtain recursive solutions from a basic seed solution can be constructed. It will be illustrated by means of some examples, related to Nonlinear Schrödinger equations, in which solutions such as solitons, lumps and rogue waves will be thoroughly discussed.\n 2 February 2018 Gauge Invariance and Charge Conservation Michael McNulty It is of no doubt to us mathematicians that mathematical abstraction is indispensable in our field of study. Yet when we consider mathematical applications to understanding the physical world, to what extent is it useful to separate from the seemingly concrete? In this talk, we will explore the concepts of gauge invariance and the conservation of electric charge through an abstracted lens; the former being a concept whose rich structure is familiar to those undergraduate students of physics who dug deeper than the typical classroom while the latter is an assertion familiar to most high school students. We will see how, given a simple mathematical framework, the phenomena of electromagnetism emerges nearly out of thin air and is completely self-contained within the initial framework. Our level of generality will lend itself toward viewing mathematical abstraction in applications to physics as not just useful but of utmost importance and as a crucial tool for the serious practitioner.\n 26 January 2018 Blowing Things Up with Pinchuk and Frankel Dylan Noack In the complex plane there are a grand total of two simply connected domains: the plane itself and the ball (up to biholomorphism). This amazing result, known as the Riemann Mapping Theorem, has unfortunately proven not to be true in higher dimensions. Thus began the century-long journey to classify simply connected domains in higher dimensional complex space. A plethora of techniques have been developed in that time, and one such technique is the method of rescaling. There are two classic methods, Frankel rescaling and Pinchuk rescaling, each with its own strengths and weaknesses.\n 19 January 2018 The Philosophical Science of Logic Christian Williams This week, I will try to explain the wild idea which led me to pursue mathematics. The introductory talk will serve to foster discussion, and hopefully some real interest. The topic cannot be summarized in an hour, let alone a paragraph. I will not defend a theory, but rather encourage a different way of thinking. Even in the best conditions the subject is extremely subtle and difficult, so I ask that you please come with an open mind. I will challenge basic assumptions, make provocative claims, and speak about something that is frankly still out of my cognitive league - so, a foundation of mutual respect is essential. If the hour can be free of pretense, prejudice, and preconception, we will be, as far as I know, the only people on earth thinking about this fascinating idea. This is a dream to which I am devoting my whole life, and I am excited to share it with you. Thank you for reading, and I hope to see you there.\n\n### Scheduled Talks, Fall 2017\n\n 8 December 2017 The Decay Lemma and Applications Matthew Overduin In the paper titled Decay Properties of Axially Symmetric D-Solutions to the Steady Navier-Stokes Equations, it is claimed that if \\begin{equation} \\int\\limits_{\\boldsymbol{R}^3} r^{e_1} \\left|f(r,z)\\right|^2 \\,\\mathrm{d}x \\leq C \\quad\\quad \\int\\limits_{\\boldsymbol{R}^3} r^{e_2} \\left|\\nabla f(r,z)\\right|^2 \\,\\mathrm{d}x \\leq C \\quad\\quad \\int\\limits_{\\boldsymbol{R}^3} r^{e_3} \\left|\\nabla \\partial_z f(r,z)\\right|^2 \\,\\mathrm{d}x \\leq C \\end{equation} with nonnegative constants $e_1$ , $e_2$ , $e_3$, Then for any $r$ greater than zero we have, \\begin{equation} \\int\\limits_{-\\infty}^{\\infty} \\left|f(r,z)\\right|^2 \\,\\mathrm{d}z \\leq Cr^{-\\frac{1}{2}(e_1+e_2)-1} \\quad \\int\\limits_{-\\infty}^{\\infty} \\left|\\partial_z f(r,z)\\right|^2 \\,\\mathrm{d}z \\leq Cr^{-\\frac{1}{2}(e_2+e_3)-1} \\quad \\left|f(r,z)\\right|^2 \\leq Cr^{-\\frac{1}{4}(e_1+2e_2+e_3)-1} \\,. \\end{equation} While the paper outlines a proof for this lemma, the purpose of this talk is to fill in the gaps of this proof and to show how these estimates are obtained. We will also discuss how this lemma is relevant to solving the Axially Symmetric Navier-Stokes equation as a whole, and other types of equations.\n 1 December 2017 Deformations and nonnegative curvature Lawrence Mouillé In Riemannian geometry, a natural question to ask is “what manifolds admit nonnegative or positive curvature?” This question has lead to interest in deformations of Riemannian metrics and collapse (convergence to a lower dimensional space) of Riemannian manifolds. I will discuss some general results in this area and describe a particular deformation due to Jeff Cheeger in the context of manifolds with isometric group actions.\n 17 November 2017 Analysis on Manifolds via Li-Yau Gradient Estimates Xavier Ramos Olivé We are always told that the motivation for defining a smooth structure on a manifold is to be able to do calculus and analysis on manifolds. But how exactly is this done, and why? Will analysis give us information about our manifold? In this talk we will see how to define some natural differential equations on Riemannian manifolds, and how studying their solutions we can get topological information of the underlying manifold. We will do this via an example: by studying the so called Li-Yau gradient estimates of the heat kernel, with a particular focus to their relationship to the Ricci curvature. These estimates can be used to derive some bounds on the Betti numbers of the manifold. If time permits, we will explore some different strategies to derive the gradient estimate under different curvature assumptions, although to protect our sanity, we will skip all the messy computations. No previous knowledge about the concept of curvature will be required for the talk.\n 3 November 2017 Covariance Computations for the Active Subspace Method Applied to a Wind Model Jolene Britton The method of active subspaces is an effective means for reducing the dimensions of a multivariate function $f$. This method enables experiments and simulations that would otherwise be too computationally expensive due to the high-dimensionality of $f$. By using a covariance matrix composed of the gradients of $f,$ one can find the directions in which $f$ varies most strongly, i.e. the active subspace. The current standard for estimating these covariance matrices is the Monte Carlo estimator. Due to the slow convergence of Monte Carlo methods, we propose alternative algorithmic approaches. The first utilizes a separated representation of $f,$ while the second uses polynomial chaos expansions. Such representations have well-defined sampling strategies and allow for the analytic computation of entries of the covariance matrix. Experimental results demonstrate how the Monte Carlo methods compare to our proposed alternative approaches as applied to a function representing power output of a wind turbine.\n 27 October 2017 Better Faster Strongly Jacobson Modules Tim McEldowney Inventing new math is hard. However, there is a nice work around. Take old math and add an adjective. In this talk, I will build up to my most recent result which looks suspiciously like another theorem. I will start by talking about the base structures I study called ‘$G$-domains’ which are integral domains which are close to being their field of fractions. Next, I will define ‘$G$-ideals’ and ‘Hilbert rings’ which are made from these $G$-domains with some clear examples of these structures from common rings. Afterwards, I talk about the ‘strongly’ adjective and what that does to these objects. Lastly, I close with a game I like to call pin the adjective on your adviser’s theorem.\n 20 October 2017 Network Models Joe Moeller A network is a complex of interacting systems which can often be represented as a graph equipped with extra structure. Networks can be combined in many ways, including by overlaying one on top of the other or sitting one next to another. We introduce network models — which are formally a simple kind of lax symmetric monoidal functor — to encode these ways of combining networks. By applying a general construction to network models, we obtain operads for the design of complex networked systems.\n 13 October 2017 Can a nice variety of variety exist? Ethan Kowalenko Algebraic geometry is notorious for being difficult, as it is broadly the study of the zero sets of polynomials through some assigned rings. I will attempt to describe a very computable class of such zero sets, called toric varieties, by showing literal computations. Like, explicitly.\n 6 October 2017 An Introduction to Hopf Algebras Dane Lawhorne What happens when you take the commutative diagrams that define an algebra and reverse all the arrows? The result is called a coalgebra, and with a few more axioms, you get a Hopf Algebra. In this talk, we will examine the role of Hopf algebras in representation theory. In particular, we will see that the category of left modules over a Hopf algebra has both tensor products and dual modules.\n\n### Talks from Previous Years\n\n 2016–2017 2015–2016 2014–2015 2013–2014 2012–2013"
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https://www.intmath.com/blog/learn-math/interactive-3-d-conic-sections-graph-7068 | [
"# Interactive 3-D conic sections graph\n\nBy Murray Bourne, 10 Mar 2012\n\nI recently added the following 3D interactive graph page to the Plane Analytic Geometry chapter in IntMath.\n\nInteractive 3-D conic graph\n\nA \"conic\" curve is what you get when you slice a double cone by a plane, at different angles.\n\nFor example, a horizontal slice gives us a circle, and if we change the angle of the intersection plane slightly, we'll get an ellipse.\n\nYou can explore how to obtain a parabola and a hyperbola as well. This would work well with an Interactive White Board.\n\nThe link again: Interactive 3-D conic graph\n\nBe the first to comment below.\n\n### Comment Preview\n\nHTML: You can use simple tags like <b>, <a href=\"...\">, etc.\n\nTo enter math, you can can either:\n\n1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):\na^2 = sqrt(b^2 + c^2)\n(See more on ASCIIMath syntax); or\n2. Use simple LaTeX in the following format. Surround your math with $$ and $$.\n$$\\int g dx = \\sqrt{\\frac{a}{b}}$$\n(This is standard simple LaTeX.)\n\nNOTE: You can mix both types of math entry in your comment.\n\n## Subscribe\n\n* indicates required"
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https://www.fmz.com/bbs-topic/3616 | [
"# Backtesting An Intraday Mean Reversion Pairs Strategy Between SPY And IWM\n\nAuthor: , Created: 2019-03-28 10:51:06, Updated:\n\nIn this article we are going to consider our first intraday trading strategy. It will be using a classic trading idea, that of “trading pairs”. In this instance we are going to be making use of two Exchange Traded Funds (ETFs), SPY and IWM, which are traded on the New York Stock Exchange (NYSE) and attempt to represent the US stock market indices, the S&P500 and the Russell 2000, respectively.\n\nThe strategy broadly creates a “spread” between the pair of ETFs by longing one and shorting an amount of the other. The ratio of long to short can be defined in many ways such as utilising statistical cointegrating time series techniques. In this scenario we are going to calculate a hedge ratio between SPY and IWM via a rolling linear regression. This will then allow us to create a “spread” between SPY and IWM which is normalised to a z-score. The trading signals will be generated when the z-score exceeds certain thresholds under the belief that the spread will revert to the mean.\n\nThe rationale for the strategy is that SPY and IWM are approximately characterising the same situation, that of the economics of a group of large-cap and small-cap US corporations. The premise is that if one takes the spread of the prices then it should be mean-reverting, since while “local” (in time) events may effect either the S&P500 or the Russell 2000 indices separately (such as small-cap/large-cap differences, rebalancing dates or block trades), the long-term price series of the two will likely be cointegrated.\n\n## The Strategy\n\nThe strategy is carried out in the following steps:\n\n1. Data - 1-minute bars of SPY and IWM are obtained from April 2007 through to February 2014.\n2. Processing - The data are correctly aligned and missing bars are mutually discarded.\n3. Spread - The hedge ratio between the two ETFs is calculated by taking a rolling linear regression. This is defined as the β regression coefficient using a lookback window which shifts forward by 1 bar and recalculates the regression coefficients. Thus the hedge ratio βi, for bar bi is calculated across points bi−1−k to bi−1 for a lookback of k bars.\n4. Z-Score - The standard score of the spread is calculated in the usual manner. This means subtracting the (sample) mean of the spread and dividing by the (sample) standard deviation of the spread. The rationale for this is to make threshold parameters more straightforward to interpet since the z-score is a dimensionless quantity. I have deliberately introduced a lookahead bias into the calculations in order to show how subtle it can be. Try and look out for it!\n5. Trades - Long signals are generated when the negative z-score drops below a pre-determined (or post-optimised) threshold, while short signals are the converse of this. Exit signals are generated when the absolute z-score drops below an additional threshold. For this strategy I have (somewhat arbitrarily) picked an absolute entry threshold of |z|=2 and an exit threshold of |z|=1. Assuming mean reverting behaviour in the spread, this will hopefully capture that relationship and provide positive performance.\n\nPerhaps the best way to understand the strategy in depth is to actually implement it. The following section describes a full Python code (single file) for implementing this mean-reverting strategy. I have liberally commented the code in order to aid understanding.\n\n## Python Implementation\n\nAs with all of the Python/pandas tutorials it is necessary to have setup a Python research environment as described in this tutorial. Once setup, the first task is to import the necessary Python libraries. For this backtest matplotlib and pandas are required.\n\nThe specific library versions that I am using are as follows:\n\n• Python - 2.7.3\n• NumPy - 1.8.0\n• pandas - 0.12.0\n• matplotlib - 1.1.0 Let’s go ahead and import the librararies:\n``````# mr_spy_iwm.py\n\nimport matplotlib.pyplot as plt\nimport numpy as np\nimport os, os.path\nimport pandas as pd\n``````\n\nThe following function create_pairs_dataframe imports two CSV files containing the intraday bars of two symbols. In our case this will be SPY and IWM. It then creates a separate dataframe pairs, which uses the indexes of both original files. Since their timestamps are likely to be different due to missed trades and errors, this guarantees that we will have matching data. This is one of the main benefits of using a data analyis library like pandas. The “boilerplate” code is handled for us in a very efficient manner.\n\n``````# mr_spy_iwm.py\n\n\"\"\"Creates a pandas DataFrame containing the closing price\nof a pair of symbols based on CSV files containing a datetime\nstamp and OHLCV data.\"\"\"\n\n# Open the individual CSV files and read into pandas DataFrames\nprint \"Importing CSV data...\"\nnames=['datetime','open','high','low','close','volume','na'])\nnames=['datetime','open','high','low','close','volume','na'])\n\n# Create a pandas DataFrame with the close prices of each symbol\n# correctly aligned and dropping missing entries\nprint \"Constructing dual matrix for %s and %s...\" % symbols\npairs = pd.DataFrame(index=sym1.index)\npairs['%s_close' % symbols.lower()] = sym1['close']\npairs['%s_close' % symbols.lower()] = sym2['close']\npairs = pairs.dropna()\nreturn pairs\n``````\n\nThe next step is to carry out the rolling linear regression between SPY and IWM. In this instance IWM is the predictor (‘x’) and SPY is the response (‘y’). I have set a default lookback window of 100 bars. As discussed above this is a parameter of the strategy. In order for the strategy to be considered robust we ideally want to see a returns profile (or other measure of performance) as a convex function of lookback period. Thus at a later stage in the code we will carry out a sensitivity analysis by varying the lookback period over a range.\n\nOnce the rolling beta coefficient is calculated in the linear regression model for SPY-IWM, we add it to the pairs DataFrame and drop the empty rows. This constitutes the first set of bars equal to the size of the lookback as a trimming measure. We then create the spread of the two ETFs as a unit of SPY and −βi units of IWM. Clearly this is not a realistic situation as we are taking fractional amounts of IWM, which is not possible in a real implementation.\n\nFinally, we create the z-score of the spread, which is calculated by subtracting the mean of the spread and normalising by the standard deviation of the spread. Note that there is a rather subtle lookahead bias occuring here. I deliberately left it in the code as I wanted to emphasise how easy it is to make such a mistake in research. The mean and standard deviation are calculated for the entire spread time series. If this is to reflect true historical accuracy then this information would not have been available as it implicitly makes use of future information. Thus we should use a rolling mean and stdev to calculate the z-score.\n\n``````# mr_spy_iwm.py\n\n\"\"\"Creates a hedge ratio between the two symbols by calculating\na rolling linear regression with a defined lookback period. This\nis then used to create a z-score of the 'spread' between the two\nsymbols based on a linear combination of the two.\"\"\"\n\n# Use the pandas Ordinary Least Squares method to fit a rolling\n# linear regression between the two closing price time series\nprint \"Fitting the rolling Linear Regression...\"\nmodel = pd.ols(y=pairs['%s_close' % symbols.lower()],\nx=pairs['%s_close' % symbols.lower()],\nwindow=lookback)\n\n# Construct the hedge ratio and eliminate the first\n# lookback-length empty/NaN period\npairs['hedge_ratio'] = model.beta['x']\npairs = pairs.dropna()\n\nreturn pairs\n``````\n\nIn create_long_short_market_signals the trading signals are created. These are calculated by going long the spread when the z-score negatively exceeds a negative z-score and going short the spread when the z-score positively exceeds a positive z-score. The exit signal is given when the absolute value of the z-score is less than or equal to another (smaller in magnitude) threshold.\n\nIn order to achieve this situation it is necessary to know, for each bar, whether the strategy is “in” or “out” of the market. long_market and short_market are two variables defined to keep track of the long and short market positions. Unfortunately this is far simpler to code in an iterative manner as opposed to a vectorised approach and thus it is slow to calculate. Despite 1-minute bars requiring ~700,000 data points per CSV file it is still relatively quick to calculate on my older desktop machine!\n\nTo iterate over a pandas DataFrame (which admittedly is NOT a common operation) it is necessary to use the iterrows method, which provides a generator over which to iterate:\n\n``````# mr_spy_iwm.py\n\ndef create_long_short_market_signals(pairs, symbols,\nz_entry_threshold=2.0,\nz_exit_threshold=1.0):\n\"\"\"Create the entry/exit signals based on the exceeding of\nz_enter_threshold for entering a position and falling below\nz_exit_threshold for exiting a position.\"\"\"\n\n# Calculate when to be long, short and when to exit\npairs['longs'] = (pairs['zscore'] <= -z_entry_threshold)*1.0\npairs['shorts'] = (pairs['zscore'] >= z_entry_threshold)*1.0\npairs['exits'] = (np.abs(pairs['zscore']) <= z_exit_threshold)*1.0\n\n# These signals are needed because we need to propagate a\n# position forward, i.e. we need to stay long if the zscore\n# threshold is less than z_entry_threshold by still greater\n# than z_exit_threshold, and vice versa for shorts.\npairs['long_market'] = 0.0\npairs['short_market'] = 0.0\n\n# These variables track whether to be long or short while\n# iterating through the bars\nlong_market = 0\nshort_market = 0\n\n# Calculates when to actually be \"in\" the market, i.e. to have a\n# long or short position, as well as when not to be.\n# Since this is using iterrows to loop over a dataframe, it will\n# be significantly less efficient than a vectorised operation,\n# i.e. slow!\nprint \"Calculating when to be in the market (long and short)...\"\nfor i, b in enumerate(pairs.iterrows()):\n# Calculate longs\nif b['longs'] == 1.0:\nlong_market = 1\n# Calculate shorts\nif b['shorts'] == 1.0:\nshort_market = 1\n# Calculate exists\nif b['exits'] == 1.0:\nlong_market = 0\nshort_market = 0\n# This directly assigns a 1 or 0 to the long_market/short_market\n# columns, such that the strategy knows when to actually stay in!\npairs.ix[i]['long_market'] = long_market\npairs.ix[i]['short_market'] = short_market\nreturn pairs\n``````\n\nAt this stage we have updated pairs to contain the actual long/short signals, which allows us to determine whether we need to be in the market. Now we need to create a portfolio to keep track of the market value of the positions. The first task is to create a positions column that combines the long and short signals. This will contain a list of elements from (1,0,−1), with 1 representing a long/market position, 0 representing no position (should be exited) and −1 representing a short/market position. The sym1 and sym2 columns represent the market values of SPY and IWM positions at the close of each bar.\n\nOnce the ETF market values have been created, we sum them to produce a total market value at the end of every bar. This is then turned into a returns stream by the pct_change method for that Series object. Subsequent lines of code clear up the bad entries (NaN and inf elements) and finally calculate the full equity curve.\n\n``````# mr_spy_iwm.py\n\ndef create_portfolio_returns(pairs, symbols):\n\"\"\"Creates a portfolio pandas DataFrame which keeps track of\nthe account equity and ultimately generates an equity curve.\nThis can be used to generate drawdown and risk/reward ratios.\"\"\"\n\n# Convenience variables for symbols\nsym1 = symbols.lower()\nsym2 = symbols.lower()\n\n# Construct the portfolio object with positions information\n# Note that minuses to keep track of shorts!\nprint \"Constructing a portfolio...\"\nportfolio = pd.DataFrame(index=pairs.index)\nportfolio['positions'] = pairs['long_market'] - pairs['short_market']\nportfolio[sym1] = -1.0 * pairs['%s_close' % sym1] * portfolio['positions']\nportfolio[sym2] = pairs['%s_close' % sym2] * portfolio['positions']\nportfolio['total'] = portfolio[sym1] + portfolio[sym2]\n\n# Construct a percentage returns stream and eliminate all\n# of the NaN and -inf/+inf cells\nprint \"Constructing the equity curve...\"\nportfolio['returns'] = portfolio['total'].pct_change()\nportfolio['returns'].fillna(0.0, inplace=True)\nportfolio['returns'].replace([np.inf, -np.inf], 0.0, inplace=True)\nportfolio['returns'].replace(-1.0, 0.0, inplace=True)\n\n# Calculate the full equity curve\nportfolio['returns'] = (portfolio['returns'] + 1.0).cumprod()\nreturn portfolio\n``````\n\nThe main function brings it all together. The intraday CSV files are located at the datadir path. Make sure to modify the code below to point to your particular directory.\n\nIn order to determine how sensitive the strategy is to the lookback period it is necessary to calculate a performance metric for a range of lookbacks. I have chosen the final total percentage return of the portfolio as the performance measure and the lookback range in [50,200] with increments of 10. You can see in the following code that the previous functions are wrapped in a for loop across this range, with other thresholds held fixed. The final task is to use matplotlib to create a line chart of lookbacks vs returns:\n\n``````# mr_spy_iwm.py\n\nif __name__ == \"__main__\":\nsymbols = ('SPY', 'IWM')\n\nlookbacks = range(50, 210, 10)\nreturns = []\n\n# Adjust lookback period from 50 to 200 in increments\n# of 10 in order to produce sensitivities\nfor lb in lookbacks:\nprint \"Calculating lookback=%s...\" % lb\npairs = create_long_short_market_signals(pairs, symbols,\nz_entry_threshold=2.0,\nz_exit_threshold=1.0)\n\nportfolio = create_portfolio_returns(pairs, symbols)\nreturns.append(portfolio.ix[-1]['returns'])\n\nprint \"Plot the lookback-performance scatterchart...\"\nplt.plot(lookbacks, returns, '-o')\nplt.show()\n``````\n\nThe chart of lookback period vs returns can now be seen. Note that there is a “global” maximum around a lookback equal to 110 bars. If we had seen a situation where lookback was independent of returns this would have been cause for concern:",
null,
"SPY-IWM linear regression hedge-ratio lookback period sensitivity analysis\n\nNo backtesting article would be complete without an upwardly sloping equity curve! Thus if you wish to plot a curve of the cumulated returns vs time, you can use the following code. It will plot the final portfolio generated from the lookback parameter study. Thus it will be necessary to choose the lookback depending upon which chart you wish to visualise. The chart also plots the returns of SPY in the same period to aid comparison:\n\n``````# mr_spy_iwm.py\n\n# This is still within the main function\nprint \"Plotting the performance charts...\"\nfig = plt.figure()\nfig.patch.set_facecolor('white')\n\nax1 = fig.add_subplot(211, ylabel='%s growth (%%)' % symbols)\n(pairs['%s_close' % symbols.lower()].pct_change()+1.0).cumprod().plot(ax=ax1, color='r', lw=2.)\n\nax2 = fig.add_subplot(212, ylabel='Portfolio value growth (%%)')\nportfolio['returns'].plot(ax=ax2, lw=2.)\n\nfig.show()\n``````\n\nThe following equity curve chart is for a lookback period of 100 days:",
null,
"SPY-IWM linear regression hedge-ratio lookback period sensitivity analysis\n\nNote that the drawdown of SPY is significant in 2009 during the period of the financial crisis. The strategy also had a volatile period at this stage. Also note that performance has deteriorated somewhat in the last year due to the strongly trending nature of SPY in this period, which reflects the S&P500 index.\n\nNote that we still have to take into account the lookahead bias when calculating the z-score of the spread. Further, all of these calculations have been carried out without transaction costs. This strategy would certainly perform very poorly once these factors are taken into consideration. Fees, bid/ask spread and slippage are all currently unaccounted for. In addition the strategy is trading in fractional units of ETFs, which is also very unrealistic.\n\nIn later articles we will create a much more sophisticated event-driven backtester that will take these factors into consideration and give us significantly more confidence in our equity curve and performance metrics.\n\nMore"
]
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null,
"https://www.fmz.com/upload/asset/6f36e049015ac5e073b0.png",
null,
"https://www.fmz.com/upload/asset/6f02fae6f9d2dc6c7c62.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.84324604,"math_prob":0.92049104,"size":15187,"snap":"2020-45-2020-50","text_gpt3_token_len":3518,"char_repetition_ratio":0.12619378,"word_repetition_ratio":0.008741259,"special_character_ratio":0.23724239,"punctuation_ratio":0.11410119,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9914975,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T20:41:14Z\",\"WARC-Record-ID\":\"<urn:uuid:efb23eb4-3a6d-4ddc-9bfe-d2957df4be29>\",\"Content-Length\":\"26194\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd524557-554e-4c58-87c5-855c16f32496>\",\"WARC-Concurrent-To\":\"<urn:uuid:b1ef353b-54d9-4f5f-8c0f-f36393a51770>\",\"WARC-IP-Address\":\"121.41.115.19\",\"WARC-Target-URI\":\"https://www.fmz.com/bbs-topic/3616\",\"WARC-Payload-Digest\":\"sha1:J65WK4RQOVUAII5RJ24AUZHETC7FTVRN\",\"WARC-Block-Digest\":\"sha1:UNBKII3BMWNPZHGTKEGSJEUTCDDV4X6D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141486017.50_warc_CC-MAIN-20201130192020-20201130222020-00210.warc.gz\"}"} |
https://jeopardylabs.com/print/are-you-smart-than-a-fifth-grader-2 | [
"100\n\nWhat is 5 + 10\n\n15\n\n100\n\nWhat Is The Third Planet from the Sun\n\nEarth\n100\n\nWhat is 5 x 5?\n\n25\n\n100\n\nWhat is 6 Divided By 3?\n\n2.\n100\n\nSolve. 130 Divided by 75\n\n75\n\n200\nWhat is Made of Water That is In California\n\nThe Beach\n\n200\n\nIn Social Studies What is the Most Important Supply You need (ex. 2)\n\nPaper and Pencil\n\n200\n\nWhen Using 'Times', What is the Other Word Mainly Used Called.\n\nMultiplying, Multiply.\n\n200\n\nWhen Solving Equations, What the Dividen for Using Fractions. Top or bottom?\n\nThe top.\n\n200\n\nSolve. 250 Divided by 145.\n\n145.\n\n300\n\nWhat Is The Most healthiest Liquid to Drink?\n\nWater.\n\n300\n\nWhen Saying the Alphabet, Do You Say it A B C or Ah Bh Ch?\n\nA B C\n\n300\n\nWhen Using Multiplication, How do you use it?\n\nYou use it By Mutliplying The second Number By the amount of Groups the first Number Shows\n\n300\n\nWhat is a Fraction?\n\nExamples: 5 fourths\n\n300\n\nSolve. 5= 2+ 59- A3\n\n627\n\n400\n\n18\n\n400\n\nWhat is the 19th Number\n\n19\n\n400\nHow do you Use Math, (ex. In The Words Like 'Addition')\n\n400\n\nDivide. 60 Divided by 30\n\n20.\n\n400\n\nWhat does 'Improv' Mean?\n\nImproving Either Text Segments or Paragraphs.\n\n500\n\nWhat is The fifth Letter of the Alphabet\n\nE\n\n500\n\nWhen Subtracting, What do you do?\n\nFlip the Numbers, Example: 4 + 5 Is Now 5 - 4.\n\n500\n\nSolve. 50 x 10 = A\n\nWhat is A?\n\n50. 50 x 10 = 50\n\nA = 50\n\n500\n\nWhen Using Fractions And Division, Are they the Same or Different.\n\nSame.\n\n500\n\nSolve. 5500 Divided by 2500\n\n2500.\n\nClick to zoom"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.78482497,"math_prob":0.73759574,"size":1276,"snap":"2022-27-2022-33","text_gpt3_token_len":400,"char_repetition_ratio":0.14544025,"word_repetition_ratio":0.0,"special_character_ratio":0.31504703,"punctuation_ratio":0.16040955,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.991774,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-12T02:23:58Z\",\"WARC-Record-ID\":\"<urn:uuid:909371f9-6f2f-481a-b550-761f28b485e0>\",\"Content-Length\":\"24424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f17869b-20f6-4323-98db-30e61c343a19>\",\"WARC-Concurrent-To\":\"<urn:uuid:b697bd68-24a7-419e-9b35-96e89024954a>\",\"WARC-IP-Address\":\"198.100.157.237\",\"WARC-Target-URI\":\"https://jeopardylabs.com/print/are-you-smart-than-a-fifth-grader-2\",\"WARC-Payload-Digest\":\"sha1:WJ65VD5Y3V77I7CQRKBMF3SDXZJT26LB\",\"WARC-Block-Digest\":\"sha1:PEOYZLOV7YFUTOA4RAFUZRT34FZNAQAU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571538.36_warc_CC-MAIN-20220812014923-20220812044923-00349.warc.gz\"}"} |
https://www.captaincoloringbook.com/2019/08/draw-flame-princess.html | [
"# Draw Flame Princess\n\nGRID STEP You can print out the base construction lines and start drawing on tracing paper or you can draw the grid layout yourself using the following steps…\n1) Draw a rectangle that will define the conditional proportions and boundaries of the chosen drawing.\n2) From the middle of the rectangle, draw one vertical and one horizontal line equally dividing the shape.\n3) Draw another horizontal line equally dividing the upper half of the rectangle. Similarly, draw a horizontal line equally dividing the bottom half of the rectangle.\n4) Draw a vertical line equally dividing the left half of the rectangle. Similarly, draw a vertical line equally dividing the right half of the rectangle.\n\nSTEP 1: Mark off the width and height of the picture. Define the general proportions for Flame Princess. Draw the shape of her head."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8166278,"math_prob":0.9769182,"size":1513,"snap":"2023-40-2023-50","text_gpt3_token_len":292,"char_repetition_ratio":0.18489066,"word_repetition_ratio":0.848,"special_character_ratio":0.19828156,"punctuation_ratio":0.09642857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9910946,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T01:47:47Z\",\"WARC-Record-ID\":\"<urn:uuid:4dcf76ec-9159-46c1-8f7a-545e5841018a>\",\"Content-Length\":\"139370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e91b24f-7ee4-42d6-ac1c-1346bc4b788e>\",\"WARC-Concurrent-To\":\"<urn:uuid:a890f0db-1a37-4d35-9893-644409de5f9f>\",\"WARC-IP-Address\":\"172.253.122.121\",\"WARC-Target-URI\":\"https://www.captaincoloringbook.com/2019/08/draw-flame-princess.html\",\"WARC-Payload-Digest\":\"sha1:ICBSZHUT7I6JH2VUBKVCVXZASKJ7XRT3\",\"WARC-Block-Digest\":\"sha1:NIOEJRAMDTGZQTQYTBCIYONC6NSM4722\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510130.53_warc_CC-MAIN-20230926011608-20230926041608-00050.warc.gz\"}"} |
https://www.hydrometronics.com/p7dop.html | [
"# 7-Parameter Dilution of Precision\n\nFrom the world of hydrography and GPS (Horizontal, Positional and Geometric Dilution of Precision):",
null,
"",
null,
"",
null,
"From the world of 7-parameter datum transformations (7-Parameter Dilution of Precision):",
null,
"Dilution of precision (DOP) is the multiplier of observation (measurement) error into parameter error in a least-squares adjustment. For example, in hydrography and GPS, HDOP is the multiplier of 1D range error into 2D horizontal position error. DOP is a consequence of the geometry (angle of cut) of the adjustment. A good DOP is a low DOP, i.e. a low multiplier of observational error into parametric (coordinate) error.\n\nSolving for the 7 parameters of a Helmert (Bursa-Wolf or similarity) datum transformation is a least-squares adjustment affected by geometry, specifically the areal distribution of the survey points in the two datums. All other things being equal, a large areal distribution (good angle of cut) produces a low P7DOP (see formula). Small areas produce high P7DOP and are not suitable for 7-parameter adjustments due to the high correlations among the parameters. Use a 3-p or 10-p (Molodensky-Badekas) datum shift instead. The chart below relates area and number of survey points to P7DOP. Avoid the red P7DOPs.",
null,
"Download the entire presentation \"Molodensky-Badekas: Reducing the Consequences of Parametric Correlation in the 7-Parameter Shift\", which describes why the derivation of a 7-parameter datum shift for a small area of the Earth results in high correlations among the parameters, whose values are consequently determined by observational \"noise\"."
]
| [
null,
"https://www.hydrometronics.com/images/hdop.png",
null,
"https://www.hydrometronics.com/images/pdop.png",
null,
"https://www.hydrometronics.com/images/gdop.png",
null,
"https://www.hydrometronics.com/images/p7dop.png",
null,
"https://www.hydrometronics.com/images/p7dop_table.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7159126,"math_prob":0.85156375,"size":1663,"snap":"2023-40-2023-50","text_gpt3_token_len":389,"char_repetition_ratio":0.12597951,"word_repetition_ratio":0.0,"special_character_ratio":0.19723392,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95684457,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T19:33:13Z\",\"WARC-Record-ID\":\"<urn:uuid:07be00e9-63bb-4216-b391-bc35d5cb4d81>\",\"Content-Length\":\"4569\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d101704-4419-44ad-88a3-7d5056f6d33d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4d9bde2-2c36-4989-b3ba-a1733bc5b540>\",\"WARC-IP-Address\":\"209.17.116.160\",\"WARC-Target-URI\":\"https://www.hydrometronics.com/p7dop.html\",\"WARC-Payload-Digest\":\"sha1:TH2ICU6MAOK5DGYO66HHNCNSVPVAA6PS\",\"WARC-Block-Digest\":\"sha1:CLMZCQVUGZRGJHEFQIVFZDZWDOHBJYEQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100135.11_warc_CC-MAIN-20231129173017-20231129203017-00036.warc.gz\"}"} |
https://lmcs.episciences.org/volume/view/id/196 | [
"# Selected Papers of the Conference \"Typed Lambda Calculi and Applications 2007\"\n\n2007\n\nEditors: Simona Ronchi Della Rocca\n\nThis special issue of Logical Methods in Computer Science (LMCS) contains extended versions of selected papers from the Conference on Typed Lambda Calculus and Applications (TLCA '07), held in Paris, France, in June 26-28, 2007.\n\nIn consultation with the TLCA '07 Program Committee, a small number of papers presented at the symposium were selected, and their authors were invited to submit full versions to this special issue. All submissions were refereed in accordance with the usual high standards of LMCS.\n\nWe are grateful to the authors of the papers for their excellent contributions, to the members of the TLCA '07 Program Committee, to the reviewers for their e orts, to the managing editors of LMCS, Dana S. Scott, Gordon D. Plotkin, Moshe Y. Vardi for proposing this special issue, to the managing editor for special issues Benjamin Pierce and to the executive editor Jiri Adamek for their help and guidance throughout this process.\n\nSimona Ronchi Della Rocca Guest editor and TLCA '07 PC Chair\nPaula Severi Guest Editor and TLCA '07 PC Member\n\n### 1. The Safe Lambda Calculus\n\nSafety is a syntactic condition of higher-order grammars that constrains occurrences of variables in the production rules according to their type-theoretic order. In this paper, we introduce the safe lambda calculus, which is obtained by transposing (and generalizing) the safety condition to the setting of the simply-typed lambda calculus. In contrast to the original definition of safety, our calculus does not constrain types (to be homogeneous). We show that in the safe lambda calculus, there is no need to rename bound variables when performing substitution, as variable capture is guaranteed not to happen. We also propose an adequate notion of beta-reduction that preserves safety. In the same vein as Schwichtenberg's 1976 characterization of the simply-typed lambda calculus, we show that the numeric functions representable in the safe lambda calculus are exactly the multivariate polynomials; thus conditional is not definable. We also give a characterization of representable word functions. We then study the complexity of deciding beta-eta equality of two safe simply-typed terms and show that this problem is PSPACE-hard. Finally we give a game-semantic analysis of safety: We show that safe terms are denoted by `P-incrementally justified strategies'. Consequently pointers in the game semantics of safe lambda-terms are only necessary from order 4 onwards.\n\n### 2. On tiered small jump operators\n\nPredicative analysis of recursion schema is a method to characterize complexity classes like the class FPTIME of polynomial time computable functions. This analysis comes from the works of Bellantoni and Cook, and Leivant by data tiering. Here, we refine predicative analysis by using a ramified Ackermann's construction of a non-primitive recursive function. We obtain a hierarchy of functions which characterizes exactly functions, which are computed in O(n^k) time over register machine model of computation. For this, we introduce a strict ramification principle. Then, we show how to diagonalize in order to obtain an exponential function and to jump outside deterministic polynomial time. Lastly, we suggest a dependent typed lambda-calculus to represent this construction.\n\n### 3. The Omega Rule is $\\mathbf{\\Pi_{1}^{1}}$-Complete in the $\\lambda\\beta$-Calculus\n\nIn a functional calculus, the so called \\Omega-rule states that if two terms P and Q applied to any closed term <i>N</i> return the same value (i.e. PN = QN), then they are equal (i.e. P = Q holds). As it is well known, in the \\lambda\\beta-calculus the \\Omega-rule does not hold, even when the \\eta-rule (weak extensionality) is added to the calculus. A long-standing problem of H. Barendregt (1975) concerns the determination of the logical power of the \\Omega-rule when added to the \\lambda\\beta-calculus. In this paper we solve the problem, by showing that the resulting theory is \\Pi\\_{1}^{1}-complete.\n\n### 4. Polynomial Size Analysis of First-Order Shapely Functions\n\nWe present a size-aware type system for first-order shapely function definitions. Here, a function definition is called shapely when the size of the result is determined exactly by a polynomial in the sizes of the arguments. Examples of shapely function definitions may be implementations of matrix multiplication and the Cartesian product of two lists. The type system is proved to be sound w.r.t. the operational semantics of the language. The type checking problem is shown to be undecidable in general. We define a natural syntactic restriction such that the type checking becomes decidable, even though size polynomials are not necessarily linear or monotonic. Furthermore, we have shown that the type-inference problem is at least semi-decidable (under this restriction). We have implemented a procedure that combines run-time testing and type-checking to automatically obtain size dependencies. It terminates on total typable function definitions.\n\n### 5. Continuation-Passing Style and Strong Normalisation for Intuitionistic Sequent Calculi\n\nThe intuitionistic fragment of the call-by-name version of Curien and Herbelin's \\lambda\\_mu\\_{\\~mu}-calculus is isolated and proved strongly normalising by means of an embedding into the simply-typed lambda-calculus. Our embedding is a continuation-and-garbage-passing style translation, the inspiring idea coming from Ikeda and Nakazawa's translation of Parigot's \\lambda\\_mu-calculus. The embedding strictly simulates reductions while usual continuation-passing-style transformations erase permutative reduction steps. For our intuitionistic sequent calculus, we even only need \"units of garbage\" to be passed. We apply the same method to other calculi, namely successive extensions of the simply-typed λ-calculus leading to our intuitionistic system, and already for the simplest extension we consider (λ-calculus with generalised application), this yields the first proof of strong normalisation through a reduction-preserving embedding. The results obtained extend to second and higher-order calculi.\n\n### 6. Observational Equivalence and Full Abstraction in the Symmetric Interaction Combinators\n\nThe symmetric interaction combinators are an equally expressive variant of Lafont's interaction combinators. They are a graph-rewriting model of deterministic computation. We define two notions of observational equivalence for them, analogous to normal form and head normal form equivalence in the lambda-calculus. Then, we prove a full abstraction result for each of the two equivalences. This is obtained by interpreting nets as certain subsets of the Cantor space, called edifices, which play the same role as Boehm trees in the theory of the lambda-calculus."
]
| [
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https://basicittopic.com/datatype-in-c-2/?utm_source=rss&utm_medium=rss&utm_campaign=datatype-in-c-2 | [
"Categories\n\n# Data type in C++\n\nThe Data type in c++ are divided into 4 types:-\n\nprimitive: bool, char, int, float, double,\n\nempty: void\n\nderived: array, pointer, reference\n\nuser-defined: enum, typedef, struct, union, and class\n\nsome basic or primitive data type with an example:-\n\n### bool:-\n\nBoolean can represent true or false.\n\nAny positive or any negative value is true except ‘0’.\n\nEvery character is true except ‘/0’ (null) character.\n\nEvery time string constant is true for Boolean.\n\nBoolean size is 1 byte because the bit is not accessible. To represent bool 7 bit are wastage.\n\nExample\n\n```#include\"iostream\"\n\nusing namespace std;\n\nmain()\n\n{\n\nbool a=10;\n\nbool b=0;\n\nbool c=\" \";\n\nbool d='/0';\n\nbool e='A';\n\nbool f=1.2;\n\ncout<<a<<b<<c<<d<<e<<f<<\\n;\n\n}```\n\nOutput:-\n\n1 0 1 0 1 1\n\n### Character:-\n\nChar does not store a string constant. Char is an integer constant.\n\nIt can not store string type. It can store an int, float, double constant.\n\nChar data type support two modifiers such as signed and unsigned.\n\nChar data type support 2 modifiers such as signed and unsigned.\n\nExample:-\n\n```#include\"iostream\"\n\nusing namespace std;\n\nmain()\n\n{\n\nchar sec;\n\nsec=\"abcdef\";\n\ncout<<sec<<endl;\n\n}\n\nOutput:- f```\n\n### Integer:-\n\nInteger is the combination of 4 char.\n\nAll the modifiers are allowed in the integer.\n\nInteger is 4 byte.\n\nExample:-\n\n```#include <iostream>\n\nusing namespace std;\n\nint main()\n\n{\n\nint x = 2, y = 3, temp;\n\ncout << \"Before swapping.\" << endl;\n\ncout << \"x = \" << x << \", y = \" << y << endl;\n\ntemp = x;\n\nx = y;\n\nx = temp;\n\ncout << \"\\nAfter swapping.\" << endl;\n\ncout << \"x = \" << x << \", y = \" << y << endl;\n\nreturn 0;\n\n}```\n\nOutput:-\n\n```Before swapping 2 3\n\nAfter swapping 3 2```\n\n### Float:-\n\nFloat is 4byte long.\n\nFloat does not support any modifiers.\n\nExample:-\n\n```#include\"iostream\"\n\nusing namespace std;\n\nmain()\n\n{\n\ncout << \"Size of float is \" << sizeof(float) << endl;\n\n}```\n\nOutput:-\n\n`Size of float is 4 byte`\n\n### Double:-\n\nDouble is 8 byte long.\n\nSigned, unsigned, shot modifier is not allowed for double.\n\nExample:-\n\n```#include\"iostream\"\n\nusing namespace std;\n\nmain()\n\n{\n\ncout << \"Size of double is \" << sizeof(double) << endl;\n\n}\n\nOutput:-\n\nSize of double is 8 byte\n\n```\n\n### What are the modifiers in C++?\n\nsigned, unsigned, short, long, long long.\n\n### What are the qualifiers in C++?\n\nconst, volatile, mutable"
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http://vinds.com.br/4j8amzv/simplify-roots-of-negative-numbers-c0df84 | [
"Then, rewrite the square root as a multiplication problem under the square root sign. Definition of square root. 3rd negative number root simplifying; Home. Types of Problems. Simplify: ⓐ ⓑ ⓒ ⓐ 10 ⓑ 2 ⓒ 3. scurtisrutherford. By using this website, you agree to our Cookie Policy. Estimate Square Roots. To calculate any root of a number use our Nth Root Calculator. The number n is called the degree of the root and a is called the radicand of the root. Rational Exponents. Example 4. then is not a real number. We use the letter i to help us when we need to simplify square roots of negative numbers. Hello, guys! 7. In this video, you'll learn how to simplify the square root of a negative number. This website uses cookies to ensure you get the best experience. If you divide 98 by 2, you get 49. melindameo TEACHER. We will apply these properties in the next two examples. Next you will simplify the square root … You just need to remember 'i' in your answer! − 169 = −13. Get step-by-step solutions to your Negative numbers problems, … not a real number. Aug 2011 3 0. FALSE this rule does not apply to negative radicands ! ... Simplify Expressions with Square Roots. Learn more Accept. 9 2.Express 8 in i notation. If we want to find the negative square root of a number, we place a negative in front of the radical sign. Get your calculator and check if you want: they are both the same value! The symbol #i# which is an imaginary number is another way to write: #sqrt(-1)# so we can rewrite the expression as: #2sqrt(color(blue)(2)) * i =># #2isqrt(color(blue)(2))# In the following exercises, simplify. then is a real number. Simplify square roots of negative numbersWhat is √-100 = ___ + ___i We write 169 = 13. Sometimes, after simplifying the square root(s), addition or subtraction becomes possible. S. Sova. Addition and subtraction of square roots after simplifying. They saw equations such as x 2 + 1 = 0, and wondered what the solution really meant. When n is an odd number, is a real number for all values of a. In the following exercises, estimate each square root between two consecutive whole numbers. Simplifying Square Roots Medium-Hard. Viewed 333 times 0 $\\begingroup$ … Algebra . Relating imaginary numbers to square roots of negative numbers Practicing how to simplify complex expression; Practice Exams . Square root is an inverse operation of the squaring a number.. 20 terms. 30 terms . The number i allows us to work with roots of all negative numbers, not just . Square Root Calculator. jessoccer06. Can we simplify $$\\sqrt{−25}$$? scurtisrutherford. These cannot be added until is simplified. Simpliflying Roots Of Negative Number - Displaying top 8 worksheets found for this concept.. Simplify and add. Fourth Roots. And here is how to use it: Example: simplify √12. It includes 6 examples. For example, if you're trying to find the square root of 98, the smallest prime number possible is 2. There are two important rules to remember: , and . Solution: For this one, we will skip some of the intermediate steps and go straight to simplifying the number by replacing the negative sign under the square root with the imaginary unit i in front of the square root sign. There is one type of problem in this exercise: Express the radical using the imaginary unit, . For example, . So let’s simplify it even more: When the exponent is: - an even number – the power is a positive number, - an odd number – the power is a negative number. Simplifying Square Roots – Techniques and Examples. High School Math / Homework Help. 20 terms. Examples: You have to be careful. To simplify a square root: make the number inside the square root as small as possible (but still a whole number): Example: √12 is simpler as 2√3. Solve Negative numbers problems with our Negative numbers calculator and problem solver. Any positive number squared is positive, and any negative number squared is also positive. Example: (-248)^1/2 = ??? Aug 22, 2011 #1 I'm trying to get up to speed with my math after being out of practice for a long time, and there is a problem that has me stumped. In the examples above, number –2 is placed in brackets. For example, − 169 = −13. Let's see some special cases: The root of degree n = 2 is known as a square root. A complex number is a number that combines a real portion with an imaginary portion. That means you have to do the operation on “whole” number: –2, as it’s presented above. Fourth root of 1 is ±1; Fourth root of 16 is ±2; Fourth root of 81 is ±3; Fourth root of 256 is ±4; Fourth root of 625 is ±5; Fourth root of 1296 is ±6 5. When n is an even number and. Simplifying Square Roots of a Negative Number. 10. not a real number. This video looks at simplifying square roots with negative numbers using the imaginary unit i. 3. Example: The square root of 9 is 3 because 3 to the power of two is 9. Multiplying Integers. Ask Question Asked 4 years, 8 months ago. Always simplify if possible. The square root of a negative number does not exist among the set of Real Numbers. In the next section, we will explore cube roots, and use the methods we have shown here to simplify them. I get the general idea like if it was: (-9)^1/2 = 3i but what if its a larger number that doesn't have a square root? 8. Active 4 years, 8 months ago. $$\\red{ \\sqrt{a} \\sqrt{b} = \\sqrt{a \\cdot b} }$$ only works if a > 0 and b > 0. Simplify: ⓐ ⓑ ⓒ ⓐ ⓑ ⓒ Simplify: ⓐ ⓑ ⓒ ⓐ 3 ⓑ 4 ⓒ 3. : This problem asks for the radical of a given number. Is there a number whose square is −25? The Simplify square roots of negative numbers exercise appears under the Algebra II Math Mission and Mathematics III Math Mission. $$\\sqrt{9} = 3$$ The root of degree n = 3 is known as a cube root… YOU MIGHT ALSO LIKE... square roots!!! Cube roots are unique from square roots in that it is possible to have a negative number under the root, such as $\\sqrt{-125}$. We write $$\\sqrt{169}=13$$. To simplify a square root, start by dividing the square root by the smallest prime number possible. Square Root of a Negative Number. Simplifying the square root of a negative number is very similar to simplifying the square root of a positive number. Here is the rule: when a and b are not negative. Simplifying nth roots Hard. Properties of . [A]8i [B]22i [C] 8i [D]22i 3.Express 80 in i notation. an odd root of a negative number because a negative number raised to an odd power is still negative. Simplifying nth Roots- Medium. Its just hard for me to factor out the i part and leave whatevers supposed to be still under the \"square root\" part I've tried going over it several times and I can't arrive at the given solution. scurtisrutherford. Roots can also be expressed as fractional exponents. Every positive number has a positive square root and a negative square root. A complex number, then, is made of a real number and some multiple of i. Example 3 – Simplify the number √-3.54 using the imaginary unit i. Algebra II Practice N.CN.A.2: Square Roots of Negative Numbers www.jmap.org NAME:_____ 1.Simplify. Imaginary is the term used for the square root of a negative number, specifically using the notation = −. Forums. 6. 169 = 13. If we want to find the negative square root of a number, we place a negative in front of the radical sign. Now, because both are alike under the radical sign, Try to simplify each one. When problems with negatives under a square root first appeared, mathematicians thought that a solution did not exist. If we want to find the negative square root of a number, we place a negative in front of the radical sign. 48 terms. For example, $$-\\sqrt{169}=-13$$. [A]45i [B] 80i [C]45i [D]80i 4.Express 75 in i notation. Why? A negative number like − 4 has no real square root because the square of a number cannot be negative. A radical sign, , is used to indicate a positive square root. $(\\;)^{2} = -25?$ None of the numbers that we have dealt with so far have a square that is −25. This exercise practices rewriting square roots of negative numbers as imaginary numbers. Just like your shadow in the shade is invisible, so is the imaginary part of complex numbers. Check out this tutorial to see how to simplify the square root of a negative number. Evaluate the Square Root of a Negative Number. See also on this page a square root chart 1 to 100. Some of the worksheets for this concept are Radical expressions radical notation for the n, Grade 9 simplifying radical expressions, Maths refresher, Simplifying radicals and complex numbers, Absolute value and roots, Simplifying radical expressions, Algebra skill, Chapter 7. Simplifying Square Roots. kmagarie. Simplifying Square Roots A square root of a number is one of two equal factors of the number. Free simplify calculator - simplify algebraic expressions step-by-step . [A]75i [B]53i [C] 75i [D]53i 5.Express 72 in i notation. Factors. Please enter a real number, the press 'Calculate': Square root result: The square root of -100 (√-100) is the imaginary number: i10 What is square root? Use the square root calculator below to find the square root of any real number, positive or negative. We know that every positive number has two square roots and the radical sign indicates the positive one. 17 terms. [A]62i [B] 72i Keywords: problem; simplify; square roots; negative numbers; imaginary numbers; i; radicals; Background Tutorials. The square root of a number x is denoted with a radical sign √x or x 1/2.A square root of a number x is such that, a number y is the square of x, simplify written as y 2 = x.. For instance, the square root of 25 is represented as: √25 = 5. For example, to simplify we are looking for a real number x so that x 2 = –1. We know that every positive number has two square roots and the radical sign indicates the positive one. Some sample complex numbers are 3+2i, 4-i, or 18+5i. Imaginary Numbers (Simplifying with Negative Squar… 12 terms. Odd roots of negative numbers are real numbers. Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. You will use these rules to rewrite the square root of a negative number as the square root of a positive number times . For complex or imaginary solutions use Simplify Radical Expressions Calculator. However, the even root of a negative number is not a real number. Square Roots of Negative Complex Numbers Simplifying Square Roots The Equation of a Circle Fractional Exponents Finding the Least Common Denominator Simplifying Square Roots That Contain Whole Numbers Solving Quadratic Equations by Completing the Square Graphing Exponential Functions Decimals and Fractions Adding and Subtracting Fractions Adding and Subtracting Rational … In other words, the product of two radicals does not equal the radical of their products when you are dealing with imaginary numbers. 47 terms. In a later section we will discuss how to work with even roots of negative numbers, but for now we state they are not real numbers. To simplify square roots with negative numbers are alike under the Algebra II Math Mission Squar… 12 terms -248 ^1/2! They saw equations such as x 2 = –1 ( simplifying with negative ;. Has a positive square root of a negative number is very similar to simplifying square! 2 ⓒ 3, to simplify a square root of a positive number squared is positive and. Simplify √12 odd roots of negative numbers problems, … Free simplify -! Not a real number and some multiple of i for complex or imaginary solutions use simplify radical expressions.! Radical of a real number i ' in your answer exist among the simplify roots of negative numbers! Smallest prime number possible numbers using the notation = − + 1 = 0, and the... Equal the radical sign inverse operation of the root 've tried going over it several times and i n't! Root chart 1 to 100 72 in i notation because both are alike under radical. Radical expressions calculator made of a real number, we place a negative number is not a real number we. Positive or negative positive one } =13\\ ) is made of a number! Simplifying the square root of a negative number raised to an odd root of degree n 2! Ⓑ 2 ⓒ 3 [ a ] 62i [ B ] 80i [ ]... Is invisible, so is the term used for the square root first appeared mathematicians. $\\begingroup$ … odd roots of negative numbers using the imaginary unit i and a called... By dividing the square root of a negative number as the square as! Such as x 2 = –1 a simplify roots of negative numbers number, is used to indicate a positive number squared positive! Number can not be negative under the Algebra II Math Mission and Mathematics III Math Mission = is. Combines a real number and some multiple of i ] 75i [ D ] 53i [ C 8i... Numbers exercise appears under the square root is an odd power is still negative =-13\\ ) above, –2... Mathematicians thought that a solution did not exist using this website uses cookies to ensure you 49... Consecutive whole numbers an odd power is still negative your negative numbers are,! Simplify ; square roots!!!!!!!!!!!!!!!. Real number simplify roots of negative numbers the square root because the square root because the of! 72I Evaluate the square root of a positive square root the number is! No real square root of 98, the even root of a positive root. Complex numbers are real numbers n is called the degree of the radical sign indicates the positive.. Is not a real number, specifically using the imaginary part of complex numbers indicates positive. Exercise practices rewriting square roots ; negative numbers are real numbers the Algebra II Math Mission Squar…... Will use these rules to rewrite the square root ( s ), addition or subtraction possible! Like... square roots of negative numbers problems, … Free simplify calculator - algebraic! On “ whole ” number: –2, as it ’ s presented above 12 terms simplify ; square of. Really meant keywords: problem ; simplify ; square roots of negative numbers exercise appears under the Algebra Math... This tutorial to see how to simplify them number does not exist the set of numbers! The given solution type of problem in this video, you 'll learn how to use it::! 53I [ C ] 75i [ B ] 80i [ C ] 45i [ B ] 53i C. Not negative, because both are alike under the square root because the square root, start by dividing square... And use the methods we have shown here to simplify we are looking for a real number is... Term used for the square root numbers are 3+2i, 4-i, or...., specifically using the notation = − \\sqrt { 169 } =-13\\ ) 2, you to! The imaginary unit, roots ; negative numbers ; imaginary numbers number does not equal the radical sign ( {. A given number simplify a square root of a negative number squared is positive... The root i ' in your answer some multiple of i solution did not exist real numbers simplify roots of negative numbers we to. The rule: when a and B are not negative equal the radical sign,, is real... 72 in i notation of degree n = 2 is known as a square root a! Get 49 false this rule does not apply to negative radicands get step-by-step solutions to negative! Simplify radical expressions calculator we know that every positive number has two square roots of negative numbers 3+2i! Numbers are real numbers the even root of degree n = 2 is known as a square root by smallest! Roots!!!!!! simplify roots of negative numbers!!!!!! 0 $\\begingroup$ … odd roots of negative numbers using the imaginary unit, you agree to Cookie! So that x 2 + 1 = 0, and wondered what the solution really.. And any negative number as the square root of a positive number times not negative roots negative... Have to do the operation on “ whole ” number: –2, as it s! We have shown here to simplify them 8 months ago smallest prime number possible is.... This rule does not apply to negative radicands because 3 to the power two... Is also positive number has two square roots of negative numbers problems with negatives under a square root of negative... Equations such as x 2 + 1 = 0, and wondered what the solution really meant –2! I to help us when we need to remember:, and any number! Let 's see some special cases: the root and a negative number is a number that combines real... Whole ” number: –2, as it ’ s presented above C ] 75i B..., mathematicians thought that a solution did not exist among the set of real numbers square of number... Be negative alike under the square root of a real number x so that x 2 =.! As a multiplication problem under the Algebra II Math Mission the examples above, number –2 is placed in.! Roots of negative numbers exercise practices rewriting square roots and the radical of their products when you are dealing imaginary... Simplifying the square root is an odd power is still negative ; radicals ; Background Tutorials to it. When n is an inverse operation of the radical of a number use Nth.... square roots and the radical using the imaginary unit i a multiplication problem under the of! The squaring a number can not be negative the best experience and i ca n't arrive at the given.. 4.Express 75 in i notation will use these rules to rewrite the square of a negative in front the... Not equal the radical using the imaginary part of complex numbers, after the. Is called the radicand of the squaring a number, we place a negative.... Of complex numbers are real numbers roots with negative numbers as imaginary numbers ; i radicals. A given number III Math Mission and Mathematics III Math Mission expressions step-by-step below to the... Keywords: problem ; simplify ; square roots of negative numbers problems with our negative numbers 3+2i! Out this tutorial to see how to simplify each one like... square roots of numbers... Complex numbers important rules to rewrite the square root of a negative number need to simplify square. Like... square roots with negative Squar… 12 terms saw equations such as x +... Shown here to simplify the square root of a negative number as square! Is an inverse operation of the radical sign however, the product of two radicals does not the! Going over it several times and i ca n't arrive at the given solution 53i [ C 75i... We have shown here to simplify each one a is called the of! Because 3 to the power of two is 9 as x 2 + 1 = 0, and wondered the! Are dealing with imaginary numbers, positive or negative to use it: example: the root and a number... A multiplication problem under the square root sign negative number i to help us when we need to them... Chart 1 to 100 the same value = 0, and simplifying square with! Next two examples numbers exercise appears under the Algebra II Math Mission Mathematics... We write \\ ( \\sqrt { 169 } =13\\ ) multiple of i Asked. In i notation the number n is called the degree of the radical of their products when you dealing! Will use these rules to remember ' i ' in your answer and check if divide!, start by dividing the square root ( s ), addition or subtraction becomes possible prime! Simpliflying roots of negative numbers ; i ; radicals ; Background Tutorials s ) addition! A is called the degree of the radical sign, Try to simplify square. Indicate a positive number has two square roots and the radical using the imaginary part of numbers... Numbers using the imaginary part of complex numbers are 3+2i, 4-i, 18+5i. Not exist among the set of real numbers - simplify algebraic expressions step-by-step ’ s presented.... The simplify square roots of negative numbers exercise appears under the Algebra II Math Mission and Mathematics III Math.... Can not be negative equal the radical of a negative in front of the radical sign radical sign an power. Estimate each square root of 9 is 3 because 3 to the of...\n\nMexican Earthquake 2018, Red Rock Cafe Napa Menu, Tipu Sultan Photos, Public Bank Moratorium Faq, How To Paint A Wooden Door, Dora The Explorer Season 4 Episode 21, Tourist Places Near Basara, Oyster Shucker Party London, Tom's Rhinoplasty Watch Online, Xiaomi Window Cleaner,"
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https://www.visionlearning.com/en/library/Math%20in%20Science/62/Linear%20Equations%20in%20Science/194 | [
"Equations\n\n# Linear Equations in Science: Relationships with Two Variables\n\nby Christine Hoekenga, Anthony Carpi, Ph.D., Anne E. Egger, Ph.D.\n\nImagine you are a forensic scientist working in the Central Identification Lab at JPAC (the Joint POW/MIA Accounting Command). Your job is to help identify human remains believed to be U.S. military personnel reported missing in action during World War II and other conflicts. A team of your colleagues recovers skeletal remains consisting of a pelvic bone, several ribs, and a femur from a 1943 military plane crash on Vanuatu.\n\nWhen the remains arrive in your lab, you photograph and measure the bones. From the shape of the pelvis, you can quickly tell that the remains most likely belong to an adult male. You note that the femur is 18.7 inches long. Bone length, especially the length of long bones like the femur, is related to an individual’s overall height. Simply put, a tall person will usually have long legs, and a short person generally has shorter legs. This relationship is so strong that you can predict an individual’s height if you know the length of one bone in the leg (Figure 1). You plug your measurement into an equation used to estimate the overall height of an adult male based on femur length:\n\nH = 1.880(L) + 32.010\n\nWhere H = height in inches and L = femur length in inches.\n\nH = (1.880 × 18.7) + 32.010\n\nH = 67.166 inches\n\nH ≅ 5 feet, 7 inches\n\nYou send the estimated height, along with the results of your other analyses, to your colleague who reviews the records of missing service members for possible matches. Based on the location of the crash and the type of plane, she had already narrowed it down to three possible missing airmen. Their service records list their heights as 5 feet, 4 inches; 6 feet even; and 5 feet, 7 inches. The third airman appears to be the closest match, and your colleague will now contact the airman’s family to request a DNA sample for confirmation.",
null,
"Figure 1: As you can see in this graph, the relationship between femur length and overall height in humans is a linear relationship. The heights and expected femur lengths of three airmen who were possible matches for the remains appear as points on the line.\n\nAlthough quite simple, this scenario is an example of how math, in this case linear algebra, is a fundamental part of science. Scientists – as well as people working in many other fields or just going about their regular routines – make use of linear equations every day. Among other things, linear equations can help us describe the relationship between two quantities or phenomena (like femur length and overall height), calculate rates (such as how quickly an object is moving), or convert from one unit of measurement to another (for example, inches to centimeters).\n\n## What are linear equations?\n\nThe formula relating femur length to estimated height in the scenario above is one example of a linear equation—a mathematical statement in which the highest exponent on any variable is 1 (none of the variables are squared, cubed, raised to the fourth power, etc.). They are also known as first-order equations.\n\nAs you might expect from the name, when graphed on the Cartesian coordinate system (the familiar x- and y-axis system), a linear equation produces a straight line (Figure 2). Other examples of linear equations include:\n\n$$y = 1.8 ( x ) + 32$$\n\nThis equation converts degrees Celsius (x) to degrees Fahrenheit (y).\n\n$$y = 2 ( x )$$\n\nAn equation that gives the proportion of oxygen atoms (y) to carbon atoms (x) in carbon dioxide.\n\n$$x = 1 4 ( y ) + 40$$\n\nThis equation relates the number of chirps per minute (y) made by a snowy tree cricket to the ambient temperature in degrees Fahrenheit (x).",
null,
"Figure 2: Three examples of linear relationships found in scientific applications.\n\n## Early history of linear equations\n\nLinear equations and other basic concepts of algebra have a long history stretching back thousands of years. The ancient Mesopotamians, Egyptians, Greeks, Chinese, and Indians all developed mathematical methods that served as early foundations for modern algebra. But most historians consider the father of algebra to be Abu Abdullah Muhammad ibn Musa al-Khwarizmi (780–850 CE), a scholar with the House of Wisdom academy in what is now Baghdad. In fact, the word algebra comes from al-jabar, a term al-Khwarizmi used to describe the technique of adding equal quantities to both sides of an equation in order to simplify it.",
null,
"Figure 3: A page from al-Khwarizmi’s most popular book Al-Kitab Al-Jabar Wa'al-Muqabelah, which roughly translates to The Book of Restoration and Balancing. Although the meaning has changed over time, the term al-jabar eventually gave rise the Latin term algebrae, and finally to the modern algebra.\n\nBut the math al-Khwarizmi and his predecessors practiced looked very different than what we think of as algebra today. Perhaps the biggest difference is that al-Khwarizmi did not use mathematical symbols. He did not use variables to stand in for unknowns or constants, nor did he use symbols to denote the operations like addition and subtraction that he performed on them. Instead of working with equations, every calculation al-Khwarizmi made was described in words—mostly everyday language with a few technical terms, like al-jabar. He usually wrote about the math needed for practical purposes, such as for dividing inheritance or digging canals.\n\nToday, the use of symbols and equations is so central to algebra that it’s logical to ask: Why are al-Khwarizmi’s word problems even considered algebra? The key features are:\n\n• solving for an unknown quantity (which separates them from simple arithmetic),\n• taking a numeric approach (rather than a purely spatial or geometric approach as many Greek scholars did), and\n• articulating general rules or techniques for working with numbers (such as al-jabar).\n\nAl-Khwarizmi also studied arithmetic, especially as it was practiced in India. Building upon early Indian scholars, he wrote one of the first known texts describing a decimal system, the operations we now call multiplication and division, and a small circle that appears to be used like a zero (Figure 3).\n\nDuring the 12th Century, parts of al-Khwarizmi’s writings were translated into Latin and read by scholars working in Europe. These scholars gradually introduced symbols for operations, numbers, and variables. This eventually led to the development of equations as we think of them today.\n\nComprehension Checkpoint\n\nWhy is Abu Abdullah Muhammad ibn Musa al-Khwarizmi considered the \"father of algebra\"?\n\n## Development of the Cartesian coordinate system\n\nIn the 17th century, another innovation helped connect algebra with geometry. René Descartes, a French philosopher and mathematician, developed a way to visualize equations with two variables by graphing them as lines (linear) or curves (nonlinear). The Cartesian coordinate system, named for Descartes, is a system of two perpendicular axes, usually labeled x and y. It is an important tool in modern math from algebra to calculus, and scientists frequently use it to visualize the relationship between two variables in their data. (For more about how scientists use the Cartesian coordinate system, see our module Using Graphs and Visual Data in Science.)\n\nDescartes’ 1637 book La géométrie described the basic idea of a coordinate system as well as an organized collection of symbols and conventions we still use today. From La géométrie, we get things like the modern square root symbol (radicand) and the convention of writing an exponent as a small, raised numeral just after its base. The practice of using lower-case letters from the beginning of the alphabet to stand for given numbers or constants and using lower-case letters from the end of the alphabet to represent variables also comes from Descartes. Today we call the equation ax + by = c the standard form of a linear equation.\n\n## Working with linear equations\n\nWhen presented with a linear equation, if we know the value of one of the variables (often x or y), we can solve the equation for the other variable. Let’s use the cricket equation shown in Figure 2 (Dolbear, 1897) as an example. If we sit outside one evening and count the snowy tree crickets chirping 80 times per minute (y = 80), we can solve for the temperature (x) as follows:\n\n$$x = 1 4 ( y ) + 40$$\n\n$$x = 1 4 ( 80 ) + 40$$\n\nPlug in 80 chirps for y.\n\n$$x = 20 + 40$$\n\nSimplified.\n\n$$x = 60 ∘ F$$\n\nThus, we know the temperature is approximately 60° F. The following evening if we observe the crickets chirping 60 times per minute (y = 60), we can plug that value in and calculate the new value for x (55° F).\n\nIn the cricket equation, x is known as the independent variable, and y is the dependent variable, since its value depends on the value of x (chirping rate depends on temperature). Sets of x and y values that make the equation true are solutions to the equation. They are generally written as ordered pairs in the form (x, y). One solution to the cricket equation is the ordered pair 60° F and 80 chirps per minute (60, 80). In theory, there is an infinite number of solutions to the equation, including (55, 60), (65, 100), and (43, 12). Each ordered pair is a point on the line described by the equation (Figure 4).",
null,
"Figure 4: A graph shows the linear relationship between the air temperature and the cadence of crickets chirping. Each point labeled on the line represents a data point gathered by observation.\n\nHowever, some ordered pairs that are solutions to a given equation may not make sense in the real world. For example, the ordered pairs (37, -12) and (200, 640) are valid solutions to the cricket equation but don’t make sense in this context. Crickets cannot chirp -12 times per minute, and they are unlikely to be alive and chirping (let alone at a pace of 640 chirps per minute) if the temperature has reached 200° F. When working with an equation in a real-world scientific context, it is important to reflect on a given solution and consider whether it makes sense in that context.\n\nIf we don’t know the value of either variable, we can still solve for one of the variables in terms of the other.\n\n$$x = 1 4 ( y ) + 40$$\n\nAgain using the equation above, we can solve for y in terms of x.\n\n$$x − 40 = 1 4 y$$\n\nSubtract 40 from both sides.\n\n$$4 ( x − 40 ) = 4 ( 1 4 y )$$\n\nMultiply both sides by 4.\n\n$$4 x − 160 = y$$\n\nSimplify both sides of the equations.\n\n$$y = 4 x − 160$$\n\nRearrange the equation so y is on the left.\n\nThus, we can see that y equals four times x, minus 160. If we are out in the field and already know the temperature (x), this equation can be used to quickly calculate the expected number of chirps (y).\n\nComprehension Checkpoint\n\nSolutions for x and y are written as\n\n## Slope-intercept form\n\nAdditionally, if we want to visualize a linear equation by graphing it, the arrangement shown above, called slope-intercept form, is often more useful. Slope-intercept form follows the general format:\n\n$$y = m x + b$$\n\nWhere x and y are the variables and m and b are constants. (Keep in mind that the m and b may be positive, negative, or equal to zero.)\n\nIn this form, m is the slope of the line – a ratio that tells us how much a line rises over a given distance. Meanwhile, b is the y-intercept – the point where the line crosses the y-axis. Keep in mind that in real-world contexts, the axes may be labeled with variables other than x and y. For example, you may see t to represent time or v to represent velocity. For this reason, it may help to think of the y-intercept as the vertical-axis-intercept.\n\nThe slope is a ratio relating the change in y to the change in x (sometimes called “rise to run”) from one point to another on a line. For example, if m = ½, each point on the line is 1 unit higher on the vertical (y) axis for every two units to the right on the horizontal (x) axis. If m = -3, each point on the line is 3 units lower on the y-axis for every one unit to the right. A positive slope means that a line trends up as one moves to the right; a negative slope occurs when a line trends down as one moves to the right (Figure 5).",
null,
"Figure 5: Two linear equations show how the slope and y-intercept of a line may be positive or negative. How does a line with a positive slope (left) look different than one with a negative slope (right)?\n\nFor linear equations used in science, b often represents the starting point. Imagine a study where a scientist measures an organism’s length as it grows over the course of a month. If he finds the growth rate is linear and writes an equation in the form y = mx + b to describe the organism’s length, b would likely indicate the organism’s length at the beginning of the study.\n\n## Describing vertical and horizontal lines with linear equations\n\nVertical and horizontal lines are also described by linear equations. In a scientific context, a horizontal or vertical line indicates that a variable is constant, regardless of changes in any other variable. In the equation above relating femur length (L) to a person’s overall height (H), the taller the person is, the longer his or her femur. But, if we consider the relationship between height (H) and number of limbs (N), we see no dependence of one upon the other. Regardless of height, N = 4 describes number of limbs in all cases.\n\nA vertical line has an undefined slope and thus cannot be written in slope-intercept form. The general equation for a vertical line is x = a, where a is a constant. On a vertical line, all points have the same x-value, and the line never crosses the y-axis (unless the equation is x = 0). A horizontal line has a slope of 0, so the slope-intercept form can be simplified to y = b, where b is the y-intercept, as well as the y-value in every ordered pair that satisfies the equation (Figure 6).",
null,
"Figure 6: A vertical line (left) represents a linear relationship in which the value of x is constant. A horizontal line (right) represents a linear relationship in which the value of y is constant.\n\nIn real world applications such as those described below, each axis—and thus each variable—represents a measurement of some factor, such as distance traveled, time elapsed, degrees Fahrenheit, etc. The linear equation describes the relationship between the two measurements. Although x and y are the default variables for the axes, you will often see other letters used in equations and on graphs that hint at what the variable represents. For example, t may be used for time, d for distance, etc. (See our module Using Graphs and Visual Data in Science for more about how graphs are used in science.)\n\nComprehension Checkpoint\n\nOn a vertical line, all points have the same value for\n\n## Using linear equations in science\n\nLinear equations can be used to describe many relationships and processes in the physical world, and thus play a big role in science. Frequently, linear equations are used to calculate rates, such as how quickly a projectile is moving or a chemical reaction is proceeding. They can also be used to convert from one unit of measurement to another, such as meters to miles or degrees Celsius to degrees Fahrenheit.\n\nIn some cases, scientists discover linear relationships during the course of research. For example, an environmental scientist analyzing data she has collected about the concentration of a certain pollutant in a lake may notice that the pollutant degrades at a constant rate. Using those data, she may develop a linear equation that describes the concentration of the pollutant over time. The equation can then be used to calculate how much of the pollutant will be present in five years or how long it will take for the pollutant to degrade entirely.\n\n### How to calculate a rate\n\nA rate is a measurement of change relative to time. Scientists often need to know how quickly or slowly (“at what rate”) a given process is occurring. A geologist, for example, may want to know the rate at which pieces of the Earth’s crust are moving in order to assess potential seismic hazards. A chemist may need to know the rate at which two substances react with one another in order to understand the products of a chemical reaction.\n\nA rate (r) is calculated by determining the amount of change (for example, distance traveled) and the time elapsed. To do this, we need two values for time (t1 and t2) and two corresponding values for the condition that is changing (d1 and d2). So for example:\n\n$$r = d 2 − d 1 t 2 − t 1$$\n\nwhere d2 is the distance traveled at time t2 and d1 is the distance traveled at time t1. The Greek letter Δ, “delta,” means change, and you will often see it used in rate calculation problems. Written using delta, our example rate equation becomes:\n\n$$r = Δ d Δ t$$\n\nThe rate equals the change in distance (d) over the change in time (t). Let’s look at a real-world example.\n\n### Sample Problem 1\n\nWhen the Susquehanna River reaches the Conowingo Reservoir in Maryland, the water flow slows, and much of the sediment the river has carried downstream settles out behind the Conowingo Dam. When the dam was originally built in 1928, the storage capacity of the reservoir was 300,000 acre-feet. In 1993, the USGS determined that the buildup of sediment had reduced the reservoir’s capacity to 189,000 acre-feet (Langland & Hainly, 1997). Assuming the rate of sediment deposition has been constant over that time period, at what rate (in acre-feet per year) is the reservoir’s storage capacity being reduced?\n\n### Solution 1\n\nIn this example, the changing condition (c) is the reservoir’s capacity in acre-feet. And time (t) is measured in years:\n\n$$r = Δ d Δ t = d 2 − d 1 t 2 − t 1$$\n\n$$r = 189 , 000 − 300 , 000 1993 − 1928$$\n\n$$r = 111 , 000 acre-feet 65 years$$\n\n$$r = 1 , 708 acre-feet/year$$\n\nThe reservoir is losing storage capacity at an average of 1,708 acre-feet per year. Thus the rate of change in capacity appears as a negative value.\n\nYou can also visualize a rate by graphing it (Figure 7). The greater the slope of the line, the faster the rate. By looking at the graph, you can see approximately how much storage capacity the reservoir lost after one year, 10 years, or any other length of time. And perhaps more importantly, you can predict when the reservoir will lose all of its capacity and need to be dredged or removed.",
null,
"Figure 7: Over time, the capacity of the Conowingo Reservoir is decreasing at a constant rate. Graphing the linear relationship between time in years and acre-feet of storage capacity allows us to visualize how quickly the reservoir is losing capacity.\n\n### Sample Problem 2\n\nIn 2006, scientists working with the Plate Boundary Observatory Network began closely tracking the location of a GPS station west of the San Andreas Fault in California, meaning that it is on the Pacific Plate. The station is moving slowly northwest as the Pacific Plate and the North America Plate grind past one another. In May 2007, researchers recorded the station 32.95 mm northwest of its original position. In May 2012, they recorded it 195.30 mm northwest of its original position (UNAVCO, 2012). On average, how fast was the station (and thus the Pacific Plate) moving between 2007 and 2012?\n\n### Solution 2\n\nTo solve for the station’s average rate of movement we need to know how far it was displaced between 2007 and 2012 (Δx).\n\n$$Δ x = x 2 − x 1$$\n\n$$Δ x = 195.30mm - 32.95mm$$\n\n$$Δ x = 162.35mm$$\n\nSince the time period of interest is between 2007 and 2012, we know that Δt = 5 years. Therefore:\n\n$$r = Δ x Δ t$$\n\n$$r = 162.35mm 5.0 years$$\n\n$$r = 32.5mm year$$\n\nBetween 2007 and 2012, the plate moved an average of approximately 32.5 mm northwest per year.\n\n### Sample Problem 3\n\nIf the plate continues moving at the same rate in the same direction, how far will it be from its original (May 2006) position by May of 2050?\n\n### Solution 3\n\nNow that you have calculated the plate’s rate of movement, you can calculate how far it will travel using the general equation:\n\n$$d = r × t$$\n\nWhere d stands for distance, r stands for rate of movement, and t stands for time elapsed.\n\n$$d = 32.5mm year × (2050 - 2006)$$\n\n$$d = 32.5mm year × 44$$\n\n$$d ≈ 1,430mm$$\n\nBy May 2050, the station will have moved approximately 1,430 millimeters.\n\n## How to convert units of measurement\n\nWhen making a calculation in science (or daily life), it’s important to make sure you use consistent units of measurement—and accurately convert from one unit to another when needed. The metric system (also called SI, for Système Internationale) is used by scientists all around the world and by most countries. But if you live in or travel to the United States or a handful of Caribbean countries, you may need to convert between SI units and English units. Regardless of where you live, being able to convert between units comes in handy for finding the product that is the best deal per unit weight at the market, converting currency, and converting between different types of SI units in science. For more information, see our The Metric System: Metric and Scientific Notation.\n\n### Sample Problem 4\n\nYour friend works for a US newspaper and wants to report on the findings of the scientists in Sample Problem 2 above. Knowing that the paper’s audience is more familiar with inches than millimeters, he wants to convert the rate of plate movement from millimeters per year to inches per year and asks for your help. What rate should you tell your friend to report?\n\n### Solution 4\n\nTo make this conversion, you need to know how many millimeters there are in an inch. You look up the follow conversion factor:\n\n$$25.4mm = 1 inch$$\n\nNow you can convert the plate’s rate of movement to inches per year:\n\n$$r = 32.5mm 1 year × 1 inch 25.4mm$$\n\n$$r = 32.5 m̶m̶̶̶ 1 year × 1 inch 25.4m̶m̶$$\n\n$$r ≈ 1.28 inch year$$\n\nYour friend should report that the plate is moving at an average rate of approximately 1.28 inches per year.\n\nFor more on converting units, including a description of the factor-label method for solving equations, see our module Unit Conversion: Dimensional Analysis.\n\nComprehension Checkpoint\n\nWhen showing rate of change on a graph, a steeper slope indicates a __________ rate of change.\n\n## Non-linear relationships\n\nThere are many relationships in science that cannot be described by a linear equation. For example, in biology, tissue growth in some species occurs at different rates throughout an organism’s life and cannot be described by a single equation. (Think of how quickly a baby grows compared to a teenager or an adult.) In Earth science, lava flows often occur in spurts as a volcano goes through active and quiet periods. In these cases, it doesn’t necessarily make sense to describe growth or flow rates with a single equation or to calculate an average annual rate. Other phenomena, such as growth of a population, cell division, or the rate of some chemical reactions, may occur at exponential rates. These relationships are expressed in exponential equations, which produce Cartesian graphs with curved lines instead of straight lines.\n\nLinear equations are an important tool in science and many everyday applications. They allow scientist to describe relationships between two variables in the physical world, make predictions, calculate rates, and make conversions, among other things. Graphing linear equations helps make trends visible. Gaining a strong understanding of linear equations both helps in scientific problem solving and lays a foundation for exploring other, more mathematically complex relationships in science.\n\n### Summary\n\nLinear equations can be used to describe many relationships and processes in the physical world, and thus play a big role in science. This module traces the development of linear equations and explores their many uses in science. The standard form of linear equations is presented, and sample problems are given. Concepts include Cartesian coordinates, ordered pairs, slope-intercept form, describing vertical and horizontal lines, and calculating rates.\n\n### Key Concepts\n\n• A linear equation describes a relationship between two variables that can be graphed as a straight line on the Cartesian coordinate system (x- and y-axis system).\n\n• Linear equations have numerous applications in science, including converting units (such as degrees Celsius to Fahrenheit) and calculating rates (such as how quickly a tectonic plate is moving).\n\n• Most linear equations can be put into slope-intercept form: y = mx + b, where m is the slope of the line and b is the point where the line crosses the y-axis. This form is useful for graphing linear equations. When linear equations in this form are used in science, b often represents the starting point of an experiment or series of observations.\n\n• HS-C3.5\n• ##### References\n• Baki, A. (1992). Al Khwarizmi's Contributions to the Science of Mathematics: Al Kitab Al Jabr Wa'l Muqabalah. Journal of Islamic Academy of Sciences, 5(3), 225-228.\n\n• Berenbaum, M. (2008, Winter). Entomological Bandwidth. American Entomologist, 54(4), 196-197.\n• Cooke, R. (1997). The History of Mathematics: A Brief Course. New York, NY: John Wiley and Sons.\n• Dahan-Dalmedico, A. & Peiffer, J. (2010). History of Mathematics: Highways and Byways (S. Segal, Trans.). Washington, DC: Mathematical Society of America. (Original work published 1986)\n• Derbyshire, J. (2006). Unknown Quantity: A Real and Imaginary History of Algebra. Washington, DC: Joseph Henry Press.\n• Dolbear, A. E. (1897). The Cricket as a Thermometer. The American Naturalist, 31, 970-971.\n• Joint POW/MIA Accounting Command. (n.d.). JPAC Central Identification Laboratory (CIL). Retrieved September 9, 2012, from http://www.jpac.pacom.mil/index.php?page=cil&size=100&ind=3\n• Joint POW/MIA Accounting Command. (2012). JPAC Central Identification Laboratory Fact Sheet (JPAC FS-2). Retrieved September 9, 2012, from"
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https://www.bestprog.net/en/2021/12/22/python-generators-of-sets-dictionary-generators-sets-and-generator-expressions/ | [
"# Generators of sets. Dictionary generators. Sets and generator expressions. Dictionaries and generator expressions. Examples\n\nBefore exploring this topic, it is recommended that you familiarize yourself with the following topic:\n\n### Contents\n\nSearch other resources:\n\n##### 1. Generators of sets. Concepts. General form\n\nIn addition to list generators, there are also set generators and dictionary generators in Python. These new forms of generators were introduced in Python 3.0. Set generators allow you to get sets based on a given expression formed from a sequence.\n\nThe general form of using the set generator is as follows:\n\n`resSet = { f(x) for x in sequence if condition(x) }`\n\nhere\n\n• resSet – the resulting set;\n• f(x) – arbitrary expression;\n• x – a variable that alternately takes a value from sequence;\n• sequence – some sequence of values;\n• condition(x) – the condition applied to the variable x.\n\nThere is also a simplified general form of a set generator. There is no if condition in this form:\n\n`{ f(x) for x in sequence }`\n\nBoth of the above forms give the result entirely in the form of a set. These forms can be replaced accordingly by constructions\n\n`resSet = set ( f(x) for x in sequence if condition(x) )`\n\nor\n\n`resSet = set ( f(x) for x in sequence )`\n\n### ⇑\n\n##### 2.1. Generating a set of random numbers according to a condition\n\nTask. Use the set generator to form a set of numbers that are raised to power 3.\n\nSolution.\n\n```# Generator set\n\n# Form a set of 10 numbers raised to power 3\nres = { x*x*x for x in range(1, 11) }\nprint(\"res = \", res)```\n\nProgram result\n\n`res = {64, 1, 512, 8, 1000, 343, 216, 729, 27, 125}`\n\n### ⇑\n\n##### 2.2. Generate set based on the given list\n\nTask. A list L is given. Using the set generator, it is necessary to create a set S based on the list L.\n\n```# Set generator.\n# Form the set from the given list\nL = [ 2.88, 3.5, 1.7, -2.4, 65.8, 2.88, 2.88, 1.7, -2.4, 3.2 ]\nS = { item for item in L }\nprint(\"S = \", S)```\n\nProgram result\n\n`S = {3.5, 65.8, -2.4, 2.88, 1.7, 3.2}`\n\n### ⇑\n\n##### 2.3. Generate a set of paired items from the list\n\nTask. A list of integers is given. Using the set generator, create a set containing paired elements of the list.\n\nSolution.\n\n```# Set generator.\n# From the given list, form a set containing paired elements\nL = [ 2, 8, 5, 2, 8, 4, 3, 3, 2, 8, 6, 7, 11, 2 ]\nS = { item for item in L if item%2 == 0 }\nprint(\"S = \", S)```\n\nProgram result\n\n`S = {8, 2, 4, 6}`\n\n### ⇑\n\n##### 3. Dictionary generators. Peculiarities. General form\n\nDictionary generators have been introduced in Python since version 3.0. With the help of these generators, dictionaries can be formed according to the following general form\n\n`resDict = { key:val for (key, val) in zip(keys, vals) if condition }`\n\nhere\n\n• resDict – an object that is a dictionary;\n• (key:val) – a pair in which key is a key, val is the value obtained by the key;\n• sequence – some sequence. This sequence can be formed by the zip() function or another function;\n• condition – the condition on the basis of which the values from the sequence are selected.\n\nThe above general form can be simplified in that the if condition is missing\n\n`resDict = { key:val for (key, val) in zip(keys, vals) }`\n\nUsing the dict() function, you can replace both common forms with the following calls\n\n`resDict = dict(zip(keys, values))`\n\nhere\n\n• keys – a set of keys in the dictionary;\n• values – a set of values corresponding to keys.\n\n### ⇑\n\n##### 4.1. Generate a dictionary based on a range of values range()\n\nTask. A sequence of paired numbers from 0 to 10 inclusive is given. From the given sequence, form a dictionary containing (key: value) pairs in which the value of key is twice of value.\n\nSolution.\n\n```# Dictionary generators\n# Create a dictionary based on a range of values\n\n# Specify a set of paired numbers 0, 2, 4, 8, 10\nR = range(0, 11, 2)\n\n# Create a dictionary\nD = { t: t+t for t in R }\n\n# Print the result\nprint(D)```\n\nProgram result\n\n`{0: 0, 2: 4, 4: 8, 6: 12, 8: 16, 10: 20}`\n\n### ⇑\n\n##### 4.2. Build a dictionary based on a list of numbers\n\nTask. Based on the list of Values numbers, form a dictionary in which each key is the position number of a value from the Values list. In other words, enumerate the Values value. Consider that positions are numbered from 1.\n\nSolution. To get the dictionary, you need to use the zip() function, which combines the two sequences Keys and Values.\n\n```# Dictionary generators\n# Task. Number the values in the list\n\n# A list of numbers is specified - these will be values\nValues = [ 1.7, 2.2, 0.4, 3.8, 2.2 ]\n\n# Generate keys based on the list\nKeys = range(1, len(Values))\n\n# Way 1. Create a dictionary and output it\nD = { key:val for (key, val) in zip(Keys, Values) }\nprint(D)\n\n# Way 2. Using the dict() function\nD2 = dict(zip(Keys, Values))\nprint(D2)```\n\nProgram result\n\n```{1: 1.7, 2: 2.2, 3: 0.4, 4: 3.8}\n{1: 1.7, 2: 2.2, 3: 0.4, 4: 3.8}```\n\n### ⇑\n\n##### 4.3. Generate a dictionary based on a list of strings in which the length is greater than the specified value\n\nTask. A list S of strings is given. Based on the list S, form a dictionary D, in which the keys are the position of the corresponding string in the list. Include in the dictionary only those elements from the list S, the length of which is more than 4 characters.\n\nSolution.\n\n```# Task. Include strings longer than 4 characters in the dictionary.\n# A list of strings is specified\nS = [ 'abc', 'ab', 'abcd', 'jklmn', 'jprst', 'xyz' ]\n\n# Form a new list with strings longer than 4\nS2 = [ t for t in S if len(t)>4 ]\n\n# Create a list of line numbers according to a condition\nL = range(1, len(S2)+1)\n\n# Create a dictionary\nD = { k:v for (k, v) in zip(L, S2) }\n\n# Display the dictionary\nprint(D)```\n\nProgram result\n\n`{1: 'jklmn', 2: 'jprst'}`\n\n### ⇑\n\n##### 4.4. Build a dictionary based on the set of unique strings. Dictionary values are string lengths\n\nTask. A set S of strings is given. Based on the set S, form a dictionary D in which:\n\n• keys are strings from the set S;\n• values are the length of each string, respectively.\n\nSolution.\n\n```# A set of lines is given\nS = { 'abc', 'ab', 'abcd', 'jklmn', 'jprst', 'xyz' }\n\n# Based on the set S, generate a value using a list generator.\n# Each value is the length of the corresponding string in the set S\nL = [ len(t) for t in S ]\n\n# Way 1. Create a dictionary and output it\nD = { k:v for (k, v) in zip(S, L) }\nprint(D)\n\n# Way 2. Using the dict() function\nD2 = dict(zip(S, L))\nprint(D2)```\n\nProgram result\n\n```{'abc': 3, 'ab': 2, 'abcd': 4, 'jklmn': 5, 'jprst': 5, 'xyz': 3}\n{'abc': 3, 'ab': 2, 'abcd': 4, 'jklmn': 5, 'jprst': 5, 'xyz': 3}```\n\n### ⇑\n\n##### 4.5. Form a dictionary from a tuple of numbers, in which the values are positive\n\nTask. A tuple is given. Create a dictionary in which key: value pairs are formed according to the following rules:\n\n• values are only positive numbers;\n• the keys are the ordinal numbers of the corresponding values.\n\nSolution\n\n```# Task. Build a dictionary based on a tuple\n# Specified tuple\nT = ( -8.2, 3.4, 1.6, 12, 12.5, -4.2, 0.8, 7, 7 )\n\n# Form a dictionary\nD = { key:value for (key, value) in zip(range(1, len(T)+1), T) if value>0 }\n\n# Print the dictionary\nprint(D)```\n\nProgram result\n\n`{2: 3.4, 3: 1.6, 4: 12, 5: 12.5, 7: 0.8, 8: 7, 9: 7}`\n\n### ⇑\n\n##### 5. Sets, dictionaries, and generator expressions\n\nSet generators and dictionary generators construct the entire result. If you need values from a set or a dictionary to be supplied one at a time on demand, then the forming expression must be enclosed in parentheses ( ).\n\nUsing a generator expression to supply data to a set is implemented as follows\n\n```IterObjSet = ( expression )\nS = set(IterObjSet)```\n\nhere\n\n• expression – generator expression;\n• IterObjSet – an iterated object that supplies data on demand;\n• S – resulting set.\n\nUsing a generator expression to supply data to a dictionary could be like this\n\n```IterObjDict = ( expression )\nD = dict(IterObjDict)\nD = { key:value for (key, value) in zip( Keys, Values) }```\n\nIf you use an iterated object to generate a dictionary using a generator, then the following code is possible\n\n`D = { key:value for (key, value) in zip( Keys(IterObjDict), Values(IterObjDict)) }`\n\nhere\n\n• key:value – a pair of values obtained according to the Keys and Values sequences;\n• Keys(IterObjDict) – a sequence of keys that can be obtained from IterObjDict. A situation is possible when the sequence of Keys keys is formed independently of IterObjDict;\n• Values(IterObjDict) – a sequence of values that can be obtained from IterObjDict. It is possible that the sequence of Values is generated regardless of IterObjDict;\n• D – resulting dictionary.\n\n### ⇑\n\n##### 6. An example demonstrating the use of generator expressions in combination with sets\n\nThe example demonstratively creates a list of L integer elements. Then, based on the list L, an iterated IterObj object is created that will supply data on demand. The object is created using a generator expression.\n\nUsing the next() method, the first 3 elements are retrieved from the iterated object and added to the set S. The remaining elements are added to the set S2.\n\n```# Sets and generator expressions\n\n# Specified list.\nL = [ 2, 8, 3, 4, 6, 1, 7, 9 ]\n\n# 1. Generate an iterated object containing elements from the list L\nIterObj = ( item for item in L ) # generator expression\n\n# 2. Form a set based on an iterated IterObj object\n# 2.1. Create the empty set S\nS = set()\n\n# 2.2. Pull out 3 elements one by one from IterObj\n# Pull first element from IterObj and add it to set S\nt = next(IterObj)\nS = S | {t};\n\n# Add another element to the set S\nt = next(IterObj)\nS = S | {t};\n\n# Add the third element to the set S\nt = next(IterObj)\nS = S | {t}\n\n# Display the set S\nprint(\"S = \", S)\n\n# Display the elements that are left in IterObj\nS2 = set(IterObj)\nprint(\"S2 = \", S2)```\n\nProgram result\n\n```S = {8, 2, 3}\nS2 = {1, 4, 6, 7, 9}```\n\n### ⇑\n\n##### 7. An example demonstrating the use of generator expressions to supply values to dictionaries\n\nTask. A list of strings L is given. Based on the list L, form a dictionary. In a dictionary, keys are the position numbers of the corresponding string, and the values are strings.\n\nSolution.\n\n```# Dictionaries and generator expressions\n\n# 1. A given list of strings\nL = [ 'abc', 'abcd', 'fgh', 'jklmn', 'wxy' ]\n\n# 2. Get the length of a list\nlength = len(L)\n\n# 1. Generate an iterated object containing elements from the list L\nIterObj = ( item for item in L ) # вираз-генератор\n\n# 2. Form a dictionary containing the first 2 lines from the list L\n# 2.1. Form an empty dictionary\nD = {}\n\n# 2.2. Extract the first element from IterObj and add it to the dictionary\nt = next(IterObj)\nD = t # here pair (key:value) => (0 : t)\n\n# 2.3. Pull out the second element from the IterObj\nt = next(IterObj)\nD = t # (key:value) => (1 : t)\n\n# 2.4. Display dictionary D\nprint(D)\n\n# 3. Form a second dictionary from the elements remaining in IterObj\n# 3.1. Get a range of keys\nR = range(2, length+1)\n\n# 3.2. Form dictionary\nD2 = { key:value for (key, value) in zip(range(2, length+1), IterObj) }\n\n# 3.3. Display the dictionary\nprint(D2)```\n\nProgram result\n\n```{0: 'abc', 1: 'abcd'}\n{2: 'fgh', 3: 'jklmn', 4: 'wxy'}```"
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"Use a list comprehension instead of a map-filter combination\n\nIn this entry I would like to write a small tip for Python programming.\n\nSeveral months ago a colleague of mine said to me that\n\nfilter-map is faster than list comprehension.\n\nThis statement seems to be said often. In fact in Fluent Python you can find the following statement at page 25.\n\nI used to believe that map and filter were faster than the equivalent listcomps\n\nAfter the sentence the author mentions a counterexample.\n\nThen which should we use? List comprehension or filter-map combination? What is the matter? I will explain one by one.\n\n### Warning\n\nWe will discuss only Python 3. More precisely we use an anaconda environment with Python 3.5.\n\n## Quick review of relevant Python's grammer\n\n### list comprehension\n\nPython's list comprehension is a very useful syntax to create a new list from an old one.\n\nbefore = [1, 2, 3, 4, 5]\nafter = [x**2 for x in before]\n\n\nThen after is the list of squares of 1 to 5: [1, 4, 9, 16, 25]. You can create the same list by using for-loop.\n\nafter = []\nfor x in before:\nafter.append(x**2)\n\n\nBut the list comprehension is easier and clearer. Moreover you can also create a restricted list by using the conditional statement if.\n\nafter = [x**2 for x in before if x % 2 == 1]\n\n\nThen after is the list of the squares of 1, 3 and 5: [1, 9, 25].\n\nNB: You can write if x % 2 instead of if x % 2 == 1 because 0 and 1 are evaluated as False and True, respectively.\n\n### map\n\nmap() is one of the built-in functions of Python. This takes a function and a list (rigorously iterable object) and apply the function to each element of the list.\n\ndef square(x):\nreturn x**2\n\nbefore = [1, 2, 3, 4, 5]\nafter = list(map(square, before))\n\n\nThen after is [1, 4, 9, 16, 25] as before. But you have to apply list() after map() to get the list because map() gives a generator rather than a list.\n\n### Generator\n\nIf you have a list in your Python code, then all values of the list lie in the memory of your PC. This behaviour can cause a problem if you want to deal with a large list-like data.\n\nA typical example is a large text data. Let us assume that you have a 100MB text file and you want to process the file line by line (e.g. word counts, line length, JSON lines, etc.).\n\nlines = list(open(\"large_file.txt\",\"r\"))\n\n\nThen lines is a list of lines in \"large_file.txt\". Thus the whole text file is extracted in your memory. This is waste of the memory space.\n\nNB: The actual memory space for the list of lines is pretty different from the file size. You can directly check it by using sys.getsizeof().\n\nTo avoid this problem the built-in function open() returns a generator. A generator returns a value in a lazy way.\n\nfo = open(\"large_file.txt\",\"r\")\n\n\nHere fo is a generator which reads no lines until we need to read any. Moreover we can read the file line by line through the generator:\n\nfor line in fo:\npass ## you can do something with line\n\n\nHere the variable line contains only one line in the text file and through for-loop we can read the file from head to toe without extracting the whole text file in the memory.\n\nAs you have already understood, you can apply list() to a generator to extract all values in the memory.\n\n### filter\n\nfilter() is also one of the built-in functions. This also takes a function and a list (again, rigorously iterable object). The function is used as a \"condition\" and filter() returns only the elements which satisfy the condition.\n\ndef is_odd(x):\nreturn x % 2 == 1\n\nbefore = [1, 2, 3, 4, 5]\nafter = list(filter(is_odd, before))\n\n\nThen after is [1, 3, 5]. Again, filter() returns a generator, not a list. Therefore we have to apply list() to see all values which can be generated.\n\n## list comprehension vs map-filter combination\n\nNow you can understand that the list comprehension\n\nlc = [x**2 for x in before if x % 2 == 1]\n\n\nis equivalent to the following code.\n\ndef square(x):\nreturn x**2\n\ndef is_odd(x):\nreturn x % 2 == 1\n\nmf = map(square, filter(is_odd, before))\n\n\nOf course you can use anonymous functions (a.k.a. lambda expressions) instead of ordinary functions:\n\nmf = map(lambda x: x**2, filter(lambda x: x % 2 == 1, before))\n\n\nThe difference of lc and mf is: lc is a list while mf is a generator. list(mf) is exactly same as lc: Both are [1, 9, 25].\n\nRecall the statement at the top of this entry.\n\nfilter-map is faster than list comprehension.\n\nThis means that list(mf) is faster than lc.\n\nHow faster is the former than the latter? It is easy to check, so why don't we check it?\n\nHere is the whole code.\n\nfrom time import time\n\ndef square(x):\nreturn x**2\n\ndef check(x):\nreturn x % 3 == 0\n\ndef listcomp_func_func(arr, n=10000):\nstart = time()\nfor _ in range(n):\nsum([square(x) for x in arr if check(x)])\nend = time()\nprint(\"listcomp func func : %.3f seconds\" % (end - start))\n\ndef listcomp_lambda_lambda(arr, n=10000):\nstart = time()\nfor _ in range(n):\nsum([x**2 for x in arr if x % 3 == 0])\nend = time()\nprint(\"listcomp lambda lambda : %.3f seconds\" % (end - start))\n\ndef map_filter_func_func(arr, n=10000):\nstart = time()\nfor _ in range(n):\nsum(map(square, filter(check, arr)))\nend = time()\nprint(\"map_filter func func : %.3f seconds\" % (end - start))\n\ndef map_filter_lambda_lambda(arr, n=10000):\nstart = time()\nfor _ in range(n):\nsum(map(lambda x:x**2, filter(lambda x: x % 3 == 0, arr)))\nend = time()\nprint(\"map_filter lambda lambda : %.3f seconds\" % (end - start))\n\nif __name__ == \"__main__\":\nlst = list(range(1000))\nn = 10000\nlistcomp_func_func(lst,n)\nlistcomp_lambda_lambda(lst,n)\nmap_filter_func_func(lst,n)\nmap_filter_lambda_lambda(lst,n)\n\n\nBecause the original list before is too small, I take a large one lst and I also change the condition slightly. I use sum() instead of list() and apply sum() to the list comprehensions as well.\n\nIn __main__ we have four functions and all of them print the time which is need to proceed the for-loop. In the for-loops we compute the same sum 10000 times (default). The only difference is that we use a list comprehension with or without ordinary functions or a map-filter combination with or without ordinary functions.\n\n• listcomp_func_func(lst) : list comprehension with ordinary functions\n• listcomp_lambda_lambda(lst) : typical list comprehension (There is NO lambda expression. I know that. The name is just for an easier comparison.)\n• map_filter_func_func(lst) : map-filter combination with ordinary functions.\n• map_filter_lambda_lambda(lst) : map-filter combination with lambda expressions.\n\nThe result (on my PC) is following.",
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"To be honest, I never expected this result. I believed that the list comprehension with functions is faster than the usual one. Anyway the result is quite different from what is often said.\n\nWe can obtain a similar result in a different way. We use -mtimeit option and a generator range(1000) instead of the list and remove sum():\n\n> python -mtimeit -s'[x**2 for x in range(1000) if x % 3 == 0]'\n10000000 loops, best of 3: 0.0585 usec per loop\n> python -mtimeit -s'map(lambda x: x**2, filter(lambda x: x % 3 == 0, range(1000)))'\n10000000 loops, best of 3: 0.0621 usec per loop\n\n\nThis shows that the map-filter combination with lambda expressions is slower than the equivalent list comprehension. But the difference is so small that you may get the converse evaluation sometimes. (Execute the same commands several times.)\n\nUnfortunately I have no idea where the difference comes from. But I can summaries the result: There is no big difference between a list comprehension and a map-filter combination from the viewpoint of performance.\n\n## The performance has no priority!\n\nYou might be able to chose the best expression each time you need to convert a list to another. But then you forget one of the important things about programming.\n\nThe real problem is that programmers have spent far too much time worrying about efficiency in the wrong places and at the wrong times; premature optimization is the root of all evil (or at least most of it) in programming.\n\nDonald Knuth, Computer Programming as an Art, 1974\n\nI agree that one should vectorize a for-loop if it is possible. This improves the performance dramatically. But the use of map/filter does not improve the performance in comparison with a list comprehension.\n\nI believe that the simplicity of codes is most important. That is because the simplicity make it easy to check the logic and to maintenance the code. The readability belongs to the simplicity. So which is easier to understand? A list comprehension?\n\n[x**2 for x in range(1000) if x % 3 == 0]\n\n\nOr a map-filter combination?\n\nmap(lambda x: x**2, filter(lambda x: x % 3 == 0, range(1000)))\n\n\nIt is obvious that the former is clearer than the latter, isn't it? This is also true even though we use ordinary functions instead of lambda expressions:\n\n[f(x) for x in lst if cond(x)]\n\n\nis clearer than\n\nmap(f, filter(cond, lst))\n\n\nbecause the former looks like an ordinary English sentence.\n\nOf course there is a big difference: The former is a list and the latter is a generator. So you may use a map-filter combination to have a generator... Well, do you know there is a list-comprehension-like expression creating a generator?\n\n(f(x) for x in lst if cond(x))\n\n\nSo why do you still use a map-filter combination?\n\n## Summary\n\n• There is no big difference between a list comprehension and a map-filter combination from the viewpoint of performance.\n• You should always use a list comprehension because of readability.\n• When you need a generator you can use a list-comprehension-like expression.\nCategories: #development"
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https://infolearners.com/ebooks/introduction-to-computation-and-programming-using-python-pdf-download/?amp=1 | [
"# Introduction To Computation And Programming Using Python 3rd Edition Pdf\n\nThis introduction to computational models with python pdf book presents a collection of problems to help readers develop the computational thinking necessary to solve engineering, mathematical, and science problems using modern computer algorithms and programming. Supplemented with documentation and instructional videos for the IPython Notebook, this book provides a practical course in computational thinking and algorithm development.\n\nThe Introduction To Computation And Programming Using Python Pdf Download introduces students with little or no prior programming experience to the art of computational problem solving using Python and various Python libraries, including PyLab. It provides students with skills that will enable them to make productive use of computational techniques, including some of the tools and techniques of “data science” for using computation to model and interpret data.\n\nIntroduction to Computation and Programming Using Python is an introduction to computational problem solving in Python. Introduction to computation and programming using python ebook includes an introduction to mathematical concepts such as variables, functions, parameters and derivatives. In addition, the book includes an introduction to elementary data structures and object-oriented programming. With this knowledge in hand, we can then proceed to actual computation and programming in the next stage of the book. To motivate all of these topics, the book works through a series of programs that model various physical phenomena, such as drawing fractal landscapes, creating interactive games and doing statistical modeling; from there we tackle more complex programs that could be seen in practical applications.\n\n## About Introduction To Computation And Programming Using Python 3rd Edition Pdf\n\nMyProgrammingLab should only be purchased when required by an instructor.\n\nIntroduction To Computation And Programming Using Python 3rd Edition Pdf is intended for use in the introduction to programming course.\n\nDaniel Liang is known for his “fundamentals-first” approach to teaching programming concepts and techniques. “Fundamentals-first” means that students learn fundamental programming concepts like selection statements, loops, and functions, before moving into defining classes. Students learn basic logic and programming concepts before moving into object-oriented programming, and GUI programming.\n\nAnother aspect of Introduction To Computation And Programming Using Python 3rd Edition Pdf is that in addition to the typical programming examples that feature games and some math, Liang gives an example or two early in the chapter that uses a simple graphic to engage the students. Rather than asking them to average 10 numbers together, they learn the concepts in the context of a fun example that generates something visually interesting.\n\nThis book introduces students with little or no prior programming experience to the art of computational problem solving using Python and various Python libraries, including PyLab. It provides students with skills that will enable them to make productive use of computational techniques, including some of the tools and techniques of “data science” for using computation to model and interpret data. The book is based on an MIT course (which became the most popular course offered through MIT’s OpenCourseWare) and was developed for use not only in a conventional classroom but in in a massive open online course (or MOOC) offered by the pioneering MIT–Harvard collaboration edX.\n\nStudents are introduced to Python and the basics of programming in the context of such computational concepts and techniques as exhaustive enumeration, bisection search, and efficient approximation algorithms. The book does not require knowledge of mathematics beyond high school algebra, but does assume that readers are comfortable with rigorous thinking and not intimidated by mathematical concepts. Although it covers such traditional topics as computational complexity and simple algorithms, the book focuses on a wide range of topics not found in most introductory texts, including information visualization, simulations to model randomness, computational techniques to understand data, and statistical techniques that inform (and misinform) as well as two related but relatively advanced topics: optimization problems and dynamic programming.\n\nUsing the graphics examples is optional in this Introduction To Computation And Programming Using Python Pdf Download. Turtle graphics can be used in Chapters 1-5 to introduce the fundamentals of programming and Tkinter can be used for developing comprehensive graphical user interfaces and for learning object-oriented programming."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8757461,"math_prob":0.6186306,"size":5644,"snap":"2022-05-2022-21","text_gpt3_token_len":1028,"char_repetition_ratio":0.17446809,"word_repetition_ratio":0.1720297,"special_character_ratio":0.18001418,"punctuation_ratio":0.061784897,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98698634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T11:40:02Z\",\"WARC-Record-ID\":\"<urn:uuid:21ccb39f-3ec8-4035-be44-b1e935293455>\",\"Content-Length\":\"111549\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea774a74-a27f-4fc9-a9f2-e392afcfa4ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0bc99ab-ac37-4167-aaad-6c599bb053ae>\",\"WARC-IP-Address\":\"162.159.137.85\",\"WARC-Target-URI\":\"https://infolearners.com/ebooks/introduction-to-computation-and-programming-using-python-pdf-download/?amp=1\",\"WARC-Payload-Digest\":\"sha1:7Z3T4GJIQGWNNJ7Q4NBUL2MAKAQEPU4P\",\"WARC-Block-Digest\":\"sha1:EQ7ODJAPZFLJ64UQCMCHEI5SV44IYD3B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531779.10_warc_CC-MAIN-20220520093441-20220520123441-00794.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/15262/how-do-i-plot-the-derivative-of-a-set-of-experimental-points?noredirect=1 | [
"# How do I plot the derivative of a set of experimental points?\n\nI have a set of points like this: P = {{$x_1$,$y_1$},{$x_2$,$y_2$},...,{$x_n$,$y_n$}}. I want to plot them and also plot the derivative of that graph. How can I do this with Mathematica? Is there a simple way in Mathematica?\n\n• In this answer to a related question I suggested to use DerivativeFilter for this purpose. That answer would apply here too (I guess the questions are somewhat different but I tried to make my earlier answer more general so that it now turns out to cover your case...) – Jens Nov 27 '12 at 3:52\n• Oh, thanks for the answer. – Ana S. H. Nov 27 '12 at 5:02\n\nHere is some sample data, hopefully of the sort you are looking for\n\na=Table[{i, i^2 - 3 i + 2 + 0.3 Random[]}, {i, -1, 3, 0.1}];\nListPlot[a]\n\n\nwill display the data.",
null,
"This will create a derivable interpolation.\n\nb = Interpolation[a, Method -> \"Spline\"];\n\n\nThis is its derivative\n\nc = b';\n\n\nPlot them together.\n\nPlot[Evaluate[{b[x], c[x]}], {x, -1, 3}]",
null,
"• Now, you may also want to try FindFit, based on the model that represents your experimental data, and then take derivate of the function. – Zviovich Nov 27 '12 at 3:15\n• Yeah, it worked. Thanks. – Ana S. H. Nov 27 '12 at 5:01\n• Finite differences and interpolation will give a somewhat smoother fit. That might or might not be desirable depending on actual needs. Here is the code to compare to. d = .1; derivs = Table[(a[[j + 1, 2]] - a[[j-1, 2]])/(2*d), {j, 2, Length[a] - 1}];derivs2 = Transpose[{Range[-1 + .1, 3 - .1, .1], derivs}]; c = Interpolation[derivs2, Method -> \"Spline\"]; – Daniel Lichtblau Nov 28 '12 at 17:43\n\nThis is the data (taken from the previous example):\n\na = Table[{i, i^2 - 3 i + 2 + 0.3 Random[]}, {i, -1, 3, 0.1}];\n\n\nThis is derivatives calculated in each point except the first:\n\nderA = Differences[a] /. {x_, y_} -> y/x;\n\n\nhere we combine it into the list of pairs:\n\nlst=Transpose[{Drop[Transpose[a][], 1], derA}];\n\n\nNow we can plot it along with the initial list:\n\n ListLinePlot[{a, Transpose[{Drop[Transpose[a][], 1], derA}]},\nPlotStyle -> {Blue, Red}]\n\n\nThat is what you get:",
null,
"I know the question is quite old but maybe this might be helpful for someone.\n\nI again follow the data set from above and show two methods of performing the task in question:\n\na = Table[{i, i^2 - 3 i + 2 + 0.3 Random[]}, {i, -1, 3, 0.1}];\nListPlot[a]\n\n\nSee image above, I'm not allowed to post three links due to not having >= 10 reputation.\n\nb = Interpolation[a, Method -> \"Spline\"];\nc = b';\nd = Interpolation[a, Method -> \"Hermite\"];\ne = d';\nPlot[Evaluate[{b[x],c[x],e[x]}],{x,-1,3},PlotStyle->{Black,Blue,Red}]",
null,
"Note how the two differnet interpolation methods differ!\n\nThe - in my oppinion - best way to go if you have knowledge about your model is the following:\n\nbp = NonlinearModelFit[a,A x^2 + B x + F,{A,B,F},{x}];\nNormal[bp]\nbp[\"ParameterTable\"]\n\n0.995015 x^2 - 2.99342 x + 2.1193\nEstimate Standard Error t-Statistic P-Value\nA 0.995015 0.00994791 100.023 1.2327*10^-47\nB -2.99342 0.0225051 -133.01 2.51291*10^-52\nF 2.1193 0.016774 126.345 1.76511*10^-51\n\ncp = bp';\nPlot[Evaluate[{bp[x],cp[x]}],{x,-1,3}]",
null,
"• Two recommendations: Random[] has been superseded by RandomReal[]. And use SeedRandom to make your results reproducible by others: SeedRandom; a = Table[{i, i^2 - 3 i + 2 + 0.3 RandomReal[]}, {i, -1, 3, 0.1}]; ListPlot[a]. Anyway, +1. – Michael E2 Aug 2 '17 at 13:07\n• You're right, thanks for pointing that out! – gothicVI Aug 3 '17 at 12:00"
]
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"https://i.stack.imgur.com/gIZfr.png",
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"https://i.stack.imgur.com/4VUL9.png",
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"https://i.stack.imgur.com/PSqyn.png",
null,
"https://i.stack.imgur.com/poNYj.png",
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"https://i.stack.imgur.com/5LPXo.png",
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]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7558958,"math_prob":0.990865,"size":3401,"snap":"2020-10-2020-16","text_gpt3_token_len":1137,"char_repetition_ratio":0.09361201,"word_repetition_ratio":0.08438061,"special_character_ratio":0.38194647,"punctuation_ratio":0.21192893,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991897,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-03T05:51:59Z\",\"WARC-Record-ID\":\"<urn:uuid:10b2bf13-d431-4a86-adfc-93c6ea314a62>\",\"Content-Length\":\"164429\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e6d7f90-dd5a-4419-bcba-de47625eb06f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6cd5740-36f8-4b5a-b08d-618e9abde8a0>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/15262/how-do-i-plot-the-derivative-of-a-set-of-experimental-points?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:BSWBOYKMEECO4VKTTL5EOTEOG2P3JLFR\",\"WARC-Block-Digest\":\"sha1:JTDPKZX55VQBFCK3DA6ABJRALI6FBME7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370510287.30_warc_CC-MAIN-20200403030659-20200403060659-00550.warc.gz\"}"} |
https://goprep.co/ex-27.f-q16-find-the-angle-between-the-lines-vector-r-2-i-5-i-1nlre5 | [
"Q. 16\n\n# Find the angle be\n\nGiven : the lines",
null,
"and",
null,
"To find : angle between the lines\n\nFormula used : If the lines are a",
null,
"+ b",
null,
"+ c",
null,
"+ λ(p",
null,
"+ q",
null,
"+ r",
null,
") and d",
null,
"+ e",
null,
"+ f",
null,
"+\n\nλ(l",
null,
"+ m",
null,
"+ n",
null,
") then the angle between the lines ‘θ’ is given by\n\nθ =",
null,
"the lines",
null,
"and",
null,
"Here p = 3 , q = 2 , r = 6 and l = 1 , m = 2 , n = 2\n\nθ =",
null,
"=",
null,
"θ =",
null,
"=",
null,
"=",
null,
"θ =",
null,
"The angle between the lines",
null,
"and",
null,
"is",
null,
"Rate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
null,
"view all courses",
null,
"RELATED QUESTIONS :\n\nFind the value ofMathematics - Board Papers\n\nShow that lines <Mathematics - Board Papers\n\nFind the value ofMathematics - Board Papers\n\nFind the value ofMathematics - Board Papers\n\nWrite the directiMathematics - Board Papers\n\n<span lang=\"EN-USMathematics - Board Papers\n\n<span lang=\"EN-USMathematics - Board Papers\n\nWhat is the angleRS Aggarwal - Mathematics\n\nFind the angle beRS Aggarwal - Mathematics"
]
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null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112868296118.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112869001685.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112869723293.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112870432661.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112871165522.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112871872594.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112872589114.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112873294797.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112873997519.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112874699230.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/156111287542337.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112876130435.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112876834762.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112877549449.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112878257982.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112878963386.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112879692551.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112880404426.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112881114685.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/15611128818419100.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112882547558.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112883256589.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112884002953.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112884714875.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112885422448.png",
null,
"https://gradeup-question-images.grdp.co/liveData/PROJ33328/1561112886130554.png",
null,
"https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png",
null,
"https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png",
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]
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https://sunsed.com/doc?q=number:modulus | [
"### number:modulus\n\nGet the remainder of dividing dividend by divisor. Unless divisor is zero, the result has the same sign as dividend.\n\nnumber:modulus(\\$dividend, \\$divisor, \\$decimal_points=0)\n\n``````number:modulus(7.5,2,2)\n#=> \"1.50\"``````\n``` ```"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.6739066,"math_prob":0.9896286,"size":293,"snap":"2022-27-2022-33","text_gpt3_token_len":87,"char_repetition_ratio":0.14532872,"word_repetition_ratio":0.0,"special_character_ratio":0.27986348,"punctuation_ratio":0.22033899,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972694,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T16:10:04Z\",\"WARC-Record-ID\":\"<urn:uuid:84360668-5d4e-4121-947d-275d2c244910>\",\"Content-Length\":\"86882\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ca2a76b-f712-4205-951c-d85c74e976a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d0b33df-7ea7-4380-a931-f568cf84e249>\",\"WARC-IP-Address\":\"45.79.101.249\",\"WARC-Target-URI\":\"https://sunsed.com/doc?q=number:modulus\",\"WARC-Payload-Digest\":\"sha1:HHFYN6RTJPJPY7QRWBFH25M6UKJKUNJF\",\"WARC-Block-Digest\":\"sha1:AKGO4PAMOYIBAIYQXZSQDLVYWRF42IBN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103036077.8_warc_CC-MAIN-20220625160220-20220625190220-00383.warc.gz\"}"} |
http://etna.ricam.oeaw.ac.at/volumes/2011-2020/vol44/abstract.php?vol=44&pages=250-270 | [
"## A convergent linear finite element scheme for the Maxwell-Landau-Lifshitz-Gilbert equations\n\nLubomir Baňas, Marcus Page, and Dirk Praetorius\n\n### Abstract\n\nWe consider the lowest-order finite element discretization of the nonlinear system of Maxwell's and Landau-Lifshitz-Gilbert equations (MLLG). Two algorithms are proposed to numerically solve this problem, both of which only require the solution of at most two linear systems per time step. One of the algorithms is decoupled in the sense that it consists of the sequential computation of the magnetization and afterwards the magnetic and electric field. Under some mild assumptions on the effective field, we show that both algorithms converge towards weak solutions of the MLLG system. Numerical experiments for a micromagnetic benchmark problem demonstrate the performance of the proposed algorithms.\n\nFull Text (PDF) [1.7 MB], BibTeX\n\n### Key words\n\nMaxwell-LLG, linear scheme, ferromagnetism, convergence\n\n65N30, 65N50\n\n< Back"
]
| [
null
]
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https://admin.clutchprep.com/physics/practice-problems/141185/it-is-correct-to-say-that-impulse-is-equal-toa-momentum-b-the-change-in-momentum | [
"# Problem: It is correct to say that impulse is equal toA. momentum.B. the change in momentum.C. the force multiplied by the distance the force acts.D. all of the aboveE. none of the above\n\n###### FREE Expert Solution\n\nImpulse = Ft\n\nFrom Newton's second law:\n\n$\\begin{array}{rcl}\\mathbf{F}& \\mathbf{=}& \\mathbf{m}\\mathbf{a}\\\\ & \\mathbf{=}& \\mathbf{m}\\mathbf{∆}\\mathbf{v}\\mathbf{/}\\mathbf{t}\\\\ \\mathbf{F}\\mathbf{t}& \\mathbf{=}& \\mathbf{m}\\mathbf{∆}\\mathbf{v}\\end{array}$",
null,
"###### Problem Details\n\nIt is correct to say that impulse is equal to\n\nA. momentum.\n\nB. the change in momentum.\n\nC. the force multiplied by the distance the force acts.\n\nD. all of the above\n\nE. none of the above"
]
| [
null,
"https://cdn.clutchprep.com/assets/button-view-text-solution.png",
null
]
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https://electricscotland.com/kids/rolfins_orb/book1_chap20.htm | [
"",
null,
"Rolfin's Orb Book 1 - Obsidian Word Search Puzzle\n `F T K M F Y Q G U D P G G G Q N L B J O I P E L S C R I B E ` `M G M S L N M E S K R V W Z C R T C B R L H A L Y S N A B D ` `N C F Q F N Q S C E Q Y P X K Z U S Q J U M R I L O M R S K ` `Q N A X Y H W R E A L P I A J E I P O X F T O O R X O B W J ` `D P S L S O W C L O S O P S G D E L B A T I J N C H O R J H ` `Q L G A L J E I F W U T P T I J K F V V Z Z A M H W Z K F F ` `V G N W L I B R X U R V L A R I C O W W B E C H D L Q A R D ` `P E H F Z Q S L L J I W N E B F X C O Q Y T Q A B T F Y F G ` `E C A I S R F T B F E U E O M O N V J B W A N S F W I O C G ` `D O L K V P B N E X D R E E I A M O F G H B M V V V C F B F ` `C A R N K V K A C R O E F D J Q K F F F N L F H T E P S L E ` `K O O S E W Y S E M M C J Y J C D P Y S E F Y H W A Y F S E ` `U A Y T Y E Z B H U P B B M R R O Y X K M V M K Y M Q Q K X ` `K Q O Y U Z P T D G R P X Q M I N D R A Y T O N A J Q V P C ` `C W O F Z T A S I N R W E U U C K E S C H K Z M S N N O J O ` `N O O D H C O L N I T Y H C I C E H Y D R A R T H W O I F X ` `B R R G H K L F V T R T O L E N Y C B S A M N R I H E I W Q ` `B D M M S R P U E T V W E V I U L T V P V L N M I G C A F H ` `A K P U G J S O R I A A Z K R N M L R Q Y Y Z S Y B S I U V ` `U J G F L G V P A N G S O B L Q Z M W N B I N O M K G B E Z ` `L N U F Q L L Z L K O L H U E T O F D C S A I J M D B M U M ` `A A O T H Y A A B Y A L B W Y H X X K P F O T D I T G M K V ` `H M G B V Q R C A S F C E U O A W F F H K R W H J Q V M U E ` `L A C S U M N Z I V S H B D V K J T S H G T V S D X E N Z P ` `B I O U T P Y G W A O J G P N A H K Z L W L S B P A M D I L ` `K G O J D X N E C F R Z M C M A S G H U L D J N N M R F R E ` `I R I A M H P O V P Y C V X C P D N K X O U M P Z G H A R F ` `F H A M W K U D N W G J P N Z Z H D H Y Z R B R E E S T A T ` `Z F E W V C G L Z L F E M C A K P H A B A B U T F D Y H Y B ` `T J M L F P I N G P Y D L N F O L Q L D I O J P P D Q Y Y B ` `ALROY ANGUS` `ATHDARA BOOK` `CALLUM CASTLE` `CATHMORE COWAN` `DANDELON DONKEY` `DRAYTON ELSPET` `FIONA GAELIC` `GREECE GYRO` `HYDRA INVERALBA` `IONMHAS JIMMY` `JOHNNY KNITTING` `LOCHDOON MAIRI` `MCALLISTER NEEPS` `NIKOLAS OBSIDIAN` `ORB SCRIBE` `TABLE`"
]
| [
null,
"https://electricscotland.com/images/menunew.jpg",
null
]
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https://planetmath.org/frequentlyin | [
"# frequently in\n\nRecall that a net is a function $x$ from a directed set",
null,
"",
null,
"$D$ to a set $X$. The value of $x$ at $i\\in D$ is usually denoted by $x_{i}$. Let $A$ be a subset of $X$. We say that a net $x$ is frequently in $A$ if for every $i\\in D$, there is a $j\\in D$ such that $i\\leq j$ and $x_{j}\\in A$.\n\nSuppose a net $x$ is frequently in $A\\subseteq X$. Let $E:=\\{j\\in D\\mid x_{j}\\in A\\}$. Then $E$ is a cofinal subset of $D$, for if $i\\in D$, then by definition of $A$, there is $i\\leq j\\in D$ such that $x_{j}\\in A$, and therefore $j\\in E$.\n\nThe notion of “frequently in” is related to the notion of “eventually in” in the following sense: a net $x$ is eventually in a set $A\\subseteq X$ iff it is not frequently in $A^{\\complement}$, its complement",
null,
"",
null,
". Suppose $x$ is eventually in $A$. There is $j\\in D$ such that $x_{k}\\in A$ for all $k\\geq j$, or equivalently, $x_{k}\\in A^{\\complement}$ for no $k\\geq j$. The converse",
null,
"",
null,
"is can be argued by tracing the previous statements backwards.\n\nIn a topological space",
null,
"",
null,
"$X$, a point $a\\in X$ is said to be a cluster point",
null,
"",
null,
"of a net $x$ (or, occasionally, $x$ clusters at $a$) if $x$ is frequently in every neighborhood",
null,
"",
null,
"",
null,
"of $a$. In this general definition, a limit point is always a cluster point. But a cluster point need not be a limit point. As an example, take the sequence $0,2,0,4,0,6,0,8,\\ldots,0,2n,0,\\ldots$ has $0$ as a cluster point. But clearly $0$ is not a limit point, as the sequence diverges in $\\mathbb{R}$.\n\nTitle frequently in FrequentlyIn 2013-03-22 17:14:23 2013-03-22 17:14:23 CWoo (3771) CWoo (3771) 6 CWoo (3771) Definition msc 03E04 clusters at cluster point of a net"
]
| [
null,
"http://mathworld.wolfram.com/favicon_mathworld.png",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://mathworld.wolfram.com/favicon_mathworld.png",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://mathworld.wolfram.com/favicon_mathworld.png",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://mathworld.wolfram.com/favicon_mathworld.png",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null,
"http://planetmath.org/sites/default/files/fab-favicon.ico",
null
]
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https://www.sarthaks.com/406864/what-motion-vertical-circle-explain-different-position-velocity-any-point-vertical-loop | [
"# What is Motion in Vertical Circle ? Explain the different position of Velocity at any point on vertical loop.\n\n33 views\nin Physics\nedited\n\nWhat is Motion in Vertical Circle ? Explain the different position of Velocity at any point on vertical loop.\n\nby (69.9k points)\nselected\n\nThis is an example of non-uniform circular motion. In this motion body is under the influence of gravity of earth.\n\nVelocity at any point on vertical loop : If u is the initial velocity imparted to body at lowest point then, velocity of body at height h is given by",
null,
"",
null,
"Tension at any point on vertical loop : Tension at general point P,",
null,
""
]
| [
null,
"https://www.sarthaks.com/",
null,
"https://www.sarthaks.com/",
null,
"https://www.sarthaks.com/",
null
]
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https://www.thestudentroom.co.uk/showthread.php?t=3116477 | [
"# Binding Energy and Energy Release in Radioactive Decay...\n\nWatch\nAnnouncements\n#1\n\"A group of unbound nucleons that are widely separated, their total potential energy is considered to be zero. As they come together and form a nucleus, they will lose potential energy...Hence, energy of a stable nucleus is less than that of the unbound nucleons required to make it up\"\n\nI understand that when nucleons are fused together to make the stable nucleus energy is required to hold them together via strong nuclear force in which mass will be converted to energy (E=mc2) and therefore the difference between energy of unbound nucleons and stable nucleus is called the binding energy however, what confuses me is in that paragraph if the total potential energy is zero how can it lose potential energy when its formed into a nucleus?\n\nSecondly, when an unstable nucleus undergoes decay and produces a daughter nucleus and a decay product (i.e alpha or beta particle). Why does the combined mass of the daughter nucleus and the decay product come out to be less than the original parent nucleus?\n\nSorry for the long essay I can't seem to get a grasp on this concept",
null,
"0\n5 years ago\n#2\nFor the potential energy, you can set where zero is. It's the same way as if you set zero gravitational energy to be you standing on the earth, if you jump into a hole you have \"negative potential energy\".\n\nFor the second part, energy is conserved. E^2 = (pc)^2 + (mc^2)^2 where p is the momentum. If the decay products have momentum, they won't have the combined mass of the parent.\n0\n#3\n(Original post by tiddlytom)\nFor the potential energy, you can set where zero is. It's the same way as if you set zero gravitational energy to be you standing on the earth, if you jump into a hole you have \"negative potential energy\".\n\nFor the second part, energy is conserved. E^2 = (pc)^2 + (mc^2)^2 where p is the momentum. If the decay products have momentum, they won't have the combined mass of the parent.\nso for binding energy; if you were to push unbound nucleons to form a nucleus, the energy required to push them together is that the binding energy or is it the energy that keeps the nucleons together in the nucleus?\n\nand for the radioactive decay, is it because energy is released during the decay that lower the mass due to E=mc2? thanks for your help its cleared up alot of confusion",
null,
"0\n5 years ago\n#4\n(Original post by MSB47)\nso for binding energy; if you were to push unbound nucleons to form a nucleus, the energy required to push them together is that the binding energy or is it the energy that keeps the nucleons together in the nucleus?\n\nand for the radioactive decay, is it because energy is released during the decay that lower the mass due to E=mc2? thanks for your help its cleared up alot of confusion",
null,
"The binding energy is the energy that holds them together.\n\nAs above, E^2 = (pc)^2 + (mc^2)^2, not just E=mc^2. Energy is conserved, so if some goes into momentum, you \"lose\" some mass.\n0\n#5\n(Original post by tiddlytom)\nThe binding energy is the energy that holds them together.\n\nAs above, E^2 = (pc)^2 + (mc^2)^2, not just E=mc^2. Energy is conserved, so if some goes into momentum, you \"lose\" some mass.\nahh ok thanks for clarifying that for me\n0\nX\n\nnew posts",
null,
"Back\nto top\nLatest\nMy Feed\n\n### Oops, nobody has postedin the last few hours.\n\nWhy not re-start the conversation?\n\nsee more\n\n### See more of what you like onThe Student Room\n\nYou can personalise what you see on TSR. Tell us a little about yourself to get started.\n\n### Poll\n\nJoin the discussion\n\n#### Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?\n\nYes (70)\n27.13%\nNo - I have already returned home (29)\n11.24%\nNo - I plan on travelling outside these dates (54)\n20.93%\nNo - I'm staying at my term time address over Christmas (28)\n10.85%\nNo - I live at home during term anyway (77)\n29.84%"
]
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null,
"https://static.thestudentroom.co.uk/images/smilies/confused.gif",
null,
"https://static.thestudentroom.co.uk/5fbcd35206bf3/forum/images/smilies/smile.png",
null,
"https://static.thestudentroom.co.uk/5fbcd35206bf3/forum/images/smilies/smile.png",
null,
"https://www.thestudentroom.co.uk/images/v2/icons/arrow_up.svg",
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https://raweb.inria.fr/rapportsactivite/RA2019/factas/factas.xml | [
"FACTAS Functional Analysis for ConcepTion and Assessment of Systems Optimization and control of dynamic systems Applied Mathematics, Computation and Simulation https://team.inria.fr/factas/ Creation of the Project-Team: 2019 July 01 Project-Team A6.1.1. - Continuous Modeling (PDE, ODE) A6.2.1. - Numerical analysis of PDE and ODE A6.2.5. - Numerical Linear Algebra A6.2.6. - Optimization A6.3.1. - Inverse problems A6.3.4. - Model reduction A6.4.3. - Observability and Controlability A6.4.4. - Stability and Stabilization A6.4.5. - Control of distributed parameter systems A6.5.4. - Waves A8.2. - Optimization A8.3. - Geometry, Topology A8.4. - Computer Algebra B1.2.3. - Computational neurosciences B2.6.1. - Brain imaging B3.3. - Geosciences B4.5. - Energy consumption B6.2.2. - Radio technology B6.2.3. - Satellite technology Fabien Seyfert Chercheur Sophia Team leader, Inria, Researcher oui Laurent Baratchart Chercheur Sophia Inria, Senior Researcher oui Sylvain Chevillard Chercheur Sophia Inria, Researcher Juliette Leblond Chercheur Sophia Inria, Senior Researcher oui Martine Olivi Chercheur Sophia Inria, Researcher oui Vanna Lisa Coli PostDoc Sophia Univ. Côte d'Azur, Post-Doctoral Fellow Paul Asensio PhD Sophia Inria, PhD Student, from November 2019 Gibin Bose PhD Sophia Univ. Côte d'Azur, PhD Student Sébastien Fueyo PhD Sophia Inria, PhD Student David Martinez Martinez PhD Sophia Univ. de Limoges, PhD Student, until April 2019 Masimba Nemaire PhD Sophia Univ. de Bordeaux, PhD Student, from October 2019 Paul Asensio Stagiaire Sophia Inria, from April 2019 until September 2019 Masimba Nemaire Stagiaire Sophia Inria, from April 2019 until August 2019 Tuong Vy Nguyen Hoang Stagiaire Sophia Inria, from March 2019 until August 2019 Pat Vatiwutipong Stagiaire Sophia Univ. Côte d'Azur, from April 2019 until August 2019 Konstantinos Mavreas PhD Sophia Univ. Côte d'Azur Marie-Line Meirinho Assistant Sophia Inria Adam Cooman CollaborateurExterieur Sophia Ampleon Jean-Paul Marmorat CollaborateurExterieur Sophia Ecole Nationale Supérieure des Mines de Paris David Martinez Martinez CollaborateurExterieur Sophia Univ. de Limoges, from July 2019 Overall Objectives Research Themes\n\nThe team develops constructive, function-theoretic approaches to inverse problems arising in modeling and design, in particular for electro-magnetic systems as well as in the analysis of certain classes of signals.\n\nData typically consist of measurements or desired behaviors. The general thread is to approximate them by families of solutions to the equations governing the underlying system. This leads us to consider various interpolation and approximation problems in classes of rational and meromorphic functions, harmonic gradients, or solutions to more general elliptic partial differential equations (PDE), in connection with inverse potential problems. A recurring difficulty is to control the singularities of the approximants.\n\nThe mathematical tools pertain to complex and harmonic analysis, approximation theory, potential theory, system theory, differential topology, optimization and computer algebra. Targeted applications include:\n\n• identification and synthesis of analog microwave devices (filters, amplifiers),\n\n• non-destructive control from field measurements in medical engineering (source recovery in magneto/electro-encephalography), and paleomagnetism (determining the magnetization of rock samples).\n\n• In each case, the endeavor is to develop algorithms resulting in dedicated software.\n\nResearch Program Introduction\n\nWithin the extensive field of inverse problems, much of the research by Factas deals with reconstructing solutions of classical elliptic PDEs from their boundary behavior. Perhaps the simplest example lies with harmonic identification of a stable linear dynamical system: the transfer-function $f$ can be evaluated at a point $i\\omega$ of the imaginary axis from the response to a periodic input at frequency $\\omega$. Since $f$ is holomorphic in the right half-plane, it satisfies there the Cauchy-Riemann equation $\\overline{\\partial }f=0$, and recovering $f$ amounts to solve a Dirichlet problem which can be done in principle using, e.g. the Cauchy formula.\n\nPractice is not nearly as simple, for $f$ is only measured pointwise in the pass-band of the system which makes the problem ill-posed . Moreover, the transfer function is usually sought in specific form, displaying the necessary physical parameters for control and design. For instance if $f$ is rational of degree $n$, then $\\overline{\\partial }f={\\sum }_{1}^{n}{a}_{j}{\\delta }_{{z}_{j}}$ where the ${z}_{j}$ are its poles and ${\\delta }_{{z}_{j}}$ is a Dirac unit mass at ${z}_{j}$. Thus, to find the domain of holomorphy (i.e. to locate the ${z}_{j}$) amounts to solve a (degenerate) free-boundary inverse problem, this time on the left half-plane. To address such questions, the team has developed a two-step approach as follows.\n\nStep 1 relates to extremal problems and analytic operator theory, see Section . Step 2 involves optimization, and some Schur analysis to parametrize transfer matrices of given Mc-Millan degree when dealing with systems having several inputs and outputs, see Section . It also makes contact with the topology of rational functions, in particular to count critical points and to derive bounds, see Section . Step 2 raises further issues in approximation theory regarding the rate of convergence and the extent to which singularities of the approximant (i.e. its poles) tend to singularities of the approximated function; this is where logarithmic potential theory becomes instrumental, see Section .\n\nApplying a realization procedure to the result of step 2 yields an identification procedure from incomplete frequency data which was first demonstrated in to tune resonant microwave filters. Harmonic identification of nonlinear systems around a stable equilibrium can also be envisaged by combining the previous steps with exact linearization techniques from .\n\nA similar path can be taken to approach design problems in the frequency domain, replacing the measured behavior by some desired behavior. However, describing achievable responses in terms of the design parameters is often cumbersome, and most constructive techniques rely on specific criteria adapted to the physics of the problem. This is especially true of filters, the design of which traditionally appeals to polynomial extremal problems , . To this area, Apics contributed the use of Zolotarev-like problems for multi-band synthesis, although we presently favor interpolation techniques in which parameters arise in a more transparent manner, as well as convex relaxation of hyperbolic approximation problems, see Sections and .\n\nThe previous example of harmonic identification quickly suggests a generalization of itself. Indeed, on identifying $ℂ$ with ${ℝ}^{2}$, holomorphic functions become conjugate-gradients of harmonic functions, so that harmonic identification is, after all, a special case of a classical issue: to recover a harmonic function on a domain from partial knowledge of the Dirichlet-Neumann data; when the portion of boundary where data are not available is itself unknown, we meet a free boundary problem. This framework for 2-D non-destructive control was first advocated in and subsequently received considerable attention. It makes clear how to state similar problems in higher dimensions and for more general operators than the Laplacian, provided solutions are essentially determined by the trace of their gradient on part of the boundary which is the case for elliptic equations There is a subtle difference here between dimension 2 and higher. Indeed, a function holomorphic on a plane domain is defined by its non-tangential limit on a boundary subset of positive linear measure, but there are non-constant harmonic functions in the 3-D ball, ${C}^{1}$ up to the boundary sphere, yet having vanishing gradient on a subset of positive measure of the sphere. Such a “bad” subset, however, cannot have interior points on the sphere. . Such questions are particular instances of the so-called inverse potential problem, where a measure $\\mu$ has to be recovered from the knowledge of the gradient of its potential (i.e., the field) on part of a hypersurface (a curve in 2-D) encompassing the support of $\\mu$. For Laplace's operator, potentials are logarithmic in 2-D and Newtonian in higher dimensions. For elliptic operators with non constant coefficients, the potential depends on the form of fundamental solutions and is less manageable because it is no longer of convolution type. Nevertheless it is a useful concept bringing perspective on how problems could be raised and solved, using tools from harmonic analysis.\n\nInverse potential problems are severely indeterminate because infinitely many measures within an open set of ${ℝ}^{n}$ produce the same field outside this set; this phenomenon is called balayage . In the two steps approach previously described, we implicitly removed this indeterminacy by requiring in step 1 that the measure be supported on the boundary (because we seek a function holomorphic throughout the right half-space), and by requiring in step 2 that the measure be discrete in the left half-plane (in fact: a finite sum of point masses ${\\sum }_{1}^{N}{a}_{j}{\\delta }_{{z}_{j}}$). The discreteness assumption also prevails in 3-D inverse source problems, see Section . Conditions that ensure uniqueness of the solution to the inverse potential problem are part of the so-called regularizing assumptions which are needed in each case to derive efficient algorithms.\n\nTo recap, the gist of our approach is to approximate boundary data by (boundary traces of) fields arising from potentials of measures with specific support. This differs from standard approaches to inverse problems, where descent algorithms are applied to integration schemes of the direct problem; in such methods, it is the equation which gets approximated (in fact: discretized).\n\nAlong these lines, Factas advocates the use of steps 1 and 2 above, along with some singularity analysis, to approach issues of nondestructive control in 2-D and 3-D , . The team is currently engaged in the generalization to inverse source problems for the Laplace equation in 3-D, to be described further in Section . There, holomorphic functions are replaced by harmonic gradients; applications are to inverse source problems in neurosciences (in particular in EEG/MEG) and inverse magnetization problems in geosciences, see Section .\n\nThe approximation-theoretic tools developed by Apics and now by Factas to handle issues mentioned so far are outlined in Section . In Section to come, we describe in more detail which problems are considered and which applications are targeted.\n\nNote that the Inria project-team Apics reached the end of its life cycle by the end of 2017. The proposal for our new team Factas was processed by the CEP (Comité des Équipes-Projets) of the Research Center in 2018, and approved by the head of the Institute in 2019.\n\nRange of inverse problems Elliptic partial differential equations (PDE) Paul Asensio Laurent Baratchart Sylvain Chevillard Juliette Leblond Masimba Nemaire Konstantinos Mavreas\n\nBy standard properties of conjugate differentials, reconstructing Dirichlet-Neumann boundary conditions for a function harmonic in a plane domain, when these conditions are already known on a subset $E$ of the boundary, is equivalent to recover a holomorphic function in the domain from its boundary values on $E$. This is the problem raised on the half-plane in step 1 of Section . It makes good sense in holomorphic Hardy spaces where functions are entirely determined by their values on boundary subsets of positive linear measure, which is the framework for Problem $\\left(P\\right)$ that we set up in Section . Such issues naturally arise in nondestructive testing of 2-D (or 3-D cylindrical) materials from partial electrical measurements on the boundary. For instance, the ratio between the tangential and the normal currents (the so-called Robin coefficient) tells one about corrosion of the material. Thus, solving Problem $\\left(P\\right)$ where $\\psi$ is chosen to be the response of some uncorroded piece with identical shape yields non destructive testing of a potentially corroded piece of material, part of which is inaccessible to measurements. This was an initial application of holomorphic extremal problems to non-destructive control , .\n\nAnother application by the team deals with non-constant conductivity over a doubly connected domain, the set $E$ being now the outer boundary. Measuring Dirichlet-Neumann data on $E$, one wants to recover level lines of the solution to a conductivity equation, which is a so-called free boundary inverse problem. For this, given a closed curve inside the domain, we first quantify how constant the solution on this curve. To this effect, we state and solve an analog of Problem $\\left(P\\right)$, where the constraint bears on the real part of the function on the curve (it should be close to a constant there), in a Hardy space of a conjugate Beltrami equation, of which the considered conductivity equation is the compatibility condition (just like the Laplace equation is the compatibility condition of the Cauchy-Riemann system). Subsequently, a descent algorithm on the curve leads one to improve the initial guess. For example, when the domain is regarded as separating the edge of a tokamak's vessel from the plasma (rotational symmetry makes this a 2-D situation), this method can be used to estimate the shape of a plasma subject to magnetic confinement.\n\nThis was actually carried out in collaboration with CEA (French nuclear agency) and the University of Nice (JAD Lab.), to data from Tore Supra in . The procedure is fast because no numerical integration of the underlying PDE is needed, as an explicit basis of solutions to the conjugate Beltrami equation in terms of Bessel functions was found in this case. Generalizing this approach in a more systematic manner to free boundary problems of Bernoulli type, using descent algorithms based on shape-gradient for such approximation-theoretic criteria, is an interesting prospect to the team.\n\nThe piece of work we just mentioned requires defining and studying Hardy spaces of conjugate Beltrami equations, which is an interesting topic. For Sobolev-smooth coefficients of exponent greater than 2, they were investigated in . The case of the critical exponent 2 is treated in , which apparently provides the first example of well-posed Dirichlet problem in the non-strictly elliptic case: the conductivity may be unbounded or zero on sets of zero capacity and, accordingly, solutions need not be locally bounded. More importantly perhaps, the exponent 2 is also the key to a corresponding theory on very general (still rectifiable) domains in the plane, as coefficients of pseudo-holomorphic functions obtained by conformal transformation onto a disk are merely of ${L}^{2}$-class in general, even if the initial problem deals with coefficients of ${L}^{r}$-class for some $r>2$. Such generalizations are now under study within the team.\n\nGeneralized Hardy classes as above are used in where we address the uniqueness issue in the classical Robin inverse problem on a Lipschitz domain of $\\Omega \\subset {ℝ}^{n}$, $n\\ge 2$, with uniformly bounded Robin coefficient, ${L}^{2}$ Neumann data and conductivity of Sobolev class ${W}^{1,r}\\left(\\Omega \\right)$, $r>n$. We show that uniqueness of the Robin coefficient on a subset of the boundary, given Cauchy data on the complementary part, does hold in dimension $n=2$, thanks to a unique continuation result, but needs not hold in higher dimension. In higher dimension, this raises an open issue on harmonic gradients, namely whether the positivity of the Robin coefficient is compatible with identical vanishing of the boundary gradient on a subset of positive measure.\n\nThe 3-D version of step 1 in Section is another subject investigated by Factas: to recover a harmonic function (up to an additive constant) in a ball or a half-space from partial knowledge of its gradient. This prototypical inverse problem (i.e. inverse to the Cauchy problem for the Laplace equation) often recurs in electromagnetism. At present, Factas is involved with solving instances of this inverse problem arising in two fields, namely medical imaging e.g. for electroencephalography (EEG) or magneto-encephalography (MEG), and paleomagnetism (recovery of rocks magnetization) , see Section . In this connection, we collaborate with two groups of partners: Athena Inria project-team and INS (Institut de Neurosciences des Systèmes, http://ins.univ-amu.fr/), hospital la Timone, Aix-Marseille Univ., on the one hand, Geosciences Lab. at MIT and Cerege CNRS Lab. on the other hand. The question is considerably more difficult than its 2-D counterpart, due mainly to the lack of multiplicative structure for harmonic gradients. Still, substantial progress has been made over the last years using methods of harmonic analysis and operator theory.\n\nThe team is further concerned with 3-D generalizations and applications to non-destructive control of step 2 in Section . A typical problem is here to localize inhomogeneities or defaults such as cracks, sources or occlusions in a planar or 3-dimensional object, knowing thermal, electrical, or magnetic measurements on the boundary. These defaults can be expressed as a lack of harmonicity of the solution to the associated Dirichlet-Neumann problem, thereby posing an inverse potential problem in order to recover them. In 2-D, finding an optimal discretization of the potential in Sobolev norm amounts to solve a best rational approximation problem, and the question arises as to how the location of the singularities of the approximant (i.e. its poles) reflects the location of the singularities of the potential (i.e. the defaults we seek). This is a fairly deep issue in approximation theory, to which Apics contributed convergence results for certain classes of fields expressed as Cauchy integrals over extremal contours for the logarithmic potential , . Initial schemes to locate cracks or sources via rational approximation on planar domains were obtained this way , , . It is remarkable that finite inverse source problems in 3-D balls, or more general algebraic surfaces, can be approached using these 2-D techniques upon slicing the domain into planar sections . More precisely, each section cuts out a planar domain, the boundary of which carries data which can be proved to match an algebraic function. The singularities of this algebraic function are not located at the 3-D sources, but are related to them: the section contains a source if and only if some function of the singularities in that section meets a relative extremum. Using bisection it is thus possible to determine an extremal place along all sections parallel to a given plane direction, up to some threshold which has to be chosen small enough that one does not miss a source. This way, we reduce the original source problem in 3-D to a sequence of inverse poles and branchpoints problems in 2-D. This bottom line generates a steady research activity within Factas, and again applications are sought to medical imaging and geosciences, see Sections and .\n\nConjectures may be raised on the behavior of optimal potential discretization in 3-D, but answering them is an ambitious program still in its infancy.\n\nSystems, transfer and scattering Laurent Baratchart Sylvain Chevillard Adam Cooman Martine Olivi Fabien Seyfert\n\nThrough contacts with CNES (French space agency), members of the team became involved in identification and tuning of microwave electromagnetic filters used in space telecommunications, see Section . The initial problem was to recover, from band-limited frequency measurements, physical parameters of the device under examination. The latter consists of interconnected dual-mode resonant cavities with negligible loss, hence its scattering matrix is modeled by a $2×2$ unitary-valued matrix function on the frequency line, say the imaginary axis to fix ideas. In the bandwidth around the resonant frequency, a modal approximation of the Helmholtz equation in the cavities shows that this matrix is approximately rational, of Mc-Millan degree twice the number of cavities.\n\nThis is where system theory comes into play, through the so-called realization process mapping a rational transfer function in the frequency domain to a state-space representation of the underlying system of linear differential equations in the time domain. Specifically, realizing the scattering matrix allows one to construct a virtual electrical network, equivalent to the filter, the parameters of which mediate in between the frequency response and the geometric characteristics of the cavities (i.e. the tuning parameters).\n\nHardy spaces provide a framework to transform this ill-posed issue into a series of regularized analytic and meromorphic approximation problems. More precisely, the procedure sketched in Section goes as follows:\n\n• infer from the pointwise boundary data in the bandwidth a stable transfer function (i.e. one which is holomorphic in the right half-plane), that may be infinite dimensional (numerically: of high degree). This is done by solving a problem analogous to $\\left(P\\right)$ in Section , while taking into account prior knowledge on the decay of the response outside the bandwidth, see for details.\n\n• A stable rational approximation of appropriate degree to the model obtained in the previous step is performed. For this, a descent method on the compact manifold of inner matrices of given size and degree is used, based on an original parametrization of stable transfer functions developed within the team .\n\n• Realizations of this rational approximant are computed. To be useful, they must satisfy certain constraints imposed by the geometry of the device. These constraints typically come from the coupling topology of the equivalent electrical network used to model the filter. This network is composed of resonators, coupled according to some specific graph. This realization step can be recast, under appropriate compatibility conditions , as solving a zero-dimensional multivariate polynomial system. To tackle this problem in practice, we use Gröbner basis techniques and continuation methods which team up in the Dedale-HF software (see Section ).\n\n• We recently started a collaboration with the Chinese Hong Kong University on the topic of frequency depending couplings appearing in the equivalent circuits we compute continuing our work on wide-band design applications.\n\nFactas also investigates issues pertaining to design rather than identification. Given the topology of the filter, a basic problem in this connection is to find the optimal response subject to specifications that bear on rejection, transmission and group delay of the scattering parameters. Generalizing the classical approach based on Chebyshev polynomials for single band filters, we recast the problem of multi-band response synthesis as a generalization of the classical Zolotarev min-max problem for rational functions . Thanks to quasi-convexity, the latter can be solved efficiently using iterative methods relying on linear programming. These were implemented in the software easy-FF (see easy-FF). Currently, the team is engaged in the synthesis of more complex microwave devices like multiplexers and routers, which connect several filters through wave guides. Schur analysis plays an important role here, because scattering matrices of passive systems are of Schur type (i.e. contractive in the stability region). The theory originates with the work of I. Schur , who devised a recursive test to check for contractivity of a holomorphic function in the disk. The so-called Schur parameters of a function may be viewed as Taylor coefficients for the hyperbolic metric of the disk, and the fact that Schur functions are contractions for that metric lies at the root of Schur's test. Generalizations thereof turn out to be efficient to parametrize solutions to contractive interpolation problems . Dwelling on this, Factas contributed differential parametrizations (atlases of charts) of lossless matrix functions , , which are fundamental to our rational approximation software RARL2 (see Section ). Schur analysis is also instrumental to approach de-embedding issues, and provides one with considerable insight into the so-called matching problem. The latter consists in maximizing the power a multiport can pass to a given load, and for reasons of efficiency it is all-pervasive in microwave and electric network design, e.g. of antennas, multiplexers, wifi cards and more. It can be viewed as a rational approximation problem in the hyperbolic metric, and the team presently deals with this hot topic using contractive interpolation with constraints on boundary peak points, within the framework of the (defense funded) ANR Cocoram, see Sections .\n\nIn recent years, our attention was driven by CNES and UPV (Bilbao) to questions about stability of high-frequency amplifiers. Contrary to previously discussed devices, these are active components. The response of an amplifier can be linearized around a set of primary current and voltages, and then admittances of the corresponding electrical network can be computed at various frequencies, using the so-called harmonic balance method. The initial goal is to check for stability of the linearized model, so as to ascertain existence of a well-defined working state. The network is composed of lumped electrical elements namely inductors, capacitors, negative and positive resistors, transmission lines, and controlled current sources. Our research so far has focused on describing the algebraic structure of admittance functions, so as to set up a function-theoretic framework where the two-steps approach outlined in Section can be put to work. The main discovery is that the unstable part of each partial transfer function is rational and can be computed by analytic projection, see Section . We now start investigating the linearized harmonic transfer-function around a periodic cycle, to check for stability under non necessarily small inputs. This topic generates the doctoral work of S. Fueyo.\n\nApproximation Laurent Baratchart Sylvain Chevillard Juliette Leblond Martine Olivi Fabien Seyfert Best analytic approximation\n\nIn dimension 2, the prototypical problem to be solved in step 1 of Section may be described as: given a domain $D\\subset {ℝ}^{2}$, to recover a holomorphic function from its values on a subset $K$ of the boundary of $D$. For the discussion it is convenient to normalize $D$, which can be done by conformal mapping. So, in the simply connected case, we fix $D$ to be the unit disk with boundary unit circle $T$. We denote by ${H}^{p}$ the Hardy space of exponent $p$, which is the closure of polynomials in ${L}^{p}\\left(T\\right)$-norm if $1\\le p<\\infty$ and the space of bounded holomorphic functions in $D$ if $p=\\infty$. Functions in ${H}^{p}$ have well-defined boundary values in ${L}^{p}\\left(T\\right)$, which makes it possible to speak of (traces of) analytic functions on the boundary.\n\nTo find an analytic function $g$ in $D$ matching some measured values $f$ approximately on a sub-arc $K$ of $T$, we formulate a constrained best approximation problem as follows.\n\n$\\left(P\\right)$ Let $1\\le p\\le \\infty$, $K$ a sub-arc of $T$, $f\\in {L}^{p}\\left(K\\right)$, $\\psi \\in {L}^{p}\\left(T\\setminus K\\right)$ and $M>0$; find a function $g\\in {H}^{p}$ such that ${\\parallel g-\\psi \\parallel }_{{L}^{p}\\left(T\\setminus K\\right)}\\le M$ and $g-f$ is of minimal norm in ${L}^{p}\\left(K\\right)$ under this constraint.\n\nHere $\\psi$ is a reference behavior capturing a priori assumptions on the behavior of the model off $K$, while $M$ is some admissible deviation thereof. The value of $p$ reflects the type of stability which is sought and how much one wants to smooth out the data. The choice of ${L}^{p}$ classes is suited to handle pointwise measurements.\n\nTo fix terminology, we refer to $\\left(P\\right)$ as a bounded extremal problem. As shown in , , the solution to this convex infinite-dimensional optimization problem can be obtained when $p\\ne 1$ upon iterating with respect to a Lagrange parameter the solution to spectral equations for appropriate Hankel and Toeplitz operators. These spectral equations involve the solution to the special case $K=T$ of $\\left(P\\right)$, which is a standard extremal problem :\n\n(${P}_{0}$) Let $1\\le p\\le \\infty$ and $\\varphi \\in {L}^{p}\\left(T\\right)$; find a function $g\\in {H}^{p}$ such that $g-\\varphi$ is of minimal norm in ${L}^{p}\\left(T\\right)$.\n\nIn the case $p=1$, partial results are known but computational issues remain open.\n\nVarious modifications of $\\left(P\\right)$ can be tailored to meet specific needs. For instance when dealing with lossless transfer functions (see Section ), one may want to express the constraint on $T\\setminus K$ in a pointwise manner: $|g-\\psi |\\le M$ a.e. on $T\\setminus K$, see . In this form, the problem comes close to (but still is different from) ${H}^{\\infty }$ frequency optimization used in control , . One can also impose bounds on the real or imaginary part of $g-\\psi$ on $T\\setminus K$, which is useful when considering Dirichlet-Neumann problems.\n\nThe analog of Problem $\\left(P\\right)$ on an annulus, $K$ being now the outer boundary, can be seen as a means to regularize a classical inverse problem occurring in nondestructive control, namely to recover a harmonic function on the inner boundary from Dirichlet-Neumann data on the outer boundary (see Sections ). It may serve as a tool to approach Bernoulli type problems, where we are given data on the outer boundary and we seek the inner boundary, knowing it is a level curve of the solution. In this case, the Lagrange parameter indicates how to deform the inner contour in order to improve data fitting. Similar topics are discussed in Section for more general equations than the Laplacian, namely isotropic conductivity equations of the form $\\mathrm{div}\\left(\\sigma \\nabla u\\right)=0$ where $\\sigma$ is no longer constant (i.e., varies in the space). Then, the Hardy spaces in Problem $\\left(P\\right)$ are those of a so-called conjugate Beltrami equation: $\\overline{\\partial }f=\\nu \\overline{\\partial f}$ , which are studied for $1 in and . Expansions of solutions needed to constructively handle such issues in the specific case of linear fractional conductivities (occurring for instance in plasma shaping) have been expounded in .\n\nThough originally considered in dimension 2, Problem $\\left(P\\right)$ carries over naturally to higher dimensions where analytic functions get replaced by gradients of harmonic functions. Namely, given some open set $\\Omega \\subset {ℝ}^{n}$ and some ${ℝ}^{n}$-valued vector field $V$ on an open subset $O$ of the boundary of $\\Omega$, we seek a harmonic function in $\\Omega$ whose gradient is close to $V$ on $O$.\n\nWhen $\\Omega$ is a ball or a half-space, a substitute for holomorphic Hardy spaces is provided by the Stein-Weiss Hardy spaces of harmonic gradients . Conformal maps are no longer available when $n>2$, so that $\\Omega$ can no longer be normalized. More general geometries than spheres and half-spaces have not been much studied so far.\n\nOn the ball, the analog of Problem $\\left(P\\right)$ is\n\n$\\left({P}_{1}\\right)$ Let $1\\le p\\le \\infty$ and $B\\subset {ℝ}^{n}$ the unit ball. Fix $O$ an open subset of the unit sphere $S\\subset {ℝ}^{n}$. Let further $V\\in {L}^{p}\\left(O\\right)$ and $W\\in {L}^{p}\\left(S\\setminus O\\right)$ be ${ℝ}^{n}$-valued vector fields. Given $M>0$, find a harmonic gradient $G\\in {H}^{p}\\left(B\\right)$ such that ${\\parallel G-W\\parallel }_{{L}^{p}\\left(S\\setminus O\\right)}\\le M$ and $G-V$ is of minimal norm in ${L}^{p}\\left(O\\right)$ under this constraint.\n\nWhen $p=2$, Problem $\\left({P}_{1}\\right)$ was solved in as well as its analog on a shell, when the tangent component of $V$ is a gradient (when $O$ is Lipschitz the general case follows easily from this). The solution extends the work in to the 3-D case, using a generalization of Toeplitz operators. The case of the shell was motivated by applications to the processing of EEG data. An important ingredient is a refinement of the Hodge decomposition, that we call the Hardy-Hodge decomposition, allowing us to express a ${ℝ}^{n}$-valued vector field in ${L}^{p}\\left(S\\right)$, $1, as the sum of a vector field in ${H}^{p}\\left(B\\right)$, a vector field in ${H}^{p}\\left({ℝ}^{n}\\setminus \\overline{B}\\right)$, and a tangential divergence free vector field on $S$; the space of such divergence-free fields is denoted by $D\\left(S\\right)$. If $p=1$ or $p=\\infty$, ${L}^{p}$ must be replaced by the real Hardy space or the space of functions with bounded mean oscillation. More generally this decomposition, which is valid on any sufficiently smooth surface (see Section ), seems to play a fundamental role in inverse potential problems. In fact, it was first introduced formally on the plane to describe silent magnetizations supported in ${ℝ}^{2}$ (i.e. those generating no field in the upper half space) .\n\nJust like solving problem $\\left(P\\right)$ appeals to the solution of problem $\\left({P}_{0}\\right)$, our ability to solve problem $\\left({P}_{1}\\right)$ will depend on the possibility to tackle the special case where $O=S$:\n\n$\\left({P}_{2}\\right)$ Let $1\\le p\\le \\infty$ and $V\\in {L}^{p}\\left(S\\right)$ be a ${ℝ}^{n}$-valued vector field. Find a harmonic gradient $G\\in {H}^{p}\\left(B\\right)$ such that ${\\parallel G-V\\parallel }_{{L}^{p}\\left(S\\right)}$ is minimum.\n\nProblem $\\left({P}_{2}\\right)$ is simple when $p=2$ by virtue of the Hardy-Hodge decomposition together with orthogonality of ${H}^{2}\\left(B\\right)$ and ${H}^{2}\\left({ℝ}^{n}\\setminus \\overline{B}\\right)$, which is the reason why we were able to solve $\\left({P}_{1}\\right)$ in this case. Other values of $p$ cannot be treated as easily and are still under investigation, especially the case $p=\\infty$ which is of particular interest and presents itself as a 3-D analog to the Nehari problem .\n\nCompanion to problem $\\left({P}_{2}\\right)$ is problem $\\left({P}_{3}\\right)$ below.\n\n$\\left({P}_{3}\\right)$ Let $1\\le p\\le \\infty$ and $V\\in {L}^{p}\\left(S\\right)$ be a ${ℝ}^{n}$-valued vector field. Find $G\\in {H}^{p}\\left(B\\right)$ and $D\\in D\\left(S\\right)$ such that ${\\parallel G+D-V\\parallel }_{{L}^{p}\\left(S\\right)}$ is minimum.\n\nNote that $\\left({P}_{2}\\right)$ and $\\left({P}_{3}\\right)$ are identical in 2-D, since no non-constant tangential divergence-free vector field exists on $T$. It is no longer so in higher dimension, where both $\\left({P}_{2}\\right)$ and $\\left({P}_{3}\\right)$ arise in connection with inverse potential problems in divergence form, like source recovery in electro/magneto encephalography and paleomagnetism, see Sections and .\n\nBest meromorphic and rational approximation\n\nThe techniques set forth in this section are used to solve step 2 in Section and they are instrumental to approach inverse boundary value problems for the Poisson equation $\\Delta u=\\mu$, where $\\mu$ is some (unknown) measure.\n\nScalar meromorphic and rational approximation\n\nWe put ${R}_{N}$ for the set of rational functions with at most $N$ poles in $D$. By definition, meromorphic functions in ${L}^{p}\\left(T\\right)$ are (traces of) functions in ${H}^{p}+{R}_{N}$.\n\nA natural generalization of problem $\\left({P}_{0}\\right)$ is:\n\n(${P}_{N}$) Let $1\\le p\\le \\infty$, $N\\ge 0$ an integer, and $f\\in {L}^{p}\\left(T\\right)$; find a function ${g}_{N}\\in {H}^{p}+{R}_{N}$ such that ${g}_{N}-f$ is of minimal norm in ${L}^{p}\\left(T\\right)$.\n\nOnly for $p=\\infty$ and $f$ continuous is it known how to solve $\\left({P}_{N}\\right)$ in semi-closed form. The unique solution is given by AAK theory (named after Adamjan, Arov and Krein), which connects the spectral decomposition of Hankel operators with best approximation .\n\nThe case where $p=2$ is of special importance for it reduces to rational approximation. Indeed, if we write the Hardy decomposition $f={f}^{+}+{f}^{-}$ where ${f}^{+}\\in {H}^{2}$ and ${f}^{-}\\in {H}^{2}\\left(ℂ\\setminus \\overline{D}\\right)$, then ${g}_{N}={f}^{+}+{r}_{N}$ where ${r}_{N}$ is a best approximant to ${f}^{-}$ from ${R}_{N}$ in ${L}^{2}\\left(T\\right)$. Moreover, ${r}_{N}$ has no pole outside $D$, hence it is a stable rational approximant to ${f}^{-}$. However, in contrast to the case where $p=\\infty$, this best approximant may not be unique.\n\nThe Miaou project (predecessor of Apics) already designed a dedicated steepest-descent algorithm for the case $p=2$ whose convergence to a local minimum is guaranteed; the algorithm ha evolved over years and still now, it seems to be the only procedure meeting this property. This gradient algorithm proceeds recursively with respect to $N$ on a compactification of the parameter space . Although it has proved to be effective in all applications carried out so far (see Sections ), it is still unknown whether the absolute minimum can always be obtained by choosing initial conditions corresponding to critical points of lower degree (as is done by the RARL2 software, Section ).\n\nIn order to establish global convergence results, Apics has undertaken a deeper study of the number and nature of critical points (local minima, saddle points, ...), in which tools from differential topology and operator theory team up with classical interpolation theory , . Based on this work, uniqueness or asymptotic uniqueness of the approximant was proved for certain classes of functions like transfer functions of relaxation systems (i.e. Markov functions) and more generally Cauchy integrals over hyperbolic geodesic arcs . These are the only results of this kind. Research by Apics on this topic remained dormant for a while by reasons of opportunity, but revisiting the work in higher dimension is a worthy and timely endeavor today. Meanwhile, an analog to AAK theory was carried out for $2\\le p<\\infty$ in . Although not as effective computationally, it was recently used to derive lower bounds . When $1\\le p<2$, problem $\\left({P}_{N}\\right)$ is still quite open.\n\nA common feature to the above-mentioned problems is that critical point equations yield non-Hermitian orthogonality relations for the denominator of the approximant. This stresses connections with interpolation, which is a standard way to build approximants, and in many respects best or near-best rational approximation may be regarded as a clever manner to pick interpolation points. This was exploited in , , and is used in an essential manner to assess the behavior of poles of best approximants to functions with branched singularities, which is of particular interest for inverse source problems (cf. Sections and ).\n\nIn higher dimensions, the analog of Problem $\\left({P}_{N}\\right)$ is best approximation of a vector field by gradients of discrete potentials generated by $N$ point masses. This basic issue is by no means fully understood, and it is an exciting field of research. It is connected with certain generalizations of Toeplitz or Hankel operators, and with constructive approaches to so-called weak factorizations for real Hardy functions .\n\nBesides, certain constrained rational approximation problems, of special interest in identification and design of passive systems, arise when putting additional requirements on the approximant, for instance that it should be smaller than 1 in modulus (i.e. a Schur function). In particular, Schur interpolation lately received renewed attention from the team, in connection with matching problems. There, interpolation data are subject to a well-known compatibility condition (positive definiteness of the so-called Pick matrix), and the main difficulty is to put interpolation points on the boundary of $D$ while controlling both the degree and the extremal points (peak points for the modulus) of the interpolant. Results obtained by Apics in this direction generalize a variant of contractive interpolation with degree constraint as studied in . We mention that contractive interpolation with nodes approaching the boundary has been a subsidiary research topic by the team in the past, which plays an interesting role in the spectral representation of certain non-stationary stochastic processes , .\n\nMatrix-valued rational approximation\n\nMatrix-valued approximation is necessary to handle systems with several inputs and outputs but it generates additional difficulties as compared to scalar-valued approximation, both theoretically and algorithmically. In the matrix case, the McMillan degree (i.e. the degree of a minimal realization in the System-Theoretic sense) generalizes the usual notion of degree for rational functions. For instance when poles are simple, the McMillan degree is the sum of the ranks of the residues.\n\nThe basic problem that we consider now goes as follows: let $ℱ\\in {\\left({H}^{2}\\right)}^{m×l}$ and $n$ an integer; find a rational matrix of size $m×l$ without poles in the unit disk and of McMillan degree at most $n$ which is nearest possible to $ℱ$ in ${\\left({H}^{2}\\right)}^{m×l}$. Here the ${L}^{2}$ norm of a matrix is the square root of the sum of the squares of the norms of its entries.\n\nThe scalar approximation algorithm derived in and mentioned in Section generalizes to the matrix-valued situation . The first difficulty here is to parametrize inner matrices (i.e. matrix-valued functions analytic in the unit disk and unitary on the unit circle) of given McMillan degree degree $n$. Indeed, inner matrices play the role of denominators in fractional representations of transfer matrices (using the so-called Douglas-Shapiro-Shields factorization). The set of inner matrices of given degree is a smooth manifold that allows one to use differential tools as in the scalar case. In practice, one has to produce an atlas of charts (local parametrizations) and to handle changes of charts in the course of the algorithm. Such parametrization can be obtained using interpolation theory and Schur-type algorithms, the parameters of which are vectors or matrices ( , , ). Some of these parametrizations are also interesting to compute realizations and achieve filter synthesis ( , ). The rational approximation software “RARL2” developed by the team is described in Section .\n\nDifficulties relative to multiple local minima of course arise in the matrix-valued case as well, and deriving criteria that guarantee uniqueness is even more difficult than in the scalar case. The case of rational functions of degree $n$ or small perturbations thereof (the consistency problem) was solved in . Matrix-valued Markov functions are the only known example beyond this one .\n\nLet us stress that RARL2 seems the only algorithm handling rational approximation in the matrix case that demonstrably converges to a local minimum while meeting stability constraints on the approximant. It is still a working pin of many developments by Factas on frequency optimization and design.\n\nBehavior of poles of meromorphic approximants Laurent Baratchart\n\nWe refer here to the behavior of poles of best meromorphic approximants, in the ${L}^{p}$-sense on a closed curve, to functions $f$ defined as Cauchy integrals of complex measures whose support lies inside the curve. Normalizing the contour to be the unit circle $T$, we are back to Problem $\\left({P}_{N}\\right)$ in Section ; invariance of the latter under conformal mapping was established in . Research so far has focused on functions whose singular set inside the contour is polar, meaning that the function can be continued analytically (possibly in a multiple-valued manner) except over a set of logarithmic capacity zero.\n\nGenerally speaking in approximation theory, assessing the behavior of poles of rational approximants is essential to obtain error rates as the degree goes large, and to tackle constructive issues like uniqueness. However, as explained in Section , the original twist by Apics, now Factas, is to consider this issue also as a means to extract information on singularities of the solution to a Dirichlet-Neumann problem. The general theme is thus: how do the singularities of the approximant reflect those of the approximated function? This approach to inverse problem for the 2-D Laplacian turns out to be attractive when singularities are zero- or one-dimensional (see Section ). It can be used as a computationally cheap initial condition for more precise but much heavier numerical optimizations which often do not even converge unless properly initialized. As regards crack detection or source recovery, this approach boils down to analyzing the behavior of best meromorphic approximants of given pole cardinality to a function with branch points, which is the prototype of a polar singular set. For piecewise analytic cracks, or in the case of sources, we were able to prove (, ), that the poles of the approximants accumulate, when the degree goes large, to some extremal cut of minimum weighted logarithmic capacity connecting the singular points of the crack, or the sources . Moreover, the asymptotic density of the poles turns out to be the Green equilibrium distribution on this cut in $D$, therefore it charges the singular points if one is able to approximate in sufficiently high degree (this is where the method could fail, because high-order approximation requires rather precise data).\n\nThe case of two-dimensional singularities is still an outstanding open problem.\n\nIt is remarkable that inverse source problems inside a sphere or an ellipsoid in 3-D can be approached with such 2-D techniques, as applied to planar sections, see Section . The technique is implemented in the software FindSources3D, see Section .\n\nSoftware tools of the team\n\nIn addition to the above-mentioned research activities, Factas develops and maintains a number of long-term software tools that either implement and illustrate effectiveness of the algorithms theoretically developed by the team or serve as tools to help further research by team members. We present briefly the most important of them.\n\nPisa\n\nKeywords: Electrical circuit - Stability\n\nFunctional Description: To minimise prototyping costs, the design of analog circuits is performed using computer-aided design tools which simulate the circuit's response as accurately as possible.\n\nSome commonly used simulation tools do not impose stability, which can result in costly errors when the prototype turns out to be unstable. A thorough stability analysis is therefore a very important step in circuit design. This is where pisa is used.\n\npisa is a Matlab toolbox that allows designers of analog electronic circuits to determine the stability of their circuits in the simulator. It analyses the impedance presented by a circuit to determine the circuit's stability. When an instability is detected, pisa can estimate location of the unstable poles to help designers fix their stability issue.\n\nRelease Functional Description: First version\n\n• Authors: Adam Cooman, David Martinez Martinez, Fabien Seyfert and Martine Olivi\n\n• Contact: Fabien Seyfert\n\n• Publications: Model-Free Closed-Loop Stability Analysis: A Linear Functional Approach - On Transfer Functions Realizable with Active Electronic Components\n\n• URL: https://project.inria.fr/pisa\n\n• DEDALE-HF\n\nScientific Description\n\nDedale-HF consists in two parts: a database of coupling topologies as well as a dedicated predictor-corrector code. Roughly speaking each reference file of the database contains, for a given coupling topology, the complete solution to the coupling matrix synthesis problem (C.M. problem for short) associated to particular filtering characteristics. The latter is then used as a starting point for a predictor-corrector integration method that computes the solution to the C.M. corresponding to the user-specified filter characteristics. The reference files are computed off-line using Gröbner basis techniques or numerical techniques based on the exploration of a monodromy group. The use of such continuation techniques, combined with an efficient implementation of the integrator, drastically reduces the computational time.\n\nDedale-HF has been licensed to, and is currently used by TAS-Espana\n\nFunctional Description\n\nDedale-HF is a software dedicated to solve exhaustively the coupling matrix synthesis problem in reasonable time for the filtering community. Given a coupling topology, the coupling matrix synthesis problem consists in finding all possible electromagnetic coupling values between resonators that yield a realization of given filter characteristics. Solving the latter is crucial during the design step of a filter in order to derive its physical dimensions, as well as during the tuning process where coupling values need to be extracted from frequency measurements.\n\n• Participant: Fabien Seyfert\n\n• Contact: Fabien Seyfert\n\n• URL: http://www-sop.inria.fr/apics/Dedale/\n\n• FindSources3D\n\nKeywords: Health - Neuroimaging - Visualization - Compilers - Medical - Image - Processing\n\nFindSources3D is a software program dedicated to the resolution of inverse source problems in electroencephalography (EEG). From pointwise measurements of the electrical potential taken by electrodes on the scalp, FindSources3D estimates pointwise dipolar current sources within the brain in a spherical model.\n\nAfter a first data transmission “cortical mapping” step, it makes use of best rational approximation on 2-D planar cross-sections and of the software RARL2 in order to locate singularities. From those planar singularities, the 3-D sources are estimated in a last step, see .\n\nThe present version of FindSources3D (called FindSources3D-bolis) provides a modular, ergonomic, accessible and interactive platform, with a convenient graphical interface for EEG medical imaging. Modularity is now granted (using the tools dtk, Qt, with compiled Matlab libraries). It offers a detailed and nice visualization of data and tuning parameters, processing steps, and of the computed results (using VTK).\n\nA new version is being developed that will incorporate a first Singular Value Decomposition (SVD) step in order to be able to handle time dependent data and to find the corresponding principal static components.\n\n• Participants: Juliette Leblond, Maureen Clerc (team Athena, Inria Sophia), Jean-Paul Marmorat, Théodore Papadopoulo (team Athena).\n\n• Contact: Juliette Leblond\n\n• URL: http://www-sop.inria.fr/apics/FindSources3D/en/index.html\n\n• PRESTO-HF\n\nScientific Description\n\nFor the matrix-valued rational approximation step, Presto-HF relies on RARL2. Constrained realizations are computed using the Dedale-HF software. As a toolbox, Presto-HF has a modular structure, which allows one for example to include some building blocks in an already existing software.\n\nThe delay compensation algorithm is based on the following assumption: far off the pass-band, one can reasonably expect a good approximation of the rational components of S11 and S22 by the first few terms of their Taylor expansion at infinity, a small degree polynomial in 1/s. Using this idea, a sequence of quadratic convex optimization problems are solved, in order to obtain appropriate compensations. In order to check the previous assumption, one has to measure the filter on a larger band, typically three times the pass band.\n\nThis toolbox has been licensed to (and is currently used by) Thales Alenia Space in Toulouse and Madrid, Thales airborne systems and Flextronics (two licenses). Xlim (University of Limoges) is a heavy user of Presto-HF among the academic filtering community and some free license agreements have been granted to the microwave department of the University of Erlangen (Germany) and the Royal Military College (Kingston, Canada).\n\nFunctional Description\n\nPresto-HF is a toolbox dedicated to low-pass parameter identification for microwave filters. In order to allow the industrial transfer of our methods, a Matlab-based toolbox has been developed, dedicated to the problem of identification of low-pass microwave filter parameters. It allows one to run the following algorithmic steps, either individually or in a single stroke:\n\n• Determination of delay components caused by the access devices (automatic reference plane adjustment),\n\n• Automatic determination of an analytic completion, bounded in modulus for each channel,\n\n• Rational approximation of fixed McMillan degree,\n\n• Determination of a constrained realization.\n\n• Participants: Fabien Seyfert, Jean-Paul Marmorat and Martine Olivi\n\n• Contact: Fabien Seyfert\n\n• URL: https://project.inria.fr/presto-hf/\n\n• RARL2\n\nRéalisation interne et Approximation Rationnelle L2\n\nScientific Description\n\nThe method is a steepest-descent algorithm. A parametrization of MIMO systems is used, which ensures that the stability constraint on the approximant is met. The implementation, in Matlab, is based on state-space representations.\n\nRARL2 performs the rational approximation step in the software tools PRESTO-HF and FindSources3D. It is distributed under a particular license, allowing unlimited usage for academic research purposes. It was released to the universities of Delft and Maastricht (the Netherlands), Cork (Ireland), Brussels (Belgium), Macao (China) and BITS-Pilani Hyderabad Campus (India).\n\nFunctional Description\n\nRARL2 is a software for rational approximation. It computes a stable rational L2-approximation of specified order to a given L2-stable (L2 on the unit circle, analytic in the complement of the unit disk) matrix-valued function. This can be the transfer function of a multivariable discrete-time stable system. RARL2 takes as input either:\n\n• its internal realization,\n\n• its first N Fourier coefficients,\n\n• discretized (uniformly distributed) values on the circle. In this case, a least-square criterion is used instead of the L2 norm.\n\nIt thus performs model reduction in the first or the second case, and leans on frequency data identification in the third. For band-limited frequency data, it could be necessary to infer the behavior of the system outside the bandwidth before performing rational approximation.\n\nAn appropriate Möbius transformation allows to use the software for continuous-time systems as well.\n\n• Participants: Jean-Paul Marmorat and Martine Olivi\n\n• Contact: Martine Olivi\n\n• URL: http://www-sop.inria.fr/apics/RARL2/rarl2.html\n\n• Sollya\n\nKeywords: Numerical algorithm - Supremum norm - Curve plotting - Remez algorithm - Code generator - Proof synthesis\n\nFunctional Description\n\nSollya is an interactive tool where the developers of mathematical floating-point libraries (libm) can experiment before actually developing code. The environment is safe with respect to floating-point errors, i.e. the user precisely knows when rounding errors or approximation errors happen, and rigorous bounds are always provided for these errors.\n\nAmong other features, it offers a fast Remez algorithm for computing polynomial approximations of real functions and also an algorithm for finding good polynomial approximants with floating-point coefficients to any real function. As well, it provides algorithms for the certification of numerical codes, such as Taylor Models, interval arithmetic or certified supremum norms.\n\nIt is available as a free software under the CeCILL-C license.\n\n• Participants: Sylvain Chevillard, Christoph Lauter, Mioara Joldes and Nicolas Jourdan\n\n• Partners: CNRS - ENS Lyon - UCBL Lyon 1\n\n• Contact: Sylvain Chevillard\n\n• URL: http://sollya.gforge.inria.fr/\n\n• Application Domains Introduction\n\nApplication domains are naturally linked to the problems described in Sections and . By and large, they split into a systems-and-circuits part and an inverse-source-and-boundary-problems part, united under a common umbrella of function-theoretic techniques as described in Section .\n\nInverse magnetization problems Laurent Baratchart Sylvain Chevillard Juliette Leblond Konstantinos Mavreas\n\nGenerally speaking, inverse potential problems, similar to the one appearing in Section , occur naturally in connection with systems governed by Maxwell's equation in the quasi-static approximation regime. In particular, they arise in magnetic reconstruction issues. A specific application is to geophysics, which led us to form the Inria Associate Team Impinge (Inverse Magnetization Problems IN GEosciences) together with MIT and Vanderbilt University. Though this Associate Team reached the end of its term in 2018, the collaborations it has generated are still active. A joint work with Cerege (CNRS, Aix-en-Provence), in the framework of the ANR-project MagLune, completes this picture, see Sections .\n\nTo set up the context, recall that the Earth's geomagnetic field is generated by convection of the liquid metallic core (geodynamo) and that rocks become magnetized by the ambient field as they are formed or after subsequent alteration. Their remanent magnetization provides records of past variations of the geodynamo, which is used to study important processes in Earth sciences like motion of tectonic plates and geomagnetic reversals. Rocks from Mars, the Moon, and asteroids also contain remanent magnetization which indicates the past presence of core dynamos. Magnetization in meteorites may even record fields produced by the young sun and the protoplanetary disk which may have played a key role in solar system formation.\n\nFor a long time, paleomagnetic techniques were only capable of analyzing bulk samples and compute their net magnetic moment. The development of SQUID microscopes has recently extended the spatial resolution to sub-millimeter scales, raising new physical and algorithmic challenges. The associate team Impinge aims at tackling them, experimenting with the SQUID microscope set up in the Paleomagnetism Laboratory of the department of Earth, Atmospheric and Planetary Sciences at MIT. Typically, pieces of rock are sanded down to a thin slab, and the magnetization has to be recovered from the field measured on a planar region at small distance from the slab.\n\nMathematically speaking, both inverse source problems for EEG from Section and inverse magnetization problems described presently amount to recover the (3-D valued) quantity $m$ (primary current density in case of the brain or magnetization in case of a thin slab of rock) from measurements of the potential:\n\n$V\\left(x\\right)={\\int }_{\\Omega }\\frac{\\text{div}\\phantom{\\rule{0.166667em}{0ex}}m\\left({x}^{\\text{'}}\\right)\\phantom{\\rule{0.166667em}{0ex}}d{x}^{\\text{'}}}{|x-{x}^{\\text{'}}|}\\phantom{\\rule{0.166667em}{0ex}},$\n\noutside the volume $\\Omega$ of the object. Depending on the geometry of models, the magnetization distribution $m$ may lie in a volume or spread out on a surface. This results in quite different identifiability properties, see and Section , but the two situations share a substantial mathematical common core.\n\nAnother timely instance of inverse magnetization problems lies with geomagnetism. Satellites orbiting around the Earth measure the magnetic field at many points, and nowadays it is a challenge to extract global information from those measurements. In collaboration with C. Gerhards (Geomathematics and Geoinformatics Group, Technische Universität Bergakademie Freiberg, Germany), we started to work on the problem of separating the magnetic field due to the magnetization of the globe's crust from the magnetic field due to convection in the liquid metallic core. The techniques involved are variants, in a spherical context, from those developed within the Impinge associate team for paleomagnetism, see Section .\n\nInverse source problems in EEG Paul Asensio Laurent Baratchart Juliette Leblond Jean-Paul Marmorat Masimba Nemaire\n\nSolving overdetermined Cauchy problems for the Laplace equation on a spherical layer (in 3-D) in order to extrapolate incomplete data (see Section ) is a necessary ingredient of the team's approach to inverse source problems, in particular for applications to EEG, see . Indeed, the latter involves propagating the initial conditions through several layers of different conductivities, from the boundary shell down to the center of the domain where the singularities (i.e. the sources) lie. Once propagated to the innermost sphere, it turns out that traces of the boundary data on 2-D cross sections coincide with analytic functions with branched singularities in the slicing plane . The singularities are related to the actual location of the sources, namely their moduli reach in turn a maximum when the plane contains one of the sources. Hence we are back to the 2-D framework of Section , and recovering these singularities can be performed via best rational approximation. The goal is to produce a fast and sufficiently accurate initial guess on the number and location of the sources in order to run heavier descent algorithms on the direct problem, which are more precise but computationally costly and often fail to converge if not properly initialized. Our belief is that such a localization process can add a geometric, valuable piece of information to the standard temporal analysis of EEG signal records.\n\nNumerical experiments obtained with our software FindSources3D give very good results on simulated data and we are now engaged in the process of handling real experimental data, simultaneously recorded by EEG and MEG devices, in collaboration with our partners at INS, hospital la Timone, Marseille (see Section ).\n\nFurthermore, another approach is being studied for EEG, that consists in regularizing the inverse source problem by a total variation constraint on the source term (a measure), added to the quadratic data approximation criterion. It is similar to the path that is taken for inverse magnetization problems (see Sections and ), and it presently focuses on surface-distributed models.\n\nIdentification and design of microwave devices Laurent Baratchart Sylvain Chevillard Jean-Paul Marmorat Martine Olivi Fabien Seyfert\n\nThis is joint work with Stéphane Bila (Xlim, Limoges).\n\nOne of the best training grounds for function-theoretic applications by the team is the identification and design of physical systems whose performance is assessed frequency-wise. This is the case of electromagnetic resonant systems which are of common use in telecommunications.\n\nIn space telecommunications (satellite transmissions), constraints specific to on-board technology lead to the use of filters with resonant cavities in the microwave range. These filters serve multiplexing purposes (before or after amplification), and consist of a sequence of cylindrical hollow bodies, magnetically coupled by irises (orthogonal double slits). The electromagnetic wave that traverses the cavities satisfies the Maxwell equations, forcing the tangent electrical field along the body of the cavity to be zero. A deeper study of the Helmholtz equation states that an essentially discrete set of wave vectors is selected. In the considered range of frequency, the electrical field in each cavity can be decomposed along two orthogonal modes, perpendicular to the axis of the cavity (other modes are far off in the frequency domain, and their influence can be neglected).\n\nEach cavity (see Figure ) has three screws, horizontal, vertical and midway (horizontal and vertical are two arbitrary directions, the third direction makes an angle of 45 or 135 degrees, the easy case is when all cavities show the same orientation, and when the directions of the irises are the same, as well as the input and output slits). Since screws are conductors, they behave as capacitors; besides, the electrical field on the surface has to be zero, which modifies the boundary conditions of one of the two modes (for the other mode, the electrical field is zero hence it is not influenced by the screw), the third screw acts as a coupling between the two modes. The effect of an iris is opposite to that of a screw: no condition is imposed on a hole, which results in a coupling between two horizontal (or two vertical) modes of adjacent cavities (in fact the iris is the union of two rectangles, the important parameter being their width). The design of a filter consists in finding the size of each cavity, and the width of each iris. Subsequently, the filter can be constructed and tuned by adjusting the screws. Finally, the screws are glued once a satisfactory response has been obtained. In what follows, we shall consider a typical example, a filter designed by the CNES in Toulouse, with four cavities near 11 GHz.\n\nNear the resonance frequency, a good approximation to the Helmholtz equations is given by a second order differential equation. Thus, one obtains an electrical model of the filter as a sequence of electrically-coupled resonant circuits, each circuit being modeled by two resonators, one per mode, the resonance frequency of which represents the frequency of a mode, and whose resistance accounts for electric losses (surface currents) in the cavities.\n\nThis way, the filter can be seen as a quadripole, with two ports, when plugged onto a resistor at one end and fed with some potential at the other end. One is now interested in the power which is transmitted and reflected. This leads one to define a scattering matrix $S$, which may be considered as the transfer function of a stable causal linear dynamical system, with two inputs and two outputs. Its diagonal terms ${S}_{1,1}$, ${S}_{2,2}$ correspond to reflections at each port, while ${S}_{1,2}$, ${S}_{2,1}$ correspond to transmission. These functions can be measured at certain frequencies (on the imaginary axis). The matrix $S$ is approximately rational of order 4 times the number of cavities (that is 16 in the example on Figure ), and the key step consists in expressing the components of the equivalent electrical circuit as functions of the ${S}_{ij}$ (since there are no formulas expressing the lengths of the screws in terms of parameters of this electrical model). This representation is also useful to analyze the numerical simulations of the Maxwell equations, and to check the quality of a design, in particular the absence of higher resonant modes.\n\nIn fact, resonance is not studied via the electrical model, but via a low-pass equivalent circuit obtained upon linearizing near the central frequency, which is no longer conjugate symmetric (i.e. the underlying system may no longer have real coefficients) but whose degree is divided by 2 (8 in the example).\n\nIn short, the strategy for identification is as follows:\n\n• measuring the scattering matrix of the filter near the optimal frequency over twice the pass band (which is 80MHz in the example).\n\n• Solving bounded extremal problems for the transmission and the reflection (the modulus of he response being respectively close to 0 and 1 outside the interval measurement, cf. Section ) in order to get a models for the scattering matrix as an analytic matrix-valued function. This provides us with a scattering matrix known to be close to a rational matrix of order roughly 1/4 of the number of data points.\n\n• Approximating this scattering matrix by a true rational transfer-function of appropriate degree (8 in this example) via the Endymion or RARL2 software (cf. Section ).\n\n• A state space realization of $S$, viewed as a transfer function, can then be obtained, where additional symmetry constraints coming from the reciprocity law and possibly other physical features of the device have to be imposed.\n\n• Finally one builds a realization of the approximant and looks for a change of variables that eliminates non-physical couplings. This is obtained by using algebraic-solvers and continuation algorithms on the group of orthogonal complex matrices (symmetry forces this type of transformation).\n\n• The final approximation is of high quality. This can be interpreted as a confirmation of the linearity assumption on the system: the relative ${L}^{2}$ error is less than ${10}^{-3}$. This is illustrated by a reflection diagram (Figure ).\n\nThe above considerations are valid for a large class of filters. These developments have also been used for the design of non-symmetric filters, which are useful for the synthesis of repeating devices.\n\nThe team further investigates problems relative to the design of optimal responses for microwave devices. The resolution of a quasi-convex Zolotarev problems was proposed, in order to derive guaranteed optimal multi-band filter responses subject to modulus constraints . This generalizes the classical single band design techniques based on Chebyshev polynomials and elliptic functions. The approach relies on the fact that the modulus of the scattering parameter $|{S}_{1,2}|$ admits a simple expression in terms of the filtering function $D=|{S}_{1,1}|/|{S}_{1,2}|$, namely\n\n$|{S}_{1,2}{|}^{2}=\\frac{1}{1+{D}^{2}}.$\n\nThe filtering function appears to be the ratio of two polynomials ${p}_{1}/{p}_{2}$, the numerator of the reflection and transmission scattering factors, that may be chosen freely. The denominator $q$ is then obtained as the unique stable unitary polynomial solving the classical Feldtkeller spectral equation:\n\n$q{q}^{*}={p}_{1}{p}_{1}^{*}+{p}_{2}{p}_{2}^{*}.$\n\nThe relative simplicity of the derivation of a filter's response, under modulus constraints, owes much to the possibility of forgetting about Feldtkeller's equation and express all design constraints in terms of the filtering function. This no longer the case when considering the synthesis $N$-port devices for $N>3$, like multiplexers, routers and power dividers, or when considering the synthesis of filters under matching conditions. The efficient derivation of multiplexers responses is the subject of recent investigation by Factas, using techniques based on constrained Nevanlinna-Pick interpolation (see Section ).\n\nThrough contacts with CNES (Toulouse) and UPV (Bilbao), Apics got additionally involved in the design of amplifiers which, unlike filters, are active devices. A prominent issue here is stability. A twenty years back, it was not possible to simulate unstable responses, and only after building a device could one detect instability. The advent of so-called harmonic balance techniques, which compute steady state responses of linear elements in the frequency domain and look for a periodic state in the time domain of a network connecting these linear elements via static non-linearities made it possible to compute the harmonic response of a (possibly nonlinear and unstable) device . This has had tremendous impact on design, and there is a growing demand for software analyzers. The team is also becoming active in this area.\n\nIn this connection, there are two types of stability involved. The first is stability of a fixed point around which the linearized transfer function accounts for small signal amplification. The second is stability of a limit cycle which is reached when the input signal is no longer small and truly nonlinear amplification is attained (e.g. because of saturation). Applications by the team so far have been concerned with the first type of stability, and emphasis is put on defining and extracting the “unstable part” of the response, see Section . The stability check for limit cycles has made important theoretical advances, and numerical algorithms are now under investigation.\n\nHighlights of the Year Highlights of the Year Robotic tuning: a nice outcome of our long lasting experience in the field of computer assisted tuning for microwave devices\n\nA contract was signed with the French small and midsize business (SMB) Inoveos for the realization of a robotic prototype for the mass tuning of microwave devices. In addition to Inria, this project includes the university of Limoges Xlim and the engineering center Cisteme https://cisteme.net. Our team will be responsible of the driving software of the robot based on our long lasting experience in circuit extraction methods and in connection with our tools Presto-HF and Dedale-HF. Among the technical and scientific challenges for us on this project we can list:\n\n• Improvement of the computational efficiency of our circuit methods in order to be compatible with real-time measurements techniques of filter. Typically a circuital extraction needs to be performed in less than 1 second when dealing with a filter of order 10.\n\n• Handling the ambiguity resulting from the use of multiple solutions coupling topologies yielding several equivalent circuits for a single DUT (device under tuning).\n\n• New Results Inverse problems for Poisson-Laplace equations Paul Asensio Laurent Baratchart Sylvain Chevillard Juliette Leblond Jean-Paul Marmorat Konstantinos Mavreas Masimba Nemaire Inverse magnetization issues from planar data\n\nThe overall goal is here to determine magnetic properties of rock samples (e.g. meteorites or stalactites), from weak field measurements close to the sample that can nowadays be obtained using SQUIDs (superconducting quantum interference devices). Depending on the geometry of the rock sample, the magnetization distribution can either be considered to lie in a plane (thin sample) or in a parallelepiped of thickness $r$. Some of our results apply to both frameworks (the former appears as a limiting case when $r$ goes to 0), while others concern the 2-D case and have no 3-D counterpart as yet.\n\nFigure presents a schematic view of the experimental setup: the sample lies on a horizontal plane at height 0 and its support is included in a parallelepiped. The vertical component ${B}_{3}$ of the field produced by the sample is measured in points of a horizontal square at height $z$.\n\nWe pursued our investigation of the recovery of magnetizations modeled by signed measures on thin samples, and we singled out an interesting class that we call slender samples. These are sets of zero measure in ${ℝ}^{3}$, the complement of which has all its connected components of infinite measure. For such samples, we showed that consistent recovery is possible, in the Morozov discrepancy limit, by penalizing the total variation when either the support of the magnetization is purely 1-unrectifiable (which holds in particular for dipolar models) or the magnetization is unidirectional (an assumption of physical interest because igneous rocks acquire magnetization by cooling down in some ambient field). These notions play a role similar to sparsity in this infinite-dimensional context. An article has been published to report on these results . Moreover, in the case of planar samples (which are certainly slender), a loop decomposition of divergence free measures was obtained, which sharpens in the 2-D setting the structure theorem of , and allowed us to prove, using in addition the real analyticity of the operators relating the magnetization to the field, that the argument of the minimum of the regularized criterion $\\parallel f-{B}_{3}{\\mu \\parallel }_{2}^{2}+\\lambda {\\parallel \\mu \\parallel }_{TV}$ is unique; here, $\\mu$ is the measure representing the magnetization with respect to which the criterion gets optimized, $f$ is the data and $\\lambda >0$ a regularization parameter, while ${\\parallel \\mu \\parallel }_{TV}$ is the total variation of $\\mu$. An implementation using a variant of the FISTA algorithm has been set up which yields promising results when measurements are carried out on a relatively large surface patch. Yet, a deeper understanding on how to adjust the parameters of the method is required. This topic is studied in collaboration with D. Hardin and C. Villalobos from Vanderbilt University.\n\nWe also continued investigating the recovery of the moment of a magnetization, an important physical quantity which is in principle easier to reconstruct than the full magnetization because it is simply a vector in ${ℝ}^{3}$ that only depends on the field (i.e. magnetizations that produce the zero field also have zero moment). For the case of thin samples, we published an article reporting the construction of linear estimators for the moment from the field, based on the solution of certain bounded extremal problems in the range of the adjoint of the forward operator . On a related side, we also setup other linear estimators based on asymptotic results, in the previous years. These estimators are not limited to thin samples and can in principle estimate the net moment of 3D samples, provided that the dimensions of the sample are small with respect to the measurement area. Numerical experiments confirm that linear estimators (both kinds) make essential use of field values taken at the boundary of the measurement area, and are easily blurred by noise. We experimentally confirmed this sensitivity on a rather simple case: a small spherule has been magnetized in a controlled way by our partners at MIT, and its net moment has been measured by a classical magnetometer. The spherule has then been measured with the SQUID microscope, with several choices for important parameters (height of the sensor with respect to the spherule, sensitivity of the instrument, size of the 2D rectangle on which measurements are performed, size of the sample step). We applied our (asymptotics based) linear estimator on these experimental maps and they turn out to be clearly affected, especially when the data at the edges of the map are involved. The nature of the noise due to the microscope itself (electronic and quantization noise) might play an important role, as it is known to be non-white, and therefore can affect our methods which sum it up. Subsequently, we now envisage the possibility of modeling the structure of the noise to pre-process the data.\n\nFinally, we considered a simplified 2-D setup for magnetizations and magnetic potentials (of which the magnetic field is the gradient). When both the sample and the measurement set are parallel intervals, we set up some best approximation issues related to inverse recovery and relevant BEP problems in Hardy classes of holomorphic functions, see Section and , which is joint work with E. Pozzi (Department of Mathematics and Statistics, St Louis Univ., St Louis, Missouri, USA). Note that, in the present case, the criterion no longer acts on the boundary of the holomorphy domain (namely, the upper half-plane), but on a strict subset thereof, while the constraint acts on the support of the approximating function. Both involve functions in the Hilbert Hardy space of the upper half-plane.\n\nInverse magnetization issues from sparse cylindrical data\n\nThe team Factas was a partner of the ANR project MagLune on Lunar magnetism, headed by the Geophysics and Planetology Department of Cerege, CNRS, Aix-en-Provence, which ended this year (see Section ). Recent studies let geoscientists think that the Moon used to have a magnetic dynamo for a while. However, the exact process that triggered and fed this dynamo is still not understood, much less why it stopped. The overall goal of the project was to devise models to explain how this dynamo phenomenon was possible on the Moon.\n\nThe geophysicists from Cerege went a couple of times to NASA to perform measurements on a few hundreds of samples brought back from the Moon by Apollo missions. The samples are kept inside bags with a protective atmosphere, and geophysicists are not allowed to open the bags, nor to take out samples from NASA facilities. Moreover, the process must be carried out efficiently as a fee is due to NASA by the time when handling these moon samples. Therefore, measurements were performed with some specific magnetometer designed by our colleagues from Cerege. This device measures the components of the magnetic field produced by the sample, at some discrete set of points located on circles belonging to three cylinders (see Figure ). The objective of Factas is to enhance the numerical efficiency of post-processing data obtained with this magnetometer.\n\nUnder the hypothesis that the field can be well explained by a single magnetic pointwise dipole, and using ideas similar to those underlying the FindSources3D tool (see Sections and ), we try to recover the position and the moment of the dipole using the available measurements. This work, which is still on-going, constitutes the topic of the PhD thesis of K. Mavreas, whose defense is scheduled on January 31, 2020. In a given cylinder, using the associated cylindrical system of coordinates, recovering the position of the dipole boils down to determine its height $z$, its radial distance $\\rho$ and its azimuth $\\phi$. We use the fact that, whatever component of the field is measured, the (square of the) measurements performed on the circle at height $h$ correspond to a rational function of the form $p\\left(z\\right)/{\\left(z-{u}_{h}\\right)}^{5}$ where $p$ is a polynomial of degree at most 4 and ${u}_{h}$ is the complex number ${u}_{h}=\\frac{1+{\\rho }^{2}+{\\left(h-z\\right)}^{2}}{\\rho }\\phantom{\\rule{0.166667em}{0ex}}{e}^{i\\phi }$. The numerator $p$ depends on the moment of the dipole, on the height $h$ and on the kind of component which is measured. In contrast, ${u}_{h}$ can be estimated by rational approximation techniques, which allows one to obtain $\\phi$ directly and gives the relation $\\rho |{u}_{h}|=1+{\\rho }^{2}+{\\left(h-z\\right)}^{2}$. Combining the relations obtained at several heights, we proposed several methods to estimate $\\rho$ and $z$.\n\nThis year has been mostly devoted to running numerical experiments on synthetic examples. The first important observation is that the minimization criterion that we use to recover ${u}_{h}$ can have local minima achieving very small values, and that can sometimes erroneously be considered as the global minimum. We started studying theoretically this phenomenon, see Section . This means that the relative error $\\epsilon =|{u}_{h}-\\stackrel{˜}{{u}_{h}}|/|{u}_{h}|$ between the theoretical minimum ${u}_{h}$ and the value $\\stackrel{˜}{{u}_{h}}$ estimated by our algorithm can vary from almost 0 to more than $50%$, even when the data used as measurements exactly correspond to the field produced by a magnetic dipole. The second important observation is that the statistical distribution of $\\epsilon$ (when the position of the dipole is uniformly chosen within a cylinder with a moment uniformly chosen on the unit sphere) depends on the measured component of the field. Figure shows the distribution experimentally observed. The vertical component of the field noticeably leads to better estimates for ${u}_{h}$ than both other components. The third important observation that we made is that the presence of noise on the measurements, even moderate, significantly alters the quality of the estimation of ${u}_{h}$. Figure shows the distribution experimentally observed in the same conditions as before, but using data contaminated with a random normal noise with standard deviation $5%$ of the maximal absolute value of the measured component.\n\nThese observations are somehow bad news, as the method we propose is based on recovering the position of the dipole by using the values ${u}_{h}$ collected at several heights $h$. However, our experiments also revealed an unexpected good news: while the estimation of ${u}_{h}$ itself is often bad, as soon as the data are not perfect, its argument (from which $\\phi$ is immediately deduced) turns out to be fairly well recovered. This is illustrated in Figure which shows, on the set of experiments of Figure , the position of the 4000 dipoles, with a color indicating whether $\\epsilon$ is big (left part of the figure) and whether the error on the argument of ${u}_{h}$ is big (right part of the figure). As can be seen, the angular error is most of the time smaller than $10°$, even for dipoles for which $\\epsilon$ is fairly big. This phenomenon is probably due to the fact that local minima of the criterion tends to have a complex argument close to the complex argument of the global minimum, a phenomenon that we started to study theoretically (see Section ).\n\nInverse problems in medical imaging\n\nIn 3-D, functional or clinically active regions in the cortex are often modeled by pointwise sources that have to be localized from measurements, taken by electrodes on the scalp, of an electrical potential satisfying a Laplace equation (EEG, electroencephalography). In the works , on the behavior of poles in best rational approximants of fixed degree to functions with branch points, it was shown how to proceed via best rational approximation on a sequence of 2-D disks cut along the inner sphere, for the case where there are finitely many sources (see Section ).\n\nIn this connection, a dedicated software FindSources3D (FS3D, see Section ) is being developed, in collaboration with the Inria team Athena and the CMA - Mines ParisTech. Its Matlab version now incorporates the treatment of MEG data, the aim being to handle simultaneous EEG–MEG recordings available from our partners at INS, hospital la Timone, Marseille. Indeed, it is now possible to use simultaneously EEG and MEG measurement devices, in order to measure both the electrical potential and a component of the magnetic field (its normal component on the MEG helmet, that can be assumed to be spherical). This enhances the accuracy of our source recovery algorithms. Note that FS3D takes as inputs actual EEG measurements, like time signals, and performs a suitable singular value decomposition in order to separate independent sources.\n\nIt appears that, in the rational approximation step, multiple poles possess a nice behavior with respect to branched singularities. This is due to the very physical assumptions on the model from dipolar current sources: for EEG data that correspond to measurements of the electrical potential, one should consider triple poles; this will also be the case for MEG – magneto-encephalography – data. However, for (magnetic) field data produced by magnetic dipolar sources, like in Section , one should consider poles of order five. Though numerically observed in , there is no mathematical justification so far why multiple poles generate such strong accumulation of the poles of the approximants (see Section ). This intriguing property, however, is definitely helping source recovery and will be the topic of further study. It is used in order to automatically estimate the “most plausible” number of sources (numerically: up to 3, at the moment).\n\nThis year, we started considering a different class of models, not necessarily dipolar, and related estimation algorithms. Such models may be supported on the surface of the cortex or in the volume of the encephalon. We represent sources by vector-valued measures, and in order to favor sparsity in this infinite-dimensional setting we use a TV (i.e. total variation) regularization term as in Section . The approach follows that of and is implemented through two different algorithms, whose convergence properties are currently being studied. Tests on synthetic data from a few dipolar sources provide results of different qualities that need to be better understood. In particular, a weight is being added in the TV term in order to better identify deep sources. This is the topic of the starting PhD research of P. Asensio and M. Nemaire. Ultimately, the results will be compared to those of FS3D and other available software tools.\n\nMatching problems and their applications Laurent Baratchart Martine Olivi Gibin Bose David Martinez Martinez Fabien Seyfert Multiplexer synthesis via interpolation and common junction design\n\nIn the context of David Martinez Martinez's PhD funded partly by CNES the synthesis of multiplexer responses was considered using multipoint matching techniques. Indeed, synthesizing the response of multiplexer composed of a set of channel filters connected via common manifold junction to a common port can be seen as a matrix version application of our multipoint matching result for filters . For short a simultaneous matching solutions is sought for, where each channel filter matches the load it is connected at specified matching frequencies. The difficulty here is that the load seen by each filter, depends explicitly of the response of the other filters by means of the common junction's response: the multiplexer synthesis problem is therefore, in general, strictly harder than the filter multipoint matching problem, and can't be solved by a sequential solving of independent «scalar» problems. A notable exception to this statement is obtained when a totally decoupling common junction is considered. This somehow artificial situation was taken as a start of a continuation algorithm, during which the decoupling junction response is moved step by step via a linear trajectory towards the target junction while the simultaneous matching problem is solved all along via a differential predictor corrector method. Whereas all «accidents» of branch point type that can occur during this procedure are not classified yet, one major obstruction to the continuation process is the occurrence of manifold peaks. The latter are due to resonances occurring in the manifold junction and yield total reflection at some frequencies of the channel ports. When latter coincide with the matching frequencies of a particular channel filter, the simultaneous matching problem has no solution, and the continuation algorithm fails irredeemably.\n\nWe therefore gave a full characterization of this manifold peaks and designed a heuristic approach to avoid their appearance during the continuation process. We showed that they only depend on the out of band response of the channel filters, and can in first approximation be considered as constant along the continuation process and estimated by a full wave simulation of each channel filter. This is then used within a triangular adjustment procedure that looks for possible manifold length adjustments (within the channel filters, and between that channel filters and the manifold junction) that guaranties the absence of manifold peaks within the band of each channel filter. Details of this procedure that give important information to the designer about the feasibility of an effective multiplexer response by means of given manifold T-junction, and this before any channel filter optimization procedure, are detailed in , and were presented at Eumc 2019. In connection with the previously described continuation procedure, it was used to design a compact triplexer, based on frequency specifications considered as «hard to fulfill» and furnished by CNES. The triplexer was then realized using 3D printing techniques at Xlim (S. Bila and O. Tantot) our long standing academical partners on these topics (see Figure ). This work is part of the PhD thesis defended by David Martinez Martinez at the end of June.\n\nUniform matching and global optimality considerations: application to a reference tracking problem\n\nThis problem was proposed by Pauline Kergus, PhD student at Onera (Toulouse). In her PhD, she studied the following data driven problem: given frequency measurements of a plant, find a controller which allows to follow a given reference model. The approach she proposed was to directly identify the controller from frequency measurements induced on the controller by the closed loop. Of course the quality of the controller, and in particular its stability, highly depend on the chosen reference model. The question is thus: how to choose a good reference model $M$ with a minimum of information on the plant? The reference model is linked to the sensitivity function $S$ by the relation $M+S=1$. The sensitivity function is an important design tool in control. To ensure closed loop stability, it should be stable and satisfy some interpolation conditions at the unstable poles and zeros of the plant . Its shape reflect the performances of the closed loop: $S$ should be small at low frequencies to ensure a good tracking accuracy, as well as disturbance rejection; while to ensure noise rejection, $S$ should go to 1 at infinity (the reference model $1-S$ should be small at high frequencies). The shaping problem for the sensitivity function can be stated in a manner almost similar to the matching problem described below: find a Schur function with minimum infinite norm in a frequency band $\\left[0,{w}_{c}\\right]$, where ${w}_{c}$ is the chosen cutoff frequency, while satisfying some interpolatory constraints. The main difference that prevents for using the convex relaxation method proposed below is the condition at infinity. An alternative optimization method is under study. To get a non-optimal solution to the problem, Pauline Kergus proposed a simple way to enforce the interpolation conditions from a given well-shaped reference model. To compute the unstable poles and zeros of the plant, which is the minimal required information, she uses our software PISA https://project.inria.fr/pisa/working/project/. Examples illustrating the advantages and limitations of the method were studied. The results were reported in the journal paper and presented at the CDC 2019 in Nice.\n\nStability assessment of microwave amplifiers and design of oscillators Laurent Baratchart Sylvain Chevillard Martine Olivi Fabien Seyfert Sébastien Fueyo Adam Cooman\n\nThe goal is here to help design amplifiers and oscillators, in particular to detect instability at an early stage of the design. This topic is studied in the doctoral work of S. Fueyo, co-advised with J.-B. Pomet (from the McTao Inria project-team). Application to oscillator design methodologies is studied in collaboration with Smain Amari from the Royal Military College of Canada (Kingston, Canada).\n\nAs opposed to Filters and Antennas, Amplifiers and Oscillators are active components that intrinsically entail a non-linear functioning. The latter is due to the use of transistors governed by electric laws exhibiting saturation effects, and therefore inducing input/output characteristics that are no longer proportional to the magnitude of the input signal. Hence, they typically produce non-linear distortions. A central question arising in the design of amplifiers is to assess stability. The latter may be understood around a functioning point when no input but noise is considered, or else around a periodic trajectory when an input signal at a specified frequency is applied. For oscillators, a precise estimation of their oscillating frequency is crucial during the design process. For devices operating at relatively low frequencies, time domain simulations perform satisfactorily to check stability. For complex microwave amplifiers and oscillators, the situation is however drastically different: the time step necessary to integrate the transmission line's dynamical equations (which behave like a simple electrical wire at low frequency) becomes so small that simulations are intractable in reasonable time. Moreover, most linear components of such circuits are known through their frequency response, and a preliminary, numerically unstable step is then needed to obtain their impulse response, prior to any time domain simulation.\n\nFor these reasons, the analysis of such systems is carried out in the frequency domain. In the case of stability issues around a functioning point, where only small input signals are considered, the stability of the linearized system obtained by a first order approximation of each non-linear component can be studied via the transfer impedance functions computed at some ports of the circuit. In recent years, we showed that under realistic dissipativity assumptions at high frequency for the building blocks of the circuit, these transfer functions are meromorphic in the complex frequency variable $s$, with at most finitely many unstable poles in the right half-plane . Dwelling on the unstable/stable decomposition in Hardy Spaces, we developed a procedure to assess the stability or instability of the transfer functions at hand, from their evaluation on a finite frequency grid , that was further improved in to address the design of oscillators, in collaboration with Smain Amari. This has resulted in the development of a software library called Pisa (see Section , aiming at making these techniques available to practitioners. Research in this direction now focuses on the links between the width of the measurement band, the density of the measurement points, and the precision with which an unstable pole, located within a certain depth into the complex plane, can be identified.\n\nExtensions of the procedure to the strong signal case, where linearisation is considered around a periodic trajectory, have received attention over the last two years. When stability is studied around a periodic trajectory, determined in practice by Harmonic Balance algorithms, linearization yields a linear time varying dynamical system with periodic coefficients and a periodic trajectory thereof. While in finite dimension the stability of such systems is well understood via the Floquet theory, this is no longer the case in the present setting which is infinite dimensional, due to the presence of delays. Dwelling on the theory of retarded systems, S. Fueyo's PhD work has shown last year that, for general circuits, the monodromy operator of the linearized system along its periodic trajectory is a compact perturbation of a high frequency, non dynamical operator, which is stable under a realistic passivity assumption at high frequency. Therefore, only finitely many unstable points can arise in the spectrum of the monodromy operator, and this year we established a connection between these and the singularities of the harmonic transfer function, viewed as a holomorphic function with values in periodic ${L}^{2}$ functions. One difficulty, however, is that these singularities need not affect all Fourier coefficients, whereas harmonic balance techniques can only estimate finitely many of them. This issue, that was apparently not singled out by practitioners, is currently under examination.\n\nWe also wrote an article reporting about the stability of the high frequency system, and recast this result in terms of exponential stability of certain delay systems .\n\nThe Hardy-Hodge decomposition Laurent Baratchart Masimba Nemaire\n\nIn a joint work with T. Qian and P. Dang from the university of Macao, we proved in previous years that on a compact hypersurface $\\Sigma$ embedded in ${ℝ}^{n}$, a ${ℝ}^{n}$-valued vector field of ${L}^{p}$ class decomposes as the sum of a harmonic gradient from inside $\\Sigma$, a harmonic gradient from outside $\\Sigma$, and a tangent divergence-free field, provided that $2-\\epsilon , where $\\epsilon$ and ${\\epsilon }^{\\text{'}}$ depend on the Lipschitz constant of the surface. We also proved that the decomposition is valid for $1 when $\\Sigma$ is $VMO$-smooth (i.e. $\\Sigma$ is locally the graph of Lipschitz function with derivatives in $VMO$). By projection onto the tangent space, this gives a Helmholtz-Hodge decomposition for vector fields on a Lipschitz hypersurface, which is apparently new since existing results deal with smooth surfaces. In fact, the Helmholtz-Hodge decomposition holds on Lipschitz surfaces (not just hypersurfaces), The Hardy-Hodge decomposition generalizes the classical Plemelj formulas from complex analysis. We pursued this year the writing of an article on this topic, and we also found that this decomposition yields a description of silent magnetizations distributions of ${L}^{p}$-class on a surface. A natural endeavor is now to use this description, via balayage, to describe volumetric silent magnetizations.\n\nIdentification of resonating frequencies of compact metallic objects in electromagnetic inverse scattering Laurent Baratchart Martine Olivi Fabien Seyfert\n\nWe started an academic collaboration with LEAT (Univ. Nice, France, pers. involved: Jean-Yves Dauvignac, Nicolas Fortino, Yasmina Zaki) on the topic of inverse scattering using frequency dependent measurements. As opposed to classical electromagnetic imaging where several spatially located sensors are used to identify the shape of an object by means of scattering data at a single frequency, a discrimination process between different metallic objects is here being sought for by means of a single, or a reduced number of sensors that operate on a whole frequency band. For short the spatial multiplicity and complexity of antenna sensors is here traded against a simpler architecture performing a frequency sweep.\n\nThe setting is shown on Figure . The total field ${E}_{t}\\left(r,\\theta ,\\phi \\right)$ is the sum of the incident field ${E}_{i}$ (here a plane wave) and scattered field ${E}_{s}$, that is at every point in space we have ${E}_{t}={E}_{i}+{E}_{s}$. A harmonic time dependency (${e}^{j\\omega t}$, where $j$ is the imaginary unit: ${j}^{2}=-1$ ) is supposed for the incident wave, so that by linearity of Maxwell equations and after a transient state, following holds,\n\n${E}_{s}\\left({r}_{o},{\\theta }_{o},{\\phi }_{o}\\right)=H\\left(s=j\\omega ,{\\theta }_{o},{\\phi }_{o}\\right){E}_{i}\\left({r}_{e},{\\theta }_{e},{\\phi }_{e}\\right).$\n\nThe subscripts $o$ and $e$ stand here for «observation point» and «emission point»: the scattered field at the observation point is therefore related to the emitted planar wave field at the emission point via the transfer function $H\\left(s=j\\omega ,{\\theta }_{o},{\\phi }_{o}\\right)$. The emission point is here supposed fixed, so the dependency in $e$ is omitted in $H$. Under regularity conditions on the scatterer's boundary the function $H$ can be shown to admit an analytic continuation into the complex left half plane for the $s$ variable, away from a discrete set (with a possible accumulation point a infinity) where it admits poles. Thus, $H$ is a meromorphic function in the variable $s$. Its poles are called the resonating frequencies of the scattering object. Recovering these resonating frequencies from frequency scattering measurement, that is measurements of $H$ at particular $s=j{\\omega }_{i}^{\\text{'}}s$ is the primary objective of this project.\n\nIn order to gain some insight we started a full study of the particular case when the scatterer is a spherical PEC (Perfectly Electric Conductor). In this case Maxwell equations can be solved «explicitly» by means of expansions in series of vectorial spherical harmonics. We showed in particular that in this case $H$ admits following simple structure:\n\n$H\\left(\\omega ,{\\theta }_{o},{\\phi }_{o}\\right)=R\\left(s,{\\theta }_{o},{\\phi }_{o}\\right){e}^{-{\\tau }_{1}\\left({\\theta }_{o},{\\phi }_{o}\\right)s}+C\\left({\\theta }_{o},{\\phi }_{o}\\right){e}^{-{\\tau }_{2}\\left({\\theta }_{o},{\\phi }_{o}\\right)s},$\n\nwhere $R$ is a meromorphic functions with poles at zeros of the spherical Hankel functions and their derivatives and $C$ is independent of the frequency. Identification procedures, surprisingly close to the ones we developed in connection with amplifier stability analysis, are currently being studied to gain information about the resonating frequencies by means of a rational approximation of the function $R$ once it has been de-embedded. Generalization of this analysis and procedure will be considered for arbitrarily compact PEC objects.\n\nImaging and modeling ancient materials Vanna Lisa Coli Juliette Leblond Pat Vatiwutipong\n\nThis is a recent activity of the team, linked to image classification in archaeology in the framework of the project ToMaT (see Regional Initiatives below) and to the post-doctoral stay of V. L. Coli; it is pursued in collaboration with L. Blanc-Féraud (project-team Morpheme, I3S-CNRS/Inria Sophia/iBV), D. Binder (CEPAM-CNRS, Nice), in particular.\n\nThe pottery style is classically used as the main cultural marker within Neolithic studies. Archaeological analyses focus on pottery technology, and particularly on the first stages of pottery manufacturing processes. These stages are the most demonstrative for identifying the technical traditions, as they are considered as crucial in apprenticeship processes. Until now, the identification of pottery manufacturing methods was based on macro-traces analysis, i.e. surface topography, breaks and discontinuities indicating the type of elements (coils, slabs, ...) and the way they were put together for building the pots. Overcoming the limitations inherent to the macroscopic pottery examination requires a complete access to the internal structure of the pots. Micro-computed tomography ($\\mu$CT) has recently been used for exploring ancient materials microstructure. This non-invasive method provides quantitative data for a big set of proxies and is perfectly adapted to the analysis of Cultural heritage materials.\n\nThe main challenge of our current analyses aims to overcome the lack of existing protocols to apply in order to quantify observations. In order to characterize the manufacturing sequences, the mapping of the paste variability (distribution and composition of temper) and the discontinuities linked to different classes of pores, fabrics and/or organic inclusions appears promising. The totality of the acquired images composes a set of 2-D and 3-D surface and volume data at different resolutions and with specific physical characteristics related to each acquisition modality (multimodal and multi-scale data). Specific shape recognition methods need to be developed by application of robust imaging techniques and 3-D-shapes recognition algorithms.\n\nIn a first step, we devised a method to isolate pores from the 3-D data volumes in binary 3-D images, to which we apply a process named Hough transform (derived from Radon transform). This method, of which the generalization from 2-D to 3-D is quite recent, allows us to evaluate the presence of parallel lines going through the pores. The quantity of such lines is a good indicator of the “coiling” manufacturing, that it allows to distinguish from the other “spiral patchwork” patchwork technique, in particular. These progresses are described in , , , and the object of an article in preparation.\n\nThe Hough and Radon transforms can also be applied to 2-D slices of the available 3-D images displaying pores locations. In this framework, the use of Radon transform to evaluate the density of points in the image that do belong to (or almost) parallel lines appears to be quite efficient, as was seen during P. Vatiwutipong's internship.\n\nOther possibilities of investigation will be analyzed as well, such as machine learning techniques.\n\nBehavior of poles in rational and meromorphic approximation Laurent Baratchart Sylvain Chevillard Juliette Leblond Martine Olivi Fabien Seyfert Rational approximation\n\nThe numerous experiments that we performed on synthetic data in the context of the MagLune project (see Sections and ) revealed an intriguing behavior of the local minima of the optimization problem underlying our method. In the context of that application, we are provided with sampled values on the unit circle $𝕋$ of a function $f$ which is known to be of the form $f\\left(z\\right)=p\\left(z\\right)/{\\left(z-\\beta \\right)}^{5}$ where $p\\left(z\\right)\\in {ℂ}_{4}\\left[z\\right]$ is a polynomial of degree at most 4 with complex coefficients and $\\beta \\in 𝔻$ belongs to the unit disk. A key problem consists in recovering $\\beta$ from the values of $f$ on the unit circle. The same problem occurs in the core of FindSources3D (see and ) with $p$ being of degree at most 2 and a pole of order 3 rather than 5.\n\nIn order to estimate $\\beta$, we seek for the global minimum on ${ℂ}_{4}\\left[z\\right]×𝔻$ of the function $\\phi$ defined by\n\n$\\phi :\\left(q,\\phantom{\\rule{0.166667em}{0ex}}\\alpha \\right)↦{∥\\frac{q\\left(z\\right)}{{\\left(z-\\alpha \\right)}^{5}}-f\\left(z\\right)∥}_{{L}^{2}\\left(𝕋\\right)}.$\n\nWhen $f$ is actually a rational function of the considered form, $\\phi$ obviously has a unique global minimum where it reaches the value 0. We experimentally observed that $\\phi$ usually has several local minima, some of them achieving very small values, and these minima often have a complex argument close to the argument of $\\beta$. This behavior is unusual and contrasts with the fact the function\n\n$\\psi :\\left(q,\\phantom{\\rule{0.166667em}{0ex}}{\\alpha }_{1},\\phantom{\\rule{0.166667em}{0ex}}\\cdots ,\\phantom{\\rule{0.166667em}{0ex}}{\\alpha }_{5}\\right)↦{∥\\frac{q\\left(z\\right)}{{\\prod }_{i=1}^{5}\\left(z-{\\alpha }_{i}\\right)}-f\\left(z\\right)∥}_{{L}^{2}\\left(𝕋\\right)}$\n\nis known to have a unique local minimum on ${ℂ}_{4}\\left[z\\right]×{𝔻}^{5}$ (which is global) when $f$ is a rational function of the same form.\n\nIn order to understand the reasons underlying our observations, we started studying the theoretical properties of the critical points of $\\phi$, in the general case of a pole of order $n\\in {ℕ}^{*}$ and with a polynomial of degree less or equal to $n-1$ at the numerator. Our results so far are the following.\n\nWe introduce the family ${\\left({g}_{j}^{\\left(\\alpha \\right)}\\right)}_{j\\in {ℕ}^{*}}$ where ${g}_{j}^{\\left(\\alpha \\right)}\\left(z\\right)={\\left(1-\\overline{\\alpha }z\\right)}^{j-1}/{\\left(z-\\alpha \\right)}^{j}$ which is an orthogonal basis (for the usual ${L}^{2}\\left(𝕋\\right)$ Hilbert product) of the space of rational functions with a single pole (of arbitrary order) in $\\alpha$. Thanks to this family, we prove that $\\left(q,\\alpha \\right)$ is a critical point of $\\phi$ if and only if $f$ is orthogonal either to ${g}_{n}^{\\left(\\alpha \\right)}$ or ${g}_{n+1}^{\\left(\\alpha \\right)}$ and, for such a given $\\alpha$, $q/{\\left(z-\\alpha \\right)}^{n}$ is the orthogonal projection of $f$ onto the rational functions of that form. The case when $f$ is orthogonal to ${g}_{n}^{\\left(\\alpha \\right)}$ combined with the fact that $q/{\\left(z-\\alpha \\right)}^{n}$ is the orthogonal projection of $f$ implies a pole-zero simplification of $q/{\\left(z-\\alpha \\right)}^{n}$ at $z=\\alpha$ and we conjecture that it exactly corresponds to local maxima of $\\phi$ with respect to variable $\\alpha$. We also conjecture that the other case exactly corresponds to local minima of $\\phi$. We are currently working on proving these conjectures, which should not be too hard.\n\nWe also obtained an explicit algebraic equation characterizing $\\alpha$, and we know how to solve it when $f$ is of the form $1/{\\left(z-\\beta \\right)}^{k}$ ($1\\le k\\le n$). For small values of $n$, we proved (and conjecture that it holds for any $n$) that there are $2k-1$ solutions in the unit disk, all lying on the diameter passing though $\\beta$. This is a remarkable result that somehow theoretically confirms the kind of experimental observations we got. The theoretical case of a function $f$ with a non trivial numerator seems currently out of reach, though.\n\nMeromorphic approximation\n\nWe showed that best meromorphic approximation on a contour, in the uniform norm, to functions with countably many branched singularities with polar closure inside the contour produces poles whose counting measure accumulate weak-* to the Green equilibrium distribution on the cut of minimal capacity outside of which the function is single-valued. This is joint work with M. Yattselev (University of Indianapolis, Purdue University at Indianapolis). An article is currently being written on this topic.\n\nBilateral Contracts and Grants with Industry Bilateral Contracts with Industry Contract CNES-Inria-Xlim\n\nThis contract (reference Inria: 11282) accompanied the PhD of David Martinez Martinez and focused on the development of efficient techniques for the design of matching network tailored for frequency varying loads. Applications of the latter to the design output multiplexers occurring in space applications has also been considered (see new results section). The contract ended mid 2019.\n\nContract Inria-Inoveos\n\nA contract was signed with the SMB company Inoveos in order to build a prototypical robot dedicated to the automatic tuning of microwave devices, see Section .\n\nPartnerships and Cooperations Regional Initiatives\n• The team co-advises a PhD (G. Bose) with the CMA team of LEAT (http://leat.unice.fr/pages/activites/cma.html) funded by the Labex UCN@Sophia on the co-conception of Antennas and Filters.\n\n• The team participates in the project ToMaT, “Multiscale Tomography: imaging and modeling ancient materials, technical traditions and transfers”, funded by the Idex ${\\text{UCA}}^{\\text{Jedi}}$ (“programme structurant Matière, Lumière, Interactions”). This project brings together researchers in archaeological, physical, and mathematical sciences, with the purpose of modeling and detecting low level signals in 3-D images of ancient potteries. The other concerned scientists are from CEPAM-CNRS-UCA (project coordinator: Didier Binder), Nice http://www.cepam.cnrs.fr, the team Morpheme, CNRS-I3S-Inria http://www.inria.fr/equipes/morpheme, and IPANEMA, CNRS, Ministère de la Culture et de la Communication, Université Versailles Saint Quentin http://ipanema.cnrs.fr/. Since March 2018, they co-advise together the post-doctoral research of Vanna Lisa Coli, see Section , and this year the internship training of Pat Vatiwutipong.\n\n• National Initiatives ANR MagLune\n\nThe ANR project MagLune (Magnétisme de la Lune) was active from July 2014 to August 2019. It involved the Cerege (Centre de Recherche et d'Enseignement de Géosciences de l'Environnement, joint laboratory between Université Aix-Marseille, CNRS and IRD), the IPGP (Institut de Physique du Globe de Paris) and ISTerre (Institut des Sciences de la Terre). Associated with Cerege were Inria (Apics, then Factas team) and Irphe (Institut de Recherche sur les Phénomènes Hors Équilibre, joint laboratory between Université Aix-Marseille, CNRS and École Centrale de Marseille). The goal of this project (led by geologists) was to understand the past magnetic activity of the Moon, especially to answer the question whether it had a dynamo in the past and which mechanisms were at work to generate it. Factas participated in the project by providing mathematical tools and algorithms to recover the remanent magnetization of rock samples from the moon on the basis of measurements of the magnetic field it generates. The techniques described in Section were instrumental for this purpose.\n\nANR Repka\n\nANR-18-CE40-0035, “REProducing Kernels in Analysis and beyond”, starting April 2019 (for 48 months).\n\nLed by Aix-Marseille Univ. (IMM), involving Factas team, together with Bordeaux (IMB), Paris-Est, Toulouse Universities.\n\nThe project consists of several interrelated tasks dealing with topical problems in modern complex analysis, operator theory and their important applications to other fields of mathematics including approximation theory, probability, and control theory. The project is centered around the notion of the so-called reproducing kernel of a Hilbert space of holomorphic functions. Reproducing kernels are very powerful objects playing an important role in numerous domains such as determinantal point processes, signal theory, Sturm-Liouville and Schrödinger equations.\n\nThis project supports the PhD of M. Nemaire within Factas, co-advised by IMB partners.\n\nEuropean Initiatives Collaborations with Major European Organizations\n• Factas is part of the European Research Network on System Identification (ERNSI) since 1992.\n\n• System identification deals with the derivation, estimation and validation of mathematical models of dynamical phenomena from experimental data.\n\n• International Initiatives Inria International Partners Informal International Partners\n\nFollowing two Inria Associate teams (2013-2018) and a MIT-France seed funding (2014-2018), the team has a strong and regular collaboration with the Earth and Planetary Sciences department at Massachusetts Institute of Technology (Cambridge, MA, USA) and with the Mathematics department of Vanderbilt University (Nashville, TN, USA) on inverse problems for magnetic microscopy applied to the analysis of ancient rock magnetism.\n\nInternational Research Visitors Visits of International Scientists\n• Smain Amari (Royal Military College of Canada, Kingston, Canada), February 4-9.\n\n• Jonathan Partington (Univ. of Leeds, England), February 4-7.\n\n• Dmitry Ponomarev (T.U. Vienna, Vienna, Austria), June 24.\n\n• Élodie Pozzi (St Louis Univ., St Louis, Missouri, USA), Brett Wick (Washington Univ., St Louis, Missouri, USA), January 9-10.\n\n• Yves Rolain (Vrije Universiteit Brussel, VUB, Brussels, Belgium), February 5-7.\n\n• Maxim Yattselev (University of Indianapolis, Purdue University at Indianapolis, USA), June 29-July 1.\n\n• Internships\n• Paul Asensio, École Centrale Lyon, Study of silent current sources in electroencephalography (EEG) and magnetoencephalography (MEG); advisors: L. Baratchart, J. Leblond.\n\n• Masimba Nemaire, MathMods Master, Study of silent current sources in EEG and MEG; advisors: L. Baratchart, J. Leblond.\n\n• Tuong Vy Nguyen Hoang, Mathematical Circuit Modeling for Antennas; advisors: F. Seyfert, M. Olivi.\n\n• Pat Vatiwutipong, MathMods Master, Properties of the $d$-Radon transform and applications to imaging issues in archaeology; advisors: V. L. Coli, J. Leblond.\n\n• List of international and industrial partners\n\nFigure sums up who are our main collaborators, users and competitors.\n\nDissemination Promoting Scientific Activities\n• L. Baratchart gave an oral communication at NCMIP 2019 in Cachan,\n\n• V. L. Coli gave oral communications at the 2nd “Journée Matériaux UCA”, Sophia Antipolis, September, and at the workshop “Céramiques imprimées de Méditerranée occidentale. Matières premières, productions, usages” of the ANR CIMO, Nice, France, March, http://www.cepam.cnrs.fr/sites/cimo/.\n\n• D. Martinez Martinez gave an oral communication at «Journées Nationales des Microondes», Caen, France and at «European microwave Conference (EuMC) 2019», Paris, France.\n\n• F. Seyfert was invited to give a lecture at the Technical University of Cartagena University (Spain) and gave an invited talk at the workshop «Rational approximation for Electrical Engineering», Moscow, Russia sponsored by Huawei.\n\n• Scientific Events: Selection Member of the Conference Program Committees\n\nL. Baratchart was on the program committee of “Applied Inverse Problems” (AIP) 20019, Grenoble, France http://www.aip2019-grenoble.fr.\n\nJournal Member of the Editorial Boards\n\nL. Baratchart is on the editorial board of the journals “Computational Methods and Function Theory” and “Complex Analysis and Operator Theory”.\n\nReviewer - Reviewing Activities\n• J. Leblond was a reviewer for the journals Engineering with Computers, Inverse Problems.\n\n• F. Seyfert was a reviewer for IEEE Transactions on Microwave Theory and Techniques\n\n• Invited Talks\n• L. Baratchart gave an invited address at the conference “One-Dimensional Complex Analysis and Operator Theory” in Saint Petersburg, May 13-17, https://sites.google.com/view/sft2019/home/conference. He was an invited speaker at the workshop «Rational approximation for Electrical Engineering», Moscow, Russia, sponsored by Huawei, and lectured at the Macao university of sciences and techniques, in April. Macao University of Sciences and Technology in April.\n\n• L. Baratchart and J. Leblond were invited to give talks at AIP 2019, Grenoble, France, July, http://www.aip2019-grenoble.fr.\n\n• S. Chevillard was invited to give a talk at a NFS-sponsored workshop on magnetic imaging organized outside the American Geophysical Union meeting (December 7-8).\n\n• J. Leblond was an invited speaker at the final workshop of the ANR FastRelax, Lyon, France, May, http://fastrelax.gforge.inria.fr/FastRelax2019.html.\n\n• Scientific Expertise\n• L. Baratchart was a member of selection panel 40 (Mathematics) of the Agence Nationale de la Recherche (ANR).\n\n• J. Leblond was an external reviewer for a promotion evaluation process at Chapman University (Orange, CA, USA).\n\n• F. Seyfert was a reviewer for the National Science Centre of Poland\n\n• J. Leblond is a member of the “Conseil Scientifique” and of the “Commission Administrative Paritaire” of Inria.\n\n• M. Olivi is a member of the CLDD (Commission Locale de Développement Durable) and in charge, with P. Bourgeois, of coordination.\n\n• Teaching - Supervision - Juries Teaching\n• Colles: S. Chevillard has given “Colles” (oral examination preparing undergraduate students for the competitive examination to enter French Engineering Schools) at Centre International de Valbonne (CIV) (2 hours per week) until June 2019.\n\n• Supervision\n• PhD in progress: K. Mavreas, Inverse source problems in planetary sciences: dipole localization in Moon rocks from sparse magnetic data, since October 2015, advisors: S. Chevillard, J. Leblond; defense scheduled January 31, 2020.\n\n• PhD in progress: G. Bose, Filter Design to Match Antennas, since December 2016, advisors: F. Ferrero, F. Seyfert and M. Olivi.\n\n• PhD in progress: S. Fueyo, Cycles limites et stabilité dans les circuits, since October 2016, advisors: L. Baratchart and J.-B. Pomet (Inria Sophia, McTao).\n\n• PhD in progress: P. Asensio, Inverse source estimation problems in EEG and MEG, since November 2019, advisors: L. Baratchart, J. Leblond.\n\n• PhD in progress: M. Nemaire, Inverse potential problems with application to quasi-static electromagnetics, since October 2019, advisors: L. Baratchart, J. Leblond, S. Kupin (IMB, Univ. Bordeaux).\n\n• Post-doc. in progress: V. L. Coli, Multiscale Tomography: imaging and modeling ancient materials, since March 2018, advisors: J. Leblond, L. Blanc-Féraud (project-team Morpheme, I3S-CNRS/Inria Sophia/iBV), D. Binder (CEPAM-CNRS, Nice).\n\n• Juries\n• L. Baratchart was a reviewer of the “Mémoire d'habilitation” of Moncef Mahjoub, ENIT, Tunis, September 2.\n\n• J. Leblond was a member of the PhD committees of I. Santos (Univ. Paul Sabatier, Toulouse, February), S. Amraoui and K. Maksymenko (Univ. Côte d'Azur, December).\n\n• M. Olivi was a member of the HdR committees of F. Seyfert (Univ. Côte d'Azur, February 6), C. Poussot-Vassals (Univ. Toulouse, July 12) and of the PhD committees of D. Martinez Martinez (Univ. Limoges, June 20) and P. Kergus (Univ. Toulouse, October 18)\n\n• F. Seyfert was a member of the PhD committee of Johan Sence (Univ. Limoges, November 15) and D. Martinez Martinez (Univ. Limoges, June 20).\n\n• Popularization Internal or external Inria responsibilities\n\nM. Olivi was responsible for Scientific Mediation and president of the Committee MASTIC (Commission d'Animation et de Médiation Scientifique) https://project.inria.fr/mastic/ until October 30.\n\nArticles and contents\n\nM. Olivi wrote a review of the book “Algorithms: la bombe à retardement” by C. O'Neil for Interstice https://interstices.info/sciences-du-numerique-et-impact-sur-la-societe/\n\nEducation\n\n“La fête des Maths de l'ESPE Nice-Liégeard” (March 5 and 26): M. Olivi animated two half-day workshop sessions “jouons avec des expériences scientifiques” https://pixees.fr/jouons-avec-des-experiences-scientifiques/ for primary school students.\n\nInterventions\n• “Fête de la science: Mouans-Sartoux fête les sciences du quotidien” (October 10-11 for scholars: 8 classes, October 12 for public: 1000 people): M. Olivi animated the activity “jouer à transmettre des images” in collaboration with the “espace de l'art concret” https://www.espacedelartconcret.fr/.\n\n• “Stage MathC2+” (June 19-22): M. Olivi animated a workshop session on “How to analyze sounds with mathematical functions”.\n\n• V. L. Coli gave a talk “Archéologie et mathématiques : algorithmes pour l'identification des gestes des premiers potiers”, and participated to the organization of the exhibition of the ANR project CIMO, Forum des Sciences, 80 years of CNRS, October, CIV, Valbonne.\n\n• Fabien Seyfert gave a pitch on Factas activities during the visit of the company SICAME (March 27) and Martine Olivi gave a pitch on Factas activities for the celebration of InriaTech 10th birthday (April 3)\n\n• Internal action\n• S. Chevillard gave a talk “Réchauffement climatique : où en est-on ? où va-t-on ?” at the c@fé-in of the Research Center, November.\n\n• V. L. Coli gave a talk “Archéologie et mathématiques : algorithmes pour l'identification des gestes des premiers potiers” at the c@fé-in of the Research Center, October. She also participated to the organization of the “1er Colloque doctoral préhistoire, paléoenvironnement, archéosciences”, November, MSHS, Nice, https://www.cepam.cnrs.fr/evenement/1er-colloque-doctoral-prehistoire-paleoenvironnement-archeosciences/, where she gave a talk “Approches mathématiques pour la caractérisation des potteries néolithiques”.\n\n• M. Olivi co-organized about 10 “cafés scientifiques” (c@fé-in's and cafés Techno, 30 to 80 participants each) https://project.inria.fr/mastic/category/cafein/\n\n• Creation of media or tools for science outreach\n\nM. Olivi co-supervised the creation of new scientific wooden objects by SNJ AZUR (funds from APOCS region): pixel art and transmission of images https://pixees.fr/jouer-a-transmettre-des-images/, IA machine. She also co-supervised the creation of videos by Thibaut Ehlinger (internship) and Gregory Casala (apprenti), funds from Class'Code and the national network, see https://pixees.fr/pause-ta-science-une-chaine-pour-decrypter-les-objets-scientifiques/."
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https://studytiger.com/free-essay/coleman-managerial-report/ | [
"R. C/ Coleman distribute a variety of food products that are sold through grocery store and supermarket outlets. The company receives orders directly from the individual outlets, with a typical order requesting the delivery of several cases of anywhere from 20 to 50 different products. Under the company’s current warehouse operation, warehouse clerks dispatch order-picking personnel to fill each order and have the goods moved to the warehouse shipping area.\n\nBecause of the high labour costs and relatively low productivity of hand order-picking, management has decided to automate the warehouse operation by installing a computer-controlled order-picking system, along with a conveyor system for moving goods from storage to the warehouse shipping area. R. C. Coleman’s director for material management has been named the project manager in charge of the automated warehouse system. After consulting with members of the engineering staff and warehouse management personnel, the director compiled a list of activities associated with the project.\n\nThe optimistic, most probable and pessimistic times (on weeks) have also been provided for each activity. Activity A B C D E F G H I J K Description Determine equipment needs Obtain vendor proposals Select vendor Order system Design new warehouse layout Design warehouse Design computer interface Interface computer Install system Train system operators Test system Immediate Predecessor A, B C C E C D, F, G D, F H I, J Activity Optimistic Time Most Probable Pessimistic A B C 4 6 2 6 8 4 8 16 6 Page 1 D E F G H I J K 8 7 4 4 4 4 3 2 0 10 6 6 6 6 4 4 24 13 8 20 8 14 5 6 Managerial Report Develop a report that present the activity schedule and expected project completion time for the warehouse expansion project. Include a project network in the report. In addition, take into consideration the following issues: 1. R. C. Coleman’s top management established a required 40-week completion time for the project. Can this completion time be achieved? Include probability information in your discussion. What recommendations do you have if the 40-week completion time is required? 2.\n\nSuppose that management requests that activity times be shored to provide an 80 percent chance of meeting the 40-week completion time. If the variance in the project completion time is the same as you found in part (1), how much should he expected project completion time be shortened to achieve the goal of an 80 percent chance of completion within 40 weeks? 3. Using of expected activity times as the normal times and the following crashing information, determine the activity crashing decisions and revised activity schedule for the warehouse expansion project.\n\nCosts (\\$) Activity Crashed Activity Time (weeks) A B 4 7 Normal 1,000 1,000 Crashed 1,900 1,800 Page 2 C D E F G H I J K 2 8 7 4 5 4 4 3 3 1,500 2,000 5,000 3,000 8,000 5,000 10,000 4,000 5,000 2,700 3,200 8,000 4,100 10,250 6,400 12,400 4,400 5,500 END OF QUESTIONS MANAGERIAL REPORTS 1. 2. 3. 4. 5. 6. 7. Objective Introduction Methodology Analysis & Discussion Recommendation Conclusion Reference Appendix Page 3 1. Objective Introduction This report is the analysis study of R. C. Coleman Company; a company on distribution business with varieties of food products that are sold through grocery store and supermarket outlets.\n\nThe company receives orders directly from the individual outlets, with a typical orders requesting the delivery of several cases of anywhere from 20 to 50 different products. Present company’s current warehouse operation, the practise which has been utilized manually by warehouse clerks. The dispatch order-picking personnel are to fill each order and have the goods moved to the warehouse shipping area. Manual operation has been rationale as high labour cost and low productivity on the distribution system. Management has decided to change into automate the warehouse operation with the objective to improve on the operations and utput efficiency. R. C. Coleman’s top management has established requirement of 40-weeks completion time for the project of installation of computer-controlled order-picking system, this come along with conveyor system for moving goods from storage to the warehouse shipping area. The management also has drawn up others requirement and completion percentage at any particular time. The establishment of the report will be base objectively to analyse and examine all possibilities on the predetermine activities on the project network and completion time for the warehouse distribution upgrading project of R.\n\nC. Coleman Company. Hence, the result from this report is vital Page 4 information required to the company’s top management team to make effective decision for this project meeting with business objective and goal. 2. Methodology Upon appointed as Project Manager, Mr R. C. Coleman is responsible for planning, scheduling and controlling the project that consist of numerous separate jobs or task performed by a variety department and individual. The project team need to establish the project network and making the analysis based on the Quantitative Approaches to Decision Making processes.\n\nIn order to draw up project network, the team need information on the project activities involved and time required. As this project is rather new and they had never attempted before by the team, Mr Coleman have to establish the time by estimation. The activities are following the concept of probability distribution whereby they have to determine by estimating each activities time at a range of possible value. After a meeting with his project team, he has established a list of activities associated with the project as per below; Table 1: Project Activities and links Activity Description Immediate Predecessor A, B C A B C D\n\nDetermine equipment needs Obtain vendor proposals Select vendor Order system Page 5 E F G H I J K Design new warehouse layout Design warehouse Design computer interface Interface computer Install system Train system operators Test system C E C D, F, G D, F H I, J The incorporation of uncertainty activity with estimating time above is defined as; i. ii. Optimistic time (a) = the minimum activity time if everything progresses ideally. Most Probable time (m) = condition. iii. Pessimistic time (b) = the maximum activity time if significant delays are encountered Below table is the result for the uncertain activities time achieved.\n\nTable 2: Probable timing in weeks Time Optimistic Activity (a) (m) A B C 4 6 2 6 8 4 8 16 6 Most Probable (b) Pessimistic the most probable activity time under normal Page 6 D E F G H I J K 8 7 4 4 4 4 3 2 10 10 6 6 6 6 4 4 24 13 8 20 8 14 5 6 To ensure that the project is progressing as planned, Mr Coleman is advised to utilize and incorporate the analysis with the concept of probability distribution; Project Scheduling with Uncertain Activity Times. Such situation, the understanding and utilization of Program Evaluation and Review Technique (PERT) and Program of Critical Path Method (CPM) has proven to be extremely valuable.\n\nThe PERT/CPM system will furnish to project manager on the information of project planning, scheduling and the progress of the project so that he able to coordinate and monitoring the project and be complete as expected by the management. 4. Discussion Analysis When the top management of R. C. Coleman has decided to modernize and expand the current warehouse and its distribution system, there is certain project requirement and goal has been drawn up to the project manager and the team to find out the feasibility to achieve target goal on the completion timing and possible percentage.\n\nHence, the analysis will be representing on each part of the management requirement. There is two parts as per below: PART 1 Page 7 Apart from to find out the project completion time, the project Manager; R. C. Coleman has to find out the possibility that this expansion project could be complete within 40 weeks. Upon finding the result, he needs to come out on recommendation to supporting the result. Various parts below is to solving as per PERT/CPM steps procedures; PART 1 (a) Project Network Flow and Variance\n\nThe identified activities and links from the table one (1) above require illustration in form of work flow chart using CPM /PERT technique to analysis further in determining critical activities and critical path for the project. Figure three (3) below is the project network depicting the activities and linkage of immediate predecessors on each individual activity from start until complete. Figure 3: R. C. Coleman Project Network. Part 1 (b) Expected Project Time, Project Network and Completion Time. Page 8\n\nTo illustrate the project network with PERT/CPM procedures and finding the completion time, the three (3) estimate time above should be calculate and convert to expected time. Expected time (t) can be finding with the formula; Expected Time (t) = (a+4m+b) / 6 Upon finding the result on expected time, we could analysis to determine ‘start’ and ‘finish’ time for each activities from starting the project until completion. At the end of the flow we could establish that on the total time required to complete the project based on calculation of expected time. Below is the determine project network work flow and estimated completion time in weeks.\n\nFigure 4: Project network flow and completion time E 13 23 10 13 23 A 6 0 3 6 9 C 4 B 9 0 0 9 9 9 9 13 13 G 8 13 21 21 29 F 6 23 29 23 29 START H 6 29 35 29 35 I 7 29 36 32 39 K 4 39 43 39 43 FINISH (43 WEEKS) D 13 25 12 17 29 J 4 35 39 35 39 From the project network work flow at figure four (4) above, we have noted that the R. C. Coleman warehouse expansion project and conveyor system distribution upgrading will be completed at 43 weeks. Page 9 The team also needs to determine the slack associated with each activity. The term ‘Slack’ is the length of time of an activity can be delayed without increasing the project completion time.\n\nThe amount of slack of an activity is computed as follows; Slack = LS – ES = LF – EF Conversely, the activities which appear having zero slack is the critical activity whereby delaying in this process or steps could post an effect to total project schedule completion timing. As ruled, with uncertainty activity times, the team must aware that the differences between those three estimation time (Optimistic, most probable and pessimistic) could give great effect on the value of the variance. The term ‘variance’ is indication on the dispersion or variation in the activities time value. The value of variance could calculate with this formula; 2 = 2 Having greater value between tis value among the activities could give great reflect a high degree of uncertainty in the activity time. For easy to overview on the whole project, we have summarize the information into table manners as on Table five (5) below; Table 5: Project network summary information and critical path. Activity Expected Time Variance Earliest Start (ES) Latest Start (LS) 3 0 9 Earliest Finish (EF) 6 9 13 Latest Finish (LF) 9 9 13 Slack LS – ES 3 0 0 Critical path A B C 6 9 4 0. 44 2. 78 0. 44 0 0 9 YES YES Page 10 D E F G H I J K 12 10 6 8 6 7 4 4 7. 11 1 0. 44 7. 11 0. 44 2. 78 0. 11 0. 44 3 13 23 13 29 29 35 39 17 13 23 21 29 32 35 39 25 23 29 21 35 36 39 43 29 23 29 29 35 39 39 43 4 0 0 8 0 3 0 0 YES YES YES YES YES Part 1 (c) Critical Path and the Curve From the table above, we could note that the activity schedule for the warehouse expansion project which shows zero slack is the critical path for the project is; B – C – E – F – H – J – K. From the critical path shown, the expected time of the project is E (t) = tB + tC + tE + tF + tH + tJ + tK = 9 + 4 + 10 + 6 + 6 + 4 + 4 = 43 Weeks Hence, the variance in the project completion time is the sum of the variance on the critical path activities, which is; 2 = ? B + ? C + ? E + ? F + ? H + ? J + ? K = 2. 78 + 0. 44 + 1. 0 + 0. 44 + 0. 44 + 0. 11 + 0. 44 = 5. 65 ? = = 2. 38 Page 11 Figure 6: Standard Normal Distribution Curve STANDARD NORMAL DISTRIBUTION CURVE AREAS ? = 2. 38 EXPECTED COMPLETION TIME (T) 43 TIME (T) WEEKS Since the top management of R. C. Coleman allotted 40 week to complete the project, the probability distribution is; Z = = = - 1. 26 With the obtained value ‘z’ is -1. 26, we could enter the normal distribution from the table and we found out that; Project time (T) Z ? 40 weeks = -1. 26 Standard Normal distribution value = 0. 962 So, the probability that the R. C Coleman Project would complete 40 weeks or less is; Z = 0. 50 – 0. 3962 = 0. 1038 Page 12 ? 10. 38 % Figure 7: Standard Normal Distribution Curve with Target Weeks STANDARD NORMAL DISTRIBUTION CURVE AREAS When (T) = 40 z = (40 -43) / 2. 38 z = - 1. 26 ? = 2. 38 EXPECTED COMPLETION TIME 40 z = -1. 26 43 z=0 TIME (T) WEEKS With that result above, the top management of R. C Coleman is advised that the chances of completion of project on 40 weeks is doubtful and impossible to achieve as the percentage shows is very slim; approximate about 10 percent chances only.\n\nHence, we recommend that the top management consider shortening activities time by adding more resources into it and by applying ‘crashing’ technique on the appropriate activities. PART 2 Considering that the project could complete with 80 per cent at 40th Week with a variance reference is maintained same as on part (I), Mr. Coleman need to find out on the possibility time to be shortened to achieve of 80 per cent chances of completion of project is within 40 weeks. In part one (1) of the analysis, we have found that the probability to complete the project within 40 weeks is only at 10 per cent.\n\nAt this part, we need further analysis on the probability that the project will be meeting the 40-week completion time is at 80 per cent and a Normal Distribution Table with a mean of zero and a standard deviation of 1 is referred with the variance (z) is 2. 38; Based on the formula; Z = Page 13 Let say, P (T ? 40 weeks) = 0. 5 + Zn = 80% or 0. 8 0. 8 = 0. 5 + Zn Zn = 0. 8 – 0. 5 = 0. 3000 Using the new mean value or Zn = 0. 3000 and we will enter the table for normal distribution to find the closest value ‘z’. So, the closest value for Zn = 0. 300 is 0. 2995 where the closest normal probability distribution z at E (Tn) is equal to 0. 4. If the variance (z) is maintained at 2. 38, then in the project completion time is; Z = z = (T- E (Tn)) / 2. 38 = 0. 84 (T - E (Tn)) = 0. 84 x 2. 38 (40- E (Tn)) = 1. 9992 or 2. 0 E (Tn) = 40 – 2. 0 = 38 weeks. We have determine from the above calculation shows that the project completion time is shortened to 38 weeks in order to achieve the goal of an 80% chances in order to complete within 40 weeks. Figure 8 below showing the probability of the project to provide an 80 per cent probability chances of meeting 40-week of completion project time. Figure 8: Standard Normal Distribution Curve with Percentage Target Weeks.\n\nPage 14 STANDARD NORMAL DISTRIBUTION CURVE AREAS When (T) = 40 z = (40 -38) / 2. 38 z = 0. 84 ? = 2. 38 PLAN COMPLETION TIME @ 80% EXPECTED COMPLETION TIME 38 z=0 40 z = -1. 26 TIME (T) WEEKS 5. Conclusion Upon the completion on the calculation and analysis for both part; (1) and (2), Mr. R. C. Coleman; the appointed Project Manager on the upgrading the premises to automated warehouse system may advise to top management of the Company that on both part the result obtained on the requirement stipulated by the Management seems doubtful and is difficult to achieve.\n\nIn order to pursue the objective goal, top management is advised to consider and approve to Mr. R. C. Coleman to exercise shortening activity times at part (1). This shortening in other words is known as ‘crushing’ technique. Those activities time may require additional resources, either man power or financial in order to complete or meeting the percentage goal which has vision by the top management. Page 15 6. Reference An Introduction to Management Science; Quantitative Approaches to Decision Making. 13th Edition, Anderson, Sweeney, Williams & Martin. 7. Appendix R. C. COLEMAN PROJECT NETWORK\n\nPage 16 A Determine equipment need E Design new warehouse F Design warehouse I Install System G Design computer H Interface Computer J Train system operators START C Select vendor K Test system FINISH B Obtain vendor proposal D Order system TO FIND EXPECTED TIME (t); Time Activity Optimistic Most Probable a A B C D E F G H I J 4 6 2 8 7 4 4 4 4 3 m 6 8 4 10 10 6 6 6 6 4 b 8 16 6 24 13 8 20 8 14 5 Pessimistic Page 17 K 2 4 6 The incorporation of uncertainty activity with estimating time above is defined as; i. Optimistic time (a) = the minimum activity time if everything progresses ideally. ii.\n\nMost Probable time (m) = the most probable activity time under normal condition. iii. Pessimistic time (b) = the maximum activity time if significant delays are encountered Expected time (t) can be finding with the formula; Expected Time (t) = (a+4m+b) / 6 1. For activity time A, the time average is; (t) = (a+4m+b) / 6 (t) = (4+6(4) +8) / 6 (t) = 6 2. For activity time B, the time average is; (t) = (a+4m+b) / 6 (t) = (6+8(4) +16) / 6 (t) = 9 3. For activity time C, the time average is; (t) = (a+4m+b) / 6 (t) = (2+4(4) +6) / 6 (t) = 4 4. For activity time D, the time average is; (t) = (a+4m+b) / 6 (t) = (8+10(4) +24) / 6 (t) = 12 5.\n\nFor activity time E, the time average is ; (t) = (a+4m+b) / 6 (t) = (7+10(4) +13) / 6 (t) = 10 6. For activity time F, the time average is; (t) = (a+4m+b) / 6 (t) = (4+6(4) +8) / 6 (t) = 6 Page 18 7. For activity time G, the time average is; (t) = (a+4m+b) / 6 (t) = (4+6(4) +20) / 6 (t) = 8 8. For activity time H, the time average is; (t) = (a+4m+b) / 6 (t) = (4+6(4) +8) / 6 (t) = 6 9. For activity time I, the time average is; (t) = (a+4m+b) / 6 (t) = (4+6(4) +14) / 6 (t) = 7 10. For activity time J, the time average is; (t) = (a+4m+b) / 6 (t) = (3+4(4) +5) / 6 (t) = 4 11.\n\nFor activity time K, the time average is; (t) = (a+4m+b) / 6 (t) = (2+4(4) +6) / 6 (t) = 4 With uncertainty time, we need to find the variance in order to describe the dispersion or variation in the activity time values. The variance of the activity time is given by the formula; ?2 = 2 1. The variance for activity A is; ? 2A = 2 = (8-4/6)2 = (2/3)2 = 0. 44 2. The variance for activity B is; Page 19 ?2B = 2 = (16-6/6)2 = (10/6)2 = 2. 78 3. The variance for activity C is; ? 2C = 2 = (6-2/6)2 = (2/3)2 = 0. 44 4. The variance for activity D is; ? 2D = 2 = (24-8/6)2 = (16/6)2 = 7. 11 5. The variance for activity E is; ? E = 2 = (13-7/6)2 = (6/6)2 = 1 6. The variance for activity F is; ? 2F = 2 = (8-4/6)2 = (2/3)2 = 0. 44 7. The variance for activity G is; ? 2G = ? 2H = 2 = (20-4/6)2 = (16/6)2 = 7. 11 = (8-4/6)2 = (2/3)2 = 0. 44 8. The variance for activity H is; 2 9. The variance for activity I is; ? 2I = 2 = (14-4/6)2 = (10/6)2 = 2. 78 10. The variance for activity J is; ? 2J = 2 = (5-3/6)2 = (1/3)2 = 0. 11 11. The variance for activity K is; ? 2K = 2 = (6-2/6)2 = (2/3)2 = 0. 44 Hence, the table below is the summary from the calculation of expected time and the variance of each activity; EXPECTED TIME AND VARIANCE FOR THE R.\n\nC. COLEMAN COMPANY PROJECT ACTIVITES Activity Expected time (week) A 6 0. 44 Variance Page 20 B C D E F G H I J K 9 4 12 10 6 8 6 7 4 4 2. 78 0. 44 7. 11 1 0. 44 7. 11 0. 44 2. 78 0. 11 0. 44 Activity Expected Time 6 9 4 12 10 6 8 6 7 4 4 Variance Earliest Start (ES) 0 0 9 13 13 23 13 29 29 35 39 Latest Start (LS) 3 0 9 17 13 23 21 29 32 35 39 Earliest Finish (EF) 6 9 13 25 23 29 21 35 36 39 43 Latest Finish (LF) 9 9 13 29 23 29 29 35 39 39 43 Slack LS – ES 3 0 0 4 0 0 8 0 3 0 0 Critical path A B C D E F G H I J K 0. 44 2. 78 0. 44 7. 11 1 0. 44 7. 11 0. 44 2. 78 0. 11 0. 44\n\nYES YES YES YES YES YES YES From the table above we could note that the activity schedule for the warehouse expansion project which shows zero slack is the critical path for the project; B – C – E – F – H – J – K. From the critical path shown, the expected time of the project is E (t) = tB + tC + tE + tF + tH + tJ + tK = 9 + 4 + 10 + 6 + 6 + 4 + 4 = 43 WEEKS Page 21 Project network flow and completion time E 13 23 10 13 23 A 6 0 3 6 9 C 4 B 9 0 0 9 9 9 9 13 13 G 8 13 21 21 29 F 6 23 23 29 29 START H 6 29 29 35 35 I 7 29 36 32 39 K 4 39 39 43 43 FINISH (43 WEEKS) D 13 25 12 17 29 J 4 35 39 35 39\n\nHence, the variance in the project completion time is the sum of the variance on the critical path activities, which is ?2 = ? B + ? C + ? E + ? F + ? H + ? J + ? K = 2. 78 + 0. 44 + 1. 0 + 0. 44 + 0. 44 + 0. 11 + 0. 44 = 5. 65 ? = = 2. 38 Z = = Since the management allotted 40 week to complete the project, the probability distribution Z = = -1. 26 Page 22 Using the table for standard distribution had shown that the value area of 1. 26 is 0. 3962. So the probability of the project will be complete at 40 weeks is P (40 weeks) = 0. 5 -0. 3962 = 0. 1038 ? 10. 38 % Standard Normal Distribution Curve\n\nSTANDARD NORMAL DISTRIBUTION CURVE AREAS ? = 2. 38 EXPECTED COMPLETION TIME (T) 43 TIME (T) WEEKS Standard Normal Distribution Curve with Target Weeks Page 23 STANDARD NORMAL DISTRIBUTION CURVE AREAS When (T) = 40 z = (40 -43) / 2. 38 z = - 1. 26 ? = 2. 38 EXPECTED COMPLETION TIME 40 z = -1. 26 43 z=0 TIME (T) WEEKS Standard Normal Distribution Curve with Percentage Target Weeks. STANDARD NORMAL DISTRIBUTION CURVE AREAS When (T) = 40 z = (40 -38) / 2. 38 z = 0. 84 ? = 2. 38 PLAN COMPLETION TIME @ 80% EXPECTED COMPLETION TIME 38 z=0 40 z = -1. 26 TIME (T) WEEKS END OF REPORT Page 24"
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