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[ "Cody\n\n# Problem 45. Make a Palindrome Number\n\nSolution 309983\n\nSubmitted on 24 Aug 2013 by JerryJames\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% a = 124; b = 545; assert(isequal(find_palindrome(a),b))\n\na = 545\n\n2   Pass\n%% a = 150; b = 303; assert(isequal(find_palindrome(a),b))\n\na = 201 a = 303\n\n3   Pass\n%% a = 291; b = 6996; assert(isequal(find_palindrome(a),b))\n\na = 483 a = 867 a = 1635 a = 6996\n\n4   Pass\n%% a = 178; b = 15851; assert(isequal(find_palindrome(a),b))\n\na = 1049 a = 10450 a = 15851\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "# [Problem Set 2 Quantitative models of behavior] Problem 2: Decision strategy for flower sampling by bees\n\nLink of the iPython notebook for the code\n\n# PROBLEM 2: Simple decision strategy for flower sampling by bees.\n\nLet us consider the following experiment: a bee lands on flowers to collect as much nectar as possible, during two day (granting that it lands on $100$ flowers a day). There are two kinds of flowers:\n\n• blue flowers, which carry a nectar reward of\n\n• $r_b = 8$ during the first day\n• $r_b = 2$ during the second day\n• yellow flowers, which carry a nectar reward of\n\n• $r_y = 2$ during the first day\n• $r_y = 8$ during the second day\n\nThe bee’s internal estimate for the rewards is $m_b$ (resp. $m_y$) for the blue (resp. yellow) flowers. The bee chooses to land on a blue flower with probability (softmax-strategy):\n\np_b = \\frac{1}{1 + \\exp(β (m_y - m_b))}\n\nwhere $β$ is called the “exploitation-exploration trade-off” parameter (we will explain why in the following section).\n\n## 1. The softmax-strategy\n\nNote that $p_b$ is a sigmoid of $-(m_y - m_b)$ (for a fixed $β$) and a sigmoid of $-β$ (for a fixed $m_y - m_b$).", null, "Figure 4.1. - Plot of $p_b$ for different values of $β$\n\nThus:\n\n• the larger $m_y - m_b$, the higher the bee’s internal estimate of the yellow flower is compared to the blue one. As a consequence, the lower the probability $p_b$ of the bee landing on the blue one.\n\n• the lower the reward difference, the worse the bee considers the yellow flower to be, compared to the blue one, and the greater the probability $p_b$ of the bee landing on the blue flower.\n\n• in between: the bee’s behavior is a mix between exploiting what the bee deems to be the most nutritious flower and exploring the other one (which is what is expected).\n\nSymetrically:", null, "Figure 4.1. - Plot of $p_b$ for different values of $m_y-m_b$\n\n### The “exploitation-exploration trade-off” parameter $β$\n\n#### For $β ≥ 0$\n\nAs seen in the previous figures: the parameter $β$ plays the same role as the inverse temperature $β ≝ \\frac{1}{k_B T}$ in statistical physics. That is, for $β ≥ 0$:\n\n• When $β ⟶ +∞$: the bigger $β$ is (which corresponds, in physics, to a low temperature/entropy), the more the bee tends to exploit the seemingly most nutritious flower. As a result:\n\n• if $m_y-m_b > 0$ (i.e. the yellow flower seems more avantageous to the bee), the probability of the bee landing on the blue flower rapidly decreases to $0$\n• if $m_y-m_b < 0$ (i.e. the blue flower seems more advantageous to the bee), the probability of the bee landing on the blue flower rapidly increases to $1$\n• When $β ⟶ 0$: the lower $β$ is (high temperature/entropy in physics), the more the bee tends to explore the flowers.\n\nIndeed: as $β ⟶ 0$, $p_b$ becomes less and less steep, to such a point that it ends up being the constant function $1/2$ when $β=0$ (at this point, the bee does nothing but exploring, since landing on either of the flowers is equiprobable, no matter how nutritious the bee deems the flowers to be)\n\n#### For $β < 0$\n\nAs\n\np_b(-β, m_y-m_b) = \\frac{1}{1 + \\exp(-β (m_y-m_b))} = \\frac{1}{1 + \\exp(β (-(m_y-m_b)))} = p_b(β, -(m_y-m_b))\n\nThe curves for $-β < 0$ are symmetric to those for $β > 0$ with respect to the y-axis, which makes no sense from a behavioral point of vue: it means that the most nutritious a flower appears to the bee, the less likely the bee is to land on it (and conversely)!\n\n## 2. Dumb bee\n\nNow, let us assume that the bee is “dumb”, i.e. it does not learn from experience: throughout all the trials, the internal reward estimates remain constant: $m_y = 5, \\; m_b = 0$.", null, "Figure 2.1. - Dumb bee: Behavior for $β=0$", null, "Figure 2.2. - Dumb bee: Behavior for $β=0.5$", null, "Figure 2.3. - Dumb bee: Behavior for $β=1$\n\nSo it appears that\n\n• for $β = 0$: the bee’s behavior is the most exploratory one: the probability of landing on a blue flower as well as a yellow flower is $1/2$, irrespective of the internal estimates $m_b$ and $m_y$\n\n• for $β = 0.5$: this $β$ corresponds to a perfectly balanced exploration-exploitation tradoff, but here: as $m_y = 5$ is significantly higher that $m_b = 0$, $p_y ≝ 1-p_b » p_b$ (which are constant, as the internal estimates are constant)\n\n• for $β = 1$: the bee’s behavior is the most exploitative one. As $m_y > m_b$, $p_y ≃ 1$: the bee keep exploiting the flower it deems the most nutritious (i.e. the yellow ones).\n\nOn the whole, this behavior can reasonably be called “dumb”, since the bee takes no account of the actual rewards of the flowers whatsoever (its internal estimates remain constant and don’t depend on these actual rewards).\n\n## 3. Smart bee\n\nNow let us suppose that the bee is “smart”, i.e. it can learn from its experiences. Whenever it visits a flower, it updates its estimated reward as follows:\n\n\\begin{align*} m_b &→ m_b + ε(r_b − m_b)\\\\ m_y &→m_y +ε(r_y − m_y) \\end{align*}\n\nGiven a learning parameter $ε = 0.2$ and the initial assumptions about flower reward from above ($m_y = 5, m_b = 0$), simulate the bees sequence of choices during the two days. How do the reward estimates change over time? Explore the case of purely explorative behavior ($β = 0$) and the case of strongly exploitative behavior ($β = 1$). What do you observe?", null, "Figure 3.1. - Smart bee: Behavior for $β=0$", null, "Figure 3.2. - Smart bee: Behavior for $β=0.5$", null, "Figure 3.3. - Smart bee: Behavior for $β=1$\n\nIt appears that\n\n• for $β = 0$: no surprise, the bee’s behavior is the most exploratory one: the probability of landing on a blue flower as well as a yellow flower is $1/2$, irrespective of the internal estimates $m_b$ and $m_y$\n\n• for $β = 1$: the bee’s behavior is the most exploitative one. Similarly to the Rescola-Wagner rule, the updating rule makes the internal estimates converge toward the actual rewards of the flowers, and as result: the exploitative behavior causes the probability $p_b$ (resp. $p_y ≝ 1-p_b$) to match the evolution of the internal estimate $m_b$ (resp. $m_y$) when $m_b > m_y$ (resp. $m_y > m_b$): the bee exploit the flower type which has the highest current estimate.\n\n• for $β = 0.5$: it is a mixed behavior bewteen exploitation and exploration: compared to the case $β=1$, there are some discrepancies in the update of the internal estimates, since the bee may not always exploit the flower type which has the best current estimate (exploratory behavior).\n\nOn the whole, this behavior can reasonably be called “smart”, since the bee, depending on the value of $β$:\n\n• takes more or less into account the actual rewards of the flowers by exploiting the seemingly most nutritious flower (exploitation, increasing with $β$)\n\n• while, from time to time, completely ignoring its internal estimates and exploring the other flower (exploration, decreasing with $β$)\n\nTags:\n\nUpdated:" ]

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[ "551\n\nThe number obtained by interchanging the two digits of a two-digit number is less than the original number by 45. If the sum of the two digits of the number so obtained is 13, then what is the original number choose which one is correct?\n\nA. 49\n\nC. 83\n\nD. Either (a) or (b)\n\nE. None of these\n\n552\\\n\nThe sum of five consecutive even numbers of set x is 440. Find the sum of a different set of five consecutive integers whose second least number is 121 less than double the least number of set x choose which one is correct?\n\nA. 248\n\nC. 228\n\nD. 236\n\nE. None of these\n\n553\n\nIn a series of six consecutive even numbers, the sum of the second and sixth numbers is 24. What is the fourth number choose which one is correct?\n\nA. 8\n\nC. 6\n\nD. 14\n\nE. None of these\n\n554\n\nThe sum of the digits of a two-digit number is 12. The difference of the digits is 6. Find the number choose which one is correct?\n\nA. 93\n\nB. 39\n\nC. 75\n\nD. 48\n\nE. Either (a) or (b) (Answer)\n\n555\n\nThe tens digit of a two-digit number is two more than its unit digit. The two-digit number is 7 times the sum of the digits. Find the units digits choose which one is correct?\n\nA. 1\n\nC. 3\n\nD. 4\n\nE. None of these\n\n556\n\nSolve the equation for x : 6x – 27 + 3x = 4 + 9 – x (Answer)\n\nB. 5\n\nC. 6\n\nD. -4\n\nE. None of these\n\n557\n\nSolve the equation for x : 19(x + y) + 17 = 19(-x + y) – 21\n\nB. -2\n\nC. -3\n\nD. -4\n\nE. None of these\n\n558\n\nThe cost of 2 chairs and 3 tables is Rs.1300. The cost of 3 chairs and 2 tables is Rs.1200. The the cost of each table is more than that of each chair by__________ choose which one is correct?\n\nA. Rs.70\n\nB. Rs.75\n\nC. Rs.50\n\nD. Rs.60\n\n559\n\nThe denominator of a fraction is 1 less than twice the numerator. If the numerator and denominators are both increased by 1, the fraction becomes 3/5. Find the fraction choose which one is correct?\n\nA. 2/3\n\nB. 3/5\n\nC. 4/7\n\nE. None of these\n\n560\n\nThe cost of 10 kg of apples is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour choose which one is correct?\n\nA. Rs.849.40\n\nC. Rs.901.60\n\nD. Rs.815.20\n\nE. None of these\n\n561\n\nA question paper consists of five problems, each problem having three interval choices. In how many ways can a candidate attempt one or more problems choose which one is correct?\n\nA. 63\n\nB. 511\n\nD. 15\n\n562\n\nIn a class there are 20 boys and 25 girls. In how many ways can a boy and a girl be selected choose which one is correct?\n\nA. 400\n\nC. 600\n\nD. 20\n\n563\n\nIn how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE” choose which one is correct?\n\nA. 25\n\nB. 13\n\nC. 40\n\n564\n\nA bag contains nine yellow balls, three white balls and four red balls. In how many ways can two balls be drawn from the bag choose which one is correct?\n\nA. ⁹C₂\n\nB. =C₂\n\nC. ;⁶C₂ (  nswer)\n\nD. ;<C₂\n\n565\n\nHow many four digit even numbers can be formed using the digits {2, 3, 5, 1, 7, 9}\n\nB. 360\n\nC. 120\n\nD. 240\n\n566\n\nThe number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is_________ choose which one is correct?\n\nA. (6!)2\n\nB. 6! * ⁷P₆\n\nC. 2(6!)\n\nD. 6! * 7\n\n567\n\nA letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most__________ choose which one is correct?\n\nA. 216\n\nB. 243\n\nD. 729\n\n568\n\nA committee has 5 men and 6 women. What are the number of ways of selecting a group of eight people choose which one is correct?\n\nB. 185\n\nC. 205\n\nD. 225\n\n569\n\nWhat are the number of ways to select 3 men and 2 women such that one man and one Women are always selected choose which one is correct?\n\nA. 100\n\nB. 60\n\nD. 20\n\n570\n\nThe number of permutations of the letters of the word ‘MESMERISE’ is___________ choose which one is correct?\n\nB. 9!/(2!)3 3!\n\nC. 9!/(2!)2 (3!)2\n\nD. 5!/(2!)2 3!\n\n571\n\nThe number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places choose which one is correct?\n\nA. 720\n\nC. 120\n\nD. 36\n\n572\n\nA boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt choose which one is correct?\n\nA. 21\n\nB. 12\n\nC. 9\n\n573\n\nThere are 30 people in a group. If all shake hands with one another , how many handshakes are possible choose which one is correct?\n\nA. 870\n\nC. 30!\n\nD. 29! + 1\n\n574\n\nIn how many different ways can the letters of the word ‘OPTIC  L’ be arranged so that the vowels always come together choose which one is correct?\n\nA. 120\n\nC. 4320\n\nD. 2160\n\n575\n\nIn how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women choose which one is correct?\n\nB. 90\n\nC. 126\n\nD. 45\n\n576\n\nIn how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position choose which one is correct?\n\nA. 9\n\nC. 14\n\nD. 18\n\n577\n\nThere are two identical red, two identical black, and two identical white balls.In how many different ways can the balls be placed in the cells (Each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells choose which one is correct?\n\nA. 15\n\nB. 18\n\nD. 30\n\n578\n\nTwelve people from a club, by picking lots. One of them will host a dinner for all once in a month. The number of dinners a particular member has to host in one year is________ choose which one is correct?\n\nA. One\n\nB. Zero\n\nC. Three\n\n579\n\nA person ordered 5 pairs of black socks and some pairs of brown socks. The price of a black pair was thrice that of a brown pair while preparing the bill the bill clerk inter-changed the number of black and brown pairs by mistake which increased the bill by 100% what was the number of pairs of brown socks in the original order choose which one is correct?\n\nA. 10\n\nB. 15\n\nC. 20\n\n580\n\nThree dice (each having six faces with each face having one number from 1 or 6) are rolled.\n\nWhat is the number of possible outcomes such that at least one dice shows the number 2 choose which one is correct?\n\nA. 36\n\nB. 81\n\nD. 116\n\n581\n\nA mixed doubles tennis game is to be played two teams(each consists of one male and one female) There are four married couples. No team consists of a husband and his wife. What is the maximum number of games that can be played choose which one is correct?\n\nA. 12\n\nB. 48\n\nC. 36\n\n582\n\nIn a question paper, there are four multiple choice type questions, each question has five choices with only one choice for it’s correct answer. What is the total number of ways in which a candidate will not get all the four answers correct choose which one is correct?\n\nA. 19\n\nB. 120\n\nD. 1024\n\n583\n\nIn how many different ways can six players be arranged in a line such that two of them, Asim and Raheem are never together choose which one is correct?\n\nA. 120\n\nB. 240\n\nC. 360\n\n584\n\nSix points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points choose which one is correct?\n\nA. 29\n\nC. 55\n\nD. 30\n\n585\n\n6 of 5/8 / 5/8 – 1/8 = ______ choose which one is correct?\n\nA. 6 1/8\n\nB. 5 3/4\n\nC. 6 3/4\n\n586\n\n(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = _________ choose which one is correct?\n\nA. 1\n\nB. 2\n\nD. 66/17\n\n587\n\n[(62.25/2.5) + (12.5/0.5)] / [(0.75 * 9) + (3.5 * 3.5)] = _________ choose which one is correct?\n\nA. 1 17/38\n\nB. 1 38/17\n\nC. 2 38/17\n\n587\n\n(180/37 of 14.8) / (17/60.1 of 180.30) = _______ choose which one is correct?\n\nB. 1.6\n\nC. 1.5\n\nD. 1.2\n\n588\n\n45/100 * 125 + 22.22 = ________ choose which one is correct?\n\nB. 76\n\nC. 74\n\nD. 72\n\n589\n\n12 3/4 * (1 2/17 / 8 1/6) / 5 4/3 = __________ choose which one is correct?\n\nA. 1\n\nC. 6/17\n\nD. 27/17\n\n590\n\nWhich of the following groups of fractions is in descending order choose which one is correct?\n\nA. 8/15, 9/13, 6/11\n\nB. 8/15, 6/11, 9/13\n\nD. 6/11, 8/15, 9/13\n\n591\n\nThe value of (2.75)3 – (2.00)3 – (0.75)3 is__________ choose which one is correct?\n\nA. 3 * 2.75 * 2.75 * 2.75\n\nB. 3 * 2.00 * 2.00 * 2.00\n\nC. 3 * 2.75 * 0.75 * 2.75\n\n592\n\n{20 – [7 – (3 -2)] + 1/3 (4.2)} / 1.4 = ________ choose which one is correct?\n\nB. 12\n\nC. 14\n\nD. 19\n\n593\n\nJameel gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Nasir gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates,then at which floor will their paths cross choose which one is correct?\n\nA. 19\n\nB. 28\n\nD. 37\n\n594\n\nA man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be_________ choose which one is correct?\n\nA. 22\n\nB. 23\n\nC. 24\n\n595\n\nFree notebooks were distributed equally among children of a class. The number of notebooks each child got one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed choose which one is correct?\n\nA. 256\n\nB. 432\n\nD. 640\n\n596\n\nIn a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hour for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for______ choose which one is correct?\n\nA. 160\n\nC. 180\n\nD. 195\n\n597\n\n105.126 * 35.201 – 90.23 * 3 + 55.11 * 27.01 = ____________ choose which one is correct?\n\nB. 40000\n\nC. 271\n\nD. 5996\n\n598\n\n(0.9 * 0.9 – 0.8 * 0.8)/1.7 = _________ choose which one is correct?\n\nB. 0.2\n\nC. 0.3\n\nD. 0.4\n\n599\n\n5/8 * 2(2/5 – 5/4) / 1 1/4 = _________ choose which one is correct?\n\nB. 4/5\n\nC. 6/5\n\nD. 8/5\n\n600\n\n12(13/18) + 16(11/27) – 10(5/9) = _________ choose which one is correct?\n\nA. 182/3\n\nC. 18\n\nD. 201/2\n\n601\n\nPositive integers indicated by x and y satisfy 3 1/x * y 2/5 = 13 3/4, the fractions being in their lowest terms, then x =  choose which one is correct? and y = ____________ choose which one is correct?\n\nA. x = 4, y = 8\n\nB. x = 4, y = 4\n\nC. x = 2, y = 4\n\nD. x = 8, y = 4 (Answer)\n\n602\n\nIf (461 + 462 + 463 + 464) is divisible by  choose which one is correct?, then  choose which one is correct? =_____________ choose which one is correct?\n\nA. 3\n\nB. 11\n\nC. 13\n\n603\n\nrectangular measure 8cm on length and it’s diagonals measures 10cm what is the perimeter of the rectangular choose which one is correct?\n\nA. 36cm\n\nB. 38cm\n\nD. 18cm\n\n604\n\nThe perimeter of a floor of a room is 18m.What is the area of four walls of the room if it; a height is 3m choose which one is correct?\n\nA. 21 Sq.m\n\nB. 42 Sq.m\n\nD. 108 Sq.m\n\n605\n\nparallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long then its area is:__________ choose which one is correct?\n\nA. 480 sq.m\n\nB. 320 sq.m\n\nC. 600√15 sq.m (  nswer)\n\nD. 450√15\n\n606\n\nIf a square and a rhombus stand on the same base then the ratio of the areas of the squares and the rhombus is:_________ choose which one is correct?\n\nA. greater than 1\n\nC. equal to 1/2\n\nD. equal to ¼\n\n607\n\nIn a rhombus whose area is 144 sq.cm one of its diagonals is twice as long as the other. The lengths of its diagonals are:_________ choose which one is correct?\n\nA. 24 cm,48cm\n\nC. 6√2cm,12√2cm\n\nD. 6cm,12cm\n\n608\n\nThe length of a rope by which a cow must be tethered in order that she may be able to graze an area of 9856 sq.m is:________ choose which one is correct?\n\nB. 64m\n\nC. 88m\n\nD. 168m\n\n609\n\nThe circumferences of two concentric circles are 176 m and 132 cm respectively. What is the difference between their radii choose which one is correct?\n\nA. 5 m\n\nC. 8 m\n\nD. 44 m\n\n610\n\nThe diameter of a circle is 105 cm less than the circumference what is the diameter of the circle choose which one is correct?\n\nA. 44 cm\n\nB. 46 cm\n\nC. 48 cm\n\n611\n\nA circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ration of 6:5 The smaller side if the rectangle is:_________ choose which one is correct?\n\nA. 30 cm\n\nC. 72 cm\n\nD. 132 cm\n\n612\n\nIf the diameter of a circle is increased by 100 % its area is increased by:_________ choose which one is correct?\n\nA. 100%\n\nB. 200%\n\nD. 400%\n\n613\n\nIf the radius of a circle be reduced by 50% its area is reduced by:_________ choose which one is correct?\n\nA. 25 %\n\nB. 50%\n\nD. 100%\n\n614\n\nArea of a square with side x is equal to the area of a triangle with base x. The altitude of the triangle is:________ choose which one is correct?\n\nA. x/2\n\nB. x\n\nD. 4x\n\n615\n\nIf the diagonal of a square is doubled. How does the area of the square change choose which one is correct?\n\nB. Becomes threefold\n\nC. Becomes twofold\n\nD. none of the above\n\n616\n\nIf the base of rectangle is increased by 10 % and the area is unchanged then the corresponding altitude must be decreased by:________ choose which one is correct?\n\nB. 10%\n\nC. 11%\n\nD. 11 1/9 %\n\n617\n\nThe length of a rectangle is twice its breadth, if length is decreased by 5 cm and the breadth is increased by 5 cm. The length of the rectangle is_________ choose which one is correct?\n\nA. 20 cm\n\nB. 30 cm\n\nD. 50 cm\n\n618\n\nD. 135m × 32m\n\n619\n\nWhich one of the following has a greater perimeter than the rest choose which one is correct?\n\nA. A square with an area of 36 sq.cm\n\nB. An Equilateral Triangle with a side of 9 cm\n\nC. A rectangle with 10 cm as length and 40 sq.cm as area (Answer)\n\nD. A circle with a radius of 4 cm\n\n620\n\nThe perimeter of a semi circle is 144 cm then the radius is_________ choose which one is correct?\n\nA. 25 cm\n\nC. 30 cm\n\nD. 35 cm\n\n621\n\nThe radius of a semicircle is 6.3 cm then its perimeter is choose which one is correct?\n\nA. 35.4 cm\n\nC. 32 cm\n\nD. 30 cm\n\n622\n\nWhat is the ratio between perimeters of two squares one having 3 times the diagonal then the other choose which one is correct?\n\nA. 4:8\n\nB. 1:4\n\nC. 6:2\n\n623\n\nA rope of which a calf is tied is increased from 12 m to 23 m, how much additional grassy ground shall it graze choose which one is correct?\n\nA. 1120 m2\n\nB. 1250 m2\n\nD. 1200 m2\n\n624\n\nThe cross-section of a cannel is a trapezium in shape. If the channel is 10 m wide at the top and 6 m wide at the bottom and the area of cross-section is 640 sq m, the depth of channel is_________ choose which one is correct?\n\nA. 20 m\n\nB. 60 m\n\nC. 40 m\n\n625\n\nThe side of a rhombus is 26 m and the length of one of its diagonals is 20 m. The area of the rhombus is_______ choose which one is correct?\n\nA. 529 sq m\n\nB. 240 sq m\n\nC. 260 sq m\n\n626\n\nAn agricultural field is in the form of a rectangle of length 20m and width 14m. A pit 6m long 3m wide and 2.5m deep is dug in a corner of the field and earth taken out of the pit is spread uniformly over. The remaining area of the field. The level of the field has been raised by________ choose which one is correct?\n\nA. 15.16cm\n\nB. 16.17cm\n\nD. 18.19m\n\n627\n\nA tank is 7m long and 4m wide at what speed should water run through pipes 5cm broad and 4cm deep so that in 6 hours and 18 minutes, the water level in the tank rises by 4.5cm choose which one is correct?\n\nA. 12km/hr\n\nC. 14km/hr\n\nD. None of these\n\n628\n\nWhat part of a ditch 48m long 16.5 broad and 4m deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4m and length 56m choose which one is correct?\n\nA. 1/9\n\nC. 7/9\n\nD. 8/9\n\n629\n\nFind the least number which when divided by 35 leaves remainder 25, when divided by 25 leaves remainder 15 and when divided by 15 leaves remainder 5 choose which one is correct?\n\nA. 420\n\nC. 435\n\nD. 518\n\nPage 285\n\n630\n\nWhat is the least number which when divided by 8, 12, 18 and 24 leaves the remainders 4, 8, 14 and 20 respectively choose which one is correct?\n\nA. 78\n\nC. 58\n\nD. None\n\n631\n\nThe wheels revolve round a common horizontal axis. They make 15, 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position choose which one is correct?\n\nB. 2 min\n\nC. 3 min\n\nD. None\n\n632\n\nTwo numbers 4242 and 2903 when divided by a certain number of three digits, leave the same remainder. Find the number choose which one is correct?\n\nA. 101\n\nC. 107\n\nD. 113\n\n633\n\nThere are four prime numbers written in ascending order of magnitude. The product of the The first three are 385 and of the last three are 1001. Find the fourth number choose which one is correct?\n\nB. 7\n\nC. 11\n\nD. 13\n\n634\n\nA room is 4 meters 37 cm long and 3 meters 23cm broad. It is required to pave the floor with minimum square slabs. Find the number of slabs required for this purpose choose which one is correct?\n\nA. 485\n\nB. 431\n\nD. 381\n\n635\n\nLCM of two numbers is 138. But their (Greatest common divisor) GCD is 23. The numbers are in a ratio 1:6. Which is the largest number amongst the two choose which one is correct?\n\nA. 46\n\nC. 69\n\nD. 23\n\n636\n\nGreatest Common Divisor of two numbers is 8 while their Least Common Multiple is 144. Find the other number if one number is 16.\n\nA. 108\n\nB. 96\n\nD. 36\n\n637\n\nThe L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is_________ choose which one is correct?\n\nA. 28\n\nB. 32\n\nD. 64\n\n638\n\nThe smallest fraction, which each of 6/7, 5/14, 10/21 will divide exactly is_________ choose which one is correct?\n\nB. 30/98\n\nC. 60/147\n\nD. 50/294\n\n639\n\nWhat will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 choose which one is correct?\n\nA. 196\n\nC. 1260\n\nD. 2520\n\n640\n\nWhich of the following fractions is the largest choose which one is correct?\n\nB. 13/16\n\nC. 31/40\n\nD. 63/80\\\n\n641\n\nThe H.C.F of two numbers is 23 and the other two factors of their L.C.M are 13 and 14. The the larger of the two numbers is___________ choose which one is correct?\n\nA. 276\n\nB. 299\n\nD. 345\n\n642\n\nFind the highest common factor of 36 and 84.\n\nA. 4\n\nB. 6\n\nD. 18\n\n643\n\n1095/1168 when expressed in simplest form is:\n\nA. 13/16\n\nC. 17/26\n\nD. 25/26\n\n644\n\nWhich of the following has the most number of divisors choose which one is correct?\n\nA. 99\n\nB. 101\n\nD. 182\n\n645\n\nCalculate the number of bricks, each measuring 25 cm * 15 cm * 8 cm required to construct a wall of dimensions 10 m * 4 m * 5 m when 10% of its volume is occupied by mortar choose which one is correct?\n\nA. 4000\n\nB. 5000\n\nD. 7000\n\nD. 1220 cm2\n\n646\n\nThe slant height of a cone is 12 cm and radius of the base is 4cm, find the curved surface of the cone.\n\nA. 74 π cm2\n\nB. 36 π cm2\n\nC. 48 π cm2 (  Answer)\n\nD. 24 π cm2\n\n647\n\nThe area of a base of a cone is 30 cm2. If the height of the cone is 6cm, find its volume choose which one is correct?\n\nA. 120 cm3\n\nB. 40 cm3\n\nC. 50 cm3\n\n648\n\nThe volumes of two cubes are in the ratio 27: 125, what shall be the ratio of their surface areas choose which one is correct?\n\nA. 6:25\n\nB. 3:5\n\nD. 16:25\n\n649\n\nI. x2 + 9x + 20 = 0,\n\nII. y2 + 5y + 6 = 0 to solve both the equations to find the values of x and y choose which one is correct?\n\nA. If x < y (Answer)\n\nB. If x > y\n\nC. If x ≤ y\n\nD. If x ≥ y\n\n650\n\nI. x2 – x – 42 = 0,\n\nII. y2 – 17y + 72 = 0 to solve both the equations to find the values of x and y choose which one is correct?\n\nA. If x < y (Answer)\n\nB. If x > y\n\nD. If x ≥ y\n\nC. If x ≤ y\n\n651\n\nFind the quadratic equations whose roots are the reciprocals of the roots of 2×2 + 5x + 3 = 0 choose which one is correct?\n\nA. 3×2 + 5x – 2 = 0\n\nB. 3×2 + 5x + 2 = 0 (Answer)\n\nC. 3×2 – 5x + 2 = 0\n\nD. 3×2 – 5x – 2 = 0\n\n652\n\nIf a and b are the roots of the equation x2 – 9x + 20 = 0, find the value of a2 + b2 + ab choose which one is correct?\n\nA. -21\n\nB. 1\n\nD. 21\n\n653\n\nThe arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of they secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of the remaining 55% is___________ choose which one is correct?\n\nA. 45\n\nB. 50\n\nD. 54.6 approx.\n\n654\n\nThe average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is___________ choose which one is correct?\n\nB. 23 years\n\nC. 28.5 years\n\nD. 29.3 years\n\n655\n\nSarfaraz obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks choose which one is correct?\n\nA. 65\n\nB. 69\n\nC. 72\n\n656\n\nA total of 3000 chocolates were distributed among 120 boys and girls such that each boy received 2 chocolates and each girl received 3 chocolates. Find the respective number of boys and girls choose which one is correct?\n\nA. 70, 50\n\nC. 50, 70\n\nD. 40, 80\n\n657\n\nWhat are Prime Numbers choose which one is correct?\n\nA. Which can be divided by Odd Numbers\n\nB. Which can be divided by Even Numbers\n\nC. Which can be divided by Number 1 & by itself Number (Answer)\n\nD. Which can be divided by any Number\n\n658\n\nTwelve inches equal to how many centimeters choose which one is correct?\n\nB. 29\n\nC. 34.12\n\nD. 24\n\n659\n\nWho is Called the Father of Calculus choose which one is correct?\n\nA. Al-Khwarizmi\n\nB. Aristotle\n\nC. Herodotus\n\n670\n\n1 inch is equal to__________ choose which one is correct?\n\nA. 2.4 cm\n\nB. 2.5 cm\n\nD. None\n\n671\n\nIf the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:\n\nA. 53.33\n\nC. 55\n\nD. None of these\n\n672\n\nMasood purchased a plot for Rs. 8,000. He sells the plot to Aftab at a profit of 20%. Aftab in turn sells that plot to Zahid at a loss of 20%. The plot costs Zahid  choose which one is correct?\n\nA. Rs. 12,000\n\nB. Rs. 10,000\n\nC. Rs. 8670" ]

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[ "## 2.2. Equation technique in CVP analysis\n\nWe just saw how to calculate the volume of minimum sales in units and dollars to break-even (i.e., have zero profits). What if we want to know how many valves need to be sold to earn a \\$30,000 profit? What about a \\$40,000 profit?\n\nThe answer can be found by using the same equation we used before:\n\nSales = Variable Costs + Fixed Costs + Profits\n\nNote that:\n\nSales = Sales Price per Unit x Unit Sales\n\nVariable Costs = Variable Cost per Unit x Unit Sales\n\nIf we replace Sales and Variable Costs in the equation, we will get the following result:\n\nSelling Price per Unit x Unit Sales = Variable Costs per Unit x Unit Sales + Fixed Costs + Profits\n\nWhen we further re-arrange the equation, we will obtain the following:\n\n Unit Sales = Fixed Costs + Profits Selling Price per Unit - Variable Costs per Unit\n\nUsing this formula, we can calculate how many units should be sold to generate \\$30,000 in profits for Friends Company:\n\n Unit Sales = \\$10,000 + \\$30,000 = 20,000 units \\$5 - \\$3\n\nThe same formula can be used when the target profit is \\$40,000:\n\n Unit Sales = \\$10,000 + \\$40,000 = 25,000 units \\$5 - \\$3\n\nThe 20,000 and 25,000 units can be converted to dollars as follows: \\$100,000 (\\$5 x 20,000 units) and \\$125,000 (\\$5 x 25,000 units), respectively.\n\nWe can check our answers for the desired \\$30,000 and \\$40,000 profits by using the same formula we applied before:\n\nProfits = Selling Price per Unit x Unit Sales - Variable Costs per Unit x Unit Sales - Fixed Costs\n\nFor the 20,000 units sold, the profits will be calculated as follows:\n\nProfits = \\$5 x 20,000 - \\$3 x 20,000 - \\$10,000 = \\$30,000\n\nAnd for the 25,000 units sold, the profits can be determined as shown below:\n\nProfits = 25,000 x \\$5 - 25,000 x \\$3 - 10,000 = \\$40,000\n\nIn both cases, we arrived at the desired level of profits and this proves that our calculations were correct.\n\nNow when we know that we've got to sell 20,000 units (or generate \\$100,000 in sales) to earn the \\$30,000 profits and 25,000 units (or \\$125,000) to make \\$40,000 of profits, Friends Company can assess (a) if it is possible considering the market situation; (b) what level of sales is more realistic in the company's situation; (c) what amount of resources the company needs; and (d) if Friends Company needs to hire more employees to generate a desired level of profits.\n\nNot a member?\nSee why people join our\nonline accounting course:\nLecture Contents:" ]

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[ "# Question 994e2\n\nMar 23, 2017\n\n$30.$g\n\n#### Explanation:\n\nA 3.0 M solution of NaOH has 3.0 moles of NaOH per liter of solution. There are 0.25 L of solution (250\"mL\" * (1\"L\")/(1000\"mL\"))#, so there are $0.25 \\text{L\" * 3.0\"mol/L\" = 0.75\"mol}$ of NaOH. The molar mass of NaOH is 40.0 g/mol, so there are $0.75 \\text{mol\" * 40.0\"g/mol\" = 30\"g}$ of NaOH, 30. g with the right number of significant figures.\n\nMar 24, 2017\n\n30 g.\n\n#### Explanation:\n\nIt's a 3.0 M solution so 1 litre of solution contains 3 moles of NaOH. 250 ml of solution therefore contains 0.25 x 3 = 0.75 moles of NaOH.\n\nThe molar mass of NaOH is 39.9971 g/mol, so 250 ml of this solution contains 0.75 x 39.9971 = 29.99 g, or if you round it up 30.0 g." ]

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[ "In logistic regression, the probability or odds of the response variable (instead of values as in linear regression) are modeled as function of the independent variables. For example, prediction of death or survival of patients, which can be coded as 0 and 1, can be predicted by metabolic markers. Logistic regression model When I use sm.Logit to predict results, do you know how I go about interpreting the results? This will create a new variable called pr which will contain the predicted probabilities. First, we try to predict probability using the regression model. Luca Massaron is a data scientist and a research director specializing in multivariate statistical analysis, machine learning, and customer insight. I used this to get the points on the ROC curve: from sklearn import metrics fpr, tpr, thresholds = metrics.roc_curve(Y_test,p) Instead we could include an inconclusive region around prob = 0.5 (in binary case), and compute the prediction table only for observations with max probabilities large enough. Conclusion: Logistic Regression is the popular way to predict the values if the target is binary or ordinal. The margins command (introduced in Stata 11) is very versatile with numerous options. About the Book Author. Logistic Regression. In the logit model the log odds of the outcome is modeled as a linear combination of the predictor variables. This page provides information on using the margins command to obtain predicted probabilities.. Let’s get some data and run either a logit model or a probit model. I ran a logistic regression model and made predictions of the logit values. Note that classes are ordered as they are in self.classes_. Exponentiating the log odds enabled me to obtain the first predicted probability obtained by the effects package (i.e., 0.1503641) when gre is set to 200, gpa is set to its observed mean value and the dummy variables rank2, rank3 and rank4 are set to their observed mean values. Logistic regression, also called a logit model, is used to model dichotomous outcome variables. Just remember you look for the high recall and high precision for the best model. Instead of two distinct values now the LHS can take any values from 0 to 1 but still the ranges differ from the RHS. His topics range from programming to home security. How can logit … You can provide multiple observations as 2d array, for instance a DataFrame - see docs.. The first column is the probability that the entry has the -1 label and the second column is the probability that the entry has the +1 label. I looked in my data set and it was 0, and that particular record had close to 0 … Prediction tables for binary models like Logit or Multinomial models like MNLogit, OrderedModel pick the choice with the highest probability. For instance, I saw a probability spit out by Statsmodels that was over 90 percent, so I was like, great! and the inverse logit formula states $$P=\\frac{OR}{1+OR}=\\frac{1.012}{2.012}= 0.502$$ Which i am tempted to interpret as if the covariate increases by one unit the probability of Y=1 increases by 50% - which I assume is wrong, but I do not understand why. For linear regression, both X and Y ranges from minus infinity to positive infinity.Y in logistic is categorical, or for the problem above it takes either of the two distinct values 0,1. After that you tabulate, and graph them in whatever way you want. Version info: Code for this page was tested in Stata 12. The precision and recall of the above model are 0.81 that is adequate for the prediction. You can provide new values to the .predict() model as illustrated in output #11 in this notebook from the docs for a single observation. John Paul Mueller, consultant, application developer, writer, and technical editor, has written over 600 articles and 97 books.\n2020 statsmodels logit predict probability" ]

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[ "## Convert fuss [German] to digit\n\n fuss digit\n\nHow many fuss in 1 digit? The answer is 0.060111364211592.\nWe assume you are converting between fuss [German] and digit.\nYou can view more details on each measurement unit:\nfuss or digit\nThe SI base unit for length is the metre.\n1 metre is equal to 3.1637560111364 fuss, or 52.631578947368 digit.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between fuss [German] and digits.\nType in your own numbers in the form to convert the units!\n\n## Quick conversion chart of fuss to digit\n\n1 fuss to digit = 16.63579 digit\n\n2 fuss to digit = 33.27158 digit\n\n3 fuss to digit = 49.90737 digit\n\n4 fuss to digit = 66.54316 digit\n\n5 fuss to digit = 83.17895 digit\n\n6 fuss to digit = 99.81474 digit\n\n7 fuss to digit = 116.45053 digit\n\n8 fuss to digit = 133.08632 digit\n\n9 fuss to digit = 149.72211 digit\n\n10 fuss to digit = 166.35789 digit\n\n## Want other units?\n\nYou can do the reverse unit conversion from digit to fuss, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## Definition: Digit\n\nunit of length equal to about 3/4 of an inch\n\n## Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]

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[ "Solutions by everydaycalculation.com\n\n## Compare 75/18 and 21/98\n\n1st number: 4 3/18, 2nd number: 21/98\n\n75/18 is greater than 21/98\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 18 and 98 is 882\n2. For the 1st fraction, since 18 × 49 = 882,\n75/18 = 75 × 49/18 × 49 = 3675/882\n3. Likewise, for the 2nd fraction, since 98 × 9 = 882,\n21/98 = 21 × 9/98 × 9 = 189/882\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 3675/882 > 189/882 or 75/18 > 21/98\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 ]

[ "# 集合的基本运算学案", null, "1. 并集\n\nA B A(B) A B B A B A\n\nA ∪\n\nB B A ?", null, "3.补集\n\nA\n\nU\n\nC U A\n\n4.求集合的并、交、补是集合间的基本运算，运算结果仍然还是集合，区分交集与并集的关键\n\n5.集合基本运算的一些结论：\n\nA∩B?A，A∩B?B，A∩A=A，A∩?=?,A∩B=B∩A\n\nA?A∪B，B?A∪B，A∪A=A，A∪?=A,A∪B=B∪A\n\n（C U A）∪A=U，（C U A）∩A=?\n\nA\n\nB\n\nB\n\nA\n\n-1 3 5 9 x", null, "【课堂练习】\n\n１.已知全集{}{}{}0,1,2,4,6,8,10,2,4,6,1U A B ===,则()U C A B ?=( ) Ａ\n\n{}0,1,8,10 Ｂ {}1,2,4,6 Ｃ {}0,8,10\n\nＤ Φ\n\n２.集合{}{}\n\n2\n\n1,4,,,1A x B x A B B ==?=且,则满足条件的实数x 的值为 ( )\n\nＡ １或０ Ｂ １,０,或２ Ｃ ０,２或－２ Ｄ １或２ 3.若{}{}{}0,1,2,1,2,3,2,3,4A B C ===???则(A B)(B C)＝ ( ) Ａ {}1,2,3 Ｂ\n\n{}2,3\n\n{}2,3,4 Ｄ {}1,2,4\n\n4.设集合{}{}|91,|32A x x B x x A B =-<<=-<\n\nA （-∞，-1）\n\nB （-1，-23）\n\nC （-2\n\n3\n\n,3）D (3,+∞)\n\n2、（广东）设集合U {1,23,4,5,6}=，，M {1,2,4}=则M U =e（ ）\n\nA ．U\n\nB ．{1,3,5}\n\nC ．{3,5,6}\n\nD ．{2,4,6}\n\n3、（湖南）设集合M={-1,0,1}，N={x|x 2≤x}，则M ∩N=（ ）\n\nA.{0}\n\nB.{0,1}\n\nC.{-1,1}\n\nD.{-1,0,0}\n\n4、（辽宁）已知全集U=｛0,1,2,3,4,5,6,7,8,9｝，集合A=｛0,1,3,5,8｝，集合B=｛2,4,5,6,8｝，则\n\n)()(B C A C U U 为（ ）\n\n(A){5,8} (B){7,9} (C){0,1,3} (D){2,4,6}\n\n5、（全国）已知集合{1,3,}A m =，{1,}B m =，A B A =，则m = （A ）0或3 （B ）0或3 （C ）1或3 （D ）1或3（ ）\n\n6、（山东）已知全集\n\n={0,1,2,3,4}，集合A={1,2,3,}，B={2,4} ，则（CuA ）\n\nB 为（ ）\n\nA {1,2,4}\n\nB {2,3,4}\n\nC {0,2,4}\n\nD {0,2,3,4} 7、（陕西）集合{|lg 0}M x x =>，2{|4}N x x =≤，则M\n\nN = （ ）\n\n（A ） (1,2) （B ） [1,2) （C ） (1,2] （D ） [1,2]", null, "8、（新课标）已知集{1,2,3,4,5}A =,{(,),,}B x y x A y A x y A =∈∈-∈；,则B 中所含元素的个数为（ ）\n\n()A 3 ()B 6 ()C 8 ()D 10\n\n9、（浙江）设集合A ＝{x|1＜x ＜4}，B ＝{x|x 2－2x －3≤0}，则A ∩(C R B)＝（ ） A ．(1，4) B ．(3，4) C ．(1，3) D ．(1，2)\n\n10、（上海）若集合}012|{>+=x x A ，}2|1||{<-=x x B ，则=B A 。 11、（四川）设全集{,,,}U a b c d =，集合{,}A a b =，{,,}B b c d =，则___________。\n\n12、（天津）已知集合={||+2|<3}A x R x ∈，集合={|()(2)<0}B x R x m x ∈--,且=(1,)A B n -，则\n\n=m ,=n .\n\n13、（江苏）已知集合{124}A =，\n\n，，{246}B =，，，则A B = ．\n\n1、（安徽）设集合A={3123|≤-≤-x x }，集合B 为函数)1lg(-=x y 的定义域，则A ?B=( ) （A ） （1，2） （B ）[1，2]（C ） [ 1，2） （D ）（1，2 ]\n\n2、（北京）已知集合{|320}A x R x =∈+>，{|(1)(3)0}B x R x x =∈+->，则A B =( )\n\n（A ）(,1)-∞- （B ）2(1,)3-- （C ）2\n\n(,3)3\n\n- （D ）(3,)+∞\n\n3、（福建）已知集合}4,3,2,1{=M ，}2,2{-=M ，下列结论成立的是（ ） A ．M N ? B ．M N M = C ．N N M = D ．}2{=N M\n\n4、（广东）设集合{1,2,3,4,5,6},{1,3,5}U M ==；则U C M =( )\n\n()A {,,}246 ()B {1,3,5 ()C {,,}124 ()D U\n\n5、 （湖北）已知集合A{x| 2x -3x +2=0,x ∈R } ， B={x|0＜x ＜5，x ∈N }，则满足条件A ?C ?B 的集合C 的个数为( )\n\nA 1\n\nB 2\n\nC 3\n\nD 4 6、（湖南）设集合M={-1,0,1}，N={x|x 2=x}，则M ∩N=( ) A.{-1，0，1} B.{0,1} C.{1} D.{0} 7、（江西） 若全集U=｛x ∈R|x 2≤4} A=｛x ∈R||x+1|≤1}的补集CuA 为( ) A |x ∈R |0＜x ＜2| B |x ∈R |0≤x ＜2| C |x ∈R |0＜x≤2| D |x ∈R |0≤x≤2| 8、（山东）已知全集{0,1,2,3,4}U =，集合{1,2,3}A =，{2,4}B =，则()U A B e为( ) (A){1,2,4} (B){2,3,4} (C){0,2,4} (D){0,2,3,4}", null, "9、（陕西）集合{|lg 0}M x x =>，2{|4}N x x =≤，则M N =\n\nA 。 (1,2)\n\nB 。 [1,2)\n\nC 。 (1,2]\n\nD 。 [1,2]（ ） 10、（四川）设集合{,}A a b =，{,,}B b c d =，则A B =（ ） A 、{}b B 、{,,}b c d C 、{,,}a c d D 、{,,,}a b c d 11、（新课标）已知集合A={x|x 2－x －2<0}，B={x|－1\n\n12.（2008北京1）已知全集U =R ，集合{}|23\n\nA x x =-≤≤，{}|14\n\nB x x x =<->或，那么集合U A\n\nC B ?=等于（ ）\n\nA ．{}\n\n|24x x -<≤ B ．{}\n\n|34x x x 或≤≥ C ．{}|21x x -<-≤\n\nD ．{}\n\n|13x x -≤≤\n\n13.已知全集U ＝｛1，2，3， 4，5｝，集合A ＝{}\n\n23Z <-∈x x ,则集合C u A 等于\n\nA {\n\n}4,3,2,1 B {}4,3,2 C {}5,1 D {}5\n\n14.已知全集U Z =，2{1,0,1,2},{|}A B x x x =-==，则\n\nU A C B ?=（ ）\n\nA ．{1,2}-\n\nB ．{1,0}-\n\nC ．{0,1}\n\nD ．{1,2}\n\n15.设集合{1\n\n2345}U =，，，，，{13}A =，， B=}{\n\n345，，，则U U C A C B ?=( )\n\nA ．{1}\n\nB ．{2}\n\nC ．{24}，\n\n}{345，\n\n， D ．{1234}，，，\n\n16.设,a b R ∈， {1,,}{0,\n\n,}b\n\na b a b a\n\n+=，则b a -=（ ） A ．1 B ．1- C ．2 D ．2-\n\n17.\n\n{123U =，，，，，\n\n2{|\n\n320\n\n}A x x x =-+=， {|2}B x x a a A ==∈，，则集合()U C A B ?中元素的个数为（ ）\n\nA ．1\n\nB ．2\n\nC ．3\n\nD ．4\n\n18.（08四川1）设集合{}{}{}1,2,3,4,5,1,2,3,2,3,4U A B ===，则()U A B =e( )\n\n（Ａ）{}2,3 （Ｂ）{}1,4,5 （Ｃ）{}4,5 （Ｄ）{}1,5", null, "19.（重庆11）设U ={1,2,3,4,5},A ={2,4},B={3,4,5},C={3,4},则(A ?B)()C ??e=\n\n19、（重庆）设函数f （x ）=x 2-4x+3，g （x ）=3x -2，集合M={x ∈R|f （g （x ））＞0}，N={x ∈R g （x ）g （x ）＜2}，则M ∩N 为( ) （A ）（1，﹢∞）（B ）（0，1）（C ）（-1，1）（D ）（-∞，1）\n\n20（浙江） 设全集U={1，2，3，4，5，6} ，设集合P={1，2，3，4} Q{3，4，5}，则P ∩（C U Q ）=( )\n\nA.{1，2，3，4，6}\n\nB.{1，2，3，4，5}\n\nC.{1，2，5}\n\nD.{1,2}\n\n21、（上海）若集合{}210A x x =->，{}\n\n1B x x =<，则A B ?＝ 17、（天津）集合\n\n{}|25A x R x =∈-≤中最小整数为 ." ]

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[ "# Grid Studies: node.js\n\nNode.js is a platform built on Chrome’s JavaScript runtime for easily building fast, scalable network applications. Node.js uses an event-driven, non-blocking I/O model that makes it lightweight and efficient, perfect for data-intensive real-time applications that run across distributed devices.\n\nNode.js is generally used to build web servers but can also be used to interface with monome devices, send and receive MIDI messages, or work with raw audio stream using the many modules available on NPM.\n\nThe `monome-grid` library has been designed for the latest stable version of Node, currently 8.9.x .\n\n## Prerequisites\n\nIf you’re very new to node.js (or JavaScript) a few tutorials and introduction videos have been provided below:\n\nCode School - Node.js\n\nThe Node Beginner Book\n\n## Initial Project Setup\n\nFirst install node.js. The official installers for your platform of choice are recommended. (This guide presumes you’re running either Mac OSX or Unix/Linux; whilst it should work on Windows, the examples are given for a Unix shell).\n\nOnce installed, open a terminal window and create a new folder:\n\n``````\\$ mkdir grid-studies\n\\$ cd grid-studies\n``````\n\nNext run `npm init` to create a `package.json` file. Just push enter on each of the prompts after running `npm init`:\n\n``````\\$ npm init\n``````\n\nNext install the `monome-grid` library:\n\n``````\\$ npm install --save monome-grid\n``````\n\nNow install the `easymidi` library:\n\n``````\\$ npm install --save easymidi\n``````\n\nYou can now copy the examples to this folder and run them using the `node` command:\n\n``````\\$ node grid_studies_2.js\n``````\n\n## 1. Connect\n\nThe `monome-grid` library facilitates easy connection and communication with grids. It uses the modern javascript idiom of ‘async/await’ to reflect that grid code can’t run until the grid has properly been initialised.\n\nFirst, require the `monome-grid` library:\n\n``````const monomeGrid = require('monome-grid');\n``````\n\nThen, connect to the grid inside an asynchronous function like so:\n\n``````async function run() {\nlet grid = await monomeGrid();\n}\n\nrun()\n``````\n\nHere the first monome device found is attached. If you want to connect to a specific grid (if you have more than one connected) you can specify a serial number, which can be found using serialosc-monitor.\n\n``````async function run() {\nlet grid = await monomeGrid('m1000011');\n}\n\nrun()\n``````\n\nThe library communicates with serialosc to discover attached devices using OSC. For a detailed description of how the mechanism and protocol work, see monome.org/docs/tech:osc.\n\n## 2. Basics\n\nSee grid_studies_2.js for this section.\n\n``````const monomeGrid = require('monome-grid');\n\nasync function run() {\nlet grid = await monomeGrid(); // optionally pass in grid identifier\n\n// initialize 2-dimensional led array\nlet led = [];\nfor (let y=0;y<8;y++) {\nled[y] = [];\nfor (let x=0;x<16;x++)\nled[y][x] = 0;\n}\n\n// refresh leds with a pattern\nlet refresh = function() {\nled = 15;\nled = 5;\nled = 5;\ngrid.refresh(led);\n}\n\n// call refresh() function 60 times per second\nsetInterval(refresh, 1000 / 60);\n\n// set up key handler\ngrid.key((x, y, s) => console.log(`key received: \\${x}, \\${y}, \\${s}`));\n}\n\nrun();\n``````\n\n### 2.1 Key input\n\nThe monome-grid library calls the function passed to `grid.key()` upon receiving input from the grid. It has three parameters.\n\n``````x : horizontal position (0-15)\ny : vertical position (0-7)\ns : state (1 = key down, 0 = key up)\n``````\n\nInside our aysnc function, we define the key function and simply print out incoming data.\n\n``````grid.key((x, y, s) => console.log(`key received: \\${x}, \\${y}, \\$[s}`));\n``````\n\nWe will, of course, do more interesting things with this function in the future.\n\n### 2.2 LED output\n\nUse the `grid.refresh()` function to update the state of the LEDs on the grid. This function accepts an array which represents the entire grid, hence the full frame is updated on each call.\n\nFirst, inside the async function, create the 2-dimensional array:\n\n``````let led = [];\nfor (let y = 0; y < 8; y++) {\nled[y] = [];\n}\n``````\n\nThis array has 8 rows which each contain an empty array representing the columns. Each LED on the grid can have a brightness range of 0-15, where 0 is off and 15 is maximum brightness. The values in the array are initially undefined, but we can set them all to 0 as follows:\n\n``````let led = [];\nfor (let y = 0; y < 8; y++) {\nled[y] = [];\nfor (let x = 0; x < 16; x++) {\nled[y][x] = 0;\n}\n}\n``````\n\nWe can draw a small pattern by setting individual elements of the array:\n\n``````led = 15;\nled = 5;\nled = 5;\n``````\n\nAnd finally, to copy this entire array to the grid:\n\n``````grid.refresh(led);\n``````\n\nAs seen in grid_studies_2.js we place this code inside the `refresh()` function - itself, part of the async function `run`. Note that we run the `refresh` function 60 times a second, using `setInterval` Upon running the sketch you will see this:", null, "### 2.3 Coupled interaction\n\nSee grid_studies_2_3.js for this section.\n\nThe previous code refreshes the grid constantly with every call of `refresh()`, which we typically do not want to do– it doesn’t make sense to have the computer spend time redrawing the same thing constantly.\n\nNext we’ll change the LED state to show which keys are being held, and only redraw when something has changed.\n\n``````const monomeGrid = require('monome-grid');\n\nasync function run() {\nlet grid = await monomeGrid(); // optionally pass in grid identifier\n\n// initialize 2-dimensional led array\nlet led = [];\nfor (let y=0;y<8;y++) {\nled[y] = [];\nfor (let x=0;x<16;x++)\nled[y][x] = 0;\n}\n\nlet dirty = true;\n\n// refresh leds with a pattern\nlet refresh = function() {\nif(dirty) {\ngrid.refresh(led);\ndirty = false;\n}\n}\n\n// call refresh() function 60 times per second\nsetInterval(refresh, 1000 / 60);\n\n// set up key handler\ngrid.key((x, y, s) => {\nled[y][x] = s * 15;\ndirty = true;\n});\n}\n\nrun();\n``````\n\nWe add a boolean variable `dirty` to indicate if the grid needs to be refreshed. We set this to `true` initially so the grid is immediately cleared upon start.\n\nNow we change the grid display upon incoming key data:\n\n``````grid.key((x, y, s) => {\nled[y][x] = s * 15;\ndirty = true;\n});\n``````\n\nSince `s` is either 0 or 1, when we multiply it by 15 we get off or full brightness. We set the LED location according to the position of the incoming key press, x and y.\n\nWe changed the `led` array, so we specify that the grid need refreshing:\n\n``````dirty = true;\n``````\n\nOnce this flag is set, the grid will be updated on the next iteration of `refresh()`:\n\n``````let refresh = function() {\nif(dirty) {\ngrid.refresh(led);\ndirty = false;\n}\n}\n``````\n\nOnce we’ve refreshed the grid, we set the `dirty` flag to `false` so we’re not needlessly refreshing.\n\nThe `refresh` function is called at 60fps unless you specify a different rate in the `setInterval(refresh, 1000 / 60)` such as `setInterval(refresh, 1000 / 10)` for 10fps.\n\nAs always, we wrap everything in an `async` function; see `grid_studies_2_3.js` for reference\n\n### 2.4 Decoupled interaction\n\nThe most basic decoupled interaction is a toggle. Turn the grid into a huge bank of toggles simply by changing line 30 (which is in the `grid.key` callback function):\n\n``````if(s == 1) led[y][x] ^= 15;\n``````\n\nNow only key downs (s = 1) do something. They use an xor operator to toggle the LED value between 0 and 15, depending on the previous state of the LED.\n\n## 3. Further\n\nNow we’ll show how basic grid applications are developed by creating a step sequencer. We will add features incrementally:\n\n• Use the top six rows as toggles.\n• Generate a clock pulse to advance the playhead from left to right, one column at a time. Wrap back to 0 at the end.\n• Display the play head on “position” (last) row.\n• Indicate the “activity” row (second to last) with a low brightness.\n• Trigger an event when the playhead reads an “on” toggle. Our “event” will be to turn on the corresponding LED in the “activity” row.\n• Jump to playback position when key pressed in the position row.\n• Adjust playback loop with two-key gesture in position row.\n\n### 3.1 Toggles\n\nSee grid_studies_3_1.js for this step.\n\nFirst we’ll create a new array called `step` that can hold 6 rows worth of step data. Note that we’ve created a new function called `create2DArray` to help us create new arrays. On key input we’ll look for key-down events in the top six rows:\n\n`````` // toggle steps\nif(s == 1 && y < 6) {\nstep[y][x] ^= 1;\ndirty = true;\n}\n``````\n\nIf this condition is true, we toggle the corresponding position in the `step` data and set the dirty flag so the grid will refresh.\n\nWe will “build” the LED display from scratch each time we need to refresh. This will be done inside of `refresh()` so we no longer need the `led` array to be global. Below we simply copy the `step` data to the `led` array, doing the proper multiplication by 15 in order to get full brightness.\n\n``````let refresh = function() {\nif(dirty) {\nlet led = create2DArray(8, 16);\n\n// display steps\nfor (let x=0;x<16;x++)\nfor (let y=0;y<6;y++)\nled[y][x] = step[y][x] * 15;\n\n// update grid\ngrid.refresh(led);\ndirty = false;\n}\n}\n``````\n\nThat’ll get us started.\n\n### 3.2 Play\n\nSee grid_studies_3_2.js for this step.\n\nFor simplicity we’re going to make a not-very-smart timer to drive our sequencer. Basically we’ll count `refresh` cycles and upon matching a specified interval, we’ll take a step forward in the sequence.\n\n``````let timer = 0;\nlet play_position = 0;\nconst STEP_TIME = 10;\n// ...\n\nlet refresh = function() {\nif(timer == STEP_TIME) {\nif(play_position == 15)\nplay_position = 0;\nelse\nplay_position++;\n\ntimer = 0;\ndirty = true;\n}\nelse timer++;\n\n// ...\n``````\n\nIn `refresh` we check `timer` against `STEP_TIME`. If they are equal, we process the next step, which in this case simply means incrementing `play_position`, which must be wrapped to 0 if it’s at the end. We reset `timer` so it can count back up, and set the dirty flag so the grid redraws.\n\nYou can change the speed by altering `STEP_TIME`. (We’ve made it a constant, for now, assuming that it won’t change during the running of the script. If you want to alter the tempo, it’ll have to become a regular variable.\n\nFor the redraw we add highlighting for the play position. Note how the multiply by 15 has been decreased to 11 to provide another mid-level brightness. We now have a series of brightness levels helping to indicate playback, lit keys, and currently active keys:\n\n``````let highlight = 0;\n\n// display steps\nfor (let x=0;x<16;x++) {\n// highlight the play position\nif(x == play_position)\nhighlight = 4;\nelse\nhighlight = 0;\n\nfor (let y=0;y<6;y++)\nled[y][x] = step[y][x] * 11 + highlight;\n}\n``````\n\nDuring this loop which copies steps to the grid, we check if we’re updating a column that is the play position. If so, we increase the highlight value. By adding this value during the copy we’ll get a nice effect of an overlaid translucent bar.\n\n### 3.3 Triggers\n\nSee grid_studies_3_3.js for this step.\n\nWhen the playhead advances to a new row we want something to happen which corresponds to the toggled-on rows. We’ll do two things: we’ll show separate visual feedback on the grid in the second-to-last (trigger) row, and we’ll print something to the console.\n\nDrawing the trigger row happens entirely in the `refresh()`:\n\n``````for(let x=0;x<16;x++) {\nled[x] = 4;\n}\n\nfor(let y=0;y<6;y++) {\nif(step[y][play_position] == 1) {\nled[y] = 15;\n}\n}\n``````\n\nFirst we create a dim row (level 4 is fairly dim). Then we search through the `step` array at the current play position, showing a bright indicator for each on state. This displays a sort of horizontal correlation of rows (or “channels”) 1-6 current state.\n\nFor the screen drawing, we create a function `trigger` which gets passed values of activated steps. This is what we do, inside `refresh` right after we change `play_position`:\n\n``````// TRIGGER SOMETHING\nfor(let y=0;y<6;y++) {\nif(step[y][play_position] == 1) {\ntrigger(y);\n}\n}\n``````\n\nAnd then `trigger` itself:\n\n``````let trigger = function(i) {\nconsole.log(`trigger at \\${i}`);\n}\n``````\n\nThis simply prints on the console. We could have this trigger a MIDI note for example.\n\n### 3.4a Cutting\n\nSee grid_studies_3_4.js for this step.\n\nWe will now use the bottom row to dynamically cut the playback position. First let’s add a position display to the last row, which will be inside `refresh()`:\n\n``````led[play_position] = 15;\n``````\n\nNow we look for key presses in the last row, in the `key` function:\n\n``````grid.key((x, y, s) => {\n// toggle steps\nif(s == 1 && y < 6) {\nstep[y][x] ^= 1;\n\ndirty = true;\n}\n// cut\nelse if(y == 7) {\nif(s == 1)\ncutting = true;\nnext_position = x;\n}\n});\n``````\n\nWe’ve added two variables, `cutting` and `next_position`. Check out the changed code where we check the timer:\n\n``````if(timer == STEP_TIME) {\nif(cutting)\nplay_position = next_position;\nelse if(play_position == 15)\nplay_position = 0;\nelse\nplay_position++;\n\ncutting = false;\n// ...\n``````\n\nNow, when pressing keys on the bottom row it will cue the next position to be played. Note that we set `cutting = false` after each cycle so that each press only affects the timer once.\n\n### 3.4b Loop\n\nLastly, we’ll implement setting the loop start and end points with a two-press gesture: pressing and holding the start point, and pressing an end point while still holding the first key. We’ll need to add a variable to count keys held, one to track the last key pressed, and variables to store the loop positions.\n\n``````let keys_held = 0\nlet key_last = 0;\nlet loop_start = 0\nlet loop_end = 15;\n``````\n\nWe set loop_end to 15 to begin with the full range. We count keys held on the bottom row thusly:\n\n``````keys_held = keys_held + (s*2) - 1;\n``````\n\nBy multiplying `s` by 2 and then subtracting one, we add one on a key down and subtract one on a key up.\n\nWe’ll then use the `keys_held` counter to do different actions:\n\n``````// cut and loop\nelse if(y == 7) {\n// track number of keys held\nkeys_held = keys_held + (s*2) - 1;\n\n// cut\nif(s == 1 && keys_held == 1) {\ncutting = true;\nnext_position = x;\nkey_last = x;\n}\n// set loop points\nelse if(s == 1 && keys_held == 2) {\nloop_start = key_last;\nloop_end = x;\n}\n}\n``````\n\nWe then modify the position change code:\n\n``````if(timer == STEP_TIME) {\nif(cutting)\nplay_position = next_position;\nelse if(play_position == 15)\nplay_position = 0;\nelse if(play_position == loop_end)\nplay_position = loop_start;\nelse\nplay_position++;\n``````\n\nDone!\n\n### 3.5a Sending MIDI Notes\n\nLet’s make it actually send some MIDI notes. The `easymidi` module can be used to create and listen to all types of MIDI events. First we’ll get some basic initialization out of the way by loading the `easymidi` module and creating a virtual MIDI input and output:\n\n``````const easymidi = require('easymidi');\n\nconst output = new easymidi.Output('grid out', true);\nconst input = new easymidi.Input('grid in', true);\n``````\n\nNext let’s implement the trigger method and make it actually do something. We’re going to add a new variable called `type` that will take the values `noteon` or `noteoff`:\n\n``````let trigger = function(type, i) {\noutput.send(type, {\nnote: 36 + i,\nvelocity: 127,\nchannel: 0\n});\n}\n``````\n\nNext we need to modify the code that calls trigger to pass it the `type` argument. We’ll need to send both “noteon” and “noteoff” messages to avoid leaving hanging notes. To do this we’ll calculate the `last_play_position` and use it to trigger “noteoff” messages while sending “noteon” messages to the current play_position:\n\n``````// TRIGGER SOMETHING\nlet last_play_position = play_position - 1;\nif(last_play_position == -1)\nlast_play_position = 15;\nfor(let y=0;y<6;y++) {\nif(step[y][last_play_position] == 1)\ntrigger('noteoff', y);\nif(step[y][play_position] == 1)\ntrigger('noteon', y);\n}\n``````\n\nNow if you open Ableton Live for example, you should see a “grid out” device that you can enable and route to an instrument. You can also route the notes to a real MIDI device by changing this line:\n\n``````let output = new easymidi.Output('grid out', true);\n``````\n\nTo something like this:\n\n``````let output = new easymidi.Output('Real Device Name');\n``````\n\nThe `true` argument means create a virtual device so don’t use that when interfacing with a real MIDI output. If you aren’t sure what the names of your devices are you can create a small script to check:\n\n``````let easymidi = require('easymidi');\nconsole.log(easymidi.getOutputs());\n``````\n\nSave this as `listmidi.js` and run it on the command line with `node listmidi.js`. You should see an array of device names.\n\n### 3.5b MIDI Clock Sync\n\nNow we’ll make the sequencer respond to MIDI clock messages. First, we can delete the STEP_TIME and timer variables as they are no longer needed. We’ll also want to move the code that handles timing out of the `refresh()` function and into a few event handlers. Here’s the main event handler to listen for midi clock messages:\n\n``````let ticks = 0;\n\ninput.on('clock', () => {\nticks++;\nif(ticks % 12 != 0)\nreturn;\n\nif(cutting)\nplay_position = next_position;\nelse if(play_position == 15)\nplay_position = 0;\nelse if(play_position == loop_end)\nplay_position = loop_start;\nelse\nplay_position++;\n\n// TRIGGER SOMETHING\nlet last_play_position = play_position - 1;\nif(last_play_position == -1)\nlast_play_position = 15;\nfor(let y=0;y<6;y++) {\nif(step[y][last_play_position] == 1)\ntrigger('noteoff', y);\nif(step[y][play_position] == 1)\ntrigger('noteon', y);\n}\n\ncutting = false;\ndirty = true;\n});\n``````\n\nThis is mostly a re-arrangement of existing code but there is a new variable called `ticks` that we’ll use to keep track of the number of MIDI clock messages we’ve received. This function will get called every time a ‘clock’ message is received on the input device (our virtual MIDI device named “grid in”).\n\nMIDI clock messages come in at a rate of 96 per measure, so on each tick we’ll check if it’s divisible evenly by 12 to provide 8th note resolution:\n\n``````ticks++;\nif(ticks % 12 != 0)\nreturn;\n``````\n\nFirst we increment `ticks` and then do the divisibility check. If it’s not divisible by 12 we’ll return out of the function and wait for the next tick. If it is divisible we’ll advance the play_position and trigger noteoff/noteon messages as needed.\n\nThis mostly works but we need to account for a few other MIDI messages. For example, this won’t trigger notes in the first `play_position`. To trigger these notes we’ll listen for the ‘start’ MIDI message:\n\n``````input.on('start', () => {\nfor(let y=0;y<6;y++)\nif(step[y][play_position] == 1)\ntrigger('noteon', y);\n});\n``````\n\nAnother issue we have is that if we reset the play position to 0 in our DAW we we still might be halfway through playing the segment on the sequencer. We should reset the `play_position` and `ticks` if this occurs:\n\ninput.on(‘position’, (data) => { if(data.value != 0) return; ticks = 0; play_position = 0; if(loop_start) play_position = loop_start; dirty = true; });\n\nAnd that’s it! We have a fully functioning MIDI sequencer that can sync to MIDI clock and trigger notes. To try it out in Ableton, simply turn on the “sync” option for the “grid in” MIDI device and turn on the “track” option for the “grid out” device (and route it to an instrument). See `grid_studies_3_5.js` for the completed MIDI implementation.\n\n## Closing\n\n### Suggested Exercises\n\n• “Record” keypresses in the “trigger” row to the toggle matrix.\n• Display the loop range on the bottom row of the grid.\n• Use the rightmost key in the “trigger” row as an “alt” key.\n• If “alt” is held while pressing a toggle, clear the entire row.\n• If “alt” is held while pressing the play row, reverse the direction of play.\n• Synchronize ticks/play_position based on the value coming in to the ‘position’ MIDI message (if it’s non-zero).\n\n## Credits\n\nNode.js is maintained by Joyent.\n\nThe monome-grid library was written and is maintained by Tom Dinchak.\n\nThis tutorial was created by Tom Dinchak and Tom Armitage for monome.org.\n\nContributions welcome. Submit a pull request to github.com/monome/docs or e-mail [email protected]." ]

[ null, "http://monome.org/docs/grid/studies/nodejs/images/grid-studies-nodejs-seen.jpg", null ]

[ "", null, "Affiliated Institutes\n\n### The MS in Computational & Applied Mathematics trains you in the analytical and numerical methods of applied mathematics to address and solve pressing challenges in all areas of science and engineering.\n\nMathematical modeling and scientific computing have become increasingly useful problem-solving tools. You’ll learn analytical and numerical methods of applied mathematics, applying them to problems in physics, biology, engineering, economics, climate science, and more. You’ll work alongside faculty who have extensive backgrounds in applied mathematics, computational science, statistics, financial engineering, industrial modeling, and related fields. Our program produces educated, capable practitioners who excel at communicating knowledge and solving complex mathematical challenges across disciplines. Join a new generation of mathematical scientists ready for leadership roles in education, industry, government laboratories, technology companies, and more.\n\n## Program At a Glance\n\nUNITS\n32 units\n\nCOMPLETION TIME\n2 years\n\nCOURSES BEGIN\nFall | Spring\n\nDEGREE AWARDED\nMS in Computational & Applied Mathematics\n\n## Featured Courses\n\nMATH 294\nMethods of Applied Mathematics\n\nStudies derivation of classical equations of applied math including quasilinear hyperbolic equations, Laplace, Poisson and Helmholtz equations, among others, in various orthogonal curvilinear coordinate systems.\n\nMATH 368\nNumerical Methods for Matrix Computations\n\nExamines numerical linear algebra and the matrix computations that arise in solving linear systems, least-squares problems, and eigenvalue problems for both dense and sparse matrices.\n\nMATH 362\nNumerical Methods for Differential Equations\n\nStudies numerical integration of ordinary and partial differential equations through methods such as Runge-Kutta and Adams formulas and more.\n\nMath 462\nMathematics of Machine Learning\n\nCovers theoretical foundations, algorithms, and methodologies for machine learning, emphasizing the role of probability and optimization and exploring a variety of real-world applications.\n\nMATH 256\nStochastic Processes\n\nAnalyzes properties of independent and dependent random variables and examines conditional expectation among topics chosen from Markov processes and others.\n\nMATH 306\nOptimization\n\nEmphasizes nonlinear programming and numerical methods for finite-dimensional optimization problems with fairly smooth functions.\n\nCurriculum\n\nDegree Requirements\nA minimum of eight courses (32 units) of graduate math coursework is required, at least five of which (20 units) must be at the 300-level or higher.\n\nStudents who lack the prerequisite undergraduate coursework may be asked to complete more than 32 units. At least 20 units of coursework must be gamma courses (300-level and above). A grade of B- or above must be earned in gamma courses.\n\nStudents in the master’s program in mathematics, computational and applied mathematics, and statistical sciences may convert one 200-level (beta) course to gamma credit. At the discretion of the IMS Director, in consultation with the student’s academic advisor, more than one conversion may be approved in exceptional cases.\n\nCore Courses (Four courses – 16 units)\n\n• Methods of Applied Mathematics (waived if the student has had equivalent preparation: complex variables, vector calculus, Fourier series/transforms, and PDEs)\n• Numerical Methods for Matrix Computations\n• Numerical Methods for Differential Equations\n• Continuous Mathematical Modeling, or Discrete Mathematical Modeling\n\nRestricted Electives (Two electives – 8 units)\n\n• Any of the core courses listed above if not taken to satisfy the core requirements\n• Mathematics Clinic\n• Stochastic Processes\n• Optimization\n• Reliability Theory\n• Deterministic and Stochastic Control\n• Mathematical Finance\n• Simulation\n• Fluid Dynamics\n• Perturbation and Asymptotic Analysis\n• Mathematical Modeling in Biology\n• Image Processing\n\nIndependent Study\nIn lieu of one formal course, the student may take Math 398 Independent Study with a research advisor leading to a publication quality technical report in an area of computational or applied math.\n\nAccelerated Degree Option\n\nIf you are an undergraduate student at the Claremont Colleges*, you can obtain a graduate degree on an accelerated track. Undergraduate students can begin working toward the master’s requirements simultaneously with the completion of their undergraduate degrees.\n\nYou can earn a Master’s Degree in Mathematics, Computational & Applied Mathematics, or Statistical Sciences from Claremont Graduate University with just one additional year. Up to 16 units of undergraduate credit may be accepted towards your graduate degree. Financial aid is available for your graduate year at CGU.\n\n*The accelerated degree option is only open to undergraduate students at Pomona, Scripps, Claremont McKenna, Harvey Mudd, and Pitzer.\n\nConsider this opportunity if:\n\n• Your longer-term plans include a PhD in a math-related field (e.g. physics, engineering, computer science)\n• You are seeking a teaching career in math or science\n• You would like to get involved in ongoing research with CGU Math faculty during your senior year\n• You would like to earn your Master’s degree in half the usual time\n\n## Resources\n\nEngineering & Computational Mathematics Clinic\nCGU’s internationally recognized Engineering & Computational Mathematics Clinic offers first-hand experience solving significant mathematical problems for industry and government clients.\n\nRecent projects include:\n\n• Optimizing Transmission of Renewable Energy–Southern California Edison\n• Hardware-Software Codesign–Los Alamos National Laboratory\n• Data Cohort Analysis–Fair Isaac\n• Optimizing Smart Power Grids–Los Alamos National Laboratory\n• Credit Risk in a Network Economy–Fitch Rating\n• Isogeometric Analysis–Boeing\n• Gate to Base Capacitance Modeling for Nanoscale MOSFETs–USC Information Sciences Institute\n• Practical Semi-Analytic Model for the Substrate Current of Short Channel MOSFETs with LDDs–USC Information Sciences Institute\n\nClaremont Center for the Mathematical Sciences\nThrough the Claremont Center for the Mathematical Sciences (CCMS), you’ll have access to one of the largest mathematical science communities in California, as well as to workshops, conferences, and seminars, including:\n\n• Southern California Analysis Seminar\n• Math-in-Industry Workshop\n• Michael E. Moody Lecture Series\n• History and Philosophy of Mathematics Seminar\n• Claremont Mathematics Weekend\n• CCMS Software Lab\n• and more\n\nFaculty & Research\n•", null, "John Angus\n\nProfessor of Mathematics\n\n##### Research Interests\n\nProbability, Statistics, Computing, Algorithms, Navigation, Systems Engineering, Mathematical Finance\n\n•", null, "Marina Chugunova\n\nProfessor of Mathematics\nProgram Director, PhD in Engineering & Computational Mathematics\n\n##### Research Interests\n\nSurfactant-driven thin film flows in biomedical applications; Nonlinear parabolic equations; Stability problems in fluid dynamics; Scientific computations; Applied operator theory; Sturm-Liouville problems\n\n•", null, "Ellis Cumberbatch\n\nProfessor of Mathematics Emeritus\n\n##### Research Interests\n\nApplied mathematics, Industrial modeling, Differential equations, Fluid mechanics, Wave propagation, Semi-conductors\n\n•", null, "Research Professor of Mathematics\n\n##### Research Interests\n\nApproximation theory, Computational harmonic analysis, Mathematics of data, Query processing\n\n•", null, "Professor of Mathematics\nJoseph H. Pengilly Chair in Mathematics\n\n##### Research Interests\n\nFluid Dynamics, Mathematical Modeling, Scientific Computing\n\n•", null, "Andrew Nguyen\n\n##### Research Interests\n\nStochastic processes, Statistics, Risk management, Financial derivatives, Actuarial sciences, Statistical software\n\n•", null, "Qidi Peng\n\nResearch Assistant Professor of Mathematics\n\n##### Research Interests\n\nStatistical inferences, Stochastic differential equations, Stochastic modeling, Simulation, Machine learning, Approximation theory, Graph theory\n\n•", null, "Allon Percus\n\nProfessor of Mathematics\n\n##### Research Interests\n\nDiscrete optimization; Network models; Statistical physics; Random combinatorial structures\n\n•", null, "Claudia Rangel-Escareño\n\n##### Research Interests\n\nProbabilistic methods in computational biology, Statistical inference of genetic networks, Bioinformatics\n\n•", null, "Henry Schellhorn\n\nProfessor of Mathematics\nDirector, Institute of Mathematical Sciences\n\n##### Research Interests\n\nFinancial engineering, Credit risk, Stochastic analysis, Traffic models\n\n## Extended Faculty\n\nIn addition to CGU core faculty, you will have access to Math faculty across the Claremont Colleges, including Pomona, Scripps, Harvey Mudd, Claremont McKenna, Pitzer, and Keck Graduate Institute, as well as faculty who are part of the W.M. Keck integrated science department.\n\nHarvey Mudd\nView Math Faculty\n\nPomona\nView Math Faculty\n\nClaremont McKenna\nView Math Faculty\n\nScripps\nView Math Faculty\n\nPitzer\nDavid Bachman\nJim Hoste\n\nW.M. Keck Integrated Science Department\n(Claremont McKenna, Pitzer, Scripps)\nJohn Milton\n\nAnimesh Ray\nJames Sterling\n\n## Where You Can Find Our Alumni\n\n• Raytheon\n\n• RAND Corporation\n\n• Memorial University of Newfoundland\n\n• Stanford University\n\n• Center for Infection and Immunity, Columbia University\n\n• Jet Propulsion Laboratory\n\n• ReResearch Nanotechnology\n\n• Cablevision\n\n• Accenture\n\n### Request information about the Computational & Applied Mathematics program\n\n• This field is for validation purposes and should be left unchanged." ]

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[ "# `applyOf()`\n\nGets the `apply()` function for the specified function.\n\nThis function is valid in v2.21.0 to v2.24.2. This function has been downloaded 18 times.\n\n### Parameters\n\n{Function|string} fn\nThe function or the function name for which to get the `apply()` function.\n{Object=} opt_owner Optional\nIf `fn` is a string then this will be used as the parent object to reference the actual function.\n{*=} opt_objThis Optional\nIf given this will be the value that is always used as the `this` keyword by `fn` (the function). If not given, the first argument passed to the returned function will be referenced as the `this` keyword.\n\n### Returns\n\n{Function}\nIf `opt_objThis` is not passed, a version of `apply()` will be returned which will use the first argument as the `this` keyword and the subsequent argument (which should be an array or an array-like object) as the normal arguments. Otherwise a function which accepts only the array of arguments (not the `this` keyword definition) will be returned.\n\n### Required Function\n\nThis function directly requires the following function which is included automatically:\n\n### Requiring Function\n\nThis function is directly required by the following function:" ]

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[ "Check if frequency of each digit is less than the digit in Python\n\nPythonServer Side ProgrammingProgramming\n\nSuppose we have a number n, we have to check whether the occurrence of each digit of n is less than or equal to digit itself.\n\nSo, if the input is like n = 5162569, then the output will be True as the digits and frequencies are (5, 2), (1, 1), (6, 2) and (9, 1), for all the frequency is either small or equal to the digit value.\n\nTo solve this, we will follow these steps −\n\n• for i in range 0 to 9, do\n• temp := n, cnt := 0\n• while temp is non-zero, do\n• if temp mod 10 is same as i, then\n• cnt := cnt + 1\n• if cnt > i, then\n• return False\n• temp := quotient of (temp / 10)\n• return True\n\nExample\n\nLet us see the following implementation to get better understanding −\n\nLive Demo\n\ndef solve(n):\nfor i in range(10):\ntemp = n\ncnt = 0\nwhile temp:\nif temp % 10 == i:\ncnt += 1\nif cnt > i:\nreturn False\ntemp //= 10\nreturn True\ns = 5162569\nprint(solve(s))\n\n5162569\n\nTrue" ]

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[ "CAT  >  Practice Questions Level 3: Probability - 2\n\n# Practice Questions Level 3: Probability - 2 - Notes | Study Level-wise Practice Questions for CAT Preparation - CAT\n\n 1 Crore+ students have signed up on EduRev. Have you?\n\nThis EduRev document offers 10 Multiple Choice Questions (MCQs) from the topic Probability (Level - 3). These questions are of Level - 3 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets a 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability of getting a 1, 2, 3 or 4 on the die?\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls one by one without replacement. Find the probability that both the balls are of different colours.\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:The probabilities that three men hit a target are 1/6, 1/4 and 1/3. Each man shoots once at the target. What is the probability that exactly one of them hits the target?\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:Three bags contain 18 balls of red and blue colours. The first bag contains 3 red and 3 blue balls. The second bag contains 4 red and 2 blue balls and the third bag contains 2 red and 4 blue balls. One ball is taken out at random from each bag. The probability of getting 2 red balls and 1 blue ball is\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:In a tailor shop, tailor A makes 500 shirts, tailor B makes 1200 shirts and tailor C makes 1600 shirts. Out of these, 200, 300 and 300 shirts are defective, respectively. One shirt is drawn at random and if it is found to be defective, then the probability that is made by tailor C is\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:Directions: Study the following information and answer the given question.\n\nThree friends A, B and C do air rifle shooting as a hobby and compete against each other in friendly competitions. On a Sunday morning, they go to an empty tract of land along with their air rifles to shoot at clay practice targets from the distances of 100 m and 200 m. A hits his target 95% of the times from a distance of 100 m and 70% times from a distance of 200 m. The same figures for B are 85% and 80%, respectively. The same figures for C are 90% and 75%, respectively. They put in numbered chits (1, 2 and 3) to be drawn to decide the order of shooting and then each of them shoots at the targets turn-wise according to the chit number, first from 100 m and then from 200 m. They decide on a point system with 1 point for hitting the target from a distance of 100 m and 2 points for hitting it from a distance of 200 m with the winner getting a round of drinks paid by the other two.\n\nQ. What is the probability that the order of shooting is ABC?\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:Directions: Study the following information and answer the given question.\n\nThree friends A, B and C do air rifle shooting as a hobby and compete against each other in friendly competitions. On a Sunday morning, they go to an empty tract of land along with their air rifles to shoot at clay practice targets from the distances of 100 m and 200 m. A hits his target 95% of the times from a distance of 100 m and 70% times from a distance of 200 m. The same figures for B are 85% and 80%, respectively. The same figures for C are 90% and 75%, respectively. They put in numbered chits (1, 2 and 3) to be drawn to decide the order of shooting and then each of them shoots at the targets turn-wise according to the chit number, first from 100 m and then from 200 m. They decide on a point system with 1 point for hitting the target from a distance of 100 m and 2 points for hitting it from a distance of 200 m with the winner getting a round of drinks paid by the other two.\n\nQ. What is the probability that all three of them will end up shooting 1 target only, with all three of them shooting the target of the same points weightage?\n\nQuestion for Practice Questions Level 3: Probability - 2\nTry yourself:A bag X contains 3 white and 2 black balls. Another bag Y contains 2 white and 4 black balls. A bag and a ball out of it are picked at random. Find the probability that the ball is of white colour.\n\nQuestion for Practice Questions Level 3: Probability - 2\n\nTwo friends A and B play a game of chance. A person tosses a coin and if the coin shows a heads, he rolls a die. He does this twice and the score is the total of the numbers that the die throws up in both the rolls, if at all. If a person ends up with a tails on tossing the coin, he does not roll the die and his score is entered as 0. The winner of the game is the person whose score is the maximum.\n\nQ. What is the probability that A wins the game with a score of 2 and B ends up with a score of 0?\n\nQuestion for Practice Questions Level 3: Probability - 2\n\nTwo friends A and B play a game of chance. A person tosses a coin and if the coin shows a heads, he rolls a die. He does this twice and the score is the total of the numbers that the die throws up in both the rolls, if at all. If a person ends up with a tails on tossing the coin, he does not roll the die and his score is entered as 0. The winner of the game is the person whose score is the maximum.\n\nQ. What is the probability that A ends up with an odd score without scoring a 0 in any of his rolls?\n\nThe document Practice Questions Level 3: Probability - 2 - Notes | Study Level-wise Practice Questions for CAT Preparation - CAT is a part of the CAT Course Level-wise Practice Questions for CAT Preparation.\nAll you need of CAT at this link: CAT\n\n## Level-wise Practice Questions for CAT Preparation\n\n277 docs\n Use Code STAYHOME200 and get INR 200 additional OFF\n\n## Level-wise Practice Questions for CAT Preparation\n\n277 docs\n\n### How to Prepare for CAT\n\nRead our guide to prepare for CAT which is created by Toppers & the best Teachers\n\nTrack your progress, build streaks, highlight & save important lessons and more!\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n;" ]

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[ "# Re:\n\nDate: Thu Mar 15 2018 - 08:24:07 CDT\n\nI think you are looking for the *mean* acceptance probability, that is, the\naverage probability with which an exchange is accepted. You can compute\nthis in two different ways which, to my knowledge, are essentially\nequivalent for any reasonably large number of REMD cycles:\n\n1) take the expectation of the Metropolis criterion P = min[1,\nexp(-Delta_ij)] where Delta_ij contains the observed energies and\ntemperatures -- this method is cumbersome and requires sifting through a\nlot of data\n\n2) just divide the number of success by the number of attempts = (50 / 450)\n= 0.11 = 11%\n\nThe reasonableness of the observed value depends on what your exchange\nscheme is. I assume that your script implements a nearest neighbor\nsampling? If the neighbors are chosen stochastically with equal\nprobabilities, then 11% is very close to optimal. If the neighbors are\nchosen in the deterministic \"up/down\" strategy then something closer to 20%\nis preferred.\n\nIf you are unhappy with your acceptance probability you have two options:\n1) assume that you have bad statistics and keep running until the\nperformance numbers change or 2) start over and choose more replicas over\nthe temperature range.\n\nHTH,\nBKR\n\nOn Thu, Mar 15, 2018 at 7:44 AM, Srijita Paul <srijitap91_at_gmail.com> wrote:\n\n> Hi,\n>\n> Can anybody explain me the output file obtained from a remd simulation\n> .job0.restart900.0.tcl.\n>\n> array set replica {index.b 4 index 3 temperature.a 283.48\n> exchanges_attempted 450 loc.a 4 temperature.b 289.07 temperature 286.26\n> exchanges_accepted 50 loc.b 18 index.a 2}\n>\n> exchanges_attempted 450\n> exchanges_accepted 50\n>\n> Is it a good result for remd? How can I find acceptance probability for my\n> system?\n>\n\nThis archive was generated by hypermail 2.1.6 : Tue Dec 31 2019 - 23:19:45 CST" ]

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[ "## #math trivia #73 solution\n\nBurt Kaliski Jr. (@modulomathy)\n3/13/12 7:42 AM\n#math trivia for #March13: #73 as a modulus gives square roots to -1. Solve x^2+1 = 0 (mod 73) where x is an integer between 0 and 72.\n\nOne way to find the square root is by trial and error, looking for the smallest integer congruent to 72 modulo 73 that is a square:\n\n– 72 — no\n– 145 — no\n– 218 — no\n– 291 — no\n– 364 — no\n– 437 — no\n– 510 — no\n– 583 — no\n– 656 — no\n– 729 — yes, 27^2\n\nSo, one of the square roots of 72 modulo 73 is 27. The other is -27, which in modular terms is 73-27 = 46. We can check this as follows:\n\n46^2 mod 73 = 2116 mod 73 = (28*73+72) mod 73\n\nAnother approach to finding the square root that may be more efficient is to take advantage of Fermat’s Little Theorem, which states that if p is a prime and x is not divisible by p, then\n\nx^(p-1) mod p = 1 .\n\nIf x is a random number modulo p, then with some probability, x^((p-1)/2) mod p will be -1. When p-1 is divisible by 4 as in the present example, x^((p-1)/4) mod p will then be a square root of -1. Concretely, we have (p-1)/4 = 18, so let’s compute x^18 mod 73 for small values of x and see what happens when we enumerate successive powers up to 18:\n\n— x = 2: 2, 4, 8, 16, 32, 64, 55, 37, 1, 2, 4, 8, 16, 32, 64, 55, 37, 1\n— x = 3: 3, 9, 27, 8, 24, 72, 69, 64, 46, 65, 49, 1, 3, 9, 27, 8, 24, 72\n\nFor x = 2, the successive powers go directly to 1; the square root is not exposed. For x = 3, the successive powers go to -1, so we still don’t have a square root of -1 at the 18th power. However, the 9th power will give is what we’re looking for: the square root of -1 mod 73 is 46, just as above. (So is 27, which is the 3rd power here; note that the 6th is also -1.)" ]

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[ "X\nX\n\n# Calculate the Greatest Common Factor or GCF of 301\n\nThe instructions to find the GCF of 301 are the next:\n\n## 1. Decompose all numbers into prime factors\n\n 301 7 43 43 1\n\n## 2. Write all numbers as the product of its prime factors\n\n Prime factors of 301 = 7 . 43\n\n## 3. Choose the common prime factors with the lowest exponent\n\nCommon prime factors: 7 , 43\n\nCommon prime factors with the lowest exponent: 71, 431\n\n## 4. Calculate the Greatest Common Factor or GCF\n\nRemember, to find the GCF of several numbers you must multiply the common prime factors with the lowest exponent.\n\nGCF = 71. 431 = 301\n\nAlso calculates the:" ]

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[ "# Yahoo Web Search\n\n1. Sort by\n\n1. ### A capacitor with low energy loss has a (A) high PF (B)low Q (C)high DF (D)low ESR?\n\nD; low ESR (effective series resistance) It will have a low PF (power factor), high Q and low DF (dissipation factor)\n\n1 Answers · Science & Mathematics · 20/08/2013\n\n2. ### derive eq: p=pf (m/(m-ma))?\n\nwhat is the variable ?\n\n1 Answers · Science & Mathematics · 10/05/2009\n\n3. ### 425 V is applied to a 7900 pF capacitor. How much energy is stored?\n\nE = (1/2)CV^2 E = (1/2)(7900e-9)(425)^2 E = 0.71346875 J\n\n3 Answers · Science & Mathematics · 15/02/2008\n\n4. ### 600 V is applied to a 7000 pF capacitor. How much energy is stored?\n\nEnergy stored = (1/2) C V^2 C = 7000pF = 7000 * 10^-9 F V = 600V = (1/2)(7000*10^-9) (600*600) =1.26*10^9*10^-9 =1.26J EDIT : Yes I am sorry... 1 picofarad = 1*10^-12 farad....He is...\n\n2 Answers · Science & Mathematics · 23/04/2008\n\n5. ### jiofw[f[ew'mmmcwopfk[wf pcwq]pf?\n\ni'm not doing your homework dude.\n\n2 Answers · Science & Mathematics · 28/01/2009\n\n6. ### 460 V is applied to a 6500 pF capacitor. How much energy is stored?\n\nE = 0.5 CV^2 = 0.5*6500*10^-12*460*460 = 0.0006877 J = 6.877*10^-4 J\n\n1 Answers · Science & Mathematics · 03/04/2013\n\n7. ### 450 V is applied to a 6700 pF capacitor. How much energy is stored?\n\nW = (1/2)∙C∙V² = (1/2) ∙ 6700×10⁻¹²F ∙ (450V)² = 1.35675×10⁻³J = 1.35675mJ\n\n1 Answers · Science & Mathematics · 22/03/2009\n\n8. ### 550 V is applied to a 6900 pF capacitor. How much energy is stored?\n\nw=(c*v^2)/2 w= (6.9*10^-9*550^2)/2= 1.043625[mJ] or 1043.625[μJ]\n\n1 Answers · Science & Mathematics · 06/02/2013\n\n9. ### You need to construct a 100 pF capacitor?\n\nok, first things first, mm's to m's .2 mm = 2*10^-4 m the formula you need is : C = EA/d ( E is constant btw, it is the permittivity of free space and it equals 8.85*10^-12) so, now u just need to plug in you values: 100*10^-12...\n\n2 Answers · Science & Mathematics · 14/02/2009\n\n10. ### A 5.00 - pF parallel-plate air-filled capacitor with circular plates is to be used in a circuit...?\n\nThe capacitance of the two plates of area A and plate separation d is given by: C = εoA/d Then, A = Cd/εo = π r² and, r = √{Cd/πεo}, The potential difference across the plates...\n\n3 Answers · Science & Mathematics · 30/09/2010" ]

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[ "# 47.29 kg to lbs - 47.29 kilograms to pounds\n\nDo you need to know how much is 47.29 kg equal to lbs and how to convert 47.29 kg to lbs? Here it is. In this article you will find everything about kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to point out that all this article is devoted to only one amount of kilograms - exactly one kilogram. So if you need to know more about 47.29 kg to pound conversion - keep reading.\n\nBefore we go to the practice - it means 47.29 kg how much lbs calculation - we want to tell you few theoretical information about these two units - kilograms and pounds. So let’s move on.\n\nHow to convert 47.29 kg to lbs? 47.29 kilograms it is equal 104.2566036998 pounds, so 47.29 kg is equal 104.2566036998 lbs.\n\n## 47.29 kgs in pounds\n\nWe will start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, that is International System of Units (in short form SI).\n\nFrom time to time the kilogram could be written as kilogramme. The symbol of this unit is kg.\n\nFirstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but impractical to use.\n\nThen, in 1889 the kilogram was described using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was substituted by another definition.\n\nNowadays the definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.\n\n## 47.29 kilogram to pounds\n\nYou learned a little bit about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to underline that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to centre only on pound-mass.\n\nThe pound is in use in the British and United States customary systems of measurements. Naturally, this unit is used also in other systems. The symbol of this unit is lb or “.\n\nThere is no descriptive definition of the international avoirdupois pound. It is equal 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.\n\nThe avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### How many lbs is 47.29 kg?\n\n47.29 kilogram is equal to 104.2566036998 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.\n\n### 47.29 kg in lbs\n\nThe most theoretical section is already behind us. In this section we are going to tell you how much is 47.29 kg to lbs. Now you know that 47.29 kg = x lbs. So it is time to know the answer. Have a look:\n\n47.29 kilogram = 104.2566036998 pounds.\n\nThis is an accurate outcome of how much 47.29 kg to pound. You may also round it off. After rounding off your outcome is as following: 47.29 kg = 104.038 lbs.\n\nYou know 47.29 kg is how many lbs, so see how many kg 47.29 lbs: 47.29 pound = 0.45359237 kilograms.\n\nObviously, in this case you can also round off the result. After rounding off your outcome is exactly: 47.29 lb = 0.45 kgs.\n\nWe also want to show you 47.29 kg to how many pounds and 47.29 pound how many kg outcomes in charts. Look:\n\nWe want to begin with a table for how much is 47.29 kg equal to pound.\n\n### 47.29 Kilograms to Pounds conversion table\n\nKilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)\n47.29 104.2566036998 104.0380\nNow see a table for how many kilograms 47.29 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n47.29 0.45359237 0.45\n\nNow you learned how many 47.29 kg to lbs and how many kilograms 47.29 pound, so we can move on to the 47.29 kg to lbs formula.\n\n### 47.29 kg to pounds\n\nTo convert 47.29 kg to us lbs you need a formula. We will show you two formulas. Let’s start with the first one:\n\nNumber of kilograms * 2.20462262 = the 104.2566036998 outcome in pounds\n\nThe first version of a formula will give you the most exact outcome. In some situations even the smallest difference could be significant. So if you need an accurate outcome - this formula will be the best solution to convert how many pounds are equivalent to 47.29 kilogram.\n\nSo go to the shorer formula, which also enables conversions to know how much 47.29 kilogram in pounds.\n\nThe shorter version of a formula is as following, let’s see:\n\nNumber of kilograms * 2.2 = the result in pounds\n\nAs you see, this version is simpler. It can be the best solution if you want to make a conversion of 47.29 kilogram to pounds in fast way, for instance, during shopping. You only have to remember that your result will be not so correct.\n\nNow we are going to learn you how to use these two formulas in practice. But before we are going to make a conversion of 47.29 kg to lbs we want to show you another way to know 47.29 kg to how many lbs totally effortless.\n\n### 47.29 kg to lbs converter\n\nAn easier way to learn what is 47.29 kilogram equal to in pounds is to use 47.29 kg lbs calculator. What is a kg to lb converter?\n\nConverter is an application. Calculator is based on longer formula which we showed you in the previous part of this article. Due to 47.29 kg pound calculator you can effortless convert 47.29 kg to lbs. You only need to enter amount of kilograms which you need to calculate and click ‘convert’ button. You will get the result in a second.\n\nSo try to calculate 47.29 kg into lbs with use of 47.29 kg vs pound converter. We entered 47.29 as an amount of kilograms. This is the result: 47.29 kilogram = 104.2566036998 pounds.\n\nAs you see, our 47.29 kg vs lbs calculator is so simply to use.\n\nNow we can move on to our chief topic - how to convert 47.29 kilograms to pounds on your own.\n\n#### 47.29 kg to lbs conversion\n\nWe will start 47.29 kilogram equals to how many pounds conversion with the first formula to get the most correct outcome. A quick reminder of a formula:\n\nNumber of kilograms * 2.20462262 = 104.2566036998 the result in pounds\n\nSo what need you do to know how many pounds equal to 47.29 kilogram? Just multiply number of kilograms, this time 47.29, by 2.20462262. It gives 104.2566036998. So 47.29 kilogram is exactly 104.2566036998.\n\nYou can also round it off, for example, to two decimal places. It is exactly 2.20. So 47.29 kilogram = 104.0380 pounds.\n\nIt is high time for an example from everyday life. Let’s calculate 47.29 kg gold in pounds. So 47.29 kg equal to how many lbs? And again - multiply 47.29 by 2.20462262. It is equal 104.2566036998. So equivalent of 47.29 kilograms to pounds, when it comes to gold, is 104.2566036998.\n\nIn this case you can also round off the result. It is the outcome after rounding off, this time to one decimal place - 47.29 kilogram 104.038 pounds.\n\nNow let’s move on to examples calculated with short formula.\n\n#### How many 47.29 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nNumber of kilograms * 2.2 = 104.038 the outcome in pounds\n\nSo 47.29 kg equal to how much lbs? And again, you have to multiply amount of kilogram, in this case 47.29, by 2.2. Let’s see: 47.29 * 2.2 = 104.038. So 47.29 kilogram is equal 2.2 pounds.\n\nMake another conversion using shorer formula. Now convert something from everyday life, for instance, 47.29 kg to lbs weight of strawberries.\n\nSo let’s calculate - 47.29 kilogram of strawberries * 2.2 = 104.038 pounds of strawberries. So 47.29 kg to pound mass is equal 104.038.\n\nIf you know how much is 47.29 kilogram weight in pounds and can calculate it with use of two different versions of a formula, let’s move on. Now we want to show you all results in charts.\n\n#### Convert 47.29 kilogram to pounds\n\nWe are aware that results shown in charts are so much clearer for most of you. We understand it, so we gathered all these results in charts for your convenience. Thanks to this you can easily compare 47.29 kg equivalent to lbs results.\n\nBegin with a 47.29 kg equals lbs table for the first version of a formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n47.29 104.2566036998 104.0380\n\nAnd now have a look at 47.29 kg equal pound table for the second version of a formula:\n\nKilograms Pounds\n47.29 104.038\n\nAs you can see, after rounding off, if it comes to how much 47.29 kilogram equals pounds, the results are not different. The bigger amount the more considerable difference. Please note it when you want to do bigger number than 47.29 kilograms pounds conversion.\n\n#### How many kilograms 47.29 pound\n\nNow you learned how to convert 47.29 kilograms how much pounds but we will show you something more. Are you curious what it is? What about 47.29 kilogram to pounds and ounces conversion?\n\nWe want to show you how you can convert it step by step. Begin. How much is 47.29 kg in lbs and oz?\n\nFirst thing you need to do is multiply number of kilograms, in this case 47.29, by 2.20462262. So 47.29 * 2.20462262 = 104.2566036998. One kilogram is exactly 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this case there are 2 pounds.\n\nTo convert how much 47.29 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.\n\nSo final result is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result will be exactly 2 pounds and 33 ounces.\n\nAs you see, conversion 47.29 kilogram in pounds and ounces quite simply.\n\nThe last conversion which we are going to show you is conversion of 47.29 foot pounds to kilograms meters. Both of them are units of work.\n\nTo convert it you need another formula. Before we show you this formula, have a look:\n\n• 47.29 kilograms meters = 7.23301385 foot pounds,\n• 47.29 foot pounds = 0.13825495 kilograms meters.\n\nNow let’s see a formula:\n\nAmount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters\n\nSo to calculate 47.29 foot pounds to kilograms meters you need to multiply 47.29 by 0.13825495. It is exactly 0.13825495. So 47.29 foot pounds is exactly 0.13825495 kilogram meters.\n\nYou can also round off this result, for example, to two decimal places. Then 47.29 foot pounds will be exactly 0.14 kilogram meters.\n\nWe hope that this calculation was as easy as 47.29 kilogram into pounds calculations.\n\nThis article was a huge compendium about kilogram, pound and 47.29 kg to lbs in calculation. Due to this conversion you know 47.29 kilogram is equivalent to how many pounds.\n\nWe showed you not only how to do a calculation 47.29 kilogram to metric pounds but also two another conversions - to check how many 47.29 kg in pounds and ounces and how many 47.29 foot pounds to kilograms meters.\n\nWe showed you also another solution to make 47.29 kilogram how many pounds calculations, that is using 47.29 kg en pound calculator. This will be the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.\n\nWe hope that now all of you are able to make 47.29 kilogram equal to how many pounds calculation - on your own or with use of our 47.29 kgs to pounds converter.\n\nSo what are you waiting for? Calculate 47.29 kilogram mass to pounds in the best way for you.\n\nDo you need to make other than 47.29 kilogram as pounds conversion? For instance, for 10 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so easy as for 47.29 kilogram equal many pounds.\n\n### How much is 47.29 kg in pounds\n\nWe want to sum up this topic, that is how much is 47.29 kg in pounds , we prepared one more section. Here we have for you all you need to know about how much is 47.29 kg equal to lbs and how to convert 47.29 kg to lbs . It is down below.\n\nHow does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 47.29 kg to pound conversion formula look? . See it down below:\n\nThe number of kilograms * 2.20462262 = the result in pounds\n\nSee the result of the conversion of 47.29 kilogram to pounds. The correct result is 104.2566036998 pounds.\n\nYou can also calculate how much 47.29 kilogram is equal to pounds with another, easier type of the formula. Have a look.\n\nThe number of kilograms * 2.2 = the result in pounds\n\nSo now, 47.29 kg equal to how much lbs ? The answer is 104.2566036998 lbs.\n\nHow to convert 47.29 kg to lbs in a few seconds? You can also use the 47.29 kg to lbs converter , which will make the rest for you and you will get an accurate result .\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.\nRead more related articles:\n 47.01 kg to lbs = 103.639 47.02 kg to lbs = 103.661 47.03 kg to lbs = 103.683 47.04 kg to lbs = 103.705 47.05 kg to lbs = 103.727 47.06 kg to lbs = 103.75 47.07 kg to lbs = 103.772 47.08 kg to lbs = 103.794 47.09 kg to lbs = 103.816 47.1 kg to lbs = 103.838 47.11 kg to lbs = 103.86 47.12 kg to lbs = 103.882 47.13 kg to lbs = 103.904 47.14 kg to lbs = 103.926 47.15 kg to lbs = 103.948 47.16 kg to lbs = 103.97 47.17 kg to lbs = 103.992 47.18 kg to lbs = 104.014 47.19 kg to lbs = 104.036 47.2 kg to lbs = 104.058 47.21 kg to lbs = 104.08 47.22 kg to lbs = 104.102 47.23 kg to lbs = 104.124 47.24 kg to lbs = 104.146 47.25 kg to lbs = 104.168\n 47.26 kg to lbs = 104.19 47.27 kg to lbs = 104.213 47.28 kg to lbs = 104.235 47.29 kg to lbs = 104.257 47.3 kg to lbs = 104.279 47.31 kg to lbs = 104.301 47.32 kg to lbs = 104.323 47.33 kg to lbs = 104.345 47.34 kg to lbs = 104.367 47.35 kg to lbs = 104.389 47.36 kg to lbs = 104.411 47.37 kg to lbs = 104.433 47.38 kg to lbs = 104.455 47.39 kg to lbs = 104.477 47.4 kg to lbs = 104.499 47.41 kg to lbs = 104.521 47.42 kg to lbs = 104.543 47.43 kg to lbs = 104.565 47.44 kg to lbs = 104.587 47.45 kg to lbs = 104.609 47.46 kg to lbs = 104.631 47.47 kg to lbs = 104.653 47.48 kg to lbs = 104.675 47.49 kg to lbs = 104.698 47.5 kg to lbs = 104.72\n 47.51 kg to lbs = 104.742 47.52 kg to lbs = 104.764 47.53 kg to lbs = 104.786 47.54 kg to lbs = 104.808 47.55 kg to lbs = 104.83 47.56 kg to lbs = 104.852 47.57 kg to lbs = 104.874 47.58 kg to lbs = 104.896 47.59 kg to lbs = 104.918 47.6 kg to lbs = 104.94 47.61 kg to lbs = 104.962 47.62 kg to lbs = 104.984 47.63 kg to lbs = 105.006 47.64 kg to lbs = 105.028 47.65 kg to lbs = 105.05 47.66 kg to lbs = 105.072 47.67 kg to lbs = 105.094 47.68 kg to lbs = 105.116 47.69 kg to lbs = 105.138 47.7 kg to lbs = 105.16 47.71 kg to lbs = 105.183 47.72 kg to lbs = 105.205 47.73 kg to lbs = 105.227 47.74 kg to lbs = 105.249 47.75 kg to lbs = 105.271\n 47.76 kg to lbs = 105.293 47.77 kg to lbs = 105.315 47.78 kg to lbs = 105.337 47.79 kg to lbs = 105.359 47.8 kg to lbs = 105.381 47.81 kg to lbs = 105.403 47.82 kg to lbs = 105.425 47.83 kg to lbs = 105.447 47.84 kg to lbs = 105.469 47.85 kg to lbs = 105.491 47.86 kg to lbs = 105.513 47.87 kg to lbs = 105.535 47.88 kg to lbs = 105.557 47.89 kg to lbs = 105.579 47.9 kg to lbs = 105.601 47.91 kg to lbs = 105.623 47.92 kg to lbs = 105.646 47.93 kg to lbs = 105.668 47.94 kg to lbs = 105.69 47.95 kg to lbs = 105.712 47.96 kg to lbs = 105.734 47.97 kg to lbs = 105.756 47.98 kg to lbs = 105.778 47.99 kg to lbs = 105.8 48 kg to lbs = 105.822" ]

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[ "# RE: st: Interpretation of regressionmodel of ln-transformed variable\n\n From Maarten buis To [email protected] Subject RE: st: Interpretation of regressionmodel of ln-transformed variable Date Wed, 5 Nov 2008 20:54:52 +0000 (GMT)\n\n```--- \"Lachenbruch, Peter\" <[email protected]> wrote:\n> You can expect differences since your model transforms the response\n> variable, while the glm transforms the mean function. The model you\n> cite below fits log(mu)=XB, while your other model fit\n> E(log(y))=XB. For non-linear functions these won't be the same.\n\nTo expand a bit on that, there are two reasons why the models give\n\n1) In case of the log transformed y, -regress- with the -eform()-\noption will give you a model for the geometric mean, while -glm- with\nthe -link(log)- option will give you a model of the arithmetic mean.\nThe two are different but the results should in most cases be pretty\nclose.\n\n2) -regress- with log transformed y will ignore all observations with\nan y equal to 0. The reason is that ln(0) is not defined so will give\nyou a missing value. -glm- models the average y, and an average of 0 is\nperfectly legal, so -glm- can handle a LOS of 0 without problem. This\ncould lead to larger differences between the two models. If you have\nobservations whose value on the dependent variable is 0, than -glm- is\nthe preferred method.\n\nHope this helps,\nMaarten\n\n-----------------------------------------\nMaarten L. Buis\nDepartment of Social Research Methodology\nVrije Universiteit Amsterdam\nBoelelaan 1081\n1081 HV Amsterdam\nThe Netherlands\n\nBuitenveldertselaan 3 (Metropolitan), room N515\n\n+31 20 5986715\n\nhttp://home.fsw.vu.nl/m.buis/\n-----------------------------------------\n\n*\n* For searches and help try:\n* http://www.stata.com/help.cgi?search\n* http://www.stata.com/support/statalist/faq\n* http://www.ats.ucla.edu/stat/stata/\n```" ]

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[ "# Physics\n\nA 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface.\n\nA) What is the speed of the block after the collision?\n\nB) What percentage of the ball's initial energy is \"lost\"?\n\n1. 👍\n2. 👎\n3. 👁\n1. .050 Vo = 1.050 v\nso\nv = .0476 Vo\n\ninitial Ke = (1/2)(.05)Vo^2\nfinal Ke = (1/2)(1.05)v^2\n\n100 (.05 Vo^2 - 1.05 v^2) /.05 Vo^2\n\n100 (.05 Vo^2 - .00238 Vo^2)/.05 Vo^2\n\n= 95.2 % of initial energy is lost\n\n1. 👍\n2. 👎\n\n## Similar Questions\n\n1. ### Physics\n\nA potter’s wheel with rotational inertia of 45 kg m2, is spinning freely at 40 rpm. The potter drops a lump of clay onto the wheel where it sticks a distance 1.2 m from the rotational axis at the wheel’s center. If the\n\n2. ### Physics\n\nA rotating uniform-density disk of radius 0.6 m is mounted in the vertical plane. The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 5.8 kg. A lump\n\n3. ### physics\n\nA pitcher throws a 0.5kg ball of clay at a 6kg block of wood. The clay sticks to the wood on impact and their joint velocity afterwards is 3m/s. What is the original speed of the clay? I know the answer is 39m/s but I have no idea\n\n4. ### physics\n\nA tennis ball has a mass of 0.057 kg. A professional tennis player hits the ball hard enough to give it a speed of 52 m/s (about 117 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the\n\n1. ### dynamics\n\nA ball is thrown so that its motion is defined by the equations: x = 5t; y = 2 + 6t - 4.9t2, where and x and y are in meters and t is in seconds. Determine: 1. The velocity of the ball at any time t 2. The initial position and\n\n2. ### Physics\n\nA potter’s wheel with rotational inertia 24 kg m2 is spinning freely with no friction at 50 rpm. A piece of clay is dropped from above the wheel and sticks 0.5 m from the center of the wheel. Afterward, the rotational speed of\n\n3. ### Physics\n\n20 g ball of clay traveling east at 3.0 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and the direction of the resulting 50 g ball of clay?\n\n4. ### Physics\n\nA bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.51s after the ball is released from his hands. What is the speed of the\n\n1. ### Physics\n\nTwo identical balls roll off the edge of a horizontal table top. One leaves traveling at twice the speed of the other. Which ball hits the floor first? Why?\n\n2. ### Physics\n\nA golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. The ball leaves her club at an angle of 37.0° above the ground, with a speed of 31.1 m/s. Find the time that\n\n3. ### Physics..please help\n\nA car is traveling at 20m/s when it pulls out to pass a truck that is traveling at only 18m/s. The car accelerates at 2.0m/s^2 for 4.0 s and then maintains this new velocity. First it asked me to calculate the distance during\n\n4. ### PHY SCI\n\nA golfer on a level fairway hits a ball at an angle of 42* to the horizontal that travels 100 yd before striking the ground. He then hits another ball from the same spot with the same speed, but at a different angle. This ball" ]

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[ "# Power Assignment in a Wireless Communication System¶\n\nby Robert Gowers, Roger Hill, Sami Al-Izzi, Timothy Pollington and Keith Briggs\n\nfrom Boyd and Vandenberghe, Convex Optimization, exercise 4.20 page 196\n\nConvex optimization can be used to maximise the minimum signal to inteference plus noise ratio (SINR) of a wireless communication system. Consider a system with $$n$$ transmitters, each with power $$p_j \\geq 0$$, transmitting to $$n$$ receivers. Let $$G_{ij} \\geq 0$$ denote the path gain from transmitter $$j$$ to receiver $$i$$. These path gains form the matrix $$G \\in \\mathbb{R}^{n \\times n}$$.\n\nEach receiver is assigned to a transmitter such that the signal power at receiver $$i$$, $$S_i = G_{ii}p_i$$ and the interefence power at receiver $$i$$ is $$I_i = \\sum_{k\\neq i} G_{ik}p_k$$. Given a noise power $$\\sigma_i$$ at each receiver, the SINR at receiver $$i$$, $$\\gamma_i = \\frac{S_i}{I_i + \\sigma_i}$$.\n\nThe objective is to maximise the minimum SINR of the system under certain power constraints. These constraints are:\n\ni - Each transmitter power $$p_j \\leq P_j^{\\text{max}}$$\n\nii - If the transmitters are partitioned into $$m$$ nonoverlapping groups, $$K_1, ..., K_m$$, which share a common power supply with total power $$P_l^{\\text{gp}}$$: $$\\sum_{k\\in K_l}p_k \\leq P_l^{\\text{gp}}$$.\n\niii - There is a maximum power that each receiver can receive $$P_i^{\\text{rc}}$$, $$\\sum_{k=1}^{n}G_{ik}p_k \\leq P_i^{\\text{rc}}$$.\n\nThe objective function can be rewritten as:\n\nminimise $$\\max_{i=1,...,n}\\frac{I_i + \\sigma_i}{S_i}$$\n\nHowever, since this is a quasiconvex objective function we cannot solve it directly using CVXPY. Instead we must use a bisection method. First we take the step of rewriting the objective, $$\\alpha = \\gamma^{-1} \\geq 0$$, as a constraint:\n\n$$I_i+\\sigma_i \\leq S_i\\alpha$$\n\nThen we choose initial lower and upper bounds $$L_0$$ and $$U_0$$ for $$\\alpha$$, which should be chosen such that $$L < \\alpha^* < U$$, where $$\\alpha^*$$ is the optimal value of $$\\alpha$$. Starting with an initial value $$\\alpha_0 = \\frac{1}{2}(L_0+U_0)$$, feasibility is checked for $$\\alpha_0$$ by using an arbitrary objective function. The new upper and lower bounds are determined from the feasibility:\n\nIf $$\\alpha_0$$ is feasible then $$L_1 = L_0$$, $$U_1 = \\alpha_0$$ and $$\\alpha_1 = \\frac{1}{2}(L_1+U_1)$$.\n\nIf $$\\alpha_0$$ is infeasible then $$L_1 = \\alpha_1$$, $$U_1 = U_0$$ and $$\\alpha_1 = \\frac{1}{2}(L_1+U_1)$$.\n\nThis bisection process is repeated until $$U_N - L_N < \\epsilon$$, where $$\\epsilon$$ is the desired tolerance.\n\n#!/usr/bin/env python3\n# @author: R. Gowers, S. Al-Izzi, T. Pollington, R. Hill & K. Briggs\n\nimport cvxpy as cp\nimport numpy as np\n\ndef maxmin_sinr(G, P_max, P_received, sigma, Group, Group_max, epsilon = 0.001):\n# find n and m from the size of the path gain matrix\nn, m = np.shape(G)\n\n# Checks sizes of inputs\nif m != np.size(P_max):\nprint('Error: P_max dimensions do not match gain matrix dimensions\\n')\nreturn 'Error: P_max dimensions do not match gain matrix dimensions\\n', np.nan, np.nan, np.nan\n\nprint('Error: P_received dimensions do not match gain matrix dimensions\\n')\nreturn 'Error: P_received dimensions do not match gain matrix dimensions', np.nan, np.nan, np.nan\n\nif n != np.size(sigma):\nprint('Error: σ dimensions do not match gain matrix dimensions\\n')\nreturn 'Error: σ dimensions do not match gain matrix dimensions', np.nan, np.nan, np.nan\n\n#I = np.zeros((n,m))\n#S = np.zeros((n,m))\n\ndelta = np.identity(n)\nS = G*delta # signal power matrix\nI = G-S # interference power matrix\n\n# group matrix: number of groups by number of transmitters\nnum_groups = int(np.size(Group,0))\n\nif num_groups != np.size(Group_max):\nprint('Error: Number of groups from Group matrix does not match dimensions of Group_max\\n')\nreturn ('Error: Number of groups from Group matrix does not match dimensions of Group_max',\nnp.nan, np.nan, np.nan, np.nan)\n\n# normalising the max power of a group so it is in the range [0,1]\nGroup_norm = Group/np.sum(Group,axis=1).reshape((num_groups,1))\n\n# create scalar optimisation variable p: the power of the n transmitters\np = cp.Variable(shape=n)\nbest = np.zeros(n)\n\n# set upper and lower bounds for sub-level set\nu = 1e4\nl = 0\n\n# alpha defines the sub-level sets of the generalised linear fractional problem\n# in this case α is the reciprocal of the minimum SINR\nalpha = cp.Parameter(shape=1)\n\n# set up the constraints for the bisection feasibility test\nconstraints = [I*p + sigma <= alpha*S*p, p <= P_max, p >= 0, G*p <= P_received, Group_norm*p <= Group_max]\n\n# define objective function, in our case it's constant as only want to test the solution's feasibility\nobj = cp.Minimize(alpha)\n\n# now check whether the solution lies between u and l\nalpha.value = [u]\nprob = cp.Problem(obj, constraints)\nprob.solve()\n\nif prob.status != 'optimal':\n# in this case the level set u is below the solution\nprint('No optimal solution within bounds\\n')\nreturn 'Error: no optimal solution within bounds', np.nan, np.nan, np.nan\n\nalpha.value = [l]\nprob = cp.Problem(obj, constraints)\nprob.solve()\n\nif prob.status == 'optimal':\n# in this case the level set l is below the solution\nprint('No optimal solution within bounds\\n')\nreturn 'Error: no optimal solution within bounds', np.nan, np.nan, np.nan\n\n# Bisection algortithm starts\nmaxLoop = int(1e7)\nfor i in range(1,maxLoop):\n# First check that u is in the feasible domain and l is not, loop finishes here if this is not the case\n# set α as the midpoint of the interval\nalpha.value = np.atleast_1d((u + l)/2.0)\n\n# test the size of the interval against the specified tolerance\nif u-l <= epsilon:\nbreak\n\n# form and solve problem\nprob = cp.Problem(obj, constraints)\nprob.solve()\n\n# If the problem is feasible u -> α, if not l -> α, best takes the last feasible value as the optimal one as\n# when the tolerance is reached the new α may be out of bounds\nif prob.status == 'optimal':\nu = alpha.value\nbest = p.value\nelse:\nl = alpha.value\n\n# final condition to check that the interval has converged to order ε, i.e. the range of the optimal sublevel set is <=ε\nif u - l > epsilon and i == (maxLoop-1):\nprint(\"Solution not converged to order epsilon\")\n\nreturn l, u, float(alpha.value), best\n\n\n## Example¶\n\nAs a simple example, we will consider a case with $$n=5$$, where $$G_{ij} = 0.6$$ if $$i=j$$ and $$0.1$$ otherwise.\n\n$$P_j^{\\text{max}} = 1$$ for all transmitters and the transmitters are split into two groups, each with $$P_l^{\\text{gp}} = 1.8$$. The first group contains transmitters 1 & 2, while the second group contains 3,4 & 5.\n\nFor all receivers $$P_i^{\\text{rc}} = 4$$ and $$\\sigma_i = 0.1$$.\n\nnp.set_printoptions(precision=3)\n\n# in this case we will use a gain matrix with a signal weight of 0.6 and interference weight of 0.1\nG = np.array([[0.6,0.1,0.1,0.1,0.1],\n[0.1,0.6,0.1,0.1,0.1],\n[0.1,0.1,0.6,0.1,0.1],\n[0.1,0.1,0.1,0.6,0.1],\n[0.1,0.1,0.1,0.1,0.6]])\n\n# in this case m=n, but this generalises if we want n receivers and m transmitters\nn, m = np.shape(G)\n\n# set maximum power of each transmitter and receiver saturation level\nP_max = np.array([1.]*n)\n\n# normalised received power, total possible would be all power from all transmitters so 1/n\n\n# set noise level\nsigma = np.array([0.1,0.1,0.1,0.1,0.1])\n\n# group matrix: number of groups by number of transmitters\nGroup = np.array([[1.,1.,0,0,0],[0,0,1.,1.,1.]])\n\n# max normalised power for groups, number of groups by 1\nGroup_max = np.array([1.8,1.8])\n\n# now run the optimisation problem\nl, u, alpha, best = maxmin_sinr(G, P_max, P_received, sigma, Group, Group_max)\n\nprint('Minimum SINR={:.4g}'.format(1/alpha))\nprint('Power={}'.format(best))\n\nMinimum SINR=1.148\nPower=[0.8 0.8 0.8 0.8 0.8]" ]

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[ "# Problem 9: An electron beam is shot into a region where there is both an electric field pointing upward of magnitude E = 9.6 N/C, and a magnetic field, B, as shown in the figure. While moving through the region the electron beam stays in a straight line with speed v = 390 m/s. Randomized Variables E = 9.6 N/C v = 390 m/s Part (a) In which direction does the electric force act on the electrons? MultipleChoice : 1) Upward. 2) To the right. 3) To the left. 4) Downward. Part (b) In order to keep the electrons in a straight line, in which direction should the magnetic force act? MultipleChoice: 1) Out of the screen. 2) To the left. 3) To the right. 4) Downward. 5) Into the screen. 6) Upward. Part (c) What should the direction of the magnetic field be in order to produce the force in part (b)? MultipleChoice : 1) Out of the screen. 2) Into the screen. Part (d) Express the magnitude of the electric force using the electric field E and the elementary charge e. Expression : FE = Select from the variables below to write your expression. Note that all variables may not be required. a, B, 0, B, d, e, E, g, h, i, j, m, P, t, v Part (e) Calculate the numerical value of the magnitude of FE, in newtons. Numeric : A numeric value is expected and not an expression. FE = Part (f) Express the magnitude of the magnetic force in terms of the elementary charge e, speed v, and B, the magnitude c the magnetic field. Expression : FB =", null, "Need a custom paper ASAP?\nWe can do it today.\nTailored to your instructions. 0% plagiarism.\n\nLooking for someone to write your essay in 2 hours? Assignment123 can help. Hire an expert writer to tackle your essay paper and get it done quickly and efficiently. You can also order a custom essay, buy essay online, or get essay help from our team of experienced writers." ]

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[ "## Working with images\n\nThe package also supports images, including hyperspectral images, however they have to be transformed into datasets. The idea is very simple, we keep information about image pixels in an unfolded form, as a matrix, and use attributes width and height to reshape the data when we need to show it as an image.\n\nThere are three methods that make this procedure easier: mda.im2data(), mda.data2im() and imshow(). The first converts an image (represented as 3-way array) to a data set, second does the opposite and the third takes dataset and shows it as an image. In the code chunk below you will see several examples how the methods work.\n\nWe will use a dataset image available in the package. It is a 3-way array of numbers, if you want to work with e.g. JPEG, PNG or other standard image files you can load them using specific packages (jpeg, png).\n\ndata(pellets)\n\n# convert image to a data matrix and add some attributed\nd = mda.im2data(pellets)\ncolnames(d) = c(\"Red\", \"Green\", \"Blue\")\nattr(d, \"name\") = \"Image\"\n\n# show data values\nmda.show(d, 10)\n## Image\n## -----\n## Red Green Blue\n## [1,] 183 80 101\n## [2,] 191 76 105\n## [3,] 187 73 99\n## [4,] 199 81 113\n## [5,] 198 81 110\n## [6,] 197 84 114\n## [7,] 191 83 109\n## [8,] 193 83 110\n## [9,] 188 83 114\n## [10,] 172 86 115\n# show separate channels and the whole image in plots\npar(mfrow = c(2, 2))\nimshow(d, 1)\nimshow(d, 2)\nimshow(d, 3)\nimshow(d, 1:3)", null, "By default image for one channels is shown using jet color palette for intensities, but you can also use gray colors, palette from colorbrewer2 as well as your own.\n\npar(mfrow = c(2, 2))\nimshow(d, 1)\nimshow(d, 1, colmap = \"gray\")\nimshow(d, 1, colmap = heat.colors(256))\nimshow(d, 1, colmap = colorRampPalette(c(\"red\", \"green\"))(256))", null, "You can work with the image values as normal dataset and show scatter, line plots, calculate statistics, etc.\n\npar(mfrow = c(1, 2))\nmdaplot(d, type = \"p\")\nmdaplot(d, type = \"l\")", null, "However, it will take some time to show these plots as this image has several hundreds of thousands pixels, a faster alternative can be the use density plot based on hexagonal binning. Use type = \"d\" for this as shown below. Number of bins can be adjusted by using parameter nbins.\n\npar(mfrow = c(1, 2))\nmdaplot(d, type = \"d\")\nmdaplot(mda.subset(d, select = c(\"Red\", \"Blue\")), type = \"d\", nbins = 40)", null, "Another useful thing is to set some of the pixels as background. The background pixels are removed from the image dataset physically, there is no way to get them back (in cotrast to excluded rows/pixels). It can be particularly useful when working with e.g. geocorrected hyperspectral images, where, often, many pixels have NA values and there is no need to keep them in memory. To set pixels as background you need to use method mda.setimbg() with either pixel indices or vector with logical values as it is shown below.\n\n# original size\nshow(dim(d))\n## 114000 3\n# set red epixels as background and show new size\nd = mda.setimbg(d, d[, \"Red\"] > 100)\nshow(dim(d))\n## 66471 3\n# show image with background pixels\npar(mfrow = c(1, 2))\nimshow(d, 1)\nimshow(d, 1:3)", null, "All image related attributes are inherited by all object/rows related results, e.g. scores, residuals, predicted values and classes, etc. This means if you provide an image to any modelling method, you can visualise the corresponding results also as an image. Some examples will be shown in chapter about PCA." ]

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[ "Cms技巧\n\n# ECSHOP模板自定义广告位的例子\n\n```/*\n\n* 调用广告位\n\n*/\n\nfunction index_ad_diy(\\$p_id)\n\n{\n\n\\$sql ='SELECT * FROM ' . \\$GLOBALS['ecs']->table('ad') . ' where position_id ='.\\$p_id.' ORDER BY ad_id desc limit 1';\n\n\\$res = \\$GLOBALS['db']->getAll(\\$sql);\n\n\\$arr = array();\n\nforeach (\\$res AS \\$idx => \\$row)\n\n{\n\n\\$arr[\\$idx]['id']          = \\$row['ad_id'];\n\n\\$arr[\\$idx]['title']       = \\$row['ad_name'];\n\n\\$arr[\\$idx]['imglink']       = \\$row['ad_link'];\n\n\\$arr[\\$idx]['imgcode']     = \\$row['ad_code'];\n\n}\n\nreturn \\$arr;\n\n}```\n\n`\\$smarty->assign('ad_diy_left', index_ad_diy(29));`\n\n```{foreach from=\\$ad_diy_left  item=diyAd}\n<a href=\"{\\$diyAd.imglink}\"><img src=\"data/afficheimg/{\\$diyAd.imgcode}\" alt=\"{\\$diyAd.title}\" title=\"{\\$diyAd.title}\" /></a>\n{/foreach}```\n\nTOP", null, "", null, "QQ客服", null, "18910140161", null, "", null, "", null, "" ]

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[ "# Unlevered Beta\n\nUnlevered beta (also called asset beta) represents the systematic risk of the assets of a company. It is the weighted average of equity beta and debt beta. It is called unlevered beta because it can be estimated by dividing the equity beta by a factor of 1 plus (1 – tax rate) times the debt-to-equity ratio of the company.\n\nUnlevered beta essentially neutralizes the effect of capital structure on a company’s systematic risk exposure. It can be used to assess the ‘true’ systematic risk of a company’s assets (not just equity). It follows from the Modigliani & Miller theories that cost of equity increases with increase in debt to equity ratio. The same applies to a company’s equity beta. Due to higher debt, the risk of the company’s stock return’s varying with reference to the market increases. It is useful to remove the systematic risk that has creeped in due to exposure to debt. Unlevered beta is the measure of systematic risk left after the additional financial risk resulting from debt is removed.\n\nUnlevered beta is an important input in the pure play method. The pure play method is an approach to estimating cost of equity which involves unlevering the publicly-available equity beta value of a comparable company using the comparable company debt-to-equity ratio to get the unlevered (asset) beta which is then relevered using the debt-to-equity ratio of the company under analysis.\n\n## Formula\n\nUnlevered beta can be obtained using the following formula:\n\n Unlevered Beta (βA) = Equity Beta (βE) 1 + (1 − t) × D/E\n\nJust like the weighted average cost of capital, we can express beta as the weighted average of the debt beta and equity beta:\n\n βA = βD × VD × (1 − t) + βE × VE VE + VD × (1 − t) VE + VD × (1 − t)\n\nWhere βA is the unlevered beta (i.e. asset beta), βE is equity beta (also called levered beta), VD is the market value of debt, VE is the market value of equity and t is the tax rate.\n\nIn many cases it is safe to assume that the debt beta is zero. This is because normally debt constitutes a lower percentage of the overall capital structure and the relationship between market risk and debt is inherently low (because debt has fixed charge).\n\nAfter such simplification, the above equation reduces as follows:\n\n βA = βE × VE VE + VD × (1 − t)\n\nDividing the numerator and denominator both by VE and taking VE as a common factor in the denominator give as the equation for unlevered beta:\n\n## Example\n\nShah works as a financial analyst at Agha Investments. His next assignment is valuation of Bolan Electric Arts (BEA), a company engaged in production of electric vehicles. He is interested in calculating cost of equity for BEA. Since BEA stock is not publically traded, he cannot estimate beta coefficient by regressing stock return on index return. He wants to unlever and relever equity beta of a similar company, Tesla Motors Inc (TSLA), and has requested you to calculate unlevered beta for TSLA.\n\nFrom a financial database you found that equity beta of Tesla Motors Inc. is 0.7, its debt to equity rate is 2 and applicable tax rate is 35%. Just plug in the figures in the following formula to calculate the unlevered beta.\n\n Unlevered Beta = 0.7 = 0.3 1 + 2 × (1 − 35%)\n\nShah must re-lever this unlevered beta (also called asset beta) in accordance with the capital structure of BEA to find applicable equity beta, which in turn can be used in CAPM to find BEA's cost of equity." ]

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[ "## GATE Exam Syllabus 2022 for Statistics\n\nGATE Statistics Syllabus: Here we bring you the Gate 2022 Exam Syllabus for Statistics subject to prepare for the Graduate Aptitude Test in Engineering (GATE), an all India level examination conducted by the GATE Committee. Check out the GATE Syllabus for Statistics and other important aspects of the GATE Exam Statistics Syllabus.\n\n### GATE Statistics Syllabus\n\nCalculus: Finite, countable and uncountable sets, Real number system as a complete ordered field, Archimedean property; Sequences and series, convergence; Limits, continuity, uniform continuity, differentiability, mean value theorems; Riemann integration, Improper integrals; Functions of two or three variables, continuity, directional derivatives, partial derivatives, total derivative, maxima and minima, saddle point, method of Lagrange’s multipliers; Double and Triple integrals and their applications; Line integrals and Surface integrals, Green’s theorem, Stokes’ theorem, and Gauss divergence theorem.\n\nLinear Algebra: Finite dimensional vector spaces over the real or complex fields; Linear transformations and their matrix representations, rank; systems of linear equations, eigenvalues and eigenvectors, minimal polynomial, Cayley-Hamilton Theorem, diagonalization, Jordan canonical form, symmetric, skew symmetric, Hermitian, skew- Hermitian, orthogonal and unitary matrices; Finite dimensional inner product spaces, Gram-Schmidt orthonormalization process, definite forms.\n\nProbability: Classical, relative frequency and axiomatic definitions of probability, conditional probability, Bayes’ theorem, independent events; Random variables and probability distributions, moments and moment generating functions, quantiles; Standard discrete and continuous univariate distributions; Probability inequalities (Chebyshev, Markov, Jensen); Function of a random variable; Jointly distributed random variables, marginal and conditional distributions, product moments, joint moment generating functions, independence of random variables; Transformations of random variables, sampling distributions, distribution of order statistics and range; Characteristic functions; Modes of convergence; Weak and strong laws of large numbers; Central limit theorem for i.i.d. random variables with existence of higher order moments.\n\nStochastic Processes: Markov chains with finite and countable state space, classification of states, limiting behaviour of n-step transition probabilities, stationary distribution, Poisson and birth-and-death processes.\n\nInference: Unbiasedness, consistency, sufficiency, completeness, uniformly minimum variance unbiased estimation, method of moments and maximum likelihood estimations; Confidence intervals; Tests of hypotheses, most powerful and uniformly most powerful tests, likelihood ratio tests, large sample test,\n\nSign test, Wilcoxon signed rank test, Mann- Whitney U test, test for independence and Chi-square test for goodness of fit.\n\nRegression Analysis: Simple and multiple linear regression, polynomial regression, estimation, confidence intervals and testing for regression coefficients; Partial and multiple correlation coefficients.\n\nMultivariate Analysis: Basic properties of multivariate normal distribution; Multinomial distribution; Wishart distribution; Hotelling’s T2 and related tests; Principal component analysis; Discriminant analysis; Clustering.\n\nDesign of Experiments: One and two-way ANOVA, CRD, RBD, LSD, 22 and 23 Factorial experiments.\n\nAlso See:" ]

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[ "# Sonometer Concept Question\n\n1. Homework Statement\n\nThis is more like a concept question.\n\nIf you have a solenoid with AC frequency f, placed in close proximity on top of a wire with both ends fixed, and projected parallel to it, why is the frequency of the standing wave observed in the wire equal to 2f?\n\n3. The Attempt at a Solution\n\nThis is about magnetic forces and fields which I haven't studied yet. This is out of curiosity, after seeing that statement without any explanation in my last lab. Thank you if you can clear this for me.\n\nRelated Introductory Physics Homework Help News on Phys.org\nandrevdh\nHomework Helper\nThe magnetic field of the solenoid causes the wire to become magnetized with an opposite polarity as that of the solenoid. This means that the wire will become a N and S pole near the solenoid during each cycle of the current in the solenoid. The wire will therefore be attracted by the solenoid twice during each cycle of the current.\n\nThanks. When you mean cycle, do you refer to current going through one coil, or going through the entire solenoid? (I think only one coil) Also, at what moments will the polarity in the solenoid change, thus inducing two magnetic forces on the wire per cycle? Thanks a lot.\n\nandrevdh\nHomework Helper\nSince the solenoid is one continuous piece of wire the current through it will be the same \"everywhere\" in it.\n\nNormally we use the right hand rule to determine the direction of the magnetic field : Curl the fingers of the right hand in the direction of conventional current flow in the coil. The thumb will then point in the direction of the magnetic field lines. This means that the field changes direction when the current changes direction (E and B fields are in phase). The induced magnetism will normally lag behind in time due to the fact that it takes a finite amount of time for the magnetic domains to orientate themselves (this is also true for the solenoid - a delayed effect).\n\nLast edited:\no I think I get it now. Great. Thank you." ]

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[ "# bitwise.slshift() function\n\nThe `bitwise.slshift()` function is experimental and subject to change at any time. By using this function, you accept the risks of experimental functions.\n\n`bitwise.slshift()` shifts bits in `a` left by `b` bits. Both `a` and `b` are integers.\n\n``````import \"experimental/bitwise\"\n\nbitwise.slshift(a: 12, b: 21)\n\n// Returns 25165824\n``````\n\n## Parameters\n\nLeft operand.\n\n### b\n\nNumber of bits to shift." ]

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[ "# A rod made up of a number of circular cross sections as shown in Fig is subjected to a tensile force of 125 . Determine the elongation of the rod. E = 205 GPa.\n\nQuestion-AnswerCategory: Strength of MaterialsA rod made up of a number of circular cross sections as shown in Fig is subjected to a tensile force of 125 . Determine the elongation of the rod. E = 205 GPa.\n\nA rod made up of a number of circular cross sections as shown in Fig is subjected to a tensile force of 125 . Determine the elongation of the rod. E = 205 GPa.\n\nnote : replace 125 KN by 121 KN", null, "", null, "" ]

[ null, "https://i0.wp.com/media.cheggcdn.com/study/347/347e049b-d694-4d0f-9501-3cbb4a1cfa69/image.png", null, "https://i0.wp.com/media.cheggcdn.com/coop/72b/72b42a11-9dd6-4a7a-b320-53e6d0ea6226/1600677118065_Document_824.png", null ]

[ "Recall the formula for R-squared from your first statistics class: \\begin{equation} R^2=\\frac{\\sum\\limits_{i=1}^N (\\hat{Y_i}-\\overline{Y})^2}{\\sum\\limits_{i=1}^N (\\hat{Y_i}-\\overline{Y})^2 + \\sum\\limits_{i=1}^N (\\hat{Y_i}-Y_i)^2}= \\frac{SS_{model}}{SS_{model}+SS_{residual}} \\end{equation} In that same class, you are taught that adding more regressors can not decrease R-squared, even if they have no relationship to the dependent variable.\nThis, however, does not apply to the constant term. At first pass, this seems hard to believe: An unconstrained model should always do at least as well as a constrained model.\nThe catch is, the variance explained by the constant term is not included in the calculation of R-squared - we subtract $\\overline{Y}$ when calculating $SS_{model}$.\nThe no constant restriction implicitly sets $\\overline{Y}$ to zero. This will increase both the model sum of squares and the residual sum of squares. The model sum of squares effect dominates, however, and $R^2$ is pushed towards one.\nI discovered this today, regressing realized GDP on forecasted GDP. Although the sum of squared errors is nearly identical for both models, the model sum of squares is much larger for the no-intercept case:", null, "My takeaway: Be careful when writing your own regress command in Matlab, or any other language. Omitting a constant term can drastically change R-squared." ]

[ null, "http://marcosammon.com/Post_Images/10_17_2016/gdp.png", null ]

[ "This is an old version, view current version.\n\n## 6.1 Numeric Literals\n\nThe simplest form of expression is a literal that denotes a primitive numerical value.\n\n### Integer Literals\n\nInteger literals represent integers of type int. Integer literals are written in base 10 without any separators. Integer literals may contain a single negative sign. (The expression --1 is interpreted as the negation of the literal -1.)\n\nThe following list contains well-formed integer literals.\n\n0, 1, -1, 256, -127098, 24567898765\n\nInteger literals must have values that fall within the bounds for integer values (see section).\n\nInteger literals may not contain decimal points (.). Thus the expressions 1. and 1.0 are of type real and may not be used where a value of type int is required.\n\n### Real Literals\n\nA number written with a period or with scientific notation is assigned to a the continuous numeric type real. Real literals are written in base 10 with a period (.) as a separator and optionally an exponent with optional sign. Examples of well-formed real literals include the following.\n\n0.0, 1.0, 3.14, -217.9387, 2.7e3, -2E-5, 1.23e+3.\n\nThe notation e or E followed by a positive or negative integer denotes a power of 10 to multiply. For instance, 2.7e3 and 2.7e+3 denote $$2.7 \\times 10^3$$, whereas -2E-5 denotes $$-2 \\times 10^{-5}$$." ]

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[ "## Search This Blog\n\n### Generalization of the Difference Equation for a Simple Harmonic Oscillator\n\nGeneralization of the Difference Equation for a Simple Harmonic Oscillator\n\n- Consider a finite set S which contains n distinct objects.\n\n- Consider a map C on S such that C(a,b)=0 or 1 where a,b are any two objects in S.\n\nWe call such a map a Connection Map. If C(a,b)=1 we say that a and b are \"connected\" and a and b are \"neighbors\".\nIn general a member of S may have many neighbors, up to a maximum of n-1.\n\nIf C(a,b)=0 we say that a and b are not connected.\n\nWe require that..\n\nAn object cannot connect to itself. So C(a,a)=0 for all a in S.\nEvery object in S must have at least 1 connection. That is, for any object a in S it's possible to find at least one other object x in S such that C(a,x)=1.\n\nNote that C is completely defined by a nxn matrix who's elements are 0 or 1 and who's diagonal elements are all 0.\n\n- We say that the matrix C imposes a connection on S. A connection completely describes the geometry of S, so we can also say that C is a geometry on S.\n\n- Clearly S can have many distinct geometries.\n\n- Now define a field over S as a function f(a) for each member of S. We'll use a scalar field, so f(a) is simply a number.\n\n- For the Simple Harmonic Oscillator (in 1 degree of freedom) we need to pick a specific geometry on S as follows..\n\nSince members of S are distinct we can label them 1,2,3,..,n. These are just labels and do not imply any arithmetic operations. We then pick a specific Geometry on S defined by C(i,i+1)=1 for i=1,2,3,..,n-1 and all other C's are zero.\n\n-The Simple Harmonic Oscillator is a field on S defined by..\n\nf(n+1)=k*f(n)-f(n-1)\n\nwhere f(n) is the value of the field for object n and k is a constant. f(1), f(2) and k are given and these 3 numbers define the amplitude, wavelength and phase of the field.\n\nWe re-write this as..\n\nf(n)=(f(n+1)+f(n-1))/k\n\nWe now interpret this equation in a more general way..\n\nf(n) is the value of the field for object n and because of the geometry we picked we notice that n+1 and n-1 are the neighbors of n.\n\nSo this equation says that the value of f for object n is determined by the value of f at the neighbors of n. This is only true because of the particular C we used. This is the generalization of the difference equation for the Simple Harmonic Oscillator.\n\nA few things to notice that differ from the usual treatment of the Simple Harmonic Oscillator..\n\nNo differential equation is used.\nThe difference equation has been generalized to a field over S.\nThere is no explicit coordinate system, and that includes the time coordinate.\n\nContent written and posted by Ken Abbott [email protected]" ]

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[ "# CALIBRATION APPARATUS, CALIBRATION SYSTEM, AND CALIBRATION METHOD\n\nIn a calibration apparatus, a captured image acquisition unit acquires data of an image of a calibration chart captured by a fish-eye imaging apparatus. A distortion correction unit makes an assumption of a parameter of the imaging apparatus, takes into account a projection system to make a correction, and adjusts the parameter until an image of the original chart is obtained. A pseudo image drawing unit uses model data of a chart to draw a pseudo captured image. A point-of-view adjustment unit compares the pseudo image and the captured image and adjusts a point-of-view position of a virtual camera according to a difference between positions of markers or intervals of patterns to specify a position of the imaging apparatus.\n\n## Latest Sony Interactive Entertainment Inc. Patents:\n\nDescription\nTECHNICAL FIELD\n\nThe present invention relates to an apparatus, a calibration system, and a calibration method for realizing calibration of an imaging apparatus.\n\nBACKGROUND ART\n\nThere is a known game, in which a camera images the body of a user or a marker, and a region of the image is replaced with another image and displayed on a display (for example, see PTL 1). A technique of detecting and analyzing an image in a captured image to acquire the position or the motion of a subject or a camera or to recognize attributes of the subject is widely implemented not only in a camera mounted on a gaming apparatus or an information terminal, but also in systems including a security camera, a vehicle-mounted camera, a robot-mounted camera, and the like.\n\nTo guarantee the processing accuracy in the systems, a calibration process of acquiring, in advance, information specific to an apparatus, such as an internal parameter of an imaging apparatus, a distortion correction coefficient, and an external parameter, is executed. The internal parameter defines a relationship between position coordinates of pixels in a captured image and position coordinates in a camera coordinate system, in which the origin is at the optical center, and the unit is length. The distortion correction coefficient is a coefficient for correcting barrel distortion or distortion in the circumferential direction caused by the lens.\n\nThe external parameter defines a relationship between the camera coordinate system and a world coordinate system, and the external parameter is particularly used to, for example, perform the calibration at the same time for a multi-lens camera, such as a stereo camera, to adjust the slopes of the images captured by the cameras. An example of a widely known calibration technique includes a Zhang method for imaging a plane chart in a checkered pattern from a plurality of points of view and optimizing parameters so that positions of feature points on the captured image and positions of the feature points in the plane of the chart in a real space correctly correspond (see NPL 1).\n\nCITATION LIST Patent Literature\n\n[PTL 1] European Patent Application Publication No. 0999518\n\nNon Patent Literature\n\n[NPL 1] Zhengyou Zhang, “A Flexible New Technique for Camera Calibration,” Microsoft Research Technical Report, MSR-TR-98-71, Dec. 2, 1998.\n\nSUMMARY Technical Problems\n\nIn the Zhang method, the imaging apparatus and the chart need to be set at some positions and postures to repeat the imaging, and the workload is large. In addition, although a technique of using a fish-eye lens to capture a wide-angle image is generally used in recent years, it is difficult to accurately correct the distortion specific to the fish-eye lens in the entire image.\n\nThe present invention has been made in view of the problems, and an object of the present invention is to provide a technique for easily and accurately performing calibration of a fish-eye camera.\n\nSolution to Problems\n\nTo solve the above problems, an aspect of the present invention relates to a calibration apparatus. The calibration apparatus includes: a captured image acquisition unit that acquires data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens; and a correction unit that corrects distortion caused by the lens based on a relationship between a position on a screen corresponding to the fish-eye lens and a position on a surface of the chart imaged at the position and that evaluates the corrected image to derive and output a value of a parameter regarding the imaging apparatus.\n\nAnother aspect of the present invention also relates to a calibration apparatus. The calibration apparatus includes: a captured image acquisition unit that acquires data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens; a parameter acquisition unit that acquires a value of a parameter regarding the imaging apparatus; a pseudo image drawing unit that arranges a virtual camera and an object model of the chart in a virtual three-dimensional space and that uses the value of the parameter to generate a pseudo image of the captured image as an image captured by the virtual camera; and a point-of-view adjustment unit that compares the generated pseudo image and the captured image to derive a position of the imaging apparatus with respect to the chart.\n\nYet another aspect of the present invention relates to a calibration system. The calibration system includes an imaging apparatus including a fish-eye lens; and the calibration apparatus.\n\nYet another aspect of the present invention relates to a calibration method. The calibration method includes: a step of acquiring data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens; a step of correcting distortion caused by the lens based on a relationship between a position on a screen corresponding to the fish-eye lens and a position on a surface of the chart imaged at the position; and a step of evaluating the corrected image to derive and output a value of a parameter regarding the imaging apparatus.\n\nNote that arbitrary combinations of the constituent elements and things obtained by converting the expressions of the present invention between a method, an apparatus, a system, a recording medium, a computer program, and the like are also effective as modes of the present invention.\n\nAccording to the present invention, the calibration of the imaging apparatus can be easily and accurately performed.\n\nBRIEF DESCRIPTION OF DRAWINGS\n\nFIG. 1 is a diagram exemplifying a configuration of a system in which the present embodiment can be applied.\n\nFIG. 2 is a diagram for describing central projection performed by a general imaging apparatus.\n\nFIG. 3 is a diagram for describing equidistant projection performed by an imaging apparatus according to the present embodiment.\n\nFIG. 4 is a diagram illustrating an internal circuit configuration of a calibration apparatus according to the present embodiment.\n\nFIG. 5 is a diagram illustrating a configuration of functional blocks of the calibration apparatus according to the present embodiment.\n\nFIG. 6 is a flow chart illustrating a procedure of specifying parameters of the imaging apparatus executed by the calibration apparatus according to the present embodiment.\n\nFIG. 7 is a diagram exemplifying a state in which an actual captured image is corrected in the present embodiment.\n\nFIG. 8 is a diagram illustrating transition of a conceptual image based on correction by a distortion correction unit of the present embodiment.\n\nFIG. 9 is a flow chart illustrating a procedure of correcting the captured image executed by the distortion correction unit in the present embodiment.\n\nFIG. 10 is a diagram for describing a process of projective transformation in S32 of FIG. 8.\n\nFIG. 11 is a diagram for describing a process of distortion correction in S30 of FIG. 8.\n\nFIG. 12 is a diagram exemplifying a relationship between an expanded screen surface of central projection and a screen surface of a fish-eye lens according to the present embodiment.\n\nFIG. 13 is a diagram for describing a method of inverse correction of the distortion correction of the lens according to the present embodiment.\n\nFIG. 14 is a diagram for describing an example of a method of searching grid lines of an image after the correction in S46 of FIG. 9.\n\nFIG. 15 is a diagram for describing a method of slightly adjusting a line representing a boundary line of the grid line in the present embodiment.\n\nFIG. 16 is a flow chart illustrating a procedure of adjusting the setting of a posture of the imaging apparatus to bring an image of the grid line after the correction into line with horizontal and vertical directions of an image plane in S48 of FIG. 9.\n\nFIG. 17 is a flow chart illustrating a procedure of adjusting an angle of view and a coefficient of the distortion correction in S50 of FIG. 9.\n\nFIG. 18 is a flow chart illustrating a procedure of drawing a pseudo image executed by a pseudo image drawing unit according to the present embodiment.\n\nFIG. 19 is a diagram exemplifying an actual pseudo image generated by the pseudo image drawing unit according to the present embodiment.\n\nFIG. 20 is a diagram illustrating an image after the correction of the pseudo image illustrated in FIG. 19.\n\nFIG. 21 is a diagram illustrating another example of a pattern of a chart that can be applied to the present embodiment.\n\nFIG. 22 is a diagram exemplifying captured images of the chart according to the present embodiment.\n\nFIG. 23 is a diagram for describing a difference between a system of a general process using an imaging apparatus including a wide-angle lens during operation and a system of the process using the present embodiment.\n\nFIG. 24 is a diagram exemplifying captured images of a chart pattern in which the position and the posture with respect to the fish-eye lens are different according to the present embodiment.\n\nFIG. 25 is a diagram for describing a procedure of generating the pseudo image in the present embodiment.\n\nFIG. 26 is a diagram for describing a procedure of generating the pseudo image in the present embodiment.\n\nFIG. 27 is a diagram for describing a procedure of generating the pseudo image in the present embodiment.\n\nFIG. 28 is a diagram illustrating examples of the pseudo image that can be generated in the present embodiment.\n\nFIG. 29 is a diagram illustrating examples of the pseudo images that can be generated in the present embodiment.\n\nFIG. 30 is a diagram illustrating definition of parameters used for calculation in the present embodiment.\n\nFIG. 31 is a diagram schematically illustrating a procedure of obtaining position coordinates in a three-dimensional space corresponding to position coordinates in the image plane in the present embodiment.\n\nFIG. 32 is a diagram schematically illustrating a procedure of obtaining the position coordinates in the three-dimensional space corresponding to the position coordinates in the image plane in the present embodiment.\n\nFIG. 33 is a flow chart illustrating a procedure of applying the distortion correction to the image of the chart pattern of the captured image executed by the distortion correction unit according to the present embodiment.\n\nFIG. 34 is a flow chart illustrating a procedure of detecting a region of the chart pattern executed by the distortion correction unit according to the present embodiment.\n\nFIG. 35 is a diagram for describing parameters to be evaluated in a posture detection process of the imaging apparatus based on orthogonality of the chart pattern according to the present embodiment.\n\nFIG. 36 is a diagram illustrating results of the correction of the image of the chart pattern in the captured image executed by the distortion correction unit according to the present embodiment.\n\nFIG. 37 is a diagram illustrating images obtained by restoring the chart of each surface from a pseudo image, in which the chart pattern is indicated inside a cube, and the images are captured with the point of view at the center, in the present embodiment.\n\nFIG. 38 is a diagram illustrating a result of restoration of the chart of a surface from a pseudo image in which the inside of the cube is imaged at an arbitrary posture in the present embodiment.\n\nDESCRIPTION OF EMBODIMENTS Embodiment 1\n\nFIG. 1 exemplifies a configuration of a system in which the present embodiment can be applied. The system includes an imaging apparatus 12, a calibration chart 200, and a calibration apparatus 10 that uses images of the chart 200 captured by the imaging apparatus 12 to carry out calibration. The calibration apparatus 10 uses captured images of the chart 200 to acquire predetermined parameters regarding the imaging apparatus 12. Here, the predetermined parameters include general camera parameters, such as a lens distortion correction coefficient, a position of a lens center, and a tangential distortion correction coefficient, as well as a position, a posture, and a horizontal/vertical angle of view (hereinafter, simply referred to as “angle of view”) of the imaging apparatus 12 at the time of imaging the chart 200.\n\nThe imaging apparatus 12 includes an imaging optical system including a fish-eye lens, an image sensor, such as a charge coupled device (CCD) sensor and a complementary metal oxide semiconductor (CMOS) sensor, and a mechanism that carries out general image processing, such as demosaicing process. The imaging apparatus 12 images a subject space at a wide angle. The imaging apparatus 12 may be a single-lens camera including only one set of a lens and an imaging sensor or may be a multi-lens camera including a plurality of sets. The configuration of the camera included in the imaging apparatus 12 may be determined according to the details of information processing executed by using captured images.\n\nFor example, in a case of acquiring the position of an object, such as a user and a controller, in a three-dimensional space from a captured image to realize an electronic game, the camera can be a stereo camera including two cameras provided on the left and right at a known interval. In addition, the mode of the imaging apparatus 12 is not limited to the illustrated one. For example, the imaging apparatus 12 may be integrated with the calibration apparatus 10. Alternatively, the imaging apparatus 12 may be installed on a robot or a head mounted display. In this case, images corresponding to the field of view of the robot or the user wearing the head mounted display can be captured at a predetermined frame rate to reflect the images on the action of the robot, or corresponding images can be used for the display on the head mounted display.\n\nThe chart 200 is a real object, such as a planar board or paper, representing a predetermined pattern (design). Alternatively, a display apparatus displaying the design, an electronic device including the display apparatus, or the like may be used. In the illustrated example, the pattern of the chart 200 includes grid lines including black horizontal lines and vertical lines of a predetermined width arranged at equal intervals on a white background and a marker 202 arranged at a predetermined position, such as on a center rectangle, in a two-dimensional array of white rectangles formed by the grid lines, as illustrated in a front view in the lower part of FIG. 1.\n\nHowever, the pattern of the chart 200 is not limited to this, and the pattern can be any combination of a design indicating the horizontal direction and the vertical direction of the chart at predetermined intervals and a design indicating a specific position on the chart. For example, a marker may be added to a checkered pattern in which white and black are reversed in adjacent rectangles among rectangular regions divided in a grid, or a barcode or the like exemplified later may be used. Although the marker 202 is a black square in the illustrated example, this is not intended to limit the marker 202 to this, and the marker 202 can have any known shape and color that allow to clearly distinguish the marker 202 from the grid lines.\n\nThe calibration apparatus 10 sets a captured image of the chart 200 as input data and acquires the parameters regarding the imaging apparatus 12. Here, the captured image set as input data is an image before correction of barrel distortion specific to the fish-eye lens. The calibration apparatus 10 uses the lens distortion to acquire, from one captured image, parameters such as a posture, a correction coefficient, and the like of the imaging apparatus 12. The calibration apparatus 10 further uses separately prepared model data of the chart to create a pseudo captured image and derives the position of the imaging apparatus 12 in the three-dimensional space based on the position and the size of the image of the marker 202.\n\nThe calibration apparatus 10 stores the obtained parameters in an internal memory or outputs the obtained parameters to an external apparatus not illustrated. Alternatively, the calibration apparatus 10 may have a function of using the obtained parameters to execute information processing of the electronic game or the like and causing an output apparatus not illustrated to display the obtained parameters. Furthermore, the calibration apparatus 10 can be provided in a robot along with the imaging apparatus 12, and the parameters may be used to analyze the captured image to create a surrounding environment map. These cases can realize an information processing apparatus having the functions of the calibration apparatus 10.\n\nFIG. 2 is a diagram for describing central projection performed by a general imaging apparatus without a fish-eye. FIG. 2 illustrates a bird's-eye view of a state in which an imaging apparatus with a focal length fa images a plane at a distance Za from a principal point 14. The image taken by the imaging apparatus corresponds to an image obtained by projecting a subject onto a screen 15 at the distance fa from the principal point 14. In addition, when the sensor size is known, the focal length fa determines an angle of view θ. In this case, ranges 13a and 13b in a same size a on the surface of the subject form images 16a and 16b in the same size (fa·a/Za) regardless of the position on the captured image.\n\nThat is, the size of the image is provided by a function of a ratio of the focal length fa and the distance Za. Therefore, if the ratio is the same, the size of the image may be the same even if the distance Za is different. In other words, the position of the point of view and the angle of view cannot be obtained at the same time from the image size of one captured image. Therefore, in the calibration of a general imaging apparatus, the chart is imaged from various directions to provide simultaneous equations, or the distance to the chart is fixed to capture an image to thereby obtain various camera parameters.\n\nFIG. 3 is a diagram for describing equidistant projection performed by the imaging apparatus according to the present embodiment. FIG. 3 also illustrates the state in which the imaging apparatus with a focal length fb images a plane at a distance Zb from a principal point 18. On the other hand, in the imaging apparatus 12 of the present embodiment, a screen 19 on which the image is projected by the fish-eye lens is spherical. In this case, the longer the distance from the center on the captured image, the smaller the images 20a and 20b of ranges 17a and 17b in a same size b on the surface of the subject.\n\nFurthermore, the change in reduction ratio with respect to the distance from the center on the captured image is dependent on the curvature of the sphere of the screen 19, that is, the focal length fb, regardless of the position of the principal point. This is used in the present embodiment to evaluate the distortion caused by the fish-eye lens to thereby specify the focal length as well as the angle of view and to evaluate the position and the size of the image to thereby specify the position of the imaging apparatus. This enables calibration by using just one captured image of the chart 200 without strict positioning.\n\nFIG. 4 illustrates an internal circuit configuration of the calibration apparatus 10. The calibration apparatus 10 includes a central processing unit (CPU) 122, a graphics processing unit (GPU) 124, and a main memory 126. The CPU 122 controls processing and signal transmission in the constituent elements inside the apparatus based on a program of an operating system, an application, or the like. The GPU 124 executes image processing. The main memory 126 includes a random access memory (RAM) and stores programs and data necessary for processing.\n\nThese components are connected to each other through a bus 130. An input-output interface 128 is further connected to the bus 130. The components connected to the input-output interface 128 include: a communication unit 132 including a peripheral device interface of a universal serial bus (USB), Institute of Electrical and Electronics Engineers (IEEE) 1394, or the like and a network interface of a wired or wireless local area network (LAN); a storage unit 134, such as a hard-disk drive and a non-volatile memory; an output unit 136 that outputs data to an output apparatus, such as a display apparatus and a printer, not illustrated; an input unit 138 that receives data from the imaging apparatus 12 or an input apparatus not illustrated; and a recording medium drive unit 140 that drives a removable recording medium, such as a magnetic disk, an optical disk, and a semiconductor memory.\n\nThe CPU 122 executes an operating system stored in the storage unit 134 to control the entire apparatus. The CPU 122 also executes various programs that are read from a removable recording medium and loaded on the main memory 126 or that are downloaded through the communication unit 132. The GPU 124 has a function of a geometry engine and a function of a rendering processor and draws an image according to a drawing command from the CPU 122.\n\nFIG. 5 illustrates a configuration of functional blocks of the calibration apparatus 10. The functional blocks illustrated in FIG. 5 can be realized by the CPU 122, the GPU 124, the main memory 126, and the like illustrated in FIG. 4 in terms of hardware and can be realized by a computer program or the like loaded from a hard disk or a recording medium to a main memory in terms of software. Therefore, it can be understood by those skilled in the art that the functional blocks can be realized in various forms, such as only hardware, only software, and combinations of hardware and software, and the functional blocks are not limited to any of these.\n\nThe calibration apparatus 10 includes: a captured image acquisition unit 30 that acquires data of a captured image; a distortion correction unit 34 that performs projective transformation of an image generated from an actual captured image and a model and that corrects lens distortion; an image storage unit 36 that stores a corrected image; a parameter storage unit 38 that stores parameters acquired in the correction; a chart image storage unit 40 that stores model data of the pattern of the chart 200; a pseudo image drawing unit 42 that uses the model data of the chart 200 to generate a distorted pseudo image; a point-of-view adjustment unit 44 that acquires a point-of-view position of the imaging apparatus 12 based on the position and the size of the image of the marker; and an output data generation unit 46 that outputs parameters obtained in relation to the imaging apparatus or that generates and outputs data indicating a result of information processing using the parameters.\n\nThe captured image acquisition unit 30 acquires, from the imaging apparatus 12, data of the captured image of the chart 200. In the case where the imaging apparatus 12 includes a multi-lens camera, the captured image acquisition unit 30 acquires all data of the image captured by each camera. The image of the chart in the captured image is distorted due to a slope corresponding to the posture of the imaging apparatus or due to the fish-eye lens. Hereinafter, the deformation from the original chart in the captured image may be collectively referred to as “distortion.”\n\nThe distortion correction unit 34 performs correction of removing the distortion from the captured image. Specifically, the distortion correction unit 34 transforms the image captured in the equidistant projection system into data of central projection and further uses a distortion correction equation to make a more detailed correction. As a result, the grid lines of the chart 200 substantially become straight lines. Furthermore, the distortion correction unit 34 performs projective transformation to eliminate the slope of the image of the chart 200 caused by the posture of the imaging apparatus 12. In these processes, the distortion correction unit 34 uses the angle of view, the posture, the lens distortion correction coefficient, the position on the image corresponding to the lens center (hereinafter, referred to as “the position of the lens center”), and the tangential distortion correction coefficient of the imaging apparatus 12.\n\nThe distortion correction unit 34 first makes an assumption of the parameters to make a correction and adjusts the parameters so that the result indicates the image of the pattern of the original chart. As a result, each parameter can be obtained. The distortion correction unit 34 further applies the same process to the pseudo captured image generated by the pseudo image drawing unit 42 using the model data of the chart. The data of the image after the correction obtained by these correction processes is stored in the image storage unit 36, and the parameters are stored in the parameter storage unit 38.\n\nThe pseudo image drawing unit 42 reads the model data of the chart 200 stored in the chart image storage unit 40 and arranges the model data in a virtual three-dimensional space along with a virtual camera. Here, the posture of the imaging apparatus 12 obtained from the actual captured image is provided to the virtual camera. The position of the imaging apparatus 12 is unknown, and therefore, a temporary value is set. In addition, equidistant projection of the chart 200 as an object model is performed to create the same state as the actual imaging environment of the imaging apparatus 12. In this case, the angle of view, the lens distortion correction coefficient, the position of the lens center, and the tangential distortion correction coefficient obtained with respect to the imaging apparatus 12 stored in the parameter storage unit 38 are used to make a correction opposite the correction for eliminating the lens distortion to thereby generate a distorted image similar to the captured image. Hereinafter, the image will be referred to as a “pseudo image.”\n\nThe point-of-view adjustment unit 44 compares the position and the size of the marker 202 in the image after the correction of the captured image and the position and the size of the marker 202 in the image after the correction of the pseudo image. If the position of the virtual camera that is set when the pseudo image drawing unit 42 generates the pseudo image is not correct, there are differences in the parameters. Therefore, the position of the point of view is adjusted based on the differences to specify the actual position of the imaging apparatus 12.\n\nThe output data generation unit 46 acquires various parameters stored in the parameter storage unit 38 and the position of the captured image specified by the point-of-view adjustment unit 44 and stores the parameters and the position as output data in the storage apparatus or outputs the output data to an external apparatus. Alternatively, the output data generation unit 46 may use the information to correct the image captured during operation or use the information for image analysis to carry out predetermined information processing to thereby generate and output the output data of a display image or the like. For example, in the case where the imaging apparatus 12 is a stereo camera, the differences between the cameras regarding the intervals of principal points and the postures are found out based on the position and the posture of each camera. Therefore, the parameters can be used to correct the stereo image obtained during operation or can be used for calculation based on triangulation to thereby accurately specify the position of the object.\n\nNext, an action of the calibration apparatus 10 that can be realized by the configuration will be described. FIG. 6 is a flow chart illustrating a procedure of specifying the parameters of the imaging apparatus executed by the calibration apparatus 10. First, the captured image acquisition unit 30 acquires the data of the captured image of the chart 200 from the imaging apparatus 12 (S10). In the case where the imaging apparatus 12 is a multi-lens camera, the captured image acquisition unit 30 acquires the data of the image captured by each camera. The data can be an image of the chart 200 captured from substantially the front, and strict positioning is not necessary at the imaging.\n\nThe posture/distortion correction unit 34 corrects the captured image to make the captured image similar to the pattern of the original chart to thereby acquire the parameters regarding the imaging apparatus (S12). As described above, the distortion of the captured image includes distortion caused by the posture of the imaging apparatus 12 and distortion caused by the structure of the imaging apparatus 12 such as a fish-eye lens. Therefore, the posture/distortion correction unit 34 makes an assumption of the parameters necessary for the correction to sequentially perform the correction for eliminating the distortion caused by the posture and the distortion caused by the structure and adjusts the parameters until the obtained data after the correction becomes substantially the same as the pattern of the chart. In this way, the posture/distortion correction unit 34 acquires the final image after the correction and the values of the parameters. However, the position of the imaging apparatus 12 is not used for the correction of the distortion, and the position is not found out at this stage. Note that in a case where the position of the imaging apparatus 12 does not have to be obtained, the process of S14 to S18 described next can be skipped.\n\nNext, the pseudo image drawing unit 42 reads the model data of the chart 200 from the chart image storage unit 40, arranges the model data in a virtual space, and draws the pseudo image based on the image of the model data captured by the virtual camera (S14). In this case, the posture of the virtual camera is set to the value acquired in S12, and a temporary value is provided for the position. In addition, the virtual camera includes the fish-eye lens and has the same angle of view, lens distortion correction coefficient, position of lens center, and tangential distortion correction coefficient as those of the imaging apparatus 12. Accordingly, the chart on the model can be projected and distorted to thereby generate an image similar to the captured image acquired in S10.\n\nNext, the posture/distortion correction unit 34 corrects the pseudo image drawn in S14, as in S12 (S16). That is, the posture/distortion correction unit 34 performs the correction of using the parameters obtained in S12 to eliminate the distortion. As a result, an image similar to the pattern of the chart is obtained, and the position and the magnification of the chart on the image are dependent on the position of the virtual camera that is set in drawing the pseudo image. The point-of-view adjustment unit 44 adjusts the position of the virtual camera so that the position of the marker and the intervals of the grid in the image after the correction of the captured image acquired in S12 and the position of the marker and the intervals of the grid in the image after the correction of the pseudo image acquired in S16 become substantially the same (S18).\n\nThe position of the virtual camera when the difference between them is equal to or smaller than a predetermined threshold is acquired as the actual position of the imaging apparatus 12. The output data generation unit 46 generates and outputs data representing the parameters obtained in the process so far or representing the results of image analysis or information processing performed by using the parameters (S20). As described above, the data may be output to an apparatus that executes other processing, such as a game and display, or to a storage apparatus. Alternatively, the data may be output to the imaging apparatus 12.\n\nNext, a correction method of the captured image performed by the distortion correction unit 34 in S12 of FIG. 6 will be described. FIG. 7 exemplifies a state in which the actual captured image is corrected. A chart in a captured image 50 has a pattern including black straight lines of a predetermined width indicated in a grid on a white background and further including a black rectangular marker indicated in one of the rectangles surrounded by the grid lines, as described in FIG. 1. The imaging apparatus 12 includes the fish-eye lens. Therefore, the longer the distance from substantially the center of the image plane corresponding to the center of the lens is, the more the image is reduced.\n\nThe distortion correction unit 34 corrects the captured image 50 to generate a state similar to the original chart pattern, that is, an image 52 after the correction in which the black straight lines are orthogonal to each other at equal intervals. As described above, the degree of reduction of the image toward the image edges as well as the condition of curve of the straight lines are mainly dependent on the internal structure of the imaging apparatus 12, such as the angle of view of the imaging apparatus 12 and the deviation of the position of the lens center. The degree of inclination of the grid in the horizontal direction and the vertical direction of the image plane is dependent on the posture of the imaging apparatus 12. Therefore, the parameters of them can be derived in the course of correction from the captured image 50 to the image 52.\n\nFIG. 8 illustrates a transition of a conceptual image based on the correction by the distortion correction unit 34. As described above, the correction is a process of eliminating the lens distortion in a captured image 60 to generate an image 62 with a linear grid (S30) and performing projective transformation of the image 62 to obtain an image 64 after correction in which the grid is orthogonal (S32). Image processing is a process of inputting the position coordinates on the captured image 60 and outputting the position coordinates on the image 64 after correction. Furthermore, in the present embodiment, the angle of view, the correction coefficient, and the like of the imaging apparatus 12 are searched so that the grid becomes a linear image in the process of S30, and the original posture of the imaging apparatus 12 is searched so that the grid becomes orthogonal in the process of S32.\n\nThat is, this includes a process of evaluating the linearity and the orthogonality in the target images, such as the images 62 and 64, and checking how the linearity and the orthogonality are improved in the case where various parameters are adjusted. In this case, the position coordinates on the images 62 and 64 to be evaluated can be input, and a processing path for outputting the original position coordinates on the captured image 60 can be prepared to make the adjustment process efficient. That is, a processing path is prepared, in which inverse projective transformation is applied to the image 64 after correction (S34) to provide lens distortion (S36).\n\nThe distortion correction unit 34 appropriately uses these processes to efficiently generate the image 64 after correction. Hereinafter, the processing path of generating the image of the chart pattern from the distorted image as in S30 and S32 will be referred to as “forward direction correction,” and the processing path of generating the distorted image from the image of the chart pattern as in S34 and S36 will be referred to as “backward direction correction.”\n\nFIG. 9 is a flow chart illustrating a procedure of correcting the captured image executed by the distortion correction unit 34. The process corresponds to S12 of FIG. 6. First, the plane of the image after correction is prepared (S40), and the parameters used for the correction are temporarily set (S42). Specifically, the angle of view, the posture (pan angle, tilt angle, and roll angle), the lens distortion correction coefficient, the position of the lens center, and the tangential distortion correction coefficient of the imaging apparatus 12 are set. In many cases, the lens distortion is corrected by the following equations using lens distortion correction coefficients k2, k4, and k6.\n\ni′=i(1+k2*r2+k4*r4+k6*r6)/(1+k2+k4+k6)\n\nj′=j(1+k2*r2+k4*r4+k6*r6)/(1+k2+k4+k6)\n\nHere, (i, j) represents the position coordinates in the distorted image, and (i′, j′) represents the position coordinates after the correction of the distorted image. In addition, r represents the distance between the position coordinates (i, j) in the image before the correction and the lens center, and r is defined by the following equation where (ic, jc) represents the position coordinates of the lens center.\n\nr=((i−ic)2+(j−jc)2)1/2\n\nNote that although the equation indicates the correction up to the sixth degree, the degree of correction is not limited. On the other hand, the tangential distortion is corrected by the following equations using tangential distortion correction coefficients p1 and p2.\n\ni′=i+(2p1*ij+p2(r2+2i2))\n\nj′=j+(p1(r2+2j2)+2p2*ij)\n\nThat is, the equations are used to make a correction dependent on only the distance from the lens center.\n\nNote that as described later, the difference in projection system is first taken into account to correct the distortion caused by the fish-eye lens, and the equations are further used to correct the distortion to improve the correction accuracy in the present embodiment. Next, the corrected image at the time of using the temporary parameters set in S42 is drawn on the image plane prepared in S40 (S44). The processing path used in the process is a processing path of the backward direction correction in which the position coordinates of each pixel in the image plane are used as input values to obtain the corresponding position coordinates on the captured image. As a result, an image similar to the chart pattern is generated, in which the temporarily set parameters are used for the correction.\n\nNext, the straight lines included in the grid of the chart pattern are searched in the image (S46). In a case where the posture of the imaging apparatus 12 set in S42 is different from the actual posture, the straight lines are not orthogonal, and the setting of the posture is adjusted until the straight lines become orthogonal to each other (S48). However, only the intersections of four straight lines that are upper, lower, left, and right straight lines detected on the extreme outside are monitored here to adjust the setting of the posture to thereby increase the processing efficiency. As a result, the posture of the imaging apparatus 12 is found out. In a case where a parameter other than the posture set in S42 is different from the actual parameter, the grid lines in the image generated in S48 are still distorted a little or inclined. Therefore, the setting of the parameter is adjusted until the distortion is eliminated (S50).\n\nIn this case, the points bisecting the width of the grid lines are sampled at predetermined intervals, and only errors with respect to approximate straight lines of the points are monitored to adjust the setting of the parameter to thereby increase the processing efficiency. As a result, the angle of view, the correction coefficient, and the position of the lens center of the imaging apparatus 12 are found out. As a result of the process of S48 and S50, the parameters set in S42 are updated. Therefore, the updated values are used to carry out the drawing process similar to S44 again to generate the final image after correction (S52). In this case, the grid lines and the markers are also detected from the image to specify the intervals of the grid lines and the size of the marker. The data of the image after correction is stored in the image storage unit 36, and the updated parameters are stored in the parameter storage unit 38.\n\nFIG. 10 is a diagram for describing the process of projective transformation in S32 of FIG. 8. In FIG. 10, (a) illustrates a three-dimensional space including the chart 200 and illustrates a point of view 205 and a line-of-sight vector 207 of the imaging apparatus 12. Note that the line-of-sight vector 207 here defines the direction of the optical axis of the imaging apparatus. In the illustrated example, position coordinates (x, y, z) of the point of view 205 are defined by a three-dimensional space formed by an XZ plane parallel to the chart 200 and a Y-axis vertically intersecting the center of the chart. An origin O of the three-dimensional space is a point of view when the Y-axis direction is set as the line-of-sight vector to display the chart pattern in the entire field of view. In other words, the image 64 after correction in FIG. 8 is obtained in a reference state, in which the point of view is at the origin O, and the line-of-sight vector overlaps the Y-axis (imaging apparatus 12 directly faces the chart 200).\n\nIn (b) of FIG. 10, the line-of-sight vector 207 is enlarged to illustrate parameters defining the posture of the imaging apparatus 12. A pan angle p and a tilt angle q are defined based on a perpendicular line L (Y-axis direction) from the point of view 205 to the chart 200, and a roll angle r is defined in the rotation direction of the line-of-sight vector 207. The change in position coordinates on the image when the point of view 205 and the line-of-sight vector 207 are displaced to the standard state can be obtained by a general projective transformation matrix using the parameters.\n\nIf the point of view 205 as a start point and the line-of-sight vector 207 are known, the image 64 after correction in FIG. 8 can be obtained by the projective transformation. In the present embodiment, the point of view 205 and the line-of-sight vector 207 are unknown. Therefore, the parameters defining the point of view 205 and the line-of-sight vector 207 are temporarily set to check the image after the transformation, and the set parameters are adjusted to obtain the target image of the chart pattern. In this way, the point of view 205 and the line-of-sight vector 207 as well as the position and the posture of the imaging apparatus 12 can be specified.\n\nFIG. 11 is a diagram for describing the process of distortion correction in S30 of FIG. 8. In the case of the fish-eye lens, particularly in the general distortion correction using the correction equations, the amount of correction may be insufficient in image edge portions at a large distance from the optical axis, and the elimination of the distortion may not be sufficient. Therefore, in the present embodiment, the projection system is taken into account to correct the distortion specific to the fish-eye lens, and then general correction equations are applied. FIG. 11 illustrates a bird's-eye view of a point of view 74, a screen surface 70 of the central projection indicated by a straight line AB, and a screen surface 72 of the equidistant projection of the fish-eye lens indicated by an arc A2B2.\n\nA and B represent left and right edges in an image with the number of horizontal pixels Nh in the central projection where θ represents the angle of view, and A2 and B2 represent left and right edges in an image of the fish-eye lens corresponding to A and B. A distance f from the point of view 74 to the image is expressed by the following equation, in which the unit is pixel.\n\nf=Nh/(2*(tan(q/2))\n\nA point P (i1,j1) on the screen surface 72 of the equidistant projection and a point Q (i2, j2) on the screen surface 70 of the central projection corresponding to the point P (i1,j1) are in the following relationships with respect to distances r1 and r2 from an image center (strictly speaking, position of the lens center) C, respectively.\n\nr1=f*atan(r2/f)\n\nr2=(i2+j2)1/2\n\nHere, atan (r2/f) represents an angle formed by the line-of-sight vector to the point P (i1, j1) and the point Q (i2, j2) before and after the correction and the optical axis. A ratio s of the distance r1 to the distance r2 is obtained as follows by using the position coordinates of the point Q (i2, j2).\n\ns=r1/r2=(f*a tan(r2/f))/r2\n\nA ratio R of the length from the center to the image edge is\n\nR=(2f*atan(Nh/2f))/Nh,\n\nand therefore, a normalized ratio s′ is as follows.\n\ns′=s/R=(Nh/2r2)*(atan(r2/f)/atan(Nh/2f))\n\nThe calculation is based on the angle of the line-of-sight vector with respect to the optical axis, unlike the general correction equation based on the distance from the image center. That is, not only the information on the plane of the image, but also the three-dimensional information considering the relationship between the information and the point of view can be used to correct the image.\n\nThis can realize a correct correction regardless of the position on the image. Note that the point Q (i2, j2) after the correction is known in obtaining the ratio s′. This is because in the actual process, the screen surface 70 of the central projection is prepared first, and the position coordinates of the pixel on the surface are multiplied by the ratio s′ to specify the referenced pixel in the image of the fish-eye lens before the correction. The pixel values of the referenced pixel can be referenced in this way to generate the image after the correction.\n\nEven after the distortion caused by the fish-eye lens is eliminated in this way, the distortion caused by the individual difference or the condition of assembly of the lens still remains, and the general correction equation based on the distance from the lens center is used to make the correction. However, in the case where the distortion of the fish-eye lens is corrected as described above, the quadratic term of the lens distortion correction equation is skipped. In the present embodiment, the set of processes can be used to perform the distortion correction to obtain an image in which the distortion is accurately eliminated up to the edges of the image. In the correction, the angle of view, the lens distortion correction coefficient, the position of the lens center, and the tangential distortion correction coefficient of the imaging apparatus are used. Therefore, an assumption of the parameters are made once to carry out the correction, and consequently, the parameters are adjusted until the grid lines become similar to straight lines with sufficient accuracy. In this way, final values of the parameters are determined.\n\nIn the illustrated example, the angle of view θ of the fish-eye lens is limited to approximately 100°, and the transformation onto the screen surface 70 of the central projection is easy. On the other hand, the camera using the fish-eye lens has an advantage that images with an angle of view of 360° relative to the point of view can be obtained at one time. To utilize the characteristic, the screen surface of the central projection can be expanded. FIG. 12 exemplifies a relationship between the expanded screen surface of the central projection and the screen surface of the fish-eye lens. As in FIG. 11, FIG. 12 is a bird's-eye view of the point of view 74 and each screen surface. The angle of view is set to 360°, and a screen surface 76 of the fish-eye lens is the entire sphere in which the focal length f is the radius.\n\nIn this case, to correct the entire image, inner walls of six surfaces of a cube circumscribing the screen surface 76 of the sphere are set as screen surfaces of the central projection (for example, screen surfaces 78a, 78b, 78c, and 78d). In addition, the space is divided into regions in a quadrangular pyramid shape, in which the point of view 74 is a vertex, and each surface of the inner wall is a bottom surface. The screen surface to be switched is selected based on the region to which the target pixel on the screen surface 76 of the sphere belongs. In FIG. 12, the screen is looked down from above, and the screen is divided into four quadrants of 90° around the point of view 74. An image belonging to quadrant I of the screen surface 76 of the sphere is corrected to an image on the screen surface 78a. Images belonging to quadrants II, III, and IV are corrected to images on the screen surfaces 78b, 78c, and 78d, respectively.\n\nThe calculation of the correction in each quadrant can be similar to the calculation described above. As for the chart 200, the chart pattern can also be indicated on the inner walls of the cube, and the imaging apparatus 12 can be arranged at the center of the chart pattern to capture images. In this way, the other processes of the present embodiment can also be similarly executed for each quadrant. As a result, the image of the fish-eye lens can be corrected to an image with almost no distortion, without a restriction on the angle of view. In addition to the case in which the angle of view is 360°, an appropriate number of planes can be prepared for arbitrary angles of view smaller than 360°, and the correction can be performed in the same way. In addition, the planes to be prepared are not limited to the surfaces of the cube, and the planes may be planes of a polyhedron circumscribing the screen surface 76 of the sphere, planes of a cylinder circumscribing the screen surface 76 of the sphere, or the like.\n\nOther than to specify the correction coefficients as described so far, the method can also be applied to a mode of generating an image without distortion from an image captured during operation and displaying the image. For example, in a case where the imaging apparatus including the fish-eye lens is provided on a head mounted display to display the captured image as it is or to draw and display a virtual object, the image is temporarily corrected to the image on the screen of the central projection. Furthermore, necessary processing can be executed on each screen, and the image can be projected again on a view screen corresponding to the line of sight of the user to thereby efficiently display the image without distortion.\n\nThe forward direction correction of S30 and S32 in FIG. 8 has been mainly illustrated in the description so far. On the other hand, the process in S44 of FIG. 9 determines the pixel values in the image plane after the correction, and the process is backward direction correction for obtaining the position coordinates of the captured image from the position coordinates. The processing efficiency of the process of adjusting the setting of the posture so that the grid lines become orthogonal to each other in S48 and the process of adjusting the correction parameters to eliminate the distortion in S50 can also be increased by using the backward direction correction to evaluate the condition of adjustment based on only the specific pixels sampled on the temporarily corrected image.\n\nThe projective transformation in S34 of FIG. 8 in the backward direction correction can be easily realized by, for example, means for using an inverse matrix of the projective transformation matrix described in relation to S32. On the other hand, the inverse correction of the distortion correction caused by the lens in S36 of FIG. 8 particularly includes a reverse process of the correction using the distortion correction equations based on the correction coefficients, and the inverse correction requires devising. In the present embodiment, the set of “the distortion correction specific to the fish-eye lens and the correction using the correction coefficients” is carried out as described above. The set is also used as it is in the inverse correction so that the correction becomes substantially opposite.\n\nFIG. 13 is a diagram for describing a method of inversion correction of the distortion correction of the lens. As illustrated in the upper part of FIG. 13, the process includes a process of using the point Q (i2, j2) on an image 82 without lens distortion to obtain the corresponding point P (i1, j1) in an image 80 with lens distortion (arrow B). The correction in the forward direction for eliminating the lens distortion (arrow A) is as described above, and the correction is also used as it is in the inverse correction.\n\nSpecifically, as illustrated in the lower part of FIG. 13, a start point is searched as follows so that an end point after the correction in the forward direction is at the point Q (i2, j2). First, an initial value P0 of the start point is temporarily set at the position of the target point Q (i2, j2), and the correction in the forward direction is performed. An end point Q0 obtained by the correction is obviously deviated from the target point Q (i2, j2). Therefore, for example, the position coordinates of the point P0 are multiplied by a predetermined proportion of the amount of deviation to shift the start point. The correction in the forward direction is applied to the start point P1 again to obtain an end point Q1.\n\nFurthermore, the position coordinates of the point P1 are multiplied by a predetermined proportion of the amount of deviation between the end point Q1 and the target point Q (i2, j2) to further shift the start point, and the correction in the forward direction is performed. The process is repeated until a position Qn of the end point falls withing a predetermined range from the target point Q (i2, j2). A start point Pn upon convergence is set as the end point P (i1, j1) of the inverse correction in the distorted image 80.\n\nFIG. 14 is a diagram for describing an example of the method of searching the grid lines of the image after the correction in S46 of FIG. 9 executed by the distortion correction unit 34. FIG. 14 illustrates an enlarged region near the center in the image after the correction. In the image region, for example, a line segment of a predetermined length around a center 84 of the image is defined, and the total of the pixel values of the line segment is acquired while changing the angle and the position of the line segment. The pixel values are minimum if the entire line segment is in a black region of the grid lines, and therefore, a line segment 86 in such a state is determined.\n\nNext, the line segment 86 is divided into equal parts at the center 84 to obtain two line segments, and the total of the pixel values of each line segment is acquired while moving the line segment in the downward direction of the image plane. When the line segment enters a white rectangle of the chart pattern, the total of the pixel values is large. Threshold determination is actually used to detect such a state. FIG. 14 illustrates line segments 88a and 88b in the state. Here, the right line segment 88b reaches the white rectangle earlier than the left line segment 88a. As a result, it is found out that the initial line segment 86 extends downward to the right with respect to the grid lines of the chart pattern. Therefore, the line segment 86 is adjusted upward to the right by a predetermined angle, and line segments obtained by dividing the line segment 86 again are moved downward to search for a white rectangle. The angle of the original line segment is adjusted based on the timing that the divided line segments reach the white rectangles.\n\nThe process is repeated for a predetermined number of times to find out approximate angle and position of the lower edge of the horizontal line of the grid line of the chart pattern. Similarly, the angle and the position are used to define the line segment, and the line segment is divided to obtain two line segments. The totals of the pixel values of the line segments are acquired while further moving the line segments in the downward direction of the image plane. The total of the pixel values is small when the line segment enters a black grid line of the chart pattern. Threshold determination is actually used to detect such a state. FIG. 14 illustrates line segments 90a and 90b in the state. In this case, a process of adjusting the angle based on which one of the line segments reaches the black grid line first and searching for a black grid line again is repeated. As a result, approximate angle and position of the upper edge of the horizontal line of the grid line of the chart pattern are found out.\n\nSubsequently, the search is performed while adjusting the angle upward and downward based on the straight lines, and all of the slopes and the positions of the upper edges and the lower edges of the horizontal lines of the grid lines can be specified. A similar process can also be applied to the vertical lines of the grid lines to specify the slopes and the positions of the left edges and the right edges. A similar process is also carried out when the final image after correction is generated in S52 of FIG. 9. In this case, a region with a black center is further searched in the white rectangular regions surrounded by the grid lines of the chart pattern to thereby detect the image of the marker. In addition, the size of the marker is obtained, and the size is used along with the intervals of the grid lines to acquire the point-of-view position of the imaging apparatus 12.\n\nThe slope and the position of the boundary line of the grid line specified as described above may be checked in a wide region and slightly adjusted to increase the accuracy. FIG. 15 is a diagram for describing a method of slightly adjusting a line representing the boundary line of the grid line. FIG. 15 illustrates a rectangle 96 detected in an image 92 to be searched, including four outermost straight lines as sides, and illustrates a line 94 to be adjusted. First, intersections 98a and 98b of the line 94 and the rectangle 96 are acquired.\n\nNext, the intersections 98a and 98b are set as start points to define, in the center direction of the image, line segments 100a and 100b of a predetermined length, with the same slope as the line 94. Furthermore, a principle similar to the method of FIG. 14 is used to acquire the totals of the pixel values of the line segments 100a and 100b while moving the line segments 100a and 100b upward or downward. In this way, the timing that the line segments 100a and 100b enter the white rectangles is detected, and the slopes of the straight lines obtained based on the positions are reflected to define the line segments again to repeat the search. According to the process, the slopes and the positions of the boundary lines can be detected at separate positions near both edges of the image, and the boundary lines are connected to provide a line after the adjustment of the line 94.\n\nFIG. 16 is a flow chart illustrating a procedure of adjusting the setting of the posture of the imaging apparatus 12 to bring the image of the grid lines after the correction into line with the horizontal and vertical directions of the image plane in S48 of FIG. 9 executed by the distortion correction unit 34. First, slope errors from the horizontal direction or the vertical direction of the image plane are obtained for four outermost straight lines included in the sides of the rectangle 96 of FIG. 15 among the boundary lines of the grid lines detected as described above (S60). Subsequently, the position coordinates of the intersections of the straight lines are acquired, and the position coordinates before the correction of the four points are specified on the captured image (S62). The backward direction correction can be applied to the process.\n\nFurthermore, the setting of the posture of the imaging apparatus 12 is adjusted in the direction for eliminating the errors obtained in S60 (S64), and an assumption of the posture is made to correct the position coordinates of the four points (S66). Furthermore, the slope errors from the horizontal direction or the vertical direction of the image plane are obtained again for each side of the rectangle with the vertex at the position coordinates after the correction (S68). In a case where the errors are equal to or greater than a predetermined threshold (N in S70), the processes from S64 to S68 are repeated. Once the errors become smaller than the threshold, it is determined that the set posture is correct, and the process ends (Yin S70). As a result, the posture of the imaging apparatus 12 can be accurately obtained by tracking the correction status of just four points.\n\nFIG. 17 is a flow chart illustrating a procedure of adjusting the angle of view and the coefficient of the distortion correction in S50 of FIG. 9 executed by the distortion correction unit 34. As described above, the distortion caused by the lens is eliminated from the captured image in the present embodiment, and the distortion correction specific to the fish-eye lens and the correction using the correction coefficient are carried out. The angle of view and the correction coefficient of the imaging apparatus are used for the correction. On the other hand, the temporarily set values are used in the process of S44 in FIG. 9, and the grid lines may be distorted or tilted. Therefore, the parameters are adjusted while the linearity of the grid lines is checked to thereby derive the correct parameters.\n\nFirst, the information of the boundary lines of the grid lines of the chart pattern detected as described in FIGS. 14 and 15 is used to extract, at predetermined intervals, a midpoint group dividing the width of the grid lines into two equal parts (S80). Next, the least-squares method or the like is used to approximate the extracted midpoint group to a straight line, and the total of the distance from each midpoint to the straight line is acquired as an error (S82). Furthermore, the position coordinates of the midpoint group before the correction are specified on the captured image (S84). The backward direction correction can be applied to the process. Furthermore, each parameter is adjusted in the direction for eliminating the errors obtained in S82 (S86), and the position coordinates of the midpoint group after the correction when the parameters are used to perform the correction are obtained (S88).\n\nFurthermore, as in S82, the errors from the approximate straight line are obtained (S90), and in a case where the errors are equal to or greater than the threshold (N in S92), the processes from S86 to S90 are repeated. Once the errors become smaller than the threshold, it is determined that correct parameters are obtained, and the process ends (Y in S92). In this case, the parameters including the angle of view of the imaging apparatus 12 can also be efficiently obtained by tracking the correction status of a limited number of points.\n\nFIG. 18 is a flow chart illustrating a procedure of drawing the pseudo image executed by the pseudo image drawing unit 42. The process corresponds to S14 of FIG. 6. The pseudo image drawing unit 42 first arranges the object model of the chart 200 in the virtual three-dimensional space as illustrated in FIG. 10 and further arranges the virtual camera with the posture and the angle of view of the imaging apparatus 12 acquired as illustrated in FIGS. 16 and 17 (S100). Temporary values are set for the position. Furthermore, the image plane of the virtual camera is prepared (S102), and each pixel in the image plane is set as a start point to obtain the corresponding position on the chart 200. The process corresponds to the forward direction correction.\n\nThat is, as described in FIG. 8, for each pixel of the image plane of the virtual camera in which the distorted image is to be originally displayed (S104), the distortion specific to the fish-eye lens is corrected, and the correction coefficient is used to perform the correction (S106). The value of each parameter of the imaging apparatus 12 illustrated in FIG. 17 is used for the correction. At this point, the pixel array of the central projection without lens distortion is provided, and the general projective transformation according to the position and the posture of the virtual camera is used to determine the position coordinates of the referenced pixel on the pattern of the chart 200 (S108).\n\nFurthermore, the pixel value is referenced from the position of the referenced pixel to determine the pixel value of the target pixel in the image plane of the virtual camera (S110). The processes from S104 to S110 are applied to all pixels of the image plane (N in S112), and the process ends once the process is completed (Y in S112). The pseudo image of the captured image generated in this way is supplied to the distortion correction unit 34. The distortion correction unit 34 applies the same process as described in FIG. 9 to the image. That is, the distortion correction unit 34 uses various parameters regarding the imaging apparatus obtained in the correction process for the captured image to carry out the correction process and generates the image without distortion.\n\nIn this case, the grid lines are also searched in the image as described in FIGS. 14 and 15, and the marker is further detected to acquire the intervals of the grid lines and the position of the marker. Furthermore, the intervals of the grid lines and the position of the marker are compared with the intervals of the grid lines and the position of the marker in the corrected image of the captured image acquired in S52 of FIG. 9, respectively. In the case where the position of the virtual camera is different from the actual position, the parameters are different from the values obtained for the captured image. Therefore, the point-of-view adjustment unit 44 adjusts the point-of-view position of the virtual camera to eliminate the difference. For example, the pseudo image drawing unit 42 is caused to draw again a pseudo image at the point-of-view position after the adjustment, and the pseudo image is corrected to make an adjustment so that the intervals of the grid lines and the position of the marker match the actual intervals of the grid lines and the position of the marker.\n\nOn the other hand, the adjustment process may be made efficient by using only the image after the correction of the pseudo image. That is, the relationship between the differences between the intervals of the grid lines and the actual intervals of the grid lines and between the position of the marker and the actual position of the marker and the amount of movement of the point-of-view position necessary for eliminating the difference is obtained in advance. The point-of-view position moved by using the relationship based on the actual difference is the position of the imaging apparatus to be obtained. Note that in the case where the imaging apparatus 12 is a multi-lens camera, the process illustrated in FIG. 6 is applied to each captured image. As a result, the position and the posture of each camera can be obtained, and the position and the posture can be used in the same way as external parameters of a general camera. That is, the position and the posture can be used to correct the captured image in extracting the corresponding point from the image captured by each camera.\n\nFIG. 19 exemplifies an actual pseudo image generated by the pseudo image drawing unit 42. That is, FIG. 19 illustrates a captured image when the chart arranged in the virtual space is to be captured by the virtual camera including the fish-eye lens. However, the position of the virtual camera is the position after the adjustment by the point-of-view adjustment unit 44. As a result, it can be recognized that an image substantially the same as the captured image 50 of FIG. 7 is obtained. FIG. 20 illustrates an image after the correction of the pseudo image illustrated in FIG. 19. The image is also substantially the same as the image 52 after the correction in FIG. 7.\n\nStrictly speaking, in the image 52 after the correction of the actual captured image, the distortion characteristics slight vary depending on the position on the image. Therefore, the grid lines may be slightly distorted. However, there is no such a variation in the image of FIG. 20. Therefore, the correction process can reproduce the chart pattern including the grid lines orthogonal to each other. Furthermore, the intervals of the grid lines and the position of the marker in the image obtained first by correcting the pseudo image are actually different from the actual intervals of the grid lines and the position of the marker depending on the setting position of the virtual camera. The position of the point of view can be adjusted so that the intervals of the grid lines and the position of the marker become the same as the actual intervals of the grid lines and the position of the marker, and an image substantially the same as the image 52 after the correction of the captured image can be obtained as illustrated in FIG. 20.\n\nFIG. 21 illustrates another example of the pattern of the chart 200 that can be applied to the present embodiment. The illustrated pattern is called a fractal barcode which is disclosed in the specification of U.S. Pat. No. 7,857,232. In the fractal barcode, a minimum unit of two-dimensional barcode is arranged in two rows and two columns, and this is further arranged in two rows and two columns. In this way, a repeated pattern of a nested layered structure is provided, and the information to be embedded can also be provided in a layered structure. In addition, each barcode includes information regarding the position with respect to the entire barcodes. Therefore, necessary information can be obtained even if only a partial region of the barcodes is imaged by bringing the imaging apparatus close.\n\nThus, the fractal barcode can be used as a marker of the chart pattern of the present embodiment, or a plurality of fractal barcodes can be arranged in the entire chart to substantially specify the position of the imaging apparatus with respect to the chart based on the captured image. Therefore, the virtual camera in the pseudo image drawing unit 42 can be accurately set compared to the original setting, and the processing efficiency can be improved. In addition, a similar effect can be obtained even if only part of the chart is imaged, and the necessity of positioning in imaging can be further reduced. For example, the calibration can be performed if the fractal barcode is in part of the wide field of view of the fish-eye lens.\n\nFIG. 22 exemplifies captured images of the chart 200, (a) illustrating a captured image in a case where the chart pattern including the grid and the marker is displayed on the display screen of a mobile terminal, (b) illustrating a captured image in a case where the chart pattern including the fractal barcode is displayed on a television receiver. The pattern of the chart can be clearly imaged even when the pattern is displayed on the screen of an electronic device in this way. Furthermore, in the case of (b) fractal barcode, the calibration can be performed even when only a partial region is imaged as illustrated. Therefore, a general user can cause an electronic device possessed by the user to display the chart at arbitrary timing during operation to thereby easily carry out the calibration.\n\nFor example, the chart can be displayed on a stand-alone monitor or television receiver, and calibration of an imaging apparatus included in a head mounted display worn by the user can be performed. In this way, the position and the posture of the head mounted display with respect to the monitor or the like are found out. This state can be set as the start point, and the images continuously captured by the imaging apparatus can be tracked to track the motion of the head mounted display. In addition, the relationship between an object in the field of view and the monitor can be found out, and an indoor environment map can be easily created. Simultaneous localization and mapping (SLAM) is known as a technique of analyzing images to create a surrounding environment map. The present embodiment can also be applied to such an image analysis technique.\n\nFIG. 23 is a diagram for describing a difference between the system of the general process using the imaging apparatus including the wide-angle lens during operation and the system of the process using the present embodiment. In the general process illustrated in (a), the two-dimensional correction information on the image plane is mainly used. As illustrated in (a), the image processing is executed, and then the correction information on the plane is provided. The distortion correction and the like can be carried out according to the correction information. As a result, the image processing can be easily executed, and an image without a sense of discomfort in displaying the image can be obtained. To attain the purpose, the image captured by the fish-eye lens can be transformed into planar information early in the calibration.\n\nOn the other hand, in the present embodiment, the distortion of the fish-eye lens can be used in the calibration to obtain three-dimensional information, such as a relationship between the position and posture of the imaging apparatus and the position and posture of the chart. The angle of view is also three-dimensional information in terms of defining the field of view with respect to the point of view. Therefore, the three-dimensional information is provided for the captured image during operation as illustrated in (b), and geometrical information in the three-dimensional space can be acquired. For example, the information can be used to create the environment map. The result can be used to significantly increase the variations of information processing. In a case of displaying the result, more attractive image world, such as virtual reality and augmented reality, can be accurately expressed.\n\nAccording to the present embodiment described above, the fish-eye lens is used to image the chart of a predetermined pattern to perform the calibration. In this case, the distortion caused by the fish-eye lens is used to obtain the angle of view of the imaging apparatus. As a result, one of the angle of view and the point-of-view position, which are difficult to isolate in one imaging, can be known. Furthermore, the obtained posture can be used to draw an image on the model of the chart as viewed from the virtual camera and compare the image with the actual image to thereby specify the point-of-view position.\n\nAs a result, general parameters, such as a correction coefficient, can also be acquired without the burden of fixing the interval between the chart and the imaging apparatus to capture an image or the burden of changing the direction of the imaging apparatus to repeatedly capture images for a plurality of times. In addition, to correct the lens distortion, the image of the equidistant projection of the fish-eye lens is first corrected to the image of the central projection, and then the general correction equation is applied. This can improve the conventionally observed situation that the longer the distance from the center of the image, the lower the correction accuracy. As a result, the information can be acquired from the entire image, and the accuracy of the calibration and the information processing using the calibration can be improved.\n\nTo obtain various parameters from the imaging apparatus, the correction is repeated while the parameters are adjusted until the target image is obtained. In this case, the optimal value of the parameter can be efficiently obtained by focusing on only the sampling point to be evaluated in the target image and tracking the change due to the adjustment. The present embodiment is none other than using the fish-eye lens to acquire the three-dimensional information from one captured image. Therefore, the present embodiment can be used for a geometrical process in the three-dimensional space, such as creating an environment map, during operation, and variations of the process using the captured image can be increased.\n\nThe present embodiment can be applied to the multi-lens camera to acquire the relative position of all cameras and the difference in posture. This can be used to adjust the position of the image captured by each camera, and the position of the subject can be easily acquired by using this. In a stitching process of connecting the captured images of the multi-lens camera to provide a display image, the difference in the relative position and the posture of the cameras can be taken into account to make a slight adjustment based on the projective transformation in the three-dimensional space, and more precise connection can be realized.\n\nIn addition, the position and the posture of the captured image with respect to the chart are also found out, and the chart and the imaging apparatus do not have to be strictly positioned in the calibration. The fractal barcode can be used as a chart pattern to specify the position of the imaging apparatus even when only a partial region is imaged, and this further increases the degree of freedom in imaging. The chart can also be displayed on an electronic device, and the user can easily carry out the calibration regardless of time or location. This can be used as a basis for the subsequent continuous acquisition of information, and the latest correction environment can be provided in relation to the temporal change in the apparatus.\n\nThe present invention has been described based on the embodiment. The embodiment is exemplary, and it can be understood by those skilled in the art that various modifications can be made regarding the combinations of the constituent elements and the processes, and the modifications are also within the scope of the present invention.\n\nFor example, the distortion correction unit of the present embodiment makes an assumption of the parameters to correct the captured image of the chart and derives the parameters by making an adjustment until the image becomes equivalent to the original chart pattern. In the time course of the process, the distortion correction unit performs distortion correction specific to the fish-eye lens based on the difference in the projective transformation. On the other hand, the correction method can be used not only for the calibration, but also for the correction of any image captured by the fish-eye camera.\n\nThat is, an image of outdoors or indoors captured by the fish-eye camera is transformed into an image of the central projection as illustrated in FIG. 11. Furthermore, a general correction equation using a correction coefficient is applied to perform the correction as necessary, and this is displayed or used for information processing. In this case, the entire image can be accurately corrected, and the distortion that often occurs in the image with a wide field of view can be minimized.\n\nFurthermore, in the present embodiment, the distortion correction unit adjusts the parameters, such as an angle of view and a correction coefficient, as described above, and the pseudo image drawing unit uses the parameters to set the virtual camera. On the other hand, other means may be used to acquire the parameters, and the pseudo image drawing unit may use the data to set the virtual camera. For example, a general calibration method is used to acquire the correction coefficient, and a motion sensor included in the imaging apparatus acquires the posture of the imaging apparatus. The result may be used to generate a pseudo image, and the point of view of the virtual camera may be adjusted as in the present embodiment to specify the position of the imaging apparatus.\n\nAlternatively, the posture may also be unknown in addition to the position of the imaging apparatus. In this case, while both the point-of-view position and the line-of-sight vector of the virtual camera are adjusted, the pseudo image obtained as a result of the adjustment can be checked. The point-of-view position and the line-of-sight vector when the pseudo image matches the captured image can be determined as the position and the posture of the imaging apparatus. For the comparison with the captured image, the grid lines and the marker in the image after the correction can be used as in the present embodiment to accurately evaluate the degree of coincidence.\n\nIn addition, although the distortion correction unit 34 in the present embodiment performs the set of the correction for transforming the project system, the general correction using the correction coefficient, and the projective transformation based on the difference in posture from the reference state, one or two of these may not be performed. For example, the correction using the correction coefficient may not be performed as long as the accuracy required only in the correction for transforming the projection system can be obtained. Furthermore, in a situation where it is apparent that the imaging apparatus directly faces the chart, the projective transformation may not be performed. An advantageous effect similar to the present embodiment can also be obtained in this way.\n\nEmbodiment 2\n\nIn Embodiment 1, the distortion specific to the fish-eye lens is basically transformed into the image of the central projection in the positional relationship in which the optical axis of the imaging apparatus matches the line perpendicular to the chart surface as illustrated in FIG. 11. According to the method, the distortion correction of the chart pattern can be realized by easy calculation. On the other hand, the wider the angle of the lens is, the lower the correction accuracy of the pattern in a region away from the optical axis may be. Therefore, in the present embodiment, accurate correction is made possible in the entire captured image regardless of the angle of view of the imaging apparatus 12. The configuration of the functional blocks of the calibration apparatus 10 and the main procedure of calibration are similar to those illustrated in FIGS. 5 and 6. Hereinafter, differences from Embodiment 1 will be mainly described.\n\nFIG. 24 exemplifies captured images of a chart pattern in which the position and the posture with respect to the fish-eye lens are different. An upper part of FIG. 24 illustrates bird's-eye views of positional relationships between the screen surface and the surface of the chart when each image is captured. In FIG. 24, (a) illustrates a captured image of a case in which a chart surface 302a directly faces a screen surface 300 as illustrated in Embodiment 1, (b) illustrates a captured image of a case in which a line perpendicular to a chart surface 302b is tilted 45° to the left with respect to an optical axis oa, and (c) illustrates a captured image of a case in which a line perpendicular to a chart surface 302c is titled 90° to the left with respect to the optical axis oa.\n\nAccording to the wide-angle fish-eye lens, a chart in a wide range can be imaged. Therefore, if the image can be corrected with the same accuracy regardless of the position and the direction of the chart as long as the pattern is imaged, the calibration that can utilize the characteristics of the fish-eye lens to the maximum extent can be realized by one imaging. Therefore, the distortion correction unit 34 and the pseudo image drawing unit 42 of the present embodiment perform calculation different from Embodiment 1 to restore the chart pattern including orthogonal grid lines from an illustrated captured image or to generate a pseudo image of the captured image.\n\nFirst, a method of generating a pseudo image executed by the pseudo image drawing unit 42 will be described. FIGS. 25, 26, and 27 are diagrams for describing the procedure of generating the pseudo image in the present embodiment. As described in FIG. 18, the pseudo image drawing unit 42 arranges the object model of the chart and the virtual camera in the virtual three-dimensional space similar to FIG. 10 and sets each pixel of the image plane corresponding to the screen surface of the virtual camera as a start point to obtain the position on the chart indicated there.\n\nIn the present embodiment, the screen surface is originally spherical in the distortion correction of the lens in S106 of FIG. 18, and based on this, the screen surface is modified and rotated in the three-dimensional space. A coordinate system illustrated in FIGS. 25, 26, and 27 is similar to FIG. 10, and the coordinate system is defined by an XZ plane parallel to the chart and a Y-axis vertically intersecting at the center of the chart. First, a region 310 in FIG. 25 represents an initial state in which a square region directly facing the chart pattern in the screen surface of the virtual camera is indicated by an XZ space.\n\nAlthough the screen surface is actually curved in the Y-axis direction, the region 310 in the XZ space corresponds to a state in which the original chart pattern is viewed from the front. The following process corresponds to obtaining a change in position on the chart pattern projected onto each pixel of the region 310 when the curve of the screen surface is taken into account. Note that the point of view matches the origin at this stage.\n\nA region 312 is obtained by applying, to the region 310, the inverse correction of the general distortion correction using the correction coefficient. The recursive method illustrated in FIG. 13 can be used for the process. However, only the general distortion correction calculation using the lens distortion correction coefficient and the tangential distortion correction coefficient is considered here. In addition, although the correction equation up to the sixth degree is illustrated in Embodiment 1, terms of a higher degree may be included. In a case where a coefficient for correcting the barrel distortion is provided, the region 312 has a pincushion shape as illustrated.\n\nAs described above, the screen surface is actually attached to a sphere around the point of view. Therefore, when the posture of the virtual camera is to be changed, the screen surface in the spherical state can be rotated according to the roll angle r, the tilt angle q, and the pan angle p to accurately reflect the original situation. The rotation in the roll direction is rotation about the Y-axis of the three-dimensional space, that is, rotation in the XZ plane. In the illustrated example, the screen surface is rotated by roll angle r=10° to provide a region 314. A lower part illustrates the state of the region 314 in the XY space when the imaging environment is looked down from above. As illustrated in the lower part, the region 314 is considered to be attached to the sphere of the lens.\n\nWith reference to FIG. 26, the rotation in the tilt direction is rotation about the X-axis in the three-dimensional space. In the illustrated example, the region 314 of FIG. 25 is rotated by tilt angle q=10° to obtain a region 316, and a bird's-eye view of the region 316 is illustrated. The rotation in the pan direction is rotation about the Z-axis in the three-dimensional space. In the illustrated example, the region 316 is rotated by pan angle p=10° to obtain a region 318, and a bird's-eye view of the region 318 is illustrated. In a case where the point of view is not at the origin, the sphere is further moved parallel in the three-dimensional space. In the illustrated example, the sphere is moved 200 mm closer to the plane of the chart to provide a region 320. Note that the order of the rotation of the pan, the tilt, and the roll and the parallel movement is not limited to the illustrated one.\n\nFIG. 27 illustrates a state of projecting the image of the chart onto the screen surface after the transformation. In FIG. 27, (a) illustrates a bird's-eye view of the region 320 of the screen surface after the transformation and a chart 322, as illustrated in FIG. 26. As a result of the illustrated transformation, the location in the three-dimensional space after the displacement of each pixel of the region 310 in the XZ space is found out. Therefore, a vector 326 from a point of view 324 of the virtual camera to the location after the displacement is obtained for each pixel. A point 328 on the chart 322 displayed as an image on the pixel exists at a location of extension of the vector 326.\n\nThat is, the vector 326 can be obtained for each pixel to acquire the position on the chart pattern displayed on the pixel. In FIG. 27, (b) illustrates a region on the chart 322 projected onto the region 320. The region has a pincushion shape as a whole as a result of the inverse correction of the barrel distortion. In addition, as a result of applying the rotation of pan, tilt, and roll in the forward direction for the posture of the virtual camera, a region with asymmetric distortion is formed as illustrated. The image of the region can be arranged as pixels of the original image plane to generate a pseudo image captured by the fish-eye lens.\n\nAs a result of the procedure, the pseudo image can be basically generated without a restriction on the angle of view. FIGS. 28 and 29 illustrate examples of the pseudo image that can be generated in the present embodiment. FIGS. 28 and 29 illustrate a situation in which the chart pattern is displayed inside each of the six surfaces of the cubes, and the point of view of the virtual camera is placed at the center. Note that the color of the chart varies according to the surface. FIG. 28 illustrates pseudo images when the posture of the virtual camera with the horizontal angle of view of 140° is changed.\n\nAmong these, as illustrated in a left bird's-eye view, (a) illustrates a pseudo image generated in a state in which the virtual camera directly faces a surface 330 of the cube, that is, all of the pan angle, the tilt angle, and the roll angle are 0° when the surface parallel to the surface is defined as an XZ plane. As illustrated in a right bird's-eye view, (b) illustrates a pseudo image in a state in which the pan angle is 30° and the tilt angle and the roll angle are 0°, and as illustrated in a left bird's-eye view, (c) illustrates a pseudo image in a state in which the tilt angle is 45° and the pan angle and the roll angle are 0°. As illustrated in a right perspective view, (d) illustrates a pseudo image in a state in which the pan angle is 60°, the tilt angle is 20°, and the roll angle is 45°.\n\nFIG. 29 illustrates pseudo images when the horizontal angle of view of the virtual camera is changed in the state in which the virtual camera directly faces one surface of the cube. The horizontal angle of view is 90° in (a). The point of view is at the center of the cube, and as a result, the breadth of the image of the chart matches the breadth of the captured image. The horizontal angle of view is 140° in (b), and the horizontal angle of view is 180° in (c). In the case of (c), the front half of the left and right surfaces of the cube is in the angle of view. The horizontal angle of view is 270° in (d), and the horizontal angle of view is 360° in (e).\n\nIn the case of (e), the surface in the backward direction of the optical axis among the six surfaces of the cube is in the peripheral region of the field of view, and the marker at the center of the chart on the surface is displayed on the circumference of the edges of the field of view. In this way, according to the method of the present embodiment, the change in the captured image dependent on the posture and the angle of view of the virtual camera can be truly reproduced. In addition, general lens distortion, such as barrel distortion, can also be reflected.\n\nNext, a method of acquiring position coordinates in the three-dimensional space from two-dimensional coordinates of a pixel defined in the XZ space when the screen surface is attached to the sphere will be described. FIG. 30 illustrates definition of parameters used for the calculation. In FIG. 30, (a) illustrates a state in which the three-dimensional space is looked down from above, and (b) illustrates the XZ space. An image plane 340 represents a region in which the chart pattern is imaged, as in the region 310 of FIG. 25. The position coordinates of each pixel are expressed by (i, j), with the origin at the lower left.\n\nThe sizes in the horizontal direction and the vertical direction of the chart 322 in the image plane 340 correspond to the number of horizontal pixels imax and the number of vertical pixels jmax of the image plane 340. In addition, when the angle range in the horizontal direction of the chart 322 from the point of view 324 is defined as θ, the number of pixels (pixel density) Dens of the image plane 340 corresponding to the angle of view of 1° in the horizontal direction is as follows.\n\ndens=i max/(2*θ)\n\nFIGS. 31 and 32 schematically illustrate a procedure of obtaining the position coordinates in the three-dimensional space corresponding to the position coordinates (i, j) in the image plane. First, (a) illustrated in an upper part of FIG. 31 illustrates the position coordinates (i, j) on the image plane 340. The position coordinates (i, j) are converted into the position coordinates in the XZ space with the origin at the center of the image plane 340, and the inverse correction of the lens distortion correction is carried out. The equations of general lens distortion and tangential distortion are used to carry out the correction as described in FIG. 13. The coordinates obtained in this way will be referred to as (x1, z1).\n\nAlthough a line 342 from the origin of the XZ space to (x1, z1) is linear in the XZ space, the line 342 is wound along a unit sphere in the three-dimensional space as illustrated on the right side of (a). Therefore, the position coordinates (x1, z1) are transformed into the position coordinates in the two-dimensional space including the longitude and the latitude as illustrated in (b) of a lower part. Assuming that a point of Y=1 on the Y-axis is the origin in the longitude and latitude space, longitude PP and latitude QQ after the transformation are as follows.\n\nPP=x1/dens\n\nQQ=z1/dens\n\nA diagram at the center of (b) illustrates a perspective view of the longitude PP and the latitude QQ in the unit sphere. Furthermore, in a diagram at the right edge, a plane 344 including the line 342 from the origin to the coordinates (PP, QQ) in the longitude and latitude plane is extracted and illustrated. As illustrated in (b), an angle R of the line 342 wound around the unit sphere is an angle of a combination of the longitude PP and the latitude QQ and is obtained as follows.\n\nR=(PP2+QQ2)1/2\n\nIn addition, an angle tt of the line 342 from the longitude axis in the longitude and latitude space corresponds to an angle of the plane 344, which includes the line 342, from the XY plane and is obtained as follows.\n\ntt=atan(QQ/PP)\n\nTo obtain position coordinates (x3, y3, z3) of the coordinates (PP, QQ) in the three-dimensional space, the plane 344 is first rotated by the angle tt to bring the plane 344 into line with the XY plane as illustrated in an upper part (a) of FIG. 32.\n\nAs a result of the transformation, the line 342 is changed to a line 346, and the following equations are used to transform the position coordinates (PP, QQ) in the longitude and latitude plane into position coordinates (x2, y2) of the XY plane.\n\nx2=sin(R)\n\ny2=cos(R)\n\nThe Y coordinate of the position coordinates (PP, QQ) in the XYZ space is maintained even after the rotation, and y2 is none other than Y coordinate y3 of the position coordinates (x3, y3, z3) in the three-dimensional space to be obtained.\n\nFurthermore, as illustrated in a lower part (b), the plane rotated in this way is rotated in the backward direction by the angle tt this time to restore the original plane 344 (however, defined in the XYZ space), and the position coordinates (x3, y3, z3) of the position coordinates (PP, QQ) in the XYZ space are eventually obtained as follows.\n\nx3=x2*cos(tt)\n\ny3=y2\n\nz3=x2*sin(tt)\n\nThe process can be applied to all pixels of the image plane 340 illustrated in FIG. 30 to transform the image plane into the screen surface of the three-dimensional space. The screen surface is further transformed according to the posture (pan angle p, tilt angle q, roll angle r) of the virtual camera. In this case, the rotation in the roll direction about the Y-axis can be carried out at the same time as the rotation by the angle tt illustrated in (b) of FIG. 32. The transformation equations in this case are as follows.\n\nx3=x2*cos(tt+r)\n\ny3=y2\n\nz3=x2*sin(tt+r)\n\nFurthermore, position coordinates (x4, y4, z4) after the rotation in the tilt direction can be obtained by the following transformation equations.\n\nx4=x3\n\ny4=y3*cos(q)+z3*sin(q)\n\nz4=y3*sin(q)+z3*cos(q)\n\nFurthermore, position coordinates (x5, y5, z5) after the rotation in the pan direction can be obtained by the following transformation equations.\n\nx5=y4*cos(p)+z4*sin(p)\n\ny5=y4*sin(p)+z4*cos(p)\n\nz5=z4\n\nFurthermore, in a case where the position coordinates of the point of view are deviated by (Δx, Δy, Δz) from the origin, position coordinates (x6, y6, z6) moved parallel are further obtained as follows.\n\nx6=x5+Dx\n\ny6=y5+Dy\n\nz6=z5+Dz\n\nThe process can realize the transformation into the region 320 illustrated in FIG. 27, and the corresponding point 328 on the chart 322 can be specified by the vector 326 from the point of view to the position coordinates after transformation. The point 328 can be set as a sampling point, and bilinear interpolation or the like can be appropriately applied to determine the pixel value. As a result, the pseudo images as illustrated in FIGS. 28 and 29 can be generated.\n\nThe transformation process from the two-dimensional space to the three-dimensional space illustrated in FIGS. 31 and 32 is reversible. That is, the procedure of obtaining the position coordinates (x3, y3, z3) of the three-dimensional space from the pixel array of (i, j) in the two-dimensional space corresponding to the original chart pattern can be followed backward to obtain the position coordinates (i, j) of the image of the original chart pattern from the position coordinates (x3, y3, z3). In other words, the two-dimensional information without the distortion can be generated from the three-dimensional information defining the distortion specific to the fish-eye lens.\n\nThe distortion correction unit 34 in the present embodiment uses this to execute a process opposite the process of generating the pseudo image executed by the pseudo image drawing unit 42 to thereby correct the image of the chart pattern in the captured image. Although the basic action of the distortion correction unit 34 is as illustrated in FIG. 9, the distortion correction unit 34 executes a process opposite the process of generating the pseudo image illustrated in FIGS. 31 and 32 in the distortion correction carried out in S44, S50, and S52. Because of the reversibility, it is only necessary to apply the recursive transformation process illustrated in FIG. 13 to the reverse process of the general distortion correction using the correction coefficient.\n\nFIG. 33 is a flow chart illustrating a procedure of applying the distortion correction to the image of the chart pattern of the captured image executed by the distortion correction unit 34. First, a target pixel in the captured image is set, and position coordinates (x7, y7, z7) in the three-dimensional space are acquired from the position coordinates of the target pixel (S200). In the captured image of the equidistant projection exemplified in FIG. 24, the position of the pixel can be expressed by polar coordinates (r, θ) with the origin at the image center. Here, the distance r from the image center is proportional to the zenith angle when the optical axis is set vertically upward in the three-dimensional space, and the angle θ from the reference axis represents the azimuth in the three-dimensional space.\n\nTherefore, the position coordinates (x7, y7, z7) in the three-dimensional space can be easily acquired from the position of the pixel in the captured image. In addition, a unit vector (x8, y8, z8) from the point of view toward the target pixel is obtained as follows (S202).\n\nx8=x7/(x72+y72+z72)1/2\n\ny8=y7/(x72+y72+z72)1/2\n\nz8=z7/(x72+y72+z72)1/2\n\nNext, a process opposite of (b) in FIG. 32 is executed with the origin at the point of view. That is, a line connecting the origin and the position coordinates (x8, y8, z8) is rotated so that the line is positioned on the XY plane (S204). A rotation angle tta in this case can be obtained as follows.\n\ntta=atan(z8/x8)\n\nIn addition, the rotation changes the position coordinates (x8, y8, z8) to the following position coordinates (x9, y9, z9).\n\nx9=x8/cos(tta)\n\ny9=y8\n\nz9=0\n\nSubsequently, a process opposite of (a) in FIG. 32 is executed. That is, an angle Ra of position coordinates (x9, y9) in the Z plane wound around the unit sphere is obtained (S206), and the angle Ra is rotated in the backward direction by the angle tta to restore the angle to thereby obtain longitude Pa1 and latitude Qa1 as follows (S208).\n\nRa=atan(x9/y9)\n\nPa1=Ra*cos(tta)\n\nQa1=Ra*sin(tta)\n\nIn addition, opposite to (b) of FIG. 31, the pixel density dens that is the number of pixels per angle of view of 1° is used to obtain position coordinates (x10, z10) in the XZ space as follows (S210).\n\nx10=Pa1*dens\n\nz10=Qa1*dens\n\nFurthermore, opposite to (a) of FIG. 31, the correction using the equations of general lens distortion and tangential distortion is applied to move the origin to the lower left to obtain the position coordinates in the original image plane (S212). The processes from S200 to S212 are executed for each pixel of the image of the chart pattern (N in S214), and the process ends once the position coordinates after transformation are obtained for all pixels (Y in S214). The process can obtain an image in which the grid of the chart pattern is substantially linear. The process of searching for the posture of the imaging apparatus that can obtain the orthogonality of the grid and searching for the angle of view and the correction coefficient that can obtain more accurate straight lines while performing the illustrated distortion correction is similar to Embodiment 1.\n\nIncidentally, the position and the posture of the chart pattern are accepted in a wide range in the present embodiment, and there are various positions for the image of the chart pattern in the captured image as illustrated in FIG. 24. Therefore, preferably, the distortion correction unit 34 searches in advance for an approximate region of the chart pattern in the captured image and an approximate direction of the chart pattern in the three-dimensional space.\n\nFIG. 34 is a flow chart illustrating a procedure of detecting the region of the chart pattern executed by the distortion correction unit 34. The distortion correction unit 34 first divides the captured image into regions (S220). For example, the angle of view is divided by a predetermined number in the horizontal direction and the vertical direction to divide the captured image into regions on the image corresponding to each angle range. For example, in a case of the imaging apparatus with the angle of view of 210°, the captured image is divided every 70° to obtain 3×3 regions. Next, one of the regions is set as a target, and the image of the region is expanded to a separately prepared image plane at a predetermined magnification (S222).\n\nSubsequently, a horizontal line of the grid lines of the chart pattern is searched from the expanded partial image (S224). The method illustrated in FIG. 14 can be used for the search. As a result of the search, in a case where an effective angle or position is not obtained (N in S226), the search for the horizontal line is repeated by changing the magnification of the image in the same region or setting again the next region as a target (S222, S224).\n\nOnce an efficient result is obtained in the search for the horizontal line (Y in S266), a vertical line is then similarly searched (S208). As a result of the search for the vertical line, in a case where an effective result is not obtained (N in S230), the search for the vertical line is performed again by changing the magnification in the same region or setting again the next region as a target (S222 to S228). In this way, once an effective result is also obtained in the search for the vertical line (Y in S230), it is determined that the region is a region in which the chart pattern is imaged (S232), and the search process ends.\n\nThe distortion correction process is applied to the region of the image of the chart pattern detected in the process, and the substantially linearly corrected chart pattern is used to obtain the posture of the imaging apparatus 12. The substantial procedure is as illustrated in FIG. 16. That is, the adjustment of the posture and the acquisition of the slope of the line are alternately repeated to bring the four outermost straight lines of the detected grid lines into line with the horizontal and vertical directions of the image plane.\n\nFIG. 35 is a diagram for describing the parameters to be evaluated in the posture detection process of the imaging apparatus based on the orthogonality of the chart pattern. An illustrated rectangle represents the image plane, and straight lines illustrated inside indicate the four outermost straight lines of the detected grid lines or indicate two straight lines facing each other.\n\nFirst, (a) illustrates a state without an error in any of the pan angle, the tilt angle, and the roll angle representing the posture of the imaging apparatus 12. In this case, four straight lines of the image of the grid are parallel to the horizontal direction and the vertical direction of the image plane as illustrated. On the other hand, (b) illustrates an image of the grid lines in a case where there is an error in the pan angle, (c) illustrates an image of the grid lines in a case where there is an error in the tilt angle, and (d) illustrates an image of the grid lines in a case where there is an error in the roll angle.\n\nFor example, in the state of (b), a difference epan between the slopes of upper and lower lines is evaluated as an error in the pan angle. The difference epan is positive when, for example, the right side is narrowed as illustrated. In the state of (c), a difference etil between the slopes of the left and right lines is evaluated as an error in the tilt angle. The difference etil is positive when, for example, the lower side is narrowed as illustrated. In the state of (d), an average value erol of the slopes of the upper, lower, left, and right straight lines is evaluated as an error in the roll angle. The average value erol is positive in, for example, the counterclockwise direction as illustrated.\n\nIn Embodiment 1, it is assumed that the chart pattern is imaged from substantially the front. Therefore, an adjustment can be made by acquiring the errors epan, etil, and erol based on the relationships between the straight lines as illustrated and subtracting the epan, etil, and erol from the set pan angle, tilt angle, and roll angle, respectively, to converge the angles into correct pan angle, tilt angle, and roll angle. On the other hand, in the case where the state that the chart pattern is parallel to the optical axis is accepted as in the present embodiment, the correspondence between the positional relationship between the straight lines and the rotation direction with an error may change.\n\nFor example, in a case where there is a chart 360a directly facing the optical axis direction as viewed from the point of view 324 as illustrated in a bird's-eye view on the right side of FIG. 35, the condition of gap between the left and right straight lines changes as illustrated in (c) as a result of rotation in the tilt direction indicated by an arrow. On the other hand, in a case where a chart 360b close to parallel to the optical axis is right next to the optical axis, there is a change similar to the original rotation in the roll direction as illustrated in (d) as a result of the same rotation in the tilt direction. On the other hand, if the chart is rotated in the roll direction in the state of the chart 360b, there is a change similar to the original rotation in the tilt direction illustrated in (c). In this way, the evaluation parameters are switched between the roll and the tilt.\n\nSimilarly, in a case where the chart is right above or right below, the evaluation parameters are switched between the roll and the pan. Therefore, the switches are taken into account depending on the case, and the errors epan, etil, and erol are appropriately combined to calculate the errors in the pan angle p, the tilt angle q, and the roll angle r. Furthermore, the errors can be subtracted from the set values of the pan angle p, the tilt angle q, and the roll angle r until the errors become smaller than thresholds, and the posture of the imaging apparatus can be accurately acquired regardless of the position of the chart.\n\nOnce the posture of the imaging apparatus 12 is obtained in this way, a procedure similar to the procedure illustrated in FIG. 17 is executed to adjust the correction coefficient and the angle of view until the image of the grid of the chart pattern indicates straight lines to thereby determine the final values of the correction coefficient and the angle of view. FIG. 36 illustrates results of the correction of the image of the chart pattern in the captured image executed by the distortion correction unit 34. Images of (a), (b), and (c) are images after correction generated from the captured images illustrated in (a), (b), and (c) of FIG. 24, respectively.\n\nThat is, (a) illustrates a correction result of the captured image of the case in which the chart surface 302a directly faces the screen surface 300, (b) illustrates a correction result of the captured image of the case in which the line perpendicular to the chart surface 302b is tilted 45° to the left with respect to the optical axis oa, and (c) illustrates a correction result of the captured image of the case in which the line perpendicular to the chart surface 302c is tilted 90° to the left with respect to the optical axis oa. It can be recognized that the chart pattern is favorably restored regardless of the position or the posture of the chart with respect to the lens.\n\nIn other words, the posture, the correction coefficient, and the angle of view of the imaging apparatus 12 can be accurately obtained without placing the chart at the center of the field of view to capture an image. The point-of-view adjustment unit 44 compares the chart pattern restored in this way and the chart pattern restored in a similar process applied to the pseudo image to acquire the position of the imaging apparatus 12. Here, a method similar to the method described in the present embodiment is also used to generate the pseudo image generated at the adjustment of the point of view.\n\nIn the mode described so far, it is assumed that the acceptable range of the position and the posture of the chart pattern of one surface is increased. This enables the calibration in the state in which part of the chart pattern is in a corner of the field of view. As a result, information processing using the captured image, such as imaging an object to progress a game according to the motion of the object and generating a display image in accordance with the field of view of the captured image, can be executed in parallel with the calibration. On the other hand, the chart pattern can include a plurality of surfaces as illustrated in FIGS. 28 and 29, and even an imaging apparatus that images the whole sky can accurately calibrate the entire field of view by one imaging.\n\nFIG. 37 illustrates images obtained by restoring the chart of each surface from a pseudo image, in which the chart pattern is indicated inside the cube, and the images are captured at the angle of view of 220°, with the point of view at the center of the chart pattern. In this case, the definition of the three-dimensional space varies in each surface of the chart. Therefore, the correction process is executed by setting a region on the image corresponding to the angle of view including each surface, and the correction process is executed on the basis of this unit. According to the angle of view of 220°, the entire front side and part of the upper, lower, left, and right surfaces are imaged as illustrated in a captured image of (a). Five three-dimensional spaces can be set for the surfaces, and the correction process can be applied to each image to generate five corrected images as illustrated in (b). As illustrated, the chart pattern of all surfaces can be restored in the range in which the images are captured.\n\nReplacing the situation illustrated in FIG. 37 with captured images, the correction process is independently applied as illustrated in (b) to each region cut out according to the angle of view in the images obtained by imaging a plurality of chart patterns at one time as illustrated in (a). As a result, the correction coefficients can be accurately obtained for all regions of the captured images. In addition, the information of the point of view and the posture can be obtained for each of the imaged surfaces. Therefore, the information can be combined to more accurately obtain the point of view, the posture, and the angle of view of the imaging apparatus.\n\nFIG. 38 illustrates a result of restoration of the chart of a surface from a pseudo image in which it is assumed that the inside of the cube is imaged at an arbitrary posture. In FIG. 38, (a) illustrates a pseudo image with the pan angle of 60°, the tilt angle of 20°, and the roll angle of 45° illustrated in (d) of FIG. 28, and (b) illustrates an image after correction of a chart surface 370 positioned in front when the pan angle is 0°, the tilt angle is 0°, and the roll angle is 0°. In this case, the part corresponding to the original chart pattern is also accurately reproduced from only the part of the chart pattern in the field of view.\n\nReplacing the situation illustrated in FIG. 38 with a captured image, an adjustment can be made until the original chart pattern as illustrated in (b) is obtained even if the posture in capturing the captured image illustrated in (a) is unknown and even if the correction coefficient and the angle of view are unknown. In this way, the parameters can be acquired.\n\nAccording to the present embodiment described above, the image plane is provided as a spherical screen surface and mapped in the three-dimensional space to accurately reproduce the distortion specific to the fish-eye lens in the calibration using the characteristics of the fish-eye lens. In addition, the sphere is rotated about each axis to accurately reflect the change in the posture of the imaging apparatus on the image. There is no constraint condition regarding the optical axis in mapping the region including the image of the chart pattern to the sphere, and a similar process can be executed regardless of the location of the chart pattern in the captured image.\n\nAs a result, the accuracy is improved by reproducing the actual situation, and in addition, the acceptable range of the position and the posture of the chart pattern with respect to the field of view in imaging can be significantly increased. This further enables the calibration in the state in which only part of the chart pattern is in the field of view, and a process in parallel with the original operation process can improve the processing efficiency of the calibration and maintain the accuracy of the operation process.\n\nIn addition, since the transformation from the image plane into the sphere is a reversible process, the process of setting the reaching point and searching for the initial state leading to the reaching point can be minimized, and the processing efficiency and the accuracy of the calibration can be improved. Furthermore, the fact that there are few restrictions on the position and the posture of the chart pattern can be utilized to use the chart pattern with a plurality of surfaces to cover the angle of view of the captured image, and accurate calibration can be realized by one imaging regardless of the angle of view. As a result of these, advantageous effects similar to the advantageous effects described in Embodiment 1 can be applied in a wider imaging environment, and the versatility and the convenience can be improved.\n\nNote that in the present embodiment, the chart pattern is also not limited to the grid with straight lines, and the chart pattern may be another pattern, such as the fractal barcode described in Embodiment 1. In addition, the modifications of the usage and the process of the calibration apparatus are similar to the modifications described in Embodiment 1.\n\nREFERENCE SIGNS LIST\n\n10 Calibration apparatus, 12 Imaging apparatus, 30 Captured image acquisition unit, 34 Distortion correction unit, 36 Image storage unit, 38 Parameter storage unit, 40 Chart image storage unit, 44 Point-of-view adjustment unit, 46 Output data generation unit, 122 CPU, 124 GPU, 126 Main memory, 200 Chart\n\nINDUSTRIAL APPLICABILITY\n\nIn this way, the present invention can be used for various information processing apparatuses, such as a calibration apparatus, an imaging apparatus, an image processing apparatus, a display apparatus, a robot, and a personal computer, and for a system and the like including any of these.\n\n## Claims\n\n1. An apparatus comprising:\n\na calibration apparatus including:\na captured image acquisition unit that acquires data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens; and\na correction unit that corrects distortion caused by the lens based on a relationship between a position on a screen corresponding to the fish-eye lens and a position on a surface of the chart imaged at the position and that evaluates the corrected image to derive and output a value of a parameter regarding the imaging apparatus.\n\n2. The apparatus according to claim 1, wherein the correction unit compares an image, which is obtained by correcting the distortion caused by the lens by making an assumption of an angle of view of the imaging apparatus, with a pattern of the chart to evaluate the image and derives the angle of view based on a result of the evaluation.\n\n3. The apparatus according to claim 1, wherein the correction unit further makes an assumption of a posture of the imaging apparatus to perform projective transformation of the corrected image and compares the image with the pattern of the chart to make an evaluation to derive the posture based on a result of the evaluation.\n\n4. The apparatus according to claim 1, further comprising:\n\na pseudo image drawing unit that arranges a virtual camera and an object model of the chart in a virtual three-dimensional space and that uses the value of the parameter to execute a process opposite the correction unit to generate a pseudo image of the captured image as an image captured by the virtual camera; and\na point-of-view adjustment unit that compares the generated pseudo image and the captured image to derive a position of the imaging apparatus with respect to the chart.\n\n5. The apparatus according to claim 4, wherein\n\nthe correction unit applies, to the generated pseudo image, the same correction as the correction applied to the captured image, and\nthe point-of-view adjustment unit compares the corrected images to derive the position of the imaging apparatus.\n\n6. The apparatus according to claim 1, wherein the correction unit makes a correction based on an angle formed by a line-of-sight vector to each pixel position in the captured image and an optical axis and then makes a correction based on a distance from a position corresponding to a lens center to each pixel position.\n\n7. The apparatus according to claim 1, wherein the correction unit obtains, for each pixel of the captured image, position coordinates on a sphere with a center at a point of view of the imaging apparatus in a three-dimensional space defined by an image plane parallel to the surface of the chart and by a line perpendicular to the surface of the chart and transforms, based on longitude and latitude coordinates of the sphere corresponding to the position coordinates, the position coordinates into position coordinates in a space of the image plane to thereby correct the distortion caused by the lens.\n\n8. The apparatus according to claim 7, wherein\n\nthe captured image acquisition unit acquires data of an image obtained by imaging, at one time, the chart displayed on each of a plurality of surfaces at different angles, and\nthe correction unit defines a corresponding three-dimensional space for each surface of the chart to correct the distortion caused by the lens to derive the value of the parameter.\n\n9. The apparatus according to claim 1, wherein the correction unit makes the correction while adjusting the parameter and repeats evaluation of a predetermined sampling point in the corrected image to determine the value of the parameter.\n\n10. The apparatus according to claim 1, wherein the correction unit individually corrects the distortion caused by the lens according to ranges of the angle of view of the imaging apparatus.\n\n11. The apparatus according to claim 1, wherein the captured image acquisition unit acquires the data of the captured image of the chart including grid lines, which include a plurality of straight lines extending in a horizontal direction and a vertical direction of the image plane, and including a marker in a predetermined shape, which is arranged in a rectangular region at a predetermined position among rectangular regions surrounded by the straight lines.\n\n12. The apparatus according to claim 1, wherein the captured image acquisition unit acquires the data of the captured image of the chart including a nested barcode array including a plurality of arrays of two-dimensional barcodes of a lower layer contained in an array of two-dimensional barcodes of an upper layer.\n\n13. An apparatus comprising:\n\na calibration apparatus including:\na captured image acquisition unit that acquires data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens;\na parameter acquisition unit that acquires a value of a parameter regarding the imaging apparatus;\na pseudo image drawing unit that arranges a virtual camera and an object model of the chart in a virtual three-dimensional space and that uses the value of the parameter to generate a pseudo image of the captured image as an image captured by the virtual camera; and\na point-of-view adjustment unit that compares the generated pseudo image and the captured image to derive a position of the imaging apparatus with respect to the chart.\n\n14. The apparatus according to claim 13, wherein the pseudo image drawing unit transforms position coordinates of pixels of an image plane parallel to a surface of the chart into position coordinates on a screen surface on a sphere with a center at a point of view of the virtual camera in a three-dimensional space defined by the image plane and by a line perpendicular to the surface of the chart and acquires, for each of the pixels, a sampling point on the surface of the chart based on the point of view and the position coordinates after the transformation to thereby generate the pseudo image of the captured image.\n\n15. The apparatus according to claim 14, wherein the pseudo image drawing unit transforms the position coordinates of the pixels of the image plane into longitude and latitude coordinates of the sphere to obtain corresponding positions on the sphere and obtains the position coordinates after the transformation in the three-dimensional space based on results of obtaining the positions.\n\n16. The apparatus according to claim 14, wherein\n\nthe parameter acquisition unit acquires information regarding a posture of the imaging apparatus, and\nthe pseudo image drawing unit rotates the sphere in the three-dimensional space based on the posture to transform the position coordinates into position coordinates corresponding to the posture.\n\n17. The apparatus of claim 1, further comprising the imaging apparatus including the fish-eye lens.\n\n18. A calibration method, comprising:\n\nacquiring data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens;\ncorrecting distortion caused by the lens based on a relationship between a position on a screen corresponding to the fish-eye lens and a position on a surface of the chart imaged at the position; and\nevaluating the corrected image to derive and output a value of a parameter regarding the imaging apparatus.\n\n19. A non-transitory, computer-readable storage medium containing a computer program, which when executed by a computer, causes the computer to carry out actions, comprising:\n\nacquiring data of a captured image of a calibration chart from an imaging apparatus including a fish-eye lens;\ncorrecting distortion caused by the lens based on a relationship between a position on a screen corresponding to the fish-eye lens and a position on a surface of the chart imaged at the position; and\nevaluating the corrected image to derive and output a value of a parameter regarding the imaging apparatus.\n\n20. The apparatus of claim 13, further comprising the imaging apparatus including the fish-eye lens.\n\nPatent History\nPublication number: 20200273205\nType: Application\nFiled: May 9, 2018\nPublication Date: Aug 27, 2020\nApplicant: Sony Interactive Entertainment Inc. (Tokyo)\nInventors: Noriyuki Yamashita (Tokyo), Akio Ohba (Kanagawa)\nApplication Number: 16/643,678\nClassifications\nInternational Classification: G06T 7/80 (20060101); H04N 17/00 (20060101); G06T 5/00 (20060101); G06T 7/70 (20060101); G06T 15/20 (20060101);" ]

[ null ]

[ "", null, "", null, "", null, "Chapter 4.2, Problem 75E\n\nChapter\nSection\nTextbook Problem\n\n# Find ∫ 1 2 x − 2 d x . Hint: Choose x i * to be the geometric mean of x i − 1 and x i (that is, x i * = x i − 1 x i ) and use the identity 1 m ( m + 1 ) = 1 m − 1 m + 1\n\nTo determine\n\nTo find:\n\nThe 12x-2dx by choosing xi* to be the geometric mean of xi-1 and xi and use the identity 1mm+1=1m-1m+1\n\nExplanation\n\n1) Concept:\n\nUse theorem (4) to find the definite integral.\n\nTheorem (4):\n\nIf f is integrable on [a, b], then\n\nabfxdx=limni=1nfxi* x\n\nWhere\n\nx= b - an and xi* is in [xi-1,xi] where xi=a+i x\n\n2) Formula:\n\ni=1n(ai-bi)= i=1nai-i=1nbi\n\n3) Given:\n\n12x-2dx\n\nAnd\n\n1mm+1=1m-1m+1\n\n4) Calculation:\n\nHere, a=1 and b=2\n\nx=b-an=2-1n=1n\n\nThus,xi=a+i x =1+ in=1+in\n\nAnd\n\nxi-1=1+i-1n\n\nChoose xi* to be the geometric mean of xi-1 and xi that is\n\nxi*=xi-1xi Thus\n\nxi*=xi-1xi= 1+i-1n1+in\n\nBy using theorem (4)\n\n12x-2dx=limni=1nfxi* x\n\n=limn1ni=1n1xi*2\n\nSubstitute,\n\nxi*= 1+\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\n#### Find more solutions based on key concepts", null, "" ]

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[ "", null, "# Php Programming Code Examples\n\n## Php > HTML and Php Code Examples\n\n### Calender in PHP\n\nC Program Codes Find Reverse of an Array - C program to read elements in an array and find its reverse. C Program to print reverse of an array. There are various ways to reverse an array. The basic three algorithms to reverse a\n\nPointer Simple Program Pass by Reference - C++ Language program sample which passes a \"Reference\" to a Function and Function has a 'pointer' as argument. Keep the return type void & display result in main body. Apply any\n\nProgram code to check symmetric matrix - C program to read elements in a matrix and check whether the given matrix is symmetric matrix or not. Symmetric matrix is a square matrix which is equal to its transpose. A is...\n\nC Operate Appropriately On The Numbers - Program take two numbers and a operator as input and store it in the variables a, b and ch respectively. Using switch statement, test the operator stored in the variable ch. Print the\n\nSum of each Row and Columns of a matrix - C program to read elements in a matrix and find the sum of elements of each row and columns of matrix. C program to calculate sum of rows and columns of matrix. Input...\n\nC Trim Trailing White Spaces from Strings - C Program code remove 'trailing white space' characters is 'lot more easier' and faster than removing leading white space characters. As in this case we need not to perform any 'char'\n\nProgram to Solve a Matching Problem For - Program to solve a matching problem. Given N men and N women, where each person has ranked all members of the opposite gender in order of preference, marry the men & women\n\nCoordinated Universal Time in C Language - C program code returns time in Coordinated Universal Time (UTC) (essentially Greenwich mean time). Printing the local time and date. Printing coordinated Universal Time and date" ]

[ null, "https://happycodings.com/images/HappyCodings.png", null ]

[ "## CHEM1109 - Syllabus", null, "# Chemistry 1 Life Sciences B\n\nThe order of topics presented here is essentially the order of presentation in the lecture course, although individual lecturers may make some variations from this. Some topics span more than one week.\n\nJump to week <1> <2> <3> <4> <5> <6> <7> <8> <9> <10> <11> <12> <13>\n\n## Topics covered\n\n1\n\nIntroduction\n- overview of course\n\nIntroduction to Chemical Energetics\n- first law of thermodynamics\n- enthalpy\n- Hess' Law\n- standard enthalpy of formation\n\n2\n\nIntroduction to Chemical Energetics (continued)\n- spontaneous processes and entropy\n- entropy and the second law of thermodynamics\n- the effect of temperature and spontaneity\n- entropy change in chemical reactions\n- free energy\n- free energy and chemical reactions\n- the dependence of free energy on pressure\n- free energy and equilibrium\n- free energy and work\n\n3\n\nGas Laws\n- effects of temperature and pressure on volume\n- partial pressures\n- ideal gas law\n\nChemical Equilibrium\n- the equilibrium condition\n- the equilibrium constant; Kc and Kp\n- heterogeneous equilibrium\n- uses of equilibrium constants - the reaction quotient\n\n4\n\nChemical Equilibrium\n- solving equilibrium problems\n- factors affecting equilibrium: Le Chatelier's principle\n- the effects of catalysts\n\n5\n\nSolutions\n- factors affecting solubility: molecular structure; pressure (Henry's Law); temperature\n- molarity and mole fraction\n- the vapour pressure of solutions\n- Raoult's law\n- colligative properties - boiling point elevation, freezing point depression\n- osmotic pressure, dialysis and the artificial kidney\n\n6\n\nAcids and Bases\n- calculations involving pH, Ka and Kb\n- acid-base properties of salt solutions\n- hydrolysis\n- the Lewis theory of acids and bases\n\n7\n\nAcids and Bases\n- the common ion effect\n- calculations involving buffer solutions\n\nSolubility\n- solubility equilibrium\n- common ion effect\n- solubility and pH\n- Qsp/Ksp criterion\n\n8\n\nOxidation Numbers\n\nComplexes\n- complex ions\n- the structure of complexes\n- chelates, biologically important metal complexes\n- nomenclature\n- isomerism\n\n9\n\nRedox Reactions and Introduction to Electrochemistry\n- electrochemical cells\n- cell EMF, standard reduction potentials\n- cell potential, electrical work and free energy\n\n10\n\nRedox Reactions and Introduction to Electrochemistry\n- dependence of cell potential on concentrations\n- the Nernst equation\n\n11\n\nChemical Kinetics\n- reaction rate\n- dependence of reaction rate on concentration\n- the rate law\n- relation between reactant concentrations and time\n- first order rate laws, half-life, second order rate laws, zero order rate laws\n- activation energy, the Arrhenius equation\n- catalysis\n- enzymes\n\n12\n\n- elementary sub-atomic particles\n- structure of nuclei, Z and A, revision\n- isotopes\n- nuclear stability and radioactive decay\n- half-life of nuclear disintegration\n- nuclear transformations\n\n13\n\nIntroduction to Colloids and Surface Chemistry\n- the colloidal state, types and properties of colloids\n- surface tension\n- detergency\n- model of biological membranes\n- micelle formation, lipid bilayers\n\n# Textbook\n\nBlackman, Bottle, Schmid, Mocerino and Wille,\nChemistry, 3rd Edition, 2015 (John Wiley) ISBN: 978-0-7303-1105-8 (paperback) and 978-0-7303-2492-8 (e-text)\n\nTextbook resources, including the answers to the odd-numbered problems of Chemistry, 3rd Edition and the web address for the publisher's site, are available via \"Textbook Resources\" on USYD eLearning." ]

[ null, "https://scilearn.sydney.edu.au/fychemistry/images/uni_logo_2.gif", null ]

[ "G05 Chapter Contents\nG05 Chapter Introduction\nNAG Library Manual\n\n# NAG Library Routine DocumentG05YMF\n\nNote:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.\n\n## 1  Purpose\n\nG05YMF generates a uniformly distributed low-discrepancy sequence as proposed by Sobol, Faure or Niederreiter. It must be preceded by a call to one of the initialization routines G05YLF or G05YNF.\n\n## 2  Specification\n\n SUBROUTINE G05YMF ( N, RCORD, QUAS, LDQUAS, IREF, IFAIL)\n INTEGER N, RCORD, LDQUAS, IREF($\\mathit{liref}$), IFAIL REAL (KIND=nag_wp) QUAS(LDQUAS,$\\mathit{tdquas}$)\n\n## 3  Description\n\nLow discrepancy (quasi-random) sequences are used in numerical integration, simulation and optimization. Like pseudorandom numbers they are uniformly distributed but they are not statistically independent, rather they are designed to give more even distribution in multidimensional space (uniformity). Therefore they are often more efficient than pseudorandom numbers in multidimensional Monte–Carlo methods.\nG05YMF generates a set of points ${x}^{1},{x}^{2},\\dots ,{x}^{N}$ with high uniformity in the $S$-dimensional unit cube ${I}^{S}={\\left[0,1\\right]}^{S}$.\nLet $G$ be a subset of ${I}^{S}$ and define the counting function ${S}_{N}\\left(G\\right)$ as the number of points ${x}^{i}\\in G$. For each $x=\\left({x}_{1},{x}_{2},\\dots ,{x}_{S}\\right)\\in {I}^{S}$, let ${G}_{x}$ be the rectangular $S$-dimensional region\n $G x = 0, x 1 × 0, x 2 ×⋯× 0, x S$\nwith volume ${x}_{1},{x}_{2},\\dots ,{x}_{S}$. Then one measure of the uniformity of the points ${x}^{1},{x}^{2},\\dots ,{x}^{N}$ is the discrepancy:\n $DN* x1,x2,…,xN = sup x∈IS SN Gx - N x1 , x2 , … , xS .$\nwhich has the form\n $DN*x1,x2,…,xN≤CSlog⁡NS+Olog⁡NS-1 for all N≥2.$\nThe principal aim in the construction of low-discrepancy sequences is to find sequences of points in ${I}^{S}$ with a bound of this form where the constant ${C}_{S}$ is as small as possible.\nThe type of low-discrepancy sequence generated by G05YMF depends on the initialization routine called and can include those proposed by Sobol, Faure or Niederreiter. If the initialization routine G05YNF was used then the sequence will be scrambled (see Section 3 in G05YNF for details).\nBratley P and Fox B L (1988) Algorithm 659: implementing Sobol's quasirandom sequence generator ACM Trans. Math. Software 14(1) 88–100\nFox B L (1986) Algorithm 647: implementation and relative efficiency of quasirandom sequence generators ACM Trans. Math. Software 12(4) 362–376\n\n## 5  Parameters\n\nNote: the following variables are used in the parameter descriptions:\n• $\\mathit{idim}={\\mathbf{IDIM}}$, the number of dimensions required, see G05YLF or G05YNF\n• $\\mathit{liref}={\\mathbf{LIREF}}$, the length of IREF as supplied to the initialization routine G05YLF or G05YNF\n• $\\mathit{tdquas}={\\mathbf{N}}$ if ${\\mathbf{RCORD}}=1$; otherwise $\\mathit{tdquas}=\\mathit{idim}$\n1:     N – INTEGERInput\nOn entry: the number of quasi-random numbers required.\nConstraint: ${\\mathbf{N}}\\ge 0$ and ${\\mathbf{N}}+\\text{previous number of generated values}\\le {2}^{31}-1$.\n2:     RCORD – INTEGERInput\nOn entry: the order in which the generated values are returned.\nConstraint: ${\\mathbf{RCORD}}=1$ or $2$.\n3:     QUAS(LDQUAS,$\\mathit{tdquas}$) – REAL (KIND=nag_wp) arrayOutput\nOn exit: contains the N quasi-random numbers of dimension idim.\nIf ${\\mathbf{RCORD}}=1$, ${\\mathbf{QUAS}}\\left(i,j\\right)$ holds the $j$th value for the $i$th dimension.\nIf ${\\mathbf{RCORD}}=2$, ${\\mathbf{QUAS}}\\left(i,j\\right)$ holds the $i$th value for the $j$th dimension.\n4:     LDQUAS – INTEGERInput\nOn entry: the first dimension of the array QUAS as declared in the (sub)program from which G05YMF is called.\nConstraints:\n• if ${\\mathbf{RCORD}}=1$, ${\\mathbf{LDQUAS}}\\ge \\mathit{idim}$;\n• if ${\\mathbf{RCORD}}=2$, ${\\mathbf{LDQUAS}}\\ge {\\mathbf{N}}$.\n5:     IREF($\\mathit{liref}$) – INTEGER arrayCommunication Array\nOn entry: contains information on the current state of the sequence.\nOn exit: contains updated information on the state of the sequence.\n6:     IFAIL – INTEGERInput/Output\nOn entry: IFAIL must be set to $0$, $-1\\text{​ or ​}1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.\nFor environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\\text{​ or ​}1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\\mathbf{1}\\text{​ or ​}\\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.\nOn exit: ${\\mathbf{IFAIL}}={\\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).\n\n## 6  Error Indicators and Warnings\n\nIf on entry ${\\mathbf{IFAIL}}={\\mathbf{0}}$ or $-{\\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).\nErrors or warnings detected by the routine:\n${\\mathbf{IFAIL}}=1$\nOn entry, either ${\\mathbf{N}}<0$ or there have been too many calls to the generator.\n${\\mathbf{IFAIL}}=2$\nOn entry, ${\\mathbf{RCORD}}\\ne 1$ or $2$.\n${\\mathbf{IFAIL}}=4$\nOn entry, ${\\mathbf{RCORD}}=1$ and ${\\mathbf{LDQUAS}}<\\mathit{idim}$.\nOn entry, ${\\mathbf{RCORD}}=2$ and ${\\mathbf{LDQUAS}}<{\\mathbf{N}}$.\n${\\mathbf{IFAIL}}=5$\nIncorrect initialization. G05YLF must be called prior to G05YMF and IREF must remain unaltered after this call.\n\nNot applicable.\n\nNone.\n\n## 9  Example\n\nThis example calls G05YLF and G05YMF to estimate the value of the integral\n $∫01 ⋯ ∫01 ∏ i=1 s 4xi-2 dx1, dx2, …, dxs = 1 .$\nIn this example the number of dimensions $S$ is set to $8$.\n\n### 9.1  Program Text\n\nProgram Text (g05ymfe.f90)\n\n### 9.2  Program Data\n\nProgram Data (g05ymfe.d)\n\n### 9.3  Program Results\n\nProgram Results (g05ymfe.r)" ]

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[ "# Motion, laws of motion\n\nMotion\n\nMotion is a change in position of an object with respect to time. Motion is typically described in terms of displacement, distance, velocity, acceleration, time and speed.\n\nThe branch of physics which deals with the study of motion of material objects is called mechanics.\n\nMechanics is divided into following branches.\n\n(i) Statics : Statics is the branch of mechanics which deals with the study of motion of objects under the effect of forces in equilibrium.\n\n(ii) Kinematics :\n\nIt is that branch of mechanics which deals with the study of motion of object without taking into account the factors (i.e. nature of forces, nature of bodies etc.) which cause motion. Here time factor plays an essential role.\n\n(iii) Dynamics :\n\nIt is that branch of mechanics which deals with the study of motion of objects taking into account the factors whichcause motion.\n\nRest : An object is said to be at rest if it does not change its position with time, with respect to its surroudings.\n\nA book lying on a table, a person sitting in a chair are the examples of rest.\n\nMotion : An object is said to be in motion if it changes its position with time, with respect to its surroundings.\n\nExample : A bird flying in air, a train moving on rails, a ship sailing on water, a man walking on road are some of the examples of motion, visible to the eye. Motion of gas molecules is an example of motion, invisible to the eye.\n\nRest & Motion are relative terms :\n\nWhen we say that an object is at rest or in motion,then this statement is incomplete and meaningless. Basically, rest & motion are relative terms. An object which is at rest can also be in motion simultaneously. This can be illustrated as follows.\n\nThe passengers sitting in a moving bus are at rest with respect to each other but they are also in motion at the same time with respect to the objects like trees, buildings on the road side. So the motion and rest are relative terms.\n\nRectilinear motion :\n\nIf a particle moves in a fixed direction, the motion of this type is called rectilinear motion or one dimensional motion.For example the motion of an ant on a wire is a rectilinear motion.\n\nTwo dimensional motion :\n\nIf the motion of a particle is in such a way that its position remains on a fixed plane, then the motion of a particle is called two dimensional motion.\n\nLaws Of Motion\n\nFirst Law of Motion\n\nA body continue to be in its state of rest or of uniform motion along a straight line, unless it is acted upon by some external force to change the state\n\n(1) If no net force acts on a body, then the velocity of the body cannot change i.e. the body cannot accelerate.\n\n(2) Newton‟s first law defines inertia and is rightly called the law of inertia. Inertia are of three types :Inertia of rest, Inertia of motion, Inertia of direction\n\n(3) Inertia of rest : It is the inability of a body to change by itself, its state of rest. This means a body atrest remains at rest and cannot start moving by its own.\n\nSecond Law of Motion\n\n(1) The rate of change of linear momentum of a body is directly proportional to the external force appliedon the body and this change takes place always in the direction of the applied force.\n\n(2) If a body of mass m, moves with velocity v then its linear momentum can be given by p= mv and if force is applied on a body, then Force = mass ? acceleration\n\nThird Law of Motion\n\nTo every action, there is always an equal (in magnitude) and opposite (in direction) reaction.\n\n(1) When a body exerts a force on any other body, the second body also exerts an equal and opposite forceon the first.\n\n(2) Forces in nature always occurs in pairs. A single isolated force is not possible.\n\n(3) Any agent, applying a force also experiences a force of equal magnitude but in opposite direction. Theforce applied by the agent is called „Action‟ and the counter force experienced by it is called „Reaction‟.\n\n(4) Action and reaction never act on the same body. If it were so the total force on a body would have always been zero i.e. the body will always remain in equilibrium.\n\n(5) If F (AB)= force exerted on body A by body B (Action) and F(BA)= force exerted on body B by body A (Reaction) Then according to Newton‟s third law of motion F (AB) = F(BA)\n\n(6) Example : (i) A book lying on a table exerts a force on the table which is equal to the weight of the book. This is the force of action.\n\nTSPSC  Notes brings Prelims and Mains programs for TSPSC  Prelims and TSPSC  Mains Exam preparation. Various Programs initiated by TSPSC  Notes are as follows:- For any doubt, Just leave us a Chat or Fill us a querry––\n\nHope we have satisfied your need for TSPSC Prelims and Mains Preparation\n\n#### Kindly review us to serve even better\n\nTSPSC Mains Test Series 2022\n\n20 Quality mock tests and GS Mains Notes\n\nMains Test Series and Notes\n\nMains Printed Notes (With COD)\n\nTSPSC Prelims Test Series 2022\n\n24 Quality mock tests and GS Prelims Notes\n\nPrelims Test Series and Notes\n\nPrelims Printed Notes (With COD)\n\n## Subscribe to TSPSC Notes\n\nNever Miss any TSPSC important update!\n\nJoin 2,500 other subscribers\n\nerror: Content is protected !!" ]

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[ "# #002a10 Color Information\n\nIn a RGB color space, hex #002a10 is composed of 0% red, 16.5% green and 6.3% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 61.9% yellow and 83.5% black. It has a hue angle of 142.9 degrees, a saturation of 100% and a lightness of 8.2%. #002a10 color hex could be obtained by blending #005420 with #000000. Closest websafe color is: #003300.\n\n• R 0\n• G 16\n• B 6\nRGB color chart\n• C 100\n• M 0\n• Y 62\n• K 84\nCMYK color chart\n\n#002a10 color description : Very dark (mostly black) cyan - lime green.\n\n# #002a10 Color Conversion\n\nThe hexadecimal color #002a10 has RGB values of R:0, G:42, B:16 and CMYK values of C:1, M:0, Y:0.62, K:0.84. Its decimal value is 10768.\n\nHex triplet RGB Decimal 002a10 `#002a10` 0, 42, 16 `rgb(0,42,16)` 0, 16.5, 6.3 `rgb(0%,16.5%,6.3%)` 100, 0, 62, 84 142.9°, 100, 8.2 `hsl(142.9,100%,8.2%)` 142.9°, 100, 16.5 003300 `#003300`\nCIE-LAB 13.787, -21.78, 12.78 0.921, 1.693, 0.768 0.272, 0.5, 1.693 13.787, 25.253, 149.597 13.787, -12.382, 11.476 13.012, -10.132, 5.607 00000000, 00101010, 00010000\n\n# Color Schemes with #002a10\n\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #2a001a\n``#2a001a` `rgb(42,0,26)``\nComplementary Color\n• #052a00\n``#052a00` `rgb(5,42,0)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #002a25\n``#002a25` `rgb(0,42,37)``\nAnalogous Color\n• #2a0005\n``#2a0005` `rgb(42,0,5)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #25002a\n``#25002a` `rgb(37,0,42)``\nSplit Complementary Color\n• #2a1000\n``#2a1000` `rgb(42,16,0)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #10002a\n``#10002a` `rgb(16,0,42)``\n• #1a2a00\n``#1a2a00` `rgb(26,42,0)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #10002a\n``#10002a` `rgb(16,0,42)``\n• #2a001a\n``#2a001a` `rgb(42,0,26)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #001106\n``#001106` `rgb(0,17,6)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #00441a\n``#00441a` `rgb(0,68,26)``\n• #005d23\n``#005d23` `rgb(0,93,35)``\n• #00772d\n``#00772d` `rgb(0,119,45)``\nMonochromatic Color\n\n# Alternatives to #002a10\n\nBelow, you can see some colors close to #002a10. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #002a06\n``#002a06` `rgb(0,42,6)``\n• #002a09\n``#002a09` `rgb(0,42,9)``\n• #002a0d\n``#002a0d` `rgb(0,42,13)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #002a14\n``#002a14` `rgb(0,42,20)``\n• #002a17\n``#002a17` `rgb(0,42,23)``\n• #002a1b\n``#002a1b` `rgb(0,42,27)``\nSimilar Colors\n\n# #002a10 Preview\n\nThis text has a font color of #002a10.\n\n``<span style=\"color:#002a10;\">Text here</span>``\n#002a10 background color\n\nThis paragraph has a background color of #002a10.\n\n``<p style=\"background-color:#002a10;\">Content here</p>``\n#002a10 border color\n\nThis element has a border color of #002a10.\n\n``<div style=\"border:1px solid #002a10;\">Content here</div>``\nCSS codes\n``.text {color:#002a10;}``\n``.background {background-color:#002a10;}``\n``.border {border:1px solid #002a10;}``\n\n# Shades and Tints of #002a10\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, #000301 is the darkest color, while #eefff5 is the lightest one.\n\n• #000301\n``#000301` `rgb(0,3,1)``\n• #001609\n``#001609` `rgb(0,22,9)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\n• #003e17\n``#003e17` `rgb(0,62,23)``\n• #00511f\n``#00511f` `rgb(0,81,31)``\n• #006526\n``#006526` `rgb(0,101,38)``\n• #00782e\n``#00782e` `rgb(0,120,46)``\n• #008c35\n``#008c35` `rgb(0,140,53)``\n• #00a03d\n``#00a03d` `rgb(0,160,61)``\n• #00b344\n``#00b344` `rgb(0,179,68)``\n• #00c74c\n``#00c74c` `rgb(0,199,76)``\n• #00db53\n``#00db53` `rgb(0,219,83)``\n• #00ee5b\n``#00ee5b` `rgb(0,238,91)``\n• #03ff63\n``#03ff63` `rgb(3,255,99)``\n• #16ff6f\n``#16ff6f` `rgb(22,255,111)``\n• #2aff7b\n``#2aff7b` `rgb(42,255,123)``\n• #3eff87\n``#3eff87` `rgb(62,255,135)``\n• #51ff93\n``#51ff93` `rgb(81,255,147)``\n• #65ffa0\n``#65ffa0` `rgb(101,255,160)``\n• #78ffac\n``#78ffac` `rgb(120,255,172)``\n• #8cffb8\n``#8cffb8` `rgb(140,255,184)``\n• #a0ffc4\n``#a0ffc4` `rgb(160,255,196)``\n• #b3ffd0\n``#b3ffd0` `rgb(179,255,208)``\n• #c7ffdc\n``#c7ffdc` `rgb(199,255,220)``\n• #dbffe8\n``#dbffe8` `rgb(219,255,232)``\n• #eefff5\n``#eefff5` `rgb(238,255,245)``\nTint Color Variation\n\n# Tones of #002a10\n\nA tone is produced by adding gray to any pure hue. In this case, #131715 is the less saturated color, while #002a10 is the most saturated one.\n\n• #131715\n``#131715` `rgb(19,23,21)``\n• #121814\n``#121814` `rgb(18,24,20)``\n• #101a14\n``#101a14` `rgb(16,26,20)``\n• #0f1b13\n``#0f1b13` `rgb(15,27,19)``\n• #0d1d13\n``#0d1d13` `rgb(13,29,19)``\n• #0b1f13\n``#0b1f13` `rgb(11,31,19)``\n• #0a2012\n``#0a2012` `rgb(10,32,18)``\n• #082212\n``#082212` `rgb(8,34,18)``\n• #062412\n``#062412` `rgb(6,36,18)``\n• #052511\n``#052511` `rgb(5,37,17)``\n• #032711\n``#032711` `rgb(3,39,17)``\n• #022810\n``#022810` `rgb(2,40,16)``\n• #002a10\n``#002a10` `rgb(0,42,16)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #002a10 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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[ "Search a number\nBaseRepresentation\nbin1101111001\n31012221\n431321\n512024\n64041\n72410\noct1571\n91187\n10889\n11739\n12621\n13535\n14477\n153e4\nhex379\n\n889 has 4 divisors (see below), whose sum is σ = 1024. Its totient is φ = 756.\n\nThe previous prime is 887. The next prime is 907. The reversal of 889 is 988.\n\nSubtracting from 889 its product of digits (576), we obtain a palindrome (313).\n\nSubtracting 889 from its reverse (988), we obtain a palindrome (99).\n\n889 is nontrivially palindromic in base 13.\n\nIt is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.\n\nIt is not a de Polignac number, because 889 - 21 = 887 is a prime.\n\nIt is a magnanimous number.\n\nIt is a Duffinian number.\n\n889 is an undulating number in base 13.\n\nIt is a plaindrome in base 10, base 14 and base 16.\n\nIt is a nialpdrome in base 12.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (881) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (7) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 57 + ... + 70.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (256).\n\nIt is an amenable number.\n\n889 is a deficient number, since it is larger than the sum of its proper divisors (135).\n\n889 is a wasteful number, since it uses less digits than its factorization.\n\n889 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 134.\n\nThe product of its digits is 576, while the sum is 25.\n\nThe square root of 889 is about 29.8161030318. The cubic root of 889 is about 9.6153977436.\n\nThe spelling of 889 in words is \"eight hundred eighty-nine\", and thus it is an aban number and an oban number.\n\nDivisors: 1 7 127 889" ]

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[ "Feeds:\nPosts\n\n## The representation theory of SU(2)\n\nToday we will give four proofs of the classification of the (finite-dimensional complex continuous) irreducible representations of", null, "$\\text{SU}(2)$ (which you’ll recall we assumed way back in this previous post). As a first step, it turns out that the finite-dimensional representation theory of compact groups looks a lot like the finite-dimensional representation theory of finite groups, and this will be a major boon to three of the proofs. The last proof will instead proceed by classifying irreducible representations of the Lie algebra", null, "$\\mathfrak{su}(2)$.\n\nAt the end of the post we’ll briefly describe what we can conclude from all this about electrons orbiting a hydrogen atom.\n\n## The quaternions and Lie algebras II\n\nWe now know what a Lie algebra is and we know they are abstractions of infinitesimal symmetries, which are given by derivations. Today we will see what we can say about associating infinitesimal symmetries to continuous symmetries: that is, given a matrix Lie group", null, "$G$, we will describe its associated Lie algebra", null, "$\\mathfrak{g}$ of infinitesimal elements and the exponential map", null, "$\\mathfrak{g} \\to G$ which promotes infinitesimal symmetries to real ones.\n\nAs in the other post, I will be ignoring some technical details for the sake of exposition. For example, I am generally not specifying how I’m topologizing various objects, and this is because of the general fact that a finite-dimensional real vector space has a unique Hausdorff topology compatible with addition and scalar multiplication. Whenever I talk about limits in such a vector space, I therefore don’t need to specify how I’m imposing a topology, although it will generally be convenient to induce it via a norm (which I am also not specifying).\n\n## Euler characteristic as homotopy cardinality\n\nLet", null, "$X$ be a finite CW complex with", null, "$c_0$ vertices,", null, "$c_1$ edges, and in general", null, "$c_i$ different", null, "$i$-cells. The Euler characteristic", null, "$\\displaystyle \\chi(X) = \\sum_{i \\ge 0} (-1)^i c_i$\n\nis a fundamental invariant of", null, "$X$, and the observation that it is homotopy invariant is the appropriate generalization of Euler’s formula", null, "$V - E + F = 2 = \\chi(S^2)$ for a convex polyhedron. But where exactly does this expression come from? The modern story involves the homology groups", null, "$H_i(X, \\mathbb{Q})$, but actually one can work on a more intuitive level characterized by the following slogan:\n\nThe Euler characteristic is a homotopy-invariant generalization of cardinality.\n\nMore precisely, the above expression for Euler characteristic can be deduced from three simple axioms:\n\n1. Cardinality:", null, "$\\chi(\\text{pt}) = 1$.\n2. Homotopy invariance: If", null, "$X \\sim Y$, then", null, "$\\chi(X) = \\chi(Y)$.\n3. Inclusion-exclusion: Suppose", null, "$X$ is the union of two subcomplexes", null, "$A, B$ whose intersection", null, "$A \\cap B$ is a subcomplex of both", null, "$A$ and", null, "$B$. Then", null, "$\\chi(X) = \\chi(A) + \\chi(B) - \\chi(A \\cap B)$.\n\nOf course, this isn’t enough to conclude that there actually exists an invariant with these properties. Nevertheless, it’s enough to motivate the search for a proof that such an invariant exists." ]

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[ "AI:EF\n\nStep-by-step explanation:\n\n1. By definition, the diameter that a one is double the radius, then:\n\nD=2r\n\nwherein D is the diameter and r is the radius:\n\n2. The AI is the lenght the the radius that the circle and also EF is the length of the diameter that the circle. Keeping this ~ above mind you have:", null, "or 1:2\n\n3. Then, AI and also EF kind the ratio:\n\n1:2\n\nAs given allude A is the center of the circle.\n\nYou are watching: In the image, point a marks the center of the circle. which two lengths must form a ratio of 1:2?\n\nWe will certainly answer this concern by considering all the options.\n\nOption 1. FA : AE < Both FA and also EA space radius of circle .So their ratio can\"t be same to 1:2.>\n\nOption 2. CB : FE < CB is a chord and FE is a Diameter as length of Chord no given, So us can\"t Predict around their Ratio.>\n\nOption 3: FE : AH\n\nOption 4 : AH : FE\n\nOption (d) AH : FE = Radius : Diameter=1:2 is true.\n\nEF:AI\n\nStep-by-step explanation:\n\nE to F is dual that that AI and also it requirements to be in the front because it is 2:1, no 1:2. Finest of luck through the rest.\n\noption b.\n\nStep-by-step explanation:\n\nwe understand that\n\nIn a circle the proportion of the diameter come the radius is equal to", null, "because", null, "", null, "so\n\nIn this difficulty\n\nEF is a diameter\n\ntherefore\n\nDraw a diameter v the center of the circle. Any radius that this circle is 1/2 the length of the diameter. Thus, if we divide the diameter right into two same parts, we have two lengths that type the ratio 2:1, where 2 represents the diameter and 1 the radius.\nI expect this helps you", null, "I would go through D) AH:EF. The easiest relationship amongst the parts of a circle the would have the proportion of 1:2 would be that of a radius to a diameter. Among the choices, the one the matches the 1:2 proportion would be AH (a radius) and EF (a diameter).\n\nStep-by-step explanation:\n\nthe the very least value the end of the data presented is 5, probably if you write-up a photo with the i could a tiny more.\n\nstep-by-step explanation:\n\n4x-10=2x+3 +10. +104x=2x+13-2x. -2x2x=13/2. /2x=6.5\n\n### Similar questions\n\nMathematics, 06.07.2019 01:40, alyssa0888\nIn the image, suggest a marks the center of the circle. Which two lengths must type a proportion of 2: 1?\ncontinue\nIn the image, suggest a marks the facility of the circle. Which 2 lengths must form a ratio of 1: 2? a)af: ae b)bc: ef c)ef: ah d)ah: ef e)ah: ad\ncontinue\nMathematics, 01.10.2019 18:30, gallla\nPoint a is the facility of this circle. The ratio of the lengths that (ef/bc) and also (ad/ih) is 2: 1.\ncontinue\nDo you recognize the exactly answer?\nIn the image, suggest a point out the facility of the circle. Which 2 lengths must type a ratio of 1 : 2?...\n\n### Questions in other subjects:\n\nBiology, 04.04.2021 04:12\nKevin started his evidence of the Pythagorean to organize by inscribing a square in ~ a square so the the outer square consists of both the within square and four triangles...\nArts, 04.04.2021 04:20\nDoes anyone know around any free drawing programs that will download on home windows 10, I gained a new drawing tablet and I desire a great program to check it out on. Please provide me a suitable a...\nMathematics, 04.04.2021 04:20\nShow how to use substitution or elimination to settle the following system that equations. Create your solution as an bespeak pair:4x - y = 10 , y = 2x - 2 ( , )\nEnglish, 04.04.2021 04:20\nWhat person(s) or groups does the work recognize as \"other\" or stranger? just how are such persons/groups described and also treated in the novel exit West by Mohsin Hamid?...\nMathematics, 04.04.2021 04:20\nThe lot of medication in the bloodstream of a patient decreases roughly exponentially. Here is the graph representing f, an exponential function that models the medicine in the bo...\nChemistry, 04.04.2021 04:20\nCool air over warmth water will:Increase in temperatureStay the exact same in temperatureDecrease in temperature...\nMathematics, 04.04.2021 04:20\nHelp plz, present your job-related too! Ty :) resolve the trouble -4(2a +3) 15 - 3(x + 2) (6x - 3) - (4x -3)...\nBiology, 04.04.2021 04:20\n1. Us have proof to support the case that the properties of antibiotic resistance has actually increased in many types of bacteria since penicillin came to be the first commercially available...\nWhich the the following would most likely use a job order costing system? a. Oil refinery b. Agency that manufactures chlorine for swim pools c. Record mill d. Swimming pool ins...\nMathematics, 04.04.2021 04:20\nWhich that these correctly rearranges the state in this polynomial so like terms are alongside each other...\nOther jobs in the category: MathematicsMore task\nTotal solved difficulties on the site: 13850511\nor\ncheck out a few ads and also unblock the prize on the site\nYou can only ask a inquiry in the app\nSee results (0)\nEmail\nToo short question\nNext\nEmail\nYou space registered.Access to her account will certainly be opened after verification and also publication of the question.\n\nSee more: According To Your Text, How Did European Colonialists In The Caribbean View Forests?\n\nNext" ]

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[ "# Discretization\n\n## Introduction", null, "To calculate flow field it is required to solve the Navier-Stokes equations inside the flow domain. This is a system of partial differential equations which cannot be solved analytically in general and needs to be converted into a system of linear equations. The process of converting partial differential equation into a linear equation system is called discretization and can use multiple different schemes.\n\nIn the Discretization panel one selects discretization schemes that are most suitable for the given simulation type. Usually, when selecting discretization scheme there is a trade-off between accuracy and stability, which may require some testing to select optimal scheme.\n\n## Time Discretization", null, "The Time discretization tab defines how terms of the form $$\\frac{\\partial\\Psi}{\\partial t}$$ need to be discretized.\n\nTime Discretization Schemes:\n\n• Steady State - terms with temporal derivatives are not discretized, which is usually used in steady-state simulation\n\n• Implicit Euler - basic Euler scheme (first-order), uses only the current and previous time-step values\n\n• Crank-Nicolson - Crank-Nicolson scheme (second-order), requires specification of an \"off-centering\" coefficient (1 - pure Crank-Nicolson, 0 - pure Implicit Euler)\n\n• Backward - backward-differencing scheme (second-order), uses the current and two previous time-step values\n\n• Local Euler - local time-step Euler scheme (first-order), local time-step field is looked-up from the database. This scheme should only be used for steady-state computations using transient codes where local time-stepping is preferable to under-relaxation for transport consistency reasons.\n\n• Bounded - option for selecting the bounded form of the underlying scheme. Boundedness is achieved by subtracting $$\\frac{\\partial \\varphi}{\\partial t} \\cdot \\Psi$$ which is non-conservative if $$\\frac{\\partial \\rho}{ \\partial t} \\neq 0$$ but conservative otherwise.\n\n## Convection (Advection) Discretization", null, "The Convection discretization tab defines how terms of the form $$\\nabla \\cdot(\\vec U\\cdot \\Psi)$$ need to be discretized. Since usually different equations require different treatment, one needs to define convection scheme for each individual equation.\n\n### Defining Convection Scheme", null, "To specify convection scheme for a given field (for example velocity), one needs to click the button next to the name of the equation. This will cause a panel with available options to expand. When the new scheme is selected, the label of the button will change and display new scheme’s name.\n\nConvection Schemes:\n\n• Upwind - upwind differencing scheme (first-order, bounded)\n\n• Linear Upwind - linear upwind differencing scheme (second-order), calculates upwind weighting factors and also applies a gradient-based explicit correction\n\n• Linear - central-differencing scheme (second-order, unbounded)\n\n• Limited Linear - TVD limited linear differencing scheme (second-order)\n\n• LUST - linear-upwind stabilised transport scheme (second-order), blended linear/linear-upwind scheme For this discretization, velocity gradients need to be specified. This scheme is used for large eddy simulation (LES).\n\n• Cubic - modified version of Linear scheme which applies an explicit correction (second-order)\n\n• van Leer - discretization scheme with van Leer flux limiter (second-order)\n\n• MUSCL - MUSCL differencing scheme (second-order)\n\n• QUICK - quadratic-upwind differencing scheme (second-order)\n\n• Interface Compression - modeled relative velocity used to compress the interface\n\n• DES Hybrid - hybrid convection scheme for DES calculations. The scheme blends between a low-dissipative linear convection scheme within the LES region and a numerically more robust upwind scheme in the RAS region. It requires you to specify responsively velocity and length scales.\n\nConvection Scheme Options:\n\n• Vector Specific - this option enables special version of the scheme dedicated for discretization of vector fields\n\n• Bounded - option for selecting bounded form of the underlying scheme. Boundedness is achieved by subtracting $$(\\nabla \\cdot \\rho U)\\Psi$$ which is non-conservative if $$\\frac{\\partial \\rho}{ \\partial t} \\neq 0$$ but conservative otherwise. Can be used for convection of bounded scalar properties in steady-state solvers to improve stability if insufficient convergence of the pressure equation causes temporary divergence of the flux field.\n\n• Skew Corrected - skewness-corrected interpolation scheme that applies an explicit correction to an underlying scheme", null, "The Gradients tab defines discretization schemes used for calculating gradients.\n\n### Defining Gradient Scheme", null, "The Gradient discretization scheme once selected is applied to all fields.\n\n• Gauss - basic gradient scheme using face-interpolation and Gauss' theorem (second-order)\n\n• Least Squares - gradient scheme using least-squares (second-order)\n\n• Fourth - gradient scheme using least-squares (fourth-order)\n\n• none - no limiting is applied\n\n• Cell Limited - scalar limiter based on limiting the extrapolated face values between the maximum and minimum cell and cell neighbour values and is applied to all components of the gradient\n\n• Face Limited - scalar limiter based on limiting the extrapolated face values between the face-neighbour cell values and is applied to all components of the gradient\n\n### Defining Surface Normal Gradient Scheme", null, "The Surface Normal Gradient once selected applies to all fields also when calculating the Laplace operator.\n\nSurface Normal Gradient Schemes:\n\n• Corrected - simple central-difference scheme with non-orthogonal correction\n\n• Uncorrected - simple central-difference scheme without non-orthogonal correction\n\n• Orthogonal - simple central-difference scheme without non-orthogonal correction\n\n• Face Corrected - simple central-difference scheme with non-orthogonal correction\n\n• Limited - scheme with limited non-orthogonal correction. The limiter is controlled by a coefficient with a value between 0 and 1 which when zero switches the limiter off and the scheme behaves as Corrected, and when set to 1 the limiter is calculated such that the non-orthogonal contribution does not exceed the orthogonal part.\n\n• Linear Fit - centred fit scheme with linear polynomial and explicit correction\n\n• Quadratic Fit - centred fit scheme with quadratic polynomial and explicit correction\n\n## Interpolation", null, "The Interpolation tab defines face interpolation scheme, which is going on be used. The selected scheme will also be used when calculating Laplace and divergence operators.\n\nInterpolation Schemes:\n\n• Linear - central-differencing interpolation\n\n• Cubic - modified version of Linear scheme which applies an explicit correction\n\n• Mid Point - mid-point interpolation (weighting factors = 0.5) scheme" ]

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[ "# PHYSICS HOMEWORK #32\n\nNow, there is the slight complication that the pulley does exert a force on your system, but it’s safe to ignore this and just picture everything in one-dimension. There are 7 identical items, each with a equally random chance of being placed in 1 of 32 containers. Right, it’s the acceleration which is constant. Verifying Newton’s 2nd Law Ask Question. First calculate the number of ways 7 items can go into n bins, none empty, using this expansion series: Extending these general cases to particular 5, 6 and 7 bins: Note that 7 items into 4 containers is the same as rolling a 4 sided die 7 times and noting the outcome, so this method can be used for permutations or probability of rolling a die several times.", null, "Physics Stack Exchange works best with JavaScript enabled. So it produces a very straight line, the data line up very well and produce a great line of best fit. Post as a guest Name. By using our site, you acknowledge that you have read and understand our Cookie Policy , Privacy Policy , and our Terms of Service. IS all that is needed to verify it?\n\n# Math 21a Homework Assignments for Spring\n\nAnd phhysics use this, in the middle paragraph, when you mention the 1st, 2nd, 3rd item etc. Reduce this to the situation where there are 7 items and 4 containers. By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.", null, "But essentially that may well have been a complication rather than a simplification – fewer steps maybe but more analysis. Note that 7 items into 4 containers is the same as ##32 a 4 sided die 7 times and noting the outcome, so this method can be used for permutations or probability of rolling a die several times. Remember we’re modelling this with bin dividers, and for 4 bins there are 3 dividers.\n\nARCANE THESIS FEAT ERRATA\n\nUnicorn Meta Zoo 3: Corrections final terms were ok, the intemediate terms for 2of2 and 3of3 were multiplied but left inside parenthesiscorrected equations: That might be appropriate for your application; I really don’t know physicz to judge. The number of ways to do this would be 32x1x1x1x31x30x I Permutation of identical elements.", null, "What are the possible permutations? Right, it’s the acceleration which is constant. Sign up using Email and Password.\n\nI can easily apply this to other numbers of bins, homewokr changing the number of events is a little slower. Looking at the case of 7 items into 4 bins with no bins empty there are 20 cases that fall into 3 patterns: Sorry rcgldr, I didn’t notice your link before.\n\nWhat is the probabilty that containers, a, b, c end up with 1 item, and container d ends up with 4 items? A thread connected the hanging mass with the one on the track.\n\n## Content is not yet loaded to the server\n\nThe distribution of balls in the pnysics is random. Amazingly similar lines of thought. Extending these general cases to particular 5, 6 and 7 bins: Hlmework, 7 items, 4 containers.\n\nESSAY ÜBER IWEIN\n\nFirst calculate the number of ways 7 items can go into n bins, none empty, using this expansion series: For example, with 4 bins used: IS all that is needed to verify it?\n\nFirst, your list of numbers are actually the number of ways of putting 7 distinct items, not identical items, into containers.", null, "I homeork have the hokework for this, but wondering if there’s some efficient way to calculate this. Odds versus number of draws to get 6 of 72 items. Is the logic in my first post treating the items or containers as distinct: So taking one of the cases, suppose we have bin allocations: So, I thought about this a little more, and I think some case-by-case work is probably inevitable.\n\nThe right-hand side is the total mass of the two-mass system multiplied by the its common acceleration. How do we grade questions?\n\n# homework and exercises – Verifying Newton’s 2nd Law – Physics Stack Exchange\n\nYour logic treats both containers and objects as distinct. These numbers are way too high for a distribution of identical objects; basically you’re homweork the long way around on this problem.\n\nThread starter rcgldr Start date Mar 20," ]

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[ "## Abstract\n\nThe paper investigates matrix-free high-order implementation of finite element discretization with p-multigrid preconditioning for the compressible Neo-Hookean hyperelasticity problem at finite strain on unstructured 3D meshes in parallel. We consider two formulations for the matrix-free action of the Jacobian in Neo-Hookean hyperelasticity: (i) working in the reference configuration to define the second Piola-Kirchhoff tensor as a function of the Green-Lagrange strain S(E) (or equivalently, the right Cauchy-Green tensor C = I+2E), and (ii) working in the current configuration to define the Kirchhoff stress in terms of the left Cauchy-Green tensor τ(b). The proposed efficient algorithm utilizes the Portable, Extensible Toolkit for Scientific Computation (PETSc), along with the libCEED library for efficient compiler optimized tensor-product-basis computation to demonstrate an efficient nonlinear solution algorithm. We utilize p-multigrid preconditioning on the high-order problem with algebraic multigrid (AMG) on the assembled linear Q1 coarse grid operator. In contrast to classical geometric multigrid, also known as h-multigrid, each level in p-multigrid is related to a different approximation polynomial order p, instead of the element size h. A Chebyshev polynomial smoother is used on each multigrid level. AMG is then applied to the assembled Q1 (trilinear hexahedral elements), which allows low storage that can be efficiently used to accelerate convergence to a solution. For the compressible Neo-Hookean hyperelastic constitutive model we exploit the stored energy density function to compute the stored elastic energy density of the Neo-Hookean material as it relates to the deformation gradient. Based on our formulation, we consider four different algorithms each with different storage strategies. Algorithms 1 and 3 are implemented in the reference and current configurations respectively and store ∇Xξ, det(∇ξX), and ∇Xu. Algorithm 2 in the reference configuration stores, ∇Xξ, det(∇ξX), ∇Xu, C−1, and λ log (J). Algorithm 4, in the current configuration, stores det(∇ξX), ∇xξ, τ, and μ – λ log(J). x refers to the current coordinates, X to the reference coordinates, and ξ to the natural coordinates. We perform 3D bending simulations of a tube composed of aluminum (modulus of elasticity E = 69 GPa, Poisson’s ratio v = 0.3) using unstructured meshes and polynomials of order p = 1 through p = 4 under mesh refinement. We explore accuracy-time-cost tradeoffs for the prediction of strain energy across the range of polynomial degrees and Jacobian representations. In all cases, Algorithm 4 using the current configuration formulation outperforms the other three algorithms and requires less storage. Similar simulations for large deformation compressible Neo-Hookean hyperelasticity as applied to the same aluminum material are conducted with ABAQUS, a commercial finite element software package which is a state-of-the-art engineering software package for finite element simulations involving large deformation. The best results from the proposed implementations and ABAQUS are compared in the case of p = 2 on an Intel system with @2.4 GHz and 128 GB RAM. Algorithm 4 outperforms ABAQUS for polynomial degree p = 2.\n\nThis content is only available via PDF.\nYou do not currently have access to this content." ]

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[ "# Efficiently summing reciprocal of polynomial over primes.\n\nFor fun, I have been trying to calculate the sum of the reciprocals of the cube-full numbers. I have managed to show that the limit is equal to $$\\frac{\\zeta(3)\\zeta(4)\\zeta(5)}{\\zeta(8)\\zeta(10)}\\prod_{p\\ \\mbox{prime}}\\left(1-\\frac1{(p^4+1)(p^5+1)}\\right)\\approx1.3397841535.$$ This product converges pretty fast, because of the order $$9$$ polynomial $$f(x)=(x^4+1)(x^5+1)$$ in the numerator. By simply taking the primes up to $$10^8$$, I already got $$64$$ digits of precision. $$1.3397841535743472465991525865148860527752422497881828066630150676$$ However, this method requires exponential time to calculate more digits. I was wondering if there is a faster, or even a polynomial time algorithm to calculate more digits.\n\nOne thing I tried is to take the logarithm of the product. $$\\log\\left(\\prod_{p\\ \\mbox{prime}}\\left(1-\\frac1{f(p)}\\right)\\right)=\\sum_{p\\ \\mbox{prime}}\\log\\left(1-\\frac1{f(p)}\\right)$$ By taking the Taylor series of the natural logarithm, we get $$\\log\\left(1-\\frac1{f(p)}\\right)=\\sum_{n=1}^\\infty\\frac{-1}{n(f(p))^n}.$$ By absolute convergence, we can interchange the sums to obtain $$\\sum_{n=1}^\\infty\\frac{-1}n\\sum_{p\\ \\mbox{prime}}\\frac1{(f(p))^n}.$$ For all $$n$$, of course $$(f(p))^n$$ is a polynomial, so the question becomes how we can efficiently sum the reciprocal of a polynomial over the primes. Is there some sort of analog for the Euler-Maclaurin formula for primes?\n\n• nice question. what a cute limit at the start! – mathworker21 Dec 29 '19 at 1:58\n\n## 1 Answer\n\n$$\\log(1-\\frac{x^{9}}{(1+x^4)(1+x^5)}) = \\sum_{m\\ge 0} a_m x^{9+m}$$\n\n$$\\log\\prod_p (1-\\frac{p^{-9}}{(1+p^{-4})(1+p^{-5})}) =\\sum_p \\sum_m a_m p^{-9-m}=\\sum_m a_m \\sum_p p^{-9-m}$$ $$=\\sum_m a_m \\sum_l \\frac{\\mu(l)}{l} \\log\\zeta(l(9+m)) = \\sum_{n\\le N}(\\sum_{l|n}\\frac{ a_{n/l-9} \\mu(l)}{l} )\\log\\zeta(n)+O(2^{-N})$$\n\n• This looks promising, but I am not experienced enough to fill in the gaps. Are we supposed to calculate the terms $a_m$ from the Taylor series? Is that possible in polynomial time? And how do we know it converges for $|x|<1$? And most importantly, how do you go from the sum over $p$ to the sum over $l$? Suddenly $\\mu$ and $\\log\\zeta$ pop up, so something must be done with inclusion exclusion and the Euler product, but this step still seems like a pretty big leap. Finally, do you have a proof that the partial sums converge exponentially in the final sum? – SmileyCraft Dec 27 '19 at 20:54" ]

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[ "• PDF\n• Cite\n• Share\nArticle Contents", null, "", null, "Article Contents\n\n# Well-posedness of a model for the growth of tree stems and vines\n\n• * Corresponding author: Prof. Alberto Bressan\n\nThe first author is supported by NSF grant DMS-1714237, \"Models of controlled biological growth\"\n\n• The paper studies a PDE model for the growth of a tree stem or a vine, having the form of a differential inclusion with state constraints. The equations describe the elongation due to cell growth, and the response to gravity and to external obstacles.\n\nThe main theorem shows that the evolution problem is well posed, until a specific \"breakdown configuration\" is reached. A formula is proved, characterizing the reaction produced by unilateral constraints. At a.e. time $t$, this is determined by the minimization of an elastic energy functional under suitable constraints.\n\nMathematics Subject Classification: Primary: 49K99, 35Q92.\n\n Citation:", null, "•", null, "• Figure 1.  Left: at any point $\\gamma(t,\\sigma)$ along the stem, an infinitesimal change in curvature is produced as a response to gravity (or stems of other plants). The angular velocity is given by the vector $\\omega(\\sigma)$. This affects the position of all higher points along the stem. Right: At a given time $t$, the curve $\\gamma(t,\\cdot)$ is parameterized by $s\\in [0,t]$. It is convenient to prolong this curve by adding a segment of length $T-t$ at its tip (dotted line, possibly entering inside the obstacle). This yields an evolution equation on a fixed functional space $H^2([0,T];\\,\\mathbb{R}^3)$.\n\nFigure 3.  For the two initial configurations on the left, the constrained growth equation (8) admits a unique solution. On the other hand, the two configurations on the right satisfy both (26) and (27) in (B). In such cases, the Cauchy problem is ill posed.\n\nFigure 2.  Left: three configurations of the stem, relative to the obstacle. Right: in an abstract space, the first two configurations are represented by points $\\gamma_1,\\gamma_2$ on the boundary of the admissible set $S$ where the corresponding cones $\\Gamma_1,\\Gamma_2$ are transversal. On the other hand, $\\gamma_3$ is a \"breakdown configuration\", satisfying all assumptions (26)-(27). Its corresponding cone $\\Gamma_3$ is tangent to the boundary of the set $S$. Here the shaded region is the complement of $S$.\n\n•", null, "## Article Metrics", null, "", null, "DownLoad:  Full-Size Img  PowerPoint" ]

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[ "# How do you convert 64.3 grams of Al_2(CO_3)_3 to moles?\n\n$\\text{Moles}$ $=$ $\\text{Mass of stuff\"/\"Molar mass of stuff}$\nSo $\\frac{64.3 \\cdot \\cancel{g}}{233.99 \\cdot \\cancel{g} \\cdot m o {l}^{-} 1}$ $\\cong$ $0.25$ $m o l$." ]

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[ "# Selected quad for the lemma: spirit_n\n\nWord A Word B Word C Word D Occurrence Frequency Band MI MI Band Prominent\nspirit_n flesh_n law_n sin_n 20,113 5 5.9622 4 true\n\n# Text snippets containing the quad\n\nID Title Author Corrected Date of Publication (TCP Date of Publication) STC Words Pages\nA31362 The moderate enquirer resolved in a plain description of several objections which are summed up together and treated upon by way of conference, concerning the contemned [sic] people commonly called Quakers who are the royal seed of God and whose innocency is here cleared in the answers to the many objections that are frequently produced by their opponents : which may be profitable for them to read that have any thing against them, and useful for all such as desire to know the certainty of those things which are most commonly reported of them / written in behalf of the brethren, in vindication of the truth, by VV. C. Caton, William, 1636-1665. 1658 (1658) Wing C1515; ESTC R29186 28,001 70\n\nThere are 2 snippets containing the selected quad. | View original text\n\nlive_v in_o the_o life_n of_o immortality_n which_o be_v make_v manifest_a in_o the_o saint_n of_o old_a so_o that_o the_o same_o light_n life_n spirit_n and_o power_n be_v now_o make_v manifest_a among_o the_o saint_n and_o to_o the_o truth_n of_o this_o these_o people_n can_v set_v their_o seal_n who_o be_v make_v partaker_n of_o the_o same_o according_a to_o their_o measure_n enq._n have_v they_o then_o revelation_n now_o by_o the_o same_o spirit_n or_o be_v there_o any_o miracle_n wrought_v among_o they_o by_o the_o same_o power_n res._n yea_o they_o have_v revelation_n by_o the_o same_o spirit_n which_o reveal_v the_o my●teries_n of_o god_n in_o former_a age_n to_o the_o saint_n for_o if_o they_o have_v not_o revelation_n how_o shall_v they_o come_v to_o the_o knowledge_n of_o the_o father_n who_o none_o know_v but_o by_o the_o revelation_n of_o the_o son_n and_o by_o the_o mighty_a power_n of_o god_n miracle_n have_v be_v wrought_v among_o they_o and_o miraculous_a thing_n have_v be_v bring_v to_o pass_v in_o a_o mystery_n which_o be_v hide_v from_o all_o vulturous_a eye_n live_v and_o they_o be_v privy_a to_o these_o thing_n who_o be_v witness_n of_o the_o power_n of_o god_n which_o have_v wrought_v they_o but_o from_o scoffer_n and_o scorner_n despiser_n and_o wonderer_n these_o thing_n be_v conceal_v enq._n but_o such_o will_v have_v a_o sign_n that_o they_o may_v believe_v and_o be_v convert●d_fw-mi also_o will_v they_o not_o res._n yea_o but_o none_o must_v be_v give_v they_o but_o that_o of_o jonah_n for_o they_o be_v of_o the_o adulterous_a generation_n that_o seek_v one_o say_v show_v we_o a_o sign_n or_o work_v we_o a_o miracle_n that_o we_o may_v believe_v that_o your_o god_n be_v the_o live_a god_n and_o that_o he_o have_v send_v you_o etc._n etc._n now_o they_o that_o will_v not_o believe_v and_o 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stand_v in_o the_o judgement_n exclaim_v out_o against_o they_o as_o if_o they_o be_v censorious_a judge_n and_o horrid_a blasphemer_n when_o they_o have_v judge_v and_o reprove_v nothing_o but_o deceit_n in_o they_o and_o that_o which_o be_v to_o be_v eternal_o condemn_v even_o the_o very_a same_o that_o be_v judge_v in_o their_o own_o particular_n enq._n well_o but_o duth_o not_o christ_n say_v judge_v not_o that_o you_o be_v not_o judge_v and_o paul_n say_v the_o man_n that_o judge_v be_v inexcusable_a whosoever_o he_o be_v res._n yea_o those_o be_v the_o word_n of_o christ_n and_o paul_n by_o the_o way_n take_v notice_n of_o the_o follow_a word_n do_v not_o christ_n there_o speak_v to_o the_o hypocrite_n that_o behold_v the_o mo●e_n that_o be_v in_o his_o brother_n eye_n but_o consider_v not_o the_o beam_n that_o be_v in_o his_o own_o eye_n and_o ●ake_v paul_n word_n also_o as_o they_o be_v thou_o be_v inexcusable_a o_o man_n say_v he_o whosoever_o thou_o be_v that_o judge_v for_o wherein_o thou_o ●●udgest_v another_o mark_v well_o you_o condem●nest_a thyself_o for_o thou_o that_o judge_v do_v the_o ●●ame_a thing_n there_o be_v now_o the_o man_n that_o be_v ●nexcusable_a who_o judge_v another_o for_o that_o which_o he_o be_v guilty_a of_o himself_o and_o if_o thou_o ●or_a any_o man_n can_v tax_v these_o people_n in_o this_o ●hen_o say_v that_o they_o be_v inexcusable_a who_o be_v ●●ound_v guilty_a in_o this_o thing_n enq._n there_o be_v another_o thing_n which_o man_n ●●ath_v against_o they_o which_o be_v this_o they_o say_v ●hat_o they_o will_v not_o respect_v person_n but_o to_o a_o prince_n or_o lord_n or_o whatsoever_o that_o he_o be_v they_o will_v say_v thou_o and_o thou_o to_o he_o and_o stand_v with_o ●heir_a hat_n on_o in_o his_o presence_n as_o if_o he_o be_v a_o ●an_n of_o no_o reputation_n be_v this_o true_a res._n he_o who_o be_v their_o god_n who_o have_v ●ade_v of_o one_o blood_n all_o nation_n of_o man_n for_o ●o_z dwell_v upon_o the_o face_n of_o the_o earth_n he_o be_v no_o respecter_n of_o person_n and_o if_o they_o who_o ●are_v his_o people_n shall_v have_v respect_n to_o person_n they_o shall_v commit_v sin_n and_o will_v be_v convince_v of_o the_o law_n as_o transgressor_n and_o for_o they_o to_o have_v respect_n of_o person_n that_o be_v not_o good_a and_o as_o for_o the_o properness_n of_o thou_o and_o thou_o i_o have_v show_v thou_o before_o and_o why_o may_v not_o they_o stand_v with_o their_o hat_n on_o in_o the_o presence_n of_o the_o great_a man_n upon_o the_o face_n of_o the_o earth_n ●e●ing_v that_o they_o stand_v daily_o cover_v in_o the_o presence_n of_o their_o god_n who_o name_n be_v the_o lord_n of_o host_n who_o fill_v heaven_n and_o earth_n in_o who_o they_o honour_v all_o man_n without_o the_o respect_n of_o any_o one_o person_n or_o the_o greatness_n and_o magnificency_n of_o any_o for_o what_o be_v man_n but_o dust_n and_o wherefore_o shall_v he_o be_v adore_v by_o his_o fellow-creature_n as_o if_o honour_n pertain_v to_o he_o which_o indeed_o be_v only_o due_a to_o his_o creator_n who_o form_v he_o of_o the_o dust_n to_o which_o he_o must_v return_v and_o then_o his_o honour_n shall_v be_v lay_v in_o the_o dust_n with_o he_o enq_n well_o but_o hear_v these_o people_n say_v that_o they_o be_v move_v of_o the_o lord_n to_o do_v this_o and_o the_o other_o and_o that_o they_o be_v send_v of_o god_n to_o publish_v his_o truth_n now_o if_o this_o be_v true_a they_o shall_v turn_v many_o from_o their_o wickedness_n shall_v they_o not_o res._n the_o truth_n they_o must_v acknowledge_v whether_o they_o before_o who_o it_o be_v acknowledge_v can_v bear_v it_o or_o no_o the_o which_o truth_n have_v be_v often_o reject_v by_o such_o as_o resist_v the_o holy_a ghost_n by_o the_o which_o these_o peo●le_n have_v be_v often_o move_v to_o declare_v un●●_n the_o son_n of_o adam_n their_o sin_n and_o to_o show_v ●hem_fw-mi their_o transgression_n and_o to_o exhort_v ●em_n to_o repentance_n and_o amendment_n of_o life_n ●nd_v many_o have_v be_v turn_v from_o darkness_n 〈◊〉_d light_n from_o the_o power_n of_o satan_n unto_o god_n by_o those_o among_o they_o that_o be_v send_v 〈◊〉_d publish_v his_o tr●th_n and_o to_o preach_v his_o e●erlasting_a gospel_n unto_o the_o spirit_n in_o pri●●n_n yea_o thousand_o be_v convert_v and_o tur●ed_v from_o iniquity_n and_o wickedness_n and_o be_v ●rought_v to_o serve_v god_n in_o purity_n and_o holy●ess_n in_o his_o own_o temple_n and_o therefore_o they_o ●ave_v not_o run_v as_o uncertain_o or_o without_o ●●●ding_v neither_o have_v they_o fight_v as_o one_o ●●at_a beat_v the_o air_n for_o they_o who_o be_v tur●ed_v from_o wickedness_n by_o they_o be_v the_o seal_n 〈◊〉_d their_o apostleship_n in_o the_o lord_n enq._n it_o be_v say_v that_o they_o permit_v woman_n to_o ●eak_v or_o prophesy_v among_o they_o be_v not_o that_o ex●es●y_n against_o the_o scripture_n for_o do_v not_o paul_n 〈◊〉_d that_o he_o will_v not_o permit_v a_o woman_n to_o speak_v 〈◊〉_d the_o church_n and_o where_o do_v they_o read_v of_o any_o ●omen_n in_o the_o new-testament_n that_o prophesy_v res._n for_o woman_n to_o prophesy_v upon_o ●hom_n the_o spirit_n of_o the_o lord_n be_v pour_v out_o ●_z not_o express_o against_o the_o scripture_n but_o a_o ●●●filling_n of_o the_o scripture_n for_o the_o scripture_n ●ith_n that_o be_v shall_v come_v to_o pass_v afterward_o ●●at_a i_o the_o lord_n will_v pour_v out_o my_o spirit_n upon_o all_o flesh_n and_o your_o son_n and_o your_o daughter_n shall_v prophesy_v now_o the_o fulfil_n of_o this_o these_o people_n do_v witness_n who_o will_n no●_z permit_v a_o woman_n upon_o who_o the_o spir●●_z of_o the_o lord_n be_v not_o pour_v out_o to_o speak_v in_o the_o church_n but_o notwithstanding_o every_o sister_n in_o christ_n upon_o who_o the_o spirit_n of_o the_o lord_n be_v pour_v out_o who_o by_o that_o spirit_n be_v move_v to_o prophesy_v they_o forbid_v not_o fo●_z the_o same_o paul_n that_o do_v not_o permit_v a_o woman_n to_o speak_v in_o the_o church_n he_o write_v un●to_o the_o church_n that_o they_o shall_v not_o quench_v the_o spirit_n nor_o despise_v prophecy_n neither_o do_v we_o read_v that_o he_o nor_o none_o of_o his_o company_n do_v forbid_v philip_n daughter_n to_o prophesy_v act_n 21._o 8_o 9_o who_o be_v virgin_n and_o four_o in_o number_n upon_o who_o the_o spirit_n of_o the_o lord_n be_v pour_v out_o and_o who_o will_v say_v that_o he_o will_v have_v quench_v the_o spirit_n in_o phoebe_n if_o she_o have_v be_v move_v by_o it_o to_o prophesy_v in_o that_o church_n of_o" ]

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[ "Question\n\n# The regression between average SAT reading scores in 2014 and average SAT reading scores in 2010 for the 51 states\n\nhas the following partial table of co-efficients:\n\nYou are testing if the slope is significantly different from zero using α = 0.001.\n\nWhat is the value of the test statistic? Answer to three decimal places.\n\nSCOICS TIT\nU TOT CHIC\n1 Stat\nEstimate\nStandard error\nIntercept\n-65.71\n30.23" ]

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[ "Cody\n\n# Problem 112. Remove the air bubbles\n\nSolution 1536512\n\nSubmitted on 22 May 2018 by Shaul Salomon\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\na = [ 1 2 3 0 4 5 6 0 0 ]; b_correct = [ 0 0 0 1 2 3 6 4 5 ]; assert(isequal(bubbles(a),b_correct))\n\nw = 0 1 6 w = 0 2 4 w = 0 3 5\n\n2   Pass\na = [ 1 0 5 0 7 0 6 ]'; b_correct = [ 0 0 0 1 5 7 6 ]'; assert(isequal(bubbles(a),b_correct))\n\nw = 0 0 0 1 5 7 6\n\n3   Pass\na = [1 0; 1 1]; b_correct = [1 0; 1 1]; assert(isequal(bubbles(a),b_correct))\n\nw = 1 1 w = 0 1\n\n4   Pass\na = [0 8 0 6 -2]'; b_correct = [0 0 8 6 -2]'; assert(isequal(bubbles(a),b_correct))\n\nw = 0 0 8 6 -2\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "", null, "CD shopping page\n\n## 2017-Sustainable Industrial Processing SummitSIPS 2017 Volume 6: Mathematics, Multiscale Mechanics, Coatings\n\n Editors: Kongoli F, Masset P, Rokicki P Publisher: Flogen Star OUTREACH Publication Year: 2017 Pages: 142 pages ISBN: 978-1-987820-71-3 ISSN: 2291-1227 (Metals and Materials Processing in a Clean Environment Series)\n\n## On the Negative-Energy 4-Spinors and Masses in the Dirac Equation\n\nValeriy Dvoeglazov1;\n1UNIVERSIDAD DE ZACATECAS, Zacatecas, Mexico;\nType of Paper: Regular\nId Paper: 36\nTopic: 38\n\n## Abstract:\n\nIt is easy to check that both algebraic equation $Det (hat p - m) =0$ and $Det (hat p + m) =0$ for $u-$ and $v-$ 4-spinors have solutions with $p_0= pm E_p =pm sqrt{{bf p}^2 +m^2}$. The same is true for higher-spin equations. Meanwhile, every book considers\nthe equality $p_0=E_p$ for both $u-$ and $v-$ spinors of the $(1/2,0)oplus (0,1/2))$ representation only, thus applying the Dirac-Feynman-Stueckelberg procedure for elimination of the negative-energy solutions. The recent Ziino works (and, independently,\nthe articles of several others) show that the Fock space can be doubled. We re-consider this possibility on the quantum field level for both $s=1/2$ and higher spin particles.\n\n## Keywords:\n\ncosmology; hadronic mechanics; physics; quantum mechanics;\n\n## Full Text:\n\nClick here to access the Full Text" ]

[ null, "http://www.flogen.org/books/imgBooks/sips2017_CD_Proceedings.jpg", null ]

[ "# Use xkcd's formula to approximate the world population\n\nIn xkcd 1047, Randall Munroe lists \"slightly wrong\" approximations of assorted quantities and numbers with varying precision and complexity, such as that the number of liters in a gallon is very close to 3 + π4. In the middle of the comic, he gives an intermission: a way to estimate the world (and United States) population based for a given year.", null, "(Cropped from xkcd: Approximations by Randall Munroe)\n\nYour task is to write a program that implements these formulas to approximate the current world and U.S. populations, replicated as follows.\n\n## World population\n\n1. Take the last two digits of the current year.\n2. Subtract the number of leap years (including the current year) since Hurricane Katrina (2005). For these purposes, any year divisible by 4 is considered a leap year.\n3. Add a decimal point between the two numbers (the same as dividing by 10).\n4. Add 6. This gives the result in billions of people.\n\n## U.S. population\n\n1. Take the last two digits of the current year.\n2. Subtract 10.\n3. Multiply by 3.\n5. Add 3 to the beginning (for this challenge, some numbers will be negative, so add 300 instead). Somehow I didn't notice that just concatenating wouldn't work because the program I used to generate the results just added 300.\n6. This gives the result in millions of people.\n\n# Details\n\nThis formula \"should stay current for a decade or two,\" but you must be able to theoretically handle any year 2000–2039 inclusive. For some cases, the leap years since Katrina will have a negative or zero value.\n\nYou are free to simplify the formula in any way, as long as all outputs match the ones below.\n\nFor the year, use the year according to the computer's clock. It must work next year and any other year this century, so you cannot simply hardcode 2015. For convenience, you might want to include a way to specify the year as a variable or input to test other years.\n\nThe output should be the approximated world population (in billions of people), followed by some delimiter (e.g. space or comma), followed by the U.S. population (in millions of people). You may also write a function that returns or prints a string or an array of numbers or strings.\n\nThis is code golf, so shortest code in bytes wins. Tiebreaker is earliest post.\n\n# Test cases\n\nThis is a list of all possible years, followed by the two outputs.\n\nYear World U.S.\n2000 6.1 280\n2001 6.2 283\n2002 6.3 286\n2003 6.4 289\n2004 6.4 292\n2005 6.5 295\n2006 6.6 298\n2007 6.7 301\n2008 6.7 304\n2009 6.8 307\n2010 6.9 310\n2011 7 313\n2012 7 316\n2013 7.1 319\n2014 7.2 322\n2015 7.3 325\n2016 7.3 328\n2017 7.4 331\n2018 7.5 334\n2019 7.6 337\n2020 7.6 340\n2021 7.7 343\n2022 7.8 346\n2023 7.9 349\n2024 7.9 352\n2025 8 355\n2026 8.1 358\n2027 8.2 361\n2028 8.2 364\n2029 8.3 367\n2030 8.4 370\n2031 8.5 373\n2032 8.5 376\n2033 8.6 379\n2034 8.7 382\n2035 8.8 385\n2036 8.8 388\n2037 8.9 391\n2038 9 394\n2039 9.1 397\n\n• Do you have to round the numbers? – Blue Sep 12 '15 at 19:35\n• @muddyfish I'm not sure I understand. If you follow the instructions in the comic exactly, there's technically no division going on, but the world population should be rounded to the nearest tenth. – NinjaBearMonkey Sep 12 '15 at 19:38\n• I'm a little confused by the United States population one. If you're concatenating a 3, shouldn't 2040 give a population of 3100? 40 - 10 = 30, 30 * 3 = 90, 90 + 10 = 100, which would give \"3\" + \"100\" = 3100 – cole Sep 12 '15 at 20:01\n• @Cole Good point, I'll make it so you only have to support years through 2039. About hardcoding the year, I don't want to allow hardcoding because that will almost always be shorter even languages that do support dates. – NinjaBearMonkey Sep 12 '15 at 20:09\n• @NinjaBearMonkey I suggest that you change the description of \"adding 3, thinking concatenation\" to a literal \"add 300\" to cover all of the edge cases that occur when the previous result isn't a nice 2-digit positive number. (For example, year 2000 gives 280 as a result of -20+300=280 and not 3 . -20= \"3-20\") – PhiNotPi Sep 13 '15 at 5:15\n\n# Pyth, 21 20 bytes\n\n-1 byte by Dennis\n\nc/J-*3.d3C\\ᙹ4T+33J\n\n\nThese have the same byte count but are ASCII-only:\n\nc/J%*3.d3 523 4T+33J\nc/-J%*3.d3*44T33 4TJ\n\n\nI don't know Pyth, so still probably possibly golfable. Using the same algorithm:\n\n# TI-BASIC, 23 bytes\n\nmax(3getDate-5753\n{.1int(Ans/4),Ans+33\n\n\ngetDate returns a list of three floats {YYYY,MM,DD} in some order depending on the date format setting (TI-84s don't have a true int datatype); the max( will be the year. Multiplying and subtracting inside the max( saves a close-paren.\n\n# Javascript (ES6), 5554 48 bytes\n\n-~((n=Date().substr(13,2)*3+280)/4-9.1)/10+' '+n\n\n\nWorks in Firefox 33; theoretically supports all years from 2000 to 2099. If programs that dump the result onto the console are not allowed, use this 51-byte function:\n\n(n=Date().substr(13,2)*3+280)=>-~(n/4-9.1)/10+' '+n\n\n\nFull program, 55 bytes:\n\nn=Date().substr(13,2)*3+280,alert(-~(n/4-9.1)/10+' '+n)\n\n\nGetting the year was quite expensive, but after using the deprecated getYear() instead of getFullYear(), all of the numbers in the equation became smaller, saving a lot of bytes. EDIT: Thanks to an eeevil trick, I skipped new and getYear() altogether. >:D\n\nSuggestions welcome!\n\n# Pyth, 30 bytes\n\n.R+*.075J%.d3C\\d6.105 1+*3J280\n\n\nMy first Pyth program!\n\nThanks @Jakube for some hints (I would never have thought of those!)\n\n• Haven't read the question, but here are a few optimizations I immediately saw. Write everything on one line. Choose a different ordering of numbers and variables (+*3Z280 instead of +*Z3 280 for instance). C\\d instead of 100 (saves spaces). Use J instead of Z (saves the =). Inline the assignment. Link – Jakube Sep 12 '15 at 20:27\n\n# Python 2, 80 bytes\n\nfrom datetime import*\ny=date.today().year%40\nprint int(61.55+.75*y)/10.,y*3+280\n\n\nNow rounds!\n\n• You have to round, as OP clarified a minute ago. ;-) – mınxomaτ Sep 12 '15 at 19:52\n• year%100 is the same as year%40. – lirtosiast Sep 12 '15 at 22:31\n\n# CJam, 28 bytes\n\net0=100%__4/(-Ad/6+S@3*280+\n\n\nTry it online\n\nTo try years other than the current one, replace the et0= at the start with the literal value of the year.\n\n• Since 2000 is divisible by 40, and you only need 2000-2039, you can use 40% to save one byte. – Andrea Biondo Sep 13 '15 at 8:34\n\n# Python 3, 134 bytes\n\nWorks well but seems a bit long\n\nfrom datetime import*\ny=str(date.today().year)\nz=int(y[2:])\nm=str(60+(z-int((int(y)-2005)/4)))\nprint(\"%s.%s\"%(m,m),310+(z-10)*3)\n\n• To shorten this, use from time import*, y=strftime('%Y'). Or copy the other python answer :P – FlipTack Nov 19 '16 at 14:31\n\n## AutoIt - 6058 56 bytes\n\n$_=@YEAR-2e3 ClipPut(($_-Int($_/4-1))/10+6&' 3'&3*$_-20)\n\n\nThe rounding eats bytes (not anymore). I have now tweaked both formulas. Some sample outputs:\n\n7.3 325 // 2015\n7.3 328\n7.4 331\n7.5 334 // 2018\n8.4 370 // 2030\n\n\nThey all seem to be accurate.\n\nA tip: Order of execution saves bytes on parentheses. For example: (a-4)/4 = a/4-1 :-).\n\n$n=date -f yy;.1*[int](61.45+.75*$n);280+3*$n Uses a similar rounding trick as muddyfish's Python 2 answer to simplify the world population calculation, since PowerShell implicitly Round()s when you cast from a [double] to an [int], rather than truncating. For the output, we assume that \"followed by some delimiter (e.g. space or comma)\" can also mean \"newline\", so we execute one result and then the second. PowerShell implicitly writes out results, so we don't need to explicitly call any printing commands. # Mathematica, 50 bytes n=Today[\"YearShort\"];{.1*(61+n-Floor[n/4]),280+3n} Note that this is dependent on having a Wolfram Engine with Version number 10+ (released 2014) due to dependence on DateObjects. # R, 64 bytes n=as.numeric(format(Sys.time(),\"%y\")) c(.1*(61+n-n%/%4),280+3*n) Direct port of Mathematica code, think I had a shorter solution but dependent on packages whereas this works with base R. • (1/10) -> .1? – lirtosiast Sep 13 '15 at 17:32 • I think you also don't need parentheses around the .1. – lirtosiast Sep 13 '15 at 17:43 # Java, 180177166152 143 bytes Thanks Thomas for helping a noob out :) class D{public static void main(String[]a){int f=java.time.LocalDate.now().getYear();System.out.printf(\"%.1f %d\\n\",(3.*f-5755)/40,3*f-5720);}} Ungolfed version: class D { public static void main(String[]a) { int f = java.time.LocalDate.now().getYear(); System.out.printf(\"%.1f %d\\n\",(3.*f-5755)/40,3*f-5720); } } Requires Java 8. • import java.time.*? 3.0->3.? You also don't need to print the year with the output. – lirtosiast Sep 20 '15 at 4:15 • Oh, I didn't realize you didn't need the year printed out... :P – a spaghetto Sep 20 '15 at 14:35 # JavaScript (ES6) 52 A function returning the output as an array. (y=(new Date).getYear())=>[(y+~(y/4)-13)/10,y*3-20] Just for testing purpose, the function accepts an input equals to current year - 1900 (e.g. 105 for 2015) That's in the line of the ETHproductions' answer (the math is the math) but avoding the evil trick it is more portable in different locales. And as a function it's 1 byte shorter. Test snippet: f=(y=(new Date).getYear())=>[(y+~(y/4)-13)/10,y*3-20] o=[]; for(a=2000;a<2050;a++)o.push(<td>${a}</td><td>${f(a-1900)}</td>); document.write(<table><tr>${o.join</tr><tr>}</tr></table>)\ntd { text-align:right; font-family:monospace }\n\n# Desmos, 140 Bytes\n\nI'm counting a newline to be a signal for a new equation. Folders on the link are just for organization.\n\n## Golfed, 140 Bytes\n\nClick add slider when prompted.\n\na=6+.1b-.1c\nd=280+3b\nc=\\sum_{n=2005}^q\\left\\{0=\\operatorname{mod}\\left(n,4\\right),0\\right\\}\nb=100\\left|\\operatorname{floor}\\left(.01q\\right)-.01q\\right|\n\n\n# Ungolfed, 261 Bytes\n\np_{world}=6+\\frac{l_{astdig}-l_{eap}}{10}\np_{us}=310+3\\left(l_{astdig}-10\\right)\ny_{ear}=2039\nl_{eap}=\\sum _{n=2005}^{y_{ear}}\\left\\{0=\\operatorname{mod}\\left(n,4\\right),0\\right\\}\nl_{astdig}=100\\left|\\operatorname{floor}\\left(\\frac{y_{ear}}{100}\\right)-\\frac{y_{ear}}{100}\\right|\n\n\n# Ruby, 57 bytes\n\ny=Time.now.year%40;puts \"#{6+(y-(y-5)/4)*0.1} #{3*y+280}\"\n\n\nUnfortunately, Time.now.year really costs some characters.\n\n# Glava, 77 bytes\n\ni|f=java.time.LocalDate.now().getYear();F(\"%.1f %d\\n\",(3.*f-5755)/40,3*f-5720\n\n\nA translation of my Java answer.\n\n# PHP, 45 bytes\n\nThe code:\n\necho(int)(($x=3*date(y)+247)/4)*.1,\" \",$x+33;\n\n\nBecause y (the argument of date()) is an undefined constant, PHP triggers a notice (that can be muted) and converts it to a string (as we need); this PHP courtesy allows saving 2 bytes.\n\nIn order to suppress the notice, the program needs to be run using the error_reporting=0 runtime directive, like this:\n\n$php -d error_reporting=0 -r 'echo(int)(($x=3*date(y)+247)/4)*.1,\" \",$x+33;' 7.3 325 ### For testing By replacing the call to date(y) with $argv (the first command line argument), the program length increases with 1 byte but it can get the year from the command line.\n\nThe expected argument is the year minus 2000; it also works for negative values (years before 2000) or values greater than 40 (after year 2040).\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 00\n6.1 280\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 01\n6.2 283\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 02\n6.3 286\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 03\n6.4 289\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 04\n6.4 292\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 15\n7.3 325\n\n$php -r 'echo(int)(($x=3*$argv+247)/4)*.1,\" \",$x+33;' 39\n9.1 397\n\n\n# APL, 2523 29 bytes\n\n{(10÷⍨⌊⍵÷4),⍵+33}⊃¯5753+3×⎕TS\n\n\nTryAPL\n\nYes, it's 29 bytes.\n\n• Save two chars: (÷∘40,+∘33)¯5753+3×⊃⎕TS – Adám Sep 30 '15 at 2:50" ]

[ null, "https://i.stack.imgur.com/DvmsW.png", null ]

[ "## I'm an IDIOT!! June 11, 2013\n\nI spent a ton of time trying to write a syntax parser for a meta-language which would be parsed by Javascript.  I just replaced a lot of confusing logic with about 40 lines of code.\n\nFirst, Javascript is dynamic and has stuff built in that can evaluate arbitrary code passed in via a string.\n\nFunction!\n\nHowever, my syntax looks like this:  Hello {{page.user == null ? 'Anonymous' : page.user.firstName }}\n\nSo, we need to know what \"page\" is in the context of the function.  I still have to parse out top level variables.  In the code above, it will get \"page\" as a top level variable.\n\nThen, I build up the function:\n\nvar f = new Function(vars, body);\n\n\"body\" is actually modified, I set it to \"return \" + body;  So that I can do {{ page.user == null ? 'Anonymous' : page.user.firstName }} and it will return the display name instead of undefined, the default behavior of a void function.\n\nI have to count up the number of variables used, and build the function accordingly.  Currently, this is a switch statement.\n\nswitch (vars.length){\ncase 0: f = new Function(body); break;\ncase 1: f = new Function(vars, body); break;\ncase 2: f = new Function(vars, vars, body); break;\n}\n\nLuckily in my code, there aren't more than 3-4 \"top level\" variables, including globals like \"Array\" and \"String\".\n\nHere's the variable parsing part:\n\n```var shared = require(\"./shared\"); require(\"strings\"); var constants = { \"true\": true, \"false\": true, \"null\": true }; var lookup = {}; this.getVariables = function(body){ if (body in lookup) return lookup[body]; var vars = []; var instr = false; var instrch = null; var buf = \"\"; var toplevel = false; var result = null; for (var i = 0; i < body.length; i++){ var ch = body.charAt(i); switch (ch){ case \"'\": case \"\\\"\": instr = ch != instrch; instrch = instrch == null ? ch : (instr ? instrch : null); break; } if ((!instr && shared.tokenSeparator.test(ch)) || i == body.length-1){ if (i == body.length-1) buf+= ch; if (!toplevel && (result = shared.variable.exec(buf)) != null && !(result in constants)){ if (!vars.some(function(d){ return d == result})){ vars.push(result); toplevel = true; } } buf = \"\"; } else if (instr) buf = \"\"; else buf += ch; if (toplevel && (instr || (shared.tokenSeparator.test(ch) && ch != \".\" && ch != \"]\"))) toplevel = false; } lookup[body] = vars; return vars; }```\n\nAnd here's the evaluation part:\n\n```var syntax = require(\"./syntax\"); var shared = require(\"./shared\"); var lookup = {}; var Evaluator = function(globals){ this.globals = globals; } Evaluator.prototype.evaluate = function(body, context){ body = shared.replaceComps(body); var vars = syntax.getVariables(body); var args = []; body = \"return \" + body; for (var i = 0; i < vars.length; i++){ if (context && vars[i] in context) args.push(context[vars[i]]); else if (this.globals && vars[i] in this.globals) args.push(this.globals[vars[i]]); } if (body in lookup){ return lookup[body].apply(null, args); } var f = null; switch (vars.length){ case 0: f = new Function(body); break; case 1: f = new Function(vars, body); break; case 2: f = new Function(vars, vars, body); break; case 3: f = new Function(vars, vars, vars, body); break; case 4: f = new Function(vars, vars, vars, vars, body); break; } var result = null; if (f != null){ result = f.apply(null, args); lookup[body] = f; } return result; } this.Evaluator = Evaluator;```\n\nshared.js has a regular expression, a map (for old syntax considerations), and a function to replace some old ways of doing things with the new, pure Javascript way of doing it.\n\n```this.replCompRegex = / (eq|ne|gt|lt|gte|lte) /; this.replCompMap = { eq: \"==\", ne: \"!=\", gt: \">\", lt: \"<\", gte: \">=\", lte: \"<=\" }; this.replaceComps = function(body){ var res = null; while ((res = this.replCompRegex.exec(body)) != null){ body = body.replace(res, this.replCompMap[res]); } return body; }```\n\n2021\n2020\n2019\n2018\n2017\n2016\n2015\n2014\n2013\n2012\n2011\n2010\n2009\n2008\n2007\n2006\n2005" ]

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[ "## 竞赛描述\n\n### 评估方法", null, "", null, "## 方案分享\n\n### AE MLP方案\n\nAE MLP引入了自动编码器进行分类特征提取。自动编码器是一种神经网络，可用于学习原始数据的压缩表示。自编码器由编码器和解码器子模型组成。编码器压缩输入，解码器尝试从编码器提供的压缩版本重新创建输入。训练后，编码器模型被保存，解码器被丢弃。然后，编码器可用作特征提取，对可用于训练不同机器学习模型的原始数据执行特征提取。", null, "``````def create_ae_mlp(num_columns,\nnum_labels,\nhidden_units,\ndropout_rates,\nls = 1e-2,\nlr = 1e-3):\n# 定义输入\ninp = tf.keras.layers.Input(shape = (num_columns, ))\nx0 = tf.keras.layers.BatchNormalization()(inp)\n# 高斯加噪\nnoise_x0 = tf.keras.layers.GaussianNoise(dropout_rates)(x0)\n# 编码器\nencoder = tf.keras.layers.Dense(hidden_units)(encoder)\nencoder = tf.keras.layers.BatchNormalization()(encoder)\nencoder = tf.keras.layers.Activation('swish')(encoder)\n# 解码器\ndecoder = tf.keras.layers.Dropout(dropout_rates)(encoder)\ndecoder = tf.keras.layers.Dense(num_columns, name = 'decoder')(decoder)\n# Classification Layer\nx_ae = tf.keras.layers.Dense(hidden_units)(decoder)\nx_ae = tf.keras.layers.BatchNormalization()(x_ae)\nx_ae = tf.keras.layers.Activation('swish')(x_ae)\nx_ae = tf.keras.layers.Dropout(dropout_rates)(x_ae)\n# Autoencoder的分类器\nout_ae = tf.keras.layers.Dense(num_labels,\nactivation='sigmoid',\nname = 'ae_action')(x_ae)\n# 拼接原始输入和编码器输入\nx = tf.keras.layers.Concatenate()([x0, encoder])\n# MLP的网络层\nx = tf.keras.layers.BatchNormalization()(x)\nx = tf.keras.layers.Dropout(dropout_rates)(x)\nfor i in range(2, len(hidden_units)):\nx = tf.keras.layers.Dense(hidden_units[i])(x)\nx = tf.keras.layers.BatchNormalization()(x)\nx = tf.keras.layers.Activation('swish')(x)\nx = tf.keras.layers.Dropout(dropout_rates[i + 2])(x)\n# MLP的分类器\nout = tf.keras.layers.Dense(num_labels, activation = 'sigmoid', name = 'action')(x)\n# 定义输入输出\nmodel = tf.keras.models.Model(inputs = inp, outputs = [decoder, out_ae, out])\n# 定义优化器，损失函数和Metrics\nloss = {'decoder': tf.keras.losses.MeanSquaredError(),\n'ae_action': tf.keras.losses.BinaryCrossentropy(label_smoothing = ls),\n'action': tf.keras.losses.BinaryCrossentropy(label_smoothing = ls),\n},\nmetrics = {'decoder': tf.keras.metrics.MeanAbsoluteError(name = 'MAE'),\n'ae_action': tf.keras.metrics.AUC(name = 'AUC'),\n'action': tf.keras.metrics.AUC(name = 'AUC'),\n},\n)\n\nreturn model``````\n\n1. 交叉验证，由于金融任务存在任务数据是时间序列，模型的训练和使用存在时间延迟 。作者使用了PurgedGroupTimeSeriesSplit进行交叉验证划分。如下图所示，PurgedGroupTimeSeriesSplit支持对时序数据进行划分，避免了验证集使用在训练集之前的数据，同时支持定义训练集和验证集之间的时间间隔。本文使用了5-折 31-间隔的交叉验证划分方式。", null, "2. 使用 swish 激活函数而不是 ReLU 来防止“死神经元”并平滑梯度； 3. 用 3 个不同的随机种子训练模型并取平均值以减少预测方差； 4. 仅使用在最后两次 交叉验证拆分中训练的模型（具有不同种子），因为它们使用了较多的数据； 5. 通过MLP 的 BCE 损失进行早停；6. 使用 Hyperopt 找到最优的超参数集。\n\n### Xgboost方案\n\nXgboost比较简单，指定超参，使用不同的随机数种子训练了三个模型进行使用。代码如下：\n\n``````X = train.loc[:, train.columns.str.contains('feature')].values\ny = train.loc[:, 'action'].astype('int').values\n\nX_ = X\ny_ = train.loc[:, 'action3'].astype('int').values\n\nclf1 = xgb.XGBClassifier(\nn_estimators=100,\nmax_depth=11,\nlearning_rate=0.05,\nsubsample=0.90,\ncolsample_bytree=0.7,\nmissing=-999,\nrandom_state=21,\ntree_method='gpu_hist', # THE MAGICAL PARAMETER\nreg_alpha=10,\nreg_lambda=10,\n)\n\nclf1.fit(X_, y_)\n\nclf2 = xgb.XGBClassifier(\nn_estimators=100,\nmax_depth=11,\nlearning_rate=0.05,\nsubsample=0.90,\ncolsample_bytree=0.7,\nmissing=-999,\nrandom_state=210,\ntree_method='gpu_hist', # THE MAGICAL PARAMETER\nreg_alpha=10,\nreg_lambda=10,\n)\n\nclf2.fit(X_, y_)\n\nclf3 = xgb.XGBClassifier(\nn_estimators=100,\nmax_depth=11,\nlearning_rate=0.05,\nsubsample=0.90,\ncolsample_bytree=0.7,\nmissing=-999,\nrandom_state=2010,\ntree_method='gpu_hist', # THE MAGICAL PARAMETER\nreg_alpha=10,\nreg_lambda=10,\n)\n\nclf3.fit(X_, y_)``````" ]

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[ "Home | | Maths 11th std | Newton Leibnitz Integral\n\n# Newton Leibnitz Integral\n\nIf F (x) is a particular antiderivative of a function f (x) on an interval I, then every antiderivative of f (x) on I is given by ∫ f ( x)dx = F ( x) + c\n\nNewton-Leibnitz Integral\n\nIntegral calculus is mainly divided into indefinite integrals and definite integrals. In this chapter, we study indefinite integration, the process of obtaining a function from its derivative.\n\nWe are already familiar with inverse operations. (+,-) (x,÷), ()n,n√ are some pairs of inverse operations. Similarly differentiation and integrations (d, ∫ ) are also inverse operations. In this section we develop the inverse operation of differentiation called ‘antidifferentiation’.", null, "## Definition 11.1\n\nA function F (x) is called an antiderivative (Newton-Leibnitz integral or primitive) of a function f (x) on an interval I if\n\nF ′(x ) = f (x) , for every value of x in I\n\nIllustration 11.1\n\nIf F (x ) = x2 + 5 then\n\nF ′(x ) = 2x .\n\nThus if f (x) is defined by", null, "f( x ) = 2x, then\n\nwe say that f (x) is the derivative of F (x) and that F (x) is an antiderivative of f (x)\n\nConsider the following table", null, "We  can see  that the derivative  of F (x) , P (x) , Q (x) and H (x) is f (x) ,  but in  reverse the antiderivatives of f (x ) = 2x is not unique. That is the antiderivatives of f (x) is a family of infinitely many functions.\n\n## Theorem 11.1\n\nIf F (x) is a particular antiderivative of a function f (x) on an interval I, then every antiderivative of f (x) on I is given by\n\n∫ f ( x)dx = F ( x) + c\n\nwhere c is called an arbitrary constant, and all antiderivatives of f (x) on I can be obtained by assigning particular value to c.\n\nThe function f(x) is called Integrand.\n\nThe variable x in dx is called variable of integration or integrator.\n\nThe process of finding the integral is called integration or antidifferentiation (Newton-Leibnitz integral).\n\nThe peculiar integral sign ∫ originates in an elongated S (like Σ ) which stands for sum.\n\nOften in applications involving differentiation it is desired to find a particular integral antiderivative that satisfies certain conditions called initial condition or boundary conditions.\n\nFor instance, if an equation involving dy/dx is given as well as the initial condition that y = y1 when x = x1, then after the set of all antiderivatives is found, if x and y are replaced by x1 and y1 , a particular value of the arbitrary constant is determined. With this value of c a particular antiderivative is obtained.\n\n## Illustration 11.2\n\nSuppose we wish to find the particular antiderivative satisfying the equation", null, "and the initial condition that y = 10 when x = 2.\n\nFrom the given equation", null, "We substitute y = 10 when x = 2 , in the above equation\n\n10 = 22 + c  â‡’  c = 6\n\nWhen this value c =6 is substituted we obtain\n\n= x2 + 6\n\nwhich gives the particular antiderivative desired.\n\nTags : Definition, Theorem , 11th Mathematics : UNIT 11 : Integral Calculus\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n11th Mathematics : UNIT 11 : Integral Calculus : Newton Leibnitz Integral | Definition, Theorem" ]

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[ "# Prism radiation at constant T\n\n## FILES\n\n C++ FLTK multiprecision desktop app source code prism_radiation_multi.cpp libgcc_s_sjlj-1.dll from MINGW-W64 [mingw-w64_8.1.0_x86_64_win32_sjlj] libgcc_s_sjlj-1.dll libstdc++-6.dll from MINGW-W64 [mingw-w64_8.1.0_x86_64_win32_sjlj] libstdc++-6.dll C++ FLTK multiprecision desktop app executable SIGNED BY xformulas.net prism_radiation_multi.exe Formulas prism_radiation.txt ZIP with ALL the files 20200424_111036_000013_prism_radiation_multi.zip\n\n## INPUT DATA\n\n M [kg] Mass z [m] Length z x [m] Length x rho [kg*m^-3] Density Tp [K] Prism effective temperature Te [K] Exterior effective temperature sigma [W*m^-2*K^-4] Stefan-Boltzmann constant t [s] Time\n\n## OUTPUT DATA\n\n V [m^3] Volume y [m] Length y S [m^2] Surface Psp [W*m^-2] Prism radiated power per each m^2 (black body approximation) Ptp [W] Prism total radiated power Etp [J] Prism total radiated energy Etp [kcal] Prism total radiated energy Pse [W*m^-2] Exterior radiated power per each m^2 (black body approximation) Pte [W] Exterior total radiated power Ete [J] Exterior total radiated energy Ete [kcal] Exterior total radiated energy Ej [J] Energy required to maintain prism temperature (basal metabolism) Ek [kcal] Energy required to maintain prism temperature (basal metabolism)\n\n### 2019-10-15\n\nGenerated by xformulas.net on 20200424_111036_000013" ]

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[ "We have over 40 years of experience.\n\n1. Home\n2. calculate\n3. Tohow To Calculate Sand Quantity In Concrete\n\n# Tohow To Calculate Sand Quantity In Concrete\n\nSand calculator,nov 23, 2020 you can also type the area of the excavation directly into our calculator if you choose an excavation of some more sophisticated shape. establish the depth of the excavation. lets say its 0.5 yd. multiply the area by the depth of the excavation to obtain its volume: 0.5 cu yd. the volume of sand required is equal to the volume\n\nChatear en línea\n\n## New Projects Of Tohow To Calculate Sand Quantity In Concrete\n\nThere are four series of products including crushing, sand making, building materials, and grinding, with excellent performance and complete models\n\n• Refined and intelligent manufacturing\n• Online and offline communication\n• Special car transportation and distribution\n• Customer acceptance and commissioning\n• Comprehensive after-sales service\n\n### Get in Touch\n\nNeed more additional information or queries? We are here to help. Please fill in the form below to get in touch.\n\nI accept the Data Protection Declaration\n\n### Sand Calculator\n\nnov 23, 2020 you can also type the area of the excavation directly into our calculator if you choose an excavation of some more sophisticated shape. establish the depth of the excavation. lets say its 0.5 yd. multiply the area by the depth of the excavation to obtain its volume: 0.5 cu yd. the volume of sand required is equal to the volume aggregate calculator concrete calculator landscape calculator cubic yards calculator real-time graphics. make the calculations and see the changes. real-time graphics. use it to calculate the volume of sand you will need, the weight of the material, and if you know the price per unit massvolume, then the total cost as well.sand calculator online estimate the sand required for your construction or landscaping project in weight and volume If you are wondering how much sand do need, our free sand calculator is here to do the math for you. information about sand density, common sand types, sand grain sizes, how much a cubic yard of sand\n\n### How To Calculate Cement Sand And Aggregate Quantity In\n\nTo calculate the individual quantity of cement, sand and aggregate and water in cum of concrete we assuming the nominal mix design concrete of grade As per IS 2000 in grade of concrete the proportion of cement, sand and aggregate is,reading time: minutequantities of materials for concrete such as cement, sand and aggregates for production of required quantity of concrete of given mix proportions such as can be calculated by absolute volume method. this method is based on the principle that the volume of fully compacted sep 07, 2017 the quantity of sand that you need 0.7608 1762 kgm 1217.28 kg of sand calculate the quantity of aggregate you need remember that from our mix ratio for part cement, you need parts of granite. that is,\n\n### How To Calculate Cement Sand And Coarse Aggregate For\n\nmake sure all the measurements are in the same unit. formula for circular volume 3.14 0.42 So the required concrete volume 0.42 now we have calculated the required volume. its time to calculate the micro-ingredients.use this paver sand calculator to solve for the amount of sand you need to create a stable layer under a pavement. whether youre planning to build an elaborate patio or a paved driveway, you must first create a strong foundation to lay your concrete or stone pavers on. this layer which supports these elements directly is usually paving sand.how to calculate cement, sand and aggregate quantity in concrete, quantity of cement sand and aggregate in concrete in this topic we know about how much cement, sand and aggregate quantity are used in of concrete. We know that concrete is mixture of cement sand and aggregate in fixed proportion, there is a several grade of concrete like and so\n\n### How To Calculate Cement Sand And Coarse\n\nmar 02, 2021 We find out the volume of concrete in wet condition So in dry conditions, we have to add to 35% sand bulkage and add 20% for wastage. accounting bulkage of sand 34%wet volume of concrete mix ratio for concrete We have to do following calculating steps:- step we have wet volume of concrete is equal to cubic metre but when we calculate actual quantity of dry ingredient cement, sand and aggregate of concrete. dry volume of concrete is equal to 1.54 1.54 jul 26, 2017 therefore volume of the block 0.02278 0.5 0.0143 step calculate the volume of the blocks If the volume of block is 0.0143, the volume of blocks 0.502 step calculate the volume sand required for the blocks the volume of bag of cement is 34.72 litres\n\n### Concrete Material Calculator Sand Cement And\n\nconcrete material calculator, just enter the volume of concrete and it will give the calculated quantity of material like sand, aggregate, and cement. read civiconcepts concrete material calculator calculate quantity of various materials of concrete, calculate concrete material per square feet consider 35% sand bulkage 1.2 calculate the volume in cubic meters. concrete of five cubic needs seven and a half cubic meters. this also depends on the size of ballast to be used and the use of the concrete. To calculate the amount of fine aggregate which is sand its important to know its properties. sand bulks when wet\n\n### Sand And Gravel Calculator Buildbase\n\nWe use cookies to help give you the best experience on our website. If you continue without changing your settings, well assume that you are happy to receive all cookies on the website.may 11, 2016 To make concrete there are four basic materials you need: portland cement, sand, aggregate and water. the ratio of aggregate to sand to cement is an important factor in determining the compressive strength of the concrete mixture. the mixing process is the fact that the better you mix the concrete, the stronger it will be.for measuring tipper sand capacity you will have to calculate the volume of the tripper, length width height, then you can calculate the total volume of the sand. tipper sand capacity in ton size of tipper vary from ft to ft, around ft ton, so tipper sand\n\n### How To Calculate Cement And Sand Quantity For Plastering\n\nadding bulking of sand & wastage. So to get 1.2 of wet cement mortar, we have to multiply the dry volume by 1.55. what is 1.55 constant in the calculation? bulking of sand If the moisture is present in the sand, then it makes the sand look bulkier, which could result in inadequate sand proportion in the concrete of concrete material cement sand aggregate and water. cement quantity for of concrete: part of cement 15.5, density of cement dry volume 1.3, weight volume density, weight of cement 15.5 1.3 1440 kg so cement is required for of grade of concrete. sand quantity for of concrete: part of sand 1.55.5, density of sand dec 16, 2020 here we calculate the quantity of cement sand and aggregate in concrete slab detailed calculation given below It is very easy to calculate how much is required for the construction of the slab. concrete slab floors come in many forms and can be used to provide great thermal comfort and lifestyle advantages.\n\n### How To Calculate Number Of Bricks Cement And Sand For\n\nrequired amount cement quantity Kg 1.26 bags required amount of sand 0.26277 cubic metre therefore, for cum of brickwork, we need. 500 numbers of bricks; kg of cement; 0.263 of sand. feel free to use the below calculator. brickwork calculatorfeb 21, 2018 assuming you are to calculate the amount of sand require for of concrete of concrete 0.637 tons of sand. of concrete tons of sand. 0.35 tons of sand this means 0.35 tons of sand will mix of concrete using mix ratio check out how to calculate the total tons of granite for a volume of concretejan 26, 2017 We will calculate quantities of materials for concrete let us assume the mix proportion is volume of wet concrete volume of dry concrete 1.54 1.54\n\n### Calculate Cement Sand And Aggregate Quantity In Concrete\n\nsep 24, 2020 how to use a concrete work calculator? steps enter required volume of concrete which is you obtain from estimate of building. steps enter concrete mix details like proportion of cement sand and coarse aggregate which is you obtain from mix design of concrete. If your mix design is that means the ratio of grade is 1.3. therefore enter cement proportion sand calculate cement concrete mix design or estimate how much cement concrete volume is required for your construction using our free online calculator. know exactly how many bags, kg and ton of cement, sand and aggregate is needed of specific cement, sand and aggregate ratio aug 03, 2018 calculation for sand quantity; consider volume of concrete dry volume of concrete 1.54 1.54 sand 1.54 0.42 1.5 is a part of sand, 5.5 is sum of ratio density of sand is for KG 0.42\n\n### How To Calculate Cement Sand And Coarse Aggregate\n\nfor cement, sand and coarse aggregate. this is a volumetric calculation. assuming we need of concrete for concrete mix, total part of the concrete 11.53 5.5 parts. therefore, cement quantity concrete volumefeb 07, 2017 the dry volume of concrete mixture is always greater than the wet volume. the ratio of dry volume to the wet volume of concrete is 1.54. So 1.54 cum of dry materials is required to produce cum of concrete. volume of cement required 1.54 1.54 0.22 cum" ]

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[ "# Number 29848781824\n\n### Properties of number 29848781824\n\nCross Sum:\nFactorization:\n2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5399 * 5399\nDivisors:\nCount of divisors:\nSum of divisors:\n59679468247\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\n6f3204400\nBase 32:\nrpi0h00\nsin(29848781824)\n-0.37465756102199\ncos(29848781824)\n-0.9271632606877\ntan(29848781824)\n0.40409017150237\nln(29848781824)\n24.119409866036\nlg(29848781824)\n10.474926611582\nsqrt(29848781824)\n172768\nSquare(29848781824)\n8.9094977637675E+20\n\n### Number Look Up\n\nLook Up\n\n29848781824 which is pronounced (twenty-nine billion eight hundred forty-eight million seven hundred eighty-one thousand eight hundred twenty-four) is a impressive figure. The cross sum of 29848781824 is 61. If you factorisate the number 29848781824 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 5399 * 5399. The figure 29848781824 has 33 divisors ( 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 5399, 10798, 21596, 43192, 86384, 172768, 345536, 691072, 1382144, 2764288, 5528576, 29149201, 58298402, 116596804, 233193608, 466387216, 932774432, 1865548864, 3731097728, 7462195456, 14924390912, 29848781824 ) whith a sum of 59679468247. The figure 29848781824 is not a prime number. The number 29848781824 is not a fibonacci number. The number 29848781824 is not a Bell Number. 29848781824 is not a Catalan Number. The convertion of 29848781824 to base 2 (Binary) is 11011110011001000000100010000000000. The convertion of 29848781824 to base 3 (Ternary) is 2212001012202020000221. The convertion of 29848781824 to base 4 (Quaternary) is 123303020010100000. The convertion of 29848781824 to base 5 (Quintal) is 442112242004244. The convertion of 29848781824 to base 8 (Octal) is 336310042000. The convertion of 29848781824 to base 16 (Hexadecimal) is 6f3204400. The convertion of 29848781824 to base 32 is rpi0h00. The sine of the number 29848781824 is -0.37465756102199. The cosine of the number 29848781824 is -0.9271632606877. The tangent of the number 29848781824 is 0.40409017150237. The root of 29848781824 is 172768.\nIf you square 29848781824 you will get the following result 8.9094977637675E+20. The natural logarithm of 29848781824 is 24.119409866036 and the decimal logarithm is 10.474926611582. I hope that you now know that 29848781824 is special number!" ]

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[ "# AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.4\n\nAP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.4 Textbook Questions and Answers.\n\n## AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.4", null, "Question 1.\nCheck whether 25110 is divisible by 45.\nSolution:\nThe given number = 25110\nIf 25110 is divisible by 45 then it should be divisible by 5 and 9.", null, "∴ The number 25110 is divisible by 45", null, "Question 2.\nCheck whether 61479 is divisible by 81.\nSolution:\nIf 61479 is divisible by 81 then it is divisible by 9.\nIf the sum of the digits of a number is dívisible by 9 then the entire number is divisible by 9.\n∴ 61479 → 6 + 1 + 4 + 7 + 9 → $$\\frac { 27 }{ 9 }$$ (R = 0)\n∴ 61479 is divisible by 81. [∵ 9 is factor of 81]\n\nQuestion 3.\nCheck whether 864 is divisible by 36? Verif,’ whether 864 is divisible by all the factors of 36 ?\nSolution:\n864 is divisible by 2 and 3.\n∴ 864 is divisible by 6.\n∴ 864 is divisible by 36 [ ∵ 6 is the factor of 36]\n∴ Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18. 36.", null, "", null, "∴ 864 is divisible by all the factor of 36.", null, "Question 4.\nCheck whether 756 is divisible by 42? Verify whether 756 is divisible by all the factors of 42?\nSolution:\n756 is divisible by 2 and 3.\n∴ 756 is divisible by 6.\n2a + 3b + c = 2 x 7 + 3 x 5 + 6 = 14 + 15 + 6 → $$\\frac { 35 }{ 7 }$$ (R = 0)\n∴ 756 is divisible by 7.\n∴ 756 is divisible by 42. [ ∵ 6, 7 are the factors of 42]\nFactors of 42 = 1, 2, 3, 6, 7, 14, 21, 42.", null, "", null, "∴ 756 is divisible by all the factor of 42.", null, "Question 5.\nCheck whether 2156 is divisible by 11 and 7? Verify whether 2156 is divisible by product of 11 and 7?\nSolution:", null, "Question 6.\nCheck whether 1435 is divisible by 5 and 7? Verify if 1435 is divisible by the product of 5 and 7?\nSolution:", null, "", null, "Question 7.\nCheck whether 456 and 618 are divisible by 6’? Also check whether 6 divides the sum of 456 and 618 ‘?\nSolution:", null, "Question 8.\nCheck whether 876 and 345 are divisible by 3. Also check whether 3 divides the difference of 876 and 345?\nSolution:\n\n Number Divisible by 3 Y/N Difference is divisible by 3 Y/N 876 8 + 7 + 6 → $$\\frac { 21 }{ 3 }$$ (R = 0) Yes 876 – 345 = 531 345 3 + 4 + 5 → $$\\frac { 12 }{ 3 }$$ (R = 0) Yes The difference of 876, 345 is divisible by 3. 531 5 + 3 + 1 → $$\\frac { 9 }{ 3 }$$ (R = 0) Yes\n\nQuestion 9.\nCheck whether 22 + 23+24 is divisible by 2 or 4 or by both 2 and 4’?\nSolution:", null, "∴ 22 + 23+24 is divisible by both 2 and 4.\n\nQuestion 10.\nCheck whether 322 is divisible by 4 or 8 or by both 4 and 8’?\nSolution:", null, "322 is divisible by 4 and 8", null, "Question 11.\nIf A679B is a 5-dit number is divisible by 72 find ‘A’ and ‘B”?\nSolution:\nIf A679B is divisible by 72 then it should be divisible by 8 and 9.\n[ ∵ 8, 9 are the factors of 72]\nA679B is divisible by 9 then\nA + 6 + 7 + 9 + B = A + B + 22 = 27 (= 9 x 3)\n=A + B = 5 ……………. (1)\nA679B → $$\\frac{79 \\mathrm{~B}}{8}$$ [From B (2,4,6,8) we take B = 2]\n= $$\\frac{792}{8}$$ (R = 0)\n∴ B = 2\nFrom (1) ⇒ A + 2 = 5\n∴ A = 3, B = 2" ]

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[ "# `if` and `if let` expressions\n\n## `if` expressions\n\nSyntax\nIfExpression :\n`if` Expressionexcept struct expression BlockExpression\n(`else` ( BlockExpression | IfExpression | IfLetExpression ) )?\n\nAn `if` expression is a conditional branch in program control. The syntax of an `if` expression is a condition operand, followed by a consequent block, any number of `else if` conditions and blocks, and an optional trailing `else` block. The condition operands must have the boolean type. If a condition operand evaluates to `true`, the consequent block is executed and any subsequent `else if` or `else` block is skipped. If a condition operand evaluates to `false`, the consequent block is skipped and any subsequent `else if` condition is evaluated. If all `if` and `else if` conditions evaluate to `false` then any `else` block is executed. An if expression evaluates to the same value as the executed block, or `()` if no block is evaluated. An `if` expression must have the same type in all situations.\n\n``````\n#![allow(unused)]\nfn main() {\nlet x = 3;\nif x == 4 {\nprintln!(\"x is four\");\n} else if x == 3 {\nprintln!(\"x is three\");\n} else {\nprintln!(\"x is something else\");\n}\n\nlet y = if 12 * 15 > 150 {\n\"Bigger\"\n} else {\n\"Smaller\"\n};\nassert_eq!(y, \"Bigger\");\n}\n``````\n\n## `if let` expressions\n\nSyntax\nIfLetExpression :\n`if` `let` Pattern `=` Scrutineeexcept lazy boolean operator expression BlockExpression\n(`else` ( BlockExpression | IfExpression | IfLetExpression ) )?\n\nAn `if let` expression is semantically similar to an `if` expression but in place of a condition operand it expects the keyword `let` followed by a pattern, an `=` and a scrutinee operand. If the value of the scrutinee matches the pattern, the corresponding block will execute. Otherwise, flow proceeds to the following `else` block if it exists. Like `if` expressions, `if let` expressions have a value determined by the block that is evaluated.\n\n``````\n#![allow(unused)]\nfn main() {\nlet dish = (\"Ham\", \"Eggs\");\n\n// this body will be skipped because the pattern is refuted\nif let (\"Bacon\", b) = dish {\nprintln!(\"Bacon is served with {}\", b);\n} else {\n// This block is evaluated instead.\nprintln!(\"No bacon will be served\");\n}\n\n// this body will execute\nif let (\"Ham\", b) = dish {\nprintln!(\"Ham is served with {}\", b);\n}\n\nif let _ = 5 {\nprintln!(\"Irrefutable patterns are always true\");\n}\n}\n``````\n\n`if` and `if let` expressions can be intermixed:\n\n``````\n#![allow(unused)]\nfn main() {\nlet x = Some(3);\nlet a = if let Some(1) = x {\n1\n} else if x == Some(2) {\n2\n} else if let Some(y) = x {\ny\n} else {\n-1\n};\nassert_eq!(a, 3);\n}\n``````\n\nAn `if let` expression is equivalent to a `match` expression as follows:\n\n``````if let PATS = EXPR {\n/* body */\n} else {\n/*else */\n}\n``````\n\nis equivalent to\n\n``````match EXPR {\nPATS => { /* body */ },\n_ => { /* else */ }, // () if there is no else\n}\n``````\n\nMultiple patterns may be specified with the `|` operator. This has the same semantics as with `|` in `match` expressions:\n\n``````\n#![allow(unused)]\nfn main() {\nenum E {\nX(u8),\nY(u8),\nZ(u8),\n}\nlet v = E::Y(12);\nif let E::X(n) | E::Y(n) = v {\nassert_eq!(n, 12);\n}\n}\n``````\n\nThe expression cannot be a lazy boolean operator expression. Use of a lazy boolean operator is ambiguous with a planned feature change of the language (the implementation of if-let chains - see eRFC 2947). When lazy boolean operator expression is desired, this can be achieved by using parenthesis as below:\n\n``````// Before...\nif let PAT = EXPR && EXPR { .. }\n\n// After...\nif let PAT = ( EXPR && EXPR ) { .. }\n\n// Before...\nif let PAT = EXPR || EXPR { .. }\n\n// After...\nif let PAT = ( EXPR || EXPR ) { .. }\n``````" ]

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[ "Amount\nFrom\nTo\n\n# 1 kilograms to stone\n\nHow many stone in 1 kilograms? 1 kilograms is equal to 0.1574730444 stone.\n\nThis page provides you how to convert between kilograms and stone with conversion factor.\n\n# How to convert 1 kg to st?\n\nTo convert 1 kg into st, follow these steps:\n\nWe know that, 1 st = 6.35029318 kg\n\nHence, to convert the value of 1 kilograms into stone, divide the weight value 1kg by 6.35029318.\n\n1 kg = 1/6.35029318 = 0.1574730444 st\n\n### Thus, 1 kg equals to 0.1574730444 st\n\nKilograms Conversion of Kilograms to Stone\n0.99 kg 0.99 kg = 0.155898314 st\n0.9 kg 0.9 kg = 0.14172574 st\n1 kg 1 kg = 0.1574730444 st\n2 kg 2 kg = 0.3149460888 st" ]

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[ "# Using Fibonacci retracements in forex trading?", null, "In forex trading, Fibonacci retracements can be used to identify possible support and resistance levels. Using Fibonacci ratios, you can get an idea of where the price will likely head next.\n\nThis article will explain how to use Fibonacci retracements in your trading strategy. We will also provide examples of how this tool can be used to generate profits.\n\n## What are Fibonacci retracements, and how do they work?\n\nIn forex trading, Fibonacci retracements identify possible support and resistance levels. The idea behind this tool is that prices tend to move in cycles. Using Fibonacci ratios, you can get an idea of where the price will likely head next.\n\nThe sequence is a set of numbers that starts with 0 and 1. The next number in the sequence is the sum of the previous two numbers. So, the sequence goes: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.\n\nAs you see, each number in the sequence is roughly 1.618 times greater than the number before. This ratio is known as the “golden ratio” and is found throughout nature. The Fibonacci sequence can be applied to forex trading to identify potential support and resistance levels.\n\nWhen applying Fibonacci retracements to a price chart, you will notice that the prices tend to bounce off the Fibonacci levels. These levels represent potential areas where the price could find support or resistance. By identifying these levels, you can make better-informed trading decisions.\n\n## The different types of Fibonacci retracements\n\nThere are three different types of Fibonacci retracements:\n\nThe first type is known as a “standard” Fibonacci retracement. To apply this tool, you must find the most recent highs and lows on your price chart. Once you have found these two points, you will draw a horizontal line from the high to the low. This line will be used to identify the Fibonacci levels. The standard Fibonacci levels are 23.6%, 38.2%, 50%, 61.8%, and 100%.\n\nThe second type of Fibonacci retracement is known as an “extended” Fibonacci retracement. To apply this tool, you must find the most recent highs and lows on your price chart. Once you have found these two points, you will draw a horizontal line from the high to the low. This line will be used to identify the Fibonacci levels. The extended Fibonacci levels are 127.2%, 161.8%, and 200%.\n\nThe third type of Fibonacci retracement is known as a “custom” Fibonacci retracement. To apply this tool, you must find the most recent highs and lows on your price chart. Once you have found these two points, you will draw a horizontal line from the high to the low. This line will be used to identify the Fibonacci levels. The custom Fibonacci levels are based on your personal preferences. You can use a forex broker to assist you with this.\n\nYou can use Fibonacci retracements in your trading strategy in a few different ways. One way is to use them to identify possible entry and exit points, and another is to set stop-loss and take-profit orders.\n\nIf you use Fibonacci retracements to identify possible entry and exit points, you will want to look for price action confirming your levels. For example, if the price is retracing back to a Fibonacci level and starts to bounce off, this could be a good sign that the level is working as support or resistance. You can then enter or exit your trade based on this price action.\n\nIf you are using Fibonacci retracements to set stop-loss and take-profit orders, you will want to place your orders around the Fibonacci levels, which will help you protect your profits and limit your losses.\n\nWhen using Fibonacci retracements, it is essential to remember that they are not an exact science, and prices can sometimes move through Fibonacci levels without bouncing off of them. However, by using Fibonacci retracements in conjunction with other technical indicators, you can increase the accuracy of your trading decisions.\n\n## Examples of how Fibonacci retracements can be used to predict price movements\n\nIn the following examples, we’ll see how Fibonacci retracements can predict price movements.\n\n### Example 1:\n\nThe price of XYZ stock is trading at \\$50 per share. The stock then starts to decline and reaches a low of \\$40 per share. From here, the stock begins to rebound and starts to move higher. As the stock moves higher, it returns to the 23.6% Fibonacci level. This level is created by drawing a line from the high of \\$50 to the low of \\$40. The 23.6% Fibonacci level is located at \\$46.40.\n\n### Example 2:\n\nThe price of ABC stock is trading at \\$100 per share. The stock then starts to decline and reaches a low of \\$80 per share. From here, the stock begins to rebound and starts to move higher. As the stock moves higher, it returns to the 38.2% Fibonacci level. This level is created by drawing a line from the high of \\$100 to the low of \\$80. The 38.2% Fibonacci level is located at \\$92.00.", null, "" ]

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[ "# Tagged: Tough number Riddl\n\n## Whatsapp Mathematical Puzzle: If 5+3+2=151022, 7+2+5=?\n\nHere is the  puzzles for Mathematical knowledge.try to solve this Whatsapp Mathematical Puzzle: If 5+3+2=151022, 7+2+5=? please share with your friends. please like and comment in website.   Whatsapp Mathematical Puzzle: If 5+3+2=151022, 7+2+5=?   solve...\n\n## whatsapp puzzles-Find The Dates on Which The Man Spent His Money\n\nHere is the logical and   mathematical whatsapp puzzles.so challenge your brain.and solve this puzzles. please solve this puzzles and share with your friends. if you like this puzzle so please comment in website. whatsapp...\n\n## New whatsapp puzzles\n\nHere is the New whatsapp puzzles:Ek Mithai Ka Naam, Dawa Ka Naam, Film Ka Naam, Ladki Ka Naam, City Ka Naam, Car Ka Naam, Place Ka Naam Aur Doggy Ka Naam in 8 sawalo ka...\n\n## whatsapp Math Riddle: How Will You Get 100 Fruits in Rs.100?\n\nHere is the whatsapp Math Riddle: How Will You Get 100 Fruits in Rs.100? this riddles is fruit  related. let’s try to solve this puzzles and share with your friends. please like and comment. whatsapp...\n\n## Why Was The Truck Driver Not Stopped By The Cops?\n\nWhy Was The Truck Driver Not Stopped By The Cops? Here we have added latest Answer to difficult riddles, Answer to IAS Exam Puzzle, answer to interview questions, Answer to whatsapp puzzles, answers to..." ]

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[ "", null, "", null, "John Zhu\n\nMean Value Theorem\n\nSlide Duration:\n\nSection 1: Functions\nDefinitions & Properties of Functions\n\n11m 26s\n\nIntro\n0:00\nDefinition\n0:28\nProperties: Vertical Line Test\n1:32\nDomain\n1:38\nRange\n1:59\nVertical Line test\n2:19\nExample 1\n2:33\nExample 2\n3:10\nProperties: Roots or Zeros\n4:04\nFinding the Root\n4:16\nProperties: Forms\n5:12\nGraphically\n5:20\nList\n5:46\nEquation\n6:11\nFunction\n6:38\nProperties: Odd & Even\n7:12\nEven Function\n7:14\nOdd Function\n8:25\nProperties: Increasing & Decreasing\n9:17\nIncreasing Function\n9:22\nDecreasing Function\n10:21\nGraphing\n\n13m 58s\n\nIntro\n0:00\nManipulating\n0:10\nA in the Equation\n0:39\nB in the Equation\n0:44\nC & D in the Equation\n0:49\nNegative values\n0:59\nExample 1\n1:17\nExample 2\n1:51\nExample 3: Absolute Value Functions\n3:43\nExample 4\n4:57\nExample 5\n6:17\nExample 6\n8:02\nExample 7\n9:10\nExample 8\n11:02\nExample 9\n11:47\nInverse Functions\n\n6m 47s\n\nIntro\n0:00\nInverse\n0:08\nDefinition\n0:18\nExample: Finding the Inverse\n1:03\nExample 2\n2:29\nExample 3\n3:12\nExample 4\n4:41\nPolynomial Functions\n\n5m 4s\n\nIntro\n0:00\nTypes of Functions: Polynomials\n0:07\nNo Domain Restrictions\n0:12\nNo Discontinuities\n0:19\nDegree Test\n0:31\nTypes of Functions: Polynomials\n1:17\n1:33\n1:54\n2:13\n2:34\n2:46\nExamples: Types of Functions: Polynomials\n3:03\nExamples: Types of Functions: Polynomials\n4:18\nTrigonometric Functions\n\n6m 45s\n\nIntro\n0:00\nTypes of Functions: Trigonometric\n0:05\n6 Functions To Be Familiar With\n0:14\nExample 1: SIN\n1:38\nExample 2: COS\n3:22\nExample 3: TAN\n4:38\nInverse Trigonometric Functions\n\n5m 58s\n\nIntro\n0:00\nTypes of Functions: Trigonometric- Inverse Trig Functions\n0:07\nExample: Inverse SIN of X\n0:45\nExample: Inverse Function\n2:30\nExample: Inverse TAN of X\n4:42\nTrigonometric Identities\n\n17m 42s\n\nIntro\n0:00\nTypes of Functions: Trigonometric- Trig Identities\n0:07\n4 Identities\n0:24\nPythagorean\n0:28\nDouble Angle\n1:10\nPower Reducing\n1:28\nSum or Difference\n1:42\nCouple More Identities\n1:59\nNegative Angle\n2:04\nProduct to Sum\n2:39\nExample 1: Prove\n3:00\nExample 2: Simplify Expression\n5:02\nExample 3: Prove\n5:56\nExample 4: Prove\n8:02\nExample 5: Prove With TAN\n12:43\nExponential Functions\n\n5m 53s\n\nIntro\n0:00\nTypes of Functions: Exponentials\n0:07\nGeneral Form\n0:10\nSpecial Exponential Function\n0:17\nExample 1: Using Exponential Properties\n0:46\nExample 2: Using Exponential Properties\n1:58\nExample 3: Using Trig Identities & Exponential Properties\n3:16\nExample 4: Using Exponential Properties\n4:37\nLogarithmic Functions\n\n7m 8s\n\nIntro\n0:00\nTypes of Functions: Logarithmic\n0:06\nGeneral Form\n0:10\n2 Special Logarithmic Func.\n0:19\nEuler's # / Natural Log\n0:27\nLogarithmic & Exponential Relationship\n0:45\nLog form\n1:56\nProperties\n2:09\nExample 1: Apply Basic Principle of Log Func.\n3:05\nExample 2: Use Properties\n3:40\nExample 3: Regular Log\n5:16\nRational Functions\n\n15m 36s\n\nIntro\n0:00\nTypes of Functions: Rational - Definition\n0:06\nExample 1: Graph Rational Func.\n0:36\nExample 2: Find Asymptotes of Func.\n7:02\nExample 3: Find Asymptotes of Func.\n8:59\nExample 4: Graph Rational Func.\n11:08\nConic Sections\n\n14m 58s\n\nIntro\n0:00\nTypes of Conic Sections\n0:06\nParabolas\n0:19\nCircles\n1:36\nEllipses\n2:40\nHyperbolas\n4:42\nComplete the Square\n6:40\nExample: Conic Sections\n9:08\nExample 2: Conic Sections\n10:59\nExample 3: Graph Conic Sections\n12:21\nSection 2: Limits and Continuity\nLimit Definition & Properties\n\n7m 15s\n\nIntro\n0:00\nDefinition\n0:06\nExample: Limit\n0:17\nProperties\n1:13\n1st Property\n1:21\n2nd Property\n1:34\nSpecial Property\n1:51\nLimits\n2:36\nExplain Example\n2:49\nLimits Example\n4:39\nLimits Example\n5:21\nSolving Limits with Algebra\n\n8m 1s\n\nIntro\n0:00\nSolving Limits with Algebra\n0:07\nExample 1: Solve Algebraically\n0:30\nSolving Limits with Algebra, Example 2\n2:28\nSolving Limits with Algebra, Example 3\n3:18\nSolving Limits with Algebra, Example 4\n4:56\nSolving Limits with Algebra, Example 5\n6:26\nRational Limit Rules\n\n3m 16s\n\nIntro\n0:00\nRational Limit Rules\n0:07\nReview of Solving Problem Algebraically\n0:08\nLimit Rules\n0:28\nRule 1\n0:35\nRule 2\n0:40\nRule 3\n0:45\nRational Limit Rules\n1:02\nApplying 1st Rule\n1:22\nRational Limit Rules\n1:50\nApplying 2nd Rule\n2:09\nRational Limit Rules\n2:26\nApplying 3rd Rule\n2:40\nOne Sided Limits\n\n9m 57s\n\nIntro\n0:00\nTypes of Limits: One-Sided Limit Rules\n0:06\nExample\n0:19\nApplying Same Rule\n0:34\nRule to Keep In Mind\n0:52\nTypes of Limits: One-Sided Limit Example 1\n1:12\nLimit of x² From Negative Side\n2:11\nTypes of Limits: One-Sided Limit, Example 2\n2:27\nTypes of Limits: One-Sided Limit, Example 3\n4:26\nTypes of Limits: One-Sided Limit, Example 4\n5:47\nOne-Sided Limit Example: X with Even Degree Polynomial\n7:00\nOne-Sided Limit Example: Entire Denominator Squared\n8:09\nSpecial Trigonometric Limits\n\n8m 28s\n\nIntro\n0:00\nTypes of Limits: Special Trig Limits\n0:07\nPre-set Rules\n0:35\nSpecial Trig Limits, Example 1\n0:58\nSpecial Trig Limits, Example 2\n2:50\nSpecial Trig Limits, Example 3\n3:55\nSpecial Trig Limits, Example 4: With More Degrees\n4:57\nSpecial Trig Limits, Example 5\n6:21\nLimits & Continuity\n\n10m 14s\n\nIntro\n0:00\nDefinition\n0:06\n3 Rules: f(x) Is Continuous…\n0:21\nExample 1: Finding Continuity\n1:06\nTypes of Discontinuity\n2:44\nJump\n2:52\nPoint\n3:24\nEssential (Asymptote)\n3:47\nRemovable\n4:17\nExample 2: Continuity Examples\n4:41\nExample 3: Continuity Examples\n6:13\nExample 4: Locate & Identify Type of Discontinuities\n8:00\nLimits: Multiple Choice Practice\n\n6m 16s\n\nIntro\n0:00\nProblem 1\n0:08\nProblem 2\n1:51\nProblem 3\n2:54\nProblem 4\n4:31\nSection 3: Derivatives\nDerivative Definition & Properties\n\n4m 11s\n\nIntro\n0:00\nDefinition\n0:09\nFormal Definition\n0:45\nDifference Quotient\n1:12\nBasic Derivatives\n1:16\nDifferentiability\n2:54\nBasic Rules of Differentiation\n\n7m 7s\n\nIntro\n0:00\nBasic Rules of Differentiation\n0:09\nConstant Rule\n0:14\nConstant Multiple Rule\n1:10\n1:40\nExample 1: Constant Rule\n2:25\nExample 2: Constant Multiple Rule\n3:01\nExample 3: Constant Multiple Rule\n3:35\nExample 4: Constant Rule\n4:34\nExample 5: Constant Multiple Rule\n5:03\nExample 6\n5:33\nPower Rule\n\n7m 14s\n\nIntro\n0:00\nPower Rule\n0:07\nPower Rule Definition\n0:30\nExample 1\n1:11\nExample 2\n2:25\nExample 3\n3:05\nExample 4\n4:18\nExample 5\n5:13\nTrigonometric Rules\n\n7m 53s\n\nIntro\n0:00\nTrigonometric Rules\n0:07\nCOS X\n0:38\nFind Derivative\n1:02\nExample 1\n2:46\nExample 2: COS Function\n3:09\nExample 3: Composite Expression\n3:54\nExample 4: Sec Function\n5:02\nExample 5: CSC\n5:33\nExample 6L COT\n6:42\nProduct Rule\n\n11m 11s\n\nIntro\n0:00\nProduct Rule\n0:07\nDefinition\n0:20\nExample 1\n0:43\nExample 2\n2:11\nExample 3\n4:24\nExample 4\n5:24\nExample 5\n6:42\nExample 6\n7:51\nQuotient Rule\n\n16m 50s\n\nIntro\n0:00\nQuotient Rule\n0:07\nDefinition\n0:30\nExample 1\n1:17\nExample 2: With No X In Numerator\n2:49\nExample 3\n4:30\nExample 4: With Decimals\n6:46\nExample 5\n8:53\nExample 6: With Trig Functions\n12:55\nChain Rule\n\n19m 48s\n\nIntro\n0:00\nChain Rule\n0:07\nDefinition\n0:17\nExample 1: Applying the Chain Rule\n1:33\nExample 2\n4:25\nExample 3\n6:02\nExample 4\n9:25\nExample 5\n12:47\nExample 6\n15:27\nHigher Order Derivatives\n\n15m\n\nIntro\n0:00\nTypes of Derivatives: Higher Order Derivatives\n0:07\n1st Derivative / F Prime\n0:19\n2nd Derivative\n0:25\n3rd Derivative\n0:32\nExample 1\n1:48\nExample 2: Find 3rd Derivative\n3:13\nExample 3: Acceleration\n4:25\nExample 4\n10:20\nExample 5: 2nd Derivative\n12:11\nDerivatives of Exponential Functions\n\n13m 14s\n\nIntro\n0:00\nTypes of Derivatives: Exponential Functions\n0:08\nDerivatives: Definition/ Formula\n0:28\nExample 1\n1:25\nExample 2\n2:47\nExample 3\n4:13\nExample 4\n7:11\nExample 5\n9:23\nExample 6\n11:06\nDerivatives of Logarithmic Functions\n\n11m 30s\n\nIntro\n0:00\nTypes of Derivatives: Logarithmic Functions\n0:06\nRule for Logarithmic Functions\n0:28\nExample 1\n0:58\nExample 2\n3:10\nExample 3\n4:38\nExample 4\n7:18\nExample 5\n8:48\nExample 6\n9:38\nDerivatives of Inverse Trigonometric Functions\n\n16m 54s\n\nIntro\n0:00\nTypes of Derivatives: Inverse Trigonometric Functions\n0:06\n6 Fundamental Properties of Inverse Trigonometric Functions\n0:38\nExample 1\n2:17\nExample 2\n3:41\nExample 3\n5:37\nExample 4\n7:24\nExample 5\n10:08\nImplicit Differentiation\n\n16m 53s\n\nIntro\n0:00\nImplicit Differentiation: First Order\n0:07\nExample 1: Setting Up\n0:45\nExample 1: Solving\n1:41\nImplicit Differentiation: Second Order (Ex. 2)\n4:55\nExample 3: Implicit Differentiation\n9:11\nExample 4: Implicit Differentiation\n9:56\nExample 5: Implicit Differentiation With Double Derivative\n12:46\nMultiple Choice Practice: Derivatives\n\n11m 7s\n\nIntro\n0:00\nPractice Problem 1\n0:09\n3:24\nPractice Problem 2\n3:36\n6:29\nPractice Problem 3\n6:42\n8:39\nPractice Problem 4\n8:43\n9:33\nPractice Problem 5\n9:41\n10:40\nSection 4: Applications of Derivatives\nTangent & Normal Lines\n\n22m 36s\n\nIntro\n0:00\nTangent and Normal Lines\n0:10\nDefinition\n0:22\nExample 1\n0:55\nTangent and Normal Lines: Example 2\n2:43\nTangent and Normal Lines\n5:21\nExample 3\n5:35\nTangent and Normal Lines: Example 4\n9:14\nTangent and Normal Lines: Example 5\n12:27\nTangent and Normal Lines: Example 6\n15:54\nTangent and Normal Lines: Example 7\n19:05\nPosition Velocity & Acceleration\n\n18m 42s\n\nIntro\n0:00\nPosition, Velocity, and Acceleration\n0:10\nPosition Function\n0:14\nVelocity Function\n0:34\nAcceleration Function\n1:01\nExample 1\n1:20\nExample 2\n6:31\nExample Continue: Velocity When Acceleration is Zero\n6:32\nExample 3: Where Is Particle Changing Directions?\n8:16\nExample 4: Total Distance Traveled From 0 to 2 Second\n11:09\nExample 5: Ball Drop Problem\n16:40\nRelated Rates\n\n26m 22s\n\nIntro\n0:00\nRelated Rates\n0:06\nFinding Rate of Change: Organization & Big Picture\n0:23\nExample 2: Area of a Circle\n1:17\nExample 3: Spherical Volume Expanding\n4:19\nExample 4: Traveling Problem\n7:57\nExample 5: Square Increase\n12:37\nExample 6: Standard Related Rates Problem\n16:59\nExample 7: Standard Related Rates Problem\n19:49\nMinimum & Maximum\n\n12m 22s\n\nIntro\n0:00\nExtrema: First Derivative Test\n0:09\nExample 1\n0:46\nExample 2: Real World Application/ Cost Function\n4:05\nExample 3: Minimums & Maximums\n7:10\nExample 4: Find Critical Points\n10:52\nConcavity\n\n11m 43s\n\nIntro\n0:00\nConcavity: Second Derivative Test\n0:06\nDefinition\n0:34\nExample 1\n0:54\nExample 2\n2:51\nExample 3\n4:08\nExample 4\n5:52\nRolles Theorem\n\n8m 28s\n\nIntro\n0:00\nRolle's Theorem\n0:07\nConditions\n0:11\nSummary\n0:41\nExample 1\n1:09\nExample 2\n3:08\nExample 3\n4:48\nMean Value Theorem\n\n9m 39s\n\nIntro\n0:00\nMean Value Theorem\n0:06\nRolle's Theorem\n0:07\nMean Value Theorem Conditions\n0:24\nMean Value Theorem Definition\n0:36\nExample 1\n0:56\nExample 2\n2:44\nExample 3\n5:28\nExample 4\n7:15\nDifferentials\n\n12m 25s\n\nIntro\n0:00\nDifferentials\n0:08\n1st Differential Formula\n0:29\n2nd Differential Formula\n0:57\nExample 1\n1:06\nExample 2\n3:21\nExample 3\n5:49\nExample 4\n7:19\nExample 5\n9:06\nApplications of Derivatives: Multiple Choice Practice\n\n13m 21s\n\nIntro\n0:00\nPractice Problem 1\n0:10\n1:57\nPractice Problem 2\n2:08\n5:39\nPractice Problem 3\n5:45\n9:59\nPractice Problem 4\n10:12\n11:49\nPractice Problem 5\n11:52\n13:00\nApplications of Derivatives: Free Response Practice\n\n10m 22s\n\nIntro\n0:00\nPractice Problem 1\n0:10\nSlope\n1:30\nTangent Line Equation\n2:17\nAbsolute Minimum\n2:24\n2 Possible X Points With Minimums\n3:15\nOne Interest Point\n4:14\nConcavity\n4:33\nPositive Value = Positive Concavity\n4:10\nMinimum Point\n5:34\nAbsolute Minimum\n6:18\nPoint(s) of Inflection\n6:31\nDefinition\n6:49\n2 Points Of Inflection\n9:59\nSection 5: Integrals\nDefinition of Integrals\n\n1m 8s\n\nIntro\n0:00\nDefinition\n0:09\nDefinition\n0:16\nExample\n0:20\nIntegrals of Power Rule\n\n8m 50s\n\nIntro\n0:00\nPower Rule\n0:06\nExample 1\n0:25\nExample 2\n2:02\nExample 3\n2:54\nExample 4\n3:45\nExample 5\n4:49\nExample 6\n6:47\nIntegrals Basic Rules of Integration\n\n9m 43s\n\nIntro\n0:00\nBasic Rules of Integration\n0:09\nConstant Rule\n0:22\nExample 1\n0:40\n1:40\nExample 2\n1:58\nExample 3: Subtraction/ Difference Rule\n2:47\nExample 4\n3:55\nExample 5\n5:19\nExample 6\n7:37\nTrigonometric Rules of Integrals\n\n8m 58s\n\nIntro\n0:00\nTrigonometric Rules\n0:09\nIntegral of SIN\n0:38\nExample 1: Integral of SIN\n1:46\nExample 2: Integral of COS\n2:38\nExample 3: With 2 terms of X\n3:06\nExample 4: Integral of SEC\n4:15\nExample 5: Integral of CSC\n5:06\nExample 6\n6:18\nChain Rule\n\n13m 59s\n\nIntro\n0:00\nChain Rule\n0:07\nExample 1\n0:37\nExample 2\n3:17\nExample 3\n5:09\nExample 4\n7:53\nExample 5\n9:40\nExample 6\n11:39\nIntegrals of Exponential Functions\n\n12m 52s\n\nIntro\n0:00\nTypes of Integrals: Exponential Functions\n0:09\nRule 1\n0:30\nRule 2\n0:49\nExample 1\n1:11\nExample 2\n2:54\nExample 3\n4:19\nExample 4\n5:19\nExample 5\n7:37\nExample 6\n9:04\nIntegrals of Natural Logarithmic Functions\n\n13m\n\nIntro\n0:00\nTypes of Integrals: Natural Log Functions\n0:09\nExample 1\n0:49\nExample 2\n2:06\nExample 3\n4:01\nExample 4\n5:37\nExample 5\n7:30\nExample 6\n9:05\nIntegrals of Inverse Trigonometric Functions\n\n8m 29s\n\nIntro\n0:00\nTypes of Integrals: Inverse Trig Functions\n0:09\nOne Property\n0:40\nExample 1\n1:19\nExample 2\n3:44\nExample 3\n4:53\nExample 4\n5:53\nIntegrals: Multiple Choice Practice\n\n15m 37s\n\nIntro\n0:00\nProblem 1\n0:09\n4:09\nProblem 2\n4:33\n5:54\nProblem 3\n5:59\n8:02\nProblem 4\n8:06\n10:27\nProblem 5\n10:43\n14:46\nSection 6: Applications of Integrals\nFundamental Theorem of Calculus\n\n15m 55s\n\nIntro\n0:00\nFundamental Theorem of Calculus: Properties\n0:10\nDefinition of Integral\n0:49\nExample 1\n1:14\nFundamental Theorem of Calculus: Properties\n2:40\nRule 1\n2:50\nRule 2\n3:14\nRule 3\n3:33\nRule 4\n3:52\nExample 2\n4:07\nExample 3\n6:17\nExample 4\n9:31\nExample 5\n10:52\nExample 6\n13:34\nArea Under A Curve\n\n18m 34s\n\nIntro\n0:00\nArea Under Curve\n0:07\nDefinition of Integral\n0:09\nLeft Endpoint\n1:17\nRight Endpoint\n1:47\nMidpoints\n2:09\nExample 1\n2:40\nExample 2\n4:59\nExample 3\n8:48\nExample 4\n10:23\nExample 5\n12:30\nExample 6\n15:32\nReimann Sums\n\n10m 35s\n\nIntro\n0:00\nReimann Sums\n0:08\nDefinition\n1:07\nExample 1\n2:48\nExample 2\n5:38\nExample 3\n7:21\nExample 4\n9:14\nTrapezoid Rule\n\n12m 46s\n\nIntro\n0:00\nThe Trapezoid Rule\n0:09\nDefinition: Area Of A Trapezoid\n0:26\nTerms of Formula\n1:35\nExample 1\n2:11\nExample 2\n4:29\nExample 3\n7:22\nExample 4\n10:01\nMean Value Theorem\n\n11m 22s\n\nIntro\n0:00\nMean Value Theorem of Integration\n0:06\nExample 1\n0:53\nExample 2\n2:29\nExample 3\n3:48\nExample 4\n6:02\nSecond Fundamental Theorem of Calculus\n\n4m 44s\n\nIntro\n0:00\nSecond Fundamental Theorem of Calculus\n0:07\nDefinition\n0:39\nExample 1\n1:08\nExample 2\n2:07\nExample 3\n2:48\nExample 4\n3:23\nArea Between Curves\n\n16m 39s\n\nIntro\n0:00\nExample 1\n0:10\nExample 2\n3:00\nExample 3\n4:46\nExample 4\n8:22\nExample 5\n11:04\nExample 6\n13:09\nRevolving Solids Washer Disk Methods\n\n21m 9s\n\nIntro\n0:00\nRevolving Solids Washer Disk Methods\n0:11\nExplanation\n0:33\nFormula\n3:12\nExample 1\n3:42\nExample 2\n6:54\nExample 3\n9:29\nExample 4\n12:16\nExample 5\n15:35\nRevolving Solids Cylindrical Shells Method\n\n26m 46s\n\nIntro\n0:00\nRevolving Solids: Cylindrical Shells Method\n0:09\nVolume Of A Solid\n0:25\nFormula\n2:51\nExample 1\n2:56\nExample 2\n7:28\nExample 3\n11:39\nExample 4\n17:36\nExample 5\n21:45\nRevolving Solids Known Cross Sections\n\n27m 41s\n\nIntro\n0:00\nRevolving Solids Known Cross Sections\n0:08\nExample 1\n0:35\nExample 2\n6:01\nExample 3\n11:03\nExample 4\n17:29\nExample 5\n22:19\nDifferential Equations Eulers Method\n\n17m 54s\n\nIntro\n0:00\nDifferential Equations\n0:08\nExample 1\n0:30\nDifferential Equations: Euler's Method\n2:33\nRules\n2:39\nExample 2\n3:00\nExample 3\n5:42\nExample 4\n9:44\nExample 5\n14:14\nDifferential Equations Slope Fields\n\n16m 30s\n\nIntro\n0:00\nSlope Fields\n0:08\nWhat Are Slope Fields\n0:21\nExample 1\n0:42\nExample 2\n6:30\nExample 3\n11:17\nApplication of Integrals: Multiple Choice Practice\n\n14m 19s\n\nIntro\n0:00\nPractice Problem 1\n0:10\n3:46\nPractice Problem 2\n3:49\n6:20\nPractice Problem 3\n6:26\n8:02\nPractice Problem 4\n8:07\n10:58\nPractice Problem 5\n11:05\n14:06\nApplication of Integrals: Free Response Practice\n\n9m 14s\n\nIntro\n0:00\nProblem 1\n0:10\nPart A\n0:24\nPart A: Solution\n2:04\nPart B\n2:10\nProblem 1, Part B Continue\n2:23\nPart B: Solution\n6:15\nProblem 1, Part C\n6:58\nPart C: Solution\n12:40\nProblem 2\n12:52\nPart A\n13:02\nPart A: Solution\n15:34\nPart B\n16:03\nPart B: Solution\n18:48\nSection 7: Sample AP Test\nAP Calculus AB Practice test: Section 1: Multiple Choice Part 1\n\n17m 50s\n\nIntro\n0:00\nProblem 1\n0:20\nProblem 2\n1:24\nProblem 3\n2:53\nProblem 4\n3:56\nProblem 5\n8:18\nProblem 6\n9:06\nProblem 7\n10:14\nProblem 8\n12:16\nProblem 9\n14:13\nAP Calculus AB Practice test: Section 1: Multiple Choice Part 2\n\n17m 32s\n\nIntro\n0:00\nProblem 10\n0:18\nProblem 11\n2:26\nProblem 12\n6:11\nProblem 13\n7:04\nProblem 14\n8:06\nProblem 15\n10:32\nProblem 16\n11:40\nProblem 17\n13:00\nProblem 18\n14:43\nAP Calculus AB Practice test: Section 1: Multiple Choice Part 3\n\n22m 14s\n\nIntro\n0:00\nProblem 19\n0:21\nProblem 20\n2:33\nProblem 21\n7:23\nProblem 22\n10:24\nProblem 23\n12:18\nProblem 24\n13:13\nProblem 25\n15:52\nProblem 26\n17:03\nProblem 27\n19:44\nAP Calculus AB Practice test: Section 1: Multiple Choice Part 4\n\n19m 35s\n\nIntro\n0:00\nProblem 28\n0:23\nProblem 29\n3:50\nProblem 30\n5:31\nProblem 31\n9:02\nProblem 32\n10:07\nProblem 33\n11:27\nProblem 34\n13:47\nProblem 35\n15:21\nProblem 36\n16:53\nAP Calculus AB Practice test: Section 1: Multiple Choice Part 5\n\n25m 43s\n\nIntro\n0:00\nProblem 37\n0:22\nProblem 38\n2:27\nProblem 39\n5:36\nProblem 40\n7:21\nProblem 41\n10:08\nProblem 42\n11:29\nProblem 43\n13:07\nProblem 44\n18:18\nProblem 45\n21:08\nAP Calculus AB Practice Test: Section 2: Free Response Part 1\n\n16m 50s\n\nIntro\n0:00\nProblem 1, Part A\n0:20\nProblem 1, Part B\n3:03\nProblem 1, Part C\n4:11\nProblem 1, Part D\n5:36\nProblem 2, Part A\n7:37\nProblem 2, Part B\n9:02\nProblem 2, Part C\n12:31\nAP Calculus AB Practice Test: Section 2: Free Response Part 2\n\n21m 36s\n\nIntro\n0:00\nProblem 3, Part A\n0:18\nProblem 3, Part B\n5:57\nProblem 4, Part A\n11:26\nProblem 4, Part B\n12:28\nProblem 4, Part C\n15:35\nProblem 4, Part D\n18:56\nAP Calculus AB Practice Test: Section 2: Free Response Part 3\n\n13m 39s\n\nIntro\n0:00\nProblem 5, Part A\n0:21\nProblem 5, Part B\n3:07\nProblem 5, Part C\n6:43\nProblem 6\n8:24\nBookmark & Share Embed\n\n## Copy & Paste this embed code into your website’s HTML\n\nPlease ensure that your website editor is in text mode when you paste the code.\n(In Wordpress, the mode button is on the top right corner.)\n×\n• - Allow users to view the embedded video in full-size.\nSince this lesson is not free, only the preview will appear on your website.\n\n• ## Related Books", null, "0 answersPost by Angela Zhang on December 7, 2014John, you are very cute! I like you <3", null, "0 answersPost by Yu Han on July 24, 2014For example 1, how can you tell if it met these two conditions?", null, "0 answersPost by Justin Powell on February 10, 2014Just for clarity:The Mean Value Theorem says that if a function f(x) is continuousand differentiable between two intervals x=a and x=b, then solving the function for these two values will give the coordinates {a,f(a)} and {b,f(b)}.Now if you draw a line between these two points, the slope will be:(f(b)-f(a))/(b-a) which is the rise over the run.The Mean Value Theoem states that at some point, {c,f(c)},on the graph of the function between the intervals x=a an x=b, the derivative or slope must be equal to (f(b)-f(a))/(b-a) at least once.So the derivative of f(x) at x=c is equal to:f'(c)=(f(b)-f(a))/(b-a)In the last example:the slope between the coordinates {-1,-1} and {5,125} equals 126/6 = 21=f'(c)=3c^2 c=sqrt(7).At f(sqrt7) the slope of the curve of the graph is parallel to the line between {-1,-1} and {5,125}.also for the MVT, f'(c)=0 is only true when f(a) and f(b) are equal (Rolles theorem).Sorry for overclarifying, but the video left out some key points.", null, "1 answerLast reply by: Andrew MuMon Jan 6, 2014 8:30 AMPost by Willie Wang on January 13, 2013AT Example 4, why c cannot be negative sqr(7)？", null, "1 answerLast reply by: Johnny ZamoraFri Jan 10, 2014 2:58 AMPost by James Xie on December 17, 2012What exactly does c stand for? Is it the Instant Rate Of Change (IROC?", null, "0 answersPost by Ryan Menezes on November 18, 2012That's exactly what i want to know......", null, "1 answerLast reply by: Siyun LiuSat Jan 24, 2015 6:57 PMPost by Steve Denton on October 18, 2012At 5:00, what happened to the 1/2 in front of (c-1)^-1/2? rewritten then 1/ (2 sq rt (c-1))?????\n\n### Mean Value Theorem\n\n• There exists a point c, between aand b where the slope is defined by change in f(x)divided by change in x.\n• Applying Mean Value Theorem\n• Check conditions\n• Take derivative\n• Set derivative function =", null, "• Replace xwith c\n• Solve for c value\n\n### Mean Value Theorem\n\nFind the value of c that satisfies the the Mean Value Theorem conditions for f(x) = ex on the interval [0, ln2]\n• f′(x) = ex\n• Continuous and differentiable everywhere\n• f′(c) = [(f(ln2) − f(0))/(ln2 − 0)]\n• f′(c) = [(eln2 − e0)/ln2]\n• f′(c) = [(2 − 1)/ln2] = [1/ln2]\n• Rolle's Theorem conditions are not met (f(a) ≠ f(b)), but the Mean Value conditions are met.\n• f′(c) = ec = [1/ln2]\n• lnec = ln[1/ln2]\nc = ln[1/ln2]\nFind the value of c that satisfies the the Mean Value Theorem conditions for f(x) = sinx on the interval [0, 2π]\n• f′(x) = cosx\n• Continuous and differentiable everywhere\n• f′(c) = [(cos2π− cos0)/(2π− 0)]\n• f′(c) = [(1 − 1)/(2π)]\n• f′(c) = 0\n• cosc = 0\n• There are many solutions, but only two in the interval\nc = [(π)/2], [(3π)/2]\nFind the value of c that satisfies the the Mean Value Theorem conditions for f(x) = x3 − x2 + 1 on the interval [0, 1]\n• f′(x) = 3x2 − 2x\n• f′(c) = [(13 − 12 + 1 − 0)/(1 − 0)]\n• f′(c) = 1\n• 3c2 − 2c = 1\n• 3c2 − 2c − 1 = 0\n• c = [(2 ±√{4 − 4(3)(−1)})/(−6)]\n• c = [(2 ±4)/(−6)]\n• c = 1, [(−1)/3]\n• Only one of these values is in the interval\nc = 1\nFind the mean value of f′(x) using the Mean Value Theorem for f(x) = 3x2 on the interval [1, 2]\n• Polynomial. Continuous and differentiable.\n• f′(x) = 6x\n• f′(c) = [(3(22) − 3(12))/(2 − 1)]\nf′(c) = 9\nFind the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [−1,1]\n• f(x) is undefined at x = 0 and that is within the interval\n• f(x) is not continuous on the interval\nMean Value Theorem cannot be applied\nFind the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [1, 2]\n• f′(x) = −[1/(x2)]\n• f′(c) = [([1/2] − [1/1])/(2 − 1)]\n• f′(c) = −[1/2]\n• −[1/2] = −[1/(c2)]\n• [1/2] = [1/(c2)]\n• c2 = 2\n• c = ±√2\n• Only one solution in the interval\nc = √2\nf(x) is continuous and differentiable on [a,b] and f(a) = f(b) and a ≠ b. Show that f′(x) has a root in the interval\n• In order for some c to be a root of f′(x), f′(c) = 0\n• Continuous and differentiable, so Mean Value Theorem can be used\n• f′(c) = [(f(b) − f(a))/(b − a)]\n• f′(c) = [(f(b) − f(b))/(b − a)]\n• f′(c) = [0/(b − a)]\n• f′(c) = 0\nf′(c) = 0, so there is a root at x = c\nFind at least one root of f′(x) on the interval [−1, 1] given f(x) = x2\n• Polynomial. Continuous and differentiable everywhere\n• f′(x) = 2x\n• f′(c) = [(12 − (−12))/(1 − (−1))]\n• f′(c) = 0\n• 2c = 0\n• c = 0\nThere's a root of f′(0)\nf(x) is continuous and differentiable on [0,5]. Also, f(0) = 1 and f′(x) ≤ 3 in the interval. Find the largest possible value of f(5).\n• f′(c) = [(f(5) − f(0))/(5 − 0)]\n• f′(c) = [(f(5) − 1)/5]\n• 5 f′(c) = f(5) − 1\n• f(5) = 5 f′(c) + 1\n• Plug in the largest value\n• f(5) = 5(3) + 1\nThe largest possible value of f(5) is 16\nFind the value of c that satisfies the the Mean Value Theorem conditions for f(x) = √x on the interval [1, 4]\n• f(x) is continuous so long as x ≥ 0\n• f′(x) = [1/(2 √x)]\n• Differentiable so long as x > 0\n• Continuous and differentiable in the interval\n• f′(c) = [(√4 − √1)/(4 − 1)]\n• f′(c) = [1/3]\n• [1/(2 √c)] = [1/3]\n• 2√c = 3\n• √c = [3/2]\nc = [9/4]\n\n*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.\n\n### Mean Value Theorem\n\nLecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.\n\n• Intro 0:00\n• Mean Value Theorem 0:06\n• Rolle's Theorem\n• Mean Value Theorem Conditions\n• Mean Value Theorem Definition\n• Example 1 0:56\n• Example 2 2:44\n• Example 3 5:28\n• Example 4 7:15\n\nOR\n\n### Start Learning Now\n\nOur free lessons will get you started (Adobe Flash® required)." ]

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[ "# Intersection Logic\n\n1 teachers like this lesson\nPrint Lesson\n\n## Objective\n\nSWBAT use conjectures about intersections to justify a claim. Students will understand how conjectures related to intersections are applied in the construction of a bisector.\n\n#### Big Idea\n\nUsing conjectures about intersections to justify a claim: students find the right argument to make their case.\n\n## Lesson Open\n\n5 minutes\n\nThe Warm Up prompt for this lesson asks whether, in the construction of a (segment) bisector, 4 arcs must all be constructed with the same radius.  The goal of the question is to get students to help each other recall the method of constructing a bisector.  I expect them to answer that, yes, all 4 arcs must be constructed with the same radius, since that is the way I taught them to construct a bisector.\n\nI present the warm-up using the slideshow for the lesson.  The lesson open follows our Team Warm-up routine, with students writing their answers to the warm-up prompt in their Learning Journals.\n\nGoal-Setting\n\nAfter reviewing the teams' answers to the warm-up problem and praising them for recalling the procedure they were taught, I reveal that the question is a setup.  Displaying the Agenda and Learning Targets, I tell the class that the goal of today's lesson of  is to use conjectures about the intersections of points, lines, and planes to reason about the properties of a geometric figure.  To start, we will analyze the construction of a bisector to understand why it works...and see that there is room to deviate from the method as we learned it.\n\n## Analyzing a Construction\n\n15 minutes\n\nIn this section, I lead the class in an analysis of the construction of a bisector, using the line of reasoning demonstrated in the video (MP3) and summarized here:\n\nReferring to the slide I point out...\n\n1. That to be a point on the bisector of A and B, we must locate point C so that it is equidistant from those points.   In other words, distance AC must equal distance BC.  We can make this happen by constructing a pair of arcs, centered at A and B, using the same compass setting.\n2. Likewise, to be a point on the bisector of A and B, point D must be located so that it is also equidistant from A and B.  That means that distance AD must equal distance BD.  It is not necessary, however, that AD=AC or that BD=BC.  (We learned to construct it that way simply because it is convenient.)  We can test this conclusion by deliberately changing our compass setting before constructing two more arcs to locate point D.\n3. As long as points C and D are both points on the bisector of A and B, we can be sure that the bisector is line CD.  Why?  Because one and only one line exists between any two points.  Since there is only one bisector of A and B, this line must be it.\n\nThis argument seems a little funny to students at first.  Speaking from personal experience, I remember finding arguments like this to be strange when I first learned geometric proofs.  With repeated exposure, however, the conventions of formal proof become more familiar, and students learn to judge what is considered a viable argument.  The purpose of this demonstration and the activity that follows is to give students experience early in the course, so that they can internalize the ways of thinking that mathematicians use to make their arguments (MP3).\n\nOf course, this argument makes a few big assumptions: that a bisector exists, and that it is unique.  We will return to those points when we use transformations to prove the properties of a bisector in a later lesson.\n\nTo check whether we have reasoned correctly, I ask every student to get out construction tools and try the alternate method of constructing a bisector.\n\n\n\n30 minutes\n\nI display the instructions and ask students to get out their Guided Notes on Intersections.  In the Activity that follows, students work together to answer questions and determine which conjecture about intersections provides the best support for thier answer.\n\nThis activity follows the Rally Coach routine, because I want students to critique each other's work.  Does the chosen conjecture really apply in this situation (MP3)?\n\n## Lesson Close and Homework\n\n4 minutes\n\nThe Lesson Close asks students to make a connection: Why, when using geometric notation, do we name a line using two points, but name a plane using three points?\n\nThis activity follows our Team Size-Up routine.\n\nHomework\n\nFor homework, I assign problems #38-40 of Homework Set 2.  Problems #38-39 ask students to use conjectures about intersections to justify claims, similar to the problems they completed in class.  Problem #40 previews the next lesson by asking students to describe the intersection of a relatively simple polyhedron with a plane." ]

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[ "## 05 January 2009\n\n### Doing my homework\n\nThe mapFilter function\n\nI recently tried to write a function using Scala which would apply a function to a list, then filter all the resulting elements according to a criteria. In my infinite wisdom I named it \"mapFilter\". Here is an example of what I expected:\n\n`mapFilter(List(1, 2), (x:Int) => if (x%2 == 0) Some(x*2) else None)// returns List(4)`\n\nOf course I wanted this function to be as efficient as possible so it had to do the job in one pass. Otherwise, I could just use a combination of \"map\" and \"filter\" to do the trick:\n\n`def mapFilter[T, U](l: List[T], f: T =>Option[U]): List[U] = {`\n`  l.map(f(_)).filter(_.isDefined).map(_.get)`\n`}`\n3 passes on the list, we can definitely do better than this!\n\nUsing flatMap\n\nSomehow I remembered that the Option class is a Monad, and my subconscious related the bind operation of monads to the flatMap operation on Iterable. I should be able to do something with it,...\n\nThen I experimented this:\n\n`List(1, 2) flatMap { (x:Int) => if (x%2 == 0) Some(x*2) else None }`\n\nBingo, it works!\n\nNow, if you read the signature of the flatMap method (on the Iterable trait), understanding how this works is not so clear:\n\n`def flatMap[B](f: A => Iterable[B]): Iterable[B]`\n\nOption is not an Iterable, this shouldn't compile! Yes, it does, because any Option can be implicitly converted to an Iterable thanks an implicit conversion:\n`object Option {  /** An implicit conversion that converts an option to an iterable value */  implicit def option2Iterable[A](xo: Option[A]): Iterable[A] = xo.toList`\n`}`\nSometimes I wish there was a way to display all the available implicit definitions for a given scope,...\n\nWhat about the \"Monadic\" stuff then? Tony Morris, on the Scala mailing-list, left me a message hinting at a purer solution:\nHi Eric,\nYou can write mapFilter using Scalaz which has a MonadEmptyPlus data\nstructure. I suggest you do not use the available subversion of the\nList.flatMap type signature from which to learn\nSo I thought that I should do my homework, have a look at it and understand why MonadEmptyPlus was a good way to write the mapFilter function.\n\nYou can find all the code from the scalaz project here.  The package I'm exploring today is the \"control\" package.\n\nA bit of Scalaz exploration\n\nLet's jump right to the MonadEmptyPlus:\n`trait MonadEmptyPlus[M[_]] extends MonadPlus[M] with MonadEmpty[M]`\nOhhh. We are high on abstraction here, because MonadPlus is:\n\n`trait MonadPlus[M[_]] extends Monad[M] with Plus[M]`\nAnd MonadEmpty is, as you can guess:\n\n`trait MonadEmpty[M[_]] extends Monad[M] with Empty[M]`\n\nLast but not least, Monad is:\n\n`trait Monad[M[_]] extends Pure[M] with Bind[M] with Functor[M] with Apply[M]`\nHonestly, I didn't suspect that my homework would take me so long! Yet, since every concept and notion of what is a Monad seems to be carefully crafted in Scalaz, this is a good idea to spend a bit of time trying to understand each of them.\n\nPure\n\nThe Pure trait is defined by:\n`/** * Project the given value into the environment that is abstracted over.  */def pure[A](a: A): P[A]`\nIt actually says that you can take any value of some type A, and using the \"pure\" function, put it in a \"Container\" of type P, called the \"environment\" here. I guess that the idea is to say that once the value is well protected and controlled in its container, it is \"pure\". That being said, this is still very abstract. Fortunately we have some implicit values in the the Pure object giving us some examples of what it means to be \"pure\":\n\n` implicit val OptionPure = new Pure[Option] {  def pure[A](a: A) = Some(a)}implicit val ListPure = new Pure[List] {  def pure[A](a: A) = List(a)}`\nA Pure[Option] of the value a is simply the value \"a\" inside the \"Some\" container. So if you think of Monads as \"Containers\" for computation then \"pure\" is what creates the container with a value inside.\n\nBind\n\nBind is the evil twin of Pure and, as far as I'm concerned, the essence of monads because it is the basis of computing with Monads:\n`/** * Binds a function through an environment (known also as sequencing). */trait Bind[BD[_]] {  /**   * Binds the given value with the given value through the environment.   */  def bind[A, B](f: A => BD[B], ba: BD[A]): BD[B]}`\nEvery word is important here. You have an \"environment\" again. This \"environment\" or \"Container\" contains values which have been put there using the pure function for example. Now, what you want to do is to apply a function to the value inside the container, without the value leaving it. So you \"bind\" a function to the environment. Again, concrete examples, given by the implicit values in the Bind object will help us understand what's going on:\n\n`implicit val OptionBind = new Bind[Option] {  def bind[A, B](f: A => Option[B], a: Option[A]) = a flatMap f}implicit val ListBind = new Bind[List] {  def bind[A, B](f: A => List[B], a: List[A]) = a flatMap f}`\nWhen I bind a function to an Option, like Some(4), it is as if I'm doing:\n`val f = x => Some(x + 2)`\n`bind(f, Some(4))`\n\n1. apply the function to the inner element\n\nSome(f(4)) === Some(Some(6))\n\n2. \"flatten\" the inner result by removing the inner \"box\"\n\nSome(6)\n\nThis brings 2 remarks here:\n\n1. \"bind\" for a monad is indeed the \"flatMap\" function of an Iterable in Scala. It maps the value then flatten the results. The big difference is in the function signature. While flatMap is a method on Iterable and accepts a function returning another Iterable, the bind function accepts a \"Container\" type of any sort and returns the same container type; Binding to an Option will return an Option, binding to a List will return a List (\"I suggest you do not use the available subversion of the List.flatMap type signature from which to learn\"...)\n\nFunctor\n\n2. why is it necessary to have a \"bind\" operation? If I want to get Some(6) as a result, something like \"map\" should be enough. True. And this even has a name, this is the \"fmap\" operation of the Functor trait (also mixed in by Monad):\n`trait Functor[F[_]] { /**  * Maps the given function across the given environment.  */  def fmap[A, B](f: A => B, fa: F[A]): F[B]}`\nThe \"fmap\" operation just means transforming the value(s) inside the Container to something else. But the bind operation offers an additional advantage. It allows to compose functions returning values in the Container. I can't compose directly 2 functions returning an Option:\n\n`val f = (x:Int) => if (x % 2 == 0) Some(x) else Noneval g = (x:Int) => if (x % 3 == 0) Some(x) else None// this doesn't work since g returns an Option and f expects an Intval composed = f . g // ??`\n\nBut I can compose them using \"bind\":\n\n`// this works as expectedval composed: (Int => Option[Int]) = (x:Int) => g(x).bind(f)`\nApply\n\nThe definition of Apply is this one:\n`trait Apply[AP[_]] {  def apply[A, B](f: AP[A => B], fa: AP[A]): AP[B]}`\nAnd a good example is provided for Lists:\n`implicit val ListApply = new Apply[List] {  def apply[A, B](f: List[A => B], a: List[A]) = f flatMap (f => a map(f(_)))`\n`}`\nWe take a list of functions, a list of values, and we return a list with the results of all the functions applied to all the values. Actually this is the essence of the \"Applicative\" model of programming, so I don't really understand why it's been added to the Monad trait. I guess it is because monad computations imply applicative computations as described here.\n\nEmpty and Plus\n\nNow that the tour of Monad is over, we can look at the Empty and Plus traits.\n\nEmpty is very easy, it just returns an empty \"Container\" (named \"Environment\" here):\n\n`trait Empty[E[_]] {  /**   * Returns the empty environment.   */  def empty[A]: E[A]}implicit val OptionEmpty = new Empty[Option] {  def empty[A] = None}implicit val ListEmpty = new Empty[List] {  def empty[A] = Nil}`Plus defines what it means to \"append\" or \"merge\" two environments together. Its result is very specific to the underlying Container type as you can see with the implicit values:\n`trait Plus[P[_]] {  /**   * Appends the two given values.   */  def plus[A](a1: => P[A], a2: => P[A]): P[A]}implicit val OptionPlus = new Plus[Option] {  def plus[A](a1: => Option[A], a2: => Option[A]) = a1 orElse a2}implicit val ListPlus = new Plus[List] {  def plus[A](a1: => List[A], a2: => List[A]) = a1 ::: a2}`\nEmpty and Plus are indeed defining some kind of addition operation on Monads, aren't they?\n\nThe real academic mapFilter function ;-)\n\nWith all that at hand I should now be able to create my mapFilter function ;-)\n`import scalaz.control._def mapFilter[T, A[_], U](f: T => A[U], m: A[T])(implicit m1: MonadEmptyPlus[A])= {`\n`  m1.bind(f, m)}`\nIs that all? Yes, let's try it:\n`// Note that Scala type inferencer doesn't infer the types heremapFilter[Int, List, Int]((x:Int) => if (x%2 == 0) List(x*2) else Nil,                          List(1, 2))`\nHowever we may wonder: is that really necessary to have a MonadEmptyPlus? No, a simple Monad is also ok:\n`def mapFilter[T, A[_], U](f: T => A[U], m: A[T])(implicit m1: Monad[A]) = { `\n`  m1.bind(f, m) `\n`}`\nAnd that's understable with the List example because we're using \"flatMap\" under the covers! But this is not totally satisfying. Somehow, I would like to enforce that if f returns the empty element for a given MonadEmptyPlus, then that element is filtered from the result.\n\nEven better with FoldRight\n\nWe can actually define this using another abstraction from the control package: FoldRight.\n`trait FoldRight[F[_]] {  /**   * Fold the given function across the given environment.   */  def foldRight[A, B](t: F[A], b: B, f: (A, => B) => B): B}`\nFoldRight works exactly as the foldRight method on Iterable in Scala and allows us to use the Empty and Plus definition from MonadEmptyPlus, accumulating the mapped values, unless they're empty:\n`def mapFilter[T, A[_], U](m: A[T], f: T => A[U])                         (implicit m1: MonadEmptyPlus[A], f1: FoldRight[A]) = {  f1.foldRight[T, A[U]](m, m1.empty, (t, result) => m1.plus(f(t), result))}`\nAnd that's the end of the assignement for today!\n\nConclusion\n\nThis exploration of Scalaz was very interesting for me because:\n1. The abstractions are nicely separated\n2. Implicit values provide lots of concrete examples\n3. The programming style is very close to the use of type classes in Haskell (which wraps my mind in a new way). The drawback is less type inference by Scala\n4. I had to think again about that \"bind\" function and why it was so special\n5. It gives me the desire to understand more deeply what are the differences between the so-called \"models of programming\": functor, monads, applicative, arrows,...\n6. I haven't been blogging for a looooonnnnng time." ]

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[ "Calculate delta values from recorded data in dual inlet files (continuous flow not yet supported). Note that these are recorded delta values of sample vs. standard in each dual inlet file and NOT normalized against any external standard or reported values of the standard. They usually need at least a frame shift correction afterwards. Also note that the delta values are only multipled by 1000 if in_permil=TRUE (the default), otherwise returns raw values without unit multiplication.\n\niso_calculate_deltas(...)\n\n# S3 method for iso_file_list\niso_calculate_deltas(\niso_files,\ndeltas,\nbracket = TRUE,\nin_permil = TRUE,\nquiet = default(quiet)\n)\n\n# S3 method for data.frame\niso_calculate_deltas(\ndf,\ndeltas,\nbracket = TRUE,\nin_permil = TRUE,\nratio_prefix = \"r\",\nquiet = default(quiet)\n)\n\n## Arguments\n\n... S3 method placeholder parameters, see class specific functions for details on parameters collection of dual inlet iso_file objects which deltas to calculate from the raw data, must be in the complete format with prefix 'd' and both numerator and denominator mass, e.g. c(\"d45/44\", \"d46/44\") to calculate both delta 45/44 and delta 46/44. Deltas can only be calculated if the corresponding ratio columns already exist. whether to display (quiet=FALSE) or silence (quiet = TRUE) information messages. a data frame of raw dual inlet data, must have at minimum the columns 'file_id', 'type' (with values \"standard\" and \"sample\") and 'cycle' (integer column), as well as all of the required ratio columns with the correct ratio_prefix. the prefix of the ratio columns. Default is \"r\" which is what iso_calculate_ratios generates.\n\n## Value\n\nthe passed in data with deltas added\n\n## Details\n\nIf bracket=TRUE (the default), the delta values are calculated from the raw data ratios of the sample and the two bracketing standards in the same cycle and cycle before it (this implicitly assumes that the pre-standard has cycle number 0) If bracket=FALSE, the delta values are calculated straight from the corresponding raw data ratios of the sample and the standard in the same cycle. Delta values are only stored in the sample rows of the raw data, not in the standard rows." ]

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[ "Heat Transfer and Temperature Distribution in a Fin\n This Demonstration calculates temperature as a function of distance from the base of a heat transfer fin of uniform cross-sectional area. The fin is attached to a base, which has the same cross-sectional area and is at a constant temperature of 100°C. Heat loss from the fin is by natural convection to the surrounding air, which is at 25°C. The rate of heat transfer is calculated for an adiabatic tip (no heat transfer through the tip surface) and for a tip that has heat transfer by convection. The temperature decreases down the fin due to conduction and to heat being lost by convection. The conduction is assumed to be one-dimensional along the length of the fin. The length and diameter of the fin can be varied, and its cross section can be either rectangular or circular. Three fin materials with different thermal conductivities can be selected. A three-dimensional representation of the fin is shown. Download the CDF file to view the simulation using the free Wolfram CDF player." ]

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[ "# A Novel Method of Radar Modeling for Vehicle Intelligence.\n\nI. INTRODUCTION\n\nAs one of the ultimate solutions to traffic safety and vehicle performance, intelligent vehicle is gaining momentum. Radar detection system plays a key role on environmental perception for intelligent vehicles. Although field experiment has been traditionally dominating in vehicle development, it is not an efficient or even safe way in developing and testing radar-based ADAS for ACC, collision avoidance, and etc. It is hard or sometimes even impossible to set up such field environment that could cover large variations, great uncertainties and complexity of real driving scenarios. Besides high cost, long duration, lack of flexibility, and worse yet, lack of guaranteed safety, limited field experiment can hardly ensure system robustness and reliability. The steadier detection performance, that is, affected slightly by environment like atmosphere, and low cost compared with Light Detection And Ranging (LiDAR), the longer detection range up to 300 meters compared to ultrasonic radar, and the ability to detect the relevant velocity makes millimeter wave radar widely used in ADAS applications by now. So this paper intends to propose a millimeter wave radar modeling method of high fidelity with the abundant driving environment provided by PanoSim aimed to maximum the meaning of virtual simulation experiment before field experiment in the minimum cost.\n\nThere are indeed lots of radar simulation solutions resulted in mature commercial products in different domains like ENDOCOM in medical domain, MultiPASS and OP2H in telecommunication, WinPROP, FERMAT and so on in environment perception . The first two solutions are based on statistical modeling and hardly meet the requirements of complex dynamic situations. This paper focuses on automotive radar simulation solutions to be discussed in detail next.\n\nThe technologies in prior research on radar modeling for automotive applications are mainly through function-based approaches, physics-based approaches or both .\n\nThe function-based modeling approach mainly focuses on the functions of radar detection and measurement with little detail on its physical components and working mechanism. In this approach, the radio wave that radar fires is treated as a confined geometrical shape, such as a cone; the targets to be detected are modeled as a collection of polygons, bounding box or just scattering center for the sake of computational efficiency. Radar detection is abstracted as a solution of the intersection between the geometrical representation of radar beam and the targets. However the simulation result is ideal and often unrealistic.\n\nThere have been mainly two ways to model radar in a physics-based approach. One approach is to model radar components with electromagnetic and electronic circuits software and concurrently to simulate the propagation channels usually by means of \"ray tracing\". This approach can be very time-consuming and often real-time prohibitive . The other approach is signal-based that synthesizes radar signal echoed from the targets and other objects in the driving environment with clutter and noise that affect radar detection with various physical factors .\n\nThen this paper presents a novel radar modeling approach, of which the architecture is depicted in Section II. This novel approach primarily consists of two modules. The first one, geometric model is discussed in Section III and the second one, physical model is described in detail in Section IV. Section V shows the independent validation of physical model in Matlab/Simulink and the integrated validation of the whole model in Matlab/Simulink and PanoSim. Finally the paper concludes in Section VI.\n\nII. DESCRIPTION OF MODEL ARCHITECTURE\n\nThe architecture of the proposed radar model is shown in figure 1 below. At first, it is necessary to clear the differences between the two conceptions, \"Radar Model\" and \"Radar System Model\". Radar, an acronym for radio detection and ranging, is an object detection system which fires electromagnetic wave into the surrounding environment, then receives and processes its echo signal reflected, so as to determine the range, range rate and angle of objects.\n\nTo abstract the function of radar, if treat the actual radar as a black box, then the input of this black box should be the surrounding environment information and the output should be the detected result. Thus the black box and its input and output composes the radar system. Accordingly in the virtual world depicted in figure 1, the red frame named radar model is equivalent to the actual radar or the black box. Note that the block of environment model refers in particular to infrastructure like buildings, ground, trees, bush and transport facilities, excluding the active transport participants like all kinds of vehicles and pedestrians, what the block of target model exactly means. The block of RCS estimation serves as the response of objects to the radar EM-wave which will be discussed more detailed later. These three blocks provide a complete input information to the radar model and the detection output which is not expressed in a block flows into the ACC controller as input. Hence all the blocks except ACC controller composes the radar system model as blue frame shows.\n\nIII. GEOMETRIC MODELING APPROACH\n\nThis paper firstly adopts a geometric modeling approach to establish the link between the object information and the radar without concerning about any details of physics factor. In this approach, the object to be detected is abstracted as a set of pre-defined Points of Shape Characteristics (PSC)illustrated in figure 2, the EM-wave radar fires is represented as a closed geometry created by an elliptic cone and a plane, and then the radar detection process is transformed into a problem to find out the PSC illuminated by the EM-wave, or radar beam. These illuminated PSC should meet the following two conditions: a) located inside of the radar beam geometry; b) not blocked by other objects, that is, there exists no objects, not PSC, on the line from the radar beam firing point to the illuminated PSC. Then among these PSC, find out the closest point to the radar of each object and transmit them to the physical model.\n\nGeometric Representation of Objects\n\nThe object in general refers to any moving or stationary things that may fall into radar RVV and generate echo signals physically. To discretize a complex object in terms of its shape, a set of points are pre-defined that capture the major characteristics of the object shape (PSC), such as vertices, center, edge, etc, as illustrated in Figure 2.\n\nAs figure 3 shows, the radar beam is assumed to be a close geometry named Radar View Volume(RVV) created by an elliptic cone and a plane and have the three property parameters: range R, azimuth angle [alpha], elevation angle [beta]. The RVV geometric representation is defined in the equation (1) where [x, y, z] denotes the coordinates of a 3D point.\n\n[x.sup.2]/[(ytan [alpha]/2).sup.2]+[z.sup.2]/[(ytan [beta/2]).sup.2] [less than or equal to]1y[less than or equal to]R (1)\n\nGeometric Representation of Detection Process\n\nIn order to determine which sets of PSC are falling into RVV and not blocked simultaneously, the premise of detection process is to unify the coordinate systems. Then find the special PSC that meet the demands quickly with a series of steps taken.\n\nA. Unify the Coordinate Systems\n\nConsidering that RVV has accurate and succinct mathematical expression and the point coordinate is easy to be converted, so transform the coordinates of PSC from world frame to vehicle frame and actually to radar frame, as shown in figure 2.\n\nFirst, to transform the coordinates of a point [P.sub.w] [[x.sub.w], [y.sub.w], [z.sub.w]] given under world frame to vehicle frame, denote the position of vehicle or the origin of vehicle coordinate system under world frame by [T.sub.V] [[X.sub.V], [Y.sub.V], [Z.sub.V]], and orientation of vehicle (in yaw-pitch-roll sequence) or the Euler angles (Gillespie, 1992) by [[psi], [theta], [phi]], then the coordinate of the point under vehicle frame can be expressed as follows\n\n[mathematical expression not reproducible] (2)\n\nWhere the rotation matrix is given by [mathematical expression not reproducible]:\n\n[mathematical expression not reproducible]\n\n[mathematical expression not reproducible]\n\n[mathematical expression not reproducible]\n\nTo obtain a homogeneous transformation matrix defined as [mathematical expression not reproducible] considering both rotation and translation for the sake of format conciseness, add a dimensionality to the point coordinate, thus equation 1 is turned as follows:\n\n[mathematical expression not reproducible]\n\n[mathematical expression not reproducible]\n\nSimilarly, the transformation from the vehicle frame to the radar frame can be represented by the homogeneous matrix [mathematical expression not reproducible]:\n\n[mathematical expression not reproducible]\n\nThus the total transformation formula of a point from world frame to sensor frame can be defined as\n\n[mathematical expression not reproducible] (3)\n\nwhere [mathematical expression not reproducible]\n\nB. Detailed Detection Process\n\nThe key problem to the radar detection process of the virtual world is how to screen PSC to get those special ones which are \"visible\" to the radar beams quickly. Consequently this paper adopts the approach of Axis Aligned Bounding Box (AABB) among others to achieve this goal. As illustrated in figure 4, AABB is defined as the minimum hexahedron that encloses object and of which each edge is parallel to the coordinate axis. The coordinate of AABB vertices can be easily calculated through the maximum and minimum values of the mesh point coordinate of objects.\n\nThe figure 5 shows the flowchart of the detailed detection procedure of which the function is accomplished primarily by four steps of judgement highlighted by four yellow diamonds. Firstly, in view of the fact that the 3D scene is far greater than the radar beam and there generally exists plenty of objects which are definitely can not be detected by radar, this paper adopts quadtree to establish and query the 3D scene. A quadtree is a tree data structure in which each internal node has exactly four children. Only those AABB located in the same node as radar beam, namely, the so-called minimum visible zone, go on with the next judgement and other AABB with the sets of PSC enclosed are excluded. Secondly, according to the different position relationship of AABB and RVV (figure 3 shows one situation that AABB lies partly in RVV), choose different measures. Then the following two procedures deal with PSC instead of AABB. Eventually the detection information (relevant range, range rate, angle) the radar geometric model outputs can be easily calculated based on geometric relationships. Note that among all the visible PSC of one object detected, only the closest PSC to the radar should be converted as detection information.\n\nIV. PHYSICAL MODELING APPROACH\n\nThis paper adopts the echo signal based physical modeling approach to take the non-ideal factors such as signal attenuation and noise resulted from the stochastic nature of the physics components, channel propagation of radio wave with absorption and scattering due to atmosphere, ground and object reflectivity, etc. The sense of physical model lies in refining the detection result of geometric model further to improve the whole model's fidelity.\n\nFigure 6 shows the FMCW radar block diagram. The CW signal with constant amplitude generated by Voltage Controlled Oscillator (VCO) is modulated in frequency to produce a linear chirp which is radiated toward targets through transceiver. The return radiation collected by transceiver is mixed with Local Oscillator (LO) reference signal and fed into the low-pass filter. The yielded beat signal is digitized by the Analogue to Digital Circuit (ADC) and processed to compute the sensing attributes.\n\nSignal Processing\n\nThe proposed physical model is based on a classic radar signal processing method, i.e., Range and Doppler Processing Method . As figure 7 shows, the basic idea of FMCW is to generate a linear frequency ramp (chirp) periodically. Conventionally one period data is often taken as a unit to process, so the instantaneous transmission frequency for one ramp can be written as:\n\n[f.sub.r](t) = [f.sub.c]+B/T t\n\nwhere [f.sub.c] is the carrier frequency, B is bandwidth and T is sweep duration. After integration, the instantaneous phase becomes\n\n[mathematical expression not reproducible]\n\nWhere [[phi].sub.0] is the initial phase. If is the time delay between the transmitted and received signal of the same target is [tau], then the instantaneous phase of received signal is\n\n([[phi].sub.R])(t) = [[phi].sub.T](t-[tau])\n\n[tau] = 2(R + vt)/c (4)\n\nwhere c is the light speed in vacuum, R and v is respectively the distance and velocity of target at the moment when radar fires EM-wave. And the instantaneous phase of the down-converted signal (beat signal) from the mixed signal is\n\n[[phi].sub.B](t) = [[phi].sub.[tau]](t) - [[phi].sub.R](t) = 2[pi]([f.sub.c][tau] + B/T t[tau] - 1B/2T [[tau].sup.2]) (5)\n\nGiving t is relatively small sufficiently, neglect the high order term and plug equation (4) into equation (5), which then becomes\n\n[[phi].sub.B](t) = 2[pi][2[f.sub.c]R/c] + ([2[f.sub.c]v/c + 2BR/Tc)t] (6)\n\nThus the instantaneous frequency of the beat signal is\n\n[f.sub.B] = d[[phi].sub.B](t)/2[pi]dt = 2[f.sub.c]v/c + 2BR/Tc (7)\n\nFrom equation (7), it can be easily seen that both R and v contribute to the frequency of beat signal. In figure 7, the black solid line denotes the transmitted signal of L periods and the black dashed line denotes the received signal reflected by a stationary target at a certain distance R from radar with the time delay [tau] (black) and the beat frequency [f.sub.B] (black). Then if the stationary target is located at a greater distance corresponding to the longer time delay [tau] (blue), the received signal (blue) will move right along the time axis and lead to a bigger beat frequency. If the target is located in the same position with a certain speed v away from radar, the received signal (red) will results in a smaller beat frequency.\n\nHowever equation (7) is ambiguous since the target's range R and range rate v cannot be determined simultaneously with one frequency ramp and multiple ramps are needed to resolve the range-Doppler ambiguity. Considering that computer can only deal with digital signal, discretize the instantaneous phase of beat signal by taking N ramps and sampling M points in every ramp (denote the sample time interval by [t.sub.s] = T/M).\n\n[[phi].sub.(n,m)] = [2[f.sub.c](R + nTv/c + 2BR/Tc)m[t.sub.s]]\n\n= 2[pi][2[f.sub.c]R/c + 2[f.sub.c]v/c nT + (2[f.sub.c]v/c + 2BR/Tc)m[t.sub.s]] (8)\n\nwhere n = 0,1...N - 1, m = 0,1...M - 1.\n\nNote two things. First, the second R is not replaced by (R + nTv) because the movement during the measurement is short compared to the distance R in terms of physics, or the increment brought by nTvt is relatively small in terms of mathematics. Second, the ramp repetition interval denoted by [T.sub.RRI] is taken to equal to the sweep duration denoted by T. Thus the discretized beat signal is\n\n[s.sub.B(n,m)] = A[e.sup.j[phi]B(n,m)] = Aexp{j[[phi].sub.B(n,m)]} = Aexp{j2[pi][2[f.sub.c]R/c + (2[f.sub.c]v/c + 2BR/Tc)m[t.sub.s]]} (9)\n\nwhere A denotes the amplitude and can be calculated from the classic Radar Equation as follows:\n\n[mathematical expression not reproducible] (10)\n\nwhere the parameter definitions are shown in figure 8.\n\nThe 2D Fourier Fast Transformation (2D-FFT) for continuous temporal signal is\n\n[mathematical expression not reproducible] (11)\n\nDiscretize the beat signal in both time and frequency domain by perform the 2D-FFT:\n\n[mathematical expression not reproducible] (12)\n\nwhere p = 0,1...N - 1, k = 0,1...M - 1.\n\nThe detailed analysis for equation (11) in respect of physics is presented as follows. Arrange the sample points of beat signal in N rows and M columns and perform 1D FFT to every M points row by row. As a result, the peak of each row appears in the same column and corresponds to the range frequency\n\n[f.sub.R] = 2[f.sub.c]v/c + 2BR/Tc [approximately equal to] 2BR/Tc (13)\n\nThen perform 1D FFT to every N computation results of the previous step column by column and consequently the peak of each column appears in the same row and corresponds to the Doppler frequency\n\n[f.sub.D] = 2[f.sub.c]v/c (14)\n\nAt last, in the whole 2D area there appears one peak of which the position corresponds to the range and range rate of target.\n\nAccording to the definition of Fourier transformation, the frequency resolution of first FFT and the range resolution are respectively:\n\n[DELTA][f.sub.R] = 1/[t.sub.s]M = 1/T [DELTA]R = cT/2B [DELTA][f.sub.R] = c/2B\n\nThen the frequency resolution of second FFT and the Doppler resolution are respectively:\n\n[DELTA][f.sub.D] = 1/TN [DELTA]V = c/2[f.sub.c][DELTA][f.sub.[DELTA]] = c/2[f.sub.cTN]\n\nIn order to fulfill the Nyquist Sampling Theorem, the sample frequency should be greater than 2 times the maximum signal frequency as follows:\n\nM/T > 2[f.sub.Rmax] = 4B/Tc [R.sub.max] 1/T > 2[f.sub.Dmax] = 4[f.sub.c]/c [v.sub.max]\n\nBased on the signal processing method mentioned above, the block diagram of FMCW radar model presented is shown in figure 8 with detailed parameter inputs.\n\nThe geometric model described in Section IV outputs the detection information (range, range rate and angle) of the closest PSC to radar of each object detected. These sets of PSC are characterized by distance from radar (and thus a path loss associated with that distance), radial velocity, incident angle and the strength of reflection, i.e., Radar Cross Section (RCS). RCS is a measure of how detectable of an object is with radar. A larger RCS indicates the object is more easily detected. This paper doesn't focus on the estimation of RCS and takes it as known. The selected sets of PSC are fed to the physical model to be screened further by actual factors. The physical model is based on the signal processing method described in the last part. First, compute the beat signal in time domain and superpose an Additive Gaussian White Noise (AGWN) on it. Then perform 2D-FFT to convert the 2D input signal from time domain to frequency domain. A Constant False Alarm Rate (CFAR) threshold is applied to the yielded frequency spectrum and only those range Doppler bins whose Signal Noise Ratio (SNR) is larger than the threshold continue to the next step. Finally calculate the refined detection information (including range, range rate and azimuth angle) according to the position of peaks not submerged by the noise and output them as radar measurements to ACC controller.\n\nV. SIMULATION RESULTS\n\nFigure 9 shows the information list of targets and the simulation results of the proposed FMCW model with the following key parameters: field of view is 30[degrees]; radar Radio Frequency (RF) [f.sub.c] = 24GHz;bandwidth B = 1GHz; sample parameter N = 256, M = 1024; transmit power [P.sub.t] = 10w; effective area of radar receive antenna [A.sub.eff] = 1[m.sup.2]; antenna gain is taken as constant 1. The designed sensing property is [-50,50]m/s for range rate and [0.1,150]m for range. The left bottom figure shows the beat signal contaminated by AGWN in time domain and the right bottom figure shows the same signal in frequency domain after performing 2D-FFT, where three peaks correspond to the three targets. It can be easily seen that none of the three targets is submerged in the noise, which leads to the detection of the three targets.\n\nFigure 10 shows an ACC simulation scheme created under PanoSim and Matlab/Simulink environment. The upper section is the block diagram of the whole system model including driver model, radar model, vehicle dynamics model and ACC algorithm in Simulink. The lower right section is the 3D animation in PanoSim. The lower left and middle section are the detection information of range and range rate respectively and the lines in red circles means the target goes out of the radar detectable range.\n\nVI. CONCLUSIONS\n\nThis paper presents a novel approach to simulate FMCW millimeter wave radar that combines both geometric and echo signal based modeling method to achieve good balance between model fidelity and computational efficiency. The mathematical derivation behind the described method is discussed in detail. The proposed model has been implemented and verified under various scenarios. Among others, a closed loop simulation with an ACC algorithm under a dynamic 3D virtual environment provided by PanoSim has been implemented and the result demonstrates the validity and effectiveness of the proposed modeling method. The proposed method further enables real-time simulation under a Hardware-Inthe-Loop (HIL) platform in the near future.\n\nREFERENCES\n\n(1.) http://www.panosim.com/\n\n(2.) Steve PECHBERTI, Dominique GRUYER and Vincent VIGNERON, \"Radar simulation in SiVIC platform for transportation issues. Antenna and propagation channel modelling\", 15th international IEEE Conference on Intelligent Transportation Systems, Anchorage, Alaska, USA, September 16-19, 2012, pp.469-474.\n\n(3.) Deng, W., Zeng, S., Zhao, Q. and Dai, J. \"Modelling and simulation of sensor-guided autonomous driving\", Int. J. Vehicle Design, Vol. 56, Nos. 1-4, 2011, pp.341-366.\n\n(4.) Dudek Manuel, et. al., \"Millimeter Wave FMCW Radar System Simulations including a 3D Ray Tracing Channel Simulator\", Proc. of IEICE, 2010, pp.1665-1668.\n\n(5.) Laet, T., Schutter, J. and Bruyninckx, H. (2008) \"Rigorously Bayesian range finder sensor model for dynamic environment\", Proceedings of 2008 IEEE International Conference on Robotics and Automation, 19-23 May, Pasadena, CA, USA, pp.2994-3001.\n\n(6.) Schuler K., Becker D., and Wiesbeck W., \"Extraction of vertical scattering centers of vehicles by ray-tracing simulations,\" IEEE Transactions on Antennas and Propagation, Vol. 56, No. 11, November 2008, pp. 3543-3551.\n\n(7.) Winkler, V., \"Range Doppler detection for automotive FMCW Radar\", Proceedings of the 4th IEEE European Radar Conference, pp. 166-169, Oct.. 2007.\n\n(8.) Rohling, H., Moller, C., \"Radar waveform for automotive radar systems and applications\", Radar Conference 2008, pp. 1-4, May 2008\n\nCONTACT INFORMATION\n\nFor questions or to contact the authors, please email to the corresponding author\n\nProf. Weiwen (Kevin) Deng State Key Lab of Automotive Simulation and Control Jilin University China 130025 [email protected]\n\nACKNOWLEDGMENTS\n\nThe authors wish to acknowledge the support of China National Natural Science Foundation under grant U1564211 and National Key Research and Development Program under grant 2016YFB0100904.\n\nJiao Guo, Weiwen Deng, Sumin Zhang, and Shiqian Qi\n\nJilin University\n\nXin Li\n\nJilin Universisty; Aviation University of Air Force\n\nAll rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of SAE International.\n\nPositions and opinions advanced in this paper are those of the author(s) and not necessarily those of SAE International. The author is solely responsible for the content of the paper.\n\ndoi:10.4271/2016-01-1892" ]

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[ "Elementary Geometry for College Students (6th Edition)\n\n1. Use formula for area of a triangle A=$\\frac{1}{2}$(base)(height) and find the area of all 3 triangles. 2. Use A=lw to find the area of the rectangle 3. Subtract the area found for all 3 triangles from the area of the rectangle. [(7$\\times$5)-(0.5$\\times$4$\\times$6)-(0.5$\\times$1$\\times$5)-(0.5$\\times$1$\\times$7)] =35-12-2.5-3.5 =17 sq units" ]

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[ "# DateTime calculated Column using Formula\n\nI have a column Birthday which contains the date of the user's birthday. I would like to create a calculated column to hold the user's next birthday.\n\nFor instance, if I wanted to know the birthday in 2014, this would be the formula:\n\n``````=DATE(2014;MONTH(Birthday);DAY(Birthday))\n``````\n\nBut I would really like to know the user's next birthday. Meaning, if the user hasn't had a birthday this year, then the year would be 2014, otherwise, it would be 2015.\n\nMore precisely:\n\n``````If Month(Birthday) > Month(Today): =DATE(2014;MONTH(Birthday);DAY(Birthday))\nIf Month(Birthday) = Month(Today) && Day(Birthday) >= Day(Today): =DATE(2014;MONTH(Birthday);DAY(Birthday))\nElse =DATE(2015;MONTH(Birthday);DAY(Birthday))\n``````\n\nHowever, a formula as simple as the following fails, due to the syntax not being right\n\n``````=IF([Birthday]>[Birthday], [Birthday], [Birthday])\n``````\n• I just updated my answer. Hope that works for you! Let me know how it goes. – Phil Greer May 23 '14 at 21:34\n\nIt looks like you are using Spanish locale, you should separate with semicolons (apart from the fact that [Cumpleaños] will never be greater than itself):\n\n``````=IF([Cumpleaños]>[Cumpleaños]; [Cumpleaños]; [Cumpleaños])\n``````\n\nFirst, you can't use \"Today\" in a calculated column in SharePoint, but there is a way around that:\n\n1) Create a temporary column called \"Today\"\n2) Setup your calculated column (with formula referencing the bogus \"Today\" column - see below)\n3) When that's working, delete the temporary \"Today\" column (you have tricked SP into using Today's date)\n\nTest thoroughly! You may need to re-create and delete your temporary \"Today\" column several times. For more details, see this blog post.\n\nSecond, the formula you need for your calculated column is as follows:\n\n``````=IF(MONTH(Birthday)>MONTH(Today),DATE(YEAR(Today),MONTH(Birthday),DAY(Birthday)),IF(MONTH(Birthday)<MONTH(Today),DATE(YEAR(Today)+1,MONTH(Birthday),DAY(Birthday)),IF(MONTH(Birthday)=MONTH(Today),IF(DAY(Birthday)>=DAY(Today),DATE(YEAR(Today),MONTH(Birthday),DAY(Birthday)),DATE(YEAR(Today)+1,MONTH(Birthday),DAY(Birthday))))))\n``````\n\nNote that the calculated column should be of type \"Date and Time\".\n\nFinally, I created a temporary column \"Today\" as suggested by Phil Greer, and ended up using the following formula:\n\n``````=IF(DATE(YEAR(Today);MONTH(Cumpleaños);DAY(Cumpleaños))>Today;DATE(YEAR(Today);MONTH(Cumpleaños);DAY(Cumpleaños));DATE(YEAR(Today)+1;MONTH(Cumpleaños);DAY(Cumpleaños)))\n``````\n\nHope this helps!" ]

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[ "# Week 11: Amortized Inference and Variational Auto Encoders¶\n\nUsing HMC to infer transmission + death rates for COVID: paper\n\n# Recap: latent variable models¶\n\nWriting down a generative model $$p_\\theta(x|z), p_\\theta(z)$$ is a simple, interpretable and powerful way to specify a complicated joint distribution $$p(x)$$.\n\n## Examples¶\n\n• Trueskill gives a joint distribution over game outcomes. Can interpret posterior over latent quantities as a belief state about skills given games.\n• Personality models (e.g. Big 5 traits) (z = personality, x is behavior), aka factor analysis.\n• Item Response Theory,\n• Latent Growth models.\n\n## What's easy?¶\n\n• Sampling $$z \\sim p(z)$$ and $$x \\sim p(x|z)$$, and $$x, z \\sim p(x|z)$$, $$x \\sim p(x)$$\n• Evaluating $$p(z)$$ and $$p(x|z)$$, and $$p(x|z)$$, including e.g. $$p(x_1, x_2|x_3, x_4)$$\n\n## What's hard?¶\n\n• Sampling $$z \\sim p(z|x)$$ or $$x_1, x_2 \\sim p(x_1, x_2|x_3, x_4)$$\n• Evaluating $$p(z|x)$$ or $$p(x_1, x_2|x_3, x_4)$$, $$p(x)$$\n• $$p(x) = \\int p(x|z)p(z) dz$$ (and simple MC will have high variance)\n• $$p(x_1, x_2|x_3, x_4) = \\int p(x_1, x_2|z)p(z|x_3, x_4) dz$$ (need posterior)\n• $$p(z|x) = \\frac{p(x, z)}{\\int p(x, z) dz}$$\n\n# Recap: Stochastic variational inference¶\n\nTo approximate $$p(x|z)$$: 1. Introduce a variational family $$q_\\phi(z|x)$$ with parameters $$\\phi$$. 2. Minimize KL divergence between $$p(z | x)$$ and $$q_\\phi(z|x)$$.\n\nTip\n\nBig distinction between model $$p(x,z)$$ and approximate inference strategy. Can use different approx. inf. for same model (such as MCMC or loopy belief propagation), even ones that weren't invented when you wrote down the model!\n\nThis is equivalent to maximizing a lower bound on the log marginal likelihood $$\\log p(x)$$:\n\n$\\log p(x) \\geq \\nabla_\\phi \\mathcal L(\\phi) = \\nabla_\\phi \\mathbb{E}_{z \\sim q_\\phi(z | x)} \\Big [ \\log p(x, z) - \\log q_\\phi(z|x) \\Big ]$\n\nWe'll want to optimize using stochastic unbiased gradients from simple Monte Carlo. We can use thereparameterization trick to bring the gradient inside the expectation:\n\n$\\nabla_\\phi \\mathcal L(\\phi) = \\nabla_\\phi \\mathbb{E}_{z \\sim q_\\phi(z | x)} \\Big [ \\log p(x, z) - \\log q_\\phi(z|x) \\Big ] \\\\ = \\nabla_\\phi \\mathbb{E}_{\\epsilon \\sim p(\\epsilon)} \\Big [ \\log p(x, T(\\phi, \\epsilon)) - \\log q_\\phi(T(\\phi, \\epsilon)|x) \\Big ] \\\\ = \\mathbb{E}_{\\epsilon \\sim p(\\epsilon)} \\nabla_\\phi \\Big [ \\log p(x, T(\\phi, \\epsilon)) - \\log q_\\phi(T(\\phi, \\epsilon)|x) \\Big ]$", null, "## Per-example latent variable models¶\n\nIn the Trueskill model, there is one big vector of $$z$$s, and one big list of game outcomes $$x$$.\nThe graphical model for Trueskill is just one big z to one big z. The graphical model for the approximate posterior is just z.\n\nWhat about our personality quiz example? E.g. there are N people each of whom takes a quiz with D questions. And we assume each person has an a priori indepedently distributed, Q-dimensional vector that specifies their personality. Then there is a separate $$z$$ vector for each $$x$$ vector. The graphical model will have a plate.\n\nThe true posterior in this model factorizes over people. But now we have N true posteriors to approximate. How could we do efficient approximate inference in this setting?\n\n### Motivation #1: SVI on a per-example LVM doesn't scale to large data¶\n\nThis argument is a bit involved, so feel free to zone out.\n\nWe could simply do SVI on everyone's $$z$$ vectors all at once on each iteration of gradient descent. Each person would have a $$\\phi_i$$. The graphical model would also have a plate. This is a good strategy for a small N. However, if N is large, we want to be able to subsample the data. However, in this case, we would have one global $$\\theta$$ and a separate approximate posterior $$q_{\\phi_i}(z_i|x_i)$$ for each person. If we subsampled one out of a thousand people each time we updated $$\\theta$$, then each $$q_{\\phi_i}(z_i|x_i)$$ would be a thousand steps out of date. Keep in mind that the true posterior will change shape as the model parameters $$\\theta$$ change. So the gradients that $$\\theta$$ would get will be for a very poor approximate posterior. We could stop and optimze each $$q_{\\phi_i}(z_i|x_i)$$ for a while before we get the gradient for $$theta$$, but this would also be slow.\n\nWe want to be able to somehow keep all the approximate posteriors in sync, without optimizing all of them whenever we update $$\\theta$$.\n\n### Motivation #2: People can learn to recognize what's going on from partial evidence¶\n\nFor example, with enough experience, doctors, plumbers, detectives, etc. can very quickly tell what is going on and what they still need more information about, if they've seen enough similar situations. Perhaps we could somehow train a neural network to look at the data for a person $$x_i$$, and then output an approximate posterior $$q_{\\phi_i}(z_i|x_i)$$?\n\n# Amortized Inference¶\n\n\"Amortize\" just means \"spread out a cost over time\". Instead of doing SVI from scratch every time we see a new datapoint, we're going to try to gradually learn a function that can look at the data for a person $$x_i$$, and then output an approximate posterior $$q_\\phi(z_i|x_i)$$. We'll call this a \"recognition model\" Instead of a separate $$\\phi_i$$ for each data example, we'll just have a single global $$\\phi$$ that specifies the parameters of the recognition model. Because the relationship between data and posteriors is complex and hard to specify by hand, we'll do this with a neural network!\n\nWe've already seen one way to specify a probability distribution given an input with neural networks. We can simply have a network take in $$x_i$$, and output the mean and variance vector for a Gaussian:\n\n$q_\\phi(z_i|x_i) = \\mathcal{N}(z_i | \\mu_\\phi(x_i), \\Sigma_\\phi(x_i) )$\n\nThe graphical model for this recognition model has the $$\\phi$$ outside the plate.\n\nThe algorithm for amortized inference looks like: 1. Sample a datapoint 2. Compute params of approximate posterior (recognition) 3. Compute gradient of Monte Carlo estimate of ELBO wrt phi. 4. Update phi\n\nThen, when we want to make predictions about a new datapoint, we can use our fast recognition model if we want! Of course, we might also want to stop and do something slower but more accurate, like per-example SVI, or MCMC.\n\n### Also optimizing model parameters.¶\n\nIf $$\\log p_\\theta(x)$$ depends on parameters $$\\theta$$, then the ELBO is a function of both, and we can optimize them together:\n\n$\\nabla_{\\theta, \\phi} \\mathcal L(\\phi) = \\nabla_{\\theta, \\phi} \\mathbb{E}_{z \\sim q_\\phi(z | x)} \\Big [ \\log p_\\theta(x, z) - \\log q_\\phi(z|x) \\Big ] \\\\ = \\mathbb{E}_{\\epsilon \\sim p(\\epsilon)} \\nabla_{\\theta, \\phi} \\Big [ \\log p_\\theta(x, T(\\phi, \\epsilon)) - \\log q_\\phi(T(\\phi, \\epsilon)|x) \\Big ]$\n\nIn the trueskill example, this would let us learn the shape of the likelihood function, for example:\n\n$p(\\textnormal{i beats j}|z_i, z_j) = \\frac{1}{1 + \\exp(-\\theta(z_i, z_j)}$\n\nThus we can jointly fit the model parameters and the recognition network, by subsampling training examples and using simple Monte Carlo with stochastic gradient descent.\n\nThis is called a variational autoencoder (VAE). we'll explain the name next.\n\nSometimes people draw the recognition graphical model on top of the generative model, giving this confusing diagram:", null, "### Example: MNIST¶\n\nLet's give an explicit model for MNIST images of handwritten digits.\n\nWe will choose our prior on $$z$$ to be the standard Gaussian with zero mean and unit variance\n\n$\\mathcal{N}(0, I)$\n\nour likelihood function to be\n\n$p_\\theta(x_i | z_i) = \\prod_{d=1}^D Ber(x_{id} | \\mu_\\theta(z_i))$\n\nand our approximate posterior to be\n\n$q_\\phi(z | x) = \\mathcal{N}(\\mu(x), \\sigma(x)I)$\n\nFinally, we use neural networks as our encoder and decoder\n\nEncoder: $$g_\\phi(x_i) = \\phi_i = [u_i, \\log \\sigma_i]$$\n\nDecoder: $$f_\\theta(z_i) = \\theta_i$$\n\nWhere $$\\mu_i$$ are the Bernoulli means for each pixel in the input. To see a \"reconstructed\" input, we can plot $$\\mu_i$$.\n\nThe entire model looks like:", null, "Where inputs $$x_i$$ are encoded to vectors $$\\mu$$ and $$\\log \\sigma_i$$, which parameterize $$q_\\phi(z | x)$$. Before decoding, we draw a sample $$z \\sim q_\\phi(z | x) = \\mathcal{N}(\\mu(x), \\sigma(x)I)$$ and evaluate its likelihood under the model with $$p_\\theta(x | z)$$. We compute the loss function $$\\mathcal L(\\theta, \\phi ; x)$$ and propagate its derivative with respect to $$\\theta$$ and $$\\phi$$, $$\\nabla_\\theta L$$, $$\\nabla_\\phi L$$, through the network during training.\n\n## Alternate derivation: Autoencoders¶\n\nAn autoencoder takes an input, encodes it into a vector, then decodes to produce something similar to the original data. Or: autoencoders reconstruct their own input using an encoder and a decoder.\n\nEncoder: $$g(x) \\rightarrow z$$\n\nDecoder: $$f(z) \\rightarrow \\hat x$$\n\nThe encoder, $$g(x)$$, takes in the input data (such as an image) and outputs a single value for each encoding dimension while the The decoder, $$f(z)$$ takes this encoding and attempts to recreate the original input.\n\nOur goal is to learn $$g, f$$ from unlabeled data, and usually we specify f and g with neural networks, and minimize squared reconstruction error.", null, "$$z$$ is the code the model attempts to compress a representation of the input, $$x$$, into. For deterministic autoencoders, it's important that this code is a bottleneck, i.e. that\n\n$\\text{dim} \\ F \\lt \\text{dim} \\ X$\n\nas this forces the encoder to reduce the dimension, for example by learning how to ignore noise (otherwise, we would just learn the identify function). The big idea is that the code contains only the most important features of the input, such that we can reconstruct the input from the code reliably\n\n$\\tilde x = f(g(x)) \\approx x$\n\n### Problems with Deterministic Autoencoders¶\n\nThere are two main problems with deterministic autoencoders.\n\n#### Problem 1: Proximity in data space does not mean proximity in feature space¶\n\nThe embeddings (or codes) learned by the model are deterministic, i.e.\n\n$g(x_1) = z_1 \\Rightarrow f(z_1) = \\tilde x_1 \\\\ g(x_2) = z_2 \\Rightarrow f(z_2) = \\tilde x_2 \\\\$\n\nbut proximity in feature space is not enforced for inputs in close proximity in data space, i.e.\n\n$z_1 \\approx z_2 \\not \\Rightarrow x_1 \\approx x_2$\n\nIf the space has regions where no data gets encoded to, and you sample/generate a variation from there, the decoder will simply generate an unrealistic output, because the decoder has no idea how to deal with that region of the latent space. During training, it never saw encoded vectors coming from that region of latent space.\n\n• Low-dim z. But what dimension?\n• Can add noise to data before encoding, reconstruct original data. But how much noise?\n• Can add noise to latents after encoding, reconstruct original data. But how much noise?\n\n## Variational Autoencoders (VAEs)¶\n\nThis stochastic generation means, that even for the same input, while the mean and standard deviations remain the same, the actual encoding will somewhat vary on every single pass simply due to sampling.\n\n### Why does a VAE solve the problems of a deterministic autoencoder?¶\n\nThe VAE generation model learns to reconstruct its inputs not only from the encoded points but also from the area around them. This allows the generation model to generate new data by sampling from an “area” instead of only being able to generate already seen data corresponding to the particular fixed encoded points.\n\n# What can we do with a latent variable model?¶\n\nIf a latent variable model has a compact prior and vector-valued $$z$$s, we can: - Sample new data to check the model - Encode (sample from the approximate posterior) of two datapoints, and interpolate between them in the latent space, then decode along the path - Learn a function to predict some property of the examples from the low-dimensional z space instead of directly from x. (Semi-supervised learning)\n\n# Tutorial¶\n\n## Consequences of using amortized inference¶\n\n• Don’t need to re-optimize phi_i each time theta changes - much faster\n• Recognition net won’t necessary give optimal phi_i\n• Can have fast test-time inference (vision)\n\n## Semi-amortized inference:¶\n\nThere's no reason why we can't use a recognition network to initialize q, then take a few steps of SVI.\n\n## Alternate forms of the ELBO:¶\n\nWe also talked about two other alternative forms or \"intuitions\" of the ELBO:\n\n$L(\\theta, \\phi ; x) = E_{z_\\phi \\sim q_\\phi} \\Big [ \\log p_\\theta({x | z}) + \\log p_\\theta({z}) - \\log {q_\\phi(z | x)} \\Big ] \\\\ = E_{z_\\phi \\sim q_\\phi} \\Big [ \\log p_\\theta({x | z}) \\Big ] - D_{KL}(q_\\phi(z | x) || p_\\theta(z))$\n\nThe second of which (intuition 3) is the loss function we use for training VAEs. Notice now that the first term corresponds to the likelihood of our input under the distribution decoded from $$z$$ and the second term the divergence of the approximate distribution posterior from the prior of the true distribution.\n\nNote\n\nThe second terms acts a regularization, by enforcing the idea that our parameterization shouldn't move us too far from the prio distribution. Also note that this term as a simple, closed form if the posterior and prior are Gaussians.\n\n## Automatically choosing latent dimension¶\n\nStandard autoencoders require choosing latent dimension What happens if a VAE has more than it needs? If q(z|x) is factorized, then KL term factorizes over dimensions, wants to make each q(z_i|x) look like p(z_i). If a dimension doesn’t help likelihood enough, it will ‘turn off’ and set q(z_i|x) = p(z_i), ignoring x. Then decoder can ignore too." ]

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[ "# Chapter 5 - Linear Functions - 5-1 Rate of Change and Slope - Standardized Test Prep - Page 298: 61\n\na) $20 < 2x +16 < 30$ b) $2 < x < 7$, please see the first graph. c) $2/3 < x < 10$, please see the second graph.", null, "a) Two of the sides of the rectangle have lengths of (x+2) and 6. Since these sides are not parallel, the other two sides have sides of the same length. $20< (x+2) + (x+2) + 6 + 6 < 30$ $20 < 2x +16 < 30$ b) $20 < 2x +16 < 30$ $20 -16 < 2x + 16 - 16 < 30-16$ $4 < 2x < 14$ $4/2 < 2x/2 < 14/2$ $2 < x < 7$ c) possible pairs of triangle legs: (6, x+4), (6, 2x), (2x,x+4) $6 + x+4 > 2x$ $10 + x > 2x$ $10 + x - x > 2x - x$ $10 > x$6 + 2x > x + 46 + 2x - x - 4 > x + 4 - x - 42 + x > 02 + x - 2 > 0 - 2x > -2 x+4 + 2x > 63x + 4 > 63x + 4 - 4 > 6 - 4 3x > 2 3x/3 > 2/3x > 2/3\\$ If we let x = 11, we would not have a triangle as the lengths would be 6, 15, and 22 (and 6 + 15 < 22). Thus, the solution set is 2/3 < x < 10.", null, "" ]

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[ "# Medical Electronics 6th MJ13\n\nB.E/B.TECH. DEGREE EXAMINATION, MAY/JUNE 2013\nSixth Semester\nElectronics and Communication Engineering\n080290046-MEDICAL ELECTRONICS\n(Regulation 2008)\n\nPART A-(10*2=20 marks)\n\n1. What is meant by perfectly polarized and non-polarized electrodes?\n2. Draw the relationship between action potential and muscle contraction.\n3. Define cardiac output.\n4. What are the factors that affect the speed of migration in electrophoresis?\n5. How to treat fibrillation problems?\n6. Draw the different types of wave forms are used in electrotherapy.\n7. What is the principle of computer axial tomography?\n8. Define Nuclear Magnetic Resonance.\n10. What is Arrhythmia?\n\nPART B-(5*16=80 marks)\n\n11. (a) (i) Explain in detail various types of bio-potential electrodes.\n(ii) Explain the characteristics of resting potential.\nOr\n(b) Explain the working of\n(i) ECG recording method.\n(ii) EMG recorder.\n\n12. (a) Explain the following cardiac output measurement techniques\n(i) Fick's Method\n(ii) Impedance Change\n(iii) Indicator dilution method.\nOr\n(b) (i) Explain the principle of Electromagnetic blood flow meter.\n(ii) Write short notes on Pulmonary function analyzers.\n\n13. (a) (i) Explain the working principle of capacitive discharge type of cardiac defibrillator with\nneat Circuit diagram.\n(ii) Explain the working of heart lung machine.\nOr\n(b) (i) With a neat block diagram, explain the operation of dialyzer machine.\n(ii) Write short notes on cutting and coagulation waveform in surgical diathermy.\n\n14. (a) Explain the operation of computer tomography by using the mathematical basis of image\nConstruction. Mention its applications.\nOr\n(b) Explain the following\n(i) MRI system\n(ii) Ultrasonic Imaging System.\n\n15. (a) (i) Explain the latest trends in diagrams and therapy in medical field.\n(ii) With neat block diagram explain the design process of pattern recognition.\nOr\n(b) (i) Explain the role of expert systems.\n(ii) Explain the VLSI design tools in medical application." ]

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[ "Disclosure: This post may contain affiliate links, meaning when you click the links and make a purchase, we receive a commission.\n\n# The Different Ways of Counting in Excel\n\nThere is a total of five functions used for counting, predominantly in Excel. These are namely the COUNT Function, the COUNTA Function, the COUNTIF Function, the COUNTIFS Function and the COUNTBLANK Function.\n\nThe COUNT Function is used to determine the number of cells that contain numbers or a number of numbers in a list. It is the simplest of the counting functions. Dates are also counted when using this function. The syntax for the COUNT Function is:", null, "The COUNTA Function is used to count the number of cells, within a particular range that is not empty. This function counts numbers, error values, logical formulas and formulas that return an empty text string. The syntax for the COUNTA Function is:", null, "The COUNTIF Function beings conditional logic into the equation and counts the number of cells that meet specific criteria. The syntax for the COUNTIF Function is:", null, "The COUNTIFS Function is similar to the COUNTIF Function, except it’s used to count the number of cells in a range that meet multiple criteria. All the criteria have to be met for COUNTIFS to count that specific value. So, if one thinks of this using logical formula logic all conditions set have to evaluate to true for a row to be counted.", null, "The COUNTBLANK Function counts the number of blank cells in a specified range. The syntax for the COUNTBLANK Function is:", null, "So let’s get started with a few simple examples illustrating the use of the different counting functions in Excel.\n\nRead More: Excel Dynamic Named Range [4 Ways]\n\n## The COUNT Function\n\nThere is a mixed list of data that has numbers, text, characters and logical values. We want to determine only the numbers in the list using the COUNT Function.", null, "1) In order to count the numbers in the cell range A5: A12, we input the following formula in Cell A13:\n=COUNT (A5: A12)", null, "2) Upon pressing CTRL-ENTER, we get a value of 4 returned.", null, "### Crossover Tip: One can also use the COUNT Function in MS Word\n\n1) We have a table in Word, with the values in numbers listed underneath the different fruits and we want to count the number of times a number value is listed.", null, "2) The way the cell references in a Table work in Word is shown below:", null, "3) In order to insert a formula in the last cell D2, with the Table Tools context sensitive menu activated, go to Table Tools>Layout>Data>Formula.", null, "4) Enter the formula\n\n=COUNT (A2: C2)\n\nSelect a number format and click Ok as shown.", null, "5) A value of 2 is delivered from the COUNT Function.", null, "6) Doing the same for the rest of the column results in the following as shown below.", null, "7) It is often necessary to manually update formulas in Word, since the formula is calculated in Word at the time of insertion and when the user opens the document containing the formula. So, in order to do this Press CTRL-A on your keyboard and the F9 key and then all the formulas in the Word document will be updated with the current most relevant results.\n\n## The COUNTA Function\n\nThere is a mixed list of data that has numbers, text, characters and logical values. We want to determine the cells that are not blank in the list using the COUNTA Function.", null, "1) In order to count the number of cells that are not blank, in the cell range A5: A12, we input the following formula in Cell A13:\n\n=COUNTA (A5: A12)", null, "2) Upon pressing CTRL-ENTER, we get a value of 8 returned.", null, "Read More: COUNTIF formula in Excel\n\n### Creating Dynamic Ranges with COUNTA\n\n1) In order to create dynamic ranges, one often has to use the COUNTA Function.\n\n2) We have a range in cell A5: A9 called Apples_Sold, which was previously created and named.", null, "3) Now let’s say we wanted to add an extra number in A10, that denotes more Apples sold or data not collected for that particular period, we would, of course, want the range to update automatically, and in order to do this, we can use the COUNTA Function in order to create a dynamic range.\n\n4) Go to Formulas>Defined Names>Name Manager as shown below:", null, "5) Make sure the Apples_Sold range is selected and choose Edit as shown.", null, "6) In the formula bar, type in the following formula:\n\n=OFFSET(\\$A\\$5,0,0,COUNTA(\\$A\\$5:\\$A\\$1048576),1)", null, "7) Press Ok and then Close.", null, "8) In Cell E5, input the following formula:\n\n=COUNTA (Apples_Sold)", null, "9) Upon pressing CTRL-ENTER, the value of 5 is delivered.", null, "10) If you type the value, Not collected in Cell A10, the formula in E5 automatically updates with the new entry since one used the COUNTA Function if you want only numbers to use COUNT instead.", null, "Read More: Count Cells that Contain Specific Text in Excel\n\n## The COUNTIF Function\n\nThere is a mixed list of data that has different fruits and the corresponding sales of the fruits. We want to count only the values that exceeded \\$ 500, this is clearly a condition or criteria attached so we will have to use the COUNTIF Function.", null, "1) In order to count the number of cells that have a value greater than \\$ 500, in the range cell B5: B11, we input the following formula in Cell D6:\n\n=COUNTIF (B5: B11, “>500”)", null, "2) Upon pressing CTRL-ENTER, we get a value of 2 is returned, so two values, in other words, exceeded the \\$ 500 threshold.", null, "## The COUNTIFS Function\n\nThere is a mixed list of data that has different fruits and the amounts sold. We want to count only the values of apples, whose sales exceeded \\$ 200, so we can use the COUNTIFS Function to accomplish this task since we now have more than one criteria that have to be fulfilled.", null, "1) In order to count the number of cells that have a value of Apples with sales exceeding \\$ 200, we input the following formula in Cell D6:\n\n=COUNTIFS(A5:A16,”Apples”,B5:B16,”>200″)", null, "2) Upon pressing CTRL-ENTER, we get a value of 2 is returned, so two values, in other words only two values fulfill the requirement of being apples with sales of greater than \\$ 200.", null, "### Using Wildcards\n\n1) One can also use Wildcards when summing, counting or using TEXT Functions in order to extend functionality in Excel.\n\n2) In the example used above, we now want count all the fruits starting with the letter A, and whose sales exceed \\$ 200.", null, "3) So in Cell D6, we enter the following formula:\n\n=COUNTIFS(A5:A16,”A*”,B5:B16,”>200″)", null, "4) Upon pressing CTRL-ENTER, we get a value of 3 is returned, so three values, in other words, only three values fulfill the requirement of starting with the letter A, with sales of greater than \\$ 200.", null, "When using COUNTIF or COUNTIFS to count certain text values, you may need to first use the CLEAN Function or the TRIM Function to clean the data, since spaces will affect the results. Neither the COUNTIF Function or the COUNTIFS Function are case sensitive so apples should be counted as well as Apples.\n\nRead More: Finding out the number of duplicate rows using COUNTIF formula\n\n## The COUNTBLANK Function\n\nThere is a mixed list of data that has different values and a few blank cells, in between and we want to use the COUNTBLANK Function in order to determine the number of blank cells in the range.", null, "1) In order to count the number of blank cells that fall within the range, we first have to turn off Enable iterative calculation. We do this by going to File>Options>Formulas and making sure that the Enable iterative calculations checkbox is clear.\n\n2) So in Cell A18, we enter the following formula:\n\n=COUNTBLANK (A5: A17)", null, "3) Upon pressing CTRL-ENTER, we get a value of 2 returned, so of the cells in the range, there were only two blank cells.", null, "### Using Conditional Formatting Example\n\n1) You can also use conditional formatting in order to highlight the blank cells.\n\n2) In order to do this select the range, you want to conditionally format and go to Home>Styles>Conditional Formatting.", null, "3) Go to Highlight Cells Rules>More rules…….", null, "4) In the New Formatting Rule dialog box, choose Format only cells that contain and then Format only cells with Blanks and choose a format, in this case, a blue fill. Click Ok.", null, "5) All the blank cells, in the range, selected now have a blue fill.", null, "And there you have it.\n\n## Conclusion\n\nIt is often necessary to count different types of values or use diverse criteria when counting. Excel provides five different counting functions that can be used on their own or in combination with other functions in order to create advanced formulas. While the Word options for adding, counting or summing values are less, Word does have a few functions that one can use to perform operations on Table data. Wildcards can also be used to extend functionality, especially when using the COUNTING Functions. The three wildcards *,? and ~ have to still be surrounded by quotation marks in formulas but when utilizing them in formulas, they allow one to encompass a greater degree of flexibility or counting options in this case.\n\nPlease feel free to comment and tell us how you utilize counting functions, in your spreadsheets or Word documents." ]

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[ "{{ message }}\n\nInstantly share code, notes, and snippets.\n\n# mblondel/lda_gibbs.py\n\nLast active Aug 10, 2021\nLatent Dirichlet Allocation with Gibbs sampler\nThis file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters\n \"\"\" (C) Mathieu Blondel - 2010 License: BSD 3 clause Implementation of the collapsed Gibbs sampler for Latent Dirichlet Allocation, as described in Finding scientifc topics (Griffiths and Steyvers) \"\"\" import numpy as np import scipy as sp from scipy.special import gammaln def sample_index(p): \"\"\" Sample from the Multinomial distribution and return the sample index. \"\"\" return np.random.multinomial(1,p).argmax() def word_indices(vec): \"\"\" Turn a document vector of size vocab_size to a sequence of word indices. The word indices are between 0 and vocab_size-1. The sequence length is equal to the document length. \"\"\" for idx in vec.nonzero(): for i in xrange(int(vec[idx])): yield idx def log_multi_beta(alpha, K=None): \"\"\" Logarithm of the multinomial beta function. \"\"\" if K is None: # alpha is assumed to be a vector return np.sum(gammaln(alpha)) - gammaln(np.sum(alpha)) else: # alpha is assumed to be a scalar return K * gammaln(alpha) - gammaln(K*alpha) class LdaSampler(object): def __init__(self, n_topics, alpha=0.1, beta=0.1): \"\"\" n_topics: desired number of topics alpha: a scalar (FIXME: accept vector of size n_topics) beta: a scalar (FIME: accept vector of size vocab_size) \"\"\" self.n_topics = n_topics self.alpha = alpha self.beta = beta def _initialize(self, matrix): n_docs, vocab_size = matrix.shape # number of times document m and topic z co-occur self.nmz = np.zeros((n_docs, self.n_topics)) # number of times topic z and word w co-occur self.nzw = np.zeros((self.n_topics, vocab_size)) self.nm = np.zeros(n_docs) self.nz = np.zeros(self.n_topics) self.topics = {} for m in xrange(n_docs): # i is a number between 0 and doc_length-1 # w is a number between 0 and vocab_size-1 for i, w in enumerate(word_indices(matrix[m, :])): # choose an arbitrary topic as first topic for word i z = np.random.randint(self.n_topics) self.nmz[m,z] += 1 self.nm[m] += 1 self.nzw[z,w] += 1 self.nz[z] += 1 self.topics[(m,i)] = z def _conditional_distribution(self, m, w): \"\"\" Conditional distribution (vector of size n_topics). \"\"\" vocab_size = self.nzw.shape left = (self.nzw[:,w] + self.beta) / \\ (self.nz + self.beta * vocab_size) right = (self.nmz[m,:] + self.alpha) / \\ (self.nm[m] + self.alpha * self.n_topics) p_z = left * right # normalize to obtain probabilities p_z /= np.sum(p_z) return p_z def loglikelihood(self): \"\"\" Compute the likelihood that the model generated the data. \"\"\" vocab_size = self.nzw.shape n_docs = self.nmz.shape lik = 0 for z in xrange(self.n_topics): lik += log_multi_beta(self.nzw[z,:]+self.beta) lik -= log_multi_beta(self.beta, vocab_size) for m in xrange(n_docs): lik += log_multi_beta(self.nmz[m,:]+self.alpha) lik -= log_multi_beta(self.alpha, self.n_topics) return lik def phi(self): \"\"\" Compute phi = p(w|z). \"\"\" V = self.nzw.shape num = self.nzw + self.beta num /= np.sum(num, axis=1)[:, np.newaxis] return num def run(self, matrix, maxiter=30): \"\"\" Run the Gibbs sampler. \"\"\" n_docs, vocab_size = matrix.shape self._initialize(matrix) for it in xrange(maxiter): for m in xrange(n_docs): for i, w in enumerate(word_indices(matrix[m, :])): z = self.topics[(m,i)] self.nmz[m,z] -= 1 self.nm[m] -= 1 self.nzw[z,w] -= 1 self.nz[z] -= 1 p_z = self._conditional_distribution(m, w) z = sample_index(p_z) self.nmz[m,z] += 1 self.nm[m] += 1 self.nzw[z,w] += 1 self.nz[z] += 1 self.topics[(m,i)] = z # FIXME: burn-in and lag! yield self.phi() if __name__ == \"__main__\": import os import shutil N_TOPICS = 10 DOCUMENT_LENGTH = 100 FOLDER = \"topicimg\" def vertical_topic(width, topic_index, document_length): \"\"\" Generate a topic whose words form a vertical bar. \"\"\" m = np.zeros((width, width)) m[:, topic_index] = int(document_length / width) return m.flatten() def horizontal_topic(width, topic_index, document_length): \"\"\" Generate a topic whose words form a horizontal bar. \"\"\" m = np.zeros((width, width)) m[topic_index, :] = int(document_length / width) return m.flatten() def save_document_image(filename, doc, zoom=2): \"\"\" Save document as an image. doc must be a square matrix \"\"\" height, width = doc.shape zoom = np.ones((width*zoom, width*zoom)) # imsave scales pixels between 0 and 255 automatically sp.misc.imsave(filename, np.kron(doc, zoom)) def gen_word_distribution(n_topics, document_length): \"\"\" Generate a word distribution for each of the n_topics. \"\"\" width = n_topics / 2 vocab_size = width ** 2 m = np.zeros((n_topics, vocab_size)) for k in range(width): m[k,:] = vertical_topic(width, k, document_length) for k in range(width): m[k+width,:] = horizontal_topic(width, k, document_length) m /= m.sum(axis=1)[:, np.newaxis] # turn counts into probabilities return m def gen_document(word_dist, n_topics, vocab_size, length=DOCUMENT_LENGTH, alpha=0.1): \"\"\" Generate a document: 1) Sample topic proportions from the Dirichlet distribution. 2) Sample a topic index from the Multinomial with the topic proportions from 1). 3) Sample a word from the Multinomial corresponding to the topic index from 2). 4) Go to 2) if need another word. \"\"\" theta = np.random.mtrand.dirichlet([alpha] * n_topics) v = np.zeros(vocab_size) for n in range(length): z = sample_index(theta) w = sample_index(word_dist[z,:]) v[w] += 1 return v def gen_documents(word_dist, n_topics, vocab_size, n=500): \"\"\" Generate a document-term matrix. \"\"\" m = np.zeros((n, vocab_size)) for i in xrange(n): m[i, :] = gen_document(word_dist, n_topics, vocab_size) return m if os.path.exists(FOLDER): shutil.rmtree(FOLDER) os.mkdir(FOLDER) width = N_TOPICS / 2 vocab_size = width ** 2 word_dist = gen_word_distribution(N_TOPICS, DOCUMENT_LENGTH) matrix = gen_documents(word_dist, N_TOPICS, vocab_size) sampler = LdaSampler(N_TOPICS) for it, phi in enumerate(sampler.run(matrix)): print \"Iteration\", it print \"Likelihood\", sampler.loglikelihood() if it % 5 == 0: for z in range(N_TOPICS): save_document_image(\"topicimg/topic%d-%d.png\" % (it,z), phi[z,:].reshape(width,-1))\n\n### cdfox commented May 28, 2012\n\nI don't think numpy/scipy are making this code slower rather faster. I did a quick test and found that a pure python implementation of sampling from a multinomial distribution with 1 trial (i.e. a discrete distribution)\n\n```import random\ndef draw(p):\nr = random.random()\nfor i in range(len(p)):\nr = r - p[i]\nif r < 0:\nreturn i\nreturn len(p) - 1```\n\nis much faster than using numpy.random's multinomial function\n\n```def draw(p):\nreturn np.random.multinomial(1,p).argmax()```\n\n### flaviotruzzi commented Dec 12, 2012\n\nI'll try to make a cython version when I have time.\n\n### marcmaxson commented Apr 3, 2013\n\nNot sure why the import failed to include the misc subfolder, but I was able to get the demonstration running with this tweak to save_document_image():\n\n``````def save_document_image(filename, doc, zoom=2):\n\"\"\"\nSave document as an image.\n\ndoc must be a square matrix\n\"\"\"\nheight, width = doc.shape\nzoom = np.ones((width*zoom, width*zoom))\n# imsave scales pixels between 0 and 255 automatically\nsp.misc.imsave(filename, np.kron(doc, zoom))\ntry:\nsp.misc.imsave(filename, np.kron(doc, zoom))\nexcept:\nimport scipy.misc\nscipy.misc.imsave(filename, np.kron(doc, zoom))\n``````\n\n### svanschalkwyk commented Apr 18, 2013\n\n@marc - still not able to find misc even though it is imported as part of sp (have verified). Is sp overwritten somewhere? Wil llook when I have time.\n@Mathiue - thanks for the code!\n\n### whille commented May 15, 2013\n\nhi @Mathiue, here const doc_length is used for training samples. According to Gregor's Parameter estimation for text analysis, Fig.7 defination, Nm(doc length) should follow Poisson distribution, E=λ.\nso in line 210, const number should better modified to\n\n``````for n in np.random.poisson(length, DOC_NUM):\n``````\n\n### elplatt commented Nov 24, 2013\n\nThanks for posting this. I'd like to use this code for a project. If that's ok, can you add a license?\n\n### corydolphin commented Dec 12, 2013\n\nQuick note to anyone struggling with the scipy.misc.imsave import, you need to have PIL installed for this import to work. Python dependency management is crazy!\n\n### jnothman commented May 10, 2014\n\nRe @cdfox's comment, you're much better off doing a `searchsorted(cumsum(p), rand())` than `np.random.multinomial(1,p).argmax()` which is efficient only when you're taking a large sample.\n\n### ChangUk commented Jun 9, 2014\n\nI implemented Gibbs sampler for standard LDA inference. My program updates alpha(vector) and beta(scalar) during the iterative sampling process by using Minka's fixed-point iteration.\nVisit here: https://gist.github.com/ChangUk/a741e0ccf5737954956e" ]

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[ "# Risk neutral pricing - Example from a book is correct?\n\nI found the following example in a book on Model Risk, while trying to explain how risk-neutral pricing takes properly into account the risk involved in different investments. The Example is this.\n\nSuppose you have a riskless asset ( a bond, say) promising you a riskless return of 50%. So, take a risk free interest rate $r=0.5$. Then, you have a security, which, under the actual probability measure can go up with probability $p=0.5$ and make a return of $125\\%$, or can go down (with probability $q=0.5$) yielding a return of $-25\\%$.\n\nUnder the actual probability the expected return of the security is then equal to $\\frac{1}{2}125\\%-\\frac{1}{2}25\\%=50\\%$. On the other hand, this is the same return promised by the bond, so the two investments produce the same expected return (using the actual probability measure, that is).\n\nThe book goes on to say that, however, the two investments are not the same because in the security's case it is present a risk that is not present in the bond case and, unless you are a special investor, you will prefer the riskless bond yielding the same expected return. That is the reason why, the book adds, you need to introduce a different measure than the actual one, the risk-neutral measure, that takes into account the presence of risk in the investments, and you shall take the expectation of the return of the risky asset with respect to this measure, and not the actual measure, in order to compare this investment to the riskless investment.\n\nNow all this makes perfect sense to me, however, if I understand well, the book implies that the two investments are not the same because in the risk neutral measure the expected return of the risky asset shall be less than the $50\\%$ arising from the actual probability.\n\nHowever, after some calculation, I realised that this is not the case: it appears that the risk-neutral probabilities $\\tilde{p}, \\tilde{q}$ in this case are the same as the actual probabilities postulated by the example, that is: $\\tilde{p}=\\tilde{q}=\\frac{1}{2}$, as it is easy to see by letting $r=0.5$, $d=0.75$ and $u=2.25$, and applying the well-known equations for the one-period model.\n\nSo, in fact one obtains that the risky asset has the same return as the bond (i.e., $50\\%$) also under the risk-neutral probability. Therefore, I have two questions:\n\n(1) Am I right to claim that, from the risk-neutral perspective, the two investments are actually equivalent (contrary to what the book seems to claim)? Or am I missing something?\n\n(2) Is the above example ill-chosen to illustrate the motivation behind risk-neutrality?\n\n• What is that book? – Robert Jul 29 '16 at 11:55\n• Morini, Understanding and Managing Model Risk. Example is on p. 24. So what is the answer in your opinion? – RandomGuy Jul 29 '16 at 13:25\n• I don't think you could use a model for derivatives as a general model. The author is saying that a \"normal\" investor will adjust the discount rate by risk, so the security will have a discount rate greater than 50% (rf + risk premium) and then, considering positive risk premiums, less value in t=0. – Robert Jul 29 '16 at 14:07\n• I disagree. The author clearly sates this example in the context of motivating the concept of risk-neutral measure as opposed to actual market measure. So it is the author himself that introduces the notion of risk-neutral measure in the context of a market with only a security and a bond (and no derivative at all). – RandomGuy Jul 29 '16 at 14:19\n• \"In order to avoid this we consider a special expectation, the risk-adjusted or risk-neutral one\". \"There may be different ways of taking risk into account\" (final of p.24). So the author is not saying how the \"risk-neutral probabilities\" are computed because those are specific for each assets class? – Robert Jul 29 '16 at 14:32" ]

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[ "# CodeSport\n\n0\n3249", null, "", null, "Welcome to CodeSport! In this month’s column, we feature a medley of questions covering C/C++, data structures, algorithms and operating systems.\n\nWe close the year by featuring a bunch of questions based on the various topics we discussed in this column over the last 12 months. These questions require you to apply some of the concepts we covered through 2010 and should help readers in preparing for software interviews. Let’s get started with some easy questions first.\n\n1. Given a stack data structure that allows these operations: push, pop, is_empty, and peek (returns the value of the topmost element without popping it), you are asked to write a code fragment to sort the elements of the stack in descending order. In other words, once the stack is sorted, the stack top should contain the maximum element, the element next to the top should be the second maximum and so on. Assume that the stack contains N integers. What is the complexity of your algorithm?\n2. Given a directed graph G(V, E) where V is the set of vertices and E is the set of edges, you are asked to find if there exists a path between two given vertices (x,y). What would be the complexity of your solution? If you are given a set of vertex pairs and asked to find if there exists a path for each pair of vertices in the set, what would your approach to the problem be?\n3. Given a DFS representation of a binary tree, with each vertex and its associated DFS number represented as (v,n) where v is the vertex ID and n is the corresponding DFS number of the vertex v, write an algorithm to give a BFS representation of the same binary tree. Assume that we number the root of the tree with the DFS number 1.\n4. Given an undirected graph G(V, E) where V is the set of vertices and E is the set of edges, give an algorithm to determine whether the graph is 2-colourable. (By 2-colourable, we mean that the graph vertices can be coloured with 2 colours such that no two adjacent vertices have the same colour.)\n5. Given a set S of N integers, write an algorithm to determine all the subsets of S. What is the complexity of your solution? If you are asked to find all the k-subsets (subsets with k members), what would your approach be?\n6. Given a two-dimensional matrix (mxn) of integer elements, in which each row of the matrix and each column of the matrix is already sorted, give an algorithm to find if a number x is present in the matrix.\n7. Write a small code snippet to find whether the system stack grows upward or downward on your system.\n8. Given a set of cities, and the distances between the cities, you are asked to visit all the cities as part of a tour. Write a program which will ensure that you visit all cities, but travel the minimum number of miles overall in the course of your journey. What is the complexity of your solution?\n9. Given an undirected graph G(V,E) where V is the set of vertices and E is the set of edges, write a program to determine whether it is a tree.\n10. In C++ programs, when would you prefer to declare your class’ constructor, copy constructor and assignment operator as private functions?\n11. We know that malloc and free are standard library functions for allocating and de-allocating memory dynamically in C programs. When we use free, we only pass the pointer to the memory to be deallocated as a parameter, and not the size of the memory block to be deallocated. How does the free library call determine what size the block to be deallocated should be? Give your implementations for malloc and free with an explanation.\n13. Let’s suppose that you have written an application, and it has been working perfectly on a 32-bit platform. Now you are asked to port your application to a 64-bit platform. What kind of coding changes would be needed?\n14. You have written an application, and it has been working well on a little-endian platform. Now you are asked to port your application to a big-endian platform. Besides, your application also reads from files stored on disks, processes the data, and writes back data to disk in output files. What, if any, kind of coding changes would be needed?\n15. We all have heard of paging and virtual memory. Does the operating system maintain a page table per process, or is there only one page table for all processes in the system? Explain your answer.\n16. In a multiprocessor system with each processor having its own cache, how are the updates made by one processor to a location in its cache made visible to other processors? Does this require hardware support? If not, can you sketch what kind of software support would be needed from the operating system?\n17. Given two threads running on two different processors, Thread A running on processor 1 updates the global variable X first, and then the global variable Y. Can thread B running on processor 2 see the updated value of Y before it sees the updated value of X? Explain your answer.\n18. You are given three arrays A, B and C, each containing N distinct integers. You are asked to find out whether there exist three such integers a€A, b€B and c€C such that a + b + c = 0. What is the complexity of your solution? Would the complexity be the same if you are asked to repeat this problem for the two arrays, A and B only, such that there exist any two integers a€A and b€B for which a + b = 0?\n19. Can you write to a memory location that is not present in the processor cache? If yes, how? If not, why not?\n20. We all have used debuggers, and we know that we can set breakpoints to stop program execution at a particular program location. Can you explain how breakpoints are implemented? Does it require underlying hardware support?\n21. What is meant by privileged instructions? In a virtualised system (system virtual machine, such as Xen/KVM/Hyper V), how are privileged instructions handled?\n22. Given an array of N integers, which is mostly sorted, with very few elements being out of sort (the number of unsorted elements is k, where k << N), what sorting algorithm would you use to sort the array, and why?\n23. Given a 32-bit representation of an integer, write a code snippet which, given a bit position k, prints the current bit value at that location, and then flips it.\n24. What is the problem with the following code snippet?\n```int* bar(int size)\n{\nint* p = alloca(size);\nreturn p;\n}\nvoid foo(int n)\n{\nint* myspace = bar(n);\n*myspace = 10;\n..\n}```\n\n## My ‘must read book’ for this month\n\nThis month, our reader Srividya K, recommends the book Concurrent Programming in Java by Doug Lea. Srividya says that this book helped him understand and apply concurrent programming techniques effectively in developing software for commercial projects.\n\nHe also cautions readers that this is not a book for beginners to Java. While on the subject of concurrency, there is an excellent article that discusses how efficiently a lock-free concurrent hash table can be implemented. It is available at as a PDF. This is a good read not just for Java programmers, but for anyone writing concurrent code.\n\nIt is also worth reading Cliff Click’s presentation on writing scalable non-blocking code for Java [PDF]. The code corresponding to the data structures discussed in his article is available on Sourceforge.\n\nIf you have a favourite programming book that you think is a must-read for every programmer, please send me a note with the book’s name, and a short note on why you think it is useful. I will mention it in this column. This will help many readers who want to improve their coding skills.\n\nPlease send me your answers to this month’s questions. If you have any favourite programming puzzles that you would like to discuss on this forum, please send them to me, along with your solutions and feedback, at sandyasm_AT_yahoo_DOT_com. Till we meet again next month, happy programming and wishing you the very best!\n\nFeature image courtesy: Valerie Everett. reused under the terms of CC-BY-SA 2.0 License." ]

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[ "# An object with a mass of 2 kg is hanging from an axle with a radius of 2 cm. If the wheel attached to the axle has a radius of 16 cm, how much work would it take to turn the wheel a length equal to the circumference of the axle?\n\nFeb 25, 2017\n\nThe work done is $= 2.46 J$\n\n#### Explanation:\n\nMass $= 2 k g$\n\nRadius of axle $= 0.02 m$\n\nRadius of wheel $= 0.16 m$\n\nCicumference of axle $= 2 \\pi r = 2 \\cdot 0.02 \\cdot \\pi = 0.126 m$\n\nWork done $= m g h = 2 \\cdot 9.8 \\cdot 0.126 = 2.46 J$" ]

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[ "C. Idempotent functions\ntime limit per test\n1 second\nmemory limit per test\n256 megabytes\ninput\nstandard input\noutput\nstandard output\n\nSome time ago Leonid have known about idempotent functions. Idempotent function defined on a set {1, 2, ..., n} is such function", null, ", that for any", null, "the formula g(g(x)) = g(x) holds.\n\nLet's denote as f(k)(x) the function f applied k times to the value x. More formally, f(1)(x) = f(x), f(k)(x) = f(f(k - 1)(x)) for each k > 1.\n\nYou are given some function", null, ". Your task is to find minimum positive integer k such that function f(k)(x) is idempotent.\n\nInput\n\nIn the first line of the input there is a single integer n (1 ≤ n ≤ 200) — the size of function f domain.\n\nIn the second line follow f(1), f(2), ..., f(n) (1 ≤ f(i) ≤ n for each 1 ≤ i ≤ n), the values of a function.\n\nOutput\n\nOutput minimum k such that function f(k)(x) is idempotent.\n\nExamples\nInput\n4\n1 2 2 4\nOutput\n1\nInput\n3\n2 3 3\nOutput\n2\nInput\n3\n2 3 1\nOutput\n3\nNote\n\nIn the first sample test function f(x) = f(1)(x) is already idempotent since f(f(1)) = f(1) = 1, f(f(2)) = f(2) = 2, f(f(3)) = f(3) = 2, f(f(4)) = f(4) = 4.\n\nIn the second sample test:\n\n• function f(x) = f(1)(x) isn't idempotent because f(f(1)) = 3 but f(1) = 2;\n• function f(x) = f(2)(x) is idempotent since for any x it is true that f(2)(x) = 3, so it is also true that f(2)(f(2)(x)) = 3.\n\nIn the third sample test:\n\n• function f(x) = f(1)(x) isn't idempotent because f(f(1)) = 3 but f(1) = 2;\n• function f(f(x)) = f(2)(x) isn't idempotent because f(2)(f(2)(1)) = 2 but f(2)(1) = 3;\n• function f(f(f(x))) = f(3)(x) is idempotent since it is identity function: f(3)(x) = x for any", null, "meaning that the formula f(3)(f(3)(x)) = f(3)(x) also holds." ]

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[ "Home / Specific Heat Capacity Conversion / Convert Kilocalorie (th)/kilogram/K to Btu (IT)/pound/°C\n\n# Convert Kilocalorie (th)/kilogram/K to Btu (IT)/pound/°C\n\nPlease provide values below to convert kilocalorie (th)/kilogram/K to Btu (IT)/pound/°C, or vice versa.\n\n From: kilocalorie (th)/kilogram/K To: Btu (IT)/pound/°C\n\n### Kilocalorie (th)/kilogram/K to Btu (IT)/pound/°C Conversion Table\n\nKilocalorie (th)/kilogram/KBtu (IT)/pound/°C\n0.01 kilocalorie (th)/kilogram/K0.0179879622 Btu (IT)/pound/°C\n0.1 kilocalorie (th)/kilogram/K0.1798796217 Btu (IT)/pound/°C\n1 kilocalorie (th)/kilogram/K1.7987962166 Btu (IT)/pound/°C\n2 kilocalorie (th)/kilogram/K3.5975924331 Btu (IT)/pound/°C\n3 kilocalorie (th)/kilogram/K5.3963886497 Btu (IT)/pound/°C\n5 kilocalorie (th)/kilogram/K8.9939810828 Btu (IT)/pound/°C\n10 kilocalorie (th)/kilogram/K17.9879621656 Btu (IT)/pound/°C\n20 kilocalorie (th)/kilogram/K35.9759243312 Btu (IT)/pound/°C\n50 kilocalorie (th)/kilogram/K89.9398108279 Btu (IT)/pound/°C\n100 kilocalorie (th)/kilogram/K179.8796216558 Btu (IT)/pound/°C\n1000 kilocalorie (th)/kilogram/K1798.7962165576 Btu (IT)/pound/°C\n\n### How to Convert Kilocalorie (th)/kilogram/K to Btu (IT)/pound/°C\n\n1 kilocalorie (th)/kilogram/K = 1.7987962166 Btu (IT)/pound/°C\n1 Btu (IT)/pound/°C = 0.5559273423 kilocalorie (th)/kilogram/K\n\nExample: convert 15 kilocalorie (th)/kilogram/K to Btu (IT)/pound/°C:\n15 kilocalorie (th)/kilogram/K = 15 × 1.7987962166 Btu (IT)/pound/°C = 26.9819432484 Btu (IT)/pound/°C" ]

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[ "# Is jump time in a continuous-time Markov chain a stopping time?\n\nAssume $$V$$ is a countable state space and $$L:V^2 \\to \\mathbb R$$ the infinitesimal generator of a continuous-time Markov chain $$(X_t)_{t \\ge 0}$$ on the probability space $$(\\Omega, \\mathcal{G}, \\mathbb{P})$$. Then we can define such random variables as first passage time, holding times, and jump times.\n\nGiven $$\\omega \\in \\Omega$$, we define a sequence of random jump times $$(\\sigma_n)$$ recursively as follows:\n\n• First, let $$\\sigma_0 := 0$$.\n\n• Second, let $$i := X_{\\sigma_n} (\\omega) \\in V$$ and $$L(i) := - L(i,i)$$ where $$X_{\\sigma_n} (\\omega) := X_{\\sigma_n (\\omega)} (\\omega)$$. Then the time until transition out of state $$i$$ is $$\\sigma_{n+1} -\\sigma_{n} \\sim \\operatorname{Exp}(L(i))$$.\n\nIt's clear that the first passage time is a stopping time. I would like to ask if jump time is a stopping time too.\n\n• Do you mean the time when the jump occurs, or the increment in time between two jumps? In the case of the latter, it is definitely not a stopping time except when $n=0$, but the reason is kind of obvious (you're comparing apples to oranges, since the definition wants you to restrict to times measured relative to the start of the process but you are really measuring times relative to the arrival at the current state).\n– Ian\nMay 4 '20 at 15:49\n• That being said, the actual time of the jump itself is a stopping time provided that you've defined your conventions consistently (i.e. all the inequalities involved are nonstrict).\n– Ian\nMay 4 '20 at 15:49\n\nDo you mean the time when the jump occurs, or the increment in time between two jumps? In the case of the latter, it is definitely not a stopping time except when $$n=0$$, but the reason is kind of obvious (you're comparing apples to oranges, since the definition wants you to restrict to times measured relative to the start of the process but you are really measuring times relative to the arrival at the current state)." ]

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[ "# Geometry Definitions Worksheet Sine And Cosine Rule 1 Higher Maths Tutorial Graphing Math Solver Website Playground", null, "geometry definitions worksheet sine and cosine rule 1 higher maths tutorial graphing math solver website playground.\n\ngeometric mean worksheets geometry definitions with answers, rectangular prism math best images of volume worksheets cube, faces edges vertices worksheet shapes chart geometry geometric,angles parallel lines geometry tutorial high school review sheet ,definition of denote math scale on a graph worksheet geometry , geometry worksheets grade basic, geometry word problems worksheets grade area and perimeter, equation for mean math, downloadable geometry worksheets for graders worksheet fun grade, 0 product property math elementary algebra practice exam solutions." ]

[ null, "http://meetamontrealer.com/wp-content/uploads/2018/11/geometry-definitions-worksheet-sine-and-cosine-rule-1-higher-maths-tutorial-graphing-math-solver-website-playground.jpg", null ]

[ "Want to Enable Key Features?\n\n# Convert Pound to Atomic Mass Unit (lb to amu)\n\nIn next fields, kindly type your value in the text box under title [ From: ] to convert from pound to atomic mass unit (lb to amu). As you type your value, the answer will be automatically calculated and displayed in the text box under title [ To: ].\n\nFrom:\nTo:\nDefinitions:\n\nPound (abbreviations: lb, or lbs, or ps): is a unit of force used in some systems of measurement including English Engineering units and the British Gravitational System. Also, it is a unit of mass used in the imperial, United States customary and other systems of measurement.\n\nAtomic Mass Unit (abbreviations: amu, or u): also known as dalton (symbol: Da), is a standard unit of mass that used on an atomic or molecular scale (atomic mass). One unified atomic mass unit is about the mass of one nucleon (either a single proton or neutron) and equivalent to 1 g/mol.\n\n## How to Convert Pounds to Atomic Mass Units\n\n### Example: How many atomic mass units are equivalent to 72.23 pounds?\n\nAs;\n\n1 pounds = 2.732E+26 atomic mass units\n\n72.23 pounds = Y atomic mass units\n\nAssuming Y is the answer, and by criss-cross principle;\n\nY equals 72.23 times 2.732E+26 over 1\n\n(i.e.) Y = 72.23 * 2.732E+26 / 1 = 1.9733236E+28 atomic mass units\n\nAnswer is: 1.9733236E+28 atomic mass units are equivalent to 72.23 pounds.\n\n### Practice Question: Convert the following units into amu:\n\nN.B.: After working out the answer to each of the next questions, click adjacent button to see the correct answer.\n\n( i ) 20.64 lb\n\n( ii ) 70.42 lb\n\n( iii ) 32.3 lb\n\nReferences\n•", null, "•", null, "•", null, "Ask questions and Share knowledge with Community\n\nFind below recent posts for automation solutions with questions and answers by community. You can search in past threads or post new question about your assignment with detailed description, and always could mark your question as request. Sharing knowledge are highly appreciated by answering on others questions, and in return awards will be decided.\n\n#\nRecent Questions\n*\ni need to know how to convert 723,413,000 sq mm into mm2.\n110\n1\n0\n*\nHow to convert kg/m3 to kn/m3\n107\n1\n0\n\nView More\n\nArticle Categories\n× Close\nResults:" ]

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[ "◄ Exodus 36:21 ►\nModern Translations\nNew International Version\nEach frame was ten cubits long and a cubit and a half wide,\n\nNew Living Translation\nEach frame was 15 feet high and 27 inches wide,\n\nEnglish Standard Version\nTen cubits was the length of a frame, and a cubit and a half the breadth of each frame.\n\nBerean Study Bible\nEach frame was ten cubits long and a cubit and a half wide.\n\nNew American Standard Bible\nTen cubits was the length of each board, and one and a half cubits the width of each board.\n\nNASB 1995\nTen cubits was the length of each board and one and a half cubits the width of each board.\n\nNASB 1977\nTen cubits was the length of each board, and one and a half cubits the width of each board.\n\nAmplified Bible\nEach board was ten cubits long and one and a half cubits wide.\n\nChristian Standard Bible\nEach support was fifteen feet long and twenty-seven inches wide.\n\nHolman Christian Standard Bible\nThe length of each plank was 15 feet, and the width of each was 27 inches.\n\nContemporary English Version\nEach frame was 4 meters high and 66 centimeters wide\n\nGood News Translation\nEach frame was 15 feet tall and 27 inches wide,\n\nGOD'S WORD® Translation\nEach frame was 15 feet long and 27 inches wide,\n\nInternational Standard Version\nEach board was ten cubits long, and one and a half cubits wide.\n\nNET Bible\nThe length of each frame was fifteen feet, the width of each frame was two and a quarter feet,\nClassic Translations\nKing James Bible\nThe length of a board was ten cubits, and the breadth of a board one cubit and a half.\n\nNew King James Version\nThe length of each board was ten cubits, and the width of each board a cubit and a half.\n\nKing James 2000 Bible\nThe length of a board was ten cubits, and the breadth of a board one cubit and a half.\n\nNew Heart English Bible\nSeventeen feet three inches was the length of a board, and two feet seven inches the breadth of each board.\n\nWorld English Bible\nTen cubits was the length of a board, and a cubit and a half the breadth of each board.\n\nAmerican King James Version\nThe length of a board was ten cubits, and the breadth of a board one cubit and a half.\n\nAmerican Standard Version\nTen cubits was the length of a board, and a cubit and a half the breadth of each board.\n\nA Faithful Version\nThe length of a board was ten cubits, and the breadth of a board was one and a half cubits.\n\nDarby Bible Translation\nten cubits the length of the boards, and one cubit and a half the breadth of one board;\n\nEnglish Revised Version\nTen cubits was the length of a board, and a cubit and a half the breadth of each board.\n\nWebster's Bible Translation\nThe length of a board was ten cubits, and the breadth of a board one cubit and a half.\n\nEarly Modern\nGeneva Bible of 1587\nThe length of a board was ten cubites, and the bredth of one board was a cubite, & an halfe.\n\nBishops' Bible of 1568\nThe length of a boorde was ten cubites, the breadth one cubite & a halfe.\n\nCoverdale Bible of 1535\neuery one ten cubytes longe, and a cubyte and a half brode,\n\nTyndale Bible of 1526\nvpright euery borde .x. cubetes longe and a cubet ad an halfe brode.\nLiteral Translations\nLiteral Standard Version\nten cubits [is] the length of the [one] board, and a cubit and a half the breadth of the [one] board;\n\nYoung's Literal Translation\nten cubits is the length of the one board, and a cubit and a half the breadth of the one board;\n\nSmith's Literal Translation\nTen cubits the length of the board, and a cubit and a half a cubit, the breadth of the one board.\n\nCatholic Translations\nDouay-Rheims Bible\nThe length of one board was ten cubits: and the breadth was one cubit and a half.\n\nCatholic Public Domain Version\nTen cubits was the length of one panel, and one and one half cubits comprised the width.\n\nTranslations from Aramaic\nPeshitta Holy Bible Translated\nTen cubits, the length of one board, and a cubit and half a cubit, the width of one board.\n\nLamsa Bible\nThe length of each board was ten cubits, and the breadth of each board one cubit and a half.\n\nOT Translations\nJPS Tanakh 1917\nTen cubits was the length of a board, and a cubit and a half the breadth of each board." ]

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[ "Search a number\nBaseRepresentation\nbin1001001010111…\n…01110010011111\n312100202110101221\n410211131302133\n5124143231312\n611344514211\n71622616214\noct445356237\n9170673357\n1076930207\n113a474892\n122191b967\n1312c27043\n14a307b0b\n156b49207\nhex495dc9f\n\n76930207 has 2 divisors, whose sum is σ = 76930208. Its totient is φ = 76930206.\n\nThe previous prime is 76930201. The next prime is 76930223. The reversal of 76930207 is 70203967.\n\nIt is a weak prime.\n\nIt is an emirp because it is prime and its reverse (70203967) is a distict prime.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 76930207 - 23 = 76930199 is a prime.\n\nIt is a congruent number.\n\nIt is not a weakly prime, because it can be changed into another prime (76930201) by changing a digit.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 38465103 + 38465104.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (38465104).\n\nAlmost surely, 276930207 is an apocalyptic number.\n\n76930207 is a deficient number, since it is larger than the sum of its proper divisors (1).\n\n76930207 is an equidigital number, since it uses as much as digits as its factorization.\n\n76930207 is an evil number, because the sum of its binary digits is even.\n\nThe product of its (nonzero) digits is 15876, while the sum is 34.\n\nThe square root of 76930207 is about 8770.9866605759. The cubic root of 76930207 is about 425.3035100736.\n\nIt can be divided in two parts, 769 and 30207, that added together give a square (30976 = 1762).\n\nThe spelling of 76930207 in words is \"seventy-six million, nine hundred thirty thousand, two hundred seven\"." ]

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[ "## Example Questions\n\n### Example Question #51 : Algebraic Functions\n\nTwo consecutive negative odd numbers have a product of 15. What is the sum of the two numbers?\n\n–6\n\n–12\n\n–8\n\n–4\n\n–10\n\n–8\n\nExplanation:\n\nDefine the variables x = first number and x + 2 = second number\n\nThe product is x(x + 2) = 15, so we need to solve the quadratic equation x2 + 2x – 15 = 0.  Factoring we get (x + 5)(x – 3) = 0, so x = –5 or x = 3.  The problem calls for negative numbers, so the answers are –5 and –3 and the sum is –8\n\n### Example Question #1 : How To Use The Quadratic Function\n\nWhat is the difference of the values of x that satisfy\n\n4x2+ 3x – 1 = 0?\n\n4/5\n\n3\n\n0\n\n5/4\n\n5/4\n\nExplanation:\n\nFactor the equation (4x – 1) (x + 1)\n\nThe solutions are x = –1 and x = 1/4.\n\nThe difference is (1/4) – (–1) = 5/4\n\n### Example Question #2702 : Act Math\n\n2x + y = 1\n\nWhat is the slope and the y-intercept of the given equation, respectively?\n\n1, 1\n\n1, 2\n\n2, 1\n\n–2, 1\n\n2, –2\n\n–2, 1\n\nExplanation:\n\nYou put the equation into slope-intercept form (y = mx b)\n\nThe new equation is y= –2x + 1\n\n= –2 and = 1\n\n### Example Question #61 : Algebraic Functions\n\nHow many solutions are there to: x– 9 = 0\n\n0\n\n1\n\nInfinitely many\n\n2", null, "" ]

[ null, "https://vt-vtwa-app-assets.varsitytutors.com/assets/problems/og_image_practice_problems-9cd7cd1b01009043c4576617bc620d0d5f9d58294f59b6d6556fd8365f7440cf.jpg", null ]

[ "×\nGet Full Access to Basic Statistics For Business And Economics - 7 Edition - Chapter 6 - Problem 53e\nGet Full Access to Basic Statistics For Business And Economics - 7 Edition - Chapter 6 - Problem 53e\n\n×\n\n# Recent crime reports indicate that 3.1 motor vehicle", null, "ISBN: 9780077384470 64\n\n## Solution for problem 53E Chapter 6\n\nBasic Statistics for Business and Economics | 7th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants", null, "Basic Statistics for Business and Economics | 7th Edition\n\n4 5 1 275 Reviews\n24\n4\nProblem 53E\n\nProblem 53E\n\nRecent crime reports indicate that 3.1 motor vehicle thefts occur each minute in the United States. Assume that the distribution of thefts per minute can be approximated by the Poisson probability distribution.\n\na. Calculate the probability exactly four thefts occur in a minute.\n\nb. What is the probability there are no thefts in a minute?\n\nc. What is the probability there is at least one theft in a minute?\n\nStep-by-Step Solution:\n\nSolution:\n\nStep 1 of 3:\n\nThe average number of motor vehicle thefts occur each minute in the United States is 3.1.\n\nThe distribution of thefts per minute can be approximated by the Poisson distribution.\n\n1. We have to find the probability that exactly four thefts occur in a minute.\n2. We have to find the probability there are no thefts occur in a minute.\n3. We have to find the probability that there is at least one theft in a minute.\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780077384470\n\nUnlock Textbook Solution" ]

[ null, "https://studysoup.com/cdn/24cover_2543290", null, "https://studysoup.com/cdn/24cover_2543290", null ]

[ "Question\n\n# Two balls are chosen randomly from an urn containing 6 red and 4 black balls, without...\n\nTwo balls are chosen randomly from an urn containing 6 red and 4 black balls, without replacement. Suppose that we win \\$2 for each black ball selected and we lose \\$1 for each red ball selected. Let X denote the amount on money we won or lost.\n(a) Find the probability mass function of X, i.e., find P(X = k) for all possible values of k.\n(b) Compute E[X].\n(c) Compute Var(X)\n\na) below is probability mass function of X :\n\nP(X=-2)=P(both red balls)=(6/10)*(5/9)=1/3\n\nP(X=1)=P(1red and 1 black ball)=2*(6/10)*(4/9)=8/15\n\nP(X=4)=P(both black balls)=(4/10)*(3/9)=2/15\n\nb)\n\n x P(x) xP(x) x2P(x) -2 1/3 -0.667 1.333 1 8/15 0.533 0.533 4 2/15 0.533 2.133 total 0.400 4.000 E(x) =μ= ΣxP(x) = 0.4000 E(x2) = Σx2P(x) = 4.0000 Var(x)=σ2 = E(x2)-(E(x))2= 3.840\n\nE(X)=0.400\n\nc)Var(X)= 3.840\n\n#### Earn Coins\n\nCoins can be redeemed for fabulous gifts.\n\n##### Need Online Homework Help?\n\nMost questions answered within 1 hours.\n\n##### Active Questions\n• In a recent baseball season, Bob hit a home run approximately once every 18.38 plate appearances....\n• Need to determine the model equation for the conservation of energy it should have 5 terms...\n• Questions 5-15 are based on the following Ricardian model. Marginal Product of Labor                           Labor Force...\n• Construct a circuit for a 1 bit ALU in Logisim ,which can do AND,.OR,NOT, Addition, Subtraction.Use...\n• Vector A is directed along the positive y-axis and has magnitude √(3.0) units. Vector B is...\n• A random variable follows the continuous uniform distribution between 45 and 135. ​a) Calculate the probability..." ]

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[ "# FDTD code that can compare different absorbing boundary conditions\n\nStatus\nNot open for further replies.\n\n##### Member level 3", null, "Hello,\nI am looking for an FDTD code 2D or 3D that can compare different\nabsorbing boundary condiditons\n\nthanks\n\n#### junction_crosstalk\n\n##### Junior Member level 2", null, "you can try CST Microwave Studio\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### peter-gr\n\n##### Member level 1", null, "FDTD code\n\nA simple 1-D FDTD code in matlab with ABC is given below.\n\n% 1-D FDTD code\nKE=200 ;%KE is he number of cells to be used\nex=zeros(1,KE);\nhy=zeros(1,KE);\nex_low_m1=0;\nex_low_m2=0;\nex_high_m1=0;\nex_high_m2=0;\n\nkc=KE/2;\n%Gaussian pulse\nt0=40;\nn=input('T=?');\nfor T=1:n\n%Main FDTD Loop\n%Ex field\n\nfor k=2:KE\nex(k)=ex(k) + 0.5*(hy(k-1)-hy(k));\nend\n\n%Gaussian pulse in the middle\n\nex(kc)=pulse;\n\n%Absorbing Boundary Conditions\n\nex(1)=ex_low_m2;\nex_low_m2=ex_low_m1;\nex_low_m1=ex(2);\n\nex(KE)=ex_high_m2;\nex_high_m2=ex_high_m1;\nex_high_m1=ex(KE-1);\n\n%Hy field\n\nfor k=1:KE-1\nhy(k)=hy(k)+ 0.5*(ex(k)-ex(k+1));\nend\n\nend\n\n%End of FDTD Loop\n\nplot(ex)\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### g86\n\n##### Full Member level 4", null, "Re: FDTD code\n\nNice piece ....\n\nCan you please teach me a bit? I have some small questions and I am not verey familier with FDTD.\n\n1. Can we replace the gaussian pulse? How?\n2. Can we MIX FDTD with some other method?\n\nthanks\n\n:!: :idea: :?:\n\npeter-gr said:\nA simple 1-D FDTD code in matlab with ABC is given below.\n\n% 1-D FDTD code\nKE=200 ;%KE is he number of cells to be used\nex=zeros(1,KE);\nhy=zeros(1,KE);\nex_low_m1=0;\nex_low_m2=0;\nex_high_m1=0;\nex_high_m2=0;\n\nkc=KE/2;\n%Gaussian pulse\nt0=40;\nn=input('T=?');\nfor T=1:n\n%Main FDTD Loop\n%Ex field\n\nfor k=2:KE\nex(k)=ex(k) + 0.5*(hy(k-1)-hy(k));\nend\n\n%Gaussian pulse in the middle\n\nex(kc)=pulse;\n\n%Absorbing Boundary Conditions\n\nex(1)=ex_low_m2;\nex_low_m2=ex_low_m1;\nex_low_m1=ex(2);\n\nex(KE)=ex_high_m2;\nex_high_m2=ex_high_m1;\nex_high_m1=ex(KE-1);\n\n%Hy field\n\nfor k=1:KE-1\nhy(k)=hy(k)+ 0.5*(ex(k)-ex(k+1));\nend\n\nend\n\n%End of FDTD Loop\n\nplot(ex)\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### loucy", null, "Re: FDTD code\n\n1. I think you can use any wave-form. How to do it would depend on the software you are using. An example would be providing a data file specifying the amplitude of pusle at a series of time steps. Now the accuracy of the result and the time to convergence would be affected by the wave-form. If your source runs for 10000 time steps before reaching \"practical\" zero, you would check carefully whether the output has stablized much earlier (say 5000). If the simulation is terminated too early, you have error. You would also think about numerical dispersion...\n\n2. Obviously you can mix FDTD with other method. Some commercial codes (e.g. WASP-net) does that. From a theoretical point of view, those methods are used to solve some differential equations, and you could appy equivalent principle to decompose a problem domain into subdomains and apply different methods in each of them.\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### peter-gr\n\n##### Member level 1", null, "Re: FDTD code\n\nOf course you can use a pulse other than gaussian.For example you can use a sinusoidal pulse.\n\ngood luck.\n\npeter-gr.\n\n#### peter-gr\n\n##### Member level 1", null, "Re: FDTD code\n\nIt is very easy to switch to switch to a sinusoidal source.Just replace the parameter pulse with the following:\n\npulse=sin(2*pi*freq_in*dt*T);\n\nThe parameter freq_in determines the frequency of the wave.\n\n#### danielwood\n\n##### Newbie level 5", null, "Re: FDTD code\n\nMaybe you can find the 2d and 3d code in the book by Taflove.\nComputational Electrodynamics: The Finite-Difference Time-Domain Method by Allen Taflove, Susan C. Hagness (Hardcover - June 2000)\n\nOr You can find another book\nElectromagnetic Simulation Using the FDTD Method by Dennis M. Sullivan (Author) (Hardcover)\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### yingyang\n\n##### Full Member level 2", null, "#### yingyang\n\n##### Full Member level 2", null, "#### fgc\n\n##### Newbie level 5", null, "FDTD code\n\nI need a 3D FDTD code implementing a point source in a free-space mesh with Mur absorbing boundaries. Would help me?\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### peter-gr\n\n##### Member level 1", null, "Re: FDTD code\n\nThe book of Dennis Sullivan gives the programs you want for 3D FDTD simulation.\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### tangvu\n\n##### Newbie level 4", null, "Re: FDTD code\n\nin the matlab source code above, when it is asking for an input of \"T\" what does that do? help is appreciated.\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### peter-gr\n\n##### Member level 1", null, "Re: FDTD code\n\nThe T indicates the time-steps.(the time the pulse will last: T=δt*n)\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### tangvu\n\n##### Newbie level 4", null, "Re: FDTD code\n\ncan someone explain the absorbing boundary condition? i'm still a bit lost on the algorithm. thanks.\n\n#### peter-gr\n\n##### Member level 1", null, "Re: FDTD code\n\nAbsorbing Boundary Conditions are necessary to keep outgoing H and E fields from being reflected back into the problem space. Normally, in calculating the E field, we know the surrounding H values;this is a fundamental assumption in the FDTD method.At the edge of the problem space we will not have the value to one side.\nSuppose we are looking for a boundary condition in the end.If the wave is going toward the boundary condition in free space, it is travelling at c, the speed of light.So in one time step of the FDTD algorithm,it travels:\ndistance=c*dt=c*(dx/2c)=dx/2\n\nThe above equation explains that it takes two time steps for a wave to cross one cell.So a common sense approach tells as that a common boundary condition would be Ex", null, "(0)=Ex(n-2)(1)\nIt is relativeley easy to implement this.Store a value of Ex(1) for two time steps and then put it in Ex(0).\n\n•", null, "labbani\n\n### labbani\n\npoints: 2\n\n#### nguyenduyvy\n\n##### Newbie level 1", null, "Re: FDTD code\n\nYes we can see this from the book of Sullivan. But, in the code you have pasted above, how were the time indexes \"n\" and \"n-2\" are presented, i.e. in\n\" ex(1) = ex_low_m2\nex_low_m2 = ex_low_m1\nex_low_m1 = ex(2) \"\nI saw just that ex(1) = ex(2).\nSorry for my zombie question. Thanks!\nAbsorbing Boundary ... So in one time step of the FDTD algorithm,it travels:\ndistance=c*dt=c*(dx/2c)=dx/2\n\nThe above equation explains that it takes two time steps for a wave to cross one cell.So a common sense approach tells as that a common boundary condition would be Ex", null, "(0)=Ex(n-2)(1)\nIt is relativeley easy to implement this.Store a value of Ex(1) for two time steps and then put it in Ex(0).\n\nLast edited:\n\nStatus\nNot open for further replies." ]

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[ "", null, "Chapter 8 Sets and Probabilities. 8.1 SETS The use of braces: { } Element (member) of a set, ,  Empty set  Distinguish: 0,  and {0} Equality of sets.\n\nPresentation on theme: \"Chapter 8 Sets and Probabilities. 8.1 SETS The use of braces: { } Element (member) of a set, ,  Empty set  Distinguish: 0,  and {0} Equality of sets.\"— Presentation transcript:\n\nChapter 8 Sets and Probabilities\n\n8.1 SETS The use of braces: { } Element (member) of a set, ,  Empty set  Distinguish: 0,  and {0} Equality of sets Set-builder notation: { x  x has property P } Universal set U\n\nSubset A set A is a SUBSET of a set B (written A  B) if every element of A is also an element of B. Proper subset For any set A:   A and A  A. {a, b} has 4 subsets, {a, b, c} has 8 subsets. Number of subsets: If set A has n elements, then A has 2 n subsets.\n\nVENN diagrams A B C U\n\nCOMPLEMENT Let A and B be any sets with U the universal set. Then: The complement of A, written A’, is A’ = { x  x  A and x  U } Example: U = {1, 2, 3, 4, 5}, A = {1, 3, 5} A’ = ? A A’\n\nINTERSECTION The intersection of A and B is: A  B = { x  x  A and x  B } Example: A = {1, 2, 3, 4, 5} B = {1, 3, 5} A  B = ? U AB\n\nDISJOINT SETS A and B are DISJOINT sets if A  B =  A B U\n\nUNION The union of two sets A and B is: A  B = { x  x  A or x  B or both} AB U\n\nHow to read set expressions x  A x  A A  B A  B A  B A  B A \\ B A  B A’ x belongs to A / x is an element of A x does not belong to A / x is not an element of A A is contained in B / A is a subset of B A contains B / B is a subset of A A cap B / A meet B / A intersection B A cup B / A join B / A union B A minus B / the difference between A and B A cross B / the cartesian product of A and B A prime\n\n8.2 APPLICATIONS OF VENN DIAGRAMS Example 1 (p. 452). Shade the region representing the sets: A’  B A’  B’\n\n8.2 APPLICATIONS OF VENN DIAGRAMS Example 2 (p. 453). Shade the region A’  (B  C’)\n\n8.2 APPLICATIONS OF VENN DIAGRAMS Example 3 (p. 454): (cont.) (a) How many students read none of the publications? (b) How many read only Fortune? (c) How many read Business Week and the Journal, but not Fortune?\n\nADDITION RULE FOR COUNTING Denote n(X): number of elements in X n(A  B) = n(A) + n(B) – n(A  B) Example 5 (p. 457): A group of 10 students meets to plan a school function. All are majoring in accounting or economics or both. Five of the students are economics major and 7 are in accounting major. How many students major in both subjects?\n\nADDITION RULE FOR COUNTING Example 6 (p. 457): Below is the result of American reading habit: (a)Find n(G  B) (b)Find n(G  B) (c)Find n( (A  C)  H’) A None B 1-5 C 6-10 D 11-50 E 51+ F No answer Total GHGH 2005 2002 16 18 38 31 14 15 25 27 6868 1111 100\n\n8.3 PROBABILITY Experiment: activity or occurrence with an observable result Trial: each repetition of an experiment. Outcomes: possible results of each trial Sample space: set of all possible outcomes for an experiment. –Example: tossing a coin, rolling a die.\n\nSample space and event Event: any subset of a sample space Example: rolling a die, Sample space S = {1, 2, 3, 4, 5, 6} Determine the following events: –The die shows an even number: E 1 – The die shows a 1: E 2 – The die shows a number less than 5: E 3 – The die shows a multiple of 3: E 4\n\nSample space and event Event: any subset of a sample space –Simple event: event with only one possible outcome –Certain event: event that equals the sample space –Impossible event: empty set Example: rolling a die, S = {1, 2, 3, 4, 5, 6} – Simple event: The die shows a 4, E = {4} – Certain event: The die shows a number less than 10, E = S – Impossible event: The die shows a 7, E = \n\nSet operations for events Let E and F be events for a sample space S. Then: –Event E  F occurs when both E and F occur; –Event E  F occurs when E or F or both occur –Event E’ occurs when E does not occur – If E  F = , E and F are disjoint events (or mutually exclusive events) Example: rolling a die, S = {1, 2, 3, 4, 5, 6} – E: The die shows an even number – F: The die shows a number greater than 2 – Determine the following events: E  F, E  F, E’, E  F’\n\nBASIC PROBABILITY PRINCIPLE (for spaces with equally likely outcomes) Suppose event E is a subset of a sample space S. Then the probability that event E occurs, written P(E), is: For any event E, 0  P(E)  1. For any sample space S, P(S) = 1 and P(  ) = 0 Example: find P(E), P(F) in example in previous slide\n\nStandard 52-card Deck Spade Club Heart Diamond Diamond King Heart Queen Diamond Jack Heart Ace\n\nExamples A single card is drawn from a standard 52- card deck, find the probability of the following events: (a) Drawing an ace (b) Drawing a face card (c) Drawing a spade (d) Drawing a spade or a heart (e) Drawing a queen (f) Drawing a diamond (g) Drawing a red card\n\nExample The table below lists the smoking habits of a group of college students: If a student is chosen at random, find the probability of getting someone who is a woman and a heavy smoker. SexNon-smokerRegular SmokerHeavy SmokerTotal Man135705210 Woman1872115223 Total3229120433\n\nADDITION RULE & COMPLEMENT RULE ADDITION RULE For any events E and F from a sample space S, P(E  F) = P(E) + P(F) – P(E  F) For mutually exclusive events E and F, P(E  F) = P(E) + P(F) COMPLEMENT RULE For any event E, P(E’) = 1 – P(E), P(E) = 1 – P(E’). 8.4 BASIC CONCEPTS OF PROBABILITY\n\nExamples Example 1: A single card is drawn from a standard 52-card deck, find the probability that it will be a red or a face card. Example 2: Two dice are rolled. Find the probability of the following events: (a) The 1 st die shows a 2 or the sum is 6 or 7 (b) The sum is 11 or the 2 nd die shows a 5\n\n1-11-21-31-41-51-6 2-12-22-32-42-52-6 3-13-23-33-43-53-6 4-14-24-34-44-54-6 5-15-25-35-45-55-6 6-16-26-36-46-56-6 Examples Example 2: Two dice are rolled. Find the probability of the following events: (a) The 1 st die shows a 2 or the sum is 6 or 7\n\n1-11-21-31-41-51-6 2-12-22-32-42-52-6 3-13-23-33-43-53-6 4-14-24-34-44-54-6 5-15-25-35-45-55-6 6-16-26-36-46-56-6 Examples Example 2: Two dice are rolled. Find the probability of the following events: (b) The sum is 11 or the 2 nd die shows a 5\n\nExamples Example 4: A die is rolled, what is the probability that any number but 5 will come up? Example 5: Two dice are rolled. Find the probability that the sum of the numbers showing is greater than 3.\n\nODDS The odds in favor of an event E are defined as the ratio of P(E) to P(E’), or,, P(E’)  0. If the odds favoring event E are m to n, then P(E) =, and P(E’) = Example 6 (p. 472) Example 7 (p. 473) Example 8 (p. 473)\n\nODDS Example 6 (p. 472) The probability of rain tomorrow is 1/3. Find the odds in favor of rain tomorrow. Example 7 (p. 473) There is a 40% chance that it will snow tomrrow. Find the odds in favor of snow tomorrow. Example 8 (p. 473) The odds that a particular bid will be low bid are 4 to 5. Find the probability that the bid will be the low bid.\n\nRelative Frequency Probability Example 9: the table lists the number of siblings indicated by respondents in a survey: (a)Find the relative frequency probability of having 0, 1, 2, …, 10 or more siblings. (b)Find the probability that a randomly chosen American has 1 or 2 siblings. Number of siblingsFrequency 0140 1505 2583 3457 4314 5224 6157 7115 877 949 10 or more137 Total2758 Number of siblingsFrequencyProbability 01400.051 15050.183 25830.211 34570.166 43140.114 52240.081 61570.057 71150.042 8770.028 9490.018 10 or more1370.050 Total27581.000\n\nPROPERTIES OF PROBABILITY Let S be a sample space consisting of n distinct outcomes s 1, s 2, …, s n. An acceptable probability assignment consists of assigning to each outcome s i a number p i (the probability of s i ) according to these rules: 1.The probability of each outcome is a number between 0 and 1 (0  p i  1, i = 1..n) 2.The sum of the probabilities of all possible outcomes is 1. (p 1 + p 2 + … + p n = 1)\n\nRelative Frequency Probability Example: two dice are rolled, and the sum is calculated. Make the probability distribution for the sum: Find the probability that the sum is at least 10. The sumProbability 2 3 4 5 6 7 8 9 10 11 12\n\nExample 10 Let L indicate the event that the respondent had a “liberal” political tendency, and let M indicate that the respondent believes that marijuana use should be legal. Below are the survey estimates: P(L) =.27, P(M) =.37, P(L  M) =.15 (a)Find the probability that a respondent does not have a liberal tendency and does not support legalizing the use of marijuana. (b)Find the probability that a respondent does not have a liberal tendency or does not support legalizing the use of marijuana.\n\n8.5 CONDITIONAL PROBABILITY The conditional probability of an event E given event F, written P(E|F), is:, P(F)  0 Example 2 (p. 483): given P(E) =.4, P(F) =.5 and P(E  F)=.7. Find P(E|F). Example 3 (p.483): 2 coins were tossed, find the probability that both were heads, if it is known that at least one was head.\n\nPRODUCT RULE OF PROBABILITY If E and F are events, then P(E  F) may be found by either of these formulas: P(E  F) = P(F)  P(E|F) or P(E  F) = P(E)  P(F|E) Example 4. (p. 484) In a class with 2/5 women and 3/5 men, 25% of the women are business majors. Find the probability that a student at random from the class is a female business major.\n\nExample 5. (p. 485) A company needs to decide between person A and B to be a new director of advertising. The research shows that A is in charge of twice as many advertising campaigns as B. And A’s campaigns have satisfactory results three out of four times, while B’s campaigns have satisfactory results only two out of five times. a)Find the probability that A runs a campaign that produces satisfactory results. b)Find the probability that B runs a campaign that produces satisfactory results. c)Find the probability that a campaign is satisfactory, unsatisfactory d)Find the probability that either A runs the campain or the results are satisfactory\n\nExample 6. (p. 486) From a box containing 1 red, 3 white and 2 green marbles, two marbles are drawn one at a time without replacing the first before the second is drawn. Find the probability that one white and one green marble are drawn.\n\nExample 7. (p. 488) Two cards are drawn without replacement from a deck. Find the probability that the first card is a heart and the second card is red.\n\nExample 8. (p. 488) Three cards are drawn without replacement from a deck. Find the probability that exactly 2 of the cards are red.\n\nINDEPENDENT EVENTS E and F are independent events if P(F|E) = P(F) or P(E|F) = P(E) PRODUCT RULE FOR INDEPENDENT EVENTS E and F are independent events if and only if P(E  F) = P(E)  P(F)\n\nExample 9. (p. 490) A calculator requires a key-stroke assembly and a logic circuit. Assume that 99% of the key-stroke assemblies are satisfactory and 97% of the circuits are satisfactory. Find the probability that a finished calculator will be satisfactory. Suppose that the failure of key- stroke assemblies and the failure of logic circuits are independent.\n\nExample 10. (p. 490) On a typical day in Saigon the probability of very hot weather is 0.10, the probability of a traffic jam is 0.80, and the probability of very hot weather or a traffic jam is 0.82. Are the event “very hot weather” and event “a traffic jam” independent?\n\n8.6 BAYES’ FORMULA BAYES’ FORMULA (Special case) Example 1 (p. 495) For a fixed length of time, the probability of worker error on a production line is 0.1, the probability that an accident will occur when there is a worker error is 0.3, and the probability that an accident will occur when there is no worker error is 0.2. Find the probability of a worker error if there is an accident.\n\nBAYES’ FORMULA (General case) Example 2 (p. 497) A survey indicated that 87% of married women have one or more children, 40% of never- married women have one or more children, and 88% of women who are divorced, separated or widowed have one or more children. The survey also indicated that 43% of women were currently married, 24% had never been married, and 33% were divorced, separated or widowed. Find the probability that a woman who have one or more children is married.\n\nUSING BAYES’ FORMULA 1.Start a probability tree with branches representing events F 1, F 2, …, F n. Label each branch with its corresponding probability. 2.From the end of each of these branches, draw a branch for event E. Label this branch with probability of getting to it, or P(E|F i ).\n\nUSING BAYES’ FORMULA 3.There are now n defferent paths that result in event E. Next to each path, put its probability – the product of the probabilities that the first branch occurs, P(F i ), and that the second branch occurs, P(E|F i ): that is P(F i )  P(E|F i ). 4.P(F i |E) is found by dividing the probability of the branch for F i by the sum of the probabilities of all the branches producing event E.\n\nExample 3 (p. 497) A manufacturer buys items from 6 different suplliers. The fraction of the total number of items obtained from each supplier, along with the probability that an item purchased from that supplier is defective, is shown in the table below. Find the probability that a defective item came from supplier 5. SupplierFraction of total supplied Probability of Defective 1.05.04 2.12.02 3.16.07 4.23.01 5.35.03 6.09.05\n\nDownload ppt \"Chapter 8 Sets and Probabilities. 8.1 SETS The use of braces: { } Element (member) of a set, ,  Empty set  Distinguish: 0,  and {0} Equality of sets.\"\n\nSimilar presentations" ]

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[ "# Starting to understand Encryption\n\nEncryption is the principal application of cryptography; it makes data incomprehensible in order to ensure its confidentiality. Encryption uses an algorithm called a cipher and a secret value called the key; if you don’t know the secret key, you can’t decrypt, nor can you learn any bit of information on the encrypted message—and neither can any attacker.\n\n## The basics of encryption\n\nWhen we’re encrypting a message, plaintext refers to the unencrypted message and ciphertext to the encrypted message. A cipher is therefore composed of two functions: encryption turns a plaintext into a ciphertext, and decryption turns a ciphertext back into a plaintext. But we’ll often say “cipher” when we actually mean “encryption.” For example, the figure below shows a cipher, E, represented as a box taking as input a plaintext, P, and a key, K, and producing a ciphertext, C, as output. I’ll write this relation as C = E(K, P). Similarly, when the cipher is in decryption mode, I’ll write D(K, C).\n\n## History of encryption\n\nOne of the earliest forms of encryption is symbol replacement, which was first found in the tomb of Khnumhotep II, who lived in 1900 B.C. Egypt. Symbol replacement encryption is “non-standard,” which means that the symbols require a cipher or key to understand. This type of early encryption was used throughout Ancient Greece and Rome for military purposes.]\n\nOne of the most famous military encryption developments was the Caesar Cipher.\n\n### The Caesar Cipher\n\nThe Caesar cipher is so named because the Roman historian Suetonius reported that Julius Caesar used it. It encrypts a message by shifting each of the letters down three positions in the alphabet, wrapping back around to A if the shift reaches Z. For example, ZOO encrypts to CRR, FDHVDU decrypts to CAESAR, and so on, as shown in the next figure.\n\nThere’s nothing special about the value 3; it’s just easier to compute in one’s head than 11 or 23.\n\nThe Caesar cipher is super easy to break: to decrypt a given ciphertext, simply shift the letters three positions back to retrieve the plaintext. That said, the Caesar cipher may have been strong enough during the time of Crassus and Cicero. Because no secret key is involved (it’s always 3), users of Caesar’s cipher only had to assume that attackers were illiterate or too uneducated to figure it out—an assumption that’s much less realistic today. (In fact, in 2006, the Italian police arrested a mafia boss after decrypting messages written on small scraps of paper that were encrypted using a variant of the Caesar cipher: ABC was encrypted to 456 instead of DEF, for example.)\n\nCould the Caesar cipher be made more secure? You can, for example, imagine a version that uses a secret shift value instead of always using 3, but that wouldn’t help much because an attacker could easily try all 25 possible shift values until the decrypted message makes sense.\n\n### The Vigenère Cipher\n\nIt took about 1500 years to see a meaningful improvement of the Caesar cipher in the form of the Vigenère cipher, created in the 16th century by an Italian named Giovan Battista Bellaso. The name “Vigenère” comes from the Frenchman Blaise de Vigenère, who invented a different cipher in the 16th century, but due to historical misattribution, Vigenère’s name stuck. Nevertheless, the Vigenère cipher became popular and was later used during the American Civil War by Confederate forces and during WWI by the Swiss Army, among others.\n\nThe Vigenère cipher is similar to the Caesar cipher, except that letters aren’t shifted by three places but rather by values defined by a key, a collection of letters that represent numbers based on their position in the alphabet. For example, if the key is DUH, letters in the plaintext are shifted using the values 3, 20, 7 because D is three letters after A, U is 20 letters after A, and H is seven letters after A. The 3, 20, 7 pattern repeats until you’ve encrypted the entire plaintext. For example, the word CRYPTO would encrypt to FLFSNV using DUH as the key: C is shifted three positions to F, R is shifted 20 positions to L, and so on. The next figure illustrates this principle when encrypting the sentence THEY DRINK THE TEA.\n\nThe Vigenère cipher is clearly more secure than the Caesar cipher, yet it’s still fairly easy to break. The first step to breaking it is to figure out the key’s length. For example, take the example in Figure 1-3, wherein THEY DRINK THE TEA encrypts to WBLBXYLHRWBLWYH with the key DUH. (Spaces are usually removed to hide word boundaries.)\n\nNotice that in the ciphertext WBLBXYLHRWBLWYH, the group of three letters WBL appears twice in the ciphertext at nine-letter intervals. This suggests that the same three-letter word was encrypted using the same shift values, producing WBL each time. A cryptanalyst can then deduce that the key’s length is either nine or a value divisible by nine (that is, three). Furthermore, they may guess that this repeated three-letter word is THE and therefore determine DUH as a possible encryption key.\n\nThe second step to breaking the Vigenère cipher is to determine the actual key using a method called frequency analysis, which exploits the uneven distribution of letters in languages. For example, in English, E is the most common letter, so if you find that X is the most common letter in a ciphertext, then the most likely plaintext value at this position is E.\n\nDespite its relative weakness, the Vigenère cipher may have been good enough to securely encrypt messages when it was used. First, because the attack just outlined needs messages of at least a few sentences, it wouldn’t work if the cipher was used to encrypt only short messages. Second, most messages needed to be secret only for short periods of time, so it didn’t matter if ciphertexts were eventually decrypted by the enemy. (The 19thcentury cryptographer Auguste Kerckhoffs estimated that most encrypted wartime messages required confidentiality for only three to four hours.)\n\n## How Ciphers Work\n\nBased on simplistic ciphers like the Caesar and Vigenère ciphers, we can try to abstract out the workings of a cipher, first by identifying its two main components: a permutation and a mode of operation. A permutation is a function that transforms an item (in cryptography, a letter or a group of bits) such that each item has a unique inverse (for example, the Caesar cipher’s three-letter shift). A mode of operation is an algorithm that uses a permutation to process messages of arbitrary size. The mode of the Caesar cipher is trivial: it just repeats the same permutation for each letter, but as you’ve seen, the Vigenère cipher has a more complex mode, where letters at different positions undergo different permutations.\n\n### The Permutation\n\nMost classical ciphers work by replacing each letter with another letter— in other words, by performing a substitution. In the Caesar and Vigenère ciphers, the substitution is a shift in the alphabet, though the alphabet or set of symbols can vary: instead of the English alphabet, it could be the Arabic alphabet; instead of letters, it could be words, numbers, or ideograms, for example. The representation or encoding of information is a separate matter that is mostly irrelevant to security. (We’re just considering Latin letters because that’s what classical ciphers use.)\n\nA cipher’s substitution can’t be just any substitution. It should be a permutation, which is a rearrangement of the letters A to Z, such that each letter has a unique inverse. For example, a substitution that transforms the letters A, B, C, and D, respectively to C, A, D, and B is a permutation, because each letter maps onto another single letter. But a substitution that transforms A, B, C, D to D, A, A, C is not a permutation, because both B and C map onto A. With a permutation, each letter has exactly one inverse.\n\nStill, not every permutation is secure. In order to be secure, a cipher’s permutation should satisfy three criteria:\n\n• The permutation should be determined by the key, so as to keep the permutation secret as long as the key is secret. In the Vigenère cipher, if you don’t know the key, you don’t know which of the 26 permutations was used; hence, you can’t easily decrypt.\n• Different keys should result in different permutations. Otherwise, it becomes easier to decrypt without the key: if different keys result in identical permutations, that means there are fewer distinct keys than distinct permutations, and therefore fewer possibilities to try when decrypting without the key. In the Vigenère cipher, each letter from the key determines a substitution; there are 26 distinct letters, and as many distinct permutations.\n• The permutation should look random, loosely speaking. There should be no pattern in the ciphertext after performing a permutation, because patterns make a permutation predictable for an attacker, and therefore less secure. For example, the Vigenère cipher’s substitution is pretty predictable: if you determine that A encrypts to F, you could conclude that the shift value is 5 and you would also know that B encrypts to G, that C encrypts to H, and so on. However, with a randomly chosen permutation, knowing that A encrypts to F would only tell you that B does not encrypt to F.\n\nWe’ll call a permutation that satisfies these criteria a secure permutation. But as you’ll see next, a secure permutation is necessary but not sufficient on its own for building a secure cipher. A cipher will also need a mode of operation to support messages of any length.\n\n### The Mode of Operation\n\nSay we have a secure permutation that transforms A to X, B to M, and N to L, for example. The word BANANA therefore encrypts to MXLXLX, where each occurrence of A is replaced by an X. Using the same permutation for all the letters in the plaintext thus reveals any duplicate letters in the plaintext. By analyzing these duplicates, you might not learn the entire message, but you’ll learn something about the message. In the BANANA example, you don’t need the key to guess that the plaintext has the same letter at the three X positions and another same letter at the two L positions. So if you know, for example, that the message is a fruit’s name, you could determine that it’s BANANA rather than CHERRY, LYCHEE, or another six-letter fruit.\n\nThe mode of operation (or just mode) of a cipher mitigates the exposure of duplicate letters in the plaintext by using different permutations for duplicate letters. The mode of the Vigenère cipher partially addresses this: if the key is N letters long, then N different permutations will be used for every N consecutive letters. However, this can still result in patterns in the ciphertext because every Nth letter of the message uses the same permutation. That’s why frequency analysis works to break the Vigenère cipher, as you saw earlier.\n\nFrequency analysis can be defeated if the Vigenère cipher only encrypts plaintexts that are of the same length as the key. But even then, there’s another problem: reusing the same key several times exposes similarities between plaintexts. For example, with the key KYN, the words TIE and PIE encrypt to DGR and ZGR, respectively. Both end with the same two letters (GR), revealing that both plaintexts share their last two letters as well. Finding these patterns shouldn’t be possible with a secure cipher.\n\nTo build a secure cipher, you must combine a secure permutation with a secure mode. Ideally, this combination prevents attackers from learning anything about a message other than its length.\n\nThis article was based on the book Serious Cryptography – A Practical Introduction to Modern Encryption by Jean-Philippe Aumasson. To learn more about encryption and cryptography in general we recommend this great book.\n\nThis is the index from the book:\n\n• Chapter 1: Encryption\n• Chapter 2: Randomness\n• Chapter 3: Cryptographic Security\n• Chapter 4: Block Ciphers\n• Chapter 5: Stream Ciphers\n• Chapter 6: Hash Functions\n• Chapter 7: Keyed Hashing\n• Chapter 8: Authenticated Encryption\n• Chapter 9: Hard Problems\n• Chapter 10: RSA\n• Chapter 11: Diffie–Hellman\n• Chapter 12: Elliptic Curves\n• Chapter 13: TLS\n• Chapter 14: Quantum and Post-Quantum\n\nEditor\n\nArticles posted after being checked by editors." ]

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[ "$$\\require{cancel}$$\n\n# 15.19: The Transverse and Oblique Doppler Effects\n\n•", null, "• Contributed by Jeremy Tatum\n• Emeritus Professor (Physics & Astronomy) at University of Victoria\n\nI pointed out in Section 15.18 that the observed Doppler effect, when the transmitted signal is electromagnetic radiation and observer or source or both are travelling at speed comparable to that of light, is a combination of two effects – the “true” Doppler effect, caused by the changing distance between source and observer, and the effect of time dilation. This raises the following questions.\n\n# Transverse Doppler effect\n\nIf a source of light is moving at right angles to (transverse to) the line joining observer to source, will the observer see a change in frequency or wavelength, even though the distance between observer and source at that instant is not changing? The answer is yes, certainly, and the effect is sometimes called the “transverse Doppler effect”, although it is the effect of relativistic time dilation rather than of a true Doppler effect.\n\nThus let us suppose that a source is moving transverse to the line of sight at a speed described by its parameter $$\\beta$$ or $$\\gamma$$, and that the period of the radiation referred to the reference frame in which the source is at rest is $$T_{0}$$ and the frequency is $$\\nu_{0}$$. The time interval between emission of consecutive wavecrests when referred to the frame in which the observer is at rest is longer by the gamma-factor, and the frequency is correspondingly less. That is, the frequency, referred to the observer’s reference frame, is\n\n$\\nu=\\frac{\\nu_{0}}{\\gamma}=\\nu_{0}\\sqrt{1-\\beta^{2}} \\label{15.19.1}$\n\nThe light from the source is therefore seen by the observer to be redshifted, even though there is no radial velocity component.\n\n# Oblique Doppler Effect\n\nThis raises a further question. Suppose a source is moving almost but not quite at right angles to the line of sight, so that it has a large transverse velocity component, and a small velocity component towards the observer. In that case, its “redshift” resulting from the time dilation might be appreciable, while its “blueshift” resulting from “true” Doppler effect (the decreasing distance between source and observer) is still very small. Therefore, even though the distance between source and observer is slightly decreasing, there is a net redshift of the spectrum. This is in fact correct, and is the “oblique Doppler effect”.\n\nIn Figure XV.30, a source S is moving at speed $$\\beta$$ times the speed of light in a direction that makes an angle $$\\theta$$ with the line of sight. It is emitting a signal of frequency $$\\nu_{0}$$ in S. (I am here using the frame “in S” as earlier in the chapter to mean “referred to a reference frame in which S is at rest.) The signal arrives at the observer O at a slightly greater frequency as a result of the decreasing distance of S from O, and at a slightly lesser frequency as a result of the time dilation, the two effects opposing each other.", null, "The frequency of the received signal at O, in O, is\n\n$\\nu=\\frac{\\nu_{0}\\sqrt{1-\\beta^{2}}}{1-\\beta\\cos\\theta}. \\label{15.19.2}$\n\nFor a given angle $$\\theta$$ the redshift is zero for a speed of\n\n$\\beta =\\frac{2\\cos\\theta}{1+\\cos^{2}\\theta} \\label{15.19.3}$\n\nor, for a given speed, the direction of motion resulting in a zero redshift is given by\n\n$\\beta\\cos^{2}\\theta-2\\cos\\theta+\\beta=0. \\label{15.19.4}$\n\nThis relation is shown in Figure XV.30. (Although Equation $$\\ref{15.19.4}$$ is quadratic in $$\\cos\\theta$$ there is only one real solution $$\\theta$$ for $$\\beta$$ between 0 and 1. Prove this assertion.) It might be noted that if the speed of the source is 99.99% of the speed of light the observer will see a redshift unless the direction of motion of S is no further than 9o 36' from the line from S to O. That is worth repeating: S is moving very close to the speed of light in a direction that is close to being directly towards the observer; the observer will see a redshift.\n\nEquation $$\\ref{15.19.2}$$, which gives $$\\nu$$ as a function of $$\\theta$$ for a given $$\\beta$$, will readily be recognized at the equation of an ellipse of eccentricity $$\\beta$$, semi minor axis $$\\nu_{0}$$ and semi major axis $$\\gamma\\nu_{0}$$. This relation is shown in Figure XV.31 for several $$\\beta$$. The curves are red where there is a redshift and blue where there is a blueshift. There is no redshift or blueshift for $$\\beta=0$$, and the ellipse for that case is a circle and is drawn in black.", null, "An alternative and perhaps more useful way of looking at Equation $$\\ref{15.19.2}$$ is to regard it as an equation that gives $$\\beta$$ as a function of $$\\theta$$ for a given Doppler ratio $$\\frac{\\nu}{\\nu_{0}}$$. For example, if the Doppler ratio of a galaxy is observed to be 0.75, the velocity vector of the galaxy could be any arrow starting at the black dot and ending on the curve marked 0.75. The curves are ellipses with semi major axis equal to $$\\frac{1}{\\sqrt{1-(\\frac{\\nu}{\\nu_{0}})^{2}}}$$ and semi minor axis $$\\frac{1}{(1-(\\frac{\\nu}{\\nu_{0}})^{2})}$$.", null, "", null, "" ]

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[ "# Automated generation of a concept map from text\n\nIn my previous post, concepts were extracted from an electronic textbook. The next step is to simulate an expert’s mind map based on the concepts extracted. For that we would need some way of computing which concepts are important, which concepts are related, how they’re related, and how important those relationships themselves are. I’m using a wikipedia article on stars for this exercise. The text was obtained with BeautifulSoup.\n\n# Concept Importance:\n\nRanking the concepts by their importance shows the student on which concepts to focus their efforts. In the absence of annotated data on which concepts are important for learning about stars, let’s evaluate the following hypotheses:\n\nHypothesis 1 — Raw counts: The more times a concept is mentioned in the text, the more important it is. Above is the distribution of raw counts of concepts in the text. Almost 1200 of the concepts extracted from the wikipedia article are mentioned<10 times. Not visible is the concept phrase ‘star’ (174 counts).\n\nZooming in further on raw counts in the image below shows a few concepts (~5) with raw counts above 20.\n\nIt should be noted that several concept phrases have already been removed based on a list of generic phrases such as ‘image’ (2 counts), ‘function’ (4 counts), ‘type’ (13 counts), ‘example’ (10 counts), group (10 counts), set (2 counts), technique (2 counts), term (5 counts) system (30 counts). These phrases are used frequently but do not add much meaning. Since this list of ‘stop concepts’ is not exhaustive, raw counts may prune some unimportant concepts, but includes other generic concepts such as ‘time’ (25 counts) and ‘year’ (17 counts).\n\nHypothesis 2 — TFIDF: If the concept is uncommon outside of this text, but frequent in this text, it is more important. (The Term Frequency-Inverse Document Frequency ratio, or TFIDF). The Inverse Document Frequency (IDF) is computed with a large collection of documents (corpus), typically in a similar subject area or genre as the text being analyzed. Assuming a typical adult readability for the documents, the Reuters corpus available from the NLTK python library was chosen. It contains 10,788 news articles and around 1.3 million words. Computing the IDF values for two word or three word concept terms was getting computationally expensive, however, so as a gross approximation, the highest IDF value of the individual words was used, with a multiplier based on the number of terms contained in the concept phrase. Below is a table of concepts ranked by TF-IDF.\n\nIn spite of a low IDF value, star has a high enough raw count of mentions to make it the most important concept. The next few concepts are sun, luminosity, core and helium. Concepts ‘helium’, ‘astronomer’, and ’supernova’ were bumped up above lower raw count concepts because of having a high IDF value — mentioned rarely in the NLTK Reuters corpus. High raw count concepts ‘year’ and ‘time’ were much lower in this ranking and not tagged as particularly important. This seems to be a reasonable way to order concept importance, although a bigger corpus in the general subject area of the analyzed text would definitely perform better.\n\nHypothesis 3 — Spread: If the concept is encountered throughout the text, it is more important than if it is only mentioned in one part of the text. A concept may have high raw count but low spread if it is only important in one subsection of the text as opposed to throughout the text. To compute spread, a numeric index was assigned to the sentences in the text. The spread was calculated as the standard deviation of the sentences that contained the concept phrase, normalized to the total length of the text in sentences (423 total sentences). Not visible is the concept ‘star’ with a spread of 0.28.\n\nThe spread may be large if an unimportant concept is mentioned three times but spread out in the text (‘stellar atmosphere’ and ‘gravitational collapse’, 2 counts each). Spread computed by sentence indices could also be affected by the ordering of subsections of text dealing with related content. So spread could be an artefact of the writing style and length of text, and is not a useful indicator by itself.\n\nThe figure below looks at the spread of the high TFIDF concepts shows which concepts are important in a subsection of the text vs important all around the text. The concepts ‘rotation’ and ‘magnetic field’ have a similar raw count to ‘radius’ and ‘white dwarf’, but are important in fewer contexts (in this particular text). Similarly, ‘massive star’ is mentioned in fewer contexts than ‘earth’, although ‘earth’ has a similar raw count. Focusing on learning about high TFIDF, high spread concepts first will help get the most out of the text.\n\n# Relationship Strength:\n\nBased on the assumption that the proximity of two concepts indicates relatedness, a preliminary graph was built. If the concepts were mentioned within a window of a number of sentences, they were considered related. To avoid an extremely dense map (capturing unimportant or distant relationships) or a map that is too sparse (relationships not captured), as a first guess, the window was set to 1 sentence before and after — if a concept y is in two sentences before or after concept x, then concept x and concept y were considered related. It should be noted that pronouns have not been included in the analysis.\n\nHow strongly related are two concepts? What are the most important connections a concept has? Three heuristic metrics were computed and compared:\n\n## Metric 1: Sentence distance count\n\nAssuming that if two concepts are mentioned in the same sentence, they are strongly related, my first guesses were that the relationship strength would be directly proportional to how often they occur closeby, and indirectly proportional to the distance between them. Accordingly, a preliminary heuristic metric for each pair of concepts x and y was calculated as:\n\nwhere sentence distance s is the difference of how many sentences separate the two concepts, starting from the same sentence (s = 0) to s = S, where S is the maximum distance at which concepts are considered related (typically set between 0 and 3).\n\nTo calculate relationship strength relative to one of the concepts, the average Sentence distance count of concept x with every related concept is subtracted from the Sentence distance count of concept x to concept y\n\nA quick look at the sorted concepts shows the limitations of this metric.\n\n1. For relatively smaller articles, with a reasonable threshold sentence distance (S= 3), this metric is not granular enough, showing over 150 concepts with the same high relationship strength. This does not help us meet our objective of screening which relationships are most important to focus on first.\n\n2. Infrequently mentioned concepts are unfairly rewarded for being mentioned in the same sentence once. This would mean that if two concepts are mentioned once in the same sentence, the relationship is as important as if the concepts are mentioned twice with a sentence distance = 2, or if the two concepts are mentioned three times with a sentence distance = 3.\n\n3. This metric does not help us understand for which concept the relationship is important. Is ‘nuclear fuel’ an important concept to ‘star’, or is it the other way around?\n\n## Metric 2: Common concept Jaccard similarity coefficient\n\nHow many concepts are both concept x and concept y related to? A Jaccard similarity metric for common concepts can be computed as:\n\nwhere xmap is the set of all concepts x is related to, and ymap is the set of all concepts to which y is related. The Jaccard similarity coefficient(cardinality of intersection divided by cardinality of union) of these two sets may provide a measure of how important the relationship is.\n\nThis metric has similar limitations to metric 1. A long sentence containing a lot of concept terms will skew this number. The following sentence from the wiki page illustrates this:\n\nAlthough concepts ‘positron’ and ‘chain reaction’ are only mentioned once, a longer sentence skews the results.\n\n## Metric 3: Co-occurence\n\nAnother heuristic metric defined to capture how closely two concepts co-occur was computed.\n\nFor every mention of concept x in the text, the distance to the nearest mention of concept y was found. The median of these shortest distances provides a measure of how closely the concept y is related to concept x. If the value is relatively small, then for most mentions of concept x, there is a concept y nearby. This would make concept y an important relationship for concept x.\n\nThis metric is a lot more granular, and since the value for concept x to concept y can be different from the value for concept y to concept x, it indicates the relative importance of the relationship for each concept.\n\nFrom the above table, we see that 50% of the mentions of ‘star’ are less than or equal to 29 sentences away from a mention of ‘light-year’, while 50% of the mentions of ‘light-year’ are less than a sentence away from the nearest mention of ‘star’. So ‘star’ is an important relationship for ‘light-year’. The concept ‘space’ is somewhat more important to ‘energy’ than the other way around.\n\nIn the absence of a multi-sourced annotated dataset on the strength of relationships, we have to make some assumptions. If we posit that language develops vocabulary to represent strong relationships succinctly, then the first metric (Sentence distance count) has merit. The relationship may be strong but not important in the context of the text being analyzed.\n\nThe second and third metrics actually answer a slightly different question — how important the relationship is in this text. In other words, while reading and comprehending the text, how often will both concepts occupy space in the reader’s working memory? The common concept Jaccard similarity index was found to skew heavily with complex sentences, unfairly rewarding infrequent concepts that were mentioned along with many others. The co-occurrence metric does not have the same limitation. It also varies with respect to concept, as the relationship may be much more important to one of the concepts.\n\nIf we restrict the concepts to the top 5% of TFIDF values and top 5% of spread (measures of concept importance), and the top 1% of co-occurring concepts (measure of relationship importance), the below partial graph is generated.\n\nThe first step to learning about stars in the article would be to understand the concepts in the above graph, and their relationship to ‘star’. The above analysis provides a good starting point for learning about any concept in a text. It becomes especially useful when the amount of text is vast.\n\n# Next steps\n\nNow that we know which concepts are important, and which relationships are important to a concept, we will look into extracting the type of relationships being extracted. This will enable us to store what we learn in a knowledge base, and for the intelligent tutoring system to ask smart questions.\n\nWe can also start constructing personalized learning paths. In the next articles, we will deal with building a personalized learning path. Setting learning objectives and figuring out what you already know is crucial while navigating a massive amount of text. The learning path will be based on individual student’s prior knowledge, and will simulate their cognitive load so the student is not bored (repeating known concepts) or overwhelmed (too many new concepts introduced at once)." ]

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