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http://jaymanntoday.ning.com/profiles/blogs/sunday-october-11-2015-early-day-stuff
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[
"It originally was scheduled for Saturday night, but heavy and persistent rains postponed the start until Sunday, when the weather forecast is superb.\n\nThe 12 remaining Chase drivers will be searching for an all-important win, which will automatically qualify them for the Eliminator Round consisting of eight drivers.\n\nTeams practiced primarily through the daylight hours, which should help now that the race has been pushed back to a 12:30 p.m. ET start. The heavy rains Saturday night washed the racing surface of any rubber build-up and giving drivers a green racetrack for Sunday's race.\n\nAs a result, there will be a competition caution on Lap 25.\n\nFive-time Charlotte winner Jeff Gordon will be making his final start at the 1.5-mile track, the track where he earned his first Sprint Cup Series win in the 1994 Coca-Cola 600. Gordon has yet to win a race this season and knows it would be something special if he can go to Victory Lane at the end of the day.\n\n\"If we walk out of here with a win, that would be extremely emotional,\" he said.\n\n((((((((((((((((((((((((((((((((((((((((((()))))))))))))))))))))))))))))))))))",
null,
"(((((((((((((((((((((((((((((((((((((((((()))))))))))))))))))))))))))))))))))))))))))))))\n\n##### Tuckerton Redmen\n\nPig Roasting for today",
null,
""
] |
[
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"http://storage.ning.com/topology/rest/1.0/file/get/395017604",
null,
"https://fbcdn-photos-e-a.akamaihd.net/hphotos-ak-xat1/v/t1.0-0/s600x600/12088068_10208085242275335_2382663820707603942_n.jpg",
null
] |
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https://algorithm.zone/blogs/find-collision-pointer.html
|
[
"# Find – collision pointer\n\nGive an integer array nums and return the index values I and j of the two numbers in the array, so that nums[i] + nums[j] is equal to a given target value, and the two indexes cannot be equal.\n\nFor example: num = [2,7,11,15], target = 9\nReturn [0,1]\n\nIdea:\n\n1. Whether the start array is in order;\n2. Is the index calculated from 0 or 1?\n3. What should I do without a solution?\n4. What if there are multiple solutions? There is guaranteed to be a unique solution.\n\nFirst try to force the solution, O(n2) traversal\n\n```class Solution:\ndef twoSum(self, nums: List[int], target: int) -> List[int]:\nlen_nums = len(nums)\nfor i in range(len_nums):\nfor j in range(i+1,len_nums):\nif nums[i] + nums[j] == target:\nreturn [i,j]\n```\n\nTry Optimization: use sorting + pointer collision (O(n) + O(nlogn) = O(nlogn))\n\nBecause the problem itself is not ordered, you need to sort the original array once. After sorting, you can solve it with O(n) pointer collision.\nBecause the index value is returned, the index value corresponding to the number should be stored before sorting\n\n```class Solution:\ndef twoSum(self, nums: List[int], target: int) -> List[int]:\nrecord = dict()\nnums_copy = nums.copy()\nsameFlag = True;\nnums.sort()\nl,r = 0,len(nums)-1\nwhile l < r:\nif nums[l] + nums[r] == target:\nbreak\nelif nums[l] + nums[r] < target:\nl += 1\nelse:\nr -= 1\nres = []\nfor i in range(len(nums)):\nif nums_copy[i] == nums[l] and sameFlag:\nres.append(i)\nsameFlag = False\nelif nums_copy[i] == nums[r]:\nres.append(i)\nreturn res\n```\n\nMore Python IC implementations:\nStart binding subscripts and values through list(enumerate(nums)), without special copy and bool judgment.\n\n```class Solution:\nnums = list(enumerate(nums))\nnums.sort(key = lambda x:x)\ni,j = 0, len(nums)-1\nwhile i < j:\nif nums[i] + nums[j] > target:\nj -= 1\nelif nums[i] + nums[j] < target:\ni += 1\nelse:\nif nums[j] < nums[i]:\nnums[j],nums[i] = nums[i],nums[j]\nreturn num[i],nums[j]\n```\n\nCopy array + bool type auxiliary variable\nLookup table – O(n)\nIn the process of traversing the array, when traversing the element V, you can only see whether the element in front of V contains the element of target-v.\n\n1. If the search is successful, the solution is returned;\n2. If the search is not successful, put v in the lookup table and continue to find the next solution.\nEven if v is placed in the previous lookup table and overwrites v, the lookup of the current v element is not affected. Because you only need to find two elements, you only need\nJust find another element of target-v.\n```class Solution:\ndef twoSum(self, nums: List[int], target: int) -> List[int]:\nrecord = dict()\nfor i in range(len(nums)):\ncomplement = target - nums[i]\n# This value has been found in the previous dictionary\nif record.get(complement) is not None:\nres = [i,record[complement]]\nreturn res\nrecord[nums[i]] = i\n```\n\nTitle Description\nGive an integer array and find all the different triples (a,b,c) in it so that a+b+c=0\nNote: the answer cannot contain duplicate triples.\nFor example: num = [- 1, 0, 1, 2, - 1, - 4],\nThe result is: [- 1, 0, 1], [- 1, - 1, 2]]\n\nProblem solving ideas\n\n1. Arrays are not ordered;\n2. The returned result is all solutions. Should the order of multiple solutions be considered There is no need to consider order\n3. What is a different triplet? Are indexes different or values different Different values are defined as triples\n4. How to return when there is no solution Empty list\n```class Solution:\ndef threeSum(self, nums: [int]) -> [[int]]:\nnums.sort()\nres = []\nfor i in range(len(nums)-2):\n# Because it is a sorted array, if the smallest is greater than 0, it can be excluded directly\nif nums[i] > 0: break\n# Exclude duplicate values of i\nif i > 0 and nums[i] == nums[i-1]: continue\nl,r = i+1, len(nums)-1\nwhile l < r:\nsum = nums[i] + nums[l] + nums[r]\nif sum == 0:\nres.append([nums[i],nums[l],nums[r]])\nl += 1\nr -= 1\n#Prevent duplication\nwhile l < r and nums[l] == nums[l-1]: l += 1\nwhile l < r and nums[r] == nums[r+1]: r -= 1\nelif sum < 0:\nl += 1\nelse:\nr -= 1\nreturn res\n```\n\nSmall routine\n\n1. Three indexes are processed in the form of for + while;\n2. When the array is not ordered, you should pay attention to the characteristics of order and what methods can be used to solve it? What's the inconvenience if it's out of order\nInside?\n3. Collision pointer routine:\n```# Colliding pointer routine\nl,r = 0, len(nums)-1\nwhile l < r:\nif nums[l] + nums[r] == target:\nreturn nums[l],nums[r]\nelif nums[l] + nums[r] < target:\nl += 1\nelse:\nr -= 1`\n```\n\nRoutine for handling duplicate values: first convert to an ordered array, and then recycle to determine whether it is duplicate with the previous value:\n\n```# 1.\nfor i in range(len(nums)):\nif i > 0 and nums[i] == nums[i-1]: continue\n# 2.\nwhile l < r:\nwhile l < r and nums[l] == nums[l-1]: l += 1\n```\n\nTitle Description\nGiven an integer array, find all the different quads (a,b,c,d) in it, so that a+b+c+d is equal to a given number target.\n\nFor example:\nnums = [1, 0, -1, 0, -2, 2]，target = 0\n\nThe result is:\n[[-1, 0, 0, 1],[-2, -1, 1, 2],[-2, 0, 0, 2]]\n\n```class Solution:\ndef fourSum(self, nums: List[int], target: int) -> List[List[int]]:\nnums.sort()\nres = []\nif len(nums) < 4: return res\nif len(nums) == 4 and sum(nums) == target:\nres.append(nums)\nreturn res\nfor i in range(len(nums)-3):\nif i > 0 and nums[i] == nums[i-1]: continue\nfor j in range(i+1,len(nums)-2):\nif j > i+1 and nums[j] == nums[j-1]: continue\nl,r = j+1, len(nums)-1\nwhile l < r:\nsum_value = nums[i] + nums[j] + nums[l] + nums[r]\nif sum_value == target:\nres.append([nums[i],nums[j],nums[l],nums[r]])\nl += 1\nr -= 1\nwhile l < r and nums[l] == nums[l-1]: l += 1\nwhile l < r and nums[r] == nums[r+1]: r -= 1\nelif sum_value < target:\nl += 1\nelse:\nr -= 1\nreturn res\n```\n\nYou can also use combinations(nums, 4) to fully arrange the four elements in the original array. After starting sort, set and de duplicate the arranged elements. However, the implementation using combinations alone will time out.\n\n```class Solution:\ndef fourSum(self, nums: List[int], target: int) -> List[List[int]]:\nnums.sort()\nfrom itertools import combinations\nres = []\nfor i in combinations(nums, 4):\nif sum(i) == target:\nres.append(i)\nres = set(res)\nreturn res\n```\n\nTitle Description\n\nGive an integer array and find the three elements a, B and C, so that the value of a+b+c is closest to another given number target.\n\nFor example: given array num = [- 1, 2, 1, - 4], and target = 1\n\nThe sum of the three numbers closest to target is 2 (-1 + 2 + 1 = 2).\n\nanalysis\nJudge whether the sum of the three numbers is equal to the target value. If it is equal to the target value, return the three values directly. If it is not equal to the target value, compare the difference between the sum of the three numbers in the array and the target, and return the three numbers with the smallest difference.\n\n```nums.sort()\nres = []\nfor i in range(len(nums)-2):\nl,r = i+1, len(nums)-1\nwhile l < r:\nsum = nums[i] + nums[l] + nums[r]\nif sum == 0:\nres.append([nums[i],nums[l],nums[r]])\nelif sum < 0:\nl += 1\nelse:\nr -= 1\n```\n```class Solution:\ndef threeSumClosest(self, nums: List[int], target: int) -> int:\nnums.sort()\ndiff = abs(nums+nums+nums-target)\nres = nums + nums + nums\nfor i in range(len(nums)):\nl,r = i+1,len(nums)-1\nt = target - nums[i]\nwhile l < r:\nif nums[l] + nums[r] == t:\nreturn nums[i] + t\nelse:\nif abs(nums[l]+nums[r]-t) < diff:\ndiff = abs(nums[l]+nums[r]-t)\nres = nums[i]+nums[l]+nums[r]\nif nums[l]+nums[r] < t:\nl += 1\nelse:\nr -= 1\nreturn res\n```\n\nTitle Description\n\nGive four shaping arrays a, B, C and D, and find how many combinations of I, J, K and l there are, so that A[i]+B[j]+C[k]+D[l]=0. his\nIn, a, B, C and d all contain the same number of elements N, and 0 < = N < = 500;\n\nInput:\n\nA = [ 1, 2]\nB = [-2,-1]\nC = [-1, 2]\nD = [ 0, 2]\n\nOutput: 2\n\nAnalysis and Implementation\nThis problem is also a variant of the Sum class problem. It changes the conditions of the same array into four arrays, which can still be realized with the idea of look-up table.\nFirst, you can consider putting the elements in the D array into the lookup table, and then traverse the first three arrays to judge whether the value of target minus each element exists in the lookup table. If so, add 1 to the result value. Then the data structure of the lookup table uses dict to store duplicate elements. The final result value plus the value of the corresponding key of dict is as follows:\n\n```from collections import Counter\nrecord = Counter()\n# First create a lookup table for array D\n`for i in range(len(D)):\nrecord[D[i]] += 1\nres = 0\nfor i in range(len(A)):\nfor j in range(len(B)):\nfor k in range(len(C)):\nnum_find = 0-A[i]-B[j]-C[k]\nif record.get(num_find) != None:\nres += record(num_find)\nreturn res\n\n```\n\nAt this time, because the time complexity of traversing the three layers is O (N3), n < = 500, if n is too large, the consumption of the algorithm is still too large. Optimization:\n\n```class Solution:\ndef fourSumCount(self, A: List[int], B: List[int], C: List[int], D:\nList[int]) -> int:\nfrom collections import Counter\nrecord = Counter()\nfor i in range(len(A)):\nfor j in range(len(B)):\nrecord[A[i]+B[j]] += 1\nres = 0\nfor i in range(len(C)):\nfor j in range(len(D)):\nfind_num = 0 - C[i] - D[j]\nif record.get(find_num) != None:\nres += record[find_num]\nreturn res\n```\n\nThe complexity of the algorithm is O(n2)\n\nThe optimized functions and python are generated as follows:\n\n```class Solution:\ndef fourSumCount(self, A: List[int], B: List[int], C: List[int], D:\nList[int]) -> int:\nrecord = collections.Counter(a + b for a in A for b in B)\nreturn sum(record.get(- c - d, 0) for c in C for d in D)\n```\n\nTitle Description\n\nGive an array of strings that group all words that can produce the same result by reversing the character order.\n\nExample:\nInput: [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"],\nOutput: [\"ate\", \"eat\", \"tea\"], [\"nat\", \"tan\"], [\"bat\"]]\n\nexplain:\nAll inputs are lowercase letters.\nThe order in which answers are output is not considered.\n\n```class Solution:\ndef groupAnagrams(self, strs: List[str]) -> List[List[str]]:\nfrom collections import defaultdict\nstrs_dict = defaultdict(list)\nres = []\nfor str in strs:\nkey = ''.join(sorted(list(str)))\nstrs_dict[key] += str.split(',')\nfor v in strs_dict.values():\nres.append(v)\nreturn res\n```\n\nThen replace the places that can be replaced by list generation. The code is as follows:\n\n```class Solution:\ndef groupAnagrams(self, strs: List[str]) -> List[List[str]]:\nfrom collections import defaultdict\nstrs_dict = defaultdict(list)\nfor str in strs:\nkey = ''.join(sorted(list(str)))\nstrs_dict[key] += str.split(',')\nreturn [v for v in strs_dict.values()]\n```\n\nTitle Description\nGiven n points on a plane, find how many triples (i,j,k) composed of these points exist, so that the distance between I and j is equal to the distance between I and K.\nWhere n is up to 500, and the range of all point coordinates is [- 1000010000].\n\nInput:\n[[0,0],[1,0],[2,0]]\n\nOutput:\n2\nExplanation:\nThe two results are: [[1,0], [0,0], [2,0]] and [[1,0], [2,0], [0,0]]\n\ndistance\nFor the calculation of distance value, if it is calculated according to the European distance method, it is easy to produce floating-point numbers. You can remove the root sign and compare the distance with the sum of squares of the difference.\nThe implementation code is as follows:\n\n```class Solution:\ndef numberOfBoomerangs(self, points: List[List[int]]) -> int:\nres = 0\nfrom collections import Counter\nfor i in points:\nrecord = Counter()\nfor j in points:\nif i != j:\nrecord[self.dis(i,j)] += 1\nfor k,v in record.items():\nres += v*(v-1)\nreturn res\ndef dis(self,point1,point2):\nreturn (point1-point2) ** 2 + (point1-point2) ** 2\n```\n\noptimization\nOptimize the implemented Code:\n\n1. Change the for loop traversal to list generation;\n2. For the operation of sum + = consider using the sum function.\n3. Use closure to embed different functions;\n```class Solution:\ndef numberOfBoomerangs(self, points: List[List[int]]) -> int:\nfrom collections import Counter\ndef f(x1, y1):\n# Sum the distance values of J and K under an i\nd = Counter((x2 - x1) ** 2 + (y2 - y1) ** 2 for x2, y2 in\npoints)\nreturn sum(t * (t-1) for t in d.values())\n# Sum the distances of each i\nreturn sum(f(x1, y1) for x1, y1 in points)\n```\n\nTitle Description\nGiven a two-dimensional plane with n points, find the maximum number of points on the same line.\n\nExample 1:\nInput: [1,1], [2,2], [3,3]]\nOutput: 3\n\nExample 2:\nInput: [1,1], [3,2], [5,3], [4,1], [2,3], [1,4]]\nOutput: 4\n\nAnalysis and Implementation\nThe requirement of this topic is to see how many points are on the same straight line, so judging whether the points are on the same straight line is actually equivalent to judging whether the slope of I and j is equal to the slope of I and K.\nReview the requirements in the previous 447 topic: make the distance between I and j equal to the distance between I and K. here, directly consider using the lookup table, that is, find the number of key s with the same slope, value.\nIn the previous question, i and j, J and i are two different situations, but in this question, they belong to the same two points. Therefore, when traversing each i to find a point with the same slope as i, we can no longer count the result number res + +, but should take the maximum value in the lookup table. If two slopes are the same, 3 points should be returned, so the result number + 1 is returned.\n\n```class Solution:\ndef maxPoints(self,points):\nif len(points) <= 1:\nreturn len(points)\nres = 0\nfrom collections import defaultdict\nfor i in range(len(points)):\nrecord = defaultdict(int)\nsamepoint = 0\nfor j in range(len(points)):\nif points[i] == points[j] and points[i] == points[j]:\nsamepoint += 1\nelse:\nrecord[self.get_Slope(points,i,j)] += 1\nfor v in record.values():\nres = max(res, v+samepoint)\nres = max(res, samepoint)\nreturn res\ndef get_Slope(self,points,i,j):\nif points[i] - points[j] == 0:\nreturn float('Infinite')\nreturn (points[i] - points[j]) / (points[i] - points[j])\n```\n\nThe time complexity of the solution is O(n2) and the space complexity is O(n)\n\nTags: Python Algorithm\n\nPosted by steven21 on Fri, 20 May 2022 16:49:37 +0300"
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[
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https://math.stackexchange.com/questions/3212209/create-a-markov-chain-that-has-a-stationary-distribution-equal-to-the-zipf-distr
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[
"# Create a Markov chain that has a stationary distribution equal to the Zipf distribution using MCMC-MH.\n\n## Background\n\nI'm trying to learn about MCMC-MH by implementing one of the examples in a textbook, specifically problem 12.1.3. A description of how to solve the problem is given, but not an implementation.\n\n## The problem\n\nLet $$M > 2$$ be an integer. An random variable $$X$$ has the zipf distribution with parameter $$a > 0$$ if its PMF is :\n\n$$P(X=k) = \\frac{\\frac{1}{k^a}}{\\sum_{j=1}^{M} \\frac{1}{j^a}}$$\n\nfor $$k = 1, 2, ...$$ (and $$0$$ other wise). Create a Markov chain $$X_0, X_1, ...$$ whose stationary distribution is the Zipf distribution, and such that $$|X_{n+1} - X_n| \\leq 1$$ for all $$n$$.\n\n## The solution\n\nWe can use the Metropolis-Hastings algorithm, after coming up with a proposal distribution. The choice of proposal distribution is a random walk on $$1, 2, ..., M$$. From state $$i$$ with $$i\\neq 1$$ or $$i\\neq M$$, we move to state $$i+1$$ or $$i-1$$ with probability $$0.5$$ each. If we are at state $$1$$, we stay there or move to state $$2$$ each with $$0.5$$ probability. If we are in state M, we stay there or move to state $$M-1$$ each with probability $$0.5$$.\n\nLet $$P$$ be the transition matrix for this chain and $$X_0$$ be the starting state. We can generate a chain $$X_0, X_1, ...$$ as follows, given that we are in state $$i$$:\n\n1. Generate a proposal state, $$j$$, according to the proposal chain P.\n\n2. Accept the proposal with probability $$min(\\frac{i^a}{j^a}, 1)$$. If the proposal is accepted, we go to state $$j$$ and stay at $$i$$ otherwise.\n\n## The implementation (Python)\n\nimport numpy as np\nfrom scipy.stats import zipf\nimport matplotlib.pyplot as plt\nfrom matplotlib.pyplot import stem\nimport seaborn\nfrom collections import Counter\n\ndef transition_prob():\n\"\"\"\n:return: (np.ndarry). For example, when M=10:\n\n[[0.5 0.5 0. 0. 0. 0. 0. 0. 0. 0. ]\n[0.5 0. 0.5 0. 0. 0. 0. 0. 0. 0. ]\n[0. 0.5 0. 0.5 0. 0. 0. 0. 0. 0. ]\n[0. 0. 0.5 0. 0.5 0. 0. 0. 0. 0. ]\n[0. 0. 0. 0.5 0. 0.5 0. 0. 0. 0. ]\n[0. 0. 0. 0. 0.5 0. 0.5 0. 0. 0. ]\n[0. 0. 0. 0. 0. 0.5 0. 0.5 0. 0. ]\n[0. 0. 0. 0. 0. 0. 0.5 0. 0.5 0. ]\n[0. 0. 0. 0. 0. 0. 0. 0.5 0. 0.5]\n[0. 0. 0. 0. 0. 0. 0. 0. 0.5 0.5]]\n\"\"\"\nP = np.array( * M * M, dtype=np.float32).reshape(M, M)\nP[0, 0] = 0.5\nP[M - 1, M - 1] = 0.5\nfor i in range(M): # 0 indexed python\nfor j in range(M):\nif j == i + 1 or j == i - 1:\nP[i, j] = 0.5\n\nreturn P\n\ndef analytical(M, a):\npmf = []\nfor k in range(1, M + 1):\npmf.append(zipf.pmf(k, a))\nreturn pmf\n\ndef process_markov_data(data):\n# count frequencies\ncount = Counter(s_rec)\n## normalise to 1\ncount = {k: v / sum(count.values()) for k, v in count.items()}\nreturn count\n\ndef plot(data):\n# obtain and plot analytical solution\nana = analytical(M, a)\nfig = plt.figure()\nax1 = plt.subplot(121)\nax1.set_title('Analytical')\nax1.stem(ana, linefmt='k-.', markerfmt='ko')\nax1.set_xlim(-1, M)\nax1.set_ylim(0, 1)\n\nax2 = plt.subplot(122)\nax2.stem(data.keys(), data.values(), linefmt='k-.', markerfmt='ko')\nax2.set_title('Simulated')\nax2.set_xlim(-1, M)\nax2.set_ylim(0, 1)\nseaborn.despine(fig=fig, top=True, right=True)\n\nplt.show()\n\nif __name__ == '__main__':\nn_iterations = 100000 # number of iterations\nburnin = 1000 # number of iterations discarded\nM = 10 # range of M for k\na = 2 # Zipf parameter a\n\nP = transition_prob() # get transition probability\nsi = 1 # initial state for markov chain\n\ns_rec = [] # initialize a vector for storing the results\n\nfor iteration in range(n_iterations):\n# retrieve probabilities relevant for state i and propose state j\npmf = P[si,]\nsj = np.random.choice(range(pmf.shape), p=pmf)\n\n# compute the acceptance probability\naij = min(1, sj * P[si, sj] / si * P[si, sj])\n\n# generare a random number r~U(0, 1)\nr = np.random.uniform()\n\n# if the random number r is smaller than acceptance probability\n# accept and go to state j\nif r < aij:\nsi = sj\n\n# Only record transitions after burn-in phase\nif iteration > burnin:\ns_rec.append(si)\n\n# turn simulated data into probabilities\ns_rec = process_markov_data(s_rec)\n\n# plot simulated vs analytical\nplot(s_rec)\n\n\n\n# Results",
null,
"## Questions\n\nWhy does my simulated PMF not resemble the Analytical PMF?\n\n# Edit: Implementing the suggestions in the comments\n\nJust to follow up from for anybody else reading this, the line:\n\n aij = min(1, sj * P[si, sj] / si * P[si, sj])\n\n\nshould be\n\n aij = min(1, (si+1)**a / (sj+1)**a)\n\n\n## Results with the change",
null,
"• Your acceptance probability does not seem consistent with what you wrote above.\n– kccu\nMay 3, 2019 at 13:32\n• Also keep in mind that since the indexing is zero-based in Python, if you want to do any calculations involving the actual value of the the state $\\texttt{si}$, you should use $\\texttt{si} + 1$.\n– kccu\nMay 3, 2019 at 13:43\n• Perfect, thanks for pointing that out. Feel free to post an answer and I'll accept. May 3, 2019 at 14:45\n\nYour acceptance probability in the code does not seem consistent with what you wrote above. Also keep in mind that since the indexing is zero-based on Python, if you want to do any calculations involving the actual value of the state $$\\texttt{si}$$, you should use $$\\texttt{si + 1}$$."
] |
[
null,
"https://i.stack.imgur.com/Ecyq6.png",
null,
"https://i.stack.imgur.com/5MZVO.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.63976526,"math_prob":0.9998351,"size":4171,"snap":"2023-40-2023-50","text_gpt3_token_len":1514,"char_repetition_ratio":0.15046796,"word_repetition_ratio":0.13261163,"special_character_ratio":0.3862383,"punctuation_ratio":0.2332696,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998846,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T18:27:22Z\",\"WARC-Record-ID\":\"<urn:uuid:d5890cce-c88c-4eb8-b29c-61ad6e805d6d>\",\"Content-Length\":\"145208\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c92ef42-22e2-4076-8ac9-cf5eca89704f>\",\"WARC-Concurrent-To\":\"<urn:uuid:0bafbedd-c040-408a-add4-75547f36db09>\",\"WARC-IP-Address\":\"104.18.10.86\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3212209/create-a-markov-chain-that-has-a-stationary-distribution-equal-to-the-zipf-distr\",\"WARC-Payload-Digest\":\"sha1:4PRPFZAK4OPLMMYOGVXGVTQTHXMPEGOF\",\"WARC-Block-Digest\":\"sha1:ENUNUEYTSQ3XZCJVH542T5SRGXKAK6HP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506029.42_warc_CC-MAIN-20230921174008-20230921204008-00509.warc.gz\"}"}
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https://isabelle.in.tum.de/repos/isabelle/file/bdc2c6b40bf2/src/HOL/Taylor.thy
|
[
"src/HOL/Taylor.thy\n author haftmann Sat Jul 05 11:01:53 2014 +0200 (2014-07-05) changeset 57514 bdc2c6b40bf2 parent 56193 c726ecfb22b6 child 58889 5b7a9633cfa8 permissions -rw-r--r--\nprefer ac_simps collections over separate name bindings for add and mult\n``` 1 (* Title: HOL/Taylor.thy\n```\n``` 2 Author: Lukas Bulwahn, Bernhard Haeupler, Technische Universitaet Muenchen\n```\n``` 3 *)\n```\n``` 4\n```\n``` 5 header {* Taylor series *}\n```\n``` 6\n```\n``` 7 theory Taylor\n```\n``` 8 imports MacLaurin\n```\n``` 9 begin\n```\n``` 10\n```\n``` 11 text {*\n```\n``` 12 We use MacLaurin and the translation of the expansion point @{text c} to @{text 0}\n```\n``` 13 to prove Taylor's theorem.\n```\n``` 14 *}\n```\n``` 15\n```\n``` 16 lemma taylor_up:\n```\n``` 17 assumes INIT: \"n>0\" \"diff 0 = f\"\n```\n``` 18 and DERIV: \"(\\<forall> m t. m < n & a \\<le> t & t \\<le> b \\<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t))\"\n```\n``` 19 and INTERV: \"a \\<le> c\" \"c < b\"\n```\n``` 20 shows \"\\<exists> t. c < t & t < b &\n```\n``` 21 f b = (\\<Sum>m<n. (diff m c / real (fact m)) * (b - c)^m) + (diff n t / real (fact n)) * (b - c)^n\"\n```\n``` 22 proof -\n```\n``` 23 from INTERV have \"0 < b-c\" by arith\n```\n``` 24 moreover\n```\n``` 25 from INIT have \"n>0\" \"((\\<lambda>m x. diff m (x + c)) 0) = (\\<lambda>x. f (x + c))\" by auto\n```\n``` 26 moreover\n```\n``` 27 have \"ALL m t. m < n & 0 <= t & t <= b - c --> DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)\"\n```\n``` 28 proof (intro strip)\n```\n``` 29 fix m t\n```\n``` 30 assume \"m < n & 0 <= t & t <= b - c\"\n```\n``` 31 with DERIV and INTERV have \"DERIV (diff m) (t + c) :> diff (Suc m) (t + c)\" by auto\n```\n``` 32 moreover\n```\n``` 33 from DERIV_ident and DERIV_const have \"DERIV (%x. x + c) t :> 1+0\" by (rule DERIV_add)\n```\n``` 34 ultimately have \"DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1+0)\"\n```\n``` 35 by (rule DERIV_chain2)\n```\n``` 36 thus \"DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)\" by simp\n```\n``` 37 qed\n```\n``` 38 ultimately\n```\n``` 39 have EX:\"EX t>0. t < b - c &\n```\n``` 40 f (b - c + c) = (SUM m<n. diff m (0 + c) / real (fact m) * (b - c) ^ m) +\n```\n``` 41 diff n (t + c) / real (fact n) * (b - c) ^ n\"\n```\n``` 42 by (rule Maclaurin)\n```\n``` 43 show ?thesis\n```\n``` 44 proof -\n```\n``` 45 from EX obtain x where\n```\n``` 46 X: \"0 < x & x < b - c &\n```\n``` 47 f (b - c + c) = (\\<Sum>m<n. diff m (0 + c) / real (fact m) * (b - c) ^ m) +\n```\n``` 48 diff n (x + c) / real (fact n) * (b - c) ^ n\" ..\n```\n``` 49 let ?H = \"x + c\"\n```\n``` 50 from X have \"c<?H & ?H<b \\<and> f b = (\\<Sum>m<n. diff m c / real (fact m) * (b - c) ^ m) +\n```\n``` 51 diff n ?H / real (fact n) * (b - c) ^ n\"\n```\n``` 52 by fastforce\n```\n``` 53 thus ?thesis by fastforce\n```\n``` 54 qed\n```\n``` 55 qed\n```\n``` 56\n```\n``` 57 lemma taylor_down:\n```\n``` 58 assumes INIT: \"n>0\" \"diff 0 = f\"\n```\n``` 59 and DERIV: \"(\\<forall> m t. m < n & a \\<le> t & t \\<le> b \\<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t))\"\n```\n``` 60 and INTERV: \"a < c\" \"c \\<le> b\"\n```\n``` 61 shows \"\\<exists> t. a < t & t < c &\n```\n``` 62 f a = (\\<Sum>m<n. (diff m c / real (fact m)) * (a - c)^m) + (diff n t / real (fact n)) * (a - c)^n\"\n```\n``` 63 proof -\n```\n``` 64 from INTERV have \"a-c < 0\" by arith\n```\n``` 65 moreover\n```\n``` 66 from INIT have \"n>0\" \"((\\<lambda>m x. diff m (x + c)) 0) = (\\<lambda>x. f (x + c))\" by auto\n```\n``` 67 moreover\n```\n``` 68 have \"ALL m t. m < n & a-c <= t & t <= 0 --> DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)\"\n```\n``` 69 proof (rule allI impI)+\n```\n``` 70 fix m t\n```\n``` 71 assume \"m < n & a-c <= t & t <= 0\"\n```\n``` 72 with DERIV and INTERV have \"DERIV (diff m) (t + c) :> diff (Suc m) (t + c)\" by auto\n```\n``` 73 moreover\n```\n``` 74 from DERIV_ident and DERIV_const have \"DERIV (%x. x + c) t :> 1+0\" by (rule DERIV_add)\n```\n``` 75 ultimately have \"DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1+0)\" by (rule DERIV_chain2)\n```\n``` 76 thus \"DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)\" by simp\n```\n``` 77 qed\n```\n``` 78 ultimately\n```\n``` 79 have EX: \"EX t>a - c. t < 0 &\n```\n``` 80 f (a - c + c) = (SUM m<n. diff m (0 + c) / real (fact m) * (a - c) ^ m) +\n```\n``` 81 diff n (t + c) / real (fact n) * (a - c) ^ n\"\n```\n``` 82 by (rule Maclaurin_minus)\n```\n``` 83 show ?thesis\n```\n``` 84 proof -\n```\n``` 85 from EX obtain x where X: \"a - c < x & x < 0 &\n```\n``` 86 f (a - c + c) = (SUM m<n. diff m (0 + c) / real (fact m) * (a - c) ^ m) +\n```\n``` 87 diff n (x + c) / real (fact n) * (a - c) ^ n\" ..\n```\n``` 88 let ?H = \"x + c\"\n```\n``` 89 from X have \"a<?H & ?H<c \\<and> f a = (\\<Sum>m<n. diff m c / real (fact m) * (a - c) ^ m) +\n```\n``` 90 diff n ?H / real (fact n) * (a - c) ^ n\"\n```\n``` 91 by fastforce\n```\n``` 92 thus ?thesis by fastforce\n```\n``` 93 qed\n```\n``` 94 qed\n```\n``` 95\n```\n``` 96 lemma taylor:\n```\n``` 97 assumes INIT: \"n>0\" \"diff 0 = f\"\n```\n``` 98 and DERIV: \"(\\<forall> m t. m < n & a \\<le> t & t \\<le> b \\<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t))\"\n```\n``` 99 and INTERV: \"a \\<le> c \" \"c \\<le> b\" \"a \\<le> x\" \"x \\<le> b\" \"x \\<noteq> c\"\n```\n``` 100 shows \"\\<exists> t. (if x<c then (x < t & t < c) else (c < t & t < x)) &\n```\n``` 101 f x = (\\<Sum>m<n. (diff m c / real (fact m)) * (x - c)^m) + (diff n t / real (fact n)) * (x - c)^n\"\n```\n``` 102 proof (cases \"x<c\")\n```\n``` 103 case True\n```\n``` 104 note INIT\n```\n``` 105 moreover from DERIV and INTERV\n```\n``` 106 have \"\\<forall>m t. m < n \\<and> x \\<le> t \\<and> t \\<le> b \\<longrightarrow> DERIV (diff m) t :> diff (Suc m) t\"\n```\n``` 107 by fastforce\n```\n``` 108 moreover note True\n```\n``` 109 moreover from INTERV have \"c \\<le> b\" by simp\n```\n``` 110 ultimately have EX: \"\\<exists>t>x. t < c \\<and> f x =\n```\n``` 111 (\\<Sum>m<n. diff m c / real (fact m) * (x - c) ^ m) + diff n t / real (fact n) * (x - c) ^ n\"\n```\n``` 112 by (rule taylor_down)\n```\n``` 113 with True show ?thesis by simp\n```\n``` 114 next\n```\n``` 115 case False\n```\n``` 116 note INIT\n```\n``` 117 moreover from DERIV and INTERV\n```\n``` 118 have \"\\<forall>m t. m < n \\<and> a \\<le> t \\<and> t \\<le> x \\<longrightarrow> DERIV (diff m) t :> diff (Suc m) t\"\n```\n``` 119 by fastforce\n```\n``` 120 moreover from INTERV have \"a \\<le> c\" by arith\n```\n``` 121 moreover from False and INTERV have \"c < x\" by arith\n```\n``` 122 ultimately have EX: \"\\<exists>t>c. t < x \\<and> f x =\n```\n``` 123 (\\<Sum>m<n. diff m c / real (fact m) * (x - c) ^ m) + diff n t / real (fact n) * (x - c) ^ n\"\n```\n``` 124 by (rule taylor_up)\n```\n``` 125 with False show ?thesis by simp\n```\n``` 126 qed\n```\n``` 127\n```\n``` 128 end\n```"
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[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.824764,"math_prob":0.9983724,"size":4530,"snap":"2020-24-2020-29","text_gpt3_token_len":1847,"char_repetition_ratio":0.18183827,"word_repetition_ratio":0.49649736,"special_character_ratio":0.5083885,"punctuation_ratio":0.083174,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998366,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-02T07:35:40Z\",\"WARC-Record-ID\":\"<urn:uuid:85790ede-4df6-43d9-8473-81216522f13b>\",\"Content-Length\":\"25107\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:645eee45-df5a-40c4-b385-afb8268181ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:dcfd31e0-617f-45e3-af14-cb225c9ecc70>\",\"WARC-IP-Address\":\"131.159.46.82\",\"WARC-Target-URI\":\"https://isabelle.in.tum.de/repos/isabelle/file/bdc2c6b40bf2/src/HOL/Taylor.thy\",\"WARC-Payload-Digest\":\"sha1:SN6RCEV7NNRFALCIWRGQGQQDVWMJLLAA\",\"WARC-Block-Digest\":\"sha1:YYTYRRWAIEVTVSLHHKEVDNWHF45VMZNR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655878519.27_warc_CC-MAIN-20200702045758-20200702075758-00095.warc.gz\"}"}
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https://www.finmath.net/finmath-lib/apidocs/net.finmath.lib/net/finmath/marketdata/products/Cap.html
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[
"# Class Cap\n\nAll Implemented Interfaces:\nAnalyticProduct, Product\nDirect Known Subclasses:\nCapShiftedVol\n\npublic class Cap extends AbstractAnalyticProduct\nImplements the valuation of a cap via an analytic model, i.e. the specification of a forward curve, discount curve and volatility surface. A cap is a portfolio of Caplets with a common strike, i.e., the strike is the same for all Caplets. The class can value a caplet with a given strike or given moneyness. If moneyness is given, the class calculates the ATM forward. Note that this is done by omitting the first (fixed) period, see getATMForward(AnalyticModel, boolean). Note: A fixing in arrears is not handled correctly since a convexity adjustment is currently not applied.\nVersion:\n1.0\nAuthor:\nChristian Fries\nTo dos:\nSupport convexity adjustment if fixing is in arrears.\n• ## Constructor Summary\n\nConstructors\nConstructor\nDescription\nCap(Schedule schedule, String forwardCurveName, double strike, boolean isStrikeMoneyness, String discountCurveName, String volatilitySurfaceName)\nCreate a Caplet with a given schedule, strike on a given forward curve (by name) with a given discount curve and volatility surface (by name).\nCap(Schedule schedule, String forwardCurveName, double strike, boolean isStrikeMoneyness, String discountCurveName, String volatilitySurfaceName, VolatilitySurface.QuotingConvention quotingConvention)\nCreate a Caplet with a given schedule, strike on a given forward curve (by name) with a given discount curve and volatility surface (by name).\n• ## Method Summary\n\nModifier and Type\nMethod\nDescription\ndouble\ngetATMForward(AnalyticModel model, boolean isFirstPeriodIncluded)\nReturn the ATM forward for this cap.\nString\ngetDiscountCurveName()\nReturns the name of the discount curve referenced by this cap.\nString\ngetForwardCurveName()\nReturns the name of the forward curve references by this cap.\ndouble\ngetImpliedVolatility(double evaluationTime, AnalyticModel model, VolatilitySurface.QuotingConvention quotingConvention)\nReturns the value of this cap in terms of an implied volatility (of a flat caplet surface).\ndouble\ngetStrike()\nReturns the strike of this caplet.\ndouble\ngetValue(double evaluationTime, AnalyticModel model)\nReturn the valuation of the product using the given model.\ndouble\ngetValueAsPrice(double evaluationTime, AnalyticModel model)\nReturns the value of this product under the given model.\nString\ntoString()\n\n### Methods inherited from class net.finmath.marketdata.products.AbstractAnalyticProduct\n\ngetValue, getValue\n\n### Methods inherited from class java.lang.Object\n\nclone, equals, finalize, getClass, hashCode, notify, notifyAll, wait, wait, wait\n\n### Methods inherited from interface net.finmath.modelling.Product\n\ngetValues\n• ## Constructor Details\n\n• ### Cap\n\npublic Cap(Schedule schedule, String forwardCurveName, double strike, boolean isStrikeMoneyness, String discountCurveName, String volatilitySurfaceName, VolatilitySurface.QuotingConvention quotingConvention)\nCreate a Caplet with a given schedule, strike on a given forward curve (by name) with a given discount curve and volatility surface (by name). The valuation is performed using analytic valuation formulas for the underlying caplets.\nParameters:\nschedule - A given payment schedule, i.e., a collection of Periods with fixings, payments and period length.\nforwardCurveName - The forward curve to be used for the forward of the index.\nstrike - The given strike (or moneyness).\nisStrikeMoneyness - If true, then the strike argument is interpreted as moneyness, i.e. we calculate an ATM forward from the schedule.\ndiscountCurveName - The discount curve to be used for discounting.\nvolatilitySurfaceName - The volatility surface to be used.\nquotingConvention - The quoting convention of the value returned by the getValue(double, net.finmath.marketdata.model.AnalyticModel)-method.\n• ### Cap\n\npublic Cap(Schedule schedule, String forwardCurveName, double strike, boolean isStrikeMoneyness, String discountCurveName, String volatilitySurfaceName)\nCreate a Caplet with a given schedule, strike on a given forward curve (by name) with a given discount curve and volatility surface (by name). The valuation is performed using analytic valuation formulas for the underlying caplets.\nParameters:\nschedule - A given payment schedule, i.e., a collection of Periods with fixings, payments and period length.\nforwardCurveName - The forward curve to be used for the forward of the index.\nstrike - The given strike (or moneyness).\nisStrikeMoneyness - If true, then the strike argument is interpreted as moneyness, i.e. we calculate an ATM forward from the schedule.\ndiscountCurveName - The discount curve to be used for discounting.\nvolatilitySurfaceName - The volatility surface to be used.\n• ## Method Details\n\n• ### getValue\n\npublic double getValue(double evaluationTime, AnalyticModel model)\nDescription copied from interface: AnalyticProduct\nReturn the valuation of the product using the given model. The model has to implement the modes of AnalyticModel.\nParameters:\nevaluationTime - The evaluation time as double. Cash flows prior and including this time are not considered.\nmodel - The model under which the product is valued.\nReturns:\nThe value of the product using the given model.\n• ### getValueAsPrice\n\npublic double getValueAsPrice(double evaluationTime, AnalyticModel model)\nReturns the value of this product under the given model.\nParameters:\nevaluationTime - Evaluation time.\nmodel - The model.\nReturns:\nValue of this product und the given model.\n• ### getATMForward\n\npublic double getATMForward(AnalyticModel model, boolean isFirstPeriodIncluded)\nReturn the ATM forward for this cap. The ATM forward is the fixed rate K such that the value of the payoffs $$F(t_i) - K$$ is zero, where $$F(t_i)$$ is the ATM forward of the i-th caplet. Note however that the is a convention to determine the ATM forward of a cap from the payoffs excluding the first one. The reason here is that for non-forward starting cap, the first period is already fixed, i.e. it has no vega.\nParameters:\nmodel - The model to retrieve the forward curve from (by name).\nisFirstPeriodIncluded - If true, the forward will be determined by considering the periods after removal of the first periods (except, if the Cap consists only of 1 period).\nReturns:\nThe ATM forward of this cap.\n• ### getImpliedVolatility\n\npublic double getImpliedVolatility(double evaluationTime, AnalyticModel model, VolatilitySurface.QuotingConvention quotingConvention)\nReturns the value of this cap in terms of an implied volatility (of a flat caplet surface).\nParameters:\nevaluationTime - The evaluation time as double. Cash flows prior and including this time are not considered.\nmodel - The model under which the product is valued.\nquotingConvention - The quoting convention requested for the return value.\nReturns:\nThe value of the product using the given model in terms of a implied volatility.\n• ### getForwardCurveName\n\npublic String getForwardCurveName()\nReturns the name of the forward curve references by this cap.\nReturns:\nthe forward curve name.\n• ### getStrike\n\npublic double getStrike()\nReturns the strike of this caplet.\nReturns:\nthe strike\n• ### getDiscountCurveName\n\npublic String getDiscountCurveName()\nReturns the name of the discount curve referenced by this cap.\nReturns:\nthe discount curve name\n• ### toString\n\npublic String toString()\nOverrides:\ntoString in class Object"
] |
[
null
] |
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|
https://hal-polytechnique.archives-ouvertes.fr/hal-00859451
|
[
"",
null,
"",
null,
"Fluid Dynamic Limits of the Kinetic Theory of Gases - Archive ouverte HAL Access content directly\nConference Papers Year : 2014\n\n## Fluid Dynamic Limits of the Kinetic Theory of Gases\n\n(1)\n1\nFrançois Golse\n• Function : Author\n• PersonId : 964158\n\n#### Abstract\n\nThese three lectures introduce the reader to recent progress on the hydrodynamic limits of the kinetic theory of gases. Lecture 1 outlines the main mathematical results in this direction, and explains in particular how the Euler or Navier-Stokes equations for compressible as well as incompressible fluids, can be derived from the Boltzmann equation. It also presents the notion of renormalized solution of the Boltzmann equation, due to P.-L. Lions and R. DiPerna, together with the mathematical methods used in the proofs of the fluid dynamic limits. Lecture 2 gives a detailed account of the derivation by L. Saint-Raymond of the incompressible Euler equations from the BGK model with constant collision frequency [L. Saint-Raymond, Bull. Sci. Math. 126 (2002), 493-506]. Finally, lecture 3 sketches the main steps in the proof of the incompressible Navier-Stokes limit of the Boltzmann equation, connecting the DiPerna-Lions theory of renormalized solutions of the Boltzmann equation with Leray's theory of weak solutions of the Navier-Stokes system, following [F. Golse, L. Saint-Raymond, J. Math. Pures Appl. 91 (2009), 508-552]. As is the case of all mathematical results in continuum mechanics, the fluid dynamic limits of the Boltzmann equation involve some basic properties of isotropic tensor fields that are recalled in Appendices 1-2.\n\n#### Domains\n\nMathematics [math] Analysis of PDEs [math.AP]\n\n### Dates and versions\n\nhal-00859451 , version 1 (08-09-2013)\n\n### Identifiers\n\n• HAL Id : hal-00859451 , version 1\n• DOI :\n\n### Cite\n\nFrançois Golse. Fluid Dynamic Limits of the Kinetic Theory of Gases. From particle systems to partial differential equations, Dec 2012, University of Minho, Braga, Portugal. viii+320 pp., ⟨10.1007/978-3-642-54271-8_1⟩. ⟨hal-00859451⟩\n\n### Export\n\nBibTeX TEI Dublin Core DC Terms EndNote Datacite\n\n483 View"
] |
[
null,
"https://piwik-hal.ccsd.cnrs.fr//matomo.php",
null,
"https://piwik-hal.ccsd.cnrs.fr//matomo.php",
null
] |
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|
http://mutshs.edu.bd/?p=2113
|
[
"## Whatever They Told You About Solve Equation Calculator with Steps Is Dead Wrong…And Here’s Why\n\nThese worksheets require students to do multiple measures to fix the equations. Math is quite a hard and complicated subject for the majority of us, but it is going to become more simple if you use our math graph solver. However, the Calculator for Trigonometry will let you execute the calculations very easily. Moreover, the equations can be very time-consuming. These equations incorporate certain imaginary numbers too.\n\n## Finding the Best Solve Equation Calculator with Steps\n\nYou can accomplish this by hand, or you may use a graphing calculator. The previous trial gives the right factorization. To acquire the reply to any quadratic calculation, it’s possible to either use manual techniques or a calculator. This calculator can allow you to do calculations easily. It can be used to factor polynomials.\n\n## The War Against Solve Equation Calculator with Steps\n\nThis means that you have to start with employing the golden rule of equation solving and the order of operations, PEMDAS, to create the expression on either side of the equals sign as easy as possible. In an equation there’s always an equality sign. The site also provides a tutorial.\n\n## Solve Equation Calculator with Steps – Overview\n\nYou may also visit our next web pages on various stuff in math. There is a plethora of distinctive choices in math calculators accessible to you on the internet. MathPapa is another site of an on-line calculator with steps. When you open the website, you will observe a search bar in which you’ll be able to put in your problem. Also, the website extends to you 6 distinct languages.\n\nThis is very natural since we increased the range of variables by one, keeping the variety of equations the exact same. So, you’ve got to be cautious when calculating the results. But since we’re older now than when we were filling in boxes, the equations can likewise be far more complicated, and so the methods we’ll utilize to fix the equations are going to be a little more advanced. Solve these inequalities.\n\n## Lies You’ve Been Told About Solve Equation Calculator with Steps\n\nPrice will come close to this level but the majority of the time won’t have the ability to break it. This internet calculator gives you the ability to address differential equations online.\n\nThe calculator or tool is the response to each of these queries. So, this tool will let you fix the equations in a short while. Time has changed since I got this program.\n\nYou may also set the Cauchy problem to the whole set of feasible solutions to choose private appropriate given initial problems. If there’s no inverse, it’s either that the 2 agents are moving parallel to one another, or are moving on the exact same line. To address a system of linear equations using Gauss-Jordan elimination you should do the subsequent steps.\n\nThe thought of the straight line is just one of the intuitive concepts of geometry. The options are in reality limitless. Recall that, as a way to boost the legitimacy of the narrative, the additional details don’t need to be actually scientifically accurate. Analyze the essence of the roots.\n\nIt lets you understand the history also. We learned a number of essential things in thelast episode. Here’s a list of some of the more popular ones. A source of information, and the question you’re attempting to reply.\n\nStandard deviation may be used to figure a minimum and maximum value within which some element of the item should fall some high proportion of the moment. Any calculation in a term is performed left to right (top to bottom), therefore the integral calculatro next step is to produce new terms depending on the order of operations. As stated by the rule, whenever there are a number of operations of the exact same rank, an individual must operate left to right. Suppose we do a full-text search on terms from 1 segment https://sites.google.com/site/mathcalculatoronline/binary-calculator across each of the segments and count the amount of common terms.\n\nThis results in the model to die or not converge whatsoever. Therefore, we must reshape it. For instance, let us examine a traditional one-variable case.\n\n## The Importance of Solve Equation Calculator with Steps\n\nIt’s tricky to understand where to begin. That’ll be a lot less difficult to calculate. This is achieved in OnRender.\n\nWith the incorrect mode, it is possible to actually become incorrect outcomes. You will discover that the moment you begin typing, beneath the input line the expression is going to be shown in the `book format’. But, you’ve got to be mindful with modes. So, choose the mode according to your selection.\n\nTeachers have zero control over if student’s use the web to help them with their homework. I’d finished each of the math and science courses provided by my public high school. The MCQ worksheets form an ideal tool to check student’s knowledge on this subject. The students who sign up for math tutoring online can choose a tutor according to their requirement. An on-line maths tutor can enable you to do practice in a handy way and to score better in your exams.\n\n## The Benefits of Solve Equation Calculator with Steps\n\nHowever, based on the function, the inverse might be tough to be defined, or might not be a function on each of the established B (only on some subset), and have many values sooner or later. EES code permits the user to input equations in any purchase and receive a solution, but in addition can contain if-then statements, which may also be nested within one another to make if-then-else statements. Conversely, a greater standard deviation indicates a larger selection of values.\n\n## The Solve Equation Calculator with Steps Chronicles\n\nAdditionally, you don’t will need to register for each site. Some (mostly recent) research that might be of interest. They might have tried opening DLR’s page to search for any announcements about the competition. A number of the sites allow you to print the calculation while some sites allow you to generate a PDF. So, prepare for the proper and accurate outcomes.\n\nIf you want to gain or get rid of weight, you’re able to also use this number for a point to eat more or less then, respectively. We are not likely to see the way that it’s derived, but in fact see the way that it works. I decided to have a chance with the Algebrator.\n\n## The Hidden Treasure of Solve Equation Calculator with Steps\n\nPlease remember to refer to a health expert if you’re searching to gain or lose a great deal of weight. Since part of the story is factual, it may be interpreted that the remainder of the story is also true, which makes it an excellent way to deceive people. This was probably the very first time I received no idea how to address current endeavor. You team was quick to reply."
] |
[
null
] |
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|
https://tutorbin.com/questions-and-answers/2-which-of-the-following-forecasting-models-is-superior-for-predicting
|
[
"Question\n\n# 2. Which of the following forecasting models is superior for predicting future sales based on the sum of squared differences criterion? a) Constant model with a constant prediction of 102 b) Last period model c) Moving average model with n = 2 d) Moving average model with n= 3 e) Weighted moving average model with n= 2, C1 =0.6 , C2 =0.4 f) Weighted moving average model with z-3, C1= 6, C2 = 4, C3= 2 g) Exponentially weighted moving average model with xo = 76 and a = 0.35 h) Exponentially weighted moving average model with xo = 76 and a = 0.7",
null,
"",
null,
"Fig: 1",
null,
"",
null,
"Fig: 2",
null,
"",
null,
"Fig: 3",
null,
"",
null,
"Fig: 4",
null,
"",
null,
"Fig: 5",
null,
"",
null,
"Fig: 6",
null,
"",
null,
"Fig: 7",
null,
"",
null,
"Fig: 8",
null,
"",
null,
"Fig: 9"
] |
[
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null
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|
https://en.wikipedia.org/wiki/Staudt-clausen_theorem
|
[
"# Von Staudt–Clausen theorem\n\n(Redirected from Staudt-clausen theorem)\n\nIn number theory, the von Staudt–Clausen theorem is a result determining the fractional part of Bernoulli numbers, found independently by Karl von Staudt (1840) and Thomas Clausen (1840).\n\nSpecifically, if n is a positive integer and we add 1/p to the Bernoulli number B2n for every prime p such that p − 1 divides 2n, we obtain an integer, i.e., $B_{2n}+\\sum _{(p-1)|2n}{\\frac {1}{p}}\\in \\mathbb {Z} .$",
null,
"This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers B2n as the product of all primes p such that p − 1 divides 2n; consequently the denominators are square-free and divisible by 6.\n\nThese denominators are\n\n6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, ... (sequence A002445 in the OEIS).\n\nThe sequence of integers $B_{2n}+\\sum _{(p-1)|2n}{\\frac {1}{p}}$",
null,
"is\n\n1, 1, 1, 1, 1, 1, 2, -6, 56, -528, 6193, -86579, 1425518, -27298230, ... (sequence A000146 in the OEIS).\n\n## Proof\n\nA proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:\n\n$B_{2n}=\\sum _{j=0}^{2n}{\\frac {1}{j+1}}\\sum _{m=0}^{j}{(-1)^{m}{j \\choose m}m^{2n}}$",
null,
"and as a corollary:\n\n$B_{2n}=\\sum _{j=0}^{2n}{\\frac {j!}{j+1}}(-1)^{j}S(2n,j)$",
null,
"where $S(n,j)$",
null,
"are the Stirling numbers of the second kind.\n\nFurthermore the following lemmas are needed:\nLet p be a prime number then,\n1. If p-1 divides 2n then,\n\n$\\sum _{m=0}^{p-1}{(-1)^{m}{p-1 \\choose m}m^{2n}}\\equiv {-1}{\\pmod {p}}$",
null,
"2. If p-1 does not divide 2n then,\n\n$\\sum _{m=0}^{p-1}{(-1)^{m}{p-1 \\choose m}m^{2n}}\\equiv 0{\\pmod {p}}$",
null,
"Proof of (1) and (2): One has from Fermat's little theorem,\n\n$m^{p-1}\\equiv 1{\\pmod {p}}$",
null,
"for $m=1,2,...,p-1$",
null,
".\nIf p-1 divides 2n then one has,\n\n$m^{2n}\\equiv 1{\\pmod {p}}$",
null,
"for $m=1,2,...,p-1$",
null,
".\nThereafter one has,\n\n$\\sum _{m=1}^{p-1}{(-1)^{m}{p-1 \\choose m}m^{2n}}\\equiv \\sum _{m=1}^{p-1}{(-1)^{m}{p-1 \\choose m}}{\\pmod {p}}$",
null,
"from which (1) follows immediately.\nIf p-1 does not divide 2n then after Fermat's theorem one has,\n\n$m^{2n}\\equiv m^{2n-(p-1)}{\\pmod {p}}$",
null,
"If one lets $\\wp =[{\\frac {2n}{p-1}}]$",
null,
"(Greatest integer function) then after iteration one has,\n\n$m^{2n}\\equiv m^{2n-\\wp (p-1)}{\\pmod {p}}$",
null,
"for $m=1,2,...,p-1$",
null,
"and $0<2n-\\wp (p-1)",
null,
".\nThereafter one has,\n\n$\\sum _{m=0}^{p-1}{(-1)^{m}{p-1 \\choose m}m^{2n}}\\equiv \\sum _{m=0}^{p-1}{(-1)^{m}{p-1 \\choose m}m^{2n-\\wp (p-1)}}{\\pmod {p}}$",
null,
"Lemma (2) now follows from the above and the fact that S(n,j)=0 for j>n.\n(3). It is easy to deduce that for a>2 and b>2, ab divides (ab-1)!.\n(4). Stirling numbers of second kind are integers.\n\nProof of the theorem: Now we are ready to prove Von-Staudt Clausen theorem,\nIf j+1 is composite and j>3 then from (3), j+1 divides j!.\nFor j=3,\n\n$\\sum _{m=0}^{3}{(-1)^{m}{3 \\choose m}m^{2n}}=3\\cdot 2^{2n}-3^{2n}-3\\equiv 0{\\pmod {4}}$",
null,
"If j+1 is prime then we use (1) and (2) and if j+1 is composite then we use (3) and (4) to deduce:\n\n$B_{2n}=I_{n}-\\sum _{(p-1)|2n}{\\frac {1}{p}}$",
null,
"where $I_{n}$",
null,
"is an integer, which is the Von-Staudt Clausen theorem."
] |
[
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/4d182fe69db4e14e3766f5af0b4b4dcddb48befa",
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|
https://www.tutorialspoint.com/java-program-to-fill-elements-in-a-byte-array
|
[
"# Java Program to fill elements in a byte array\n\nElements can be filled in a byte array using the java.util.Arrays.fill() method. This method assigns the required byte value to the byte array in Java. The two parameters required are the array name and the value that is to be stored in the array elements.\n\nA program that demonstrates this is given as follows −\n\n## Example\n\nLive Demo\n\nimport java.util.Arrays;\npublic class Demo {\npublic static void main(String[] argv) throws Exception {\nbyte[] byteArray = new byte;\nbyte byteValue = 5;\nArrays.fill(byteArray, byteValue);\nSystem.out.println(\"The byte array content is: \" + Arrays.toString(byteArray));\n}\n}\n\n## Output\n\nThe byte array content is: [5, 5, 5, 5, 5]\n\nNow let us understand the above program.\n\nFirst the byte array byteArray[] is defined. Then the value 5 is filled in the byte array using the Arrays.fill() method. Finally, the byte array is printed using the Arrays.toString() method. A code snippet which demonstrates this is as follows −\n\nbyte[] byteArray = new byte;\nbyte byteValue = 5;\nArrays.fill(byteArray, byteValue);\nSystem.out.println(\"The byte array content is: \" + Arrays.toString(byteArray));"
] |
[
null
] |
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|
http://onlinetutorpakistan.com/gradient-slope-of-straight-line-grade-6-quiz/
|
[
"##",
null,
"Find Slope (gradient) of Straight line passes through the points (4,2) and (6,3)\n\nFind Slope (gradient) of Straight line passes through the points (-4,5) and (1,2)\n\nFind Slope (gradient) of Straight line passes through the points (-3,4) and (7,-6)\n\nFind Slope (gradient) of Straight line passes through the points (3a, -2a ) and (7a, 2a)\n\nFind Slope (gradient) of Straight line passes through the points (0, 5x) and (10x, 0)\n\nThe Line joining (3, -5) to (6, x) has a gradient 4. Find the value of x\n\nThe line joining (5, x) to (8, 3 ) has gradient -3 . Find the value of y.\n\nThe line joining (x, 4) to (7, 6) has a gradient 3/4, find x\n\nThe line joining the points (-3, -2) to (2b, 5) has a gradient 2. Find b\n\nThe line joining the points (3,-4) to (-b, 2b) has gradient -3, find the value of b\n\nFor More Quiz Visit Math Quiz\n\nOnline Tuition Pakistan\n\nSite Protection is enabled by using WP Site Protector from Exattosoft.com"
] |
[
null,
"http://onlinetutorpakistan.com/wp-content/uploads/2018/08/Gradient-Slope-of-Straight-Line-Grade-6-Quiz.jpeg",
null
] |
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|
https://www.hepdata.net/search/?observables=SIG&page=1&author=Flechl%2C+Martin
|
[
"Showing 25 of 844 results\n\n#### Measurement of the $t\\bar{t}$ production cross-section in the lepton+jets channel at $\\sqrt{s}=13\\;$TeV with the ATLAS experiment\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\n2020.\nInspire Record 1802524\n\nThe $t\\bar{t}$ production cross-section is measured in the lepton+jets channel using proton$-$proton collision data at a centre-of-mass energy of $\\sqrt{s}=13$ TeV collected with the ATLAS detector at the LHC. The dataset corresponds to an integrated luminosity of 139 fb$^{-1}$. Events with exactly one charged lepton and four or more jets in the final state, with at least one jet containing $b$-hadrons, are used to determine the $t\\bar{t}$ production cross-section through a profile-likelihood fit. The inclusive cross-section is measured to be ${\\sigma_{\\text{inc}} = 830 \\pm 0.4~\\text{(stat.)}\\pm 36~\\text{(syst.)}\\pm 14~\\text{(lumi.)}~\\mathrm{pb}}$ with a relative uncertainty of 4.6%. The result is consistent with theoretical calculations at next-to-next-to-leading order in perturbative QCD.\n\n5 data tables\n\nThe results of fitted inclusive and fiducial ${t\\bar{t}}$ cross-sections\n\nRanking of the systematic uncertainties on the measured cross-section, normalised to the predicted value, in the inclusive fit to data. The impact of each nuisance parameter, $\\Delta \\sigma_{\\text{inc}}/\\sigma^{\\text{pred.}}_{\\text{inc}}$, is computed by comparing the nominal best-fit value of $\\sigma_{\\text{inc}}/\\sigma^{\\text{pred}}_{\\text{inc}}$ with the result of the fit when fixing the considered nuisance parameter to its best-fit value, $\\theta$, shifted by its pre-fit (post-fit) uncertainties $\\pm \\Delta \\theta$ ($\\pm \\Delta \\hat{\\theta}$). The figure shows the effect of the ten most significant uncertainties.\n\nRanking of the systematic uncertainties on the measured cross-section, normalised to the predicted value, in the fiducial fit to data. The impact of each nuisance parameter, $\\Delta \\sigma_{\\text{fid}}/\\sigma^{\\text{pred.}}_{\\text{fid}}$, is computed by comparing the nominal best-fit value of $\\sigma_{\\text{fid}}/\\sigma^{\\text{pred}}_{\\text{fid}}$ with the result of the fit when fixing the considered nuisance parameter to its best-fit value, $\\theta$, shifted by its pre-fit (post-fit) uncertainties $\\pm \\Delta \\theta$ ($\\pm \\Delta \\hat{\\theta}$). The figure shows the effect of the ten most significant uncertainties.\n\nMore…\n\n#### Dependence of inclusive jet production on the anti-$k_\\mathrm{T}$ distance parameter in pp collisions at $\\sqrt{s} =$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nJHEP and tables can be found at http://cms-results.web.cern.ch/cms-results/public-results/publications/SMP-19-003 (CMS Public Pages), 2020.\nInspire Record 1795080\n\nThe dependence of inclusive jet production in proton-proton collisions with a center-of-mass energy of 13 TeV on the distance parameter $R$ of the anti-$k_\\mathrm{T}$ algorithm is studied using data corresponding to integrated luminosities up to 35.9 fb$^{-1}$ collected by the CMS experiment in 2016. The ratios of the inclusive cross sections as functions of transverse momentum $p_\\mathrm{T}$ and rapidity $y$, for $R$ in the range 0.1 to 1.2 to those using $R=$ 0.4 are presented in the region 84 $\\lt p_\\mathrm{T} \\lt$ 1588 GeV and $|y|\\lt$ 2.0. The results are compared to calculations at leading and next-to-leading order in the strong coupling constant using different parton shower models. The variation of the ratio of cross sections with $R$ is well described by calculations including a parton shower model, but not by a leading-order quantum chromodynamics calculation including nonperturbative effects. The agreement between the data and the theoretical predictions for the ratios of cross sections is significantly improved when next-to-leading order calculations with nonperturbative effects are used.\n\n88 data tables\n\nRatio of differential cross section of AK1 jets with respect to AK4 jets a function of jet PT in the rapidity range |y|<0.5. The nonperturbative correction can be used to scale fixed-order theory prediction to compare to data at particle level.\n\nRatio of differential cross section of AK1 jets with respect to AK4 jets a function of jet PT in the rapidity range 0.5<|y|<1.0. The nonperturbative correction can be used to scale fixed-order theory prediction to compare to data at particle level.\n\nRatio of differential cross section of AK1 jets with respect to AK4 jets a function of jet PT in the rapidity range 1.0<|y|<1.5. The nonperturbative correction can be used to scale fixed-order theory prediction to compare to data at particle level.\n\nMore…\n\n#### Observation of electroweak production of two jets and a $Z$-boson pair with the ATLAS detector at the LHC\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\n2020.\nInspire Record 1792133\n\nElectroweak symmetry breaking explains the origin of the masses of elementary particles via their interactions with the Higgs field. Besides the measurements of the Higgs boson properties, the study of the scattering of massive vector bosons (with spin one) at the Large Hadron Collider allows to probe the nature of electroweak symmetry breaking with an unprecedented sensitivity. Among all processes related to vector-boson scattering, the electroweak production of two jets and a $Z$-boson pair is a rare and important one. This article reports on the first observation of this process using proton-proton collision data corresponding to an integrated luminosity of 139 fb$^{-1}$ recorded at a centre-of-mass energy of 13 TeV by the ATLAS detector. Two different final states originating from the decays of the $Z$-boson pair, one containing four charged leptons and the other containing two charged leptons and two neutrinos, are considered. The hypothesis of no electroweak production is rejected with a statistical significance of 5.5 $\\sigma$, and the measured cross-section for electroweak production is consistent with the Standard Model prediction. In addition, cross-sections for inclusive production of a $Z$-boson pair and two jets are reported for the two final states.\n\n1 data table\n\nMeasured and predicted fiducial cross-sections in both the lllljj and ll$\\nu\\nu$jj channels for the inclusive ZZjj processes. Uncertainties due to different sources are presented\n\n#### Measurement of the cross section for electroweak production of a Z boson, a photon and two jets in proton-proton collisions at $\\sqrt{s} =$ 13 TeV and constraints on anomalous quartic couplings\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nJHEP 06 (2020) 076, 2020.\nInspire Record 1781935\n\nA measurement is presented of the cross section for electroweak production of a Z boson and a photon in association with two jets (Z$\\gamma$jj) in proton-proton collisions. The Z boson candidates are selected through their decay into a pair of electrons or muons. The process of interest, electroweak Z$\\gamma$jj production, is isolated by selecting events with a large dijet mass and a large pseudorapidity gap between the two jets. The measurement is based on data collected at the CMS experiment at $\\sqrt{s} =$ 13 TeV, corresponding to an integrated luminosity of 35.9 fb$^{-1}$. The observed significance of the signal is 3.9 standard deviations, where a significance of 5.2 standard deviations is expected in the standard model. These results are combined with published results by CMS at $\\sqrt{s} =$ 8 TeV, which leads to observed and expected respective significances of 4.7 and 5.5 standard deviations. From the 13 TeV data, a value is obtained for the signal strength of electroweak Z$\\gamma$jj production and bounds are given on quartic vector boson interactions in the framework of dimension-eight effective field theory operators.\n\n3 data tables\n\nThe measured EWK Zgamma+2j fiducial cross section. The uncertainty is the combined stastical uncertianty and the systematic uncertainty including experimental and theortical sources\n\nThe measured combined QCD-induced and EWK Zgamma+2j fiducial cross section. The uncertainty is the combined stastical uncertianty and the systematic uncertainty including experimental and theortical sources\n\naQGC limits on effective field theory parameters in EWK Zgamma events\n\n#### Measurement of the $\\Upsilon$(1S) pair production cross section and search for resonances decaying to $\\Upsilon$(1S)$\\mu^+\\mu^-$ in proton-proton collisions at $\\sqrt{s} =$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nPhys.Lett.B 808 (2020) 135578, 2020.\nInspire Record 1780982\n\nThe fiducial cross section for $\\Upsilon$(1S) pair production in proton-proton collisions at a center-of-mass energy of 13 TeV in the region where both $\\Upsilon$(1S) mesons have an absolute rapidity below 2.0 is measured to be 79 $\\pm$ 11 (stat) $\\pm$ 6 (syst) $\\pm$ 3 ($\\mathcal{B}$) pb assuming the mesons are produced unpolarized. The last uncertainty corresponds to the uncertainty in the $\\Upsilon$(1S) meson dimuon branching fraction. The measurement is performed in the final state with four muons using proton-proton collision data collected in 2016 by the CMS experiment at the LHC, corresponding to an integrated luminosity of 35.9 fb$^{-1}$. This process serves as a standard model reference in a search for narrow resonances decaying to $\\Upsilon$(1S)$\\mu^+\\mu^-$ in the same final state. Such a resonance could indicate the existence of a tetraquark that is a bound state of two b quarks and two $\\bar{\\mathrm{b}}$ antiquarks. The tetraquark search is performed for masses in the vicinity of four times the bottom quark mass, between 17.5 and 19.5 GeV, while a generic search for other resonances is performed for masses between 16.5 and 27 GeV. No significant excess of events compatible with a narrow resonance is observed in the data. Limits on the production cross section times branching fraction to four muons via an intermediate $\\Upsilon$(1S) resonance are set as a function of the resonance mass.\n\n9 data tables\n\nThe fiducial cross section measured in bins of the absolute rapidity difference between the mesons for events in the fiducial region with 2 Y(1S) with absolute rapidity less than 2.0.\n\nThe fiducial cross section measured in bins of the invariant mass of the two mesons for events in the fiducial region with 2 Y(1S) with absolute rapidity less than 2.0.\n\nThe fiducial cross section measured in bins of the transverse momentum of the meson pair for events in the fiducial region with 2 Y(1S) with absolute rapidity less than 2.0.\n\nMore…\n\n#### Measurement of the top quark pair production cross section in dilepton final states containing one $\\tau$ lepton in pp collisions at $\\sqrt{s}=$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nJHEP 02 (2020) 191, 2020.\nInspire Record 1767671\n\nThe cross section of top quark pair production is measured in the $\\mathrm{t\\bar{t}}\\to (\\ell\\nu_{\\ell})(\\tau_\\mathrm{h}\\nu_{\\tau})\\mathrm{b\\bar{b}}$ final state, where $\\tau_\\mathrm{h}$ refers to the hadronic decays of the $\\tau$ lepton, and $\\ell$ is either an electron or a muon. The data sample corresponds to an integrated luminosity of 35.9 fb$^{-1}$ collected in proton-proton collisions at $\\sqrt{s}=$ 13 TeV with the CMS detector. The measured cross section is $\\sigma_{\\mathrm{t\\bar{t}}} =$ 781 $\\pm$ 7 (stat) $\\pm$ 62 (syst) $\\pm$ 20 (lum) pb, and the ratio of the partial width $\\Gamma($t$\\to\\tau\\nu_{\\tau}$b) to the total decay width of the top quark is measured to be 0.1050 $\\pm$ 0.0009 (stat) $\\pm$ 0.0071 (syst). This is the first measurement of the $\\mathrm{t\\bar{t}}$ production cross section in proton-proton collisions at $\\sqrt{s}=$ 13 TeV that explicitly includes $\\tau$ leptons. The ratio of the cross sections in the $\\ell\\tau_\\mathrm{h}$ and $\\ell\\ell$ final states yields a value $R_{\\ell\\tau_\\mathrm{h}/\\ell\\ell}=$ 0.973 $\\pm$ 0.009 (stat) $\\pm$ 0.066 (syst), consistent with lepton universality.\n\n3 data tables\n\nThe measured inclusive top quark pair production cross section in the dilepton final state with one tau lepton.\n\nThe ratio between top quark production cross sections measured in lepton-tau and light dilepton final states.\n\nThe ratio of the partial width to the total decay width of the top quark.\n\n#### Version 2 rivet Analysis Measurement of the $Z(\\rightarrow\\ell^+\\ell^-)\\gamma$ production cross-section in $pp$ collisions at $\\sqrt{s} =13$ TeV with the ATLAS detector\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nJHEP 03 (2020) 054, 2020.\nInspire Record 1764342\n\nThe production of a prompt photon in association with a $Z$ boson is studied in proton-proton collisions at a centre-of-mass energy $\\sqrt{s} =$ 13 TeV. The analysis uses a data sample with an integrated luminosity of 139 fb$^{-1}$ collected by the ATLAS detector at the LHC from 2015 to 2018. The production cross-section for the process $pp \\rightarrow \\ell^+\\ell^-\\gamma+X$ ($\\ell = e, \\mu$) is measured within a fiducial phase-space region defined by kinematic requirements on the photon and the leptons, and by isolation requirements on the photon. An experimental precision of 2.9% is achieved for the fiducial cross-section. Differential cross-sections are measured as a function of each of six kinematic variables characterising the $\\ell^+\\ell^-\\gamma$ system. The data are compared with theoretical predictions based on next-to-leading-order and next-to-next-to-leading-order perturbative QCD calculations. The impact of next-to-leading-order electroweak corrections is also considered.\n\n7 data tables\n\nThe measured fiducial cross section. \"Uncor\" uncertainty includes all systematic uncertainties that are uncorrelated between electron and muon channels such as the uncertainty on the electron identification efficiency and the uncorrelated component of the background uncertainties. The parton-to-particle correction factor $C_{theory}$ is the ratio of the cross-section predicted by Sherpa LO samples at particle level within the fiducial phase-space region defined in Table 4 to the predicted cross-section at parton level within the same fiducial region but with the smooth-cone isolation prescription defined above replacing the particle-level photon isolation criterion, and with Born-level leptons in place of dressed leptons. This correction should be applied on fixed order parton-level calculations. The systematic uncertainty is evaluated from a comparison with the correction factor obtained using events generated with SHERPA 2.2.2 at NLO. In the case that the calculations are valid for dressed leptons, a modified correction factor excluding the Born-to-dressed lepton correction should be applied instead. This correction only takes into account the particle-level isolation criteria, and is provided separately here. The Sherpa 2.2.8 NLO cross-sections given below include a small contribution from EW $Z\\gamma jj$ production of 4.57 fb.\n\nThe measured fiducial cross section vs $E_{\\mathrm{T}}^\\gamma$. The central values are provided along with the statistical and systematic uncertainties together with the sign information. The statistical and \"Uncor\" uncertainty should be treated as uncorrelated bin-to-bin, while the rest are correlated between bins, and they are written as signed NP variations. The parton-to-particle correction factor $C_{theory}$ is the ratio of the cross-section predicted by Sherpa LO samples at particle level within the fiducial phase-space region defined in Table 4 to the predicted cross-section at parton level within the same fiducial region but with the smooth-cone isolation prescription defined above replacing the particle-level photon isolation criterion, and with Born-level leptons in place of dressed leptons. This correction should be applied on fixed order parton-level calculations. The systematic uncertainty is evaluated from a comparison with the correction factor obtained using events generated with SHERPA 2.2.2 at NLO. In the case that the calculations are valid for dressed leptons, a modified correction factor excluding the Born-to-dressed lepton correction should be applied instead. This correction only takes into account the particle-level isolation criteria, and is provided separately here. The Sherpa 2.2.8 NLO cross-sections given below include a small contribution from EW $Z\\gamma jj$ production.\n\nThe measured fiducial cross section vs $|\\eta^\\gamma|$. The central values are provided along with the statistical and systematic uncertainties together with the sign information. The statistical and \"Uncor\" uncertainty should be treated as uncorrelated bin-to-bin, while the rest are correlated between bins, and they are written as signed NP variations. The parton-to-particle correction factor $C_{theory}$ is the ratio of the cross-section predicted by Sherpa LO samples at particle level within the fiducial phase-space region defined in Table 4 to the predicted cross-section at parton level within the same fiducial region but with the smooth-cone isolation prescription defined above replacing the particle-level photon isolation criterion, and with Born-level leptons in place of dressed leptons. This correction should be applied on fixed order parton-level calculations. The systematic uncertainty is evaluated from a comparison with the correction factor obtained using events generated with SHERPA 2.2.2 at NLO. In the case that the calculations are valid for dressed leptons, a modified correction factor excluding the Born-to-dressed lepton correction should be applied instead. This correction only takes into account the particle-level isolation criteria, and is provided separately here. The Sherpa 2.2.8 NLO cross-sections given below include a small contribution from EW $Z\\gamma jj$ production.\n\nMore…\n\n#### Measurement of differential cross sections for single diffractive dissociation in $\\sqrt{s} = 8$ TeV $pp$ collisions using the ATLAS ALFA spectrometer\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nJHEP 02 (2020) 042, 2020.\nInspire Record 1762584\n\nA dedicated sample of Large Hadron Collider proton-proton collision data at centre-of-mass energy $\\sqrt{s}=8$ TeV is used to study inclusive single diffractive dissociation, $pp \\rightarrow Xp$. The intact final-state proton is reconstructed in the ATLAS ALFA forward spectrometer, while charged particles from the dissociated system $X$ are measured in the central detector components. The fiducial range of the measurement is $-4.0 < \\log_{10} \\xi < -1.6$ and $0.016 < |t| < 0.43 \\ {\\rm GeV^2}$, where $\\xi$ is the proton fractional energy loss and $t$ is the squared four-momentum transfer. The total cross section integrated across the fiducial range is $1.59 \\pm 0.13 \\ {\\rm mb}$. Cross sections are also measured differentially as functions of $\\xi$, $t$, and $\\Delta \\eta$, a variable that characterises the rapidity gap separating the proton and the system $X$. The data are consistent with an exponential $t$ dependence, ${\\rm d} \\sigma / {\\rm d} t \\propto \\text{e}^{Bt}$ with slope parameter $B = 7.65 \\pm 0.34 \\ {\\rm GeV^{-2}}$. Interpreted in the framework of triple Regge phenomenology, the $\\xi$ dependence leads to a pomeron intercept of $\\alpha(0) = 1.07 \\pm 0.09$.\n\n3 data tables\n\nHadron-level differential SD cross section as a function of Delta Eta.\n\nHadron-level differential SD cross section as a function of t.\n\nHadron-level differential SD cross section as a function of log_10 xi.\n\n#### Underlying Event properties in pp collisions at $\\sqrt{s}$ = 13 TeV\n\nThe collaboration Acharya, Shreyasi ; Adamova, Dagmar ; Adler, Alexander ; et al.\nJHEP 04 (2020) 192, 2020.\nInspire Record 1762350\n\nThis article reports measurements characterizing the Underlying Event (UE) associated with hard scatterings at midrapidity in pp collisions at $\\sqrt{s}=13$ TeV. The hard scatterings are identified by the leading particle, the charged particle with the highest transverse momentum ($p_{\\rm T}^{\\rm leading}$) in the event. Charged-particle number and summed transverse-momentum densities are measured in different azimuthal regions defined with respect to the leading particle direction: Toward, Transverse, and Away. The Toward and Away regions contain the fragmentation products of the hard scatterings in addition to the UE contribution, whereas particles in the Transverse region are expected to originate predominantly from the UE. The study is performed as a function of $p_{\\rm T}^{\\rm leading}$ with three different $p_{\\rm T}$ thresholds for the associated particles, $p_{\\rm T}^{\\rm min} >$ 0.15, 0.5, and 1.0 GeV/$c$. The charged-particle density in the Transverse region rises steeply for low values of $p_{\\rm T}^{\\rm leading}$ and reaches a plateau. The results confirm the trend that the charged-particle density in the Transverse region shows a stronger increase with $\\sqrt{s}$ than the inclusive charged-particle density at midrapidity. The UE activity is increased by approximately 20% when going from 7 to 13 TeV. The plateau in the Transverse region ($5 < p_{\\rm T}^{\\rm leading} < ~ 40$ GeV/$c$ ) is further characterized by the probability distribution of its charged-particle multiplicity normalized to its average value (relative transverse activity, $R_{T}$) and the mean transverse momentum as a function of $R_{T}$. Experimental results are compared to model calculations using PYTHIA 8 and EPOS LHC. The overall agreement between models and data is within 30%. These measurements provide new insights on the interplay between hard scatterings and the associated UE in pp collisions.\n\n5 data tables\n\nFig. 3: Number density $N_{ch}$ (left) and $\\\\Sigma p_{T}$ (right) distributions as a function of $p_{T}^{leading}$ in Toward, Transverse, and Away regions for $p_{T}^{track} >$ 0.15 GeV/$c$. The shaded areas represent the systematic uncertainties and vertical error bars indicate statistical uncertainties.\n\nFig. 9: R_T probability distribution in the Transverse region for $p_{T}^{track} >$ 0.15 GeV/$c$ and $|\\\\eta|<$ 0.8. The result (solid circles) is compared to the PYTHIA 8 and EPOS LHC calculations (lines). The red line represents the result of the NBD fit, where the multiplicity is scaled by its mean value, m. The parameter k is related to the standard deviation of the distribution via $\\\\sigma$ = $\\\\sqrt{ \\\\frac{1}{m} + \\\\frac{1}{k} }$. The open boxes represent the systematic uncertainties and vertical error bars indicate statistical uncertainties. No uncertainties are shown for the MC calculations. The bottom panel shows the ratio between the NBD fit, as well as those of the MC to the data.\n\nFig. 10: $<p_{T}>$ in the Transverse region as a function of $R_{T}$ for $p_{T}^{track} >$ 0.15 GeV/$c$ and $|\\\\eta|<$ 0.8. Data (solid circles) are compared to the results of PYTHIA 8 and EPOS LHC calculations (lines). The open boxes represent the systematic uncertainties and vertical error bars indicate statistical uncertainties. No uncertainties are shown for the MC calculations. The bottom panel shows the ratio of the MC to data.\n\nMore…\n\n#### Search for a heavy pseudoscalar Higgs boson decaying into a 125 GeV Higgs boson and a Z boson in final states with two tau and two light leptons at $\\sqrt{s}=$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nJHEP 03 (2020) 065, 2020.\nInspire Record 1761088\n\nA search is performed for a pseudoscalar Higgs boson, A, decaying into a 125 GeV Higgs boson h and a Z boson. The h boson is specifically targeted in its decay into a pair of tau leptons, while the Z boson decays into a pair of electrons or muons. A data sample of proton-proton collisions collected by the CMS experiment at the LHC at $\\sqrt{s} =$ 13 TeV is used, corresponding to an integrated luminosity of 35.9 fb$^{-1}$. No excess above the standard model background expectations is observed in data. A model-independent upper limit is set on the product of the gluon fusion production cross section for the A boson and the branching fraction to Zh$\\to\\ell\\ell\\tau\\tau$. The observed upper limit at 95% confidence level ranges from 27 to 5 fb for A boson masses from 220 to 400 GeV, respectively. The results are used to constrain the extended Higgs sector parameters for two benchmark scenarios of the minimal supersymmetric standard model.\n\n1 data table\n\nThe expected and observed 95% CL model-independent upper limits on the product of the cross section and branching fraction for the A boson (pseudoscalar Higgs boson).\n\n#### Search for new resonances in mass distributions of jet pairs using 139 fb$^{-1}$ of $pp$ collisions at $\\sqrt{s}=13$ TeV with the ATLAS detector\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nJHEP 03 (2020) 145, 2020.\nInspire Record 1759712\n\nA search for new resonances decaying into a pair of jets is reported using the dataset of proton-proton collisions recorded at $\\sqrt{s}=13$ TeV with the ATLAS detector at the Large Hadron Collider between 2015 and 2018, corresponding to an integrated luminosity of 139 fb$^{-1}$. The distribution of the invariant mass of the two leading jets is examined for local excesses above a data-derived estimate of the Standard Model background. In addition to an inclusive dijet search, events with jets identified as containing $b$-hadrons are examined specifically. No significant excess of events above the smoothly falling background spectra is observed. The results are used to set cross-section upper limits at 95% confidence level on a range of new physics scenarios. Model-independent limits on Gaussian-shaped signals are also reported. The analysis looking at jets containing $b$-hadrons benefits from improvements in the jet flavour identification at high transverse momentum, which increases its sensitivity relative to the previous analysis beyond that expected from the higher integrated luminosity.\n\n24 data tables\n\nThe probability of an event to pass the b-tagging requirement after the rest of the event selection, shown as a function of the resonance mass and for the 1b and 2b analysis categories.\n\nDijet invariant mass distribution for the inclusive category with |y*| < 0.6.\n\nDijet invariant mass distribution for the inclusive category with |y*| < 1.2.\n\nMore…\n\n#### Measurement of the $\\mathrm{t\\bar{t}}\\mathrm{b\\bar{b}}$ production cross section in the all-jet final state in pp collisions at $\\sqrt{s} =$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nPhys.Lett.B 803 (2020) 135285, 2020.\nInspire Record 1753720\n\nA measurement of the production cross section of top quark pairs in association with two b jets ($\\mathrm{t\\bar{t}}\\mathrm{b\\bar{b}}$) is presented using data collected in proton-proton collisions at $\\sqrt{s} =$ 13 TeV by the CMS detector at the LHC corresponding to an integrated luminosity of 35.9 fb$^{-1}$. The cross section is measured in the all-jet decay channel of the top quark pair by selecting events containing at least eight jets, of which at least two are identified as originating from the hadronization of b quarks. A combination of multivariate analysis techniques is used to reduce the large background from multijet events not containing a top quark pair, and to help discriminate between jets originating from top quark decays and other additional jets. The cross section is determined for the total phase space to be 5.5 $\\pm$ 0.3 (stat)${}^{+1.6}_{-1.3}$ (syst) pb and also measured for two fiducial $\\mathrm{t\\bar{t}}\\mathrm{b\\bar{b}}$ definitions. The measured cross sections are found to be larger than theoretical predictions by a factor of 1.5-2.4, corresponding to 1-2 standard deviations.\n\n1 data table\n\nThe measured cross sections. The first uncertainty is statistical, the second uncertianty is the systematic.\n\n#### Search for light long-lived neutral particles produced in $pp$ collisions at $\\sqrt{s} =$ 13 TeV and decaying into collimated leptons or light hadrons with the ATLAS detector\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nEur.Phys.J.C 80 (2020) 450, 2020.\nInspire Record 1752519\n\nSeveral models of physics beyond the Standard Model predict the existence of dark photons, light neutral particles decaying into collimated leptons or light hadrons. This paper presents a search for long-lived dark photons produced from the decay of a Higgs boson or a heavy scalar boson and decaying into displaced collimated Standard Model fermions. The search uses data corresponding to an integrated luminosity of 36.1 fb$^{-1}$ collected in proton-proton collisions at $\\sqrt{s} =$ 13 TeV recorded in 2015-2016 with the ATLAS detector at the Large Hadron Collider. The observed number of events is consistent with the expected background, and limits on the production cross section times branching fraction as a function of the proper decay length of the dark photon are reported. A cross section times branching fraction above 4 pb is excluded for a Higgs boson decaying into two dark photons for dark-photon decay lengths between 1.5 mm and 307 mm.\n\n19 data tables\n\nUpper limits at 95% CL on the cross section times branching fraction for the process $H \\to 2\\gamma_d + X$ with $m_H$ = 125 GeV in the muon-muon final state.\n\nUpper limits at 95% CL on the cross section times branching fraction for the process $H \\to 4\\gamma_d + X$ with $m_H$ = 125 GeV in the muon-muon final state.\n\nUpper limits at 95% CL on the cross section times branching fraction for the process $H \\to 2\\gamma_d + X$ with $m_H$ = 800 GeV in the muon-muon final state.\n\nMore…\n\n#### Search for supersymmetry in proton-proton collisions at 13 TeV in final states with jets and missing transverse momentum\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nJHEP 10 (2019) 244, 2019.\nInspire Record 1749379\n\nResults are reported from a search for supersymmetric particles in the final state with multiple jets and large missing transverse momentum. The search uses a sample of proton-proton collisions at $\\sqrt{s} =$ 13 TeV collected with the CMS detector in 2016-2018, corresponding to an integrated luminosity of 137 fb$^{-1}$, representing essentially the full LHC Run 2 data sample. The analysis is performed in a four-dimensional search region defined in terms of the number of jets, the number of tagged bottom quark jets, the scalar sum of jet transverse momenta, and the magnitude of the vector sum of jet transverse momenta. No significant excess in the event yield is observed relative to the expected background contributions from standard model processes. Limits on the pair production of gluinos and squarks are obtained in the framework of simplified models for supersymmetric particle production and decay processes. Assuming the lightest supersymmetric particle to be a neutralino, lower limits on the gluino mass as large as 2000 to 2310 GeV are obtained at 95% confidence level, while lower limits on the squark mass as large as 1190 to 1630 GeV are obtained, depending on the production scenario.\n\n90 data tables\n\nObserved yields and pre-fit background predictions for Njets 2-3.\n\nObserved yields and pre-fit background predictions for Njets 4-5.\n\nObserved yields and pre-fit background predictions for Njets 6-7.\n\nMore…\n\n#### Multiplicity dependence of (multi-)strange hadron production in proton-proton collisions at $\\sqrt{s}$ = 13 TeV\n\nThe collaboration Acharya, Shreyasi ; Adamova, Dagmar ; Adhya, Souvik Priyam ; et al.\nEur.Phys.J.C 80 (2020) 167, 2020.\nInspire Record 1748157\n\nThe production rates and the transverse momentum distribution of strange hadrons at mid-rapidity ($\\ |y\\ | < 0.5$) are measured in proton-proton collisions at $\\sqrt{s}$ = 13 TeV as a function of the charged particle multiplicity, using the ALICE detector at the LHC. The production rates of $\\rm{K}^{0}_{S}$, $\\Lambda$, $\\Xi$, and $\\Omega$ increase with the multiplicity faster than what is reported for inclusive charged particles. The increase is found to be more pronounced for hadrons with a larger strangeness content. Possible auto-correlations between the charged particles and the strange hadrons are evaluated by measuring the event-activity with charged particle multiplicity estimators covering different pseudorapidity regions. When comparing to lower energy results, the yields of strange hadrons are found to depend only on the mid-rapidity charged particle multiplicity. Several features of the data are reproduced qualitatively by general purpose QCD Monte Carlo models that take into account the effect of densely-packed QCD strings in high multiplicity collisions. However, none of the tested models reproduce the data quantitatively. This work corroborates and extends the ALICE findings on strangeness production in proton-proton collisions at 7 TeV.\n\n59 data tables\n\n$K^{0}_{S}$ transverse momentum spectrum - V0M multiplicity classes. Total systematic uncertainties include both correlated and uncorrelated uncertainties across multiplicity. Uncorrelated systematic originating from the multiplicity dependence of the efficiency (2%) is not included.\n\n$\\Lambda+\\bar{\\Lambda}$ transverse momentum spectrum - V0M multiplicity classes. Total systematic uncertainties include both correlated and uncorrelated uncertainties across multiplicity. Uncorrelated systematic originating from the multiplicity dependence of the efficiency (2%) is not included.\n\n$\\Xi^{-}+\\bar{\\Xi^{+}}$ transverse momentum spectrum - V0M multiplicity classes. Total systematic uncertainties include both correlated and uncorrelated uncertainties across multiplicity. Uncorrelated systematic originating from the multiplicity dependence of the efficiency (2%) is not included.\n\nMore…\n\n#### Search for light pseudoscalar boson pairs produced from decays of the 125 GeV Higgs boson in final states with two muons and two nearby tracks in pp collisions at $\\sqrt{s}=$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nPhys.Lett.B 800 (2020) 135087, 2020.\nInspire Record 1744267\n\nA search is presented for pairs of light pseudoscalar bosons, in the mass range from 4 to 15 GeV, produced from decays of the 125 GeV Higgs boson. The decay modes considered are final states that arise when one of the pseudoscalars decays to a pair of tau leptons, and the other one either into a pair of tau leptons or muons. The search is based on proton-proton collisions collected by the CMS experiment in 2016 at a center-of-mass energy of 13 TeV that correspond to an integrated luminosity of 35.9 fb${-1}$. The 2$\\mu$2$\\tau$ and 4$\\tau$ channels are used in combination to constrain the product of the Higgs boson production cross section and the branching fraction into 4$\\tau$ final state, $\\sigma\\mathcal{B}$, exploiting the linear dependence of the fermionic coupling strength of pseudoscalar bosons on the fermion mass. No significant excess is observed beyond the expectation from the standard model. The observed and expected upper limits at 95% confidence level on $\\sigma\\mathcal{B}$, relative to the standard model Higgs boson production cross section, are set respectively between 0.022 and 0.23 and between 0.027 and 0.19 in the mass range probed by the analysis.\n\n1 data table\n\nExpected and observed 95% CL upper limits on (sigma(pp->h)/sigma(pp->hSM)) * B(h -> aa -> tautautautau) as a function of m(a) obtained from the 13 TeV data, where h(SM) is the Higgs boson of the standard model, h is the observed particle with mass of 125 GeV, and (a) denotes a light Higgs-like state.\n\n#### Measurement of $W^{\\pm }$-boson and Z-boson production cross-sections in pp collisions at $\\sqrt{s}=2.76$ TeV with the ATLAS detector\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nEur.Phys.J.C 79 (2019) 901, 2019.\nInspire Record 1742785\n\nThe production cross-sections for $W^{\\pm}$ and $Z$ bosons are measured using ATLAS data corresponding to an integrated luminosity of 4.0 pb$^{-1}$ collected at a centre-of-mass energy $\\sqrt{s}=2.76$ TeV. The decay channels $W \\rightarrow \\ell \\nu$ and $Z \\rightarrow \\ell \\ell$ are used, where $\\ell$ can be an electron or a muon. The cross-sections are presented for a fiducial region defined by the detector acceptance and are also extrapolated to the full phase space for the total inclusive production cross-section. The combined (average) total inclusive cross-sections for the electron and muon channels are: \\begin{eqnarray} \\sigma^{\\text{tot}}_{W^{+}\\rightarrow \\ell \\nu}& = & 2312 \\pm 26\\ (\\text{stat.})\\ \\pm 27\\ (\\text{syst.}) \\pm 72\\ (\\text{lumi.}) \\pm 30\\ (\\text{extr.})\\text{pb} \\nonumber, \\\\ \\sigma^{\\text{tot}}_{W^{-}\\rightarrow \\ell \\nu}& = & 1399 \\pm 21\\ (\\text{stat.})\\ \\pm 17\\ (\\text{syst.}) \\pm 43\\ (\\text{lumi.}) \\pm 21\\ (\\text{extr.})\\text{pb} \\nonumber, \\\\ \\sigma^{\\text{tot}}_{Z \\rightarrow \\ell \\ell}& = & 323.4 \\pm 9.8\\ (\\text{stat.}) \\pm 5.0\\ (\\text{syst.}) \\pm 10.0\\ (\\text{lumi.}) \\pm 5.5 (\\text{extr.}) \\text{pb} \\nonumber. \\end{eqnarray} Measured ratios and asymmetries constructed using these cross-sections are also presented. These observables benefit from full or partial cancellation of many systematic uncertainties that are correlated between the different measurements.\n\n28 data tables\n\nMeasured fiducial cross section times leptonic branching ratio for W+ production in the W+ -> e+ nu final state.\n\nMeasured fiducial cross section times leptonic branching ratio for W+ production in the W+ -> mu+ nu final state.\n\nMeasured fiducial cross section times leptonic branching ratio for W- production in the W- -> e- nu final state.\n\nMore…\n\n#### Version 2 Search for long-lived particles using nonprompt jets and missing transverse momentum with proton-proton collisions at $\\sqrt{s} =$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nPhys.Lett.B 797 (2019) 134876, 2019.\nInspire Record 1740108\n\nA search for long-lived particles decaying to displaced, nonprompt jets and missing transverse momentum is presented. The data sample corresponds to an integrated luminosity of 137 fb$^{-1}$ of proton-proton collisions at a center-of-mass energy of 13 TeV collected by the CMS experiment at the CERN LHC in 2016-2018. Candidate signal events containing nonprompt jets are identified using the timing capabilities of the CMS electromagnetic calorimeter. The results of the search are consistent with the background prediction and are interpreted using a gauge-mediated supersymmetry breaking reference model with a gluino next-to-lightest supersymmetric particle. In this model, gluino masses up to 2100, 2500, and 1900 GeV are excluded at 95% confidence level for proper decay lengths of 0.3, 1, and 100 m, respectively. These are the best limits to date for such massive gluinos with proper decay lengths greater than $\\sim$0.5 m.\n\n28 data tables\n\nSummary of the estimated number of background events.\n\nThe timing distribution of the backgrounds predicted to contribute to the signal region, compared to those for a representative signal model. The time is defined by the jet in the event with the largest $t_{\\mathrm{jet}}$ passing the relevant selection. The distributions for the major backgrounds are taken from control regions and normalized to the predictions. The observed data is shown by the black points.\n\nThe product of the acceptance and efficiency in the $c\\tau_{0}$ vs. $m_{\\tilde{g}}$ plane for the GMSB model, after all requirements.\n\nMore…\n\n#### Search for a heavy charged boson in events with a charged lepton and missing transverse momentum from $pp$ collisions at $\\sqrt{s} = 13$ TeV with the ATLAS detector\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nPhys.Rev.D 100 (2019) 052013, 2019.\nInspire Record 1739784\n\nA search for a heavy charged-boson resonance decaying into a charged lepton (electron or muon) and a neutrino is reported. A data sample of 139 fb$^{-1}$ of proton-proton collisions at $\\sqrt{s} = 13$ TeV collected with the ATLAS detector at the LHC during 2015-2018 is used in the search. The observed transverse mass distribution computed from the lepton and missing transverse momenta is consistent with the distribution expected from the Standard Model, and upper limits on the cross section for $pp \\to W^\\prime \\to \\ell\\nu$ are extracted ($\\ell = e$ or $\\mu$). These vary between 1.3 pb and 0.05 fb depending on the resonance mass in the range between 0.15 and 7.0 TeV at 95% confidence level for the electron and muon channels combined. Gauge bosons with a mass below 6.0 TeV and 5.1 TeV are excluded in the electron and muon channels, respectively, in a model with a resonance that has couplings to fermions identical to those of the Standard Model $W$ boson. Cross-section limits are also provided for resonances with several fixed $\\Gamma / m$ values in the range between 1% and 15%. Model-independent limits are derived in single-bin signal regions defined by a varying minimum transverse mass threshold. The resulting visible cross-section upper limits range between 4.6 (15) pb and 22 (22) ab as the threshold increases from 130 (110) GeV to 5.1 (5.1) TeV in the electron (muon) channel.\n\n14 data tables\n\nTransverse mass distribution for events satisfying all selection criteria in the electron channel.\n\nTransverse mass distribution for events satisfying all selection criteria in the muon channel.\n\nUpper limits at the 95% CL on the cross section for SSM $W^\\prime$ production and decay to the electron+neutrino channel as a function of the $W^\\prime$ pole mass.\n\nMore…\n\n#### Measurement of prompt D$^{0}$, D$^{+}$, D$^{*+}$, and ${\\mathrm{D}}_{\\mathrm{S}}^{+}$ production in p–Pb collisions at $\\sqrt{{\\mathrm{s}}_{\\mathrm{NN}}}$ = 5.02 TeV\n\nThe collaboration Acharya, Shreyasi ; Adamova, Dagmar ; Adhya, Souvik Priyam ; et al.\nJHEP 12 (2019) 092, 2019.\nInspire Record 1738950\n\nThe measurement of the production of prompt D$^0$, D$^+$, D$^{*+}$, and D$^+_s$ mesons in proton$-$lead (p$-$Pb) collisions at the centre-of-mass energy per nucleon pair of $\\sqrt{s_{\\rm NN}}$ = 5.02 TeV, with an integrated luminosity of $292\\pm 11$ $\\mu$b$^{-1}$, are reported. Differential production cross sections are measured at mid-rapidity ($-0.96<y_{\\rm cms}<0.04$) as a function of transverse momentum ($p_{\\rm T}$) in the intervals $0< p_{\\rm T} < 36$ GeV/$c$ for D$^0$, $1< p_{\\rm T} <36$ GeV/$c$ for D$^+$ and D$^{*+}$, and $2< p_{\\rm T} <24$ GeV/$c$ for D$^+_s$ mesons. For each species, the nuclear modification factor $R_{\\rm pPb}$ is calculated as a function of $p_{\\rm T}$ using a proton-proton (pp) reference measured at the same collision energy. The results are compatible with unity in the whole $p_{\\rm T}$ range. The average of the non-strange D mesons $R_{\\rm pPb}$ is compared with theoretical model predictions that include initial-state effects and parton transport model predictions. The $p_{\\rm T}$ dependence of the D$^0$, D$^+$, and D$^{*+}$ nuclear modification factors is also reported in the interval $1< p_{\\rm T} < 36$ GeV/$c$ as a function of the collision centrality, and the central-to-peripheral ratios are computed from the D-meson yields measured in different centrality classes. The results are further compared with charged-particle measurements and a similar trend is observed in all the centrality classes. The ratios of the $p_{\\rm T}$-differential cross sections of D$^0$, D$^+$, D$^{*+}$, and D$^+_s$ mesons are also reported. The D$^+_s$ and D$^+$ yields are compared as a function of the charged-particle multiplicity for several $p_{\\rm T}$ intervals. No modification in the relative abundances of the four species is observed with respect to pp collisions within the statistical and systematic uncertainties.\n\n27 data tables\n\n$p_{\\rm{T}}$ differential cross section of prompt D0 mesons obtained from the analysis without vertexing reconstruction in p-Pb collisions at $\\mathbf{\\sqrt{{\\textit s}_{\\rm NN}}~=~5.02~TeV}$.\n\n$p_{\\rm{T}}$ differential cross section of inclusive D0 mesons from the analysis without vertexing reconstruction in p-Pb collisions at $\\mathbf{\\sqrt{{\\textit s}_{\\rm NN}}~=~5.02~TeV}$.\n\n$p_{\\rm{T}}$ differential cross section of inclusive D0 mesons from the analysis without vertexing reconstruction in pp collisions at $\\mathbf{\\sqrt{{\\textit s}}~=~5.02~TeV}$ multiplied by A=208.\n\nMore…\n\n#### Observation of electroweak production of a same-sign $W$ boson pair in association with two jets in $pp$ collisions at $\\sqrt{s}=13$ TeV with the ATLAS detector\n\nPhys.Rev.Lett. 123 (2019) 161801, 2019.\nInspire Record 1738841\n\nThis Letter presents the observation and measurement of electroweak production of a same-sign $W$ boson pair in association with two jets using 36.1 fb$^{-1}$ of proton-proton collision data recorded at a center-of-mass energy of $\\sqrt{s}=13$ TeV by the ATLAS detector at the Large Hadron Collider. The analysis is performed in the detector fiducial phase-space region, defined by the presence of two same-sign leptons, electron or muon, and at least two jets with a large invariant mass and rapidity difference. A total of 122 candidate events are observed for a background expectation of $69 \\pm 7$ events, corresponding to an observed signal significance of 6.5 standard deviations. The measured fiducial signal cross section is $\\sigma^{\\mathrm {fid.}}=2.89^{+0.51}_{-0.48} \\mathrm{(stat.)} ^{+0.29}_{-0.28} \\mathrm{(syst.)}$ fb.\n\n6 data tables\n\nMeasured fiducial cross section.\n\nThe $m_{jj}$ distribution for events meeting all selection criteria for the signal region. Signal and individual background distributions are shown as predicted after the fit. The last bin includes the overflow. The highest value measured in a candidate event in data is $m_{jj}=3.8$ TeV.\n\nThe $m_{ll}$ distribution for events meeting all selection criteria for the signal region as predicted after the fit. The fitted signal strength and nuisance parameters have been propagated, with the exception of the uncertainties due to the interference and electroweak corrections for which a flat uncertainty is assigned. The last bin includes the overflow. The highest value measured in a candidate event in data is $m_{ll}=824$ GeV.\n\nMore…\n\n#### Combination of CMS searches for heavy resonances decaying to pairs of bosons or leptons\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nPhys.Lett.B 798 (2019) 134952, 2019.\nInspire Record 1737724\n\nA statistical combination of searches for heavy resonances decaying to pairs of bosons or leptons is presented. The data correspond to an integrated luminosity of 35.9 fb$^{-1}$ collected during 2016 by the CMS experiment at the LHC in proton-proton collisions at a center-of-mass energy of 13 TeV. The data are found to be consistent with expectations from the standard model background. Exclusion limits are set in the context of models of spin-1 heavy vector triplets and of spin-2 bulk gravitons. For mass-degenerate W' and Z' resonances that predominantly couple to the standard model gauge bosons, the mass exclusion at 95% confidence level of heavy vector bosons is extended to 4.5 TeV as compared to 3.8 TeV determined from the best individual channel. This excluded mass increases to 5.0 TeV if the resonances couple predominantly to fermions.\n\n14 data tables\n\nObserved and expected 95% CL upper limits on the product of the cross section and branching fraction of a spin-1 resonance decaying to a pair of SM bosons.\n\nObserved and expected 95% CL upper limits on the product of the cross section and branching fraction of a spin-2 resonance decaying to a pair of SM bosons.\n\nObserved and expected 95% CL upper limits on the product of the cross section of a W' resonance decaying to a pair of SM bosons.\n\nMore…\n\n#### Search for the electroweak diboson production in association with a high-mass dijet system in semileptonic final states in $pp$ collisions at $\\sqrt{s}=13$ TeV with the ATLAS detector\n\nThe collaboration Aad, Georges ; Abbott, Brad ; Abbott, Dale Charles ; et al.\nPhys.Rev.D 100 (2019) 032007, 2019.\nInspire Record 1735560\n\nThis paper reports on a search for the electroweak diboson ($WW/WZ/ZZ$) production in association with a high-mass dijet system, using data from proton-proton collisions at a center-of-mass energy of $\\sqrt{s}=13$ TeV. The data, corresponding to an integrated luminosity of 35.5 fb$^{-1}$, were recorded with the ATLAS detector in 2015 and 2016 at the Large Hadron Collider. The search is performed in final states in which one boson decays leptonically, and the other boson decays hadronically. The hadronically decaying $W/Z$ boson is reconstructed as either two small-radius jets or one large-radius jet using jet substructure techniques. The electroweak production of $WW/WZ/ZZ$ in association with two jets is measured with an observed (expected) significance of 2.7 (2.5) standard deviations, and the fiducial cross section is measured to be $45.1 \\pm 8.6(\\mathrm{stat.}) ^{+15.9} _{-14.6} (\\mathrm{syst.})$ fb.\n\n2 data tables\n\nSummary of predicted and measured fiducial cross sections for EW $VVjj$ production. The three lepton channels are combined. For the measured fiducial cross sections in the merged and resolved categories, two signal-strength parameters are used in the combined fit, one for the merged category and the other one for the resolved category; while for the measured fiducial cross section in the inclusive fiducial phase space, a single signal-strength parameter is used. For the SM predicted cross section, the error is the theoretical uncertainty (theo.). For the measured cross section, the first error is the statistical uncertainty (stat.), and the second error is the systematic uncertainty (syst.).\n\nSummary of predicted and measured fiducial cross sections for EW $VVjj$ production. in the three lepton channels. The measured values are obtained from a simultaneous fit where each lepton channel has its own signal-strength parameter, and in each lepton channel the same signal-strength parameter is applied to both the merged and resolved categories. For the SM predicted cross section, the error is the theoretical uncertainty (theo.). For the measured cross section, the first error is the statistical uncertainty (stat.), and the second error is the systematic uncertainty (syst.).\n\n#### Search for the production of W$^\\pm$W$^\\pm$W$^\\mp$ events at $\\sqrt{s} =$ 13 TeV\n\nThe collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al.\nPhys.Rev.D 100 (2019) 012004, 2019.\nInspire Record 1734235\n\nA search for the production of events containing three W bosons predicted by the standard model is reported. The search is based on a data sample of proton-proton collisions at a center-of-mass energy of 13 TeV recorded by the CMS experiment at the CERN LHC and corresponding to a total integrated luminosity of 35.9 fb$^{-1}$. The search is performed in final states with three leptons (electrons or muons), or with two same-charge leptons plus two jets. The observed (expected) significance of the signal for W$^\\pm$W$^\\pm$W$^\\mp$ production is 0.60 (1.78) standard deviations, and the ratio of the measured signal yield to that expected from the standard model is 0.34 $^{+0.62}_{-0.34}$. Limits are placed on three anomalous quartic gauge couplings and on the production of massive axionlike particles.\n\n9 data tables\n\nLost-lepton and three-lepton background contributions.\n\nNon-prompt lepton background estimates.\n\nSummary of typical systematic uncertainties of estimated background contributions.\n\nMore…\n\n#### rivet Analysis Measurement of fiducial and differential $W^+W^-$ production cross-sections at $\\sqrt{s}=13$ TeV with the ATLAS detector\n\nEur.Phys.J.C 79 (2019) 884, 2019.\nInspire Record 1734263\n\nA measurement of fiducial and differential cross-sections for $W^+W^-$ production in proton-proton collisions at $\\sqrt{s}=$13 TeV with the ATLAS experiment at the Large Hadron Collider using data corresponding to an integrated luminosity of $36.1$ fb$^{-1}$ is presented. Events with one electron and one muon are selected, corresponding to the decay of the diboson system as $WW\\rightarrow e^{\\pm}\\nu\\mu^{\\mp}\\nu$. To suppress top-quark background, events containing jets with a transverse momentum exceeding 35 GeV are not included in the measurement phase space. The fiducial cross-section, six differential distributions and the cross-section as a function of the jet-veto transverse momentum threshold are measured and compared with several theoretical predictions. Constraints on anomalous electroweak gauge boson self-interactions are also presented in the framework of a dimension-six effective field theory.\n\n43 data tables\n\nMeasured fiducial cross-section as a function of the jet-veto $p_{T}$ threshold. The value at the jet-veto $p_{T}$ threshold of 35GeV corresponds to the nominal fiducial cross section measured in this publication.\n\nStatistical correlation between bins in data for the measured fiducial cross-section as a function of the jet-veto $p_{T}$ threshold. The value at the jet-veto $p_{T}$ threshold of 35GeV corresponds to the nominal fiducial cross section measured in this publication.\n\nTotal correlation between bins in data for the measured fiducial cross-section as a function of the jet-veto $p_{T}$ threshold. The value at the jet-veto $p_{T}$ threshold of 35GeV corresponds to the nominal fiducial cross section measured in this publication.\n\nMore…"
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https://www.osh.net/calc/feet-to-inches/31
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[
"# What is 31 feet in inches?\n\n31 feet = 372 inches\n\nSaid another way, 31 feet = 31 feet 0 inches\n\nConvert another measurement\n\n## Formula for converting feet to inches\n\nThe formula for converting feet to inches is feet x 12. So for a length of 31 feet, the formula would be 31 x 12, with a result of 372 inches.\n\nThe process for converting feet to feet and inches is by leaving the measurement in feet alone and multiplying everything after the decimal point by 12. So for a length of 31 feet, you would keep the 31 and multiply 0 * 12, with a result of 31 feet 0 inches.\n\n## Convert 31.01 - 31.99 feet\n\n62626262626262626262626262626262626262626262626262\n Feet\n62626262626262626262626262626262626262626262626262\n Feet\n62626262626262626262626262626262626262626262626262\n Feet\n626262626262626262626262626262626262626262626262\n Feet\n\n## Look up numbers near 31\n\n← Prev num Next num →\n30 32"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.9136831,"math_prob":0.9871291,"size":583,"snap":"2023-40-2023-50","text_gpt3_token_len":167,"char_repetition_ratio":0.18998273,"word_repetition_ratio":0.10526316,"special_character_ratio":0.30874786,"punctuation_ratio":0.09375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9824128,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T05:18:42Z\",\"WARC-Record-ID\":\"<urn:uuid:50ab59f0-7029-4646-a7b6-b28216897a0a>\",\"Content-Length\":\"7674\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:797896cd-fa77-4ac5-89bf-9112752c1273>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad74368f-c09e-4d48-bce7-f0d57a72772f>\",\"WARC-IP-Address\":\"35.209.4.189\",\"WARC-Target-URI\":\"https://www.osh.net/calc/feet-to-inches/31\",\"WARC-Payload-Digest\":\"sha1:ECHR7F3HOWVBR4XZQIBZOJDKMONDAPDQ\",\"WARC-Block-Digest\":\"sha1:VOCSVP6GGA4EMFRDQ55C2I2BIWYLHT5L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510259.52_warc_CC-MAIN-20230927035329-20230927065329-00463.warc.gz\"}"}
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https://swiflearn.com/ncert-solutions/class-10/science/chapter-10/
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[
"> > > > NCERT Solution for Class 10 Science Chapter 10 : Light Reflection and Refraction\n\nNCERT Solution for Class 10 Science Chapter 10 : Light Reflection and Refraction",
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"Click to rate this post!\n[Total: 22 Average: 4.6]\n\nChapter 10 of NCERT Class 10 Science is Light Reflection and Refraction, In this chapter, you will study about Light Reflection and Refraction with the help of laws of light reflection and refraction and identify the nature of light after striking on different shaped structures like curve, plane, etc. This chapter has 5 questions that will strengthen your concepts regarding the nature of light.\n\nNCERT Solutions for Class 10 Science Chapter 10 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 10 Science Chapter 10 are designed as per the CBSE Class 10th Science Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 10 Science Chapter 10.",
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"NCERT Solution for Class 10 Science Chapter 10 Light Reflection and Refraction PDF\n\nQuestion 1:Define the principal focus of a concave mirror.\n\nSolution:\nLight rays parallel to the principal axis of a concave mirror after reflecting from its surface converge at a specific point on its principal axis. This point of convergence of parallel light\nrays is called the principal focus of the concave mirror.\n\nQuestion 2.The radius of curvature of a spherical mirror is 20 cm. What is its focallength?\n\nSolution:\nGiven:\nRadius of curvature of spherical mirror, R = 20 cm\nRadius of curvature of a spherical mirror = 2 × Focal length (f)\nR = 2f\nf = R/2\n= 20/2\n= 10 cm\nHence, the focal length of the given spherical mirror is 10 cm\n\n.Question 3.Name the mirror that can give an erect and enlarged image of an object.\n\nSolution:\nA concave mirror forms a virtual, erect, and enlarged image of an object placed between its pole and the principal focus. Neither plain nor convex mirror can make a virtual, erect, and\nenlarged image.\n\nQuestion 4:Why do we prefer a convex mirror as a rear-view mirror in vehicles?\n\nSolution 4:\nWhen an object is placed in front of the convex mirror it gives a virtual, erect, and diminished image of the object. We need to see as many areas as possible behind the vehicle also we need all the images erect because of this criterion convex mirrors are preferred as a rear-view mirror in vehicles because and they give a wider field of view, this type of view allows the driver to see most of the traffic behind him.\n\nQuestion 1:Find the focal length of a convex mirror whose radius of curvature is 32cm.\n\nSolution 1:\nGiven:\nRadius of curvature of the mirror = R = 32 cm\nRadius of curvature of the mirror = 2 × Focal length (f)\nR = 2f\nf = R/2\n= 32/2\n= 16cm\nHence, the focal length of the given convex mirror is 16 cm.\n\nQuestion 2:A concave mirror produces three times magnified (enlarged) real image ofobject placed at 10 cm in front of it. Where is the image located?\n\nSolution 2:\nMagnification produced by a spherical mirror is given by the relation,\nm = Height of the image/ Height of the Object = – Image Distance/Object distance\nm = h1/h0 = -v/u\nLet’s assume the height of the object = h0 = h\nThen, the height of the image, h1 = −3h (because the Image formed is real)\nAccording to relation,\n-3h/h = -v/u\nv/u = 3\nGiven:\nObject distance, u = −10 cm\nv = 3 × (−10) = −30 cm\nImage is formed at a distance of 30cm the mirror. Here, the negative sign indicates that an inverted image is formed.\n\nQuestion 1:A ray of light travelling in air enters obliquely into water. Does the lightray bend towards the normal or away from the normal? Why?\n\nSolution 1:\nLight refracts towards the normal while travelling from a rare medium to a dense medium. This is the reason why light bends towards normal while travelling from air into water.\n\nQuestion 2:Light enters from air to glass having refractive index 1.50. What is thespeed of light in the glass? The speed of light in vacuum is 3 × 108 m s−1.\n\nSolution 2:\nGiven: c = 300000000m/s\nμ =1.5\nSpeed of light in glass = speed of light in vacuum / Reflective index of glass\n1.5 = 3 x 108\nvg\nvg= 2 x 108 m/s\nSo, the speed of light in glass is 20000000 m/s.\n\nQuestion 3:Find out, from Table, the medium having highest optical density. Also findthe medium with lowest optical density.\n\nSolution 3:\nHighest optical density = Diamond with μ=2.42\nLowest optical density = Air with μ=1.0003\nThe optical density increases with an increase in the refractive index of the medium. A material which has the highest refractive index will have the highest optical density and viceversa.\n\nSolution 4:\n\nRefer pdf.\n\nSolution 5:\n\nRefer pdf.\n\nQuestion 1:Define 1 dioptre of power of a lens.\n\nSolution 1:\nDioptre is the SI unit of power of lens is denoted by the letter D. 1 dioptre can be defined as the power of a lens of focal length 1 metre.\n\nSolution 2:\n\nRefer pdf.\n\nQuestion 3:Find the power of a concave lens of focal length 2 m.\n\nSolution 3:\nFocal length of concave lens, f = -2 m\nNow, we know that,\nP= 1/f\nP=1/ (-2)\nHence, the power of the given lens is -0.5 D.\nHere, negative sign arises due to the divergent nature of concave lens.\n\nQuestion 1:Which one of the following materials cannot be used to make a lens?(a) Water(b) Glass(c) Plastic(d) Clay\n\nSolution 1:\n(d) Clay. A lens must allow the light rays to pass through it. Since clay does not have such property, it can’t be used in making lens.\n\nQuestion 2:The image formed by a concave mirror is observed to be virtual, erect andlarger than the object. Where should be the position of the object?(a) Between the principal focus and the center of curvature(b) At the center of curvature(c) Beyond the center of curvature(d) Between the pole of the mirror and its principal focus.\n\nSolution:\n(d) Between the pole of the mirror and its principal focus. The image formed is virtual, erect, and larger than the object, when an object is placed between the pole and principal focus of a concave mirror.\n\nQuestion 3:Where an object should be placed in front of a convex lens to get a realimage of the size of the object?(a) At the principal focus of the lens(b) At twice the focal length(c) At infinity(d) Between the optical centre of the lens and its principal focus.\n\nSolution 3:\n(b) At twice the focal length. The image is formed at the centre of curvature on the other side of the lens when an object is placed at the centre of curvature in front of a convex lens. It forms an image that is real, inverted, and of the same size as the object.\n\nQuestion 4:A spherical mirror and a thin spherical lens have each a focal length of −15cm. The mirror and the lens are likely to be(a) Both concave(b) Both convex(c) The mirror is concave and the lens is convex(d) The mirror is convex, but the lens is concave\n\nSolution 4:\n(a) Both concave.\nThe primary focus of the concave lens in on the same side as the object while the same for a concave mirror is in front of the mirror. So we can see that both are negative because as per convention the focal length of a convex mirror and a concave lens are taken as negative. Hence, both the thin spherical lens and the spherical mirror are concave in nature.\n\nQuestion 5:No matter how far you stand from a mirror, your image appears erect. Themirror is likely to be(a) Plane(b) Concave(c) Convex(d) Either plane or convex\n\nSolution 5:\n(d)Both plane and convex mirror produce erect image for all object positions.\n\nQuestion 6:Which of the following lenses would you prefer to use while reading smallletters found in a dictionary?(a) A convex lens of focal length 50 cm(b) A concave lens of focal length 50 cm(c) A convex lens of focal length 5 cm(d) A concave lens of focal length 5 cm\n\nSolution 6:\n(c)When the object is between F and O, a convex lens will make a magnified and erect image. Also, convex lenses having shorter focal length produce more magnification. Therefore, for reading small letters, using a convex lens of focal length 5 cm would be best.\n\nQuestion 7:We wish to obtain an erect image of an object, using a concave mirror offocal length 15 cm. What should be the range of distance of the object fromthe mirror? What is the nature of the image? Is the image larger or smallerthan the object? Draw a ray diagram to show the image formation in thiscase.\n\nSolution 7:\nWhen an object is placed between the principal focus (F) and pole (P), a concave mirror gives an erect image.. The image formed will be virtual, erect, and magnified in nature if the object is placed between the pole and principal focus of the concave mirror.\n\nQuestion 8:Name the type of mirror used in the following situations.(a) Headlights of a car(b) Side/rear-view mirror of a vehicle(c) Solar furnaceSupport your answer with reason.\n\nSolution 8:\n(a) Concave mirrors can produce a strong parallel beam of light when the light source is placed at their principal focus F. Hence, a concave mirror is used in the headlights of a car.\n(b) A convex mirror is usually used in side /rear-view mirror of a vehicle. Convex mirrors are capable of giving a virtual, erect, and diminished image of the objects placed in front of it, as a result, a wider field of view is obtained. This wider field of view enables the driver to see most of the traffic behind him/her.\n(c) In the solar furnace, concave mirrors are used. Concave mirrors are converging mirrors. Due to this property of concave mirrors they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point of the focus known as the principal focus of the mirror. Hence, they produce a large amount of heat at their principal focus.\n\nQuestion 9:One-half of a convex lens is covered with a black paper. Will this lensproduce a complete image of the object? Verify your answerexperimentally. Explain your observations.\n\nSolution 9:\nA full image but with lesser intensity is produced. We can understand these using two cases.\nCase I: When the upper half of the lens is covered:\nAs the upper half of the lens is covered, the light rays coming from the object is refracted by the lower half of the lens only. After refraction, these rays meet at the other side of the lens to form the image of the given object.\nCase II: When the lower half of the lens is covered:\nAs the lower half of the lens is covered, the light rays coming from the object is refracted by the upper half of the lens only. After refraction, these rays meet at the other side of the lens to form the image of the given object.\nIn both the cases some rays are blocked. But the other rays pass through the lens and form a complete image but the intensity of the image is less.\n\nSolution 10:\n\nRefer pdf.\n\nSolution 11:\n\nRefer pdf.\n\nSolution 12:\n\nRefer pdf.\n\nSolution 13:\n\nRefer pdf.\n\nSolution 14:\n\nRefer pdf.\n\nSolution 15:\n\nRefer pdf.\n\nSolution 16:\n\nRefer pdf.\n\nQuestion 17:A doctor has prescribed a corrective lens of power +1.5 D. Find the focallength of the lens. Is the prescribed lens diverging or converging?\n\nSolution 17:\n\nRefer pdf.",
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https://www.element14.com/community/docs/DOC-92119/l/element14-essentials-voltage-references?ICID=learningctr-mobile
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[
"# element14 Essentials: Voltage References\n\nVersion 17\n\n1. Introduction\n\nMost analog circuits, such as comparators and A/D & D/A converters, need a voltage reference to achieve the desired accuracy and performance. The precision of these circuits depends on a reference that must be independent of manufacturing process tolerances, temperature changes, power supply variations, and more. Primary voltage standards such as the Weston cell are implemented using electrochemical cells or batteries, which have severe limitations in size, cost, complexity, and stability over temperature and time. Advances in precision voltage references have resulted in semiconductor voltage references that can be quite accurate and stable. This learning module covers the essentials of voltage references, focusing on measurement parameters, types, and applications.\n\n2. Objectives\n\nAfter completing this learning module, you will be able to:",
null,
"Understand the basic concepts and operation of a voltage reference",
null,
"Explain the different types of voltage references",
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"Discuss the quality parameters that influence voltage references",
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"Describe how voltage references are used in different applications\n\n3. Basic Concepts\n\nA voltage reference is an electronic circuit or device that ideally produces a constant voltage irrespective of the variation in input and loading on the device. A simple example of a voltage reference is a Zener diode. When current passes through the Zener diode, a voltage develops across its terminals, and as the current continues to increase, this voltage remains relatively constant. This voltage drop depends on the design of the device and materials used. Let us expand our discussion to three different technologies that voltage references are built upon —Zener breakdown, Buried Zener, and bandgap voltage.\n\n- 3.1 Zener-based Voltage Reference\n\nZener diodes are PN junction diodes with controlled reverse-biased properties, making them useful as voltage reference devices. The V-I characteristics of the ideal Zener diode are shown in Figure 1.\n\nThe reverse characteristic shows that, at the breakdown point, the knee voltage (Vz) where the current through the PN junction starts increasing rapidly is relatively independent of the diode current. This knee voltage, also known as Zener voltage, is controlled by the amount of doping applied during the manufacturing process. Practical Zener diodes typically have tolerances of a few percent, so high levels of precision can be difficult to achieve. In addition, while the breakdown voltage is ideally constant as a function of reverse current, in practice this voltage will vary with current, adding further uncertainty to the value. In a simple Zener reference circuit, with the diode driven by a resistor connected to a power supply voltage, the breakdown voltage remains relatively constant as the supply voltage – and therefore the reverse current – varies. If the input voltage increases, the diode maintains a nominally constant voltage across the load by absorbing the extra current and keeping the load current constant. If the load resistance decreases, the current drawn by the Zener diode decreases and the extra current through the load maintains a nominally constant load voltage.\n\nA voltage reference requires stability of the output voltage relative to temperature variations. The Zener voltage varies with temperature, and its variation depends on Zener and avalanche breakdown. The Zener breakdown voltage decreases with the increase in temperature, creating a negative temperature coefficient (NTC). Avalanche breakdown voltage rises with the increase in temperature, creating a positive temperature coefficient. Near 5.6V, the negative temperature coefficient from Zener breakdown is balanced by the positive temperature coefficient from avalanche breakdown, creating a device with a low-temperature coefficient.\n\nIn the manufacturing of a practical Zener-based voltage reference IC, a conventional forward biased diode is used in series with a Zener operating in the avalanche mode to avoid the temperature effect. A forward-biased diode has a negative temperature coefficient, which cancels the positive temperature coefficient of the Zener diode operating in avalanche mode.\n\nA Zener-based voltage reference IC implementation is shown in Figure 2. R4 provides the startup current for the diodes, thus setting the positive input of the op-amp at V2. R3 sets the desired bias current for the diodes. Manufacturers set the output voltage through the ratio of R1 and R2. By trimming these resistors, the output voltage can be set to the desired value. We can optimize bias current to a point where a minimum temperature coefficient is obtained by trimming R3 – but note that any variations in the 15V power supply will affect the current, and therefore the output voltage and the temperature coefficient.\n\n- 3.2 Buried Zener Voltage References\n\nZener diodes are used to make high-quality voltage references. However, they are very noisy and unstable due to surface impurities and crystal imperfections. In a buried Zener reference, the Zener junction is covered by a protective diffusion below the surface, which makes the device stable and protects it from the noise generated by surface impurities and crystal imperfections.\n\n- 3.3 Bandgap Voltage Reference\n\nA bandgap voltage reference produces the sum of two voltages having opposite temperature coefficients. Similar to Zener-based references, this approach cancels out the temperature effect. Figure 3 shows a bandgap voltage reference circuit where one voltage is the forward voltage (VBE) of the conventional diode (the base-emitter junction) in transistor Q1, which has a negative temperature coefficient; the other is ΔVBE, the difference between the forward voltages of two diodes (the base-emitter junction of Q1 and Q2) with the same current but operating at two different current densities in the ratio of n:1. This is achieved by fabricating Q1 with a larger emitter area than Q2. The negative dependence of VBE is compensated by the positive dependency of ΔVBE and dropped across R1.\n\nBandgap references typically provide a voltage ranging from 1.2V to 10V. The benefit of bandgap devices is the ability to not only provide reference voltages below 5V but also to enable operating currents down to milliamps and microamps.\n\n4. Types of Voltage References\n\nWe will now discuss different types of commercially available integrated voltage reference devices and their circuit implementations.\n\n- 4.1 Series Voltage Reference\n\nA series voltage reference is a three-terminal device connected in series with its loads, as shown in Figure 4. This mode regulates the output by creating a voltage drop between its input and output. It can be viewed as a voltage-controlled resistance where the output controls an internal resistance between the input and output terminals of the reference device. In the absence of any load, the series reference draws a small amount of current through the internal resistance and drops a voltage between the input and output that is necessary to produce the correct output.\n\nWhen load current increases, the reference changes its internal resistance to provide the correct voltage drop between the input and output to maintain the specified output voltage.\n\n- 4.2 Shunt Voltage Reference\n\nA shunt voltage reference is a two-terminal device connected in parallel with its load, as shown in Figure 5. The complete circuit can be viewed as a voltage-controlled current sink where the controlling voltage is applied to its input terminal. With no load applied, enough current flows through the shunt reference so that the voltage drop across R1 produces the desired output voltage. For example, if the applied input voltage is 8.0V, and the desired reference voltage is 5.0V, the reference IREF creates a 3.0V drop across R1.\n\nWhen we apply a load to the reference, IREF is not equal to IR1, because load current produces part of the voltage drop across R1. In this case, the reference automatically adjusts its sink current to maintain a constant voltage across the reference. Therefore, the total current through R1 does not change. IR1 is shunted between the reference and load, and hence is called a \"shunt reference.\" A shunt reference controls the output voltage by adjusting its sink current and opposes load current changes.\n\n5. Measurement Units for a Voltage Reference\n\nThe units for voltage reference parameters, such as accuracy, differ among manufacturers. Commonly used units of accuracy include parts per million (ppm), the percentage of full scale (%), decibels (dB), and voltage (V/ µV). Let us discuss the relationships for converting one unit to another.\n\n- 5.1 Accuracy Percentage of Full Scale\n\nPercentage of the nominal value is a common means for stating reference accuracy. It follows the convention for expressing tolerance on resistors, capacitors, and inductors. The percent accuracy specifications for voltage references are 1%, 1.5%, 2%, and 5%.\n\nWe should multiply the output voltage of the reference device by the percent accuracy and divide by 100 to determine the voltage deviation of a reference. For example, a 3.5V reference accurate to ±1.5% has a deviation of: ± (3.5V × 1.5)/100 = ±0.0525V, or ±52.5mV\n\n- 5.2 Accuracy in Parts per Million\n\nParts per Million (ppm) is used to specify temperature coefficients and other parameters that change very little under varying conditions. For a 3.5V reference, 1ppm is one-millionth of 3.5V, or 3.5µV. If the reference is accurate to within 10ppm, its output tolerance is: 3.5V × 10/10-6 = 35µV\n\nConversion in voltage accuracy:\n3.5V ± 35µV = 3.499965 V, or 3.500035 V\n\nConversion in percent:\n± (35E - 6V) × 100/3.5V = ±0.001%\n\n- 5.3 Accuracy in Bits\n\nAccuracy in bits is specified as a ± number of bits Least Significant Bit (LSB) or Most Significant Bit (MSB). For example, a 3.5V reference voltage that claims to be 16-bit accurate must not deviate by more than one part in 216 (65536). So, 1 bit is 1/65536 volt of the total voltage value. In this case, it is 3.5/65536 ≈ 53µV. If we consider 1-bit accuracy (±1 LSB), the output voltage can be 1 bit higher or lower than nominal, i.e., ±53µV.\n\nConverting to voltage accuracy:\n3.5V ±53µV = 3.499947V, or 3.500053V\n\nConverting to percent:\n(±53E - 6V/3.5V) × 100 = ±0.0015%\n\n6. Performance Parameters for Voltage References\n\n- 6.1 Initial Accuracy\n\nThe initial accuracy of a voltage reference is defined as its worst-case tolerance after the complete production process. We can measure the output voltage of the reference using automatic test equipment (ATE). This specification is generally given for room temperature, with defined load current and input voltage. Package stress can affect the initial accuracy tolerance, so control of the solder temperature profile is required.\n\n- 6.2 Noise\n\nNoise in the reference voltage is a random signal produced by active and passive devices inside the IC. It affects the accuracy. The MAX6250 and MAX6350 are good choices for low noise applications with 3µVp-p noise from 0.1Hz to 10Hz.\n\n- 6.3 Temperature Drift\n\nThe terms temperature drift and temperature coefficient are generally used interchangeably. Temperature drift is the change in output voltage with regard to temperature and is broadly used to evaluate voltage reference performance. Temperature drift is caused by imperfections and nonlinearities in the circuit elements and specified by ppm/°C. Figure 6 shows how the output of one voltage reference drifts with temperature. The plot below is data for the MAX6005, which is specified as a 100ppm/°C (max) reference.\n\n- 6.4 Thermal Hysteresis\n\nThermal hysteresis is the change in output voltage when the temperature is cycled - usually between room temperature and the operating temperature limits. As an example, consider a reference that operates over a temperature range of -40°C to +85°C. Thermal hysteresis can be measured by first measuring the output voltage at room temperature (+25°C), then cooling the reference to -40°C, heating it to +85°C, and finally returning it to 25°C. We measure and re-record the output voltage. The temperture hysteresis is the difference in these measurments. This parameter is related to stress on the die. The MAX6001 device is a 3-pin SOT23 with a typical temperature hysteresis of 130ppm. However, a larger, more stable package, like the MAX6190 in the 8-pin SO, has a temperature hysteresis of only 75ppm.\n\n- 6.5 Dropout voltage\n\nDropout voltage is the difference between input and output voltage that allows the reference to maintain its specified accuracy. It is a critical factor in low-voltage and battery-operated equipment, and applies only to the series reference. Battery voltage decreases with the number of discharges, and the reference must maintain an accurate output voltage by the lowest possible battery voltage to maximize its useful life. Hence, a lower dropout voltage allows continued operation at a lower battery voltage.\n\n- 6.6 Line Regulation\n\nLine regulation is a change in output voltage due to a change in input voltage. It is a DC parameter and is typically specified at DC. It is generally specified by %/V, ppm/V and µV/V of the input voltage. Line regulation is inversely related to the rate at which the line voltage changes, and can be specified by PSRR (power supply rejection ratio). Input capacitors are generally recommended to minimize high-frequency input noise.\n\nLoad regulation is a change in output voltage due to a change in the reference load current, and it is specified by ppm/mA, µA/mA, %/mA or percent change from no load to full load. It includes any self-heating due to power dissipation with load current. This parameter is essential if the reference load current fluctuates while the reference is working, e.g. when a reference drives some types of resistive ladder-type digital-to-analog converter (DAC) with no reference buffer. If the ladder impedance changes significantly with DAC code, the variations in output current can affect the reference’s output voltage. The load regulation is inversely proportional to the rate of change of the load current. There must be output capacitors to stabilize the output voltage.\n\n- 6.8 Line Transient Response\n\nLine transient response is the deviation of the voltage reference output in response to a momentary or transient change in input voltage. For example, a voltage reference can be designed for an output voltage deviation less than 5mV for a 200mV input voltage step in a range of 3.2V to 3.4V in 10 microseconds. It should be noted that the input and output capacitors have a remarkable effect on the performance of the reference.\n\nLoad transient response, or output settling time, is the response characteristic to an abrupt load variation. It is the time required for the output voltage to return to its specified value after having fallen or risen. Proper capacitor values connected to the reference’s input and output have a significant impact on the performance of the reference.\n\n- 6.10 Turn-on/Turn-off Settling Time\n\nThe turn-on settling time is the time to stabilize the output voltage of the reference after an initial power-up. The output is only required to be stable, not necessarily to have reached the specified accuracy. It is dependent on several factors, including the reference’s internal architecture, the input and output capacitor values used, and the applied load. Turn-off settling time is the time for the output voltage of the reference to reach virtually zero volts. This parameter is also dependent on the input and output capacitor values and the applied load.\n\n- 6.11 Long-Term Drift (LTD)\n\nLong-term drift is the variation in output voltage over a long period at some specified condition of steady-state operation. It is a measure of the maximum and minimum output-voltage deviations over an extended period. All other conditions such as temperature, input voltage, and load current must be constant if this measurement is to reflect drift in the reference accurately. It is specified in ppm per 1000 hours.\n\nMany factors can affect a voltage reference's Long-Term Drift, including package stress due to package size, mold compound, PCB stress, and external environmental factors such as temperature and humidity.\n\nThe LTD (ppm) can be described by the following function:\nLTD (ppm) = f (V, package stress, PCB design, PCB assembly quality, compound settling time, Temperature, humidity)\n\n7. Use Cases of Voltage References\n\nIt is necessary to understand the voltage reference specifications and its contribution to error in order to select the correct reference for the application. Applications such as data acquisition, sensor driving, and comparators need precise voltage references. Let us look at a few design examples.\n\n- 7.1 Data Converters\n\nWhen used with an ADC or a DAC, the voltage reference determines the input or output range, and therefore has a role in setting the value of the LSB. The digital output of the ADC denotes the ratio of the input voltage and the reference voltage, whereas the digital input to a DAC defines the ratio of its analog output voltage to its reference voltage.\n\nMost data-converters use a sample & hold technique, where a sampling clock processes a large number of equally-spaced analog samples. The quantization value of the analog sample is inversely proportional to the reference voltage. Any instability in the reference will reflect in the precision of the ADC digital value. A voltage reference connected with a successive-approximation-register (SAR) ADC is shown in Figure 7.\n\n- 7.2 Sensors\n\nA precise voltage reference is useful for driving some sensor circuits. A voltage reference precisely drives sensors such as resistance temperature detectors, thermistors, or thermocouples in a Wheatstone circuit configuration, which is the first choice for front-end sensors, as shown in Figure 8. A Wheatstone bridge is used to measure an unknown electrical resistance by balancing two legs of the bridge. It has a single impedance-variable element that is inherently nonlinear when away from the balance point.\n\nSince bridge circuits are simple, they are useful for monitoring temperature, pressure, humidity, light, and other analog properties in industrial and medical applications.\n\n- 7.3 Comparators\n\nA voltage reference is used with comparators to make a precise and accurate set trip point. A comparator circuit with Zener voltage reference is shown in Figure 9. The behavior and accuracy of the comparator circuit depends highly on the voltage reference. Comparators are widely used in ADC, level detection, Schmitt triggers, clock-recovery circuits, window detectors, and on-off controls.\n\n8. Application Examples\n\nVoltage references form significant circuit blocks in systems such as data conversion systems, power supply monitors, test instruments, and sensor drivers. Let us look at some applications of Voltage reference ICs designed by Maxim.\n\n- 8.1 Analog Current/Voltage Output for Industrial Automation and Control\n\nThe Carmel subsystem reference design board, shown in figure 10, is an analog current/voltage output board that uses a MAX6126 voltage reference IC. This IC features curvature-correction circuitry and high-stability, laser-trimmed, thin-film resistors that result in 3ppm/°C (max) temperature coefficients and ±0.02% (max) initial accuracy. This design board is suitable for programmable logic controllers (PLCs), distributed control systems (DCSs), and other industrial applications to produce voltages and currents.\n\n- 8.2 Simple Power Monitor using MAX6025A, SOT23-3 Voltage References\n\nThe power monitoring circuit is used to monitor a system supply voltage. It is implemented using a MAX6025, which is a low power, low-dropout voltage reference IC. This circuit allows an ADC to monitor the system supply voltage through the reference input of the ADC.\n\nPower monitor circuits often use a reference voltage that is lower than the supply voltage. An external resistor-divider may be used to decrease the voltage applied to the ADC input, but external resistors produce errors that may be objectionable in many applications. The modified circuit shown in Figure 11 eliminates the voltage divider, and applies the supply voltage directly to the ADC’s reference input. In this case, the ADC must be capable of accepting an external reference as high as the supply voltage, which is a common feature of most general-purpose ADCs.\n\n- 8.3 4-20mA Input Isolated Analog Front End (AFE) for Industrial Sensor and Process Control\n\nThe Campbell (MAXREFDES4#) subsystem reference design, as shown in Figure 12, is mainly targeted for industrial sensors, automation, process control, programmable logic controllers (PLCs), and medical applications where a 16-bit high-accuracy industrial analog front end (AFE) is required. It accepts a 4–20mA current loop or a 0.2V-to-4.096V voltage input signal and features isolated power and data. The Campbell design integrates an ultra-high-precision 4.096V voltage reference (MAX6126), a precision low-noise buffer (MAX44250), 600VRMS data isolation (MAX14850), a high-accuracy ADC (MAX11100), and isolated/regulated 5V power rails (MAX256/MAX1659).\n\n- 8.4 Application of LM404 in Base Station\n\nThe power amplifier (PA) system is one of the critical components in a Base station for wireless communications. The PA system performance is optimized by combining components such as an ADC, DAC, temperature sensor, a current sense amplifier, precision reference voltage, and a microcontroller unit as shown in Figure 13. The accuracy and performance of all components are highly dependent on the voltage reference block used.\n\nThe LM4040 is a precision, two-terminal shunt mode, bandgap voltage reference. It is available in fixed reverse breakdown voltages of 2.048V, 2.500V, 3.000V, 3.3V, 4.096V, and 5.000V. The Laser-trimmed resistors ensure precise initial accuracy. With a 100ppm/°C temperature coefficient, the device has four grades of initial accuracy ranging from 0.1% to 1%. There is no capacitor required on the output, and it ensures stability with any capacitive load and operates over the temperature range of -40°C to +125°C.\n\nFigure: 13: Current Monitoring and Control in the PA System and LM4040 Operating Circuit\n\n- 8.5 80mA Precision Reference\n\nLarge systems with many loads on the reference voltage may require more current than the voltage reference IC is capable of sourcing. A buffer circuit at the reference-IC output is a standard solution, but the offset voltage of the buffer tends to decrease reference accuracy. A better solution is to incorporate the buffer inside the feedback loop of the reference error amplifier. The MAX6033A voltage reference, as shown in Figure 14, offers output force and sense pins for this purpose. This device features a SOT23 package, 0.04% initial accuracy, and a 7ppm/°C temperature coefficient.\n\nThe power dissipation limit of the output transistor restricts the output current to 80mA.\n\n*Trademark. Maxim Integrated is a trademark of Maxim Integrated Inc. Other logos, product and/or company names may be trademarks of their respective owners.",
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"Shop our wide range of voltage references, including series, shunt, fixed, or adjustable, evaluation boards, and other accessories.",
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"Are you ready to demonstrate your Voltage Reference knowledge? Then take a quick 15-question multiple choice quiz to see how much you've learned from this learning module.\n\nTo earn the Voltage Reference Badge, read through the module, attain 100% in the quiz at the bottom, leave us some feedback in the comments section, and give the module a star rating.\n\n## 15) Load regulation in a voltage reference is a change in____ due to a change in the reference _____and specified by ppm/mA, µA/mA, and %/mA.\n\nAlas, you didn't quite meet the grade. You only got %. Have another look through the course, and try again.\nYou nailed it, and scored %! To earn the Voltage References Badge, leave us some feedback in the comments section and give the module a star rating. You may also download the pdf for future reference. Other topics you want to learn? Send a suggestion."
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https://courses.ansys.com/index.php/courses/turbulent-pipe-flow-les/lessons/pre-analysis-start-up-lesson-2-13/
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[
"# Pre-Analysis & Start-Up — Lesson 2\n\n### Preliminary Analysis\n\nIn Large-Eddy Simulations, the instantaneous velocity U(x,t) is decomposed into a filtered component Ū(x,t) and a residual component u'(x,t). The filtered velocity component represents the large-scale unsteady motions. In LES, the large-scale turbulent motions are directly represented, whereas the effects of small-scale turbulent motions are modeled. The filtered equations for the filtered velocity can be obtained from the Navier-Stokes equations. The non linear convective term in the momentum equation introduces a residual stress tensor which is due to the residual motions. Closure is needed for this residual stress tensor and hence it requires modeling. There is a range of models, from simple to complex, in Fluent which we will use.\n\nSince we are solving for Ū(x,t), the LES is an unsteady simulation where we march in time. In order to collect statistics like the mean and the root mean square (rms) velocities, we need to first reach a statistically stationary state. In comparison, simulation using the k-ε model solves only for the mean velocity.\n\nMore information on the LES can be found in Turbulent Flows book by Pope .\n\n### Start Ansys Fluent\n\nSince LES is a three-dimensional unsteady simulation, the computational domain is a full pipe.\n\nPrior to opening Ansys Workbench, create a folder called turbulent_pipe_LES in a convenient location. We'll use this as the working folder in which files created during the session will be stored. For this simulation Fluent will be run within the Ansys Workbench Interface. Start Ansys Workbench:\n\nStart > All Programs > Ansys > Workbench\n\nThe following figure shows the Workbench window.",
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https://aaamazingphoenix.wordpress.com/2020/09/09/covid-odyssey-dont-underestimate-ro-how-many-people-can-one-person-infect/
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[
"# COVID Odyssey: Don’t underestimate Ro ~ How many people may one person infect on average?\n\nRo\n\nStill forever be fraught\nWill we still wonder\nIt’ll be under\n‘Til true value be sought\n\nAlan Grace\n9 September 2020\n\nIn New Zealand we find Ro ~ 4.1 (4.143238) with a 10-day infectious period. We find 10-day moving averages work well to help confirm the value for Ro (~ 4.1) given a daily increase in new case numbers close to 40% over several days (r ~ 1.4) in theory and using actual NZ cases.\n\nIf a case may be infectious for 15 days (three 5-day cycles) we find that Ro may be 6.04, close to the value we previously considered for a 10-day (two 5-day cycles) infectious period.\n\nRo is by definition the reproduction number in a totally naïve population (with no quarantine or isolation). e.g. If Ro = 4 then one person with COVID-19 will on average infect four other people.\n\nWe also revise estimates for Ro worldwide. We assume a 10-day infectious period.\n\nThe estimation of Ro for COVID-19 is fraught with underestimation. All estimations of Ro taken over a long period of time that involve averages (means), moving averages, least-squares approximation or rely on actual case numbers over a long period are likely to underestimate Ro.\n\nCumulative case numbers are used (here) in analysing Ro. A daily cumulative rate also means the same rate for daily new case numbers. We have already proved this.\n\nLet\n\n• r denote the effective Reproduction rate of COVID-19 for one day\n• Ro (R0; R-Zero; R-Nought) denote the Reproduction rate without any quarantine or isolation\n• Re denote the effective Reproduction rate of COVID-19\n(Re assumes isolation/quarantine is happening)\n• a case be defined as a person diagnosed as having COVID-19\n\nIn New Zealand we have found a daily rate of increase of r = 1.4 (40% increase per day) in March and assume a 10-day infectious period (when a case can infect others).\n\nWe assume a daily increase of r (e.g. r = 1.4) and a daily decay of infectivity of 1/r over a 10-day infectious period for each case.\n\nCOVID Odyssey: [Pre-]Summer Summary~ How many people may one person infect on average?\n\nWe have r ~ 1.4 (a = 1.4) giving a good fit 16 March to 26 March (the first day of Lockdown Level 4) :",
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"",
null,
"Since cases each day produce new cases over a 10-day period, the number of new cases on each successive day is contributed towards by cases over ten days. With a daily increase of r and a decay of 1/r, we end up with the same contribution from each day (r*1/r = 1).\n\nThe table below relates the various values for r we have used, and estimates for Ro using the formula at the top of the table, and similar formulae",
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"We estimated in New Zealand a value for r of 1.4 to calculate Ro.\n\nSince r*r is 1.96 (close to 2) we also consider r = SQRT(2).\n\nOur original values for Ro are in the last column. Hence when you see values near these elsewhere on this site, you will need to divide by r to get our current estimate for Ro ~ 4.1.\n\nPreviously we considered Ro ~ 5.8 and 6 (see last column in the above table).\n\nWe note that a median value for Ro of 5.7 was found in this article:\n20-0282\n\nFor various estimates for Ro (some over 6), you may also like to look at:\ntaaa021",
null,
"An estimate for r in the article, r ~ 1.22, was estimated by us as r ~ 1.4 (see graph above).\n\nNote that r = SQRT(2) is also included on the graph.\n\nWith r ~ 1.22 we would have estimated Ro ~ 2.5;\nwith r ~ 1.4 we estimate Ro ~ 4.1.\n\nSince cases are usually isolated or quarantined once they are identified, our estimate of r ~ 1.4 may be low and maybe r ~ SQRT(2) should be considered. This would make Ro almost 4.3 (4.28).\n\nThe last row in the table above shows that a 2% increase in r (from r ~ 1.4 to r = SQRT(2) may produce a 3% increase in estimating Ro.\n\nSince some will consider our estimate for Ro ~ 4.1 high, we retain this estimate and do not increase it.\n\nWe have previously estimated Ro ~ 3, 4, and 6 in New Zealand.\n\nThese values correspond to (note the change in the first power of r below):\nRo=10 * r^9 * (r-1)/(r^10 -1)\nRo=10 * r^10 * (r-1)/(r^10 -1)\nRo=10 * r^11 * (r-1)/(r^10 -1)\n\nusing r = 1.4 or r = SQRT(2).\n\nPreviously we saw that\n\nRo=10 * r^10 * (r-1)/(r^10 -1)\n\ngave exact values for case numbers when r = 1.4 using the calculations below (first adding the totals for the previous 10 days, then multiplying the result by Ro/10):",
null,
"Hence we adopt Ro = 4.1\n[4.1432 (4 d.p.)].\n\nr = SQRT(2) also works (and other values):",
null,
"We can estimate the total number of Cases (C) for a day (the next day) as\n\nC = AV * Ro\n\nwhere AV is the average number of cases over the preceding n days (e.g. n = 10).\n\nAV is a moving average.\n\nIn the penultimate table(s) above, Sum10*Ro/10 can be re-arranged as (Sum10/10) * Ro = AV * Ro.\n\nThis is only likely to work reasonably accurately over many weeks near our ideal situation where the number of cases increases by close to r every day.\n\nWe obtain the estimates below using actual NZ cases:",
null,
"Usually the above calculations using r = 1.4 give a better estimate.\nThe estimates in the last three columns are all acceptable.\n\nBelow are some runs of our CSAW v3 simulation using r = 1.4:\nCSAWv3Samples\n\nWe conclude that in New Zealand (and elsewhere where r ~ 1.4), one person with COVID-19 may infect on average 4.1 other people (and maybe up to 4.3 (4.27575) using r = SQRT(2)).\n\nBelow is a table estimating Ro for other values of r.\nSee Re#2* (last column):",
null,
"We conjecture the following estimates for Ro for 9-day and 11-day infectious periods for r = 1.4 and r = SQRT(2):\nNZRoTable\n\nIn general, for an n-day infectious period, we have:\n\nRo=n * r^n * (r-1)/(r^n -1)\n\nSince a person may infect others up to two days before becoming symptomatic (showing symptoms of COVID-19), we consider infectious periods of n = 10, 11, and 12 days, and get these estimates for Ro:",
null,
"We get these estimates for Ro and for NZ case numbers:",
null,
"We continue up to 15 infectious days (three 5-day cycles):",
null,
"We note that in the last line above, we could have expected close to 400 cases. Clearly isolation/ quarantine was having an effect before the first full day of Lockdown Level 4.\n\nThe curve had already started to flatten before Lockdown Level 4 started.\n\nIn New Zealand we adopt Ro ~ 4.1 (4.143238) with a 10-day infectious period since cases are infectious for at least 10 days, and all Ro values match the actual number of cases equally well. This means the estimate for Ro is likely to be lower than it should be. i.e. A person infected with COVID-19 may infect on average at least 4.1 other people over a 10-day (two 5-day cycles) infectious period.\nNote: Allowing for an extra two infectious days (12 days), we would have obtained Ro ~ 4.9 (4.886185).\n\nAll the values for Ro give good estimates for the actual case numbers for\nn = 10 to 15.\n\nIf a case may be infectious for 15 days (three 5-day cycles) we find that Ro may be 6.04, close to the value we previously considered for a 10-day (two 5-day cycles) infectious period.\n\nWhen the number of cases each day is exactly r times the previous number, we get an exact match:",
null,
"The top line is the number of infectious days (n) with each column using an n-day moving average. We see that when n = 15, Ro ~ 6.04, close to the value we previously considered. In this scenario a case may be infectious for three 5-day cycles.\n\nWe conclude that a person infected with COVID-19 may infect on average at least 4.1 other people over a 10-day (two 5-day cycles) infectious period.\n\nIf a case may be infectious for 15 days (three 5-day cycles) we find that Ro may be 6.04, close to the value we previously considered for a 10-day (two 5-day cycles) infectious period.\n\nWe compare the 10 and 15-day values (scaled to 1) for a decay rate of 1/1.4. We see that less than 3% (~2.8%) of infections come from Days 11-15 with almost 1% coming from Day#11; infections from Days 1-10 are comparable although almost 3% (~2.8%) less for each day.",
null,
"Epidemiologists will need to consider the likelihood of infections over 15 days.\n\nRo ~ 4.1 (4.143238) is close to the means we have seen in 10-Day infection simulations. See:\nCOVID Odyssey: CSAW simulation ~ form[ul]ation explanation\n\nWe have a range for Ro of 4.1 to 6.\n\nAll these estimates are within the range in this graph:",
null,
"The above graph was obtained from this article: taaa021 (Click to view PDF):\nThe reproductive number of COVID-19 is higher compared to SARS coronavirus\npublished 13 February 2020, obtained from here:\n(Journal of Travel Medicine, Volume 27, Issue 2, March 2020)\n\nConservatively we adopt Ro = 4.1.\n\nRo = 6 may still be possible.\n\nWe conclude that our estimates for Ro are likely to be underestimates particularly since the “totally naïve population” requirements are unlikely to have been met. We may need to consider r = SQRT(2).\n\nRo\n\nStill forever be fraught\nWill we still wonder\nIt’ll be under\n‘Til true value be sought\n\nAlan Grace\n9 September 2020\n\nIn New Zealand we find …\n\n========\nAppendix\n========\n\nThe CSAW (COVID-19 Sampling Analysis Worksheets; “SeeSaw”) simulations produced summaries like this for a 10-day infectious period in CSAW v2\n(See CSAW v3 below these):",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Maximum Infected means the number of people directly infected by 1 case (i.e. not by another person).\n\nWe end with a sample runs of the CSAW v3 Excel simulation:\n10-day period of infectivity; r = 1.4 and the daily rate of decay in infectivity is P = 1/1.4 i.e. 1/r:\nCSAWsimV3Sampling\n\nBelow are some of our worldwide estimates for Ro:\nCOVIDWorldAvNewRanked4r\nCOVIDWorldAvNewAlpha4r\n\nI share my posts at:\nhttps://guestdailyposts.wordpress.com/guest-pingbacks/"
] |
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|
http://iesbdc.ddns.us/corey/43-Solving-Linear-Quadratic-System-Homework-Answers.html
|
[
"# Unit 4 Solving Quadratic Equations Homework 9.\n\nRight from quadratic equations to equations, we have got every part discussed. Come to Algebra-equation.com and read and learn about equation, formula and loads of other algebra topics.",
null,
"Solving Quadratic Systems. In a linear-quadratic system, use substitution to solve. In a quadratic-quadratic system, use elimination to solve. When graphing inequalities, remember the conventions about graphing boundaries using either solid or dotted lines. If possible, check your solutions to systems of equations by graphing.",
null,
"We tried to locate some good of Linear Quadratic Systems Worksheet 1 together with Algebra 1 Worksheets image to suit your needs. Here it is. It was from reliable on line source and that we love it. We hope this graphic will likely be one of excellent reference.",
null,
"Rectangular shape related questions in linear equations. 9. System of linear equations. 10. System of linear-quadratic equations. 11. System of quadratic-quadratic equations. 12. Solving 3 variable systems of equations by substitution. 13. Solving 3 variable systems of equations by elimination. 14.",
null,
"Linear Quadratic Systems. Linear Quadratic Systems - Displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are Systems of quadratic equations, Unit 2 solving systems of linear and quadratic equations, Mcr 3u date linear quadratic systems of equations, Linear quadratic systems line parabola intersection, Systems of two equations, Solving quadratic systems.",
null,
"Students begin to work with Linear Quadratic Systems in a series of math worksheets, lessons, and homework. A quiz and full answer keys are also provided.",
null,
"Section 4.3 Solving Systems of Linear Inequalities A2.5.4 Solve systems of linear equations and inequalities in two variables by substitution, graphing, and use matrices with three variables; Packet.\n\n## Solving Linear Systems by Substitution - 2012.",
null,
"Start studying Solving Quadratic-Linear Systems Algebraically. Learn vocabulary, terms, and more with flashcards, games, and other study tools.",
null,
"This homework assignment is a bit of a confidence builder, but as we approach the end of the unit, confidence is important: Solve systems of linear and quadratic functions algebraically 14 - 8 HW157 Solve systems of linear and quadratic functions algebraically.docx.",
null,
"Linear Quadratic Systems. Showing top 8 worksheets in the category - Linear Quadratic Systems. Some of the worksheets displayed are Systems of quadratic equations, Unit 2 solving systems of linear and quadratic equations, Mcr 3u date linear quadratic systems of equations, Linear quadratic systems line parabola intersection, Systems of two equations, Solving quadratic systems, Chapter 5.",
null,
"Homework Practice Workbook. 3-2 Solving Linear Equations by Graphing .37 3-3 Rate of Change and Slope. 9-4 Solving Quadratic Equations by Completing the Square .121 9-5 Solving Quadratic Equations by Using the Quadratic Formula.",
null,
"Linear and Quadratic Graphs Exercises STUDYSmarter Question 4 For each of the following quadratic equations, nd 1.whether it is concave up (smile) or concave down (frown).",
null,
"Solve a system containing a linear equation and a quadratic equation in two variables (conic sections possible) graphically and symbolically. Common Core: HSA-REI.C.7 Solving Linear and Quadratic Systems How to find the solutions to linear and quadratric systems given a graph, equations, or a context?",
null,
"Determine the number of solutions to the given linear-quadratic system. Linear Quadratic Systems of Equations DRAFT. 9th grade. 213 times. Mathematics. 70% average accuracy. a year ago. yreyes. 0. Save. Edit. Edit. Linear Quadratic Systems of Equations DRAFT. a year ago. by yreyes. Played 213 times. 0.. When solving linear-quadratic systems.\n\n## Chapter 5: Graphing Quadratics Systems of Equations.\n\nNow is the time to redefine your true self using Slader’s free enVision Algebra 1 answers. Shed the societal and cultural narratives holding you back and let free step-by-step enVision Algebra 1 textbook solutions reorient your old paradigms.Linear Quadratic Systems Five Pack Math Worksheets Land from linear quadratic systems worksheet 1, source:yumpu.com 45 Free Download Linear Quadratic Systems Worksheet 1 By Frank Webb Posted on September 21, 2017 July 28, 2018 1 views.Task 2: Non-Linear System of Equations Create a system of equations that includes one linear equation and one quadratic equation. Part 1: Show all work to solving your system of equations algebraically. Part 2: Graph your system of equations, and show the. asked by lauren on June 7, 2018; college geomrtry.\n\nSolving Systems of Equations by Substitution Method. Substitution is the most elementary of all the methods of solving systems of equations. Substitution method, as the method indicates, involves substituting something into the equations to make them much simpler to solve.A fun foldable to teach or review solving nonlinear systems graphically and algebraically. Each system has a quadratic function and a linear function which models 1 solution, 2 solutions, and no real solution.\n\nessay service discounts do homework for money"
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https://grass.osgeo.org/grass78/manuals/i.topo.corr.html
|
[
"",
null,
"## NAME\n\ni.topo.corr - Computes topographic correction of reflectance.\n\n## KEYWORDS\n\nimagery, terrain, topographic correction\n\n## SYNOPSIS\n\ni.topo.corr\ni.topo.corr --help\ni.topo.corr [-is] [input=name[,name,...]] output=name basemap=name zenith=float [azimuth=float] [method=string] [--overwrite] [--help] [--verbose] [--quiet] [--ui]\n\n### Flags:\n\n-i\nOutput sun illumination terrain model\n-s\nScale output to input and copy color rules\n--overwrite\nAllow output files to overwrite existing files\n--help\nPrint usage summary\n--verbose\nVerbose module output\n--quiet\nQuiet module output\n--ui\nForce launching GUI dialog\n\n### Parameters:\n\ninput=name[,name,...]\nName of reflectance raster maps to be corrected topographically\noutput=name [required]\nName (flag -i) or prefix for output raster maps\nbasemap=name [required]\nName of input base raster map (elevation or illumination)\nzenith=float [required]\nSolar zenith in degrees\nazimuth=float\nSolar azimuth in degrees (only if flag -i)\nmethod=string\nTopographic correction method\nOptions: cosine, minnaert, c-factor, percent\nDefault: c-factor\n\n## DESCRIPTION\n\ni.topo.corr is used to topographically correct reflectance from imagery files, e.g. obtained with i.landsat.toar, using a sun illumination terrain model. This illumination model represents the cosine of the incident angle i, i.e. the angle between the normal to the ground and the sun rays.\n\nNote: If needed, the sun position can be calculated for a given date with r.sunmask.",
null,
"Figure showing terrain and solar angles\n\nUsing the -i flag and given an elevation basemap (metric), i.topo.corr creates a simple illumination model using the formula:\n\n• cos_i = cos(s) * cos(z) + sin(s) * sin(z) * cos(a - o)\nwhere, i is the incident angle to be calculated, s is the terrain slope angle, z is the solar zenith angle, a the solar azimuth angle, o the terrain aspect angle.\n\nFor each band file, the corrected reflectance (ref_c) is calculate from the original reflectance (ref_o) using one of the four offered methods (one lambertian and two non-lambertian).\n\n### Method: cosine\n\n• ref_c = ref_o * cos_z / cos_i\n\n### Method: minnaert\n\n• ref_c = ref_o * (cos_z / cos_i) ^k\nwhere, k is obtained by linear regression of\nln(ref_o) = ln(ref_c) - k ln(cos_i/cos_z)\n\n### Method: c-factor\n\n• ref_c = ref_o * (cos_z + c)/ (cos_i + c)\nwhere, c is a/m from ref_o = a + m * cos_i\n\n### Method: percent\n\nWe can use cos_i to estimate the percent of solar incidence on the surface, then the transformation (cos_i + 1)/2 varied from 0 (surface in the side in opposition to the sun: infinite correction) to 1 (direct exhibition to the sun: no correction) and the corrected reflectance can be calculated as\n• ref_c = ref_o * 2 / (cos_i + 1)\n\n## NOTES\n\n1. The illumination model (cos_i) with flag -i uses the actual region as limits and the resolution of the elevation map.\n2. The topographic correction use the full reflectance file (null remain null) and its resolution.\n3. The elevation map to calculate the illumination model should be metric.\n\n## EXAMPLES\n\nFirst, make a illumination model from the elevation map (here, SRTM). Then make perform the topographic correction of e.g. the bands toar.5, toar.4 and toar.3 with output as tcor.toar.5, tcor.toar.4, and tcor.toar.3 using c-factor (= c-correction) method:\n\n```# first pass: create illumination model\ni.topo.corr -i base=SRTM zenith=33.3631 azimuth=59.8897 output=SRTM.illumination\n\n# second pass: apply illumination model\ni.topo.corr base=SRTM.illumination input=toar.5,toar.4,toar.3 output=tcor \\\nzenith=33.3631 method=c-factor\n```\n\n## REFERENCES\n\n• Law K.H. and Nichol J, 2004. Topographic Correction For Differential Illumination Effects On Ikonos Satellite Imagery. International Archives of Photogrammetry Remote Sensing and Spatial Information, pp. 641-646.\n• Meyer, P. and Itten, K.I. and Kellenberger, KJ and Sandmeier, S. and Sandmeier, R., 1993. Radiometric corrections of topographically induced effects on Landsat TM data in alpine terrain. Photogrammetric Engineering and Remote Sensing 48(17).\n• Riaño, D. and Chuvieco, E. and Salas, J. and Aguado, I., 2003. Assessment of Different Topographic Corrections in Landsat-TM Data for Mapping Vegetation Types. IEEE Transactions On Geoscience And Remote Sensing, Vol. 41, No. 5\n• Twele A. and Erasmi S, 2005. Evaluating topographic correction algorithms for improved land cover discrimination in mountainous areas of Central Sulawesi. Göttinger Geographische Abhandlungen, vol. 113."
] |
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null,
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null,
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|
https://docs.snowflake.net/manuals/sql-reference/functions/grouping_id.html
|
[
"Categories:\n\nAggregate Functions (General)\n\n# GROUPING_ID¶\n\nDescribes which of a list of expressions are grouped in a row produced by a GROUP BY query.\n\nAlias for GROUPING.\n\n## Syntax¶\n\n```GROUPING_ID( <expr1> [ , <expr2> , ... ] )\n```\n\n## Usage Notes¶\n\nGROUPING_ID is not an aggregate function, but rather a utility function that can be used alongside aggregation, to determine the level of aggregation a row was generated for:\n\n• GROUPING_ID(`expr`) returns 0 for a row that is grouped on `expr`, and 1 for a row that is not grouped on `expr`.\n\n• GROUPING_ID(`expr1`, `expr2` , … , `exprN`) returns the integer representation of a bit-vector containing GROUPING_ID(`expr1`) , GROUPING_ID(`expr2`) , … , GROUPING_ID(`exprN`).\n\n## Examples¶\n\nThe examples use the following table and data:\n\n```CREATE OR REPLACE TABLE aggr2(col_x int, col_y int, col_z int);\nINSERT INTO aggr2 VALUES (1, 2, 1),\n(1, 2, 3);\nINSERT INTO aggr2 VALUES (2, 1, 10),\n(2, 2, 11),\n(2, 2, 3);\n```\n\nThis example groups on col_x. Calling `GROUPING_ID(col_x)` returns 0, indicating that col_x is indeed one of the grouping columns.\n\n```SELECT col_x, sum(col_z), GROUPING_ID(col_x)\nFROM aggr2\nGROUP BY col_x\nORDER BY col_x;\n+-------+------------+--------------------+\n| COL_X | SUM(COL_Z) | GROUPING_ID(COL_X) |\n|-------+------------+--------------------|\n| 1 | 4 | 0 |\n| 2 | 24 | 0 |\n+-------+------------+--------------------+\n```\n\nThis query groups by sets:\n\n```SELECT col_x, col_y, sum(col_z),\nGROUPING_ID(col_x),\nGROUPING_ID(col_y),\nGROUPING_ID(col_x, col_y)\nFROM aggr2\nGROUP BY GROUPING SETS ((col_x), (col_y), ())\nORDER BY col_x;\n+-------+-------+------------+--------------------+--------------------+---------------------------+\n| COL_X | COL_Y | SUM(COL_Z) | GROUPING_ID(COL_X) | GROUPING_ID(COL_Y) | GROUPING_ID(COL_X, COL_Y) |\n|-------+-------+------------+--------------------+--------------------+---------------------------|\n| 1 | NULL | 4 | 0 | 1 | 1 |\n| 2 | NULL | 24 | 0 | 1 | 1 |\n| NULL | NULL | 28 | 1 | 1 | 3 |\n| NULL | 2 | 18 | 1 | 0 | 2 |\n| NULL | 1 | 10 | 1 | 0 | 2 |\n+-------+-------+------------+--------------------+--------------------+---------------------------+\n```",
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""
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[
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"https://docs.snowflake.net/manuals/_static/img/snowflake-logo-reversed.png",
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http://managedstudy.com/micro/measurement-of-price-elasticity-of-demand.htm
|
[
"# Measurement of price elasticity of demand\n\nThree popular methods for measuring price elasticity of demand are discussed below:\n\n## 1. Percentage method\n\nThe percentage method measures price elasticity of demand by dividing the percentage change in amount demand by percentage change in price of commodity. The elasticity of demand is unity, greater than unity and less than unity. Demand is unity if change in demand is proportionate to the change in price. Demand is greater than unity when change in demand is more than proportionate change in price. The demand is less than unity if change is less than proportionate change in price. The coefficient of price elasticity of demand is always negative because change in price brings a change in demand in opposite direction. Negative signs are usually disregarded.\n\nThe following formula is used for the measurement of price elasticity of demand:\n\nEy = Δq/Δp × p/q\n\nElasticity less than unity:\nA consumer buys 5 kg of commodity at \\$10. If the price falls to \\$6 the consumer buys 6 kilograms. Elasticity in this case is less than 1.\n\nEy = Δq/Δp × p/q = 1/4 × 10/5 = 1/2 < 1 (less elastic)\n\nElasticity is unity:\nA consumer buys 5 kg of commodity at \\$10. If the price rises to \\$12 the consumers buys 4 kilograms. Elasticity in this case is1.\n\nEy = Δq/Δp × p/q = 1/2 × 10/5 = 1 (unity)\n\nElasticity is more than unity:\nA consumer buys 5 kg of commodity at \\$10. if the price rises to \\$11 the consumers buys 4 kilograms. Elasticity in this case is more than 1. Thus\n\nEy = Δ q /Δ p × p/q = 1/1 × 10/5 = 2 > 1 (elastic)\n\n## 2. Total expenditure method\n\nMarshal evolved total expenditure (outlay) of consumer or total revenue of seller as measure of elasticity. Total expenditure of a producer is compared before and after change in price. Total outlay is equal to price multiplied by quantity demanded. This method of the measurement of price elasticity of demand tells us whether demand for a commodity is elastic or greater than unity. When fall in its price leads to increase in total expenditure in it and a rise the price causes decrease in total expenditure. Demand is equal to unity when total expenditure remains the same whether there is increase in price or decrease in price of goods. Demand is said to be inelastic or less than unity when total expenditure decrease with decrease in price and increase with increase in price.\n\nSchedule for expenditure elastic demand\n\n Price Quantity demanded outlay 12 2 24 6 4 24 3 8 24",
null,
"The total expenditure remains unchanged at \\$24 in unitary elastic demand. There is fall in price from \\$12 to \\$6 and \\$3, the total expenditure remains unchanged at \\$24. When there is rise in price from \\$3 to \\$6 and \\$12, the total expenditure remains unchanged at \\$24. The elasticity of price is = 1 or unitary elastic demand whether there is increase in price or decrease in price.\n\nSchedule of expenditure elastic demand\n\n Price Quantity demanded Outlay 12 2 24 6 6 36 3 16 48",
null,
"The total expenditure rises from \\$24 to \\$36 and \\$48 respectively with the fall in price from \\$12 to \\$6 and \\$3. The total expenditure falls from \\$48 to \\$36 and \\$24 respectively with the rise in price from \\$3 to \\$6 and \\$12. The elasticity of price is > 1 in this case.\n\nSchedule of expenditure inelastic demand\n\n Price Quantity demanded Outlay 12 2 24 6 3 18 3 4 12",
null,
"The total expenditure falls from \\$24 to \\$18 and \\$12 respectively with fall in price from \\$12 to \\$6 and \\$3. The total expenditure rises from \\$12 to \\$18 and \\$24 respectively with the rise in price from \\$3 to \\$6 and \\$12. The elasticity price is < 1. There is inelastic demand in such case.\n\nThis method is simple as it classifies the elasticity into three categories. Its drawback is that it fails to measure elasticity in exact figures.\n\n## 3. Point elasticity method\n\nThe point elasticity of demand is proportionate change in quantity demand in response to very small proportionate change in price. The point elasticity concept is useful when change in price and the consequent change in quantity demanded is very small. Marshal suggested this method. It is also called geometrical method because graph is drawn to show the demand curve. This method is explained with the help of (1) straight line demand curve and (2) convex demand curve.\n\n(1). Straight line demand curve:\nThe diagram shows a straight line demand curve. We join both sides of the straight line demand curve with the two axes at points D and C. Elasticity at any points is equal to the ratio of the distance from the point P to the X axis and the distance to the Y axis. In the diagram the point N is half way between D and C. The elasticity of demand is = Ep = NC / ND. The elasticity is equal to 1 at this point. Similarly at point M elasticity is greater than 1. The elasticity at point P is less than 1.",
null,
"(2). Convex demand curve:\nThere is convex demand curve DD. Suppose we want to check price elasticity at point A. we can draw a tangent RS at the point A. the elasticity is found as AM / AR. Similarly for finding out elasticity at point B we draw a tangent at this point to the demand curve. The elasticity at this point is given by the ratio of the distance along the tangent to the X-axis divided by the distance of the Y-axis.",
null,
"## ARC method:\n\nThe concept of ARC elasticity was provided by Dalton and than it was further developed by Lerner. This method for the measurement of price elasticity of demand is applied when the change in price is somewhat large or the price elasticity over an ARC of demand is provided. ARC elasticity of demand is the elasticity between distinct points on the demand curve. It is an increase of average responsiveness to price change shown by a demand curve. Any two points on demand curve make an ARC. The area between A and B on the DD curve is an ARC that measures elasticity over a certain range of price and quantities.",
null,
"The diagram shows quantity demanded on OX-axis and price on OY-axis. The demand curve DD is passing through two points A and B for quantity demanded between Q and Q1. The formula for measuring ARC elasticity is Δq ÷ Δp × (p + p1) ÷ (q + q1). Suppose the price of a commodity falls from \\$10 to \\$5 the quantity demanded increases from 100 to 200 units. The ARC elasticity is = 100 ÷ 5 × (10 + 5) ÷ (100 + 200) = 100 ÷ 5 × 15 ÷ 300 = 1"
] |
[
null,
"http://managedstudy.com/micro/images/measurement-of-price-elasticity-of-demand1.png",
null,
"http://managedstudy.com/micro/images/mopeod2.png",
null,
"http://managedstudy.com/micro/images/mopeod3.png",
null,
"http://managedstudy.com/micro/images/straight-line-demand-curve.png",
null,
"http://managedstudy.com/micro/images/convex-demand-curve.png",
null,
"http://managedstudy.com/micro/images/arc-method-of-measuring-elasticity.png",
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] |
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https://www.bartleby.com/questions-and-answers/2-5-4-3-4-5-the-function-graphed-above-is-increasing-on-the-intervals-preview-decreasing-on-the-inte/bf08ea1a-1c2f-4247-8a0c-1ef239ad5e77
|
[
"# 2-5 -4 -3 - -4 5The function graphed above is:Increasing on the interval(s)PreviewDecreasing on the interval(s)PreviewGet help: Videoer\n\nQuestion\n1 views",
null,
"help_outlineImage Transcriptionclose2 -5 -4 -3 - - 4 5 The function graphed above is: Increasing on the interval(s) Preview Decreasing on the interval(s) Preview Get help: Video er fullscreen\ncheck_circle\n\nStep 1\n\nGiven graph is:\n\nFrom the graph, the critical points are at x = -1 and 2 as the graph changes here.\n\nNow the domain is divided into three intervals based on the critical points i.e. (-∞,-1),(-1,2),(2,∞).\n\nConsider the interval (-∞,-1):\n\nAs we move from left to right in the interval, the graph is moving downwards i.e. the value of the function is decreasing in this interval.\n\nCon...\n\n### Want to see the full answer?\n\nSee Solution\n\n#### Want to see this answer and more?\n\nSolutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*\n\nSee Solution\n*Response times may vary by subject and question.\nTagged in\n\n### Calculus",
null,
""
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[
null,
"https://prod-qna-question-images.s3.amazonaws.com/qna-images/question/fdaf9d65-ef37-47e0-9965-c2cbe1b7de49/bf08ea1a-1c2f-4247-8a0c-1ef239ad5e77/zg5ejhr.png",
null,
"https://www.bartleby.com/static/logo.svg",
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https://www.us-b.fun/dsa/2020/1166/43/11/3510.431035241111662020
|
[
"# LeetCode-1024. 视频拼接\n\nUS-B.Ralph\n2020-10-24\n\n## 问题地址\n\nLeetCode每日一题/2020-10-24\n\n## 问题描述\n\n### 示例\n\n• 示例1\n输入：clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10\n\n\n• 示例2\n输入：clips = [[0,1],[1,2]], T = 5\n\n\n• 示例3\n输入：clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9\n\n\n• 示例4\n输入：clips = [[0,4],[2,8]], T = 5\n\n\n\n### 进阶\n\n• 1 <= clips.length <= 100\n• 0 <= clips[i] <= clips[i] <= 100\n• 0 <= T <= 100\n\n1. 时间复杂度\n2. 空间复杂度\n\n## 官方解法\n\n### 方法一 动态规划\n\n#### 思路及算法\n\n• 比较容易想到的方法是动态规划，我们令 \\textit{dp}[i] 表示将区间 [0,i) 覆盖所需的最少子区间的数量。由于我们希望子区间的数目尽可能少，因此可以将所有 \\textit{dp}[i] 的初始值设为一个大整数，并将 \\textit{dp}（即空区间）的初始值设为 0。\n\n• 我们可以枚举所有的子区间来依次计算出所有的 \\textit{dp} 值。我们首先枚举 i，同时对于任意一个子区间 [a_j,b_j)，若其满足d： a_j < i ≤ b_j 。\n\n• 那么它就可以覆盖区间 [0, i) 的后半部分，而前半部分则可以用 \\textit{dp}[a_j] 对应的最优方法进行覆盖，因此我们可以用 dp[a_j] + 1 来更新 \\textit{dp}[i]。状态转移方程如下：dp[i] = \\min{dp[a_j]}+ 1\n\n• 最终的答案即为 \\textit{dp}[T]\n\n#### 复杂度分析：\n\n1. 时间复杂度:O(T×N)，其中 T 是区间的长度，N 是子区间的数量。对于任意一个前缀区间 [0,i) ，我们都需要枚举所有的子区间，时间复杂度 O(N)。总时间复杂度为 O(T) \\times O(N) = O(T \\times N)\n2. 空间复杂度：O(T)，其中 T 是区间的长度。我们需要记录每一个前缀区间 [0,i) 的状态信息。\n\n#### 编码实现\n\nclass Solution {\npublic int videoStitching(int[][] clips, int T) {\nint[] dp = new int[T + 1];\nArrays.fill(dp, Integer.MAX_VALUE - 1);\ndp = 0;\nfor (int i = 1; i <= T; i++) {\nfor (int[] clip : clips) {\nif (clip < i && i <= clip) {\ndp[i] = Math.min(dp[i], dp[clip] + 1);\n}\n}\n}\nreturn dp[T] == Integer.MAX_VALUE - 1 ? -1 : dp[T];\n}\n}\n\n\n### 方法二 贪心\n\n#### 思路\n\n• 注意到对于所有左端点相同的子区间，其右端点越远越有利。且最佳方案中不可能出现两个左端点相同的子区间。于是我们预处理所有的子区间，对于每一个位置 i，我们记录以其为左端点的子区间中最远的右端点，记为 \\textit{maxn}[i]\n• 我们可以参考「55. 跳跃游戏的官方题解」，使用贪心解决这道题。\n• 具体地，我们枚举每一个位置，假设当枚举到位置 i 时，记左端点不大于 i 的所有子区间的最远右端点为 \\textit{last}。这样 \\textit{last} 就代表了当前能覆盖到的最远的右端点。\n• 每次我们枚举到一个新位置，我们都用 \\textit{maxn}[i] 来更新 \\textit{last}。如果更新后 last == ilast==i，那么说明下一个位置无法被覆盖，我们无法完成目标。\n• 同时我们还需要记录上一个被使用的子区间的结束位置为 \\textit{pre}，每次我们越过一个被使用的子区间，就说明我们要启用一个新子区间，这个新子区间的结束位置即为当前的 \\textit{last}。也就是说，每次我们遇到 i == pre，则说明我们用完了一个被使用的子区间。这种情况下我们让答案加 1，并更新 \\textit{pre} 即可。\n\n#### 复杂度分析：\n\n1. 时间复杂度:O(T+N)，其中 T 是区间的长度，N 是子区间的数量。我们都需要枚举每一个位置，时间复杂度 O(T)。总时间复杂度为 O(T) + O(N) = O(T + N)\n\n2. 空间复杂度：O(T)，其中 T 是区间的长度。我们需要记录以其为左端点的子区间的最右端点。\n\n#### 编码实现\n\nclass Solution {\npublic int videoStitching(int[][] clips, int T) {\nint[] maxn = new int[T];\nint last = 0, ret = 0, pre = 0;\nfor (int[] clip : clips) {\nif (clip < T) {\nmaxn[clip] = Math.max(maxn[clip], clip);\n}\n}\nfor (int i = 0; i < T; i++) {\nlast = Math.max(last, maxn[i]);\nif (i == last) {\nreturn -1;\n}\nif (i == pre) {\nret++;\npre = last;\n}\n}\nreturn ret;\n}\n}\n\n\n## 精彩评论\n\n### 跳转地址1：【Java】1024节，不准不“贪心”\n\n#### 思路\n\n(本题解 借鉴了 官方题解 的思想)\n– 很纯粹的 贪心 思想：\n– 1、首先来初始化一个 maxEnd数组，用于保存 以当前数字(下标)为起点 的区间的 最大的结束位置\n– 2、遍历clips，初始化maxEnd数组(每个元素开头的区间的最大结束位置)\n– 3、定义三个变量，辅助完成之后的操作：\n– 1、pre 记录 结果中上一次的最大结束位置(本轮的最小开始位置)\n– 2、last 记录 当前遍历到的 区间最大结束位置\n– 3、count 记录 结果\n– 4、根据maxEnd数组，计算最终结果\n\n• 因为maxEnd[i]数组为最大结束位置，因此：\n• 当前元素 == 本区间最大元素，即：区间断开，无法连续到后续位置，返回-1;\n• 当前元素 == 上一个区间的最大结束元素，即：到达了上一个满足条件的区间的结束位置\n这时的last为当前最大的结束位置，我们将其放入满足条件的区间集合之中(因为本题只需要我们记录 满足条件的区间个数，因为只需要 更新count和pre 即可)\n\n#### 编码实现\n\nclass Solution {\npublic int videoStitching(int[][] clips, int T) {\nif (clips == null) {\nreturn 0;\n}\n\nint[] maxEnd = new int[T]; // 用于保存 以当前数字(下标)为起点 的区间的 最大的结束位置\n\n/*\n遍历clips，初始化maxEnd数组(每个元素开头的区间的最大结束位置)\n*/\nfor (int[] clip : clips) {\nif (clip < T) {\nmaxEnd[clip] = Math.max(maxEnd[clip], clip);\n}\n}\n\n/*\n根据maxEnd数组，计算最终结果\n因为maxEnd[i]数组为最大结束位置，因此：\n1、当前元素 == 本区间最大元素，\n即：区间断开，无法连续到后续位置，返回-1\n2、当前元素 == 上一个区间的最大结束元素，\n即：到达了上一个满足条件的区间的结束位置\n这时的last为当前最大的结束位置，我们将其放入满足条件的区间集合之中\n(因为本题只需要我们记录 满足条件的区间个数，因为只需要 更新count和pre 即可)\n*/\nint pre = 0; // 记录 结果中上一次的最大结束位置(本轮的最小开始位置)\nint last = 0; // 记录当前遍历到的 区间最大结束位置\nint count = 0; // 记录结果\nfor (int i = 0; i < T; i++) {\nlast = Math.max(maxEnd[i], last);\nif (i == last) { // 当前元素 == 本区间最大元素(无法到达后续位置)\nreturn -1;\n}\n\nif (i == pre) { // 当前元素 == 上一个区间的最大元素\ncount++;\npre = last;\n}\n}\nreturn count;\n}\n}\n}\n\n//反转链表\nListNode prev = null;\n}\nreturn prev;\n}\n\n\n### 跳转地址2：「手画图解」动态规划 思路 | 分情况讨论\n\n#### 思路",
null,
"• 怎么求dp[j]？即，递推关系怎么找？我们画图看看，有哪些情况：",
null,
"#### 编码实现\n\nconst videoStitching = (clips, T) => {\nclips.sort((a, b) => a - b); // 按开始时间从小到大排序\n// dp[j]：涵盖[0:j]需要的clip的最少个数，目标是求dp[T]，初始值为101，大于所有可能值\nconst dp = new Array(T + 1).fill(101);\ndp = 0; // base case\n\nfor (let i = 0; i < clips.length; i++) { // 遍历每个片段\nconst [start, end] = clips[i]; // 获取当前片段的开始和结束时间\nfor (let j = start + 1; j <= end; j++) { // 计算当前片段上每个点的dp[j]\ndp[j] = Math.min(dp[j], dp[start] + 1);\n}\n}\nif (dp[T] == 101) { // 如果dp值为101，说明覆盖不了[0:T]，否则返回dp[T]\nreturn -1;\n}\nreturn dp[T];\n};\n\n\n#### 优化\n\nconst videoStitching = (clips, T) => {\nclips.sort((a, b) => a - b); // 按开始时间从小到大排序\n// dp[j]：涵盖[0:j]需要的clip的最少个数，目标是求dp[T]，初始值为101，大于所有可能值\nconst dp = new Array(T + 1).fill(101);\ndp = 0; // 涵盖[0:0]不需要clip，所以为0\n\nfor (let i = 0; i < clips.length; i++) { // 遍历每个片段\nconst [start, end] = clips[i]; // 获取当前片段的开始和结束时间\nif (dp[start] == 101) { // 该片段上都覆盖不了，直接退出循环，保持为101\nbreak;\n}\nfor (let j = start + 1; j <= end; j++) { // 计算当前片段上每个点的dp[j]\ndp[j] = Math.min(dp[j], dp[start] + 1);\n}\n}\n\nif (dp[T] == 101) {\nreturn -1;\n}\nreturn dp[T];\n};",
null,
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|
https://avpnftq.web.app/bazelais65714wos/traceless-stress-energy-tensor-xifi.html
|
[
"Gravitational Waves - USU\n\nThis is the classical stress tensor. Even if it is traceless, the quantum stress tensor might have a non-zero trace, for two di↵erent reasons. First, the UV regulator introduces a scale, and may introduce a trace. In fact, in a renormalizable theory, Tµ µ (x)= X i g i O i(x) , (23.20) where O i … Electromagnetism II, Lecture Notes 9 symmetric tensor, because it is just a number. (I am using. S. for symmetric tensors, while reserving. C. for traceless symmetric tensors.) It takes 3 numbers to specify. S (1) i, since the 3 values. S (1) (1) (1) 1, S (2) 2,and. S. 3. can each be specified independently. For. S. ij, however, weseetheconstraintsofsymmetry: S (2) hastoequal (2 Cosmological Dynamics - E. Bertschinger We have introduced two 3-scalar fields (x, ) and (x, ), one 3-vector field w(x, ) = w i e i, and one symmetric, traceless second-rank 3-tensor field h(x, ) = h ij e i e j.No generality is lost by making h ij traceless since any trace part can be put into .The factors of 2 and signs have been chosen to simplify later expressions. Quantum Field Theory: Is there any physical interpretation\n\n## Regularization of the stress-energy tensor for vector and\n\non the use of the traceless stress tensor (TST). It is shown that it naturally leads to the appearance of a modified viscosity given by C. =3/ tr.˝/ where is the shear-viscosity coefficient, the relaxation time and tr(˝) the trace of the extra stress tensor. This modified … CHAPTER 6 EINSTEIN EQUATIONS - WordPress.com\n\n### Feb 19, 2019\n\nJun 23, 2019 · However, which only coincides with his final (correct) equation if the stress-energy tensor T (and hence also R) is traceless, i.e. that the sum of the elements on the main diagonal of the matrix We worked out the stress-energy tensor Tij for the case of a perfect fluid in its rest frame, so now we want to generalize this to the case where the fluid is viewed from some more general coordinate system, possibly in curved spacetime. The result is simply stated in Moore’s equation 20.16 and in pretty well every other source I looked at. the gravitational contribution to the total energy is small, then this expression can be treated in the weak field limit, and reduces to the famous quadrupole formula: hTT jk = 2G c4 1 r I¨TT jk (t− r/c) → 2 r I¨TT jk (t− r) Here I jk is the reduced (trace-free) quadrupole moment tensor , given by Ijk = Ijk − 1 3 δjkδ lmI lm where Casimir energy of the CFT on a toroidal geometry. Finally, we present a discussion of our results in section 4. While this paper was in preparation, ref. appeared which discusses calculating the CFT stress-energy using techniques similar to those in section 2.2. 2 Stress-Energy Tensor The improved stress-energy tensor defines the same field energy-momentum and angular momentum as the conventional tensor, and it is traceless for a non-interacting field theory when all coupling constants are physically dimensionless. 9. Stress-Energy-Momentum Tensor of Matter Fields 10. Stress-Energy-Momentum Tensors of Gauge Potentials 11. Stress-Energy-Momentum Tensors of Proca Fields 12. Topological Gauge Theories Part 2. 13. Reduced Second Order Lagrangian Formalism 14. Conservation Laws in Einstein’s Gravitation Theory 15. First Order Palatini Formalism 16. Stress"
] |
[
null
] |
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|
https://ivov.dev/notes/graphs
|
[
"Graphs\n\n# Graphs\n\n2020-02-13T23:46:37.003Z\n\nGraphs represent relationships between entities—graphs convey how data is connected. Graphs have no root, only nodes (or vertices) and edges. If two nodes are directly connected, we say they are adjact or neighbors. If the edges have a direction, we say it is a directed graph. Connections may have a weight, i.e. how strong the connection is.",
null,
"Graph",
null,
"Directed and undirected graph\n\nTrees as graphs\n\nA tree is actually a special case of a graph—a tree is simply a graph without a cycle. A tree is a special case of a graph where each node (except the root) has exactly one node referencing it (its parent) and there are no cycles.\n\nUnlike tree nodes, graph nodes (or vertices) can have multiple parents. In addition, the links between nodes may have values or weights. These links are called \"edges\" because they may contain more information than just a pointer. In a graph, edges can be one-way or two-way. A graph with one-way edges is called a \"directed graph\". A graph with only two-way edges is called an \"undirected graph\". Graphs may have cycles, so when these algorithms are applied to graphs, some mechanism is needed to avoid re-traversing parts of the graph that have already been visited.\n\nGraphs can be implemented with an adjacency matrix:",
null,
"When we do time complexity analysis on graphs, we use `V` (for vertices) and `E` (for edges). For example, adding an item to and removing an item from an adjacency matrix is `O(V^2)` because we have to copy all the other items around. Adding and removing an edge as well as querying an edge is `O(1)` because we use a hash table to keep the vertices indexed. Finding the neighbors of a node runs in `O(V)`.",
null,
"Graphs can also be implemented with an adjacency list (an array of linked lists):",
null,
"A dense graph is one where all vertices are connected to all other vertices. If you are dealing with a dense graph, use an adjacency matrix; otherwise, use an adjacency list in most cases.\n\nA graph may be directed or undirected. In a directed graph, each edge has a direction, which indicates that you can move from one vertex to the other through the edge. In an undirected graph, you can move in both directions between vertices.\n\nTwo vertices in a graph are said to be adjacent if they are connected by the same edge. Similarly, two edges are said to be adjacent if they are connected to the same vertex. An edge in a graph that joins two vertices is said to be incident to both vertices. The degree of a vertex is the number of edges incident to it.\n\nTwo vertices are called neighbors if they are adjacent. Similarly, two edges are called neighbors if they are adjacent.\n\nYou can represent a graph using an adjacency matrix or adjacency lists. Which one is better? If the graph is dense (i.e., there are a lot of edges), using an adjacency matrix is preferred. If the graph is very sparse (i.e., very few edges), using adjacency lists is better, because using an adjacency matrix would waste a lot of space.\n\nThere are two types of weighted graphs: vertex weighted and edge weighted. In a vertexweighted graph, each vertex is assigned a weight. In an edge-weighted graph, each edge is assigned a weight. Of the two types, edge-weighted graphs have more applications.\n\n## Implementation\n\n``````public class Graph {\nprivate class Node {\nprivate String label;\nNode(String label) {\nthis.label = label;\n}\n\n@Override\npublic void toString() {\nreturn label;\n}\n}\nprivate Map<String, Node> nodes = new HashMap<>();\nprivate Map<Node, List<Node>> adjacencyList = new HashMap<>();\n// this implementation uses maps, but still considered an adjacency list\n\npublic void addNode(String label) {\nvar node = new Node(label);\nnodes.putIfAbsent(label, node);\n}\n\npublic void addEdge(String from, String to) {\nvar fromNode = nodes.get(from);\nif (fromNode == null)\nthrow new IllegalArgumentException();\n\nvar toNode = nodes.get(to);\nif (toNode == null)\nthrow new IllegalArgumentException();\n\n}\n\npublic void print() {\nfor (var source : adjecencyList.keySet()) {\nvar targets = adjacencyList.get(source)\nif (!targets.isEmpty()) {\nSystem.out.println(source + \" is connected to \" + targets);\n}\n}\n}\n\npublic void removeNode(String label) {\nvar node = nodes.get(label);\nif (node == null)\nreturn;\n\n// iterate over every node in this graph\nfor (var n : adjacencyList.keySet()) {\n}\nnodes.remove(node);\n}\n\npublic void removeEdge(String from, String to) {\nvar fromNode = nodes.get(from);\nvar toNode = nodes.get(to);\n\nif (fromNode == null || toNode == null)\nreturn;"
] |
[
null,
"https://ivov.dev/posts/graphs/graph-1.png",
null,
"https://ivov.dev/posts/graphs/graph-2.png",
null,
"https://ivov.dev/posts/graphs/adjacency-matrix.png",
null,
"https://ivov.dev/posts/graphs/adjacency-matrix-complexity.png",
null,
"https://ivov.dev/posts/graphs/adjacency-list.png",
null
] |
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|
https://www.cienciasinseso.com/en/the-error-of-confidence/
|
[
"# The error of confidence\n\nOur life is full of uncertainty. There’re many times that we want to know information that is out of our reach, and then we have to be happy with approximations. The problem is that approximations are subject to error, so we can never be completely sure that our estimates are true. But do we can measure our degree of uncertainty.\n\nThis is one of the things statistics is responsible for, quantifying uncertainty. For example, let’s suppose we want to know the mean cholesterol levels of adults between 18 and 65 years from the city where I live in. If we want the exact number we have to call them all, convince them to be analyzed (most of them are healthy and won’t want to know anything about analysis) and make the determination to every one of them to calculate the average we want to know.\n\nThe problem is that I live in a big city, with about five million people, so it’s impossible from a practical point of view to determine serum cholesterol to all adults from the age range that we are interested in. What can we do?. We can select a more affordable sample, calculate the mean cholesterol of their components and then estimate what is the average value in the entire population.",
null,
"So, I randomly pick out 500 individuals and determine their cholesterol levels, in milligrams per deciliter, getting a mean value of 165, a standard deviation of 25, and an apparently normal distribution, as it’s showed in the graph attached.\n\nLogically, as the sample is large enough, the average value of the population will probably be close to that of 165 obtained from the sample, but it’s also very unlikely to be exactly that. How can we know the value of the population? The answer is that we cannot know the exact value, but we can know what the approximate value is. In other words, we can calculate a range within which the value of my unaffordable population is, always with a certain level of confidence (or uncertainty) that can be settled by us.\n\nLet’s consider for a moment what would happen if we repeat the experiment many times. We would get a slightly different value every time, but all of them should be similar and close to the actual value of the population. If we repeat the experiment a hundred times and get a hundred mean values, these values will follow a normal distribution with a specific mean and standard deviation.\n\nNow, we know that, in a normal distribution, about 95% of the sample is located in the interval enclosed by the mean plus minus two standard deviations. In the case of the distribution of means of our experiments, the standard deviation of the means distribution is called the standard error, but its meaning is the same of any standard deviation: the range between the mean plus minus two standard errors contains 95% of the means of the distribution. This implies, roughly, that the actual mean of our population will be included in that interval 95% of the times, and that we don’t need to repeat the experiment a hundred times, it’s enough to compute the interval as the obtained mean plus minus two standard errors. And how can we get the mean’s standard error? Very simple, using the following expression:\n\nstandard error = standard deviation / square root of sample size\n\nIn our example, the standard error equals 1.12, which means that the mean value of cholesterol in our population is within the range 165 – 2.24 to 165 + 2.24 or, what is the same, 162.76 to 167.24, always with a probability of error of 5% (a level of confidence of 95%).\n\nWe have thus calculated the 95% confidence interval of our mean, which allow us to estimate the values between which the true population mean is. All confidence intervals are calculated similarly, varying in each case how to calculate the standard error, which will be different depending on whether we are dealing with a mean, a proportion, a relative risk, etc.\n\nTo finish this post I have to tell you that the way we have done this calculation is an approximation. When we know the standard deviation in the population we can use a standard normal distribution to calculate the confidence interval. If we don’t know it, which is the usual situation, and the sample is large enough, we can make an approximation using a normal distribution committing little error. But if the sample is small the distribution of our means of repetitive experiments won’t follow a normal distribution, but a Student’s t distribution, so we should use this distribution to calculate the interval. But that’s another story…\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] |
[
null,
"https://www.cienciasinseso.com/wp-content/uploads/2014/06/colesterol_normal_en-300x196.jpeg",
null
] |
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|
https://www.instron.cn/zh-cn/our-company/library/test-types/tensile-test
|
[
"# 拉伸\n\n## What is Tensile Testing?\n\nA tensile test, also known as tension test, is probably the most fundamental type of mechanical test you can perform on material. Tensile tests are simple, relatively inexpensive, and fully standardized. By pulling on something, you will very quickly determine how the material will react to forces being applied in tension. As the material is being pulled, you will find its strength along with how much it will elongate.\n\n## Why Perform a Tensile Test or Tension Test?\n\nYou can learn a lot about a substance from tensile testing. As you continue to pull on the material until it breaks, you will obtain a good, complete tensile profile. A curve will result showing how it reacted to the forces being applied. The point of failure is of much interest and is typically called its \"Ultimate Strength\" or UTS on the chart.",
null,
"### Hooke's Law\n\nFor most tensile testing of materials, you will notice that in the initial portion of the test, the relationship between the applied force, or load, and the elongation the specimen exhibits is linear. In this linear region, the line obeys the relationship defined as \"Hooke's Law\" where the ratio of stress to strain is a constant, or",
null,
". E is the slope of the line in this region where stress (σ) is proportional to strain (ε) and is called the \"Modulus of Elasticity\" or \"Young's Modulus\".",
null,
"### Modulus of Elasticity\n\nThe modulus of elasticity is a measure of the stiffness of the material, but it only applies in the linear region of the curve. If a specimen is loaded within this linear region, the material will return to its exact same condition if the load is removed. At the point that the curve is no longer linear and deviates from the straight-line relationship, Hooke's Law no longer applies and some permanent deformation occurs in the specimen. This point is called the \"elastic, or proportional limit\". From this point on in the tensile test, the material reacts plastically to any further increase in load or stress. It will not return to its original, unstressed condition if the load were removed.\n\n### Yield Strength\n\nA value called \"yield strength\" of a material is defined as the stress applied to the material at which plastic deformation starts to occur while the material is loaded.\n\n### Offset Method\n\nFor some materials (e.g., metals and plastics), the departure from the linear elastic region cannot be easily identified. Therefore, an offset method to determine the yield strength of the material tested is allowed. These methods are discussed in ASTM E8 (metals) and D638 (plastics). An offset is specified as a % of strain (for metals, usually 0.2% from E8 and sometimes for plastics a value of 2% is used). The stress (R) that is determined from the intersection point \"r\" when the line of the linear elastic region (with slope equal to Modulus of Elasticity) is drawn from the offset \"m\" becomes the Yield Strength by the offset method.\n\n### Alternate Moduli\n\nThe tensile curves of some materials do not have a very well-defined linear region. In these cases, ASTM Standard E111 provides for alternative methods for determining the modulus of a material, as well as Young's Modulus. These alternate moduli are the secant modulus and tangent modulus.\n\n### Strain\n\nYou will also be able to find the amount of stretch or elongation the specimen undergoes during tensile testing This can be expressed as an absolute measurement in the change in length or as a relative measurement called \"strain\". Strain itself can be expressed in two different ways, as \"engineering strain\" and \"true strain\". Engineering strain is probably the easiest and the most common expression of strain used. It is the ratio of the change in length to the original length,",
null,
". Whereas, the true strain is similar but based on the instantaneous length of the specimen as the test progresses,",
null,
", where Li is the instantaneous length and L0 the initial length.\n\n### Ultimate Tensile Strength\n\nOne of the properties you can determine about a material is its ultimate tensile strength (UTS). This is the maximum load the specimen sustains during the test. The UTS may or may not equate to the strength at break. This all depends on what type of material you are testing. . .brittle, ductile or a substance that even exhibits both properties. And sometimes a material may be ductile when tested in a lab, but, when placed in service and exposed to extreme cold temperatures, it may transition to brittle behavior."
] |
[
null,
"https://www.instron.cn/-/media/images/instron/landing-page-images/test-types/tension_test_graph1.gif",
null,
"https://www.instron.cn/-/media/images/instron/landing-page-images/test-types/hookes_law.gif",
null,
"https://www.instron.cn/-/media/images/instron/landing-page-images/test-types/tension_test_graph2.gif",
null,
"https://www.instron.cn/-/media/images/instron/landing-page-images/test-types/engineering_strain_formula.gif",
null,
"https://www.instron.cn/-/media/images/instron/landing-page-images/test-types/true-strain-formula.gif",
null
] |
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|
https://scicomp.stackexchange.com/questions/10005/inverting-many-small-matrices-in-parallel
|
[
"# Inverting many small matrices in parallel\n\nI am trying to find a good way to handle the following problem: Let C be an N by 3 array (corresponding to points in $\\mathbb{R}^3$). There is a method I am interested in testing which requires constructing, for i=1....N, a matrix $A_i$ of size $n_i <<N$ and a vector $b_i$ of size $n_i$ using only data from the row i of C. Then, we seek the solution $x_i = A_i^{-1} b_i$.\n\nI currently am running this in MATLAB as a large for loop going from i=1 to i=N, and doing the above operations. For N = 30,000, this takes a while and I was hoping to find some way to speed things up. Since each iteration is independent of the other, I was hoping I could parallelize the for loop. However, I am unfamiliar with parallel programming.\n\nI do not have access to the matlab parallel toolbox, so I cannot simply invoke parfor to parallelize the loop. If it helps, I am familiar with Python+Numpy+Scipy, and also a bit with C/C++ and eager to learn MPI or CUDA or whatever would be useful.\n\nAm I correct in thinking that this for loop could be parallelized for faster performance? Can anyone point me to some resources to get started for a parallel (or, faster) implementation?\n\nThanks!\n\n• What you need is probably OpenMP with MEX files. Nov 10, 2013 at 22:47\n• MATLAB is rather slow with small matrices. A simple MEX file (without parallelisation) - Eigen is good for this since it is header only - yields nearly an order of magnitude speed-up in my experience. Nov 11, 2013 at 11:51"
] |
[
null
] |
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|
https://www.answerswave.com/ExpertAnswers/there-are-seven-photocopiers-and-three-of-these-photocopiers-are-defective-the-buyer-begins-to-test--aw339
|
[
"# (Solved): There Are Seven Photocopiers, And Three Of These Photocopiers Are Defective. The Buyer Begins To Tes...\n\nThere are seven photocopiers, and three of these photocopiers are defective. The buyer begins to test them one at a time to figure out which three are defective.\n\na. What is the probability that the first defective photocopier is found on the fifth test?\n\nb. What is the probability that the second defective photocopier is found on the fifth test?\n\nc. What is the probability that the third defective photocopier is found on the fifth test?\n\nd. What is the probability that no more than five tests are needed to find the three defective photocopiers?\n\ne. Given that exactly one of the defective photocopiers was found within the first three tests, what is the probability that the other two defective photocopiers are found within the next three tests?\n\nf. Given that exactly two of the defective photocopiers were found within the first three tests, what is the probability the last defective photocopier is found within the next two tests?\n\ng. Given that exactly two of the defective photocopiers were found within the tests 1, 3, 5, what is the probability the other defective photocopier was found on tests 6 or 7?\n\nh. Given the last defective photocopier was found within the last two tests, what is the probability that the first two defective photocopiers were found within the first three tests?\n\nWe have an Answer from Expert"
] |
[
null
] |
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|
https://web2.0calc.com/questions/help-please-urgent_4
|
[
"+0\n\n# Help please urgent\n\n+1\n469\n3\n\nFind all real numbers t such that $$\\frac{2}{3} t - 1 < t + 7 \\le -2t + 15$$. Give your answer as an interval.\n\nJul 17, 2019\n\n### Best Answer\n\n#1\n+2\n\nFind all real numbers t such that $$\\frac{2}{3} t - 1 < t + 7 \\le -2t + 15$$. Give your answer as an interval.\n\nHello Guest!\n\n$$\\frac{2}{3} t - 1 < t + 7 \\le -2t + 15\\\\ -8 <\\frac{1}{3}t\\\\ \\color{blue}t<24$$\n\n$$ $$\n\n$$t+7\\le-2t+15\\\\ -8\\le -3t\\\\\\color{blue}t\\le24$$\n\n$$t\\in \\{ \\mathbb{R}|t<24\\}$$",
null,
"!\n\nJul 17, 2019\nedited by asinus Jul 17, 2019\nedited by asinus Jul 17, 2019\n\n### 3+0 Answers\n\n#1\n+2\nBest Answer\n\nFind all real numbers t such that $$\\frac{2}{3} t - 1 < t + 7 \\le -2t + 15$$. Give your answer as an interval.\n\nHello Guest!\n\n$$\\frac{2}{3} t - 1 < t + 7 \\le -2t + 15\\\\ -8 <\\frac{1}{3}t\\\\ \\color{blue}t<24$$\n\n$$ $$\n\n$$t+7\\le-2t+15\\\\ -8\\le -3t\\\\\\color{blue}t\\le24$$\n\n$$t\\in \\{ \\mathbb{R}|t<24\\}$$",
null,
"!\n\nasinus Jul 17, 2019\nedited by asinus Jul 17, 2019\nedited by asinus Jul 17, 2019\n#2\n+5\n\nIn order from oldest to newest:\n\n Answers by CPhill and me: https://web2.0calc.com/questions/help_49470 Second answer by CPhill: https://web2.0calc.com/questions/help_85516 Answer by heureka: https://web2.0calc.com/questions/help_71861 Second answer by me: https://web2.0calc.com/questions/help_31003 Found from first page of search results here.\nJul 18, 2019\n#3\n+2\n\nLOL!!!!!",
null,
"",
null,
"",
null,
"CPhill Jul 18, 2019"
] |
[
null,
"https://web2.0calc.com/img/emoticons/smiley-laughing.gif",
null,
"https://web2.0calc.com/img/emoticons/smiley-laughing.gif",
null,
"https://web2.0calc.com/img/emoticons/smiley-cool.gif",
null,
"https://web2.0calc.com/img/emoticons/smiley-cool.gif",
null,
"https://web2.0calc.com/img/emoticons/smiley-cool.gif",
null
] |
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|
http://www.biotech2012.org/2019/09/10/supplementary-materialssupp-datas1-discrete-particles-of-the-membrane-cytoskeletal-assembly-and/
|
[
"# Supplementary MaterialsSupp DataS1. discrete particles of the membrane, cytoskeletal assembly, and\n\nSupplementary MaterialsSupp DataS1. discrete particles of the membrane, cytoskeletal assembly, and the cytoplasm are described using the Lennard-Jones potential with empirical constants. By exploring the parameter space of this CGMD model, we have successfully simulated the dynamics of varied filopodia formations. Comparative analyses of length and thickness of filopodia show that our numerical simulations are in agreement with experimental measurements of flow-induced activated platelets. Experiments below), resulting in the numeric mapping correspondence between the parameters and the measurements. Accordingly, PA-824 manufacturer we construct an inverse mapping scheme in which the observed geometric measurements can be converted to the space of model parameters. After carefully studying several alternatives, we introduce the linear function fr between r and L and quadratic function f between and T: r =?fr(L) =?L +?r0 L??[0,?Lmax] (2) 98%). The constants and are dependent on the coarse-graining level of filament bundles used to simulate the filopod formation. Thus, the constant for both of the filopodia is determined by fitting a linear function fr as shown in Fig. 7a. Similarly, the constant for both of the filopodia is determined by fitting a quadratic function f as shown in Fig. 7b. In this framework, depending on the model structure and the coarsening level, the constants [0.95 * 10?2,5.26 * 10?2] and [0.8 * 10?5,3.0 * 10?5]. This represents the physiologically possible range of filopodia formation patterns that can be simulated by this model, within the current limits of the membrane stability and structural integrity of the model. In Fig. 8, simulation snapshots of a medium sized filopod are shown. The linear function fr and f for this filopod are plotted in Fig. 7a and Fig. 7b (shown in red). The corresponding and values fall in the range shown above. Open in a separate window Figure 7 Correlation between model parameter space and experimental measurements. Plots for the (a) linear function f_r between the model parameter r and the PA-824 manufacturer noticed geometric dimension L from the filopod. (b) Quadratic function f_ between your model parameter as well as the noticed geometric dimension T^2 from the filopod for both representative instances in Fig. 5 and Fig. 6 (demonstrated in dark). Also the outcomes for the filopod of moderate length (demonstrated in reddish colored) are plotted. The products of r,,L,T are in Angstroms. Open up in another window Shape 8 Full platelet membrane after filopod development using an intermediate filament package. A thorough experimental data source of platelet activation comprising geometrical measurements of filopodia for different shear stress-exposure period combinations will increase this parameter space. With such data, one can use the framework established above that gives independent FGF18 inverse mapping functions fr and f. Given a desired experimental measurement L for a filopod, the linear function fr will generate corresponding model parameter r. The maximum length of the filopod that can be simulated by the model is limited by the coarsening level and elasticity of the model membrane. Similarly, given a desired experimental measurement T, the quadratic function f will generate corresponding model parameter . With such a framework, we can simulate the dynamic growth of filopodia of desired lengths and thicknesses observed in platelets that are exposed to varying levels of shear stress-exposure time combinations. 3.3. Model Verification In Fig. 9, three examples of the dynamic simulation results achieved by the end of the simulation time (corresponding to the experimental exposure time of the platelets to a prescribed level of shear stress), are compared to the geometric features of the measurements of filopodia formation (length and thickness) processed from SEM images of the exposed platelets. These images were obtained from the flow-induced shear stress PA-824 manufacturer experiments conducted using the HSD. While the simulations depict the dynamic formation of the filopodia, the comparisons are of snapshots from the dynamic simulations corresponding to the experimental endpoints. Open in a separate window Figure 9 Visual comparisons of experimental and simulated filopod formation. (a) SEM images at (i) 1 dyne cm^-2 – 4 min (ii) 70 dyne cm^-2 – 4 min (iii) 70 dyne cm^-2 – 1 min. Tracings of the images in panel (a) illustrate the retained ellipsoidal shape of the platelet and filopod formation. The length (L) and thickness (T) are observed to be (i) 0.68 and 0.29 m (ii) 1.39 and 0.35 m and (iii) 0.29 and 0.32 m, respectively. (b) Simulated filopod formation on model platelet (c-i to c-vi) The model parameters for the three simulations. The Y-axis units are Angstroms, X-axis units are number of simulation steps. As shown in Supporting Data S1, when the platelets undergo shear stress (1."
] |
[
null
] |
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|
https://www.dadisp.com/webhelp/mergedProjects/dspguide/CHAP6A/Replacing_Outliers.htm
|
[
"## Replacing Outliers\n\nTo identify and replace the outliers with a specific value, use binary series for a quick and efficient result. (W1 > 10.0) yields a series which is 1.0 wherever W1 is greater than 10.0, and 0.0 elsewhere. Multiply this by the desired replacement value:\n\n(W1 > 10.0) * -1.5\n\nto obtain a series which is -1.5 at each replacement location and 0.0 elsewhere.\n\nThis is \"close\", but is only half of the solution. The remaining part of the problem is to retain those points in W1 that are not outliers (i.e. W1 <= 10.0).\n\n(not(W1 > 10.0) * W1)\n\nreturns a series that is the same as W1 everywhere W1 does not exceed 10.0, and 0.0 elsewhere. Adding the two expressions together produces the desired result:\n\n((W1 > 10.0) * (-1.5)) + (not(W1 > 10.0) * W1))\n\nTo generalize with a macro:\n\n#define replace(s, cond, val) (((cond)*(val)) + ((not(cond))*(s)))\n\nwhere s is the input series, cond is the condition for replacement, and val is the desired replacement value.\n\nWhat happens if NAVALUE is used for the replacement value? Try:\n\n((W1 > 10.0) * (navalue)) + (not(W1 > 10.0) * W1))\n\nPoints greater than 10.0 are \"dropped out\" of the display of the series, but the x-axis is retained because the NAVALUE serves as a placeholder.\n\nFinally, to replace outliers with a linear interpolation of the surrounding points, consider the following:\n\ndelete(W1, W1 > 10.0)\n\nreturns a series of the y-values which are not outliers, i.e. W1 <= 10.0.\n\ndelete(xvals(w1), w1 > 10.0)\n\nreturns a series of the x-values where the y-values of W1 are not outliers.\n\nxy(delete(xvals(w1), w1 > 10.0), delete(w1, w1 > 10.0))\n\ncreates an XY plot where the outliers are removed from the x- and y-values. DADiSP will graphically \"connect the dots\" in the XY plot, so that it appears to have interpolated between the points on either side of the outliers, but, no extra points have been added. To insert the interpolated points, use:\n\nxyinterp(delete(xvals(w1), w1 > 10.0),delete(w1, w1 > 10.0))\n\nFinally, the OUTLIER function incorporates these ideas into one, simple outlier removal function."
] |
[
null
] |
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|
https://www.jobilize.com/physics-k12/section/evaluating-relative-velocity-by-making-reference-object-stationary?qcr=www.quizover.com
|
[
"Relative velocity in one dimension (Page 3/4)\n\n Page 3 / 4\n\n$\\begin{array}{l}⇒{v}_{C}={v}_{B}+{v}_{CB}\\\\ ⇒{v}_{CB}={v}_{C}-{v}_{B}\\end{array}$\n\nThis is an important relation. This is the working relation for relative motion in one dimension. We shall be using this form of equation most of the time, while working with problems in relative motion. This equation can be used effectively to determine relative velocity of two moving objects with uniform velocities (C and B), when their velocities in Earth’s reference are known. Let us work out an exercise, using new notation and see the ease of working.\n\nProblem : Two cars, initially 100 m distant apart, start moving towards each other with speeds 1 m/s and 2 m/s along a straight road. When would they meet ?\n\nSolution : The relative velocity of two cars (say 1 and 2) is :\n\n$\\begin{array}{l}{v}_{21}={v}_{2}-{v}_{1}\\end{array}$\n\nLet us consider that the direction ${v}_{1}$ is the positive reference direction.\n\nHere, ${v}_{1}$ = 1 m/s and ${v}_{2}$ = -2 m/s. Thus, relative velocity of two cars (of 2 w.r.t 1) is :\n\n$\\begin{array}{l}⇒{v}_{21}=-2-1=-3\\phantom{\\rule{2pt}{0ex}}m/s\\end{array}$\n\nThis means that car \"2\" is approaching car \"1\" with a speed of -3 m/s along the straight road. Similarly, car \"1\" is approaching car \"2\" with a speed of 3 m/s along the straight road. Therefore, we can say that two cars are approaching at a speed of 3 m/s. Now, let the two cars meet after time “t” :\n\n$\\begin{array}{l}t=\\frac{\\mathrm{Displacement}}{\\mathrm{Relative velocity}}=\\frac{100}{3}=33.3\\phantom{\\rule{2pt}{0ex}}s\\end{array}$\n\nOrder of subscript\n\nThere is slight possibility of misunderstanding or confusion as a result of the order of subscript in the equation. However, if we observe the subscript in the equation, it is easy to formulate a rule as far as writing subscript in the equation for relative motion is concerned. For any two subscripts say “A” and “B”, the relative velocity of “A” (first subscript) with respect to “B” (second subscript) is equal to velocity of “A” (first subscript) subtracted by the velocity of “B” (second subscript) :\n\n$\\begin{array}{l}{v}_{AB}={v}_{A}-{v}_{B}\\end{array}$\n\nand the relative velocity of B (first subscript) with respect to A (second subscript) is equal to velocity of B (first subscript) subtracted by the velocity of A (second subscript):\n\n$\\begin{array}{l}{v}_{BA}={v}_{B}-{v}_{A}\\end{array}$\n\nEvaluating relative velocity by making reference object stationary\n\nAn inspection of the equation of relative velocity points to an interesting feature of the equation. We need to emphasize that the equation of relative velocity is essentially a vector equation. In one dimensional motion, we have taken the liberty to write them as scalar equation :\n\n$\\begin{array}{l}{v}_{BA}={v}_{B}-{v}_{A}\\end{array}$\n\nNow, the equation comprises of two vector quantities ( ${v}_{B}$ and $-{v}_{A}$ ) on the right hand side of the equation. The vector “ $-{v}_{A}$ ” is actually the negative vector i.e. a vector equal in magnitude, but opposite in direction to “ ${v}_{A}$ ”. Thus, we can evaluate relative velocity as following :\n\n• Apply velocity of the reference object (say object \"A\") to both objects and render the reference object at rest.\n• The resultant velocity of the other object (\"B\") is equal to relative velocity of \"B\" with respect to \"A\".\n\nThis concept of rendering the reference object stationary is explained in the figure below. In order to determine relative velocity of car \"B\" with reference to car \"A\", we apply velocity vector of car \"A\" to both cars. The relative velocity of car \"B\" with respect to car \"A\" is equal to the resultant velocity of car \"B\".\n\nwhat's lamin's theorems and it's mathematics representative\nif the wavelength is double,what is the frequency of the wave\nWhat are the system of units\nA stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained\n58asagravitasnal firce\nAmar\nwater boil at 100 and why\nwhat is upper limit of speed\nwhat temperature is 0 k\nRiya\n0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale\nMustapha\nHow MKS system is the subset of SI system?\nwhich colour has the shortest wavelength in the white light spectrum\nhow do we add\nif x=a-b, a=5.8cm b=3.22 cm find percentage error in x\nx=5.8-3.22 x=2.58\nwhat is the definition of resolution of forces\nwhat is energy?\nAbility of doing work is called energy energy neither be create nor destryoed but change in one form to an other form\nAbdul\nmotion\nMustapha\nhighlights of atomic physics\nBenjamin\ncan anyone tell who founded equations of motion !?\nn=a+b/T² find the linear express\nأوك\nعباس\nQuiklyyy",
null,
"",
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"By Anonymous User",
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"",
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[
null,
"https://www.jobilize.com/quiz/thumb/social-dances-and-dance-mixers-1-1-test-by-marion-cabalfin.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://www.jobilize.com/quiz/thumb/r5-family-quizzes.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://www.jobilize.com/quiz/thumb/lbst-1104-midterm-exam-quiz-by-jonathan-long.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://farm4.staticflickr.com/3780/11322953266_db29ce0659_t.jpg",
null,
"https://www.jobilize.com/quiz/thumb/negotiations-conflict-management-mcq-quiz-by-charles-jumper.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://www.jobilize.com/quiz/thumb/measurement-experimentation-lab-mcq-quiz-by-dr-steve-gibbs.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://www.jobilize.com/quiz/thumb/ch-01-the-cranial-nerves-and-the-circle-of-willis-quiz-by-university.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://farm4.staticflickr.com/3217/3007857588_60b3c3be76_t.jpg",
null,
"https://www.jobilize.com/quiz/thumb/classical-music-quiz-by-marion-cabalfin-5081012.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null,
"https://www.jobilize.com/quiz/thumb/summer-kitchens-quiz-by-heather-mcavoy.png;jsessionid=UDqVH_uRW5geyjLUlI_ZwRrzyIZB9cHHwGrOifdp.condor3363",
null
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|
http://modules.xjtlu.edu.cn/MOD_CAT.aspx?mod_code=MTH315
|
[
"Module Catalogues, Xi'an Jiaotong-Liverpool University\n\nModule Code: MTH315\nModule Title: Geometry of Curves and Surfaces\nModule Level: Level 3\nModule Credits: 5.00\nSemester: SEM1\nOriginating Department: Mathematical Sciences\nPre-requisites: MTH207\n\n Aims • To introduce the ideas and methods of the classical differential geometry of curves and surfaces in three dimensional Euclidean space.• To translate intuitive geometrical ideas into mathematical concepts that allow for quantitative study.• To illustrate geometrical concepts through examples.This module builds on concepts introduced in vector analysis, and has a wide range of applications in science and technology. It provides a foundation for further study of topics such as Riemannian geometry, continuum mechanics and general relativity.\n Learning outcomes A Apply techniques from calculus and linear algebra with confidence to the study of geometrical objectsB Define and interpret the meaning of core concepts in the differential geometry of curves and surfaces in 3-dimensional spaceC Calculate various quantities (e.g. length, curvature, torsion; fundamental forms, area) for given examples and interpret their geometrical significanceD Discriminate between intrinsic and extrinsic geometrical properties\n Method of teaching and learning This module will be delivered through a combination of formal lectures and tutorials.\n Syllabus 1. Curves in the plane: tangent vectors, arc-length, parametrisation, curvature. 2. Curves in 3-space: curvature and torsion, Frenet-Serret coordinate frame.3. Geometry of the 2-sphere: great circles, parallel transport, stereographic projection.4. Surfaces in 3-space: surfaces of rotation, quadrics and other examples, parametrisation, tangent plane, normal vector and orientation.5. Intrinsic metric quantities: first fundamental form, distance, angle, area.6. Curvature: the Gauss map, second fundamental form, principal curvatures and vectors, mean and Gaussian curvatures, Gauss’s Theorema Egregium, geodesics.7. Global properties of surfaces: Euler characteristic and genus, Gauss-Bonnet Theorem.\nDelivery Hours\n Lectures Seminars Tutorials Lab/Prcaticals Fieldwork / Placement Other(Private study) Total Hours/Semester 39 13 98 150\n\n## Assessment\n\n Sequence Method % of Final Mark 1 Coursework 15.00 2 Final Exam 85.00\n Module Catalogue generated from SITS CUT-OFF: 12/10/2019 12:15:04 AM"
] |
[
null
] |
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|
http://info.bakercharters.org/teaching-textbooks-math/
|
[
"# Teaching Textbooks Math",
null,
"## Summary\n\nTeaching Textbooks is a quality math program that provides computer based lessons and workbooks. A student can use one or both to move through the lessons. Instruction is clear and easy to understand and there are plenty of practice problems.\n\n### Pros:\n\n• Student can do both workbook and/or online.\n• All instruction is given on the computer\n• Lessons provide clear instruction and quality practice.\n• Computer will auto grade each lesson, quiz and test\n• When an answer is wrong a pop up will appear that tells the student why\n\n### Cons:\n\n• Not aligned to the SBAC standards used for state testing of grades 3 – 8.\n• Not being standards-aligned makes it hard to change to a new curriculum later without having gaps in knowledge.\n• Numbering system is not intuitive. Teaching Textbooks 3 = 2nd grade curriculum; Teaching Textbooks 4 = 3rd grade curriculum; Teaching Textbooks 5 = 4th grade curriculum; etc.\n\n## Learn More:\n\nCathy Duffy Review\n\nTT Website\n\nTeachingTextbooks Math Placement"
] |
[
null,
"http://quizzical-monitor.flywheelsites.com/wp-content/uploads/2017/11/Pic043.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.93178254,"math_prob":0.5481244,"size":904,"snap":"2021-04-2021-17","text_gpt3_token_len":197,"char_repetition_ratio":0.12,"word_repetition_ratio":0.0,"special_character_ratio":0.21238938,"punctuation_ratio":0.0931677,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95708823,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-20T22:34:25Z\",\"WARC-Record-ID\":\"<urn:uuid:5e4c2096-9a9d-4213-9902-6025eeac2d30>\",\"Content-Length\":\"28137\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eda3fd0b-130d-4eb8-b2cd-b17f64cfd1e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:53fce45a-7788-43ea-979e-e47cdf4e2c27>\",\"WARC-IP-Address\":\"104.131.165.133\",\"WARC-Target-URI\":\"http://info.bakercharters.org/teaching-textbooks-math/\",\"WARC-Payload-Digest\":\"sha1:5GFWEWJEL2M3F6BOXXDQ7NFM36DXZXBO\",\"WARC-Block-Digest\":\"sha1:UQINLW6EWTYTB3U4GJ5SQVWD6UCWUAJJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039491784.79_warc_CC-MAIN-20210420214346-20210421004346-00574.warc.gz\"}"}
|
https://en.sfml-dev.org/tutorials/2.0/system-time.php
|
[
"# Handling time\n\n## Time in SFML\n\nUnlike many other libraries where time is a uint32 number of milliseconds, or a float number of seconds, SFML doesn't impose any specific unit or type for time values. Instead it leaves this choice to the user through a flexible class: `sf::Time`. All SFML classes and functions that manipulate time values use this class.\n\n`sf::Time` represents a time period (in other words, the time that elapses between two events). It is not a date-time class which would represent the current year/month/day/hour/minute/second as a timestamp, it's just a value that represents a certain amount of time, and how to interpret it depends on the context where it is used.\n\n## Converting time\n\nA `sf::Time` value can be constructed from different source units: seconds, milliseconds and microseconds. There is a (non-member) function to turn each of them into a `sf::Time`:\n\n``````sf::Time t1 = sf::microseconds(10000);\nsf::Time t2 = sf::milliseconds(10);\nsf::Time t3 = sf::seconds(0.01f);\n``````\n\nNote that these three times are all equal.\n\nSimilarly, a `sf::Time` can be converted back to either seconds, milliseconds or microseconds:\n\n``````sf::Time time = ...;\n\nsf::Int64 usec = time.asMicroseconds();\nsf::Int32 msec = time.asMilliseconds();\nfloat sec = time.asSeconds();\n``````\n\n## Playing with time values\n\n`sf::Time` is just an amount of time, so it supports arithmetic operations such as addition, subtraction, comparison, etc. Times can also be negative.\n\n``````sf::Time t1 = ...;\nsf::Time t2 = t1 * 2;\nsf::Time t3 = t1 + t2;\nsf::Time t4 = -t3;\n\nbool b1 = (t1 == t2);\nbool b2 = (t3 > t4);\n``````\n\n## Measuring time\n\nNow that we've seen how to manipulate time values with SFML, let's see how to do something that almost every program needs: measuring the time elapsed.\n\nSFML has a very simple class for measuring time: `sf::Clock`. It only has two functions: `getElapsedTime`, to retrieve the time elapsed since the clock started, and `restart`, to restart the clock.\n\n``````sf::Clock clock; // starts the clock\n...\nsf::Time elapsed1 = clock.getElapsedTime();\nstd::cout << elapsed1.asSeconds() << std::endl;\nclock.restart();\n...\nsf::Time elapsed2 = clock.getElapsedTime();\nstd::cout << elapsed2.asSeconds() << std::endl;\n``````\n\nNote that `restart` also returns the elapsed time, so that you can avoid the slight gap that would exist if you had to call `getElapsedTime` explicitly before `restart`.\nHere is an example that uses the time elapsed at each iteration of the game loop to update the game logic:\n\n``````sf::Clock clock;\nwhile (window.isOpen())\n{\nsf::Time elapsed = clock.restart();\nupdateGame(elapsed);\n...\n}\n``````"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8486224,"math_prob":0.9810835,"size":2625,"snap":"2023-14-2023-23","text_gpt3_token_len":646,"char_repetition_ratio":0.15986265,"word_repetition_ratio":0.0,"special_character_ratio":0.25980952,"punctuation_ratio":0.25301206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9917804,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T11:56:47Z\",\"WARC-Record-ID\":\"<urn:uuid:177d4626-eb04-402a-a95f-142e838213cd>\",\"Content-Length\":\"8800\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ccf52520-16fe-4fb3-b1d4-0e3679794a42>\",\"WARC-Concurrent-To\":\"<urn:uuid:18204ee8-9737-4fb4-8a9b-a5287f4003a6>\",\"WARC-IP-Address\":\"78.47.82.133\",\"WARC-Target-URI\":\"https://en.sfml-dev.org/tutorials/2.0/system-time.php\",\"WARC-Payload-Digest\":\"sha1:D24EKIRXQIOWFHMSA4ST6KJSAWP5KA2U\",\"WARC-Block-Digest\":\"sha1:FOPWZMV47IV66FOQ3CWO3HUQNS4NF2T7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647810.28_warc_CC-MAIN-20230601110845-20230601140845-00342.warc.gz\"}"}
|
https://wiki.scar-divi.com/TPAGroup
|
[
"# GroupTPA\n\n(Redirected from TPAGroup)\n\n## Definition\n\n`function GroupTPA(const TPA: TPointArray; const Dist: Integer): T2DPointArray;`\n\n## Availability\n\nSCAR Divi 3.28 > Current\n\n• Contained a bug in 3.26-3.27 that caused the initial element in every subarray of the result to be duplicated inside of that subarray.\n• In 3.26-3.27 the algorithm was implemented incorrectly and split every point of TPA up into a subarray and then included every other point within a distance of the first point in that subarray.\n\n### Aliases\n\n• TPAGroup (SCAR Divi 3.26 > 3.34)\n• TPAToATPA (SCAR Divi 3.26 > 3.34)\n\n## Description\n\nThis function splits a given TPointArray TPA into separate TPointArrays, by grouping points that are within the given distance Dist of the first point of each subarray. If a point isn't within the given distance of any first point of the subarrays, a new subarray is created containing that point as first one. The function returns a T2DPointArray containing all the resulting TPointArrays. An extended function with additional functionality is available as GroupTPAEx.\n\n## Example\n\n```var\nTPA: TPointArray;\nATPA: T2DPointArray;\ni: Integer;\n\nbegin\nClearDebug;\nTPA := [Point(2, 5), Point(6, 9), Point(0, 0), Point(5, 5), Point(1, 1), Point(5, 7), Point(-2, 0)];\nATPA := GroupTPA(TPA, 2);\nfor i := 0 to High(ATPA) do\nWriteLn(TPAToStr(ATPA[i]));\nend.```\n\nOutput:\n\n```(2,5)\n(6,9)\n(0,0);(1,1);(-2,0)\n(5,5);(5,7)\n```"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.7672409,"math_prob":0.9561396,"size":1399,"snap":"2022-27-2022-33","text_gpt3_token_len":430,"char_repetition_ratio":0.14910394,"word_repetition_ratio":0.00952381,"special_character_ratio":0.28377414,"punctuation_ratio":0.20401338,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746615,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T12:35:31Z\",\"WARC-Record-ID\":\"<urn:uuid:dd8f272b-933d-4a79-8eff-1f7ce991083a>\",\"Content-Length\":\"17385\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc91c7f2-3682-41c5-ac83-f6bcf94bbef2>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1db7355-12e2-43c1-8361-53d15473cc4a>\",\"WARC-IP-Address\":\"178.18.82.165\",\"WARC-Target-URI\":\"https://wiki.scar-divi.com/TPAGroup\",\"WARC-Payload-Digest\":\"sha1:T47VPML4NAOG44T5H7QBSLPMEODD2V52\",\"WARC-Block-Digest\":\"sha1:HBSFBPD2GLSW3BK6PF6ASA6HXLWWRIDX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104576719.83_warc_CC-MAIN-20220705113756-20220705143756-00384.warc.gz\"}"}
|
https://nl.mathworks.com/matlabcentral/cody/problems/23-finding-perfect-squares/solutions/1416225
|
[
"Cody\n\n# Problem 23. Finding Perfect Squares\n\nSolution 1416225\n\nSubmitted on 14 Jan 2018 by Jafar Mortadha\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 = [2 3 4]; assert(isequal(isItSquared(a),true))\n\nsorted = 2 3 4\n\n2 Pass\na = [20:30]; assert(isequal(isItSquared(a),false))\n\nsorted = 20 21 22 23 24 25 26 27 28 29 30\n\n3 Pass\na = ; assert(isequal(isItSquared(a),true))\n\nsorted = 1\n\n4 Pass\na = [6 10 12 14 36 101]; assert(isequal(isItSquared(a),true))\n\nsorted = 6 10 12 14 36 101\n\n5 Pass\na = [6 10 12 14 101]; assert(isequal(isItSquared(a),false))\n\nsorted = 6 10 12 14 101"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6272006,"math_prob":0.9934481,"size":746,"snap":"2019-51-2020-05","text_gpt3_token_len":264,"char_repetition_ratio":0.14150943,"word_repetition_ratio":0.07751938,"special_character_ratio":0.42627347,"punctuation_ratio":0.09150327,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9747733,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T13:21:44Z\",\"WARC-Record-ID\":\"<urn:uuid:b673f54e-0565-4094-b925-ddb947e674c6>\",\"Content-Length\":\"75959\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f62bea7-9e76-44e6-b094-357bcef86fee>\",\"WARC-Concurrent-To\":\"<urn:uuid:e93661a5-a581-4088-9b22-e5fcfbefdd61>\",\"WARC-IP-Address\":\"104.117.0.182\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/23-finding-perfect-squares/solutions/1416225\",\"WARC-Payload-Digest\":\"sha1:5GLKEQ5UC3WGBZL2TSO7VGJWITZITGHT\",\"WARC-Block-Digest\":\"sha1:W2PQO3HUQ2QURPKG5727AMBDMVFA4XYH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540488620.24_warc_CC-MAIN-20191206122529-20191206150529-00277.warc.gz\"}"}
|
https://astronomy.stackexchange.com/questions/40363/how-to-find-the-distance-between-two-galaxies-from-ra-and-dec-and-redshift
|
[
"# How to find the distance between two galaxies from ra and dec and redshift [duplicate]\n\nI have ra and dec and corresponding redshift of galaxies. I would like to calculate the distance between two galaxies.\n\nI did like this to find the nearest neighbour galaxies.\n\nSuppose I am trying to find the nearest neighbour of galaxy A(which has RA,DEC) and the target galaxy has ra1,dec1,\n\nso to find the distance between two galaxy, I did like this\n\ndist_sqd= sqrt( (RA-ra1)**2 + (DEC-dec1)**2 )\n#RA, DEC are in degree\n\n\nIs this the correct way to find the distance.\n\nAs there is projection in 2D, Since, I already have redshift, Is there any best way to find the distance between two galaxies properly?\n\n• That won't work, consider two galaxies near the north celestial pole, one with an ra of 0 and one with an ra of 180... Dec 13, 2020 at 17:05\n• You mean the angular separation? What has it got to do with the redshifts? Dec 13, 2020 at 17:26\n• Dec 13, 2020 at 17:28\n• Dec 13, 2020 at 17:39\n\nSince you are presumably using Hubble's law to get the distance to each of the galaxies, the approximate estimate would the \"the greater distance ± the smaller distance\".\n\nBut the geometric solution is useful too. Your strategy appears to be something like:\n\n1. Calculate angular separation $$\\Theta$$ from RA and DEC.\n2. By trigonometry, the distance is then $$d_{a,b} = \\sqrt{d_a^2\\sin(\\Theta)^2 + (d_b - d_a\\cos(\\Theta))^2}$$\n\nThe problem is that you can not calculate the angular separation on a sphere by the Pythagorean formula, instead, you need the haversine formula. To keep this post self-contained, that is:\n\n$$\\Theta = 2\\sin^{-1}\\left(\\sqrt{\\sin^2\\left(\\frac{\\phi_2 - \\phi_1}{2}\\right) + \\cos(\\phi_1)\\cos(\\phi_2)\\sin^2\\left(\\frac{\\lambda_2 - \\lambda_1}{2}\\right)}\\right)$$\n\nWhere $$\\phi$$ is declination and $$\\lambda$$ is right ascension.\n\nPeople are also linking this answer from the comments, which deals with the problem by converting everything to Cartesian coordinates first, which may or may not be easier for your use case.\n\n• what is this d_a and d_b? Apr 19, 2021 at 10:26\n• @astronerd Distance to galaxy A and galaxy B Apr 19, 2021 at 10:56\n• Thank you. Is it comoving distance or proper distance? Apr 20, 2021 at 5:17"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9612176,"math_prob":0.9951553,"size":601,"snap":"2023-40-2023-50","text_gpt3_token_len":148,"char_repetition_ratio":0.16750419,"word_repetition_ratio":0.05882353,"special_character_ratio":0.23793676,"punctuation_ratio":0.10483871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99886537,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T07:58:04Z\",\"WARC-Record-ID\":\"<urn:uuid:aebe4ab7-349d-4dd3-8b2f-9b5d6703f164>\",\"Content-Length\":\"148033\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f90e769-029c-42bc-bf21-5410a9df430f>\",\"WARC-Concurrent-To\":\"<urn:uuid:b17190fd-8adb-4c1b-8937-0271f8b4f892>\",\"WARC-IP-Address\":\"104.18.43.226\",\"WARC-Target-URI\":\"https://astronomy.stackexchange.com/questions/40363/how-to-find-the-distance-between-two-galaxies-from-ra-and-dec-and-redshift\",\"WARC-Payload-Digest\":\"sha1:BQPURIWRWVW5JELBTIJL4HSKFYBYMA6N\",\"WARC-Block-Digest\":\"sha1:KV2HDI5UAUDIELDOZE2Y3ENPRJ7KOLHD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100525.55_warc_CC-MAIN-20231204052342-20231204082342-00431.warc.gz\"}"}
|
https://de.mathworks.com/matlabcentral/cody/problems/42460-the-cake-is-a-lie/solutions/1449767
|
[
"Cody\n\n# Problem 42460. The cake is a lie...\n\nSolution 1449767\n\nSubmitted on 26 Feb 2018 by cokakola\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\nx = 1;y_correct = 0; assert(isequal(birthday_cake(x),y_correct))\n\n2 Pass\nx = 2;y_correct = 1; assert(isequal(birthday_cake(x),y_correct))\n\n3 Pass\nx = 4;y_correct = 2; assert(isequal(birthday_cake(x),y_correct))\n\n4 Pass\nx = 7;y_correct = 3; assert(isequal(birthday_cake(x),y_correct))\n\n5 Pass\nx = 12;y_correct = 4; assert(isequal(birthday_cake(x),y_correct))\n\n6 Pass\nx = 27;y_correct = 6; assert(isequal(birthday_cake(x),y_correct))\n\n7 Pass\nx = 127;y_correct = 9; assert(isequal(birthday_cake(x),y_correct))\n\n8 Pass\nx = 2015;y_correct = 23; assert(isequal(birthday_cake(x),y_correct))\n\n9 Pass\nx = 4060225;y_correct = 290; assert(isequal(birthday_cake(x),y_correct))\n\n10 Pass\nx = 1234567890;y_correct = 1950; assert(isequal(birthday_cake(x),y_correct))\n\n11 Pass\nx = 1362067890;y_correct = 2015; assert(isequal(birthday_cake(x),y_correct))\n\n12 Pass\ny=arrayfun(@(x) birthday_cake(x),1:1000); assert(isequal(sum(y),13965)) [x1,y1]=hist(y,unique(y)); [m1,m2]=max(x1); assert(isequal(m1,154)) assert(isequal(x1(isprime(x1)),[2 7 11 29 37 67 79 137]))"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.62894493,"math_prob":0.9997726,"size":1562,"snap":"2019-51-2020-05","text_gpt3_token_len":518,"char_repetition_ratio":0.26379976,"word_repetition_ratio":0.0,"special_character_ratio":0.37195903,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995334,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-07T17:07:48Z\",\"WARC-Record-ID\":\"<urn:uuid:1a04552c-1446-45c0-a3cf-4b411c585d06>\",\"Content-Length\":\"80326\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60221aa4-1d4d-4f07-a47e-e5cf253607bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:810f22d9-1783-4ec2-a51d-b1c5fa52696b>\",\"WARC-IP-Address\":\"104.117.0.182\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/cody/problems/42460-the-cake-is-a-lie/solutions/1449767\",\"WARC-Payload-Digest\":\"sha1:43FPLOFIIAHXFNC56FTAALCZ3VLLY2PZ\",\"WARC-Block-Digest\":\"sha1:KMA3EM5G77QSIHLKBTJFDSXR2OMCRBMC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540500637.40_warc_CC-MAIN-20191207160050-20191207184050-00110.warc.gz\"}"}
|
https://www.includehelp.com/python/how-to-convert-floats-to-ints-in-pandas.aspx
|
[
"# How to convert floats to ints in Pandas?\n\nLearn how to convert floats to ints in Pandas?\nSubmitted by Pranit Sharma, on May 14, 2022\n\nPandas is a special tool which allows us to perform complex manipulations of data effectively and efficiently. Inside pandas, we mostly deal with a dataset in the form of DataFrame. DataFrames are 2-dimensional data structure in pandas. DataFrames consists of rows, columns and the data. The data inside the rows and columns can be of any type like int, float, string etc.\n\nIn this article, we are going to learn how to convert floats to ints in pandas DataFrame? For this purpose, we are going to use pandas.DataFrame.astype() method.\n\n## pandas.DataFrame.astype()\n\nThis method is used for type casting. It cast an object to a specified dtype, it is very easy to understand when it comes to cast the data type of a particular column.\n\nSyntax:\n\n```DataFrame.astype(\ndtype,\ncopy=True,\nerrors='raise'\n)\n\n# or\nDataFrame.astype(dtype='')\n```\n\nTo work with pandas, we need to import pandas package first, below is the syntax:\n\n```import pandas as pd\n```\n\nLet us understand with the help of an example:\n\n```# Importing pandas package\nimport pandas as pd\n\n# Create a dictionary for the DataFrame\ndict = {\n'Name': ['Sudhir', 'Pranit', 'Ritesh','Sanskriti', 'Rani','Megha','Suman','Ranveer'],\n'Age': [16, 27, 27, 29, 29,22,19,20],\n'Height (cm)': [140.23, 143.6, 157.4, 148.78, 133.23,149.22,139.32,148.65]\n}\n\n# Converting Dictionary to Pandas Dataframe\ndf = pd.DataFrame(dict)\n\n# Display original DataFrame\nprint(\"Original DataFrame:\\n\",df,\"\\n\")\n\n# Converting float to int\ndf['Height (cm)'] = df['Height (cm)'].astype(int)\n\n# Display modified DataFrame\nprint(\"Modified DataFrame\\n\",df)\n```\n\nOutput:",
null,
"Preparation"
] |
[
null,
"https://www.includehelp.com/python/images/convert-floats-to-ints-in-pandas.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.71151584,"math_prob":0.8671967,"size":1766,"snap":"2022-05-2022-21","text_gpt3_token_len":461,"char_repetition_ratio":0.13564132,"word_repetition_ratio":0.03816794,"special_character_ratio":0.2774632,"punctuation_ratio":0.200542,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9902266,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T23:48:50Z\",\"WARC-Record-ID\":\"<urn:uuid:b7bb7868-fe9f-4b4b-ba1b-9ee57a286772>\",\"Content-Length\":\"167167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e3561a0-63db-4cc2-9b41-c55aeeb66153>\",\"WARC-Concurrent-To\":\"<urn:uuid:21a0c971-bf84-48d1-b547-d38ed67f7f40>\",\"WARC-IP-Address\":\"104.21.23.12\",\"WARC-Target-URI\":\"https://www.includehelp.com/python/how-to-convert-floats-to-ints-in-pandas.aspx\",\"WARC-Payload-Digest\":\"sha1:JGVFJ5VZU3MDPQBW5Q2LM2YAZWP45QET\",\"WARC-Block-Digest\":\"sha1:KL7LCGOSL7YHL5KZH5OLKBEMMX6Y7IVM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662534693.28_warc_CC-MAIN-20220520223029-20220521013029-00190.warc.gz\"}"}
|
http://bartleylawoffice.com/faq/what-is-pascals-law.html
|
[
"# What is pascal’s law\n\n## What is Pascal’s law in simple terms?\n\nPascal’s principle, also called Pascal’s law, in fluid (gas or liquid) mechanics, statement that, in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container.\n\n## What is Pascal’s law for kids?\n\nPascal’s law (also known as Pascal’s principle) is the statement that in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container. The law was first stated by French scientist and religious philosopher Blaise Pascal.\n\n## What is Pascal law class 9?\n\nPascal’s law states that the pressure exerted anywhere in a confined incompressible liquid is transmitted equally in all directions irrespective of the area on which it acts and it always acts at right angles to the surface of containing vessel.\n\n## What is Pascal law and its application?\n\nApplications of Pascal’s Law\n\nThis is the principle of working of hydraulic lift. It works based on the principle of equal pressure transmission throughout a fluid (Pascal’s Law). … Few more applications include a hydraulic jack and hydraulic press and forced amplification is used in the braking system of most cars.22 мая 2020 г.\n\n## What is Pascal’s formula?\n\nPascal’s Identity is a useful theorem of combinatorics dealing with combinations (also known as binomial coefficients). It can often be used to simplify complicated expressions involving binomial coefficients. Pascal’s Identity is also known as Pascal’s Rule, Pascal’s Formula, and occasionally Pascal’s Theorem.\n\nYou might be interested: How Much Tax Is Taken Out Of Stocks? (Perfect answer)\n\n## How do you prove Pascal’s law?\n\nThe pressure of liquid exerts the force normal to the surface. Let us assume pressure P1exerts the force F on the surface ABFE, pressure P2 exerts the force F2 on the surface ABDC and pressure P3 exerts force on the surface CDFE. Hence the Pascal’s law is proved.\n\n## What does Pascal’s mean?\n\nThe pascal (symbol: Pa) is the SI derived unit of pressure used to quantify internal pressure, stress, Young’s modulus and ultimate tensile strength. The unit, named after Blaise Pascal, is defined as one newton per square metre.\n\n## Where do you observe Pascal’s principle in daily life give a few examples?\n\nPascal’s principle can be observed in our daily life as in vehicle braking system and new electronic parking system in which cars are directly moved to next floor.\n\n## How a hydraulic lift can raise an object as heavy as a car?\n\nIf a car sits on top of the large piston, it can be lifted by applying a relatively small force to the smaller piston, the ratio of the forces being equal to the ratio of the areas of the pistons. … The work required to lift the heavy object equals the work done by the small force.\n\n## Why is Pascal’s law important?\n\nPascal’s law states that a change in pressure at any point in an enclosed fluid is transmitted equally throughout the fluid. The ability of fluids to transmit pressure in this way can be very useful—from getting toothpaste out of a tube to applying the brakes on a car.\n\nYou might be interested: When Is California Sales Tax Due? (Solution found)\n\n## Where is Pascal’s law used?\n\nA typical application of Pascal’s principle for gases and liquids is the automobile lift seen in many service stations (the hydraulic jack). Increased air pressure produced by an air compressor is transmitted through the air to the surface of oil in an underground reservoir.\n\n## What uses Bernoulli’s principle?\n\nAn example of Bernoulli’s principle is the wing of an airplane; the shape of the wing causes air to travel for a longer period on top of the wing, causing air to travel faster, reducing the air pressure and creating lift, as compared to the distance traveled, the air speed and the air pressure experienced beneath the …"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.93128294,"math_prob":0.83808535,"size":3727,"snap":"2023-40-2023-50","text_gpt3_token_len":777,"char_repetition_ratio":0.16062315,"word_repetition_ratio":0.08832808,"special_character_ratio":0.2004293,"punctuation_ratio":0.07777778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9725293,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T20:37:12Z\",\"WARC-Record-ID\":\"<urn:uuid:01414d25-1396-4613-a48e-91442433c31e>\",\"Content-Length\":\"67622\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:40b39416-c5c3-4b17-8c25-6eb6dc35642b>\",\"WARC-Concurrent-To\":\"<urn:uuid:35909331-ff23-4c40-a386-861d0e777e14>\",\"WARC-IP-Address\":\"104.21.73.37\",\"WARC-Target-URI\":\"http://bartleylawoffice.com/faq/what-is-pascals-law.html\",\"WARC-Payload-Digest\":\"sha1:GKQRHZOYCILYARO5U7D6YMYTAB5B3XGY\",\"WARC-Block-Digest\":\"sha1:W6LV3OESREPMGO6E5YLY2IKGQH3C3QUR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510454.60_warc_CC-MAIN-20230928194838-20230928224838-00136.warc.gz\"}"}
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https://www.convertunits.com/from/kilogram/meter+hour/to/kilogram/meter-second
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[
"## ››Convert kilogram/meter-hour to kilogram/meter-second\n\n kilogram/meter hour kilogram/meter-second\n\nHow many kilogram/meter hour in 1 kilogram/meter-second? The answer is 3600.\nWe assume you are converting between kilogram/meter-hour and kilogram/meter-second.\nYou can view more details on each measurement unit:\nkilogram/meter hour or kilogram/meter-second\nThe SI derived unit for dynamic viscosity is the pascal second.\n1 pascal second is equal to 3600 kilogram/meter hour, or 1 kilogram/meter-second.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between kilograms/meter hour and kilograms/meter second.\nType in your own numbers in the form to convert the units!\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from kilogram/meter-second to kilogram/meter hour, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\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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https://www.arxiv-vanity.com/papers/1603.09257/
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[
"# Characterization of hyperfine interaction between an NV electron spin and a first shell 13C nuclear spin in diamond\n\nK. Rama Koteswara Rao and Dieter Suter Fakultät Physik, Technische Universität Dortmund\nMarch 24, 2022\n###### Abstract\n\nThe Nitrogen-Vacancy (NV) center in diamond has attractive properties for a number of quantum technologies that rely on the spin angular momentum of the electron and the nuclei adjacent to the center. The nucleus with the strongest interaction is the C nuclear spin of the first shell. Using this degree of freedom effectively hinges on precise data on the hyperfine interaction between the electronic and the nuclear spin. Here, we present detailed experimental data on this interaction, together with an analysis that yields all parameters of the hyperfine tensor, as well as its orientation with respect to the atomic structure of the center.\n\nNV center, hyperfine interaction\n###### pacs:\n03.67.Lx, 76.70.Hb, 33.35.+r, 61.72.J-\n\nNitrogen-Vacancy (NV) centers in diamond have interesting properties for applications in room-temperature metrology, spectroscopy, and Quantum Information Processing (QIP) Dutt et al. (2007); Balasubramanian et al. (2008); Neumann et al. (2008); Dolde et al. (2011); Staudacher et al. (2013); Waldherr et al. (2014). Nuclear spins, coupled by hyperfine interaction to the electron spin of the NV-center, are important for many of these applications Childress et al. (2006); Dutt et al. (2007); Neumann et al. (2008, 2010); Smeltzer et al. (2011); Waldherr et al. (2014); Taminiau et al. (2014); Álvarez et al. (2015); Fuchs et al. (2011); Maurer et al. (2012). For example, in QIP applications, the nuclear spins can hold quantum information Fuchs et al. (2011); Maurer et al. (2012), serving as part of a quantum register Dutt et al. (2007). Accurate knowledge of the hyperfine interaction is necessary, e.g., for designing precise and fast control sequences for the nuclear spins Smeltzer et al. (2009); Chen et al. (2015). With a known Hamiltonian, control sequences can be tailored by optimal control techniques to dramatically improve the speed and precision of multi-qubit gates Khaneja et al. (2005); Khaneja (2007); Chou et al. (2015); Rong et al. (2015).\n\nThe C nuclear spin of the first coordination shell is a good choice for a qubit due to its large hyperfine coupling to the electronic spin of the NV center, which can be used to implement fast gate operations Jelezko et al. (2004); Neumann et al. (2008); Mizuochi et al. (2009) or high-speed quantum memories Shim et al. (2013a). Full exploitation of this potential requires accurate knowledge of the hyperfine interaction, including the anisotropic (tensor) components. The interaction tensor has been calculated by DFT Łuszczek et al. (2004); Gali et al. (2008); Nizovtsev et al. (2010); Szász et al. (2013), but only a limited number of experimental studies of this hyperfine interaction exist to date Loubser and van Wyk (1978); Felton et al. (2009); Shim et al. (2013b). In this work, we present a detailed analysis of this hyperfine interaction. The experiments were carried out on single NV centers of a diamond crystal with a natural abundance of C and a nitrogen concentration of ppb using a home-built confocal microscope and microwave electronics for excitation Shim et al. (2013a).",
null,
"Figure 1: Symmetry of the system along with the energy level diagram and spectra. (a) Structure of the NV center in diamond with a first shell 13C. The mirror plane and the x, y, and z- axes of the NV frame are also shown. (b) Energy level diagram considering only the electron and 13C nuclear spins. The red arrows and green wave represent electron- and nuclear spin transitions, respectively. (c) Experimental ESR spectrum for a particular orientation of the magnetic field vector. The spectrum consists of 24 resonance lines due to the interaction with the 14N nuclear spin.\n\nThe presence of a nuclear spin in the first coordination shell reduces the symmetry of the NV center from to , a single mirror plane. This symmetry plane passes through the NV symmetry axis and the C nuclear spin as illustrated in Fig. 1(a). The NV symmetry axis defines the -axis of the NV frame of reference, the -axis lies in the symmetry plane of the center, and the -axis is perpendicular to both of them. Due to the symmetry of the system, only those elements of the hyperfine tensor that are invariant with respect to the inversion of the -coordinate can be non-zero. Hence, the hyperfine Hamiltonian in the NV frame of reference can be written as\n\n Hhf=AzzSzIz+AxxSxIx+AyySyIy+Axz(SxIz+SzIx). (1)\n\nHere, and represent components of the electron and nuclear spin angular momenta, respectively, and the components of the hyperfine tensor. The full Hamiltonian of the system consisting of an NV electron spin and a C nuclear spin can be written as\n\n H=DS2z+γeB⋅S+γnB⋅I+Hhf. (2)\n\nHere, represents the zero-field splitting of the electron spin, and are the gyromagnetic ratios of the electron and nuclear spins respectively,\n\nWe start the determination of the hyperfine tensor by measuring ESR spectra for different orientations of the magnetic field vector and then fitting the resonance frequencies to obtain all relevant Hamiltonian parameters, including the orientation of the NV center with respect to the laboratory frame. As a first step we determined the orientation of the NV-axis of the center, which corresponds to the -axis of the Hamiltonian of Eq. (2). For this purpose, we rotated the magnetic field with a fixed magnitude around two orthogonal rotation axes crossing at the NV center. ESR spectra were measured as Fourier transforms of Ramsey fringes for nine non-coplanar orientations of the magnetic field. The transition frequencies of these spectra were fitted numerically to determine the orientation of the -axis of the center. This orientation was reconfirmed by also measuring spectra of a neighboring NV center with no C nucleus in the first coordination shell. The two data sets yielded the same direction for the -axis. If the field is aligned with the -axis, the splitting between the transitions =0 =1 reaches a maximum. The numerical fit also provided estimates of the Hamiltonian parameters MHz, MHz, MHz, which give the dominant contribution to the transition frequencies when .\n\nThe and components of the hyperfine tensor contribute only in second order to the energies and spectral positions. Accordingly, their uncertainty is quite large. Their influence on the transition frequencies is maximised if the magnetic field is close to the -plane. The highest precision is obtained by measuring the transitions with the smallest linewidth. In the NV system, this is the nuclear spin transition between the states. As a nuclear spin transition, the width of this resonance line is about one order of magnitude smaller ( kHz) than that of the ESR transitions. This transition can be excited not only by resonant radio-frequency pulses, but also by the Raman excitation scheme shown in Fig. 1(b): A microwave pulse that drives the transitions from both states to one of the states creates nuclear spin coherence and can also probe this nuclear spin coherence by converting it back into population of the state. The optimal duration of the microwave excitation pulse for the nuclear spin coherence is in general twice as long as that of the optimal pulse for the ESR transitions. Apart from the narrower linewidth, the nuclear spin transition can be identified in the ESR spectrum as a zero-quantum transition: Its position does not change with the carrier frequency used for the excitation and readout pulses. Using second-order perturbation theory, its frequency can be written as Shim et al. (2013b)\n\n Δ≈2|γeBsinθ|D(√A2xx+A2xzcos2ϕ+|Ayy|sin2ϕ). (3)\n\nTherefore, by measuring these zero-quantum frequencies for different orientations of the magnetic field for a fixed , it is possible to obtain estimates of the quantities and . Fig. 2 shows a graphical representation of the experimental data, measured for magnetic fields oriented at an angle from the -axis, together with the numerical fit.",
null,
"Figure 2: Zero-quantum transition frequencies for an azimuthal (ϕ) rotation of the magnetic field for θ=84.5∘. The experimental data (blue stars) were fitted to the equation κ1cos2ϕ+κ2sin2ϕ. The fit parameters are κ1 = 8.5 MHz and κ2 = 5.88 MHz.\n\nFrom the fitted curve, we find that the parameters and must have the values 193.0 and 133.5 MHz. The quantity is larger (smaller) than if the maximum (minimum) of the zero-quantum frequencies corresponds to the -axis. Combining these data with those for , , and , we obtain several possible parameter sets, with different signs, two different orientations for the - and -axes, and different ratios .\n\nTo eliminate the remaining ambiguities and to determine the orientation of the and -axes, it is important to use not only the transition frequencies, but also the transition amplitudes (dipole moments). For this purpose, we measured the transition amplitudes of the spectral lines for different orientations of the magnetic field. The experimental amplitudes depend on the orientation of the microwave magnetic field. Since absolute amplitudes are hard to measure, we determined ratios of transitions amplitudes for two different data sets: First, we measured ratios of Rabi frequencies for pairs of transitions whose frequencies differ by the nuclear spin transition frequency discussed above. These transitions connect the two states with the same state. In the second set of data, we compared the transition amplitudes of spectra in the -plane. The first set of data indicated that the -axis corresponds to the maximum of the zero-quantum frequency (see Fig. 2). This implies that MHz and MHz. These data also indicate that if and have the same sign, the ratio must be .",
null,
"Figure 3: Transition probabilities of low frequency ESR lines (ν<2870 MHz) when B is oriented in the xy-plane. (a) and (b) correspond to different sign combinations of the hyperfine parameters. In (a), det(A) > 0 and in (b), det(A) < 0.\n\nFig. 3 shows the numerically calculated transition probabilities of the low-frequency spectral lines ( MHz) as a function of the azimuthal angle of the magnetic field in the -plane. Here, we observe two qualitatively different cases, depending on the relative signs of the hyperfine parameters. Labeling the transition probabilities in ascending frequency order as , we observe strong variations when , with and in phase and , shifted by . However, for , the amplitudes are almost constant, with and much smaller than and . As shown in Fig. 4, the experimental data show large variations of the transition probabilities, which is well compatible with the case and excludes the case . This allows us to disregard the parameter sets with in the following. Also, the difference in the position of the peak of the quantity in Figs. 3 and 4 allows us to determine the direction of the transverse component of the microwave field with respect to the -axis of the NV frame.",
null,
"Figure 4: Ratios of amplitudes of the low-frequency spectral lines (ν<2870 MHz) in the xy-plane as a function of ϕ. The experimental data (red stars and blue triangles) were fitted to a lorentzian ab/((ϕ−ϕ1)2+b2) (red solid and blue dashed lines). For the red solid line, the fit parameters are a=178.5, b=16.3, and ϕ1=-73.3∘ and for the blue dashed line a=270.5, b=13.8, and ϕ1=9.9∘.\n\nFrom a combined fit of the available experimental data, we thus obtained values for all components of the hyperfine tensor. Table 1 lists the different parameter sets, in the NV frame of reference, that are compatible with the experimental data. The different solutions all have the same principal components, except for the signs. The signs of can be chosen positive or negative; the sign change corresponds to a -rotation around the -axis.\n\nIt is useful to consider also the principal axis representation of the hyperfine tensor. We write , , and for the principal components of the hyperfine tensor in increasing magnitude. Table 2 lists the possible values for the principal components. All four parameter sets result in identical transition frequencies and amplitudes and are therefore experimentally indistinguishable. The -axes of the principal axis system (PAS) and the NV frame of reference coincide and the angle between the NV symmetry axis and the -axis of the PAS is =109.3. As a second-rank tensor, the hyperfine tensor is invariant under -rotations around the principal axes. Accordingly, orientations with the angles and are equivalent solutions. All of these solutions are compatible with the experimental data. Since DFT calculations Gali et al. (2008); Simanovskaia et al. (2013) indicate that the largest component of the hyperfine tensor, which we write as , points in the direction of the C atom, the solution =109.3 appears to be the most meaningful one, since it agrees very well with the theoretical value obtained from the geometry.\n\nThe hyperfine tensor components determined here are in reasonable agreement with other values found in the literature. The values from the earlier ensemble EPR measurements Felton et al. (2009); Simanovskaia et al. (2013) are =199.7 and =120.3 MHz, and those from the DFT calculations Gali et al. (2008); Szász et al. (2013) are =114.0, =114.1, and =198.4 MHz. However, the earlier experimental studies assumed uni-axial symmetry of the hyperfine tensor and there was no information about the signs of the hyperfine components. Here, we measured the deviation from the uni-axial symmetry and found four equivalent sign combinations.\n\nThe parameter set determined in Ref. Shim et al. (2013b), which was given as =166.9, =122.9, =90.0, and =90.3 MHz in the NV frame of reference, is comparable in magnitude with the solutions and of Table 1. However, the signs of the parameters and the ratio together are not compatible with the measured ratios of Rabi frequencies of ESR transitions as discussed earlier.\n\nIn conclusion, we have performed a detailed analysis of the hyperfine interaction between an NV electron spin and a C nuclear spin of the first shell. This analysis yielded accurately the hyperfine tensor and its PAS. The present study will be helpful for implementing precise control operations in quantum registers containing the first-shell C nuclear spin of the NV center. This nuclear spin is particularly attractive because of its strong coupling to the NV electron spin, which is necessary for implementing fast gate operations in hybrid quantum registers consisting of electron and nuclear spins.\n\nWe gratefully acknowledge experimental assistance from J. Zhang and useful discussions with F. D. Brandão, J. H. Shim, and T. S. Mahesh. This work was supported by the DFG through grant Su 192/31-1."
] |
[
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"https://media.arxiv-vanity.com/render-output/5987962/x1.png",
null,
"https://media.arxiv-vanity.com/render-output/5987962/x2.png",
null,
"https://media.arxiv-vanity.com/render-output/5987962/x3.png",
null,
"https://media.arxiv-vanity.com/render-output/5987962/x4.png",
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https://www.geeksforgeeks.org/edge-detection-using-prewitt-scharr-and-sobel-operator/?ref=leftbar-rightbar
|
[
"# Edge detection using Prewitt, Scharr and Sobel Operator\n\n• Last Updated : 16 Oct, 2021\n\nThe discontinuity in the image brightness is called an edge. Edge detection is the technique used to identify the regions in the image where the brightness of the image changes sharply. This sharp change in the intensity value is observed at the local minima or local maxima in the image histogram, using the first-order derivative.\n\nThe techniques used here in this article are first-order derivative techniques namely:\n\n## Prewitt Operator\n\nThe Prewitt operator was developed by Judith M. S. Prewitt. Prewitt operator is used for edge detection in an image. Prewitt operator detects both types of edges, these are:\n\n• Horizontal edges or along the x-axis,\n• Vertical Edges or along the y-axis.\n\nWherever there is a sudden change in pixel intensities, an edge is detected by the mask. Since the edge is defined as the change in pixel intensities, it can be calculated by using differentiation. Prewitt mask is a first-order derivate mask. In the graph representation of Prewitt-mask’s result, the edge is represented by the local maxima or local minima.\n\n• More weight means more edge detection.\n• The opposite sign should be present in the mask. (+ and -)\n• The Sum of the mask values must be equal to zero.\n\nPrewitt operator provides us two masks one for detecting edges in the horizontal direction and another for detecting edges in a vertical direction.\n\nPrewitt Operator [X-axis] = [ -1 0 1; -1 0 1; -1 0 1]\n\nPrewitt Operator [Y-axis] = [1 1 1; 1 1 1; 1 1 1]\n\nSteps:\n\n• Convert into grayscale if it is colored.\n• Convert into the double format.\n• Define the mask or filter.\n• Detect the edges along X-axis.\n• Detect the edges along Y-axis.\n• Combine the edges detected along the X and Y axes.\n• Display all the images.\n\nImtool() is the inbuilt function in Matlab. It is used to display the image. It takes 2 parameters; the first is the image variable and the second is the range of intensity values. We provide an empty list as the second argument which means the complete range of intensity has to be used while displaying the image.\n\nExample:\n\n## Matlab\n\n `% MATLAB code for prewitt``% operator edge detection``k=imread(``\"logo.png\"``);``k=rgb2gray(k);``k1=double(k);``p_msk=[-1 0 1; -1 0 1; -1 0 1];``kx=conv2(k1, p_msk, ``'same'``);``ky=conv2(k1, p_msk``', '``same');``ked=sqrt(kx.^2 + ky.^2);` `% display the images.``imtool(k,[]);`` ` `% display the edge detection along x-axis.``imtool(abs(kx), []);` `% display the edge detection along y-axis.``imtool(abs(ky),[]);`` ` `% display the full edge detection.``imtool(abs(ked),[]);`\n\nOutput:",
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"",
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null,
"",
null,
"## Scharr Operator\n\nThis is a filtering method used to identify and highlight gradient edges/features using the first derivative. Performance is quite similar to the Sobel filter.\n\nScharr Operator [X-axis] = [-3 0 3; -10 0 10; -3 0 3];\n\nScharr Operator [Y-axis] = [ 3 10 3; 0 0 0; -3 -10 -3];\n\nExample:\n\n## Matlab\n\n `%Scharr operator -> edge detection`` ``k=imread(``\"logo.png\"``);`` ``k=rgb2gray(k);`` ``k1=double(k);`` ``s_msk=[-3 0 3; -10 0 10; -3 0 3];`` ``kx=conv2(k1, s_msk, ``'same'``);`` ``ky=conv2(k1, s_msk``', '``same');`` ``ked=sqrt(kx.^2 + ky.^2);`` ``%display the images.`` ``imtool(k,[]);`` ``%display the edge detection along x-axis.`` ``imtool(abs(kx), []);`` ``%display the edge detection along y-axis.`` ``imtool(abs(ky),[]);`` ``%display the full edge detection.`` ``imtool(abs(ked),[]);`\n\nOutput:",
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"",
null,
"## Sobel Operator\n\nIt is named after Irwin Sobel and Gary Feldman. Like the Prewitt operator Sobel operator is also used to detect two kinds of edges in an image:\n\n• Vertical direction\n• Horizontal direction\n\nThe difference between Sobel and Prewitt Operator is that in Sobel operator the coefficients of masks are adjustable according to our requirement provided they follow all properties of derivative masks.\n\nSobel-X Operator = [-1 0 1; -2 0 2; -1 0 1]\n\nSobel-Y Operator = [-1 -2 -1; 0 0 0; 1 2 1]\n\nExample:\n\n## Matlab\n\n `% MATLAB code for Sobel operator``% edge detection`` ``k=imread(``\"logo.png\"``);`` ``k=rgb2gray(k);`` ``k1=double(k);`` ``s_msk=[-1 0 1; -2 0 2; -1 0 1];`` ``kx=conv2(k1, s_msk, ``'same'``);`` ``ky=conv2(k1, s_msk``', '``same');`` ``ked=sqrt(kx.^2 + ky.^2);`` ` ` ``%display the images.`` ``imtool(k,[]);`` ` ` ``%display the edge detection along x-axis.`` ``imtool(abs(kx), []);`` ` ` ``%display the edge detection along y-axis.`` ``imtool(abs(ky),[]);`` ` ` ``%display the full edge detection.`` ``imtool(abs(ked),[]);`` `\n\nOutput:",
null,
"",
null,
"",
null,
"",
null,
"My Personal Notes arrow_drop_up"
] |
[
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927125134/1-300x268.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927125135/2-300x280.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927125136/3-140x200.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927125137/4-139x200.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927125139/5-236x300.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927125140/6-232x300.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124844/1-300x268.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124846/2-300x244.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124847/3-140x200.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124848/4-147x200.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124851/6-236x300.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124849/5-233x300.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124443/1-300x130.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124445/2-300x215.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124446/31-228x300.png",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210927124448/32-230x300.png",
null
] |
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|
https://www.exceldemy.com/absolute-cell-references-in-excel/
|
[
"What is and How to Do Absolute Cell Reference in Excel?\n\nIn this article, I will discuss what is an absolute reference and how to do absolute reference in Excel? Then I will show how you can shift from one type of reference to another type.\n\nBefore understanding absolute cell references, you have to understand cell reference in a formula. Say I have a formula like this ‘=B3*C3+D3’. In this formula there are three cell references, they are B3, C3, and D3. Every reference in a formula links with a cell in excel worksheet. In another word, every cell has a unique reference. B3 can link to only one cell in the worksheet.\n\nThere are two types of references in excel-relative references and absolute references. Relative and absolute references behave differently when they are copied and filled with other cells. If you use relative references in a formula, the formula will change if you copy the formula to another cell. On the other hand, absolute references will remain constant, no matter in which they are copied.\n\nWhat is absolute cell references\n\nSometimes you may need that a cell reference will not change when you will fill cells. In relative references, when you copy your formula to another cell, the formula will be changed according to rows or columns. In contrast, in absolute references when you copy your formula to another cell, the whole formula will not change according to rows or columns.\n\nRead More: Relative cell references in Excel\n\nSay, for example, you have a formula like this ‘=D4*F2’ in cell E4. If E4 is active cell then click Fill Handle, hold it and drag over the cells you want to fill. See the following example. We have filled the cell ranges E5: E14 in our example. Check the formula in every cell. For example, we just double-click on cell E8 to check formula in it. You will find that the formula is ‘=D8*F6’. Row numbers have been increased by 4, D4 has been D8 and F2 has been F6. This is relative references.",
null,
"A formula using relative cell references.\n\nRead More: How to reference a cell from a different worksheet in Excel\n\nA formula using absolute references\n\nBut we want that F2 cell’s value will be unchanged, and its value will be as it is in the mother formula. For this, we have to use absolute references. We shall write newly our formula in cell E4 as ‘=D4*\\$F\\$2’. We have used a dollar sign(\\$) before column letter and row number. This is the system how you can make a cell reference absolute.",
null,
"Absolute cell references in Excel. Formula with absolute cell references.\n\nThree types of absolute references\n\nThere are three types of absolute references where you can make absolute the whole cell in a formula, only making the column absolute in a formula or only making the row absolute in a formula. The following table describes three types of absolute references.\n\nTABLE: Three types of absolute references\n\nAbsolute cell referencesDescription\n\\$F\\$5Using these type of absolute references tell us that both column and row is fixed in the formula. It means that cell F5’s value is fixed in every formula in the filled cells.\nF\\$5Using this type of absolute references tell us that only row is fixed in the formula.\n\\$F5Using this type of absolute references tell us that only column is fixed in the formula.\n\nNote: We shall mainly use \\$F\\$5 type of absolute reference in the formula. Last two types of absolute references are rarely used.\n\nCreating and Copying formula using absolute cell references in Excel\n\nWe shall create a formula in cell E4 to calculate the VAT for all items in column E. In our example the VAT rate is 15% and the value is set in cell F2. We shall use \\$F\\$2 absolute reference in our formula. As our every formula is using the same VAT rate, we want that reference will remain constant when the formula is copied and filled to other cells in column E.\n\nRead More: Creating and Copying formula using relative references in Excel\n\nStep 1:\n\nSelect the cell that will contain your formula. In our example, the cell is E4 where we shall use the formula.",
null,
"Absolute cell references in Excel. E4 cell will contain the formula.\n\nStep 2:\n\nEnter your formula to calculate the desired value in the selected cell. We shall type ‘=D4*\\$F\\$2’ in cell E4.",
null,
"Absolute cell references in Excel. E4 cell is containing ‘=D4*\\$F\\$2’ formula.\n\nStep 3:\n\nNow press Enter on your keyboard. The formula will be calculated internally and you will see the calculated result in cell E4.\n\nStep 4:\n\nBy default when you pressed Enter, the active cell is now E5. Use up arrow(↑) key to make E4 as active cell again. Now locate the fill handle in the lower-right corner of the cell E4.",
null,
"Absolute cell references in Excel. Locating Fill Handle in cell E4.\n\nStep 5:\n\nNow click the fill handle, hold it and drag over the cells you wish to fill. We are selecting cells E5: E14 in our example.",
null,
"Absolute cell references in Excel. Selecting cells dragging Fill Handle.\n\nStep 6:\n\nNow release the mouse. The formula, we made in cell E4, will be copied to every cell we have selected with relative references. Calculated values will be visible in every cell.",
null,
"Absolute cell references in Excel. Cells are filled with values.\n\nNote: To check whether your formula is correct, just double-click on any cell from the filled cells. You will find that every formula has \\$F\\$2 absolute references, relative references have changed with the row numbers. For the row number 8, the formula is ‘=D8*\\$F\\$2’, just the row numbers have been changed.",
null,
"Absolute cell references in Excel. The formula has both relative and absolute cell references. Relative references have changed with the row numbers but absolute cell references are same in every formula.\n\nRead More: Excel Reference Cell in Another Sheet Dynamically\n\nMixed Cell Reference in Excel\n\nMixed cell reference means either an absolute row and relative column or an absolute column and relative row. When we add the \\$ sign before a row number, we create an absolute row or if we add the \\$ sign before a column letter we create an absolute column.\n\nIn the following image, \\$A1 is a mixed cell reference.",
null,
"Switching between references\n\nTyping \\$ sign manually before the column letter and/or row number is very painful and boring work. I’m going to show here how you can shift a relative cell reference to absolute and then change absolute cell reference to mixed, and finally changing a mixed cell reference to a relative.\n\n1) Making a cell reference absolute\n\nHere I want to show you the shortcut way to make an absolute cell reference with an example.\n\nI want to make A1 cell reference as an absolute cell reference in the cell C1.\n\nAt first, I take the C1 cell to edit mode. Now I keep the cursor on the A1 relative reference.",
null,
"When the cursor is on the A1 cell reference, I just press the F4 key on the keyboard. A1 cell reference will be converted into \\$A\\$1 absolute reference.",
null,
"2) Shifting an Absolute Cell Reference to Mixed Cell Reference\n\nContinuing from above picture…\n\nNow I press the F4 key again. \\$A\\$1 will be converted to a mixed cell reference: A\\$1. The column will be relative but row number will be absolute.",
null,
"Now if we again press the F4 function key again, then A\\$1 will be converted into \\$A1, that means Column letter will be absolute and row number will be relative.",
null,
"3) Shifting a Mixed Cell Reference to Relative Cell Reference\n\nContinuing from the above procedure……\n\nAgain if we press F4 function key again (Fourth times), then \\$A1 will be converted into A1 relative cell reference.",
null,
""
] |
[
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-01.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-active-cell.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-formula.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-fill-handle.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-dragging-cells.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-filled-cells.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/absolute-cell-references-formula-check.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/3.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/7.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/1.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/2.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/6.png",
null,
"https://www.exceldemy.com/wp-content/uploads/2013/12/4-1.png",
null
] |
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|
https://www.stumblingrobot.com/2015/08/19/calculate-the-average-power-in-an-electrical-circuit/
|
[
"Home » Blog » Calculate the average power in an electrical circuit\n\n# Calculate the average power in an electrical circuit\n\nLet",
null,
"and",
null,
"denote the current and voltage, respectively, in a circuit at time",
null,
". Define",
null,
"Then, we define the average power by the integral equation",
null,
"where",
null,
"is the period of the voltage and current. Find",
null,
"and calculate the average power.\n\nFirst, since the voltage is given by",
null,
"we know it is periodic with period",
null,
", so",
null,
". Then, we compute the average power,",
null,
""
] |
[
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-100265804c406efe22459c7cc833e453_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b92853c3ac8db0e1e00333103653d462_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-541210184f5a4d2187a101d5f5f9b52d_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-588f50e409a827881640b118254ee034_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-807301dcfb221b89320e977e783f27eb_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-df1e4c1dafc9d1aaaebfa7064c3abd03_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-df1e4c1dafc9d1aaaebfa7064c3abd03_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-47f86063226ec940607e9e4008a78d6a_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2755ee5187c69ebcefda1e608f750d14_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cf73a837dcbbe96cb5ef1301912eb155_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3ecf5609fd7df5ad4d87e3d4babca986_l3.png",
null
] |
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|
https://www.dummies.com/education/math/business-statistics/how-to-find-the-cumulative-frequency-of-a-class-and-all-prior-classes/
|
[
"",
null,
"How to Find the Cumulative Frequency of a Class and All Prior Classes - dummies\n\n# How to Find the Cumulative Frequency of a Class and All Prior Classes\n\nCumulative frequency refers to the total frequency of a given class and all prior classes in a graph. For example, say that you have researched the price of gas at several gas stations in your area, and you broke down the price ranges into classes. Using a class range of \\$0.25, you might find results similar to those in the first two columns of the following table.\n\nNow, say you wanted to find out the cumulative frequencies for the gas station data. To figure out the cumulative frequency of each class, you simply add its frequency to the frequency of the previous class.\n\nCumulative Frequency of Prices at 20 Gas Stations\nGas Prices\n(\\$/Gallon)\nNumber of Gas Stations Cumulative Frequency Cumulative Frequency\n(percent)\n\\$3.50–\\$3.74 6 6 30%\n\\$3.75–\\$3.99 4 6 + 4 = 10 50%\n\\$4.00–\\$4.24 5 6 + 4 + 5 = 15 75%\n\\$4.25–\\$4.49 5 6 + 4 + 5 + 5 = 20 100%\n\nIn this example, for the \\$3.75 to \\$3.99 class, you add its class frequency (4) to the frequency of the previous class (\\$3.50 to \\$3.74, which is 6), so 6+4 = 10. This result shows you that ten gas stations’ prices are between \\$3.50 and \\$3.99. Because 20 gas stations were used in the sample, the percentage of all gas stations with prices between \\$3.50 and \\$3.99 is 10/20 or 50 percent of the total."
] |
[
null,
"https://www.facebook.com/tr",
null
] |
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|
https://www.physicsforums.com/threads/calculating-the-molar-concentration-of-h3o-and-ph-of-solutions.869598/
|
[
"# Calculating the molar concentration of H3O+ and pH of solutions\n\nNYK\n\n## Homework Statement\n\nCalculate the the molar concentration of H3O+ ions and the pH of the following solutions:\na) 25.0 cm3 of 0.144 M HCl(aq) was added to 25.0 cm3 of 0.125 M NaOH(aq)\nb) 25.0 cm3 of 0.15 M HCl(aq) was added to 35.0 cm3 of 0.15 M KOH(aq)\nc) 21.2 cm3 of 0.22 M HNO3(aq) was added to 10.0 cm3 of 0.30 M NaOH(aq)\n\n## Homework Equations\n\npH = -log[H3O+] Handerson - Hasselbach eqn.\n\n## The Attempt at a Solution\n\n[/B]\nI have only been able to solve part a)\n\nI did that by multiplying the molairty of the strong acid/base by the total volume\n\nthen: [H3O+] = [HCl] - [NaOH] = 9.5 x 10-3mol\n\nthen took the negative log of that number to find the pH = 2.0\n\nwhich are the correct answers, but when I do that process for part b) and c) which are also strong acid/base combos, the answers are no where near correct.\n\nAny help will be appreciated.\n\nMentor\nIn the second case - what is the limiting reagent? What is left in the solution after the neutralization reaction took place?\n\nNYK\nIn the second case - what is the limiting reagent? What is left in the solution after the neutralization reaction took place?\n\nIn the second case the LR is the acid. When i do that calculation I find that the [H3O+] = .04875 mol/L\n\nthe answer is 21 mmol/L and a pH = 12.3\n\nMentor\nIn the second case the LR is the acid.\n\nGood.\n\nWhen i do that calculation I find that the [H3O+] = .04875 mol/L\n\nthe answer is 21 mmol/L and a pH = 12.3\n\nI am afraid neither of these numbers is correct.\n\nFirst of all - you said acid was the limiting reagent. If so, how come there is mo much H3O+ left?\n\n21 mmol/L of what?\n\npH of 12.3 is quite close - but it is possible to easily give a better answer.\n\nHomework Helper\nGold Member\nStrong acids and bases, so Henderson - Hasselbach eqn. never came into it."
] |
[
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https://imparquet.ru/calculate-1-of-7
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[
"# 1% of 7\n\nWhat is 1 percent of 7? What is 7 plus 1% and 7 minus 1%?\n\n 1% of 7 = 0.07 7 + 1% = 7.07 7 − 1% = 6.93\n\nCalculate online:\n % of\n\nMore calculations with 7:\n 13% of 7 = 0.91 7 + 13% = 7.91 7 − 13% = 6.09\n 33% of 7 = 2.31 7 + 33% = 9.31 7 − 33% = 4.69\n 75% of 7 = 5.25 7 + 75% = 12.25 7 − 75% = 1.75\n 276% of 7 = 19.32 7 + 276% = 26.32 7 − 276% = -12.32\nMore calculations with 1%:\n 1% of 1 = 0.01 1 + 1% = 1.01 1 − 1% = 0.99\n 1% of 346 = 3.46 346 + 1% = 349.46 346 − 1% = 342.54\n 1% of 525 = 5.25 525 + 1% = 530.25 525 − 1% = 519.75\n 1% of 2316 = 23.16 2316 + 1% = 2339.16 2316 − 1% = 2292.84"
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[
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https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/4/lesson/4.2.4/problem/4-120
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[
"",
null,
"",
null,
"### Home > A2C > Chapter 4 > Lesson 4.2.4 > Problem4-120\n\n4-120.\n\nThe quadratic formula can be used to help solve $4x^{3} + 23x^{2} − 2x = 0$. Show or explain how.\n\nFactor out the x.\n\n$x\\left(4x^{2} + 23x − 2\\right) = 0$\n\nBy Zero Product Property:\n\n$x = 0$ or\n$4x^{2} + 23x − 2 = 0$\n\n$\\textit{x}=\\frac{-23\\pm\\sqrt{23^{2}-4(4)(-2)}}{2(4)}$\nNow solve for $x$."
] |
[
null,
"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
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",
null
] |
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https://www.texasgateway.org/resource/91-work-power-and-work%E2%80%93energy-theorem?binder_id=292571
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[
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"Learning Objectives\n\nLearning Objectives\n\nBy the end of this section, you will be able to do the following:\n\n• Describe and apply the work–energy theorem\n• Describe and calculate work and power\n energy gravitational potential energy joule kinetic energy mechanical energy potential energy power watt work work–energy theorem\n\nThe Work–Energy Theorem\n\nThe Work–Energy Theorem\n\nIn physics, the term work has a very specific definition. Work is application of force, $ff$, to move an object over a distance, d, in the direction that the force is applied. Work, W, is described by the equation\n\n$W=fd.W=fd.$\n\nSome things that we typically consider to be work are not work in the scientific sense of the term. Let’s consider a few examples. Think about why each of the following statements is true.\n\n• Homework is not work.\n• Lifting a rock upwards off the ground is work.\n• Carrying a rock in a straight path across the lawn at a constant speed is not work.\n\nThe first two examples are fairly simple. Homework is not work because objects are not being moved over a distance. Lifting a rock up off the ground is work because the rock is moving in the direction that force is applied. The last example is less obvious. Recall from the laws of motion that force is not required to move an object at constant velocity. Therefore, while some force may be applied to keep the rock up off the ground, no net force is applied to keep the rock moving forward at constant velocity.\n\nWork and energy are closely related. When you do work to move an object, you change the object’s energy. You (or an object) also expend energy to do work. In fact, energy can be defined as the ability to do work. Energy can take a variety of different forms, and one form of energy can transform to another. In this chapter we will be concerned with mechanical energy, which comes in two forms: kinetic energy and potential energy.\n\n• Kinetic energy is also called energy of motion. A moving object has kinetic energy.\n• Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy.\n\nLet’s examine how doing work on an object changes the object’s energy. If we apply force to lift a rock off the ground, we increase the rock’s potential energy, PE. If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground.\n\nThe force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g.\n\n$f=w=mgf=w=mg$\n\nThe work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential energy, PEe.\n\n$W=PEe=fmgW=PEe=fmg$\n\nKinetic energy depends on the mass of an object and its velocity, v.\n\n$KE=12mv2KE=12mv2$\n\nWhen we drop the rock the force of gravity causes the rock to fall, giving the rock kinetic energy. When work done on an object increases only its kinetic energy, then the net work equals the change in the value of the quantity$12mv212mv2$. This is a statement of the work–energy theorem, which is expressed mathematically as\n\nThe subscripts 2 and 1 indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule, shown in Figure 9.2.\n\nDoes the name Joule sound familiar? The joule (J) is the metric unit of measurement for both work and energy. The measurement of work and energy with the same unit reinforces the idea that work and energy are related and can be converted into one another. 1.0 J = 1.0 N∙m, the units of force multiplied by distance. 1.0 N = 1.0 k∙m/s2, so 1.0 J = 1.0 k∙m2/s2. Analyzing the units of the term (1/2)mv2 will produce the same units for joules.\n\nFigure 9.2 The joule is named after physicist James Joule (1818–1889). (C. H. Jeens, Wikimedia Commons)\n\nWatch Physics\n\nWork and Energy\n\nThis video explains the work energy theorem and discusses how work done on an object increases the object’s KE.\n\nGrasp Check\n\nTrue or false—The energy increase of an object acted on only by a gravitational force is equal to the product of the object's weight and the distance the object falls.\n\n1. True\n2. False\n\nCalculations Involving Work and Power\n\nCalculations Involving Work and Power\n\nIn applications that involve work, we are often interested in how fast the work is done. For example, in roller coaster design, the amount of time it takes to lift a roller coaster car to the top of the first hill is an important consideration. Taking a half hour on the ascent will surely irritate riders and decrease ticket sales. Let’s take a look at how to calculate the time it takes to do work.\n\nRecall that a rate can be used to describe a quantity, such as work, over a period of time. Power is the rate at which work is done. In this case, rate means per unit of time. Power is calculated by dividing the work done by the time it took to do the work.\n\n$P=WtP=Wt$\n\nLet’s consider an example that can help illustrate the differences among work, force, and power. Suppose the woman in Figure 9.3 lifting the TV with a pulley gets the TV to the fourth floor in two minutes, and the man carrying the TV up the stairs takes five minutes to arrive at the same place. They have done the same amount of work $(fd)(fd)$ on the TV, because they have moved the same mass over the same vertical distance, which requires the same amount of upward force. However, the woman using the pulley has generated more power. This is because she did the work in a shorter amount of time, so the denominator of the power formula, t, is smaller. (For simplicity’s sake, we will leave aside for now the fact that the man climbing the stairs has also done work on himself.)\n\nFigure 9.3 No matter how you move a TV to the fourth floor, the amount of work performed and the potential energy gain are the same.\n\nPower can be expressed in units of watts (W). This unit can be used to measure power related to any form of energy or work. You have most likely heard the term used in relation to electrical devices, especially light bulbs. Multiplying power by time gives the amount of energy. Electricity is sold in kilowatt-hours because that equals the amount of electrical energy consumed.\n\nThe watt unit was named after James Watt (1736–1819) (see Figure 9.4). He was a Scottish engineer and inventor who discovered how to coax more power out of steam engines.\n\nFigure 9.4 Is James Watt thinking about watts? (Carl Frederik von Breda, Wikimedia Commons)\n\nWatch Physics\n\nWatt's Role in the Industrial Revolution\n\nThis video demonstrates how the watts that resulted from Watt's inventions helped make the industrial revolution possible and allowed England to enter a new historical era.\n\nGrasp Check\n\nWhich form of mechanical energy does the steam engine generate?\n\n1. Potential energy\n2. Kinetic energy\n3. Nuclear energy\n4. Solar energy\n\nBefore proceeding, be sure you understand the distinctions among force, work, energy, and power. Force exerted on an object over a distance does work. Work can increase energy, and energy can do work. Power is the rate at which work is done.\n\nWorked Example\n\nApplying the Work–Energy Theorem\n\nAn ice skater with a mass of 50 kg is gliding across the ice at a speed of 8 m/s when her friend comes up from behind and gives her a push, causing her speed to increase to 12 m/s. How much work did the friend do on the skater?\n\nStrategy\n\nThe work–energy theorem can be applied to the problem. Write the equation for the theorem and simplify it if possible.\n\nSolution\n\nIdentify the variables. m = 50 kg,\n\n9.1$v2=12ms, andv1=8msv2=12ms, andv1=8ms$\n\nSubstitute.\n\n9.2\nDiscussion\n\nWork done on an object or system increases its energy. In this case, the increase is to the skater’s kinetic energy. It follows that the increase in energy must be the difference in KE before and after the push.\n\nTips For Success\n\nThis problem illustrates a general technique for approaching problems that require you to apply formulas: Identify the unknown and the known variables, express the unknown variables in terms of the known variables, and then enter all the known values.\n\nPractice Problems\n\nPractice Problems\n\nHow much work is done when a weightlifter lifts a $200N$ barbell from the floor to a height of $2m$?\n1. $0J$\n2. $100J$\n3. $200J$\n4. $400J$\n\nIdentify which of the following actions generates more power. Show your work.\n\n• carrying a $100N$ TV to the second floor in $50s$ or\n• carrying a $24N$ watermelon to the second floor in $10s$?\n1. Carrying a $100N$ TV generates more power than carrying a $24N$ watermelon to the same height because power is defined as work done times the time interval.\n2. Carrying a $100N$ TV generates more power than carrying a $24N$ watermelon to the same height because power is defined as the ratio of work done to the time interval.\n3. Carrying a $24N$ watermelon generates more power than carrying a $100N$ TV to the same height because power is defined as work done times the time interval.\n4. Carrying a $24N$ watermelon generates more power than carrying a $100N$ TV to the same height because power is defined as the ratio of work done and the time interval.\n\nExercise 1\nIdentify two properties that are expressed in units of joules.\n1. work and force\n2. energy and weight\n3. work and energy\n4. weight and force\nExercise 2\n\nWhen a coconut falls from a tree, work W is done on it as it falls to the beach. This work is described by the equation\n\n9.5\n\nIdentify the quantities F, d, m, v1, and v2 in this event.\n\n1. F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the earth, v1 is the initial velocity, and v2 is the velocity with which it hits the beach.\n2. F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the coconut, v1 is the initial velocity, and v2 is the velocity with which it hits the beach.\n3. F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the earth, v1 is the velocity with which it hits the beach, and v2 is the initial velocity.\n4. F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the coconut, v1 is the velocity with which it hits the beach, and v2 is the initial velocity."
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https://support.nag.com/numeric/nl/nagdoc_28.4/flhtml/d03/d03faf.html
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[
"# NAG FL Interfaced03faf (dim3_ellip_helmholtz)\n\n## ▸▿ Contents\n\nSettings help\n\nFL Name Style:\n\nFL Specification Language:\n\n## 1Purpose\n\nd03faf solves the Helmholtz equation in Cartesian coordinates in three dimensions using the standard seven-point finite difference approximation. This routine is designed to be particularly efficient on vector processors.\n\n## 2Specification\n\nFortran Interface\n Subroutine d03faf ( xs, xf, l, bdxs, bdxf, ys, yf, m, bdys, bdyf, zs, zf, n, bdzs, bdzf, ldf, ldf2, f, w, lwrk,\n Integer, Intent (In) :: l, lbdcnd, m, mbdcnd, n, nbdcnd, ldf, ldf2, lwrk Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: xs, xf, bdxs(ldf2,n+1), bdxf(ldf2,n+1), ys, yf, bdys(ldf,n+1), bdyf(ldf,n+1), zs, zf, bdzs(ldf,m+1), bdzf(ldf,m+1), lambda Real (Kind=nag_wp), Intent (Inout) :: f(ldf,ldf2,n+1) Real (Kind=nag_wp), Intent (Out) :: pertrb, w(lwrk)\n#include <nag.h>\n void d03faf_ (const double *xs, const double *xf, const Integer *l, const Integer *lbdcnd, const double bdxs[], const double bdxf[], const double *ys, const double *yf, const Integer *m, const Integer *mbdcnd, const double bdys[], const double bdyf[], const double *zs, const double *zf, const Integer *n, const Integer *nbdcnd, const double bdzs[], const double bdzf[], const double *lambda, const Integer *ldf, const Integer *ldf2, double f[], double *pertrb, double w[], const Integer *lwrk, Integer *ifail)\nThe routine may be called by the names d03faf or nagf_pde_dim3_ellip_helmholtz.\n\n## 3Description\n\nd03faf solves the three-dimensional Helmholtz equation in Cartesian coordinates:\n $∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 +λu=f(x,y,z).$\nThis subroutine forms the system of linear equations resulting from the standard seven-point finite difference equations, and then solves the system using a method based on the fast Fourier transform (FFT) described by Swarztrauber (1984). This subroutine is based on the routine HW3CRT from FISHPACK (see Swarztrauber and Sweet (1979)).\nMore precisely, the routine replaces all the second derivatives by second-order central difference approximations, resulting in a block tridiagonal system of linear equations. The equations are modified to allow for the prescribed boundary conditions. Either the solution or the derivative of the solution may be specified on any of the boundaries, or the solution may be specified to be periodic in any of the three dimensions. By taking the discrete Fourier transform in the $x$- and $y$-directions, the equations are reduced to sets of tridiagonal systems of equations. The Fourier transforms required are computed using the multiple FFT routines found in Chapter C06.\nSwarztrauber P N (1984) Fast Poisson solvers Studies in Numerical Analysis (ed G H Golub) Mathematical Association of America\nSwarztrauber P N and Sweet R A (1979) Efficient Fortran subprograms for the solution of separable elliptic partial differential equations ACM Trans. Math. Software 5 352–364\n\n## 5Arguments\n\n1: $\\mathbf{xs}$Real (Kind=nag_wp) Input\nOn entry: the lower bound of the range of $x$, i.e., ${\\mathbf{xs}}\\le x\\le {\\mathbf{xf}}$.\nConstraint: ${\\mathbf{xs}}<{\\mathbf{xf}}$.\n2: $\\mathbf{xf}$Real (Kind=nag_wp) Input\nOn entry: the upper bound of the range of $x$, i.e., ${\\mathbf{xs}}\\le x\\le {\\mathbf{xf}}$.\nConstraint: ${\\mathbf{xs}}<{\\mathbf{xf}}$.\n3: $\\mathbf{l}$Integer Input\nOn entry: the number of panels into which the interval $\\left({\\mathbf{xs}},{\\mathbf{xf}}\\right)$ is subdivided. Hence, there will be ${\\mathbf{l}}+1$ grid points in the $x$-direction given by ${x}_{\\mathit{i}}={\\mathbf{xs}}+\\left(\\mathit{i}-1\\right)×\\delta x$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{l}}+1$, where $\\delta x=\\left({\\mathbf{xf}}-{\\mathbf{xs}}\\right)/{\\mathbf{l}}$ is the panel width.\nConstraint: ${\\mathbf{l}}\\ge 5$.\n4: $\\mathbf{lbdcnd}$Integer Input\nOn entry: indicates the type of boundary conditions at $x={\\mathbf{xs}}$ and $x={\\mathbf{xf}}$.\n${\\mathbf{lbdcnd}}=0$\nIf the solution is periodic in $x$, i.e., $u\\left({\\mathbf{xs}},y,z\\right)=u\\left({\\mathbf{xf}},y,z\\right)$.\n${\\mathbf{lbdcnd}}=1$\nIf the solution is specified at $x={\\mathbf{xs}}$ and $x={\\mathbf{xf}}$.\n${\\mathbf{lbdcnd}}=2$\nIf the solution is specified at $x={\\mathbf{xs}}$ and the derivative of the solution with respect to $x$ is specified at $x={\\mathbf{xf}}$.\n${\\mathbf{lbdcnd}}=3$\nIf the derivative of the solution with respect to $x$ is specified at $x={\\mathbf{xs}}$ and $x={\\mathbf{xf}}$.\n${\\mathbf{lbdcnd}}=4$\nIf the derivative of the solution with respect to $x$ is specified at $x={\\mathbf{xs}}$ and the solution is specified at $x={\\mathbf{xf}}$.\nConstraint: $0\\le {\\mathbf{lbdcnd}}\\le 4$.\n5: $\\mathbf{bdxs}\\left({\\mathbf{ldf2}},{\\mathbf{n}}+1\\right)$Real (Kind=nag_wp) array Input\nOn entry: the values of the derivative of the solution with respect to $x$ at $x={\\mathbf{xs}}$. When ${\\mathbf{lbdcnd}}=3$ or $4$, ${\\mathbf{bdxs}}\\left(\\mathit{j},\\mathit{k}\\right)={u}_{x}\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$, for $\\mathit{j}=1,2,\\dots ,{\\mathbf{m}}+1$ and $\\mathit{k}=1,2,\\dots ,{\\mathbf{n}}+1$.\nWhen lbdcnd has any other value, bdxs is not referenced.\n6: $\\mathbf{bdxf}\\left({\\mathbf{ldf2}},{\\mathbf{n}}+1\\right)$Real (Kind=nag_wp) array Input\nOn entry: the values of the derivative of the solution with respect to $x$ at $x={\\mathbf{xf}}$. When ${\\mathbf{lbdcnd}}=2$ or $3$, ${\\mathbf{bdxf}}\\left(\\mathit{j},\\mathit{k}\\right)={u}_{x}\\left({\\mathbf{xf}},{y}_{\\mathit{j}},{z}_{\\mathit{k}}\\right)$, for $\\mathit{j}=1,2,\\dots ,{\\mathbf{m}}+1$ and $\\mathit{k}=1,2,\\dots ,{\\mathbf{n}}+1$.\nWhen lbdcnd has any other value, bdxf is not referenced.\n7: $\\mathbf{ys}$Real (Kind=nag_wp) Input\nOn entry: the lower bound of the range of $y$, i.e., ${\\mathbf{ys}}\\le y\\le {\\mathbf{yf}}$.\nConstraint: ${\\mathbf{ys}}<{\\mathbf{yf}}$.\n8: $\\mathbf{yf}$Real (Kind=nag_wp) Input\nOn entry: the upper bound of the range of $y$, i.e., ${\\mathbf{ys}}\\le y\\le {\\mathbf{yf}}$.\nConstraint: ${\\mathbf{ys}}<{\\mathbf{yf}}$.\n9: $\\mathbf{m}$Integer Input\nOn entry: the number of panels into which the interval $\\left({\\mathbf{ys}},{\\mathbf{yf}}\\right)$ is subdivided. Hence, there will be ${\\mathbf{m}}+1$ grid points in the $y$-direction given by ${y}_{\\mathit{j}}={\\mathbf{ys}}+\\left(\\mathit{j}-1\\right)×\\delta y$, for $\\mathit{j}=1,2,\\dots ,{\\mathbf{m}}+1$, where $\\delta y=\\left({\\mathbf{yf}}-{\\mathbf{ys}}\\right)/{\\mathbf{m}}$ is the panel width.\nConstraint: ${\\mathbf{m}}\\ge 5$.\n10: $\\mathbf{mbdcnd}$Integer Input\nOn entry: indicates the type of boundary conditions at $y={\\mathbf{ys}}$ and $y={\\mathbf{yf}}$.\n${\\mathbf{mbdcnd}}=0$\nIf the solution is periodic in $y$, i.e., $u\\left(x,{\\mathbf{yf}},z\\right)=u\\left(x,{\\mathbf{ys}},z\\right)$.\n${\\mathbf{mbdcnd}}=1$\nIf the solution is specified at $y={\\mathbf{ys}}$ and $y={\\mathbf{yf}}$.\n${\\mathbf{mbdcnd}}=2$\nIf the solution is specified at $y={\\mathbf{ys}}$ and the derivative of the solution with respect to $y$ is specified at $y={\\mathbf{yf}}$.\n${\\mathbf{mbdcnd}}=3$\nIf the derivative of the solution with respect to $y$ is specified at $y={\\mathbf{ys}}$ and $y={\\mathbf{yf}}$.\n${\\mathbf{mbdcnd}}=4$\nIf the derivative of the solution with respect to $y$ is specified at $y={\\mathbf{ys}}$ and the solution is specified at $y={\\mathbf{yf}}$.\nConstraint: $0\\le {\\mathbf{mbdcnd}}\\le 4$.\n11: $\\mathbf{bdys}\\left({\\mathbf{ldf}},{\\mathbf{n}}+1\\right)$Real (Kind=nag_wp) array Input\nOn entry: the values of the derivative of the solution with respect to $y$ at $y={\\mathbf{ys}}$. When ${\\mathbf{mbdcnd}}=3$ or $4$, ${\\mathbf{bdys}}\\left(\\mathit{i},\\mathit{k}\\right)={u}_{y}\\left({x}_{\\mathit{i}},{\\mathbf{ys}},{z}_{\\mathit{k}}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{l}}+1$ and $\\mathit{k}=1,2,\\dots ,{\\mathbf{n}}+1$.\nWhen mbdcnd has any other value, bdys is not referenced.\n12: $\\mathbf{bdyf}\\left({\\mathbf{ldf}},{\\mathbf{n}}+1\\right)$Real (Kind=nag_wp) array Input\nOn entry: the values of the derivative of the solution with respect to $y$ at $y={\\mathbf{yf}}$. When ${\\mathbf{mbdcnd}}=2$ or $3$, ${\\mathbf{bdyf}}\\left(\\mathit{i},\\mathit{k}\\right)={u}_{y}\\left({x}_{\\mathit{i}},{\\mathbf{yf}},{z}_{\\mathit{k}}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{l}}+1$ and $\\mathit{k}=1,2,\\dots ,{\\mathbf{n}}+1$.\nWhen mbdcnd has any other value, bdyf is not referenced.\n13: $\\mathbf{zs}$Real (Kind=nag_wp) Input\nOn entry: the lower bound of the range of $z$, i.e., ${\\mathbf{zs}}\\le z\\le {\\mathbf{zf}}$.\nConstraint: ${\\mathbf{zs}}<{\\mathbf{zf}}$.\n14: $\\mathbf{zf}$Real (Kind=nag_wp) Input\nOn entry: the upper bound of the range of $z$, i.e., ${\\mathbf{zs}}\\le z\\le {\\mathbf{zf}}$.\nConstraint: ${\\mathbf{zs}}<{\\mathbf{zf}}$.\n15: $\\mathbf{n}$Integer Input\nOn entry: the number of panels into which the interval $\\left({\\mathbf{zs}},{\\mathbf{zf}}\\right)$ is subdivided. Hence, there will be ${\\mathbf{n}}+1$ grid points in the $z$-direction given by ${z}_{\\mathit{k}}={\\mathbf{zs}}+\\left(\\mathit{k}-1\\right)×\\delta z$, for $\\mathit{k}=1,2,\\dots ,{\\mathbf{n}}+1$, where $\\delta z=\\left({\\mathbf{zf}}-{\\mathbf{zs}}\\right)/{\\mathbf{n}}$ is the panel width.\nConstraint: ${\\mathbf{n}}\\ge 5$.\n16: $\\mathbf{nbdcnd}$Integer Input\nOn entry: specifies the type of boundary conditions at $z={\\mathbf{zs}}$ and $z={\\mathbf{zf}}$.\n${\\mathbf{nbdcnd}}=0$\nif the solution is periodic in $z$, i.e., $u\\left(x,y,{\\mathbf{zf}}\\right)=u\\left(x,y,{\\mathbf{zs}}\\right)$.\n${\\mathbf{nbdcnd}}=1$\nif the solution is specified at $z={\\mathbf{zs}}$ and $z={\\mathbf{zf}}$.\n${\\mathbf{nbdcnd}}=2$\nif the solution is specified at $z={\\mathbf{zs}}$ and the derivative of the solution with respect to $z$ is specified at $z={\\mathbf{zf}}$.\n${\\mathbf{nbdcnd}}=3$\nif the derivative of the solution with respect to $z$ is specified at $z={\\mathbf{zs}}$ and $z={\\mathbf{zf}}$.\n${\\mathbf{nbdcnd}}=4$\nif the derivative of the solution with respect to $z$ is specified at $z={\\mathbf{zs}}$ and the solution is specified at $z={\\mathbf{zf}}$.\nConstraint: $0\\le {\\mathbf{nbdcnd}}\\le 4$.\n17: $\\mathbf{bdzs}\\left({\\mathbf{ldf}},{\\mathbf{m}}+1\\right)$Real (Kind=nag_wp) array Input\nOn entry: the values of the derivative of the solution with respect to $z$ at $z={\\mathbf{zs}}$. When ${\\mathbf{nbdcnd}}=3$ or $4$, ${\\mathbf{bdzs}}\\left(\\mathit{i},\\mathit{j}\\right)={u}_{z}\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{l}}+1$ and $\\mathit{j}=1,2,\\dots ,{\\mathbf{m}}+1$.\nWhen nbdcnd has any other value, bdzs is not referenced.\n18: $\\mathbf{bdzf}\\left({\\mathbf{ldf}},{\\mathbf{m}}+1\\right)$Real (Kind=nag_wp) array Input\nOn entry: the values of the derivative of the solution with respect to $z$ at $z={\\mathbf{zf}}$. When ${\\mathbf{nbdcnd}}=2$ or $3$, ${\\mathbf{bdzf}}\\left(\\mathit{i},\\mathit{j}\\right)={u}_{z}\\left({x}_{i},{y}_{j},{\\mathbf{zf}}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{l}}+1$ and $\\mathit{j}=1,2,\\dots ,{\\mathbf{m}}+1$.\nWhen nbdcnd has any other value, bdzf is not referenced.\n19: $\\mathbf{lambda}$Real (Kind=nag_wp) Input\nOn entry: the constant $\\lambda$ in the Helmholtz equation. For certain positive values of $\\lambda$ a solution to the differential equation may not exist, and close to these values the solution of the discretized problem will be extremely ill-conditioned. If $\\lambda >0$, d03faf will set ${\\mathbf{ifail}}={\\mathbf{3}}$, but will still attempt to find a solution. However, since in general the values of $\\lambda$ for which no solution exists cannot be predicted a priori, you are advised to treat any results computed with $\\lambda >0$ with great caution.\n20: $\\mathbf{ldf}$Integer Input\nOn entry: the first dimension of the arrays f, bdys, bdyf, bdzs and bdzf as declared in the (sub)program from which d03faf is called.\nConstraint: ${\\mathbf{ldf}}\\ge {\\mathbf{l}}+1$.\n21: $\\mathbf{ldf2}$Integer Input\nOn entry: the second dimension of the array f and the first dimension of the arrays bdxs and bdxf as declared in the (sub)program from which d03faf is called.\nConstraint: ${\\mathbf{ldf2}}\\ge {\\mathbf{m}}+1$.\n22: $\\mathbf{f}\\left({\\mathbf{ldf}},{\\mathbf{ldf2}},{\\mathbf{n}}+1\\right)$Real (Kind=nag_wp) array Input/Output\nOn entry: the values of the right-side of the Helmholtz equation and boundary values (if any).\n $f(i,j,k) = f(xi,yj,zk) i=2,3,…,l, j=2,3,…,m and k=2,3,…,n.$\nOn the boundaries f is defined by\n lbdcnd ${\\mathbf{f}}\\left(1,j,k\\right)$ ${\\mathbf{f}}\\left({\\mathbf{l}}+1,j,k\\right)$ 0 $f\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$ $f\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$ 1 $u\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$ $u\\left({\\mathbf{xf}},{y}_{j},{z}_{k}\\right)$ 2 $u\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$ $f\\left({\\mathbf{xf}},{y}_{j},{z}_{k}\\right)$ $j=1,2,\\dots ,{\\mathbf{m}}+1$ 3 $f\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$ $f\\left({\\mathbf{xf}},{y}_{j},{z}_{k}\\right)$ $k=1,2,\\dots ,{\\mathbf{n}}+1$ 4 $f\\left({\\mathbf{xs}},{y}_{j},{z}_{k}\\right)$ $u\\left({\\mathbf{xf}},{y}_{j},{z}_{k}\\right)$ mbdcnd ${\\mathbf{f}}\\left(i,1,k\\right)$ ${\\mathbf{f}}\\left(i,{\\mathbf{m}}+1,k\\right)$ 0 $f\\left({x}_{i},{\\mathbf{ys}},{z}_{k}\\right)$ $f\\left({x}_{i},{\\mathbf{ys}},{z}_{k}\\right)$ 1 $u\\left({x}_{i},{\\mathbf{ys}},{z}_{k}\\right)$ $u\\left({x}_{i},{\\mathbf{yf}},{z}_{k}\\right)$ 2 $u\\left({x}_{i},{\\mathbf{ys}},{z}_{k}\\right)$ $f\\left({x}_{i},{\\mathbf{yf}},{z}_{k}\\right)$ $i=1,2,\\dots ,{\\mathbf{l}}+1$ 3 $f\\left({x}_{i},{\\mathbf{ys}},{z}_{k}\\right)$ $f\\left({x}_{i},{\\mathbf{yf}},{z}_{k}\\right)$ $k=1,2,\\dots ,{\\mathbf{n}}+1$ 4 $f\\left({x}_{i},{\\mathbf{ys}},{z}_{k}\\right)$ $u\\left({x}_{i},{\\mathbf{yf}},{z}_{k}\\right)$ nbdcnd ${\\mathbf{f}}\\left(i,j,1\\right)$ ${\\mathbf{f}}\\left(i,j,{\\mathbf{n}}+1\\right)$ 0 $f\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$ $f\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$ 1 $u\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$ $u\\left({x}_{i},{y}_{j},{\\mathbf{zf}}\\right)$ 2 $u\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$ $f\\left({x}_{i},{y}_{j},{\\mathbf{zf}}\\right)$ $i=1,2,\\dots ,{\\mathbf{l}}+1$ 3 $f\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$ $f\\left({x}_{i},{y}_{j},{\\mathbf{zf}}\\right)$ $j=1,2,\\dots ,{\\mathbf{m}}+1$ 4 $f\\left({x}_{i},{y}_{j},{\\mathbf{zs}}\\right)$ $u\\left({x}_{i},{y}_{j},{\\mathbf{zf}}\\right)$\nNote: if the table calls for both the solution $u$ and the right-hand side $f$ on a boundary, the solution must be specified.\nOn exit: contains the solution $u\\left(\\mathit{i},\\mathit{j},\\mathit{k}\\right)$ of the finite difference approximation for the grid point $\\left({x}_{\\mathit{i}},{y}_{\\mathit{j}},{z}_{\\mathit{k}}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{l}}+1$, $\\mathit{j}=1,2,\\dots ,{\\mathbf{m}}+1$ and $\\mathit{k}=1,2,\\dots ,{\\mathbf{n}}+1$.\n23: $\\mathbf{pertrb}$Real (Kind=nag_wp) Output\nOn exit: ${\\mathbf{pertrb}}=0$, unless a solution to Poisson's equation $\\left(\\lambda =0\\right)$ is required with a combination of periodic or derivative boundary conditions (lbdcnd, mbdcnd and ${\\mathbf{nbdcnd}}=0$ or $3$). In this case a solution may not exist. pertrb is a constant, calculated and subtracted from the array f, which ensures that a solution exists. d03faf then computes this solution, which is a least squares solution to the original approximation. This solution is not unique and is unnormalized. The value of pertrb should be small compared to the right-hand side f, otherwise a solution has been obtained to an essentially different problem. This comparison should always be made to ensure that a meaningful solution has been obtained.\n24: $\\mathbf{w}\\left({\\mathbf{lwrk}}\\right)$Real (Kind=nag_wp) array Output\n25: $\\mathbf{lwrk}$Integer Input\nOn entry: the dimension of the array w as declared in the (sub)program from which d03faf is called. w is no longer used as workspace, lwrk can, therefore, be safely set to zero.\n26: $\\mathbf{ifail}$Integer Input/Output\nOn entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.\nA value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.\nIf halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $0$ is recommended. When the value $-\\mathbf{1}$ 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## 6Error Indicators and Warnings\n\nIf on entry ${\\mathbf{ifail}}=0$ or $-1$, 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, ${\\mathbf{l}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{l}}\\ge 5$.\nOn entry, ${\\mathbf{lbdcnd}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{lbdcnd}}\\ge 0$ and ${\\mathbf{lbdcnd}}\\le 4$.\nOn entry, ${\\mathbf{ldf2}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{m}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{ldf2}}\\ge {\\mathbf{m}}+1$.\nOn entry, ${\\mathbf{ldf}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{l}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{ldf}}\\ge {\\mathbf{l}}+1$.\nOn entry, ${\\mathbf{m}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{m}}\\ge 5$.\nOn entry, ${\\mathbf{mbdcnd}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{mbdcnd}}\\ge 0$ and ${\\mathbf{mbdcnd}}\\le 4$.\nOn entry, ${\\mathbf{n}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{n}}\\ge 5$.\nOn entry, ${\\mathbf{nbdcnd}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{nbdcnd}}\\ge 0$ and ${\\mathbf{nbdcnd}}\\le 4$.\nOn entry, ${\\mathbf{xs}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{xf}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{xs}}<{\\mathbf{xf}}$.\nOn entry, ${\\mathbf{ys}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{yf}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{ys}}<{\\mathbf{yf}}$\nOn entry, ${\\mathbf{zs}}=⟨\\mathit{\\text{value}}⟩$ and ${\\mathbf{zf}}=⟨\\mathit{\\text{value}}⟩$.\nConstraint: ${\\mathbf{zs}}<{\\mathbf{zf}}$.\n${\\mathbf{ifail}}=3$\n$\\lambda >0.0$ in Helmholtz's equation – a solution may not exist.\n${\\mathbf{ifail}}=-99$\nSee Section 7 in the Introduction to the NAG Library FL Interface for further information.\n${\\mathbf{ifail}}=-399$\nYour licence key may have expired or may not have been installed correctly.\nSee Section 8 in the Introduction to the NAG Library FL Interface for further information.\n${\\mathbf{ifail}}=-999$\nDynamic memory allocation failed.\nSee Section 9 in the Introduction to the NAG Library FL Interface for further information.\n\nNot applicable.\n\n## 8Parallelism and Performance\n\nd03faf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.\nd03faf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.\nPlease consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.\n\nThe execution time is roughly proportional to ${\\mathbf{l}}×{\\mathbf{m}}×{\\mathbf{n}}×\\left({\\mathrm{log}}_{2}{\\mathbf{l}}+{\\mathrm{log}}_{2}{\\mathbf{m}}+5\\right)$, but also depends on input arguments lbdcnd and mbdcnd.\n\n## 10Example\n\nThis example solves the Helmholz equation\n $∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 +λ u=f(x,y,z)$\nfor $\\left(x,y,z\\right)\\in \\left[0,1\\right]×\\left[0,2\\pi \\right]×\\left[0,\\frac{\\pi }{2}\\right]$, where $\\lambda =-2$, and $f\\left(x,y,z\\right)$ is derived from the exact solution\n $u(x,y,z)=x4sinycosz.$\nThe equation is subject to the following boundary conditions, again derived from the exact solution given above.\n• $u\\left(0,y,z\\right)$ and $u\\left(1,y,z\\right)$ are prescribed (i.e., ${\\mathbf{lbdcnd}}=1$).\n• $u\\left(x,0,z\\right)=u\\left(x,2\\pi ,z\\right)$ (i.e., ${\\mathbf{mbdcnd}}=0$).\n• $u\\left(x,y,0\\right)$ and ${u}_{z}\\left(x,y,\\frac{\\pi }{2}\\right)$ are prescribed (i.e., ${\\mathbf{nbdcnd}}=2$).\n\n### 10.1Program Text\n\nProgram Text (d03fafe.f90)\n\n### 10.2Program Data\n\nProgram Data (d03fafe.d)\n\n### 10.3Program Results\n\nProgram Results (d03fafe.r)"
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[
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https://www.michaelchimenti.com/2015/05/how-not-to-use-ipython-parallel-on-a-laptop/
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[
"How not to use IPython.parallel on a laptop\n\nIn this post I want to focus on an aspect of using the IPython.parallel implementation that may be confusing to new users.\n\nIn the IPython.parallel documentation, one of the first things you do to show that you have started the parallel python engines is a call to python’s “map” method with a lambda function that takes x to the 10th power over a range of x.\n\nIn serial (non-parallel) form that is as follows:\n\nserial_result = map(lambda x:x**10, range(100))\n\nThen, you do the same in parallel with the python engines you’ve started:\n\nparallel_result = lview.map(lambda x:x**10, range(100))\n\nThen, you assert that the results are the same:\n\nassert serial_result == parallel_result\n\nThis works fine, but there is a problem. You would probably never actually use an IPython.parallel client for work like this. Given that the documentation is aimed at introducing new users, it is a bit confusing to present this simple example without the caveat that this is not a typical use case.\n\nHere is why you’d never actually code this calculation in parallel:\n\nIn : %timeit map(lambda x:x**10, range(3000))\n100 loops, best of 3: 9.91 ms per loop\n\nIn : %timeit lview.map(lambda x:x**10, range(3000))\n1 loops, best of 3: 22.8 s per loop\n\nNotice that the parallel version of this calculation over a range of just 3000, took 22 secs to complete! That is 2,300 times slower than just using one core and the built-in map method.\n\nApparently, this surprising result is because there is a huge amount of overhead associated with distributing the 3000 small, very fast jobs in the way I’ve written statement above. Every time the job is distributed to an engine, the function and data have to be serialized and deserialized (“pickled”), if my understanding is correct.\n\nIn response to my StackOverflow question on this issue, Univerio helpfully suggested the following more clever use of parallel resources (he is using 6 cores in this example):\n\nIn : %timeit map(lambda x:x**10, range(3000))\n100 loops, best of 3: 3.17 ms per loop\n\nIn : %timeit lview.map(lambda i:[x**10 for x in range(i * 500)], range(6)) # range(6) owing to 6 cores available for work\n100 loops, best of 3: 11.4 ms per loop\n\nIn : %timeit lview.map(lambda i:[x**10 for x in range(i * 1500)], range(2))\n100 loops, best of 3: 5.76 ms per loop\n\nNote that what Univerio is doing in line is to distribute equal shares of the work across 6 cores. Now the time to complete the task is within the same order of magnitude as the single-threaded version. If you use just two tasks in example , the time is cut in half again owing to less overhead.\n\nThe take-home message is that if you’re going to expend the overhead necessary to setup and start multiple IPython.parallel engines and distribute jobs to them, the jobs need to be more resource-consuming than just a few ms each. And you should try to make as few function calls as possible. Each call should do as much work as possible."
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https://www.chegg.com/homework-help/definitions/polar-moment-of-inertia-5
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[
"# Polar Moment Of Inertia\n\nPolar moment of inertia is defined as a measurement of a round bar's capacity to oppose torsion. It is required to compute the twist of a beam subjected to a torque. It suggests that to turn the shaft at an angle, more torque is required, which means more polar moment of inertia is required. Polar moment of inertia is denoted by",
null,
".\n\nWrite the general formula for the three types of cross-section polar moment of inertia:\n\n1. Solid circular shaft\n\nExpress the polar moment of inertia of solid circular shaft.",
null,
"Here, radius of a solid circular shaft is R.\n\n2. Hollow shaft\n\nExpress the polar moment of inertia of hollow shaft.",
null,
"Here, the inner and outer radius of a hollow shaft is",
null,
"respectively.\n\n3. Thin-walled shaft\n\nExpress the polar moment of inertia of thin-walled shaft.",
null,
"Here, thickness of thin-walled shaft is t, and inner and outer radius of a hollow shaft is respectively.\n\nSee more Mechanical Engineering topics",
null,
"1:00\ntutorial\nWave Equation",
null,
"1:00\ntutorial\nMach Number",
null,
"1:00\ntutorial\nHeat Transfer\n\n## Need more help understanding polar moment of inertia?\n\nWe've got you covered with our online study tools\n\n### Q&A related to Polar Moment Of Inertia\n\nExperts answer in as little as 30 minutes\n• Q:\nProblem #3 [40 Points] Rankine Cycle Analysis Water is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the 1st stage turbine at P = 8.0 MPa, T,=480°C, and then expands to P2 =700 ...\nA:\n• Q:\nProblem # 2 [30 Points] Rankine Cycle Analysis Water is the working fluid in an ideal Rankine cycle. Superheated vapor enters the turbine inlet at P, = 10 MP, T, = 480°C, and the condenser at P, = 6 kPa. Determine t...\nA:\n• Q:\nProblem 4 (12 pts) A flat plate with length 1m is maintained at a uniform surface temperature of 150°C by using independently controlled, heat-generating rectangular modules of thickness a = 10mm. Each module is ins...\nA:\n• Q:\nProblem 10 Consider the steel beam shown. Use the diagrams and formulas in the appropriate Appendix to help answer part or all of the following: Draw the Free-Body Diagram (FBD) a. b. Say whether the beam is statical...\nA:\n• Q:\nAnswer the following questions related to capillary forces and vadose zone flow. a) Estimate the height of capillary rise in a very thin tube having a diameter of 2.5 um. Assume the fluid is perfectly wetting contact...\nA:\n• Q:\n3) A stream of water of diameter d -0.1 m flows steadily shown in Figure. Determine the flow rate, , remains constant, h = 2.0 m. ne the flow rate. O. needed from the inflow pipe hly from a tank of diameter D=1.0 m a...\nA:\n• Q:\nCM 425 Midterm Exam - Summer 2019 Problem 2 (25 points) For the truss below: - Using the Method of Joints, determine the internal forces in members AB, AH, UB and BC only. State if the members are in tension or compr...\nA:\n• Q:\nProblem #1 [30 Points] Rankine Cycle Analysis Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at P, = 10 MPa and the condenser pressure is P2 = 6 kPa. Determine the thermal ef...\nA:\n• Q:\nIn an isolated subbasin within Mexico City, pumping of a consisting of a 25 m thick aquifer unit overlain by a 35 m thick aquitard unit has resulted in a uniform (same everywhere) 17 m decline in hydraulic head throu...\nA:\n• Q:\n2. Find the maximum deflection of a uniformly loaded beam using Castigliano's theorem. The fictitious load is placed in the middle of the beam with following parameters: 112 M²dx Simplify the maximum deflection with...\nA:\n• Q:\n100 -100 Problem 7-26 Dimensions in millimeters. - 350- 2.4 KN 3.8 kN 22.4 kN T = 540 Nm\nA:\n• Q:\n2. Find the maximum deflection of a uniformly loaded beam using Castigliano's theorem. The fictitious load is placed in the middle of the beam with following parameters: 112 M²dx Simplify the maximum deflection with...\nA:\n• Q:\nINSTRUCTIONS: Show all work, including equation numbers, tables and figures used. Also show all necessary steps, including algebraic manipulations, derivation of equations, integration, etc. Write UNITS. 1. The steel...\nA:\n• Q:\nPage 2. June 6, 2019 4. A stress element is shown. (a)Draw the Mohr Circle and completely label it.Using MOHR's Circle determine (b) Principal Stresses, (c) Max inplane shear stress, and the (d) Angles op and s. Show...\nA:\n• Q:\nProblem 4: (20 pts) 0.15 m of steam at 320 °C and 20 bar expands adiabatically to final pressure of 2 bar. Find the maximum work that can be extracted from this process? |! water water V,= 0,15 m T; - 320°c P= 20 b...\nA:\n• Q:\nRequired information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The two blocks shown are originally at rest. Neglect the masses of the pulleys and the...\nA:\n• Q:\n5. A mass of 4 kg of air is contained in a piston-cylinder device at 150 kPa and 120°C. Heat is transferred to the air, and the piston, which is resting on stops, starts moving when the pressure reaches 220 kPa. Hea...\nA:\n• Q:\nProblem 1 5.9 Two resistors are to be combined to form an equivalent resistance of 1000 . The resistors are taken from available stock on hand as acquired over the years. Readily available rated at 500 50 and two com...\nA:\n• Q:\nProblem 6: (20 pts) An isolated tank is divided into two parts, connected by a valve. Initially, one part contains steam at 1 MPa and 500 °C and other part is empty. Valve is opened and steam fills the tank and reac...\nA:"
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[
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"https://cramster-image.s3.amazonaws.com/definitions/DC-11V1.png",
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"https://cramster-image.s3.amazonaws.com/definitions/DC-11V2.png",
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"https://img.youtube.com/vi/j5r0F6VhV8c/0.jpg",
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{"ft_lang_label":"__label__en","ft_lang_prob":0.9252777,"math_prob":0.94778055,"size":1560,"snap":"2019-26-2019-30","text_gpt3_token_len":314,"char_repetition_ratio":0.14267352,"word_repetition_ratio":0.1007752,"special_character_ratio":0.19294871,"punctuation_ratio":0.08754209,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98867166,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T22:07:34Z\",\"WARC-Record-ID\":\"<urn:uuid:f6d96dc9-4254-4ed1-bc0e-399dc0b1314f>\",\"Content-Length\":\"235806\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17590bf7-8cb1-4f7f-888a-990182f9f99c>\",\"WARC-Concurrent-To\":\"<urn:uuid:be2fe024-cd91-4c9a-ac50-af516aa42d8d>\",\"WARC-IP-Address\":\"99.84.216.60\",\"WARC-Target-URI\":\"https://www.chegg.com/homework-help/definitions/polar-moment-of-inertia-5\",\"WARC-Payload-Digest\":\"sha1:MYRL6OQTKMFTP6KVB6GUVU36GAQ3HPJX\",\"WARC-Block-Digest\":\"sha1:WOYA54E5SIC3JHJB2J2YVV4DSDL37BWO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525829.33_warc_CC-MAIN-20190718211312-20190718233312-00135.warc.gz\"}"}
|
https://ard.bmj.com/highwire/markup/196033/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed
|
[
"Table 2\n\nComparison of patient-reported and work-related outcomes by week 26 CDC achievement*†\n\nPatient-reported outcomes in the pooled population\nCDC achievers\n(A)\nNon-achievers\n(B)\nDifference\n(C)=(A)–(B)\nN=200N=1267\nMean (95% CI)Mean (95% CI)Mean (95% CI)p Value\nWeek 26\nVAS-Pain change, N=1463−46.91 (−49.83 to −43.98)−26.88 (−28.04 to −25.72)−20.03 (−23.18 to −16.88)<0.0001\nn=200n=1263\nFACIT-Fatigue change, N=145213.29 (12.03 to 14.54)7.51 (7.02 to 8.00)5.78 (4.42 to 7.13)<0.0001\nn=196n=1256\nSF-36 mental change, N=1086§8.06 (6.45 to 9.66)4.98 (4.39 to 5.57)3.07 (1.36 to 4.78)0.0004\nn=131n=955\nSF-36 physical change, N=1086§19.66 (18.13 to 21.19)8.91 (8.35 to 9.47)10.76 (9.12 to 12.39)<0.0001\nn=131n=955\nWeek 52\nVAS-Pain change, N=1306−45.68 (−48.76 to −42.61)−28.83 (−30.08 to −27.57)−16.86 (−20.18 to −13.53)<0.0001\nn=189n=1117\nFACIT-Fatigue change, N=132612.98 (11.66 to 14.30)8.05 (7.52 to 8.57)4.93 (3.51 to 6.36)<0.0001\nn=186n=1140\nSF-36 mental change, N=1001§7.26 (5.61 to 8.92)5.12 (4.49 to 5.74)2.15 (0.38 to 3.91)0.0174\nn=126n=875\nSF-36 physical change, N=1001§19.37 (17.73 to 21.01)10.03 (9.42 to 10.65)9.34 (7.57 to 11.10)<0.0001\nn=126n=875\nWork-related outcomes in PREMIER/DE032¶\nCDC achievers\n(A)\nNon-achievers\n(B)\nDifference\n(C)=(A)−(B)\nN=98N=563\nMean (95% CI)Mean (95% CI)Mean (95% CI)p Value\nWeek 26\nP-HEQ absenteeism among employed patients, N=2849.69 (3.36 to 16.02)11.87 (8.82 to 14.93)−2.19 (−9.22 to 4.85)0.5414\nn=54n=230\nP-HEQ presenteeism among employed patients, N=278−44.23 (−50.30 to −38.17)−29.94 (−32.84 to −27.04)−14.30 (−21.03 to −7.56)<0.0001\nn=52n=226\nP-HEQ absenteeism among homemakers, N=1822.30 (−5.88 to 10.48)8.51 (5.32 to 11.69)−6.21 (−14.99 to 2.58)0.1650\nn=24n=158\nP-HEQ presenteeism among homemakers, N=179−52.00 (−61.89 to −42.11)−31.52 (−35.16 to −27.87)−20.49 (−31.07 to −9.90)0.0002\nn=22n=157\nWeek 52\nP-HEQ absenteeism among employed patients, N=2396.80 (−2.71 to 16.32)14.03 (9.09 to 18.98)−7.23 (−17.96 to 3.51)0.1860\nn=51n=188\nP-HEQ presenteeism among employed patients, N=237−46.29 (−51.49 to −41.10)−33.65 (−36.33 to −30.97)−12.64 (−18.50 to −6.78)<0.0001\nn=50n=187\nP-HEQ absenteeism among homemakers, N=1542.94 (−12.62 to 18.49)12.97 (6.46 to 19.47)−10.03 (−26.90 to 6.84)0.2420\nn=23n=131\nP-HEQ presenteeism among homemakers, N=152−51.12 (−60.83 to −41.41)−34.16 (−38.09 to −30.24)−16.96 (−27.49 to −6.42)0.0018\nn=22n=130\nWork-related outcomes in OPTIMA**\nCDC achievers\n(A)\nNon-achievers\n(B)\nDifference\n(C)=(A)−(B)\nN=58N=255\nMean (95% CI)Mean (95% CI)Mean (95% CI)p Value\nWeek 26\nWPAI activity impairment, N=308−28.89 (−34.12 to −23.67)−10.94 (−13.37 to −8.51)−17.96 (−23.81 to −12.10)<0.0001\nn=58n=250\nWPAI work impairment, N=131−25.47 (−32.88 to −18.06)−7.77 (−12.30 to −3.24)−17.70 (−26.59 to −8.80)0.0001\nn=37n=94\nWPAI absenteeism, N=131−8.20 (−14.42 to −1.99)−1.53 (−5.42 to 2.35)−6.67 (−14.02 to 0.68)0.0750\nn=37n=94\nWPAI presenteeism, N=144−23.19 (−29.04 to −17.34)−9.41 (−12.99 to −5.82)−13.78 (−20.82 to −6.74)0.0002\nn=41n=103\nRA-WIS, N=153−5.72 (−7.45 to −3.99)−2.09 (−3.00 to −1.18)−3.63 (−5.61 to −1.65)0.0004\nn=34n=119\nWeek 52\nWPAI activity impairment, N=284−29.85 (−35.69 to −24.01)−10.60 (−13.37 to −7.84)−19.24 (−25.81 to −12.68)<0.0001\nn=55n=229\nWPAI work impairment, N=125−26.09 (−32.86 to −19.33)−11.63 (−16.00 to −7.27)−14.46 (−22.72 to −6.20)0.0007\nn=38n=87\nWPAI absenteeism, N=125−8.93 (−14.10 to −3.76)−4.56 (−7.96 to −1.15)−4.37 (−10.59 to 1.84)0.1659\nn=38n=87\nWPAI presenteeism, N=134−24.31 (−30.07 to −18.56)−9.81 (−13.39 to −6.23)−14.50 (−21.48 to −7.53)0.0001\nn=39n=95\nRA-WIS, N=146−6.38 (−8.40 to −4.36)−1.74 (−2.81 to −0.67)−4.64 (−6.96 to −2.32)0.0001\nn=33n=113\n• Boldface indicates statistical significance at the 0.05 level.\n\n• *The adjusted analysis compared the mean change from baseline for each outcome by CDC achievement using an analysis of covariance model in which the impact of CDC on change in the outcome was estimated adjusting for the baseline value of the outcome.\n\n• †For VAS-Pain, negative change from baseline indicates an improvement in the outcome. For FACIT-Fatigue, SF-36 Mental Health and SF-36 Physical Health, positive change from baseline indicates an improvement in the outcome.\n\n• ‡Indicates difference exceeds the respective MCID for the outcome.\n\n• §SF-36 Physical and Mental Health Component Scores were not measured in OPTIMA.\n\n• ¶For P-HEQ absenteeism, the mean cumulative number of work days missed from baseline to week 26 and from baseline to week 52 was compared by CDC achievement. For P-HEQ presenteeism, the mean change from baseline to week 26 and from baseline to week 52 was compared by CDC achievement adjusting for the baseline P-HEQ presenteeism.\n\n• **For WPAI measures and RA-WIS, the mean change from baseline to week 26 and from baseline to week 52 was compared by CDC achievement adjusting for the baseline value.\n\n• CDC, comprehensive disease control; FACIT, Functional Assessment of Chronic Illness Therapy; MCID, minimal clinically important difference; P-HEQ, Patient Health Economic Questionnaire; RA-WIS, Rheumatoid Arthritis-Work Instability Scale; SF-36, Short Form 36; VAS, Visual Analogue Scale; WPAI, Work Productivity and Activity Impairment."
] |
[
null
] |
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|
https://www.savemyexams.co.uk/notes/as-chemistry-cie/1-physical-chemistry-as/1-2-atoms-molecules-stoichiometry-as/1-2-1-relative-masses-as/
|
[
"# 1.2.1 Relative Masses\n\n### Atomic Mass Unit\n\n• The mass of a single atom is so small that it is impossible to weigh it directly\n• Atomic masses are therefore defined in terms of a standard atom which is called the unified atomic mass unit\n• This unified atomic mass is defined as one-twelfth of the mass of a carbon-12 isotope\n• The symbol for the unified atomic mass is u (often Da, Dalton, is used as well)\n• 1 u = 1.66 x 10-27 kg",
null,
"### Relative Masses\n\n#### Relative atomic mass, Ar\n\n• The relative atomic mass (Ar) of an element is the ratio of the average mass of the atoms of an element to the unified atomic mass unit\n• The relative atomic mass is determined by using the average mass of the isotopes of a particular element\n• The Ar has no units as it is a ratio and the units cancel each other out",
null,
"#### Relative isotopic mass\n\n• The relative isotopic mass is the mass of a particular atom of an isotope compared to the value of the unified atomic mass unit\n• Atoms of the same element with a different number of neutrons are called isotopes\n• Isotopes are represented by writing the mass number as 20Ne, or neon-20 or Ne-20\n• To calculate the average atomic mass of an element the percentage abundance is taken into account\n• Multiply the atomic mass by the percentage abundance for each isotope and add them all together\n• Divide by 100 to get average relative atomic mass\n• This is known as the weighted average of the masses of the isotopes",
null,
"#### Relative molecular mass, Mr\n\n• The relative molecular mass (Mr) is the ratio of weighted average mass of a molecule of a molecular compound to the unified atomic mass unit\n• The Mr has no units",
null,
"• The Mr can be found by adding up the relative atomic masses of all atoms present in one molecule\n• When calculating the Mr the simplest formula for the compound is used, also known as the formula unit\n• Eg. silicon dioxide has a giant covalent structure, however the simplest formula (the formula unit) is SiO2\n\n#### Relative formula mass, Mr\n\n• The relative formula mass (Mr) is used for compounds containing ions\n• It has the same units and is calculated in the same way as the relative molecular mass\n• In the table above, the Mr for potassium carbonate, calcium hydroxide and ammonium sulfates are relative formula masses",
null,
"### Author: Francesca\n\nFran has taught A level Chemistry in the UK for over 10 years. As head of science, she used her passion for education to drive improvement for staff and students, supporting them to achieve their full potential. Fran has also co-written science textbooks and worked as an examiner for UK exam boards.\nClose",
null,
"Close\n\n#",
null,
"## Download all our Revision Notes as PDFs\n\nTry a Free Sample of our revision notes as a printable PDF.\n\nAlready a member?"
] |
[
null,
"data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTI4MCA4OCIgd2lkdGg9IjEyODAiIGhlaWdodD0iODgiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyI+PC9zdmc+",
null,
"data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTI4MCA4NCIgd2lkdGg9IjEyODAiIGhlaWdodD0iODQiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyI+PC9zdmc+",
null,
"data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTI4MCAxMTgiIHdpZHRoPSIxMjgwIiBoZWlnaHQ9IjExOCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIj48L3N2Zz4=",
null,
"data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTMzNCA4OCIgd2lkdGg9IjEzMzQiIGhlaWdodD0iODgiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyI+PC9zdmc+",
null,
"data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgNDAwIDQwMCIgd2lkdGg9IjQwMCIgaGVpZ2h0PSI0MDAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyI+PC9zdmc+",
null,
"https://www.savemyexams.co.uk/notes/as-chemistry-cie/1-physical-chemistry-as/1-2-atoms-molecules-stoichiometry-as/1-2-1-relative-masses-as/",
null,
"data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMjk3MCAyMTAwIiB3aWR0aD0iMjk3MCIgaGVpZ2h0PSIyMTAwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==",
null
] |
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|
https://planetmath.org/crazydice
|
[
"# crazy dice\n\nIt is a standard exercise in elementary combinatorics",
null,
"",
null,
"to the number of ways of rolling any given value with 2 fair 6-sided dice (by taking the of the two rolls). The below table the number of such ways of rolling a given value $n$:\n\n$n$ # of ways\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 5\n9 4\n10 3\n11 2\n12 1\n\nA somewhat (un?)natural question is to ask whether or not there are any other ways of re-labeling the faces of the dice with positive integers that these sums with the same frequencies. The surprising answer to this question is that there does indeed exist such a re-labeling, via the labeling\n\n $\\displaystyle\\mbox{Die }1$ $\\displaystyle=\\{1,2,2,3,3,4\\}$ $\\displaystyle\\mbox{Die }2$ $\\displaystyle=\\{1,3,4,5,6,8\\}$\n\nand a pair of dice with this labeling are called a set of crazy dice. It is straight-forward to verify that the various possible occur with the same frequencies as given by the above table.\n\nTitle crazy dice CrazyDice 2013-03-22 15:01:43 2013-03-22 15:01:43 mathcam (2727) mathcam (2727) 7 mathcam (2727) Definition msc 05A15"
] |
[
null,
"http://dlmf.nist.gov/style/DLMF-16.png",
null,
"http://mathworld.wolfram.com/favicon_mathworld.png",
null
] |
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|
https://www.code-saturne.org/forum/viewtopic.php?f=2&t=2277&p=12430
|
[
"## derive residual exact calculation\n\nQuestions and remarks about Code_Saturne usage\nForum rules\ndaniele\nPosts: 82\nJoined: Wed Feb 01, 2017 11:42 am\n\n### derive residual exact calculation\n\nHello,\n\nI looked for the posts on this subject, but could not find a precise answer: what is the actual meaning of the \"derive\" residual and how is it calculated? Is it the lhs minus rhs of the equation? More exactly max(lhs - rhs) among all cells?\nI have a converged calculation, with derive residuals all lower than 10^-3, except for the enthalpy one which is as high as 10^+5.\nI have a source term of 10^8 W (total source term integrated on the calculation domain) for the enthalpy equation, I would like to know if the 10^5 derive residual can be acceptable considered the order of magnitude of the source term.\nDoes it mean that I have a cell where lhs minus rhs of the enthalpy equation gives 10^5 W?"
] |
[
null
] |
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|
http://sacs.aeronomie.be/nrt/index.php?Year=2017&Month=12&Day=08&InstruGOME2=4&InstruOMI=0&InstruOMPS=0&InstruTROPOMI=0&InstruIASI=2&InstruIASIb=3&InstruAIRS=1&InstruSCIA=&obsVCD=1&obsAAI=2&obsCCF=0&obsHGT=0&Region=000
|
[
"",
null,
"SACS home > Nrt > near real-time",
null,
"Royal Belgian Institute for Space Aeronomy",
null,
"NEAR REAL-TIME NOTIFICATIONS PRODUCTS",
null,
"",
null,
"",
null,
"InfraRed\n\nInstruments\n\nGOME 2 [A&B] OMI OMPS\n\nIASI [A] IASI [B] AIRS\n\nTime of observations\n< day\n< month\n< year\n\n08 December 2017\n\nday >\nmonth >\nyear >\n Select a date today 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 NRT Either click on a region in the map to submit or select a region from the list-menu and click 'submit' 000 == World view 001 == North pole 002 == South pole ----------------- 101 == -150.0 60.0 102 == -120.0 60.0 103 == -90.0 60.0 104 == -60.0 60.0 105 == -30.0 60.0 106 == 0.0 60.0 107 == 30.0 60.0 108 == 60.0 60.0 109 == 90.0 60.0 110 == 120.0 60.0 111 == 150.0 60.0 112 == 180.0 60.0 ----------------- 201 == -150.0 30.0 202 == -120.0 30.0 203 == -90.0 30.0 204 == -60.0 30.0 205 == -30.0 30.0 206 == 0.0 30.0 207 == 30.0 30.0 208 == 60.0 30.0 209 == 90.0 30.0 210 == 120.0 30.0 211 == 150.0 30.0 212 == 180.0 30.0 ----------------- 301 == -150.0 0.0 302 == -120.0 0.0 303 == -90.0 0.0 304 == -60.0 0.0 305 == -30.0 0.0 306 == 0.0 0.0 307 == 30.0 0.0 308 == 60.0 0.0 309 == 90.0 0.0 310 == 120.0 0.0 311 == 150.0 0.0 312 == 180.0 0.0 ----------------- 401 == -150.0 -30.0 402 == -120.0 -30.0 403 == -90.0 -30.0 404 == -60.0 -30.0 405 == -30.0 -30.0 406 == 0.0 -30.0 407 == 30.0 -30.0 408 == 60.0 -30.0 409 == 90.0 -30.0 410 == 120.0 -30.0 411 == 150.0 -30.0 412 == 180.0 -30.0 ----------------- 501 == -150.0 -60.0 502 == -120.0 -60.0 503 == -90.0 -60.0 504 == -60.0 -60.0 505 == -30.0 -60.0 506 == 0.0 -60.0 507 == 30.0 -60.0 508 == 60.0 -60.0 509 == 90.0 -60.0 510 == 120.0 -60.0 511 == 150.0 -60.0 512 == 180.0 -60.0 (region defined by the centre longitute and latitude)\n\nVOLCANOES in this region",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Status of the NRT processing",
null,
"",
null,
"",
null,
"",
null,
"obs. of SO2 SO2 height Ash / AAI Cloud"
] |
[
null,
"http://sacs.aeronomie.be/include/images/spacer_orange.jpg",
null,
"http://sacs.aeronomie.be/include/images/spacer_grey.jpg",
null,
"http://sacs.aeronomie.be/include/images/spacer_grey.jpg",
null,
"http://sacs.aeronomie.be/nrt/lastSO2notif.png",
null,
"http://sacs.aeronomie.be/nrt/lastASHnotif.png",
null,
"http://sacs.aeronomie.be/nrt/linkSUBSCRIBE.png",
null,
"http://sacs.aeronomie.be/nrt/volcanoes/picts/volcanoes_000_sm.gif",
null,
"http://sacs.aeronomie.be/archive/GOME2_O3MSAF.png",
null,
"http://sacs.aeronomie.be/archive/IASIalertDATA.png",
null,
"http://sacs.aeronomie.be/archive/OMIdataRequest.png",
null,
"http://sacs.aeronomie.be/archive/linkKNMI.png",
null,
"http://sacs.aeronomie.be/archive/AIRSalertVCD.png",
null,
"http://sacs.aeronomie.be/archive/GOME2archiveAAI.png",
null,
"http://sacs.aeronomie.be/archive/IASIalertASH.png",
null,
"http://sacs.aeronomie.be/archive/OMIarchiveAAItemis.png",
null,
"http://sacs.aeronomie.be/archive/AIRSalertASH.png",
null
] |
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|
https://stat.ethz.ch/CRAN/web/packages/bang/vignettes/bang-d-ppc-vignette.html
|
[
"# Posterior Predictive Checking\n\n#### 2020-02-24\n\nThis short vignette illustrates the use of the pp_check method pp_check.hef, which provides an interface to the posterior predictive checking graphics in the bayesplot package (Gabry and Mahr 2017). For details see the bayesplot vignette Graphical posterior predictive checks. and/or Chapter 6 of Gelman et al. (2014). bayesplot functions return a ggplot object that can be customised using the gglot2 package (Wickham 2016).\n\nWe revisit the examples presented in the vignettes Hierarchical 1-way Analysis of Variance and Conjugate Hierarchical Models. In the code below hef and hanova1 have been called with the extra argument nrep = 50. This results in 50 simulated replicates of the data, returned in object$data_rep, on which posterior predictive checking can be based. The general idea is that if the model fits well then the observed data should not appear unusual when compared to replicates from the posterior predictive distribution. library(bang) # Beta-binomial rat tumor example rat_res <- hef(model = \"beta_binom\", data = rat, nrep = 50) # Gamma-Poisson pump failure example pump_res <- hef(model = \"gamma_pois\", data = pump, nrep = 50) # 1-way Hierarchical ANOVA global warming example RCP26_2 <- temp2[temp2$RCP == \"rcp26\", ]\ntemp_res <- hanova1(resp = RCP26_2[, 1], fac = RCP26_2[, 2], nrep = 50)\n\nWe show some examples of the graphical posterior predictive checks that are available from bayesplot, but make no comments on their content. The commented lines above the calls to pp_check describe briefly the type of plot produced.\n\n## Beta-binomial model\n\nThe aspect of the data that appears in these plots is the proportion of successful trials.\n\nlibrary(bayesplot)\n#> This is bayesplot version 1.7.0\n#> - Online documentation and vignettes at mc-stan.org/bayesplot\n#> - bayesplot theme set to bayesplot::theme_default()\n#> * Does _not_ affect other ggplot2 plots\n#> * See ?bayesplot_theme_set for details on theme setting\nlibrary(ggplot2)\n# Overlaid density estimates\npp_check(rat_res)\n# Overlaid distribution function estimates\npp_check(rat_res, fun = \"ecdf_overlay\")",
null,
"",
null,
"# Multiple histograms\npp_check(rat_res, fun = \"hist\", nrep = 8)\n#> stat_bin() using bins = 30. Pick better value with binwidth.\n# Multiple boxplots\npp_check(rat_res, fun = \"boxplot\")",
null,
"",
null,
"# Predictive medians vs observed median\npp_check(rat_res, fun = \"stat\", stat = \"median\")\n#> stat_bin() using bins = 30. Pick better value with binwidth.\n# Predictive (mean, sd) vs observed (mean, sd)\npp_check(rat_res, fun = \"stat_2d\", stat = c(\"mean\", \"sd\"))",
null,
"",
null,
"## Gamma-Poisson model\n\nThe aspect of the data that appears in these plots is the exposure-adjusted rate $$y_j / e_j$$, where $$y_j$$ is the observed count and $$e_j$$ a measure of exposure. See the Conjugate Hierarchical Models vignette for more detail.\n\n# Overlaid density estimates\npp_check(pump_res)\n# Predictive (mean, sd) vs observed (mean, sd)\npp_check(pump_res, fun = \"stat_2d\", stat = c(\"mean\", \"sd\"))",
null,
"",
null,
"## One-way Hierarchical ANOVA\n\nThe raw responses appear in these plots.\n\n# Overlaid density estimates\npp_check(temp_res)\n# Predictive (mean, sd) vs observed (mean, sd)\npp_check(temp_res, fun = \"stat_2d\", stat = c(\"mean\", \"sd\"))",
null,
"",
null,
""
] |
[
null,
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",
null,
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",
null,
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",
null,
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",
null,
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",
null,
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",
null,
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",
null,
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",
null,
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MFQa/nwkXeuZ+3VNMpJ5siTVY8W/u81c4nh1HddpNUpunfmc0Y408lF6EGN0MMQtHWfiYlHXwxTI7qUI1hvvtSLMRRlnDX2dMx1TRYynT15+wUYYQfmJ7MSSqSgPiJAa7N1q3V3shWUEj50vAyCPu0s7o7WhGSBTtuZqVZyCfZlut8cfMh7LEgEboXrVKsNXeCW96Jz86cHQ4Ul+oLGTQtjvlQNgr4RAz0/SbHeudDxeOlyYCP0MAR1ZUvhON2m+i3WWBwj7ibjMZS2pZQs6G79lXjEhiUvpo/Y1DFvwITLZRthKYCL/1J5I4l0dpYhQg3U8eKtHwc1Qo9BUBeg1TNFFUM/6IlYijO6jzApcQ45+roYBjHZxVaYYF8FQyM26gM+WQafPJXeBOiHCRo+F0ryRSxTig3IvuLzg4K4LxyKoJRPDMLTTCxucLbOINfS6hXyJkPNceK4u+qtb04k1ZC3XtGBOiT1KtD2kq+WiVw6T/fkKNvNTcq5uu92TB1/EIJS9sSet1ozmykk+LBIpjO9FBMR3ZPXyHFC29ITahItJKjtmIHFkJ1YswDVD7jw84m6n8Zik+Ldc2KtSlcwvR8ciKAW06SGt+hN24fZjqpemyHtWG/r1PYqGibp4qVtjiPjqBpFIhHXlOto5UKXuxTgJrF1mV3M175SJzgEQZuG9xhosjJl8lNWZKs2mROGn1pWfhmBNJdKVwNHIgNVrKi5aOOnZ5GuU8xhOjDYj+45DsHV4osVBkG7hGt48SqRzDE0kT9fHsgQORLpUTlqGaV5NYJX2i/qmcyGpW8/vWymYfDiItst0OtoZfUlrAXFrDvNFgo9qhF6HIKa3qYABaXgxtfS2oCr54fQ7gZvPYuV2dNRipFi1avililOkrUw1S6skyvWeWcavtSrSWW2prLeBBKUlVHiDXSDoF3CDEIz6vCO0+G20GU7ImbPWIscY09sCrLC5tnsjFFzaoPs1b2Yypo9hIRYq6eeunqlFelNeIkedXzmYGYn6Lzk5kA4AkGbAHULlLV1sppqoxLDvm1Igu2dq6KRFy1za3yULml4U56WZ4ie8z2j7iMGKl95umQoGpeF82x964dGQ9WxcBiCMn0otlPYkumLgcgSq4VrNYZagEgisE8t7uaAJXnmyFQUgnCTTLZtxi445XMbOqykhS1zEKGJhoMcNFR/AIKGD8/yixckeqDhS22dcKXEABE7XCJJaVzl5g2Lm/rrzSmkBWeQ87Fth2GBVu9nvvqMbREiA00TQTFFRw4baDoCQVv4ko1IaFMvOft6Y43x9DaAm8e7nRmZJu9EljbaluRs9UzhPSVxEW1Skxp+nuF8FShCzUvK1PFMew2CdkpQK6mzDCTaMYRBHK2+i6gpeq3R4+FK3tOeVvphfZ7ajAG1UrgWf0ruUb3oUsN/jqDVqkBpNGD/Ajd7SdSn3uq67ANHISiy7IWpbbz5ScMxD6EY5XNWryyRcbTnWp+nWos73HbLd4rX5gHchlSafjf1Xz7JT494sQQ1m1Uig6C9EtQYI9nyQgmb3qSIb0EKb7y2n2KYWFiYpASV/DItGoXLzmoEK6sVcN5CgLoI1exLFRI3KFrZ3SBobzANn6jhMwfRCPs98JhGMcjibReZF8wFQy0rH3O67Zl1JrVazdNKv7PCo9ZPEpQ2RM2oW3Y/3hg6CNofjH+JIfRkZR2crVA9WV5PY+neE+QRT49kmpZfLISZJy+b2rd4lNRZgH5ew1cXoRy7l21Ank9cPpyOPwhBMTGegz5AsmLTGmptofp1yxBz6xo8qy0r/yJtiFj8C4cpa/UaO42owcLl/wQ4sYGx18wJeUx9DYJ2CKNLsomGNlV7VvDLLUir51QOnY3iN4pNi3B6MaiJT4kupcmyfTkVoC5hP0snilCUBUqC6WxdVHK8epHuCRomqGl4Gp3J45MRqV+YlvEkEZvbHYki9XmiiyFz8SREK8nPGolNqRGP+oK4owjN7sej5g4Dlw9ohPZPUDNBGSuiQOKE2Vpr1MS7Vp7HMsRd3JgZDG0Tw2XNZZDYAwQeRg8Nb7NEPs8mhAUYaEq2yZs6vpxJ+95xDIImc4atICh555u2SKhHlppdWc0OtUFgLVLqduhSx6pF1LXNug8BGotov0AmJlcRfEjCdKcboUcTob0TNExQdR+eb7i6Bq2zoq4LtWw/VZsyYuMdPOj04vMcIn3E5gxL8atTp41e8Ff5/JlLtWU2QjfJmpQGQXuDpYISSZRYoDRp+FrDXtTYNMNjZUVX85JiCDfJZstqajSum7/0EvbnuQD9EpdQZIdROpMBbfn4wjaVQdCuQGrlDA2frWgplbnOKCqY/FbUcRq8mpMtShEr9U0gL63nQ0OmRo6zWu/HF12kOCcUtQhq+LjLphzQCD0CQTEHlpPiOKEjeSvGYt0xYNVHUZxn9xQyrrBNvXpOqTpFJzkat3Sudp+nf9lLfOnc1XYtFRaZFm6fL4OgfcEcH2p4iy4hclNrmYNM7UB8DbEYNqaGnM0tq4Q7Xpbwwjx/SJcC9KvKWCjvTcdn+vHHM0I7JygFKFKGFEGIIiYbFydrCi2c92ac1tigBM5lI5+7/iFHxb0tDQHqIjek3NcJmtHemdUMFFbgDYJ2BBKlsFGIgzthggrl52V+uqZv9xRX2WG0xlAH4bAm9WhAqS2ez4OK3kaAVmuCxi4FGqFcotxcuoOgb4LS78lGUBZ0YJG7xl6DC/y0u6KYyYQhX8etAl8WE7eq15U0AWrPaW30Xz5/ukhIGKBiBMmkMg8pPQYOQ9BsI2uS2Bg6e/wSjzx4b7eKhDxVTyhps0ytLs/JHI9W7xS58MKfOn/hajB+wqjjy9HGMHZNUNO49OFf2LTOKL1onQVo1H8sn1VbwlMjBtqmJDfH37x+17cLC1TldgKUxEeIPj5hrLQ+lhHaPUHRv5nNReIPbHB3fs4l7xHXbA6S8y2mgK8Y2hjs8+xjoIIlBcoiXfplCCNNhW5SMX4OgvYDxnxysWYkqHbsiFn46HZQ5DtbgXxrFjYpyUynLANTjcrilfUSEXrLPN2QoOjg5+xcziplyd2xjNCuCepDZdHXw4lg2BHjI2hbGf1cGb8INpmpWU2yziK0LjYqVWsOsaNnfi4IegsW4QXRPZqLDdYvHEA6CNoHqOEx7NvWZbB3PZsBGjLydEGMLqqMnQiYJcYfIwwvZf7Og1v+CU8ot+Rn5SB9LLYptryxCHtLD6TjuydoMgFa0nRHyhZwr7PtebYfZsHQaJoTK4ASn4BjfrQu6/CCn9y3dGOCIgOG/Be33x7PCO2aoHgjJ6FZkYhHHDOn3JqGPDF5QV829oaMZQVUKcl2wLtr721NrSVELSg0i+bbcEiVLpKkYvu5uZrkSDq+d4JamvOFGr7kpAst7jOW61lupqXj25HV5tOSnj7su/HTskhelS835me1yn703aOmj5sby6F0fMcEJe+ymHIENRP0vHs6NbZwtmPXT13I0Rplob7ttRmoEYpqOzu969OfdCNwWg8mooSOP5YR2jtBUeiLSibYnilJSxJF+dHq8NVPi5i80Vmi4M6rSGmROs1NfC4L+G5HUA4YyQw00cjIZRC0C1DDg0PcKFhrhhg1Aaq6EJDzE9YMbWxzu8B2J1q1p7vvMf/Ttfs9+MnBuBMzEzcq5OyjTQdB9w+1dZw2MowaPmuzL2ddvHqKrp9fG0NtczwLPzngw/NIsbzTKGPjSmq9+NqfxqTfJ0slMxRasq0UPY4I7Zeg6kOVxUaGVUlpzc9L7/FaMonMPn0wNJJHxfcnqU8V9a2HN8witXPi4JvEOXcWqj+Sju+YoCiEr/Qt2GksyWopPUf5GodO9HyzBcKlymV+FR8bz3D+3QQod9wIjFD0rSAvliMbdgR0S1AfUMuZx4Iq5ekt1tiMfckCbU9cMdQz8OGwRwo0HrUgVK1tQvjtBSgLTCd20o9XS8zLIOjusdDwbPPIhZpRWiT01Xf41Ju3BnnbY4C7YmR4q7Bzf2wR278tQfl3lISEA/c0aDqQl9Q3QYsaQYul4SWcI4lBta88V1e3IzsUk+VE2lHNh1/PsrstezAkFDqeI8wxhOJIkdCOCWotEq7hc6JbYzMbFuH0yyQ9LWDWqL7zR7zhoy6C9Uvb88bc8f13GW5StdKR4xih/RKUvW1siGdDJHs9PIxZlxy6SNHzoP1sulKYWpFo1EVFzd5dBKitYMKcJgxiZIX9gSKhvRJUq0+nYbvcJEtTNEQuUkRx7EWKrn4IVqsfz5mLXjyv89zlOxGUO2nEAk0cznegmtBuCapWVvFiU2MoQH0jgv9/cviFV1j+IOGya20SM+Y3amOuH3yHP2ZiaKooaao26E4GQfcNKzpCmhNdndYaGcu06wUOvcfQVlyynE0f309KQG9PHdbST39CspImzVqO4iV1S9BiyR72k7OYs7UaXRCgfMY7pPU+kBYH1XbU3flZbZ9CkmaEsnf6Dr9ne+iUoByACLcIM2MYpLdpyW8Q9CJFVz+E87/yslo69K4ERcGdaEqWFoPNchQjtFuCZq5cR2jbEoWcq6xv8fNdhs6pouZEx4SRxZPvwhvqeMnQ8aiu96rlI6BXgkLDc01bsUy2pNn8fOOdvdz/EWgTcRYD7k/l550IKk3HswFFomOve3RLUNEIMqGfyJs96psC1J55RtHlY7pQ8nUZup8Puw9vEJ5HRattvEcF80GM0F4Jyo0DlKAFJU3eLVffEaB86htCNGJKF/h5VwFqRujEymRVy4OgOwfrNr0bqdgbG7T6QHjm9J3XJfs0mppq61zyR05u3BbU8QiGwgjFSPOj1IR2SlAkOasNNIJCTKV8UMP7888oGjdq25+oumbuyY1bA+nNQiMUIlRtFtq9ftmG0ClBMwP13K6ZWEopHxeg9gqvUDSSm+uO+vvzE2N4Fba0GEFFBkF3DOW4byYGJ62oOckVGt5f4qKijwFirSbUHjm/dXOgujUXr1rGyNODGKFdEpRJpEoTFKGZmkq5lp/2MicHy1xHuvKc5ptfO+83gfENwu5jLmksBwnV90lQxLGDoNObmfRTBK2tJ6kuY/PzY4t7Vw/dBZjJNBmhjNVzjOQxsp19EpS1yWyXQwS0cE08H/gMiVRfo+GpDLsvQTltObMo1HZ/DILuFczCi43bQgi0hENzE7/i1Ze4M13YLjdREzpecpFj6PguCcqJNGIaPtl2jJimfIu39JXXuDdbWJVlEhRp+XK6GbxP9EhQtSXabDjOGKpcS5uNfJtfcOll7k4W7KDJiDTZIMa8Ws7cLbokKGOGWJ4tGCeShdMT6w2jQBde5/7SrGAnTbF0vGZs8TyCCO2QoLaZw0cSJwywFb2pAK0XGPoIqhQbFUodX1gxMgi6RwiXvxWr+5lMUKjG27lIjlP//RFMEVaMZGY7CytGBkH3iML9cCimZwj0Dhr+7MUeRBTULNsInMomFp9m0jV6JWixdqSc6p0IuqwUeRRNuMprco/QOpcxKnSVce0TPRIUtXY0Qac3keHCcnMNb9DTGP6dwfkNkj3QlPQIgab+COomKFezpUINr1Eft3NwYbxVLXNq/RECTf0RlBKTYxfRiMQB7ydF7/sF4mfJCkYoQQ9ghHZIUJv8DYKWpJxx0IkApfUC/z1jBk5OMgi6R0AN0gRNCg2PMcsf6kXaAxDgnfRCydbLIgfwkrojKH14mqAFg14zOj9wfxcEVRQZJPrxTMcfINDUH0Gp0jO7kRIWec0tRB0A9ks2CTp9/NIg6P7AsUUotcvQ8MV2G/XCT1sSigpC6vh0ACO0N4KyG96ioOImaEcCFNPAC9faWqGBoKz+2ed0X/RGUOaQqiVbMLM2ZEw/BEWI3qK8inEUvYvQ3giKqt5KE7SmVBM2XeLuTvjJ0bnFJuDgRlLt3Y/vjqDQ8AiCFtPwUIK1H4LSCMXuZvPjuV/n2ad0X3RGUMtzcu9q4RxN6UrDU4QyPGHj6gdB94bZBIWYSRNX7f3rRg+yGtuXc1eU3PVuhPZFUCwkdILmkmvCgCbc340A5ejogq15rFqezGw1G6ZbdEZQG2vHBccoPhftzAStvj6eySRVeEmDoDuCaXiYoJOCx3wYV4DdaPjQ8cJBoRNBS9W+jdC+CGqJeMwoTFzN1p2GrywKLUUTs53YMNt5MqkrgtpURMFsg5Sxnc193I4EqOXKQFAmk0DQvnV8bwR1E5Rdub0SFOtLmEyavnXvJfVFUDRBVDEfXjlShKWgXelAGqGe7oQROgi6G5iCrwjBYFxD6lKAmhGqVtE0EbR3I7QngnIgE0xPSNDSupH6EqDU8fAAkxmhpdgU7e8AACAASURBVPNAU18ExeoE8+ELeiO0RwHqOj6M0MkeHQTdCYQangTFfvUiXWp4JJMQYEqYnTJ584OgewEb4CdSotbOhriToL1peFuEQwcevZ0F3ce9fQYX6Iugk/xkCBvbD7mrs/YnQC3QhLl97PvovSa0J4IqG44xGSZDhGq3BK0FazvhJWEaY9+Bpp4IShO0ZF8gK0ofqT9+TgRFEiKbEToRFWX2zz6nu6EfgoYJignZGYl4EywdEjSSSdlaA5Nox0ZoVwTFwkNfI4ShBjTNenzroCgw+hSfxZowRKw/TzDQG0FLbGkplmHp8p1zHY9h4LlCgkqXn0OiI4IyEc+G44K5RbVXE5RjfNj3gUCTwp3vWMd3Q1AzQbliQBbF9H2+cSjKzkW58KNAxw+Cbh7Q8JMkYbvcbIJ2qeFXOh5lI+Whc54fi14IKjbdlX6DTR9mqV2nggU6Hp1J/GspQbsVof0QlAOxkYgvlQRV0V4FKAiqudQMI7Tk3LMR2g9BqdRJUG0maK/vGnQ8Vyah+xi7oPrV8R0RlCIUBBXbYyl9E9SMUOTjE5blDoJuGmKzXTlwA/lppWPb65tWuQEnqQWaQNB+I6HdEJTyk25t1uLJlU7fMwCJXIywJUHRIdirEdoTQUuFTWYN8d0TdFEwkhMDGH2qi14Iym4kSRYGhQnasQ8P4K/DDHDX8exmffY53QV9EFSsYY5ppCZBuyYo1zCm0PGcxD8Iul1MPjyEKLdYTi68z63t8x1zwIBJWZDtxEqv0qsR2hVBE2tB4dNK3z58dR2f0RePOT/o7ezz7+2EoJxpJ8mioBYR7Z+gNfukUNg0uVOfsAuCtiATCcrEn3T6fs1AlCIV+oWI+5ZOVUZPBLVuj8kGpQna5du1AFREykzuJkjQThNnnRDUGo6doLbSsvcNV9TxOVeIUE7r7VNn9EFQyE9WMnHDgGrXXToO9LNghBiTEwm1I10qjR4IushzVu4H1H4TKwugKLQZoWUQdLsQjg2jhsc2dY85Pfus7g7q+MTPZcqpVyO0A4LijcGKYxI0Fe6v7F/DG0Gh22l6J5ls70HQTUIqZ964CZrLcQgKHY/ZN0bQ0qWO74KgnFRLAYp+Y7VSkf7BQJNvpEkFe/U6FKHdENSCTJMzWw5igrqOzxmqI6ekucsCw/0TVLhg1UxQFtOrdD1NawEOcEj045Pp+EHQ7QEEzfBnU8wFrfj/2af1CEDHT9YnKppAUOkx29kBQdH9oJJtalhmlrPrvRczlK1JkwzNXK2n+/Hjf/Xtj35Z669//HfvHrl7gqLMbCIkp2xgc8JxTNDKghHk403Hq+yHoPWHP/uTWr//0d+8e+DuCSrWEp+MoJrlEHlOh+n4TILa2oi9/OV/+MVPJpb++fsH9kFQqU5Qa+c8iIY3P14SpoBDx2veFEFfzrB89Fc/qfV/vy9AOyDo9JZkZTDQhtoVrYci6KQ8Mpfr0b7ZEEPfIeiP/+4f/vUHXmXvBHUTNLGmh6s9DqTh3QgNP77uqWDk1z/+v//5fRdp/wQNE9QImlV86M1BYEYoezuT9X3sxU369Y/+6y8/clwHBOWGeHY3Ck3Qspf36AbgkPOJoJiLiprQ/VQ0ff/tTz503O4Jiil24t1IJKgcSMM7Qa11jn0fu9Hx3//Fx47bOUHNAp0EqHV7iByjVnkGu4+TERSz+fdC0H/4Zx88cOcENRNUku0/PFAlU8CSSdla58pOjNDvf/x//ulHHCRg3wRF/Q52cnKSK+Zl2gyxAwGEpI63bGfZxUCV7799IMfp2DdBEZjGqko68bY6oRzIh69uhOak3VYtd0BQNMR7GkkOR9CoaIIEzSBo7azmbu8EtbGgsYJb6sFM0NDxbEyaLsMg6KZAExTzQG3sYrZVXs8+qweDo2zNCk8cCtCXjt81QVlMHwQ9pAlaPdBkcYxkOn4QdCvAtBfOEgFBNXOZ7CEJin2IDIVK6W30+c4JinYkG64hHLuoRzNBq+l4FIWiYAT12oOgWwErmQTzsyYXSWGC+gbZYwEEnfxEEnT6kHam4/dMUGGpCLb+Yv/h5MUfq5IpgN5O9AyypGkQdENoGh4maM224bij9+aDYKAJXhJFqHQ2u3ffBJ0kaDaCkqkYgHM8gpoRWlJi3weCwdtXI4fo6uSKX3bkcGIDtyPV45mgtvADBV26o0DTEbo6IS7hGjGNRILqEU1QDzSFjtfJadTtj5c+Qlen2C4rhfuKkiYmOrf+xtwF1PE5FQ4DzxwwsvnrcICuTnbHwQQFQXPNcqh2uSXgxwti9QjVZxihW3CT0hmWj/bf1anU6JlppIIVx3K4QpEA/fjC0QAort9IwcjbBI2uzh9+9l/+6Je/+vbjv/vhZ//j24U+pf0SFM4qhmImi4IeWMN7oClb2XbZh473rs4//GIi5fd/Un/1z3/x7Sf1H/7JWafSjgk6OUQIqnB+a0ZR01E1fBihTPlmbO7cAUGjq/OHn/1N/fW3b99+ghv112cidNcEVRQwmQlqWaQNKLanYGLjdCE4ZDonEHT7odDo6iRB/8Jv9ERQmqBY84sMn2Yf2fDss3oSQFCoE65TSHkHBG1dneDl9z+pP/z5D//qlxX/TrBbgjLPiVkFLLWj4VU2r9fuBup4sTF3snkdv+zqhH80ffnR3/zwZ9+YXjrBXgnKNFJOlXaX2or4w5qgTlBl1TLXx9dN53wvdnVSxZ9jrwSFD4+B9GgIV7V96Yc1QSsYivkV1vixDyP0FN0RlEFQ2384maAo092y0Lg3FIOli+n4kkS3LUIv4A+/uNwsv1uCss9BuYIbt0SPbIKajscAC8TqdfM6/grslKBiRbrCyAoKRUo9sgnqVcslFwTd8HHV/en4y9grQdkuN70RVqssiV7Cs8/qmaCOz6bjU0aqs4/Kw90SVBGlF5k1/KFNUCMomjuZj0/ajY7fJ0ERY6IPT4JSlA6C0ggtxQpGONPi2ed0C+yToCwF5awsNhxPXkE91ljQC1DWJmTLJkk3InS/BIWGL75Als1IhzZBW2cSrPKUsTJpEPR5YCUTVqVnFtMnqvzDE9RSFzRCOa2+CzdplwS1hcbe7YHpduyQ70FefAVwkzBlmmMsMCi0i0uyS4Jy3zZMUDQyFjNBu3g3vgROmzaCJhbPdKFUdkpQOEbqU+0E3vy2y3ceAvrxmR0G1h4/CPos2GoksYkNnIk5TNAwQjVZbyc20vRghO6RoLbKI1u3B/Q8o05HF6BuhE7U5DhflM/08KndI0E5ib5gbi1mikiqg6AG9ZUn6EzqxQjdIUHpIYlHQQvGLyIMuvlZGg8Ax0+aZY5QaO1Bx++QoNyUgC54TrVDg0M3eb0vQlF0yB6Dl2RG6P5F6B4JCg2PSJ+tTqAJOggKmB9vOj6hCWYQ9AlgzhlbD/E+1DJJijp8eAc4ifZ4WqHSRWxjfwRllD5L9HPijRgC1GEEzVZzV2CB7v7C7JCg6C/GoAaYWmrr5fb/PtwGkJlo0DI/HvP6dy9C90lQ7KTku2C7ZwZBA2rJJO5FLNTxew+/7Y6grBPBwoRiQ8Oo4TsIp9wGoORkhBaupKk91NztjqBiM5nMR4Ku76Vs5yYwgmbOWs7cTL53EbpDgtrYWlaV2fauQdAZGFCfmQNO3MszCPpoTJysk/mfWZjLgP0g6AKgZFKUIdpKmrr33bJ7Iyi3xU4C1LZ3IVxfhwm6AIoSvGo52ad35yJ0bwQ1DY/NxpOninz8EKBr2NRUyTZhZP86fncELco0J01QBJsGQdfgcjN6kInR4r27STsjKDQ8EMuntYwg0wkKjFAWKhS4SXvX8TsjKCuVa7HdM5gs0klZ7g2BWpqiOUbZDoI+EqhjUm7eZrtcoYQYBF2BTmRFMqkkTBjRfdfK7ougEKCI7mUKCASbdq/Cbg/T8TZFTHev4/dH0MxItNAE5RS3IUDXIEHNTZLYmbRfhu6KoJAHikrQAh9puvhZhw9/BpbcmZLJydu19nuJ9kXQjJYGUV+bahp++PCnwDBwMR2fi3tJu2XorggqGRzFbiQEmWrOWnvoarg1MCi0sGMrZdsePwj6GGTW6igTeSU0/CDoKVBInxhoQsWIt2TvlaF7Iig0PJRXKeyHRxi0DhP0HCwYEbPTTcfLIOgjMGn4OvmnJhxyLWysHSboGcBHEFTcCN21jt8TQZNSw3MEJrZ3jSj9K0B7fFYS1LqPWXX37LP6HHZEUERAFZXKQhfJNfwg6DlQMIJ8PP14LYOgj8Gk4THZGnW40PDoPK7DBL0E6nhrOrBk0o7dpB0RdPLhMScssdhRp3dgaPhXAD5iQbkt7ty3m7QfgmqqIOjknMrQ8O8AZqcP+E1FvK1wEPS+KEyQ1DJJTg5hl0HQVwGdnsRGB8xLk3bJ0P0Q1DT85MrniaCI0ZcRZHoNpuMr/His7xSfbLFHhu6HoNDw3JcqCDLlIUDfAPx2zE1FMglRDy+pGQS9IzALS6jhIRYmg3QQ9A24jvdZtrnuV8fvhaCQAokZEqSRsKeKI67rIOhlCC8XrlVN1vjBz/OzT+t67IWgVidSoa8y59aaAN1zrfhdwcJZgRGKMWJlvyJ0JwRFjEmsFDQWJwwN/xawbBe9c1jUM+v4HTJ0JwSdZABdpGJ2P7fHDgH6BmCElsmPz5FN2quO3wdBtZiGn6ypzM0eOgToO8D4b4Tkko8K3auO3wlBXcNj5s2ks4aGfx8c0hQ6fscidB8EdReppJosCKoM0I9Ckdfh+Xgkk9A7B4LWPSbkd0FQaHgIUFj9nPutw4d/D6hRho6nTYR1u2oxub1dsl0QFIV2mMpkGr5MCp9pzqHh3wIIyQmMcCqz7FXH74GgGGVtdSIoIHuBPM24f2j4twAROun4XF+8YqTusuhuBwTFcmNOtSvFqh9kaPgPAARF4wcn/WKvwj5F6B4ISgGq7PVGIVPSoeE/AM60mEzQip3lSHDEkKZ9MXT7BEXEWY2gaKRFpZ1p+EHQtwEjNGPE7wsjTbZPqg6C3hpQ8NicUKSg9qEg5IQg09Dw7wDXJ4GgXLq7Vx2/eYLCAsVQ61pKwUw7bRp+EPRtIPKJ2QGorlGZZ4zsi6HbJygieEIficNcYEsNDf8RkKClsK5+styLD7MdBL0pYIFiN5JgrLJr+OHDfwzU8RI6Hm2eO3STtk9QzgSln8T120PDfxzKCSPThbOEfJUd6viNE1QZBIWGVy6ZllYnsvMFag8B/XhO9Gfnh9gUnEHQG4LDbuDATzzFUFAWhQ4B+kEgVs8JAqbjkY3bnQjdNkFNgCpiJabh2W6MR3a9ueJRgJuEWRfFiu5QZGNz7nZ07TZNUOhxF6BFE3uRSrhIe7rIT0PT8ZkZuKTiptF+Lt6WCaqWRWKiM7MSlI3HpuF3dI2fB9bRF5v4Oyn4TB2/LxG6aYIq8/DF/qVFncjQ8B8Dl6Dkyb+EFYpc8f5E6IYJyg0JmRTVwsJb1ImMIOgVsEuIQm8a8MVXI9ZB0FugCVDGmtm6ULBtug6Cfhi4To2gNWHkXd0XQ7dLUH7U4SLBcZdUuX67Dg1/FULH00KqeYc6frMEpXbCZmNUg04CtNpIOx0+/DVgq1wRm7aMcobSNirs5ApulaBaQ4DCMeJyYxTfDg1/JZiPVyMo5oklLqkZBP0yKEBRpYxSZcm2fbuYAB0a/uNgXb2IpztpjvpGhZ1cwo0S1AQoq0DhIyU2y7HaoQ4Nfw1Mx6utN0WiI1fZlY7fJkH5AbcAE8x6mz5AE1RHnch1oI6fHCVBJg47pnRfInSTBOVHfCFAQdBCY3QI0GvBsmXEQrLpeHTR1R2J0G0SlB9wuu/IJFe68CFAdznA5XkwHV8lFZtmK6xdrLsRoRskqC+d4oUUFopAwwuVfR0C9EpYMgnbZdnRxVWSe9LxWySo/c9pDSzFsYFMmM80CHo9mo5PHIKDtJzofhi6PYKadYRCUAhPiFD2yunQ8J8E4snT5Uw05Sv2e4UI3cOF3BxB/boJ+ImgnY0TyeiMDwG6h+u6IbDmDjtO2X88iVA2IA+Cfg5qEVD0wottnqEA9bXndWj469HcJGG0Dp5n6PgdXMqNETTkI6eJcJscW5FMgHr4fvsXdVvgFcto/kisGNmXCN0WQSNJDJVUWOMwCVBmOXMdAvSzsHQniu0K10R7b9I+3KRNEVT9C/x14cRA6+V8aQJ0uEifgPvxylj93kTolgga/GQrvGKXh3KrB8tEhob/NELHS0rhJsGv34cI3RBB1b+qNRubX5Q5+SqHBVoHQT8B1/GVBOVsAfQm7IOhmyGoNn5iR1cujDGp7e2CMRoVTpu/ohuE6/iayVA48VTyg6BXQNs35TQ7ekhaMbK2VApQr2La+gXdInjZMsaMJFxQjgD2S7r567kNgkb9nJqCR/WillrqEKC3QYjQiaCorAc7s+5DhG6CoLqQn8gXCRf7cLAyi2zF11PsbfDVZhAELZlWaJaFCN34Fd0AQVv5sdo2NPhHaOUEQTl6VYcA/SJCx1fT8ZqwAGQfbtLTCbqojucEYG7pzay1KxCg6r3GQ4B+BRShaKJJ2WuaeF13IEKfTNBl8waNIpaAZsY9RVkHmrgzQec008D1IEGhnXLi3iR8/HPdg5/0TIKuu4eNn7A/WaUsvj/FWzmHAP0aZjepeKQJ00J3IEKfRVA97W3nyAuWiKAKFMFQfbGR32VYoF+Ht3GHm8RBwLsQoc8gqJ6xs1pDMfxMBJiEyXgfNWBLkQZBvwjPJqm1dyIhj3CoT1N/9sm9gccRVBsuPBi1yCygh7U0iVB80CcXvkSJfR0E/QLCTZpsUN/Xp8wlb12E3pWgusRbx4kfjQRnptVZMLC2oovbBOhuemi2itDxUkqySZZqNWIbF6F3Iuj7pFweHAOX0LWNEXZY68FZIsJeuSFAb4FFsN4GCU1iYQfppHsQ9OPUrM0gxVe4R6xhEquyQ7ajhHCtW76KewCzIBAFiZOEMA58FqHbvba3JuhV5AxJ6+YnZ9lh56nYMDYkO9wClUHQr4IsDBEa+c7tW6G3Jej17HTdbQVMitEsjNC90I6XlQDd7kXcB8JN0uTDAk2E+rirZ5/da7glQa9X7LXxk1umkvFTuZSC5QwuOY2mXzy9o8N7Fao3J70k3YMjfzuCfoSeZz6925a+KyHF3kMYoOiYk0WIabOXcC9oOl5SZp0t5gFvfjHNrQh6Sk+9iEtPUluGhJYZWytnNUxYcFzqEKA3xMJN4rirYiJ026PEbkPQk5qPD5qi4b0z9475a9zTJSAo6WkCdFigNwMnCikvNQY1YQRODMTYrBV6E4KuKuY++oc270idn5MZilQnDNCXwuVIRkwZ/LwRZjeJMxwgQqGzwpB69uldxA0I2urhaTR+WHq6+EQ+A0UMmJ/OANOL8ZMOJo4c/LwZ6CbhyiehCIUFWlQ2LUK/TNBmR7pCrpeLQU6fpO1JhY6R1SgLCcpok4bkFHvGZ85t4ARhhWIkm00EV9Y9bFiEfpWg0eZ2wQF6laQ6P0kY6Jj4yVUp4CprQBlrGgL05mBCXtyRf7FYaNEYhLPJi/xFgjoNL3PxIkddeArZqdTjkjOHiEBscqUxr5ks+LnNa7dDzI68xUKLabE2FWN7+BpB36Ln4oDFT8ZOcetz+jDr9Cme+GlNnPoS65DEn1jsiVef2cBFkKBKEZrgjc4idKvB0K8QVBijEFuf+RpaxsgTm5Sb1rLpExqwyhhCszo/c622bapWWw6/yQu3T1iwno68wk0KEbpdP+mTBGV0vditi17RMjxv4lDbpbDQJ9veMe4CFaDk50vwU9rCmeEh3RjmyFvJCEoe2N+psTtpgxf6UwQlHX1x+3xnY1/QdUVRNzr5TNYsQXpaDT2bOKv1wCMWKuIvELJ5g9dtr1iJUDQtFE5fNsN0iwy9nqBud7p6X3EQxFsgTE6KW06jpcOerUQJBgLmWeVCUxTiE8UibI03Zurg5+1hJSOV7XPwSVkygn/R97UxXElQyzyImAh0FwnEZB1nnd0mi2naIy4wxWLyHt0kKznNiiap8ROqBvevdhpv76LtGRFD4eYka5xlzZ1uVIReQVB3cFw6knwuM51I9kNG14YdMaOgnTjnlnNKdNWnQ3M47y+S2W2soeCbWbu5a7ZrkIa8xJNvGn4SlPxGReiHCRqujalp42ZJABiJNbqZKzlQ0llwH392ohYJq5O8RtgTS+BTov/+Yg3wxdx8YfWStlnKm7tkOwd1POLLyMh769csQjfH0A8S1OjpuwmpzRUDpQvaB86RmfdU0+55aZfCd8yo9sqZ9HTl/qLJF81gr4dYhcjg532gIUKnS56QWp5F6AYXSX+IoDzxYrWaHsyc6MUGVkrSlxk8CtybHld3lSTcKMjdRK9IrT2u2pMyQ/UkpQ+6kRY13toF2z9MybP1CyXiFKEWZNmikv8IQZt2t6HxFJr4siTmBeQcCsO9JWM4Xo/mazssK0xRZjUHPx+AJkKx26vUJkJ9vde2Lvn7BKX0y7Q7S+NmWlHRLBiR1xgrzXWn9yNLiTvRE3QlKZmEb/n9Wgc/74ImQicl7yLUkimeUHn2+a3wLkFR/sJgpTYrsrFzNjp5tx1SqOLlnKZnIOfh90ffQakRyw9+buti9YJWFypZyNBcrb1zg578OwRVc8wtMcmyo2AnPfWEUUrTt4wf4EV5ukjMd8qxavNMpNLV55gb+vjOT7XgKH9x+zJwc6xEKPWeNBGqNudyM3iboBBumEZj/nWygTT4e0qLIgn9HgY/EWmKID3sTLrqefbzbXqypY4wxQ5elgegQM8kc05/yM+7YilCc/EmxbpJJf8WQQVlRjBT6IHnUNsMxkcmU0r46GAsMkPZ/CALeBay14KmFllCiQkDohC8NDt5rQrrRub2kfZl4A4IEYpQk9JPYpV4dU9+SyL0dYLabGPKNcsYkZ1UxlTKVPu5JeNBW8pJErEUj9AXZyr9HwmapuJVn2zLZshUfUao/e72ZeAuMBHKmeDT24ZRl8I2effkN8TQ1whK8ghj7pjp5VEhC6IX8fR7Lq0YpNXTUfmby2/hqWIln8XEp9mcQteoeFEoCvfcJvJf3rz4gTvBO5AxnqmYCHVhYZr/2ac34zJBNZRyYeaIu0sYjGB7MO1IJ2F2i7K0SibmjuJBC8rH3VwKW32PHOSmJY1soPKCkFuzgzqEJTzVlv44QzOV2Mb8pEsEVWsKSmxkm/6AbLVwJGBOnrwsoZBF5JRNrWo+Kp2YdPcYVbL6kJC6smzwsGcPfj4AXrnMjrDEd1izxaC3ZYaeEdT6hzEinl6PeiYzMYEetSE5G8VOC48CrvGtaKZ4iUirEPXnFq/aq7pm+ODnI3BS1TQztHo6+9kn6DglqH+qoKOz7dwQpyeMxkSD0iWgWZPuIp1w1H40xd5CR0ZmCX7OidD1U7dzcbqGyxCxnpvMlQBJvSO8tdM+HWuCqs/umXx3oyd3lkz8NDIWBu3DrNT4a7TVJ+sMO6y24uVF75IulLqettxtr+CrU5i0CB2XLBiaq4ubzXjyK4KqlWmIOUFQ6cZPhupZJldW7FrObNBGRL/vhGgeZbL+uLkV5IL5es8/d2CG1ugjm+Rnsup6DxvKdvTYiQQ1fsJqbgl1y1giyFlCm8uZPtfl7VU0c3mUfbfGj1D769N5u4V54JZoIpSRxGQJpexKcTMMXUtQphzBT4pLsz6xkwyTvRAiaty84BadkNS0++KOxaGzrj99iU1ck6PAGBr7/SwpX0o4FRsxtk5UfK3ILDAv7s57YltqsRNf+jPnp78mmJump6yb5evKZL1kFQzcFxopechQjBqBRMrNt9hGaeiJF6+IIjEH7+IzsXkDtcf1bIbIBT6dCcFLAag48vzJn/oTBj4Ni0RzhltGUxgJuoggbuAdWRNU4BcJ1meZcwT9TiNa1tIzcPEPOBWkl46/9MQNXI2jYVbyOkkiTmtCQklqlN49/z1ZEZT8nJwkb1OHfq+cTWNJ80vPv/wXrMzJM1ZeNm6efy0OCHeU8P4mVPva+KHqWfktKPkVQVkVD/1u9LSAEAOf8vqZvmZLL4JIi3TTa47QoOdT4PkUpGYwQcMCN0nCeZDnM/SEoNb2a8GlqHLPep5tX+Etd8/9n5ZzevWwz5z8wJdBoeHbe1VtjedLinz3BsoiVgQtsQTmhZkknrrVI71zku8EiC5F5NePX3PKA7dEmKEIT0OEFmeoB5qeztCTQL3x05rkYH1WNqx/5AxfpeBa0V8+4LpzHrghtMXrQ8nTlzeGclDjcxm6joOW7KUhMEOwlSzn98Vnw7mcPLnnAokHO58O0/KsQ0cC0TylxJliVU6HbD4caxWfTX4WRpeKIkN7LYPWoffLD1+4PfA06MxQdDJmG8iRhf1iUfDztLNbE5Rnlq2/EiWh6T4MepW+A8+AJa+LFVmqT6+fZGitVltZn8nQFUERprU8JxmahgI+Bvgmo1INTT6Nobm4f6/tmCfghKCoCjF+InI7+HkQBEMZq0chm2aToVzy64sCnsOFNUHFh4ggcTT4eSAYQ2F2ZqyeRu/4iy/sbiUYz2HD2ovPUrz8StIruc2BLhF2KHtxlRNh6JDYtrVg6BMYsXaSYk6I5PKx6OdAL3CGlgJDVGryVvNMfVpaFcXDz2tNUFZ9iuo82nPgKAiG4t1HaVOKcJNYwMkOejgr1pkk7sXmmKVBz8NhlqGcWhgMxSCuWkqUAj+aoie5eBrKuQzz84jwmPzkJ2XhGISSYo5rRc/88rCH4aSiHrtF0+DnUdGEqG26KD5KG7NxJJd21CMpepKLnwyQWZoPHA5eOiIc3WrTCk3NF9QFrw97DNYV9ZXVdQ/75QObQ0vMc3w2Bx3ZUNjMbZaL4x7E0ZNyuzzE58FhJaCTsORk90nJSw4humLoo6ToSaB+8HOAFLXEPBkaWzNSUStjXx14b6wImvLg54AJR07gmCO4NgAAAzxJREFUxOBMaZYoEvW56GMpupagg58DRMxttbnuOYSo7cp4aJDnbLLIwAAhrB1Bx3xOs57nWM5H+tFX7osfOA4s2iQ2FqetF7Q80+MoOgg68BpQNsSZxTa8ONYHTo4TikYfpOgHQQdehdXaGT+5lC38eay6lsdQdBB04HVwPBMXWlb17ZaNokyI3p+jg6ADb8BiolxYqeJbh8pLzDVGb8i9z2AQdOAt2LDQZAFQdCshOPrSKDok6MCTYYNsE1exYoOWbbAORZ/v/vsHQQfeAYVoseVYlZvWueLaxOiQoANPh6l57qfm4FDuDazu0d/7lw+CDrwLG7WMjXTU89hCyCxoHgQd2AY4hbFw/7VtaUdSHhwdKn5gE7DFH1y2nrIbpSIjDjqwGdjSmlLEKYpZnYsukHthEHTgg7Bae25sTdTviDnd/bcOgg58HLZVA4u0CkuaHlDUNAg6cAVcihbWMT+k6m4QdOAqhKK3hqVB0IHtwTqRzKO/+y8bBB24HjbAHV7SCDMNbBSm6+/+a1YEvQP++B4veiXGORjucQ4PJeg98Md3fv2PYJyDYQvncDUGQR+DcQ6fxCDoYzDO4ZMYjtHApjEIOrBpDIIObBqDoAObxiDowKYxCDqwadyNoL/7yz/9a974/b/77l/8r3v9lnfwm5+enMwTz+F51+Ef/9t3/uc/8zp8Evci6O///V//7t/w/fh/z7skv/nupycn87xzeOJ1mH7z3/7L39bnXofP4l4E/fufTh/cn1cKjr+60+94Hy692sk88Ryeeh2q8fKp1+GTuBdB8bb8rb0jv/vLp70zTo7FyTztHJ56Herv/u1v65OvwydxN4L+1Xwtfv8fnmyDLk/mWedQn3kd6t/zr3/qdfgkHkHQ+j8HQYmnXYff/8ff8lQGQRuW5s4//vcnE3QLNmh95nXwT8awQWfAYaTdU0O/PAO/mb34OJlnnUN94nX4zc/r7/5TffJ1+CTuGwf93V/+/O+/++6n7x99H/B3T+fwzPhfnMMTr8Pffvfdd3/610++Dp/FyCQNbBqDoAObxiDowKYxCDqwaQyCDmwag6ADm8Yg6MCmMQg6sGkMgg5sGoOgA5vG/wcB+POtlTQtIwAAAABJRU5ErkJggg==",
null,
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",
null
] |
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|
https://mathematica.stackexchange.com/questions/7879/cant-compute-definite-integral
|
[
"# Can't compute definite integral\n\nConsider a scalar field (in polar co-ordinates), $f(r) = l-r$.\n\nNow I want to evaluate the field integral over a circular region of radius $b$, centered at a distance of $x$ from the origin. According the law of cosines, The arc of radius $r$ from the origin subtends an angle of $\\theta = 2 \\arccos(\\frac{r^2+x^2-b^2}{2 r x})$ in the circular region.",
null,
"Assuming $x>b$, the integral over the region should be: $$\\int_{x-b}^{x+b} r f(r) 2\\cos^{-1}[\\frac{r^2+x^2-b^2}{2 r x} ] dr$$ Or, $$\\int_{x-b}^{x+b} r (l-r) 2\\cos^{-1}[\\frac{r^2+x^2-b^2}{2 r x} ] dr$$\n\nSince this is a real valued integral with real limits, we should be able to compute the integral. In Mathematica, I have:\n\nIntegrate[\n2 ArcCos[(r^2 + x^2 - b^2)/(2 r x)] r (l - r), {r, x - b, x + b},\nAssumptions -> l > 0 && x > b && b > 0 ]\n\n\nwhich never evaluates!\n\nIf, however, I try to do the indefinite integral:\n\nexpr = Integrate[2 ArcCos[(r^2 + x^2 - b^2)/(2 r x)] r (l - r), r,\nAssumptions -> l > 0 && x > b && b > 0 && x - b <= r <= x + b ]\n\n\nI get the answer\n\n1/18 r ((-9 l + 4 r) x Sqrt[-((\nb^4 + (r^2 - x^2)^2 - 2 b^2 (r^2 + x^2))/(r^2 x^2))] -\n6 r (-3 l + 2 r) ArcCos[(-b^2 + r^2 + x^2)/(\n2 r x)]) - (2 I (b - x)^2 (7 b^2 + x^2) Sqrt[(\nb^2 - r^2 - 2 b x + x^2)/(b - x)^2] Sqrt[(\nb^2 - r^2 + 2 b x + x^2)/(b + x)^2]\nEllipticE[\nI ArcSinh[r Sqrt[-(1/(b + x)^2)]], (b + x)^2/(b - x)^2] +\nb (9 b l Sqrt[-(1/(b + x)^2)]\nSqrt[-b^4 - (r^2 - x^2)^2 + 2 b^2 (r^2 + x^2)]\nArcTan[(b^2 - r^2 + x^2)/\nSqrt[-b^4 - (r^2 - x^2)^2 + 2 b^2 (r^2 + x^2)]] -\n4 I (b - x)^2 (3 b - x) Sqrt[(\nb^2 - r^2 - 2 b x + x^2)/(b - x)^2] Sqrt[(\nb^2 - r^2 + 2 b x + x^2)/(b + x)^2]\nEllipticF[\nI ArcSinh[r Sqrt[-(1/(b + x)^2)]], (b + x)^2/(b -\nx)^2]))/(9 r x Sqrt[-(1/(b + x)^2)]\nSqrt[-((b^4 + (r^2 - x^2)^2 - 2 b^2 (r^2 + x^2))/(r^2 x^2))])\n\n\nwhich looks ugly, but since there should be no discontinuities, I should be able to evaluate the definite integral as\n\n(expr /. {r -> x + b}) - (expr /. {r -> x - b})\n\n\nWhich gives another ugly answer. Now, if I try to evaluate for some numerical values:\n\nN[((expr /. {r -> x + b}) - (expr /. {r -> x - b})) /. {x -> 5,\nb -> 1, l -> 10}]\n\n\nI get errors such as\n\nPower::infy: Infinite expression 1/Sqrt encountered.\n\n\nQuestions:\n\n1. Why can't we evaluate the definite integral directly?\n2. Since the integrand is real and finite, why does the integral give infinity expressions?\n• FWIW: it is my opinion that Mathematica's handling of elliptic integrals is very far from optimal. More often than not, I've managed to produce better results from manual evaluation than from letting Mathematica lose at my integral. – J. M.'s torpor Jul 5 '12 at 2:11\n• Keep in mind that ReplaceAll is not a mathematical operation, but just pure term replacement. To see the difference try Limit[ArcTan[(b^2-r^2+x^2)/Sqrt[-b^4-(r^2-x^2)^2+2b^2(r^2+x^2)]],r->x-b] and Simplify[(b^2-r^2+x^2)/Sqrt[-b^4-(r^2-x^2)^2+2b^2(r^2+x^2)]]/.r->x-b] (where the argument is the term pointed out by @SjoerdC.deVries below). I think (but haven't tried) that by taking the limit you will be able to evaluate the resulting expression. However, the approaches outlined by Simon Woods and b.gatessucks below are most probably better. – sebhofer Jul 5 '12 at 8:17\n• You might be able to get a result by using Limit instead of ReplaceAll. But it could be incorrect if there are jump discontinuities in the indefinite integral. This can happen even if integrand is continuous along integration path, reason being integrand may have branch points. To make matters worse, presence or absence of branch cuts may depend on actual values later used for your symbolic parameters. Which quite possibly is why Integrate punted on that definite integral. – Daniel Lichtblau Jul 5 '12 at 13:52\n• @J.M., perhaps that means Mathematica can improve it. Do they take suggestions like this? Does someone from Wolfram read this? – highBandWidth Jul 5 '12 at 17:08\n• Maybe, but bear in mind that what's easily done by hand is not always easy to do for a computer following a strict algorithm, and vice-versa... – J. M.'s torpor Jul 5 '12 at 17:12\n\n## 4 Answers\n\nIf the circle is centered at {x0, y0} and has a radius b then it can be parametrized with\n\n$$x = x0 + b \\cos(\\theta')$$\n\n$$y = y0 + b \\sin(\\theta')$$\n\n$$\\theta' \\in [0, 2 \\pi]$$\n\nThen your integral is\n\n$$int = \\int_{0}^{2 \\pi} (l- \\sqrt{(x0 + b \\cos(\\theta'))^2 + (y0 + b \\sin(\\theta'))^2}) d\\theta '$$\n\nint = Integrate[(l - Sqrt[(x0 + b Cos[tp])^2 + (y0 + b Sin[tp])^2 // Expand]), {tp, 0, 2 \\[Pi]}, Assumptions -> {x0 > 0, y0 > 0, b > 0}]\n\n(* ConditionalExpression[2 l \\[Pi]-4 Sqrt[b^2+x0^2+y0^2-2 b Sqrt[x0^2+y0^2]] EllipticE[-((4 b Sqrt[x0^2+y0^2])/(b^2+x0^2+y0^2-2 b Sqrt[x0^2+y0^2]))],Sqrt[x0^2+y0^2]!=b] *)\n\n\nThe simplest check seems ok :\n\nint /. {x0 -> 0, y0 -> 0}\n\n(* ConditionalExpression[-2 Sqrt[b^2] \\[Pi] + 2 l \\[Pi], b != 0] *)\n\n• You have integrated along the circle boundary, rather than the circular disk. Perhaps we need a second integral Integrate[ . , {r,0,b}]? – highBandWidth Jul 5 '12 at 17:06\n• @highBandWidth Added another answer, if that is what you need I'll delete this one. – b.gates.you.know.what Jul 5 '12 at 20:03\n\nMy version of Mathematica has no problems calculating analytically the integral\n\n$\\int_{x-b}^{b+x} 2 r (l-r) \\cos ^{-1}\\left(\\frac{s^2+x^2-b^2}{2 s x}\\right) \\, dr$\n\nThe result looks nice (IMHO) rather than \"ugly\".\n\n$\\frac{1}{9} b^3 \\left(9 L \\pi -2 (2+z) (8+z (2+z)) \\text{EllipticE}\\left[\\frac{4 (1+z)}{(2+z)^2}\\right]+2 z (2+z)^2 \\text{EllipticK}\\left[-\\frac{4 (1+z)}{z^2}\\right]\\right)/.\\left\\{z\\to \\frac{x}{b},L\\to \\frac{l}{b}\\right\\}$\n\n$Version \"8.0 for Microsoft Windows (64-bit) (October 7, 2011)\" The idea is to remove as many symbols from the expression as possile by making it dimensionless, i.e. letting s -> r/b, z -> x/b-1, L-> l/b, the transformed integral is easily and quickly calculated by MMA: g[z_, L_] = b^3 Integrate[ 2 ArcCos[(s^2 + z (2 + z))/(2 s (1 + z))] s (L - s), {s, z, z + 2}, Assumptions -> {z > 0, L > 0}] (* 1/9 b^3 (9 L \\[Pi] - 2 (2 + z) (8 + z (2 + z)) EllipticE[(4 (1 + z))/(2 + z)^2] + 2 z (2 + z)^2 EllipticK[-((4 (1 + z))/z^2)]) *) Checking with NIntegrate shows that the result is correct. By the way, the g as a function of z looks very simple. But observe that the two elliptic integrals don't. Obviously, their non-trivial parts cancel in a \"peaceful\" manner. Best regards, Wolfgang • Your result may be further simplified to b^2*l*Pi - (4/ 9)*(x*(7*b^2 + x^2)* EllipticE[b^2/x^2] + ((4*b^4 - (b^2 + x^2)^2)/x)* EllipticK[b^2/x^2]) – Andreas Dec 2 '20 at 21:23 • @ Andreas There seems to be something missing in your comment after the minus sign. – Dr. Wolfgang Hintze Dec 4 '20 at 10:05 • The OPs integral:b = 1.7; x = 2; l = 2.7; NIntegrate[ r*(l - r)*2*ArcCos[(r^2 + x^2 - b^2)/(2*r*x)], {r, x - b, x + b}] and the simplified result isb^2*l*Pi - (4/ 9)*(x*(7*b^2 + x^2)* EllipticE[b^2/x^2] + ((4*b^4 - (b^2 + x^2)^2)/x)* EllipticK[b^2/x^2]) /. {b -> 1.7, x -> 2, l -> 2.7}. Both agree. I think the space is an artifact of the comment function – Andreas Dec 4 '20 at 14:29 I don't have an answer for your questions; below is just a way to calculate the integral. You want to integrate$f(r)=l-r$over the disk of radius$b$centered at$\\{x_0,y_0\\}$. If you imagine a line from the origin through the center of the disk then, since$f(r)$is symmetric about that line, we can consider the top half-disk only and multiply the corresponding integral by$2$. The top half-disk is described by$\\{x = x0 + r' \\cos(\\theta'), y = y0 + r' \\sin(\\theta ')\\}$(thanks to Heike who pointed this out to me) where $$r' \\in \\left[0, b\\right]$$ $$\\theta' \\in [\\theta_0, \\theta_0 + \\pi]$$ $$\\tan(\\theta_0) = \\frac{y0}{x0}$$",
null,
"Your integral is then $$\\int_{Disc(\\{ x0, y0\\}, b)} d\\theta \\ dr \\ r \\ f(r) = 2 \\int_{\\theta_0}^{\\theta_0+\\pi} d\\theta' \\ \\int_{0}^{b} dr' r |J| (l - r) \\equiv one - two$$ where $$J = \\frac{1}{r} (r' + x0 \\cos(\\theta') + y0 \\sin(\\theta'))$$ $$r = \\sqrt{(x0 + r' \\cos(\\theta'))^2 + (y0 + r' \\sin(\\theta '))^2}$$ th0 = Arctan[x0, y0]; one = 2 Integrate[ l (rp + x0 Cos[tp] + y0 Sin[tp]), {tp, th0, th0 + \\[Pi]}, {rp, 0, b}] /. {y0 Cos[th0] -> x0 Sin[th0]} (* b^2 l \\[Pi] *) aux = Integrate[(1 + \\[Beta]^2 + 2 \\[Beta] Cos[tp - \\[Alpha]])^(3/2), tp, Assumptions -> \\[Beta] \\[Element] Reals] auxTwo[\\[Alpha]_, \\[Beta]_] = Simplify[(aux /. (tp -> \\[Alpha] + \\[Pi])) - (aux /. (tp -> \\[Alpha])), Assumptions -> {\\[Alpha] \\[Element] Reals, \\[Beta] \\[Element] Reals}] two = Simplify[2/3 ((x0^2 + y0^2)^(3/2) (auxTwo[th0, b/Sqrt[x0^2 + y0^2]]) - \\[Pi] (x0^2 + y0^2)^(3/2)), Assumptions -> {b > 0, x0 \\[Element] Reals, y0 \\[Element] Reals}] result = one - two /. {y0 Cos[Arctan[x0, y0]] -> x0 Sin[Arctan[x0, y0]]} (* b^2 l \\[Pi] - 2/9 (x0^2 + y0^2)^(3/2) (-3 \\[Pi] + 2 (1 + b/Sqrt[x0^2 + y0^2]) (4 (1 + b^2/(x0^2 + y0^2)) EllipticE[( 4 b Sqrt[x0^2 + y0^2])/(b + Sqrt[x0^2 + y0^2])^2] - (-1 + b/ Sqrt[x0^2 + y0^2])^2 EllipticK[( 4 b Sqrt[x0^2 + y0^2])/(b + Sqrt[x0^2 + y0^2])^2])) *) One simple check is that it depends only on x0^2 + y0^2 as it should by symmetry. Using this we can simplify the result; with x0^2 + y0^2 -> r0^2 : Simplify[result /. {x0^2 + y0^2 -> r0^2}, Assumptions -> {r0 > 0}] (* b^2 l \\[Pi]-2/9 r0^3 (-3 \\[Pi]+(2 (b+r0) (4 (b^2+r0^2) EllipticE[(4 b r0)/(b+r0)^2]- (b-r0)^2 EllipticK[(4 b r0)/(b+r0)^2]))/r0^3) *) Another simple check is to compare to the case when the circle is centered at the origin : Integrate[r (l - r), {r, 0, b}, {\\[Theta], 0, 2 \\[Pi]}] - Limit[Limit[result, y0 -> 0], x0 -> 0] (* 0 *) • Your domain of integration doesn't actually describe a circle. Try for example With[{x0 = 5, y0 = 0, b = 1}, ParametricPlot[{r Cos[th], r Sin[th]}, {r, Sqrt[x0^2 + y0^2] - b, Sqrt[x0^2 + y0^2] + b}, {th, 0, ArcSin[b/Sqrt[x0^2 + y0^2]]}]] to see what it looks like. To describe the circle, the bounds for r would have to depend on theta (or vice versa if you swap the order of integration). – Heike Jul 6 '12 at 7:09 • @Heike You are right, I messed up my notation. Will fix it soon. Many thanks for checking. – b.gates.you.know.what Jul 6 '12 at 8:24 If I do a FullSimplify on the output of the indefinite Integrate I get this:",
null,
"Please examine the part of this expression marked in red and you'll see why the substitution of r -> x + b and r -> x - b will cause some problems. • Thanks for the insight! Now if we go back to the problem however, the integrand is always finite, even if$r=x+b$or$r=x-b\\$. So shouldn't the integral also be finite? I think the terms in red actually cancel with the other such terms in the limit. How should this be evaluated? – highBandWidth Jul 4 '12 at 17:48\n• @highBandWidth Well Limit[(b^2 - r^2 + x^2)/( Sqrt[b - r - x] Sqrt[(-b - r + x) (b - r + x) (b + r + x)]), r -> x + b, Assumptions -> {x > b && b > 0}] gives me DirectedInfinity[I]. r -> x - b yields -Infinity. – Sjoerd C. de Vries Jul 4 '12 at 18:55\n• @SjoerdC.deVries But of course the limit of the ArcTan does exist. So your comment may be misleading... – sebhofer Jul 5 '12 at 8:08"
] |
[
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"https://i.stack.imgur.com/Uf7RR.png",
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"https://i.stack.imgur.com/XnzQ6.png",
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"https://i.stack.imgur.com/zUWPD.png",
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https://www.geeksforgeeks.org/php-dechex-function/?ref=lbp
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[
"Related Articles\n\n# PHP | dechex( ) Function\n\n• Last Updated : 09 Mar, 2018\n\nThe dechex() is a built-in function in PHP and is used to convert a given decimal number to an equivalent hexadecimal number. The word ‘dechex’ in the name of function stands for decimal to hexadecimal. The dechex() function works only with unsigned numbers. If the argument passed to it is negative then it will treat it as an unsigned number.\nThe largest number that can be converted is 4294967295 in decimal resulting to “ffffffff”.\n\nSyntax:\n\n`string dechex(\\$value)`\n\nParameters: This function accepts a single parameter \\$value. It is the decimal number you want to convert in hexadecimal representation.\n\nReturn Value: It returns the hexadecimal string representation of number passed to it as argument.\n\nExamples:\n\n```Input : dechex(10)\nOutput : a\n\nInput : dechex(47)\nOutput : 2f\n\nInput : dechex(4294967295)\nOutput : ffffffff\n```\n\nBelow programs illustrate dechex() function in PHP:\n\n• Passing 10 as a parameter:\n ` `\n\nOutput:\n\n`a`\n• Passing 47 as a parameter:\n ` `\n\nOutput:\n\n`2f`\n• When the largest possible decimal number is passed as a parameter:\n ` `\n\nOutput:\n\n`ffffffff`\n\nMy Personal Notes arrow_drop_up"
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[
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https://kmath.cn/math/detail.aspx?id=5874
|
[
"",
null,
"$\\text{A.}$ 匀强磁场的方向垂直导轨平面向下 $\\text{B.}$ $M N$ 刚开始运动时的加速度 $\\frac{B\\left(E_1+E_2\\right) L}{m R}$ $\\text{C.}$ $M N$ 离开导轨后的最大速度为 $\\frac{B\\left(E_1-E_2\\right) L C}{m}$ $\\text{D.}$ $M N$ 离开导轨后的最大速度为 $\\frac{B\\left(E_1+E_2\\right) L C}{m}$"
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[
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"https://kmath.cn/uploads/2023-04/63ffc2.jpg",
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https://www.sparknotes.com/cs/searching/linearsearch/problems_2/
|
[
"Problem : You are a teacher for a class of high school computer science students and want to praise those students who do well in the class. As such, you need to find out who they are. Given an array of n doubles where each value represents a student's grade, write a function to find the highest grade and return the index it sits at.\n\nint find_highest_grade(int arr[], int n) { int i, largest = 0; for(i=1; i<n; i++) { if (arr[i] > arr[largest]) largest=i; } return largest; }\n\nProblem : Given an array of n strings, write a function that returns the first string whose length is greater than 15 characters.\n\nchar *find_big_string(char *arr[], int n) { int i; for(i=0; i<n; i++) { if (strlen(arr[i]) > 15) return arr[i]; } return NULL; }\n\nProblem : A friend tells you that he has come up with a function that implements linear search for arrays in O(logn) time. Do you congratulate him, or call him a liar? Why?\n\nYou call him a liar. Linear search requires you to look at, on average, half the elements in the list. Therefore it is O(n) by definition and can't be done in O(logn) time.\n\nProblem : Write a function that takes an array of n integers and returns the number of integers that are a power of two. Challenge: determining whether a number is a power of two can be done in a single line of code.\n\nint num_power_two(int arr[], int n) { int i, num_powers = 0; for(i=0; i<n; i) { /* Use bitwise operators to check for powers of 2. */ if (!(arr[i] & (arr[i] - 1))) num_powers; } return num_powers; }\n\nProblem : Challenge (this is tricky): Write a function that takes an array of integers (and its length) and returns the largest consecutive sum found in the array. In other words, if the array were: -1 10 -1 11 100 -1000 20 It would return 120 (10 + -1 + 11 + 100).\n\nvoid find_big_seq(int numbers[], int n) { int maxsofar = numbers; int maxendhere = numbers; int i, a, b; for (i = 1; i < n; i++) { a = maxendhere + numbers[i]; b = numbers[i]; maxendhere = a > b ? a : b; if (maxendhere > maxsofar) maxsofar = maxendhere; } return maxsofar; }\n\nProblem : You are given a two-dimensional array of integers: int arr; Write a function that returns the largest integer in the array.\n\nint find_large_int(int arr) { int i,j,largest_x=0, largest_y=0; for(i=0; i<100; i) { for(j=0; j<50; j) { if (arr[i][j] > arr[largest_x][largest_y]) { largest_x = i; largest_y = j; } } } return arr[largest_x][largest_y]; }\n\nProblem : Linear search uses an exhaustive method of checking each element in the array against a key value. When a match is found, the search halts. Will sorting the array before using the linear search have any effect on its efficiency?\n\nNo.\n\nProblem : In a best case situation, the element will be found with the fewest number of comparisons. Where in the list would the key element be located?\n\nIt will be located at the beginning of the list."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8838712,"math_prob":0.9688022,"size":2131,"snap":"2022-40-2023-06","text_gpt3_token_len":471,"char_repetition_ratio":0.12458862,"word_repetition_ratio":0.11825193,"special_character_ratio":0.2322853,"punctuation_ratio":0.101149425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99665636,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T05:05:49Z\",\"WARC-Record-ID\":\"<urn:uuid:2d8e8bed-cbe8-4564-bbfc-cb662f32b09f>\",\"Content-Length\":\"220677\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0386d3da-ba2b-4e35-893f-2051b1f0f28a>\",\"WARC-Concurrent-To\":\"<urn:uuid:a35483d5-d70f-4de4-888a-f67d82726879>\",\"WARC-IP-Address\":\"23.54.211.37\",\"WARC-Target-URI\":\"https://www.sparknotes.com/cs/searching/linearsearch/problems_2/\",\"WARC-Payload-Digest\":\"sha1:EP5VA4EVRQQCRQUI347BGT74JF2727HL\",\"WARC-Block-Digest\":\"sha1:JF5SSOWJFXM76RO72L6UXPZ5X673242S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500215.91_warc_CC-MAIN-20230205032040-20230205062040-00169.warc.gz\"}"}
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https://www.physicsforums.com/threads/what-is-an-integral.712501/
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[
"# What is an integral\n\nI was explaining to a friend about some differences between the Riemann integral and Lebesgue integral. He asked what an integral is and the best I could come up with was \"it defines a notion of area\" but I don't think that's good enough. There's Riemann integral, Lebesgue, Darboux, Haar, line, surface, double, multiple, etc. So in its most general case, what is an integral? I haven't been able to find widely accepted axiomatic treatment of integration. Could I just say that the integral of all functions is 0 and call it the \"johnqwertyful integral\"? Or call any object or function I can think of an integral?\n\nSimon Bridge\nHomework Helper\nAn integral is a fancy way of doing a sum.\n\nIt does not \"define a notion of area\" - it is a way of calculating areas by dividing them up into arbitrarily small bits.\n\npwsnafu\nI haven't been able to find widely accepted axiomatic treatment of integration.\n\nSee the Daniell integral.\n\nAn integral is a fancy way of doing a sum.\n\nIt does not \"define a notion of area\" - it is a way of calculating areas by dividing them up into arbitrarily small bits.\n\nMy real analysis professor said integrals define area. If not, then how IS area defined? What if two different integrals give you two different areas, which one is the \"true\" area?\n\nAlso, \"fancy way of doing sum\" is incredibly handwavey. Can I call any fancy way of doing sum an integral? Johnqwertyful integral: add up the natural numbers 1 through 100 then subtract 4. That's an integral because it's a fancy way of doing a sum.\n\nLast edited:\nSee the Daniell integral.\n\nInteresting. I like the axiomatic treatment of it. How widely are these axioms accepted/used? Are there any interesting examples of integrals that follow these axioms?\n\nWannabeNewton\nAn integral is a fancy way of doing a sum.\n\nIt does not \"define a notion of area\" - it is a way of calculating areas by dividing them up into arbitrarily small bits.\n\nActually a sum is a special type of integral.\n\nmathwonk\nHomework Helper\n2020 Award\nan integral is essentially a way to assign a number m(f) to each function f in some linear space of functions which is linear, i.e. such that m(af+bg) = a.m(f) + b.m(g), and non negative, i.e. such that m(f) ≥ 0 whenever f ≥ 0. I.e. it is a non negative linear functional defined on a linear space of functions.\n\nDifferent ways of assigning this number m(f) to f are valid for different spaces of functions. E.g. the Riemann and Lebesgue definitions of the number m(f) agree for those functions where both assignments make sense, but the Lebesgue definition applies to a larger space of functions.\n\nIf you want to pin it down more you can assume, for functions defined on subsets of the real line, that if f is a function equal to 1 on any interval [a,b] and equal to zero elsewhere, that m(f) = b-a. Just these simple requirements already force the definition of the integral for Riemann integrable functions, i.e. those with sufficiently good approximations from above and below by step functions.\n\nIf we add a condition on the behavior of the number m(f) under taking pointwise limits of functions, it seems we can also force the definition of the number m(f) for all Lebesgue integrable functions.\n\nLast edited:\nSimon Bridge\nHomework Helper\nMy real analysis professor said integrals define area. If not, then how IS area defined? What if two different integrals give you two different areas, which one is the \"true\" area?\nIf two different integrals give you different areas - then how can an integral define \"area\"?\n(Your real-analysis prof was probably being glib.)\n\nNotice that the concept of area in geometry predates the invention of integral calculus and can be taught to students before they know about integration. OTOH: See WannabeNewton's comment below - maybe grade-schoolers have \"actually\" been doing a form of integration when they thought they were \"merely\" finding areas?\n\nAlso, \"fancy way of doing sum\" is incredibly handwavey. Can I call any fancy way of doing sum an integral? Johnqwertyful integral: add up the natural numbers 1 through 100 then subtract 4. That's an integral because it's a fancy way of doing a sum.\nNaturally. There are lots of different ways to do sums, integration is one of them, but not all ways of doing sums are integration. Integration is closer to grade-school summation than, say, multiplication.\n\nI realize you want some sort of axiomatic approach - you have been given one.\n\nActually a sum is a special type of integral.\n\nChickens and eggs.\n\nIf two different integrals give you different areas - then how can an integral define \"area\"?\n(Your real-analysis prof was probably being glib.)\nYou said that integrals calculate area, but what if two different integrals give you two different areas? You're saying that there is one true area that every integral should give you. If that is true, why are their different integrals?\n\nIt's the same idea as different metrics. A metric on a set doesn't calculate distance, it defines distance. And two different metrics will give you two different distances.\n\nNotice that the concept of area in geometry predates the invention of integral calculus and can be taught to students before they know about integration. OTOH: See WannabeNewton's comment below - maybe grade-schoolers have \"actually\" been doing a form of integration when they thought they were \"merely\" finding areas?\n\nAnd I'm sure grade-schoolers have been doing addition and multiplication without the slightest idea of what rings, fields, groups, etc. are. Newton did calculus incredibly hand wavey, and when the founders of analysis came through, they formalized certain concepts. Everything gets done hand wavey at first, then gets formalized later.\n\nNaturally. There are lots of different ways to do sums, integration is one of them, but not all ways of doing sums are integration. Integration is closer to grade-school summation than, say, multiplication.\n\nI realize you want some sort of axiomatic approach - you have been given one.\nMhmm, and I appreciate it.\n\nSimon Bridge\nHomework Helper\nYou said that integrals calculate area, but what if two different integrals give you two different areas?\nNo, I didn't.\nI said that integrals are a way of calculating areas.\n\nYou're saying that there is one true area that every integral should give you. If that is true, why are their different integrals?\nBecause there are different size areas and different ways of applying the concept of area.\n\nIt's the same idea as different metrics. A metric on a set doesn't calculate distance, it defines distance. And two different metrics will give you two different distances.\nIf you walk the distance - does your experience depend on which metric you applied? i.e. what is the empirical experience of the metric?\n\nThere will be cases where the different metrics change the result of an experiment - what does that mean?\n\nLook - I get it. We are approaching this from different world-views ... it is entirely a philosophical difference here.\n\natyy\n\nIs there such a thing as a sum of an unmeasuable set, and would that constitute an example where the sum is not a special instance of integration?\n\nLast edited:\nSimon Bridge\nHomework Helper\nI think that shows that there exist sums where integration is not a useful strategy...\n\nJust like there are sums where multiplication is not a useful strategy.\n\nIs there such a thing as a sum of an unmeasuable set, and would that constitute an example where the sum is not a special instance of integration?\nNot with respect to the counting measure. The measurable sets are the power set.\n\nchiro"
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https://dsp.stackexchange.com/questions/15680/how-to-interpret-results-of-fourier-transform-by-using-python
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[
"# How to interpret results of Fourier transform by using Python\n\nConnected with this topic , I would like to have some clarifications about the Fourier Transform and the Python tools that make it. Having my data set whose plot is showed in the above link, my array of data is stored in rate. Then, if I check the result from the Fourier transform:\n\nyfft = np.fft.rfft(rate)\ny_smooth = np.fft.irfft(yfft)\nprint y_smooth\n\n\nI get:\n\n[ 3.20000028e+01+0.j -7.92633533e+00-7.75032655j\n-5.34245688e+00-3.82390808j -2.39880935e+00-0.60037314j\n-8.71249166e-01-0.04362804j -8.52507194e-02+1.09973189j\n-1.58332047e-01+1.15256295j -3.38909974e-01+0.96327143j\n5.51323015e-02+0.43936749j 3.81205886e-01-0.02489853j\n2.99110153e-01-0.26319101j 3.28283652e-01+0.11668814j\n1.08849205e-02-0.11317937j 3.79952011e-01+0.22278952j\n7.37720998e-02-0.26810083j -2.14247586e-01-0.25950865j\n-1.32242152e-01+0.j ]\n\n\nSo, I am quite confused about this result, and I would need some tips how to interpret these numbers.\n\n1) The array contains $17$ complex number : does it mean that my signal can be approximated by a superimposition of $17$ harmonics?\n\n2) Are these numbers the power of each harmonic, or what?\n\n3) are they ordered by the frequency order, or by power? Let's say I want to put to zero the higher orders, which ones should I put to zero?\n\n## 2 Answers\n\nYour data consist of 32 real values. Its discrete Fourier transform (DFT), which you computed using an FFT (Fast Fourier Transform) algorithm also has 32 (complex) values. However, since your data are real, the Fourier transform is symmetric and the last 15 values are redundant (i.e. can be computed from the other values). This is why you see 17 remaining non-redundant values. These values are no approximation but they are an exact representation of your data, just in a different form.\n\nThe DFT is given by\n\n$$X_k=\\sum_{n=0}^{N-1}x_ne^{-jnk2\\pi/N},\\quad k=0,1,\\ldots,N-1$$\n\nwhere $N$ is the number of data points (32 in your case). If $x_n$ is real-valued you can show that\n\n$$X_k=X^*_{N-k},\\quad k=1,\\ldots, N-1$$\n\nwhich means that all information about your data is contained in the values $X_k$ for $k=0,1,\\ldots,N/2$. Note that $X_0$ and $X_{N/2}$ are real-valued. So we have $N/2+1$ non-redundant complex number, two of which are real-valued. So you have $N/2-1$ real and imaginary parts, and 2 real numbers, which in total are exactly $N$ real numbers. No surprise that they are sufficient (and necessary) to represent your $N$ data points.\n\nIn order to see how you can interpret the values $X_k$ you need to look at the inverse transform\n\n$$x_n=\\frac{1}{N}\\sum_{n=0}^{N-1}X_ke^{jnk2\\pi/N}$$\n\nSo the values $X_k$ are the complex amplitudes of the complex exponentials representing your signal. They are ordered according to frequency. $X_0$ represents the DC value, and $X_{N/2}$ (assuming that $N$ is even) represents the value at half the sampling rate.\n\n• Thanks! Put in this way, it seems that I need to do as many DFT as much are the component of my signal, instead of just one DFT that gives back all the components. Is it correct to say the amplitudes are the square root of the Power of each considered harmonic? – Py-ser Apr 16 '14 at 7:24\n• @Py-ser The amplitudes are complex, so you could say that the squared magnitude of the amplitudes are equivalent to the energy in the $k^{th}$ frequency component. I do not understand the first part of your comment. – Matt L. Apr 16 '14 at 8:24\n• Forget about the first part. Why do you use the term \"energy\" instead of power? – Py-ser Apr 16 '14 at 8:27\n• @Py-ser $\\sum_n|x_n|^2=\\frac{1}{N}\\sum_k|X_k|^2$ is the total energy of the signal $x_n$ (look up Parseval's theorem). So $|X_k|^2$ can be considered the energy in bin $k$. – Matt L. Apr 16 '14 at 10:21\n• ... I should have said proportional to the energy, because I forgot the factor $1/N$. – Matt L. Apr 16 '14 at 13:25\n\nThe array contains 17 complex number : does it mean that my signal can be approximated by a superimposition of 17 harmonics?\n\n16.\n\nThe first component is the DC component.\n\nNote that this is not specific to your signal. Any real sequence of length 32 can be decomposed as 1 real DC component, 16 complex amplitudes, and 1 real amplitude (32 real number ins, 32 real numbers out).\n\nAre these numbers the power of each harmonic, or what?\n\nThey are the complex amplitudes.\n\nare they ordered by the frequency order, or by power? Let's say I want to put to zero the higher orders\n\nThey are ordered by frequency.\n\nwhich ones should I put to zero?\n\nThe last ones, but why would you want to do that?\n\n• Thanks. Because I want to smooth a signal leaving off the higher orders. – Py-ser Apr 16 '14 at 7:26\n• Then you are probably doing it wrong: dsp.stackexchange.com/questions/6220/… . Design a filter with all the properties you want, and apply it to the time-domain data. – pichenettes Apr 16 '14 at 7:47\n• I think in my case (see the linked discussion in the main topic) it is fine. At least, people more experienced than me do it in such cases. If you can, please, take a look at it and tell me if you still think it is not the case to proceed this way. I will not use it anyway (since I will use the Savitzky-Golay), but this post serves me to understand the principles, so it would be great to understand if the next time I don't even have to think about FFT smoothing. – Py-ser Apr 16 '14 at 8:08"
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https://coursesxrem.web.app/tantum610quv/need-help-with-a-math-problem-2701.html
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[
"# Need help with a math problem\n\nMathway | Algebra Problem Solver For a new problem, you will need to begin a new tutoring session. You can contact support with any questions regarding your current subscription. You will be able to enter math problems once our session is over.\n\nWord Problems Calculator - Math Celebrity If you cannot find what you need, post your word problem in our calculator forum ... Solves various basic math and algebra word problems with numbers Features: ... 15 Techniques to Solve Math Problems Faster | Prodigy For more fresh approaches to teaching math in your classroom, consider treating your students to a range of fun math activities. Final Thoughts About these Ways to Solve Math Problems Faster. Showing these 15 techniques to students can give them the confidence to tackle tough questions.\n\n## Need Help With Math ... Interconnected ideas that would reinforce the theme and opens up a new vision of the problem. But the study is not the meaning of life. Time ...\n\nWord problems in mathematics often pose a challenge because they require that students read and comprehend the text of the problem, identify the question that needs to be answered, and finally create and solve a numerical equation. Many ELLs may have difficulty reading and understanding the written ... Algebra Homework Help, Algebra Solvers, Free Math Tutors Pre-Algebra, Algebra I, Algebra II, Geometry: homework help by free math tutors, solvers, lessons.Each section has solvers (calculators), lessons, and a place where you can submit your problem to our free math tutors. Algebra I | Math | Khan Academy\n\n### I Need Help Math Problems - Your Personal Essay Writing Website With Professional Team Of Writers. We Do An Excellent Job Of Making Our Customers Happy.\n\nSecret to Solving Math Word Problems. Hint: It's Not about ... You don't need to be an artist. For many math students, there's truth to the saying \"a picture is worth a thousand words\". Pictures help most students visualize the problem. It clarifies what they are given and what they need to find. This strategy is particularly good for students with strong visual memory. Solving Math Problems & Assignments - Expert Math Help Email Based Math Analysis Homework Help. Experienced tutors are aware of which kind of assistance in math analysis you might need, and the best way for addressing your problem. There are some online math experts that offer email based math analysis homework assistance and you may just email them with your problems to get a helping hand. Need Help With Math - from \\$ 6 - s3.amazonaws.com Need Help With Math ... Interconnected ideas that would reinforce the theme and opens up a new vision of the problem. But the study is not the meaning of life. Time ... Illustration: The Key to Helping Your Child Solve Math Word ...\n\n### Free Statistics Help - StatisticsHelp\n\nFor a new problem, you will need to begin a new ... I am only able to help with one math problem per ... from the menu located in the upper left corner of the Mathway ... Need help with a GRE Problem? May 19, 2019 (Cool math ... The trick is to minimize the perimeter as much as possible Math Homework Help - Answers to Math Problems - Hotmath Math homework help. Hotmath explains math textbook homework problems with step-by-step math answers for algebra, geometry, and calculus. Online tutoring available for ... Math Help: Do My Math Homework for Me | Homeworkforschool"
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https://www.numbersaplenty.com/7054
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[
"Search a number\nBaseRepresentation\nbin1101110001110\n3100200021\n41232032\n5211204\n652354\n726365\noct15616\n910607\n107054\n115333\n1240ba\n133298\n1427dc\n152154\nhex1b8e\n\n7054 has 4 divisors (see below), whose sum is σ = 10584. Its totient is φ = 3526.\n\nThe previous prime is 7043. The next prime is 7057. The reversal of 7054 is 4507.\n\nIt is a semiprime because it is the product of two primes.\n\nIt is an alternating number because its digits alternate between odd and even.\n\nIt is a nialpdrome in base 11.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (7057) by changing a digit.\n\n7054 is an untouchable number, because it is not equal to the sum of proper divisors of any number.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 1762 + ... + 1765.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (2646).\n\n27054 is an apocalyptic number.\n\n7054 is a deficient number, since it is larger than the sum of its proper divisors (3530).\n\n7054 is a wasteful number, since it uses less digits than its factorization.\n\n7054 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 3529.\n\nThe product of its (nonzero) digits is 140, while the sum is 16.\n\nThe square root of 7054 is about 83.9880943944. The cubic root of 7054 is about 19.1783755383.\n\nThe spelling of 7054 in words is \"seven thousand, fifty-four\".\n\nDivisors: 1 2 3527 7054"
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https://blog.i-ll.cc/archives/64
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[
"Description\n\nAn important topic nowadays in computer science is cryptography. Some people even think that\ncryptography is the only important field in computer science, and that life would not matter at all\nwithout cryptography.\nAlvaro is one of such persons, and is designing a set of cryptographic procedures for cooking paella. ´\nSome of the cryptographic algorithms he is implementing make use of big prime numbers. However,\nchecking if a big number is prime is not so easy. An exhaustive approach can require the division of the\nnumber by all the prime numbers smaller or equal than its square root. For big numbers, the amount\nof time and storage needed for such operations would certainly ruin the paella.\nHowever, some probabilistic tests exist that offer high confidence at low cost. One of them is the\nFermat test.\nLet a be a random number between 2 and n−1 (being n the number whose primality we are testing).\nThen, n is probably prime if the following equation holds:\na^n mod n = a\nIf a number passes the Fermat test several times then it is prime with a high probability.\nUnfortunately, there are bad news. Some numbers that are not prime still pass the Fermat test\nwith every number smaller than themselves. These numbers are called Carmichael numbers.\nIn this problem you are asked to write a program to test if a given number is a Carmichael number.\nHopefully, the teams that fulfill the task will one day be able to taste a delicious portion of encrypted\npaella. As a side note, we need to mention that, according to Alvaro, the main advantage of encrypted ´\npaella over conventional paella is that nobody but you knows what you are eating.\n\nInput\n\nThe input will consist of a series of lines, each containing a small positive number n (2\nA number n = 0 will mark the end of the input, and must not be processed.\n\nOutput\n\nFor each number in the input, you have to print if it is a Carmichael number or not, as shown in the\nsample output.\n\nSample Input\n\n1729\n17\n561\n1109\n431\n0\n\nSample Output\n\nThe number 1729 is a Carmichael number.\n17 is normal.\nThe number 561 is a Carmichael number.\n1109 is normal.\n\nSolution\n\nan mod n=(a mod n)n mod n=a\n\nCode\n\n#include\n#include\n#define MAXN 65000\ntypedef long long ll;\nll mod, ans;\nint n;\n\nint isprime(int n)\n{\nfor (int i = 2; i*i if (n%i == 0)\nreturn 0;\nreturn 1;\n}\n\nll power(ll a, ll n, ll m)\n{\nll ans = 1;\nwhile (n)\n{\nif (n & 1) ans = ans*a%m;\na = a * a % m;\nn >>= 1;\n}\nreturn ans;\n}\n\nint main()\n{\n\nwhile (scanf(\"%d\", &n) && n) {\nint flag = 1;\nif (isprime(n)) flag = 0;\nelse\n{\nfor (ll i = 2; i if (power(i, n, n) != i)\n{\nflag = 0;\nbreak;\n}\n}\nif (flag) printf(\"The number %d is a Carmichael number.\\n\",n);\nelse printf(\"%d is normal.\\n\",n);\n\n}\nreturn 0;\n\n}"
] |
[
null
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8087952,"math_prob":0.9945431,"size":2936,"snap":"2021-31-2021-39","text_gpt3_token_len":915,"char_repetition_ratio":0.12039564,"word_repetition_ratio":0.0,"special_character_ratio":0.26873296,"punctuation_ratio":0.120065786,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924196,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T09:33:49Z\",\"WARC-Record-ID\":\"<urn:uuid:ee9bfbdc-daa8-486c-8a6b-437f8db74bbb>\",\"Content-Length\":\"59392\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9960bf5-68bc-48fa-945a-c8957766f508>\",\"WARC-Concurrent-To\":\"<urn:uuid:498c554e-77d0-44a8-8414-b3487e86d54a>\",\"WARC-IP-Address\":\"47.94.205.243\",\"WARC-Target-URI\":\"https://blog.i-ll.cc/archives/64\",\"WARC-Payload-Digest\":\"sha1:ZSV5P64SMWSBBGUD2GGOOHEJFMDCKSMW\",\"WARC-Block-Digest\":\"sha1:T4WJPB4I473RWCEUJJH4CTE4PE5QKE3D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154457.66_warc_CC-MAIN-20210803092648-20210803122648-00107.warc.gz\"}"}
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https://www.aqua-calc.com/what-is/energy/megawatt-hour
|
[
"# What is a megawatt hour (unit)\n\n## The megawatt hour is a unit of measurement of energy\n\nMegawatt hour (MWh) is a derived unit of energy equal to 3.6 gigajoules (GJ), and expressed as power [megawatt (MW)] multiplied by time [hour (h)].\n\n• What is energyInstant conversionsConversion tables\n• 1 MWh = 3 412 141.63 BTU\n• 1 MWh = 859 845 228 cal\n• 1 MWh = 2.2469432850375×10+28 eV\n• 1 MWh = 2 655 223 730 ft-lb\n• 1 MWh = 0.8598452279 GCal\n• 1 MWh = 2.2469432850375×10+19 GeV\n• 1 MWh = 3.6 GJ\n• 1 MWh = 0.001 GWh\n• 1 MWh = 3 600 000 000 J\n• 1 MWh = 859 845.228 kcal\n• 1 MWh = 2.2469432850375×10+25 keV\n• 1 MWh = 1 000 kWh\n• 1 MWh = 3 600 000 kJ\n• 1 MWh = 859.845228 MCal\n• 1 MWh = 2.2469432850375×10+22 MeV\n• 1 MWh = 3 600 MJ\n• 1 MWh = 3.6×10+15 µJ\n• 1 MWh = 8.59845227859×10+14 mCal\n• 1 MWh = 3 600 000 000 000 mJ\n• 1 MWh = 3.6×10+18 nJ\n• 1 MWh = 22 469 432 850 375 PeV\n• 1 MWh = 3.6×10+21 pJ\n• 1 MWh = 2.2469432850375×10+16 TeV\n• 1 MWh = 0.0036 TJ\n• 1 MWh = 1 000 000 Wh\n\n#### Foods, Nutrients and Calories\n\nBOLOGNA MADE WITH CHICKEN AND PORK, UPC: 073890002683 contain(s) 304 calories per 100 grams or ≈3.527 ounces [ price ]\n\nCRISPY ALL BUTTER COOKIES, UPC: 816059012709 contain(s) 481 calories per 100 grams or ≈3.527 ounces [ price ]\n\n#### Gravels, Substances and Oils\n\nCaribSea, Marine, CORALine, Aruba Puka Shell weighs 1 121.29 kg/m³ (69.99985 lb/ft³) with specific gravity of 1.12129 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nFaserton [Al2O3] weighs 3 965 kg/m³ (247.52686 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]\n\nVolume to weightweight to volume and cost conversions for Refrigerant R-408A, liquid (R408A) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)\n\n#### Weights and Measurements\n\nA zeptometer is a derived metric measurement unit of length\n\nThe surface density of a two-dimensional object, also known as area or areal density, is defined as the mass of the object per unit of area.\n\ntroy/cm³ to dwt/fl.oz conversion table, troy/cm³ to dwt/fl.oz unit converter or convert between all units of density measurement.\n\n#### Calculators\n\nVolume of a rectangular box calculator. Definition and formulas."
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[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.5261634,"math_prob":0.9946615,"size":1570,"snap":"2020-24-2020-29","text_gpt3_token_len":747,"char_repetition_ratio":0.30970627,"word_repetition_ratio":0.07462686,"special_character_ratio":0.4490446,"punctuation_ratio":0.05775076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98680407,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-27T03:35:28Z\",\"WARC-Record-ID\":\"<urn:uuid:f8175831-8612-4275-924f-aec402ba8b57>\",\"Content-Length\":\"24315\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c774dc74-e5d8-4bfd-8069-a9b6c044e8d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d7c5384-b8f1-4125-839d-fa46f257cac8>\",\"WARC-IP-Address\":\"69.46.0.75\",\"WARC-Target-URI\":\"https://www.aqua-calc.com/what-is/energy/megawatt-hour\",\"WARC-Payload-Digest\":\"sha1:AKQTBIA6TCSJFLPPRJ7YWGXQZCRSQR2C\",\"WARC-Block-Digest\":\"sha1:KRTHYQ77EU74GJTQHKHK7ADDLDR2IQAA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347392057.6_warc_CC-MAIN-20200527013445-20200527043445-00591.warc.gz\"}"}
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http://derminelift.info/free-decimal-worksheets/
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[
"# Free Decimal Worksheets\n\nlong division practice worksheets grade free math worksheets for 3rd grade pdf.\n\ndecimal division worksheets collection of free math worksheets for 4th graders to print.\n\nkindergarten adding and subtracting decimals worksheets image free printable decimal worksheets.\n\ndecimals free math worksheets for 7th grade geometry.\n\nmultiplying decimals worksheets grade 7 free math worksheets for 5th grade volume.\n\nfraction decimal worksheet pack classroom caboodle free math worksheets for 3rd graders geometry.\n\nfree printable decimal place value worksheets free printable decimal worksheets.\n\nfree printable reading passages for grade download page free math worksheets decimal multiplication.\n\npercent to decimal worksheet free worksheets percent to decimal free math worksheets for 3rd grade place value.\n\nprintable decimal worksheets math worksheet division grade fractions free printable worksheets decimals to fractions.\n\ndecimal fraction and number of the day worksheets free common core math worksheets for 7th grade.\n\nmath worksheet decimals worksheets grade common core decimal free printable math worksheets for 6th grade exponents.\n\nyear 4 maths worksheets printable free new best decimal free fun math worksheets for 7th grade.\n\nmultiplication with decimals worksheets free image collections free math worksheets for 4th graders to print.\n\ndecimals free math worksheets for 5th grade mean median mode.\n\nmath decimals worksheets free printable grade for rounding free math worksheets for 4th grade.\n\nmultiplying decimals worksheets free printable decimal free math worksheets 7th grade fractions.\n\nfree worksheets library download and print worksheets free on free math worksheets for 4th grade geometry.\n\ndecimal addition hundredths worksheet free math worksheets for 6th grade word problems.\n\naccounting long division with decimals worksheet worksheets free math worksheets decimal multiplication.\n\ndecimal worksheets free decimal place value worksheets 5th grade.\n\neasy decimal worksheets multiplying decimals printable worksheets free math worksheets for 5th grade volume.\n\naddition and subtraction decimals worksheets free the best free printable math worksheets for 6th grade exponents."
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[
null
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https://optimization-online.org/2014/05/4371/
|
[
"# Strong duality in Lasserre’s hierarchy for polynomial optimization\n\nA polynomial optimization problem (POP) consists of minimizing a multivariate real polynomial on a semi-algebraic set \\$K\\$ described by polynomial inequalities and equations. In its full generality it is a non-convex, multi-extremal, difficult global optimization problem. More than an decade ago, J.~B.~Lasserre proposed to solve POPs by a hierarchy of convex semidefinite programming (SDP) relaxations of increasing size. Each problem in the hierarchy has a primal SDP formulation (a relaxation of a moment problem) and a dual SDP formulation (a sum-of-squares representation of a polynomial Lagrangian of the POP). In this note, when the POP feasibility set \\$K\\$ is compact, we show that there is no duality gap between each primal and dual SDP problem in Lasserre's hierarchy, provided a redundant ball constraint is added to the description of set \\$K\\$. Our proof uses elementary results on SDP duality, and it does not assume that \\$K\\$ has an interior point."
] |
[
null
] |
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|
https://pandas.pydata.org/pandas-docs/version/0.14.0/generated/pandas.Index.html
|
[
"# pandas.Index¶\n\nclass pandas.Index\n\nImmutable ndarray implementing an ordered, sliceable set. The basic object storing axis labels for all pandas objects\n\nParameters : data : array-like (1-dimensional) dtype : NumPy dtype (default: object) copy : bool Make a copy of input ndarray name : object Name to be stored in the index tupleize_cols : bool (default: True) When True, attempt to create a MultiIndex if possible\n\nNotes\n\nAn Index instance can only contain hashable objects\n\nAttributes\n\n T Same as self.transpose(), except that self is returned if self.ndim < 2. base Base object if memory is from some other object. ctypes An object to simplify the interaction of the array with the ctypes module. data Python buffer object pointing to the start of the array’s data. flags flat A 1-D iterator over the array. imag The imaginary part of the array. is_monotonic itemsize Length of one array element in bytes. names nbytes Total bytes consumed by the elements of the array. ndim Number of array dimensions. nlevels real The real part of the array. shape Tuple of array dimensions. size Number of elements in the array. strides Tuple of bytes to step in each dimension when traversing an array. values\n asi8 dtype inferred_type is_all_dates is_unique name\n\nMethods\n\n all([axis, out]) Returns True if all elements evaluate to True. any([axis, out]) Returns True if any of the elements of a evaluate to True. append(other) Append a collection of Index options together argmax([axis, out]) Return indices of the maximum values along the given axis. argmin([axis, out]) Return indices of the minimum values along the given axis of a. argpartition(kth[, axis, kind, order]) Returns the indices that would partition this array. argsort(*args, **kwargs) See docstring for ndarray.argsort asof(label) For a sorted index, return the most recent label up to and including the passed label. asof_locs(where, mask) where : array of timestamps astype(dtype) byteswap(inplace) Swap the bytes of the array elements choose(choices[, out, mode]) Use an index array to construct a new array from a set of choices. clip(a_min, a_max[, out]) Return an array whose values are limited to [a_min, a_max]. compress(condition[, axis, out]) Return selected slices of this array along given axis. conj() Complex-conjugate all elements. conjugate() Return the complex conjugate, element-wise. copy([names, name, dtype, deep]) Make a copy of this object. cumprod([axis, dtype, out]) Return the cumulative product of the elements along the given axis. cumsum([axis, dtype, out]) Return the cumulative sum of the elements along the given axis. delete(loc) Make new Index with passed location(-s) deleted diagonal([offset, axis1, axis2]) Return specified diagonals. diff(other) Compute sorted set difference of two Index objects dot(b[, out]) Dot product of two arrays. drop(labels) Make new Index with passed list of labels deleted dump(file) Dump a pickle of the array to the specified file. dumps() Returns the pickle of the array as a string. equals(other) Determines if two Index objects contain the same elements. factorize([sort, na_sentinel]) Encode the object as an enumerated type or categorical variable fill(*args, **kwargs) This method will not function because object is immutable. flatten([order]) Return a copy of the array collapsed into one dimension. format([name, formatter]) Render a string representation of the Index get_duplicates() get_indexer(target[, method, limit]) Compute indexer and mask for new index given the current index. get_indexer_for(target, **kwargs) guaranteed return of an indexer even when non-unique get_indexer_non_unique(target, **kwargs) return an indexer suitable for taking from a non unique index get_level_values(level) Return vector of label values for requested level, equal to the length get_loc(key) Get integer location for requested label get_value(series, key) Fast lookup of value from 1-dimensional ndarray. get_values() getfield(dtype[, offset]) Returns a field of the given array as a certain type. groupby(to_groupby) holds_integer() identical(other) Similar to equals, but check that other comparable attributes are insert(loc, item) Make new Index inserting new item at location intersection(other) Form the intersection of two Index objects. Sortedness of the result is is_(other) More flexible, faster check like is but that works through views is_floating() is_integer() is_lexsorted_for_tuple(tup) is_mixed() is_numeric() is_type_compatible(typ) isin(values) Compute boolean array of whether each index value is found in the item(*args) Copy an element of an array to a standard Python scalar and return it. itemset(*args, **kwargs) This method will not function because object is immutable. join(other[, how, level, return_indexers]) Internal API method. Compute join_index and indexers to conform data map(mapper) max() The maximum value of the object mean([axis, dtype, out]) Returns the average of the array elements along given axis. min() The minimum value of the object newbyteorder([new_order]) Return the array with the same data viewed with a different byte order. nonzero() Return the indices of the elements that are non-zero. nunique() Return count of unique elements in the object. order([return_indexer, ascending]) Return sorted copy of Index partition(kth[, axis, kind, order]) Rearranges the elements in the array in such a way that value of the element in kth position is in the position it would be in a sorted array. prod([axis, dtype, out]) Return the product of the array elements over the given axis ptp([axis, out]) Peak to peak (maximum - minimum) value along a given axis. put(*args, **kwargs) This method will not function because object is immutable. ravel([order]) Return a flattened array. reindex(target[, method, level, limit, ...]) For Index, simply returns the new index and the results of rename(name[, inplace]) Set new names on index. repeat(repeats[, axis]) Repeat elements of an array. reshape(shape[, order]) Returns an array containing the same data with a new shape. resize(new_shape[, refcheck]) Change shape and size of array in-place. round([decimals, out]) Return a with each element rounded to the given number of decimals. searchsorted(v[, side, sorter]) Find indices where elements of v should be inserted in a to maintain order. set_names(names[, inplace]) Set new names on index. set_value(arr, key, value) Fast lookup of value from 1-dimensional ndarray. setfield(val, dtype[, offset]) Put a value into a specified place in a field defined by a data-type. setflags([write, align, uic]) Set array flags WRITEABLE, ALIGNED, and UPDATEIFCOPY, respectively. shift([periods, freq]) Shift Index containing datetime objects by input number of periods and slice_indexer([start, end, step]) For an ordered Index, compute the slice indexer for input labels and slice_locs([start, end]) For an ordered Index, compute the slice locations for input labels sort(*args, **kwargs) squeeze([axis]) Remove single-dimensional entries from the shape of a. std([axis, dtype, out, ddof]) Returns the standard deviation of the array elements along given axis. sum([axis, dtype, out]) Return the sum of the array elements over the given axis. summary([name]) swapaxes(axis1, axis2) Return a view of the array with axis1 and axis2 interchanged. sym_diff(other[, result_name]) Compute the sorted symmetric difference of two Index objects. take(indexer[, axis]) Analogous to ndarray.take to_datetime([dayfirst]) For an Index containing strings or datetime.datetime objects, attempt to_native_types([slicer]) slice and dice then format to_series([keep_tz]) Create a Series with both index and values equal to the index keys tofile(fid[, sep, format]) Write array to a file as text or binary (default). tolist() Overridden version of ndarray.tolist tostring([order]) Construct a Python string containing the raw data bytes in the array. trace([offset, axis1, axis2, dtype, out]) Return the sum along diagonals of the array. transpose(*axes) Returns a view of the array with axes transposed. union(other) Form the union of two Index objects and sorts if possible unique() Return array of unique values in the object. value_counts([normalize, sort, ascending, bins]) Returns object containing counts of unique values. var([axis, dtype, out, ddof]) Returns the variance of the array elements, along given axis. view(*args, **kwargs)"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.59603596,"math_prob":0.9309559,"size":8381,"snap":"2020-24-2020-29","text_gpt3_token_len":1847,"char_repetition_ratio":0.16616927,"word_repetition_ratio":0.054187194,"special_character_ratio":0.22204988,"punctuation_ratio":0.13263296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9775082,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-09T18:53:07Z\",\"WARC-Record-ID\":\"<urn:uuid:c79c9b89-960d-4d23-886f-51dbe3064cd3>\",\"Content-Length\":\"49877\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac496240-2a23-45cd-b9ae-cdba836ec5ac>\",\"WARC-Concurrent-To\":\"<urn:uuid:864cae39-1005-4cb5-ae68-302fd640c82e>\",\"WARC-IP-Address\":\"104.26.1.204\",\"WARC-Target-URI\":\"https://pandas.pydata.org/pandas-docs/version/0.14.0/generated/pandas.Index.html\",\"WARC-Payload-Digest\":\"sha1:WJGM6KTXTG5NUQCGRKFOAUGIGXV55ISG\",\"WARC-Block-Digest\":\"sha1:WW3U7KHNPGMQFSVJIFQ5NKKGTCZF7I6N\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655900614.47_warc_CC-MAIN-20200709162634-20200709192634-00112.warc.gz\"}"}
|
https://www.physicsforums.com/threads/equivalent-forces-in-statics.567830/
|
[
"# Equivalent forces in Statics.\n\n#### Huumah\n\n1. Homework Statement",
null,
"Replace the force for by the equivalent of two forces paralel at the dot and the other perpendicular to the beam.\n\nThe force P is 2000 N\n\nDetermine the forces!\n\n2. Homework Equations\n\nFx= P*cos(x) =\nFy= P*sin(x) =\n\n3. The Attempt at a Solution\n\nFx= 2000 N* cos(75)= 0.517 kN\nFy= 2000 N* sin(75) = 1.93 kN\n\nAm I on the right track? It's just the angle that i'm not sure about.\n\nRelated Introductory Physics Homework Help News on Phys.org\n\n#### Spinnor\n\nGold Member\nDoes this look right?\n\n#### Attachments\n\n• 12.3 KB Views: 286\n\n#### Huumah\n\nYes. This looks better. I cant believe i didn't think to check on what Pythagoras thought about my solution.\n\nThank you!"
] |
[
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"data:image/svg+xml;charset=utf-8,%3Csvg xmlns%3D'http%3A%2F%2Fwww.w3.org%2F2000%2Fsvg' width='316' height='239' viewBox%3D'0 0 316 239'%2F%3E",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8010939,"math_prob":0.8583565,"size":772,"snap":"2019-43-2019-47","text_gpt3_token_len":261,"char_repetition_ratio":0.13541667,"word_repetition_ratio":0.9139073,"special_character_ratio":0.35362694,"punctuation_ratio":0.09550562,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9766689,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T19:34:31Z\",\"WARC-Record-ID\":\"<urn:uuid:97e3ce41-629c-46cf-8bf4-41d0c2517db3>\",\"Content-Length\":\"71991\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2555848e-0f62-4686-8ed5-eae0d9b849f9>\",\"WARC-Concurrent-To\":\"<urn:uuid:903b2439-938a-4264-996b-6e12ee98a15d>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/equivalent-forces-in-statics.567830/\",\"WARC-Payload-Digest\":\"sha1:76PIHBHENODU6KIPPBEX3LRWQGC2IUBH\",\"WARC-Block-Digest\":\"sha1:QBLLZ2PPPKOYEYEOYZ3MMTZTFMCXMC37\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670948.64_warc_CC-MAIN-20191121180800-20191121204800-00209.warc.gz\"}"}
|
https://hotel-gurko.com/rational-or-irrational-worksheet-coloring.html
|
[
"# Rational Or Irrational Worksheet Coloring\n\n### Worksheets are Concept 13 rational irrational numbers, Irrational numbers, Work 1 rational and irrational numbers, Rational numbers irrational numbers, Rational and irrational numbers, Numbers rational and irrational, Adding and subtracting positive and negative numbers date, Identifying rational and irrational numbers.\n\nRational or irrational worksheet coloring. Displaying all worksheets related to - Irrational Number. Percent Increase And Decrease Word Problems Worksheet. Displaying top 8 worksheets found for - Rational Or Irrational. Students will be coloring or shading in the boxes that have rational numbers.\n\nHere is the Rational Or Irrational Worksheet section. Classifying Numbers Classifying Numbers Math. Some of the worksheets for this concept are Concept 13 rational irrational numbers, Identifying rational and irrational numbers, Work 1 rational and irrational numbers, Irrational and imaginary root theorems, Radicals and rational exponents, Descending order of rational numbers work, Rational and irrational. Some of the worksheets below are Rational and Irrational Numbers Worksheets, Identifying Rational and Irrational Numbers, Determine if the given number is rational or irrational, Classifying Numbers, Distinguishing between rational and irrational numbers and tons of exercises.\n\nComplete 2 of the following tasks IXL Practice Worksheets Creating D1 (8th) All the way to 100 Score = _____ Level 2 Worksheet Rational & Irrational #s Vocabulary Poster for the term Rational or Irrational # 3. This would make a great practice worksheet or review on rational and irrational numbers. Rational or irrational worksheet. For example, 12 is a natural, whole, integer, and rational number.\n\nRational Numbers activities for 5th grade and middle school. Color by Classifying worksheet, blue, green, yellow, and black coloring utensils Directions: Rational and Irrational Numbers Worksheet for 6th. Students should be familiar with various sets that numbers can be classified in (natural, whole, integer, rational, irrational, real, not real).\n\nAnswers 1 Answer Key. Displaying all worksheets related to - Classifying Rational Numbers. 7329 - rational because this number is a natural, whole, integer . Classify these numbers as rational or irrational and give your reason.\n\nTo link to this page, copy the following code to your site: It can be used as cyclical review of the difference between rational and irrational numbers. Students should be familiar with various sets that numbers can be classified in (natural, whole, integer, rational, irrational, real, not real). For instance there are many worksheet that you can print here, and if you want to preview the Rational Or Irrational Worksheet simply click the link or image and you will take to save page section.\n\nStudents will be coloring or shading in the boxes that have rational numbers. Students will classify numbers. Some of the worksheets for this concept are Concept 13 rational irrational numbers, Work 1 rational and irrational numbers, Numbers rational and irrational, Irrational and imaginary root theorems, Add subtract multiply divide rational numbers date period, Irrational numbers.\n\nDisplaying top 8 worksheets found for - 8th Rational Number. Color by Classifying worksheet, blue, green, yellow, and black coloring utensils Directions: 1-10 95 90 85 80 75 70 65 60 55 50 11-20 45 40 35 30 25 20 15 10 5 0 1) 6. Rational Irrational Identifying - Displaying top 8 worksheets found for this concept..\n\nHere you will find all we have for Rational Or Irrational Worksheet. Watch the video (Level 2: Once you find your worksheet(s), you can either click on the pop-out icon or download button to print or download your. The boxes with irrational numbers will be left white.\n\nHow To Write The Letters Of The Alphabet Worksheet. The boxes with irrational numbers will be left white. This worksheet will help students identify natural, whole, integer, rational, and irrational numbers. Students will classify numbers.\n\nRational And Irrational Numbers. Answer key is included. 49 1 3 5 √-Directions: Each box will be colored in with the matching description of the number.\n\nStudents will be coloring or shading in the boxes that have rational numbers. Color by Classifying Objective: -- Color all rational numbers blue - Leave all irrational numbers white √ Rationals include: This would make a great practice worksheet or review on rational and irrational numbers.\n\nCollege Kids Worksheet Tutorial. Rational And Irrational Numbers - Displaying top 8 worksheets found for this concept.. Memorial Coloring Printable Write The Beginning Sound Of Rational And Irrational Numbers Worksheet 8th Grade Handwriting Worksheets For Toddlers Alphabets Preschool Decimal Questions.\n\nIrrational Numbers Sorting Activity is a quick and engaging way for students to practice identifying rational and irrational numbers quickly. Rational Numbers Worksheet Pdf - Rational Numbers Worksheet Pdf , Classifying Rational & Irrational Numbers.algebra 1 Worksheets.pare and order Rational Numbers Reteach 22 2.identifying Rational and Irrational Numbers Quiz by.identifying Rational & Irrational Numbers Grade 8 Free.class 8 Math Worksheets and Problems Rational Numbers Tracing Letters For Kids. Color by Classifying Objective:\n\nWorksheets are Work classifying numbers which, Concept 13 rational irrational numbers, Classifying rational numbers 72a, Rational numbers module 3, Sets of real numbers date period, Classifying real numbers work, Cclassifying real numberslassifying real numbers, Descending order of rational. The Rational Number System Worksheet Solutions .",
null,
"### Classifying Rational Numbers Rational numbers, Teaching",
null,
"### Algebra Worksheets Identifying Rational and Irrational",
null,
"### Identifying Rational and Irrational Numbers Irrational",
null,
"### Rational Roots Theorem Algebra Color by Number Algebra",
null,
"### Classify Rational and Irrational Numbers Coloring Activity\n\nSource : pinterest.com"
] |
[
null,
"https://i.pinimg.com/originals/43/3d/6a/433d6aa7308f269c020f663fcc2470b3.jpg",
null,
"https://i.pinimg.com/736x/af/9d/4c/af9d4cda699f61d1bdde95e3cff1596a.jpg",
null,
"https://i.pinimg.com/originals/15/d4/bc/15d4bc4f73e25b7c5105fb4f83c66175.jpg",
null,
"https://i.pinimg.com/originals/56/62/b7/5662b7d560e74279565a3eeb98d372cd.jpg",
null,
"https://i.pinimg.com/474x/6a/55/e4/6a55e44f010fcf58c2b891fa680c2d0f.jpg",
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|
https://socratic.org/questions/5853a209b72cff6a4ce758dc
|
[
"Question 758dc\n\nDec 16, 2016\n\nThis is $F {e}^{3 +}$ ion.........., i.e. $\\text{ferric ion, Fe(III)}$\n${\\left[F e {\\left(C \\equiv N\\right)}_{6}\\right]}^{3 -}$ is the complex ion, that is composed of a $F {e}^{3 +}$ centre, and 6xx\"\"N-=C^(-) ligands.\nOf course, the resultant complex preserves the ionic charge of its consituent ligands and metal centre, and the net charge is 6xx(-1)+3=??#\nIf the counterion were potassium, i.e. ${K}^{+}$, what salt would likely precipitate from solution?"
] |
[
null
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http://www.365yrb.com/laliga/index_25.html
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[
"/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()\n\n/ ()阅读()"
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[
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{"ft_lang_label":"__label__zh","ft_lang_prob":0.9727911,"math_prob":0.9988749,"size":1699,"snap":"2022-05-2022-21","text_gpt3_token_len":2077,"char_repetition_ratio":0.08672567,"word_repetition_ratio":0.0,"special_character_ratio":0.20306063,"punctuation_ratio":0.33333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999616,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T12:44:56Z\",\"WARC-Record-ID\":\"<urn:uuid:0f81f8dd-f831-4374-b34e-31f8ad5ef4ea>\",\"Content-Length\":\"34430\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2bc59cd-5513-490a-92b2-ef7960cd2c8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd0cfa24-2295-42ba-a107-913208bd89da>\",\"WARC-IP-Address\":\"101.36.117.191\",\"WARC-Target-URI\":\"http://www.365yrb.com/laliga/index_25.html\",\"WARC-Payload-Digest\":\"sha1:Z77HDFQVO3YPKDAGCPQOQEKLWDXIZ3HJ\",\"WARC-Block-Digest\":\"sha1:3SU4K2YFSHHA3XX73XAD46MBMOF6LY7L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662587158.57_warc_CC-MAIN-20220525120449-20220525150449-00771.warc.gz\"}"}
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https://bugs.ghostscript.com/show_bug.cgi?id=697497
|
[
"Bug 697497 - Integer Overflow and crash in 'js_pushstring' function in jsrun.c\nInteger Overflow and crash in 'js_pushstring' function in jsrun.c\n Status: RESOLVED FIXED None MuJS Unclassified general (show other bugs) unspecified PC Windows NT P4 normal Tor Andersson Bug traffic\n\n Reported: 2017-01-23 04:44 UTC by op7ic 2017-01-24 06:30 UTC (History) 0 users ---\n\nAttachments\n\n Note You need to log in before you can comment on or make changes to this bug.\n op7ic 2017-01-23 04:44:38 UTC ```Howdy there is a int overflow in js_pushstring function in jsrun.c source when parsing corrupted JS input. Version: latest from git as of 23/01/2017 Vulnerable Source file: mujs/jsrun.c Function: js_pushstring Compile Flags CFLAGS += -g3 -ggdb -O0 Compile Command: make Valgrind shows it pretty clear: valgrind ./mujs-clean/build/mujs PoC.js ==1340== Memcheck, a memory error detector ==1340== Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al. ==1340== Using Valgrind-3.10.0 and LibVEX; rerun with -h for copyright info ==1340== Command: ./mujs-clean/build/mujs PoC.js ==1340== ==1340== Invalid read of size 1 ==1340== at 0x4C2C1A2: strlen (vg_replace_strmem.c:412) ==1340== by 0x403E80: js_pushstring (jsrun.c:114) ==1340== by 0x41611A: Ap_join (jsarray.c:138) ==1340== by 0x416DA7: Ap_toString (jsarray.c:398) ==1340== by 0x406F77: jsR_callcfunction (jsrun.c:1023) ==1340== by 0x4072A0: js_call (jsrun.c:1065) ==1340== by 0x409CCE: jsV_toString (jsvalue.c:56) ==1340== by 0x409EBB: jsV_toprimitive (jsvalue.c:103) ==1340== by 0x40A4B7: jsV_tonumber (jsvalue.c:209) ==1340== by 0x4047B9: js_tonumber (jsrun.c:261) ==1340== by 0x408C6F: jsR_run (jsrun.c:1484) ==1340== by 0x406EBD: jsR_callscript (jsrun.c:1006) ==1340== Address 0x0 is not stack'd, malloc'd or (recently) free'd ==1340== ==1340== ==1340== Process terminating with default action of signal 11 (SIGSEGV) ==1340== Access not within mapped region at address 0x0 ==1340== at 0x4C2C1A2: strlen (vg_replace_strmem.c:412) ==1340== by 0x403E80: js_pushstring (jsrun.c:114) ==1340== by 0x41611A: Ap_join (jsarray.c:138) ==1340== by 0x416DA7: Ap_toString (jsarray.c:398) ==1340== by 0x406F77: jsR_callcfunction (jsrun.c:1023) ==1340== by 0x4072A0: js_call (jsrun.c:1065) ==1340== by 0x409CCE: jsV_toString (jsvalue.c:56) ==1340== by 0x409EBB: jsV_toprimitive (jsvalue.c:103) ==1340== by 0x40A4B7: jsV_tonumber (jsvalue.c:209) ==1340== by 0x4047B9: js_tonumber (jsrun.c:261) ==1340== by 0x408C6F: jsR_run (jsrun.c:1484) ==1340== by 0x406EBD: jsR_callscript (jsrun.c:1006) ==1340== If you believe this happened as a result of a stack ==1340== overflow in your program's main thread (unlikely but ==1340== possible), you can try to increase the size of the ==1340== main thread stack using the --main-stacksize= flag. ==1340== The main thread stack size used in this run was 8388608. ==1340== ==1340== HEAP SUMMARY: ==1340== in use at exit: 129,584 bytes in 1,269 blocks ==1340== total heap usage: 1,343 allocs, 74 frees, 136,545 bytes allocated ==1340== ==1340== LEAK SUMMARY: ==1340== definitely lost: 0 bytes in 0 blocks ==1340== indirectly lost: 0 bytes in 0 blocks ==1340== possibly lost: 0 bytes in 0 blocks ==1340== still reachable: 129,584 bytes in 1,269 blocks ==1340== suppressed: 0 bytes in 0 blocks ==1340== Rerun with --leak-check=full to see details of leaked memory ==1340== ==1340== For counts of detected and suppressed errors, rerun with: -v ==1340== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0) Segmentation fault PoC (base64 encoded): KyBBcnJheSggbnVsbC00KQ== PoC execution: base64 -d /tmp/b64PoC.poc > /tmp/proof.js valgrind ./mujs-clean/build/mujs /tmp/proof.js``` Tor Andersson 2017-01-24 06:30:50 UTC ```Fixed in commit 4006739a28367c708dea19aeb19b8a1a9326ce08 Author: Tor Andersson Date: Tue Jan 24 14:42:36 2017 +0100 Fix 697497: Ensure array length is positive. As a side effect when changing to using regular integers (and avoid the nightmare of mixing signed and unsigned) we accidentally allowed negative array lengths.```"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.6106184,"math_prob":0.7337176,"size":3516,"snap":"2022-40-2023-06","text_gpt3_token_len":1299,"char_repetition_ratio":0.20928246,"word_repetition_ratio":0.28539824,"special_character_ratio":0.42861205,"punctuation_ratio":0.1958457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789633,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T03:31:24Z\",\"WARC-Record-ID\":\"<urn:uuid:1b95ebf8-d988-460c-92db-484842e63f55>\",\"Content-Length\":\"29071\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:83ce7eda-faf3-4756-9ebb-312aa677f169>\",\"WARC-Concurrent-To\":\"<urn:uuid:a1acf992-cfee-4d50-b686-830f7bfe3ac8>\",\"WARC-IP-Address\":\"35.172.177.77\",\"WARC-Target-URI\":\"https://bugs.ghostscript.com/show_bug.cgi?id=697497\",\"WARC-Payload-Digest\":\"sha1:DMYQGBJ32T7JVGGOEAPGUCDSZVMA67TH\",\"WARC-Block-Digest\":\"sha1:SZDRIEDYOIGIUCHU6VSBUE776AWLZTII\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337531.3_warc_CC-MAIN-20221005011205-20221005041205-00075.warc.gz\"}"}
|
https://mathematica.stackexchange.com/questions/181256/is-it-possible-to-parallelize-a-code-with-this-structure
|
[
"# Is it possible to parallelize a code with this structure?\n\nIs it possible to parallelize a code with this structure?\n\nn=0;\nDo[If[condition, n++],{100}]\nn\n\n\nParallelDo always returns n=0.\n\nMathematica implements distributed memory parallelism. Memory cannot be shared between parallel threads. Thus, strictly speaking, the answer is no. Each parallel thread will have a separate n.\n\nSetSharedVariable will let you pretend that n is shared between the parallel kernels, but this is not what really happens. n will instead be always evaluated in on the main kernel. This means that every time n is read or written, a callback to the main kernel is necessary. This will often have a very large performance impact, and is not typically a workable solution (unless your condition takes a very long time to evaluate).\n\nWhat one normally does is think not in terms of a specific algorithm, but in terms of the problem that needs to be solved. Try to come up with a solution that requires no communication between the parallel kernels. It seems to me that you want to count how many times of 100 is a condition satisfied. Well, break that 100 down into 4 groups, each having 25 iterations, and count separately in each group. The add up the counts at the end. ParallelCombine does this quite directly, but it can be implemented with other parallel constructs as well.\n\nTo give a specific example, the following code counts how many times the condition RandomReal[] < 0.3 is satisfied out of 10000 trials.\n\nTotal@ParallelTable[\nIf[RandomReal[] < 0.3, 1, 0],\n{10000}\n]\n\n• Is something like this correct? ParallelCombine[Do[If[condition, n++],{100}]] – mattiav27 Sep 5 '18 at 10:06\n• @mattiav27 I'm sorry to have to ask this, but have you looked at the ParallelCombine documentation? I don't understand how you came to the conclusion that this would be correct based on the documentation page. – Szabolcs Sep 5 '18 at 10:09\n• @mattiav27 Also, if you don't understand parts of my answer, that's fine. Just ask (be specific), and I'll try to clarify. But please do follow the links and do read through all of the answer. I tried to explain why it is not possible to use the n variable the way you were trying to. In this comment you are trying to use it the same way. – Szabolcs Sep 5 '18 at 10:11"
] |
[
null
] |
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|
https://www.bystudin.com/the-circumference-of-the-chord-is-72-and-the-distance-from-the-center-of-the-circle-to-this-chord-is-27-find-the-diameter-of-the-circle/
|
[
"# The circumference of the chord is 72, and the distance from the center of the circle to this chord is 27. Find the diameter of the circle",
null,
"Denote the key points as shown. Draw a segment of AO.\nConsider the triangle AOB.\nThis triangle is right-angled, since the distance OB is the height (shortest distance).\nAB is equal to half the length of the chord (according to the third property of the chord).\nThen, by the Pythagorean theorem:\nAO2 = OB2 + AB2\nAO ^ 2 = 27 ^ 2 + (72/2) ^ 2\nAO ^ 2 = 729 + 1296 = 2025\nAO = 45 is the radius of the circle, therefore, the diameter D = 2 * AO = 2 * 45 = 90",
null,
"Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position."
] |
[
null,
"https://www.bystudin.com/wp-content/uploads/2020/05/biol-19.jpg",
null,
"https://www.bystudin.com/ximg.jpg.pagespeed.ic.EFrGzzeaQT.jpg",
null
] |
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https://cdv.dei.uc.pt/visualizing-urban-mobility/
|
[
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"# Visualizing Urban Mobility\n\nThe goal of this research is understanding urban mobility through the visualization of the use of public transport systems. We focus on the visualization of anomalies regarding the number of passengers. To find patterns of use we analyze the raw data, which contains people counts for every bus stop in Coimbra city with half hour interval. For each stop, and for each day of the week, we calculate the average number of passengers and its standard deviation in each 30 minute interval. This allows us to identify situations that deviate from the norm.\n\nTo further promote visibility, we resort to exaggeration…\n\n### Visualization Model\n\nTo produce the visual artifacts we rely on the Metaballs technique. In order to augment performance all the data is pushed to video card and the visual output calculated using fragment and vertex shaders written in GLSL. Our representation of urban dynamics is embodied by two visualization models. The first represents the deviations and the city map on separate layers; The second embeds the deviations into the vectors of the city map. These complex and computationally heavy representations have the goal of simplifying the identification of patterns in large quantities of data. To further promote visibility, we resort to exaggeration, which allows the visualization of small details, even in the global view, which would be invisible by direct mapping. Thus, the goal is to convey the truth using visual distortion.",
null,
"Figure 2Visualizations of the urban mobility created by applying the Metaball technique to colorize the pixels with general view, image on the left, and zoomed view, image on the right. Displayed the data for 13 of April 2012 at 14:00h.\nWe generate an isosurface where the position of each charge point is the GPS position transformed to the screen coordinates, and the force is the absolute value of the deviation. A positive deviation, i.e. an abnormally high number of passengers, is represented in red, while a negative deviation is represented in green. Forces superior to 1 are represented with an alpha component of 0.5, forces superior to 0.5 are represented with an alpha component of 0.3. The justification for the use of a two-band threshold, can be understood by analyzing the global view of the city and the zoomed area. As it can be observed, the forces superior to 1 tend to result in circles since the influence of other charge points tends to be negligible by comparison. This contrasts with the areas of transparent green and red generated by forces in the [0.5,1] interval, which assume a more organic nature filling in the gaps among strong forces and highlighting areas where anomalies are occurring. When we zoom in, the exaggerations caused by the representation of these small forces are reduced and the visualization becomes more rigorous.",
null,
"Figure 3Detail of visualization where we can see the behavior of Metaball technique with two classes of charge points\nThe second approach is based on the same technique. The difference is that the output of the function is applied directly to the shape vertices of the map. First, the road map is retrieved from OpenStreeMap, unnecessary objects are filtered, and all lines are parsed into shapes. Then, the Metaball technique is applied to determine the color for each vertex of the map. Although, this approach is less computationally expensive, in our opinion, the visual output is not as clear and informative.\n\nThe outcome of this process is an interactive visualization application with two models of visualization, one presents a clear and informative, albeit exaggerated, view of the anomalies, while the other sacrifices visibility for the sake of rigor.\n\n### Application\n\nSo, while validating our model with TfL data we noticed that the amount of exaggeration was too high. In order to change the exaggeration we added one variable that controls the slope of the equation. As we increase the slope the resulted blobs tend to the shape of circle with radius equal to the target radius. While the low values for the slope makes the blobs to be more expanded/exaggerated. The presented visualization model was embedded in an exploratory application along with a bar graph that represents the variation of network’s load through the time. The application allows the user to navigate in time, as well as zoom and pan the visualization. The model transforms according to zoom level and morphs from state to state regarding to the selected date and time.",
null,
"Figure 5Visualisation application of TfL data — standard deviations from normal, with controlled slope of metaball function. Image on the left shows a network load at 9.00h and image on the right shows the deviations at 21.00h. Eight bus routes were represented.\n\nAuthor\n\nEvgheni Polisciuc"
] |
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null
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|
https://keysenvironmentalcalendar.org/rutherford/grouped-frequency-distribution-example-problems.php
|
[
"# Rutherford Grouped Frequency Distribution Example Problems\n\n## Chapter 2 Frequency Distributions\n\n### SOLUTION Construct a grouped frequency distribution for",
null,
"Solved Grouped-Data Formulas. When data are grouped in a. Grouped Mean Median Mode 1. Image , and better understand the current problem situation. It is the biggest frequency group and forms tallest column., Example: Problem. Stephen has been Multiply each midpoint by the frequency for “Measures of Central Tendency: Mean, Median, and Mode” by the National.\n\n### SOLUTION Construct a grouped frequency distribution for\n\nDeciles for grouped data VrcAcademy. Know how to calculate the Mean Deviation for Grouped Data and Continuous Frequency Distribution. CLASSES. CLASSES 4-5; for example, if we need to, CHAPTER 1. Introduction to Statistics and Frequency Distributions. 3. should complete all of the practice problems. Most students benefit from a few repetitions.\n\nFor Example: Following is the we will form cumulative frequency column. Quartiles for grouped data will be from the frequency distribution of weights For example- we are given data about people according to the need of the problem. mean and median for the continuous frequency distribution of grouped\n\nChapter 2: Frequency Distributions and Graphs Ungrouped frequency distributions Grouped frequency di Example: Use the frequency distribution for the ages of ... frequency distribution GROUPED data problem.5 17.Formulas for Grouped Data Problems **Please remember how to distinguish between Grouped and Ungrouped Data\n\nGrouped and Ungrouped Data. Example: The sale of shoes of various sizes at a shop, Types of Grouped Frequency Distribution 1. Free math problem solver answers your algebra, Finding the Relative Frequency of the Frequency Creating a Grouped Frequency Distribution Table; Finding the\n\nMean Median Mode for Grouped Data. Grouped MEAN = ( Total of Frequency x Midpoints ) Worked Examples for Grouped Mean Median Mode. Understanding Frequency Distributions. by looking at a chart of its frequency distribution. For example, Figure 1.15 uses a grouped frequency distribution,\n\nExample Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as For example- we are given data about people according to the need of the problem. mean and median for the continuous frequency distribution of grouped\n\nMATH 10043 CHAPTER 2 EXAMPLES & DEFINITIONS A grouped frequency distribution is commonly displayed visually as a Referring to practice problem 1 above, ... from a grouped frequency table, examples and Grouped Frequency Distribution examples, or type in your own problem and check your\n\nFrequency Distributions. Ungrouped. Example - Example - Rules. Practice Problems Rules for creating a grouped frequency distribution. There are many ways to group this data. For example, This would change the frequency distribution a lot, Difference Between Grouped and Ungrouped Data Related\n\nRead all the 5 types of Frequency distribution with examples and details. Grouped Frequency Distributions. Introductory Statistics: Concepts, Models, and Applications 2nd edition INTRODUCTORY STATISTICS: CONCEPTS, MODELS, AND\n\nSolutions for Chapter 3 Problem 94E. Problem 94E: Grouped-Data Formulas. When data are grouped in a frequency distribution, we use the following formulas to obtain Grouped frequency distributions are more common than ungrouped and the authors use the IQ as an example of a hypothetical distribution with many Problems. You\n\nCHAPTER 1. Introduction to Statistics and Frequency Distributions. 3. should complete all of the practice problems. Most students benefit from a few repetitions Example Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as\n\nThere are many ways to group this data. For example, This would change the frequency distribution a lot, Difference Between Grouped and Ungrouped Data Related Grouped Mean Median Mode 1. Image , and better understand the current problem situation. It is the biggest frequency group and forms tallest column.\n\nStatistics: Grouped Frequency Distributions. Guidelines for classes. There should be between 5 and 20 classes. Creating a Grouped Frequency Distribution. The normal distribution is the most important of all probability distributions. It is applied directly to many practical problems, grouped frequency distribution\n\nAllan G. Allan G. Bluman Bluman that are organized in a grouped frequency distribution Relative frequency example: The Most Common Graphs Frequency distribution tables can be used for In this example, the cumulative frequency is 1 and the while discrete variables can be grouped into class\n\n... examples, exercises and problems with the same frequency and that frequency is the maximum, the distribution is bimodal of the Mode for Grouped Chapter 2: Frequency Distributions and Graphs Ungrouped frequency distributions Grouped frequency di Example: Use the frequency distribution for the ages of\n\nCreate a grouped frequency distribution; Create a histogram based on a grouped frequency in the chapter \" Summarizing Distributions.\") In our example, Chapter 2: Frequency Distribution and Graphs How to construct a grouped frequency Distribution Example\nThis frequency distribution shows the number of\n\nRead all the 5 types of Frequency distribution with examples and details. Example: Problem. Stephen has been Multiply each midpoint by the frequency for “Measures of Central Tendency: Mean, Median, and Mode” by the National\n\nExample Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as What is a Relative frequency distribution? Relative Frequency Distribution: Definition and Examples . Need help NOW with a homework problem?\n\nThe problem is that so much The process of drawing grouped frequency distributions can be An alternative frequency table for the example data with SOLUTION: Construct a grouped frequency distribution for the following 28 scores using a class width of 4: 15 32 23 31 29 26 21 15 34 25 26 33 27\n\nMAT 142 College Mathematics Module #3 then make a frequency distribution. Example 7. Grouped Data. In our examples we will use frequency distributions Example of Categorical Frequency Distribution A sample of 20 M&M’s is observed and their colors are recorded. Example of Group Frequency Distribution\n\nSolved problem examples and problems for the reader to solve The graphical representations of grouped frequency distributions are usually more readily understood Example Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as\n\n### SOLUTION Construct a grouped frequency distribution for",
null,
"Solved Grouped-Data Formulas. When data are grouped in a. Measure of relative standing of an observation in Grouped Data. Example: Deciles estimate estimation Frequency Distribution graph Heteroscedasticity problems, ... we use the grouped frequency distribution table. For example a frequency distribution table is shown below. Frequency Distribution Table with Intervals..\n\n### SOLUTION Construct a grouped frequency distribution for",
null,
"Grouped Frequency Distributions Calculation of the. 24/03/2012В В· frequency distribution is the same thing as grouping data. choose which one would be appropriate for any given problem. Mode of Group Data Example: https://en.wikipedia.org/wiki/Frequency_probability Mean Example Problems. Example 1. Mean can also be found for grouped data, but before we see an example on that, let us first define frequency..",
null,
"What is a Grouped Frequency Distribution Table Grouped Frequency Distribution Table Example Problems A survey was conducted by a group of Example Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as\n\nFrequency distribution tables can be used for In this example, the cumulative frequency is 1 and the while discrete variables can be grouped into class Chapter 2: Frequency Distributions and Graphs Ungrouped frequency distributions Grouped frequency di Example: Use the frequency distribution for the ages of\n\nFrequency Distributions. Ungrouped. Example - Example - Rules. Practice Problems Rules for creating a grouped frequency distribution. frquency distribution or grouped Grouped Data Standard Deviation Calculator, This below solved example problem for frequency distribution standard\n\n10/10/2016 · This short video details the steps to follow in order to calculate the Variance and Standard Deviation of a Grouped Frequency Distribution Grouped frequency distributions are more common than ungrouped and the authors use the IQ as an example of a hypothetical distribution with many Problems. You\n\nFrequency distribution tables can be used for In this example, the cumulative frequency is 1 and the while discrete variables can be grouped into class Learn how to represent data in the form of a frequency distribution table Grouped Frequency Distribution. down the frequency in the frequency column. Example\n\nStatistics and probability problems with Construct a frequency distribution for the data. b) If a person is selected randomly from the group of twenty Example of Categorical Frequency Distribution A sample of 20 M&M’s is observed and their colors are recorded. Example of Group Frequency Distribution\n\n10/10/2016 · This short video details the steps to follow in order to calculate the Variance and Standard Deviation of a Grouped Frequency Distribution Free math problem solver answers your algebra, Finding the Relative Frequency of the Frequency Creating a Grouped Frequency Distribution Table; Finding the\n\nExample Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as Measure of relative standing of an observation in Grouped Data. Example: Deciles estimate estimation Frequency Distribution graph Heteroscedasticity problems\n\nThe problem is that so much The process of drawing grouped frequency distributions can be An alternative frequency table for the example data with What is a Grouped Frequency Distribution Table Grouped Frequency Distribution Table Example Problems A survey was conducted by a group of",
null,
"The problem is that so much The process of drawing grouped frequency distributions can be An alternative frequency table for the example data with Statistics and probability problems with Construct a frequency distribution for the data. b) If a person is selected randomly from the group of twenty\n\n## Solved Grouped-Data Formulas. When data are grouped in a",
null,
"Statistics and Probability for Engineering Applications. For Example: Following is the we will form cumulative frequency column. Quartiles for grouped data will be from the frequency distribution of weights, ... from a grouped frequency table, examples and Grouped Frequency Distribution examples, or type in your own problem and check your.\n\n### SOLUTION Construct a grouped frequency distribution for\n\nMATH 10043 CHAPTER 2 EXAMPLES & DEFINITIONS. What Is the Difference Between Grouped and Ungrouped Frequency Distributions? In a grouped frequency distribution, data is sorted and separated into groups called, Solutions for Chapter 3 Problem 94E. Problem 94E: Grouped-Data Formulas. When data are grouped in a frequency distribution, we use the following formulas to obtain.\n\nChapter 2: Frequency Distributions and Graphs Ungrouped frequency distributions Grouped frequency di Example: Use the frequency distribution for the ages of Variance and standard deviation for grouped data: The sample standard deviation: Example 5.6-1 The following gives the frequency distribution of the daily\n\n... examples, exercises and problems with the same frequency and that frequency is the maximum, the distribution is bimodal of the Mode for Grouped Measure of relative standing of an observation in Grouped Data. Example: Deciles estimate estimation Frequency Distribution graph Heteroscedasticity problems\n\nExample: Problem. Stephen has been Multiply each midpoint by the frequency for “Measures of Central Tendency: Mean, Median, and Mode” by the National deciles formula, calculating deciles, decile calculator for grouped data\n\ndeciles formula, calculating deciles, decile calculator for grouped data In worksheet on frequency distribution the questions are based on arranging data in Construct the grouped frequency table with the class Math Problem Ans;\n\nSolved problem examples and problems for the reader to solve The graphical representations of grouped frequency distributions are usually more readily understood Frequency Distributions. Ungrouped. Example - Example - Rules. Practice Problems Rules for creating a grouped frequency distribution.\n\nStatistics: Grouped Frequency Distributions. Guidelines for classes. There should be between 5 and 20 classes. Creating a Grouped Frequency Distribution. Variance and standard deviation for grouped data: The sample standard deviation: Example 5.6-1 The following gives the frequency distribution of the daily\n\nExample: Problem. Stephen has been Multiply each midpoint by the frequency for “Measures of Central Tendency: Mean, Median, and Mode” by the National Mean Median Mode for Grouped Data. Grouped MEAN = ( Total of Frequency x Midpoints ) Worked Examples for Grouped Mean Median Mode.\n\nWhat Is the Difference Between Grouped and Ungrouped Frequency Distributions? In a grouped frequency distribution, data is sorted and separated into groups called Grouped and Ungrouped Data. Example: The sale of shoes of various sizes at a shop, Types of Grouped Frequency Distribution 1.\n\nSolutions for Chapter 3 Problem 94E. Problem 94E: Grouped-Data Formulas. When data are grouped in a frequency distribution, we use the following formulas to obtain Example: Problem. Stephen has been Multiply each midpoint by the frequency for “Measures of Central Tendency: Mean, Median, and Mode” by the National\n\nExample of Categorical Frequency Distribution A sample of 20 M&M’s is observed and their colors are recorded. Example of Group Frequency Distribution In worksheet on frequency distribution the questions are based on arranging data in Construct the grouped frequency table with the class Math Problem Ans;\n\nExample: Problem. Stephen has been Multiply each midpoint by the frequency for “Measures of Central Tendency: Mean, Median, and Mode” by the National Grouped Mean Median Mode 1. Image , and better understand the current problem situation. It is the biggest frequency group and forms tallest column.\n\nRead all the 5 types of Frequency distribution with examples and details. Variance and standard deviation for grouped data: The sample standard deviation: Example 5.6-1 The following gives the frequency distribution of the daily\n\n... frequency distribution GROUPED data problem.5 17.Formulas for Grouped Data Problems **Please remember how to distinguish between Grouped and Ungrouped Data Grouped frequency distributions are more common than ungrouped and the authors use the IQ as an example of a hypothetical distribution with many Problems. You\n\nFor Example: Following is the we will form cumulative frequency column. Quartiles for grouped data will be from the frequency distribution of weights Frequency distribution tables can be used for In this example, the cumulative frequency is 1 and the while discrete variables can be grouped into class\n\nFree math problem solver answers your algebra, Finding the Relative Frequency of the Frequency Creating a Grouped Frequency Distribution Table; Finding the Allan G. Allan G. Bluman Bluman that are organized in a grouped frequency distribution Relative frequency example: The Most Common Graphs\n\nStatistics: Grouped Frequency Distributions. Guidelines for classes. There should be between 5 and 20 classes. Creating a Grouped Frequency Distribution. Mean Example Problems. Example 1. Mean can also be found for grouped data, but before we see an example on that, let us first define frequency.\n\n... frequency distribution GROUPED data problem.5 17.Formulas for Grouped Data Problems **Please remember how to distinguish between Grouped and Ungrouped Data Example of Categorical Frequency Distribution A sample of 20 M&M’s is observed and their colors are recorded. Example of Group Frequency Distribution\n\n10/10/2016 · This short video details the steps to follow in order to calculate the Variance and Standard Deviation of a Grouped Frequency Distribution Grouped frequency distributions are more common than ungrouped and the authors use the IQ as an example of a hypothetical distribution with many Problems. You\n\n10/10/2016 · This short video details the steps to follow in order to calculate the Variance and Standard Deviation of a Grouped Frequency Distribution ... we use the grouped frequency distribution table. For example a frequency distribution table is shown below. Frequency Distribution Table with Intervals.\n\n7/10/2016 · This video details the steps to be followed in order to construct a Grouped Frequency Distribution from a Raw Data Set. The details four instructive steps. Know how to calculate the Mean Deviation for Grouped Data and Continuous Frequency Distribution. CLASSES. CLASSES 4-5; for example, if we need to\n\nLearn how to represent data in the form of a frequency distribution table Grouped Frequency Distribution. down the frequency in the frequency column. Example What Is the Difference Between Grouped and Ungrouped Frequency Distributions? In a grouped frequency distribution, data is sorted and separated into groups called\n\n### Chapter 2 Frequency Distributions",
null,
"Statistics and Probability for Engineering Applications. calculation of median of grouped problem regarding the second part of my question-We have to find out the median score from the following frequency distribution, Know how to calculate the Mean Deviation for Grouped Data and Continuous Frequency Distribution. CLASSES. CLASSES 4-5; for example, if we need to.\n\n### Median of Discrete Frequency Distribution Statistic",
null,
"Grouped Frequency Distributions Constructing a Grouped. Grouped frequency distributions are more common than ungrouped and the authors use the IQ as an example of a hypothetical distribution with many Problems. You https://simple.wikipedia.org/wiki/Grouped_data Solutions for Chapter 3 Problem 94E. Problem 94E: Grouped-Data Formulas. When data are grouped in a frequency distribution, we use the following formulas to obtain.",
null,
"• Quartiles Deciles and Percentiles В» Economics Tutorials\n• Chapter 2 Frequency Distributions\n• Chapter 2 Frequency Distributions\n\n• For Example: Following is the we will form cumulative frequency column. Quartiles for grouped data will be from the frequency distribution of weights calculation of median of grouped problem regarding the second part of my question-We have to find out the median score from the following frequency distribution\n\nfrquency distribution or grouped Grouped Data Standard Deviation Calculator, This below solved example problem for frequency distribution standard 10/10/2016В В· This short video details the steps to follow in order to calculate the Variance and Standard Deviation of a Grouped Frequency Distribution\n\nGrouped frequency distributions are more common than ungrouped and the authors use the IQ as an example of a hypothetical distribution with many Problems. You What is a Grouped Frequency Distribution Table Grouped Frequency Distribution Table Example Problems A survey was conducted by a group of\n\nSolutions for Chapter 3 Problem 94E. Problem 94E: Grouped-Data Formulas. When data are grouped in a frequency distribution, we use the following formulas to obtain deciles formula, calculating deciles, decile calculator for grouped data\n\nFree math problem solver answers your algebra, Finding the Relative Frequency of the Frequency Creating a Grouped Frequency Distribution Table; Finding the For Example: Following is the we will form cumulative frequency column. Quartiles for grouped data will be from the frequency distribution of weights\n\nWhat is a Relative frequency distribution? Relative Frequency Distribution: Definition and Examples . Need help NOW with a homework problem? MATH 10043 CHAPTER 2 EXAMPLES & DEFINITIONS A grouped frequency distribution is commonly displayed visually as a Referring to practice problem 1 above,\n\n... from a grouped frequency table, examples and Grouped Frequency Distribution examples, or type in your own problem and check your Create a grouped frequency distribution; Create a histogram based on a grouped frequency in the chapter \" Summarizing Distributions.\") In our example,\n\nExample Calculate the median from the Median of Discrete Frequency Distribution. To calculate the median of a grouped frequency distribution we proceed as Problem Set 1: Frequency Distribution Table. For Exercises in Measure of Central Tendency-Grouped and PSY24 Professor: Ms. Nhelda Nacion Problem Set no\n\nGrouped and Ungrouped Data. Example: The sale of shoes of various sizes at a shop, Types of Grouped Frequency Distribution 1. Construct a grouped frequency distribution for a The solution to this problem is to create a grouped distributions. An example of a normal\n\n... examples, exercises and problems with the same frequency and that frequency is the maximum, the distribution is bimodal of the Mode for Grouped There are many ways to group this data. For example, This would change the frequency distribution a lot, Difference Between Grouped and Ungrouped Data Related",
null,
"... from a grouped frequency table, examples and Grouped Frequency Distribution examples, or type in your own problem and check your What Is the Difference Between Grouped and Ungrouped Frequency Distributions? In a grouped frequency distribution, data is sorted and separated into groups called\n\nView all posts in Rutherford category"
] |
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https://www.arxiv-vanity.com/papers/hep-ph/0104073/
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"# Opening the Crystalline Color Superconductivity Window\n\nAdam K. Leibovich1 Theory Group, Fermilab\nP.O. Box 500, Batavia, IL 60510\n\nKrishna Rajagopal2, Eugene Shuster3 Center for Theoretical Physics\nMassachusetts Institute of Technology\nCambridge, MA 02139\n\nMIT-CTP-3108, FERMILAB-Pub-01/041-T, hep-ph/0104073, April 4, 2001\n###### Abstract\n\nCold dense quark matter is in a crystalline color superconducting phase wherever pairing occurs between species of quarks with chemical potentials whose difference lies within an appropriate window. If the interaction between quarks is modeled as point-like, this window is rather narrow. We show that when the interaction between quarks is modeled as single-gluon exchange, the window widens by about a factor of ten at accessible densities and by much larger factors at higher density. This striking enhancement reflects the increasingly -dimensional nature of the physics at weaker and weaker coupling. Our results indicate that crystalline color superconductivity is a generic feature of the phase diagram of cold dense quark matter, occurring wherever one finds quark matter which is not in the color-flavor locked phase. If it occurs within the cores of compact stars, a crystalline color superconducting region may provide a new locus for glitch phenomena.\n\n## I Introduction\n\nAt asymptotic densities, the ground state of QCD with three quarks with equal masses is expected to be the color-flavor locked (CFL) phase [1, 2, 3, 4]. This phase features a condensate of Cooper pairs of quarks which includes , , and pairs. Quarks of all colors and all flavors participate equally in the pairing, and all excitations with quark quantum numbers are gapped.\n\nThe CFL phase persists for unequal quark masses, so long as the differences are not too large [5, 6]. In the absence of any interaction (and thus in the absence of CFL pairing) a quark mass difference pushes the Fermi momenta for different flavors apart, yielding different number densities for different flavors. In the CFL phase, however, the fact that the pairing energy is maximized when , , and number densities are equal enforces this equality . This means that if one imagines increasing the strange quark mass , all quark number densities remain equal until a first order phase transition, at which CFL pairing is disrupted, (some) quark number densities spring free under the accumulated tension, and a less symmetric state of quark matter is obtained . This state of matter, which results when pairing between two species of quarks persists even once their Fermi momenta differ, is a crystalline color superconductor.\n\nWe can study crystalline color superconductivity more simply by focusing just on pairing between massless up and down quarks whose Fermi momenta we attempt to push apart by turning on a chemical potential difference, rather than a quark mass difference. That is, we introduce\n\n μu = ¯μ−δμ μd = ¯μ+δμ . (1)\n\nIf is nonzero but less than some , the ground state is precisely that obtained for [8, 9, 10]. In this state, red and green up and down quarks pair, yielding four quasiparticles with superconducting gap [11, 12, 13, 14]. Furthermore, the number density of red and green up quarks is the same as that of red and green down quarks. As long as is not too large, this BCS state remains unchanged (and favored) because maintaining equal number densities, and thus coincident Fermi surfaces, maximizes the pairing and hence the gain in interaction energy. As is increased, the BCS state remains the ground state of the system only as long as its negative interaction energy offsets the large positive free energy cost associated with forcing the Fermi seas to deviate from their normal state distributions. In the weak coupling limit, in which , the BCS state persists for [8, 10].\n\nThese conclusions are the same (as long as ) whether the interaction between quarks is modeled as a point-like four-fermion interaction or is approximated by single-gluon exchange . The latter analysis [15, 16, 17, 18, 19, 20, 21, 22, 3, 4] is of quantitative validity at densities which are so extremely high that the QCD coupling . Applying these asymptotic results at accessible densities nevertheless yields predictions for the magnitude of the BCS gap , and thus for , which are in qualitative agreement with those made using phenomenologically normalized models with simpler point-like interactions [19, 3, 4].\n\nAbove the BCS state, in a range , the crystalline color superconducting phase occurs. We shall demonstrate in this paper that applying asymptotic results at accessible densities yields a window that is more than a factor of ten wider than in models with point-like interactions. The crystalline color superconductivity window in parameter space may therefore be much wider than previously thought, making this phase a generic feature of the phase diagram for cold dense quark matter. The reason for this qualitative increase in can be traced back to the fact that gluon exchange at weaker and weaker coupling is more and more dominated by forward-scattering, while point-like interactions describe -wave scattering. What is perhaps surprising is that even at quite large values of , gluon exchange yields an order of magnitude increase in .\n\nThe crystalline color superconducting state is the analogue of a state first explored by Larkin and Ovchinnikov and Fulde and Ferrell in the context of electron superconductivity in the presence of magnetic impurities. Translating LOFF’s results to the case of interest, the authors of Ref. found that for it is favorable to form a state in which the and Fermi momenta are given by and as in the absence of interactions, and are thus not equal, but pairing nevertheless occurs. Whereas in the BCS state, obtained for , pairing occurs between quarks with equal and opposite momenta, when it is favorable to form a condensate of Cooper pairs with nonzero total momentum. This is favored because pairing quarks with momenta which are not equal and opposite gives rise to a region of phase space where each quark in a Cooper pair can be close to its own Fermi surface, even when the up and down Fermi momenta differ, and such pairs can be created at low cost in free energy.444LOFF condensates have also recently been considered in two other contexts. In QCD with , and , one has equal Fermi momenta for antiquarks and quarks, BCS pairing occurs, and consequently a condensate forms [26, 27]. If and differ, and if the difference lies in the appropriate range, a LOFF phase with a spatially varying condensate results [26, 27]. Our conclusion that the LOFF window is much wider than previously thought applies in this context also. Suitably isospin asymmetric nuclear matter may also admit LOFF pairing, as discussed recently in Ref. . Condensates of this sort spontaneously break translational and rotational invariance, leading to gaps which vary periodically in a crystalline pattern. If in some shell within the quark matter core of a neutron star (or within a strange quark star) the quark chemical potentials are such that crystalline color superconductivity arises, as we now see occurs for a wide range of reasonable parameter values, rotational vortices may be pinned in this shell, making it a locus for glitch formation . Rough estimates of the pinning force suggest that it is comparable to that for a rotational vortex pinned in the inner crust of a conventional neutron star, and thus may yield glitches of phenomenological interest .\n\nAs in Refs. [10, 29], we shall restrict our attention here to the simplest possible “crystal” structure, namely that in which the condensate varies like a plane wave:\n\n ⟨ψ(x)ψ(x)⟩∝Δe2iq⋅x . (2)\n\nWherever this condensate is favored over the homogeneous BCS condensate and over the state with no pairing at all, we expect that the true ground state of the system is a condensate which varies in space with some more complicated spatial dependence. The phonon associated with the plane wave condensate (2) has recently been analyzed . A full analysis of all the phonons in the crystalline color superconducting phase must await the determination of the actual crystal structure.\n\nThe authors of Refs. [10, 29] studied crystalline color superconductivity in a model in which two flavors of quarks with chemical potentials (I) interact via a point-like four-fermion interaction with the quantum numbers of single gluon exchange. In the LOFF state, each Cooper pair has total momentum with . The direction of is chosen spontaneously. The LOFF phase is characterized by a gap parameter and a diquark condensate, but not by an energy gap: the quasiparticle dispersion relations vary with the direction of the momentum, yielding gaps which range from zero up to a maximum of . The condensate is dominated by those “ring-shaped” regions in momentum space in which a quark pair with total momentum has both members of the pair within approximately of their respective Fermi surfaces.\n\nThe gap equation which determines was derived in Ref. using variational methods, along the lines of Refs. [25, 31]. It has since been rederived using now more familiar methods, namely a diagrammatic derivation of the one-loop Schwinger-Dyson equation with a Nambu-Gorkov propagator modified to describe the spatially varying condensate . This gap equation can then be used to show that crystalline color superconductivity is favored over no pairing for . If quarks interact via a weak point-like four-fermion interaction, [24, 25, 31, 10]. For a stronger four-fermion interaction, tends to decrease, closing the crystalline color superconductivity window .\n\nCrystalline color superconductivity is favored for . As increases, one finds a first order phase transition from the ordinary BCS phase to the crystalline color superconducting phase at and then a second order phase transition at at which decreases to zero. Because the condensation energy in the LOFF phase is much smaller than that of the BCS condensate at , the value of is almost identical to that at which the naive unpairing transition from the BCS state to the state with no pairing would occur if one ignored the possibility of a LOFF phase, namely . For all practical purposes, therefore, the LOFF gap equation is not required in order to determine . The LOFF gap equation is used to determine and the properties of the crystalline color superconducting phase .\n\nIn this paper, we generalize the diagrammatic analysis of Ref. to analyze the crystalline color superconducting phase which occurs when quarks interact by the exchange of a propagating gluon, as is quantitatively valid at asymptotically high densities. At weak coupling, quark-quark scattering by single-gluon exchange is dominated by forward scattering. In most scatterings, the angular positions of the quarks on their respective Fermi surfaces do not change much. As a consequence, the weaker the coupling the more the physics can be thought of as a sum of many -dimensional theories, with only rare large-angle scatterings able to connect one direction in momentum space with others . Suppose for a moment that we were analyzing a truly -dimensional theory. The momentum-space geometry of the LOFF state in one spatial dimension is qualitatively different from that in three. Instead of Fermi surfaces, we would have only “Fermi points” at and . The only choice of which allows pairing between and quarks at their respective Fermi points is . In dimensions, in contrast, is favored because it allows LOFF pairing in “ring-shaped” regions of the Fermi surface, rather than just at antipodal points [24, 25, 10]. Also, in striking contrast to the -dimensional case, it has long been known that in a true -dimensional theory with a point-like interaction between fermions, in the weak-interaction limit .\n\nWe therefore expect that in -dimensional QCD with the interaction given by single-gluon exchange, as and the which characterizes the LOFF phase should become closer and closer to and should diverge. We shall demonstrate both effects, and shall furthermore show that both are clearly in evidence already at the rather large coupling , corresponding to MeV using the conventions of Refs. [19, 23]. At this coupling, , meaning that , which is much larger than . If we go to much higher densities, where the calculation is under quantitative control, we find an even more striking enhancement: when we find ! We see that (relative to expectations based on experience with point-like interactions) the crystalline color superconductivity window is wider by more than four orders of magnitude at this weak coupling, and is about one order of magnitude wider at accessible densities if weak-coupling results are applied there.\n\nIn Section II, we present the gap equation derived from a Schwinger-Dyson equation as in Ref. , but now assuming that quarks interact via single-gluon exchange. As our goal is the evaluation of , the value of at which the crystalline color superconducting gap vanishes, in Section III we take the limit of the gap equation. Analysis of the resulting equation yields . In Section IV we look ahead to implications of our central result that the crystalline color superconductivity window is wider by (an) order(s) of magnitude than previously thought.\n\n## Ii The Gap Equation for Crystalline Color Superconductivity at Weak Coupling\n\nIn the crystalline color superconducting phase [24, 25, 10, 29], the condensate contains pairs of and quarks with momenta such that the total momentum of each Cooper pair is given by , with the direction of chosen spontaneously. Such a condensate varies periodically in space with wavelength as in (2). Wherever there is an instability towards (2), we expect that the true ground state will be a crystalline condensate which varies in space like a sum of several such plane waves with the same . As in Refs. [10, 29], we focus here on finding the region of where an instability towards (2) occurs, leaving the determination of the favored crystal structure to future work.\n\nWe begin by reviewing the field theoretic framework for the analysis of crystalline color superconductivity presented in Ref. . In order to describe pairing between quarks with momentum and quarks with momentum , we must use a modified Nambu-Gorkov spinor defined as\n\n Ψ(p,q)=⎛⎜ ⎜ ⎜ ⎜ ⎜⎝ψu(p+q)ψd(p−q)¯ψTd(−p+q)¯ψTu(−p−q)⎞⎟ ⎟ ⎟ ⎟ ⎟⎠ . (3)\n\nNote that by we mean the four-vector . The Cooper pairs have nonzero total momentum, and the ground state condensate (2) is static. The momentum dependence of (3) is motivated by the fact that in the presence of a crystalline color superconducting condensate, anomalous propagation does not only mean picking up or losing two quarks from the condensate. It also means picking up or losing momentum . The basis (3) has been chosen so that the inverse fermion propagator in the crystalline color superconducting phase is diagonal in -space and is given by\n\n S−1(p,q)=⎡⎢ ⎢ ⎢ ⎢⎣p/+q/+μuγ00−¯Δ(p,−q)00p/−q/+μdγ00¯Δ(p,q)−Δ(p,−q)0(p/−q/−μdγ0)T00Δ(p,q)0(p/+q/−μuγ0)T⎤⎥ ⎥ ⎥ ⎥⎦ , (4)\n\nwhere and is a matrix with color (, ) and Dirac indices suppressed. Note that the condensate is explicitly antisymmetric in flavor. is the relative momentum of the quarks in a given pair and is different for different pairs. In the gap equation below, we shall integrate over and . As desired, the off-diagonal blocks describe anomalous propagation in the presence of a condensate of diquarks with momentum . The choice of basis we have made is analogous to that introduced previously in the analysis of a crystalline quark-antiquark condensate .\n\nWe obtain the gap equation by solving the one-loop Schwinger-Dyson equation given by\n\n S−1(k,q)−S−10(k,q)=−g2∫d4p(2π)4ΓAμS(p,q)ΓBνDμνAB(k−p), (5)\n\nwhere is the gluon propagator, is the full quark propagator, whose inverse is given by (4), and is the fermion propagator in the absence of interaction, given by with . The vertices are defined as follows:\n\n ΓAμ=⎛⎜ ⎜ ⎜ ⎜ ⎜⎝γμλA/20000γμλA/20000−(γμλA/2)T0000−(γμλA/2)T⎞⎟ ⎟ ⎟ ⎟ ⎟⎠. (6)\n\nIn previous analyses [10, 29], a point-like interaction was introduced by replacing by times a constant. Here, instead, we analyze the interaction between quarks given by the exchange of a medium-modified gluon and thus use a gluon propagator given by\n\n Dμν(p)=PTμνp2−G(p)+PLμνp2−F(p)−ξpμpνp4 , (7)\n\nwhere is the gauge parameter, and are functions of and , and the projectors are defined as follows:\n\n PTij=δij−^pi^pj, PT00=PT0i=0, PLμν=−gμν+pμpνp2−PTμν. (8)\n\nThe functions and describe the effects of the medium on the gluon propagator. If we neglect the Meissner effect (that is, if we neglect the modification of and due to the gap in the fermion propagator) then describes Thomas-Fermi screening and describes Landau damping and they are given in the hard dense loop (HDL) approximation by \n\n F(p) = m2p2|p|2(1−ip0|p|Q0(ip0|p|)), with Q0(x)=12log(x+1x−1), G(p) = 12m2ip0|p|⎡⎣⎛⎝1−(ip0|p|)2⎞⎠Q0(ip0|p|)+ip0|p|⎤⎦ , (9)\n\nwhere is the Debye mass for .\n\nThe method of solving the Schwinger-Dyson equation has been outlined in Ref. . One obtains coupled integral equations for the gap parameters , which describes particle-particle and hole-hole pairing, and , which describe particle-antiparticle and antiparticle-antiparticle pairing. Because only describes pairing of quarks near their respective Fermi surfaces, this is the gap parameter which describes the gap in the quasiparticle spectrum and is therefore the gap parameter of physical interest. For the same reason, the gap equation is dominated by the terms containing . Upon dropping and renaming , the gap equation derived in Ref. can be written as:\n\n Δ(k0) = −ig23sin2β(k,k)2∫d4p(2π)4Δ(p0)(p0+E1)(p0−E2) (10) ×[CF(k−p)2−F(k−p)+CG(k−p)2−G(k−p)+Cξξ(k−p)2],\n\nwhere\n\n CF = cos2β(k,p)2cos2β(p,k)2−cos2β(k,−p)2cos2β(−p,k)2−sin2β(k,k)2sin2β(p,p)2, CG = cosβ(k,−p)cosβ(−p,k)−cosβ(k,p)cosβ(p,k)2−2sin2β(k,k)2sin2β(p,p)2 −cosα(k,p)(cosα(p,k)sin2β(k,−p)2+cosα(−p,−k)sin2β(k,p)2) −cosα(−k,−p)(cosα(p,k)sin2β(p,k)2+cosα(−p,−k)sin2β(−p,k)2), Cξ = sin2β(k,p)2sin2β(p,k)2−sin2β(k,k)2sin2β(p,p)2−sin2β(k,−p)2sin2β(−p,k)2 (11) +cosα(k,p)(cosα(p,k)sin2β(k,−p)2+cosα(−p,−k)sin2β(k,p)2) +cosα(−k,−p)(cosα(p,k)sin2β(p,k)2+cosα(−p,−k)sin2β(−p,k)2),\n\nand\n\n E1(p)= +δμ+12(|p+q|−|p−q|)+12√(|p+q|+|p−q|−2¯μ)2+4Δ2sin2β(p,p)2, E2(p)= −δμ−12(|p+q|−|p−q|)+12√(|p+q|+|p−q|−2¯μ)2+4Δ2sin2β(p,p)2, (12)\n\nwith\n\n cosα(k,p)=ˆ(k−q)⋅ˆ(k−p) , cosβ(k,p)=ˆ(q+k)⋅ˆ(q−p) . (13)\n\nIn the next section, we shall use this gap equation to obtain , the upper boundary of the crystalline color superconductivity window.\n\n## Iii The Zero-Gap Curve and the Calculation of δμ2\n\nSolving the full gap equation, Eq. (10), is numerically challenging. Fortunately, our task is simpler. As , the gap . Therefore, in order to determine we need only analyze the limit of Eq. (10). Upon taking this limit, we shall find a “zero-gap curve” relating and , as obtained for a point-like interaction in Refs. [25, 31, 10]. The largest value of on this curve is , and the at this point on the curve is the favored value of in the crystalline color superconducting phase for .\n\nTo take the limit, we first divide both sides of Eq. (10) by . We must be careful, however: simply cancelling the which now occurs on the right-hand side would yield an integral which diverges in the ultraviolet. This divergence is in fact regulated by the -dependence of itself. While the precise shape of remains unknown unless we solve the full gap equation (10), we assume has a similar dependence as in the BCS case [19, 23], where it is relatively flat for and then decreases rapidly to zero as increases to several times . This rapid decrease regulates the integral. This means that for , we can approximate by a step function which imposes an ultraviolet cutoff on the integral at , where is of order several times . We shall investigate the dependence of our results on the choice of . After the factor has been replaced by a cutoff, it is safe to simply set to zero everywhere else in the integrand.\n\nAfter rotating to Euclidean space we obtain\n\n 1 = g23sin2β(k,k)2∫Λd4p(2π)41(p0−iE1)(p0+iE2) (14) ×[CF(k−p)2+F(k−p)+CG(k−p)2+G(k−p)+Cξξ(k−p)2],\n\nwhere the energies can be obtained from Eq. (II) by setting equal to 0. We also simplify the functions and to\n\n F(p)=m2,G(p)=π4m2p0|p|, (15)\n\nvalid for and to leading order in perturbation theory. We shall work in the gauge with , as this will allow us to compare our results for to values of obtained in the same gauge in the analysis of the version of Eq. (10) in Ref. . As the gauge dependence in the calculation of decreases as is reduced below , we expect that the same is true here for . We leave an analysis of the -dependence of to the future. In evaluating the right-hand side of Eq. (14), we should set , but instead choose for simplicity as we expect to be almost constant for [19, 23]. We must choose in the region of momentum space which dominates the pairing: for , for example, one takes [19, 23]. Now, we choose on the ring in -space which describes pairs of quarks both of whose momenta ( and ) lie on their respective noninteracting Fermi surfaces, as this is where pairing is most important [24, 25, 10]. For , as opposed to , the ring degenerates to the point in momentum space with and antiparallel to .\n\nWe do the integral analytically. There are six poles in the complex plane: two from the quark propagator located at at and four from the gluon propagator located at , . The poles from the gluon propagator have residues which are smaller than those from the quark propagator by factors of . We therefore keep only the poles at and . Furthermore, we can drop the pieces in the gluon propagators because at the poles they are negligible. Upon doing the contour integral over , we notice that we only obtain a non-zero result if both and are positive. This defines the “pairing region” of Refs. [10, 29]. We get\n\n 1=g23sin2β(k,k)2∫ΛPp2dpdΩ(2π)31(E1+E2)(CF|k−p|2+F+CG|k−p|2+G), (16)\n\nwhere the integral is taken over the pairing region with an upper cutoff on the integral of . The remaining integrals are done numerically.\n\nUpon doing the integrals, Eq. (16) becomes an equation relating and . We explore the plane, evaluating the right-hand side of Eq. (16) at each point, and in so doing map out the “zero-gap curves” along which Eq. (16) is satisfied. Several zero-gap curves are shown in Fig. 1, in order to exhibit the dependence of the curve on the coupling and on the parameter . The top panel is for : if we use one-loop running with MeV and , this corresponds to for MeV. The lower panel is for corresponding in the same sense to MeV. The three curves in each plot are drawn with , 2.5, and 3.0. The choice of makes some difference to the scale of the zero-gap curves, but does not change the qualitative behavior. We are only interested in the qualitative behavior, since the couplings which are appropriate at accessible densities are much too large for the analysis to be under quantitative control anyway. In subsequent calculations, we shall take as our canonical choice. Removing the artificially introduced parameter would require solving the full gap equation (10).\n\nWe only show the zero-gap curve for , because the favored value of must lie in this region [24, 25, 31, 10]. Each zero-gap curve in Fig. 1 begins somewhere on the line , moves up and to the right following the line very closely, and then turns around sharply and heads downward, eventually reaching the axis. This behavior is most clearly seen in the upper panel. The smaller the value of , the more sharply the curve turns around and the more closely the downward-going curve hugs the line. Solutions to the full gap equation (10) with occur in the region bounded by the zero-gap curve, between it and the line. As is decreased, this region becomes a sharper and sharper sliver, squeezed closer and closer to the line. Since in dimensions is favored, the change in the zero-gap curves from the upper panel to the lower vividly demonstrates how the physics embodied in the gap equation (10) becomes effectively -dimensional as is reduced.\n\nThe largest value of on the zero-gap curve is . For , a range of values of can be found for which a crystalline color superconducting phase with exists. The within this range which results in a phase with the lowest free energy is favored, and the corresponding obtained by solving Eq. (10) at that is favored. At , however, . For each curve in Fig. 1, the location of is marked. In all cases, at is very close to .\n\nWe see from Fig. 1 that decreases as decreases, but this is not the important comparison. Recall that the crystalline color superconducting phase occurs within a window where at weak coupling, with the BCS gap obtained at . To evaluate the width of the window, we must therefore also compare to . We obtain from Ref. , and in Fig. 2 we plot the zero-gap curves for equal to MeV, MeV, and MeV, in each case scaled by the corresponding BCS gap . In contrast, the star marks the location of obtained in a theory with a weak point-like interaction: and .\n\nThe effect of the single-gluon exchange interaction, then, is to increase dramatically. Because this interaction favors forward scattering at weak coupling, it causes to approach the line and to diverge, as occurs in dimensions where there are only “Fermi points”. In Table 1, we show the values for different values of , with all calculations done for . Already at , corresponding to MeV, is within 1% of the line and . Although this result will depend somewhat on , it means that the crystalline color superconducting window is about ten times wider than for a point-like interaction. Once , corresponding to MeV, is closer than one part in a thousand to the line and , meaning that the crystalline color superconducting window is about four hundred times wider than for a point-like interaction.\n\n## Iv Implications and Future Work\n\nWe have found that diverges in QCD as the weak-coupling, high-density limit is taken. Applying results valid at asymptotically high densities to those of interest in compact stars, namely MeV, we find that even here the crystalline color superconductivity window is an order of magnitude wider than that obtained previously upon approximating the interaction between quarks as point-like.\n\nThis discovery has significant implications for the QCD phase diagram and may have significant implications for compact stars. At high enough baryon density the CFL phase in which all quarks pair to form a spatially uniform BCS condensate is favored. Suppose that as the density is lowered the nonzero strange quark mass induces the formation of some less symmetrically paired quark matter before the density is lowered so much that baryonic matter is obtained. In this less symmetric quark matter, some quarks may yet form a BCS condensate. Those which do not, however, will have differing Fermi momenta. These will form a crystalline color superconducting phase if the differences between their Fermi momenta lie within the appropriate window. In QCD, the interaction between quarks is forward-scattering dominated and the crystalline color superconductivity window is consequently wide open. This phase is therefore generic, occurring almost anywhere there are some quarks which cannot form BCS pairs. Evaluating the critical temperature above which the crystalline condensate melts requires solving the nonzero temperature gap equation obtained from Eq. (10) as described in Ref. , but we expect that all but the very youngest compact stars are colder than . This suggests that wherever quark matter which is not in the CFL phase occurs within a compact star, rotational vortices may be pinned resulting in the generation of glitches as the star spins down.\n\nSolidifying the implications of our results requires further work in several directions. First, we must confirm that pushing Fermi surfaces apart via quark mass differences has the same effect as pushing them apart via a introduced by hand. Second, we must extend the analysis to the three flavor theory of interest. Third, we need to re-evaluate by comparing the crystalline color superconducting phase to a BCS phase in which spin-1 pairing between quarks of the same flavor is allowed . The results of Ref. suggest that this can be neglected, but confirming this in the present case requires solving the full gap equation (10). And, fourth, before evaluating the pinning force on a rotational vortex and making predictions for glitch phenomena, we need to understand which crystal structure is favored.\n\n###### Acknowledgements.\nWe are grateful to J. Bowers, J. Kundu, D. Son, and F. Wilczek for helpful conversations. The work of AL is supported in part by the U.S. Department of Energy (DOE) under grant number DE-AC02-76CH03000. The work of KR and ES is supported in part by the DOE under cooperative research agreement DE-FC02-94ER40818. The work of KR is supported in part by a DOE OJI grant and by the Alfred P. Sloan Foundation."
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{"ft_lang_label":"__label__en","ft_lang_prob":0.9019044,"math_prob":0.98628485,"size":31607,"snap":"2023-40-2023-50","text_gpt3_token_len":7926,"char_repetition_ratio":0.14201184,"word_repetition_ratio":0.035782263,"special_character_ratio":0.25601923,"punctuation_ratio":0.15150064,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98663557,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T05:20:09Z\",\"WARC-Record-ID\":\"<urn:uuid:0b514240-0b7b-4051-8ae4-2da216535445>\",\"Content-Length\":\"744565\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:25aea9bb-5d26-41dc-83eb-c1ea3b3f0cbd>\",\"WARC-Concurrent-To\":\"<urn:uuid:7bd519ed-9c7f-437c-8ecc-9f3cb0fba6cb>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/hep-ph/0104073/\",\"WARC-Payload-Digest\":\"sha1:AF3Y36KDISKSQL6URHOUTHCWTSRO2TMP\",\"WARC-Block-Digest\":\"sha1:NQHOW7AINL2OZ2LAPBK5UTOF2BGVSDX2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510149.21_warc_CC-MAIN-20230926043538-20230926073538-00670.warc.gz\"}"}
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http://boxbase.org/entries/2015/mar/30/minitex/
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[
"# MiniTeX\n\nMiniTeX is a typesetting system implemented as a library in a programming language.\n\nThe document produced by MiniTeX is a box, which can be rendered on screen. The user may pick individual boxes inside the rendered document.\n\nDocuments are composed from combinators. They're functions that take in environment and produce a list of boxes. Here's an implementation of `par` -combinator:\n\n``````def par(env):\n```` return [vglue(env.font_size)]````\n\nCombinators are passed to functions that produce larger combinators, for exapmle `vbox` or `table`:\n\n``````vbox([\"first paragraph\", par, \"second paragraph\"])\n``````\n``````table([\n`````` [\"a\", \"b\", \"c\"],\n`````` [\"d\", \"f\", \"g\"],\n````])````\n\nThe user can produce his own combinators to supplement additional needs. Here's an implementation of `scope` -combinator:\n\n``````def scope(contents, values={}):\n`````` def _scope(env):\n`````` env = Environ.let(env, values)\n`````` for item in contents:\n`````` for box in fold(item, env):\n`````` yield box\n```` return _scope````\n\nThe `fold` function invokes a combinator. It allows strings literals to be to be treated as combinators, eg.\n\n``````fold(\"string\", env) -> [s t r i n g]\n````fold(par, env) -> [vglue(env.font_size)]````\n\nThe user may adjust rendering by giving parameters to the environment. For example scale and font size may be changed.\n\nFinally the combinators can be gathered into a list to produce the document and display it:\n\n``````env = Environ.root(width=screen.width)\n``````box = toplevel([\n`````` \"Hello World\"\n``````], env)\n``````print box.width, box.height, box.depth\n````display(box, x, y)````\n\nI've got one of these in textended-editor, but it's not suitable for a library release. I don't have good enough rendering infrastructure in place."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.75994027,"math_prob":0.76338655,"size":1581,"snap":"2021-21-2021-25","text_gpt3_token_len":379,"char_repetition_ratio":0.10779962,"word_repetition_ratio":0.0,"special_character_ratio":0.23908919,"punctuation_ratio":0.16721311,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.955501,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T15:36:24Z\",\"WARC-Record-ID\":\"<urn:uuid:8d3d674b-727e-4f61-b838-b402fcd563ec>\",\"Content-Length\":\"5143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7f0c4a26-40bb-408b-a4c7-da0f99adbf02>\",\"WARC-Concurrent-To\":\"<urn:uuid:39fcb6fe-30ca-4ca5-b56e-bd0ecbcfdf73>\",\"WARC-IP-Address\":\"185.179.239.7\",\"WARC-Target-URI\":\"http://boxbase.org/entries/2015/mar/30/minitex/\",\"WARC-Payload-Digest\":\"sha1:VHQD6WHO4K6C6G5Z4NN6YZWNQVQIUTU4\",\"WARC-Block-Digest\":\"sha1:BLVMGZLEPSZCV3OW56VZQCABJ4NSZG7H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991224.58_warc_CC-MAIN-20210516140441-20210516170441-00576.warc.gz\"}"}
|
https://www.physicsforums.com/threads/terminology-if-ds-2-s-t-interval-what-to-call-ds.952294/
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[
"# Terminology: if ds^2=s-t interval, what to call ds\n\n• I\nGold Member\n\n## Main Question or Discussion Point\n\nIn the usual relativistic equation, ds2 = (cdt)2 - dx2 - dy 2 -dz2 or dx2 + dy 2 + dz2 - (cdt)2, depending on the convention of your choice, and ds2 is called the spacetime interval between the corresponding events, the square being used to avoid nasty ambiguities and irritating imaginary numbers. Fine. But if one wants to refer to ds, is there any standard term for it besides the clumsy \"the square root of the spacetime interval\" or \"that which when squared gives the spacetime interval\"?\nThanks.\n\nRelated Special and General Relativity News on Phys.org\nPeroK\nHomework Helper\nGold Member\nIn the usual relativistic equation, ds2 = (cdt)2 - dx2 - dy 2 -dz2 or dx2 + dy 2 + dz2 - (cdt)2, depending on the convention of your choice, and ds2 is called the spacetime interval between the corresponding events, the square being used to avoid nasty ambiguities and irritating imaginary numbers. Fine. But if one wants to refer to ds, is there any standard term for it besides the clumsy \"the square root of the spacetime interval\" or \"that which when squared gives the spacetime interval\"?\nThanks.\nThe (infinitesimal) spacetime interval is the four-vector ##(c dt, dx, dy, dz)##. That's essentially what ##ds## is, although you should really use a vector notation, such as ##\\textbf{ds}##.\n\n##ds^2 = \\textbf{ds} \\cdot \\textbf{ds}##\n\nSo, ##ds^2## is really the inner product of the spacetime interval with itself. And, in a different context, this is the definition of the (in this case flat) spacetime metric.\n\nThe quantity ##\\sqrt{\\pm ds^2}## is the proper time or proper length of the spacetime interval, depending on whether ##ds^2## is positive or negative. If it's ##0## you have a null spacetime interval.\n\nLast edited:\n•",
null,
"Gold Member\nAh, thank you, PeroK. The explanation clears up the confusion. This confusion was caused by the Wikipedia site https://en.wikipedia.org/wiki/Spacetime#Spacetime_interval which calls the scalar quantity ds2 the spacetime interval, but your pointing out that the spacetime interval is the vector ds, and the scalar quantity the inner product ds⋅ds and so forth, makes sense.\n\nAndrew Mason\nHomework Helper\nIn the usual relativistic equation, ds2 = (cdt)2 - dx2 - dy 2 -dz2 or dx2 + dy 2 + dz2 - (cdt)2, depending on the convention of your choice, and ds2 is called the spacetime interval between the corresponding events, the square being used to avoid nasty ambiguities and irritating imaginary numbers. Fine. But if one wants to refer to ds, is there any standard term for it besides the clumsy \"the square root of the spacetime interval\" or \"that which when squared gives the spacetime interval\"?\nThanks.\nGenerally, s2 is a useful not because of what it represents physically but because it is a metric associated with two events that is the same in all reference frames (and because its sign tells us whether the interval is timelike or spacelike). It has physical significance only in certain frames: If they took place at the same location in a reference frame, s represents the time interval between the events (multiplied by the factor c); if the two events took place at the same time but at different locations in a particular reference frame, s = ##\\sqrt{s^2}## is the distance between the two events in that frame multiplied by the imaginary number i. For this reason, we refer to s2 rather than s as the metric for the space-time interval.\n\nAM\n\n•",
null,
"Gold Member\nThanks, Andrew Mason. That is a good elaboration of the last sentence of PeroK's last sentence when adopting the convention [+,-,-,-].\n\nrobphy\nHomework Helper\nGold Member\nFor ##ds^2##, I prefer “[spacetime] square-interval”.\nFor the 4-vector ##d\\vec s## (or ##d\\tilde s##, if I need to use ##d\\vec V## for 3-vectors), I prefer “[spacetime] displacement vector”.\nBoth may have to be prefixed by infinitesimal, if needed for clarity.\n\n•",
null,
"Gold Member\nThanks, robphy. Interesting suggestions. Are they standard (or at least one of the standards)?\n\nrobphy\nHomework Helper\nGold Member\nI haven’t surveyed the literature to see if my terms are standard. But one of my motivations is asking what their Euclidean or Galilean analogues would sound like.\n\nIn addition, my possibly annoying use of the hyphen suggests that it’s a single term “square-interval”, not to be mistakenly separated into an adjective “square” modifying an object noun “interval”.\n\n•",
null,
"Gold Member\nThanks, robphy. If your terms are acceptable and understood, then since there seems to be no set term, yours are also good.\n\nI'll add a lot more to this discussion, but I'll keep it without time for now, and i'll put it in spherical form. After typing this all out, i recognize this might not be what you were looking for, and moreso just wanted to know what to call this vector, but I'm not sure so might as well post.\n\nLet ##ds^2=dr^2+r^2d\\theta^2+r^2\\sin^2\\theta^2 d\\phi^2## then, as stated above, you can recognize some vector ##d\\vec{r} = dr\\hat{r}+rd\\theta \\hat{\\theta} + r\\sin\\theta d\\phi \\hat{\\phi}## In other words, ##d\\vec{r}## is a vector-valued one-form! Which is a great introduction into the wonderful world of differential forms.\n\nFrom here, you can recognize your basis as ##\\omega = \\{dr, rd\\theta, r\\sin\\theta d\\phi \\} = dr \\wedge rd\\theta \\wedge r\\sin\\theta d\\phi ##\n\nSo, what's the big deal? By putting things in this format, it's a lot easier to find your connections, curvature, etc which I will do for you so you can recognize how to do this in practice (also, i've been lazy this summer so, i might as well refresh myself).\n\nThis post would be way too long if I had to type out motivation for the formulas, but you can ask questions if you're unsure where the formulas come from!\n\nLet ##d\\hat{e_i} = a^j_i \\hat{e_j}## where ##a^j_i## is called a \"connection\" and the derivative in this post is called an \"exterior derivative\". Let's type out how our basis vectors change (Recognize that 1 refers to the first coordinate in my basis, 2 as second, 3 as third)\n##d\\hat{e_1} = a^1_1 \\hat{e_1} + a^2_1 \\hat{e_2} + a^3_1 \\hat{e_3} = 0 +a^2_1 \\hat{\\theta} + a^3_1 \\hat{\\phi} ##. Now, from the levi-cievta condition, we recognize that ##a^j_j = 0##\n##d\\hat{e_2} = a^1_2 \\hat{e_1} + a^2_2 \\hat{e_2} + a^3_2 \\hat{e_3} = a^1_2 \\hat{r} + 0 + a^3_2 \\hat{\\phi} ##\n##d\\hat{e_3} = a^1_3 \\hat{e_1} + a^2_3 \\hat{e_2} + a^3_3 \\hat{e_3} = a^1_3 \\hat{r} + a^2_3 \\hat{\\theta} + 0 ##\n\nNow, we must use our levi-cievta connection to complete how to solve for the connection. The Levi-cievta connection has two properties which are metric compatibility (If you're interested in these, I believe they come from your structure equations, but someone feel free to correct me!), which looks like ##a^j_i +a^i_j = 0## or ##a^j_i = -a^i_j## in orthongonal basis, but i believe in a general basis, this looks like ##a^j_i+a^i_j = dg_{ij}## and torsion free which looks like ##d{b^i} + a^i_j \\wedge b^j = 0## where ##b^i## is your ACTUAL components of your basis! That is, you have to recognize that ##b^2 = rd\\theta##. Since i decided to work in orthogonal basis, I have to take the whole thing. I think if you decide to not take the r in this, you'd be in a \"coordinate basis\" (once again, more math-y people, feel free to correct me).\n\nGreat, now using these two we can finally see how our basis changes from point to point, which will lead us to curvature!\nSo, going through the equation we get:\n##0 = d(b^1) + a^1_1 \\wedge b^1 + a^1_2 \\wedge b^2 + a^1_3 \\wedge b^3 = 0 + 0 + a^1_2 \\wedge rd\\theta + a^1_3 \\wedge r\\sin\\theta d\\phi## So here you need to know a little more about wedge products, and I'll supply the fact that if you have two of the same one forms wedged together, the equal zero. That is, ##dr \\wedge dr = d\\theta \\wedge d\\theta = d\\phi \\wedge d\\phi = 0## Which we must utilize in order to solve the equations we will have ahead! The other one is a property of the exterior derivative, which is: the second exterior derivative is equal to 0. Which is why ##d(b^1) = d(dr) = 0## in this case.\n\nSo just from my first one, I know that my ##a^1_2## MUST contain a ##d\\theta## term, otherwise I can't satisfy my torsion free condition. Like wise, ##a^1_3## MUST contain a ##d\\phi##. Once again, a natural question may arise if you haven't seen wedge products before! What about the ##r\\sin\\theta## term before the ##d\\phi##?\n\nWell, wedge products can be thought of as multiplication, so we can just bypass them by just wedging together the one forms. For example, if ##a^1_3 = d\\phi## we'd have ##d\\phi \\wedge r\\sin\\theta d\\phi = r\\sin\\theta d\\phi \\wedge d\\phi = 0## Or, if that doesn't satisfy you, you can also use the property of wedge products that switching the order of the placements will give you a minus sign (this is known as an \"anti-symmetric\" property). So, using this, you can see that ##d\\phi \\wedge r\\sin\\theta d\\phi = - r\\sin\\theta d\\phi \\wedge d\\phi = 0## We will be using the anti-symmetric property on the other two torsion free equations!\n\n##0 = d(b^2) + a^2_1 \\wedge b^1 + a^2_2 \\wedge b^2 + a^2_3 \\wedge b^3 = d(rd\\theta) + a^2_1 \\wedge dr + 0 + a^2_3 \\wedge r\\sin\\theta d\\phi = dr \\wedge d\\theta +a^2_1 \\wedge dr + a^2_3 + \\wedge r\\sin\\theta d\\phi = dr \\wedge d\\theta - dr \\wedge a^2_1 + a^2_3 + \\wedge r\\sin\\theta d\\phi = 0##\n\nNow, let's pause here! Where did that ##dr \\wedge rd\\theta## term come from? It came from the product rule with derivatives (yes, exterior derivatives still follow this rule!) ##d(rd\\theta) = dr \\wedge d\\theta + rd^2\\theta = dr \\wedge d\\theta + 0## and the ##- dr \\wedge a^2_1## came from the anti-symmetric property of wedge products, and I used it because I want to get rid of that first term! I have to recognize that ##a^2_1 = d\\theta## WHICH IS GREAT! Recall from my first torsion free equation, I recognized that ##a^1_2## HAD to have ##d\\theta## in it! Even better, this is basically just the metric compatibility condition! That is, if ##a^1_2 = -a^2_1##, or in our case, ##a^2_1 = d\\theta## therefore ##a^1_2 = -d\\theta##\n\nThe only thing left, is we must recognize that ##a^2_3## HAS to have a term ##d\\phi## in it to satisfy the torsion free condition.\n\nFinally, we have: ## 0 = d(b^3) + a^3_1 \\wedge b^1 + a^3_2 \\wedge b^2 + a^3_3 \\wedge b^3 = d(r\\sin\\theta d\\phi) + a^3_1 \\wedge dr + a^3_2 \\wedge rd\\theta + 0 = dr \\wedge \\sin\\theta d\\phi + r\\cos\\theta d\\theta \\wedge d\\phi - dr \\wedge a^3_1 - rd\\theta \\wedge a^3_2 = 0##\n\nOnce again, we used product rule for derivatives, and used the anti symmetric property of wedge products. Here, we recognize that ##a^3_1 = \\sin\\theta d\\phi## where we NEED all the stuff out front because in order to completely get rid of the first term, we need all of those (You can see this by imagining if I factored out a dr \\wedge d\\phi from the equation, what would you have left?) and then let's try to get ##a^3_2##.\n\nIf recall from my 2nd torsion free equation that ##a^2_3## had to have something with ##d\\phi## in it. So, let's see if it was just ##d\\phi##.\n\nI'd get ##rcos\\theta d\\theta \\wedge d\\phi - rd\\theta \\wedge d\\phi## if I factor out a ##d\\theta \\wedge d\\phi## term, I'll be left with ##d\\theta \\wedge d\\phi(r\\cos\\theta - r)## Well, that is obviously not 0! So I know that I have to have a ##\\cos\\theta## term in my ##a^3_2##. That is, ##a^3_2 = \\cos\\theta d\\phi## which would make ##a^2_3 = -\\cos\\theta d\\phi##\n\nNow, you can plug these all in, and check if you get zero. So finally, we can see how our basis changes!\n\nThat is:\n##d\\hat{r} = d\\theta \\hat{\\theta} + \\sin\\theta d\\phi \\hat{\\phi}##\n##d\\hat{\\theta} = -d\\theta \\hat{r} + \\cos\\theta d\\phi \\hat{\\phi}##\n##d\\hat{\\phi} = -\\sin\\theta d\\phi \\hat{r} -\\cos\\theta d\\phi \\hat{\\theta}##\n\nCurvature is then found with the equation: ##\\Omega^i_j = d(a^i_j) + a^i_k \\wedge a^k_j## (This is called the curvature 2-form, and it also has the property ##\\Omega^i_j = -\\Omega^j_1##) This post is long enough, so you can try this out on your own!\n\nAnd last, but not least, with that vector valued one form ##d\\vec{r}## you can calculate your geodesics! First recognize that a geodesic is when we have no acceleration, that is ##\\vec{a} = 0##. How do we get this from ##d\\vec{r}##?\n\nOnce again, long enough post, but the way you do it is: Start with ##d\\vec{r}## then divide by ##dt##. To make this one explicit, I'll do the first step so you can see what i mean by \"divide\".\n\nLet ##d\\vec{r} = dr\\hat{r}+rd\\theta \\hat{\\theta} + r\\sin\\theta d\\phi \\hat{\\phi}## If I want to find velocity, that is just the change of my position with respect to time, or \"dividing\" by dt. So, let's do that ##\\vec{v} = \\frac{d\\vec{r}}{dt} = \\dot{r}\\hat{r}+r\\dot{\\theta} \\hat{\\theta} + r\\sin\\theta \\dot{\\phi} \\hat{\\phi}## Where the dots over the variables are derivatives with respect to time as usual!\n\nFrom here, find ##d\\vec{v}## then \"divide\" by dt, then you have ##\\vec{a}## now set it equal to zero, and solve your differential equations!\n\nSo, long post is done! Sorry if this was more than you wanted, but there is so much that can be done with that ##d\\vec{r}##! The math doesn't change by adding another dimension, but it just gets longer. Which is why I did a 3D one!\n\nSorry for any typos.\n\n•",
null,
"Gold Member\nThanks, romsofia. Although this was more than I asked for, it is not superfluous. There is a lot in your post that I must slowly digest, and perhaps I will come back to you later to ask questions on specific points, if this is OK. In the meantime, a question concerning post #2 of PeroK (I am slow to react)\nThe quantity √±ds2\\sqrt{\\pm ds^2} is the proper time or proper length of the spacetime interval, depending on whether ds2ds^2 is positive or negative. If it's 00 you have a null spacetime interval.\nFirst, I thought that whether ±ds^2 is positive or negative depended not only on the separation of the events (i.e., whether (ct)2 is greater or less than the sum of the squares of the spatial coordinate changes) but also upon which convention one used [+, -, -, -] or [-, + ,+, +], so it would seem to say that whether ±ds^2 corresponds to a proper time or a proper distance would depend on which convention one selected? That sounds strange, so I presume I am missing something here. Secondly, if the same quantity [hence the same units] can be either time or length, I am presuming that one is setting c=1? Furthermore, in Post #4, Andrew Mason states that such a correspondence only works in the special cases -- or did I misunderstand that?\n\nPeroK\nHomework Helper\nGold Member\nThanks, romsofia. Although this was more than I asked for, it is not superfluous. There is a lot in your post that I must slowly digest, and perhaps I will come back to you later to ask questions on specific points, if this is OK. In the meantime, a question concerning post #2 of PeroK (I am slow to react)\n\nFirst, I thought that whether ±ds^2 is positive or negative depended not only on the separation of the events (i.e., whether (ct)2 is greater or less than the sum of the squares of the spatial coordinate changes) but also upon which convention one used [+, -, -, -] or [-, + ,+, +], so it would seem to say that whether ±ds^2 corresponds to a proper time or a proper distance would depend on which convention one selected? That sounds strange, so I presume I am missing something here. Secondly, if the same quantity [hence the same units] can be either time or length, I am presuming that one is setting c=1? Furthermore, in Post #4, Andrew Mason states that such a correspondence only works in the special cases -- or did I misunderstand that?\nWhether it is proper time or proper length depends on the sign and the convention. In any case, Timelike means that the time component is greater than the spatial component. Or, alternatively, that there is a reference frame in which the spatial component is zero. Spacelike means the opposite, and there is reference frame where the time component is zero.\n\nIf you multiply a time by a speed (of light) you get a length. You can either have ##(t, x, y, z)## or ##(ct, x, y, z)## or, use units where ##c =1##.\n\nBut, you have to be clear which one you are using.\n\nGold Member\nThanks very much, PeroK. That makes sense.\n\nIf you multiply a time by a speed (of light) you get a length.\nMaybe I'm being pedantic, but I think it's better to say: \"If you multiply a time by the speed of light, you get that same time expressed in length-units\" (assuming the time was originally expressed in time-units).\n\n•",
null,
"Gold Member\nThanks, SiennaTheGr8. You are of course correct; that is probably what PeroK meant by\nBut, you have to be clear which one you are using.\n\nAndrew Mason\nHomework Helper\nFirst, I thought that whether ±ds^2 is positive or negative depended not only on the separation of the events (i.e., whether (ct)2 is greater or less than the sum of the squares of the spatial coordinate changes) but also upon which convention one used [+, -, -, -] or [-, + ,+, +], so it would seem to say that whether ±ds^2 corresponds to a proper time or a proper distance would depend on which convention one selected? That sounds strange, so I presume I am missing something here.\nWhat the sign signifies depends on the convention. If you use ## s^2 = (ct)^2 - x^2 - y^2 - z^2## as the space-time interval, then the interval between the events will be timelike if positive and spacelike if negative. It is the other way around if you use the convention ## s^2 = x^2 + y^2 + z^2 -(ct)^2##.\n\nFurthermore, in Post #4, Andrew Mason states that such a correspondence only works in the special cases -- or did I misunderstand that?\nWhat I said was that ##s = \\sqrt{s^2}## corresponds to something physical only in certain reference frames where the interval is either purely time or purely distance. Otherwise, s2 is just a useful metric (because it is has the same value in all reference frames).\n\nAM\n\n•",
null,
""
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https://en.m.wikipedia.org/wiki/Fermat_prime
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"# Fermat number\n\n(Redirected from Fermat prime)\n\nIn mathematics a Fermat number, named after Pierre de Fermat who first studied them, is a positive integer of the form\n\nNamed after Pierre de Fermat 5 5 Fermat numbers 3, 5, 17, 257, 65537 65537 A019434\n$F_{n}=2^{2^{n}}+1,$",
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"where n is a nonnegative integer. The first few Fermat numbers are:\n\n3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, … (sequence A000215 in the OEIS).\n\nIf 2k + 1 is prime, and k > 0, it can be shown that k must be a power of two. (If k = ab where 1 ≤ a, bk and b is odd, then 2k + 1 = (2a)b + 1 ≡ (−1)b + 1 = 0 (mod 2a + 1). See below for a complete proof.) In other words, every prime of the form 2k + 1 (other than 2 = 20 + 1) is a Fermat number, and such primes are called Fermat primes. As of 2019, the only known Fermat primes are F0, F1, F2, F3, and F4 (sequence A019434 in the OEIS).\n\n## Basic properties\n\nThe Fermat numbers satisfy the following recurrence relations:\n\n$F_{n}=(F_{n-1}-1)^{2}+1$\n\nfor n ≥ 1,\n\n$F_{n}=F_{n-1}+2^{2^{n-1}}F_{0}\\cdots F_{n-2}$\n$F_{n}=F_{n-1}^{2}-2(F_{n-2}-1)^{2}$\n$F_{n}=F_{0}\\cdots F_{n-1}+2$\n\nfor n ≥ 2. Each of these relations can be proved by mathematical induction. From the last equation, we can deduce Goldbach's theorem (named after Christian Goldbach): no two Fermat numbers share a common integer factor greater than 1. To see this, suppose that 0 ≤ i < j and Fi and Fj have a common factor a > 1. Then a divides both\n\n$F_{0}\\cdots F_{j-1}$\n\nand Fj; hence a divides their difference, 2. Since a > 1, this forces a = 2. This is a contradiction, because each Fermat number is clearly odd. As a corollary, we obtain another proof of the infinitude of the prime numbers: for each Fn, choose a prime factor pn; then the sequence {pn} is an infinite sequence of distinct primes.\n\n## Primality of Fermat numbers\n\nFermat numbers and Fermat primes were first studied by Pierre de Fermat, who conjectured (but admitted he could not prove) that all Fermat numbers are prime. Indeed, the first five Fermat numbers F0,...,F4 are easily shown to be prime. However, the conjecture was refuted by Leonhard Euler in 1732 when he showed that\n\n$F_{5}=2^{2^{5}}+1=2^{32}+1=4294967297=641\\times 6700417.$\n\nEuler proved that every factor of Fn must have the form k2n+1 + 1 (later improved to k2n+2 + 1 by Lucas).\n\nThe fact that 641 is a factor of F5 can be easily deduced from the equalities 641 = 27×5+1 and 641 = 24 + 54. It follows from the first equality that 27×5 ≡ −1 (mod 641) and therefore (raising to the fourth power) that 228×54 ≡ 1 (mod 641). On the other hand, the second equality implies that 54 ≡ −24 (mod 641). These congruences imply that −232 ≡ 1 (mod 641).\n\nFermat was probably aware of the form of the factors later proved by Euler, so it seems curious why he failed to follow through on the straightforward calculation to find the factor. One common explanation is that Fermat made a computational mistake.\n\nThere are no other known Fermat primes Fn with n > 4. However, little is known about Fermat numbers with large n. In fact, each of the following is an open problem:\n\nAs of 2014, it is known that Fn is composite for 5 ≤ n ≤ 32, although amongst these, complete factorizations of Fn are known only for 0 ≤ n ≤ 11, and there are no known prime factors for n = 20 and n = 24. The largest Fermat number known to be composite is F3329780, and its prime factor 193 × 23329782 + 1, a megaprime, was discovered by the PrimeGrid collaboration in July 2014.\n\n### Heuristic arguments for density\n\nThere are several probabilistic arguments for the finitude of Fermat primes.\n\nAccording to the prime number theorem, the \"probability\" that a number n is prime is about 1/ln(n). Therefore, the total expected number of Fermat primes is at most\n\n{\\begin{aligned}\\sum _{n=0}^{\\infty }{\\frac {1}{\\ln F_{n}}}&={\\frac {1}{\\ln 2}}\\sum _{n=0}^{\\infty }{\\frac {1}{\\log _{2}\\left(2^{2^{n}}+1\\right)}}\\\\&<{\\frac {1}{\\ln 2}}\\sum _{n=0}^{\\infty }2^{-n}\\\\&={\\frac {2}{\\ln 2}}.\\end{aligned}}\n\nThis argument is not a rigorous proof. For one thing, the argument assumes that Fermat numbers behave \"randomly\", yet we have already seen that the factors of Fermat numbers have special properties.\n\nIf (more sophisticatedly) we regard the conditional probability that n is prime, given that we know all its prime factors exceed B, as at most A ln(B) / ln(n), then using Euler's theorem that the least prime factor of Fn exceeds 2n + 1, we would find instead\n\n{\\begin{aligned}A\\sum _{n=0}^{\\infty }{\\frac {\\ln 2^{n+1}}{\\ln F_{n}}}&=A\\sum _{n=0}^{\\infty }{\\frac {\\log _{2}2^{n+1}}{\\log _{2}\\left(2^{2^{n}}+1\\right)}}\\\\&\n\n### Equivalent conditions of primality\n\nLet $F_{n}=2^{2^{n}}+1$ be the nth Fermat number. Pépin's test states that for n > 0,\n\n$F_{n}$ is prime if and only if $3^{(F_{n}-1)/2}\\equiv -1{\\pmod {F_{n}}}.$\n\nThe expression $3^{(F_{n}-1)/2}$ can be evaluated modulo $F_{n}$ by repeated squaring. This makes the test a fast polynomial-time algorithm. However, Fermat numbers grow so rapidly that only a handful of Fermat numbers can be tested in a reasonable amount of time and space.\n\nThere are some tests that can be used to test numbers of the form k2m + 1, such as factors of Fermat numbers, for primality.\n\nProth's theorem (1878). Let N = k2m + 1 with odd k < 2m. If there is an integer a such that\n$a^{(N-1)/2}\\equiv -1{\\pmod {N}}$\nthen N is prime. Conversely, if the above congruence does not hold, and in addition\n$\\left({\\frac {a}{N}}\\right)=-1$ (See Jacobi symbol)\nthen N is composite.\n\nIf N = Fn > 3, then the above Jacobi symbol is always equal to −1 for a = 3, and this special case of Proth's theorem is known as Pépin's test. Although Pépin's test and Proth's theorem have been implemented on computers to prove the compositeness of some Fermat numbers, neither test gives a specific nontrivial factor. In fact, no specific prime factors are known for n = 20 and 24.\n\n## Factorization of Fermat numbers\n\nBecause of the size of Fermat numbers, it is difficult to factorize or even to check primality. Pépin's test gives a necessary and sufficient condition for primality of Fermat numbers, and can be implemented by modern computers. The elliptic curve method is a fast method for finding small prime divisors of numbers. Distributed computing project Fermatsearch has successfully found some factors of Fermat numbers. Yves Gallot's proth.exe has been used to find factors of large Fermat numbers. Édouard Lucas, improving the above-mentioned result by Euler, proved in 1878 that every factor of Fermat number $F_{n}$ , with n at least 2, is of the form $k\\times 2^{n+2}+1$ (see Proth number), where k is a positive integer. By itself, this makes it easy to prove the primality of the known Fermat primes.\n\nFactorizations of the first twelve Fermat numbers are:\n\n F0 = 21 + 1 = 3 is prime F1 = 22 + 1 = 5 is prime F2 = 24 + 1 = 17 is prime F3 = 28 + 1 = 257 is prime F4 = 216 + 1 = 65,537 is the largest known Fermat prime F5 = 232 + 1 = 4,294,967,297 = 641 × 6,700,417 (fully factored 1732) F6 = 264 + 1 = 18,446,744,073,709,551,617 (20 digits) = 274,177 × 67,280,421,310,721 (14 digits) (fully factored 1855) F7 = 2128 + 1 = 340,282,366,920,938,463,463,374,607,431,768,211,457 (39 digits) = 59,649,589,127,497,217 (17 digits) × 5,704,689,200,685,129,054,721 (22 digits) (fully factored 1970) F8 = 2256 + 1 = 115,792,089,237,316,195,423,570,985,008,687,907,853,269,984,665,640,564,039,457,584,007,913,129,639,937 (78 digits) = 1,238,926,361,552,897 (16 digits) × 93,461,639,715,357,977,769,163,558,199,606,896,584,051,237,541,638,188,580,280,321 (62 digits) (fully factored 1980) F9 = 2512 + 1 = 13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,030,073,546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,097 (155 digits) = 2,424,833 × 7,455,602,825,647,884,208,337,395,736,200,454,918,783,366,342,657 (49 digits) × 741,640,062,627,530,801,524,787,141,901,937,474,059,940,781,097,519,023,905,821,316,144,415,759,504,705,008,092,818,711,693,940,737 (99 digits) (fully factored 1990) F10 = 21024 + 1 = 179,769,313,486,231,590,772,930...304,835,356,329,624,224,137,217 (309 digits) = 45,592,577 × 6,487,031,809 × 4,659,775,785,220,018,543,264,560,743,076,778,192,897 (40 digits) × 130,439,874,405,488,189,727,484...806,217,820,753,127,014,424,577 (252 digits) (fully factored 1995) F11 = 22048 + 1 = 32,317,006,071,311,007,300,714,8...193,555,853,611,059,596,230,657 (617 digits) = 319,489 × 974,849 × 167,988,556,341,760,475,137 (21 digits) × 3,560,841,906,445,833,920,513 (22 digits) × 173,462,447,179,147,555,430,258...491,382,441,723,306,598,834,177 (564 digits) (fully factored 1988)\n\nAs of 2018, only F0 to F11 have been completely factored. The distributed computing project Fermat Search is searching for new factors of Fermat numbers. The set of all Fermat factors is A050922 (or, sorted, A023394) in OEIS.\n\nIt is possible that the only primes of this form are 3, 5, 17, 257 and 65,537. Indeed, Boklan and Conway published in 2016 a very precise analysis suggesting that the probability of the existence of another Fermat prime is less than one in a billion.\n\nThe following factors of Fermat numbers were known before 1950 (since the '50s, digital computers have helped find more factors):\n\nYear Finder Fermat number Factor\n1732 Euler $F_{5}$ $5\\cdot 2^{7}+1$\n1732 Euler $F_{5}$ (fully factored) $52347\\cdot 2^{7}+1$\n1855 Clausen $F_{6}$ $1071\\cdot 2^{8}+1$\n1855 Clausen $F_{6}$ (fully factored) $262814145745\\cdot 2^{8}+1$\n1877 Pervushin $F_{12}$ $7\\cdot 2^{14}+1$\n1878 Pervushin $F_{23}$ $5\\cdot 2^{25}+1$\n1886 Seelhoff $F_{36}$ $5\\cdot 2^{39}+1$\n1899 Cunningham $F_{11}$ $39\\cdot 2^{13}+1$\n1899 Cunningham $F_{11}$ $119\\cdot 2^{13}+1$\n1903 Western $F_{9}$ $37\\cdot 2^{16}+1$\n1903 Western $F_{12}$ $397\\cdot 2^{16}+1$\n1903 Western $F_{12}$ $973\\cdot 2^{16}+1$\n1903 Western $F_{18}$ $13\\cdot 2^{20}+1$\n1903 Cullen $F_{38}$ $3\\cdot 2^{41}+1$\n1906 Morehead $F_{73}$ $5\\cdot 2^{75}+1$\n1925 Kraitchik $F_{15}$ $579\\cdot 2^{21}+1$\n\nAs of March 2019, 349 prime factors of Fermat numbers are known, and 305 Fermat numbers are known to be composite. Several new Fermat factors are found each year.\n\n## Pseudoprimes and Fermat numbers\n\nLike composite numbers of the form 2p − 1, every composite Fermat number is a strong pseudoprime to base 2. This is because all strong pseudoprimes to base 2 are also Fermat pseudoprimes - i.e.\n\n$2^{F_{n}-1}\\equiv 1{\\pmod {F_{n}}}$\n\nfor all Fermat numbers.\n\nIn 1904, Cipolla showed that the product of at least two distinct prime or composite Fermat numbers $F_{a}F_{b}\\dots F_{s},a>b>\\dots >s>1$ will be a Fermat pseudoprime to base 2 if and only if $2^{s}>a$ .\n\n## Other theorems about Fermat numbers\n\nLemma. — If n is a positive integer,\n\n$a^{n}-b^{n}=(a-b)\\sum _{k=0}^{n-1}a^{k}b^{n-1-k}.$\n\nTheorem — If $2^{k}+1$ is an odd prime, then $k$ is a power of 2.\n\nTheorem — A Fermat prime cannot be a Wieferich prime.\n\nTheorem — Any prime divisor p of $F_{n}=2^{2^{n}}+1$ is of the form $k2^{n+2}+1$ whenever n > 1.\n\n## Relationship to constructible polygons\n\nCarl Friedrich Gauss developed the theory of Gaussian periods in his Disquisitiones Arithmeticae and formulated a sufficient condition for the constructibility of regular polygons. Gauss stated without proof that this condition was also necessary, but never published his proof. A full proof of necessity was given by Pierre Wantzel in 1837. The result is known as the Gauss–Wantzel theorem:\n\nAn n-sided regular polygon can be constructed with compass and straightedge if and only if n is the product of a power of 2 and distinct Fermat primes: in other words, if and only if n is of the form n = 2kp1p2ps, where k is a nonnegative integer and the pi are distinct Fermat primes.\n\nA positive integer n is of the above form if and only if its totient φ(n) is a power of 2.\n\n## Applications of Fermat numbers\n\n### Pseudorandom Number Generation\n\nFermat primes are particularly useful in generating pseudo-random sequences of numbers in the range 1 … N, where N is a power of 2. The most common method used is to take any seed value between 1 and P − 1, where P is a Fermat prime. Now multiply this by a number A, which is greater than the square root of P and is a primitive root modulo P (i.e., it is not a quadratic residue). Then take the result modulo P. The result is the new value for the RNG.\n\n$V_{j+1}=\\left(A\\times V_{j}\\right){\\bmod {P}}$ (see linear congruential generator, RANDU)\n\nThis is useful in computer science since most data structures have members with 2X possible values. For example, a byte has 256 (28) possible values (0–255). Therefore, to fill a byte or bytes with random values a random number generator which produces values 1–256 can be used, the byte taking the output value − 1. Very large Fermat primes are of particular interest in data encryption for this reason. This method produces only pseudorandom values as, after P − 1 repetitions, the sequence repeats. A poorly chosen multiplier can result in the sequence repeating sooner than P − 1.\n\n## Other interesting facts\n\nA Fermat number cannot be a perfect number or part of a pair of amicable numbers. (Luca 2000)\n\nThe series of reciprocals of all prime divisors of Fermat numbers is convergent. (Křížek, Luca & Somer 2002)\n\nIf nn + 1 is prime, there exists an integer m such that n = 22m. The equation nn + 1 = F(2m+m) holds in that case.\n\nLet the largest prime factor of Fermat number Fn be P(Fn). Then,\n\n$P(F_{n})\\geq 2^{n+2}(4n+9)+1.$ (Grytczuk, Luca & Wójtowicz 2001)\n\n## Generalized Fermat numbers\n\nNumbers of the form $a^{2^{\\overset {n}{}}}+b^{2^{\\overset {n}{}}}$ with a, b any coprime integers, a > b > 0, are called generalized Fermat numbers. An odd prime p is a generalized Fermat number if and only if p is congruent to 1 (mod 4). (Here we consider only the case n > 0, so 3 = $2^{2^{0}}+1$ is not a counterexample.)\n\nAn example of a probable prime of this form is 12465536 + 5765536 (found by Valeryi Kuryshev).\n\nBy analogy with the ordinary Fermat numbers, it is common to write generalized Fermat numbers of the form $a^{2^{\\overset {n}{}}}+1$ as Fn(a). In this notation, for instance, the number 100,000,001 would be written as F3(10). In the following we shall restrict ourselves to primes of this form, $a^{2^{\\overset {n}{}}}+1$ , such primes are called \"Fermat primes base a\". Of course, these primes exist only if a is even.\n\nIf we require n > 0, then Landau's fourth problem asks if there are infinitely many generalized Fermat primes Fn(a).\n\n### Generalized Fermat primes\n\nBecause of the ease of proving their primality, generalized Fermat primes have become in recent years a topic for research within the field of number theory. Many of the largest known primes today are generalized Fermat primes.\n\nGeneralized Fermat numbers can be prime only for even a, because if a is odd then every generalized Fermat number will be divisible by 2. The smallest prime number $F_{n}(a)$ with $n>4$ is $F_{5}(30)$ , or 3032+1. Besides, we can define \"half generalized Fermat numbers\" for an odd base, a half generalized Fermat number to base a (for odd a) is ${\\frac {a^{2^{n}}+1}{2}}$ , and it is also to be expected that there will be only finitely many half generalized Fermat primes for each odd base.\n\n(In the list, the generalized Fermat numbers ($F_{n}(a)$ ) to an even a are $a^{2^{n}}+1$ , for odd a, they are ${\\frac {a^{2^{n}}+1}{2}}$ . If a is a perfect power with an odd exponent (sequence A070265 in the OEIS), then all generalized Fermat number can be algebraic factored, so they cannot be prime)\n\n(For the smallest number $n$ such that $F_{n}(a)$ is prime, see )\n\n$a$ numbers $n$\nsuch that\n$F_{n}(a)$ is prime\n$a$ numbers $n$\nsuch that\n$F_{n}(a)$ is prime\n$a$ numbers $n$\nsuch that\n$F_{n}(a)$ is prime\n$a$ numbers $n$\nsuch that\n$F_{n}(a)$ is prime\n2 0, 1, 2, 3, 4, ... 18 0, ... 34 2, ... 50 ...\n3 0, 1, 2, 4, 5, 6, ... 19 1, ... 35 1, 2, 6, ... 51 1, 3, 6, ...\n4 0, 1, 2, 3, ... 20 1, 2, ... 36 0, 1, ... 52 0, ...\n5 0, 1, 2, ... 21 0, 2, 5, ... 37 0, ... 53 3, ...\n6 0, 1, 2, ... 22 0, ... 38 ... 54 1, 2, 5, ...\n7 2, ... 23 2, ... 39 1, 2, ... 55 ...\n8 (none) 24 1, 2, ... 40 0, 1, ... 56 1, 2, ...\n9 0, 1, 3, 4, 5, ... 25 0, 1, ... 41 4, ... 57 0, 2, ...\n10 0, 1, ... 26 1, ... 42 0, ... 58 0, ...\n11 1, 2, ... 27 (none) 43 3, ... 59 1, ...\n12 0, ... 28 0, 2, ... 44 4, ... 60 0, ...\n13 0, 2, 3, ... 29 1, 2, 4, ... 45 0, 1, ... 61 0, 1, 2, ...\n14 1, ... 30 0, 5, ... 46 0, 2, 9, ... 62 ...\n15 1, ... 31 ... 47 3, ... 63 ...\n16 0, 1, 2, ... 32 (none) 48 2, ... 64 (none)\n17 2, ... 33 0, 3, ... 49 1, ... 65 1, 2, 5, ...\n b known generalized (half) Fermat prime base b 2 3, 5, 17, 257, 65537 3 2, 5, 41, 21523361, 926510094425921, 1716841910146256242328924544641 4 5, 17, 257, 65537 5 3, 13, 313 6 7, 37, 1297 7 1201 8 (not possible) 9 5, 41, 21523361, 926510094425921, 1716841910146256242328924544641 10 11, 101 11 61, 7321 12 13 13 7, 14281, 407865361 14 197 15 113 16 17, 257, 65537 17 41761 18 19 19 181 20 401, 160001 21 11, 97241, 1023263388750334684164671319051311082339521 22 23 23 139921 24 577, 331777 25 13, 313 26 677 27 (not possible) 28 29, 614657 29 421, 353641, 125123236840173674393761 30 31, 185302018885184100000000000000000000000000000001 31 32 (not possible) 33 17, 703204309121 34 1336337 35 613, 750313, 330616742651687834074918381127337110499579842147487712949050636668246738736343104392290115356445313 36 37, 1297 37 19 38 39 761, 1156721 40 41, 1601 41 31879515457326527173216321 42 43 43 5844100138801 44 197352587024076973231046657 45 23, 1013 46 47, 4477457, 46512+1 (852 digits: 214787904487...289480994817) 47 11905643330881 48 5308417 49 1201 50\n\n(See for more information (even bases up to 1000), also see for odd bases)\n\n(For the smallest prime of the form $F_{n}(a,b)$ (for odd $a+b$ ), see also )\n\n$a$ $b$ numbers $n$ such that\n${\\frac {a^{2^{n}}+b^{2^{n}}}{\\gcd(a+b,2)}}(=F_{n}(a,b))$\nis prime\n2 1 0, 1, 2, 3, 4, ...\n3 1 0, 1, 2, 4, 5, 6, ...\n3 2 0, 1, 2, ...\n4 1 0, 1, 2, 3, ...\n4 3 0, 2, 4, ...\n5 1 0, 1, 2, ...\n5 2 0, 1, 2, ...\n5 3 1, 2, 3, ...\n5 4 1, 2, ...\n6 1 0, 1, 2, ...\n6 5 0, 1, 3, 4, ...\n7 1 2, ...\n7 2 1, 2, ...\n7 3 0, 1, 8, ...\n7 4 0, 2, ...\n7 5 1, 4, ...\n7 6 0, 2, 4, ...\n8 1 (none)\n8 3 0, 1, 2, ...\n8 5 0, 1, 2, ...\n8 7 1, 4, ...\n9 1 0, 1, 3, 4, 5, ...\n9 2 0, 2, ...\n9 4 0, 1, ...\n9 5 0, 1, 2, ...\n9 7 2, ...\n9 8 0, 2, 5, ...\n10 1 0, 1, ...\n10 3 0, 1, 3, ...\n10 7 0, 1, 2, ...\n10 9 0, 1, 2, ...\n11 1 1, 2, ...\n11 2 0, 2, ...\n11 3 0, 3, ...\n11 4 1, 2, ...\n11 5 1, ...\n11 6 0, 1, 2, ...\n11 7 2, 4, 5, ...\n11 8 0, 6, ...\n11 9 1, 2, ...\n11 10 5, ...\n12 1 0, ...\n12 5 0, 4, ...\n12 7 0, 1, 3, ...\n12 11 0, ...\n13 1 0, 2, 3, ...\n13 2 1, 3, 9, ...\n13 3 1, 2, ...\n13 4 0, 2, ...\n13 5 1, 2, 4, ...\n13 6 0, 6, ...\n13 7 1, ...\n13 8 1, 3, 4, ...\n13 9 0, 3, ...\n13 10 0, 1, 2, 4, ...\n13 11 2, ...\n13 12 1, 2, 5, ...\n14 1 1, ...\n14 3 0, 3, ...\n14 5 0, 2, 4, 8, ...\n14 9 0, 1, 8, ...\n14 11 1, ...\n14 13 2, ...\n15 1 1, ...\n15 2 0, 1, ...\n15 4 0, 1, ...\n15 7 0, 1, 2, ...\n15 8 0, 2, 3, ...\n15 11 0, 1, 2, ...\n15 13 1, 4, ...\n15 14 0, 1, 2, 4, ...\n16 1 0, 1, 2, ...\n16 3 0, 2, 8, ...\n16 5 1, 2, ...\n16 7 0, 6, ...\n16 9 1, 3, ...\n16 11 2, 4, ...\n16 13 0, 3, ...\n16 15 0, ...\n\n(For the smallest even base a such that $F_{n}(a)$ is prime, see )\n\n$n$ bases a such that $F_{n}(a)$ is prime (only consider even a) OEIS sequence\n0 2, 4, 6, 10, 12, 16, 18, 22, 28, 30, 36, 40, 42, 46, 52, 58, 60, 66, 70, 72, 78, 82, 88, 96, 100, 102, 106, 108, 112, 126, 130, 136, 138, 148, 150, ... A006093\n1 2, 4, 6, 10, 14, 16, 20, 24, 26, 36, 40, 54, 56, 66, 74, 84, 90, 94, 110, 116, 120, 124, 126, 130, 134, 146, 150, 156, 160, 170, 176, 180, 184, ... A005574\n2 2, 4, 6, 16, 20, 24, 28, 34, 46, 48, 54, 56, 74, 80, 82, 88, 90, 106, 118, 132, 140, 142, 154, 160, 164, 174, 180, 194, 198, 204, 210, 220, 228, ... A000068\n3 2, 4, 118, 132, 140, 152, 208, 240, 242, 288, 290, 306, 378, 392, 426, 434, 442, 508, 510, 540, 542, 562, 596, 610, 664, 680, 682, 732, 782, ... A006314\n4 2, 44, 74, 76, 94, 156, 158, 176, 188, 198, 248, 288, 306, 318, 330, 348, 370, 382, 396, 452, 456, 470, 474, 476, 478, 560, 568, 598, 642, ... A006313\n5 30, 54, 96, 112, 114, 132, 156, 332, 342, 360, 376, 428, 430, 432, 448, 562, 588, 726, 738, 804, 850, 884, 1068, 1142, 1198, 1306, 1540, 1568, ... A006315\n6 102, 162, 274, 300, 412, 562, 592, 728, 1084, 1094, 1108, 1120, 1200, 1558, 1566, 1630, 1804, 1876, 2094, 2162, 2164, 2238, 2336, 2388, ... A006316\n7 120, 190, 234, 506, 532, 548, 960, 1738, 1786, 2884, 3000, 3420, 3476, 3658, 4258, 5788, 6080, 6562, 6750, 7692, 8296, 9108, 9356, 9582, ... A056994\n8 278, 614, 892, 898, 1348, 1494, 1574, 1938, 2116, 2122, 2278, 2762, 3434, 4094, 4204, 4728, 5712, 5744, 6066, 6508, 6930, 7022, 7332, ... A056995\n9 46, 1036, 1318, 1342, 2472, 2926, 3154, 3878, 4386, 4464, 4474, 4482, 4616, 4688, 5374, 5698, 5716, 5770, 6268, 6386, 6682, 7388, 7992, ... A057465\n10 824, 1476, 1632, 2462, 2484, 2520, 3064, 3402, 3820, 4026, 6640, 7026, 7158, 9070, 12202, 12548, 12994, 13042, 15358, 17646, 17670, ... A057002\n11 150, 2558, 4650, 4772, 11272, 13236, 15048, 23302, 26946, 29504, 31614, 33308, 35054, 36702, 37062, 39020, 39056, 43738, 44174, 45654, ... A088361\n12 1534, 7316, 17582, 18224, 28234, 34954, 41336, 48824, 51558, 51914, 57394, 61686, 62060, 89762, 96632, 98242, 100540, 101578, 109696, ... A088362\n13 30406, 71852, 85654, 111850, 126308, 134492, 144642, 147942, 150152, 165894, 176206, 180924, 201170, 212724, 222764, 225174, 241600, ... A226528\n14 67234, 101830, 114024, 133858, 162192, 165306, 210714, 216968, 229310, 232798, 422666, 426690, 449732, 462470, 468144, 498904, 506664, ... A226529\n15 70906, 167176, 204462, 249830, 321164, 330716, 332554, 429370, 499310, 524552, 553602, 743788, 825324, 831648, 855124, 999236, 1041870, ... A226530\n16 48594, 108368, 141146, 189590, 255694, 291726, 292550, 357868, 440846, 544118, 549868, 671600, 843832, 857678, 1024390, 1057476, 1087540, ... A251597\n17 62722, 130816, 228188, 386892, 572186, 689186, 909548, 1063730, 1176694, 1361244, 1372930, 1560730, 1660830, 1717162, 1722230, 1766192, ... A253854\n18 24518, 40734, 145310, 361658, 525094, 676754, 773620, 1415198, 1488256, 1615588, 1828858, 2042774, 2514168, 2611294, 2676404, 3060772, ... A244150\n19 75898, 341112, 356926, 475856, 1880370, 2061748, 2312092, ... A243959\n20 919444, 1059094, ... A321323\n\nThe smallest base b such that b2n + 1 is prime are\n\n2, 2, 2, 2, 2, 30, 102, 120, 278, 46, 824, 150, 1534, 30406, 67234, 70906, 48594, 62722, 24518, 75898, 919444, ... (sequence A056993 in the OEIS)\n\nThe smallest k such that (2n)k + 1 is prime are\n\n1, 1, 1, 0, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 0, 4, 1, ... (The next term is unknown) (sequence A079706 in the OEIS) (also see and )\n\nA more elaborate theory can be used to predict the number of bases for which $F_{n}(a)$ will be prime for fixed $n$ . The number of generalized Fermat primes can be roughly expected to halve as $n$ is increased by 1.\n\n### Largest known generalized Fermat primes\n\nThe following is a list of the 5 largest known generalized Fermat primes. They are all megaprimes. As of September 2018 the whole top-5 was discovered by participants in the PrimeGrid project.\n\nRank Prime rank Prime number Generalized Fermat notation Number of digits Found date ref.\n1 14 10590941048576 + 1 F20(1059094) 6,317,602 Nov 2018 \n2 15 9194441048576 + 1 F20(919444) 6,253,210 Sep 2017 \n3 26 2877652524288 + 1 F19(2877652) 3,386,397 Jun 2019 \n4 27 2788032524288 + 1 F19(2788032) 3,379,193 Apr 2019 \n5 28 2733014524288 + 1 F19(2733014) 3,374,655 Mar 2019 \n\nOn the Prime Pages you can perform a search yielding the current top 100 generalized Fermat primes."
] |
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null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/9c168e5c8ec2740a7e6eab3f171c979166f1c96d",
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http://www.e-booksdirectory.com/details.php?ebook=4565
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[
"",
null,
"# Introduction to Statistical Theory of Fluid Turbulence",
null,
"Introduction to Statistical Theory of Fluid Turbulence\nby\n\nPublisher: arXiv\nNumber of pages: 40\n\nDescription:\nFluid and plasma flows exhibit complex random behaviour at high Reynolds numbers; this phenomena is called turbulence. This text is a brief introduction to the statistical theory of fluid turbulence, with emphasis on field-theoretic treatment of renormalized viscosity and energy fluxes.\n\n(570KB, PDF)\n\n## Similar books",
null,
"The Physics of Ocean Waves\nby - SurfLibrary.org\nThis text is intended to be a general but comprehensive overview on physical phenomena associated with the marine environment. The author provides his own perspective on this field from his particular experiences and training.\n(4813 views)",
null,
"A Practical Introduction to Numerical Hydrodynamics\nby - Leiden University\nAn introduction to the field of numerical hydrodynamics. It will give you some insight in what is involved in such calculations. Numerical hydrodynamics is used in many parts of astrophysics. The applications we consider in this exercise are stellar.\n(11263 views)",
null,
"The Secret of Sailing\nby\nThis book presents a mathematical theory of sailing based on a combination of analysis and computation. This new theory is fundamentally different from that envisioned in the classical theories for lift in inviscid flow and for drag in viscous flow.\n(10006 views)",
null,
"An Introduction to the Mechanics of Fluids\nby - Longmans, Green\nIn writing this book, while preserving the usual rigour, the endeavour has been made to impart to it by the character of the illustrations and examples, a modern and practical flavour which will render it more widely useful. The calculus is not used.\n(6371 views)"
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"http://www.e-booksdirectory.com/images/4565.jpg",
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https://www.colorhexa.com/cd56b1
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[
"# #cd56b1 Color Information\n\nIn a RGB color space, hex #cd56b1 is composed of 80.4% red, 33.7% green and 69.4% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 58% magenta, 13.7% yellow and 19.6% black. It has a hue angle of 314.1 degrees, a saturation of 54.3% and a lightness of 57.1%. #cd56b1 color hex could be obtained by blending #ffacff with #9b0063. Closest websafe color is: #cc6699.\n\n• R 80\n• G 34\n• B 69\nRGB color chart\n• C 0\n• M 58\n• Y 14\n• K 20\nCMYK color chart\n\n#cd56b1 color description : Moderate magenta.\n\n# #cd56b1 Color Conversion\n\nThe hexadecimal color #cd56b1 has RGB values of R:205, G:86, B:177 and CMYK values of C:0, M:0.58, Y:0.14, K:0.2. Its decimal value is 13457073.\n\nHex triplet RGB Decimal cd56b1 `#cd56b1` 205, 86, 177 `rgb(205,86,177)` 80.4, 33.7, 69.4 `rgb(80.4%,33.7%,69.4%)` 0, 58, 14, 20 314.1°, 54.3, 57.1 `hsl(314.1,54.3%,57.1%)` 314.1°, 58, 80.4 cc6699 `#cc6699`\nCIE-LAB 54.878, 57.725, -25.746 36.44, 22.811, 44.076 0.353, 0.221, 22.811 54.878, 63.206, 335.962 54.878, 62.419, -47.402 47.761, 52.607, -21.283 11001101, 01010110, 10110001\n\n# Color Schemes with #cd56b1\n\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #56cd72\n``#56cd72` `rgb(86,205,114)``\nComplementary Color\n• #ae56cd\n``#ae56cd` `rgb(174,86,205)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #cd5676\n``#cd5676` `rgb(205,86,118)``\nAnalogous Color\n• #56cdae\n``#56cdae` `rgb(86,205,174)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #76cd56\n``#76cd56` `rgb(118,205,86)``\nSplit Complementary Color\n• #56b1cd\n``#56b1cd` `rgb(86,177,205)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #b1cd56\n``#b1cd56` `rgb(177,205,86)``\n• #7256cd\n``#7256cd` `rgb(114,86,205)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #b1cd56\n``#b1cd56` `rgb(177,205,86)``\n• #56cd72\n``#56cd72` `rgb(86,205,114)``\n• #a6318a\n``#a6318a` `rgb(166,49,138)``\n• #b9379b\n``#b9379b` `rgb(185,55,155)``\n• #c742a8\n``#c742a8` `rgb(199,66,168)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #d36aba\n``#d36aba` `rgb(211,106,186)``\n• #d97dc3\n``#d97dc3` `rgb(217,125,195)``\n• #de91cc\n``#de91cc` `rgb(222,145,204)``\nMonochromatic Color\n\n# Alternatives to #cd56b1\n\nBelow, you can see some colors close to #cd56b1. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #cb56cd\n``#cb56cd` `rgb(203,86,205)``\n• #cd56c5\n``#cd56c5` `rgb(205,86,197)``\n• #cd56bb\n``#cd56bb` `rgb(205,86,187)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #cd56a7\n``#cd56a7` `rgb(205,86,167)``\n• #cd569d\n``#cd569d` `rgb(205,86,157)``\n• #cd5693\n``#cd5693` `rgb(205,86,147)``\nSimilar Colors\n\n# #cd56b1 Preview\n\nThis text has a font color of #cd56b1.\n\n``<span style=\"color:#cd56b1;\">Text here</span>``\n#cd56b1 background color\n\nThis paragraph has a background color of #cd56b1.\n\n``<p style=\"background-color:#cd56b1;\">Content here</p>``\n#cd56b1 border color\n\nThis element has a border color of #cd56b1.\n\n``<div style=\"border:1px solid #cd56b1;\">Content here</div>``\nCSS codes\n``.text {color:#cd56b1;}``\n``.background {background-color:#cd56b1;}``\n``.border {border:1px solid #cd56b1;}``\n\n# Shades and Tints of #cd56b1\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, #0d040b is the darkest color, while #fefdfe is the lightest one.\n\n• #0d040b\n``#0d040b` `rgb(13,4,11)``\n• #1c0817\n``#1c0817` `rgb(28,8,23)``\n• #2b0d24\n``#2b0d24` `rgb(43,13,36)``\n• #3a1130\n``#3a1130` `rgb(58,17,48)``\n• #49163d\n``#49163d` `rgb(73,22,61)``\n• #581a4a\n``#581a4a` `rgb(88,26,74)``\n• #671f56\n``#671f56` `rgb(103,31,86)``\n• #772363\n``#772363` `rgb(119,35,99)``\n• #862870\n``#862870` `rgb(134,40,112)``\n• #952c7c\n``#952c7c` `rgb(149,44,124)``\n• #a43189\n``#a43189` `rgb(164,49,137)``\n• #b33595\n``#b33595` `rgb(179,53,149)``\n• #c239a2\n``#c239a2` `rgb(194,57,162)``\n• #c947aa\n``#c947aa` `rgb(201,71,170)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #d165b8\n``#d165b8` `rgb(209,101,184)``\n• #d674bf\n``#d674bf` `rgb(214,116,191)``\n• #da83c6\n``#da83c6` `rgb(218,131,198)``\n• #df93cd\n``#df93cd` `rgb(223,147,205)``\n• #e3a2d4\n``#e3a2d4` `rgb(227,162,212)``\n• #e8b1db\n``#e8b1db` `rgb(232,177,219)``\n• #ecc0e2\n``#ecc0e2` `rgb(236,192,226)``\n• #f1cfe9\n``#f1cfe9` `rgb(241,207,233)``\n• #f5def0\n``#f5def0` `rgb(245,222,240)``\n• #faedf7\n``#faedf7` `rgb(250,237,247)``\n• #fefdfe\n``#fefdfe` `rgb(254,253,254)``\nTint Color Variation\n\n# Tones of #cd56b1\n\nA tone is produced by adding gray to any pure hue. In this case, #929192 is the less saturated color, while #f72cc7 is the most saturated one.\n\n• #929192\n``#929192` `rgb(146,145,146)``\n• #9a8996\n``#9a8996` `rgb(154,137,150)``\n• #a3809b\n``#a3809b` `rgb(163,128,155)``\n• #ab789f\n``#ab789f` `rgb(171,120,159)``\n• #b46fa4\n``#b46fa4` `rgb(180,111,164)``\n• #bc67a8\n``#bc67a8` `rgb(188,103,168)``\n``#c55ead` `rgb(197,94,173)``\n• #cd56b1\n``#cd56b1` `rgb(205,86,177)``\n• #d54eb5\n``#d54eb5` `rgb(213,78,181)``\n• #de45ba\n``#de45ba` `rgb(222,69,186)``\n• #e63dbe\n``#e63dbe` `rgb(230,61,190)``\n• #ef34c3\n``#ef34c3` `rgb(239,52,195)``\n• #f72cc7\n``#f72cc7` `rgb(247,44,199)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #cd56b1 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] |
[
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https://www.studymode.com/subjects/probability-theory-page1.html
|
[
"# \"Probability theory\" Essays and Research Papers\n\nSort By:\nSatisfactory Essays\nGood Essays\nBetter Essays\nPowerful Essays\nBest Essays\nPage 1 of 50 - About 500 Essays\n• Good Essays\n\nI. Probability Theory * A branch of mathematics concerned with the analysis of random phenomena. The outcome of a random event cannot be determined before it occurs‚ but it may be any one of several possible outcomes. The actual outcome is considered to be determined by chance. * The word probability has several meanings in ordinary conversation. Two of these are particularly important for the development and applications of the mathematical theory of probability. One is the interpretation\n\nPremium Probability theory Statistical hypothesis testing\n\n• 1874 Words\n• 8 Pages\nGood Essays\n• Powerful Essays\n\nBBA (Fall - 2014) Business Statistics Theory of Probability Ahmad Jalil Ansari Business Head Enterprise Solution Division Random Process In a random process we know that what outcomes or events could happen; but we do not know which particular outcome or event will happen. For example tossing of coin‚ rolling of dice‚ roulette wheel‚ changes in valuation in shares‚ demand of particular product etc. Probability It is the numeric value representing the chance‚ likelihood‚ or possibility\n\n• 5153 Words\n• 54 Pages\nPowerful Essays\n• Good Essays\n\nP(S) The symbol for the probability of success P(F) The symbol for the probability of failure p The numerical probability of a success q The numerical probability of a failure P(S) = p and P(F) = 1 - p = q n The number of trials X The number of successes The probability of a success in a binomial experiment can be computed with the following formula. Binomial Probability Formula In a binomial experiment\n\n• 751 Words\n• 4 Pages\nGood Essays\n• Powerful Essays\n\nStudy Guide for Probability Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. Which inequality represents the probability‚ x‚ of any event happening? a.||c.|| b.||d.|| ____ 2. Which event has a probability of zero? a.|choosing a letter from the alphabet that has line symmetry|c.|choosing a pair of parallel lines that have unequal slopes| b.|choosing a number that is greater than 6 and is even|d.|choosing a triangle that is both\n\n• 5784 Words\n• 24 Pages\nPowerful Essays\n• Satisfactory Essays\n\nProbability theory Probability: A numerical measure of the chance that an event will occur. Experiment: A process that generates well defined outcomes. Sample space: The set of all experimental outcomes. Sample point: An element of the sample space. A sample point represents an experimental outcome. Tree diagram: A graphical representation that helps in visualizing a multiple step experiment. Classical method: A method of assigning probabilities that is appropriate when all the experimental\n\n• 311 Words\n• 2 Pages\nSatisfactory Essays\n• Good Essays\n\nguys‚ this is the probability Assignment. Last date for submission is 10 aug... Q1. What is the probability of picking a card that was either red or black? Q2. A problem in statistics is given to 5 students A‚ B‚ C‚ D‚ E. Their chances of solving it are ½‚1/3‚1/4‚1/5‚1/6. What is the probability that the problem will be solved? Q3. A person is known to hit the target in 3 out of 4 shots whereas another person is known to hit the target in 2 out of 3 shots. Find the probability that the target\n\n• 970 Words\n• 4 Pages\nGood Essays\n• Powerful Essays\n\nbe able to ONEDefine probability. TWO Describe the classical‚ empirical‚ and subjective approaches to probability. THREEUnderstand the terms experiment‚ event‚ outcome‚ permutation‚ and combination. FOURDefine the terms conditional probability and joint probability. FIVE Calculate probabilities applying the rules of addition and multiplication. SIXUse a tree diagram to organize and compute probabilities. SEVEN Calculate a probability using Bayes theorem. What is probability There is really no answer\n\n• 2902 Words\n• 4 Pages\nPowerful Essays\n• Good Essays\n\nWeek Four Discussion 2 1. In your own words‚ describe two main differences between classical and empirical probabilities. The differences between classical and empirical probabilities are that classical assumes that all outcomes are likely to occur‚ while empirical involves actually physically observing and collecting the information. 2. Gather coins you find around your home or in your pocket or purse. You will need an even number of coins (any denomination) between 16 and 30. You do not\n\nPremium Probability theory Theory Quantum mechanics\n\n• 541 Words\n• 3 Pages\nGood Essays\n• Good Essays\n\nt) P (X > s + t) P (X > t) e−λ(s+t) e−λt e−λs P (X > s) – Example: Suppose that the amount of time one spends in a bank is exponentially distributed with mean 10 minutes‚ λ = 1/10. What is the probability that a customer will spend more than 15 minutes in the bank? What is the probability that a customer will spend more than 15 minutes in the bank given that he is still in the bank after 10 minutes? Solution: P (X > 15) = e−15λ = e−3/2 = 0.22 P (X > 15|X > 10) = P (X > 5) = e−1/2 =\n\n• 1600 Words\n• 7 Pages\nGood Essays\n• Good Essays\n\nKaran negi 12.2 12.3 We use equation 2 to find out probability: F(t)=1 – e^-Lt 1-e^-(0.4167)(10) = 0.98 almost certainty. This shows that probability of another arrival in the next 10 minutes. Now we figure out how many customers actually arrive within those 10 minutes. If the mean is 0.4167‚ then 0.4167*10=4.2‚ and we can round that to 4. X-axis represents minutes (0-10) Y-axis represents number of people. We can conclude from this chart that the highest point with the most visitors\n\nPremium Probability theory Arithmetic mean Poisson distribution\n\n• 592 Words\n• 3 Pages\nGood Essays\nPrevious\nPage 1 2 3 4 5 6 7 8 9 50"
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[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.88990575,"math_prob":0.9369278,"size":5323,"snap":"2023-40-2023-50","text_gpt3_token_len":1159,"char_repetition_ratio":0.1750329,"word_repetition_ratio":0.048122067,"special_character_ratio":0.22055233,"punctuation_ratio":0.08220603,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99789375,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T17:06:26Z\",\"WARC-Record-ID\":\"<urn:uuid:37fb5d0a-7f54-4cc2-bd8e-ecfcede1001d>\",\"Content-Length\":\"79301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c54353d-b497-44fc-ba0f-0aa4c9be0fbc>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7aa1c76-d60e-4a59-8595-a0f4b1df81a1>\",\"WARC-IP-Address\":\"13.32.151.99\",\"WARC-Target-URI\":\"https://www.studymode.com/subjects/probability-theory-page1.html\",\"WARC-Payload-Digest\":\"sha1:GXUNBP2MLNYVLUATXLYUG57ILDBHTRPR\",\"WARC-Block-Digest\":\"sha1:SQQDBCESVJW52ITWU6TKJLFNF5DGZXWB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511386.54_warc_CC-MAIN-20231004152134-20231004182134-00742.warc.gz\"}"}
|
https://blog.51cto.com/u_15736437/6454193
|
[
"## 信道容量\n\n• 无噪无损信道\n• 有噪无损信道\n• 无噪有损信道\n• 二进制对称信道\n• AWGN信道\n\n### 信道容量的定义\n\n#### 信息传输率 R",
null,
"#### 信息传输速率",
null,
"#### 平均互信息",
null,
"#### 信道容量",
null,
"",
null,
"### 三种特殊信道的容量\n\n#### 无噪无损信道",
null,
"",
null,
"#### 有噪无损信道",
null,
"",
null,
"#### 无噪有损信道",
null,
"### 典型信道的信道容量\n\n#### BSC信道容量",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"I(X, Y) 对",
null,
"存在一个极大值,该极大值为信源的压缩极限。\n\nBSC 信道容量 $C=1-H(p)$\n\n•",
null,
", $C=1-0=1 bit =H(X)$ (信道无噪声)\n•",
null,
", $C=1-H(\\frac{1}{2}, \\frac{1}{2})=0$ (信道强噪声)",
null,
"",
null,
"",
null,
"",
null,
"BSC信道如下图,",
null,
"符号/秒,错误传递概率",
null,
"求:信道容量和实际信息传输速率。",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
", 有",
null,
"",
null,
"#### 连续信道的信道容量\n\n##### 单符号高斯连续信道",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"##### 限频、限功率高斯信道的容量",
null,
"$Y(t), X(t), n(t)$ 的带宽为 B , 以 2B 采样，得",
null,
", $Y(t_{n}), \\ldots, Y(t_{L}) \\ldots, X(t_{1}), X(t_{2}), \\ldots, X(t_{n}), \\ldots, X(t_{L}) \\ldots, n(t_{1}), n(t_{2}), \\ldots , n(t_{n}), \\ldots, n(t_{L}) \\ldots$ 。时刻",
null,
"",
null,
"##### 单符号信号一>多符号多维信道",
null,
"分别表示 L 个抽样",
null,
"的 L 维向量, 则对多符号信道",
null,
"",
null,
"统计独立时",
null,
"T 时间内抽样数 L=2BT , 则信道传输最大信息量",
null,
"",
null,
"",
null,
"为限带高斯白噪声 n(t) 的单边功率谱密度。",
null,
"时,",
null,
"; 当 $B arrow \\infty$ 时,",
null,
"一确定值。",
null,
"",
null,
",有",
null,
",即带宽不受限制时, 传输1bit信息, 信噪比最低只需要-1.6dB, 这是加性高斯噪声信道信息传输速率的极限值, 是一切编码方式所能达到的理论极限。",
null,
"-- 单位频带的信息传输速率（频带利用率）。",
null,
"时,",
null,
", 此时信道完全丧失通信能力。",
null,
"",
null,
"##### Shannon信道编码定理",
null,
"",
null,
"### 信源与信道的匹配",
null,
", 信源与信道匹配。",
null,
", 信源与信道不匹配, 信道有冗余",
null,
"",
null,
"1. Proakis, John G., et al. Communication systems engineering. Vol. 2. New Jersey: Prentice Hall, 1994.\n2. Proakis, John G., et al. SOLUTIONS MANUAL Communication Systems Engineering. Vol. 2. New Jersey: Prentice Hall, 1994.\n3. 周炯槃. 通信原理（第3版）\\[M]. 北京：北京邮电大学出版社, 2008.\n4. 樊昌信, 曹丽娜. 通信原理（第7版） \\[M]. 北京：国防工业出版社, 2012."
] |
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null,
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"https://s2.51cto.com/images/blog/202306/10101034_6483db9a08a2c50695.png",
null,
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null,
"https://math-api.51cto.com/",
null,
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null,
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null,
"https://math-api.51cto.com/",
null,
"https://math-api.51cto.com/",
null,
"https://math-api.51cto.com/",
null,
"https://math-api.51cto.com/",
null,
"https://math-api.51cto.com/",
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null,
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|
https://math.stackexchange.com/questions/555363/lower-bound-on-a-number-theoretic-function
|
[
"# Lower bound on a number theoretic function\n\nLet $n$ be a positive odd integer, let $$n_j = \\Bigl\\{\\frac{n}{2^{j+1}}\\Bigr\\}\\,,$$ where $\\{x\\}$ denotes the fractional part of $x$, and finally let $k = \\lceil \\log_2 n\\rceil$. Consider the function $$f(n) = \\prod_{j=1}^{k} \\bigl[1- 4n_j(1-n_j)\\bigr]\\,.$$ Each term in this product is bounded in the interval $(0,1)$, so $f(n)$ will tend to $0$ for large $n$. My question is how one can\n\nProve that $f(n) \\ge \\frac{c}{n^a}$ for some absolute positive constants $a$, $c$.\n\nIt is easy to show (using bounds on the fractional part) that there exist $a,c>0$ such that $$f(n) \\ge \\frac{c}{n^{a \\log n}} \\,,$$ so that the product decays only subexponentially slowly, but a heuristic argument as well as numerics indicates that the true scaling is actually a much slower inverse polynomial. The heuristic argument is, roughly, that the terms in the product are highly correlated, and if one term is small, the following term must be large. I have been unable to leverage this into a proof, however.\n\nI indicated in the title that this is a \"number theoretic\" function because the behavior depends heavily on the bit-wise structure of $n$, and it has some interesting fractal features as a result.\n\nNumerical experimentation suggests that the fastest-decaying values occur at $n_k=[2^k/3]$, where $[\\cdot]$ indicates rounding off to the nearest integer. These integers have binary expansion $101010\\cdots$. It probably is possible to work out very good bounds for $f(n_k)$. Numerically, we seem to have $f(n_k)$ asymptotic to a constant times $1/n_k^\\alpha$ where $\\alpha = \\log 9/\\log 2$.\n(Similarly, the slowest-decaying values are almost surely $m_k = 2^k-1$, for which we seem to have $f(m_k) \\sim c/m_k^2$ for some constant $c$.)\n• Thanks Greg! Yes, I have also noticed that the binary expansion is alternating at the local minima, i.e. those in the interval $(2^{k-1},2^k)$, and the asymptotic constant follows. What I'm stuck on is how to prove that this is the right decay. The problem is that I don't know how to bound the behavior of sequences which are close in Hamming distance in a satisfactory way. – Steve Flammia Nov 7 '13 at 21:55\n• I imagine you would use bounds on the derivative of $\\log(1-4x(1-x))$ to show that changing a single bit of $n$ only changes each term in the sum defining $\\log f(n)$ by a controllable amount, and hence that $f(n)$ changes only by a controllable amount (since these various error form a quickly converging sequence). Exponentiating, this shows that changing one bit of $n$ only changes $f(n)$ by a controllable percentage. And on from there, hopefully...? – Greg Martin Nov 7 '13 at 22:10\n• My fear is that there are no good bounds on that derivative. If you change the last bit to make $n$ an even number, then $f(n)$ immediately goes to zero. So the function is very delicate, it seems. I'll have to give it a try though. – Steve Flammia Nov 7 '13 at 22:54\n• Yes, certainly the derivative is not bounded on $(0,1)$, and a number ending in the bits $\\dots011111111$ or $\\dots1000000001$ will hit this function in the scary region. Presumably the contribution from the bad-derivative part is compensated for by the fact that all the stuff that led up to it was really nice, as for $2^k-1$. But yeah, it's not obvious to me how to proceed (otherwise I'd've done it!).... – Greg Martin Nov 8 '13 at 1:10"
] |
[
null
] |
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|
https://www.tutorialspoint.com/find-maximum-power-of-a-number-that-divides-a-factorial-in-cplusplus
|
[
"# Find maximum power of a number that divides a factorial in C++\n\nC++Server Side ProgrammingProgramming\n\nSuppose we have two numbers n and fact. We have to find the largest power of n, that divides fact! (factorial of fact). So if fact = 5, and n = 2, then output will be 3. So 5! = 120, and this is divisible by 2^3 = 8.\n\nHere we will use the Legendre’s formula. This finds largest power of a prime, that divides fact!. We will find all prime factors of n, then find largest power of it, that divides fact!.\n\nSo if fact is 146, and n = 15, then prime factors of n are 5 and 3. So\n\nfor 3, it will be [146/3] + [48/3] + [16/3] + [5/3] + [1/3] = 48 + 16 + 5 + 1 + 0 = 70.\n\nfor 5, it will be [146/5] + [29/5] + [5/5] + [1/3] = 29 + 5 + 1 + 0 = 35.\n\n## Example\n\n#include<iostream>\n#include<cmath>\nusing namespace std;\nint getPowerPrime(int fact, int p) {\nint res = 0;\nwhile (fact > 0) {\nres += fact / p;\nfact /= p;\n}\nreturn res;\n}\nint findMinPower(int fact, int n) {\nint res = INT_MAX;\nfor (int i = 2; i <= sqrt(n); i++) {\nint cnt = 0;\nif (n % i == 0) {\ncnt++;\nn = n / i;\n}\nif (cnt > 0) {\nint curr = getPowerPrime(fact, i) / cnt;\nres = min(res, curr);\n}\n}\nif (n >= 2) {\nint curr = getPowerPrime(fact, n);\nres = min(res, curr);\n}\nreturn res;\n}\nint main() {\nint fact = 146, n = 5;\ncout << \"Minimum power: \" << findMinPower(fact, n);\n}\n\n## Output\n\nMinimum power: 35"
] |
[
null
] |
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|
https://physics.stackexchange.com/questions/381889/generalised-coordinates-in-3d-rotation
|
[
"# Generalised Coordinates in 3D Rotation\n\nIf you have N particles on a surface of a rigid body and the rigid body is rotating about some axis, we say there are six generalised coordinates for the system (N particles on the surface) and set up the lagrangian.\n\nThe constraints we know are\n\n1. The distance between any two particles is invariant\n\n2. The angles between the line joining any particles is invariant as well and that's it.\n\nWe also know a formula for finding the number of genralised coordinates i.e difference between number of degrees of freedom (3N) and number of constraints (which is N(N-1)/2). Clearly using this formula doesn't give 6 generalised coordinates.\n\nWhere's the mistake and how to count the number of generalised coordinates?\n\nNote : Even I know the argument as to if you know 3 points on surface, you can determine the position of any other particle on the surface. But my question is about counting the generalised coordinates using the above formula.\n\nYour mistake is to assume that the number of constraints is $N(N-1)/2$. This number is only true for $N\\leq 4$.\nLet us label the particles by $1,2,\\ldots,N$ and say that $\\bar{ij}$ denotes the constraint between particles $i$ and $j$. For one particle there is no constraint. For two particles there is one constraint (see the diagrams bellow). If $N=3$ we have three constraints: $\\bar{12},\\bar{13},\\bar{23}$. Add a fourth particle and we gain three others constraints: $\\bar{14}$ and $\\bar{24}$ fix the distance from particle $4$ to particles $1$ and $2$ and $\\bar{34}$ which forbids the fourth particle to rotate about the axis joining $1$ and $2$. This fixes the distance from $4$ to $3$. This gives a total of six constraints. Finally consider adding one more particle. Now things get different. The new constraints $\\bar{15},\\bar{25}$ fix the distance to $1$ and $2$. The only freedom the sixth particle still have is to rotate about the line joining $1$ and $2$. Therefore just one more constraint, say $\\bar{13}$, is enough to rigidly fix the sixth particle. The total number of constraints in this case is six.",
null,
"As you can see in this construction, the first particle added zero constraints, the second one added one constraint, the third one added two constraints and the nth ($n\\geq 4$) one added three constraints. The total number of constraints for $N\\geq 5$ particles is therefore $1+2+3(N-3)=3(N-2)$. Hence there are $3N-3(N-2)=6$ degrees of freedom."
] |
[
null,
"https://i.stack.imgur.com/7yDsf.jpg",
null
] |
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|
https://stats.stackexchange.com/questions/242004/why-do-neural-network-researchers-care-about-epochs
|
[
"# Why do neural network researchers care about epochs?\n\nAn epoch in stochastic gradient descent is defined as a single pass through the data. For each SGD minibatch, $k$ samples are drawn, the gradient computed and parameters are updated. In the epoch setting, the samples are drawn without replacement.\n\nBut this seems unnecessary. Why not draw each SGD minibatch as $k$ random draws from the whole data set at each iteration? Over a large number of epochs, the small deviations of which samples are seen more or less often would seem to be unimportant.\n\n• Anecdotal evidence, but I recently fitted a one layer neural network using SGD on the MNIST data, which are 50000 in training size. After one random run-through the classification accuracy was not much higher than 30-40%, and the log-likelihood clearly had not converged. So I repeated the procedure for 30 more epochs leading to over 90% accuracy. At least by counterexample this showed to me they can be necessary. – tomka Oct 24 '16 at 14:30\n• @tomka That seems to provide evidence that multiple passes over the data are necessary, which is consistent with the method proposed here: keep drawing $k$ samples per training iteration ad nauseam. – Sycorax Oct 24 '16 at 14:36\n• Another interesting question would be: will mini batch order also has an impact on overfitting? – Kh40tiK Oct 25 '16 at 8:58\n• @Kh40tiK I think someone asked that question today but I can't find the link. – Sycorax Oct 26 '16 at 2:58\n• @Pinocchio The standard SGD practice is sampling without replacement (until the pool of samples is depleted, at which point a new epoch starts again with all the data). My question is why it does not use sampling with replacement. It turns out that one answer is that sampling without replacement improves the rate of convergence for the model. – Sycorax Jan 20 '17 at 4:34\n\nIn addition to Franck's answer about practicalities, and David's answer about looking at small subgroups – both of which are important points – there are in fact some theoretical reasons to prefer sampling without replacement. The reason is perhaps related to David's point (which is essentially the coupon collector's problem).\n\nIn 2009, Léon Bottou compared the convergence performance on a particular text classification problem ($n = 781,265$).\n\nBottou (2009). Curiously Fast Convergence of some Stochastic Gradient Descent Algorithms. Proceedings of the symposium on learning and data science. (author's pdf)\n\nHe trained a support vector machine via SGD with three approaches:\n\n• Random: draw random samples from the full dataset at each iteration.\n• Cycle: shuffle the dataset before beginning the learning process, then walk over it sequentially, so that in each epoch you see the examples in the same order.\n• Shuffle: reshuffle the dataset before each epoch, so that each epoch goes in a different order.\n\nHe empirically examined the convergence $\\mathbb E[ C(\\theta_t) - \\min_\\theta C(\\theta) ]$, where $C$ is the cost function, $\\theta_t$ the parameters at step $t$ of optimization, and the expectation is over the shuffling of assigned batches.\n\n• For Random, convergence was approximately on the order of $t^{-1}$ (as expected by existing theory at that point).\n• Cycle obtained convergence on the order of $t^{-\\alpha}$ (with $\\alpha > 1$ but varying depending on the permutation, for example $\\alpha \\approx 1.8$ for his Figure 1).\n• Shuffle was more chaotic, but the best-fit line gave $t^{-2}$, much faster than Random.\n\nThis is his Figure 1 illustrating that:",
null,
"This was later theoretically confirmed by the paper:\n\nGürbüzbalaban, Ozdaglar, and Parrilo (2015). Why Random Reshuffling Beats Stochastic Gradient Descent. arXiv:1510.08560. (video of invited talk at NIPS 2015)\n\nTheir proof only applies to the case where the loss function is strongly convex, i.e. not to neural networks. It's reasonable to expect, though, that similar reasoning might apply to the neural network case (which is much harder to analyze).\n\n• This is a very insightful answer. Thank you very much for your contribution. – Sycorax Oct 24 '16 at 20:19\n• sorry for the ignorance, but do you mind explaining a bit more what the difference between the three is? In particular I am confused about Random, when you say \"sample\", what do you mean? I know this is not what you are referencing but the standard Neural Net mini batch SGD usually samples batches without replacement at each iteration. Is that what Random does? If it is, how is that different from Shuffle? – Pinocchio Jan 20 '17 at 4:32\n• Now that I re-read it all three seem the same algorithm, whats the difference if the data set is shuffled or not and how often if the batches for SGD are always random anyway? – Pinocchio Jan 20 '17 at 4:33\n• @Pinocchio Imagine a four-lament dataset. Random might go ACADBBCA; each entry is completely random. Cycle might go BDAC BDAC BDAC; it chooses one order for each epoch and then repeats. Shuffle could be BDAC ADCB CBAD; it goes in epochs, but each one is random. This analysis doesn't use minibatches, just one-element-at-a-time SGD. – djs Jan 20 '17 at 11:07\n• This is a great answer. Thnx you! – DankMasterDan Jan 26 '18 at 17:22\n\nIt is indeed quite unnecessary from a performance standpoint with a large training set, but using epochs can be convenient, e.g.:\n\n• it gives a pretty good metric: \"the neural network was trained for 10 epochs\" is a clearer statement than \"the neural network was trained for 18942 iterations\" or \"the neural network was trained over 303072 samples\".\n• there's enough random things going on during the training phase: random weight initialization, mini-batch shuffling, dropout, etc.\n• it is easy to implement\n• it avoids wondering whether the training set is large enough not to have epochs\n\n gives one more reason, which isn't that much relevant given today's computer configuration:\n\nAs for any stochastic gradient descent method (including the mini-batch case), it is important for efficiency of the estimator that each example or minibatch be sampled approximately independently. Because random access to memory (or even worse, to disk) is expensive, a good approximation, called incremental gradient (Bertsekas, 2010), is to visit the examples (or mini-batches) in a fixed order corresponding to their order in memory or disk (repeating the examples in the same order on a second epoch, if we are not in the pure online case where each example is visited only once). In this context, it is safer if the examples or mini-batches are first put in a random order (to make sure this is the case, it could be useful to first shuffle the examples). Faster convergence has been observed if the order in which the mini-batches are visited is changed for each epoch, which can be reasonably efficient if the training set holds in computer memory.\n\n Bengio, Yoshua. \"Practical recommendations for gradient-based training of deep architectures.\" Neural Networks: Tricks of the Trade. Springer Berlin Heidelberg, 2012. 437-478.\n\n• These seem like good points, but regarding your update, it seems that sampling $k$ per epoch is dependent sampling (because the probability of a sample being seen twice in an epoch is 0). So I'm not sure how the authors can claim that the epoch construction is independent, unless their meaning of \"approximately independently\" is \"not at all independently.\" – Sycorax Oct 24 '16 at 14:19\n• @Sycorax Sampling without replacement, despite of course being not independent, is \"approximately independent\" in the sense that it is exchangeable. From the point of view of training a classifier which doesn't care too much about any one data point, this exchangeability is definitely fairly close to \"approximately independent.\" – djs Oct 24 '16 at 17:58\n\nI disagree somewhat that it clearly won't matter. Let's say there are a million training examples, and we take ten million samples.\n\nIn R, we can quickly see what the distribution looks like with\n\nplot(dbinom(0:40, size = 10 * 1E6, prob = 1E-6), type = \"h\")",
null,
"Some examples will be visited 20+ times, while 1% of them will be visited 3 or fewer times. If the training set was chosen carefully to represent the expected distribution of examples in real data, this could have a real impact in some areas of the data set---especially once you start slicing up the data into smaller groups.\n\nConsider the recent case where one Illinois voter effectively got oversampled 30x and dramatically shifted the model's estimates for his demographic group (and to a lesser extent, for the whole US population). If we accidentally oversample \"Ruffed Grouse\" images taken against green backgrounds on cloudy days with a narrow depth of field and undersample the other kinds of grouse images, the model might associate those irrelevant features with the category label. The more ways there are to slice up the data, the more of these subgroups there will be, and the more opportunities for this kind of mistake there will be.\n\n• I don't think it would make a large difference in practice for a large training set, but definitely I expect it would with a smaller training set. – Franck Dernoncourt Oct 24 '16 at 14:19\n• @FranckDernoncourt well, the whole point was that it can matter for large datasets if you start looking at small sub-groups. Which is not an uncommon procedure in large datasets, – dimpol Oct 24 '16 at 14:38\n• pretty sure you should have used a uniform distribution, not a binomial – lahwran Aug 23 '17 at 18:00\n• @lahwran We're sampling $10^7$ times from $10^6$ elements with replacement. In R, this would be samples = sample(1:1E6, size = 1E7, replace = TRUE). From there, you can plot the frequency distribution with plot(table(table(samples)) / 1E7). It looks just like the binomial distribution I plotted above. – David J. Harris Aug 23 '17 at 18:35\n• aha! I was wrong, then. – lahwran Aug 23 '17 at 20:15"
] |
[
null,
"https://i.stack.imgur.com/bl3eQ.png",
null,
"https://i.stack.imgur.com/3iOMa.png",
null
] |
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https://www.alfredforum.com/topic/12028-simplify-fractions/
|
[
"# Simplify fractions\n\n## Recommended Posts\n\nI write a workflow for converting a given decimal to a simplified fraction as well as simplifying a given fraction.\n\nIt saves me a lot of time and I hope it is useful for you too.\n\nExamples",
null,
"• `frac .11` ==> `.11 = 11 / 100`\n• `frac 4/6` ==> `4/6 = 2 / 3`\n• `frac -1.4/2.2` ==> `-1.4/2.2 = -7 / 11`\n• `frac 1.2/0` ==> `Error: divided by zero.`\n• `frac b/3` ==> `Error: Invalid input format`\n\n-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\n\nUpdate\n\n7 additional math tools are added and the new workflow is renamed to MathTools\n\nQuote\n\n@CJK @Thomas_U Thank you for your feedback and suggestions.\n\n#### 1) Simplify fractions\n\n##### Examples:\n• `frac .11` ==> `.11 = 11 / 100`\n• `frac 4/6` ==> `4/6 = 2 / 3`\n• `frac -1.4/2.2` ==> `-1.4/2.2 = -7 / 11`\n\n#### 4) Simplify surds\n\n##### Examples:\n• `sqrt .0144` ==> `√(.0144) = 3/25 = 0.12`\n• `sqrt 4 8/81` ==> `³√(8/81) = (2/3) ³√(1/3)`\n##### Note:\n\nNegative numbers are not supported.\n\n#### 5) Log functions\n\n##### Examples:\n• `log 5` ==> `log₁₀(5) = 0.698970004336`\n• `log2 1.0001` ==> `log₂(1.0001) = 0.000144262291095`\n• `ln e` ==> `ln(e) = 1.0`\n##### Note:\n\ne=2.718281828... can be used only in some simple cases, and in most cases it is used as the scientific notation.\n\n#### 6) Prime factorization\n\n##### Examples:\n• `factor 100` ==> `factor(100) = [1, 2, 2, 5, 5]`\n• `factor 31` ==> `factor(31) = [1, 31]`\n##### Note:\n\nThe maximum value of input integer is limited in case of memory overflow.\n\n#### 7) Permutations and Combinations\n\n##### Examples:\n• `C( 4 2` ==> `C(4, 2) = 6 `\n• `c( 1000 3` ==> `C(1000, 3) = 166167000 `\n• `P( 4 2` ==> `P(4, 2) = 12`\n• `p( 1000 3` ==> `P(1000, 3) = 997002000 `\n##### Note:\n\nThe maximum value of m is limited for both permutations and combinations.\n\nEdited by Wangyou Zhang\nUpdate\n\nNice idea and workflow. You could maybe think about expanding its capabilities to become a more general maths tool. From simplifying fractions, it's not a big leap to incorporate GCD and LCM. And then, maybe, handling/simplifying surds, expressing compound fractions, obtaining the integer or fractional portions of a number, and coping with modular arithmetic.\n\nJust thinking out loud. Thanks for the sharing this.\n\nWhat a little jewel! Thank you so much!\n\nWould it be possible to simplify german comma instead of dot/period?\n\n15 hours ago, CJK said:\n\nNice idea and workflow. You could maybe think about expanding its capabilities to become a more general maths tool. From simplifying fractions, it's not a big leap to incorporate GCD and LCM. And then, maybe, handling/simplifying surds, expressing compound fractions, obtaining the integer or fractional portions of a number, and coping with modular arithmetic.\n\nJust thinking out loud. Thanks for the sharing this.\n\n@CJK Thank you for the suggestions! I will consider adding these features later :)\n\n13 hours ago, Thomas_U said:\n\nWhat a little jewel! Thank you so much!\n\nWould it be possible to simplify german comma instead of dot/period?\n\nHi, I added support for German comma as a decimal point just now. You can try the updated workflow here.\n\nEdited by Wangyou Zhang\n\nThanks - works perfect!"
] |
[
null,
"https://content.invisioncic.com/r229491/monthly_2018_11/demo.png.85fcc8703ab6b07174d6c840773db1f8.png",
null
] |
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|
https://external.codecademy.com/courses/learn-php/lessons/php-numbers/exercises/modulo
|
[
"Learn\nPHP Numbers\nModulo\n\nPHP also provides an operator that might be less familiar: modulo (`%`). The modulo operator returns the remainder after the left operand is divided by the right operand.\n\n``echo 7 % 3; // Prints: 1``\n\nIn the code above, `7 % 3` returns `1`. Why? We’re trying to fit `3` into `7` as many times as we can. `3` fits into `7` at most twice. What’s left over—the remainder—is `1`, since `7` minus `6` is `1`.",
null,
"The modulo operator will convert its operands to integers before performing the operation. This means `7.9 % 3.8` will perform the same calculation as `7 % 3`—both operations will return `1`.\n\nLet’s look at another example of the modulo operator in action:\n\n``````\\$num_cookies = 27;\n\nLet’s practice using modulo!\n\n### Instructions\n\n1.\n\nWe have 82 students going on a class trip. We want to divide the students into groups of 6. Use the modulo operator to `echo` how many students will be left without groups."
] |
[
null,
"https://content.codecademy.com/courses/php-strings-variables/PHP_m2l2.gif",
null
] |
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|
https://wiki.tcl-lang.org/page/A+little+math+language+revisited?R=0
|
[
"## A little math language revisited\n\nArjen Markus (7 may 2004) Recent interest in the possibilities of using Tcl for numerical analysis, brought back the page A little math language. I decided to explore this a bit further ... It is not complete yet, but you get the drift, I think, from the examples/test code.\n\nAM (10 may 2004) One thing I'd like to add is support for such mathematical tools like:\n\n``` y = lim x->inf x*(atan(2x)-atan(x))\ny = int x=0 to 1 exp(x**2)\ny = deriv x=0 atan(x+sqrt(x))```\n\n(oh, none of these have any practical value as far as I am concerned, I just made them up whilst typing ...). This did bring me to consider a Simple method for computing mathematical limits.\n\nAM (8 july 2004) A slight update to the original script: I added the possibility to use array elements. This does assume a fixed list of math functions in [expr] and it does not yet work with:\n\n` x(i+1) = x(i) * a`\n\nand such ... (Makes me wonder: how does Tcl handle:\n\n``` set i 1\nset a [expr {\\$b(\\$i-1)+\\$b(\\$i)]```\n\nassuming b(0) and b(1) are defined ... After all these years still some puzzles to solve and a little interactive tclsh session led to the answer:\n\n``` % set b(0) 1\n1\n% set b(1) 2\n2\n% set i 1\n1\n% set a [expr {\\$b(\\$i-1)+\\$b(\\$i)}]\ncan't read \"b(1-1)\": no such element in array\n% ```\n\n)\n\nThe reason I added support for array elements is that Shen-Yeh Chen, who is working on a package called OpTcl - optimisation with Tcl - asked for this. For his customers, Tcl is a way to use the optimisation package (written in Fortran) without having to program in C or Fortran themselves.\n\nBecause statements like\n\n``` G1=0.52*X1*X1-3.12*X1+0.72*X1*X2-21.6*X2+0.73*X2*X2-3.68\nG1=G1/100.0\nG2=225.0-pow(X1+30.0,2.0)-pow(X2-30.0,2.0)\nG2=G2/225.0 ```\n\nare much more natural for them than the ordinary Tcl syntax, this little script helps a lot.\n\n(His project provides me with a nice opportunity to further develop my Ftcl library that makes the combination of Fortran and Tcl possible - see: Combining Fortran and Tcl in one program).\n\n## Code\n\n```# mathlang.tcl --\n# Provide commands that allow a more usual mathematical syntax:\n#\n# mathfunc {x} {\n# sinc = sin(x)/x if x != 0\n# sinc = 0 otherwise\n# }\n# math {\n# a = x*x + y*y\n# }\n#\n# Still to do: mathfunc\n#\n# (7 july 2004) Small improvement:\n# recognise array elements\n#\n\nnamespace eval ::mathsyntax {\nnamespace export math\nvariable cached_calcs {}\nvariable func_names \\\n{abs acos asin atan atan2 ceil cos cosh double\nexp floor fmod hypot int log log10 pow rand round\nsin sinh sqrt srand tan tanh wide}\n}\n\n# ToExpr --\n# Transform an expression to the form expr wants\n# Arguments:\n# expression A right-hand side of an assignment\n# Result:\n# Valid Tcl expression\n#\nproc ::mathsyntax::ToExpr { expression } {\nvariable func_names\n\nset rhs [string map {\" \" \"\"} \\$expression]\nset indices [regexp -inline -all -indices {[a-zA-Z][a-zA-Z0-9_]*} \\$rhs]\nset offset 0\nforeach idx \\$indices {\nforeach {start stop} \\$idx {break}\nset start [expr {\\$start+\\$offset}]\nset stop [expr {\\$stop+\\$offset}]\nset next [expr {\\$stop+1}]\n\nif { [string index \\$rhs \\$next] != \"(\" } {\nset char [string index \\$rhs \\$start]\nset rhs [string replace \\$rhs \\$start \\$start \"\\\\$\\$char\" ]\nincr offset\n} else {\nset char [string index \\$rhs \\$start]\nset name [string range \\$rhs \\$start \\$stop]\nif { [lsearch \\$func_names \\$name] < 0 } {\nset rhs [string replace \\$rhs \\$start \\$start \"\\\\$\\$char\" ]\n}\n}\n}\nreturn \\$rhs\n}\n\n# Transform --\n# Transform a series of mathematical expressions into Tcl code\n# Arguments:\n# id ID to use\n# calc One or more mathematical assignments\n# Result:\n# None\n# Side effects:\n# A private procedure is created\n# Note:\n# No conditions yet\n#\nproc ::mathsyntax::Transform { id calc } {\nset calc [split \\$calc \"\\n\"]\nset body {\"uplevel 2 \\{\"}\n\nforeach assign \\$calc {\nset assign [string trim \\$assign]\nif { \\$assign != \"\" } {\nregexp {([a-zA-Z][a-zA-Z0-9_()]*) *= *(.*)} \\$assign ==> lhs rhsfull\n\n#\n# Is there a condition?\n#\nset cond1 [string first \" if\" \\$rhsfull]\n\n# PM: set cond2 [string first \" otherwise\" \\$rhsfull]\n\nset cond \"\"\nif { \\$cond1 > 0 } {\nset rhs [string range \\$rhsfull 0 [expr {\\$cond1-1}]]\nset cond [string range \\$rhsfull [expr {\\$cond1+3}] end]\nlappend body \"if { [ToExpr \\$cond] } \\{\"\n} else {\nset rhs \\$rhsfull\n}\n\n# If the left-hand side refers to an array element,\n# we need to add a dollar-sign\n#\nset lhs [string map {\"(\" \"(\\$\"} \\$lhs]\n\n#\n# Prepare the assignment\n#\nset rhs [ToExpr \\$rhs]\n\nlappend body \"set \\$lhs \\[expr {\\$rhs}\\]\"\n\nif { \\$cond != \"\" } {\nlappend body \"\\}\"\n}\n}\n}\n\nlappend body \"\\}\"\n\nproc Cached\\$id {} [join \\$body \"\\n\"]\n}\n\n# math --\n# Allow mathematical expressions inside Tcl code\n# Arguments:\n# calc One or more mathematical assignments\n# Result:\n# None\n# Side effects:\n# As the code is executed in the caller's scope, variables\n# in the calling procedure are set\n# The code is transformed into a procedure that is cached\n#\nproc ::mathsyntax::math { calc } {\nvariable cached_calcs\n\nset id [lsearch \\$cached_calcs \\$calc]\nif { \\$id < 0 } {\nlappend cached_calcs \\$calc\nset id [expr {[llength \\$cached_calcs]-1}]\nTransform \\$id \\$calc\n}\n\n::mathsyntax::Cached\\$id\n}\n\n#\n# Simple test\n#\nnamespace import ::mathsyntax::math\n\nset a 1\nset b 1\nset c \"\"\nset d \"\"\nset sinc \"\"\nmath {\nc = a + b\nd = a + cos(b+c)\n}\nputs \"\\$c \\$d\"\n\nfor {set i 0} {\\$i < 20} {incr i} {\nmath {\nx = 0.1*i\nsinc = 1 if x == 0\nsinc = sin(x)/x if x != 0\ny(i) = sinc*sinc\n}\nputs \"\\$i \\$x \\$sinc \\$y(\\$i)\"\n}\n#\n# Just to check\n#\nparray y```\n\n## Discussion\n\nFM maybe we should return a list, because it should be sometime interesting to get all the values. To illustrate, here is what how I did that :\n\n`variable Res [list]`\n\nin namespace eval ::mathsyntax { ... }\n\nIn proc ::mathsyntax::Transform { id calc } { ... }\n\n`set calc [split \\$calc \"\\n\\;\"] `\n\ninstead of : set calc [split \\$calc \"\\n\"]\n\n`lappend body \"lappend ::mathsyntax::Res \\[set \\$lhs \\[expr {\\$rhs}\\]\\]\"`\n\ninstead of lappend body \"[[set \\$lhs [[expr {\\$rhs}\\]\\]\"\n\nAdding at the end of proc ::mathsyntax::math { calc } { ... }\n\n`lindex \\$::mathsyntax::Res`\n``` ## to test :\npack [canvas .c]\n.c create line {*}[set L [list]; for {set i -200} {\\$i < 200} {incr i} {\nmath {\nx = 0.1*i\nsinc = 1 if x == 0\nsinc = sin(x)/x if x != 0\n}\nlappend L {*}[math {X=200+10*x;Y=200-100*sinc}]\n}; set L]```\n\ngold9/20/2020, added appendix and pix, but above text and code unchanged.\n\n### Figure 1. A little math language revisited screenshot one",
null,
"### Figure 2. A little math language revisited screenshot two",
null,
"### figure 3. A little math language revisited screenshot 3",
null,
""
] |
[
null,
"https://wiki.tcl-lang.org/image/A+little+math+language+revisited+screenshot",
null,
"https://wiki.tcl-lang.org/image/A+little+math+language+revisited+screenshot+two",
null,
"https://wiki.tcl-lang.org/image/Testbed+for+Little%5FMath%5FLanguage+screenshot",
null
] |
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|
https://math.stackexchange.com/questions/1143982/combinatorics-how-many-ways-are-there-to/1143997
|
[
"# Combinatorics: How many ways are there to\n\nHow many ways are there for a child to take 10 pieces of candy with 4 types of candy if the child takes at most 5 pieces of any type of candy?\n\n• This can be rephrased: How many solutions does $X_1+X_2+X_3+X_4 = 10,\\quad X_i \\leq5$ have? – Frank Vel Feb 11 '15 at 19:22\n\n## 1 Answer\n\n146 ways.\n\nFound programatically:\n\nvoid enumerate() {\nint a1, a2, a3, a4;\nint count = 0;\n\nfor (a1 = 0; a1 < 6; a1++)\nfor (a2 = 0; a2 < 6; a2++)\nfor (a3 = 0; a3 < 6; a3++)\nfor (a4 = 0; a4 < 6; a4++)\n{\nif (a1 + a2 + a3 + a4 == 10)\n{\ncount ++;\n}\n}\ncout << count << endl;\n}"
] |
[
null
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|
http://www.raymaps.com/index.php/tag/beamforming/page/2/
|
[
"# Massive MIMO Fundamentals and Code\n\n###### Background\n\nJust like different frequency bands and time slots can be used to multiplex users, spatial domain can also be exploited to achieve the same result. It is well known that if there are 4 transmit antennas and 4 receive antennas then four simultaneous data streams can be transmitted over the air. This can be scaled up to 8 x 8 or in the extreme case to 128 x 128. When the number of transmit or receive antennas is greater than 100 we typically call it a Massive MIMO scenario and we need specialized signal processing techniques to handle this case. Computationally complex techniques like Maximum Likelihood (ML) become quite difficult to implement in real-time and we have to resort to some simplified linear array processing techniques. This is the topic of this blog post.\n\n###### Description of the Scenario\n\nTo understand the scenario that we will discuss here, please look at our previous post. Also note that when we talk about nT x nR case we do not necessarily mean nT transmit antennas and nR receive antennas, it could also mean nT users with 1 antenna each and co-located nR receive antennas, such as at a base station. We typically assume that the number of receive antennas is greater than the number of users. When the number of users is greater than the number of receive antennas we call it overloaded case and this is not discussed here. Here the number of users is fixed at 16 (randomly distributed between 0 and 360 degrees within the cell) and the number of receive antennas is varied from 20 to 100.\n\n###### Linear Signal Processing Techniques for Massive MIMO\n\nThe four signal processing techniques that are applied at the receive array are:\n\n1. Matched Filtering (MF)\n2. Moore Penrose Pseudo-Inverse without controlling the threshold (PINV)\n3. Moore Penrose Pseudo-Inverse with a specified threshold (PINV with tol)\n4. Minimum Mean Squared Error (MMSE)\n###### Simulation Results\n\nThere are some other techniques that we experimented with but are omitted here for the sake of brevity. The MATLAB code and simulation results showing bit error rate as a function of receive array size (nR) are given below. It is seen that simple Matched Filter works quite well when the receive array size is small but with increasing nR the performance improvement is not that great. Least Squares (LS) technique using Moore-Penrose Pseudo Inverse shows improved performance with increasing nR and this can be further improved by controlling the threshold (tol). We found that a threshold of 0.1 gave significantly improved results as compared to no threshold case. Lastly we implemented MMSE and found that it gave us the best results. It must be noted that we also implemented ML for a limited size of receive array and found that its BER performance was far superior than any other technique.\n\n###### MATLAB Code for Massive MIMO Scenario\n```%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n% MASSIVE MIMO BEAMFORMING\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nclear all\nclose all\n\n% SETTING THE PARAMETERS FOR THE SIMULATION\nf=1e9; %Carrier frequency\nc=3e8; %Speed of light\nl=c/f; %Wavelength\nd=l/2; %Rx array spacing\nM=16; %Transmit array size (users)\ntheta=2*pi*(rand(1,M)); %Angular separation of users\nEbNo=10; %Energy per bit to noise PSD\nsigma=1/sqrt(2*EbNo); %Standard deviation of noise\n\nn=1:N; %Rx array number\nn=transpose(n); %Row vector to column vector\n\ns=2*(round(rand(M,1))-0.5); %BPSK signal of length M\nH=exp(-i*(n-1)*2*pi*d*cos(theta)/l); %Channel matrix of size NxM\nwn=sigma*(randn(N,1)+i*randn(N,1)); %AWGN noise of length N\nx=H*s+wn; %Receive vector of length N\n\n% LINEAR ARRAY PROCESSING - METHODS\n% 1-MATCHED FILTER\ny=H'*x;\n\n% 2-PINV without tol\n% y=pinv(H)*x;\n\n% 3-PINV with tol\n% y=pinv(H,0.1)*x;\n\n% 4-Minimum Mean Square Error (MMMSE)\n% y=(H'*H+(2*sigma^2)*eye([M,M]))^(-1)*(H')*x;\n\n% DEMODULATION AND BER CALCULATION\ns_est=sign(real(y)); %Demodulation\nber=sum(s!=s_est)/length(s); %BER calculation\n\n%Note: Please select the array processing technique\n%you want to implement (1-MF, 2-LS1, 3-LS2, 4-MMSE)```\n\nNote:\n\n1. In the code above, N=nR and M=nT\n2. What is labelled here as Matched Filter (MF) is strictly speaking Maximal Ratio Combining (MRC)\n3. The case we discuss above is categorized as Multiuser MIMO (MU-MIMO) for the uplink\n4. MU-MIMO for the downlink is not that straight forward and will be the subject of some future post\n5. We have considered a deterministic channel model as opposed to a probabilistic channel model\n6. Probabilistic channel model can be easily implemented by assuming that channel coefficients are independent and identically distributed (IID) complex Gaussian random variables with mean zero and variance of 0.5 per dimension\n7. The initial results we have obtained using this probabilistic channel model are much better than the results shown above, but the question remains which is the more accurate representation of a real channel\n###### Update: Simulation Using a Probabilistic Channel\n\nSince most of the literature in Massive MIMO uses a probabilistic channel instead of a deterministic channel, we decided to investigate this further. To implement such a channel model we simply need to change one line of the MATLAB code shown above. Instead of defining H as:\n\nH=exp(-i*(n-1)*2*pi*d*cos(theta)/l);\n\nWe define H as:\n\nH=(1/sqrt(2))*randn(N,M)+i*(1/sqrt(2))*randn(N,M);\n\nThe results are shown below. It is seen that the BER performance is orders of magnitude better. We would next investigate the performance degradation if the channel coefficients are not independent and identically distributed (IID) but have some correlation. This is closely tied to inter-element separation of the antenna array."
] |
[
null
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|
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