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https://numbermatics.com/n/8476480/	[ "# 8476480\n\n## 8,476,480 is an even composite number composed of three prime numbers multiplied together.\n\nWhat does the number 8476480 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 28 divisors.\n\n8476480 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twenty-eight divisors.\n\n## Prime factorization of 8476480:\n\n### 26 × 5 × 26489\n\n(2 × 2 × 2 × 2 × 2 × 2 × 5 × 26489)\n\nSee below for interesting mathematical facts about the number 8476480 from the Numbermatics database.\n\n### Names of 8476480\n\n• Cardinal: 8476480 can be written as Eight million, four hundred seventy-six thousand, four hundred eighty.\n\n### Scientific notation\n\n• Scientific notation: 8.47648 × 106\n\n### Factors of 8476480\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 8\n• Sum of prime factors: 26496\n\n### Divisors of 8476480\n\n• Number of divisors d(n): 28\n• Complete list of divisors:\n• Sum of all divisors σ(n): 20185380\n• Sum of proper divisors (its aliquot sum) s(n): 11708900\n• 8476480 is an abundant number, because the sum of its proper divisors (11708900) is greater than itself. Its abundance is 3232420\n\n### Bases of 8476480\n\n• Binary: 1000000101010111010000002\n• Base-36: 51OHS\n\n### Squares and roots of 8476480\n\n• 8476480 squared (84764802) is 71850713190400\n• 8476480 cubed (84764803) is 609041133344161792000\n• The square root of 8476480 is 2911.4395064985\n• The cube root of 8476480 is 203.8943448787\n\n### Scales and comparisons\n\nHow big is 8476480?\n• 8,476,480 seconds is equal to 14 weeks, 2 hours, 34 minutes, 40 seconds.\n• To count from 1 to 8,476,480 would take you about twenty-one weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 8476480 cubic inches would be around 17 feet tall.\n\n### Recreational maths with 8476480\n\n• 8476480 backwards is 0846748\n• The number of decimal digits it has is: 7\n• The sum of 8476480's digits is 37\n• More coming soon!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84738994,"math_prob":0.9507989,"size":2873,"snap":"2021-31-2021-39","text_gpt3_token_len":781,"char_repetition_ratio":0.13105612,"word_repetition_ratio":0.057777777,"special_character_ratio":0.34145492,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9927085,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T20:46:17Z\",\"WARC-Record-ID\":\"<urn:uuid:506e10c8-b4ec-4282-a3b8-0c54206c10d9>\",\"Content-Length\":\"19038\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08bab654-e007-46ec-adcd-84ba7af40798>\",\"WARC-Concurrent-To\":\"<urn:uuid:de8b62a4-538f-4f7e-b89e-4fa2114fe895>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/8476480/\",\"WARC-Payload-Digest\":\"sha1:UZX6FJXQEPFYI2H7KKOGBN47Y7XYWUBR\",\"WARC-Block-Digest\":\"sha1:FHASXKXAGVJNGFDSMNY4MWZBMX3VK7SH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154471.78_warc_CC-MAIN-20210803191307-20210803221307-00233.warc.gz\"}"}
https://docs.nvidia.com/cuda/cuquantum/python/api/generated/cuquantum.custatevec.abs2sum_array.html	[ "# cuquantum.custatevec.abs2sum_array¶\n\ncuquantum.custatevec.abs2sum_array(intptr_t handle, intptr_t sv, int sv_data_type, uint32_t n_index_bits, intptr_t abs2sum, bit_ordering, uint32_t bit_ordering_len, mask_bit_string, mask_ordering, uint32_t mask_len)[source]\n\nCalculates the sum of squared absolute values for a given set of index bits.\n\nParameters\n• handle (intptr_t) – The library handle.\n\n• sv (intptr_t) – The pointer address (as Python `int`) to the statevector (on device).\n\n• sv_data_type (cuquantum.cudaDataType) – The data type of the statevector.\n\n• n_index_bits (uint32_t) – The number of index bits.\n\n• abs2sum (intptr_t) – The pointer address (as Python `int`) to the array (on either host or device) that would hold the sums.\n\n• bit_ordering\n\nA host array of index bit ordering. It can be\n\n• an `int` as the pointer address to the array\n\n• a Python sequence of index bit ordering\n\n• bit_ordering_len (uint32_t) – The length of `bit_ordering`.\n\nA host array for specifying mask values. It can be\n\n• an `int` as the pointer address to the array\n\n• a Python sequence of mask values\n\n• an `int` as the pointer address to the array\n• mask_len (uint32_t) – The length of `mask_ordering`." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72311306,"math_prob":0.96486783,"size":1277,"snap":"2023-14-2023-23","text_gpt3_token_len":347,"char_repetition_ratio":0.145326,"word_repetition_ratio":0.3121693,"special_character_ratio":0.24510571,"punctuation_ratio":0.119617224,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9930406,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T13:07:50Z\",\"WARC-Record-ID\":\"<urn:uuid:8b38c5eb-0bda-40ab-951a-bcb72fc64930>\",\"Content-Length\":\"65325\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d385925b-aac3-4dd8-ab8d-8157f7aa642f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e5bd3d8-2d35-484f-b05c-6829b0ae0be3>\",\"WARC-IP-Address\":\"23.222.16.104\",\"WARC-Target-URI\":\"https://docs.nvidia.com/cuda/cuquantum/python/api/generated/cuquantum.custatevec.abs2sum_array.html\",\"WARC-Payload-Digest\":\"sha1:R7MZEXL2QYVW44BP3G2GN6KC6JVM5EG7\",\"WARC-Block-Digest\":\"sha1:URDVSGHV6ZWRMFJQTD2KIHYKIYGAC7QX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649986.95_warc_CC-MAIN-20230604125132-20230604155132-00360.warc.gz\"}"}
https://stats.stackexchange.com/questions/28170/clustering-a-dataset-with-both-discrete-and-continuous-variables	[ "# Clustering a dataset with both discrete and continuous variables\n\nI have a dataset X which has 10 dimensions, 4 of which are discrete values. In fact, those 4 discrete variables are ordinal, i.e. a higher value implies a higher/better semantic.\n\n2 of these discrete variables are categorical in the sense that for each of these variables, the distance e.g. from 11 to 12 is not the same as the distance from 5 to 6. While a higher variable value implies a higher in reality, the scale is not necessarily linear (in fact, it is not really defined).\n\nMy question is:\n\n• Is it a good idea to apply a common clustering algorithm (e.g. K-Means and then Gaussian Mixture (GMM)) to this dataset which contains both discrete and continuous variables?\n\nIf not:\n\n• Should I remove the discrete variables and focus only on the continuous ones?\n• Should I better discretize the continuous ones and use a clustering algorithm for discrete data?\n• You need to find a good distance measure (often the most difficult task in clustering): if you can find a distance measure that correctly and accurately describes how similar (or not) your data items are, then you should not have any problems. – Andrew May 10 '12 at 10:59\n• Speaking about those 2 categorical variables you in effect described them as ordinal. Now, what's about the rest 2 \"ordinal\" variables? How are they different from those? – ttnphns May 10 '12 at 14:12\n• They are also discrete, but both of them have a meaningful distance function, i.e. they are interval-based (if I am not messing up the definition of interval-based). – ptikobj May 10 '12 at 14:43\n\nSo you've been told you need an appropriate distance measure. Here are some leads:\n\nand, of course: Mahalanobis distance.\n\nI've had to deal with this kind of problem in the past, and I think there could be 2 interesting approaches:\n\n• Continuousification: transform symbolic attributes with a sequence of integers. There are several ways to do this, all of which described in this paper. You can try NBF, VDM and MDV algorithms.\n\n• Discretization: transform continuous attributes into symbolic values. Again, many algorithms, and a good lecture on this would be this article. I believe the most commonly used method is Holte's 1R, but the best way to know for sure is to look at the ROC curves against algorithms like EWD, EFD, ID, LD or NDD.\n\nOnce you have all your features in the same space, it becomes an usual clustering problem.\n\nChoosing between continuousification or discretization depends on your dataset and what your features look like, so it's a bit hard to say, but I advise you to read the articles I gave you on that topic.\n\nK-means obviously doesn't make any sense, as it computes means (which are nonsensical). Same goes for GMM.\n\nYou might want to try distance-based clustering algorithms with appropriate distance functions, for example DBSCAN.\n\nThe main challenge is to find a distance function!\n\nWhile you could put a different distance function into k-means, it will still compute the mean which probably doesn't make much sense (and probably messes with a distance function for discrete values).\n\nAnyway, first focus on define what \"similar\" is. Then cluster using this definition of similar!\n\nIf you are comfortable working with a distance matrix of size num_of_samples x num_of_samples, you could use random forests, as well.\n\nClick here for a reference paper titled Unsupervised learning with random forest predictors.\n\nThe idea is creating a synthetic dataset by shuffling values in the original dataset and training a classifier for separating both. During classification you will get an inter-sample distance matrix, on which you could test your favorite clustering algorithm.\n\nMixed approach to be adopted: 1) Use classification technique (C4.5 decision tree) to classify the data set into 2 classes. 2) Once it is done, leave categorical variables and proceed with continuous variables for clustering.\n\n• I could not follow your suggestion. Which two classes, and how will that help? – KarthikS Jun 11 '15 at 10:07\n• I think what Swapnil Soni needs to say is that once we use the classification technique to classify it into two class. We can then use the label of classification output as a binary variable. So instead of all the categorical variable you get an indicative binary variable and then your clustering algorithm can proceed with the data ( consisting of all continuous plus 1 binary variable). My interpretation can be wrong though. – Tusharshar Jun 23 '15 at 11:33\n• perfectly fine! – Swapnil Soni Aug 17 '15 at 6:18" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.920322,"math_prob":0.8795368,"size":860,"snap":"2020-10-2020-16","text_gpt3_token_len":192,"char_repetition_ratio":0.1495327,"word_repetition_ratio":0.0,"special_character_ratio":0.2255814,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.973353,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-02T23:14:10Z\",\"WARC-Record-ID\":\"<urn:uuid:3dd08e6b-9e6e-4eb3-b8ee-98f9d66664bd>\",\"Content-Length\":\"178550\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a7e2460a-265c-43e8-bf11-a2a01152c4fc>\",\"WARC-Concurrent-To\":\"<urn:uuid:8de71450-9f5c-4a36-a345-dc862a271e24>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/28170/clustering-a-dataset-with-both-discrete-and-continuous-variables\",\"WARC-Payload-Digest\":\"sha1:4BVUB23BBUJFHAT3K57CMNNJFSANCO33\",\"WARC-Block-Digest\":\"sha1:RWHJK3QQQXRXK7LNGLC3KWQFPIHYZIWX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370508367.57_warc_CC-MAIN-20200402204908-20200402234908-00373.warc.gz\"}"}
https://practicaldev-herokuapp-com.global.ssl.fastly.net/janvanryswyck/prevent-domain-knowledge-from-sneaking-into-solitary-tests-16he	[ "## DEV Community is a community of 700,720 amazing developers\n\nWe're a place where coders share, stay up-to-date and grow their careers.\n\n# Prevent domain knowledge from sneaking into solitary tests", null, "Jan Van Ryswyck\nOwner @ Principal IT, husband, father of three, geek, enjoys running, fan boy of many things but nothing in particular.\nOriginally published at principal-it.eu on ・3 min read\n\nPreviously we discussed why solitary tests should be easy to read. Sometimes, the readability of solitary tests is affected by those developers who overcomplicate or overengineer things. Well intentioned no doubt, but in the end quite harmful nonetheless. Complex solitary tests can cause some serious headaches for other members of the team.\n\nOne example of this is an issue that I see popping up from time to time. It’s the case where domain logic sneaks into the implementation of solitary tests. This seems to occur most often in solitary tests that exercise algorithms or business logic in domain objects.\n\nLet’s have a look at an example to see this in action.\n\n``````public class SolarPanelInstallation\n{\npublic IEnumerable<SolarPanel> SolarPanels { get; }\n\npublic SolarPanelInstallation(IEnumerable<SolarPanel> solarPanels)\n{\nSolarPanels = solarPanels;\n}\n\npublic Watts CalculateTheoreticalCapacity()\n{\nreturn SolarPanels.Aggregate(Watts.Of(0), (accumulator, solarPanel)\n=> accumulator + solarPanel.Capacity);\n}\n}\n\npublic class SolarPanel\n{\npublic Watts Capacity { get; }\n\npublic SolarPanel(Watts capacity)\n{\nCapacity = capacity;\n}\n}\n\n{\npublic int Value { get; }\n\nprivate Watts(int value)\n{\nValue = value;\n}\n\npublic static Watts Of(int value)\n{\nreturn new Watts(value);\n}\n\npublic static Watts operator +(Watts a, Watts b)\n{\nreturn Of(a.Value + b.Value);\n}\n\npublic override string ToString()\n{\nreturn \\$\"{Value} Watts\";\n}\n}\n\n``````\n\nHere we’ve entered the realm of generating green energy using solar panels. A solar panel installation can be comprised of one or multiple solar panels. Each solar panel has its own capacity which is expressed in watts. The SolarPanelInstallation class provides a method that calculates the theoretical capacity of the entire installation.\n\nLet’s have a look at the test code.\n\n``````[Specification]\npublic class When_calculating_the_theoretical_capacity_of_a_solar_panels_installation\n{\n[Establish]\npublic void Context()\n{\nvar solarPanels = new[]\n{\nnew SolarPanel(Watts.Of(368)),\nnew SolarPanel(Watts.Of(368)),\nnew SolarPanel(Watts.Of(278))\n};\n\n_sut = new SolarPanelInstallation(solarPanels);\n}\n\n[Because]\npublic void Of()\n{\n_theoreticalCapacity = _sut.CalculateTheoreticalCapacity();\n}\n\n[Observation]\npublic void Then_it_should_yield_the_total_capacity_of_all_solar_panels_of_the_installation()\n{\nvar expectedCapacity = _sut.SolarPanels\n.Select(solarPanel => solarPanel.Capacity.Value)\n.Sum();\n\n_theoreticalCapacity.Should_be_equal_to(Watts.Of(expectedCapacity));\n}\n\nprivate SolarPanelInstallation _sut;\nprivate Watts _theoreticalCapacity;\n}\n\n``````\n\nNotice how the value of the expectedCapacity variable is being calculated. This is very similar to the calculation in the CalculateTheoreticalCapacity method of the SolarPanelInstallation class. Although the way their respective implementation calculates the capacity is slightly different, we can conclude that in this example the test code contains the same knowledge as the production code. Sometimes, I even encounter tests where the developer just “borrowed” from the production code directly.\n\nThis is also a nice example where state verification tests are too tightly coupled to the production code. So when the implementation of the CalculateTheoreticalCapacity method is refactored, chances are quite high that the test code needs to be modified as well.\n\nIt’s also more difficult to read this test and figure out what the expected value should be. For an easy example like this it doesn’t require that much additional brain cycles. However, with more complex algorithms or business logic, developers often execute the test in debug mode just to figure out what the expected value should be. How’s that for readability?\n\nLet’s have a look at an improved version of this test.\n\n``````[Specification]\npublic class When_calculating_the_theoretical_capacity_of_a_solar_panels_installation\n{\n[Establish]\npublic void Context()\n{\nvar solarPanels = new[]\n{\nnew SolarPanel(Watts.Of(368)),\nnew SolarPanel(Watts.Of(368)),\nnew SolarPanel(Watts.Of(278))\n};\n\n_sut = new SolarPanelInstallation(solarPanels);\n}\n\n[Because]\npublic void Of()\n{\n_theoreticalCapacity = _sut.CalculateTheoreticalCapacity();\n}\n\n[Observation]\npublic void Then_it_should_yield_the_total_capacity_of_all_solar_panels_of_the_installation()\n{\n_theoreticalCapacity.Should_be_equal_to(Watts.Of(1014));\n}\n\nprivate SolarPanelInstallation _sut;\nprivate Watts _theoreticalCapacity;\n}\n\n``````\n\nHere we just provided the value that we expect to be the result of the calculation. That’s it! No more duplicate domain knowledge, no more tight coupling of the test and no more debugging. The expected value is just right there. Simplicity can be a beautiful thing.\n\nDomain knowledge that sneaks into your tests is something to be avoided. Be very mindful about this. Don’t use the Subject Under Test itself for determining the outcome of a test." ]	[ null, "https://res.cloudinary.com/practicaldev/image/fetch/s--Z2zYXPlP--/c_fill,f_auto,fl_progressive,h_50,q_auto,w_50/https://dev-to-uploads.s3.amazonaws.com/uploads/user/profile_image/222059/a845194f-1a28-45e5-b119-6148b9cfe2e2.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.805881,"math_prob":0.82372546,"size":4859,"snap":"2021-31-2021-39","text_gpt3_token_len":1028,"char_repetition_ratio":0.17590113,"word_repetition_ratio":0.14751554,"special_character_ratio":0.2082733,"punctuation_ratio":0.124492556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95051473,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T12:08:35Z\",\"WARC-Record-ID\":\"<urn:uuid:0e0fb24b-b84c-4e94-86c0-9305de25c490>\",\"Content-Length\":\"114316\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a74a348-9396-43da-bc0c-8c8554c9fd8c>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1138b34-548d-4f84-a90c-61577ea17e9a>\",\"WARC-IP-Address\":\"151.101.201.194\",\"WARC-Target-URI\":\"https://practicaldev-herokuapp-com.global.ssl.fastly.net/janvanryswyck/prevent-domain-knowledge-from-sneaking-into-solitary-tests-16he\",\"WARC-Payload-Digest\":\"sha1:2BGE66G7JYMZ6TPCL4ENBBXA76RCVS3D\",\"WARC-Block-Digest\":\"sha1:UGAI6JRHWX24TALQ7YSCM5MVCIFBVSMK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056856.4_warc_CC-MAIN-20210919095911-20210919125911-00599.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/42-10-plus-70-90	[ "Solutions by everydaycalculation.com\n\n1st number: 4 2/10, 2nd number: 70/90\n\n42/10 + 70/90 is 224/45.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 10 and 90 is 90\n2. For the 1st fraction, since 10 × 9 = 90,\n42/10 = 42 × 9/10 × 9 = 378/90\n3. Likewise, for the 2nd fraction, since 90 × 1 = 90,\n70/90 = 70 × 1/90 × 1 = 70/90\n378/90 + 70/90 = 378 + 70/90 = 448/90\n5. 448/90 simplified gives 224/45\n6. So, 42/10 + 70/90 = 224/45\nIn mixed form: 444/45\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8011613,"math_prob":0.99505085,"size":724,"snap":"2020-10-2020-16","text_gpt3_token_len":298,"char_repetition_ratio":0.1625,"word_repetition_ratio":0.0,"special_character_ratio":0.5359116,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99686176,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-28T05:11:09Z\",\"WARC-Record-ID\":\"<urn:uuid:8f857265-7393-4959-ad1d-0508cd97d50f>\",\"Content-Length\":\"7568\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64b801f4-967d-4d28-99d3-2150a39a1a5c>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea0482d3-bf97-43d5-877d-499f80d856a7>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/42-10-plus-70-90\",\"WARC-Payload-Digest\":\"sha1:7KQW3Y44GGN6NZMW572HUF4NWOI3YLW5\",\"WARC-Block-Digest\":\"sha1:VRMOSB23EKZUMNRBX66NBGEETXWH4ZFP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875147054.34_warc_CC-MAIN-20200228043124-20200228073124-00558.warc.gz\"}"}
https://www.portfolioprobe.com/2012/03/12/the-quality-of-variance-matrix-estimation/	[ "# The quality of variance matrix estimation\n\nA bit of testing of the estimation of the variance matrix for S&P 500 stocks in 2011.\n\n## Previously\n\nThere was a plot in “Realized efficient frontiers” showing the realized volatility in 2011 versus a prediction of volatility at the beginning of the year for a set of random portfolios. A reader commented to me privately that they expected the plot not to be so elliptical — that they expected there to be higher dispersion for high volatility portfolios.\n\nMy initial hypothesis was that maybe the Ledoit-Wolf estimate is better than the factor models that the reader is used to. We’ll test that and a few other hypotheses.\n\nThe realized period in this post is all of 2011.\n\nFigure 1 shows the realized and predicted volatility of 10,000 random portfolios.\n\nFigure 1: Ledoit-Wolf estimate with linearly decreasing weights on a 1 year lookback for portfolios with risk fraction constraints.", null, "This shows the correlation among the random portfolios and the line where predicted and realized are equal. Many tests of variance matrices use one portfolio and look at the vertical distance between that portfolio and the equality line. That’s not really much of a test.\n\nThe future is very likely to be either more volatile or less volatile than the sample period. The plots here suggest that getting the level right is just a matter of a single scaling parameter.\n\nWhat is more important — certainly in the context of optimization — is knowing which of two portfolios will have the larger volatility. That is, the correlation between realized and predicted is really of interest.\n\n## Estimators\n\nMy hypothesis was that perhaps factor models wouldn’t be as good as Ledoit-Wolf shrinkage. Figure 2 is the same as Figure 1 except that a statistical factor model is used instead of Ledoit-Wolf to predict volatility.\n\nFigure 2: Statistical factor estimate with linearly decreasing weights on a 1 year lookback for portfolios with risk fraction constraints.", null, "Figures 1 and 2 are close to identical with Ledoit-Wolf nosing out the statistical factor model. It would be interesting to see results for other factor models.\n\n## Constraints\n\nThe random portfolios in Figures 1 and 2 were generated with the constraints:\n\n• long-only\n• 90 to 100 names\n• no asset can contribute more than 3% to the portfolio variance\n\nFigure 3 shows what happens when we replace the last constraint with:\n\n• no asset can contribute more than 3% weight\n\nFigure 3: Ledoit-Wolf estimate with linearly decreasing weights on a 1 year lookback for portfolios with weight constraints.", null, "Figure 3 has a much lower correlation than Figure 1. However, that is not a general result. The presentation “Portfolio optimisation inside out” shows a case where the weight constraints have a higher correlation than the corresponding risk fraction constraints.\n\nAn important difference in the present case is that the range of predicted volatilities is much narrower for the weight constraints, so it is natural for the correlation to be lower.\n\n## Time weights\n\nIt is possible to give more weight to some observations than others. The default weighting in the variance estimators (Ledoit-Wolf and statistical factor) in the BurStFin package is linearly decreasing. The most recent observation has 3 times the weight of the first observation in the sample period. That is the weighting used in Figures 1 through 3.\n\nFigure 4 uses equal weights for all the sample period, which is 2010.\n\nFigure 4: Ledoit-Wolf estimate with equal weights on a 1 year lookback for portfolios with risk fraction constraints.", null, "This is barely worse than using the linearly decreasing weights.\n\n## Lookback\n\nThe size of the sample period is an extreme form of controlling time weights. The time weights of observations too far back in time are set to zero.\n\nFigure 5: Ledoit-Wolf estimate with linearly decreasing weights on a 5 year lookback for portfolios with risk fraction constraints.", null, "The five-year sample period is worse than the one-year period.\n\n## Caveat\n\nYou shouldn’t think that we’ve really learned anything here. In order to get a firm sense of what works best we need to do this sort of thing over multiple times. The market changes. Different periods are different. What we’ve seen here is just for one period.\n\nHowever, for this one period the observations are:\n\n• risk fraction constraints give higher correlation than weight constraints (this is known to be time-dependent, and in this case not a valid comparison of accuracy)\n• 1-year lookback is better than 5-year lookback\n• linearly decreasing weights are slightly better than equal weights\n• Ledoit-Wolf shrinkage is slightly better than the statistical factor model\n\n## Summary\n\nUsing random portfolios to test variance matrix estimators provides vastly more useful information than the alternative.\n\nThe details of variance estimation seem to be fairly unimportant. I of course mean within reasonable limits — the sample variance isn’t going to do when there are more assets than observations.\n\n## Questions\n\nIf we wanted to get a subset of the risk fraction random portfolios that had approximately the same distribution of predicted volatility as the weight constraint random portfolios, what would be a good way of selecting the subset?\n\n## Epilogue\n\nIt turned out to be the howling of a dog\nOr a wolf, to be exact.\nThe sound sent shivers down my back\n\nfrom “Furr” by Eric Earley\n\n## Appendix R\n\nAn explanation of the R computations is given in “Realized efficient frontiers” but a few extra things were done here.\n\nThe statistical factor model as well as the Ledoit-Wolf estimator is from the BurStFin package.\n\n`require(BurStFin)`\n`var.sf.10.lw <- factor.model.stat(sp5.ret[1007:1258, ])`\n\nThe command to get equal weights is:\n\n`var.lw.10.qw <- var.shrink.eqcor(sp5.ret[1007:1258, ], weight=rep(1,252))`\n\n#### safety tip\n\nThis analysis involved two sets of random portfolios. It would be easy to mix objects concerning the two sets. But there is a trick to reduce that possibility.\n\nThe first set of random portfolios contained 10,000 portfolios. The second was created to have 10,001 portfolios. Mixing up objects in this case is very likely to produce either an error or a warning.\n\nSubscribe to the Portfolio Probe blog by Email\n\nThis entry was posted in Quant finance, R language and tagged , , . Bookmark the permalink.\n\n### 7 Responses to The quality of variance matrix estimation\n\n1.", null, "Stanislav Lem says:\n\nThe plots look terrific when plotted with anti-aliased device such as\n\nlibrary(cairoDevice)\nCairo()\nCairo_png()\n\nJust a humble suggestion. Really enjoying your work.\n\n•", null, "Pat says:\n\nStanislav,\n\nThanks. Some day when I’m not being lazy I’ll try it out.\n\n2.", null, "eran says:\n\nHi\nI just wanted to ask why:\n” look at the vertical distance between that portfolio and the equality line.” is not really much of a test? isnt the distribution of a distance defined properly?\nThanks\n\n•", null, "Pat says:\n\nEran,\n\nSorry for the late reply — somehow I didn’t see the notice of your comment.\n\nThe problem isn’t with the distance — the point clouds imply that the distance is quite well defined.\n\nOne problem is that it is a one-dimensional test of a multi-dimensional model. In particular, that one dimension is pretty much completely irrelevant for portfolio optimization. What matters there is knowing which portfolios will have lower volatility, not really the value of the volatility.\n\nThe phenomenon of variance clustering is ubiquitous in markets — volatility goes up and down over time. And the moves are to some extent unpredictable. The general level of volatility is going to either be lower or higher than the risk model says. We don’t know which, but we can be pretty sure the model will have the level wrong.\n\nYou can test how close it is over multiple time periods. It will take a long time to get significant results doing this. The methodology of creating the model may change over time, so you’ll have data snooping problems. You also have to decide on a time horizon — you can make the model look better or worse by changing the horizon you assume.\n\n3.", null, "eran says:\n\nHi Pat,\n\nDespite the fact that you can conclude on the distance, since you test only one portfolio, its not really meaningful for future portfolio construction, think that is what I missed.\n\nThanks for the post and the reply." ]	[ null, "https://portfoliolive-wpengine.netdna-ssl.com/wp-content/uploads/2012/03/samp1_lewolf_10_linw.png", null, "https://portfoliolive-wpengine.netdna-ssl.com/wp-content/uploads/2012/03/samp1_fac_10_linw.png", null, "https://portfoliolive-wpengine.netdna-ssl.com/wp-content/uploads/2012/03/samp2w_lewolf_10_linw.png", null, "https://portfoliolive-wpengine.netdna-ssl.com/wp-content/uploads/2012/03/samp1_lewolf_10_eqw.png", null, "https://portfoliolive-wpengine.netdna-ssl.com/wp-content/uploads/2012/03/samp1_lewolf_my_linw.png", null, "https://secure.gravatar.com/avatar/db26fa651bead519747438a4f4322c29", null, "https://secure.gravatar.com/avatar/1f1538e8d3f3619915c032101bb80a62", null, "https://secure.gravatar.com/avatar/ecec66f18445b116cd2ebba3ad7a6343", null, "https://secure.gravatar.com/avatar/1f1538e8d3f3619915c032101bb80a62", null, "https://secure.gravatar.com/avatar/2fbd15caa53245ee6de9ece21aaafb82", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9226928,"math_prob":0.95100266,"size":8172,"snap":"2021-21-2021-25","text_gpt3_token_len":1705,"char_repetition_ratio":0.13222331,"word_repetition_ratio":0.06945482,"special_character_ratio":0.20349975,"punctuation_ratio":0.08560053,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9878254,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T10:20:23Z\",\"WARC-Record-ID\":\"<urn:uuid:f21fe433-07b0-412e-aa9a-b3660b4729f3>\",\"Content-Length\":\"97617\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b00eda6-edc8-46f7-9ea3-0462dd737a52>\",\"WARC-Concurrent-To\":\"<urn:uuid:6003a1ab-a96f-40f3-81be-68bb23dcebdc>\",\"WARC-IP-Address\":\"35.197.229.230\",\"WARC-Target-URI\":\"https://www.portfolioprobe.com/2012/03/12/the-quality-of-variance-matrix-estimation/\",\"WARC-Payload-Digest\":\"sha1:FLFOXQQA2M5NHVHE6UM7O5DFVGR7BLDK\",\"WARC-Block-Digest\":\"sha1:GXR6RGHMOQJJII3Z2UPGDAFVXRGERN6S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487630081.36_warc_CC-MAIN-20210625085140-20210625115140-00077.warc.gz\"}"}
https://mathemerize.com/asymptotes-of-hyperbola/	[ "# Equation of Asymptotes of Hyperbola – Director Circle\n\nHere, you will learn equation of asymptotes of hyperbola and how to find the asymptotes of the hyperbola and the director circle of hyperbola.\n\nLet’s begin –\n\n## Equation of Asymptotes of hyperbola\n\nIf the length of the perpendicular let fall from the point on the hyperbola to a straight line tends to zero as the point on the hyperbola moves to infinity along the hyperbola, then the straight line is called the Asymptote of the hyperbola.\n\nHow to find the asymptotes of the hyperbola :\n\nLet y = mx + c is the asymptote of the hyperbola $$x^2\\over a^2$$ – $$y^2\\over b^2$$ = 1.\n\nSolving these two we get the quadratic as $$(b^2 – a^2m^2)$$$$x^2$$ – 2$$a^2$$mcx – $$a^2(b^2 + c^2)$$ = 0 …….(1)\n\nIn order that y = mx + c be an asymptote, both roots of equation (1) must approach infinity, the condition for which are :\n\ncoefficient of $$x^2$$ = 0 & coefficient of x = 0.\n\n$$\\implies$$ $$(b^2 – a^2m^2)$$ = 0 or m = $$\\pm b\\over a$$ & $$a^2$$mc = 0 $$\\implies$$ c = 0.\n\n$$\\therefore$$ equations of asymptote are $$x\\over a$$ + $$y\\over b$$ = 0 and $$x\\over a$$ – $$y\\over b$$ = 0\n\ncombined equation to the asymptotes $$x^2\\over a^2$$ – $$y^2\\over b^2$$ = 0\n\nWhen b = a, the asymptotes of the rectangular hyperbola.\n\n$$x^2 – y^2$$ = $$a^2$$ are y = $$\\pm$$x which are at right angles.\n\nExample : Find the asymptotes of the hyperbola $$2x^2 + 5xy + 2y^2 + 4x + 5y$$ = 0\n\nSolution : Let $$2x^2 + 5xy + 2y^2 + 4x + 5y + k$$ = 0 be asymptotes. This will represent two straight line\n\nso $$abc + 2fgh – af^2 – bg^2 – ch^2$$ = 0 $$\\implies$$ 4k + 25 – $$25\\over 2$$ – 8 – $$25\\over 4$$k = 0\n\n$$\\implies$$ k = 2\n\n$$\\implies$$ $$2x^2 + 5xy + 2y^2 + 4x + 5y + 2$$ = 0 are asymptotes\n\n$$\\implies$$ (2x+y+2) = 0 and (x+2y+1) = 0 are asymptotes" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80478376,"math_prob":1.00001,"size":1707,"snap":"2023-14-2023-23","text_gpt3_token_len":661,"char_repetition_ratio":0.19671169,"word_repetition_ratio":0.093655586,"special_character_ratio":0.41007617,"punctuation_ratio":0.05232558,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T14:38:10Z\",\"WARC-Record-ID\":\"<urn:uuid:33d98def-e665-4da7-a3c1-e1de60c3690f>\",\"Content-Length\":\"217755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:87422033-373f-4079-b5c6-9c274b7ba16d>\",\"WARC-Concurrent-To\":\"<urn:uuid:386fa482-c1f6-4e7e-9ed4-455e25434e02>\",\"WARC-IP-Address\":\"50.16.223.119\",\"WARC-Target-URI\":\"https://mathemerize.com/asymptotes-of-hyperbola/\",\"WARC-Payload-Digest\":\"sha1:H6IVFBYEDYIV7MZIZXS3I2MNBPDS45X3\",\"WARC-Block-Digest\":\"sha1:PAKNIYA5GKLVXWLIU6ND6SGDAURZA6F5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656737.96_warc_CC-MAIN-20230609132648-20230609162648-00412.warc.gz\"}"}
https://adv-math.com/optimal-expected-value-assets-black-scholes-equation-transaction-costs/	[ "Optimal Expected Value of Assets Under Black-Scholes Equation with Transaction Costs Joy Ijeoma Adindu-Dick\n\n# Optimal Expected Value of Assets Under Black-Scholes Equation with Transaction CostsJoy Ijeoma Adindu-Dick\n\nAbstract: This paper deals with optimal expected value of assets under Black-Scholes equation with transaction costs. The partial differential equation for option pricing with transaction costs on the domain $(P,T)\\in(0,\\infty)\\times (0,T)$ with terminal condition $C(P,T)=max(P-E,0),P\\in(0,\\infty)$ for European call options with strike price $E$, and a suitable terminal condition for European puts was obtained and then solved using Euler’s substitution method.\n\nKeywords: Black-Scholes equation with transaction costs, Optimal value, Euler’s substitution method, Option pricing." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7809575,"math_prob":0.97728515,"size":917,"snap":"2020-34-2020-40","text_gpt3_token_len":208,"char_repetition_ratio":0.13800657,"word_repetition_ratio":0.104347825,"special_character_ratio":0.21046892,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98326266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T15:38:55Z\",\"WARC-Record-ID\":\"<urn:uuid:0325963a-cef4-455a-8c9a-5bced3d38ccf>\",\"Content-Length\":\"54866\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6c34df03-bf64-4295-93da-3ed54b6d0b73>\",\"WARC-Concurrent-To\":\"<urn:uuid:3967fb3c-d9f0-489f-9d0f-cdb2ac07f6af>\",\"WARC-IP-Address\":\"103.212.121.63\",\"WARC-Target-URI\":\"https://adv-math.com/optimal-expected-value-assets-black-scholes-equation-transaction-costs/\",\"WARC-Payload-Digest\":\"sha1:JHWSSV75CFF4RQ6REUN6MB3NUDZHFBR7\",\"WARC-Block-Digest\":\"sha1:D4ZML5I4CG2YXC5DAW3SSVVLDCDMMPAN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400283990.75_warc_CC-MAIN-20200927152349-20200927182349-00563.warc.gz\"}"}
https://shkafy-vkupe.ru/how-to-solve-a-division-problem-6274.html	[ "# How To Solve A Division Problem", null, "This tutorial shows you how to take a word problem and translate it into a mathematical equation involving a complex fraction.\n\nLuckily, there's some key words to look out for in a word problem that help tell you what math operation to use!\n\nThis tutorial shows you some of these key words to look for in a word problem. This tutorial shows you how to take a word problem and translate it into a mathematical equation involving fractions.\n\nFor example, the nine in the dividend 59 is brought down and placed next to the 1 to form the number 19.\n\nRepeat this process until all of the numbers in the dividend have been brought down.\n\nDividing large numbers is a complex process that can become difficult for some students.\n\nThe division process involves many different steps that must be completed in the correct order, and this process must be practiced to ensure mastery.\n\nIf the subtracted answer is larger than the divisor, the student needs to find and fix the mistake in either the division or multiplication step.\n\nBring down the number to the right in the dividend and place it next to the subtracted answer.\n\n\" and using it as a step-by-step guide when dividing large numbers.\n\nDivide the first number of the dividend by the divisor." ]	[ null, "https://shkafy-vkupe.ru/rurohajikn/how-to-solve-a-division-problem-9639.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9263003,"math_prob":0.9095799,"size":1271,"snap":"2021-21-2021-25","text_gpt3_token_len":251,"char_repetition_ratio":0.12786108,"word_repetition_ratio":0.12669684,"special_character_ratio":0.19590874,"punctuation_ratio":0.0661157,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953395,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-14T09:15:41Z\",\"WARC-Record-ID\":\"<urn:uuid:309f120a-3b97-4f80-b05b-0eda1413a5f0>\",\"Content-Length\":\"27687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc10228e-f908-46ec-9dc5-e4315552c10c>\",\"WARC-Concurrent-To\":\"<urn:uuid:fcd38ec2-4806-41a1-a304-637b2c92c33e>\",\"WARC-IP-Address\":\"104.21.12.50\",\"WARC-Target-URI\":\"https://shkafy-vkupe.ru/how-to-solve-a-division-problem-6274.html\",\"WARC-Payload-Digest\":\"sha1:R3YYYIOZBD443JWWIECRKTQKXKUF2JUR\",\"WARC-Block-Digest\":\"sha1:54AJ5AWGJO3HGQKVGDEH2FT34R3FJ224\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487611641.26_warc_CC-MAIN-20210614074543-20210614104543-00173.warc.gz\"}"}
https://www.physicsforums.com/threads/bounding-the-volume-distortion-of-a-manifold.876101/	[ "# Bounding the volume distortion of a manifold\n\n• I\neyenir\nLet $U$ be a compact set in $\\mathbb{R}^k$ and let $f:U\\to\\mathbb{R}^n$ be a $C^1$ bijection. Consider the manifold $M=f(U)$.\n\nIts volume distortion is defined as $G=det(DftDf).$ If $n=1$, one can deduce that $G=1+\|\\nabla f\|^2$.\n\nWhat happens for $n>1$? Can one bound from below this $G$? If so: under which assumptions?\n\nLet $U$ be a compact set in $\\mathbb{R}^k$ and let $f:U\\to\\mathbb{R}^n$ be a $C^1$ bijection.\nSo you suggest taking a finite sub-cover ${U_i}$ such that $M\|_{U_i}=\\{(x_1,\\dots,x_{n-1},g_i\\}$ where $g_i:\\mathbb{R^{n-1}\\to R}$ is a $C^1$ function on $U_i$?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77286416,"math_prob":1.0000006,"size":533,"snap":"2022-27-2022-33","text_gpt3_token_len":205,"char_repetition_ratio":0.07183365,"word_repetition_ratio":0.63529414,"special_character_ratio":0.35084426,"punctuation_ratio":0.13492064,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-14T13:04:29Z\",\"WARC-Record-ID\":\"<urn:uuid:5425081a-9df5-4674-9568-ee2923dfcaff>\",\"Content-Length\":\"67342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:da2cf710-062a-43b0-8b42-eff916849dd8>\",\"WARC-Concurrent-To\":\"<urn:uuid:6e0e42bd-e609-4b5b-8461-de2ea18981c4>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/bounding-the-volume-distortion-of-a-manifold.876101/\",\"WARC-Payload-Digest\":\"sha1:F35OOFHQZ3FZ5ZJDY6L4IV6DXNENKTVP\",\"WARC-Block-Digest\":\"sha1:OILKSPSM6NXH6H4VZUR3IA2K4NSYRSRI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572033.91_warc_CC-MAIN-20220814113403-20220814143403-00454.warc.gz\"}"}
https://forum.seeedstudio.com/t/groove-temprature-sensor-calculation/17383	[ "", null, "# Groove Temprature Sensor Calculation\n\nHi,\n\nI am using Groove Temperature sensor and code posted on wiki (seeedstudio.com/wiki/Grove_- … ure_Sensor) seems to be working fine.\n\nHowever I am not able to understand the equation used in the calculation. Wiki says that equation is taken from Data sheet. However I am unable to find the equation in the data sheet. Data sheet only mentioned B value. I am not able to co-relate equation with Steinhart–Hart Equation either as Steinhart–Hart Equation uses three coefficients but this equation is using only B value.\n\nCan someone please enlighten me for following two statements.\n\nresistance=(float)(1023-a)*10000/a; //get the resistance of the sensor;\ntemperature=1/(log(resistance/10000)/B+1/298.15)-273.15;//convert to temperature via datasheet ;\n\nData Sheet attached.\nTTC03.pdf (68.6 KB)\n\nHi.\n\nI agree we might improve the sample code. I found a fairly elaborate article about the Steinhart–Hart Equation and how to use it when measuring the output from thermistors. I can’t post the link directly, but I suggest you look at the examples on the main Arduino site, and search for “Thermistor2”.\n\nI followed that article all the way down to the section titled “The Elaborate Code (cleaned up a bit)”.\n\nI hope this helps.\n\nBest wishes,\n\nBryce\n\nI suspect that there is a problem with my sensor. Even using the updated code suggested from the Arduino site, I am getting unexpected results:\n\nADC: 270/1024, vcc: 4.91, pad: 9.850 Kohms, Volts: 1.295, Resistance: 27507 ohms, Celsius: 3.4, Fahrenheit: 38.1\nADC: 270/1024, vcc: 4.91, pad: 9.850 Kohms, Volts: 1.295, Resistance: 27507 ohms, Celsius: 3.4, Fahrenheit: 38.1\nADC: 270/1024, vcc: 4.91, pad: 9.850 Kohms, Volts: 1.295, Resistance: 27507 ohms, Celsius: 3.4, Fahrenheit: 38.1\nADC: 270/1024, vcc: 4.91, pad: 9.850 Kohms, Volts: 1.295, Resistance: 27507 ohms, Celsius: 3.4, Fahrenheit: 38.1\n\nit’s 70 degrees Fahrenheit in my home during these tests.\n\nI found the issue. The shield switch had been moved to the 3 volt position." ]	[ null, "http://forum.seeedstudio.com/uploads/default/original/2X/c/c1add9aca880b030ab7e0129311ecb6b9c16568c.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86498034,"math_prob":0.8252047,"size":789,"snap":"2021-43-2021-49","text_gpt3_token_len":192,"char_repetition_ratio":0.124840766,"word_repetition_ratio":0.018348623,"special_character_ratio":0.24968314,"punctuation_ratio":0.11764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96015936,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T16:58:49Z\",\"WARC-Record-ID\":\"<urn:uuid:7c77e1e7-5f87-4021-90dc-f2b0edd8ee33>\",\"Content-Length\":\"21157\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f501dede-bbf0-4c4b-bc41-05361cbdc44e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e6582644-26cc-468e-b273-3ed3da1b0d4a>\",\"WARC-IP-Address\":\"168.62.217.176\",\"WARC-Target-URI\":\"https://forum.seeedstudio.com/t/groove-temprature-sensor-calculation/17383\",\"WARC-Payload-Digest\":\"sha1:QZEZ64VDFVQ6EH3FX46Z336CRKKJ765H\",\"WARC-Block-Digest\":\"sha1:7D7FQ74PELBCEN22TGEJQRH5HLRI74EC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363215.8_warc_CC-MAIN-20211205160950-20211205190950-00030.warc.gz\"}"}
https://centumacademy.com/questions/question/qod-olympiad-track-020722	[ "# CENTUM\n\n### CENTUM\n\n0\n0", null, "• You must to post comments\n0\n0\n0\n0\n\nx + y = 1/2^y-x\nx + y = 2^x-y\n(x + y)^x – y = 2\n2^(x – y)^2 = 2\n(x – y)^2 = 1\nx – y = +-1\nx + y = 2^+-1 = 2 or 1/2\n\n• You must to post comments\n0\n0\n• You must to post comments\nShowing 4 results" ]	[ null, "https://centumacademy.com/caassets/uploads/2022/07/QOD1-1024x248.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5494316,"math_prob":0.9999182,"size":557,"snap":"2022-27-2022-33","text_gpt3_token_len":244,"char_repetition_ratio":0.17179024,"word_repetition_ratio":0.023529412,"special_character_ratio":0.33752245,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9970863,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T09:23:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ebd9b281-e305-4430-8f8b-c14a8ce16692>\",\"Content-Length\":\"108965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d96fa30d-37cf-40dc-8f8a-4ff39b870b3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3c5c524-b675-4fc7-90bc-6365aae7029c>\",\"WARC-IP-Address\":\"143.110.248.157\",\"WARC-Target-URI\":\"https://centumacademy.com/questions/question/qod-olympiad-track-020722\",\"WARC-Payload-Digest\":\"sha1:TD2SX5CJOW4LELCWCDS2WDSEHDVQF6AQ\",\"WARC-Block-Digest\":\"sha1:RYU6HAXJKROPONK4CY43LCIOYMFBTH2M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571246.56_warc_CC-MAIN-20220811073058-20220811103058-00157.warc.gz\"}"}
https://pub.uni-bielefeld.de/record/2304016	[ "# Coding theorems of quantum information theory\n\nWinter A (1999)\nBielefeld (Germany): Bielefeld University.\n\nDownload", null, "Bielefelder E-Dissertation \| Englisch", null, "Autor\nBetreuer\nEinrichtung\nAbstract / Bemerkung\nCoding theorems and (strong) converses for memoryless quantum communication channels and quantum sources are proved: for the quantum source the coding theorem is reviewed, and the strong converse proven. For classical information transmission via quantum channels we give a new proof of the coding theorem, and prove the strong converse, even under the extended model of nonstationary channels. As a by-product we obtain a new proof of the famous Holevo bound. Then multi-user systems are investigated, and the capacity region for the quantum multiple access channel is determined. The last chapter contains a preliminary discussion of some models of compression of correlated quantum sources, and a proposal for a program to obtain operational meaning for quantum conditional entropy. An appendix features the introduction of a notation and calculus of entropy in quantum systems.\nStichworte\nJahr\nPUB-ID\n\n### Zitieren\n\nWinter A. Coding theorems of quantum information theory. Bielefeld (Germany): Bielefeld University; 1999.\nWinter, A. (1999). Coding theorems of quantum information theory. Bielefeld (Germany): Bielefeld University.\nWinter, A. (1999). Coding theorems of quantum information theory. Bielefeld (Germany): Bielefeld University.\nWinter, A., 1999. Coding theorems of quantum information theory, Bielefeld (Germany): Bielefeld University.\nA. Winter, Coding theorems of quantum information theory, Bielefeld (Germany): Bielefeld University, 1999.\nWinter, A.: Coding theorems of quantum information theory. Bielefeld University, Bielefeld (Germany) (1999).\nWinter, Andreas. Coding theorems of quantum information theory. Bielefeld (Germany): Bielefeld University, 1999.\nAlle Dateien verfügbar unter der/den folgenden Lizenz(en):\nCopyright Statement:\nThis Item is protected by copyright and/or related rights. [...]\nVolltext(e)\nName\nAccess Level", null, "Open Access\nZuletzt Hochgeladen\n2015-12-11T14:38:17Z\n\nOpen Data PUB" ]	[ null, "https://pub.uni-bielefeld.de/images/access_open.png", null, "https://pub.uni-bielefeld.de/thumbnail/2304016", null, "https://pub.uni-bielefeld.de/images/access_open.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73130995,"math_prob":0.7261784,"size":1968,"snap":"2019-13-2019-22","text_gpt3_token_len":454,"char_repetition_ratio":0.20977597,"word_repetition_ratio":0.18604651,"special_character_ratio":0.19512194,"punctuation_ratio":0.18012422,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9527092,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-27T11:44:17Z\",\"WARC-Record-ID\":\"<urn:uuid:d0ffb94b-bda2-469d-b3f1-c4d4dd4024f8>\",\"Content-Length\":\"31174\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4811c57-c13f-44eb-9d00-13e47380c61d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e007e53-1814-4715-a925-31bb7be87cff>\",\"WARC-IP-Address\":\"129.70.43.67\",\"WARC-Target-URI\":\"https://pub.uni-bielefeld.de/record/2304016\",\"WARC-Payload-Digest\":\"sha1:3YMKOUP3K26IYI5U4IVFPDH3OAHGQOIK\",\"WARC-Block-Digest\":\"sha1:2DFEQZVUYPGXD25XLJHGIJ6P3KYTNRJ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232262369.94_warc_CC-MAIN-20190527105804-20190527131804-00306.warc.gz\"}"}
https://www.asiabookkeeping.com/23-3-break-even-point-calculation/	[ "# Break-even Point Calculation\n\n1. Break-even points can be calculated based on fixed costs, variable costs per unit, and margin contribution per unit using the following formula.\n\nEXAMPLE\n\nThe Yang Yang Manufacturing Company is a factory that manufactures badminton racket. Details of the expenditure are as follows:\n\n2. The Break-even Point calculation assumes that:\n\na) the total cost consists of fixed costs and cost changed.\n\nb) Fixed costs consist of overhead costs.\n\nc) Cost changed consist of material cost and labor cost.\n\nd) firms only produce a type of release only\n\n3. The firm’s fixed cost is the same despite changes in the amount of release. For example rent, and insurance.\n\nThe cost changed in the firm’s firms changes just as the change in production. For example, the cost of raw materials and direct labor costs.\n\n5. The break-even point can be represented in the form of graphs or through a contribution margin formula." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8799504,"math_prob":0.90268993,"size":898,"snap":"2021-43-2021-49","text_gpt3_token_len":183,"char_repetition_ratio":0.13199106,"word_repetition_ratio":0.0,"special_character_ratio":0.2026726,"punctuation_ratio":0.114285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9884725,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T06:53:56Z\",\"WARC-Record-ID\":\"<urn:uuid:e60fa940-0fed-4fe8-84d2-2ab9a7c0d4b5>\",\"Content-Length\":\"43303\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3dff6a8-188f-48b0-966c-cc6ebcc270f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:b33659cb-2c37-404b-8bb8-b7647c492fbf>\",\"WARC-IP-Address\":\"35.213.146.187\",\"WARC-Target-URI\":\"https://www.asiabookkeeping.com/23-3-break-even-point-calculation/\",\"WARC-Payload-Digest\":\"sha1:3YTJYSCJUPUFT7QRS2NXLUBBS2I67CJ6\",\"WARC-Block-Digest\":\"sha1:ZVRNBKSGJW7SO5Z4KKZSC3HU22MYO5MS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363445.41_warc_CC-MAIN-20211208053135-20211208083135-00603.warc.gz\"}"}
https://voiceofwave.com/what-is-the-best-wave-period	[ "# What is the best wave period?\n\n1", null, "Date created: Tue, Sep 7, 2021 8:13 AM\nDate updated: Fri, May 27, 2022 10:30 AM\n\nContent\n\n## Top best answers to the question «What is the best wave period»\n\n• 1-5 seconds: Local wind swells with bumpy and disordered waves…\n• 6-8 seconds: Regional and local wind swells with average surfing conditions…\n• 8-10 seconds: Medium-distance swells improve the local surfing conditions.\n\nFAQ\n\nThose who are looking for an answer to the question «What is the best wave period?» often ask the following questions:\n\n### 👋 What is period wave?\n\nThe period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years.\n\n### 👋 Which is the best definition of wave period?\n\n• Wave period Wave period: the time it takes to finish one wave cycle Crests/Peaks: the highest points of a wave Troughs: the lowest points of a wave Wavelength: the measurement in meters from one peak to the next peak of a wave Wave cycles: one completion of a wave's repeating up-and-down pattern\n\n### 👋 What does wave period mean?\n\nWave period is the distance between two waves passing through a stationary point, measured in seconds. When it comes reading forecast graphs, swell period is definitely the magic number… The shorter the period, the weaker and slower the swell, and the closer to the surface it travels.\n\nWe've handpicked 25 related questions for you, similar to «What is the best wave period?» so you can surely find the answer!\n\nWhat does wave period mean in science?\n\n#### What is an example of a wave period?\n\n• Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. Regardless of the source of the wave, the relationship between waveform frequency and the period is the same. A wave period is the time in seconds between two wave peaks and is inversely proportional to frequency.\nWhat happens when a wave period decreases?\n• We see that frequency is inversely proportional to the period: therefore, if the period decreases, the frequency of the wave increases. What happens when the frequency of a wave decreases? As the frequency decreases, the wavelength gets longer. There are two basic types of waves: mechanical and electromagnetic.\nWhat is measured by a wave period?\n• The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.\nWhat is period in a wave function?\n\nThe period of a wave is the time it takes between two points of the wave with the same amplitude.\n\nWhat is period in characteristics of wave?\n• Wave Period = time for one full wavelength to pass a given point (s) These characteristics are important in determining the size of waves, the speed at which they travel, how they break on shore, and much more. We will refer back to them throughout the following unit. Wave Development and Movement. Waves typically propagate from the centre of a storm.\nWhat is period of a sine wave?\n\nThe period of the sine curve is the length of one cycle of the curve. The natural period of the sine curve is . So, a coefficient of b=1 is equivalent to a period of 2π.\n\nWhat is the definition of period wave?\n• The period of the wave is the time between wave crests. The period is measured in time units such as seconds. The period is usually represented by the upper case \"T.\". The period and frequency are closely related to each other. The period equals 1 over the frequency and the frequency is equal to one over the period.\nWhat is the period of wave equation?\n• The wave period is the time for two consecutive crests to pass a fixed point. The wave speed, C, can be calculated by dividing the wavelength by the wave period (C=L/T) since a wave travels one wave length each wave period.\nWhat is the wave period and frequency?\n• The wave period is the time taken by the medium's particle to complete one full vibrational cycle. Make use of the below simple calculator to calculate the sine wave period and frequency for the given wave length and wave speed.\nWhat is time period of sound wave?\n\nf is the number of waves produced by a source per second, it is measured in hertz (Hz). T is the time it takes for one complete oscillation , it is measured in seconds… A sound wave has a time period of 0.0001 seconds.\n\nWhat is wavelength divided by wave period?\n• The wave speed, C, can be calculated by dividing the wavelength by the wave period (C=L/T) since a wave travels one wave length each wave period. Waves are classified as deep water waves when the water depth is greater than half the wavelength and, for deep water waves, speed is determined by the wavelength.\nWhich is the best description of the period of a wave?\n• In general, it is the distance from the equilibrium midpoint of the wave to its maximum displacement, or it is half the total displacement of the wave. period (T) - is the time for one wave cycle (two pulses, or from crest to crest or trough to trough), in SI units of seconds (though it may be referred to as \"seconds per cycle\").\nA period of a wave?\n\nThe period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.\n\nHow to find wave period?\n\n#### What is the equation for the period of a wave?\n\n• Period of wave is the time it takes the wave to go through one complete cycle, = 1/f, where f is the wave frequency. Wavelength Frequency formula: λ = v/f. where: λ: Wave length, in meter.\nWhat is the difference between wave speed and wave period?\n\n#### What is the relation between wavelength and period of a wave?\n\n• Since speed is distance traveled / time spent, and a wave moves a distance of one wavelength in a time of one period, its speed must be: speed = distance / time = wavelength / period v = λ / p We more often use frequency = 1 / period.\nWhat does the period of a wave determine?\n• The period of a wave is the time for a particle on a medium to make one complete vibrational cycle . Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.\nWhat does wave period mean in science terms?\n\n#### What is the formula to calculate the period of a wave?\n\n• The period of the wave depends on how fast it's moving and on its wavelength ( λ ). The wave moves a distance of one wavelength in a time of one period, so the wave speed formula is v = λ / T , where v is the velocity. Reorganizing to express period in terms of the other quantities, you get:\nWhat is a good wave period for boating?\n\nFor boaters, long wave periods (e.g., 12 seconds) are better for sailing because it typically means most of the waves will be swells which means a smoother ride for small boats.\n\nWhat is an example of a wave period?\n• Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. Regardless of the source of the wave, the relationship between waveform frequency and the period is the same. A wave period is the time in seconds between two wave peaks and is inversely proportional to frequency.\nWhat is the dominant period of a wave?\n• Dominant Wave Period. The dominant wave period (in seconds) is a wave period associated with highest energetic waves at a specific point or area in the total wave spectrum and is always either the swell period or the wind- wave period. Dominant wave period is also known as the “peak” period.\nWhat is the period of a light wave?\n\nThe wave period is the time it takes to complete one cycle. The standard unit of a wave period is in seconds, and it is inversely proportional to the frequency of a wave, which is the number of cycles of waves that occur in one second. In other words, the higher the frequency of a wave, the lower the wave period.\n\nWhat is the period of a longitudinal wave?\n• In a longitudinal wave, the distance from the equilibrium position in the medium to compression or rarefaction is the amplitude. The time taken by the wave to move one wavelength is known as the period.\nWhat is the period of a sinusoidal wave?\n\nThe period of a sinusoid is the length of a complete cycle. For basic sine and cosine functions, the period is . This length can be measured in multiple ways… In the image above, the top red line would represent a regular cosine wave. The center red line would represent a regular sine wave with a horizontal shift.\n\nWhat is the period of a sound wave?\n\nA sound wave has a time period of 0.0001 seconds.\n\nWhat is the period of a square wave?\n• x = square(t) generates a square wave with period 2 π for the elements of the time array t. square is similar to the sine function but creates a square wave with values of –1 and 1." ]	[ null, "https://voiceofwave.com/storage/qr/a1e7c4eb7ce611d848cc7a629f510caf.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.934207,"math_prob":0.9902812,"size":8603,"snap":"2022-05-2022-21","text_gpt3_token_len":1891,"char_repetition_ratio":0.21863008,"word_repetition_ratio":0.29867676,"special_character_ratio":0.22306173,"punctuation_ratio":0.10131654,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99647194,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T10:46:43Z\",\"WARC-Record-ID\":\"<urn:uuid:5ac05875-a1ff-4fad-b445-6ace8fc3232c>\",\"Content-Length\":\"99596\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f20ec7d3-a9d9-4e18-93c1-5ac8a02ae831>\",\"WARC-Concurrent-To\":\"<urn:uuid:d677aa46-5a8a-49a4-98dd-f3b469d5724f>\",\"WARC-IP-Address\":\"172.67.189.86\",\"WARC-Target-URI\":\"https://voiceofwave.com/what-is-the-best-wave-period\",\"WARC-Payload-Digest\":\"sha1:QVAETORIWRAN6HCWEZZ2YKZ7UWSDO6WC\",\"WARC-Block-Digest\":\"sha1:I5VWBBPDLCOM37R3QN53MHIXLO5GNIZT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016373.86_warc_CC-MAIN-20220528093113-20220528123113-00620.warc.gz\"}"}
https://www.physicsforums.com/threads/remainder-theory-proof.6010/	[ "# Remainder theory proof\n\nIf p^2 is exactly divisible by p+q, then proof q^2 is exactly divisible by p+q.\n\nHow do I proof this, and how do I apply the remainder theorem?\n\nI know if f(x) = x^2 + 2x + 1, since f(-1) = 0 there fore (x + 1) is a factor of f(x).\n\nSo in this case\np^2 = p x p or p^2 x 1...\n\nHallsofIvy\nHomework Helper\nSince you specifically mention x2+ 2x+ 1, haven't you looked at (p+q)<sup>2</sup>= p<sup>2</sup>+ 2pq+ q<sup>2</sup>?\n\nThis is what I've done afterwards, but I'm not sure if it is right.\n\nLet p^2 = x^2 + 2xy + y^2.\n\nwhere y^2 = y.y or -y.-y or x^2 = x.x or -x.-x.\n\nWhen y = -x, or x = -y, p^2 = 0\n\nTherefore (x + y) is a factor of p^2, also (x + y) = (p + q)\n\nTherefore p^2 = p^2 + 2pq + p^2. p^2 = 0 when p = -q,\n\nbut (p)^2 = (-q)^2\np^2 = q^2\ntherefore (p+q)mod q^2 = 0\n\nAre there any flaw in my argument or have I made any calcuation error?\n\nSorry, it's q^2mod(p+q) = 0\n\nMathematicalPhysicist\nGold Member\nOriginally posted by Hyperreality\nThis is what I've done afterwards, but I'm not sure if it is right.\n\nLet p^2 = x^2 + 2xy + y^2.\n\nwhere y^2 = y.y or -y.-y or x^2 = x.x or -x.-x.\n\nWhen y = -x, or x = -y, p^2 = 0\n\nTherefore (x + y) is a factor of p^2, also (x + y) = (p + q)\n\nTherefore p^2 = p^2 + 2pq + p^2. p^2 = 0 when p = -q,\n\nbut (p)^2 = (-q)^2\np^2 = q^2\ntherefore (p+q)mod q^2 = 0\n\nAre there any flaw in my argument or have I made any calcuation error?\np^2=P^2+2pq+q^2\n-2pq=q^2\n-2p=q\np=-q/2\nnot p=-q" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88248837,"math_prob":0.99997234,"size":480,"snap":"2021-21-2021-25","text_gpt3_token_len":188,"char_repetition_ratio":0.10504202,"word_repetition_ratio":0.779661,"special_character_ratio":0.43125,"punctuation_ratio":0.11594203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000093,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-17T21:27:07Z\",\"WARC-Record-ID\":\"<urn:uuid:8e200627-8508-4c77-a2fe-16ae65aa5720>\",\"Content-Length\":\"69143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d8754eee-c7af-4a2f-bede-4dc6ad3f1de5>\",\"WARC-Concurrent-To\":\"<urn:uuid:699f5122-dd3f-4758-b0e0-4da32b611de7>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/remainder-theory-proof.6010/\",\"WARC-Payload-Digest\":\"sha1:B3JSH3BU3FBCBNRPRIDMNWM23KFMJIJ4\",\"WARC-Block-Digest\":\"sha1:FDUEAGUPATKGUH33QCW7UWJXKGKLPOV5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991870.70_warc_CC-MAIN-20210517211550-20210518001550-00092.warc.gz\"}"}
https://mathhelpboards.com/threads/root-calculations.7447/	[ "# [SOLVED]Root calculations\n\n#### wishmaster\n\n##### Active member\nIm just wondering what is the easiest way to deal with calculations where roots are involved?\nFor example how do you solve this one?\n\n$$\\displaystyle \\frac{\\frac{1}{2}}{1-\\frac{\\sqrt{2}}{2}}$$\n\nThank you for replies!\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nIm just wondering what is the easiest way to deal with calculations where roots are involved?\nFor example how do you solve this one?\n\n$$\\displaystyle \\frac{\\frac{1}{2}}{1-\\frac{\\sqrt{2}}{2}}$$\n\nThank you for replies!\nA rule of thumb for fractions is to get a common denominator and to always attempt to rationalise denominators.\n\nWhat have you tried so far?\n\n#### wishmaster\n\n##### Active member\nA rule of thumb for fractions is to get a common denominator and to always attempt to rationalise denominators.\n\nWhat have you tried so far?\nTo multiply the fraction with 2.\n\n#### MarkFL\n\nStaff member\nTo multiply the fraction with 2.\nI assume you mean to multiply by:\n\n$$\\displaystyle 1=\\frac{2}{2}$$\n\nThis is a good first step.", null, "What did you get in doing so?\n\n#### wishmaster\n\n##### Active member\nI assume you mean to multiply by:\n\n$$\\displaystyle 1=\\frac{2}{2}$$\n\nThis is a good first step.", null, "What did you get in doing so?\nYes,Mark,i did it so as you said. Actualy i understand this,but i am wondering if there some other way to deal with this.....\n\nAnd i got:\n\n$$\\displaystyle \\frac{1}{2-\\sqrt{2}}$$\n\n#### MarkFL\n\nStaff member\nYes,Mark,i did it so as you said. Actualy i understand this,but i am wondering if there some other way to deal with this.....\n\nAnd i got:\n\n$$\\displaystyle \\frac{1}{2-\\sqrt{2}}$$\nOkay, now you want to rationalize the denominator. Think of the difference of squares formula...\n\n#### wishmaster\n\n##### Active member\nOkay, now you want to rationalize the denominator. Think of the difference of squares formula...\nI multiply fraction with $$\\displaystyle (2+\\sqrt{2})$$\n\nSo i got:\n\n$$\\displaystyle \\frac{2+\\sqrt{2}}{2}$$\n\n#### MarkFL\n\nStaff member\nI multiply fraction with $$\\displaystyle (2+\\sqrt{2})$$\n\nSo i got:\n\n$$\\displaystyle \\frac{2+\\sqrt{2}}{2}$$\nGood! You could choose to leave it like that, or express it as:\n\n$$\\displaystyle 1+\\frac{\\sqrt{2}}{2}$$\n\n#### wishmaster\n\n##### Active member\nGood! You could choose to leave it like that, or express it as:\n\n$$\\displaystyle 1+\\frac{\\sqrt{2}}{2}$$\n\nthank you!" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7864414,"math_prob":0.990503,"size":409,"snap":"2021-43-2021-49","text_gpt3_token_len":128,"char_repetition_ratio":0.11851852,"word_repetition_ratio":0.3508772,"special_character_ratio":0.34718826,"punctuation_ratio":0.078947365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990921,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T09:58:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ffe211b5-48c4-46de-8aab-621366ce57bf>\",\"Content-Length\":\"95681\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1dbf8152-1b13-44b7-8605-d617708af936>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5cecf2d-161d-4296-b7d3-ed56d1f7971e>\",\"WARC-IP-Address\":\"50.31.99.218\",\"WARC-Target-URI\":\"https://mathhelpboards.com/threads/root-calculations.7447/\",\"WARC-Payload-Digest\":\"sha1:C3BPCU6AVGJBF5J6DPRVR2OFG3WPCSEJ\",\"WARC-Block-Digest\":\"sha1:6LWB5ATOZQLU47WYAVW5OLCJFTYYNDFC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585916.29_warc_CC-MAIN-20211024081003-20211024111003-00122.warc.gz\"}"}
https://www.slideserve.com/semah/chapter-4-fluid-kinematic	[ "# Chapter 4 – Fluid Kinematic - PowerPoint PPT Presentation", null, "Download Presentation", null, "Chapter 4 – Fluid Kinematic\n\nChapter 4 – Fluid Kinematic", null, "Download Presentation", null, "## Chapter 4 – Fluid Kinematic\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Chapter4 – Fluid Kinematic • Concept • Position • Rate of motion\n\n2. Chapter4 – Fluid Kinematic • Lagrangian / Eulerian (chap. 4.1)\n\n3. Chapter4 – Fluid Kinematic • Eulerian description • Velocity • Accelerator\n\n4. Chapter4 – Fluid Kinematic • Eulerian description • Gradient operator • Material Derivative\n\n5. Chapter4 – Fluid Kinematic • Deformation of control volume (chap. 4.4 p 139) • Four examples • Linear strain rate (normal force) • Volume strain rate\n\n6. Chapter4 – Fluid Kinematic • Deformation of control volume • Four examples • Shear strain (shear force) • Strain rate tensor\n\n7. Chapter4 – Fluid Kinematic • Deformation of control volume • Vorticity vector • Rotationality\n\n8. Chapter4 – Fluid Kinematic • Deformation of control volume • Rotationality Rotational or Irrotational ?\n\n9. Chapter4 – Fluid Kinematic • Motion of control volume (4.5) • Reynolds control theorem • Total volume • Alternative RTT • Small element volume Properties ‘b’ small volume ‘B’ total volume\n\n10. Chapter4 – Fluid Kinematic • Motion of control volume • RTT with steady flow • Total volume not change • Properties ‘b’ is constant\n\n11. Chapter4 – Fluid Kinematic • Flow visualization • Problems: • Non visualization • Can’t measurement/sensor • Validation of modelling\n\n12. Chapter4 – Fluid Kinematic • Flow visualization • Steamline and Steamtube • Pathline Tracer particle\n\n13. Chapter4 – Fluid Kinematic • Flow visualization • Pathline Particle Image Vilocimetry (PIV) Andrian, 1991\n\n14. Chapter4 – Fluid Kinematic • Flow visualization • Streakline Andrian, 1988\n\n15. Chapter4 – Fluid Kinematic • Flow visualization • Timeline Bippes, H. 1972\n\n16. Chapter4 – Fluid Kinematic • Flow visualization • Shadowgraph (schlieren technique) Settles 2001 Shadowgraph Steakline\n\n17. Chapter4 – Fluid Kinematic • Flow visualization • Profile plots Vector Contour" ]	[ null, "https://www.slideserve.com/img/replay.png", null, "https://thumbs.slideserve.com/1_1768040.jpg", null, "https://www.slideserve.com/photo/32367.jpeg", null, "https://www.slideserve.com/img/output_cBjjdt.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.550276,"math_prob":0.80479217,"size":1775,"snap":"2021-21-2021-25","text_gpt3_token_len":419,"char_repetition_ratio":0.26877472,"word_repetition_ratio":0.26102942,"special_character_ratio":0.23549296,"punctuation_ratio":0.049327355,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9845438,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-10T03:03:51Z\",\"WARC-Record-ID\":\"<urn:uuid:5e2b7719-a7a4-4b75-9a5a-0cb1e76ecc3c>\",\"Content-Length\":\"87974\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9e8f292-8272-476c-b575-474e7469d50d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9b24104-db9a-4a91-9a21-31b0f777678b>\",\"WARC-IP-Address\":\"52.25.140.216\",\"WARC-Target-URI\":\"https://www.slideserve.com/semah/chapter-4-fluid-kinematic\",\"WARC-Payload-Digest\":\"sha1:MKMT5TBBI3LP2BLGHK6OKKHUTV7D2X6O\",\"WARC-Block-Digest\":\"sha1:DF2AGRQTYRC7M3UVG76G66APZET43TTH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989030.65_warc_CC-MAIN-20210510003422-20210510033422-00421.warc.gz\"}"}
https://community.rstudio.com/t/help-with-code-how-to-find-the-length-of-gaps-with-missing-values-in-a-matrix/165345	[ "# Help with code. How to find the length of gaps with missing values in a matrix\n\nThis is the code and is not working as intended, the code is suposed to tell me if there is any adjacente NA values in a given column and how many. This is done by checking if a current value is NA and if the previous is NA as well, if this is the case, the will be a counter that will be registering this and saving the count in a new matrix, so at the end i will only search for the biggest value in every column and i will know how long is the longest chain of NA values.\n\n#datos is the name of a matrix from excel\n\n# this part is to obtain the dimensions of the matrix\n\nfilas <- nrow(datos)\ncolumnas <- ncol(datos)\n\n#Creating a matrix to save the results\n\n# counter at 0\n\n#going for every column\nfor (i in 1:columnas) {\n\n# going down at the current column\n\nfor (j in 1:filas) {\n# verifying if the current value is NA\nif (is.na(datos[j,i])) {\n# verifying if the preious value is NA\nif (j > 1 && is.na(datos[j-1,i])) {\n# if the previous value is NA, +1 to the counter\n} else {\n# if is not, counter restarts at 1\n}\n} else {\n# if the value is not NA, restart the counter at cero\n}\n# savinf every value to the results matrix\n}\n}\n\n# show the matrix with the results\n\nThe problem is that you do not store the maximum value of `contador` as you move down the column. Imagine that the column starts with 5 consecutive NA values but then there are no more NAs. At the bottom of the column, `contador` will have a value of zero and that will be stored in `resultado`." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70718056,"math_prob":0.94263494,"size":1410,"snap":"2023-14-2023-23","text_gpt3_token_len":408,"char_repetition_ratio":0.15576102,"word_repetition_ratio":0.0074626864,"special_character_ratio":0.29219857,"punctuation_ratio":0.089108914,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9689856,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T18:23:05Z\",\"WARC-Record-ID\":\"<urn:uuid:fe8357f1-d916-4c7d-9377-e84a234fcc67>\",\"Content-Length\":\"24696\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24666de7-fd81-41d9-a7a3-7309ee7888f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:213727eb-e230-4784-aa82-eda67ee997cb>\",\"WARC-IP-Address\":\"167.99.20.217\",\"WARC-Target-URI\":\"https://community.rstudio.com/t/help-with-code-how-to-find-the-length-of-gaps-with-missing-values-in-a-matrix/165345\",\"WARC-Payload-Digest\":\"sha1:IW25XXMYERULB6SARJNNVLLJN3EVTEDN\",\"WARC-Block-Digest\":\"sha1:AJPM32UAIKUWA2ZFZUP5WVN36XBBUOWP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654012.67_warc_CC-MAIN-20230607175304-20230607205304-00224.warc.gz\"}"}
https://ifyoufeedme.com/question/4633/	[ "# Which of the following graphs could represent a quartic function?\n\nStudents were asked to answer a question at institution and to tell what is most important for them to succeed. One which response stood out from the rest was practice. People who generally are successful do not become successful by being born. They work hard and dedication their lives to succeeding. This is how you can accomplish your goals. followed below are one of the answer and question example that you could simply utilize to practice and improve your information and also give you insights that could just guide you to preserve your study in school.\n\n## Question:\n\nWhich of the following graphs could represent a quartic function?", null, "D. Graph D\n\nStep-by-step explanation:\n\nAs we know,\n\nQuartic function is a function of the form", null, ", where a≠0.\n\nThat is, a quartic function is a polynomial of degree 4 called a quartic polynomial.\n\nBy the ‘Fundamental Theorem of algebra’, we have, ‘An n-degree polynomial will have ‘n’ number of real roots’.\n\nThus, we get,\n\nA quartic polynomial will have 4 real roots and so, a quartic function has 4 zeros.\n\nSince, a quartic function has 4 zeros. Its graph will cut the y-axis at 4 points.\n\nMoreover, when a graph touches the axis, it is known to be a repeated zero.\n\nSo, from the options, we get,\n\nOption D represents a quartic function as one of its root is repeated and it cuts the the graph 4 times.\n\nFrom the answer and question examples above, hopefully, they can definitely guide the student resolve the question they had been looking for and keep in mind of each and every thing declared in the answer above. You may then have a discussion with your classmate and continue the school learning by studying the problem alongside one another.\n\nREAD MORE Which story premise most clearly contains a supernatural element?\nKata Kunciwhich of the following graphs could represent a quartic function brainly -" ]	[ null, "https://us-static.z-dn.net/files/df9/3b0f6b8ccbc309f1d9ad4d56fe4c67b7.jpg", null, "https://tex.z-dn.net/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9686626,"math_prob":0.9062475,"size":1761,"snap":"2022-40-2023-06","text_gpt3_token_len":375,"char_repetition_ratio":0.13488902,"word_repetition_ratio":0.0397351,"special_character_ratio":0.20499717,"punctuation_ratio":0.109195404,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96060467,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T22:39:49Z\",\"WARC-Record-ID\":\"<urn:uuid:ea57a1b9-9840-411b-a557-5e70f58f0e57>\",\"Content-Length\":\"49177\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:293c8ae0-6ff6-48d7-b6fe-ab1fca1141fb>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0e818e6-6fe2-473f-ba87-3aa8e6a6ef20>\",\"WARC-IP-Address\":\"172.67.165.182\",\"WARC-Target-URI\":\"https://ifyoufeedme.com/question/4633/\",\"WARC-Payload-Digest\":\"sha1:HPITWBY43TCBV3QH5DTJOUQI33RJEHAL\",\"WARC-Block-Digest\":\"sha1:7IFD6S4QE3X4LECT2UAJDIDEXSAWIJIF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500365.52_warc_CC-MAIN-20230206212647-20230207002647-00808.warc.gz\"}"}
http://ros.fei.edu.br/roswiki/eigen(2f)Cookbook.html	[ "", null, "[Documentation] [TitleIndex] [WordIndex]\n\n## Cookbook/Common issues\n\nEigen mostly achieves a pleasantly readable mathematical syntax, but quirks of the C++ language sometimes show through. This section lists some common gotchas you may encounter when using Eigen.\n\n### Structures containing Eigen types as members\n\nSuppose you are using a fixed-size vectorizable Eigen type in one of your own structures, which you allocate dynamically:\n\n``` 1 class Foo\n2 {\n3 double x;\n4 Eigen::Vector2d v;\n5 }\n6\n7 Foo foo; // Fine\n8 Foo foo_ptr = new Foo; // PROBLEM!\n9\n```\n\nEigen ensures that Eigen::Vector2d v is 128-bit aligned with respect to the start of Foo. Stack-allocating an instance of Foo will also respect the alignment. The problem comes when dynamically allocating an instance of Foo; the default operator new is not required to allocate a 128-bit aligned block of data, so member foo_ptr->v may be misaligned.\n\nTo address this, Eigen provides a macro which overloads Foo's operator new to Do The Right Thing:\n\n``` 1 class Foo\n2 {\n3 double x;\n4 Eigen::Vector2d v;\n5 public:\n6 EIGEN_MAKE_ALIGNED_OPERATOR_NEW\n7 }\n8\n9 Foo foo; // Still fine\n10 Foo foo_ptr = new Foo; // Now OK!\n11\n```\n\nFull explanation: http://eigen.tuxfamily.org/dox/StructHavingEigenMembers.html\n\n### Using Eigen types with STL containers\n\nWhen using fixed-size vectorizable Eigen types in STL containers, you must use an aligned allocator. Eigen provides one:\n\n``` 1 USING_PART_OF_NAMESPACE_EIGEN\n2\n4 std::map<int, Vector4f> m; // WRONG\n5 // Use\n6 std::map<int, Vector4f, std::less<int>, Eigen::aligned_allocator<Vector4f> > m;\n```\n\nUnfortunately the situation with std::vector is complicated by a defect in the current C++ language definition; Eigen provides a header specifically to make std::vector work with aligned types:\n\n``` 1 #define EIGEN_USE_NEW_STDVECTOR\n2 #include <Eigen/StdVector>\n3\n4 std::vector<Eigen::Vector4f, Eigen::aligned_allocator<Eigen::Vector4f> > v;\n```\n\nThe macro EIGEN_USE_NEW_STDVECTOR will avoid some problems with the original version of Eigen/StdVector and make your code forward-compatible with future versions of Eigen.\n\nFull explanation: http://eigen.tuxfamily.org/dox/StlContainers.html\n\n### Syntax for calling member templates\n\nEigen's matrix types are class templates, allowing you to configure them on element type, storage order (e.g. row-major or col-major), or specify fixed sizes for one or both dimensions. Some member functions of these matrix types have template parameters of their own which the user must specify explicitly; these are member templates.\n\n``` 1 template<typename T>\n2 class Foo\n3 {\n4 void bar(); // Normal member function\n5 template<int N> void baz(); // Member template\n6 };\n```\n\nFor example, Eigen's MatrixBase (inherited by Matrix) has a useful member template block, which returns a read-write view of a sub-block of a matrix:\n\n``` 1 template<int BlockRows, int BlockCols>\n2 ReturnType block(int startRow, int startCol);\n```\n\nIf we know the dimensions of the block at compile time, specifying them as template parameters is a nice win; it avoids unnecessary dynamic memory allocations, and may allow Eigen to perform optimizations such as vectorization and loop unrolling. Unfortunately, calling a member template can be tricky:\n\n``` 1 // This works:\n2 void transform(Eigen::Matrix<double,3,4>& m,\n3 const Eigen::Matrix<double,3,3> trans,\n4 const Eigen::Quaternion<double> qrot)\n5 {\n6 m.block<3,3>(0,0) = qrot.toRotationMatrix().transpose();\n7 m.block<3,1>(0,3) = -m.block<3,3>(0,0) * trans;\n8 }\n9\n10 // But when we template on the element type, this fails to compile!\n11 template<typename T>\n12 void transform(Eigen::Matrix<T,3,4>& m,\n13 const Eigen::Matrix<T,3,3> trans,\n14 const Eigen::Quaternion<T> qrot)\n15 {\n16 m.block<3,3>(0,0) = qrot.toRotationMatrix().transpose();\n17 m.block<3,1>(0,3) = -m.block<3,3>(0,0) * trans;\n18 }\n```\n\nIn the second (templated) case, not only will it (correctly) fail to compile, gcc will misparse the expression so thoroughly that you get error messages along the lines of \"warning: left-hand operand of comma has no effect\", \"error: lvalue required as left operand of assignment\", \"error: invalid operands of types '<unresolved overloaded function type>' and 'int' to binary 'operator<'\".\n\nWhat happened? In the second case, the matrix types are dependent on the template parameter T. Dependent types require us to give the compiler a little extra help. We need to preface the name of the member template with the keyword template:\n\n``` 1 // Correct!\n2 template<typename T>\n3 void transform(Eigen::Matrix<T,3,4>& m,\n4 const Eigen::Matrix<T,3,3> trans,\n5 const Eigen::Quaternion<T> qrot)\n6 {\n7 m.template block<3,3>(0,0) = qrot.toRotationMatrix().transpose();\n8 m.template block<3,1>(0,3) = -m.template block<3,3>(0,0) * trans;\n9 }\n```\n\nUsing other Eigen functions, below is another (perhaps cleaner) way to write the function body. Note that col is not a member template.\n\n``` 1 m.template corner<3,3>(Eigen::TopLeft) = qrot.toRotationMatrix().transpose();\n2 m.col(3) = -m.template corner<3,3>(Eigen::TopLeft) * trans;\n```\n\nMember functions of Eigen::MatrixBase which are member templates include block, cast, corner, end, lpNorm, part, segment, and start. Many of these functions have overloads which are not member templates; those replace the template arguments with regular function arguments, which is more flexible but introduces runtime overhead.\n\n### Creating typedefs for Eigen types\n\nCreating a typedef for a fully-defined Eigen type is easy:\n\n``` 1 typedef Eigen::Matrix<float,3,1> Point;\n2\n3 Point pt;\n```\n\nBut what if we want to parameterize our Point type over floats and doubles? Then we want something like:\n\n``` 1 // NOT LEGAL C++!\n2 template<typename T>\n3 typedef Eigen::Matrix<T,3,1> Point;\n4\n5 Point<float> pt;\n```\n\nUnfortunately C++ does not currently allow template typedefs (they will be in C++0x with different syntax). The standard workaround is to make Point a \"meta-function\" that returns the desired type through its member typedef type:\n\n``` 1 template<typename T>\n2 struct Point\n3 {\n4 typedef Eigen::Matrix<T,3,1> type;\n5 };\n6\n7 Point<float>::type pt;\n```\n\nThis approach is still fairly simple, but introduces a bit of syntactic noise since the user has to remember to add ::type. Another approach is to use inheritance:\n\n``` 1 // WARNING: Has some unexpected behavior, see below\n2 template<typename T>\n3 class Point : public Eigen::Matrix<T,3,1> {};\n4\n5 Point<float> pt;\n```\n\nThis looks like what we want; unfortunately the derived class Point does not inherit the constructors or assignment operators of the base Eigen type, so Point will not inter-operate nicely with regular Eigen matrices. Here is a more complex definition of Point that fixes those issues:\n\n``` 1 template<typename T>\n2 class Point : public Eigen::Matrix<T,3,1>\n3 {\n4 typedef Eigen::Matrix<T,3,1> BaseClass;\n5\n6 public:\n7 // 3-element constructor, delegates to base class constructor\n8 Point(const T& x, const T& y, const T& z)\n9 : BaseClass(x, y, z)\n10 {}\n11\n12 // Copy constructor from any Eigen matrix type\n13 template<typename OtherDerived>\n14 Point(const Eigen::MatrixBase<OtherDerived>& other)\n15 : BaseClass(other)\n16 {}\n17\n18 // Reuse assignment operators from base class\n19 using BaseClass::operator=;\n20 };\n21\n22 // Now all of these operations work\n23 Point<float> pt(1.0, 2.5, -3.1);\n24 Eigen::Vector3f v = pt;\n25 pt = v;\n26 Point<float> pt2(v);\n```\n\n2019-12-14 12:35" ]	[ null, "http://ros.fei.edu.br/roswiki/logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7540907,"math_prob":0.9085143,"size":5564,"snap":"2019-51-2020-05","text_gpt3_token_len":1270,"char_repetition_ratio":0.12356115,"word_repetition_ratio":0.0024183798,"special_character_ratio":0.23759885,"punctuation_ratio":0.16411181,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9627676,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T19:49:09Z\",\"WARC-Record-ID\":\"<urn:uuid:c35044bd-c4ef-4882-996a-b20b8659e39b>\",\"Content-Length\":\"66210\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:47297db9-9a2f-43ff-8944-6959e2644ef0>\",\"WARC-Concurrent-To\":\"<urn:uuid:551356e7-7f9c-4415-a7f6-fd700010fe9d>\",\"WARC-IP-Address\":\"200.232.90.144\",\"WARC-Target-URI\":\"http://ros.fei.edu.br/roswiki/eigen(2f)Cookbook.html\",\"WARC-Payload-Digest\":\"sha1:TMICPRHCFPM44SHQG47K4D2RQEMXSW5P\",\"WARC-Block-Digest\":\"sha1:QPUI3ON4QO27VBTL7AXJ3CPVCTXCQGDA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541309137.92_warc_CC-MAIN-20191215173718-20191215201718-00066.warc.gz\"}"}
https://gonitsora.com/experience-of-an-m-math-interview-at-indian-statistical-institute/?shared=email&msg=fail	[ "", null, "## 25 Jul Experience of an M.Math Interview at Indian Statistical Institute\n\nGetting into ISI was a dream. I appeared for ISI B.Math exam twice but couldn’t make it and appeared for M.Math in 2019, I attempted 18 questions in the objective part and fully solved 5 questions from the subjective part in the written exam held on 5th May 2019. My name was shortlisted for interview that was scheduled to be held on 20th may at 3:30 pm.\n\nOn the day of interview, I reached ISI at 1 pm, had my lunch there, took a walk around the Campus and walked into the Stat-Math unit at 2:30pm. The person sitting in the office asked me to produce my original documents for verification and wait in the lobby. My turn came at 3:15 as I was asked to enter room S18 where interviews were taking place.\n\nThere were 5 professors in the interview panel. Later, (after checking ISI Bangalore website), I came to know their names: Ramesh Sreekantan (P1), Parthanil Roy (P2), Yogeswaran D (P3), Manish Kumar (P4) and Mathew Joseph (P5).\n\nBelow is a transcript of the interview from my memory.\n\nP2: Hi!\n\nI: Hi Sir.\n\nI: I am Pijush Pratim Sarmah and I am completing my Bachelor of Science this year from Tezpur University.\n\nP2: Which college?\n\nI: There are no colleges under Tezpur University. My program is 5-year Integrated M.Sc in Mathematics.\n\nP5: You are allowed to leave after 3 years?\n\nI: Yes Sir.\n\nP2: Fine, other than ISI what entrances have you appeared for?\n\nI: I qualified IIT JAM with AIR 242 and passed the written exam of TIFR but couldn’t pass the interviews.\n\nP1: What are your interests in Mathematics?\n\nI: Real Analysis and Linear Algebra.\n\nP5: Please go to the board and take a piece of chalk.\n\nI do as they command\n\nP5: Suppose A is a", null, "$3\\times 2$ real matrix and B is a", null, "$2\\times 3$ real matrix. What will be the determinant of the product matrix", null, "$AB$?\n\nI: The determinant will be", null, "$0$.\n\nP5: How?\n\nI: We know that", null, "$rank(AB)\\leq rank(A)$ and", null, "$rank(A)$ is at most", null, "$2$. So", null, "$AB$ is a", null, "$3\\times 3$ matrix whose rank is less than", null, "$3$. Hence, it is not invertible and its determinant is", null, "$0$.\n\nP3: Can you show that", null, "$rank(AB)\\leq \\min\\{rank(A),rank(B)\\}$?\n\n I showed this using the fact that column space of", null, "$AB$ is a subspace of column space of", null, "$A$ and row space of", null, "$AB$ is a subspace of row space of", null, "$B$\n\nP2: Okay, good. If", null, "$A$ is a real symmetric matrix, can we compare the ranks of", null, "$A$ and", null, "$A^2$?\n\nNow, this is a question that I solved just the night before my day of interview, so I showed that", null, "$rank(A)=rank(A^2)$. My confidence grew and I knew that I could do very well from there.\n\nP3: Your solution is correct. Do you know that real symmetric matrices have a beautiful property?\n\nI: Yes, Sir. Real Symmetric matrices are diagonalizable.\n\nP3: Good. Now using the fact that real symmetric matrices are diagonalizable, can you show that", null, "$rank(A)=rank(A^2)$ when", null, "$A$ is a real symmetric matrix?\n\nI: There are many useful properties of diagonalizable matrices. Where should I start?\n\nP5: Just use the definition of diagonalizable matrices.\n\nI started solving the problem and got stuck in a step where I needed to use the fact that if", null, "$D$ is a Diagonal matrix then", null, "$rank(D)=rank(D^2)$, but I hesitated to say that\n\nP2: Yes, you are in the right direction. What can you say about ranks of", null, "$D$ and", null, "$D^2$?\n\nP1: What is the rank of a diagonal matrix?\n\nI: It is precisely the number of non-zero diagonal entries. — (The moment I said that, all my confusion disappeared and I proceeded with my explanation)– Now, since,", null, "$D^2$ is obtained by squaring the diagonal entries of", null, "$D$, rank of", null, "$D$ and", null, "$D^2$ are equal and so are the ranks of", null, "$A$ and", null, "$A^2$.\n\nP2: Correct! Lets move onto Analysis. Suppose", null, "$f:\\mathbb{R} \\rightarrow \\mathbb{R}$ is continuous and", null, "$\\{x_n\\}$ be a Cauchy sequence in R. Will", null, "$\\{f(x_n)\\}$ be a Cauchy sequence?\n\nI: Since", null, "$\\mathbb{R}$ is complete,", null, "$\\{x_n\\}$ converges to some point, say", null, "$x$. By sequential criteria of continuity,", null, "$\\{f(x_n)\\}$ converges to", null, "$f(x)$. Therefore,", null, "$\\{f(x_n)\\}$ is a Cauchy sequence.\n\nP2: Very good. Now, if we restrict the domain of the function to", null, "$(0, +\\infty)$, will the same result hold?\n\nI: No. If we consider the function", null, "$f(x)=1/x$ which is continuous on the given domain and the Cauchy sequence", null, "$\\{x_n\\}=\\{1/n\\}$ in the domain, then", null, "$\\{f(x_n)\\}=n$, which is unbounded and hence not a Cauchy sequence.\n\nP2: Under what condition does a continuous function maps Cauchy sequences to Cauchy sequences?\n\nI: When the function is uniformly continuous.\n\nP5: If a function is continuous on", null, "$\\mathbb{R}$, can you impose some condition on the function which will make the function uniformly continuous?\n\nI: If limit of the function at", null, "$+\\infty$ and", null, "$-\\infty$ exists in", null, "$\\mathbb{R}$.\n\nP1: Okay, then it has to be bounded. Does there exist any real-valued function which is continuous and bounded on", null, "$\\mathbb{R}$ but not uniformly continuous?\n\nI:", null, "$\\sin(x^2)$.\n\nP2: How will you show that?\n\nI: I can give two sequences such that the limit of the sequence obtained by subtracting the corresponding terms of the sequences is zero but the limit of the sequence obtained by subtracting the corresponding terms of the image sequences is non-zero.\n\nP3: Okay. You have the idea. You don’t need to do it further.\n\nAfter this, asked me a few questions about my life at Tezpur University and said that the interview was over. From their encouraging response throughout the interview, which lasted for 25-30 minutes, I was quite sure that I would get in. And I did get in.\n\nI would like to thank Prof. Anjan Kumar Chakrabarty (IIT-Ghy), Kuldeep Sarma (Research Scholar, Tezpur University) and Sourav Koner (M.Sc, Tezpur University) for guiding me in my preparation and making this possible for me.\n\nDownload this post as PDF (will not include images and mathematical symbols).\n\n•", null, "Posted at 07:37h, 27 July Reply\n\nHappy to know about you,can i contact you\n\n•", null, "##### Vikas Rajpoot\nPosted at 19:51h, 23 March Reply\n\nHello\nMy name is Vikas\nI am 3rd year student At IEHE bhopal Pursuing bacgelor of science\nI am Preparing for ISI M.Math\nrecenlty i have Qyalified IIT JAM Mathematics exam\n\n[email protected]\n\n•", null, "##### Manjil Saikia\nPosted at 14:51h, 15 April Reply\n\nAn email was sent to you answering your query, hope it helps.\n\n•", null, "##### Rishabh Jain\nPosted at 15:18h, 02 May Reply\n\nHello\nI am Rishabh\nI am graduated from Ramanujan College , Delhi university in Bsc Maths hons\nI am Preparing for ISI M.Math\nrecenlty i have Qyalified IIT JAM Mathematics exam with AIR 173\n\nI need your guidence for the exam !!!!!\n\n•", null, "##### Manjil Saikia\nPosted at 18:06h, 16 August Reply\n\nHi, which exam do you mean? 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https://science.howstuffworks.com/math-concepts/imaginary-numbers.htm	[ "# What Are Imaginary Numbers?\n\nBy: Patrick J. Kiger \|", null, "Renaissance mathematicians were the first to come up with the idea of imaginary numbers. imagestockdesign/shutterstock\n\nIn Dan Brown's mega-bestselling 2003 mystery thriller \"The Da Vinci Code,\" there's a bit of repartee in the book between the book's hero, Robert Langdon, and cryptographer Sophie Neveu, in which she expresses skepticism about value \"of religious believers living by faiths that include miraculous occurrences. It appears their reality is false,\" she sneers.\n\nLangdon laughs, and says that those beliefs are no more bogus \"than that of a mathematical cryptographer who believes in the imaginary number 'i' because it helps her break codes.\"\n\nFor those of us who aren't mathematically inclined, Langdon's joke was a bit puzzling. What in the heck is he talking about when he says that a number is imaginary? How could that be?\n\nAs it turns out, though, an imaginary number – basically, a number that, when squared, results in a negative number – really is a thing in mathematics, first discovered back in the 1400s and 1500s as a way to solve certain bedeviling equations. While initially thought of as sort of a parlor trick, in the centuries since, they've come to be viewed as a tool for conceptualizing the world in complex ways, and today are useful in fields ranging from electrical engineering to quantum mechanics.\n\n\"We invented imaginary numbers for some of the same reasons that we invented negative numbers,\" explains Cristopher Moore. He's a physicist at the Santa Fe Institute, an independent research institution in New Mexico, and co-author, with Stephan Mertens, of the 2011 book \"The Nature of Computation.\"\n\n\"Start with ordinary arithmetic,\" Moore continues. \"What is two minus seven? If you've never heard of negative numbers, that doesn't make sense. There's no answer. You can't have negative five apples, right? But think of it this way. You could owe me five apples, or five dollars. Once people started doing accounting and bookkeeping, we needed that concept.\" Similarly, today we're all familiar with the idea that if we write big checks to pay for things, but don't have enough money to cover them, we could have a negative balance in our bank accounts.\n\n## Creative Thinking Goes a Long Way\n\nAnother way to look at negative numbers — and this will come in handy later — is to think of walking around in a city neighborhood, Moore says. If you make a wrong turn and in the opposite direction from our destination — say, five blocks south, when you should have gone north — you could think of it as walking five negative blocks to the north.\n\n\"By inventing negative numbers, it expands your mathematical universe, and enables you to talk about things that were difficult before,\" Moore says.\n\nImaginary numbers and complex numbers — that is, numbers that include an imaginary component – are another example of this sort of creative thinking. As Moore explains it: \"If I ask you, what is the square root of nine, that's easy, right? The answer is three – though it also could be negative three,\" since multiplying two negatives results in a positive.\n\nBut what is the square root of negative one? Is there a number, when multiplied by itself, that gives you in negative one? \"At one level, there is no such number,\" Moore says.\n\nBut Renaissance mathematicians came up with a clever way around that problem. \"Before we invented negative numbers there was no such number that was two minus seven,\" Moore continues. \"So maybe we should invent a number that is square root of negative one. Let's give it a name. i.\"\n\nOnce they came up with the concept of an imaginary number, mathematicians discovered that they could do some really cool stuff with it. Remember that multiplying a positive by a negative number equals a negative, but multiplying two negatives by one another equals a positive. But what happens when you start multiplying i times seven, and then times i again? Because i times i is negative one, the answer is negative seven. But if you multiply seven times i times i times i times i, suddenly you get positive seven. \"They cancel each other out,\" Moore notes.\n\nNow think about that. You took an imaginary number, plugged it into an equation multiple times, and ended up with an actual number that you commonly use in the real world.\n\n## Imaginary Numbers Are Points on a Plane\n\nIt wasn't until few hundred years later, in the early 1800s, that mathematicians discovered another way of understanding imaginary numbers, by thinking of them as points on a plane, explains Mark Levi. He's a professor and head of the mathematics department at Penn State University and author of the 2012 book \"Why Cats Land on Their Feet: And 76 Other Physical Paradoxes and Puzzles.\"\n\nWhen we think of numbers as points on a line, and then add a second dimension, \"the points on that plane are the imaginary numbers,\" he says.\n\nEnvision a number line. When you think of a negative number, it’s 180 degrees away from the positive numbers on the line. \"When you multiply two negative numbers, you add their angles, 180 degrees plus 180 degrees, and you get 360 degrees. That's why it's positive,\" Levi explains.", null, "The Y axis is helpful when you're thinking about imaginary numbers since you can't put the square root of -1 on the X axis. zizou7/shutterstock\n\nBut you can't put the square root of negative one anywhere on the X axis. It just doesn't work. However, if you create a Y axis that's perpendicular to the X, you now have a place to put it.\n\nAnd while imaginary numbers seem like just a bunch of mathematical razzle-dazzle, they're actually very useful for certain important calculations in the modern technological world, such as calculating the flow of air over an airplane wing, or figuring out the drain in energy from resistance combined with oscillation in an electrical system. And the fictional Robert Langdon wasn't pulling our legs when he mentioned that they're also used in cryptography.\n\nComplex numbers with imaginary components also are useful in theoretical physics, explains Rolando Somma, a physicist who works in quantum computing algorithms at Los Alamos National Laboratory.\n\n\"Due to their relation with trigonometric functions, they are useful for describing, for example, periodic functions,\" Somma says via email. \"These arise as solutions to the wave equations, so we use complex numbers to describe various waves, such an electromagnetic wave. Thus, as in math, complex calculus in physics is an extremely useful tool for simplifying calculations.\"\n\nComplex numbers also have a role in quantum mechanics, a theory that describes the behavior of nature at the scale of atoms and subatomic particles.\n\n\"In quantum mechanics 'i' appears explicitly in Schrödinger's equation,\" Somma explains. \"Thus, complex numbers appear to have a more fundamental role in quantum mechanics rather than just serving as a useful calculational tool.\"\n\n\"The state of a quantum system is described by its wave function,\" he continues. \"As a solution to Schrodinger's equation, this wave function is a superposition of certain states, and the numbers appearing in the superposition are complex. Interference phenomena in quantum physics, for example, can be easily described using complex numbers.\"" ]	[ null, "https://media.hswstatic.com/eyJidWNrZXQiOiJjb250ZW50Lmhzd3N0YXRpYy5jb20iLCJrZXkiOiJnaWZcL2ltYWdpbmFyeS1udW1iZXJzLnBuZyIsImVkaXRzIjp7InJlc2l6ZSI6eyJ3aWR0aCI6ODI4fX19", null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABAAAAAJCAQAAACRI2S5AAAAEElEQVR42mNkIAAYRxWAAQAG9gAKqv6+AwAAAABJRU5ErkJggg==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9680558,"math_prob":0.9384989,"size":2101,"snap":"2023-14-2023-23","text_gpt3_token_len":452,"char_repetition_ratio":0.093466856,"word_repetition_ratio":0.0,"special_character_ratio":0.20990005,"punctuation_ratio":0.122137405,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9687179,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T22:44:38Z\",\"WARC-Record-ID\":\"<urn:uuid:2ef54889-7c16-4c6a-8e08-bfccc3f68b21>\",\"Content-Length\":\"182704\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80bd6f83-f4ff-4263-8ff3-e1cb0efb9c0e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2c5ef62-67a2-4a8c-a2a9-fdc361e816d4>\",\"WARC-IP-Address\":\"18.160.41.11\",\"WARC-Target-URI\":\"https://science.howstuffworks.com/math-concepts/imaginary-numbers.htm\",\"WARC-Payload-Digest\":\"sha1:USZ6LINOFAA5HLQBB3P66WIIGYAQDHH5\",\"WARC-Block-Digest\":\"sha1:QPHVKDSEMDIHGA2F7IFOTPGV4CUQXGJB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655143.72_warc_CC-MAIN-20230608204017-20230608234017-00274.warc.gz\"}"}
https://doc-snapshots.qt.io/qt5-5.10/qrect.html	[ "QRect Class\n\nThe QRect class defines a rectangle in the plane using integer precision. More...\n\n Header: #include qmake: QT += core\n\nNote: All functions in this class are reentrant.\n\nPublic Functions\n\n QRect() QRect(const QPoint &topLeft, const QPoint &bottomRight) QRect(const QPoint &topLeft, const QSize &size) QRect(int x, int y, int width, int height) void adjust(int dx1, int dy1, int dx2, int dy2) QRect adjusted(int dx1, int dy1, int dx2, int dy2) const int bottom() const QPoint bottomLeft() const QPoint bottomRight() const QPoint center() const bool contains(const QPoint &point, bool proper = false) const bool contains(const QRect &rectangle, bool proper = false) const bool contains(int x, int y) const bool contains(int x, int y, bool proper) const void getCoords(int x1, int y1, int x2, int y2) const void getRect(int x, int y, int width, int height) const int height() const QRect intersected(const QRect &rectangle) const bool intersects(const QRect &rectangle) const bool isEmpty() const bool isNull() const bool isValid() const int left() const QRect marginsAdded(const QMargins &margins) const QRect marginsRemoved(const QMargins &margins) const void moveBottom(int y) void moveBottomLeft(const QPoint &position) void moveBottomRight(const QPoint &position) void moveCenter(const QPoint &position) void moveLeft(int x) void moveRight(int x) void moveTo(int x, int y) void moveTo(const QPoint &position) void moveTop(int y) void moveTopLeft(const QPoint &position) void moveTopRight(const QPoint &position) QRect normalized() const int right() const void setBottom(int y) void setBottomLeft(const QPoint &position) void setBottomRight(const QPoint &position) void setCoords(int x1, int y1, int x2, int y2) void setHeight(int height) void setLeft(int x) void setRect(int x, int y, int width, int height) void setRight(int x) void setSize(const QSize &size) void setTop(int y) void setTopLeft(const QPoint &position) void setTopRight(const QPoint &position) void setWidth(int width) void setX(int x) void setY(int y) QSize size() const CGRect toCGRect() const int top() const QPoint topLeft() const QPoint topRight() const void translate(int dx, int dy) void translate(const QPoint &offset) QRect translated(int dx, int dy) const QRect translated(const QPoint &offset) const QRect transposed() const QRect united(const QRect &rectangle) const int width() const int x() const int y() const QRect operator&(const QRect &rectangle) const QRect & operator&=(const QRect &rectangle) QRect & operator+=(const QMargins &margins) QRect & operator-=(const QMargins &margins) QRect operator\|(const QRect &rectangle) const QRect & operator\|=(const QRect &rectangle)\n bool operator!=(const QRect &r1, const QRect &r2) QRect operator+(const QRect &rectangle, const QMargins &margins) QRect operator+(const QMargins &margins, const QRect &rectangle) QRect operator-(const QRect &lhs, const QMargins &rhs) QDataStream & operator<<(QDataStream &stream, const QRect &rectangle) bool operator==(const QRect &r1, const QRect &r2) QDataStream & operator>>(QDataStream &stream, QRect &rectangle)\n\nDetailed Description\n\nThe QRect class defines a rectangle in the plane using integer precision.\n\nA rectangle is normally expressed as a top-left corner and a size. The size (width and height) of a QRect is always equivalent to the mathematical rectangle that forms the basis for its rendering.\n\nA QRect can be constructed with a set of left, top, width and height integers, or from a QPoint and a QSize. The following code creates two identical rectangles.\n\nQRect r1(100, 200, 11, 16);\nQRect r2(QPoint(100, 200), QSize(11, 16));\n\nThere is a third constructor that creates a QRect using the top-left and bottom-right coordinates, but we recommend that you avoid using it. The rationale is that for historical reasons the values returned by the bottom() and right() functions deviate from the true bottom-right corner of the rectangle.\n\nThe QRect class provides a collection of functions that return the various rectangle coordinates, and enable manipulation of these. QRect also provide functions to move the rectangle relative to the various coordinates. In addition there is a moveTo() function that moves the rectangle, leaving its top left corner at the given coordinates. Alternatively, the translate() function moves the rectangle the given offset relative to the current position, and the translated() function returns a translated copy of this rectangle.\n\nThe size() function returns the rectange's dimensions as a QSize. The dimensions can also be retrieved separately using the width() and height() functions. To manipulate the dimensions use the setSize(), setWidth() or setHeight() functions. Alternatively, the size can be changed by applying either of the functions setting the rectangle coordinates, for example, setBottom() or setRight().\n\nThe contains() function tells whether a given point is inside the rectangle or not, and the intersects() function returns true if this rectangle intersects with a given rectangle. The QRect class also provides the intersected() function which returns the intersection rectangle, and the united() function which returns the rectangle that encloses the given rectangle and this:", null, "", null, "intersected() united()\n\nThe isEmpty() function returns true if left() > right() or top() > bottom(). Note that an empty rectangle is not valid: The isValid() function returns true if left() <= right() and top() <= bottom(). A null rectangle (isNull() == true) on the other hand, has both width and height set to 0.\n\nNote that due to the way QRect and QRectF are defined, an empty QRect is defined in essentially the same way as QRectF.\n\nFinally, QRect objects can be streamed as well as compared.\n\nRendering\n\nWhen using an anti-aliased painter, the boundary line of a QRect will be rendered symmetrically on both sides of the mathematical rectangle's boundary line. But when using an aliased painter (the default) other rules apply.\n\nThen, when rendering with a one pixel wide pen the QRect's boundary line will be rendered to the right and below the mathematical rectangle's boundary line.\n\nWhen rendering with a two pixels wide pen the boundary line will be split in the middle by the mathematical rectangle. This will be the case whenever the pen is set to an even number of pixels, while rendering with a pen with an odd number of pixels, the spare pixel will be rendered to the right and below the mathematical rectangle as in the one pixel case.", null, "", null, "Logical representation One pixel wide pen", null, "", null, "Two pixel wide pen Three pixel wide pen\n\nCoordinates\n\nThe QRect class provides a collection of functions that return the various rectangle coordinates, and enable manipulation of these. QRect also provide functions to move the rectangle relative to the various coordinates.\n\nFor example the left(), setLeft() and moveLeft() functions as an example: left() returns the x-coordinate of the rectangle's left edge, setLeft() sets the left edge of the rectangle to the given x coordinate (it may change the width, but will never change the rectangle's right edge) and moveLeft() moves the entire rectangle horizontally, leaving the rectangle's left edge at the given x coordinate and its size unchanged.", null, "Note that for historical reasons the values returned by the bottom() and right() functions deviate from the true bottom-right corner of the rectangle: The right() function returns left() + width() - 1 and the bottom() function returns top() + height() - 1. The same is the case for the point returned by the bottomRight() convenience function. In addition, the x and y coordinate of the topRight() and bottomLeft() functions, respectively, contain the same deviation from the true right and bottom edges.\n\nWe recommend that you use x() + width() and y() + height() to find the true bottom-right corner, and avoid right() and bottom(). Another solution is to use QRectF: The QRectF class defines a rectangle in the plane using floating point accuracy for coordinates, and the QRectF::right() and QRectF::bottom() functions do return the right and bottom coordinates.\n\nIt is also possible to add offsets to this rectangle's coordinates using the adjust() function, as well as retrieve a new rectangle based on adjustments of the original one using the adjusted() function. If either of the width and height is negative, use the normalized() function to retrieve a rectangle where the corners are swapped.\n\nIn addition, QRect provides the getCoords() function which extracts the position of the rectangle's top-left and bottom-right corner, and the getRect() function which extracts the rectangle's top-left corner, width and height. Use the setCoords() and setRect() function to manipulate the rectangle's coordinates and dimensions in one go.\n\nConstraints\n\nQRect is limited to the minimum and maximum values for the int type. Operations on a QRect that could potentially result in values outside this range will result in undefined behavior.\n\nMember Function Documentation\n\nQRect::QRect()\n\nConstructs a null rectangle.\n\nQRect::QRect(const QPoint &topLeft, const QPoint &bottomRight)\n\nConstructs a rectangle with the given topLeft and bottomRight corners.\n\nQRect::QRect(const QPoint &topLeft, const QSize &size)\n\nConstructs a rectangle with the given topLeft corner and the given size.\n\nQRect::QRect(intx, inty, intwidth, intheight)\n\nConstructs a rectangle with (x, y) as its top-left corner and the given width and height.\n\nAdds dx1, dy1, dx2 and dy2 respectively to the existing coordinates of the rectangle.\n\nQRect QRect::adjusted(intdx1, intdy1, intdx2, intdy2) const\n\nReturns a new rectangle with dx1, dy1, dx2 and dy2 added respectively to the existing coordinates of this rectangle.\n\nint QRect::bottom() const\n\nReturns the y-coordinate of the rectangle's bottom edge.\n\nNote that for historical reasons this function returns top() + height() - 1; use y() + height() to retrieve the true y-coordinate.\n\nQPoint QRect::bottomLeft() const\n\nReturns the position of the rectangle's bottom-left corner. Note that for historical reasons this function returns QPoint(left(), top() + height() - 1).\n\nQPoint QRect::bottomRight() const\n\nReturns the position of the rectangle's bottom-right corner.\n\nNote that for historical reasons this function returns QPoint(left() + width() -1, top() + height() - 1).\n\nQPoint QRect::center() const\n\nReturns the center point of the rectangle.\n\nbool QRect::contains(const QPoint &point, boolproper = false) const\n\nReturns true if the given point is inside or on the edge of the rectangle, otherwise returns false. If proper is true, this function only returns true if the given point is inside the rectangle (i.e., not on the edge).\n\nbool QRect::contains(const QRect &rectangle, boolproper = false) const\n\nReturns true if the given rectangle is inside this rectangle. otherwise returns false. If proper is true, this function only returns true if the rectangle is entirely inside this rectangle (not on the edge).\n\nbool QRect::contains(intx, inty) const\n\nReturns true if the point (x, y) is inside this rectangle, otherwise returns false.\n\nbool QRect::contains(intx, inty, boolproper) const\n\nReturns true if the point (x, y) is inside or on the edge of the rectangle, otherwise returns false. If proper is true, this function only returns true if the point is entirely inside the rectangle(not on the edge).\n\nvoid QRect::getCoords(int x1, int y1, int x2, int y2) const\n\nExtracts the position of the rectangle's top-left corner to x1 and y1, and the position of the bottom-right corner to x2 and y2.\n\nvoid QRect::getRect(int x, int y, int width, int height) const\n\nExtracts the position of the rectangle's top-left corner to x and y, and its dimensions to width and height.\n\nint QRect::height() const\n\nReturns the height of the rectangle.\n\nQRect QRect::intersected(const QRect &rectangle) const\n\nReturns the intersection of this rectangle and the given rectangle. Note that r.intersected(s) is equivalent to r & s.", null, "This function was introduced in Qt 4.2.\n\nbool QRect::intersects(const QRect &rectangle) const\n\nReturns true if this rectangle intersects with the given rectangle (i.e., there is at least one pixel that is within both rectangles), otherwise returns false.\n\nThe intersection rectangle can be retrieved using the intersected() function.\n\nbool QRect::isEmpty() const\n\nReturns true if the rectangle is empty, otherwise returns false.\n\nAn empty rectangle has a left() > right() or top() > bottom(). An empty rectangle is not valid (i.e., isEmpty() == !isValid()).\n\nUse the normalized() function to retrieve a rectangle where the corners are swapped.\n\nbool QRect::isNull() const\n\nReturns true if the rectangle is a null rectangle, otherwise returns false.\n\nA null rectangle has both the width and the height set to 0 (i.e., right() == left() - 1 and bottom() == top() - 1). A null rectangle is also empty, and hence is not valid.\n\nbool QRect::isValid() const\n\nReturns true if the rectangle is valid, otherwise returns false.\n\nA valid rectangle has a left() <= right() and top() <= bottom(). Note that non-trivial operations like intersections are not defined for invalid rectangles. A valid rectangle is not empty (i.e., isValid() == !isEmpty()).\n\nint QRect::left() const\n\nReturns the x-coordinate of the rectangle's left edge. Equivalent to x().\n\nReturns a rectangle grown by the margins.\n\nThis function was introduced in Qt 5.1.\n\nQRect QRect::marginsRemoved(const QMargins &margins) const\n\nRemoves the margins from the rectangle, shrinking it.\n\nThis function was introduced in Qt 5.1.\n\nvoid QRect::moveBottom(inty)\n\nMoves the rectangle vertically, leaving the rectangle's bottom edge at the given y coordinate. The rectangle's size is unchanged.\n\nvoid QRect::moveBottomLeft(const QPoint &position)\n\nMoves the rectangle, leaving the bottom-left corner at the given position. The rectangle's size is unchanged.\n\nvoid QRect::moveBottomRight(const QPoint &position)\n\nMoves the rectangle, leaving the bottom-right corner at the given position. The rectangle's size is unchanged.\n\nvoid QRect::moveCenter(const QPoint &position)\n\nMoves the rectangle, leaving the center point at the given position. The rectangle's size is unchanged.\n\nvoid QRect::moveLeft(intx)\n\nMoves the rectangle horizontally, leaving the rectangle's left edge at the given x coordinate. The rectangle's size is unchanged.\n\nvoid QRect::moveRight(intx)\n\nMoves the rectangle horizontally, leaving the rectangle's right edge at the given x coordinate. The rectangle's size is unchanged.\n\nvoid QRect::moveTo(intx, inty)\n\nMoves the rectangle, leaving the top-left corner at the given position (x, y). The rectangle's size is unchanged.\n\nvoid QRect::moveTo(const QPoint &position)\n\nMoves the rectangle, leaving the top-left corner at the given position.\n\nvoid QRect::moveTop(inty)\n\nMoves the rectangle vertically, leaving the rectangle's top edge at the given y coordinate. The rectangle's size is unchanged.\n\nvoid QRect::moveTopLeft(const QPoint &position)\n\nMoves the rectangle, leaving the top-left corner at the given position. The rectangle's size is unchanged.\n\nvoid QRect::moveTopRight(const QPoint &position)\n\nMoves the rectangle, leaving the top-right corner at the given position. The rectangle's size is unchanged.\n\nQRect QRect::normalized() const\n\nReturns a normalized rectangle; i.e., a rectangle that has a non-negative width and height.\n\nIf width() < 0 the function swaps the left and right corners, and it swaps the top and bottom corners if height() < 0.\n\nReturns the x-coordinate of the rectangle's right edge.\n\nNote that for historical reasons this function returns left() + width() - 1; use x() + width() to retrieve the true x-coordinate.\n\nvoid QRect::setBottom(inty)\n\nSets the bottom edge of the rectangle to the given y coordinate. May change the height, but will never change the top edge of the rectangle.\n\nvoid QRect::setBottomLeft(const QPoint &position)\n\nSet the bottom-left corner of the rectangle to the given position. May change the size, but will never change the top-right corner of the rectangle.\n\nvoid QRect::setBottomRight(const QPoint &position)\n\nSet the bottom-right corner of the rectangle to the given position. May change the size, but will never change the top-left corner of the rectangle.\n\nvoid QRect::setCoords(intx1, inty1, intx2, inty2)\n\nSets the coordinates of the rectangle's top-left corner to (x1, y1), and the coordinates of its bottom-right corner to (x2, y2).\n\nvoid QRect::setHeight(intheight)\n\nSets the height of the rectangle to the given height. The bottom edge is changed, but not the top one.\n\nvoid QRect::setLeft(intx)\n\nSets the left edge of the rectangle to the given x coordinate. May change the width, but will never change the right edge of the rectangle.\n\nEquivalent to setX().\n\nvoid QRect::setRect(intx, inty, intwidth, intheight)\n\nSets the coordinates of the rectangle's top-left corner to (x, y), and its size to the given width and height.\n\nvoid QRect::setRight(intx)\n\nSets the right edge of the rectangle to the given x coordinate. May change the width, but will never change the left edge of the rectangle.\n\nvoid QRect::setSize(const QSize &size)\n\nSets the size of the rectangle to the given size. The top-left corner is not moved.\n\nvoid QRect::setTop(inty)\n\nSets the top edge of the rectangle to the given y coordinate. May change the height, but will never change the bottom edge of the rectangle.\n\nEquivalent to setY().\n\nvoid QRect::setTopLeft(const QPoint &position)\n\nSet the top-left corner of the rectangle to the given position. May change the size, but will never change the bottom-right corner of the rectangle.\n\nvoid QRect::setTopRight(const QPoint &position)\n\nSet the top-right corner of the rectangle to the given position. May change the size, but will never change the bottom-left corner of the rectangle.\n\nvoid QRect::setWidth(intwidth)\n\nSets the width of the rectangle to the given width. The right edge is changed, but not the left one.\n\nvoid QRect::setX(intx)\n\nSets the left edge of the rectangle to the given x coordinate. May change the width, but will never change the right edge of the rectangle.\n\nEquivalent to setLeft().\n\nvoid QRect::setY(inty)\n\nSets the top edge of the rectangle to the given y coordinate. May change the height, but will never change the bottom edge of the rectangle.\n\nEquivalent to setTop().\n\nQSize QRect::size() const\n\nReturns the size of the rectangle.\n\nCGRect QRect::toCGRect() const\n\nCreates a CGRect from a QRect.\n\nThis function was introduced in Qt 5.8.\n\nint QRect::top() const\n\nReturns the y-coordinate of the rectangle's top edge. Equivalent to y().\n\nQPoint QRect::topLeft() const\n\nReturns the position of the rectangle's top-left corner.\n\nQPoint QRect::topRight() const\n\nReturns the position of the rectangle's top-right corner.\n\nNote that for historical reasons this function returns QPoint(left() + width() -1, top()).\n\nvoid QRect::translate(intdx, intdy)\n\nMoves the rectangle dx along the x axis and dy along the y axis, relative to the current position. Positive values move the rectangle to the right and down.\n\nvoid QRect::translate(const QPoint &offset)\n\nMoves the rectangle offset.x() along the x axis and offset.y() along the y axis, relative to the current position.\n\nQRect QRect::translated(intdx, intdy) const\n\nReturns a copy of the rectangle that is translated dx along the x axis and dy along the y axis, relative to the current position. Positive values move the rectangle to the right and down.\n\nQRect QRect::translated(const QPoint &offset) const\n\nReturns a copy of the rectangle that is translated offset.x() along the x axis and offset.y() along the y axis, relative to the current position.\n\nQRect QRect::transposed() const\n\nReturns a copy of the rectangle that has its width and height exchanged:\n\nQRect r = {15, 51, 42, 24};\nr = r.transposed(); // r == {15, 51, 24, 42}\n\nThis function was introduced in Qt 5.7.\n\nQRect QRect::united(const QRect &rectangle) const\n\nReturns the bounding rectangle of this rectangle and the given rectangle.", null, "This function was introduced in Qt 4.2.\n\nint QRect::width() const\n\nReturns the width of the rectangle.\n\nint QRect::x() const\n\nReturns the x-coordinate of the rectangle's left edge. Equivalent to left().\n\nint QRect::y() const\n\nReturns the y-coordinate of the rectangle's top edge. Equivalent to top().\n\nQRect QRect::operator&(const QRect &rectangle) const\n\nReturns the intersection of this rectangle and the given rectangle. Returns an empty rectangle if there is no intersection.\n\nQRect &QRect::operator&=(const QRect &rectangle)\n\nIntersects this rectangle with the given rectangle.\n\nQRect &QRect::operator+=(const QMargins &margins)\n\nAdds the margins to the rectangle, growing it.\n\nThis function was introduced in Qt 5.1.\n\nQRect &QRect::operator-=(const QMargins &margins)\n\nReturns a rectangle shrunk by the margins.\n\nThis function was introduced in Qt 5.1.\n\nQRect QRect::operator\|(const QRect &rectangle) const\n\nReturns the bounding rectangle of this rectangle and the given rectangle.\n\nQRect &QRect::operator\|=(const QRect &rectangle)\n\nUnites this rectangle with the given rectangle.\n\nRelated Non-Members\n\nbooloperator!=(const QRect &r1, const QRect &r2)\n\nReturns true if the rectangles r1 and r2 are different, otherwise returns false.\n\nQRectoperator+(const QRect &rectangle, const QMargins &margins)\n\nReturns the rectangle grown by the margins.\n\nThis function was introduced in Qt 5.1.\n\nQRectoperator+(const QMargins &margins, const QRect &rectangle)\n\nReturns the rectangle grown by the margins.\n\nThis function was introduced in Qt 5.1.\n\nQRectoperator-(const QRect &lhs, const QMargins &rhs)\n\nReturns the lhs rectangle shrunken by the rhs margins.\n\nThis function was introduced in Qt 5.3.\n\nQDataStream &operator<<(QDataStream &stream, const QRect &rectangle)\n\nWrites the given rectangle to the given stream, and returns a reference to the stream." ]	[ null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-intersect.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-unite.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-diagram-zero.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-diagram-one.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-diagram-two.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-diagram-three.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-coordinates.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-intersect.png", null, "https://doc-snapshots.qt.io/qt5-5.10/images/qrect-unite.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67157626,"math_prob":0.9813694,"size":25575,"snap":"2022-05-2022-21","text_gpt3_token_len":5862,"char_repetition_ratio":0.2306128,"word_repetition_ratio":0.2906391,"special_character_ratio":0.23476051,"punctuation_ratio":0.15317611,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9893242,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,4,null,4,null,2,null,2,null,2,null,2,null,2,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-25T11:39:49Z\",\"WARC-Record-ID\":\"<urn:uuid:476cc10a-c47b-4307-bcb5-5539c5bbe38d>\",\"Content-Length\":\"94727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23379372-70c4-4ea9-8934-45149947fce3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f16d3dbf-91d9-42fd-acbd-9a09d2751ec2>\",\"WARC-IP-Address\":\"18.203.89.116\",\"WARC-Target-URI\":\"https://doc-snapshots.qt.io/qt5-5.10/qrect.html\",\"WARC-Payload-Digest\":\"sha1:6NHUO3YNJ67WH3QACN7SIRBBELZ6VNO7\",\"WARC-Block-Digest\":\"sha1:PMTV4YB6VMWUSA6SL4ZK7OUWHXUYP4KN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304810.95_warc_CC-MAIN-20220125100035-20220125130035-00078.warc.gz\"}"}
https://discuss.pytorch.org/t/nan-in-torch-norm-if-input-is-zero/6844	[ "# NaN in torch.norm if input is zero\n\nHello !\n\nThis code prints an array of nan", null, ":\n\n``````a = Variable(torch.zeros(3,3), requires_grad=True)\nb = a.norm()\nb.backward()\n``````\n\nHave I done anything wrong ? Looks rather like a formula bug …\n\nI have found a similar issue here, may be related …\n\nThe problem is that you are trying to get the derivative of the square root function at 0. Which is + infinity. The gradient of a is then `+infinity * 0 = Nan`.\n\n1 Like\n\nOk, got it.\n\nI guess the fact that is also arises for `b = a.norm(p=1)` is because derivative of abs is not defined in 0.\n\nMay i know how to deal with this problem? Is there any version fix this problem?\n\nThis has been fixed in master, now the norm return the subgradient with value 0 at 0.\n\nHi,\nHow can i find infinity values in my pytorch tensor.\n\nYou should be able to compare it to inf. Something like:\n\n`tensor == float('inf')`\n\nthanks\nyour solution is correct. I solved it in this way\n\n``````def get_new_weights(weights):\nnw=weights.view(-1)\nPINFINITY = float('inf')\nNINFINITY = -PINFINITY\n# if there is -inf --> -1e10, +inf-->max(weights), nan --> 1\nnw[nw != nw] = 1\nnw[nw == PINFINITY] = max(weights)\nnw[nw == NINFINITY] = -1e10\nreturn nw``````" ]	[ null, "https://discuss.pytorch.org/images/emoji/apple/cry.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85472715,"math_prob":0.97279555,"size":1233,"snap":"2022-05-2022-21","text_gpt3_token_len":368,"char_repetition_ratio":0.085435316,"word_repetition_ratio":0.0,"special_character_ratio":0.297648,"punctuation_ratio":0.12062257,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983067,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-16T08:10:17Z\",\"WARC-Record-ID\":\"<urn:uuid:a770cc73-6269-46cd-be5d-4d235efe388f>\",\"Content-Length\":\"28847\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5570ed53-90e9-42ba-b018-aad1852147d9>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6f0e65b-f7bc-46ab-8f0a-aa0db754d42d>\",\"WARC-IP-Address\":\"159.203.145.104\",\"WARC-Target-URI\":\"https://discuss.pytorch.org/t/nan-in-torch-norm-if-input-is-zero/6844\",\"WARC-Payload-Digest\":\"sha1:6WG3XFDOMDCFFY5ZZEO2NABC2LXWFX2I\",\"WARC-Block-Digest\":\"sha1:4TDQKKXQ32EZE5WYC4WHVZRPGUJIGO7X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662510097.3_warc_CC-MAIN-20220516073101-20220516103101-00080.warc.gz\"}"}
https://studymaterialz.in/advanced-engineering-mathematics-by-h-k-dass-nw6/	[ "0\n582\n\nDownload Advanced Engineering Mathematics By H K Dass – Advanced Engineering Mathematics, compiled by H. K. Dass, is a comprehensive book for engineering students of all branches. It discusses topics like Partial Differentiation, Multiple Integral, Differential Equations, Vectors, Special Functions, Determinants and Matrices, Complex Numbers, Statistics, Probability, Fourier Series, Laplace Transforms, Z-Transforms, Fuzzy Sets, Hilbert Transform, Infinite Series, Tensor Analysis, Calculus of Variations and Linear Programming among many other concepts.\n\nKey Featuures\n\n• Two New Chapters -Transformation and Taylor’s and Laurent’s Series have been included\n• Every topic relating to the subject has been provided with ample coverage.\n• Close to 1400 solved examples (including previous year questions of different universities) on various topics have been incorporated for the better understanding and to make familiar with the standard and trend of questions set in the examinations.\n• More than 260 exercise sets and book-end solved question papers provide apt practice of all concepts explained.\n• Useful formulae and 10 years question papers have been provided, which will be helpful for students while preparing for examinations\n\n### Book Details:\n\n Title Of The Book Advanced Engineering Mathematics Author’s Name H K Dass Publishers H K Dass File Size – MB File Type PDF\n\n### Table Of Content:\n\n1. Partial Differentiation\n2. Multiple Integral\n3. Differential Equations\n4. Determinants and Matrices\n5. Vectors\n6. Complex Numbers\n7. Functions of a Complex Variable\n8. Special Functions\n9. Partial Differential Equations\n10. Statistics\n11. Probability\n12. Fourier Series\n13. Laplace Transformation\n14. Integral Transforms\n15. Numerical Techniques\n16. Numerical Method for Solution of Partial Differential Equation\n17. Calculus of Variations\n18. Tensor Analysis\n19. Z-transforms\n20. Infinite Series\n21. Gamma, Beta Functions, Differentiation under the Integral Sign\n22. Chebyshev Polynomials\n23. Fuzzy Sets\n24. Hankel Transform\n25. Hilbert Transform\n26. Empirical Laws and Curve Fitting (Method of Least Squares)\n27. Linear Programming\nUseful Formulae\nSolved Questions Paper\nQuestion Papers, 2006, 2005, 2004\nIndex" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75603485,"math_prob":0.7233269,"size":2514,"snap":"2021-31-2021-39","text_gpt3_token_len":546,"char_repetition_ratio":0.13426295,"word_repetition_ratio":0.053072624,"special_character_ratio":0.20843278,"punctuation_ratio":0.14489311,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98195285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T11:03:42Z\",\"WARC-Record-ID\":\"<urn:uuid:9b5a1afa-6724-45d6-a525-0fff499505ca>\",\"Content-Length\":\"348728\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:22b820d4-eb6b-4e36-9b0b-817901bda2bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e8e848d-7822-456d-8265-58735049bd5b>\",\"WARC-IP-Address\":\"104.21.41.206\",\"WARC-Target-URI\":\"https://studymaterialz.in/advanced-engineering-mathematics-by-h-k-dass-nw6/\",\"WARC-Payload-Digest\":\"sha1:BZUQN3LUJVM7WZQ5G5ERW6UYTXLUWW7S\",\"WARC-Block-Digest\":\"sha1:ALICTXXBDM3H2ZY2C2YZCDYNRKCEATHJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058415.93_warc_CC-MAIN-20210927090448-20210927120448-00395.warc.gz\"}"}
https://physics.stackexchange.com/questions/24560/how-does-gravity-force-get-transmitted	[ "# How does gravity force get transmitted?\n\nHow does gravity force get transmitted?\n\nIt is not transmitted by particles I guess. Because if it was, then its propagation speed would be limited by the speed of light. If it is not transmitted by particles how is it transmitted then?\n\n• – Qmechanic Apr 28 '12 at 18:01\n\nWhen you learn about gravity at college you're almost always considering situations that are static i.e. they don't change with time. For example you'll learn early on that the gravitational potential of the Earth is given by Newton's equation and this doesn't have any dependance on time. Later on you'll learn that General Relativity gives a more accurate description of the gravity of the Earth (or more usually of a star) called the Schwarzchild metric. This does have a time variable in it, but the metric itself it is independant of time just like Newton's equation.\n\nTo address your question about the propogation of gravity you have to ask questions like \"what would happen to satellites if the Earth suddenly disappeared?\". When the Earth is present the satellites are happily orbiting because of Earth's gravity. If the Earth suddenly disappeared would the satellites instantly veer off in a straight line, or would it take some time before they reacted to the Earth's disappearance.\n\nThe answer is that they would take some time to react because the change in the Earth's gravitational field would propagate at the speed of light. The change propagates by gravitational waves and these travel at the speed of light.\n\nA gravitational wave is basically a disturbance in the curvature of space. Consider this analogy. A water wave is a disturbance in the surface of water. Suppose you had a model of the Earth floating on a pond and you suddenly pulled it out to leave a hemispherical dimple. Waves would flow into the dimple then spread out across the pond. A duck floating some distance away wouldn't know immediately that the model Earth was gone: it would only know when the waves reached it.\n\nThis is a somewhat dodgy analogy (I can hear the general relativists screaming already!) so don't take it too seriously. Apart from anything else gravity waves are mathematically very different from water waves. Still, I hope it gives the general idea. Changes in gravitational fields propagate by gravitational waves, and these move at the speed of light.\n\nYou mention particles. The description about is a classical one, and you might ask how quantum mechanics views the situation. After all, radio waves are a classical description and quantum mechanics views them as made up from particles called photons. Well you can describe a quantum gravitational field as being made up of particles called gravitons. However it is not at all clear that gravitons are a good description of quantum gravity. No-one has ever observed them, but then it would take energies far far higher than those attainable at the LHC to see gravitons, so it's no surprise they haven't been observed yet. If gravitons do exist they will travel at the speed of light just like photons.\n\n• As an aside, gravitons are pretty speculative, and having been through any sort of testing but to support the gravitational waves idea, the general theory of relativity predicts these ripples as a way of dissipating energy and this has (implicitly) been observed in different ways. One of the most convincing observations was the prediction of the slowing of PSR J0737-3039 - a double pulsar. What you have here is two radiating emitting neutron stars (The densest objects known other than black holes) orbiting around the pair's centre of mass. – Light Matters Apr 28 '12 at 19:34\n• The orbit causes huge distortions in space-time and the predicted energy emitted in gravitational waves was used to estimate the rate of slowing in the orbit and when compared to observations, there was only a discrepancy found in the range of ~0.03%! – Light Matters Apr 28 '12 at 19:36\n• Yes indeed, though wasn't PSR B1913+16 (en.wikipedia.org/wiki/PSR_B1913%2B16) the first pulsar used to detect gravity waves? Pretty much everyone believes gravity waves exist, though they haven't been detected on Earth yet. – John Rennie Apr 28 '12 at 19:38\n• It was suggestive, but PSR J0737-3039 is the closest we've ever got to astrophysical proof. PSR B1913+16, for those who don't know, is made of at least a pulsar and a neutron star, but PSR J0737-3039 is made up of two pulsars whose irradiation beams both cross the Earth: which allows for accurate measurements of Shapiro delays (Time dilation effect). – Light Matters Apr 28 '12 at 19:53\n• You can't make the Earth disappear, you can only shake it. If you make it disappear, the gravitational field becomes inconsistent, since it implies the conservation of mass-energy. – Ron Maimon Apr 30 '12 at 5:03\n\nPropagation of gravitational force is limited by the speed of light. In fact transmission of any kind of information is restricted by the speed of light. According to the general theory of relativity gravitational force propagates at the speed of light\n\nIt depends on how you think about gravity. In the framework of general relativity (the most complete, accepted paradigm), then gravity isn't a 'force' in the classical sense---but is instead the results of the geometry of space-time. Energy/mass curve spacetime; other bodies react to that curvature in their motion. Thus there is no force-carrier.\n\nIf you consider gravity in a particle-physics framework (which we don't have a complete model for, but many people are working on models of such a 'quantum gravity'), then gravity is believed to be conveyed by the spin-2 graviton.\n\nIn both cases changes in gravity propagate at the speed of light.\n\n• The two pictures, force and geometry, are dual, and believing one is right does not require one to deny the other. – Ron Maimon Apr 30 '12 at 5:04\n• Absolutely. That's why I used the terms 'how you think about gravity', 'framework of GR', and 'consider in a particle-physics framework'. – DilithiumMatrix Apr 30 '12 at 14:16\n\nIn some respects, static gravity is similar to static electricity in the sense that both follow the inverse square law. The major difference between the two is that similar charges attract in gravity, but repel in electricity. Also as far as we know, there is only positive mass (charge) in gravity. If negative mass existed, it would in fact be repelled by (normal) positive mass according to the inverse square law anyway, and will not be found around. The similar charges attract issue can in fact be easily overcome- by putting im for the mass m, where i is the imaginary number unit.\n\nIn the dynamic case where we have motion, we have another similarity between the two forces- the effects of the forces propagate at the fixed speed of light. I think this is a big hint that the two are closely related. If we then take such delay into consideration, then look to modify the forces accordingly, we can use the retarded potential integral. If we do this for electric charges, we can recover the complete set of Maxwell equations, from the static Coulomb force alone. As we know, these equations are relativistic.\n\nThe same can be done to gravity and Newton's law to obtain the so called gravito-magnetic equations. These are similar to Maxwell equations in form, and shown to be derivable from GR in the weak limit or linear gravity. These equations work not in 4D space-time, but like Maxwell in 3+1 dimensions(flat space and time). In this case, it is possible in principle to introduce gravitons very much like the photons.\n\nThe question now is that GR is non-linear and the gravito-magnetic equations, like Maxwell equations, are linear. This could stop the flow of this argument from going any further. In my opinion however, the reason GR is nonlinear is the same as Maxwell equations becoming non-linear, or the Schrodinger equation becoming nonlinear, in material media. It's a result of double counting.. we take the energy in a field and mix it with that in a mass- resulting in a multiplicity of counting and a nonlinearity in the variables of the resulting equations.\n\nHaving said all that, it must also be recalled that the photon is a quantity of energy representing the way electrically charged bodies exchange energy, and as such, is closely related to the energy levels in an atom. Thus, for the graviton to have a use and a meaning, similar energy exchanges and levels are needed in gravitating systems. Gravitating systems mainly exchange mass-energy much more than any other form of energy." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95979506,"math_prob":0.9493317,"size":6819,"snap":"2020-10-2020-16","text_gpt3_token_len":1381,"char_repetition_ratio":0.12795304,"word_repetition_ratio":0.077391304,"special_character_ratio":0.19460331,"punctuation_ratio":0.088074826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9837683,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-27T20:16:51Z\",\"WARC-Record-ID\":\"<urn:uuid:9616e461-892d-4ffc-b69b-e9bd4746a5bd>\",\"Content-Length\":\"173292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6615e6f7-32c5-4150-b1f2-c85f084b751a>\",\"WARC-Concurrent-To\":\"<urn:uuid:29f55b56-f887-44d0-acf7-09da23315bd3>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/24560/how-does-gravity-force-get-transmitted\",\"WARC-Payload-Digest\":\"sha1:KB4IBE63R76STBWLCVRAVJDB6XM46XU5\",\"WARC-Block-Digest\":\"sha1:U4W3CBOQ3OEEH57RYNA6WQT6GM7ADDAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146809.98_warc_CC-MAIN-20200227191150-20200227221150-00384.warc.gz\"}"}
https://indiascreen.ir/post/if-the-strength-of-the.p155608	[ "if you want to remove an article from website contact us from top.\n\n# if the strength of the current flowing through a wire is increased the strength of the magnetic field produced by it\n\nCategory :\n\n### Mohammed\n\nGuys, does anyone know the answer?\n\nget if the strength of the current flowing through a wire is increased the strength of the magnetic field produced by it from screen.\n\n## How is the strength of magnetic field affected when current increases?\n\nHow is the strength of magnetic field affected when current increases?", null, "Magnetic Field Due to Current through a Circular Loop\n\nHow is the st... Question\n\nHow is the strength of magnetic field affected when current increases?\n\nOpen in App Solution\n\nThe strength of magnetic field is always proportional to the magnitude of current flowing. Hence, when the current increases, the magnetic field also increases.\n\nSuggest Corrections 7", null, "SIMILAR QUESTIONS\n\nQ.\n\nA magnetic field is produced by the current passing through a long straight wire. When will the strength of the magnetic field increase?\n\nQ.\n\nThe magnetic field strength of a current carrying solenoid increases with\n\nQ. Draw the pattern of magnetic field lines through and around a current-carrying wire. Make the direction of (i) electric current in the wire (ii) magnetic field lines.\n\nHow would the strength of magnetic field due to current, carrying loop be affected if\n\n(a) radius of the loop is reduced to half its original value?\n\n(b) strength of current through the wire is doubled?\n\nQ. How can the strength of a magnetic field be increased?Q.\n\nThe magnetic field strength of a current carrying solenoid increases with a rise in its temperature.\n\nView More RELATED VIDEOS", null, "Magnetic Field Due to a Current Carrying Conductor\n\nPHYSICS Watch in App EXPLORE MORE\n\nMagnetic Field Due to Current through a Circular Loop\n\nStandard X Physics\n\nस्रोत : byjus.com\n\n## When the strength of the current flowing through a coil is increased, which of the following statements is true for it?\n\nClick here👆to get an answer to your question ✍️ When the strength of the current flowing through a coil is increased, which of the following statements is true for it?", null, "Question\n\nA\n\nB\n\nC\n\nD\n\n## Strength of the magnetic field remains constant\n\nMedium Open in App\n\nUpdated on : 2022-09-05\n\nSolution Verified by Toppr\n\nCorrect option is B)\n\n## The strength of the magnetic field of a coil is directly proportional to the strength of the current flowing through it.\n\nSolve any question of Moving Charges and Magnetism with:-\n\nPatterns of problems\n\n>\n\n18 1\n\nस्रोत : www.toppr.com\n\n## How does an increase in current cause the magnetic field strength to increase?\n\nAnswer (1 of 5): Most likely is due to a larger number of electrons are part of the current. Thus a larger number of atoms participating in generating the magnetic field. From Wikipedia “More precisely, the term magnetic moment normally refers to a system's magnetic dipole moment, which produce...", null, "How does an increase in current cause the magnetic field strength to increase?\n\nWhat is Aspose.OCR for C++ library?\n\nOCR API capable of extracting text from BMP, JPEG and other images having different fonts and styles.\n\nSort Peter Webb\n\nI might first point out that however deep you go into the laws of physics, you always end up with “well, that’s just how it is”. I can peel a few of layers off.\n\nLayer 1\n\nRunning a current through even a straight wire generates a magnetic field around the wire. This is called Ampère's circuital law - Wikipedia\n\nand dates from 1861. It is, as they say, an observational fact.\n\nLayer 2\n\nThis was bundled up with 3 other laws about electrical and magnetic fields to form Maxwell's equations - Wikipedia\n\n. This doesn’t help answer your question directly, because Ampere’s Law is effectively just one of the laws.\n\nRelated questions\n\nHow does increasing current increase the magnetic field strength, based on the formula?\n\nHow does an increase in magnetic field cause current?\n\nWhy does increasing current increase a magnetic field?\n\nWhat magnetic field strength is dangerous?\n\nWhy is current produced in a magnetic field?\n\nDavid W. Vogel\n\nA2A: I don't think any theory yet explains how. They only accommodate the observation that it does. You are likely to receive the answer “by induction\" or “because the Maxwell-Ampere equation says it must”.\n\nIt is an interesting question, as is “Why are Maxwell's equations almost symmetrical?”\n\nI'm personally working on the assumption that the right postulate will eventually clarify these mysteries. I believe I have an answer, but my model is not sufficiently developed for disclosure.\n\nYou can get 200 safe backlinks from major news sites within 10 days using a news publishing service.\n\nAmerico Perez\n\nMost likely is due to a larger number of electrons are part of the current. Thus a larger number of atoms participating in generating the magnetic field.\n\nFrom Wikipedia\n\n“More precisely, the term magnetic moment normally refers to a system's magnetic dipole moment, which produces the first term in the multipole expansion of a general magnetic field. The dipole component of an object's magnetic field is symmetric about the direction of its magnetic dipole moment, and decreases as the inverse cube of the distance from the object.”\n\nMagnetic moment - Wikipedia\n\nThe keywords are “multipole expansion”\n\nI do Time Traveler\n\nWorked at Institute of Electrical and Electronics EngineersAuthor has 9.6K answers and 1.4M answer views8mo\n\nThe more electrons flowing then the stronger the field.\n\nRelated questions\n\nHow does increasing magnetic field strength affect current flow in a conductor?\n\nWhat are the two methods to increase the strength of a magnetic field by a current carrying wire?\n\nWhat is the relationship between magnetic field strength and current?\n\nDo current carrying wires experience a force due to the magnetic field?\n\nWhat are some effects of variation in magnetic field strength?\n\nSteven J Greenfield\n\nRelated\n\nWhy does a changing magnetic field induce a current?\n\nOriginally Answered: why does a changing magnetic field induce a current?\n\nHow do moving charges, such as electric current, produce a magnetic field?\n\nThere is no such thing as a magnetic field.\n\nWhat appears to be a magnetic field is really just changing electric fields. If you analyze the situation using Special Relativity, you can see that it has exactly the effects we see experimentally.\n\nMaxwell, a brilliant man, working with the knowledge we had then and experimental observations, came up with four equations that exactly describe electric and magnetic fields. But no one had discovered relativity, just yet.\n\nPicture two wires parallel to each other, a short distance\n\nSponsored by Synapse: New Drug Intelligence Platform\n\nHow do you facilitate pharmaceutical R&D on new therapeutics?\n\nWith Synapse can get technical intelligence of competitors and whole pharmaceuticals' data in 10 seconds.\n\nAshwin Sharma\n\nFormer Mechanical EngineerUpvoted by\n\nAjith Varghese\n\n, Msc Physics, University of Mysore (2017) and\n\nNaba Prakash Nayak\n\n, PhD Physics, Indian Institute of Technology, Bombay (2024)Updated 2y\n\nRelated\n\nWhy does a moving charge produce a magnetic field around it?\n\nHow special theory of relativity deals with electromagnetism?\n\nHow the magnetic field and electric field depend on the observer's frames of reference?", null, "In this answer, I am not talking about permanent magnets they are a different thing to understand.\n\nwe all know that a current-carrying conductor generates a magnetic field around itself, and we can experience that magnetic field by moving another charge around it, as we know a current-carrying conductor exerts a force on the moving charge and it is calculated by the equation F= qv x b. We all know this by Lenz law but what does relativity has to do\n\nस्रोत : www.quora.com\n\nDo you want to see answer or more ?\nMohammed 11 day ago\n\nGuys, does anyone know the answer?" ]	[ null, "https://indiascreen.ir/red", null, "https://indiascreen.ir/red", null, "https://indiascreen.ir/red", null, "https://indiascreen.ir/red", null, "https://indiascreen.ir/red", null, "https://indiascreen.ir/red", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9248662,"math_prob":0.85663027,"size":8296,"snap":"2023-14-2023-23","text_gpt3_token_len":1758,"char_repetition_ratio":0.21008201,"word_repetition_ratio":0.19019316,"special_character_ratio":0.19527483,"punctuation_ratio":0.09325397,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9522442,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T17:48:12Z\",\"WARC-Record-ID\":\"<urn:uuid:f5a66edc-3c9c-4621-b314-0488081f0a21>\",\"Content-Length\":\"30004\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ccdd278-72a8-4450-a4c2-19bcaee1faed>\",\"WARC-Concurrent-To\":\"<urn:uuid:61b357d4-d00b-44c2-bef9-23e8ddcd070c>\",\"WARC-IP-Address\":\"185.141.107.57\",\"WARC-Target-URI\":\"https://indiascreen.ir/post/if-the-strength-of-the.p155608\",\"WARC-Payload-Digest\":\"sha1:RGNDVGP5UFSRAGBTIVN2SWEGDKYNNQVF\",\"WARC-Block-Digest\":\"sha1:FLKJYZ5WAT3G36X437OE67WIZCTXKG2K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948868.90_warc_CC-MAIN-20230328170730-20230328200730-00718.warc.gz\"}"}
https://ask.learncbse.in/t/for-any-positive-integer-n-prove-that-n3-n-is-divisible-by-6/63753	[ "", null, "# For any positive integer n prove that n3 - n is divisible by 6\n\nFor any positive integer n prove that n3 - n is divisible by 6.\n\nn3 - n = n (n2 - 1) = n (n - 1) (n + 1)\n\nWhenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.\n∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.\nIf n = 3p, then n is divisible by 3.\nIf n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.\nIf n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.\nSo, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.\n⇒ n (n – 1) (n + 1) is divisible by 3.\n\nSimilarly, whenever a number is divided 2, the remainder obtained is 0 or 1.\n∴ n = 2q or 2q + 1, where q is some integer.\nIf n = 2q, then n is divisible by 2.\nIf n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.\nSo, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.\n⇒ n (n – 1) (n + 1) is divisible by 2.\nSince, n (n – 1) (n + 1) is divisible by 2 and 3.\n\n∴ n (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)" ]	[ null, "https://ask.learncbse.in/images/discourse-logo-sketch.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9384208,"math_prob":1.000009,"size":1123,"snap":"2020-34-2020-40","text_gpt3_token_len":488,"char_repetition_ratio":0.23413762,"word_repetition_ratio":0.3057325,"special_character_ratio":0.46482635,"punctuation_ratio":0.11034483,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000046,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-24T05:08:36Z\",\"WARC-Record-ID\":\"<urn:uuid:b48ec7ef-1c03-4b35-982c-4610ff66047a>\",\"Content-Length\":\"11603\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5fac77e-fa4f-4ee8-91e7-d7b8d522f245>\",\"WARC-Concurrent-To\":\"<urn:uuid:137b445d-4f83-4bbb-8fbd-c158529e5c6e>\",\"WARC-IP-Address\":\"46.165.252.193\",\"WARC-Target-URI\":\"https://ask.learncbse.in/t/for-any-positive-integer-n-prove-that-n3-n-is-divisible-by-6/63753\",\"WARC-Payload-Digest\":\"sha1:G7VZIWIUDJO6ZYCEXHSJGXCT5M7SMLWH\",\"WARC-Block-Digest\":\"sha1:2M3OFOE6SPITMF3JXLAGONMZWXAYSY2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400213454.52_warc_CC-MAIN-20200924034208-20200924064208-00196.warc.gz\"}"}
https://tomopt.com/docs/propt/tomlab_propt104.php	[ "## 103 Singular Control 3\n\nITERATIVE DYNAMIC PROGRAMMING, REIN LUUS\n\n10.2.3 Example 3\n\nCHAPMAN & HALL/CRC Monographs and Surveys in Pure and Applied Mathematics\n\n### 103.1 Problem Formulation\n\nFind u over t in [0; 5 ] to minimize\n\n J = x3(tF)\n\nsubject to:\n\n dx1 dt\n= x2\n dx2 dt\n= u\n dx3 dt\n= x12 + x22\n\nThe initial condition are:\n\n x(0) = [0 1 0]\n −1 <= u <= 1\n\nReference: \n\n### 103.2 Problem setup\n\n```toms t\np = tomPhase('p', t, 0, 5, 60);\nsetPhase(p);\n\ntomStates x1 x2 x3\ntomControls u\n\n% Initial guess\nx0 = {icollocate({x1 == 0; x2 == 1; x3 == 0})\ncollocate(u == 0)};\n\n% Box constraints\ncbox = {-1 <= collocate(u) <= 1};\n\n% Boundary constraints\ncbnd = initial({x1 == 0; x2 == 1; x3 == 0});\n\n% ODEs and path constraints\nceq = collocate({dot(x1) == x2\ndot(x2) == u; dot(x3) == x1.^2 + x2.^2});\n\n% Objective\nobjective = final(x3);\n```\n\n### 103.3 Solve the problem\n\n```options = struct;\noptions.name = 'Singular Control 3';\nsolution = ezsolve(objective, {cbox, cbnd, ceq}, x0, options);\nt = subs(collocate(t),solution);\nu = subs(collocate(u),solution);\n```\n```Problem type appears to be: lpcon\nStarting numeric solver\n===== * * * =================================================================== * * *\nTOMLAB - Tomlab Optimization Inc. Development license 999001. Valid to 2011-02-05\n=====================================================================================\nProblem: --- 1: Singular Control 3 f_k 0.753994561590098700\nsum(\|constr\|) 0.000000015978054113\nf(x_k) + sum(\|constr\|) 0.753994577568152800\nf(x_0) 0.000000000000000000\n\nSolver: snopt. EXIT=0. INFORM=1.\nSNOPT 7.2-5 NLP code\nOptimality conditions satisfied\n\nFuncEv 1 ConstrEv 43 ConJacEv 43 Iter 34 MinorIter 366\nCPU time: 0.671875 sec. Elapsed time: 0.688000 sec.\n```\n\n### 103.4 Plot result\n\n```figure(1)\nplot(t,u,'+-');\nlegend('u');\ntitle('Singular Control 3 control');\n```", null, "" ]	[ null, "https://tomopt.com/docs/propt/pngs/singularControl3_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5666903,"math_prob":0.98011744,"size":1644,"snap":"2021-43-2021-49","text_gpt3_token_len":539,"char_repetition_ratio":0.17926829,"word_repetition_ratio":0.025316456,"special_character_ratio":0.4908759,"punctuation_ratio":0.20860927,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9898463,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T14:56:52Z\",\"WARC-Record-ID\":\"<urn:uuid:a104b5c9-2f6f-4130-8765-3523cc4dbeb0>\",\"Content-Length\":\"22638\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5c00f403-84ef-40cd-bf5e-4a523729481e>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb21b5e9-fa88-4ce0-a5da-ecc9053b9dad>\",\"WARC-IP-Address\":\"74.208.236.155\",\"WARC-Target-URI\":\"https://tomopt.com/docs/propt/tomlab_propt104.php\",\"WARC-Payload-Digest\":\"sha1:JRSLMAWF5JNEKWUMZ3XINCX5TFDECRWN\",\"WARC-Block-Digest\":\"sha1:3LJ44W5A23W3IOVNCHFCURQWZJKM3QFE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584886.5_warc_CC-MAIN-20211016135542-20211016165542-00203.warc.gz\"}"}
https://www.rdocumentation.org/packages/elliptic/versions/1.4-0/topics/P.laurent	[ "# P.laurent\n\n0th\n\nPercentile\n\n##### Laurent series for elliptic and related functions\n\nLaurent series for various functions\n\nKeywords\nmath\n##### Usage\nP.laurent(z, g=NULL, tol=0, nmax=80)\nPdash.laurent(z, g=NULL, nmax=80)\nsigma.laurent(z, g=NULL, nmax=8, give.error=FALSE)\nzeta.laurent(z, g=NULL, nmax=80)\n##### Arguments\nz\n\nPrimary argument (complex)\n\ng\n\nVector of length two with g=c(g2,g3)\n\ntol\n\nTolerance\n\ngive.error\n\nIn sigma.laurent(), Boolean with default FALSE meaning to return the computed value and TRUE to return the error (as estimated by the sum of the absolute values of the terms along the minor long diagonal of the matrix)\n\nnmax\n\nNumber of terms used (or, for sigma(), the size of matrix used)\n\n##### Aliases\n• P.laurent\n• Pdash.laurent\n• sigma.laurent\n• zeta.laurent\n• e18.5.1\n• e18f.5.3\n• e18.5.4\n• e18.5.5\n• e18.5.6\n##### Examples\n# NOT RUN {\nsigma.laurent(z=1+1i,g=c(0,4))\n\n# }\n\nDocumentation reproduced from package elliptic, version 1.4-0, License: GPL-2\n\n### Community examples\n\nLooks like there are no examples yet." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.51709247,"math_prob":0.900944,"size":842,"snap":"2020-10-2020-16","text_gpt3_token_len":259,"char_repetition_ratio":0.16825776,"word_repetition_ratio":0.0,"special_character_ratio":0.2624703,"punctuation_ratio":0.17127071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9530504,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T20:41:43Z\",\"WARC-Record-ID\":\"<urn:uuid:466d9497-5fe2-4818-816a-8240419c7b3e>\",\"Content-Length\":\"14161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f725ac92-6751-47c4-952e-a81a85a2b3e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:e15fc562-4ec9-4cb1-82db-9d92025965d1>\",\"WARC-IP-Address\":\"3.226.88.111\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/elliptic/versions/1.4-0/topics/P.laurent\",\"WARC-Payload-Digest\":\"sha1:DIQF5IHPNFAVKM5LDB4YPE2U4MUBLFNK\",\"WARC-Block-Digest\":\"sha1:HRQYMA7SQIRMWTI3PIWHXBAIVGNDR7QP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146485.15_warc_CC-MAIN-20200226181001-20200226211001-00228.warc.gz\"}"}
https://math.stackexchange.com/questions/456273/sum-of-all-positive-integers-less-than-n-and-relatively-prime-to-n	[ "sum of all positive integers less than $n$ and relatively prime to $n$\n\ncould any one tell me how to find the sum of all positive integers less than $n$ and relatively prime to $n$? $n>1$\n\nFor $n=7$, I have $\\phi(n)=6$ and the sum is ${6(6+1)\\over 2}={n\\over 2}.\\phi(n)$, is that true for any $n$?\n\n• This would be true for any prime $n$ (like your $7$), since all numbers less than $n$ would in that case be relatively prime. For any other number, not so much. Jul 31 '13 at 10:28\n\nIt’s true for all $n>2$. The reason is that if $k\\in\\{1,\\ldots,n-1\\}$ is relatively prime to $n$, so is $n-k$, so the integers that you’re adding can be combined into $\\frac{\\varphi(n)}2$ pairs whose members sum to $n$. If $n>2$, $\\frac{n}2$ is never relatively prime to $n$, so you really do get pairs $\\{k,n-k\\}$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8741762,"math_prob":0.999992,"size":1000,"snap":"2022-05-2022-21","text_gpt3_token_len":317,"char_repetition_ratio":0.11646587,"word_repetition_ratio":0.43529412,"special_character_ratio":0.336,"punctuation_ratio":0.097457625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000061,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T12:03:46Z\",\"WARC-Record-ID\":\"<urn:uuid:849ba219-ec5c-4989-a7d7-9c8df0236e99>\",\"Content-Length\":\"136237\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fa00dc64-3dbb-4cf3-bfa2-3eb1e3f0c360>\",\"WARC-Concurrent-To\":\"<urn:uuid:440e1975-1ce7-4c35-824a-0eb43fa28253>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/456273/sum-of-all-positive-integers-less-than-n-and-relatively-prime-to-n\",\"WARC-Payload-Digest\":\"sha1:YQSFECAVM25UPKUKZCC67GOKH2ZOWSJD\",\"WARC-Block-Digest\":\"sha1:7JJHA7JWL3JUKF3L6LYLT52NAXQPASB7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301309.22_warc_CC-MAIN-20220119094810-20220119124810-00202.warc.gz\"}"}
https://eyqr.xn----8sbelb9aup5ak9a.xn--p1ai/70047.php	[ "# Fibonacci Fan Forex Strategy", null, "", null, "This trading system is based on the “ Fibonacci Fan ” to identify trades that have a relatively small target. The “ Fibonacci Fan ” produces 3 lines set at the main Fibonacci retracement numbers, %, %, and %.More often than not the main support line on the “Fibonacci Fan” is the %.\n\nThe Fibonacci Fan forex trading strategy is forex strategy that utilizes the eyqr.xn----8sbelb9aup5ak9a.xn--p1ai4 forex indicator. This indicator is based on the Fibonacci numbers and can be used to create several strategies around these Fibonacci numbers.\n\nThe strategy discussed here uses the Fibonacci fan component of the indicator to create the strategy. · A Fibonacci fan is a method of plotting support and resistance levels based on the ratios provided by the Fibonacci series.\n\nTrendlines are drawn at intervals of, 50, and percent.", null, "· Fibonacci Fan buy strategy You can buy an asset in an uptrend if the price reaches or level. Take confirmation from 50 SMA. The price must be above the 50 SMA. The Fibonacci Fan Trading Strategy is a combination of Metatrader 4 (MT4) indicator (s) and template.\n\nThe essence of this forex system is to transform the accumulated history data and trading signals. · The most basic use of Fibonacci fan (or just fibo fan) is to mark out lines of support and resistance within a trend channel.\n\n## Fibonacci Fan Definition \| Forexpedia by BabyPips.com\n\nTo set up the indicator you simply mark two points on a forming trend. Fan lines drawn by the indicator then show “zones” where support or resistance is likely to occur. · The one I would like to address here in this thread, is the Fibonacci Fan: On an '' up '' Fibo Fan; do you short when price breaks below the Fibo trend line, or do you wait for a breach of the level to short?\n\nDo you place your Stop loss at the previous Swing high? · Here is another strategy called The PPG Forex Trading Strategy. Let’s get started The Best Gann Fan Trading Strategy (Rules for BUY Trade) Step #1: Pick a significant High, Draw Gann Fan Angles and Wait For the 1/1 Line to Break to the Upside.\n\nThe best Gann fan trading strategy works the same in every time frame. The price projection is based on fan-like trend lines that represent already familiar to us Fibonacci numbers:and That how it looks on the sketch: To place Fibonacci Fan on the chart, choose Fibonacci Fan tool from menu on your trading platform and then look for the highest high and lowest low points (in other words two.\n\n· Forex traders use Fibonacci retracements to pinpoint where to place orders for market entry, taking profits and stop-loss orders. Fibonacci levels are commonly used in forex trading. View Full Webinar at eyqr.xn----8sbelb9aup5ak9a.xn--p1ai?id=cabd6-a7eb7c-8eef-ffce7f15c Summary: Fibonacci ratios are a very popular.\n\n## Fibonacci Fan Forex Strategy - Forex Fibonacci Fan Indicator - Forexprofitindicators.com\n\n· In the stock market, the Fibonacci trading strategy traces trends in stocks. When a stock is trending in one direction, some believe that there will be a pullback, or decline in prices.\n\n## Fibonacci Forex trading strategy (system)\n\nFibonacci traders contend a pullback will happen at the Fibonacci retracement. · Select Chart and Time frame where you want to test your Forex system. Right-click on your trading chart and hover on “Template”. Move right to select “eyqr.xn----8sbelb9aup5ak9a.xn--p1ai-THV4PivotSystem” trading system and strategy; You will see “RVM Fractals Level and Fibonacci Fan MT4 Trading Indicator” is available on your Chart.\n\n· Fibonacci Fan Trading System combines Moving Averages with Fibonacci and adds Auto Fibonacci level to come up with a strategy which is able to deliver trade signals within the trend directions. These trade signals are demonstrated in a very simple manner so that even a newbie trader can be benefited from this trading system. Fibonacci Fan is the default indicator on MetaTrader 4 (MT4) and MetaTrader 5 (MT5), and you can assess it directly.\n\nOnce the price breaks the % level, it will usually go to the % level. We can make an entry ataiming for This rule works best in a trending environment, but it can also be used in a countertrend. · The basic idea behind a Fibonacci trading strategy is to look for a retracement to lose inertia and turn back to the initial trend direction, so you buy into the dips and exit at the higher highs on an uptrend and the reverse on a downtrend.\n\n· The Fibonacci Fans binary options trading strategy discussed here aims to spot opportunities to initiate Call or Put trades using and indicator. · So yes, aside from forex, that includes you stock, options, and futures people too! but for those who have yet to adopt a strategy in their arsenal, consider the Fibonacci Channel Strategy!\n\nIf you are a fan of Fibonacci don't forget to check out our fibonacci trendline strategy! The Forex Trading Strategies Series is comprised of three separate modules, each teaching one specific strategy. As no one video is a prerequisite for any others, they can be purchased and utilized separately or together as a set if the trader so chooses.\n\n· This Fibonacci Fan indicator display as a fan. There are Fibonacci retracement lines. Once, you install this forex indicator, you can see, how user-friendly indicators is that.\n\nFibonacci Forex strategy traditionally means that the first max/min is not the most optimum point to start setting up Fibo grid. It is recommended to find at least small double top or a double bottom in a zone wh ere the current trend begins, and it is necessary to construct Fibo levels from the second key point. Why Expert Traders Use Fibonacci Trading Strategy in Forex? The main and big reason to use the Fibonacci tools in the forex trading strategy is that- it works.\n\nAs we know, Fibonacci is everywhere and there are many real examples of the golden ratio in nature. Therefore, traders believe that the % retracement and extensions may give. · Fibonacci Arcs trading strategy.", null, "Our basic Fibonacci Arcs strategy is based on pullback from Fibonacci Arcs levels. Fibonacci Arcs buy trading signal. You can buy an asset if the price reachesor level in an uptrend.\n\nStop-loss can be placed slightly below the recent low.\n\n• Fibonacci Fan Trading System - Forex Strategies - Forex ...\n• Fibonacci forex strategy - Live otc charts - dateccoating.com\n• Fibonacci Archives - Free Forex Trading Strategies And ...\n• Fibonacci Fans - How Traders Use Them To Construct Support ...\n• Fibonacci Fans Binary Options Trading Strategy\n\nSubmit by Ketang Fibonacci Trend Strategy is an strategy suitable for day trader and swing trader based on Finacci indicators bur following the direction of retracement. Time Frame 15 min, 30 min, 60 min, min. Currency pairs: major, minor, Gold and Indices.\n\n· Fibonacci fans is a tool that will help you to analyze trends. Fibonacci fans are sets of trendlines drawn from a high or a low of the price chart through a set of points dictated by Fibonacci retracements.\n\nJust like with Fibonacci retracement, you need to choose the tool and connect a swing high and a swing low with a base line. Retracement as an important tool to predict forex market.\n\n## Forex Strategies That Use Fibonacci Retracements\n\nIn this article I have included some graphic formats such as Fibonacci arcs, fan, channel, expansion, wich are created also with Fibonacci retracement and also rules to perfect chart plotting.\n\nI have analyzed some examples of Fibonacci retracements pattern in a downtrend and in an uptrend. · Basically, all you need is the Fibonacci retracement tool + Fibonacci Fan tool (both on mt4) and instead of taking any trade who gets to a fibonacci retracement level, use the fibonacci fan for double confirmation to find high probality trades.\n\nI made this trade today. Fibonacci Fan. Fibonacci Fan as a line instrument is built as follows: a trendline — for example from a trough to the opposing peak is drawn between two extreme points. Then, an \"invisible\" vertical line is automatically drawn through the second extreme point.\n\nFibonacci numbers, when applied in technical analysis through Fibonacci retracement and Fibonacci extension, are one of the most prolific techniques traders use to qualify or disqualify forex. · Forex strategies that use Fibonacci levels include: If you place a stop-loss order just below the 50% level, then it is possible to buy near the % retracement level.\n\nBy placing the stop-loss order just below the % level, the trader can by near the 50% level. Fibonacci Fan ini secara kasat mata dapat diartikan sebagai trendline yang dibuat berjajar berdasarkan level tertentu.\n\n## Fibonacci Fan - Fibonacci Tools - MetaTrader 5 Help\n\nKalau dalam pembuatan trendline kita menarik ujung suatu titik ke titik tertentu maka dalam penggunaan Fibonacci Fan ini kita tinggal mencari titik tertinggi atau terendahnya lalu garis – garis akan muncul secara otomatis. Falling Fibonacci Fan. Fan Line 1: Peak to % retracement.", null, "Fan Line 2: Peak to 50% retracement. Fan Line 3: Peak to % retracement. Chart 2 shows the S&P ETF with falling Fibonacci Fan lines.\n\nThe lines are based on the April peak (high) and the July trough (low). The horizontal pink lines show the Fibonacci Retracements. The Fibonacci fan also predicts the range of a market for a short period of time, as prices tend to “bounce” between the lowest and highest of the three Fibonacci fan lines, occasionally hovering or rebounding from the 50% line at the middle of the projection.\n\nSeveral traders also use Fibonacci fans in conjunction with Fibonacci arcs.\n\n## Fibonacci Fan? How do you trade with it? \| Forex Factory\n\n· After we discuss about Fibonacci Retracemet dan Fibonacci Expansion, so now we will discuss one of the other types, namely Fibonacci Fibonacci eyqr.xn----8sbelb9aup5ak9a.xn--p1aicci Fan is in plain view can be interpreted as the trendline created lined by a certain level. If we draw a trendline in the manufacture of the end of a point to a certain point in the use of Fibonacci Fan then we stayed seek the highest or.\n\nExtensions use Fibonacci numbers and patterns to determine profit taking points. Extensions continue past the % mark and indicate possible exits in line with the trend.\n\n## Using the Fibonacci Fans to determine the nature of a pullback\n\nFor the purposes of using Fibonacci numbers for day trading forex, the key extension points consist of %, % and %. Fibonacci Forex Trading Strategies In Action. · Fibonacci Fans use Fibonacci ratios based on time and price to construct support and resistance trendlines.\n\nconcepts, and strategies including Momentum, Elliot Waves, Market Thrust, Moving Averages, and more Fibonacci Patterns. Also see our guides on Forex, Crypto, Stock, CFDs, and Options brokers to find out which tools brokerages offer. · The Fibonacci fans tool is listed on the MT4 among the Fibonacci group of indicators. To attach it to the MT4 chart, click on Insert -> Fibonacci -> Fans The modifications to the indicator can be either to increase or reduce the retracement levels as well as changing the colours and the thickness of each line for better visibility.\n\n· The sky is once again falling on the U.S. stock market as COVID fallout continues to dominate sentiment. At the halfway point of the American session, the DJIA DOW (), S&P SPX (), and NASDAQ () are plummeting into the red.\n\nAt press time, it appears that the March E-mini DOW has rejected a key Fibonacci resistance level.\n\n## 3 Simple Fibonacci Trading Strategies [Infographic]\n\n· forex market session; tarjeta neteller chile; forex trading time zones chart; cheap options broker; perfect iq band app; triple bottom chart; traderush usa; Fibonacci forex strategy. Option strategy calculator. Compra a descoberto. · Fibonacci arcs, time zones, and fans are for specific tasks and are rarely used by traders.\n\n## How to setup Gann Fans And Fib Retracements!\n\nA Fibonacci line, on the other hand, is considered to be the primary tool and is used together with different trading strategies – when confirming existing signals or searching for existing signals. Adding and adjusting Fibonacci levels to your chart. Fibonacci is one of the most powerful tool for predicting price movement on the Forex and Stock Market.\n\n## Cryptocurrencies And Blockchains Their Importance In The Future\n\n Forex trading system profitability higher than 80 Forex trading tips for beginners pdf Day trading forex tools Is it worth it to learn options trading Day trading platform south africa First state super growth investment option Danove priznanie z forexu Straddle option strategy calculator How to make monthly income trading options\n\nThroughout this course you will be learning about Fibonacci numbers, Fibonacci Ratios, Fibonacci retracement and extension levels, Fibonacci as support and resistance levels, Fibonacci clusters, additional Fibonacci tools, how to combine Fibonacci with other tools, I will give you some.\n\n· Fibonacci trading with MACD is a price action strategy based on Fibonacci retracemet and MACD as trend filter. This System is based on Fibonacci System, but it's not necessary to know all about the Fibonacci Sequencesand Numbers, just follow these simple rules. All. · Manual forex trading strategies,Forex trading using fibonacci and elliott wave todd gordon eyqr.xn----8sbelb9aup5ak9a.xn--p1ai By December 6, Uncategorized. No Comments.\n\nManual Forex Trading Strategies. How many more years do you have to go? É, mais ou menos, como um plano pré-pago de celular. Homem gol! Forex Trend Strategy with Fibonacci Retracement is trend following strategy but it is based on the lines of the support and resistance of Fibonacci.\n\nTime Frame 60 min. Currency pairs:any also metals and Oil. Even so, I hope that reviews about it Technical Forex Trading Strategies And Fibonacci Fan Forex Strategy will be useful/10(K). The fibonacci fan trading strategy is a combination of metatrader 4 mt4 indicator s and template. Indicators mt4 user manual template https www ebay com itm fibonacci fans automatic this system is based on the fibonacci fan to identify the transaction auto assembly.\n\nThe fibonacci fans tool is listed on the mt4 among the fibonacci group of." ]	[ null, "https://www.metatrader5.com/i/help/terminal/en/fibo_fan.png", null, "https://lh3.googleusercontent.com/proxy/B3AFSV8YXCrlw26mDSDTO_U5FShwxf02orvOJ3_McGh5l4wgEr0yiw9G_0JWquki27GHNHI8eLG4YO6xoHeQtoT74jTA2Ge5QSP_8hz9aTTrP-3317r026dizA=s0-d", null, "https://2.bp.blogspot.com/-1iJO6R9bpuw/TmHy01IvkII/AAAAAAAALnY/bMvNmwecl-4/s1600/fibonacci.GIF", null, "https://www.metatrader5.com/i/help/terminal/en/fibo_fan.png", null, "https://i.ytimg.com/vi/VGeA_N-FzuY/hqdefault.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8287215,"math_prob":0.6718329,"size":14681,"snap":"2021-21-2021-25","text_gpt3_token_len":3162,"char_repetition_ratio":0.19588472,"word_repetition_ratio":0.0041254126,"special_character_ratio":0.20291533,"punctuation_ratio":0.10052317,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96230376,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,9,null,1,null,2,null,9,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T20:35:13Z\",\"WARC-Record-ID\":\"<urn:uuid:3ea0c93f-48b5-49b1-bfd6-4a8f6f6c20f2>\",\"Content-Length\":\"21865\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:564f689c-1625-4e9a-a63b-9985f6257e2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7e467e9-7f9c-4162-a0b6-18f5a883c4d3>\",\"WARC-IP-Address\":\"172.67.194.66\",\"WARC-Target-URI\":\"https://eyqr.xn----8sbelb9aup5ak9a.xn--p1ai/70047.php\",\"WARC-Payload-Digest\":\"sha1:QSM2DQDU7SJEBON2QJ2Q2RMSMYKCVFD4\",\"WARC-Block-Digest\":\"sha1:CETFIF6EWIFRFISWNVDCTYPJ7LOQ5V5R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991207.44_warc_CC-MAIN-20210514183414-20210514213414-00529.warc.gz\"}"}
https://scialert.net/fulltext/?doi=jas.2010.2805.2813&org=11	[ "Research Article\n\n# Effects of Uncertain Inflationary Conditions on an Inventory Model for Deteriorating Items with Shortages", null, "Abolfazl Mirzazadeh\n\nABSTRACT\n\nThis study proposes an inventory model with stochastic internal and external inflation rates for deteriorating items and allowable shortages. The many economic, political, social and cultural variables affect the inflation rates. For instance, economic factors such as changes in the world inflation rate, demand level, labor cost, cost of raw materials, exchange rates, unemployment rate, productivity level, tax, liquidity, etc are effective in this direction. Therefore, the assumption of constant inflation rates is not valid, especially, when the time horizon is long. This model considers stochastic inflationary conditions. Numerical examples are used to illustrate the theoretical results, which are further clarified through a sensitivity analysis on the model parameters. It has been shown that the optimal solution is highly sensitive to considerable uncertainty of the inflation rates.\n\n Services Related Articles in ASCI Similar Articles in this Journal Search in Google Scholar View Citation Report Citation Science Alert\n\n How to cite this article: Abolfazl Mirzazadeh , 2010. Effects of Uncertain Inflationary Conditions on an Inventory Model for Deteriorating Items with Shortages. Journal of Applied Sciences, 10: 2805-2813. DOI: 10.3923/jas.2010.2805.2813 URL: https://scialert.net/abstract/?doi=jas.2010.2805.2813\n\nReceived: July 26, 2010; Accepted: September 03, 2010; Published: October 11, 2010\n\nINTRODUCTION\n\nThe classical inventory models have not been considered the inflation and time value of money. However, consequence of high inflation, it is important to investigate how time-value of money influences various inventory policies. Since, 1975 a series of related papers appeared that considered the effects of inflation on the inventory system. Before the 1990s, the earlier efforts have been considered simple situations. The first attempt in this field has been reported by Buzacott (1975) that dealt with an Economic Order Quantity (EOQ) model with inflation subject to different types of pricing policies. Misra (1979) developed a discounted-cost model and included internal (company) and external (general economy) inflation rates for various costs associated with an inventory system. Sarker and Pan (1994) surveyed the effects of inflation and the time value of money on order quantity with finite replenishment rate. Some efforts were extended to consider variable demand, such as Uthayakumar and Geetha (2009), Maity (2010), Vrat and Padmanabhan (1990), Datta and Pal (1991), Hariga (1995), Hariga and Ben-Daya (1996) and Chung (2003).\n\nIn above cases, it has been implicitly assumed that the rate of inflation is known with certainty. Yet, inflation enters the inventory picture only because it may have an impact on the future inventory costs and the future rate of inflation is inherently uncertain and unstable. Horowitz (2000) discussed an EOQ model with a normal distribution for the inflation rate.\n\nCertain types of commodities either deteriorate or become obsolete throughout course of time and hence are unstable. For example, the commonly used goods like fruits, vegetables, meat, foodstuffs, perfumes, alcohol, gasoline, radioactive substances, photographic films, electronic components, etc. where deterioration is usually observed during their normal storage period. Inventoried goods can be broadly classified into four meta-categories (Goyal and Giri, 2001):\n\n • Obsolescence refers to items that lose their value through time because of rapid changes of technology or the introduction of a new product by a competitor • Deterioration refers to the damage, spoilage, dryness, vaporization, etc. of the products • Amelioration refers to items whose value or utility or quantity increase with time • No obsolescence/deterioration/amelioration\n\nThere are several studies of deteriorating inventory models under inflationary conditions. Chung and Tsai (2001) presented an inventory model for deteriorating items with the demand of linear trend considering the time-value of money. Wee and Law (2001) derived a deteriorating inventory model under inflationary conditions when the demand rate is a linear decreasing function of the selling price. Chen and Lin (2002) discussed an inventory model for deteriorating items with a normally distributed shelf life, continuous time-varying demand and shortages under an inflationary and time discounting environment. Chang (2004) established a deteriorating EOQ model when the supplier offers a permissible delay to the purchaser if the order quantity is greater than or equal to a predetermined quantity. Yang (2004) discussed the two-warehouse inventory problem for deteriorating items with a constant demand rate and shortages. Moon et al. (2005) considered ameliorating/deteriorating items with a time-varying demand pattern. Maiti et al. (2006) proposed an inventory model with stock-dependent demand rate and two storage facilities under inflation and time value of money where the planning horizon is stochastic in nature and follows the exponential distribution with a known mean. Lo et al. (2007) developed an integrated production-inventory model with assumptions of varying rate of deterioration, partial backordering, inflation, imperfect production processes and multiple deliveries. A Two storage inventory problem with dynamic demand and interval valued lead-time over a finite time horizon under inflation and time-value of money considered by Dey et al. (2008). Maity and Maiti (2008) developed a numerical approach to a multi-objective optimal inventory control problem for deteriorating multi-items under fuzzy inflation and discounting.\n\nMirzazadeh (2010) assumed the inflation is time-dependent and demand rate is assumed to be inflation-proportional. Roy et al. (2009) prepared an inventory model for a deteriorating item with displayed stock dependent demand under fuzzy inflation and time discounting over a random planning horizon. Another research has been performed by Ameli et al. (2011) with considering an economic order quantity model for imperfect items under fuzzy inflationary conditions. Other efforts on inflationary inventory systems for deteriorating items have been made by Hsieh and Dye (2010), Sana (2010), Su et al. (1996), Chen (1998), Wee and Law (1999), Sarker et al. (2000), Yang et al. (2001, 2009), Liao and Chen (2003), Balkhi (2004a, b), Hou and Lin (2006), Shah (2006), Hou (2006), Jaggi et al. (2006), Yang (2006) and Chern et al. (2008).\n\nIn this study, a detailed analysis has been done for surveying the effect of uncertain inflationary conditions on the optimal ordering policy under stochastic inflationary conditions and arbitrary probability density functions (pdfs) for the internal and external inflation rates. Deteriorating items and shortages have been considered. A numerical example and a sensitivity analysis are used to illustrate the model.\n\nThis study concluded that the No. of replenishments and the expected value of cost are sensitive to the external inflation rate and the optimal solution is sensitive to the uncertainty of the inflation rates when the standard deviations of the inflation rates are sufficiently high. Particular cases of the problem, which follow the main problem and correspond to the situation of (1) a single inflation rate for all cost components, (2) no shortages, (3) no deterioration and (4) all the three previous cases together are discussed.\n\nTHE MATHEMATICAL MODEL AND ANALYSIS\n\nThe following assumptions are used throughout this study:\n\n • The internal and external inflation rates are random variables with known distribution • The demand rate is known and constant • Shortages are allowed and fully backlogged except for the final cycle • The replenishment is instantaneous and lead time is zero • The system operates for prescribed time-horizon of length H • A constant fraction of the on-hand inventory deteriorates per unit time\n\nThe cost components may be divided into internal and external classes. Van Hees and Monhemius (1972) have given the breakdown of the various costs of inventory system. In the real world, internal and external costs exhibit different behaviours, so that the internal cost changes by the current inflation rate of the company and the external cost varies with the inflation rate of the general economy (or of the supplier company). Therefore, two different pdfs for the inflation rates can be used in this model. The notations described as follows:\n\n im : Internal (for m = 1) and external (for m = 2) inflation rates f(im) : The pdf of im r : The discount rate D : The demand rate per unit time A : The ordering cost per order at time zero clm : The internal (for m = 1) and external (for m = 2) inventory carrying cost (for l = 1) and shortage cost (for l = 2) per unit per unit time at time zero p : The external purchase cost at time zero θ : The constant deterioration rate Mim(Y) : The moment generating function of im for m = 1 and 2 H : The fixed time horizon\n\nOther notations will be introduced later. It is assumed that the length of the planning horizon is H = nT, where, n is an integer for the number of replenishments to be made during period H and T is an interval of time between replenishment. The unit of time can be considered as a year, a month, a week, etc. and k (0<k≤1) is the proportion of time in any given inventory cycle which orders can be filled from the existing stock. Thus, during the time interval [(j-1)T,jT], the inventory level leads to zero and shortages occur at time (j+k-1)T. Shortages are accumulated until jT before they are backordered and are not allowed in the last replenishment cycle. The optimal inventory policy yields the ordering and shortage points, which minimize the total expected inventory cost over the time horizon.\n\nThe mathematical formulation: Let ECP, ECH, ECS and ECR denote the expected present values of the purchasing, carrying, shortage and replenishment costs, respectively. The detailed analysis of each cost function is given as follows:\n\nExpected present value of the purchasing cost: The expected present value of the purchasing cost for the j-th period (j = 1, 2, …, n-1), as shown in Appendix A, is equal to:", null, "", null, "(1)\n\nand for the last period is given by Appendix A:", null, "(2)\n\nTherefore, the total purchase cost for all cycles can be written as follows:", null, "(3)\n\nExpected present value of the inventory cost: The inventory carrying cost is divided into internal (for m = 1) and external (for m = 2) classes. From Appendix B the expected present value of the inventory carrying cost for the j-th cycle for the m-th class can be written as:", null, "(4)\n\nIn the last period the inventory level comes to zero at the end of period. Therefore:", null, "(5)\n\nThe total internal and external carrying costs for all cycles can be given as follows:", null, "(6)\n\nExpected present value of the shortages cost: The shortages cost may be divided to internal and external classes similar to the holding cost. The expected present value of the shortages cost for the j-th cycle for the m-th class can be formulated as follows (Appendix C):", null, "(7)\n\nThe total shortages cost during the entire planning horizon H can be written as follows:", null, "(8)\n\nExpected present value of the ordering cost: The expected present value of the ordering cost for replenishment at time (j-1)T for the j-th cycle is:", null, "(9)\n\nThe total replenishment cost can be given as follows:", null, "(10)\n\nHence, the total expected inventory cost of the system during the entire planning horizon H is given by:", null, "(11)\n\nTHE SOLUTION PROCEDURE\n\nThe problem is determining n and k to lead the minimum of the total expected inventory system cost. For a given value of n, the necessary condition of optimality is as follows:", null, "(12)\n\nThe iterative methods such as Newton method can be used for calculating k. By increasing n, the objective function decreases to lead to minimum. The second-order condition for a minimum is:", null, "(13)\n\nNUMERICAL EXAMPLE\n\nAccording to the results, the following example is providing. Let r = \\$0.2/\\$/year, D = 1000 units/year, A = \\$100/order, c11 = \\$0.2/unit/year, c12 = \\$0.4/unit/year, c21 = \\$0.8/unit/year, c22 = \\$0.6/unit/year, p = \\$5/unit, H = 10 years, θ = 0.01. The internal and external inflation rates have the normal distribution function with means of μ1 = 0.08 and μ2 = 0.14, standard deviations of σ1 = 0.04 and σ2 = 0.06, respectively. The results are shown in Table 1. The minimum expected cost over the planning horizon is 44 537.26 for n* = 41 and k* = 0.664623. Optimal interval of time between replenishment, T, equals to H/n = 0.244 year.\n\nSENSITIVITY ANALYSIS\n\nTo study the effects of system parameters changes D, H, θ, r, μ1, μ2, σ1, σ2, A, p, c11, c12, c21 and c22 on the optimal cost, the replenishment time and k* which is derived by the proposed method, a sensitivity analysis was performed. This fact is done by increasing the parameters by 20, 50, 100% and decreasing the parameters to 20, 50, 90%, taking each one at a time and keeping the remaining parameters at their original values. The following conclusion can be derived from the sensitivity analysis based on Table 2:\n\n • As the mean of the internal inflation rate increases, the number of replenishments (n) decreases and k increases. By increasing the mean of external inflation rate, increase the number of replenishments (n) and k. The optimal expected present value of cost (ETVC) increases when μ1 and μ2 increase but is highly sensitive to μ2. Induction of this result is the purchase cost increasing by external inflation rate is more than other cost components, i.e., purchase cost constitutes considerable portion of total cost. Hence, the total cost has a similar role as the purchase cost • Table 2 shows that the optimal value of k and number of replenishments (n) are insensitive to changes in the standard deviations of inflation rates. But, the operating doctrine is highly sensitive, to high values of σ1 and σ2. For example, considering σ1 = 0.04 and σ2 = 0.4, Table 3 shows that the optimal number of replenishments equals 1 and we have no shortages (n* = k* = 1). The numerical example has solved by considering σ1 = 0.04, 0.1, 0.2, 0.3, 0.4, 0.5 and σ2 = 0.06, 0.1, 0.2, 0.3, 0.4 and 0.5. The results are shown in Table 3. The derived model is sufficiently sensitive if the uncertainty of inflation rates is higher than a certain magnitude. In that situation, larger order quantity should be placed • High uncertainty inflation rate is occurred in the real world, for instants, international financial Statistical Yearbook, 2004, shows that the mean change of wholesale prices indices in Russia in 1993-2001 was 1.91 and it’s the standard deviation was 3.05. For another case, in Norway in 1997-2001 the mean change of wholesale prices indices were 0.08 and the standard deviation was 0.2 • The number of replenishments (n) is highly sensitive to the change of the parameters D, A and H, is slightly sensitive to changes in c12 and insensitive to changes in r, θ, p, c11, c21 and c22 • The optimal value of k is highly sensitive to the change of the parameters c12, c21 and c22, is moderately sensitive to r and c11 and is insensitive to D, θ, H, p and A • The total expected inventory cost of the system is highly sensitive to the changes in the parameters D, r, H and p and insensitive to θ, A, c11, c12, c21 and c22\n\n Table 1: Optimal solution for numerical example", null, "Table 2: Effects of changes in model parameters on n, k and optimal expected system cost", null, "Table 3: Effects of changes in the standard deviations of inflation rates on n, k and optimal expected system cost", null, "SOME PARTICULAR CASES\n\nHere, an attempt has been made to study some important special cases of the model.\n\nCase (I): If the internal and external inflation rates have the same pdf, the expected present value of the total cost ETVC (n,k) can be obtained by deleting:", null, "in Eq. 11 and substituting:", null, "(14)\n\nThe previous numerical example assumes that the inflation rate has the normal distribution function with the mean of μ = 0.11 and the standard deviation of σ = 0.05. The optimal solution in this case is as follows: n* = 38, k* = 0.666099, ETVC(n,k) = 39 296.36 and T* = 0.263 year. The number of replenishments and inventory system cost decrease and k increases.\n\nCase (II): If shortages are not allowed, k = 1 can be substituted in expression (11) and the expected present worth of the total variable cost ETVC(n) can be obtained. The minimum solution of ETVC(n) for the discrete variable of n must satisfy the following equation:", null, "(15)\n\nwhere, ETVC(n) = ETVC(n)-ETVC(n-1). In the numerical example, using the above inequality, the following solution is obtained: n* = 50, ETVC(n) = 45 613.73 and T* = 0.2 year. It shows that n and ETVC increase in the without shortages case.\n\nCase (III): If available inventory has no deterioration (θ = 0) over time the cost function, after modeling, may be rewritten as follows:", null, "(16)\n\nThe cost function can be minimized by the methods indicated in this study. For a given value of n, the necessary condition of optimality is:", null, "(17)\n\nand the sufficient condition is the second derivative is positive. In this case, the numerical result is obtained as follows: n* = 41, k* = 0.667940, ETVC(n,k) = 44 513.44 and T* = 0.244 year. Thus ETVC is decreased, k is increased and n is not changed in comparison to the main model.\n\nCase (IV): Now assume that the internal and external inflation rates have the same pdf, no shortages allowed and θ = 0. This may be solved by using Eq. 16 and 17, substituting k = 1 and considering (14). Therefore, the optimal solution is as follows: n* = 46, ETVC(n,k) = 40 391.48 and T* = 0.217.\n\nCONCLUSIONS\n\nUsually, in the inventory systems under inflationary conditions, it has been assumed that the inflation rates are constant over the planning horizon. But, many economic, political, social and cultural variables may also affect the future changes in the inflation rate. Therefore, assuming constant inflation rates is not valid, especially when the time horizon is more than two years. The current study considers stochastic inflation rates where any distribution function is applicable.\n\nThis model incorporates some realistic features that are likely to be associated with the inventory of certain types of goods. First, deterioration over time is a natural feature for goods. Second, occurrence of inventory shortages is a natural real situation phenomenon. In the solution process of the problem, the subject of moment generating function has been used. The numerical examples have been given and sensitivity analysis has been conducted to illustrate the theoretical results. The results of sensitivity analysis indicate that which of the parameters are more sensitive to the optimal solution. Also, the results indicate the importance of taking into account stochastic inflation, especially when there is considerable uncertainty associated with the inflation rates. Finally, four special cases have been discussed: identical inflation rates, no shortages situation, no deterioration and considering all the three cases simultaneously. These cases are compared with the main model through the numerical example.\n\nAPPENDIX A\n\nDuring any given period, the order quantity is consisting of both demand and deterioration for the relevant period excluding shortage part of the period and the amount required to satisfy the demand during the shortage period in the preceding time interval. For the j-th cycle (j = 1, 2, …, n-1) the expected present value of the purchase cost can be formulated as follows:", null, "(A1)\n\nThe above equation can be rewritten as Eq. 1. In the last period shortages are not allowable, therefore the expected present value of the purchase cost is:", null, "(A2)\n\nwhere R2 = r-i2. It can similarly be rewritten as Eq. 2.\n\nAPPENDIX B\n\nThe expected present value of the inventory carrying cost for the j-th cycle (j = 1, 2, …, n-1) for the m-th class (m = 1,2) is:", null, "(B1)\n\nwhere, Rm = r- im and Eq. B 1 can be rewritten as Eq. 4. In the last period for the m-th class (m = 1,2) from similar machinations we have:", null, "(B2)\n\nAPPENDIX C\n\nThe expected present value of the shortages cost for the j-th cycle (j = 1, 2, …, n-1) for the m-th class (m = 1,2) can be computed as:", null, "(C1)\n\nwhere, Rm = r- im . It can be rewritten as Eq. 7.\n\nREFERENCES\n1: Ameli, M., A. Mirzazadeh and M.A. Shirazi, 2011. Economic order quantity model with imperfect items under fuzzy inflationary conditions. Trends Applied Sci. Res., 6: 294-303.\nCrossRef \| Direct Link \|\n\n2: Balkhi, Z.T., 2004. An optimal solution of a general lot size inventory model with deteriorated and imperfect products, taking into account inflation and time value of money. Int. J. Syst. Sci., 35: 87-96.\nCrossRef \|\n\n3: Balkhi, Z.T., 2004. On the optimality of inventory models with deteriorating items for demand and on-hand inventory dependent production rate. IMA J. Manage. Math., 15: 67-86.\nCrossRef \|\n\n4: Buzacott, J.A., 1975. Economic order quantities with inflation. Operat. Res. Q., 26: 553-558.\nDirect Link \|\n\n5: Chang, C.T., 2004. An EOQ model with deteriorating items under inflation when supplier credits linked to order quantity. Int. J. Prod. Econ., 88: 307-316.\nCrossRef \|\n\n6: Chen, J.M., 1998. An inventory model for deteriorating items with time-proportional demand and shortages under inflation and time discounting. Int. J. Prod. Econ., 55: 21-30.\nCrossRef \|\n\n7: Chen, J.M. and C.S. Lin, 2002. An optimal replenishment model for inventory items with normally distributed deterioration. Produc. Plan. Control, 13: 470-480.\nDirect Link \|\n\n8: Chern, M.S., H.L. Yang, J.T. Teng and S. Papachristos, 2008. Partial backlogging inventory lot-size models for deteriorating items with fluctuating demand under inflation. Eur. J. Operat. Res., 191: 127-141.\nDirect Link \|\n\n9: Chung, K.J., 2003. An algorithm for an inventory model with inventory-level-dependent demand rate. Comput. Operat. Res., 30: 1311-1317.\nDirect Link \|\n\n10: Chung, K.J. and S.F. Tsai, 2001. Inventory systems for deteriorating items with shortage and a linear trend in demand-taking account of time value. Comput. Operat. Res., 28: 915-934.\nCrossRef \|\n\n11: Datta, T.K. and A.K. Pal, 1991. Effects of inflation and time-value of money on an inventory model with linear time-dependent demand rate and shortages. Eur. J. Operat. Res., 52: 326-333.\nCrossRef \|\n\n12: Dey, J.K., S.K. Mondal and M. Maiti, 2008. Two storage inventory problem with dynamic demand and interval valued lead-time over finite time horizon under inflation and time-value of money. Eur. J. Operat. Res., 185: 170-194.\nCrossRef \|\n\n13: Goyal, S.K. and B.C. Giri, 2001. Recent trends in modeling of deteriorating inventory. Eur. J. Operat. Res., 134: 1-16.\nDirect Link \|\n\n14: Hariga, M.A., 1995. Effects of inflation and time-value of money on an inventory model with time-dependent demand rate and shortages. Eur. J. Operat. Res., 81: 512-520.\nCrossRef \|\n\n15: Hariga, M. and M. Ben-Daya, 1996. Optimal time-varying lot-sizing models under inflationary conditions. Eur. J. Operat. Res., 89: 313-325.\nCrossRef \|\n\n16: Horowitz, I., 2000. EOQ and inflation uncertainty. Int. J. Prod. Econ., 65: 217-224.\nCrossRef \|\n\n17: Hou, K.L., 2006. An inventory model for deteriorating items with stock-dependent consumption rate and shortages under inflation and time discounting. Eur. J. Operat. Res., 168: 463-474.\nCrossRef \|\n\n18: Hou, K.L. and L.C. Lin, 2006. An EOQ model for deteriorating items with price- and stock-dependent selling rates under inflation and time value of money. Int. J. Syst. Sci., 37: 1131-1139.\nDirect Link \|\n\n19: Hsieh, T.P. and C.Y. Dye, 2010. Pricing and lot-sizing policies for deteriorating items with partial backlogging under inflation. Expert Syst. Appl. Int. J., 37: 7234-7242.\nDirect Link \|\n\n20: Jaggi, C.K., K.K. Aggarwal and S.K. Goel, 2006. Optimal order policy for deteriorating items with inflation induced demand. Int. J. Prod. Econ., 103: 707-714.\nCrossRef \|\n\n21: Liao, H.C. and Y.K. Chen, 2003. Optimal payment time for retailer's inventory system. Inter. J. Syst. Sci., 34: 245-253.\nCrossRef \|\n\n22: Lo, S.T., H.M. Wee and W.C. Huang, 2007. An integrated production-inventory model with imperfect production processes and Weibull distribution deterioration. Int. J. Prod. Econ., 106: 248-260.\nCrossRef \|\n\n23: Maiti, A.K., M.K. Maiti and M. Maiti, 2006. Two storage inventory model with random planning horizon. Applied Math. Comput., 183: 1084-1097.\nCrossRef \|\n\n24: Maity, A.K., 2010. One machine multiple-product problem with production-inventory system under fuzzy inequality constraint. Applied Soft Comput., 10.1016/j.asoc.2009.12.029\n\n25: Maity, K. and M. Maiti, 2008. A numerical approach to a multi-objective optimal inventory control problem for deteriorating multi-items under fuzzy inflation and discounting. Comput. Math. Appl., 55: 1794-1807.\nCrossRef \|\n\n26: Mirzazadeh, A., 2010. Effects of variable inflationary conditions on an inventory model with inflation-proportional demand rate. J. Applied Sci., 10: 551-557.\nCrossRef \| Direct Link \|\n\n27: Misra, R.B., 1979. A note on optimal inventory management under inflation. Naval Res. Logist., 26: 161-165.\nCrossRef \|\n\n28: Moon, I., B.C. Giri and B. Ko, 2005. Economic order quantity models for ameliorating/deteriorating items under inflation and time discounting. Eur. J. Operat. Res., 162: 773-785.\n\n29: Roy, A., M.K. Maiti, S. Kar and M. Maiti, 2009. An inventory model for a deteriorating item with displayed stock dependent demand under fuzzy inflation and time discounting over a random planning horizon. Applied Math. Modell., 33: 744-759.\nCrossRef \|\n\n30: Sana, S.S., 2010. Demand influenced by enterprises initiatives-A multi-item EOQ model of deteriorating and ameliorating items. Math. Comput. Modell., 52: 284-302.\nCrossRef \|\n\n31: Sarker, B.R., A.M.M. Jamal and S. Wang, 2000. Supply chain models for perishable product under inflation and permissible delay in payments. Comput. Operat. Res., 27: 59-75.\nCrossRef \|\n\n32: Sarker, B.R. and H. Pan, 1994. Effects of inflation and the time value of money on order quantity and allowable shortage. Int. J. Prod. Econ., 34: 65-72.\nCrossRef \|\n\n33: Shah, N.H., 2006. Inventory model for deteriorating items and time value of money for a finite time horizon under the permissible delay in payments. Int. J. Syst. Sci., 37: 9-15.\nDirect Link \|\n\n34: Su, C.T., L.I. Tong and H.C. Liao, 1996. An inventory model under inflation for stock dependent demand rate and exponential decay. Operat. Res., 33: 72-82.\n\n35: Vrat, P. and G. Padmanabhan, 1990. An inventory model under inflation for stock-dependent consumption rate items. Eng. Costs Prod. Econ., 19: 379-383.\nCrossRef \|\n\n36: Wee, H.M. and S.T. Law, 1999. Economic production lot size for deteriorating items taking account of the time-value of money. Comput. Operat. Res., 26: 545-558.\nDirect Link \|\n\n37: Wee, H.M. and S.T. Law, 2001. Replenishment and pricing policy for deteriorating items taking into account the time-value of money. Int. J. Prod. Econ., 71: 213-220.\nCrossRef \|\n\n38: Yang, H.L., 2004. Two-warehouse inventory models for deteriorating items with shortages under inflation. Eur. J. Operat. Res., 157: 344-356.\nCrossRef \|\n\n39: Yang, H.L., 2006. Two-warehouse partial backlogging inventory models for deteriorating items under inflation. Int. J. Prod. Econ., 103: 362-370.\nCrossRef \|\n\n40: Yang, H.L., J.T. Teng and M.S. Chern, 2001. Deterministic inventory lot-size models under inflation with shortages and deterioration for fluctuating demand. Naval Res. Logistics, 48: 144-158.\nDirect Link \|\n\n41: Yang, H.L., J.T. Teng and M.S. Chern, 2009. An inventory model under inflation for deteriorating items with stock-dependent consumption rate and partial backlogging shortages. Int. J. Prod. Econ., 123: 8-19.\nCrossRef \| Direct Link \|\n\n42: Uthayakumar, R. and K.V. Geetha, 2009. Replenishment policy for single item inventory model with money inflation. Opsearch, 46: 345-357.\nDirect Link \|\n\n43: Van Hees, R.N. and W. Monhemius, 1972. Production and Inventory Control: Theory and Practice. Barnes and Noble, New York.", null, "© 2021 Science Alert. 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https://cran.opencpu.org/web/packages/distributional/news/news.html	[ "distributional 0.3.0\n\nNew features\n\nProbability distributions\n\n• Added dist_categorical() for the Categorical distribution.\n• Added dist_lognormal() for the log-normal distribution. Mathematical conversion shortcuts have also been added, so exp(dist_normal()) produces dist_lognormal().\n\nGenerics\n\n• Added parameters() generic for obtaining the distribution’s parameters.\n• Added family(<distribution>) for getting the distribution’s family name.\n• Added covariance() to return the covariance of a distribution.\n• Added support() to identify the distribution’s region of support (#8).\n• Added log_likelihood() for computing the log-likelihood of observing a sample from a distribution.\n\nImprovements\n\n• variance() now always returns a variance. It will not default to providing a covariance matrix for matrices. This also applies to multivariate distributions such as dist_multivariate_normal(). The covariance can now be obtained using the covariance() function.\n• dist_wrap() can now search for distribution functions in any environment, not just packages. If the package argument is NULL, it will search the calling environment for the functions. You can also provide a package name as before, and additionally an arbitrary environment to this argument.\n• median() methods will now ignore the na.rm option when it does not apply to that distribution type (#72).\n• dist_sample() now allows for missing values to be stored. Note that density(), quantile() and cdf() will remove these missing values by default. This behaviour can be changed with the na.rm argument.\n• <hilo> objects now support non-numeric and multivariate distributions. <hilo> vectors that have different bound types cannot be mixed (#74).\n• Improved performance of default methods of mean() and variance(), which no longer use sampling based means and variances for univariate continuous distributions (#71, @mjskay)\n• dist_binomial() distributions now return integers for quantile() and generate() methods.\n• Added conditional examples for distributions using functions from supported packages.\n\nBug fixes\n\n• Fixed fallback format() function for distributions classes that have not defined this method (#67).\n\nBreaking changes\n\n• variance() on a dist_multivariate_normal() will now return the diagonal instead of the complete variance-covariance matrix.\n• dist_bernoulli() will now return logical values for quantile() and generate().\n\ndistributional 0.2.2\n\nNew features\n\n• Added is_distribution() to identify if an object is a distribution.\n\nImprovements\n\n• Improved NA structure of distributions, allowing it to work with is.na() and vctrs vector resizing / filling functionality.\n• Added as.character(<hilo>) method, allowing datasets containing hilo() objects to be saved as a text file (#57).\n\nBug fixes\n\n• Fixed issue with hdr() range size incorrectly being treated as 100-size, giving 5% ranges for 95% sizes and vice-versa (#61).\n\ndistributional 0.2.1\n\nA small performance and methods release. Some issues with truncated distributions have been fixed, and some more distribution methods have been added which improve performance of common tasks.\n\nNew features\n\nProbability distributions\n\n• Added dist_missing() for representing unknown or missing (NA) distributions.\n\nImprovements\n\n• Documentation improvements.\n• Added cdf() method for dist_sample() which uses the emperical cdf.\n• dist_mixture() now preserves dimnames() if all distributions have the same dimnames().\n• Added density() and generate() methods for sample distributions.\n• Added skewness() method for dist_sample().\n• Improved performance for truncated Normal and sample distributions (#49).\n• Improved vectorisation of distribution methods.\n\nBug fixes\n\n• Fixed issue with computing the median of dist_truncated() distributions.\n• Fixed format method for dist_truncated() distributions with no upper or lower limit.\n• Fixed issue with naming objects giving an invalid structure. It now gives an informative error (#23).\n• Fixed documentation for Negative Binomial distribution (#46).\n\ndistributional 0.2.0\n\nNew features\n\nProbability distributions\n\n• Added dist_wrap() for wrapping distributions not yet added in the package.\n\nMethods\n\n• Added likelihood() for computing the likelihood of observing a sample from a distribution.\n• Added skewness() for computing the skewness of a distribution.\n• Added kurtosis() for computing the kurtosis of a distribution.\n• The density(), cdf() and quantile() methods now accept a log argument which will use/return probabilities as log probabilities.\n\nImprovements\n\n• Improved documentation for most distributions to include equations for the region of support, summary statistics, density functions and moments. This is the work of @alexpghayes in the distributions3 package.\n• Documentation improvements\n• Added support for displaying distributions with View().\n• hilo() intervals can no longer be added to other intervals, as this is a common mistake when aggregating forecasts.\n• Incremented d for numDeriv::hessian() when computing mean and variance of transformed distributions.\n\nDeprecated features\n\n• Graphics functionality provided by autoplot.distribution() is now deprecated in favour of using the ggdist package. The ggdist package allows distributions produced by distributional to be used directly with ggplot2 as aesthetics.\n\ndistributional 0.1.0\n\nFirst release.\n\nNew features\n\nObject classes\n\n• distribution: Distributions are represented in a vectorised format using the vctrs package. This makes distributions suitable for inclusion in model prediction output. A distribution is a container for distribution-specific S3 classes.\n• hilo: Intervals are also stored in a vector. A hilo consists of a lower bound, upper bound, and confidence level. Each numerical element can be extracted using \\$, for example my_hilo\\$lower to obtain the lower bounds.\n• hdr: Highest density regions are currently stored as lists of hilo values. This is an experimental feature, and is likely to be expanded upon in an upcoming release.\n\nGeneric functions\n\nValues of interest can be computed from the distribution using generic functions. The first release provides 9 functions for interacting with distributions:\n\n• density(): The probability density/mass function (equivalent to d...()).\n• cdf(): The cumulative distribution function (equivalent to p...()).\n• generate(): Random generation from the distribution (equivalent to r...()).\n• quantile(): Compute quantiles of the distribution (equivalent to q...()).\n• hilo(): Compute probability intervals of probability distribution(s).\n• hdr(): Compute highest density regions of probability distribution(s).\n• mean(): Obtain the mean(s) of probability distribution(s).\n• median(): Obtain the median(s) of probability distribution(s).\n• variance(): Obtain the variance(s) of probability distribution(s).\n\nGraphics\n\n• Added an autoplot() method for visualising the probability density function ([density()]) or cumulative distribution function ([cdf()]) of one or more distribution.\n• Added geom_hilo_ribbon() and geom_hilo_linerange() geometries for ggplot2. These geoms allow uncertainty to be shown graphically with hilo() intervals.\n\nProbability distributions\n\n• Added 20 continuous probability distributions: dist_beta(), dist_burr(), dist_cauchy(), dist_chisq(), dist_exponential(), dist_f(), dist_gamma(), dist_gumbel(), dist_hypergeometric(), dist_inverse_exponential(), dist_inverse_gamma(), dist_inverse_gaussian(), dist_logistic(), dist_multivariate_normal(), dist_normal(), dist_pareto(), dist_student_t(), dist_studentized_range(), dist_uniform(), dist_weibull()\n• Added 8 discrete probability distributions: dist_bernoulli(), dist_binomial(), dist_geometric(), dist_logarithmic(), dist_multinomial(), dist_negative_binomial(), dist_poisson(), dist_poisson_inverse_gaussian()\n• Added 3 miscellaneous probability distributions: dist_degenerate(), dist_percentile(), dist_sample()\n\nDistribution modifiers\n\n• Added dist_inflated() which inflates a specific value of a distribution by a given probability. This can be used to produce zero-inflated distributions.\n• Added dist_transformed() for transforming distributions. This can be used to produce log distributions such as logNormal: dist_transformed(dist_normal(), transform = exp, inverse = log)\n• Added dist_mixture() for producing weighted mixtures of distributions.\n• Added dist_truncated() to impose boundaries on a distribution’s domain via truncation." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7277789,"math_prob":0.9293981,"size":6600,"snap":"2022-05-2022-21","text_gpt3_token_len":1377,"char_repetition_ratio":0.18814433,"word_repetition_ratio":0.009569378,"special_character_ratio":0.21257576,"punctuation_ratio":0.1521099,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99134743,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T22:50:02Z\",\"WARC-Record-ID\":\"<urn:uuid:9f5c9291-f7b2-4059-9b70-90cd493aeac0>\",\"Content-Length\":\"13568\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e2539ca-8f24-4c28-b969-ce6c714f1628>\",\"WARC-Concurrent-To\":\"<urn:uuid:e42febce-079f-4ed7-a964-9d61cb5a6445>\",\"WARC-IP-Address\":\"104.21.78.204\",\"WARC-Target-URI\":\"https://cran.opencpu.org/web/packages/distributional/news/news.html\",\"WARC-Payload-Digest\":\"sha1:2NDYIOPNZQDXGASULSHYI7EZDXD2MXRV\",\"WARC-Block-Digest\":\"sha1:4U6PDJ5VZBX5CS32N7JXOX73S4HR334V\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303917.24_warc_CC-MAIN-20220122224904-20220123014904-00224.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/1408.0184/	[ "# Double Higgs production at LHC, see-saw type II and Georgi-Machacek model\n\nS.I. Godunov Institute for Theoretical and Experimental Physics, Moscow, 117218, Russia Novosibirsk State University, Novosibirsk, 630090, Russia M.I. Vysotsky Institute for Theoretical and Experimental Physics, Moscow, 117218, Russia Moscow Institute of Physics and Technology, 141700, Dolgoprudny, Moscow Region, Russia Moscow Engineering Physics Institute, 115409, Moscow, Russia E.V. Zhemchugov Institute for Theoretical and Experimental Physics, Moscow, 117218, Russia\n###### Abstract\n\nThe double Higgs production in the models with isospin-triplet scalars is studied. It is shown that in the see-saw type II model the mode with an intermediate heavy scalar, , may have the cross section which is compatible with that in the Standard Model. In the Georgi-Machacek model this cross section could be much larger than in SM since the vacuum expectation value of the triplet can be large.\n\nThis paper is our present to Valery Anatolievich Rubakov on his anniversary. Many students (and not only students) in the world are studying Physics reading his excellent books, papers and listening his brilliant lectures.\n\n## I Introduction\n\nAfter the discovery of the Higgs-BE boson at LHC Higgs2012 the next steps to check the Standard Model (SM) are: the measurement of the coupling constants of the Higgs boson with other SM particles () with better accuracy and the measurement of the Higgs self-coupling which determines the shape of the Higgs potential. In the SM the triple and quartic Higgs couplings are predicted in terms of the known Higgs mass and vacuum expectation value. Deviations from these predictions would mean the existence of New Physics in the Higgs potential. The triple Higgs coupling can be measured at LHC in double Higgs production, in which the gluon fusion dominates: . However, the production cross section is very small. According to deFlorian2014 at the cross section with accuracy. For the final states with the reasonable signal/background ratios (such as ) only at HL-LHC with integrated luminosity double Higgs production will be found and triple Higgs coupling will be measured Baglio2014 111The decays into and final states can be even more promising for the measurement of triple Higgs coupling Englert2012 ; Zurita2012 .. We are looking for the extensions of the SM Higgs sector in which the double Higgs production is enhanced.\n\nOne of the well-motivated examples of non-minimal Higgs sector is provided by the see-saw type II mechanism of the neutrino mass generation Schechter1980 . In this mechanism a scalar isotriplet with hypercharge () is added to the SM. The vacuum expectation value (vev) of the neutral component generates Majorana masses of the left-handed neutrinos. There are two neutral scalar bosons in the model: the light one in which the doublet Higgs component dominates and which should be identified with the particle discovered at LHC (), and the heavy one in which the triplet Higgs component dominates (). The neutrino masses equal , where () originates from Yukawa couplings of Higgs triplet with the lepton doublets. If neutrinos are light due to a small value of while are of the order of one, then decays into the neutrino pairs. Three states (or ), , and are almost degenerate in the model considered in Sect. II and the absence of the same-sign dileptons at LHC from decays provides the lower bound HHiggs2012 . We are interested in the opposite case: reaches the maximum allowed value while neutrinos are light because of small values of . In this case can be the dominant decay mode of a heavy neutral Higgs. In this way we get an additional mechanism of the double production at LHC.\n\nThe bound HHiggs2012 cannot be applied now since mainly decays into the same-sign diboson Yagyu1 . We only need to be heavy enough for decay to occur. This case is analyzed in Sect. II. The invariant mass of additionally produced state peak at which is a distinctive feature of the proposed mechanism, see also Englert2012_2 ; Barger2014 .\n\ncontains a small admixture of the isodoublet state which makes gluon fusion a dominant mechanism of production at LHC. The admixture of the isodoublet component in equals approximately , where is the vacuum expectation value of the neutral component of isodoublet, and in Sect. II for and we will get . Taking into account that is about , we obtain enhancement of double Higgs production in comparison with SM.\n\nSince the nonzero value of violates the well checked equality of the strength of charged and neutral currents at tree level,\n\n g2/M2W¯g2/M2Z=1+2v2Δv2, (1)\n\nshould be less than (see Sect. II). The numerical estimate of cross section was made for maximum allowed value when the isodoublet admixture is about .\n\nThe bound is removed in the Georgi-Machacek model Georgi , in which in addition to a scalar isotriplet with is introduced. If the vev of the neutral component of this additional field equals then we get just one in the r.h.s. of (1): correction proportional to is cancelled. Thus can be much larger than . The bounds on come from the measurement of the Higgs boson couplings to vector bosons and fermions, which would deviate from their SM values: .\n\nThe consideration of an enhancement of production in GM variant of see-saw type II model is presented in Sect. III. Since at the moment the accuracy of the measurement of values in production and decay is poor, as large as is allowed and can reach value which makes it accessible with the integrated luminosity prior to HL-LHC run. We summarize our results in Conclusions.\n\n## Ii Double h production in H decays at LHC\n\n### ii.1 Scalar sector of the see-saw type II model\n\nIn this subsection we will present the necessary formulas; for a detailed description see Accomando . In addition to the SM isodoublet field ,\n\n Φ≡[Φ+Φ0]≡[Φ+1√2(v+φ+iχ)], (2)\n\nin see-saw type II an isotriplet is introduced:\n\n ≡ [δ+/√2δ++δ0−δ+/√2], (3) δ0=1√2(vΔ+δ+iη).\n\nHere are the Pauli matrices.\n\nThe scalar sector kinetic terms are\n\n Lkinetic=∣∣DμΦ∣∣2+Tr[(DμΔ)†(DμΔ)], (4)\n\nwhere\n\n DμΦ = ∂μΦ−ig2AaμσaΦ−ig′2BμΦ, (5) DμΔ = [∂μΔa+gεabcAbμΔc−ig′BμΔa]σa√2= (6) = ∂μΔ−ig2[Aaμσa,Δ]−ig′BμΔ.\n\nHypercharge was substituted for isodoublet and for isotriplet. The terms quadratic in vector boson fields are the following:\n\n LV2=g2∣∣δ0∣∣2W+W−+12g2∣∣Φ0∣∣2W+W−+¯g2∣∣δ0∣∣2Z2+14¯g2∣∣Φ0∣∣2Z2. (7)\n\nVector boson masses are\n\n (8)\n\nFor the ratio of vector boson masses neglecting the radiative corrections from isotriplet (not a bad approximation as far as the heavy triplet decouples) we get:\n\n MWMZ≈(MWMZ)SM(1−v2Δv2). (9)\n\nComparing the result of SM fit (PDG, , p.145), , with the experimental value, , at level we get the following upper bound:\n\n vΔ<5 GeV, (10)\n\nand since the cross sections we are interested in are proportional to we will use an upper bound for numerical estimates in this section.\n\nFrom the numerical value of Fermi coupling constant in muon decay we obtain:\n\n v2+2v2Δ=(246 GeV)2, (11)\n\nso for the value can be safely used in deriving (10).\n\nThe scalar potential looks like:\n\n V(Φ,Δ) = −12m2Φ(Φ†Φ)+λ2(Φ†Φ)2+ (12) + M2ΔTr[Δ†Δ]+μ√2(ΦTiσ2Δ†Φ+h.c.),\n\nwhich is a truncated version of the most general renormalizable potential (see for example Dev , eq. (2.6)). We may simply suppose that the coupling constants which multiply the omitted terms in the potential (, and ) are small. In the case of SM only the first line in (12) remains; mass of the Higgs boson equals while its expectation value , .\n\nSince at the minimum of (12) the following equations are valid:\n\n ⎧⎪⎨⎪⎩12m2Φ=12λv2−μvΔ,M2Δ=12μv2vΔ, (13)\n\nfor vev’s of isodoublet and isotriplet we obtain:\n\n v2 = m2ΦM2ΔλM2Δ−μ2, (14) vΔ = μm2Φ2λM2Δ−2μ2=μ2v2M2Δ. (15)\n\nQuadratic in terms according to (12) are\n\n V(φ,δ)=12m2Φφ2+12M2Δδ2−μvφδ. (16)\n\nHere and below the terms suppressed as are omitted.\n\nDenoting the states with the definite masses as and we obtain:\n\n [φδ]=[cosα−sinαsinαcosα][hH], tan2α=2μvM2Δ−m2Φ, (17)\n M2h = 12(m2Φ+M2Δ−√(M2Δ−m2Φ)2+4μ2v2)≈m2Φ, (18) M2H = 12(m2Φ+M2Δ+√(M2Δ−m2Φ)2+4μ2v2)≈M2Δ. (19)\n\nSince , mass eigenstate consists mostly of and consists mostly of . We suppose that the particle observed by ATLAS and CMS is , so is about .\n\nThe scalar sector of the model in addition to the massless goldstone bosons, which are eaten up by the vector gauge bosons, contains one double charged field , one single charged field , and three real neutral fields , and . is mostly with small admixture, is mostly with small admixture. All these particles except are heavy; their masses equal with small corrections proportional to .\n\n### ii.2 H decays\n\nThe second and fourth terms in potential (12) contribute to decays:\n\n λ2(Φ†Φ)2 → λv2φ3, (20) μ√2(ΦTiσ2Δ†Φ+h.c.) → −μ2δ(φ2−χ2), (21)\n\nwhere in the second line is dominantly a goldstone state which forms the longitudinal polarization.\n\nWith the help of (17) we obtain the expression for the effective lagrangian which describes decay:\n\n LHhh=μ2⎡⎢ ⎢ ⎢⎣1+3(MHMh)2−1⎤⎥ ⎥ ⎥⎦Hh2=vΔM2Hv2⎡⎢ ⎢ ⎢⎣1+3(MHMh)2−1⎤⎥ ⎥ ⎥⎦Hh2. (22)\n\nIn the see-saw type II model neutrino masses are generated by the Yukawa couplings of isotriplet with lepton doublets. These couplings generate decays as well. As it was noted in Yagyu1 for diboson decays dominate. It happens because the amplitude of diboson decay is proportional to , while Yukawa couplings are inversely proportional to it, . That is why for leptonic decays are completely negligible.\n\nThe amplitudes of and decays are contained in (7):\n\n LHVV = g2(vΔcosα−12vsinα)W+W−H+¯g2(vΔcosα−14vsinα)Z2H (23) ≈ −g2M2h/M2H1−M2h/M2HvΔW+W−H+¯g221−2M2h/M2H1−M2h/M2HvΔZ2H,\n\nand we see that decay is suppressed (see, for example, Perez ).\n\ndecay occur through admixture:\n\n LHt¯t=sinαmtvt¯tH=2vΔ/v1−M2h/M2Hmtvt¯tH, (24)\n\nas well as decay into two gluons:\n\n LHgg=αs12πsinαG2μν. (25)\n\nLet us note that all the amplitudes of decays are proportional to triplet vev .\n\nFor the decay probabilities we obtain:\n\n ΓH→hh = v2Δv4M3H8π⎡⎢ ⎢ ⎢⎣1+2(MhMH)21−(MhMH)2⎤⎥ ⎥ ⎥⎦2 ⎷1−4M2hM2H, (26) ΓH→ZZ = v2Δv4M3H8π⎡⎢ ⎢ ⎢⎣1−2(MhMH)21−(MhMH)2⎤⎥ ⎥ ⎥⎦2(1−4M2ZM2H+12M4ZM4H) ⎷1−4M2ZM2H, (27) ΓH→WW = v2Δv4M3H4π⎡⎢ ⎢ ⎢⎣M2h/M2H1−(MhMH)2⎤⎥ ⎥ ⎥⎦2(1−4M2WM2H+12M4WM4H) ⎷1−4M2WM2H, (28) ΓH→t¯t = v2Δv4Ncm2tMH2π1(1−M2h/M2H)2(1−4m2tM2H)3/2, (29)\n\nwhere is the number of colors. Finally for the width of decay into two gluon jets we obtain:\n\n ΓH→gg=v2Δv4M3H2π(αs3π)2(1−M2hM2H)−2, (30)\n\nand it is always negligible.\n\nIn what follows we suppose that and the decay is forbidden kinematically. Let us note that even for the branching ratio of decay is large, however production cross section becomes small due to the large mass.\n\nThe lighter the larger its production cross section, however, for the decay is kinematically forbidden. That is why for numerical estimates we took the value for which and decays dominate222The decay provides great opportunity for the discovery of heavy Higgs . and . Thus (or a little bit lighter) mostly decays to two Higgs bosons.\n\nA technical remark: the equality in the limit follows from the equality (up to the sign) of and decay amplitudes, see (21).\n\n### ii.3 H production at LHC\n\nThe dominant mechanism of production is the gluon fusion, cross section of which equals that of SM Higgs production multiplied by . In Table 1 the relevant numbers are presented. All the numbers correspond to LHC energy.\n\nThe subdominant mechanisms of production are fusion and associative production. Comparing and vertices we will recalculate the cross sections of SM processes of production into that of production. In SM we have\n\n LhZZ=14¯g2vZ2h. (31)\n\nFrom (23) we get:\n\n σZZ→H=(2vΔv1−2M2h/M2H1−M2h/M2H)2×(σZZ→h)SM≈10−3×(σZZ→h)SM, (32)\n\nthe same relation holds for associative production cross section.\n\nWe separate VBF cross section of SM Higgs production into that in fusion (which dominates) and in fusion (which is the one that matters for production) with the help of the computer code HAWK HAWK . The obtained results are presented in Table 2.\n\nIn Table 3 the results for the associative production cross sections are presented.\n\nWe see that gluon fusion dominates production at LHC. Using model parameters and , we obtain that the branching ratio of decay equals . Thus, decays of provide of double production cross section in addition to coming from SM. However, unlike SM in which invariant mass is spread along rather large interval, in the case of decays invariant mass equals .\n\n## Iii H production enhancement in Georgi–Machacek variant of see-saw type II model\n\nThe amplitudes of production both via fusion and VBF are proportional to the triplet vev and due to the upper bound these amplitudes and the corresponding cross sections are severely suppressed.\n\nThe triplet vev should be small in order to avoid the noticeable violation of custodial symmetry which guarantees the degeneracy of and bosons in the SM at tree level in the limit . The vacuum expectation value of the complex isotriplet with hypercharge violates the custodial symmetry, see (8). The custodial symmetry is preserved when two isotriplets (complex and real with ) are added to SM and when vev’s of their neutral components are equal Georgi . Thus in GM variant of see-saw type II model is not bounded by (10) and can be considerably larger. Instead of (8) in GM model we have:\n\n (33)\n\nand instead of (11):\n\n v2+4v2Δ=(246 GeV)2. (34)\n\nNote that our is by bigger than what is usually used in the papers devoted to GM model; our is also usually denoted by , while the value is denoted by .\n\nThe scalar particles are conveniently classified in GM model by their transformation properties under the custodial . Two singlets which mix to form mass eigenstates and are:\n\n ⎧⎨⎩H01=φ,H02=√23δ+√13ξ0, (35)\n\nsee, for example, Logan2014 . Due to considerable admixture of in the coupling constant is not suppressed and three modes of decays are essential: .\n\nThe recently discovered Higgs boson should be identified with . The deviations of couplings to vector bosons and fermions from their values in SM lead to the upper bound on . These deviations in the limit of heavy scalar triplets were studied in a recent paper Logan2014 (see also Englert2013 ). From equations (59) and (61) of Logan2014 we get the following estimates for the ratios of the (here ) and coupling constants to that in SM:\n\n ⎧⎪ ⎪⎨⎪ ⎪⎩kV≈1+3(vΔv)2,kf≈1−(vΔv)2. (36)\n\nSince at LHC the Higgs boson is produced mainly in gluon fusion through -quark triangle, for the ratio of the cross sections to that in SM we get:\n\n ⎧⎪ ⎪⎨⎪ ⎪⎩μτ¯τ≈1−(2vΔv)2,μVV≈1+(2vΔv)2. (37)\n\nSince decay is studied in associative production, , we get\n\n μb¯b≈1+(2vΔv)2. (38)\n\nFinally in case of decay SM factor in the amplitude is modified in the following way:\n\n 169−7 → [1−(vΔv)2][169(1−(vΔv)2)−7(1+3(vΔv)2)]= (39) = 169(1−2(vΔv)2)−7(1+2(vΔv)2),\n\nwhere the first factor in the first line takes into account damping of production in gluon fusion.333We take into account only -quark and -boson loops omitting the loops with charged Higgses.\n\nLet us suppose that is ten times larger than the number used in Section II, . Then from (34) we get , and , while . From (39) we get: . With the up-to-date level of the experimental accuracy one can not exclude these deviations of the quantities from their SM values .\n\nOne order of magnitude growth of leads to two orders of magnitude growth of production cross section. Hence heavy Higgs boson can be produced at LHC with cross section which should be large enough for it to be discovered prior to HL-LHC. The search strategy should be the same as for the SM Higgs boson: decay is a golden discovery mode, the cross section of which can be as large as , where depends on the model parameters, see Logan2014 .\n\n## Iv Conclusions\n\nThe case of extra isotriplet(s) provides rich Higgs sector phenomenology with additional to SM Higgs boson charged and neutral scalar particles. With the growth of triplet vev, production cross section of new scalar grows and the dominant decays of new particles become decays to gauge and lighter scalar bosons. The charged scalars () are produced through electroweak interactions. The bounds on the model parameters from nondiscovery of and with the LHC data and the prospects of their discovery at LHC are discussed in particular in Yagyu2 . In the present paper we have discussed the neutral heavy Higgs production at LHC in which the gluon fusion dominates. decay contributes significantly to the double Higgs production and even may dominate in the GM variant of the see-saw type II model. The best discovery mode for is the “golden mode” , and its cross section can be only few times smaller than for the heavy SM Higgs.\n\nAfter this paper had been written, paper BhCh2014 appeared in arXiv in which the enhancement of double Higgs production due to heavy Higgs decay is considered in the framework of MSSM model with two isodoublets. resonant decay in MSSM at small was previously analyzed in Englert2012_2 .\n\nWe are grateful to A. Denner for the clarifications concerning HAWK code and to C. Englert and J. Zurita for providing us with relevant references. The authors are partially supported under the grants No. 12-02-00193, 14-02-00995, and NSh-3830.2014.2. SG was also supported by Dynasty Foundation and by the Russian Federation Government under grant No. 11.G34.31.0047.\n\n## References\n\nWant to hear about new tools we're making? Sign up to our mailing list for occasional updates.\n\nIf you find a rendering bug, file an issue on GitHub. Or, have a go at fixing it yourself – the renderer is open source!\n\nFor everything else, email us at [email protected]." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8861937,"math_prob":0.94766265,"size":17266,"snap":"2021-04-2021-17","text_gpt3_token_len":4411,"char_repetition_ratio":0.14106129,"word_repetition_ratio":0.017974421,"special_character_ratio":0.25906405,"punctuation_ratio":0.14882506,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97577876,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T04:20:03Z\",\"WARC-Record-ID\":\"<urn:uuid:a47183f7-9396-4b93-99b7-282725f6fcf2>\",\"Content-Length\":\"788163\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c0165ec-5e8e-4640-ba2d-75a6916d50ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9e5884f-3906-4b36-97e8-e865a995593b>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/1408.0184/\",\"WARC-Payload-Digest\":\"sha1:3ANZNSNCN4EUG5YULCQQSTRU3QKDP23P\",\"WARC-Block-Digest\":\"sha1:NUFIROGYZDN5HUGYTL4AVIUMQPC7XRV2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038101485.44_warc_CC-MAIN-20210417041730-20210417071730-00326.warc.gz\"}"}
http://www.simplescience.info/physics/answerpage-physics-2	[ "Physics‎ > ‎\n\n### Answer page - L to Z.\n\nposted Oct 16, 2015, 8:44 PM by Upali Salpadoru [ updated Jul 31, 2019, 3:20 PM ]\nGet the topics according to Alphabet.\n\nL\n• Light - Introduction.\n• Q.1.0\n\n• Phenomenon Can the wave theory explain? Can the particle theory explain? Reflection Yes. Yes. Refraction Yes Yes Interference Yes No. Diffraction Yes No Photo electric effect. No Yes.\n\n Q. No. Answers Q. No. Answers 2.03.04.0 1 – B. 2- C, 3 – D 4 – B , 5 – A , 1 42˚, 2 – Image. 3 - O˚ 4 – Red and blue, 5 – Red.1 Refractive index (1.3) = sine i. / sine r. Sine r. = sine i. / 1.3 = sne 20˚ / 1.3 = 0.3420 /1.3 = 0.2630 r. = 15.3˚ 2 Refractive index = Sine 90˚./ sine r 1.3 = 1/Sin r. Sine r. = 1/1.3 = 0.7602Critical angle r. = 50.28˚ 5.06.0 1 Image, 2 Real, 3 Inverted , 4 Usually smaller, 5. Same size.1 – A , 2 – C, 3 – A, 4 - C, 5 – D,\n\n### 1.0", null, "i) and ii) 2x2+1=5 marks\n\n2.0", null, "i)\n\nii) Let the sun’s rays fall on the lens perpendicular to the plane as shown. The distance between the lens and the converging point will be the focal length.\n\n5x2=10 marks\n\n3.0", null, "Show how the 5 rays will emerge through the lens.10 marks.\n\n4.0", null, "Show how the 5 rays will emerge through the lens.\n\n10 marks.", null, "5.0Red and blue spots are 2 light emitting diodes.\n\ni) Where will be the 2 sharpest images be?\n\nii) If the lens is moved to F what changes can be observed in the images?\n\n5x3=15 marks.", null, "6.0\n\ni) Draw a ray diagram to show the image of the object shown.\n\nii) Real,Inverted, smaller.\n\n10x2=20 marks.", null, "7.0\n\nThe diagram shows an image formed by a double convex lens. Use ray diagrams to find the position of the object.\n\n10 marks.\n\n8.0 Anobject 2cm high is placed 15 cm away from a convex lens whose focal length is 5cm.\n\ni) Position of the image .ii) Size of the image.\n\n 1/v + 1/u = 1/f1/v + 1/15 = 1/ 51/v= 1/5 - 1/151 = 3 -1 v 15v= 15/4 =3 .75 cm. M= Height of image = Image distance Height of object Object distance. h/2 = 3.8 / 15 H = 2x3.8 /15\n\n10 marks.\n\n9.0\n\nWhat lenses are normally used for the following.\n\n.i) Camera, - Convex lens ( sometimes convex+ concave)\n\nii) Flash light, - Concave to spread the light.\n\niii) Door viewers (peepholes.)- concave to see a large area.\n\niv) Near nearsightedness, concave\n\nv) Slide projector. convex.\n\n10 marks\n\nM\n\nMachines.\n\n 1.0 W x 0.1 = 5 x 0.3 W = 15 N 2.0(W+5) x 0.5 = 16 x (0.5 + 0.5) W+5 = 16/0.5 W = 32 -5 = 27 N\n\n3.0\n Taking L as Pivot ACW moments = R x 6 CW moments = 3000x3 + 400 x 4 6 R = 9000 + 1600 R = 10600 / 6 =1767 N Taking R as pivotCW moments = L x 6ACW moments = 3000 x3 +40 x 2 6L = 9000 + 800 = 9800/6 =1633Verification:-Downward forces = 3000 + 400 = 3400Up ward = 1767 + 1633 = 3400\n\n4.0 4.1 A couple. 4.2 Torque from one hand = 100 x 0.3 Nm.\n\nFor two hands = 30x2 = 60 Nm.\n\nMirrors.\n\n 1.0 Image is virtual, (behind the mirror) laterally inverted ( not inverted) same size.", null, "If you rotate the mirror by 90 deg. you will know why we give the above answer. 2.0", null, "3.0", null, "Image is virtual behind the mirror same size laterally inverted. 4.0i) A piece of paper - diffuse reflection.ii) A water surface - Some rays will go through while some will undergo regular reflection producing a parallel beam.", null, "Outside surface Inside surface\n 5.0", null, "Rear view mirror of a car - diverging. 6.0", null, "Virtual small.\n\n7.0\n\nA 2-cm object is placed at a distance of 25 cm from a concave mirror having a focal length of 15 cm. Calculate (i) the image distance (ii) the image size.\n\nFocal length = 15 cm., Object distance = 25 cm., Object height = 2cm.\n\nFormula:- 1/ object distance + 1/ image distance = 1/focal length.\n\n1/25 + 1/ di = 1/ 15\n\n1/di = 1/15 - 1/25\n\ndi = 75 / 5 - 3\n\n= 37.5\n\nMachines.\n\n# Machines - Inclined plane.\n\n7.1 W= weight x height\nW = 1000 x 30 = 30,000 J\n7.2 W = mgh\n= 70 x 10 x 30 J\n= 21000 J\n7.3 Work done on cart = Force x displacement.\n30,000 = F x 50\nF = 600 N\n7.4 Efficiency =useful work/ total work\n=30,000 / 30,000 +2100.\n= 30,000 / 32100\n=0. 9968 (As a percentage 99.7 %) (10 x 4 = 40 mark\n\nMachines - Levers.\n\n 1.0", null, "Effort x Effort arm = Load x Load arm. 10 x W= 5 x 40 W = 200/10 = 20 N 2.0", null, "The centre of gravity of the ruler has to be at 50 cm. mark. Therefore the distance from G to pivot is 0.2 m.ACW Moments = CW Moments W x 30 = 30 x 80 W = 80 N\n 3.0", null, "3.1 Clockwise moments= anti clockwise moments R 1 x 6 = 3000 x 3 + 400 x 2 R 1 = (9000 + 800) / 6 = 9800/6 = 1633 N\n\n1\n\n1\n\n3.1\n\nClockwise moments= anti clockwise moments\n\nR 1 x 6 = 3000 x 3 + 400 x 2\n\nR 1 = (9000 + 800) / 6\n\nR 1 = 9800/ 6 = 1633 N 2\n\n3.2\n\nTotal up force = Total down forces\n\nR 1 + R 2 = 3000 + 400\n\nR 2 = 3400 - R 1\n\nR 2 = 3400 - 1633 = 1767 N 1\n\n4.0\n\n4.1 A couple. 1\n4.2 Torque of a couple = One of the forces x diameter.\n100 x 0.6 = 60. Nm. 1\n\n5.0\n5.1 Fulcrum is in the centre of the bolt.\nRotating of the nut forms a couple.\nMoment of a couple = F x Diameter\nClockwise moments = Anticlockwise moments\n)\n(F x 0.07) + 10 x0.1 = 40 x 0.35\n0.07 F = (14 – 1) / 0.07\nF = 186 N 1\n\n5.2 .Anticlockwise moments = Clockwise moments.\nF x 0.07 = (10 x 0.1) + (40 x 0.07)\n0.07 F = 1 + 14\nF = 15/0.07 = 214 N 1\n\n5.3 When lifting the weight of spanner will be acting against you. 1\n\nMachines - Pulleys and Wheels.\n\n1.1 ….. 2\n\n1.2 Velocity ratio = Load / Effort\n\n2 = L / 15\n\nL = 2 x 15 = 30 N\n\nWeight of pineapple = Load - weight of pulley.\n\n= 30 – 2 = 28 N.\n\nTherefore mass of pineapple = 2.8 Kg.\n\n= 28 / 15 = 1.9\n\n1.4 Efficiency = Mechanical advantage / Velocity ratio\n\n= 1.9 / 2 = 0.95\n\nE % = 95 %.\n\n80/100 = MA / 2\n\nMA = 0.8 x 2 = 1.6\n\nUsing equation MA = Load / Effort\n\n1.6 = 700/ Effort.\n\nEffort = 700/1.6 = 437.5 N\n\n3.1 ……10 m\n\nVelocity ratio = Distance Effort moves / Distance Load moves.\n\n= 20 / 10 = 2.\n\n3.2 Method 1. Total Load = Weight of log + weight of hanging pulley\n\n= 1200 + 5 = 1205 N\n\nAs the load is supported by 2 strands each strand wwill get Load / 2\n\n= b1205 / 2 = 602.5 N\n\nAs the single fixed pulley does not alter the magnitude the man will have to give a minimum force of 602.5 N.\n\nMethod 2.\n\nVelocity ratio = Ideal Mechanical advantage\n\n2 = 1205 / Effort\n\nE = 602.5 N\n\n3.3\n\n= 1200 / 602.5 = 1.99\n\nEfficiency = Mechanical advantage / Velocity ratio.\n\nE = 1.99/2 = 0.996\n\n= 99.6 %\n\n3.4 ……………Lower.\n\n3.5………….. Lower\n\n4.0\n\n Question A B 4.1 How many strands are supporting the load? 4 5 4.2 What is the velocity ratio? 4 5 4.3 What will be the effort required to lift a load of 482.0? (482+18)/ 4= 125 N (482+18)/5= 100 N 4.4 What is the mechanical advantage? 482/125= 3.9 482/100= 4.8 4.5 What is the percentage efficiency? 3.9 / 4x100= 97.5% 4.8/5= 96 %\n\nV.R = rw / r a = 18/3 = 6\n\nConsidering the Velocity ratio as the Ideal Mechaanical advantage, we get\n\n6 = 750/ E\n\n6 E = 750\n\nE = 750 /6 = 125 N\n\n7.0\n7.1 Cogwheel ratio = 16 : 64\n= 1 : 4\nAs the rare wheel is attached to the rare cog wheel it will make 4 turns.\n7.2 For 1 turn of pedal the rare wheel will rotate 4 times.\nThe circumference of wheel = 2pi r\n= 2 x 3.14 x0.4 m\nThe distance bike will travel = 2 x 3.14 x 0.4 x 4\n\n= 10.05 m\n\n• ## Momentum\n\nQ. 1.0\n\n1.1 A = 4000 x 6 = 24,000. Kg.ms-1\n\n1.2 B = 5500 x4 = - 22000 Kg.ms-1 .\n\n1.3 A + B Combined momentum 24000 + (– 22000) = 2000\n\nCombined mass = 9500\n\nAs momentum is conserved\n\n2000 = 9500 . V\n\nV = 2000/9500 = 0.21 ms-1 to Right.\n\nQ. 2.0\n\n2.1 Using Pythagoras theorem;\n\nR2 = 8 2 + 6 2 = 64 + 36 = 100\n\nR = 10 kg,ms-1\n\n2.2 Momentum = Mass x Velocity\n\n6 kg.ms-1 = m x 13.5\n\nM = 6 / 13.5 = 0.44 kg.\n\n2.3 Force x time = Impulse = Change in momentum\n\nF x t = 10 kg,ms-1\n\nF = 10 /0.06 N\n\n= 166.67 N. 36.9 west of the original direction.\n\nQ. 3.0\n\n3.1 ..0.\n\n3.3 Momentum of Part 3, = √ 22 + 12\n\n= 2.24kgms-1\n\n3.4 Momentum = mv\n\n∴ v =2.24÷ o.1\n\n= 22.4ms-\n\nUsing trigonometry to get the angle:-\n\n### Tan 0.5 =27°\n\n3.0\n\ni. 90\n\nii. 20gms-1.\n\niii. 15gms-1\n\niv. As the direction is 90 we can use the Pythagoras theorem to get the resultant.\n\nR2 = 202 + 152\n\nR = 25g.ms-1\n\nv. 37\n\nMotion Linear\n\nQ.1.0\n\n i. Velocity = distance/ timeV = 50/ 2V= 25 ms-1. ii.Time = 50 + 30 /25 = 80/25 = 3.2 s. iii. a = Change in velocity / time = 25 /5 = - 5 m.s-2\n i. Velocity = distance/ timeV = 50/ 2V= 25 ms-1. ii.Time = 50 + 30 /25 = 80/25 = 3.2 s. iii. a = Change in velocity / time = 25 /5 = - 5 m.s-2\n\nQ.2.0\n\n i. V = d / t. 150 = d x 5 ∴ d = 150 x 5 = 750 m. ii. a =(V f – Vi ) / t. a =(0 - 150) / 10 = -15 m.s-2 iii. method 1.d = V1t + ½ at2 d = (150 x 10) + -15x100 d= 1500 -1500/2 = = 1500 - 750 = 750 mMethod 2. V.av = (Vi + V.f) / 2 = 75 m/s d= V xt = 750 m.\n\n### i. What is the maximum velocity attained? ii. Find the positive acceleration .. iii. Find the acceleration during the reducing speed. iv. Find the height freely fallen before the chord tightens.\n\n i.40 ms-1 ii.a =(V 2 – Vi ) / t. = (40 - 0) / 4.5 = 8.9 ms-2 iii.a =(V 2 – Vi ) / t = (0- 40) /4.5 = - 8.9 ms-2 iv. d =V.av x t V.av = 1/2 (V f +Vi ) = 1/2 (40 + 0) = 20 ms-1 d =V.av x t = 20 x 4.5 =4.4 m.\n\n### Motion in a circle\n\n1.0\n\na. The distance of the circular path is = 2 πr = 2x 3.14x 4 m\n\nPeriod of rotation = 3 s\n\nSpeed = d/t = (2x 3.14x 4) / 3\n\n= 25.12 ms-1\n\nb. Acceleration = v2/r = v2 /r = 25.12 x 25.12 / 4\n\n=157.8 ms-2\n\nc. Acceleration is towards the centre of the circle.\n\nd. The tension is due to centripetal force and the centrifugal force.\n\nForce = mass x acceleration.\n\n= 0.25 x 157.8 N\n\n=132.8 N\n\n2.0\n\na, Acceleration = v2/r = 100 x 100/ 10 = 1000 ms-2 to the centre.\n\nb, F= ma = 800 x 1000 = 800,000 N\n\nc, 800,000 N\n\n3.0\n\nRadius of circle is the sum of half width of each\n\nRadius = 1.2/2 + 1.4 /2 = 0.6 + 0.7 =1.3 m\n\nAcceleration = 4π2 r / T = 4 x (3.14)2 x 1.3 / T\n\n= 4x 9.86 x 1.3 / T\n\n= 51.3 /T\n\nForce = mass x acceleration = 40 x a\n\n380 = 40 x 51.3 / T\n\n320 = /T\n\nT = 2050/ 320 = 6.4 seconds\n\na. Accceleration = 51.3 / 6.4 = 8 ms-2\n\na= v2/r\n\nV2 = ar\n\nV = ar\n\nb. Velocity = 8 x 1.3 = 10.4\n\n= 3 ms-1\n\nc. Shell moves at a tangent to the circle.\n\n• # Newton's Laws.\n\n Newton’s Laws 1 and 2 only i. ii. iii. iv. v. Marks 1.0 100m/s2 310 m/s. 3,100kg.m/s 3,000 kg.m/s 1000 kg.m/s2 5 2.0 15 kg.m/s 0 15 kg.m/s 15kg.m/s 75 N 5 3.0 . 0.75m 20m/s 0.0375s 48,000kg.m/s 1,280,000 N 5 4.0 Equal. A. 2000m/s2 500m/s2 1.6 m 5\n\nAsessment\n\n18 - 20 Excellent 13 - 17 Very good 10 - 12 Good 5 - 9 Fai\n\nNuclear Changes\n\n1.0\n\n1.C 2. A 3. B, 4. E, 5 D.\n\n2.0\n\n1.236. 2. 56. 3. No charge, 4. Inside the nucleus. 5. C.\n\n3.0\n\n1. H+ + Cl- HCl ……………….. Chemical change.\n\n2. Mg24 + n1 Na24 + H1 ………..Artificial transmutation.\n\n3. 6C14 7N14+ -1e0……(β)……….. .Beta decay. / Beta emission.\n\n4. 86 Rn222 = 2H4 +84 Po218 ………...Alpha decay /Alpha emission.\n\n5. 92U235 + 0n1 + 92U236 ………….Nuclearreaction.\n\nQ.4.0\n\n1. Isotopes. 2. A neutron. 3. Ba, Kr and three 0n1.\n\n4. Forming U-236 and splitting them forming three Ba, threeKr and emitting six 0n1.\n\n5. A chain reaction.\n\n5.0\n\n1. 124.\n\n2. Excess neutrons over a stable ratio.\n\n3. 3.7 x2 = 7.4 s.\n\n4. 84Y215\n\n5. Alpha and Gamma.\n\n# Motion - Projectiles.\n\nQ.1.0 1.1 ….At 45°\n\n1.2…..At 90°\n\nQ.2.0 2.1….Yes.\n\n2.2….If you select the stone it would be because it experience less air friction .The stone has a high density and therefore a small surface area .\n\n…..If you select the stick it would be because it has a greater surface area and therefore the chance of hitting the target is greater.\n\nQ 3.0\n\n3.1\n\n0 = V1 + 2x -10 x 30\n\nV12 = 600\n\nV = √ 600 = 24.5 m upward.\n\n3.2\n\nIt will come down at the same speed but in the opposite direction.\n\n24.5 m downward\n\nQuestion 4.0\n\n4.1.\n\nUsing formula:- a = V2-V1 /t\n\n-10 = 0 – 150 /t\n\n-10 = -150/t\n\nt. = 15 s.\n\n4.2.\n\nDistance = velocity x time.\n\nd.= 150 x 15 m = 2250 m\n\nQuestion 5.0\n\n(5.1)\n\nVertical component of velocity = R sin 45°\n\n=14 x 0.707 = 9.9 ms-1\n\nUsing formula a= Д V/t\n\nt = Д V/a\n\n=9.9 / -10 = 0.99 seconds.\n\nAssuming that the horizontal velocity does not change and the total time is double that for the projectile to rise we get the range from this.\n\nRange = R cos Θ x0.99x2 = 9.9 x1.98\n\n= 19.6 m\n\n(5.2) Velocity of the put when coming down will be the same as the starting velocity at the same level. This is R cos Θ\n\nUsing the formula v22 =V1+ 2ad.\n\nV2 2 = 9.92 + (2x 10 x 2 )\n\nV2 = 138\n\nV = 11.7 ms-1\n\nAverage velocity while falling the lat 2m\n\nVav = V1 + V2 / 2\n\nVav = (9.9 + 11.7) / 2 = 10.8ms-1\n\nTime = distance / velocity\n\nt. = 2/ 10.8 = 0.19 seconds\n\nAs the horizontal speed remans the same\n\nd = R cos Θ t\n\n= 9.9 x 0.19 = 1.9 m.\n\nThe total displacement = 19.6 + 1.9 = 21.5 meters\n\nQ\n\nR\n\nRotatory Motion.\n\n1.0\n\n1.1 b. 1.2 a. 1.3 d. 1.4 c 1.5 d.\n\n2.0\n\n2.1 40 rev in 120 s\n\n1 rev. in 120/40 = 3 s.\n\n2.2 One rev is 2π radians.\n\nTime taken = 3 s.\n\nAngular velocity = 2π / 3 = 6.28/3 = 2.1 rad s-1.\n\n3.0\n\n3.1 . τ = F r.\n\n= 25 x0.16 Nm = 4.0 Nm.\n\n3.2 τ = Ft x r\n\nFt = τ / r = 4.0 / 0.08 = 50 Nm.\n\n3.3 Angular displacement of large cog wheel = 2 π\n\nCircumference =2 π r 1\n\nRotations of small wheel = 2 π r 1 / 2 π r 2 = r1/r2 = 0.08/0.04 = 2\n\nRotations of rare wheel = 2\n\n3.4 Distance cycle will travel == 2 x 2 π r 3 = 3.77 m\n\n4.0\n\n4.1 . α = ( ω f – ω i ) / time\n\n= ( 7 - 0 ) / 4 = 1.75 rad s-2.\n\n4.2 0. ( horizontal line shows uniform velocity )\n\n4.3 ω ( av) = ( ω f + ω i ) / 2\n\n= 8 / 2 = 4 Rad s =1.\n\nAngular displacement in 5 seconds 5 x 4 = 20 rad.\n\n2 π radians will be 1 rev.\n\n20 rad will be 20/ 2π = 20/ 6.28 = 3.18 turns.\n\nS\n\n• Simple Harmonic Motion.\n\nQ 1.0\n\n1.1- a, 1.2- c, 1.3 - a, 1.4-b. 1.5- c . 1.6- b\n\nQ. 2.0\n\ni. Swinging of a chandelier.\n\niv. Plucking a string in a guitar.\n\nThese cannot be taken as SHM.\n\nii. A palm can get pushed by the wind and it can come back only the wind speed decrease.\n\niii. When a child jumps up it accelerates down by the gravitational force. In SHM the the whole time acceleration has to be directed towards the centre of motion.\n\nv. The same answer as for above.\n\nQ. 3.0.\n\ni. 10g= 0.01kg.\n\nPotential gravitational energy that can be released = mgh.\n\n= 9.8x0.01 x0.2 J\n\n= 0.0196 J\n\nii. Potential energy changes to kinetic energy.\n\nKinetic energy= ½ mv2.\n\n½ mv2. = 0.0196\n\nv2.= 0.0196/ 0.5x0.01\n\nv = 3.92 = 1.98 ms-1\n\niii. Zero. Iv. Up (push by the water in the other arm)\n\niv. Va= v1 +v2 / 2\n\n= 0.99 ms-1\n\nv. The distance particle has to cover in one cycle.\n\nd = vavx t.\n\nPeriod is t = d/va\n\nt =0.4 /0.99\n\n= 0.4 seconds.\n\n0.4s ………..= 1cycle\n\n1. = 1/0.4 = 2.5 Hz.\n\nVi . Throughout the oscillations the force acts towards the equilibrium position.\n\n• # Speed Velocity and Acceleration.\n\nQ. 1.0\n\nThe distance train has to travel = 50 m\n\nTime taken = 2 s\n\nThere fore speed = 50m / 2s = 25 m.s-i\n\nIn passing the platform the train travels the length of the train plus the length pf the platform.\n\n= 59m + 30 m\n\nUsing formula S = V . t\n\nMaking t subject t = S / V\n\nt = 80 m / 25 ms -1\n\n= 3.2 s. \\\n\nQ. 2.0\n\ni. The velocity at cruising speed is shown by the blue section of the graph.\nThe area of this part = 150 x 15\n= 2250 (The area below graph)\nThe displacement at cruising speed = 2250m.\n\nii. Change in velocity = V2 - V 1\nTime taken = 10 s\nTherefore acceleration = ( 0 – 150) m.s-1 / 10 s\n= - 15 m.s -2\n\niii. The displacement during landing = Area below red graph\n= Area of green triangle\n= ½ base x height\n= (150 x 10) / 2\n= 750\nThe displacement = 750 m\n\nQ. 3.00\n\ni. Maximum speed = 35 ms-1\nii. This is only for three seconds.\nUsing formula a = ( V2 – V1 ) / t\na = ( 30 – 0 ) / 3 m.s -1\na = 10 m.s -2\n\nlll. Using the same formula\na = ( V2 – V1 ) / t\na = ( 0 – 35m ) / 4 m.s -1\n= 8.75 m.s -2\n\nIv, Area of the green part. 45 m.\n\nQ.4.0\n\nEquation 1. acceleration = change in velocity /time\n\na = (V2 - V1 ) /\n\nTherefore t = (V2 - V1 ) / a\nt = 30 s-1 / 6 ms-1\n= 5 s\nl.\n\nl.\n\n Period First second Second second Third second Fourth second Velocity avg. 35 25 15 5\n1\n\nii.\n\n Period First second Second second Third second Fourth second Displacement 35x 1 =35 m 25X1=25m 15x1 =15m 5x1 =5\n\niii.\n\n Time / s 1 2 3 4 Displacement 35.m 60.m 75.m 80.m\n\niv.\n\n## T\n\n• TORQUE\n\n1.0\n\nCW Torque = ACW Torque\n\nJ X2 = 225 x 2.5\n\nJ = 562.5 / 2 …………………..= 281.25 N.\n\nTotal downward force = 281.25 + 225 ……………..= 506.25\n\nSupport force = 506.26 N.\n\n2.0\n\nTaking A as Fulcrum\n\nCW Torques= 8 xB\n\nACW = 2X10) +4.5 X 25\n\nAS the torques cause no rotation 8B = 20 + 112.5\n\nB= 132.5 / 8………….16.6 N\n\nTotal downward force = 10 + 25 = 35\n\nThen A gets 35 - 16.6 …………………………………….18.4 N\n\n3.0\n\n1. T = Fxd\n\n= 50 x sine30 x 0.25\n\n= 6.25 Nm.\n\nB. At 90 .\n\nC. The weight of spanner will act down through tnhe centre of mass.\n\n4.0\n\n4,0A\n\n1. Weight.\n\n2. Normal reaction\n\n3. Foreward force.\n\n4. Vertical component of upward force.\n\n5. Friction.\n\nB\n\n1 Weight cos 30 = R\n\nR = 40x 0.87\n\nR = 34.8 N\n\n2. Foreward force = W sine 30\n\n= 40 x0.5\n\n= 20 N\n\n3. If the friction stops sliding the torque = foreward force x radius.\n\nT = 20 x 0.35\n\n= 7 Nm.\n\n## U\n\nUnits of Measurement\n\n1. The volume of a solid block is 20 m3 . The density is 6 kg m3. What is the weight of the mass?\n\nAnswer: Mass = volume x density. 20 x 6 = 120 kg.\n\nWeight - mass x gravitational force. 120 x 9.8 = 1176 N.\n\n2. John getting a standing start runs 100m. in 15 s. gradually increasing velocity . Find the final velocity?\n\nAverage velocity = (Vi + Vf ) / 2\n\n100/15 = 0 + Vf / 2\n\nVf = 2x100/ 15 …… V13.33ms-1.\n\n3. Jane has a mass of 50 kg and her cycle 20 kg. On riding at a speed of 10ms-1 a snake decides to cross her path. She brings the bike to a stop in 2 seconds. What was her braking power.\n\nKinetic energy of cycle and rider = ½ mv2.\n\nKe. = 70/2 x 10= 3500 J.\n\nPower = J/t\n\n= 3500/2 ……= 1750 W.\n\n4. 1.0 kW. Steel kettle of 250g. is used to heat 2 litres of water at 20C . Find the minimum time it will take to boil the water.\n\nHeat absorbed by water = mct ………….2kgx 2400 x (100-20)\n\n= 2x2400 x80\n\n= 432000 J\n\nHeat absorbed by kettle = mct ………….0.25 x490 x 80\n\n= 9800. J\n\nTotal heat absorbed = 441800 J\n\nPower = J/t ……….t= J / p ……..= 441800 / 1000\n\nT= 441.8 s……. 7.4 minutes.\n\n=\n\n5. A car and passengers having a mass of 1500 kg. speeds up from 15m-s 30 m-s in 3 seconds. Find the force car has used.\n\nAcceleration = (vf – vi ) / t …………….(30 – 15) / 3\n\n= 5 ms-2\n\nForce = mass x acceleration…………..= 1500 x 5 N.\n\n= 7500 N.\n\n• ## 1.01. - d. 2. - b. 3. - c 4 – a,\n\n2.0 c. 3.0- a&b.\n4.0\n\n.to 8. Obtain the answers from this diagram.\n\nFig. 5. Free body force diagram.\n\n9. 860 N (Approx)\n\nTrigonometry solution\n\nSine 60° = h/6 = 0.866\n\nh = 6x 0.866 = 5.196\n\nTan 60° = 5.196/ x = 1.7321\n\nX = 5.196 /1.7321 = 3\n\nTherefore base = 4+3= 7\n\nTan Θ = 5.196 / 7 =0.7423\n\n=37°\n\nSine 37° = 5.196/ R =0.6018\n\nR = 5.196/ 0.6018 = 8.6\n\nActual force is 860 N\n\n10. A force equal in magnitude to resultant in the opposite direction.\n\n## W\n\n• WAVE PROPERTIES\n• Q.1.0 D\n\nQ.2.0\n\n6.5x10-7/ 1x10-4 = …..x / 3)\n\n0.02m\n\n3.0\n\n1.\n\nLoud sounds are a result of constructive interference while faint sounds are at lines of destructive interference.\n\n2.\n\nV =fλ…………………………λ= V/f\n\n= 330/264 m.\n\nλ = 1.25m.\n\n3. Tan θ = 2λ/3\n\n= 2x 1.25/3…………=2.5/3 ….=0.833\n\nΘ = 39.8°\n• Q.4.0\n\n1. W…………...500Hz.\n\n2. N………….. f a. = Vw x f w\n\nVw-Vs\n\n= 330 x 500 / 330-20\n\n= 165000 / 310………..= 632.26 Hz.\n\n3. E…………...500 Hz.\n\n4 S ………...f a. = Vw x f w\n\nVw+ Vs\n\nfa = 330x500 / 330 + 20\n\n= 165000 / 350……...471.43 Hz.\n\n2. As she would be getting closer, she will receive more waves. Apparent frequency would go higher and pitch will increase.\n\n• ## WORK , POWER and Efficiency\n\n1.1 F = ma, F= 45x9.8 = 441 N. down.\n\n1.2 Reaction = 441 up.\n\n1.3 w= fd., W= 441+ (25x 9.8) x 3000\n\n= 441 + 245\n\n= 686 x3000 = 2058000 J\n\n1.4 P = w/t. P = 2058000/ 30x60 = 60 W.\n\n 2.1 2500 x 8 = 20000 J. 2.2. 2500 W. 2.3 Power = Work/ time 2500 = 20000 / t t. = 20000 / 2500 . t.= 8 s. Q 3.0 Answers 3.1 3 (800 + 600) = 4200 N. , 3.2 3 ( 800+600)/30 = 140 W. , 3.3 800 x 3 = 2400 J. 3.4 (800 x 3) ÷ 5 =480 N. , 3.5 512 – 480 = 32 N. Q.4.0 Answers. 4.1 Work= 5000 x 9.8 x 25 = 1225000 J ( 1225 kJ.). 4.2 746 x 2 = 1492 J. 4.3 P = W/t , t. = W/p . t. = 1225000 / 1492. = 821 s = 13.7 minutes. 4.4 According to the given HP of the pump in 20 mts. ( 20 x 600 = 12000 s) it should perform 1492 x 20x60 J. Actual work done was = 1250000. Efficiency = 1225000 / 1790400 = 0.68 Percentage efficiency = 0.68 x100 = 68. %. Q. 5.0 5.1 400 N. 5.2. 400 x 12 x 0.2 = 480. J. 5.3. 480/ 5 = 96 W , 5.4 P.E g. = mgh = 40 x 10 x 2.4 = 96 J. 5.5 All the potential energy gets converted to Kinetic energy = 96 J. 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https://www.band1.org/paper/?id=6632	[ "# 中五數學試卷 (F5 Maths Past Paper)\n\n6632\n\npdf\n\n20\n\nF5 maths _MY_M1_16_17\n\n▼ 圖片只作預覽, 如欲下載整份卷, 請按「免費成為會員」 ▼", null, "▲ 圖片只作預覽, 如欲下載整份卷, 請按「免費成為會員」 ▲\n\n## 中五數學試卷 PDF 下載\n\nLa Salle College\nMid-Year Examination 2016-2017\nMathematics\nExtended Part\n(Calculus and Statistics)\nTime allowed: 2 hours 30 minutes\nInstructions\nWrite your examination number in the spaces provided on\nthe top right corner of the cover page.\nThis paper consists of Section A and Section B. Each section\ncarries 50 marks.\nAnswer ALL questions in this paper.\nAll working must be clearly shown.\nUnless otherwise specified, numerical answers should be\neither exact or correct to 4 decimal places.\nThe diagrams in this paper are not necessarily drawn to scale.\nMid-Year Examination 2016-2017\nF.5 Module 1 - Calculus and Statistics\nThe curve y = x³ - 3x² - 10x+24 intersects the x-axis at points A, B and C. Find the shaded area\nenclosed by the curve and the x-axis as shown in the figure.\nMid-Year Examination 2016-2017\nand Statistics\ny=x² − 3x² −10x + 24\n(a) Show that [(x+1)+¹(x−1)″-¹] = 2(nx − 1)(x+1)″ (x−1)″-², where n is a constant.\nGiven that the slope at any point (x, y) of a curve C is given by\n2(2017x-1)(x+1) 2017(x-1) 20¹5. If the y-intercept of C is 2, find the equation of C.\nMid-Year Examination 2016-2017\nand Statistics\nSection B [50 marks]\nA researcher models the rate of change of antibodies in the bloodstream of a patient t hours after\ninjection of a vaccine by the following formula:\n(a) Prove that A = 40ln\nwhere A units is the amount of antibodies in the bloodstream. When t = 0, A = 300.\ndt t² - 6t+10\nMid-Year Examination 2016-2017\nand Statistics\nt² - 6t+10\n0≤t≤300,\nProve that t²-6t+10>0 for all real values of t.\nWhen will the amount of antibodies in the bloodstream reach the minimum?\nWhat is the minimum amount of antibodies in the bloodstream?\nThe patient will be protected when the amount of antibodies is not less than 500 units. The\nresearcher claims that the patient can be protected within 40 hours after the injection of the\nMid-Year Examination 2016-2017\nand Statistics\nThe market research department of a smartphone manufacturer estimates the total sales (in\nafter t months of its launch,\nthousands) of a new phone model will increase at a rate S'(t)=\nwhere 0(a) The researcher uses the trapezoidal rule with 5 sub-intervals to estimate the total sales for the\nfirst 5 months after its launch.\n(i) Find the estimate.\nThe researcher re-estimates the total sales by expanding\npowers of t.\nMid-Year Examination 2016-2017\nand Statistics\nin ascending powers of t as far as the term in 1³.\n(ii) Using the result of (b)(i), re-estimate the total sales for the first 5 months after its\nas a series in ascending\nMid-Year Examination 2016-2017\nand Statistics\nLet f(x)= x² Inx.\nFind the range of value(s) of x such that f(x)≥0.\nUsing the result of (b)(i), or otherwise, show that\nx + C, where C is a constant.\n(x³ In x).\n√ x² In x dx\n(iii) Hence, show that the area bounded by the x-axis, the lines x = 1 and x = 2 as well as the\ncurve y = x ln x is\nMid-Year Examination 2016-2017\nand Statistics\nUsing the results of (b)(iii) and (c)(i), show that\n1.014 In 1.01 +1.024 In 1.02+1.034 In 1.03+...+In 1.994 In 1.99 > 632 ln 2-124.\nMid-Year Examination 2016-2017\nand Statistics\nA researcher models the number of people infected by bird flu x days after the outbreak of the\ndisease in a region with the following formula:\n1+ a[e¹(x-1)]\nwhere a and b are constants.\nMid-Year Examination 2016-2017\nand Statistics\nas a linear function of x.\nIt is found that the slope and the x-intercept are −3 and\nFind a and b.\nShow that y = P(x) is an increasing function.\n(iii) Will the number of infected people exceed 500? Explain your answer.\n+1 respectively.\nLet a be the root of P\"(x)=0. Find a.\nBriefly describe the behaviour of P'(x) before and after x = a.\nMid-Year Examination 2016-2017\nand Statistics\nSection A [50 marks]\nSolve the following equations.\n(a) ex (e*+2)=8\n(b) In(x+2)+In(x-1)=0\nMid-Year Examination 2016-2017\nand Statistics\nMid-Year Examination 2016-2017\nand Statistics\n- End of Paper\n2. Evaluate lim\nMid-Year Examination 2016-2017\nand Statistics\nIn the expansion of\n(a) Find the value of n.\nMid-Year Examination 2016-2017\nand Statistics\nHence, find the constant term.\nin ascending powers of x, the coefficient of the 3rd term is 420.\n4. Find the equation of the tangent to the curve y = (x\ny=(x² − 3)e¹-x at x=2.\nMid-Year Examination 2016-2017\nand Statistics\nIt is given that f(x)= x³ − 2x² − 4x+5. Find the range of values of x such that the curve\ny = f(x) is\n(a) decreasing;\n(b) concave upwards.\nMid-Year Examination 2016-2017\nand Statistics\nAnson deposited a certain amount of money in a bank at the beginning of 2016. The manager\nclaimed that Anson's profit, P(t) (in thousand dollars) after t months, followed the equation\n= 3e 100. Find the profit obtained from March 2016 to August 2016, correct to the nearest\nthousand dollars.\nMid-Year Examination 2016-2017\nand Statistics\nAdrian goes from Town A to Town B. He first rides on boat with a speed of 10 km/h to cross the\nriver to reach C and then rides on bicycle with a speed of 20 km/h to Town B as shown in the\nfigure. It is given that AD L BD with AD = 5 km and BD = 30 km. Let x km be the length of CD\nand t hours be the time taken for Adrian to travel from Town A to Town B.\nProve that t=\nMid-Year Examination 2016-2017\nand Statistics\n2√x² +25-x+30\nAdrian claims that he can travel from Town A to Town B within 2 hours. Is the claim correct?" ]	[ null, "https://examjpg.sgp1.cdn.digitaloceanspaces.com/06632.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8138213,"math_prob":0.9583387,"size":6008,"snap":"2023-14-2023-23","text_gpt3_token_len":1894,"char_repetition_ratio":0.1937042,"word_repetition_ratio":0.07743154,"special_character_ratio":0.302763,"punctuation_ratio":0.08675079,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9943296,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T19:58:14Z\",\"WARC-Record-ID\":\"<urn:uuid:f4515fd5-4a1b-4c42-bd0c-136024eb08ad>\",\"Content-Length\":\"22366\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:600a191f-f642-440d-8432-29a0d7917738>\",\"WARC-Concurrent-To\":\"<urn:uuid:0075f6a3-6dc6-4c1b-8389-6fa884ec6354>\",\"WARC-IP-Address\":\"13.215.98.233\",\"WARC-Target-URI\":\"https://www.band1.org/paper/?id=6632\",\"WARC-Payload-Digest\":\"sha1:T3TXAMMMZSMHBO4ZRF2LIBUU5JEONBRA\",\"WARC-Block-Digest\":\"sha1:464MAOBJPXWGDWPBCOYSNA7H2XIGVNVH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945372.38_warc_CC-MAIN-20230325191930-20230325221930-00425.warc.gz\"}"}
https://www.tessshebaylo.com/writing-linear-equations-2-quizlet/	[ "# Writing Linear Equations 2 Quizlet\n\nBy \| December 2, 2017\n\nWriting linear equations study set diagram quizlet graphing flashcards checkpoint 2 of lines chapter 5 3 04 review quiz write go live final test the year slope intercept form standard", null, "Writing Linear Equations Study Set Diagram Quizlet", null, "Graphing Writing Linear Equations Flashcards Quizlet", null, "Checkpoint 2 Writing Equations Of Lines Flashcards Quizlet", null, "Writing Linear Equations Flashcards Quizlet", null, "Chapter 5 Linear Equations Diagram Quizlet", null, "3 04 Review Quiz Write Equations Of Lines Diagram Quizlet", null, "Go Live Writing Linear Equations Flashcards Quizlet", null, "Final Test Of The Year Flashcards Quizlet", null, "Writing Equations Slope Intercept Form Flashcards Quizlet", null, "Writing Linear Equations Flashcards Quizlet", null, "Linear Equations Graphing Standard Form Flashcards Quizlet", null, "Topic 2 Linear And Nar Graphs Tables Equations Flashcards Quizlet", null, "Math 1 Unit 5 Voary Flashcards Quizlet", null, "Writing Linear Equations Flashcards Quizlet", null, "Graphing Linear Equations Flashcards Quizlet", null, "Linear Equations Systems Algebra 2 Trig Diagram Quizlet", null, "Systems Of Equations Flashcards Quizlet", null, "Slope And Linear Equations Part 1 Flashcards Quizlet", null, "Linear Equations And Graphing Flashcards Quizlet", null, "12 21 Writing Linear Equations 2 Flashcards Quizlet", null, "Sem 2 Systems Of Linear Equations Inequalities Voary Unit 4 Diagram Quizlet", null, "Review Slope Or Rate Of Change Y Intercept Flashcards Quizlet", null, "Linear Equations Eoc Blitz Flashcards Quizlet\n\nWriting linear equations study set graphing of lines flashcards diagram quizlet 3 04 review quiz write final test the year slope intercept form standard\n\nThis site uses Akismet to reduce spam. 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https://teclado.com/30-days-of-python/python-30-day-24-project/	[ "# Day 24 Project: Dice Roller\n\nWelcome to the day 24 project in the 30 Days of Python series! In this project we're going to be using a module in the standard library called `argparse` to take in configuration from a user before the program even runs.\n\nThe program we're going to be writing to demonstrate this is a command line dice roller that can simulate the rolls of various configurations of dice.\n\nBefore we can really do this, we need to learn a little bit about what repl.it does when we press the \"Run\" button, and we need to learn a little bit about the `argparse` module itself.\n\nAlso, we've got a video walkthrough of this entire blog post available!\n\n## Running Python code\n\nThose of you who have been working in local development environments may have already learnt how to run a Python program yourselves, but for the rest of us, this step has been hidden away behind repl.it's \"Run\" button.\n\nSo, what actually happens when we press this button?\n\nWhen we press the button, repl.it runs a command which looks like this:\n\n``````python main.py\n``````\n\nThis is the reason that repl.it always runs `main.py`. It runs `main.py` because this is the file it specifies as part of the default run command.\n\nWe can actually configure repl.it to use a different run command if we want to, and we do this by creating a special file in the repl called `.replit`. This file is written in a format called TOML (which stands for Tom's Obvious, Minimal Language), which is a common format for configuration files, since it's so easy to write and read.\n\nLet's have a go at changing the run command, since this is something we're going to be doing a fair bit in this project.\n\nFirst, create a file called `app.py` and put some code in there so that you can verify when it runs. Something like this would do:\n\napp.py\n``````print(\"Hello from app.py\")\n``````\n\nNow create a file called `.replit` and put the following code inside:\n\n``````run = \"python app.py\"\n``````\n\nNow press the \"Run\" button. If everything worked, your `app.py` should have run instead of `main.py`.\n\n## Running a program with flags and arguments\n\nOne thing we can do with many real world console applications is run them with flags and arguments. These flags are used to configure how the program is run, either by turning on certain settings, or by providing values for various options.\n\nFor example, if we wanted to run our `app.py` file with the `help` flag, we could do something like this:\n\n``````python app.py --help\n``````\n\nThis `--help` flag is generally used to find out information regarding how to use the program.\n\nHowever, at the moment we can't use this flag. Our program has no idea what it means. This is where `argparse` comes in: it let's us specify which flags and arguments we're going to accept, and it gives us a way to access the values the user specified when calling our application.\n\n## A quick look at `argparse`\n\nHere we're going to create a small program that returns a number raised to a certain power to learn about some `argparse` concepts.\n\n### Creating a parser\n\nIn order to use `argparse` we first need to import the module and create an `ArgumentParser` like this:\n\napp.py\n``````import argparse\n\nparser = argparse.ArgumentParser()\n``````\n\nIf we like, we can specify a description for our program by passing in a value when creating this `ArgumentParser`.\n\napp.py\n``````parser = argparse.ArgumentParser(description=\"Returns a number raised to a specified power.\")\n``````\n\n### Specifying positional arguments\n\nNow that we have this `parser`, it's time to start specifying arguments. To start with, let's make it so that we can accept a positional argument when the user calls our application.\n\nTo accept this argument, we need to write the following:\n\napp.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"Returns a number raised to a specified power.\")\nparser.add_argument(\"base\", help=\"A number to raise to the specified power\")\n``````\n\nHere we've called the `add_argument` method on our `parser`, passing in two values.\n\nThe first, `\"base\"` is the name of the parameter which is going to accept the argument from the user. We're going to use this name to get hold of the value later on.\n\nThe second value we specified using a keyword argument, and it's going to be used by `argparse` to create documentation for our program.\n\nIn order to process the values the user passes in, we need one more thing: we need to parse the arguments the user passed in.\n\napp.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"Returns a number raised to a specified power.\")\nparser.add_argument(\"base\", help=\"A number to raise to the specified power\")\n\nargs = parser.parse_args()\n\nprint(args.base) # access the value of the base argument\n``````\n\nNow we can change our `.replit` file to something like this:\n\n``````run = \"python app.py --help\"\n``````\n\nJust make sure to change the file name to wherever you wrote all of your code. If everything worked, you should be able to press the run button and get output like this:\n\n``````usage: app.py [-h] base\n\nReturns a number raised to a specified power.\n\npositional arguments:\nbase A number to raise to the specified power\n\noptional arguments:\n-h, --help show this help message and exit\n``````\n\nThis is the documentation that `argparse` created for us.\n\nNow let's change the `.replit` file to something like this instead:\n\n``````run = \"python app.py 23\"\n``````\n\nNow we should get the number `23` printed to the console, as we specified in our file.\n\n### Specifying optional arguments\n\nNow let's take a quick look at optional arguments, which are passed in using flags. We create these in just the same way, but we use a `--` in front of the name.\n\nWe're going to specify an exponent using an optional argument, and we're going to set a default value of `2` if the user doesn't provide a value.\n\napp.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"Returns a number raised to a specified power.\")\n\nparser.add_argument(\"base\", help=\"A number to raise to the specified power\")\nparser.add_argument(\"--exponent\", help=\"A power to raise the provided base to\")\n\nargs = parser.parse_args()\n\nprint(args.base)\nprint(args.exponent)\n``````\n\nYou may have noticed from the help output, that `--help` has a shortcut form: `-h`. We can do the same thing for our optional arguments by providing a second name with a single `-`.\n\napp.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"Returns a number raised to a specified power.\")\n\nparser.add_argument(\"base\", help=\"A number to raise to the specified power\")\nparser.add_argument(\"-e\", \"--exponent\", help=\"A power to raise the provided base to\")\n\nargs = parser.parse_args()\n\nprint(args.base)\nprint(args.exponent)\n``````\n\nThere are a couple of final things we can do to improve our program. First, we should set a default value for `exponent`, and we should specify the types we expect for each value.\n\napp.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"Returns a number raised to a specified power.\")\n\nparser.add_argument(\"base\", type=float, help=\"A number to raise to the specified power\")\n\"-e\",\n\"--exponent\",\ntype=float,\ndefault=2,\nhelp=\"A power to raise the provided base to\"\n)\n\nargs = parser.parse_args()\n\nprint(args.base ** args.exponent)\n``````\n\nNow we can change our `.replit` file to something like this:\n\n``````run = \"python app.py 2 -e 5\"\n``````\n\nAnd our program outputs `32.0`, with is 2⁵.\n\nIf you want to look into `argparse` in more detail, you can find a really good tutorial in the documentation here.\n\n## The brief\n\nNow that we've learnt a little bit about `argparse` we can get to the meat of the project. For this project we're going to be creating a dice roller for n-sided dice.\n\nThe user is going to be able to specify a selection of dice using the following syntax, where the number before the `d` represents the number of dice, and the number after the `d` represents how many sides those dice have.\n\n``````python main.py 3d6\n``````\n\nIn this case, the user has requested three six-sided dice.\n\nUsing the `random` module, we're going to simulate the dice rolls the user requested, and we're going to output some results in the console, like this:\n\n``````Rolls: 1, 2, 4\nTotal: 7\nAverage: 2.33\n``````\n\nHere we have the numbers rolled, the sum of the values, and the average of the rolls.\n\nIn addition to printing this result to the console, we're also going to keep a permanent log of the rolls in a file called `roll_log.txt`. The user can specify a different log file if they wish with an option argument called `--log`.\n\n``````python main.py 2d10 --log rolls.txt\n``````\n\nIn addition to specifying a custom log file, the user can specify a number of times to roll the dice set using a `--repeat` flag.\n\n``````python main.py 6d4 --repeat 2\n``````\n\nBoth `--repeat` and `--log` should have appropriate documentation, and the user should be able to use `-r` and `-l` as short versions of the flags. The user can also use both the `--repeat` and `--log` flags if they want to.\n\nGood luck!\n\n## Our solution\n\nFirst things first, let's set up our parser. I'm going to put this in its own `parser.py` file along with any code that deals with parsing the arguments.\n\nparser.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"A command line dice roller\")\n\nargs = parser.parse_args()\n``````\n\nWe need to register three arguments for our application: one positional and two optional.\n\nThe position argument is going to catch the dice configuration that the user specifies using our `xdy` syntax.\n\nThe two optional parameters are going to catch the number of repetitions for the roll, and the place to log the rolls. Both of these arguments are going to need default values.\n\nLet's start with the positional argument, which I'm just going to call `dice`.\n\nparser.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"A command line dice roller\")\n\nparser.add_argument(\"dice\", help=\"A representation of the dice you want to roll\")\n\nargs = parser.parse_args()\n``````\n\nWe don't really have to do anything special here. The input is going to be a string, and we don't need to specify any flags or default values. The only thing we need to do is specify some help text for the program documentation.\n\nThe two optional arguments are a fair bit more complicated. First let's tackle the `--repeat` argument.\n\n`--repeat` should have a default value of `1`, because if the user doesn't specify a repeat value, it's probably because they don't want to repeat the roll. It's also important that we make sure we get an integer here, and not a float, or something which can't be represented as an integer. It doesn't make much sense to repeat a roll `2.6` times, for example.\n\nWith this in mind, I think a decent implementation for this argument would be something like this:\n\nparser.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"A command line dice roller\")\n\nparser.add_argument(\"dice\", help=\"A representation of the dice you want to roll\")\n\"-r\",\n\"--repeat\",\nmetavar=\"number\",\ndefault=1,\ntype=int,\nhelp=\"How many times to roll the specifed set of dice\"\n)\n\nargs = parser.parse_args()\n``````\n\nOne thing that I've added here is a value for the `metavar` parameter. This is just going to change what shows up as a placeholder for the value in the program documentation.\n\nNow let's add the `--log` argument configuration.\n\nIn this case we want to specify a default file name for the logs, which can be whatever you want. I'm going to use `roll_log.txt`.\n\nWe also probably want to make sure that the value we get is a string, so I'm going to specify a type of `str` for this argument.\n\nparser.py\n``````import argparse\n\nparser = argparse.ArgumentParser(description=\"A command line dice roller\")\n\nparser.add_argument(\"dice\", help=\"A representation of the dice you want to roll\")\n\"-r\",\n\"--repeat\",\nmetavar=\"number\",\ndefault=1,\ntype=int,\nhelp=\"How many times to roll the specifed set of dice\"\n)\n\"-l\",\n\"--log\",\nmetavar=\"path\",\ndefault=\"roll_log.txt\",\ntype=str,\nhelp=\"A file to use to log the result of the rolls\"\n)\n\nargs = parser.parse_args()\n``````\n\nLooking good!\n\nThe only thing left to do in this `parser.py` file is to actually parse the dice specification. Assuming everything is okay with the the user's value, this should be as simple as splitting the string by the character `\"d\"` and converting the values in the resulting list to `integers`.\n\nparser.py\n``````def parse_roll(args):\nquantity, die_size = [int(arg) for arg in args.dice.split(\"d\")]\n\nreturn quantity, die_size\n``````\n\nHowever, there's always the change that the user enters and invalid configuration, so we need to do a bit of exception handling.\n\nWe're going to catch a `ValueError` first, which is going to catch cases where the user enters something like `fd6`, `d6`, or `6d`.\n\n`d6` and `6d` are going to be caught because if nothing features before or after the `\"d\"`, `\"\"` will be in the list returned by `strip`. If we try to pass `\"\"` to `int`, we get a `ValueError`.\n\nparser.py\n``````def parse_roll(args):\ntry:\nquantity, die_size = [int(arg) for arg in args.dice.split(\"d\")]\nexcept ValueError:\nraise ValueError(\"Invalid dice specification. Rolls must be in the format of 2d6\") from None\n\nreturn quantity, die_size\n``````\n\nIn this case we can't really do anything to properly handle the error, since we don't know what the user intended, but I'm raising a new `ValueError` with a more helpful exception for the user. I've decided to raise using `from None` so that the user gets a trimmed down version of the traceback.\n\nThe `ValueError` is actually also catching another issue for us as well: having too many values to unpack.\n\nAttempting to do something like the example below results in a `ValueError`:\n\n``````x, y = [1, 2, 3]\n``````\n\nIf we wanted to provide more helpful feedback to the user, we could break this comprehensions up into different steps, but I think this is good enough for our case.\n\nNow let's turn to `main.py` where we're going to make use of these things we defined in `parser.py`.\n\n`main.py` is going to be very short and is really just here to compose the various functions we define in our other files into a useful application.\n\nFirst, we're going to import `parser` and `random`, and we're going to get hold of the `args` variable we defined in `parser.py`.\n\nmain.py\n``````import parser\nimport random\n\nargs = parser.args\n``````\n\nNow that we have hold of this, we can call `parser.parse_rolls`, passing in this `args` value. We can also get hold of the specified number of repetitions, and the specified log file, assigning them to nicer names.\n\nmain.py\n``````import parser\nimport random\n\nargs = parser.args\n\nquantity, die_size = parser.parse_roll(args)\nrepetitions = args.repeat\nlog_file = args.log\n``````\n\nNow we have all the information we need, we can start actually simulating the dice rolls. For a single roll, the logic is going to look something like this:\n\n``````rolls = [random.randint(1, die_size) for _ in range(quantity)]\ntotal = sum(rolls)\naverage = total / len(rolls)\n``````\n\nWe use a list comprehension to call `randint` once for each die the user specified. So if we got `3d6`, we're going to generate a list of 3 results.\n\n`randint` chooses a number for us from an inclusive range, so we just need to specify `1` to the size of the die.\n\nOnce we have our results stored in `rolls`, we can calculate the `total` and `average`.\n\nAll of this logic is going to go in a loop, however, since we may want to perform several repetitions of the roll.\n\nmain.py\n``````import parser\nimport random\n\nargs = parser.args\n\nquantity, die_size = parser.parse_roll(args)\nrepetitions = args.repeat\nlog_file = args.log\n\nfor _ in range(repetitions):\nrolls = [random.randint(1, die_size) for _ in range(quantity)]\ntotal = sum(rolls)\naverage = total / len(rolls)\n``````\n\nAt the moment we're not really doing anything with any of the results, so let's fix that by writing some functions to take care of the formatting of our results, and the writing to our log file.\n\nI'm going to keep all of this functionality in a third file called `output.py`. Feel free to name it whatever you like.\n\nThe content of this file is very easy to understand, so we can breeze through it.\n\noutput.py\n``````roll_template = \"\"\"Rolls: {}\nTotal: {}\nAverage: {}\n\"\"\"\n\ndef format_result(rolls, total, average):\nrolls = \", \".join(str(roll) for roll in rolls)\nreturn roll_template.format(rolls, total, average)\n\ndef log_result(rolls, total, average, log_file):\nwith open(log_file, \"a\") as log:\nlog.write(format_result(rolls, total, average))\nlog.write(\"-\" * 30 + \"\\n\")\n``````\n\nFirst I'm defining a template which we can populate with values. Using a multi-line string like this helps avoid lots of `\"\\n\"` characters.\n\nI've then defined a function called `format_results` which actually populates this template with values. It also takes care of joining the rolls together so that we don't have any square brackets in our ouput.\n\nThe `log_results` function is entirely concerned with writing to the log file. It takes in a log file as an argument, and it uses a context manager to open this file in append mode. If the file does not exist, this will create it.\n\nAfter opening the file, `log_results` then formats the rolls and writes the result to the file, followed by 30 `-` characters on a new line. This is going to serve as a separator in the file.\n\nWith that, we just have to import `output` in `main.py` and call our functions.\n\nmain.py\n``````import output\nimport parser\nimport random\n\nargs = parser.args\n\nquantity, die_size = parser.parse_roll(args)\nrepetitions = args.repeat\nlog_file = args.log\n\nfor _ in range(repetitions):\nrolls = [random.randint(1, die_size) for _ in range(quantity)]\ntotal = sum(rolls)\naverage = total / len(rolls)\n\nprint(output.format_result(rolls, total, average))\noutput.log_result(rolls, total, average, log_file)\n``````\n\nNow it's time to test our program with various test cases. Here is what things look like for a valid set of arguments:\n\n``````> python main.py 8d10 -r 3\nRolls: 3, 3, 1, 3, 8, 8, 8, 9\nTotal: 43\nAverage: 5.375\n\nRolls: 10, 7, 5, 6, 10, 2, 3, 6\nTotal: 49\nAverage: 6.125\n\nRolls: 10, 6, 10, 6, 2, 4, 6, 4\nTotal: 48\nAverage: 6.0\n``````\n\nAnd here is the content of `roll_log.txt`:\n\n``````Rolls: 3, 3, 1, 3, 8, 8, 8, 9\nTotal: 43\nAverage: 5.375\n------------------------------\nRolls: 10, 7, 5, 6, 10, 2, 3, 6\nTotal: 49\nAverage: 6.125\n------------------------------\nRolls: 10, 6, 10, 6, 2, 4, 6, 4\nTotal: 48\nAverage: 6.0\n------------------------------\n``````\n\nWe can also use the `-h` or `--help` flags to see our lovely generated documentation:\n\n``````usage: main.py [-h] [-r number] [-l path] dice\n\nA command line dice roller\n\npositional arguments:\ndice A representation of the dice you want to roll\n\noptional arguments:\n-h, --help show this help message and exit\n-r number, --repeat number\nHow many times to roll the specifed set of dice\n-l path, --log path A file to use to log the result of the rolls\n``````\n\nHopefully you were able to tackle this on your own, even if you did it in a very different way to me. We'd love to see some of your solutions over on our Discord server if you feel like sharing!\n\nIf you want to dig into `argparse` further, then you should definitely check out the tutorial and main documentation page." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7684265,"math_prob":0.62087584,"size":18401,"snap":"2023-40-2023-50","text_gpt3_token_len":4368,"char_repetition_ratio":0.14649127,"word_repetition_ratio":0.13812155,"special_character_ratio":0.24183469,"punctuation_ratio":0.13560687,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95908415,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T20:52:12Z\",\"WARC-Record-ID\":\"<urn:uuid:66b3e95c-adb7-406b-99cb-5d5f25c6b91b>\",\"Content-Length\":\"181201\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71801ff8-a666-435e-bc3c-18ff3e42dfd0>\",\"WARC-Concurrent-To\":\"<urn:uuid:595a86e1-5770-4bdb-8b1c-6c6b2522ed22>\",\"WARC-IP-Address\":\"216.24.57.1\",\"WARC-Target-URI\":\"https://teclado.com/30-days-of-python/python-30-day-24-project/\",\"WARC-Payload-Digest\":\"sha1:XRL33GHQBY26QCEWOSMGEODCWCCNQGQN\",\"WARC-Block-Digest\":\"sha1:FAIAAXXA2VGNUDJJTNMMAQX2CY2O57X2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100568.68_warc_CC-MAIN-20231205204654-20231205234654-00788.warc.gz\"}"}