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https://stacks.math.columbia.edu/tag/0E72
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[
"Lemma 107.21.5. The morphisms $\\mathcal{C}\\! \\mathit{urves}^{semistable} \\to \\mathop{\\mathrm{Spec}}(\\mathbf{Z})$ and $\\mathcal{C}\\! \\mathit{urves}^{semistable}_ g \\to \\mathop{\\mathrm{Spec}}(\\mathbf{Z})$ are smooth.\n\nProof. Since $\\mathcal{C}\\! \\mathit{urves}^{semistable}$ is an open substack of $\\mathcal{C}\\! \\mathit{urves}^{nodal}$ this follows from Lemma 107.18.2. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)."
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https://cstheory.stackexchange.com/questions/7351/genetic-algorithm-to-draw-a-graph-position-assignment-problem
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[
"# Genetic Algorithm to Draw a Graph? Position assignment problem\n\nI have an assignment problem at hand and am wondering how suitable it would be to apply local search techniques to reach a desirable solution (the search space is quite large).\n\nI have a directed graph (a flow-chart) that I would like to visualize on 2-D plane in a way that it is very clear, understandable and easy to read by human-eye. Therefore; I will be assigning (x,y) positions to each vertex. I'm thinking of solving this problem using simulated annealing, genetic algorithms, or any such method you can suggest\n\nInput: A graph G = (V,E)\nOutput: A set of assignments, {(xi, yi) for each vi in V}. In other words, each vertex will be assigned a position (x, y) where the coordinates are all integers and >= 0.\n\nThese are the criteria that I will use to judge a solution (I welcome any suggestions):\n\n• Number of intersecting edges should be minimal,\n• All edges flow in one direction (i.e from left to right),\n• High angular resolution (the smallest angle formed by two edges incident on the same vertex),\n• Small area - least important.\n\nFurthermore; I have an initial configuration (assignment of positions to vertices), made by hand. It is very messy and that's why I'm trying to automate the process.\n\nMy questions are,\n\n• How wise would it be to go with local search techniques? How likely would it produce a desired outcome?\n\n• And what should I start with? Simulated annealing, genetic algorithms or something else?\n\n• Should I seed randomly at the beginning or use the initial configuration to start with?\n\n• Or, if you already know of a similar implementation/pseudo-code/thing, please point me to it.\n\nAny help will be greatly appreciated. Thanks.\n\n• Michele Sebag (lri.fr/~sebag), Marc Schoenauer (lri.fr/~marc) and Alejandro Rosete-Suarez (from Cuba, I could not find back his webpage) studied how to use genetic algorithms to generate improved representation of graphs, based on user input. – Jeremy Jul 12 '11 at 13:58\n\nI don't think it is wise to use local search techniques. There are many (fast) methods which work pretty good for many classes of graphs. Of course you can also test local search methods.\n\n• Number of intersecting edges should be minimal\n\nIs your graph planar? If yes, there are many good methods for planar graphs (e.g. dominance drawings).\n\nIs it a general digraph? In this case a hierarchical approach may be suitable (more information later).\n\n• All edges flow in one direction (i.e from left to right)\n\nDoes it contain cycles? If yes, this is not possible.\n\nA great source for graph drawing algorithms is Graph Drawing: Algorithms for the Visualization of Graphs.\n\nIn my opinion a good first approach to this problem is the hierarchical approach. Check the paper by Sugiyama et. al Methods for Visual Understanding of Hierarchical System Structures (chapter 9 in 1 describes such methods).\n\nIf you go with the local optimization methods, I would definitely use the initial configuration to start with (since it should be close to a good local optimum).\n\nI was given task similar to yours on my job. I used two things given above. Both the things work and should give you general idea. In the end you will have to do some tweaks that are specific to your graphs.\n\ngraphviz contains tools to make attractive planar layouts. the papers used to inform the programs are listed at http://www.graphviz.org/Documentation.php .\n\nforce-based methods are probably the most common for graph visualization but its possible that GAs (more recently investigated for this purpose) are effective & currently underexplored & hybrid GA-force based solutions are possible via force analysis in the fitness function. here are some other refs not cited so far"
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https://www.colorhexa.com/01fe48
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[
"# #01fe48 Color Information\n\nIn a RGB color space, hex #01fe48 is composed of 0.4% red, 99.6% green and 28.2% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 71.7% yellow and 0.4% black. It has a hue angle of 136.8 degrees, a saturation of 99.2% and a lightness of 50%. #01fe48 color hex could be obtained by blending #02ff90 with #00fd00. Closest websafe color is: #00ff33.\n\n• R 0\n• G 100\n• B 28\nRGB color chart\n• C 100\n• M 0\n• Y 72\n• K 0\nCMYK color chart\n\n#01fe48 color description : Vivid cyan - lime green.\n\n# #01fe48 Color Conversion\n\nThe hexadecimal color #01fe48 has RGB values of R:1, G:254, B:72 and CMYK values of C:1, M:0, Y:0.72, K:0. Its decimal value is 130632.\n\nHex triplet RGB Decimal 01fe48 `#01fe48` 1, 254, 72 `rgb(1,254,72)` 0.4, 99.6, 28.2 `rgb(0.4%,99.6%,28.2%)` 100, 0, 72, 0 136.8°, 99.2, 50 `hsl(136.8,99.2%,50%)` 136.8°, 99.6, 99.6 00ff33 `#00ff33`\nCIE-LAB 87.657, -82.961, 69.007 36.622, 71.354, 17.973 0.291, 0.567, 71.354 87.657, 107.91, 140.246 87.657, -81.648, 96.708 84.471, -70.437, 46.515 00000001, 11111110, 01001000\n\n# Color Schemes with #01fe48\n\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #fe01b7\n``#fe01b7` `rgb(254,1,183)``\nComplementary Color\n• #39fe01\n``#39fe01` `rgb(57,254,1)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #01fec7\n``#01fec7` `rgb(1,254,199)``\nAnalogous Color\n• #fe0139\n``#fe0139` `rgb(254,1,57)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #c701fe\n``#c701fe` `rgb(199,1,254)``\nSplit Complementary Color\n• #fe4801\n``#fe4801` `rgb(254,72,1)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #4801fe\n``#4801fe` `rgb(72,1,254)``\n• #b7fe01\n``#b7fe01` `rgb(183,254,1)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #4801fe\n``#4801fe` `rgb(72,1,254)``\n• #fe01b7\n``#fe01b7` `rgb(254,1,183)``\n• #01b232\n``#01b232` `rgb(1,178,50)``\n• #01cb3a\n``#01cb3a` `rgb(1,203,58)``\n• #01e541\n``#01e541` `rgb(1,229,65)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #1afe5a\n``#1afe5a` `rgb(26,254,90)``\n• #34fe6d\n``#34fe6d` `rgb(52,254,109)``\n• #4dfe7f\n``#4dfe7f` `rgb(77,254,127)``\nMonochromatic Color\n\n# Alternatives to #01fe48\n\nBelow, you can see some colors close to #01fe48. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01fe09\n``#01fe09` `rgb(1,254,9)``\n• #01fe1e\n``#01fe1e` `rgb(1,254,30)``\n• #01fe33\n``#01fe33` `rgb(1,254,51)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #01fe5d\n``#01fe5d` `rgb(1,254,93)``\n• #01fe72\n``#01fe72` `rgb(1,254,114)``\n• #01fe87\n``#01fe87` `rgb(1,254,135)``\nSimilar Colors\n\n# #01fe48 Preview\n\nThis text has a font color of #01fe48.\n\n``<span style=\"color:#01fe48;\">Text here</span>``\n#01fe48 background color\n\nThis paragraph has a background color of #01fe48.\n\n``<p style=\"background-color:#01fe48;\">Content here</p>``\n#01fe48 border color\n\nThis element has a border color of #01fe48.\n\n``<div style=\"border:1px solid #01fe48;\">Content here</div>``\nCSS codes\n``.text {color:#01fe48;}``\n``.background {background-color:#01fe48;}``\n``.border {border:1px solid #01fe48;}``\n\n# Shades and Tints of #01fe48\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #001406 is the darkest color, while #ffffff is the lightest one.\n\n• #001406\n``#001406` `rgb(0,20,6)``\n• #00270b\n``#00270b` `rgb(0,39,11)``\n• #003b11\n``#003b11` `rgb(0,59,17)``\n• #004e16\n``#004e16` `rgb(0,78,22)``\n• #00621c\n``#00621c` `rgb(0,98,28)``\n• #007521\n``#007521` `rgb(0,117,33)``\n• #018927\n``#018927` `rgb(1,137,39)``\n• #019c2c\n``#019c2c` `rgb(1,156,44)``\n• #01b032\n``#01b032` `rgb(1,176,50)``\n• #01c337\n``#01c337` `rgb(1,195,55)``\n• #01d73d\n``#01d73d` `rgb(1,215,61)``\n• #01ea42\n``#01ea42` `rgb(1,234,66)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\n• #15fe56\n``#15fe56` `rgb(21,254,86)``\n• #28fe64\n``#28fe64` `rgb(40,254,100)``\n• #3cfe72\n``#3cfe72` `rgb(60,254,114)``\n• #4ffe80\n``#4ffe80` `rgb(79,254,128)``\n• #63fe8e\n``#63fe8e` `rgb(99,254,142)``\n• #76fe9c\n``#76fe9c` `rgb(118,254,156)``\n• #8affab\n``#8affab` `rgb(138,255,171)``\n• #9dffb9\n``#9dffb9` `rgb(157,255,185)``\n• #b1ffc7\n``#b1ffc7` `rgb(177,255,199)``\n• #c4ffd5\n``#c4ffd5` `rgb(196,255,213)``\n• #d8ffe3\n``#d8ffe3` `rgb(216,255,227)``\n• #ebfff1\n``#ebfff1` `rgb(235,255,241)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nTint Color Variation\n\n# Tones of #01fe48\n\nA tone is produced by adding gray to any pure hue. In this case, #77887c is the less saturated color, while #01fe48 is the most saturated one.\n\n• #77887c\n``#77887c` `rgb(119,136,124)``\n• #6d9277\n``#6d9277` `rgb(109,146,119)``\n• #639c73\n``#639c73` `rgb(99,156,115)``\n• #59a66f\n``#59a66f` `rgb(89,166,111)``\n• #4fb06a\n``#4fb06a` `rgb(79,176,106)``\n• #46b966\n``#46b966` `rgb(70,185,102)``\n• #3cc362\n``#3cc362` `rgb(60,195,98)``\n• #32cd5e\n``#32cd5e` `rgb(50,205,94)``\n• #28d759\n``#28d759` `rgb(40,215,89)``\n• #1ee155\n``#1ee155` `rgb(30,225,85)``\n• #15ea51\n``#15ea51` `rgb(21,234,81)``\n• #0bf44c\n``#0bf44c` `rgb(11,244,76)``\n• #01fe48\n``#01fe48` `rgb(1,254,72)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01fe48 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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[
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https://gis.stackexchange.com/questions/329453/how-to-get-all-the-sub-rectangles-subgeohashes-of-a-given-geohash
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[
"# How to get all the sub-rectangles (subgeohashes) of a given geohash\n\nI am trying to use geohash to generate a grid of rectangles to cover an area. But I am not very familiar with the coding system of geohash. My question is:\n\nIf I have a specific geohash `gh` of 3 characters, how do I get a list of all 4,5,6-character geohashes that are contained in the `ST_GeomFromGeohash` of the original geohash `gh`?\n\nI am using PostGIS (2.5, 3.0alpha; PostgreSQL 11), and can manage the basics of getting the geometries of a geohash etc. I am just not sure how to get all the \"sub-geohashes\", e.g. by string operations. If there is a built-in function in PostGIS/PostgreSQL, it is even better.\n\nRelated:\n\nWhat is the precision of a Geohash\n\n## 1 Answer\n\nFirst of all, a geohash is easier to explain with referencee to points, but the logic can easily be extended to a grid, using two points, for opposite corners, similar to how ST_MakeBox2D works. A geohash in made up of interwoven bits, where each even bit represents increasing precision (powers of two in longitude) and each odd bit represents increasing precision in latitude, as explained very well in the wikipedia article. In other words the first bit, 0 or 1, determines that the range is either from -180 to 0 or 0 to 180, so the range for each possible bit value is 180, the mean value is +/- 90, and the maximum error is 90 degrees. The second bit determines the range as either -90 to 0 or 0 to 90, so half the range, mean value and maximum error of the corresponding longitude values. The third bit then splits the range in half, so from +/-180 to +/-90 or +/-90 to 0, depending on whether the previous odd bit was a 0 or 1.\n\nOnce you have interweaved bit values, you convert to a geohash using a modified base32 encoding, as illustrated in Wikipedia. So, a geohash of 3 letters, corresponds three 5 bit words, ie, 15 digits, which means 8 digits of precision in longitude and 7 digits in latitude, ie, the maximum error is now 0.352 in both directions.\n\nSo, for example, I am sitting at 41.48N, 2.17E. This corresponds to a geohash of sp3e3kupuxj6bggjfdc4, as you can see from:\n\n``````SELECT ST_GeoHash(ST_SetSRID(ST_MakePoint(2.17, 41.38), 4326));\n``````\n\nDoing the inverse transform, we get a polygon (box) representing the bounds of the precision -- as determined by the bit length of the base32 encoding.\n\nSo, with a precision of 20, this gives,\n\n`````` SELECT ST_Astext(ST_GeomFromGeoHash(ST_GeoHash(ST_SetSRID(ST_MakePoint(2.17, 41.38), 4326), 20)));\n``````\n\nPOLYGON((2.16999999999999 41.3799999999999,2.16999999999999 41.3800000000001,2.17000000000031 41.3800000000001,2.17000000000031 41.3799999999999,2.16999999999999 41.3799999999999))\n\nwhich is an extremely tight box around the original point. If we change the precision to 3, for example, then we get a bounding box of\n\nPOLYGON((1.40625 40.78125,1.40625 42.1875,2.8125 42.1875,2.8125 40.78125,1.40625 40.78125))\n\nie, a range of 1.406 degrees, which corresponds to +/-0.703, potential error around the exact point.\n\nSo, going back to your original question, if you have a 3 letter geohash, you can't get beyond 6 decimal places, no matter what precision you request, as there are only 15 bits in a 3 letter word.\n\n``````SELECT ST_AsText(ST_PointFromGeoHash('sp3'), 7);\n``````\n\nyields:\n\nPOINT(2.109375 41.484375)\n\nas does\n\n`````` SELECT ST_AsText(ST_PointFromGeoHash('sp3'), 10);\n``````\n\nas you can't get any more precision that that which is contained in a 3 letter base32 encoding.\n\nAs you are interested in bounding boxes, you can use the function, ST_Box2dFromGeoHash, which will return the bounding box around a point, based on the rules outlined above.\n\nIf you run this function, you will be able to determine the width of your bounding box, and to make the grid denser, you would simply subdivide this box into smaller boxes, and increase the ST_GeoHash precision to get tighter bounds each time.\n\nI do not believe that there is a a built in function, but this would be easy enough to do by chaining together ST_Box2dFromGeoHash, ST_XMin, ST_Ymin, ST_Xmax, ST_Ymax, ST_GeoHash and generate_series twice for the x and y direction and a final generate_series for the precision, ie, conceptually a triple for loop, and each time you increase the precision, you will get a longer hash code.\n\n• Thanks for the detailed answer. What I was looking for is a way to enumerate e.g. 4-letter geohash codes that \"bleong to\" the 3-letter code geographically. In your example, `sp3` has `sp3e` obviously. But what are the others? Is it `sp3[0-9,a-z]`? Then, similarly what are the 5-letter codes. – tinlyx Jul 21 at 12:57\n• I am uncertain because going down one level of a quad-tree should add two-bits, but adding one letter in geohash seems to add 5-bit to it. Again, this is because I don't know much about geohash, and whether its hierarchical approximated code systems maps to quad-trees. – tinlyx Jul 21 at 13:00\n• Yes, because one letter of a base32 encoding is 5 bits. From the wikipedia article, the first \"letter\" of the base32 alphabet is 0, which corresponds to 00000 and the last letter is z, which corresponds to 11111, ie, 31. So, one level down in the quad tree, will add 10 bits, 5 for each direction. – John Powell Jul 21 at 13:07"
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{"ft_lang_label":"__label__en","ft_lang_prob":0.83850586,"math_prob":0.94274676,"size":3492,"snap":"2019-35-2019-39","text_gpt3_token_len":978,"char_repetition_ratio":0.116972476,"word_repetition_ratio":0.0036363637,"special_character_ratio":0.31586483,"punctuation_ratio":0.1755102,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98626566,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T15:48:14Z\",\"WARC-Record-ID\":\"<urn:uuid:60498012-39f5-465c-8694-2fa066e749f5>\",\"Content-Length\":\"139354\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:960ceaaa-fd47-4b2d-924c-e6d1cb287110>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b0aedf5-3644-4b0c-9610-c94e9b319c47>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://gis.stackexchange.com/questions/329453/how-to-get-all-the-sub-rectangles-subgeohashes-of-a-given-geohash\",\"WARC-Payload-Digest\":\"sha1:LHCMX5KYERVULPMQDX5CNCH5GCLCBJAG\",\"WARC-Block-Digest\":\"sha1:EY2FC2XZ6PZM5VLRYV5QB45VRKHJHOAP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313936.42_warc_CC-MAIN-20190818145013-20190818171013-00270.warc.gz\"}"}
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https://dataplatform.cloud.ibm.com/exchange/public/entry/preview?url=https://raw.githubusercontent.com/IBMDataScience/sample-notebooks/master/Cloud/HTML/Predicting%2Bchurn%2Bwith%2BSPSS%2Brandom%2Btree%2Balgorithm.html
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[
"# Predicting churn with the SPSS random tree algorithm¶\n\nThis shows you how to create a predictive model of churn rate by using IBM SPSS Algorithm on Apache Spark version 2.0. You'll learn how to create an SPSS random tree model by using the IBM SPSS Machine Learning API, and how to view the model with IBM SPSS Model Viewer.\n\nBecause it consists of multiple classification and regression trees (CART), you can use random tree algorithms to generate accurate predictive models and solve complex classification and regression problems. Each tree develops from a bootstrap sample that is produced by resampling the original data points with replacement data. During the resampling phase, the best split variable is selected for each node from a specified smaller number of variables that are drawn randomly from the full set of variables. Each tree grows without pruning and then, during the scoring phase, the random tree algorithm aggregates tree scores by majority voting (for classification) or average (for regression).\n\nIn this notebook, you'll create a model with telecommunications data to predict when its customers will leave for a competitor, so that you can take some action to retain the customer.\n\nTo get the most out of this notebook, you should have some familiarity with the Scala programming language.\n\nThis notebooks runs on Scala 2.11 with Spark 2.0. Some familiarity with Scala is recommended.\n\n## Contents¶\n\nThis notebook contains the following main sections:\n\n## 1. Load the Telco Churn data to the cloud data repository¶\n\nTelco Churn is a hypothetical data file that concerns a telecommunications company's efforts to reduce turnover in its customer base. Each case corresponds to a separate customer and it records various demographic and service usage information. Before you can work with the data, you must use the URL to get the telco.csv and telco_Feb.csv files from the GitHub repository.\n\nIn :\nval link_telco = \"https://raw.githubusercontent.com/AlgorithmDemo/SampleData/master/telco.csv\"\n\nimport sys.process._\nimport java.net.URL\nimport java.io.File\nnew URL(link_telco) #> new File(\"telco.csv\") !!\n\nval link_telco_Feb = \"https://raw.githubusercontent.com/AlgorithmDemo/SampleData/master/telco_Feb.csv\"\n\nimport sys.process._\nimport java.net.URL\nimport java.io.File\nnew URL(link_telco_Feb) #> new File(\"telco_Feb.csv\") !!\n\n\n## 2. Prepare the data¶\n\nAfter uploading the CSV files that contain the data, you must create a SQLContext, put the data from the telco.scv file into a Spark DataFrame, and show the first row in the DataFrame.\n\nIn :\nval sqlContext = new org.apache.spark.sql.SQLContext(sc)\n\nval dfTelco = sqlContext.\nread.\nformat(\"com.databricks.spark.csv\").\noption(\"header\", \"true\").\noption(\"inferschema\", \"true\").\nload(\"telco.csv\")\n\ndfTelco.show(1)\n\n+------+------+---+-------+-------+------+---+------+------+------+------+--------+-----+--------+--------+-------+-------+--------+-------+-------+-------+-------+--------+-------+-------+--------+-----+-----+--------+------+--------+-------+------+-----+----------------+-------+-------+----------------+-------+----------------+-------+-----+\n|region|tenure|age|marital|address|income| ed|employ|retire|gender|reside|tollfree|equip|callcard|wireless|longmon|tollmon|equipmon|cardmon|wiremon|longten|tollten|equipten|cardten|wireten|multline|voice|pager|internet|callid|callwait|forward|confer|ebill| loglong|logtoll|logequi| logcard|logwire| lninc|custcat|churn|\n+------+------+---+-------+-------+------+---+------+------+------+------+--------+-----+--------+--------+-------+-------+--------+-------+-------+-------+-------+--------+-------+-------+--------+-----+-----+--------+------+--------+-------+------+-----+----------------+-------+-------+----------------+-------+----------------+-------+-----+\n| 2| 13| 44| 1| 9| 64| 4| 5| 0| 0| 2| 0| 0| 1| 0| 3.7| 0.0| 0.0| 7.5| 0.0| 37.45| 0.0| 0.0| 110.0| 0.0| 0| 0| 0| 0| 0| 0| 1| 0| 0|1.30833281965018| | |2.01490302054226| |4.15888308335967| 1| 1|\n+------+------+---+-------+-------+------+---+------+------+------+------+--------+-----+--------+--------+-------+-------+--------+-------+-------+-------+-------+--------+-------+-------+--------+-----+-----+--------+------+--------+-------+------+-----+----------------+-------+-------+----------------+-------+----------------+-------+-----+\nonly showing top 1 row\n\n\n\nReview the data. Print the schema of the DataFrame to look at what kind of data you have.\n\nIn :\ndfTelco.printSchema\n\nroot\n|-- region: integer (nullable = true)\n|-- tenure: integer (nullable = true)\n|-- age: integer (nullable = true)\n|-- marital: integer (nullable = true)\n|-- address: integer (nullable = true)\n|-- income: integer (nullable = true)\n|-- ed: integer (nullable = true)\n|-- employ: integer (nullable = true)\n|-- retire: integer (nullable = true)\n|-- gender: integer (nullable = true)\n|-- reside: integer (nullable = true)\n|-- tollfree: integer (nullable = true)\n|-- equip: integer (nullable = true)\n|-- callcard: integer (nullable = true)\n|-- wireless: integer (nullable = true)\n|-- longmon: double (nullable = true)\n|-- tollmon: double (nullable = true)\n|-- equipmon: double (nullable = true)\n|-- cardmon: double (nullable = true)\n|-- wiremon: double (nullable = true)\n|-- longten: double (nullable = true)\n|-- tollten: double (nullable = true)\n|-- equipten: double (nullable = true)\n|-- cardten: double (nullable = true)\n|-- wireten: double (nullable = true)\n|-- multline: integer (nullable = true)\n|-- voice: integer (nullable = true)\n|-- pager: integer (nullable = true)\n|-- internet: integer (nullable = true)\n|-- callid: integer (nullable = true)\n|-- callwait: integer (nullable = true)\n|-- forward: integer (nullable = true)\n|-- confer: integer (nullable = true)\n|-- ebill: integer (nullable = true)\n|-- loglong: double (nullable = true)\n|-- logtoll: string (nullable = true)\n|-- logequi: string (nullable = true)\n|-- logcard: string (nullable = true)\n|-- logwire: string (nullable = true)\n|-- lninc: double (nullable = true)\n|-- custcat: integer (nullable = true)\n|-- churn: integer (nullable = true)\n\n\n\nCreate a DataFrame for the telco_Feb.csv data. You'll use this year's data to build the model, and use the February data for accuracy value.\n\nIn :\nval dfTelcoFeb = sqlContext.\nread.\nformat(\"com.databricks.spark.csv\").\noption(\"header\", \"true\").\noption(\"inferschema\", \"true\").\nload(\"telco_Feb.csv\")\n\n\n## 3. Configure the RandomTrees model¶\n\nBy running this portion of the code, you create the random trees estimator, import the libraries, and set the ordinal and nominal variables. Because no inputFieldList value is set, all fields except the target, frequency, and analysis weight fields are treated as input fields. To make the random tree model build faster, set the number of trees to 10 instead of the default value, which is 100. Finally, you must specify the churn target field.\n\nIn :\nimport com.ibm.spss.ml.classificationandregression.ensemble.RandomTrees\nimport com.ibm.spss.ml.utils.DataFrameImplicits.DataFrameEnrichImplicitsClass\n\nval ordinal = Array(\"ed\")\nval nominal = Array(\"region\",\n\"marital\",\n\"retire\",\n\"gender\",\n\"tollfree\",\n\"equip\",\n\"callcard\",\n\"wireless\",\"multline\",\n\"voice\",\"pager\",\"internet\",\"callid\",\"callwait\",\"forward\",\"confer\",\n\"ebill\",\n\"custcat\",\n\"churn\"\n)\nval srf = RandomTrees().setTargetField(\"churn\").setNumTrees(10)\nval srfModel = srf.fit(dfTelco.setNominalMeasure(nominal,true).setOrdinalMeasure(ordinal,true))\n\n\nDo the prediction and get your results.\n\nIn :\nval predResult = srfModel.transform(dfTelcoFeb)\nval predResultNew = predResult.withColumn(\"prediction\", predResult(\"prediction\").cast(\"double\")).\nwithColumn(\"churn\", predResult(\"churn\").cast(\"double\"))\n\n\nTo get the accuracy result, use the Apache Spark MulticlassClassificationEvaluator function. Notice that the accuracy is above 90%.\n\nIn :\nimport org.apache.spark.ml.evaluation.MulticlassClassificationEvaluator\nval evaluator = new MulticlassClassificationEvaluator().setLabelCol(\"churn\").setMetricName(\"accuracy\")\nval acc_result = evaluator.evaluate(predResultNew)\n\n# Summary and next steps¶\n\nYou have created a predictive model of churn rate by using IBM SPSS Algorithm on Apache Spark. Now you can create a different model to compare model evaluations, such as the test of model effects, residuals, and so on. See SPSS documentation.\n\n## Authors¶\n\nWang Zhiyuan and Yu Wenpei are SPSS Algorithm Engineers at IBM.\n\nCopyright © 2017 IBM. This notebook and its source code are released under the terms of the MIT License."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.76228845,"math_prob":0.76870066,"size":5857,"snap":"2019-35-2019-39","text_gpt3_token_len":1303,"char_repetition_ratio":0.10302409,"word_repetition_ratio":0.04260985,"special_character_ratio":0.20710261,"punctuation_ratio":0.18224299,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.970573,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T01:44:45Z\",\"WARC-Record-ID\":\"<urn:uuid:1e98def0-d5cb-4ecf-9651-c9b1e2cc0492>\",\"Content-Length\":\"315330\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0ca500e6-8a3b-42e5-ac4c-ed20c4fdffb8>\",\"WARC-Concurrent-To\":\"<urn:uuid:2eab0717-681b-47f7-9c5a-f1ba977275bf>\",\"WARC-IP-Address\":\"169.61.138.198\",\"WARC-Target-URI\":\"https://dataplatform.cloud.ibm.com/exchange/public/entry/preview?url=https://raw.githubusercontent.com/IBMDataScience/sample-notebooks/master/Cloud/HTML/Predicting%2Bchurn%2Bwith%2BSPSS%2Brandom%2Btree%2Balgorithm.html\",\"WARC-Payload-Digest\":\"sha1:O57FH47BYNO24NPUH3CHLWCOKTOKOXIS\",\"WARC-Block-Digest\":\"sha1:GPB3BAWBCN5A3ADX4I2D4CR3F7JW4EH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027322160.92_warc_CC-MAIN-20190825000550-20190825022550-00363.warc.gz\"}"}
|
https://math.stackexchange.com/questions/1519239/proving-that-if-ab-a-and-ba-b-then-both-matrices-are-idempotent/1519244
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[
"# Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent\n\nLet $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\\cdot A=A$ and $B\\cdot B=B$?\n\nSteps I took:\n\n1. Let $A= \\left[\\begin{array}{rr} a & b \\\\ c & d \\\\ \\end{array}\\right]$ and let $B= \\left[\\begin{array}{rr} w & x \\\\ y & z \\\\ \\end{array}\\right]$\n\n2. $\\left[\\begin{array}{rr} a & b \\\\ c & d \\\\ \\end{array}\\right] \\cdot \\left[\\begin{array}{rr} w & x \\\\ y & z \\\\ \\end{array}\\right] = \\left[\\begin{array}{rr} aw+by & ax+bz \\\\ cw+dy & cx+dz \\\\ \\end{array}\\right]$ (Which is also equal to A)\n\n3. $\\left[\\begin{array}{rr} w & x \\\\ y & z \\\\ \\end{array}\\right] \\cdot \\left[\\begin{array}{rr} a & b \\\\ c & d \\\\ \\end{array}\\right] = \\left[\\begin{array}{rr} wa+xc & wb+xd \\\\ ya+zc & yb+zd \\\\ \\end{array}\\right]$ (Which is also equal to B)\n\nAt this point I am stuck. I don't know how to proceed and I imagine that I started off on the wrong track to begin with. I'd like a hint to guide me in the right direction.\n\nMy proof after consultation with answerers below:\n\nProof:\n\n1) Since $AB=A$, we can say that: $(AB)A=AA$ which is equal to $A^2$\n\n2) Then, (by associativity of matrix multiplication), we can say that $A(BA) = AB$ (since $BA=B$)\n\n3) Then, $AB=A$ (since $AB=A$ was given)\n\n4) Therefore, $AA$ is equal to $A$\n\n5) Since $BA=B$, we can also say that: $(BA)B=BB$ which is equal to $B^2$\n\n6) Then, (by associativity of matrix multiplication) we can say that $B(AB)=BA$ (since $AB=A$)\n\n7) Then, $BA=B$\n\n8) Therefore, $BB$ is equal to $B$\n\n9) Thus, $AA$ is equal to $A$ and $BB$ is equal to $B$\n\nQ.E.D.\n\n• Are you assuming they're $2 \\times 2$ matrices? – Marcus M Nov 8 '15 at 16:52\n• @MarcusM Yes, I am. Is that wrong? – Cherry_Developer Nov 8 '15 at 16:53\n• @Cherry_Developer Well, it is not sufficient to look at the case of $2\\times 2$ matrices if you want to prove something for general matrices. If you want to prove a statement of $2\\times 2$ matrices, then it is not wrong. – Eff Nov 8 '15 at 16:54\n• This is not a good way to solve the exercise. The only thing you need is associativity of matrix product. – Crostul Nov 8 '15 at 16:58\n• One comment on the edit: associativity allows us to say $A(B)A = A(BA)$ or $(BA)B = B(AB)$. It is transitivity of equality that then allows us to say $AA = (AB)A = A(BA) = AB = A$, or $BB = B(AB) = B(AB) = BA = B$. So you're \"combining steps\" at step 2) and 6), – David Wheeler Nov 9 '15 at 23:50\n\nIf $AB = A$, and $BA = B$, then:\n\n$A^2 = AA = (AB)A$ (since we can replace $A$ with $AB$ since they are equal)\n\n$= A(BA)$ (by associativity of matrix multiplication)\n\n$= AB$ (since $BA = B$)\n\n$= A$ (since $AB = A$).\n\nNow you can try to use the same reasoning to prove $B^2 = B$.\n\nThis works for any two $n \\times n$ matrices, $A,B$ that satisfy $AB = A$ and $BA = B$, not just $2 \\times 2$ matrices.\n\n• I wrote up the proof. Would you be willing to look it over for me? – Cherry_Developer Nov 8 '15 at 21:10\n• @Cherry_Developer you are always welcome to post your own answer to a question here, even when it's your question. This happens a lot, actually. – David Wheeler Nov 9 '15 at 11:43\n• Would you be willing to look it over though? – Cherry_Developer Nov 9 '15 at 11:58\n• Sure, i don't see why not. – David Wheeler Nov 9 '15 at 12:07\n• I added it to my original post. Please look it over, if you can. Let me know what you think. – Cherry_Developer Nov 9 '15 at 12:40\n\nNotice that $AA = (AB)(AB) = A(BA)B = ABB = AB = A$.\n\n• As noted in the comments above, associativity of the matrix product is crucial. From my vantage point, I immediately see the truth, here, but a beginner might not understand why we can play fast and loose with the parentheses. – David Wheeler Nov 8 '15 at 17:53\n\nHint: Start with $AA = (AB)A$ (because of the first equality)\n\n$$A \\cdot B = A \\\\ \\implies (A \\cdot B) \\cdot A = A \\cdot A \\\\ \\implies A \\cdot (B \\cdot A) = A^2 \\\\ \\implies A \\cdot B = A^2 = A$$"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.84352803,"math_prob":0.99992025,"size":1492,"snap":"2019-26-2019-30","text_gpt3_token_len":537,"char_repetition_ratio":0.16801076,"word_repetition_ratio":0.17490494,"special_character_ratio":0.38270777,"punctuation_ratio":0.073619634,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998844,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-21T20:57:14Z\",\"WARC-Record-ID\":\"<urn:uuid:9623f525-4018-4a55-8781-02f803a5ead5>\",\"Content-Length\":\"166845\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c73f146-eb09-4538-8303-6937a0f0f62f>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f6b425f-c3ae-4be1-a99a-c7f7f5a75250>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1519239/proving-that-if-ab-a-and-ba-b-then-both-matrices-are-idempotent/1519244\",\"WARC-Payload-Digest\":\"sha1:RWCIBLIM2YPKZYTC5RHYV6FC45YGXFE6\",\"WARC-Block-Digest\":\"sha1:Y6LSNLFF4DEHD3ZFNACICUTLLBSMZERV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527204.71_warc_CC-MAIN-20190721205413-20190721231413-00475.warc.gz\"}"}
|
https://tensorflow.google.cn/tutorials/customization/custom_training_walkthrough?hl=zh-CN
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[
"TF 2.0 is out! Get hands-on practice at TF World, Oct 28-31. Use code TF20 for 20% off select passes.\n\n自定义训练: 演示\n\nTensorFlow 编程\n\n1. 数据集的导入与解析\n2. 选择模型类型\n3. 对模型进行训练\n4. 评估模型效果\n5. 使用训练过的模型进行预测\n\n环境的搭建\n\n配置导入\n\nfrom __future__ import absolute_import, division, print_function, unicode_literals\n\nimport os\nimport matplotlib.pyplot as plt\nimport tensorflow as tf\nprint(\"TensorFlow version: {}\".format(tf.__version__))\nprint(\"Eager execution: {}\".format(tf.executing_eagerly()))\nTensorFlow version: 2.0.0-rc2\nEager execution: True\n\n鸢尾花分类问题\n\n• 山鸢尾\n• 维吉尼亚鸢尾\n• 变色鸢尾",
null,
"Figure 1. 山鸢尾 (by Radomil, CC BY-SA 3.0), 变色鸢尾, (by Dlanglois, CC BY-SA 3.0), and 维吉尼亚鸢尾 (by Frank Mayfield, CC BY-SA 2.0).\n\n导入和解析训练数据集\n\n下载数据集\n\ntrain_dataset_fp = tf.keras.utils.get_file(fname=os.path.basename(train_dataset_url),\norigin=train_dataset_url)\n\nprint(\"Local copy of the dataset file: {}\".format(train_dataset_fp))\n8192/2194 [================================================================================================================] - 0s 0us/step\nLocal copy of the dataset file: /home/kbuilder/.keras/datasets/iris_training.csv\n\n检查数据\n\n120,4,setosa,versicolor,virginica\n6.4,2.8,5.6,2.2,2\n5.0,2.3,3.3,1.0,1\n4.9,2.5,4.5,1.7,2\n4.9,3.1,1.5,0.1,0\n\n• 共有 120 个样本。每个样本都有四个特征和一个标签名称，标签名称有三种可能。\n• 后面的行是数据记录，每个样本各占一行，其中：\n• 前四个字段是特征: 这四个字段代表的是样本的特点。在此数据集中，这些字段存储的是代表花卉测量值的浮点数。\n• 最后一列是标签:即我们想要预测的值。对于此数据集，该值为 0、1 或 2 中的某个整数值（每个值分别对应一个花卉名称）。\n\n# CSV文件中列的顺序\ncolumn_names = ['sepal_length', 'sepal_width', 'petal_length', 'petal_width', 'species']\n\nfeature_names = column_names[:-1]\nlabel_name = column_names[-1]\n\nprint(\"Features: {}\".format(feature_names))\nprint(\"Label: {}\".format(label_name))\nFeatures: ['sepal_length', 'sepal_width', 'petal_length', 'petal_width']\nLabel: species\n\n• 0 : 山鸢尾\n• 1 : 变色鸢尾\n• 2 : 维吉尼亚鸢尾\n\nclass_names = ['Iris setosa', 'Iris versicolor', 'Iris virginica']\n\n创建一个 tf.data.Dataset\n\nTensorFlow的 Dataset API 可处理在向模型加载数据时遇到的许多常见情况。这是一种高阶 API ，用于读取数据并将其转换为可供训练使用的格式。如需了解详情，请参阅数据集快速入门指南\n\nbatch_size = 32\n\ntrain_dataset = tf.data.experimental.make_csv_dataset(\ntrain_dataset_fp,\nbatch_size,\ncolumn_names=column_names,\nlabel_name=label_name,\nnum_epochs=1)\nWARNING:tensorflow:From /tmpfs/src/tf_docs_env/lib/python3.5/site-packages/tensorflow_core/python/data/experimental/ops/readers.py:521: parallel_interleave (from tensorflow.python.data.experimental.ops.interleave_ops) is deprecated and will be removed in a future version.\nInstructions for updating:\nUse `tf.data.Dataset.interleave(map_func, cycle_length, block_length, num_parallel_calls=tf.data.experimental.AUTOTUNE)` instead. If sloppy execution is desired, use `tf.data.Options.experimental_determinstic`.\n\nmake_csv_dataset 返回一个(features, label) 对构建的 tf.data.Dataset ，其中 features 是一个字典: {'feature_name': value}\n\nfeatures, labels = next(iter(train_dataset))\n\nprint(features)\nOrderedDict([('sepal_length', <tf.Tensor: id=68, shape=(32,), dtype=float32, numpy=\narray([7. , 4.9, 5.2, 5.7, 6.1, 4.9, 5.7, 5.1, 5.1, 4.4, 5.4, 6.4, 5.8,\n6.5, 4.9, 5.8, 5.2, 5.1, 6. , 6.8, 7.3, 5. , 7.7, 6.7, 7.2, 6.4,\n6.6, 6.5, 7.6, 5. , 5. , 6.5], dtype=float32)>), ('sepal_width', <tf.Tensor: id=69, shape=(32,), dtype=float32, numpy=\narray([3.2, 2.4, 3.4, 3.8, 2.8, 3. , 2.9, 3.8, 3.7, 2.9, 3.9, 3.2, 2.6,\n3. , 3.1, 4. , 3.5, 3.8, 2.2, 3.2, 2.9, 3. , 2.8, 3.1, 3.2, 3.1,\n3. , 2.8, 3. , 3.4, 3.3, 3. ], dtype=float32)>), ('petal_length', <tf.Tensor: id=66, shape=(32,), dtype=float32, numpy=\narray([4.7, 3.3, 1.4, 1.7, 4.7, 1.4, 4.2, 1.9, 1.5, 1.4, 1.7, 4.5, 4. ,\n5.8, 1.5, 1.2, 1.5, 1.5, 5. , 5.9, 6.3, 1.6, 6.7, 5.6, 6. , 5.5,\n4.4, 4.6, 6.6, 1.5, 1.4, 5.2], dtype=float32)>), ('petal_width', <tf.Tensor: id=67, shape=(32,), dtype=float32, numpy=\narray([1.4, 1. , 0.2, 0.3, 1.2, 0.2, 1.3, 0.4, 0.4, 0.2, 0.4, 1.5, 1.2,\n2.2, 0.1, 0.2, 0.2, 0.3, 1.5, 2.3, 1.8, 0.2, 2. , 2.4, 1.8, 1.8,\n1.4, 1.5, 2.1, 0.2, 0.2, 2. ], dtype=float32)>)])\n\nplt.scatter(features['petal_length'],\nfeatures['sepal_length'],\nc=labels,\ncmap='viridis')\n\nplt.xlabel(\"Petal length\")\nplt.ylabel(\"Sepal length\")\nplt.show()",
null,
"def pack_features_vector(features, labels):\n\"\"\"将特征打包到一个数组中\"\"\"\nfeatures = tf.stack(list(features.values()), axis=1)\nreturn features, labels\n\ntrain_dataset = train_dataset.map(pack_features_vector)\n\nDataset 的特征元素被构成了形如 (batch_size, num_features) 的数组。我们来看看前几个样本:\n\nfeatures, labels = next(iter(train_dataset))\n\nprint(features[:5])\ntf.Tensor(\n[[6.2 3.4 5.4 2.3]\n[7.4 2.8 6.1 1.9]\n[5. 3.5 1.3 0.3]\n[6.7 3.1 4.4 1.4]\n[7.2 3.2 6. 1.8]], shape=(5, 4), dtype=float32)\n\n选择模型类型\n\n选择模型",
null,
"图 2. 包含特征、隐藏层和预测的神经网络\n\n使用 Keras 创建模型\n\nTensorFlow tf.keras API 是创建模型和层的首选方式。通过该 API，您可以轻松地构建模型并进行实验，而将所有部分连接在一起的复杂工作则由 Keras 处理。\n\ntf.keras.Sequential 模型是层的线性堆叠。该模型的构造函数会采用一系列层实例；在本示例中，采用的是 2 个密集层（各自包含10个节点）,以及 1 个输出层（包含 3 个代表标签预测的节点。第一个层的 input_shape 参数对应该数据集中的特征数量，它是一项必需参数：\n\nmodel = tf.keras.Sequential([\ntf.keras.layers.Dense(10, activation=tf.nn.relu, input_shape=(4,)), # 需要给出输入的形式\ntf.keras.layers.Dense(10, activation=tf.nn.relu),\ntf.keras.layers.Dense(3)\n])\n\n使用模型\n\npredictions = model(features)\npredictions[:5]\n<tf.Tensor: id=231, shape=(5, 3), dtype=float32, numpy=\narray([[ 0.764444 , -1.514788 , 4.705434 ],\n[-0.01505657, -1.4119253 , 4.742084 ],\n[ 0.4466586 , -0.7942972 , 3.0395765 ],\n[ 0.22428769, -1.1655877 , 4.2541385 ],\n[ 0.06556055, -1.5437315 , 4.723908 ]], dtype=float32)>\n\ntf.nn.softmax(predictions[:5])\n<tf.Tensor: id=236, shape=(5, 3), dtype=float32, numpy=\narray([[0.01902157, 0.0019471 , 0.9790314 ],\n[0.00849907, 0.00210242, 0.9893985 ],\n[0.06822292, 0.01972377, 0.91205335],\n[0.01739081, 0.00433216, 0.978277 ],\n[0.00937544, 0.00187536, 0.98874927]], dtype=float32)>\n\nprint(\"Prediction: {}\".format(tf.argmax(predictions, axis=1)))\nprint(\" Labels: {}\".format(labels))\nPrediction: [2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2]\nLabels: [2 2 0 1 2 1 1 2 1 1 2 0 0 2 1 0 0 0 2 0 0 0 1 1 2 2 2 2 2 2 0 0]\n\n训练模型\n\n定义损失和梯度函数\n\nloss_object = tf.keras.losses.SparseCategoricalCrossentropy(from_logits=True)\ndef loss(model, x, y):\ny_ = model(x)\n\nreturn loss_object(y_true=y, y_pred=y_)\n\nl = loss(model, features, labels)\nprint(\"Loss test: {}\".format(l))\nLoss test: 2.253187417984009\n\ndef grad(model, inputs, targets):\nwith tf.GradientTape() as tape:\nloss_value = loss(model, inputs, targets)\nreturn loss_value, tape.gradient(loss_value, model.trainable_variables)\n\n创建优化器",
null,
"图 3. 优化算法在三维空间中随时间推移而变化的可视化效果。(来源: 斯坦福大学 CS231n 课程，MIT 许可证，Image credit: Alec Radford)\n\nTensorFlow有许多可用于训练的优化算法。此模型使用的是 tf.train.GradientDescentOptimizer ， 它可以实现随机梯度下降法（SGD）。learning_rate 被用于设置每次迭代（向下行走）的步长。 这是一个 超参数 ，您通常需要调整此参数以获得更好的结果。\n\nprint(\"Step: {}, Initial Loss: {}\".format(optimizer.iterations.numpy(),\nloss_value.numpy()))\n\nprint(\"Step: {}, Loss: {}\".format(optimizer.iterations.numpy(),\nloss(model, features, labels).numpy()))\nStep: 0, Initial Loss: 2.253187417984009\nStep: 1, Loss: 1.8874703645706177\n\n训练循环\n\n1. 迭代每个周期。通过一次数据集即为一个周期。\n2. 在一个周期中，遍历训练 Dataset 中的每个样本，并获取样本的特征x）和标签y）。\n3. 根据样本的特征进行预测，并比较预测结果和标签。衡量预测结果的不准确性，并使用所得的值计算模型的损失和梯度。\n4. 使用 optimizer 更新模型的变量。\n5. 跟踪一些统计信息以进行可视化。\n6. 对每个周期重复执行以上步骤。\n\nnum_epochs 变量是遍历数据集集合的次数。与直觉恰恰相反的是，训练模型的时间越长，并不能保证模型就越好。num_epochs 是一个可以调整的超参数。选择正确的次数通常需要一定的经验和实验基础。\n\n## Note: 使用相同的模型变量重新运行此单元\n\n# 保留结果用于绘制\ntrain_loss_results = []\ntrain_accuracy_results = []\n\nnum_epochs = 201\n\nfor epoch in range(num_epochs):\nepoch_loss_avg = tf.keras.metrics.Mean()\nepoch_accuracy = tf.keras.metrics.SparseCategoricalAccuracy()\n\n# Training loop - using batches of 32\nfor x, y in train_dataset:\n# 优化模型\n\n# 追踪进度\nepoch_loss_avg(loss_value) # 添加当前的 batch loss\n# 比较预测标签与真实标签\nepoch_accuracy(y, model(x))\n\n# 循环结束\ntrain_loss_results.append(epoch_loss_avg.result())\ntrain_accuracy_results.append(epoch_accuracy.result())\n\nif epoch % 50 == 0:\nprint(\"Epoch {:03d}: Loss: {:.3f}, Accuracy: {:.3%}\".format(epoch,\nepoch_loss_avg.result(),\nepoch_accuracy.result()))\nEpoch 000: Loss: 1.626, Accuracy: 35.000%\nEpoch 050: Loss: 0.093, Accuracy: 96.667%\nEpoch 100: Loss: 0.063, Accuracy: 99.167%\nEpoch 150: Loss: 0.052, Accuracy: 99.167%\nEpoch 200: Loss: 0.081, Accuracy: 98.333%\n\n可视化损失函数随时间推移而变化的情况\n\nfig, axes = plt.subplots(2, sharex=True, figsize=(12, 8))\nfig.suptitle('Training Metrics')\n\naxes.set_ylabel(\"Loss\", fontsize=14)\naxes.plot(train_loss_results)\n\naxes.set_ylabel(\"Accuracy\", fontsize=14)\naxes.set_xlabel(\"Epoch\", fontsize=14)\naxes.plot(train_accuracy_results)\nplt.show()",
null,
"评估模型的效果\n\n5.93.04.31.511\n6.93.15.42.122\n5.13.31.70.500\n6.0 3.4 4.5 1.6 12\n5.52.54.01.311\n\n建立测试数据集\n\ntest_fp = tf.keras.utils.get_file(fname=os.path.basename(test_url),\norigin=test_url)\n8192/573 [============================================================================================================================================================================================================================================================================================================================================================================================================================================] - 0s 0us/step\ntest_dataset = tf.data.experimental.make_csv_dataset(\ntest_fp,\nbatch_size,\ncolumn_names=column_names,\nlabel_name='species',\nnum_epochs=1,\nshuffle=False)\n\ntest_dataset = test_dataset.map(pack_features_vector)\n\n根据测试数据集评估模型\n\ntest_accuracy = tf.keras.metrics.Accuracy()\n\nfor (x, y) in test_dataset:\nlogits = model(x)\nprediction = tf.argmax(logits, axis=1, output_type=tf.int32)\ntest_accuracy(prediction, y)\n\nprint(\"Test set accuracy: {:.3%}\".format(test_accuracy.result()))\nTest set accuracy: 96.667%\n\ntf.stack([y,prediction],axis=1)\n<tf.Tensor: id=115075, shape=(30, 2), dtype=int32, numpy=\narray([[1, 1],\n[2, 2],\n[0, 0],\n[1, 1],\n[1, 1],\n[1, 1],\n[0, 0],\n[2, 1],\n[1, 1],\n[2, 2],\n[2, 2],\n[0, 0],\n[2, 2],\n[1, 1],\n[1, 1],\n[0, 0],\n[1, 1],\n[0, 0],\n[0, 0],\n[2, 2],\n[0, 0],\n[1, 1],\n[2, 2],\n[1, 1],\n[1, 1],\n[1, 1],\n[0, 0],\n[1, 1],\n[2, 2],\n[1, 1]], dtype=int32)>\n\n使用经过训练的模型进行预测\n\n• 0: 山鸢尾\n• 1: 变色鸢尾\n• 2: 维吉尼亚鸢尾\npredict_dataset = tf.convert_to_tensor([\n[5.1, 3.3, 1.7, 0.5,],\n[5.9, 3.0, 4.2, 1.5,],\n[6.9, 3.1, 5.4, 2.1]\n])\n\npredictions = model(predict_dataset)\n\nfor i, logits in enumerate(predictions):\nclass_idx = tf.argmax(logits).numpy()\np = tf.nn.softmax(logits)[class_idx]\nname = class_names[class_idx]\nprint(\"Example {} prediction: {} ({:4.1f}%)\".format(i, name, 100*p))\nExample 0 prediction: Iris setosa (99.9%)\nExample 1 prediction: Iris versicolor (99.9%)\nExample 2 prediction: Iris virginica (99.1%)"
] |
[
null,
"https://tensorflow.google.cn/images/iris_three_species.jpg",
null,
"https://tensorflow.google.cn/tutorials/customization/custom_training_walkthrough_files/output_me5Wn-9FcyyO_0.png",
null,
"https://tensorflow.google.cn/images/custom_estimators/full_network.png",
null,
"https://cs231n.github.io/assets/nn3/opt1.gif",
null,
"https://tensorflow.google.cn/tutorials/customization/custom_training_walkthrough_files/output_agjvNd2iUGFn_0.png",
null
] |
{"ft_lang_label":"__label__zh","ft_lang_prob":0.63466644,"math_prob":0.9855499,"size":14816,"snap":"2019-43-2019-47","text_gpt3_token_len":8176,"char_repetition_ratio":0.13725358,"word_repetition_ratio":0.04833837,"special_character_ratio":0.36190605,"punctuation_ratio":0.31778815,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95125324,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,3,null,null,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-17T03:02:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f582be7c-7590-4374-b04f-c9fd056d2ac1>\",\"Content-Length\":\"104425\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec5613ab-eeb7-45d0-b9b6-d20eeaff8998>\",\"WARC-Concurrent-To\":\"<urn:uuid:6952c010-6489-4cca-942c-4242b5817485>\",\"WARC-IP-Address\":\"172.217.164.131\",\"WARC-Target-URI\":\"https://tensorflow.google.cn/tutorials/customization/custom_training_walkthrough?hl=zh-CN\",\"WARC-Payload-Digest\":\"sha1:RJF47P2FFZXVC5EH7NFKWJOPZCGPP3QM\",\"WARC-Block-Digest\":\"sha1:6MCSQVJBNYBTGIMLNRQFULTBRYPY3JS6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986672548.33_warc_CC-MAIN-20191017022259-20191017045759-00519.warc.gz\"}"}
|
https://stackoverflow.com/questions/10256122/using-random-to-assign-chance
|
[
"# Using Random To Assign Chance\n\nIs there a way to set up an if statement like this where 1 if statement covers multiple integers?\n\n``````variable = random.randrange(1,10)\n\nif variable is between 1 - 3\nthen do this\nif variable is between 4-5\nthen do this\nif variable is between 6-9\nthen do this\n``````\n\nor maybe something like this\n\n``````a = 1,2,3,4,5,6,7,8,9,10\n\nvariable = random.randrange(1,10)\n\nif variable == a:\nthen do this\n``````\n• You should probably just post the segment of code that is relevant to using random to assign chance. – jamylak Apr 21 '12 at 4:14\n• Okay. How does it work? And what's the actual question (you are just telling me that you have troubling applying `random` module) to your game design. Describe the difficulty. What do you want to achieve? No one will read the entire program.... and do Sherlock Holmes lol. – CppLearner Apr 21 '12 at 4:19\n• part i'm having a problem with is at the botton (starts with the variable attack_player) – RandomPhobia Apr 21 '12 at 4:48\n• pay attention to @jamylak's comment, there's a reason 4 or more people up-voted it. This post only needs the last 10 lines, and telling people you're writing a strategy game is a good way to get your post moved off SO and onto gamedev.stackexchange.com – Shep Apr 21 '12 at 4:58\n• @RandomPhobia much better... although... what's the code on top add? – Shep Apr 21 '12 at 5:17\n\n## 1 Answer\n\nWhat about choosing a random number for each player, then the player with the largest value gets to attack first.\n\nOr, if this isn't turn based, then set a threshold, and if the random number is over the threshold, then the player can attack.\n\nThe threshold could also be a random number. For example:\n\n``````player1.attack = randn ()\n[player2 etc]\nminval = randn ()\nfor player in players:\nif player.attack > minval:\n[...attack...]\n``````\n\nThe question at the bottom is easy to implement, basically exactly as you've written it:\n\n``````variable = random.randrange(1,10)\n\nif 0 <= variable < 3:\nthen do this\nif 3 <= variable < 5:\nthen do this\nif 5 <= variable < 9:\nthen do this\n``````\n• Turn based would be easy enough to set up, the game is more like this. Each player picks a class and each class has a different chance of attacking. Class 1(1 attack chance), Class 2(2 attack chances), Class 3(3 attack chances)... and the the players get the choice to choose between these classes so the code to choose who attacks must be dynamic. – RandomPhobia Apr 21 '12 at 4:25\n• I don't quite understand what you mean in your comment here - strategically speaking, why would anyone choose Class 1 or Class 2? – lxop Apr 21 '12 at 4:53\n• With lower attack chance comes great delt damage so depending on how the player wants to go they can pick a class with high damage ,but low chance of actually attacking, low damage but a very good chance of attacking or something in between (a \"balance class\" if you will) – RandomPhobia Apr 21 '12 at 5:02\n• that will works perfectly lxop. – RandomPhobia Apr 21 '12 at 5:19"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.87754303,"math_prob":0.9271329,"size":562,"snap":"2019-13-2019-22","text_gpt3_token_len":167,"char_repetition_ratio":0.16487455,"word_repetition_ratio":0.083333336,"special_character_ratio":0.30071175,"punctuation_ratio":0.1764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.953023,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T07:58:17Z\",\"WARC-Record-ID\":\"<urn:uuid:122cb51b-ba36-43d0-88f3-79bf5bceafbc>\",\"Content-Length\":\"128145\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6693cb66-0a81-49a6-ba4c-5b0f1bffabbb>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca895369-220e-44df-8276-37e517fc524b>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/10256122/using-random-to-assign-chance\",\"WARC-Payload-Digest\":\"sha1:BC3TQUAN7FQQ3FCXRB4NK3STV7EJHHZY\",\"WARC-Block-Digest\":\"sha1:MKOIVHCHAJLKAC5AWQNQJ7PGQL54LA3Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256764.75_warc_CC-MAIN-20190522063112-20190522085112-00051.warc.gz\"}"}
|
http://techiemathteacher.com/2014/05/21/sum-of-arithmetic-progression/
|
[
"# Sum of Arithmetic Progression\n\nSum of arithmetic progression is very interesting because of its application in real life from basic arithmetic to Olympiad math levels. When dealing with the sum of arithmetic series the most important number is the middle term. I have read a story about a little boy how he easily sum up the sum of the first 100 numbers when he was 10 years old. That kid is no other than Carl Friedrich Gauss.\n\nFormula:\n\nLet,",
null,
"$S_n$– Sum of series",
null,
"$a_m$– middle term",
null,
"$d$– common difference",
null,
"$n$– number of terms",
null,
"$a_1$– first term",
null,
"$a_n$– last term\n\nFormula Derivation:\n\nGiven the middle term:",
null,
"$S_n=a_m\\cdot n$ – 1st formula\n\nMiddle term can be calculated by the formula",
null,
"$a_m=\\displaystyle\\frac{a_1+a_n}{2}$",
null,
"$S_n=a_m\\cdot n$",
null,
"$S_n=\\displaystyle\\frac{n(a_1+a_n)}{2}$ – 2nd formula\n\nWe also learned from the previous post that",
null,
"$a_n=a_1+(n-1)d$\n\nFrom 2nd formula,",
null,
"$S_n=\\displaystyle\\frac{n(a_1+a_n)}{2}$",
null,
"$S_n=\\displaystyle\\frac{n}{2}[a_1+a_1+(n-1)d]$",
null,
"$S_n=\\displaystyle\\frac{n}{2}[2a_1+(n-1)d]$3rd formula\n\nSample Problem 1:\n\nThe first term of an arithmetic sequence is 4, the last term is 60. The second term is obtained by adding 4. Find the sum of the series.\n\nSolution:\n\nWe have three formula available, we need to find the best formula according to the given. Better to list down the given.",
null,
"$a_1=4$,",
null,
"$a_n=60$,",
null,
"$d=4$\n\nThe three formulas need the number of terms and that is what we need to find out.\n\nUsing the formula for nth term,",
null,
"$a_n=a_1+(n-1)d$",
null,
"$60=4+(n-1)4$",
null,
"$n=15$\n\nSolving for sum of the series, we can solve one of the three formulas.",
null,
"$S_n=\\displaystyle\\frac{n(a_1+a_n)}{2}$",
null,
"$S_n=\\displaystyle\\frac{15(4+60)}{2}$",
null,
"$S_n=480$\n\nSample Problem 2:\n\nThe nth term of an arithmetic series is given by the formula",
null,
"$a_n=5+2n$. Find the sum of the first 100 terms.\n\nSolution:\n\nWe’re doomed right? Where is our a1,an,d,n?\n\nWith the previous post there is the same problem and we know that the coefficient of n is the common difference. Thus, d=2.\n\nSolving for",
null,
"$a_1$ :\n\nLet n=1",
null,
"$a_n=5+2n$",
null,
"$a_1=5+2(1)$",
null,
"$a_1=7$\n\nSolving for",
null,
"$a_{100}$\n\nLet n=100",
null,
"$a_n=5+2n$",
null,
"$a_{100}=5+2(100)$",
null,
"$a_{100}=205$\n\nSolving for the sum of the first 100th terms:",
null,
"$S_n=\\displaystyle\\frac{n(a_1+a_n)}{2}$",
null,
"$S_{100}=\\displaystyle\\frac{100(7+205)}{2}$",
null,
"$S_{100}=10,600$\n\nWorked Problem 3:\n\nFind the sum of the first 50 positive multiples of 3.\n\nSolution:\n\nThe first multiple of 3 is obviously 3, then 6, 9, 12,…\n\nThe last term then is 3(50)=150.",
null,
"$S_n=\\displaystyle\\frac{n(a_1+a_n)}{2}$",
null,
"$S_{50}=\\displaystyle\\frac{50(3+150)}{2}$",
null,
"$S_{50}=3825$\n\nWorked Problem 4:\n\nThe logs of lumber are piled in a shape of a triangle where the tip of the pile has one lumber. The next row has two, three and so on. If the bottom part of the pile have 15 lumbers. How many logs of lumber are there?\n\nSolution:\n\nThe solution is like finding the sum of first 15 counting numbers. There is a special formula for that which is",
null,
"${n+1\\choose 2}$",
null,
"$S_n={n+1\\choose 2}=\\displaystyle\\frac{(n+1)n}{2}$",
null,
"$S_n=\\displaystyle\\frac{(15+1)15}{2}$",
null,
"$S_n=120$\n\nWorked Problem 5: Adapted from Engr. Joselito Torculas of Adamson University Team\n\nFind the value of",
null,
"$a_2+a_4+a_6+\\ldots + a_{96}+a_{98}$\n\nIf",
null,
"$a_1,a_2,a_3,\\ldots$is an arithmetic progression with common difference 1 and",
null,
"$a_1+a_2+a_3+\\dots+a_{97}+a_{98}=137$\n\nSolution:",
null,
"$a_1+a_2+a_3+\\dots+a_{97}+a_{98}=137$\n\nWe know that",
null,
"$a_2=a_1+1 \\to a_1=a_2-1$, generally",
null,
"$a_{n-1}=a_n-1$",
null,
"$(a_2-1)+a_2+(a_4-1)+a_4+\\dots+(a_{98}-1)+a_{98}=137$",
null,
"$2(a_2+a_4+a_6+\\dots+a_{96}+a_{98})-49=137$",
null,
"$2(a_2+a_4+a_6+\\dots+a_{96}+a_{98})=186$",
null,
"$a_2+a_4+a_6+\\dots+a_{96}+a_{98}=93$"
] |
[
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
"http://s0.wp.com/latex.php",
null,
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{"ft_lang_label":"__label__en","ft_lang_prob":0.92554444,"math_prob":0.9999405,"size":2311,"snap":"2021-04-2021-17","text_gpt3_token_len":572,"char_repetition_ratio":0.15041178,"word_repetition_ratio":0.020833334,"special_character_ratio":0.24231935,"punctuation_ratio":0.118609406,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000079,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T01:54:49Z\",\"WARC-Record-ID\":\"<urn:uuid:d693e80d-898f-4827-9542-00af64ac442a>\",\"Content-Length\":\"60957\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:921eba47-4c1c-4369-a921-5f2ed60208ad>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a4bd6f1-c801-4c9f-a9d6-8e3c1cfc3ec4>\",\"WARC-IP-Address\":\"184.168.164.1\",\"WARC-Target-URI\":\"http://techiemathteacher.com/2014/05/21/sum-of-arithmetic-progression/\",\"WARC-Payload-Digest\":\"sha1:YILLGRQAIIUSLTKW32FAQW7BLJANCE62\",\"WARC-Block-Digest\":\"sha1:HNHGWU3SVXTF27HJG7R33QKH64U6YLMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703531702.36_warc_CC-MAIN-20210123001629-20210123031629-00082.warc.gz\"}"}
|
https://www.bartleby.com/solution-answer/chapter-101-problem-33e-mathematical-applications-for-the-management-life-and-social-sciences-11th-edition/9781305108042/in-problems-31-36-both-a-function-and-its-derivative-are-given-use-them-to-find-critical-values/432b46a7-655c-11e9-8385-02ee952b546e
|
[
"",
null,
"",
null,
"",
null,
"Chapter 10.1, Problem 33E",
null,
"### Mathematical Applications for the ...\n\n11th Edition\nRonald J. Harshbarger + 1 other\nISBN: 9781305108042\n\n#### Solutions\n\nChapter\nSection",
null,
"### Mathematical Applications for the ...\n\n11th Edition\nRonald J. Harshbarger + 1 other\nISBN: 9781305108042\nTextbook Problem\n\n# In Problems 31-36, both a function and its derivative are given. Use them to find critical values, critical points, intervals on which the function is increasing and decreasing, relative maxima, relative minima, and horizontal points of inflection; sketch the graph of each function. y = x 3 ( x − 5 ) 2 27 d y d x = 5 x 2 ( x − 3 ) ( x − 5 ) 27\n\nTo determine\n\nTo calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of the function y=x3(x5)227 and the derivative dydx=5x2(x3)(x5)27 and also sketch the graph.\n\nExplanation\n\nGiven Information:\n\nThe provided function is y=x3(x5)227 and the derivative is dydx=5x2(x3)(x5)27.\n\nFormula Used:\n\nThe simple power rule of derivative,\n\nddx(xn)=nxn1\n\nTo obtain the relative maxima and minima of a function,\n\n1. Find the first derivative of the function.\n\n2. Set the derivative equal to 0 to obtain the critical points.\n\n3. Evaluate f(x) at values of x to the left and one to the right of each critical point to develop a sign diagram.\n\n(a) If f(x)>0 to the left and f(x)<0 to the right of the critical value, the critical point is a relative maximum point.\n\n(b) If f(x)<0 to the left and f(x)>0 to the right of the critical value, the critical point is a relative minimum point.\n\nCalculation:\n\nConsider the provided function y=x3(x5)227, and the derivative dydx=5x2(x3)(x5)27,\n\nNow, equate the derivative to 0.\n\ndydx=05x2(x3)(x5)27=0\n\nEvaluate the values of x from the equation:\n\n5x2(x3)(x5)27=0x=0,5,3\n\nHence, the value(s) of x are x=0,5,3.\n\nEvaluate the values of the original functions with the critical values:\n\nSubstitute 0 for x in the function y=x3(x5)227.\n\ny=(0)3(05)227=027=0\n\nHence, (0,0) is a critical point.\n\nPut x=3 in the equation y=x3(x5)227,\n\ny=(3)3(35)227=27(4)27=4\n\nHence, (3,4) is a critical point.\n\nPut x=5 in the equation y=x3(x5)227,\n\ny=(5)3(55)227=125(0)27=0\n\nHence, (5,0) is a critical point\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\n#### Find more solutions based on key concepts",
null,
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|
https://topfuturepoint.com/prime-numbers-of-20-2/
|
[
"# Prime Numbers of 20\n\nThere are 20 prime numbers under 8: 2, 3, 5, 7, 11, 13, 17 and 19 .\n\nSimilarly, how do you find the factorization of 20? 20 = 1 x 20, 2 x 10 , or 4 x 5. Factors of 20: 1, 2, 4, 5, 10, 20. Prime factors: 20 = 2 x 2 x 5, also written as 20 = 2² x 5 can go. When 20 is a clue in the Find the Factor puzzle, use 4 x 5 or 2 x 10.\n\nWhat is the prime factorization for 20 using the exponent? The prime factorization of 20 using the exponent is 22u22175 2 2 u2217 5 . First, we create prime factors for 20.\n\nWhat are the multiples of 20? Multiples of 20 are 20, 40, 60 80, 100, 120, 140 and so on.\n\nWhy is the second 20 not a prime number? No, 20 is not a prime number . The number 20 is divisible by 1, 2, 4, 5, 10, 20. For a number to be classified as a prime number, it must have exactly two factors. Since 20 has more than two factors, i.e. 1, 2, 4, 5, 10, 20, it is not a prime number.\n\n## What is the divisible by 20?\n\nWhen we list them in this way it becomes easy to see that the numbers divisible by 20 are 1, 2, 4, 5, 10 and 20 .\n\nSo is 20 a factor or a multiple? Table of Factors and Multiples\n\nHow many factors of 20 are there? The factors of 20 are 1, 2, 4, 5, 10 and 20 .\n\n## How do you find prime numbers?\n\nTo prove whether a number is a prime number, try dividing it by 2 first, and see if you get a whole number . If you do, it cannot be a prime number. If you don’t get a whole number, try dividing it by primes: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below) ).\n\nWhat is prime factor? The prime factors are the factors of a number which are, themselves, prime numbers . There are several methods of finding the prime factors of a number, but one of the most common is to use a prime factor tree.\n\nIs 20 a correct number?\n\nThe number 20 is not a whole number . This can be demonstrated by finding its proper divisor and showing that their sum is not equal to 20.\n\nIs every even number divisible by 20? As 20 is an even number, any number divisible by 20 is an even number.\n\n## Is 20 divisible by 3 yes or no?\n\nSince division does not give a whole number, this shows us that 20 3 . is not divisible by .\n\nWhat are the multiples of 4 to 20?\n\nMultiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 , …\n\nHow do you find a factor? How to find the number of factors?\n\n1. Find its prime factorization, that is, express it as the product of primes.\n2. Write the prime factorization in exponentiation form.\n3. Add 1 to each exponent.\n4. Multiply all the resulting numbers.\n5. This product will give the number of factors of the given number.\n\nWhich of the following is not a factor of 20? 20! In order not to be a factor of , a number must not lie between 1 and 20 and should be called 20! Cannot be calculated as a product of numbers contained in . That means 29, which is prime and greater than 20 and 121 which is 11*11 and in 20! We just have n’t a factor of 11 20! Answer D and H.\n\n## Which of the following is not a factor of 20?\n\n20! In order not to be a factor of , a number must not lie between 1 and 20 and should be called 20! Cannot be calculated as a product of numbers contained in . That means 29, which is prime and greater than 20 and 121 which is 11*11 and in 20! We just have n’t a factor of 11 20! Answer D and H.\n\nWhat are prime numbers in mathematics? Prime numbers are special numbers greater than 1, which have exactly two factors, themselves and 1 . 19 is a prime number. It can only be divided by 1 and 19. … the prime numbers below 20 are: 2, 3, 5, 7, 11, 13, 17, 19. Don’t forget: the number 1 is not a prime number because it is only a factor.\n\nWhat are the first 20 mixed numbers?\n\nThe composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, and 20 .\n\nWhat are prime number charts? prime number chart\n\n### How do you solve prime factors?\n\nPrime Factorization Methods\n\n1. Step 1: Divide the given number by the smallest prime number. ,\n2. Step 2: Again, divide the quotient by the smallest prime number.\n3. Step 3: Repeat the process until the quotient becomes 1.\n4. Step 4: Lastly, multiply all the prime factors.\n\nHow do you factorize? To fully factorize an expression, find the highest common factor (HCF) of all terms . For example, the HCF of 4 x 2 is and 2 is the largest number that will be divisible into 4 and 6 and is the largest variable that will be divisible by and.\n\nWhat is the whole number of 20?\n\nList of Mersenne Prime Numbers and Whole Numbers\n\nWhich is the best number in the world? More Videos on YouTube\n\n## Is Armstrong a number?\n\nAn Armstrong number, also known as a narcissistic number, is a number that is equal to the sum of the cubes of its own digits . For example, 370 is an Armstrong number because 370 = 3*3*3 + 7*7*7 + 0*0*0 ."
] |
[
null
] |
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|
http://scholarpedia.org/article/Canards
|
[
"# Canards\n\nPost-publication activity\n\nCurator: Martin Wechselberger",
null,
"Figure 1: A canard (with hat) of the van der Pol oscillator\nCanards were discovered and first analyzed by French mathematicians (Benoît et al. ) who studied 2D relaxation oscillators, in particular the van der Pol oscillator. There the classical canard phenomenon explains the very fast transition upon variation of a parameter from a small amplitude limit cycle via canard cycles to a large amplitude relaxation cycle. This very fast transition called canard explosion happens within an exponentially small range of a control parameter. Thus this phenomenon is very hard to detect and it seems to be more like a canard in a newspaper. Furthermore, the shape of a canard cycle in the phase space resembles that of a duck (Figure 1). So the notion canard was born and the chase on these creatures began with either nonstandard (Benoît et al ) or standard (Eckhaus ) methods.\n\n## Methods\n\nThe main mathematical methods to analyze canards are nonstandard analysis (see, e.g., Benoît et al. , Diener ), matched asymptotic expansions (see, e.g., Eckhaus , Mishchenko et al. ) and the blow-up technique (see, e.g., Dumortier and Roussarie , Krupa and Szmolyan , Szmolyan and Wechselberger ), which extends geometric singular perturbation theory known as Fenichel theory to non-hyperbolic points. Moreover, with complex analysis, the canard phenomenon can be understood with either a geometrical point of view (see, e.g. Callot or Benoît et al. ) or the Gevrey theory which is convenient to study the divergent asymptotic expansions of canards (see, e.g. Canalis-Durand et al. or Benoît ).\n\n## Definition\n\nCanards are a phenomenon occurring in singularly perturbed systems (also known as slow-fast systems), i.e. they occur in systems of the form\n\n$$\\begin{array}{rcl} \\varepsilon\\dot{x}&=&f(x,z,\\varepsilon)\\\\ \\dot{z}&=&g(x,z,\\varepsilon)\\,, \\end{array}$$\n\nwith fast variable $$x\\in\\mathbb{R}^n\\ ,$$ slow variable $$z\\in\\mathbb{R}^m\\ ,$$ sufficiently smooth functions $$g\\ ,$$ $$f$$ and small parameter $$0\\!<\\!\\varepsilon\\!\\ll\\! 1\\ .$$ This system evolves on a slow time scale $$t=\\varepsilon \\tau\\ .$$ The limiting problem $$\\varepsilon\\to 0$$ on this slow time scale $$t$$ is called the reduced problem and describes the evolution of the slow variable $$z\\in\\mathbb{R}^m\\ .$$ The phase space of the reduced problem is the critical manifold $$S$$ defined by $$S:=\\{(x,z)\\in\\mathbb{R}^n\\times\\mathbb{R}^m\\,:\\,f(x,z,0)=0\\}\\ .$$ On the other hand, the limiting problem $$\\varepsilon\\to 0$$ on the fast time scale $$\\tau$$ is called the layer problem and describes the evolution of the fast variable $$x\\in\\mathbb{R}^n$$ for fixed $$z\\in\\mathbb{R}^m\\ .$$\n\nBy Fenichel theory (Fenichel ), normally hyperbolic subsets of $$S$$ perturb to nearby slow invariant manifolds $$S_\\varepsilon$$ of the singularly perturbed system with the flow given approximately by the flow of the reduced system. The most common case where normal hyperbolicity breaks down is given by a folded critical manifold $$S=S_a\\cup L \\cup S_r$$ where $$S_a$$ denotes the attracting part of the critical manifold $$S\\ ,$$ $$S_r$$ the repelling part of $$S\\ ,$$ and $$L$$ the fold of $$S$$ along which normal hyperbolicity is lost via a saddle-node bifurcation of the layer problem. Of course, any bifurcation of the layer problem leads to a loss of normal hyperbolicity of $$S\\ .$$ Canards are a special class of solutions of singularly perturbed systems where normal hyperbolicity is lost.",
null,
"Figure 2: Examples of canard solutions (green) near critical manifold S; normal hyperbolicity is lost at L, e.g., via saddle-node bifurcation of layer problem (Fig. A), or via transcritical bifurcation of layer problem (Fig. B)\n\nDefinition: A canard is a solution of a singularly perturbed system which follows an attracting slow manifold $$S_a\\ ,$$ passes close to a bifurcation point $$p\\in L$$ of the critical manifold, and then follows a repelling slow manifold $$S_r$$ for a considerable amount of time.\n\nIn geometric terms a canard solution corresponds to the intersection of an attracting and repelling slow manifold $$S_{a,\\varepsilon}\\cap S_{r,\\varepsilon}$$ near a non-hyperbolic point $$p\\in L\\ .$$ This geometric object is called a maximal canard.\n\nIn the case of a Hopf bifurcation of the layer problem, the phenomenon is quite different: there is a delayed loss of stability but there exist no canards (see Wallet , Neishtadt ). This case will not be explained here.\n\n## Codimension of the Canard Phenomenon\n\nCanards in singularly perturbed systems with just one slow variable ($$z\\in\\mathbb{R}$$) and one fast variable ($$x\\in\\mathbb{R}$$) are non generic, since maximal canards in such systems occur only for discrete values of a control parameter (Krupa and Szmolyan ), i.e. a one parameter family of singularly perturbed systems is needed to unfold this canard problem. In general, if canards are considered in singular perturbation problems with $$z\\in\\mathbb{R}$$ and $$x\\in\\mathbb{R}^n$$ where normal hyperbolicity is lost via a saddle-node bifurcation of the layer problem, then the singularly perturbed system can be reduced to a one slow and one fast variable system by a center manifold reduction.\n\n## The Classical Canard Phenomenon\n\n### Canard Cycles and their Explosion\n\nThe classical canard phenomenon (discovered by Benoît et al. ) occurs in singularly perturbed systems with $$x\\in\\mathbb{R}$$ and $$z\\in\\mathbb{R}\\ .$$ Its prototypical example is the van der Pol oscillator given by\n\n$$\\begin{array}{rcl} \\varepsilon\\dot{x}&=& z-x^3/3+x\\\\ \\dot{z}&=& a-x\\,, \\end{array}$$\n\nwhere $$a\\in\\mathbb{R}$$ denotes a control parameter (external forcing). The critical manifold $$S$$ is cubic shaped consisting of two attracting outer branches and a middle repelling branch connected via two fold-points where normal hyperbolicity is lost.",
null,
"Figure 3: Canard explosion of the van der Pol oscillator ($$\\varepsilon=0.01$$) within an exponentially small neighbourhood of a=0.998740451245 where the transition from relaxation oscillations to small amplitude limit cycles happens via canard cycles. The small amplitude limit cycles then terminate at a=1 via a Hopf bifurcation.\n\nHere, the nature of the classical canard phenomenon is the transition from a small amplitude oscillatory state to a (large amplitude) relaxation oscillatory state within an exponentially small range $$O(\\exp \\left(-1/\\varepsilon \\right))$$of the control parameter $$a$$ (see Figure 3). This transition, also called canard explosion, occurs through a sequence of canard cycles which can be asymptotically stable, but they are hard to observe in an experiment or simulation because of sensitivity to the control parameter and also because of sensitivity to noise. This is well known in chemical literature where a canard explosion is classified as a hard transition, because, for practical purposes, the transition from a small amplitude oscillation to a relaxation oscillation occurs immediately (see, e.g., Brøns and Bar-Eli , Peng et al. ).\n\n## Generic Canards\n\nCanards in singularly perturbed systems with two or more slow variables ($$z\\in\\mathbb{R}^m,\\, m\\ge 2$$) and one fast variable ($$x\\in\\mathbb{R}$$) are robust, since maximal canards generically persist under small parameter changes (Benoît , Mishchenko et al. , Szmolyan and Wechselberger , Benoît , Wechselberger ). In general, if canards are considered in singular perturbation problems with $$z\\in\\mathbb{R}^m,\\,m\\ge 2$$ and $$x\\in\\mathbb{R}^n$$ where normal hyperbolicity is lost via a saddle-node bifurcation of the layer problem, then the singularly perturbed system can be reduced to an $$m$$ slow and one fast variable system by a center manifold reduction. Therefore, a generic example with minimal dimension is given by a 3D singularly perturbed system with a 2D folded critical manifold.",
null,
"Figure 4: A: The reduced flow near a saddle singularity, B-C: The corresponding reduced flow near a folded saddle singularity (B) and the same reduced flow shown on the 2D folded critical manifold (C)\n\n### Classification\n\nThere exist different types of canards in 3D systems with 2D folded critical manifolds (Benoît , Szmolyan and Wechselberger ). The classification of these canards is based on the analysis of the corresponding 2D reduced system. There exist (discrete) folded singularities on the fold $$F\\ ,$$ called canard points, where the reduced flow crosses from the attracting to the repelling branch of the critical manifold. Generically, these folded singularities are either folded saddles (see Figure 4) or folded nodes (see Figure 5), resembling similar phase portraits as ordinary saddles or nodes in 2D phase space, besides that the reduced flow allow trajectories to cross the fold $$F$$ at these canard points. Note that in the folded saddle case (Figure 4B-4C) two solutions of the reduced flow cross via the folded saddle from the attracting branch to the repelling branch of the critical manifold or vice versa. More strikingly, a whole family of solutions (Figure 5B, shadowed sector) crosses via the folded node singularity from the attracting to the repelling branch. This is possible since existence and uniqueness results of ordinary differential equations are violated along the fold $$F\\ .$$",
null,
"Figure 5: The reduced flow near a node singularity, B-C: The corresponding reduced flow near a folded node singularity (B) and the same reduced flow shown on the 2D folded critical manifold (C)\n\nThe folded singularity corresponding to a saddle-node bifurcation of a folded saddle and a folded node is called a folded saddle-node type I. Another possible bifurcation scenario is a transcritical bifurcation of a folded singularity and an ordinary singularity. In this case, the ordinary singularity crosses via a transcritical bifurcation from the attracting branch $$S_a$$ to the repelling branch $$S_r$$ or vice versa. The corresponding folded singularity at this transcritical bifurcation is called a folded saddle-node type II.\n\n## The Generalized Canard Phenomenon\n\n### Mixed Mode Oscillations (MMOs)\n\nMMOs correspond to switching between small amplitude oscillations and relaxation oscillations. These patterns were first discovered in the famous Belousov-Zhabotinsky reaction and, since then, have been frequently observed in experiments and models of chemical and biological rhythms. One way to explain these patterns is based on canards of folded node type. The reason is that canards of folded node type can be responsible for small amplitude oscillations (Wechselberger ). A good intuition for MMOs is that a system moves dynamically from a small amplitude oscillatory state to a relaxation oscillatory state and the feature of the large relaxation oscillation is to bring the system back to the basin of attraction of the small amplitude oscillatory state. Other proposed mechanisms for MMOs are break-up/loss of stability of a Shilnikov homoclinic orbit (Koper ), break-up of an invariant torus (Larter and Steinmetz ) or slow passage through a delayed Hopf bifurcation (Larter et al. ).",
null,
"Figure 6: Explanation of MMOs via generalized canard phenomenon: the upper center figure shows a time trace of a $$1^4$$ MMO pattern ($$\\epsilon=0.01$$); the lower right figure summarizes Assumptions 1-3 in the phase space such that a singularly perturbed system possesses MMOs; the lower left figure shows the 4 small amplitude oscillations in the phase space near the corresponding folded node singularity.\n\nFigure 6 shows a $$1^4$$ MMO pattern consisting of 1 large amplitude oscillation and 4 small amplitude oscillations. In general, the symbol $$L^s$$ is assigned to a MMO pattern with $$L$$ large and $$s$$ small oscillations. The observed MMO pattern in Figure 6 can be explained as follows. Given a singularly perturbed system:\n\n$$\\begin{array}{rcl} \\varepsilon\\dot{x}&=&f(x,y,z,\\varepsilon)\\\\ \\dot{y}&=&g_1(x,y,z,\\varepsilon)\\\\ \\dot{z}&=&g_2(x,y,z,\\varepsilon) \\end{array}$$\n\n• Assumption 1: The critical manifold $$S$$ of the singularly perturbed system is (locally) a folded surface.\n• Assumption 2: The corresponding reduced problem possesses a folded node singularity.\n• Assumption 3: There exist a singular periodic orbit (Figure 6, lower right) which consists of a segment on $$S_a$$ (blue) within the singular funnel (shadowed region) with the folded node singularity (black circle) as an endpoint, fast fibers (red) of the layer problem and a global return mechanism (green). The global return mechanism has to be specified.\n\nTheorem (Brøns et al. ): Given the above singularly perturbed system under Assumptions 1-3. Then, for sufficiently small $$\\varepsilon\\ ,$$ there exist MMOs of type $$1^s\\ .$$ It is possible to calculate the number $$s$$ of small oscillations.\n\nThe number of small oscillations is given by $$s=[(\\mu+1)/(2\\mu)]$$ with $$\\mu=\\lambda_1/\\lambda_2\\le 1$$ where $$\\lambda_1$$ and $$\\lambda_2$$ are the corresponding eigenvalues of the folded node singularity whenever $$1/\\mu \\notin \\mathbb{N}$$ (except resonance cases)."
] |
[
null,
"http://scholarpedia.org/w/images/5/52/Van_canard.jpg",
null,
"http://scholarpedia.org/w/images/thumb/f/f9/2d-canards.jpg/400px-2d-canards.jpg",
null,
"http://scholarpedia.org/w/images/thumb/8/8d/Cexplosion-new3.jpg/600px-Cexplosion-new3.jpg",
null,
"http://scholarpedia.org/w/images/thumb/0/0f/Fsaddle-scholar.jpg/300px-Fsaddle-scholar.jpg",
null,
"http://scholarpedia.org/w/images/thumb/a/a3/Fnode-scholar.jpg/300px-Fnode-scholar.jpg",
null,
"http://scholarpedia.org/w/images/thumb/4/4a/Mmo-scholar.jpg/600px-Mmo-scholar.jpg",
null
] |
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|
https://lbs-to-kg.appspot.com/47.3-lbs-to-kg.html
|
[
"Pounds To Kg\n\n# 47.3 lbs to kg47.3 Pounds to Kilograms\n\nlbs\n=\nkg\n\n## How to convert 47.3 pounds to kilograms?\n\n 47.3 lbs * 0.45359237 kg = 21.454919101 kg 1 lbs\nA common question is How many pound in 47.3 kilogram? And the answer is 104.278650013 lbs in 47.3 kg. Likewise the question how many kilogram in 47.3 pound has the answer of 21.454919101 kg in 47.3 lbs.\n\n## How much are 47.3 pounds in kilograms?\n\n47.3 pounds equal 21.454919101 kilograms (47.3lbs = 21.454919101kg). Converting 47.3 lb to kg is easy. Simply use our calculator above, or apply the formula to change the length 47.3 lbs to kg.\n\n## Convert 47.3 lbs to common mass\n\nUnitMass\nMicrogram21454919101.0 µg\nMilligram21454919.101 mg\nGram21454.919101 g\nOunce756.8 oz\nPound47.3 lbs\nKilogram21.454919101 kg\nStone3.3785714286 st\nUS ton0.02365 ton\nTonne0.0214549191 t\nImperial ton0.0211160714 Long tons\n\n## What is 47.3 pounds in kg?\n\nTo convert 47.3 lbs to kg multiply the mass in pounds by 0.45359237. The 47.3 lbs in kg formula is [kg] = 47.3 * 0.45359237. Thus, for 47.3 pounds in kilogram we get 21.454919101 kg.\n\n## 47.3 Pound Conversion Table",
null,
"## Alternative spelling\n\n47.3 Pounds to kg, 47.3 Pounds in kg, 47.3 lbs to kg, 47.3 lbs in kg, 47.3 Pound to Kilogram, 47.3 Pound in Kilogram, 47.3 lbs to Kilogram, 47.3 lbs in Kilogram, 47.3 lb to Kilogram, 47.3 lb in Kilogram, 47.3 lb to kg, 47.3 lb in kg, 47.3 lb to Kilograms, 47.3 lb in Kilograms, 47.3 Pound to kg, 47.3 Pound in kg, 47.3 Pounds to Kilogram, 47.3 Pounds in Kilogram"
] |
[
null,
"https://lbs-to-kg.appspot.com/image/47.3.png",
null
] |
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|
https://www.tutorialspoint.com/how-to-draw-an-arrowed-line-on-an-image-in-opencv-python
|
[
"# How to draw an arrowed line on an image in OpenCV Python?\n\nOpenCV provides the function cv2.arrowedLine() to draw an arrowed line on an image. This function takes different arguments to draw the line. See the syntax below.\n\ncv2.arrowedLine(img, start, end, color, thickness, line_type, shift, tip_length)\n\n• img − The input image on which the line is to be drawn.\n\n• Start − Start coordinate of the line in (width, height) format.\n\n• End − End coordinate of the line in (width, height) format.\n\n• Color − Color of the line. For a red color in BGR format we pass (0, 0, 255)\n\n• Thickness − Thickness of the line in pixels.\n\n• line_type − Type of the line.\n\n• shift − The number of fractional bits.\n\n• tip_length − The length of the arrow tip in relation to the arrow length.\n\nOutput − It returns the image with lines drawn on it.\n\n## Steps\n\nFollow the steps given below to draw arrowed lines on an image −\n\nImport the required library. In all the following Python examples, the required Python library is OpenCV. Make sure you have already installed it.\n\nimport cv2\n\n\nimage = cv2.imread('cabinet.jpg')\n\n\nDraw the arrowed lines on image using cv2.arrowedLine() passing the required arguments.\n\ncv2.arrowedLine(image, (50, 100), (300, 450), (0,0,255), 3, 5, 0, 0.1)\n\n\nDisplay the image with arrowed lines drawn on it.\n\ncv2.imshow(\"ArrowedLine\",image)\ncv2.waitKey(0)\ncv2.destroyAllWindows()\n\n\nLet's have a look at some examples for a clear understanding of how it is done.\n\n## Example 1\n\nIn this program, we draw a red line on the image with the following line properties −\n\n• start_point = (50, 100),\n\n• end_point = (300, 450),\n\n• color = (0,0,255),\n\n• thickness = 3,\n\n• line_type = 5,\n\n• shift = 0, and\n\n• tip_length = 0.1\n\n# import required libraries\nimport cv2\n\n# Draw the arrowed line passing the arguments\ncv2.arrowedLine(image, (50, 100), (300, 450), (0,0,255), 3, 5, 0, 0.1)\n\ncv2.imshow(\"ArrowedLine\",image)\ncv2.waitKey(0)\ncv2.destroyAllWindows()\n\n\n## Output\n\nWhen you run the above program, it will produce the following output −",
null,
"## Example 2\n\nIn this program, we draw three different lines with different line properties −\n\nimport cv2\n\ncv2.arrowedLine(image, (50, 50), (200, 150), (0,0,255), 3, 7, 0, 0.2)\ncv2.arrowedLine(image, (300, 120), (50, 320), (0,255,255), 3, 7, 0, 0.2)\ncv2.arrowedLine(image, (50, 200), (500, 400), (255,0,255), 3, 7, 0, 0.05)\n\ncv2.imshow(\"ArrowedLines\",image)\ncv2.waitKey(0)\ncv2.destroyAllWindows()\n\n\n## Output\n\nOn execution, it will produce the following output window −",
null,
""
] |
[
null,
"https://www.tutorialspoint.com/assets/questions/media/394091-1664277066.jpg",
null,
"https://www.tutorialspoint.com/assets/questions/media/394091-1664277092.jpg",
null
] |
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|
https://www.cheenta.com/rectangle-pattern-amc-10a-2016-problem-10/
|
[
"",
null,
"Cheenta is joining hands with Aditya Birla Education Academy for AMC Training.\n\n# Rectangle Pattern | AMC-10A, 2016 | Problem 10\n\nTry this beautiful problem from Geometry based on Rectangle Pattern from AMC 10A, 2016, Problem 10.\n\n## Rectangle Pattern- AMC-10A, 2016- Problem 10\n\nA rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?\n\n• $1$\n• $2$\n• $4$\n• $6$\n• $8$\n\n### Key Concepts\n\nGeometry\n\nRectangle\n\nsquare\n\nAnswer: $2$\n\nAMC-10A (2016) Problem 10\n\nPre College Mathematics\n\n## Try with Hints\n\nGiven that length of the inner rectangle be $x$. Therefore the area of that rectangle is $x \\cdot 1=x$\nThe second largest rectangle has dimensions of $x+2$ and 3 , Therefore area $3 x+6$. Now area of the second shaded rectangle= $3 x+6-x=2 x+6$\n\ncan you finish the problem........\n\nNow the dimension of the largest rectangle is $x+4$ and 5 , and the area= $5 x+20$. The area of the largest shaded region is the largest rectangle- the second largest rectangle, which is $(5 x+20)-(3 x+6)=2 x+14$\n\ncan you finish the problem........\n\nNow The problem states that $x, 2 x+6,2 x+14$ is an arithmetic progression,i.e the common difference will be same . So we can say $(2 x+6)-(x)=(2 x+14)-(2 x+6) \\Longrightarrow x+6=8 \\Longrightarrow x=2$\n\nTherefore the side length =$2$\n\n## Subscribe to Cheenta at Youtube\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\n# Knowledge Partner",
null,
"",
null,
""
] |
[
null,
"https://www.facebook.com/tr",
null,
"https://www.cheenta.com/wp-content/uploads/2020/01/Cheenta-small.png",
null,
"https://www.cheenta.com/wp-content/uploads/2021/07/abea.png",
null
] |
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|
https://beta.geogebra.org/m/saau9kNx
|
[
"# Sign diagram\n\nBelow is the sign diagram for three linear expressions, denoted as , and (colour coded), where are constant whose value you can change by operating with your mouse on the sliders. Points , and indicate the points where each expression equals zero; continuous lines identify values of which render each expression positive, while the dashed lines refer to values of making an expression negative.]\nExperiment with various values of and determine the sign of the expressions and algebraically with pen and paper. Then use the construction to check your results."
] |
[
null
] |
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|
https://bookdown.org/animestina/intro_stats_rms/vectors.html
|
[
"# Chapter 5 Vectors\n\n## 5.1 Numeric Data\n\nLet’s imagine that someone in the audience works for R, likes the look of this workbook, and decides to sign me up to write textbooks for them. Imagine that they then start to send me my sales data monthly. Let’s suppose I have 50 sales in June, 20 in July, 10 in August, and 70 in September, but then no other sales for the rest of the year.\n\nTask: I want to create a variable, called monthly.sales that stores this data. The first number should be 50, the second 20, and so on. We again want to use the combine function c() to help us do this. To create our vector, we should write:\n\nmonthly.sales <- c(50, 20, 10, 70, 0, 0, 0)\nmonthly.sales\n## 50 20 10 70 0 0 0\n\nTo summarise, we have created a single variable called monthly.sales, and this variable is a vector with 7 elements.\n\nSo, now that we have our vector, how do we get information out of it? What if I wanted to know how many sales I made in August, for example. Since we started in June, August was the 3rd month of sales, so lets try:\n\nmonthly.sales\n## 10\n\nTurns out that the numbers I received for the August sales were wrong, and I actually had 100 sold, not 10! How can I fix this in my monthly.sales variable? I could make the whole vector again, but thats a lot of typing and wasteful, given that I only need to change one value.\n\nWe can just tell R to change that one specific value:\n\nmonthly.sales <- 100\nmonthly.sales\n## 50 20 100 70 0 0 0\n\nYou could also use the edit() and fix() functions, but we won’t be covering these in this session. You should check them out in your own time.\n\nYou can also ask R to return multiple values at once by indexing. For example, say I wanted to know how much I earned between July (2nd element) and October (5th element). The first way to ask for an element is to simply provide the numeric position of the desired element in the structure (vector, list…) in a set of square brackets [ ] at the end of the object name. I would ask R:\n\nmonthly.sales[2:5]\n## 20 100 70 0\n# equivalent to\nmonthly.sales[c(2, 3, 4, 5)]\n## 20 100 70 0\n\nNotice that the order matters here. If I asked for it in the reverse order, then R would output the data in the reverse too.\n\nmonthly.sales[5:2]\n## 0 70 100 20\n# equivalent to\nmonthly.sales[c(5, 4, 3, 2)]\n## 0 70 100 20\n\nNext I want to figure out how much money I’ll be making each month (given that the end of the year isn’t looking too good, I hope the next few months are!). Since I earn £5 per book, I can just multiply each element of monthly.sales by 5. Sounds pretty easy, and it is!\n\nmonthly.sales * 5\n## 250 100 500 350 0 0 0\n\n## 5.2 Text/Character Data\n\nAlthough you will mostly be dealing with numeric data, this isn’t always the case. Sometimes, you’ll use text. Let’s create a simple variable:\n\ngreeting <- \"hello\"\ngreeting\n## \"hello\"\n\nIt is important to note the use of quotation marks here. This is because R recognises this as a “character”, a string of characters, no matter how long. It can be a single letter, 'g', but it can equally well be a sentence, \"Descriptive statistics can be like online dating profiles: technically accurate and yet pretty darn misleading.\"\n\nBack to my R book example, I might want to create a variable that includes the names of the months. To do so, I could tell R:\n\nmonths <- c(\"June\", \"July\", \"August\", \"September\", \"October\", \"November\", \"December\")\n\nIn simple terms, you have now created a character vector containing 7 elements, each of which is the name of a month. Lets say I wanted to know how many what the 5th month was. What would I type?\n\nmonths\n## \"October\"\n\n## 5.3 Logical Data\n\nA logical element can take one of two values, TRUE or FALSE. Logicals are usually the output of logical operations (anything that can be phrased as a yes/no question, e.g., is x equal to y?). In formal logic, TRUE is represented as 1 and FALSE as 0. This is also the case in R.\n\nIf we ask R to calculate 2 + 2, it will always give the same answer\n\n2+2\n## 4\n\nIf we want R to judge whether something is a TRUE statement, we have to explicitly ask. For example:\n\n2+2 == 4\n## TRUE\n\nBy using the equality operator == , R is being forced to make a TRUE or FALSE judgement.\n\n2+2 == 3\n## FALSE\n\nWhat if we try to force R to believe some fake news (aka incorrect truths)?\n\n2+2 = 3\n## Error in 2 + 2 = 3: target of assignment expands to non-language object\n\nR cannot be convinced that easily. It understands that the 2+2 is not a variable (“non-language object”), and it won’t let you change what 2+2 is. In other words, it wont let you change the ‘definition’ of the value of 2.\n\nThere are several other logical operators that you can use, some of which are detailed in the below table.\n\nOperation R code Example Input Example Output\nLess than < 1 < 2 TRUE\nGreater than > 1 > 2 FALSE\nLess than or equal to <= 1 <= 2 TRUE\nGreater than or equal to <= 1 >= 2 FALSE\nEqual to == 1 == 2 FALSE\nNot equal to != 1 != 2 TRUE\nNot ! !(1==1) FALSE\nOr | (1==1) (1==2)\nAnd & (1==1) (1==2)\n\nLets apply some of these logical operators to our vectors. Lets use our monthly.sales vector, and ask R when I actually sold a book:\n\nmonthly.sales > 0\n## TRUE TRUE TRUE TRUE FALSE FALSE FALSE\n\nI can then store this into a vector:\n\nany.sales <- monthly.sales > 0\nany.sales \n## TRUE TRUE TRUE TRUE FALSE FALSE FALSE\n\nTo summarise, we have created a new logical vector called any.sales, whose elements are TRUE only if the corresponding sale is > 0.\n\nBut this output isn’t very helpful, as a big list of TRUE and FALSE values don’t give me much insight to which months I’ve sold my book in.\n\nWe can use logical indexing to ask for the names of the months where sales are > 0. Ask R:\n\nmonths[ any.sales > 0 ]\n## \"June\" \"July\" \"August\" \"September\"\n\nYou can apply the same logic to find the actual sales numbers for these months too:\n\nmonthly.sales [monthly.sales > 0]\n## 50 20 100 70\n\nYou could also do the same thing with text. It turns out that the one store that sold my R book didn’t always have books in stock. Let’s create a variable called stock.levels to have a look at this:\n\nstock.levels <- c(\"high\", \"high\", \"low\", \"high\", \"low\", \"out\", \"out\")\nstock.levels\n## \"high\" \"high\" \"low\" \"high\" \"low\" \"out\" \"out\"\n\nNow, apply the same logical indexing trick, but with the character vector instead, to see when the book was not in stock.\n\nmonths[stock.levels == \"out\"]\n## \"November\" \"December\"\n\nThat explains the lack of sales anyway! But what if I wanted to know when the shop either had low or no copies? You could ask R one of two things:\n\nmonths[stock.levels == \"out\" | stock.levels == \"low\"]\n## \"August\" \"October\" \"November\" \"December\"\n#Alternatively\nmonths[stock.levels != \"high\"]\n## \"August\" \"October\" \"November\" \"December\"\n\n### 5.3.1 Exercise\n\nTry to create a few variables of your own, and ask R to return you specific elements that are [blah]."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.90997607,"math_prob":0.86625963,"size":6915,"snap":"2020-24-2020-29","text_gpt3_token_len":1949,"char_repetition_ratio":0.12762263,"word_repetition_ratio":0.035633054,"special_character_ratio":0.3065799,"punctuation_ratio":0.12879789,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9567144,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-05T22:06:34Z\",\"WARC-Record-ID\":\"<urn:uuid:dc8c4d6d-1fb3-47b9-9183-155bf4ded846>\",\"Content-Length\":\"38835\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e9cbc1b-f6f1-4da9-bcf7-c700a00b03bf>\",\"WARC-Concurrent-To\":\"<urn:uuid:539b46af-3d6b-4fd2-b82e-1d16ce05a59f>\",\"WARC-IP-Address\":\"3.227.40.207\",\"WARC-Target-URI\":\"https://bookdown.org/animestina/intro_stats_rms/vectors.html\",\"WARC-Payload-Digest\":\"sha1:6G27A2OVIEH63EW6OXVZVDMVXEYXCCIN\",\"WARC-Block-Digest\":\"sha1:HLNNF2XEMOKC7VEFMA2G77GZM2D37GCG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655889877.72_warc_CC-MAIN-20200705215728-20200706005728-00461.warc.gz\"}"}
|
https://forum.getodk.org/t/use-calculate-function-to-determine-resulting-points-based-on-weighted-answer-options/29211
|
[
"",
null,
"# Use 'calculate' function to determine resulting points based on weighted answer options\n\nHello community,\n\nI'm new to ODK, using ODK Build (in browser) and I'm creating an XLSForm that should be able to calculate the underlying values of 'choose one' - answer options (always 2,1 or 0) x 'weight field' (priority), store the results as 'points' in an extra field (- not necessarily shown, can be hidden) and give the final results at the end of the survey or a section.\n\nHow this can be done?\n\nI know in the advanced-section of any question I can use 'calculate'. I would store the priority value (always changing) there or do the calculation directly there?\n\nHow to adress the values of the answer options with \\${'data name of question'}?\n\nWhere to store the results and give the final result?\n\nIs it possible to 'colorize' a question, based on the answer option chosen (like 2,1,0 -> green, yellow, red)?\n\nYes this is almost certainly possible, but I'm a bit unclear on exactly where this (these?) weights come from. Could you perhaps give a couple of specific examples to clarify, and I'm sure we can come up with something.\n\nsi es posible hacerlo,\ncon el ejemplo mas especifico como lo dice @Xiphware será mucho más fácil ayudarte\n\nThanks for your responses. I will try to make my approach more clear with a screenshot...\n\nSo basically, this is an example of the excel sheet my ODK form should be based on.\n\nEvery row here represents a question which you find in box 1.\nIn box 2 we'll have the answer options in the ODK form:\n\n• Existing / fulfilled (2 points),\n• Partly existing / partly fulfilled (1 point),\n• Not existing / not fulfilled (0 points).\n\nThe box behind circle 2 shows the answer options and the corresponding colors:\n2 = green, 1 = yellow, 0 = red\n\nCircle 3 contains the weight field (priority) for every question and in circle 4 the calculation should be done which is:\n\nI want to know how to implement the calculation in my form and where to store the values.\n\nFurther on, I want to know if its possible to applicate the 'traffic light' in my form, so when answer option 'existing' is chosen the question or a box appears green and so on...\n\nI hope the issue is more clear now."
] |
[
null,
"https://getodk-a3b1.kxcdn.com/uploads/default/original/2X/4/4d7c4387ae5b12f125b1ca9817f66f128a59da5b.png",
null
] |
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|
https://techcommunity.microsoft.com/t5/azure-observability/log-analytics-alert-filtered-query/td-p/3595386
|
[
"",
null,
"# Log Analytics Alert Filtered Query\n\nOccasional Visitor\n\n# Log Analytics Alert Filtered Query\n\nI made this query:\n\n``````AzureDevOpsAuditing\n| where ActorUPN != \"Azure DevOps Service\"\n| where Area == \"Release\"\n| summarize Count = count()\nby\nOperationName,\nbin(TimeGenerated, 1440min),\nActorUPN,\nDetails,\nScopeDisplayName\n| summarize sum(Count)\n| where sum_Count > 0``````\n\nAnd I made an alert that takes that query, evaluates the results of one day and sends an email.\n\nBut when the filtered results come out it doesnt show the other columns that im looking for (ActorUPN, details ...)\n\nThe query:\n\n``````AzureDevOpsAuditing\n| where ActorUPN != \"Azure DevOps Service\"\n| where Area == \"Release\"\n| summarize Count = count()\nby\nOperationName,\nbin_at(TimeGenerated, 1440min, datetime(2022-08-09T20:09:14.0000000Z)),\nActorUPN,\nDetails,\nScopeDisplayName\n| summarize sum(Count)\n| where sum_Count > 0\n| extend TimeGenerated = column_ifexists('TimeGenerated', datetime(2022-08-08T20:09:14.0000000Z))\n| summarize AggregatedValue = sum(sum_Count) by bin_at(TimeGenerated, 1440m, datetime(2022-08-09T20:09:14.0000000Z))``````\n\nand show TimeGenerated and AggregatedValue, nothing else.\n\n# Re: Log Analytics Alert Filtered Query\n\nWhen you do line #12, you essential drop the other columns, so they are not available for any other lines after #12 to process.\n\nOne way is to union the results, it does mean you process the data twice, once to get the count and once for the sum. See this simple example, as a place to start.\n\n``````union\n(\nUsage\n| where DataType ==\"AzureActivity\"\n| summarize count() by DataType, IsBillable, Solution, Quantity\n),\n(\nUsage\n| where DataType ==\"AzureActivity\"\n| summarize count() by IsBillable, Solution, Quantity\n| summarize Total = sum(count_), DataType=\"This is you Total\"\n) ``````",
null,
""
] |
[
null,
"https://www.facebook.com/tr",
null,
"https://techcommunity.microsoft.com/t5/image/serverpage/image-id/396514i53820367F979D99F/image-size/medium",
null
] |
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|
https://vestnik.utmn.ru/eng/energy/vypuski-arhiv/2013/112882/
|
[
"# Temperature waves in the ground near the base of heat construction\n\nTyumen State University Herald. Physical and Mathematical Modeling. Oil, Gas, Energy\n\nRelease:\n\nReleases Archive. Вестник ТюмГУ. Физико-математические науки. Информатика (№7, 2013)\n\nTitle:\nTemperature waves in the ground near the base of heat construction\n\nAuthors:\n\nAbout the authors:\n\nMaria B. Atmanskikh, Postgraduate Student of Mathematical Modeling Department, Tyumen State University\nIlya P. Rilo, Head of Research Department of Scientific and Production Association «Fundamentstroyarkos», Public Joint Stock Company\nAlexey V. Tatosov, Dr. Sci. (Phys.-Math.), Professor, Department of Fundamental Mathematics and Mechanics, University of Tyumen; [email protected]\n\nAbstract:\n\nNumerical investigation of nonstationary conductivity in the ground near the pile is presented in this paper. The ground and the pile are exposed to temperature waves – seasonal variations of air temperature. The process of heat conduction in the ground is described by a two-dimensional nonstationary heat conduction equation with a variable conductivity coefficient, without the source term in the axisymmetric coordinate system. Algorithm CONDUCT is used for numerical solution of the problem. “Stationary periodic” mode is achieved during five periods in the present problem. Heat exchange with the environment is more intensive in the pile near the surface than on the surface of the ground. Penetration depth of temperature waves decreases exponentially with the depth of the construction, therefore their greatest influence on temperature distribution is found near the upper boundary. Thermal inertia demonstrates hyperbolic properties of solution for heat conduction equation with time-periodic boundary conditions. It contributes to phase difference between temperature oscillations at different depths.\n\nKeywords:\n\nReferences:\n\n1. Lykov, A.V. Teorija teploprovodnosti [Theory of Heat Conductivity] M.: Vysshaja shkola, 1967. 599 p. (in Russian).\n\n2. Sivuhin D.V. Obshhij kurs fiziki: V 3 tt. T. 2. Termodinamika i molekuljarnaja fizika [Guidelines to Physics. In 3 vol. Vol. 2. Thermodynamics and Molecular Physics]. M.: Nauka, 1975. 519 p. (in Russian).\n\n3. Tihonov, A.N. Samarskij, A.A. Uravnenija matematicheskoj fiziki [Equations of Mathematical Physics]. M.: Nauka, 1972. 736 p. (in Russian).\n\n4. Kutateladze, S.S. Osnovy teorii teploobmena [Fundamentals of Heat Exchange Theory]. M.: Atomizdat, 1979. 416 p. (in Russian).\n\n5. Eger, D., Karslou, G. Teploprovodnost' tverdyh tel [Thermal Conductivity of Solids]. M.: Nauka, 1964. 488 p. (in Russian).\n\n6. Atmanskih, M.B., Zubkov, P.T. The Influence of Temperature and Thermoacoustic Waves on Heat Transmission in a Layer of Perfect Viscous Gas. Dinamika sploshnoj sredy — Continuous-Medium Dynamics. 2010. №. 126. Pp. 37-41. (in Russian).\n\n7. Kalitkin, N.N. Chislennye metody [Numerical Techniques]. M.: Nauka, 1978. 512 p.\n\n8. Fletcher, K. Vychislitel'nye metody v dinamike zhidkostej. T. 1. [Computational Techniques for Fluid Dynamics. Vol. 1]. M.: Mir, 1991. 504 p. (in Russian).\n\n9. Patankar, S.V. Computation of Conduction and Duct Flow Heat Transfer. New York: Hemisphere, 1990. 250 p.\n\n10. Patankar, S.V. Numerical Heat Transfer and Fluid Flow. New York: Hemisphere, 1980. 200 p.\n\n"
] |
[
null
] |
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|
https://ounces-to-grams.appspot.com/18700-ounces-to-grams.html
|
[
"Ounces To Grams\n\n# 18700 oz to g18700 Ounce to Grams\n\noz\n=\ng\n\n## How to convert 18700 ounce to grams?\n\n 18700 oz * 28.349523125 g = 530136.082438 g 1 oz\nA common question is How many ounce in 18700 gram? And the answer is 659.623088457 oz in 18700 g. Likewise the question how many gram in 18700 ounce has the answer of 530136.082438 g in 18700 oz.\n\n## How much are 18700 ounces in grams?\n\n18700 ounces equal 530136.082438 grams (18700oz = 530136.082438g). Converting 18700 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 18700 oz to g.\n\n## Convert 18700 oz to common mass\n\nUnitMass\nMicrogram5.30136082438e+11 µg\nMilligram530136082.438 mg\nGram530136.082438 g\nOunce18700.0 oz\nPound1168.75 lbs\nKilogram530.136082438 kg\nStone83.4821428572 st\nUS ton0.584375 ton\nTonne0.5301360824 t\nImperial ton0.5217633929 Long tons\n\n## What is 18700 ounces in g?\n\nTo convert 18700 oz to g multiply the mass in ounces by 28.349523125. The 18700 oz in g formula is [g] = 18700 * 28.349523125. Thus, for 18700 ounces in gram we get 530136.082438 g.\n\n## 18700 Ounce Conversion Table",
null,
"## Alternative spelling\n\n18700 Ounces in Grams, 18700 oz to Grams, 18700 oz to Gram, 18700 oz in Gram, 18700 Ounces to Gram, 18700 Ounces in Gram, 18700 Ounce to Gram, 18700 Ounce in Gram, 18700 Ounce in g,"
] |
[
null,
"https://ounces-to-grams.appspot.com/image/18700.png",
null
] |
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|
https://snippets.snyk.io/python/dimension-of-pandas-dataframe
|
[
"# How to use 'dimension of pandas dataframe' in Python\n\nEvery line of 'dimension of pandas dataframe' code snippets is scanned for vulnerabilities by our powerful machine learning engine that combs millions of open source libraries, ensuring your Python code is secure.",
null,
"## All examples are scanned by Snyk Code\n\nBy copying the Snyk Code Snippets you agree to\n``947def dframe(self, dimensions=None, multi_index=False):948 \"\"\"Convert dimension values to DataFrame.949950 Returns a pandas dataframe of columns along each dimension,951 either completely flat or indexed by key dimensions.952953 Args:954 dimensions: Dimensions to return as columns955 multi_index: Convert key dimensions to (multi-)index956957 Returns:958 DataFrame of columns corresponding to each dimension959 \"\"\"960 if dimensions is None:961 dimensions = [d.name for d in self.dimensions()]962 else:963 dimensions = [self.get_dimension(d, strict=True).name for d in dimensions]964 df = self.interface.dframe(self, dimensions)965 if multi_index:966 df = df.set_index([d for d in dimensions if d in self.kdims])967 return df``"
] |
[
null,
"https://snippets.snyk.io/images/vuln-safe.svg",
null
] |
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|
https://rdrr.io/cran/EDISON/man/NetworkRatioExp.html
|
[
"# NetworkRatioExp: Calculates the ratio of exponential network prior... In EDISON: Network Reconstruction and Changepoint Detection\n\n## Description\n\nThis function calculates the ratio of exponential network information sharing prior probabilities.\n\n## Usage\n\n `1` ```NetworkRatioExp(network.info) ```\n\n## Arguments\n\n `network.info` Network information collected using the function `CollectNetworkInfo`. Note that `network.info\\$new.nets` has to be set.\n\n## Value\n\nReturns the ratio [prior of new network]/[prior of old network].\n\n## Author(s)\n\nFrank Dondelinger\n\n## References\n\nFor information about the exponential information sharing prior, see:\n\nHusmeier et al. (2010), \"Inter-time segment information sharing for non-homogeneous dynamic Bayesian networks\", NIPS.\n\nDondelinger et al. (2012), \"Non-homogeneous dynamic Bayesian networks with Bayesian regularization for inferring gene regulatory networks with gradually time-varying structure\", Machine Learning.\n\n`NetworkProbExp`, `CalculatePriorRatio`"
] |
[
null
] |
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http://lyjunan.com/corpnews.html
|
[
"",
null,
"",
null,
"",
null,
"```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`"
] |
[
null,
"http://lyjunan.com/skin/default/images/temp_13.jpg",
null,
"http://lyjunan.com/skin/default/images/temp_16.jpg",
null,
"http://lyjunan.com/skin/default/images/innerbanner/news.jpg",
null
] |
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|
https://www.questarter.com/q/-0-1-is-commonly-called-the-unit-interval-is-there-a-similar-term-for-1-1-21_3295648.html
|
[
"# $[0,1]$ is commonly called the unit interval - is there a similar term for $[-1,1]$?\n\nby Candid Moon _Max_ Last Updated July 17, 2019 12:20 PM - source\n\nThe interval $$[0, 1]$$ is commonly called the 'unit interval'. Is there something similar for $$[-1, 1]$$? Like a pre-defined name.\n\nTags :\n\nI don't think so. Personally speaking, I have never found a particular definition for intervals of the form $$[a,b]$$ in general. I believe that the reason of that is that any closed interval is homeomorphic to $$[0,1]$$ and then it shares the same topological properties of $$[0,1]$$.\n\nAny pair of closed intervals $$[a,b]$$ and $$[c,d]$$ are homeomorphic, for any choice of $$a,b,c,d\\in\\Bbb R$$. In fact the function $$f:[a,b]\\to [c,d]$$ defined as $$f(x)=\\frac{x-a}{b-a}(d-c)+c$$ is a possible homeomorphism.\n\nBy taking $$c=0$$ and $$d=1$$, you find an explicit homeomorphism between $$[a,b]$$ and $$[0,1]$$.\n\nAlso, the unit interval appears in more contexts than a general closed interval. For instance in algebraic topology an homotopy is parametrised by $$[0,1]$$. Another example comes from probability theory. The probability in measure by a real number in the unit interval $$[0,1]$$. So it makes sense that $$[0,1]$$ has a privileged role and has an own definition.\n\nIt's the closed unit ball of $$\\mathbb{R}$$ in its usual absolute value norm.\n\nIn any normed space $$(X, \\|\\cdot\\|)$$, we have a closed unit ball (or disk) $$D=\\{x \\in X: \\|x\\| \\le 1\\}$$ and a unit sphere\n\n$$S=\\{x \\in X: \\|x\\|=1\\}$$ The unit sphere in the reals is $$\\{-1,1\\}$$ of course, the boundary of the unit ball $$[-1,1]$$.\n\nThey often get indexed by their dimension $$n$$ if it is finite, so $$D^1$$ and $$S^0$$ in the above case. $$\\partial D^n = S^{n-1}$$ for all $$n\\ge 1$$.\n\n## Vocabulary needed 2\n\nUpdated March 18, 2017 00:20 AM\n\n## Terminology of Rectangle\n\nUpdated August 19, 2019 14:20 PM"
] |
[
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|
http://www.romannumerals.co/number-converter/457-in-roman-numerals/
|
[
"## What is 457 in Roman Numerals?\n\n### A: CDLVII\n\n457 = CDLVII\n\nYour question is, \"What is 457 in Roman Numerals?\", and the answer is 'CDLVII'. Here we will explain how to convert, write and read the number 457 in the correct Roman numeral figure format.\n\n## How is 457 converted to Roman numerals?\n\nTo convert 457 to Roman Numerals the conversion involves you to split it up into place values (ones, tens, hundreds, thousands), like this:\n\nPlace ValueNumberRoman Numeral\nConversion400 + 50 + 7CD + L + VII\nHundreds400CD\nTens50L\nOnes7VII\n\n## How to write 457 in Roman numerals?\n\nTo write 457 in Roman numerals correctly you combine the values together. The highest numerals should always precede the lower numerals in order of precedence to give you the correct written combination, like in the table above (top to bottom). like this:\n\nCD+L+VII = CDLVII\n\n## How do you read 457 as Roman numerals\n\nTo correctly read the number 457 as the Roman numeral CDLVII, It must be read as it is written; from left to right and from high to low numbers.\n\nIt is incorrect to use the Roman symbol CDLVII in a text, unless it represents an ordinal value. In any other usage case it should be written in the normal format (arabic number) 457.\n\n## More from Roman Numerals.co\n\n458 in Roman numerals\n\nNow you understand how to read and write 457 in Roman Numerals, see how the number 458 is written.\n\nConvert Another Number"
] |
[
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http://www.sci-pi.org.uk/maths/fern.html
|
[
"# An IFS Fern\n\nThe examples on the Logistic Map and the Mandelbrot Set have shown surprisingly complex behaviour from simple quadratic equations. Here something simpler is considered: a set of linear equations.\n\nConsider the following four linear transformations of a point in 2D space:\n\n1. x= 0;y= 0.16y\n2. x= 0.849x+0.037y; y=-0.037x+0.849y+1.6\n3. x= 0.197x-0.226y; y= 0.226x+0.197y+1.6\n4. x=-0.150x+0.283y; y= 0.260x+0.238y+0.44\n\nWhat would happen if one started at a random point, say (0.5,0.5), picked one of the above four transformations at random and applied it, and then repeatedly selected one of the above transformations at random and applied it? It sounds like a recipe for a random scatter of dots.",
null,
"Not a random scatter at all!\n\n```#!/usr/bin/python3\nfrom random import randint\nimport numpy as np\nimport matplotlib.pyplot as plt\n\ndef ifs(x,y):\nr=randint(1,4)\nif (r==1):\nx=0\ny=0.16*y\nif (r==2):\nnx=0.849*x+0.037*y\ny=-0.037*x+0.849*y+1.6\nx=nx\nif (r==3):\nnx=0.197*x-0.226*y\ny=0.226*x+0.197*y+1.6\nx=nx\nif (r==4):\nnx=-0.15*x+0.283*y\ny=0.26*x+0.238*y+0.44\nx=nx\nreturn(x,y)\n\nx=.5\ny=.5\n\nfor i in range (1,100):\n(x,y)=ifs(x,y)\n\nplt.axes().set_aspect('equal')\nplt.xlim(-3,3)\nplt.ylim(0,10)\nplt.ion()\nplt.show()\n\nfor i in range (200):\nvx=[]\nvy=[]\nfor j in range (100):\n(x,y)=ifs(x,y)\nvx=vx+[x]\nvy=vy+[y]\nplt.plot(vx,vy,'k,')\nplt.pause(0.0001)\n\nprint(\"Done\")\nplt.ioff()\nplt.show()\n```\n\nThe above code bundles points into collections of one hundred for efficiency before calling the plot routine. It plots 20,000 points, after discarding an initial 100 iterations. It should run in a little under half a minute.\n\nThe result is similar to the picture on this page, but distinctly sparser. The trick for making a better quality image reasonably quickly is to change the random selection of the four possible transformations. Whilst almost any scheme will ultimately produce the same result, if one biases in favour of the second transform then a well filled in fern is produced much faster.\n\nThe function `ifs` was modified as below. The probabilities used here are not optimal, but are a lot better than the previous uniform distribution.\n\n```def ifs(x,y):\nr=randint(1,10)\nif (r==1):\nx=0\ny=0.16*y\nelif (r<=8):\nnx=0.849*x+0.037*y\ny=-0.037*x+0.849*y+1.6\nx=nx\nelif (r==9):\nnx=0.197*x-0.226*y\ny=0.226*x+0.197*y+1.6\nx=nx\nelif (r==10):\nnx=-0.15*x+0.283*y\ny=0.26*x+0.238*y+0.44\nx=nx\nreturn(x,y)\n```\n\n## IFS?\n\nIterated Function System.\n\nThe fern given above was found by Michael Barnsley, and looks remarkably natural. Other 2D and 3D geometric patterns can also be produced from a small number of transformations. In the book The Science of Fractal Images (Springer-Verlag, 1988), Barnsley descibes the above 2D fern as being reduced from a 3D fern, again using just four linear transformations, and states that \"on a 68020 based workstation a single view of the 3D fern required one minute of computation.\" The Motorola 68020 was released in 1984 and used in some Apple Macintoshes and the Sun 3. His code must have been better optimsed than the code here."
] |
[
null,
"http://www.sci-pi.org.uk/maths/fern.png",
null
] |
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|
https://www.statsmedic.com/apstats-chapter3-day4
|
[
"top of page\n\n## Chapter 3 - Day 4 - Lesson 3.2\n\n##### Learning Targets\n• Make predictions using regression lines, keeping in mind the dangers of extrapolation.\n\n• Calculate and interpret a residual.\n\n• Interpret the slope and y intercept of a regression line.\n\n• Determine the equation of a least-squares regression line using technology or computer output.\n\n##### Activity: How good are the predictions for Barbie?\n\nStudents use the online applet to find the line of best fit for some Barbie Bungee data collected by one of the groups. The group forgot to record a value for 5 rubber bands, so students will use the line of best fit to make a prediction. It is then revealed that the group found their measurement for 5 rubber bands, leading students to think about how close their prediction was to the actual value (residual!).\n\nNotice that in the activity, we avoided \"formal\" language (residual, extrapolation). This is done intentionally to keep the activity accessible to all students. We always layer on the formality when we debrief the activity. Remember: Experience first, formalize later.\n\nWe let students write their own interpretation for slope and y-intercept when they were working on the activity. During the debrief, we formalize by dialing in the necessary components of interpreting slope and y-intercept.\n\n##### Teaching Tip:\n\nStudents are most familiar with slope intercept form for equations of lines (y = mx + b). So why do statisticians prefer\n\ny = a + bx? The answer is that most often in the real world, there are more than one explanatory variables that can help predict the response variable. Statisticians might create a model that had three explanatory variables (x1, x2, x3) that looks like this:\n\ny = a + bx1 + b2x2 + b3x3. The y-intercept (a) is a starting point for making a prediction for the response variable and then each time we add one more explanatory variable we are refining that prediction. This process is called multiple regression (not part of the AP® Statistics course).\n\nAlways have students use context rather than x and y when writing out a regression equation. Also, make sure they don’t forget the “hat” on the response variable, or the word “predicted” in front. This will help them with calculating residuals and interpreting slope and y-intercept.\n\nStudents will often mix up the order in a calculation of a residual by taking (predicted y – actual y). An easy way to remember the correct order of subtraction is to think AP = Actual − Predicted. They should be able to remember this one.\n\nbottom of page"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9115774,"math_prob":0.9605901,"size":2556,"snap":"2023-40-2023-50","text_gpt3_token_len":551,"char_repetition_ratio":0.120297804,"word_repetition_ratio":0.004672897,"special_character_ratio":0.20931143,"punctuation_ratio":0.0982906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.988858,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T22:48:20Z\",\"WARC-Record-ID\":\"<urn:uuid:1eaf5ed4-8d02-407d-b38c-8e754ea7bc30>\",\"Content-Length\":\"1050555\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36ec784e-ff0f-4faa-96b7-c5f0b9e4d562>\",\"WARC-Concurrent-To\":\"<urn:uuid:82cc6168-5127-4d54-9803-302bcb652169>\",\"WARC-IP-Address\":\"146.75.33.84\",\"WARC-Target-URI\":\"https://www.statsmedic.com/apstats-chapter3-day4\",\"WARC-Payload-Digest\":\"sha1:JFOOFUUKCUCYKKVYJUZYMB47GRQWRMLX\",\"WARC-Block-Digest\":\"sha1:7THKJTFQRS7DO2FMBEJQWOW2PRU7CG2H\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510326.82_warc_CC-MAIN-20230927203115-20230927233115-00556.warc.gz\"}"}
|
https://socratic.org/questions/how-do-you-find-a-power-series-representation-for-x-1-x-2-and-what-is-the-radius
|
[
"# How do you find a power series representation for x/(1-x^2) and what is the radius of convergence?\n\nOct 24, 2015\n\nUse the Maclaurin series for $\\frac{1}{1 - t}$ and substitution to find:\n\n$\\frac{x}{1 - {x}^{2}} = {\\sum}_{n = 0}^{\\infty} {x}^{2 n + 1}$\n\nwith radius of convergence $1$.\n\n#### Explanation:\n\nThe Maclaurin series for $\\frac{1}{1 - t}$ is ${\\sum}_{n = 0}^{\\infty} {t}^{n}$\n\nsince $\\left(1 - t\\right) {\\sum}_{n = 0}^{\\infty} {t}^{n} = {\\sum}_{n = 0}^{\\infty} {t}^{n} - t {\\sum}_{n = 0}^{\\infty} {t}^{n} = {\\sum}_{n = 0}^{\\infty} {t}^{n} - {\\sum}_{n = 1}^{\\infty} {t}^{n} = {t}^{0} = 1$\n\nSubstitute $t = {x}^{2}$ to get:\n\n$\\frac{1}{1 - {x}^{2}} = {\\sum}_{n = 0}^{\\infty} {\\left({x}^{2}\\right)}^{n} = {\\sum}_{n = 0}^{\\infty} {x}^{2 n}$\n\nMultiply by $x$ to get:\n\n$\\frac{x}{1 - {x}^{2}} = x {\\sum}_{n = 0}^{\\infty} {x}^{2 n} = {\\sum}_{n = 0}^{\\infty} {x}^{2 n + 1}$\n\nThis is a geometric series with common ratio ${x}^{2}$, so it will converge if $\\left\\mid {x}^{2} \\right\\mid < 1$, which is when $\\left\\mid x \\right\\mid < 1$. That is, the radius of convergence is $1$."
] |
[
null
] |
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|
http://www.mpim-bonn.mpg.de/node/10573
|
[
"# Discrete subgroups of small critical exponent\n\nPosted in\nSpeaker:\nShi Wang\nAffiliation:\nMPIM\nDate:\nThu, 07/01/2021 - 16:30 - 18:00\n\nIn this talk, I will present recent joint work with Beibei Liu. Let X be a simply connected, pinched negatively curved manifold, G be a finitely generated, torsion free, discrete subgroup of Isom(X). The critical exponent \\delta(G) is defined to be the exponential growth rate of the number of G-orbit points inside a ball in X with respect to the radius. We show that if the critical exponent of G is small enough, then G is convex cocompact.\n\n © MPI f. Mathematik, Bonn Impressum & Datenschutz"
] |
[
null
] |
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|
https://www.javatpoint.com/star-numbers-in-java
|
[
"# Star Numbers in Java\n\nIn this section, we are going to learn about stars numbers in Java. The stars number resembles the board of the Chinese checkers game. A star number is a hexagram. Here, a hexagram means a six-pointed star. Observe the following diagram.",
null,
"Mathematically, the number is represented by\n\nSn = 6 x n x (n - 1) + 1, where n >= 1\n\nThus,\n\nS1 = 6 x 1 x (1 - 1) + 1 = 6 x 1 x 0 + 1 = 0 + 1 = 1\n\nS2 = 6 x 2 x (2 - 1) + 1 = 6 x 2 x 1 + 1 = 12 + 1 = 13\n\nS3 = 6 x 3 x (3 - 1) + 1 = 6 x 3 x 2 + 1 = 36 + 1 = 37\n\nS4 = 6 x 4 x (4 - 1) + 1 = 6 x 4 x 3 + 1 = 72 + 1 = 73\n\nS5 = 6 x 5 x (5 - 1) + 1 = 6 x 5 x 4 + 1 = 120 + 1 = 121\n\nS6 = 6 x 6 x (6 - 1) + 1 = 6 x 6 x 5 + 1 = 180 + 1 = 181\n\n## Implementation\n\nThere are three ways to write the code for the star numbers in Java. One way is using the formula defined above, the other is recursive, and the last one is iterative. Let's start with the formula-based approach.\n\n### Formula based Approach\n\nFileName: StarNumbers.java\n\nOutput:\n\n```The first 10 star numbers are:\n1 13 37 73 121 181 253 337 433 541\n```\n\nTime-Complexity: The time-complexity of the above program is O(n), where n is the total number of star numbers that have to be found.\n\nSpace Complexity: The above program is not using any extra space; therefore, the space complexity of the above program is O(1).\n\n### Recursive Approach\n\nUsing recursion also, one can compute the star numbers. However, before implementing the recursion, one must know the recursive formula for computing the star numbers. The recursive formula is defined as:\n\nSn = Sn - 1 + 12 x (n - 1), where n >= 2 and S1 = 1\n\nThus,\n\nS2 = S2 - 1 + 12 x (2 - 1) = = S1 + 12 x 1 = 1 + 12 = 13\n\nS3 = S3 - 1 + 12 x (3 - 1) = = S2 + 12 x 2 = 13 + 24 = 37\n\nS4 = S4 - 1 + 12 x (4 - 1) = = S3 + 12 x 3 = 37 + 36 = 73\n\nS5 = S5 - 1 + 12 x (5 - 1) = = S4 + 12 x 4 = 73 + 48 = 121\n\nS6 = S6 - 1 + 12 x (6 - 1) = = S5 + 12 x 5 = 121 + 60 = 181\n\nObserve the following Java code.\n\nFileName: StarNumbers1.java\n\nOutput:\n\n```The first 10 star numbers are:\n1 13 37 73 121 181 253 337 433 541\n```\n\nTime-Complexity: For finding the nth star number, one has to recursive go from n to 1. Therefore, the time complexity of the above program is O(n2), where n is the total number of star numbers that have to be computed.\n\nSpace Complexity: If we ignore the implicit memory consumption due to recursion, the space complexity of the above program is O(1).\n\nIn order to optimize the above program, one has to use the memorization technique. The following program shows the same.\n\nFileName: StarNumbers2.java\n\nOutput:\n\n```The first 10 star numbers are:\n1 13 37 73 121 181 253 337 433 541\n```\n\nTime Complexity: Because of the memorization technique, the time complexity of the above program is O(n), where n is the total number of star numbers that have to be computed.\n\nSpace Complexity: The space complexity of the above program is O(n), where n is the total number of star numbers that have to be computed.\n\n### Iterative Approach\n\nUsing a loop also, one can easily compute the value of the star numbers. The following program shows the same.\n\nFileName: StarNumbers3.java\n\nOutput:\n\n```The first 10 star numbers are:\n1 13 37 73 121 181 253 337 433 541\n```\n\nTime Complexity: Because of the for-loop used in the calculation of finding each star number, the time complexity of the above program is O(n2), where n is the total number of star numbers one has to find.\n\nSpace Complexity: The space complexity of the above program is O(n), where n is the total number of star numbers that have to be computed.\n\nWe can optimize the above code to reduce the time complexity. See the following program.\n\nFileName: StarNumbers4.java\n\nOutput:\n\n```The first 10 star numbers are:\n1 13 37 73 121 181 253 337 433 541\n```\n\nTime Complexity: As there is only one loop for computing the star numbers; therefore, the time complexity of the above program is O(n), where n is the total number of star numbers one has to find.\n\nSpace Complexity: The space complexity of the above program is O(n), where n is the total number of star numbers that have to be computed.\n\nWe can still do some optimization to reduce the space complexity of the program. If we observe the recursive formula, we find that the nth star number is dependent on the (n- -1)th star number, and (n - 1)th star number is dependent on the (n - 2)th star number, and so on. That means the current star number is dependent on the previous star number. Therefore, we do not need an array for storing the star numbers. All we have to do is to store the previous star number in a variable to compute the current star number. The following program shows the same.\n\nFileName: StarNumbers5.java\n\nOutput:\n\n```The first 10 star numbers are:\n1 13 37 73 121 181 253 337 433 541\n```\n\nTime Complexity: As there is only one loop for computing the star numbers; therefore, the time complexity of the above program is O(n), where n is the total number of star numbers one has to find.\n\nSpace Complexity: As we are using only two variables for computing the star numbers; therefore, the space complexity of the above program is O(1).\n\n## Relation with the Other Numbers\n\n• There are many star numbers that are also triangular numbers. Numbers such as 1, 253, 49141, and many more are the star as well as the triangular number.\n• There are many star numbers that are also square numbers. Numbers such as 1 = 12, 121 = 112, 11881 = 1092, and many more are the star as well as the square number.\n• A star prime number is a number that is a prime as well as a star number. For example: 13, 37, 73, and many more are the star as well as the prime numbers.\n\n### Feedback",
null,
"",
null,
"",
null,
""
] |
[
null,
"https://static.javatpoint.com/core/images/star-numbers-in-java.png",
null,
"https://www.javatpoint.com/images/facebook32.png",
null,
"https://www.javatpoint.com/images/twitter32.png",
null,
"https://www.javatpoint.com/images/pinterest32.png",
null
] |
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|
http://foldoc.org/algebraic+data+type
|
[
"",
null,
"## algebraic data type\n\n(Or \"sum of products type\") In functional programming, new types can be defined, each of which has one or more constructors. Such a type is known as an algebraic data type. E.g. in Haskell we can define a new type, \"Tree\":\n\n``` data Tree = Empty | Leaf Int | Node Tree Tree\n\n```\nwith constructors \"Empty\", \"Leaf\" and \"Node\". The constructors can be used much like functions in that they can be (partially) applied to arguments of the appropriate type. For example, the Leaf constructor has the functional type Int -> Tree.\n\nA constructor application cannot be reduced (evaluated) like a function application though since it is already in normal form. Functions which operate on algebraic data types can be defined using pattern matching:\n\n``` depth :: Tree -> Int\ndepth Empty\t = 0\ndepth (Leaf n)\t = 1\ndepth (Node l r) = 1 + max (depth l) (depth r)\n\n```\nThe most common algebraic data type is the list which has constructors Nil and Cons, written in Haskell using the special syntax \"[]\" for Nil and infix \":\" for Cons.\n\nSpecial cases of algebraic types are product types (only one constructor) and enumeration types (many constructors with no arguments). Algebraic types are one kind of constructed type (i.e. a type formed by combining other types).\n\nAn algebraic data type may also be an abstract data type (ADT) if it is exported from a module without its constructors. Objects of such a type can only be manipulated using functions defined in the same module as the type itself.\n\nIn set theory the equivalent of an algebraic data type is a discriminated union - a set whose elements consist of a tag (equivalent to a constructor) and an object of a type corresponding to the tag (equivalent to the constructor arguments).\n\nLast updated: 1994-11-23\n\n### Nearby terms:\n\nTry this search on Wikipedia, Wiktionary, Google, OneLook."
] |
[
null,
"http://foldoc.org/foldoc.gif",
null
] |
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|
http://mrpt.ual.es/reference/devel/_c_point3_d_8h_source.html
|
[
"MRPT 1.9.9\nCPoint3D.h\nGo to the documentation of this file.\n1 /* +------------------------------------------------------------------------+\n2 | Mobile Robot Programming Toolkit (MRPT) |\n3 | https://www.mrpt.org/ |\n4 | |\n5 | Copyright (c) 2005-2019, Individual contributors, see AUTHORS file |\n8 +------------------------------------------------------------------------+ */\n9 #pragma once\n10\n11 #include <mrpt/math/CVectorFixed.h>\n12 #include <mrpt/poses/CPoint.h>\n14\n15 namespace mrpt::poses\n16 {\n17 /** A class used to store a 3D point.\n18 *\n19 * For a complete description of Points/Poses, see mrpt::poses::CPoseOrPoint,\n20 * or refer\n21 * to the <a href=\"http://www.mrpt.org/2D_3D_Geometry\" >2D/3D Geometry\n22 * tutorial</a> in the wiki.\n23 *\n24 * <div align=center>\n25 * <img src=\"CPoint3D.gif\">\n26 * </div>\n27 *\n28 * \\ingroup poses_grp\n29 * \\sa CPoseOrPoint,CPose, CPoint\n30 */\n31 class CPoint3D : public CPoint<CPoint3D, 3>,\n33 {\n36 public:\n37 /** [x,y,z] */\n39\n40 public:\n41 /** Constructor for initializing point coordinates. */\n42 inline CPoint3D(const double x = 0, const double y = 0, const double z = 0)\n43 {\n44 m_coords = x;\n45 m_coords = y;\n46 m_coords = z;\n47 }\n48\n49 /** Constructor from a XYZ 3-vector */\n50 explicit inline CPoint3D(const mrpt::math::CVectorFixedDouble<3>& xyz)\n51 : m_coords(xyz)\n52 {\n53 }\n54\n55 /** Constructor from an CPoint2D object. */\n56 explicit CPoint3D(const CPoint2D& p);\n57\n58 /** Constructor from an CPose3D object. */\n59 explicit CPoint3D(const CPose3D& p);\n60\n61 /** Constructor from an CPose2D object. */\n62 explicit CPoint3D(const CPose2D& p);\n63\n64 /** Constructor from lightweight object. */\n65 inline explicit CPoint3D(const mrpt::math::TPoint3D& p)\n66 {\n67 m_coords = p.x;\n68 m_coords = p.y;\n69 m_coords = p.z;\n70 }\n72\n73 /** Returns this point as seen from \"b\", i.e. result = this - b */\n74 CPoint3D operator-(const CPose3D& b) const;\n75\n76 /** Returns this point minus point \"b\", i.e. result = this - b */\n77 CPoint3D operator-(const CPoint3D& b) const;\n78\n79 /** Returns this point plus point \"b\", i.e. result = this + b */\n80 CPoint3D operator+(const CPoint3D& b) const;\n81\n82 /** Returns this point plus pose \"b\", i.e. result = this + b */\n83 CPose3D operator+(const CPose3D& b) const;\n84\n85 /** Return the pose or point as a 3x1 vector [x y z]' */\n86 void asVector(vector_t& v) const { v = m_coords; }\n87\n88 enum\n89 {\n91 };\n92 static constexpr bool is_3D() { return is_3D_val != 0; }\n93 enum\n94 {\n96 };\n97 static constexpr bool is_PDF() { return is_PDF_val != 0; }\n98 /** @name STL-like methods and typedefs\n99 @{ */\n100 /** The type of the elements */\n101 using value_type = double;\n102 using reference = double&;\n103 using const_reference = double;\n104 using size_type = std::size_t;\n105 using difference_type = std::ptrdiff_t;\n106\n107 // size is constant\n108 enum\n109 {\n111 };\n112 static constexpr size_type size() { return static_size; }\n113 static constexpr bool empty() { return false; }\n114 static constexpr size_type max_size() { return static_size; }\n115 static inline void resize(const size_t n)\n116 {\n117 if (n != static_size)\n118 throw std::logic_error(format(\n119 \"Try to change the size of CPoint3D to %u.\",\n120 static_cast<unsigned>(n)));\n121 }\n122 /** @} */\n123\n124 void setToNaN() override;\n125\n126 }; // End of class def.\n127\n128 /** Dumps a point as a string (x,y,z) */\n129 std::ostream& operator<<(std::ostream& o, const CPoint3D& p);\n130\n131 } // namespace mrpt::poses\nA compile-time fixed-size numeric matrix container.\nDefinition: CMatrixFixed.h:33\nstatic constexpr bool is_3D()\nDefinition: CPoint3D.h:92\nvoid setToNaN() override\nSet all data fields to quiet NaN.\nDefinition: CPoint3D.cpp:158\nstd::string std::string format(std::string_view fmt, ARGS &&... args)\nDefinition: format.h:26\nmrpt::math::CVectorFixedDouble< 3 > m_coords\n[x,y,z]\nDefinition: CPoint3D.h:38\nstd::ostream & operator<<(std::ostream &o, const CPoint2D &p)\nDumps a point as a string (x,y)\nDefinition: CPoint2D.cpp:102\nvoid asVector(vector_t &v) const\nReturn the pose or point as a 3x1 vector [x y z]'.\nDefinition: CPoint3D.h:86\nCPoint3D(const mrpt::math::TPoint3D &p)\nConstructor from lightweight object.\nDefinition: CPoint3D.h:65\nstatic constexpr bool empty()\nDefinition: CPoint3D.h:113\n#define DEFINE_SCHEMA_SERIALIZABLE()\nThis declaration must be inserted in all CSerializable classes definition, within the class declarati...\nstatic constexpr size_type size()\nDefinition: CPoint3D.h:112\nCPoint3D(const mrpt::math::CVectorFixedDouble< 3 > &xyz)\nConstructor from a XYZ 3-vector.\nDefinition: CPoint3D.h:50\nA base class for representing a point in 2D or 3D.\nDefinition: CPoint.h:24\ndouble x() const\nCommon members of all points & poses classes.\nDefinition: CPoseOrPoint.h:143\nA class used to store a 2D point.\nDefinition: CPoint2D.h:32\nA class used to store a 3D point.\nDefinition: CPoint3D.h:31\nClasses for 2D/3D geometry representation, both of single values and probability density distribution...\nmrpt::math::TPoint3D asTPoint() const\nDefinition: CPoint3D.cpp:164\nstd::size_t size_type\nDefinition: CPoint3D.h:104\nT x\nX,Y,Z coordinates.\nDefinition: TPoint3D.h:23\nThis is the global namespace for all Mobile Robot Programming Toolkit (MRPT) libraries.\nA class used to store a 2D pose, including the 2D coordinate point and a heading (phi) angle...\nDefinition: CPose2D.h:39\nA class used to store a 3D pose (a 3D translation + a rotation in 3D).\nDefinition: CPose3D.h:85\nCPoint3D operator-(const CPose3D &b) const\nReturns this point as seen from \"b\", i.e.\nDefinition: CPoint3D.cpp:110\ndouble value_type\nThe type of the elements.\nDefinition: CPoint3D.h:101\nCPoint3D(const double x=0, const double y=0, const double z=0)\nConstructor for initializing point coordinates.\nDefinition: CPoint3D.h:42\nCVectorFixed< double, N > CVectorFixedDouble\nSpecialization of CVectorFixed for double numbers.\nDefinition: CVectorFixed.h:32\nThe virtual base class which provides a unified interface for all persistent objects in MRPT...\nDefinition: CSerializable.h:30\nstatic void resize(const size_t n)\nDefinition: CPoint3D.h:115\n#define DEFINE_SERIALIZABLE(class_name, NS)\nThis declaration must be inserted in all CSerializable classes definition, within the class declarati...\nstatic constexpr bool is_PDF()\nDefinition: CPoint3D.h:97\nCPoint3D operator+(const CPoint3D &b) const\nReturns this point plus point \"b\", i.e.\nDefinition: CPoint3D.cpp:141\nstd::ptrdiff_t difference_type\nDefinition: CPoint3D.h:105\nstatic constexpr size_type max_size()\nDefinition: CPoint3D.h:114\n\n Page generated by Doxygen 1.8.14 for MRPT 1.9.9 Git: 24b95e159 Thu Jan 23 01:15:46 2020 +0100 at jue ene 23 01:30:10 CET 2020",
null,
""
] |
[
null,
"http://c29.statcounter.com/2835288/0/862cd600/0/",
null
] |
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|
https://discourse.julialang.org/t/how-to-evaluate-substitute-numerical-values-of-params-n-states-a-jacobian-in-modelingtoolkit-using-generate-jacobian-or-calculate-jacobian/71466
|
[
"# How to evaluate(substitute numerical values of params n states) a jacobian in modelingtoolkit using generate_jacobian or calculate_jacobian?\n\ni found that `ModelingToolkit.calculate_jacobian()` function symbolically generates Jacobian of ODESystem in MTK docs.\nHow does one substitute numerical values of parameters and state varables numerical values in this matrix to evaluate ?\n\nTry `Symbolics.substitute`\n\nthanks @baggepinnen\ni am doing currently following\n\n``````x₀map = states(odesys) .=> x0\npmap = parameters(odesys) .=> ps\njac = substitute.( ModelingToolkit.calculate_jacobian(odesys), (Dict([x₀map;pmap]),) )\njac = Symbolics.value.(jac)\n``````\n\nHope i am doing this right!\n\nLooks about right to me, just make sure the states and parameters are returned in the order you expect.",
null,
""
] |
[
null,
"https://emoji.discourse-cdn.com/twitter/smiley.png",
null
] |
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|
https://wiki.frictionalgames.com/hpl3/community/scripting/classes/cvector3f
|
[
"#",
null,
"Frictional Game Wiki\n\n### Sidebar\n\nhpl3:community:scripting:classes:cvector3f\n\n## cVector3f\n\n### Fields\n\nField Name Type Description\nx float The x value of the vector.\ny float The y value of the vector.\nz float The z value of the vector.\n\n### Functions\n\nReturn Type Function Name Parameters Description\nfloat GetElement uint64 alIdx,\nconst\nGets the value at the given index. (Indices 0, 1, and 2 are equal to x, y, and z, respectively.)\nvoid SetElement uint64 alIdx,\nfloat,\nconst\nSets the value at the given index to the given value. (Indices 0, 1, and 2 are equal to x, y, and z, respectively.)\nfloat SqrLength const Returns the length-squared of this vector.\nfloat Length const Returns the length of this vector.\nfloat Normalize Returns the normalization factor for this vector. (See Remarks.)\n\n### Remarks\n\nA normalized vector is a vector whose length is equal to one, otherwise known as a unit vector. To convert a vector into a unit vector, get the normalization factor by calling the `Normalize` function, then divide each of the vector's x, y, and z coordinates by that factor.\n\n```cVector3f vBaseVector(2.0, 5.0, 3.0);\nfloat fNormFactor = vBaseVector.Normalize();\ncVector3f vNormalizedVector(vBaseVector.x / fNormFactor,\nvBaseVector.y / fNormFactor,\nvBaseVector.z / fNormFactor);```",
null,
""
] |
[
null,
"https://wiki.frictionalgames.com/lib/tpl/dokuwiki/images/logo.png",
null,
"https://wiki.frictionalgames.com/lib/exe/indexer.php",
null
] |
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|
http://icpc.njust.edu.cn/Problem/Pku/1650/
|
[
"# Integer Approximation\n\nTime Limit: 1000MS\n\nMemory Limit: 65536K\n\n## Description\n\nThe FORTH programming language does not support floating-point arithmetic at all. Its author, Chuck Moore, maintains that floating-point calculations are too slow and most of the time can be emulated by integers with proper scaling. For example, to calculate the area of the circle with the radius R he suggests to use formula like R * R * 355 / 113, which is in fact surprisingly accurate. The value of 355 / 113 ≈ 3.141593 is approximating the value of PI with the absolute error of only about 2*10-7. You are to find the best integer approximation of a given floating-point number A within a given integer limit L. That is, to find such two integers N and D (1 <= N, D <= L) that the value of absolute error |A - N / D| is minimal.\n\n## Input\n\nThe first line of input contains a floating-point number A (0.1 <= A < 10) with the precision of up to 15 decimal digits. The second line contains the integer limit L. (1 <= L <= 100000).\n\n## Output\n\nOutput file must contain two integers, N and D, separated by space.\n\n## Sample Input\n\n3.14159265358979\n10000\n\n\n## Sample Output\n\n355 113\n\nN/A"
] |
[
null
] |
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|
https://derive-it.com/category/trigonometry/
|
[
"## The Radial Unit Vector\n\nIn this post, I find an expression for the radial unit vector, $\\vec{e}_r$. The three unit vectors in the following digram form a right-handed spherical coordinate system. This unit vector is easier to find than the other two unit vectors because all that is needed is vector addition. The Radial Unit Vector in Terms of Spherical Coordinates Suppose $r=1$. Using vector addition, $\\vec{r} = r’ \\cos\\phi \\vec{e}_x + r’ \\sin\\phi \\vec{e}_y + \\cos \\theta \\vec{e}_z$. Since $r=1$, the expression on the right is equal to $\\vec{e}_r$: $\\vec{e}_r = r’ \\cos\\phi \\vec{e}_x + r’ \\sin\\phi \\vec{e}_y + \\cos \\theta \\vec{e}_z$. […]\n\n## Proof of the Triangle Inequality for Real Numbers\n\nThe triangle inequality for real numbers is $|a+b| \\le |a| + |b|$ in which $a$ is a variable for a real number, and $b$ is a variable for a real number. Proof: I use four cases. Case 1: If $a=0$ and $b$ is any real number, then the left side of the triangle inequality is $|b|$. And the right side is $|b|$. So the left side is equal to the right side. Similarly, if $b=0$ and $a$ is any real number, then the left side is $|a|$ and the right side is $|a|$. Case 2: If $a$ and\n\n## Do the angle addition identities only work for positive angles?\n\nThe Angle Addition Identities were derived using $\\sin\\theta \\equiv \\frac{opp}{hyp}$ and $\\cos\\theta \\equiv \\frac{adj}{hyp}$. In addition, $\\theta$ was constrained to the interval of $[0, \\frac{\\pi}{2})$. The question addressed in this post is, do the Angle Addition Identities work if the arguments of the sine function and cosine functions are generalized to include negative numbers? In the following, $\\theta$ will still be constrained to the interval of $[0, \\frac{\\pi}{2})$. The generalization of choice is $\\displaystyle \\sin(-\\theta) = -\\sin\\theta$ and $\\displaystyle \\cos(-\\theta) = \\cos\\theta$ This generalization is applicable to the unit circle, specifically in the 1st and 4th quadrants. The 2nd\n\n## The Pythagorean Theorem and the Unit Circle\n\nThis following diagram shows that $\\displaystyle \\sin^2 \\theta + \\cos^2 \\theta = 1$, given $\\displaystyle \\cos \\theta \\equiv \\frac{adj}{hyp}$ and $\\displaystyle \\sin \\theta \\equiv \\frac{opp}{hyp}$ and the Pythagorean theorem. Here, $opp$, $hyp$, and $adj$ are variables for lengths. The values corresponding to these lengths are positive numbers or zero.\n\n## Double Angle Formulas\n\nIn the Angle Addition Identities post it was shown that $\\displaystyle \\sin(x+y) = \\sin x \\cos y + \\sin y \\cos x$ and $\\displaystyle \\cos(x+y) = \\cos x \\cos y – \\sin x \\sin y$. These identities are valid if $\\sin \\theta \\equiv \\frac{opp}{hyp}$ and $\\cos \\theta \\equiv \\frac{adj}{hyp}$, which implies that $\\theta \\in [0,\\frac{\\pi}{2})$ in order for $\\theta$ to be an acute angle of a right triangle. These constraints are imposed because the two Angle Addition Identities were derived using these constraints. The next step, following the steps in reference , is setting $y$ equal to\n\n## The sine function divided by its angle, and a certain limit\n\nWhat is $\\lim_{x\\rightarrow0} \\frac{\\sin x}{x}$ ? Recall the definition of a limit, repeated here for reference . A function $f(x)$ approaches a limit $A$ as $x$ approaches $a$ if, and only if, for each positive number $\\epsilon$ there is another, $\\delta$, such that whenever $0 < |x-a| < \\delta$ we have $|f(x) – A| < \\epsilon$. That is, when $x$ is near $a$ (within a distance $\\delta$ from it), $f(x)$ is near $A$ (within a distance $\\epsilon$ from it). In symbols we write $\\lim_{x \\rightarrow a} f(x) = A$. Using this definition, $a$ is $0$ in this case. Now\n\n## The Radian\n\nIn Chapter 2 of Reference , one of the footnotes has a definition of a radian. For a unit circle with a radius of one unit of length called $u$, one radian is the angle corresponding to an arc-length of $u$. In degrees, one radian is approximately 57 degrees, which is perhaps easier to visualize. A diagram of one radian is included here for reference. References Konrad Knopp. Theory and Application of Infinite Series. Dover Publications. 1990."
] |
[
null
] |
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|
https://www.experts-exchange.com/questions/26871049/In-MS-EXCEL-VBA-I-want-to-dynamically-calculate-count-sum-avg-for-a-row-as-its-added.html
|
[
"",
null,
"# In MS EXCEL VBA, I want to dynamically calculate count,sum,avg for a row as its added\n\nIn column A I have a list of names. Columns B,C,D have AVG,SUM,COUNT. COUNT and SUM get their values from column E,F,G,H and so on. Column B calculates the AVG from SUM and COUNT. From time to time I add info to the sheet which includes names that already exist in column A and add new names if they don't exist in column A. I would like to dynamically add the sum,avg,count to B,C,D as the sheet is updated. I know this can be done with a cell formula but I don't want cells to show #DIV/0! or zeros if there is no information for them. Can this be done with VBA instead? here is sample vba:\n\nFor y = 8 To last_cell\nIf player_name = Worksheets(\"Current Year\").Cells(y, 1) Then\nWorksheets(\"Current Year\").Cells(y, 5).Value = position\n\n' what i would like to calculate\nWorksheets(\"Current Year\").Cells(y, 2).Value = AVG\nWorksheets(\"Current Year\").Cells(y, 3).Value = SUM\nWorksheets(\"Current Year\").Cells(y, 4).Value = COUNT of range ' E \"y\" ' to so on\n\nplayer_exists = True\nEnd If\nMicrosoft Excel",
null,
"Last Comment\nmartyk\n\n8/22/2022 - Mon\nEric Zwiekhorst\n\ndear Dmalovich,\n\nit can be done with formula and without error like #DIV/0\n\ncellAVG = =IF( ISERROR(AVERAGE(E1:H1)),0,AVERAGE(E1:H1))\ncellSUM= =SUM(E1:H1)\ncell Count= =COUNTIF(E1:H1;\">0\")\n\nroyhsiao\n\nBecome an EE member today7-DAY FREE TRIAL\nMembers can start a 7-Day Free trial then enjoy unlimited access to the platform\nor\nLearn why we charge membership fees\nWe get it - no one likes a content blocker. Take one extra minute and find out why we block content.\nNot exactly the question you had in mind?\nSign up for an EE membership and get your own personalized solution. With an EE membership, you can ask unlimited troubleshooting, research, or opinion questions.\nroyhsiao\n\nThe cell shows #DIV/0 in avg. when the count is 0. I could add an if statement to avoid that error. here is an example.\n``````Sub Test()\nDim Player_total_count As Integer\nDim i As Integer\nDim y As Integer\n\nPlayer_total_count = Application.WorksheetFunction.CountA(Range(\"A1:A2000\"))\ni = 2\n\nFor i = 2 To Player_total_count\ny = Application.WorksheetFunction.CountA(Range(\"E\" & i & \":\" & \"H\" & i))\n\nIf y = 0 Then\nWorksheets(\"Current Year\").Cells(i, 3).Value = \"=sum(E\" & i & \":\" & \"H\" & i & \")\"\nWorksheets(\"Current Year\").Cells(i, 4).Value = \"=count(E\" & i & \":\" & \"H\" & i & \")\"\nElseIf y <> 0 Then\nWorksheets(\"Current Year\").Cells(i, 2).Value = \"=average(E\" & i & \":\" & \"H\" & i & \")\"\nWorksheets(\"Current Year\").Cells(i, 3).Value = \"=sum(E\" & i & \":\" & \"H\" & i & \")\"\nWorksheets(\"Current Year\").Cells(i, 4).Value = \"=count(E\" & i & \":\" & \"H\" & i & \")\"\nEnd If\n\nNext i\nEnd Sub\n``````\ndmang\n\nHi there\nI have attached a workbook that contains macro code that should provide what you're looking for.\nThere are sample columns and values provided.\nEach row's avg, sum, count are calculated based on the row contents..that is, in one row you can have 6 cols of values, in another - 8, etc\n\nRun the macro calcall.\n\nRegards\ndmang 26871049.xls\nExperts Exchange has (a) saved my job multiple times, (b) saved me hours, days, and even weeks of work, and often (c) makes me look like a superhero! This place is MAGIC!\nWalt Forbes\ndmalovich\n\nIt worked. Thanks.....\nmartyk\n\nYou can use formulas and avoid the display of #DIV/0 and zeros without resorting to a macro.\n\nIn column D, the COUNT column, use\n=if(count(e1:z1)>0,count(e1:z1),\"\")\n\nColumn D will be blank if all the columns E to Z are empty.\n\nThen columns B and C can use the D column result to avoid the unwanted display.\n\nIn Column B, the AVG column, use the formula\n=if(and(d1<>\"\" , d1>0),c1/d1,\"\")\n\nAnd in column C, the SUM column\n=if(and(d1<>\"\" , d1>0),sum(e1:z1),\"\")\n\nNote you must check both d1<>\"\" and d1>0 for the formula to work."
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"https://cdn.experts-exchange.com/images/experts-exchange/avatar-01-large.gif",
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"https://filedb.experts-exchange.com/files/public/2013/11/18/9b3cdc25-0086-4058-8dc3-233c37b01772.png",
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http://cran.itam.mx/web/packages/keras/vignettes/training_callbacks.html
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[
"Training Callbacks\n\nOverview\n\nA callback is a set of functions to be applied at given stages of the training procedure. You can use callbacks to get a view on internal states and statistics of the model during training. You can pass a list of callbacks (as the keyword argument callbacks) to the fit() function. The relevant methods of the callbacks will then be called at each stage of the training.\n\nFor example:\n\nlibrary(keras)\n\n# generate dummy training data\ndata <- matrix(rexp(1000*784), nrow = 1000, ncol = 784)\nlabels <- matrix(round(runif(1000*10, min = 0, max = 9)), nrow = 1000, ncol = 10)\n\n# create model\nmodel <- keras_model_sequential()\n\nmodel %>%\nlayer_dense(32, input_shape = c(784)) %>%\nlayer_activation('relu') %>%\nlayer_dense(10) %>%\nlayer_activation('softmax') %>%\ncompile(\nloss='binary_crossentropy',\noptimizer = optimizer_sgd(),\nmetrics='accuracy'\n)\n\n# fit with callbacks\nmodel %>% fit(data, labels, callbacks = list(\ncallback_model_checkpoint(\"checkpoints.h5\"),\ncallback_reduce_lr_on_plateau(monitor = \"val_loss\", factor = 0.1)\n))\n\nBuilt in Callbacks\n\nThe following built-in callbacks are available as part of Keras:\n\n callback_progbar_logger() Callback that prints metrics to stdout. callback_model_checkpoint() Save the model after every epoch. callback_early_stopping() Stop training when a monitored quantity has stopped improving. callback_remote_monitor() Callback used to stream events to a server. callback_learning_rate_scheduler() Learning rate scheduler. callback_tensorboard() TensorBoard basic visualizations callback_reduce_lr_on_plateau() Reduce learning rate when a metric has stopped improving. callback_csv_logger() Callback that streams epoch results to a csv file callback_lambda() Create a custom callback\n\nCustom Callbacks\n\nYou can create a custom callback by creating a new R6 class that inherits from the KerasCallback class.\n\nHere’s a simple example saving a list of losses over each batch during training:\n\nlibrary(keras)\n\n# define custom callback class\nLossHistory <- R6::R6Class(\"LossHistory\",\ninherit = KerasCallback,\n\npublic = list(\n\nlosses = NULL,\n\non_batch_end = function(batch, logs = list()) {\nself$losses <- c(self$losses, logs[[\"loss\"]])\n}\n))\n\n# define model\nmodel <- keras_model_sequential()\n\nmodel %>%\nlayer_dense(units = 10, input_shape = c(784)) %>%\nlayer_activation(activation = 'softmax') %>%\ncompile(\nloss = 'categorical_crossentropy',\noptimizer = 'rmsprop'\n)\n\n# create history callback object and use it during training\nhistory <- LossHistory$new() model %>% fit( X_train, Y_train, batch_size=128, epochs=20, verbose=0, callbacks= list(history) ) # print the accumulated losses history$losses\n 0.6604760 0.3547246 0.2595316 0.2590170 ...\n\nFields\n\nCustom callback objects have access to the current model and it’s training parameters via the following fields:\n\nself$params Named list with training parameters (eg. verbosity, batch size, number of epochs…). self$model\n\nReference to the Keras model being trained.\n\nMethods\n\nCustom callback objects can implement one or more of the following methods:\n\non_epoch_begin(epoch, logs)\n\nCalled at the beginning of each epoch.\n\non_epoch_end(epoch, logs)\n\nCalled at the end of each epoch.\n\non_batch_begin(batch, logs)\n\nCalled at the beginning of each batch.\n\non_batch_end(batch, logs)\n\nCalled at the end of each batch.\n\non_train_begin(logs)\n\nCalled at the beginning of training.\n\non_train_end(logs)\n\nCalled at the end of training.\n\non_train_batch_begin\n\nCalled at the beginning of every batch.\n\non_train_batch_end\n\nCalled at the end of every batch.`\n\non_predict_batch_begin\n\nCalled at the beginning of a batch in predict methods.\n\non_predict_batch_end\n\nCalled at the end of a batch in predict methods.\n\non_predict_begin\n\nCalled at the beginning of prediction.\n\non_predict_end\n\nCalled at the end of prediction.\n\non_test_batch_begin\n\nCalled at the beginning of a batch in evaluate methods. Also called at the beginning of a validation batch in the fit methods, if validation data is provided.\n\non_test_batch_end\n\nCalled at the end of a batch in evaluate methods. Also called at the end of a validation batch in the fit methods, if validation data is provided.\n\non_test_begin\n\nCalled at the beginning of evaluation or validation.\n\non_test_end\n\nCalled at the end of evaluation or validation."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.73226845,"math_prob":0.89431924,"size":4243,"snap":"2022-05-2022-21","text_gpt3_token_len":994,"char_repetition_ratio":0.165841,"word_repetition_ratio":0.14691152,"special_character_ratio":0.25760075,"punctuation_ratio":0.1223565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9769572,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-23T06:10:30Z\",\"WARC-Record-ID\":\"<urn:uuid:a5ecbb3f-f11b-421d-9b24-c9f36736ef4c>\",\"Content-Length\":\"21249\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c71f2ee3-ecdc-48c0-8476-387a269100e2>\",\"WARC-Concurrent-To\":\"<urn:uuid:4044a3bc-6d20-4d1b-9a41-2c1a4a8e4aad>\",\"WARC-IP-Address\":\"148.205.148.16\",\"WARC-Target-URI\":\"http://cran.itam.mx/web/packages/keras/vignettes/training_callbacks.html\",\"WARC-Payload-Digest\":\"sha1:DTYFQO6HDWCYXVMOD33O2PNYNTQBEBD6\",\"WARC-Block-Digest\":\"sha1:KYM5V2PHL6PUUOBGTNJDFGPE2F4LB77M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304134.13_warc_CC-MAIN-20220123045449-20220123075449-00684.warc.gz\"}"}
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https://www.freepatentsonline.com/y2012/0002887.html
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[
"Title:\nSYSTEMS, METHODS, AND COMPUTER-READABLE MEDIA FOR DETERMINING BASIC PROBABILITY NUMBERS FOR DATA FUSION\nKind Code:\nA1\n\nAbstract:\nSystems, devices, methods, and computer-readable media relating to determining basic probability numbers for use within data fusion are disclosed. A method of defining a basic probability number may comprise measuring an intensity of a pixel of a radiograph wherein the pixel is associated with an interrogation space of an interrogation volume. The method may also include calculating an assumed intensity of the pixel for each possible configuration of a plurality of possible configurations for the interrogation space. Further, the method may include classifying each possible configuration as a possible target configuration if the measured intensity is within an error-factor of the assumed intensity. The method may further include defining a basic probability number of the pixel as a ratio of a number of possible target configurations to a number of possible configurations.\n\nInventors:\nGerts, David W. (Idaho Falls, ID, US)\nRoney, Timothy J. (Idaho Falls, ID, US)\nApplication Number:\n12/829462\nPublication Date:\n01/05/2012\nFiling Date:\n07/02/2010\nExport Citation:\nAssignee:\nBATTELLE ENERGY ALLIANCE, LLC (Idaho Falls, ID, US)\nPrimary Class:\nOther Classes:\n702/181\nInternational Classes:\nG06F17/18; G06K9/62\nView Patent Images:\nRelated US Applications:\n 20080008391 Method and System for Document Form Recognition January, 2008 Geva et al. 20150030241 METHOD AND SYSTEM FOR DATA IDENTIFICATION AND EXTRACTION USING PICTORIAL REPRESENTATIONS IN A SOURCE DOCUMENT January, 2015 Kakkar et al. 20050259878 Motion estimation algorithm November, 2005 Drezner et al. 20090097734 CELLULAR THICKNESS MEASUREMENT METHOD April, 2009 Fukuda et al. 20100149210 IMAGE CAPTURING APPARATUS HAVING SUBJECT CUT-OUT FUNCTION June, 2010 Matsunaga et al. 20150010243 METHOD FOR ENCODING/DECODING HIGH-RESOLUTION IMAGE AND DEVICE FOR PERFORMING SAME January, 2015 Yie et al. 20070036453 Image quality control in an imaging system February, 2007 Lindquist et al. 20110130096 OPERATION CONTROL AND DATA PROCESSING IN AN ELECTRONIC PEN June, 2011 Dunkars 20090147843 METHOD AND APPARATUS FOR QUANTIZATION, AND METHOD AND APPARATUS FOR INVERSE QUANTIZATION June, 2009 Han et al. 20140037162 SYSTEM AND METHOD TO AID DIAGNOSES USING CROSS-REFERENCED KNOWLEDGE AND IMAGE DATABASES February, 2014 Papier et al. 20080100716 Estimating A Point Spread Function Of A Blurred Digital Image Using Gyro Data May, 2008 Fu et al.\n\nOther References:\nWang et al, Adaptive Particle Filter for Data Fusion of Multiple Cameras, Dec. 2007, J. VLSI Signal Process. Syst., Vol 49 No 3, pp. 363-376\nWolfram Mathworld, Multiple Integral, Jul. 12, 2007,\nDataDirect Technologies, Using Indexes, May 2004,\nMoral et al, A Monte Carlo Algorithm for Combining Dempster-Shafer Belief Based on Approximate Pre-computation, 1999, LNAI 1638, pp. 305-315\nPrimary Examiner:\nLULTSCHIK, WILLIAM G\nAttorney, Agent or Firm:\nTraskBritt / Battelle Energy Alliance, LLC (Salt Lake City, UT, US)\nClaims:\nWhat is claimed is:\n\n1. A method of defining a basic probability number for use within data fusion, comprising: measuring an intensity of a pixel of a radiograph, wherein the pixel is associated with an interrogation space of an interrogation volume; calculating an assumed intensity of the pixel for each possible configuration of a plurality of possible configurations for the interrogation space; classifying each possible configuration as a possible target configuration if the measured intensity of the pixel is within an error-factor of the assumed intensity; and defining a basic probability number of the pixel as a ratio of a number of possible target configurations to a number of possible configurations.\n\n2. The method of claim 1, wherein calculating comprises calculating the assumed intensity of the pixel for each possible configuration of the plurality of possible configurations by solving a multi-dimensional integral.\n\n3. The method of claim 2, wherein solving the multi-dimensional integral comprises evaluating the multi-dimensional integral with a Monte Carlo integration technique.\n\n4. The method of claim 3, wherein evaluating the multi-dimensional integral with a Monte Carlo integration technique comprises generating a random number to identify at least one variable at which to evaluate at least one integrand of the multi-dimensional integral.\n\n5. The method of claim 2, wherein solving the multi-dimensional integral comprises solving a multi-dimensional integral having one integral constrained by a value equal to a distance between a source and an associated detector of a detection system configured for measuring the intensity of the pixel of the radiograph.\n\n6. The method of claim 1, further comprising generating a random number to determine each of a number of materials assumed to be present within the interrogation space, a type of material for each of the number of materials assumed to be present within the interrogation space, and a number of electrons associated with each type of material present within the interrogation space.\n\n7. The method of claim 1, wherein classifying each possible configuration as a possible target configuration if the measured intensity is within an error-factor of the assumed intensity comprises classifying each possible configuration as a possible target configuration if the measured intensity is substantially equal to the assumed intensity.\n\n8. The method of claim 1, further comprising measuring a mass of the interrogation volume interrogated to generate the radiograph.\n\n9. The method of claim 8, further comprising determining a mass of each interrogation space of the interrogation volume by dividing the mass of the interrogation volume by a number of pixels in the radiograph.\n\n10. The method of claim 1, wherein classifying each possible configuration as a possible target configuration comprises classifying each possible configuration as a possible explosive threat configuration if the measured intensity is within an error-factor of the assumed intensity.\n\n11. The method of claim 1, wherein defining the basic probability number comprises defining a basic probability of the pixel as a ratio of a number of possible explosive threat configurations to a number of possible configurations.\n\n12. A method, comprising: detecting an intensity of each pixel of a radiograph generated by interrogating a target object with a first interrogation method; detecting an intensity of each pixel of another radiograph generated by interrogating the target object with at least one other interrogation method; determining basic probability numbers for each pixel of the radiograph and for each associated pixel of the another radiograph, wherein a basic probability number for a pixel is determined by: measuring an intensity of the pixel; solving a multi-dimensional integral for each possible configuration of a plurality of possible configurations of an interrogation space associated with the pixel to calculate an assumed intensity of the pixel; classifying each possible configuration as a possible target configuration if the intensity is substantially equal to the assumed intensity; and defining a basic probability number of the pixel as a ratio of a number of possible target configurations to a number of possible configurations; and combining associated basic probability numbers from the radiograph and the another radiograph using Dempster-Shafer's orthogonal rule of combination.\n\n13. The method of claim 12, wherein interrogating the target object with the first interrogation method comprises interrogating the target object with a first detection modality, and interrogating the target object with the at least one other interrogation method comprises interrogating the target object with a second, different detection modality.\n\n14. The method of claim 12, wherein interrogating the target object with the first interrogation method comprises interrogating the target object with a detection modality at a first energy, and interrogating the target object with the at least one other interrogation method comprises interrogating the target object with the detection modality at a second energy, different from the first energy.\n\n15. A computer-readable storage medium for storing instructions that when executed by a processor cause the processor to perform instructions for determining a basic probability number for use within a data fusion method, the instructions comprising: measuring an intensity of a pixel of a radiograph; calculating an assumed intensity of the pixel for each possible configuration of a plurality of possible configurations of an interrogation space; classifying each possible configuration as a possible target configuration if the measured intensity of the pixel is substantially equal to the assumed intensity of the pixel; and defining a basic probability number of the pixel as a ratio of possible target configurations to possible configurations.\n\n16. The computer-readable storage medium of claim 15, wherein calculating comprises calculating the assumed intensity of the pixel for each possible configuration of the plurality of possible configurations by solving an associated multi-dimensional integral.\n\n17. The computer-readable storage medium of claim 15, wherein measuring the intensity of the pixel of the radiograph comprises measuring an intensity of a pixel of a radiograph generated by one of an x-ray detection system and a neutron detection system.\n\n18. A method of defining a basic probability number of a pixel of a radiograph for use within the Dempster-Shafer theory, comprising: generating a plurality of possible configurations for an interrogation space associated with a pixel of a radiograph; determining an assumed intensity of the pixel for each possible configuration of the plurality of configurations; comparing the assumed intensity of each possible configuration to a measured intensity of the pixel to determine whether each possible configuration comprises a target configuration; and dividing a number of target configurations by a number of possible configurations to define a basic probability number for the pixel.\n\n19. The method of claim 18, wherein generating the plurality of possible configurations comprises generating each possible configuration of the plurality of possible configurations using Monte Carlo integration.\n\n20. The method of claim 18, wherein determining the assumed intensity of the pixel for each possible configuration of the plurality of configurations comprises solving a multi-dimensional integral for each possible configuration to determine the assumed intensity.\n\n21. The method of claim 20, wherein solving the multi-dimensional integral for each possible configuration to determine the assumed intensity comprises solving a multi-dimensional integral for each possible configuration having at least one integral constrained by known data.\n\n22. The method of claim 20, wherein solving the multi-dimensional integral for each possible configuration to determine the assumed intensity comprises solving a multi-dimensional integral for each possible configuration having at least one integral constrained by at least one randomly generated variable.\n\n23. A system, comprising: a source configured to transmit a first signal into a target object; a detector configured to receive a second signal emitted from the target object and responsive to the first signal being transmitted into the target object; and a computer operably coupled to each of the source and the detector and configured to: determine an assumed intensity of a pixel of a radiograph for each possible configuration of a plurality of possible configurations of an interrogation space associated with the pixel; compare the assumed intensity of each possible configuration to a measured intensity of the pixel to determine whether each possible configuration comprises a target configuration; and define a basic probability number of the pixel as a ratio of a number of target configurations to a number of possible configurations.\n\n24. The system of claim 23, wherein the computer is further configured to generate at least one random number to identify at least one variable of a possible configuration of the plurality of configurations.\n\n25. The system of claim 23, wherein the computer is further configured to generate a random number to determine a number of materials associated with each possible configuration of the plurality of configurations.\n\n26. The system of claim 23, wherein the computer is further configured to generate a random number to determine a material type of each material associated with each possible configuration of the plurality of configurations.\n\nDescription:\n\n# GOVERNMENT RIGHTS\n\nThis invention was made with government support under Contract No. DE-AC07-051D14517 awarded by the United States Department of Energy. The government has certain rights in the invention.\n\n# TECHNICAL FIELD\n\nEmbodiments of the present invention relate generally to data fusion and, more specifically, to systems, methods, and computer-readable media for determining basic probability numbers for use within data fusion methods.\n\n# BACKGROUND\n\nSensor systems incorporating a plurality of data sources (e.g., multi-sensor systems) are widely used for a variety of military applications including ocean surveillance, air-to-air and surface-to-air defense (e.g., self-guided munitions), battlefield intelligence, surveillance and target detection (classification), and strategic warning and defense. In addition, multi-sensor systems are used for a plurality of civilian applications including condition-based maintenance, robotics, automotive safety, remote sensing, weather forecasting, medical diagnoses, and environmental monitoring (e.g., weather forecasting).\n\nTo obtain the full advantage of a multi-source system, an efficient data fusion method (or architecture) may be selected to optimally combine the received data from the multiple data sources (e.g., sensors). For military applications, especially target recognition, a sensor-level fusion process is widely used wherein data received by each individual sensor is fully processed at each sensor before being output to a data fusion processor.\n\nThe well-known Dempster-Shafer theory of evidential reasoning provides a means of combining information from different data sources to allow a system to make proper decisions. The Dempster-Shafer theory uses explicit representations of ignorance and conflict to avoid the shortcomings of classical Bayesian probability calculus. The Dempster-Shafer theory uses basic probability numbers, which represent the distribution of probability mass in a system (i.e., how strongly something is believed). The Dempster-Shafer theory, therefore, is based on obtaining degrees of belief for one question from subjective probabilities for a related question and uses Dempster-Shafer's rule for combining such degrees of belief when they are based on independent items or evidence.\n\nA limitation of the Dempster-Shafer theory is determination of the basic probability numbers. Conventionally, basic probability numbers have been determined via expert knowledge, which may not be based on measured data and, therefore, may be subjective.\n\nThere is a need for enhanced methods and systems for determination of basic probability numbers for use within data fusion. Specifically, there is a need for methods, systems, and computer-readable media for determining basic probability numbers, for use within data fusion methods, based on measured data.\n\n# BRIEF SUMMARY\n\nAn embodiment of the present invention comprises a method of defining a basic probability number for use within data fusion. The method may comprise measuring an intensity of a pixel of a radiograph wherein the pixel is associated with an interrogation space of an interrogation volume. The method may also include calculating an assumed intensity of the pixel for each possible configuration of a plurality of possible configurations for the interrogation space. Further, the method may include classifying each possible configuration as a possible target configuration if the measured intensity of the pixel is within an error-factor of the assumed intensity. The method may further include defining a basic probability number of the pixel as a ratio of a number of possible target configurations to a number of possible configurations.\n\nAnother embodiment of the present invention includes a method of defining a basic probability number of a pixel of a radiograph for use within the Dempster-Shafer theory. The method may include generating a plurality of possible configurations for an interrogation space associated with a pixel of a radiograph and determining an assumed intensity of the pixel for each possible configuration of the plurality of configurations. The method may further include comparing the assumed intensity of each possible configuration to a measured intensity of the pixel to determine whether each possible configuration comprises a target configuration. Moreover, the method may include dividing a number of target configurations by a number of possible configurations to define a basic probability number for the pixel.\n\nAnother embodiment of the present invention includes a method comprising detecting an intensity of each pixel of a radiograph generated by interrogating a target object with a first interrogation method and detecting an intensity of each pixel of another radiograph generated by interrogating the target object with at least one other interrogation method. Additionally, the method may comprise determining basic probability numbers for each pixel of the radiograph and each associated pixel of the other radiograph. A basic probability number for a pixel may be determined by measuring an intensity of the pixel and solving a multi-dimensional integral for each possible configuration of a plurality of possible configurations of an interrogation space associated with the pixel to calculate an assumed intensity of the pixel. Further, the basic probability number for the pixel may be determined by classifying each possible configuration as a possible target configuration if the intensity is substantially equal to the assumed intensity and defining a basic probability number of the pixel as a ratio of a number of possible target configurations to a number of possible configurations. Moreover, the method may include combining associated basic probability numbers from the radiograph and the other radiograph using Dempster-Shafer's orthogonal rule of combination.\n\nAnother embodiment of the present invention includes a system. The system may comprise a source configured to transmit a first signal into a target object and a detector configured to receive a second signal emitted from the target object and responsive to the first signal being transmitted into the target object. The system may further include a computer operably coupled to each of the source and the detector. The computer may be configured to determine an assumed intensity of a pixel of a radiograph for each possible configuration of a plurality of possible configurations of an interrogation space associated with the pixel. The computer may also be configured to compare the assumed intensity of each possible configuration to a measured intensity of the pixel to determine whether each possible configuration comprises a target configuration. Additionally, the computer may be configured to define a basic probability number of the pixel as a ratio of a number of target configurations to a number of possible configurations.\n\nYet another embodiment of the present invention includes a computer-readable storage medium storing instructions that, when executed by a processor, cause the processor to perform instructions for determining a basic probability number for use within a data fusion method according to an embodiment of the present invention.\n\n# BRIEF DESCRIPTION OF THE DRAWINGS\n\nFIG. 1 illustrates a block diagram of a conventional detection system;\n\nFIG. 2 depicts a block diagram of a system including a computer, a target object and a plurality of detection systems in accordance with an embodiment of the present invention;\n\nFIG. 3 illustrates a object including an explosive stimulant;\n\nFIG. 4 illustrates an x-ray radiograph of the object illustrated in FIG. 3 taken at 300 keV;\n\nFIG. 5 illustrates an x-ray radiograph of the object illustrated in FIG. 3 taken at 450 keV;\n\nFIGS. 6A-6C illustrate basic probability numbers for the analysis window 240 of the radiograph 400 in FIG. 4;\n\nFIGS. 7A-7C illustrate basic probability numbers for the analysis window 240 of the radiograph 500 in FIG. 5; and\n\nFIGS. 8A-8C illustrate fused basic probability numbers for the radiographs illustrated in FIGS. 6A-6C and 7A-7C.\n\n# DETAILED DESCRIPTION\n\nIn the following detailed description, reference is made to the accompanying drawings, which form a part hereof and in which is shown by way of illustration specific embodiments in which the invention may be practiced. These embodiments are described in sufficient detail to enable those of ordinary skill in the art to practice the invention, and it is to be understood that other embodiments may be utilized, and that structural, logical, and electrical changes may be made within the scope of the disclosure.\n\nIn this description, functions may be shown in block diagram form in order not to obscure the present invention in unnecessary detail. Furthermore, specific implementations shown and described are only examples and should not be construed as the only way to implement the present invention unless specified otherwise herein. Block definitions and partitioning of logic between various blocks represent a specific, non-limiting implementation. It will be readily apparent to one of ordinary skill in the art that the various embodiments of the present invention may be practiced by numerous other partitioning solutions. For the most part, details concerning timing considerations, and the like, have been omitted where such details are not necessary to obtain a complete understanding of the present invention in its various embodiments and are within the abilities of persons of ordinary skill in the relevant art.\n\nWhen executed as firmware or software, instructions for performing the methods and processes described herein may be stored on a computer-readable medium. A computer-readable medium includes, but is not limited to, magnetic and optical storage devices such as disk drives, magnetic tape, CDs (compact disks), DVDs (digital versatile discs or digital video discs), and semiconductor devices such as RAM, DRAM, ROM, EPROM, and Flash memory.\n\nReferring in general to the following description and accompanying drawings, various embodiments of the present invention are illustrated to show its structure and method of operation. Common elements of the illustrated embodiments are designated with like numerals. It should be understood that the figures presented are not meant to be illustrative of actual views of any particular portion of the actual structure or method but are merely idealized representations, which are employed to more clearly and fully depict the present invention.\n\nA mathematical “inverse problem” may be defined as a problem wherein the answer is known, but the question is unknown. The embodiments described herein may be used to solve an “inverse problem.” More specifically, given a measurement (e.g., an intensity measurement of a pixel of a radiograph) taken in response to interrogation of a target object, the embodiments described herein may be used to assist in determining what one or more materials within the target object caused the measurement.\n\nMoreover, as described more fully below, various embodiments of the present invention include systems, methods, and computer-readable media for determining basic probability numbers for use within a data fusion method (e.g., the Dempster-Shafer theory). More specifically, various embodiments of the present invention are related to systems, methods, and computer-readable media for calculating basic probability numbers for each individual pixel in a radiograph by determining a ratio of possible target configurations to possible configurations of an interrogation space associated with each pixel.\n\nBefore describing various embodiments of the present invention related to determining basic probability numbers for use within the Dempster-Shafer theory, a brief, general description of the Dempster-Shafer theory will first be provided. A conventional detection system and a detection system, according to an embodiment of the present invention, will each then be described. Thereafter, a method of determining basic probability numbers for use within a data fusion method (e.g., the Dempster-Shafer theory) in accordance with one or more embodiments of the present invention will be described.\n\nThe Dempster-Shafer theory is well known in the art and, therefore, in the following description thereof, some details may be omitted in order to avoid unnecessarily obscuring the invention. The Dempster-Shafer theory is a mathematical theory that allows data from multiple sources to be combined to arrive at a degree of belief. The Dempster-Shafer theory may be based on a) obtaining degrees of belief for one question from subjective probabilities of another, related question; and b) combining such degrees of belief based on independent items of evidence. A method of using the Dempster-Shafer theory may begin by determining a “question of interest.” For example, in the context of a method of detecting one or more “threat materials” (e.g., explosives or illicit drugs), a “question of interest” may comprise “is (are) there a threat(s) (e.g., an explosive material(s)) in a pixel of a radiograph of a target object?” Thereafter, a set of possible answers (also referred to as “hypotheses”) to the “question of interest” may be determined. These hypotheses may also be referred to as a “frame of discernment.” A hypothesis may include, for example, “explosive threat material,” “organic material,” “illicit drugs,” “electrical components,” “machine parts,” “paper products,” or “air.” In the context of explosive detection, a hypothesis may include, for example only, “explosive threat material.”\n\nFurthermore, basic probability numbers for each data source may be determined. A set of possible subsets of Θ, which may be identified as 2Θ, may be defined by the following equation:\n\n2Θ={A|AΘ}. (1)\n\nFurthermore, the belief function, Bel functions, satisfy the following three axioms:\n\nAxiom 1:Bel(φ)=0; (2)\n\nAxiom 2:Bel(Θ)=1; (3)\n\nwherein φ is the null set.\n\nMoreover, for any whole number n, and subsets A1 . . . An, and all subsets of Θ:\n\n$Axiom3:Bel(⋃i=1nAi)≥∑i⋐{1,2,…,n≠φ}(-1)|I|+1Bel(⋃i∈IAi).(4)$\n\nBelief may be divided into one or more basic probability numbers (b(A)) and allocated to one or more subsets A, such that:\n\nΣ{b(A)|AΘ}=1. (5)\n\nFurthermore, basic probability numbers may be related to belief functions by the following equation:\n\nBel(A)−Σ{b(B)|BA}; (6)\n\nwherein B represents proper subsets of A, and the summation is over all sets B.\n\nTwo or more pieces of evidence with basic probability numbers assigned thereto may be combined using Dempster-Shafer's orthogonal rule of combinations according to the following equations:\n\n$b=b1⊕b2;and(7)b(Ai)=∑Ap⋂Aq=Aib1(Ap)b2(Aq)1-∑Ap⋂Aq=Øb1(Ap)b2(Aq).(8)$\n\nwherein the numerator indicates the summation of the combinations of basic probabilities for a given set of overlapping hypotheses and the denominator indicates one minus the summation of basic probabilities for the combinations of all the considered hypotheses.\n\nMoreover, combined basic probability numbers may be evaluated to determine the strength of evidence of a specific hypothesis.\n\nA conventional detection system will now be described. FIG. 1 illustrates a block diagram of a conventional detection system 100, which may be used for detection of one or more materials within a target object 112. Detection system 100, which includes a source 110 and a detector 114, may be configured for transmission radiography using, for example only, neutrons, bremsstrahlung x-rays, or gamma rays. As an example, source 110 may comprise an x-ray source configured to emit a signal 116 (e.g., an x-ray beam) toward target object 112. Furthermore, in this example, detector 114 may comprise an x-ray detector configured to detect a signal 117 (e.g., x-rays) emitted from target object 112. As another example, source 110 may comprise a neutron source configured to emit signal 116 (e.g., a stream of neutrons) toward target object 112. Furthermore, in this example, detector 114 may comprise a gamma ray detector configured to detect signal 117 (e.g., gamma rays) emitted from target object 112. FIG. 1 illustrates functional interactions. Actual physical placements of elements 110, 112, and 114 as well as directions of signals 116 and 117 may be in different configurations from what is illustrated in FIG. 1.\n\nFIG. 2 illustrates a detection system 200 in accordance with an embodiment of the present invention. Detection system 200 includes a computer 102, target object 112, and a plurality of detection systems 100 (i.e., a plurality of sources 110, wherein each source 110 is coupled to computer 102 and is associated with a detector 114, which is also coupled to computer 102). Although detection system 200 is illustrated as having three detection systems (i.e., three sources with each source having an associated detector), a system including any number of detection systems 100 is within the scope of the present invention. Moreover, as with FIG. 1, FIG. 2 illustrates functional interactions among elements and is not intended to convey actual physical placement of elements.\n\nTarget object 112, which may also be referred to herein as an “object of interest” or an “interrogation volume” is, or includes, a material with respect to which a determination is being made regarding its elemental components. For example, target object 112 may be any item capable of transporting or smuggling explosives. As a more specific example, target object 112 may be a vehicle, a bag, a storage drum, a box, a container, or any combination thereof.\n\nComputer 102 may include a processor 106 and a memory 104. Memory 104 may include a computer-readable medium (e.g., data storage device 108), which may include, but is not limited to, magnetic and optical storage devices, such as disk drives, magnetic tape, CDs (compact disks), DVDs, and semiconductor devices such as RAM, DRAM, ROM, EPROM, and Flash memory. As illustrated, computer 102 may be operably coupled to each source 110 and each detector 114 within detection system 200. According to an embodiment of the present invention, computer 102 may be configured to control operation of each source 110 and receive an output from each detector 114. Memory 104 may include one or more software applications configured for performing various methods described herein. For example, memory 104 may include one or more software applications configured for determining basic probability numbers for each pixel of each generated radiograph according to the methods described herein.\n\nDuring operation of detection system 200, two or more sources 110 may transmit an associated signal 116 toward target object 112 and a response may be detected by each associated detector 114. Furthermore, each detector 114 may transmit associated data to computer 102. Upon receipt of data from one or more detectors 114, computer 102 may generate associated radiographs and determine basic probability numbers for each pixel of each generated radiograph in accordance with embodiments described herein. Thereafter, computer 102 may fuse data (i.e., combine associated basic probability numbers) according to Demspter's orthogonal rule of combination.\n\nA method of determining a basic probability number for a pixel in a radiograph in accordance with an embodiment of the present invention will first be generally described. Thereafter, the method of determining a basic probability number for a pixel in a radiograph will be described in more detail. Generally, a method may include measuring an intensity of a pixel of a radiograph. Additionally, the method may include calculating, based on known data (e.g., the measured intensity of the pixel, a mass of an interrogation volume (e.g., mass of target object 112), a distance from a source (e.g., source 110) to an associated detector (e.g., detector 114), the number density of electrons associated with the pixel, or a combination thereof), and an assumed intensity of the pixel for each possible configuration of a plurality of possible configurations of an interrogation space (i.e., a space within an interrogation volume that is associated with a pixel of a radiograph). Furthermore, the method may include comparing the assumed intensity calculated for each possible configuration to the measured intensity of the pixel to determine whether each possible configuration may be considered a “possible target configuration.” Moreover, the method may include calculating a ratio of all “possible target configurations” to all “possible configurations” for the pixel.\n\nMore specifically, possible configurations of an interrogation space (i.e., a space within an interrogation volume that is associated with a pixel of a radiograph) may be generated by varying, through random number generation, the number of materials presumed to be present within the interrogation space, the types of material assumed to be present within the interrogation space, and the number of electrons associated with each material assumed to be present within interrogation space. Furthermore, each possible configuration may be integrated over known data, assumed data, or both, to calculate an assumed intensity of the associated pixel. For each specific possible configuration, the calculated assumed intensity of the pixel may then be compared to the measured intensity of the pixel and, if the intensities are within a user-defined error factor (e.g., substantially equal within the error factor), the specific possible configuration may be considered a “possible target configuration” and a “possible configuration.” If the intensities are not within a user-defined error factor, the specific possible configuration may be considered as only a “possible configuration.” After each “possible configuration” has been generated, and each calculated assumed intensity has been compared to the measured intensity, a basic probability number (“BPN”) for the pixel may be defined as the ratio of “possible target configurations” to all “possible configurations,” as discussed in more detail below and as provided in the following equation:\n\n$BPN=possibletargetconfigurationspossibleconfigurations.(9)$\n\nA method of determining a basic probability number for a pixel in a radiograph will now be described in more detail. As will be appreciated by a person having ordinary skill in the art, data, which is associated with a detection system (e.g., detection system 200), may be measured, or known, prior to determining basic probability numbers for each pixel within a radiograph. For example, intensities of each pixel within a radiograph may be measured. Furthermore, a mass of an interrogation volume (e.g., target object 112) may be measured according to known methods. Moreover, according to one embodiment, a mass of an interrogation volume may be divided by a number of pixels within a radiograph to determine an assumed mass associated with each pixel (i.e., an assumed mass of an interrogation space). For example, for a 3×3 pixel array having a total mass of 0.9 kg, even distribution of the mass would result in each pixel having 0.1 kg.\n\nAccording to another embodiment, masses of each pixel may be at least partially related to the distribution of intensities of the pixels. More specifically, by comparing relative intensities of the pixels in a radiograph, a mass may be apportioned to each pixel. For example, if 90% of the measured attenuation is in the middle 10% of pixels of a radiograph, the middle 10% of pixels may be apportioned 90% of measured mass of the interrogation volume. More specifically, in the following example in which a pixel array has a total mass of 0.9 kg, a first pixel of the array has an intensity ratio of 0.0, a second pixel has an intensity ratio of 0.1, a third pixel has an intensity ratio of 0.2, a fourth pixel has an intensity ratio of 0.4, a fifth pixel has an intensity ratio of 0.3, and every other pixel in a radiograph has an intensity ratio of 0.0. Accordingly, in this example, the pixels with intensity ratios of 0.0 would be assigned a mass of 0 kg, the second pixel would be assigned a mass of 0.09 kg, the third pixel would be assigned a mass of 0.18 kg, the fourth pixel would be assigned a mass of 0.36 kg, and the fifth pixel would be assigned a mass of 0.27 kg.\n\nFurthermore, a distance from a source (e.g., source 110) to an associated detector (e.g., detector 114) may also be measured. In addition, a ratio of an intensity of a transmitted signal (e.g., signal 116) to a detected signal (e.g., signal 117) may be determined. Moreover, a number density of electrons within an interrogation space associated with each pixel may be determined from the angle-integrated Klein-Nishina formula, as will be understood by a person having ordinary skill in the art.\n\nAccording to an embodiment of the present invention, the above-described measured or known data, which may also be referred to as “known variables,” may be used to limit “configurations” to only “possible configurations.” For example, because the mass of the interrogation space may be known, only configurations that may have a mass substantially equal to the mass of the interrogation space are “possible configurations.” Furthermore, because a measured intensity of a pixel and a distance from a source (e.g., source 110) to an associated detector (e.g., detector 114) may be known, only configurations that exhibit these attributes may be “possible configurations.” As a more specific example, if an intensity measurement of a pixel is relatively high and a distance from a source (e.g., source 110) to a detector (e.g., detector 114) is relatively short, it may not be possible for an associated configuration to include only organic products (e.g., fish and flowers). Therefore, in this example, the configuration including only organic products would not be considered a “possible configuration” and, as a result, would not be generated and tested.\n\nThe following multi-dimensional integral of equation (10) may be solved for each possible configuration of an interrogation space to calculate an “Assumed Intensity” of a pixel associated with the interrogation space:\n\n$∫NumberofMaterials(∫MaterialType[∫MaterialCharacteristic{∫Distance-σxx}])=AssumedIntensity.(10)$\n\nMonte Carlo integration may be used for evaluation of multi-dimensional integrals. Monte Carlo integration is well known in the art and, therefore, will not be described in detail. However, a method of randomly identifying variables for generating a configuration of an interrogation space to be evaluated by equation (10) is described below.\n\nOne or more integrals of the multi-dimensional integral of equation (10) may be constrained (i.e., limited) by a variable (e.g., the “Number of Materials,” “Material Type,” or both), as described more fully below. With reference to equation (10), initially, a random number may be generated to determine an assumed “Number of Materials” present within an interrogation space. For example, if a number “3” is randomly generated, it may be assumed that the interrogation space includes “3” materials.\n\nThereafter, random numbers may be generated for each assumed material to determine a “Material Type” of each material. For example, if a number “2,” which is associated with organic materials, is randomly generated for the first of the three materials assumed to be present within the interrogation space, the first material may be assumed to be an organic material (e.g., flowers). As another example, if a number “5,” which is associated with “explosive threat” materials, is randomly generated for the second of the three materials assumed to be present within the interrogation space, the second material may be assumed to be an explosive threat material. It is noted that in the context of “threat material” detection, at least one “threat material” (e.g., explosive or illicit drug) may be assumed to be present within the interrogation space. The process of randomly generating numbers to determine the types of materials assumed to be present within the interrogation space may be repeated for each material assumed to be present to define a material type of each material. Accordingly, as will be appreciated by a person having ordinary skill in the art, the “Number of Materials” and “Material Type” variables may be used to constrain at least two integrals of equation (10).\n\nIt may be desirable to interrogate a target object (e.g., target object 112) for a specific material of interest (e.g., Nitroglycerin). Accordingly, in one embodiment of the present invention, a specific material of interest (e.g., Nitroglycerin) may be assumed to be present within the interrogation space. Therefore, in this embodiment, a random number may be generated to determine an assumed “Number of Materials” present within the interrogation space. Thereafter, for each material other than the specific material of interest, random numbers may be generated to determine a “Material Type” of the materials. Accordingly, in this example, the evaluation of equation (10) is not entirely random and, therefore, the variance associated with evaluation thereof may be reduced.\n\nOne or more integrals of the multi-dimensional integral of equation (10) may be constrained (i.e., limited) by known or measured data (e.g., “Material Characteristic,” “Distance,” or both), as described more fully below. With continued reference to equation (10), after determining an assumed number of materials present within an interrogation space and a material type for each material assumed to be present, at least one “Material Characteristic” may determined. The term “Material Characteristic” may include a combined mass of the one or more materials presumed to be present within the interrogation space, a number of electrons apportioned to each material assumed to be present within the interrogation space, or a combination thereof.\n\nA number density of electrons in the interrogation space may be determined via use of the angle-integrated Klein-Nishina formula. More specifically, because the mass of the interrogation space, the distance from a source (e.g., source 110) to a detector (e.g., detector 114), and a ratio of an intensity of a transmitted signal (e.g., signal 116) to a detected signal (e.g., signal 117) is known, the angle-integrated Klein-Nishina formula may be used to calculate the number density of electrons in the interrogation space, as will be understood by a person having ordinary skill in the art. Furthermore, because each material assumed to be present within the interrogation volume has been identified as a type of material (e.g., organic, threat, machine part, etc.); each material may be assumed to have a known chemical composition and a known density. Accordingly, a number of electrons may be apportioned to each material assumed to be present within the interrogation space and a number density of electrons may be used as the “Material Characteristic” variable to constrain at least two integrals of equation (10). Moreover, a “Distance” as identified in equation (10) may represent a distance from a source (e.g., source 110) to an associated detector (e.g., detector 114) of a detection system 200 and may be used to constrain at least one integral of equation (10).\n\nFor each individual possible configuration, the calculated “Assumed Intensity” of the pixel may then be compared to the measured intensity of the pixel and, if the intensities are within a user-defined error factor (e.g., substantially equal), the individual possible configuration may be considered a “possible target configuration” and a “possible configuration.” If the intensities are not within the user-defined error factor, the individual possible configuration may be considered as only a “possible configuration.” After each possible configuration for a pixel has been generated and a calculated “Assumed Intensity” has been compared to a measured intensity, a basic probability number for the pixel may be defined as the ratio of all “possible target configurations” to all “possible configurations.” It is noted that in the context of explosive detection, a basic probability number for the pixel may be defined as the ratio of all “possible explosive threat configurations” to all “possible configurations.” This method of defining a basic probability number may be repeated for each pixel within a radiograph to determine a basic probability number for each pixel.\n\nIt is noted that basic probability numbers associated with different detection modalities (e.g., neutrons, gamma ray, and x-ray), or similar detection modalities at different energies, may be added together using Dempster-Shafer's orthogonal rule of combination as explained above.\n\nFIG. 3 illustrates an object 210 including a material 202 that models explosive stimulant. Elements 212 illustrate the flaps of an open corrugated cardboard box. Elements 214 model printed circuit boards, such as, for example, dielectric layers and conduction layers interleaved with an epoxy resin or other suitable material. Foam packaging material 216 is interposed between circuit boards 214 and around the perimeter of the box. Object 210 is used as one example of possible objects that may be used with embodiments of the present invention to illustrate an example of how the object may be processed by embodiments of the present invention. Of course, depending on the materials to be analyzed and configurations of the materials, other objects may be very different from object 210.\n\nFIGS. 4 and 5 respectively illustrate radiographs 400 and 500 of object 210 illustrated in FIG. 3. FIG. 4 illustrates a radiograph 400 of object 210 taken at 300 keV and FIG. 5 illustrates a radiograph 500 of object 210 taken at 450 keV. Element 203 indicates the region where the modeled explosive stimulant material 202 is present in FIG. 3 and region 245 illustrates the regions where the foam packaging material 216 and circuit boards 214 are placed in FIG. 3. Analysis window 240 indicates a region of radiographs 400 and 500 that is used for additional basic probability analysis as is shown in FIGS. 6A-8C.\n\nFIGS. 6A-6C illustrate basic probability numbers for the analysis window 240 of the radiograph 400 in FIG. 4. FIG. 6A is a top view 610 of the analysis window 240 and FIGS. 6B and 6C are different perspective views (620 and 630, respectively) of the analysis window 240 to better illustrate the various amplitudes of the various voxels within the analysis window 240. The amplitudes of the individual voxels indicate the relative likelihood of there being explosive stimulant at that location determined by basic probabilities according to an embodiment of the present invention. In other words, the taller the voxel, the more likely that there is explosive stimulant at that location. Regions 612, 622, and 632 illustrate a region generally corresponding to the flap 212 of the corrugated cardboard box near the top of analysis window 240 in FIG. 4. Regions 614, 624, and 634 illustrate a region generally corresponding to the foam packaging material 216 between the flap 212 of the corrugated cardboard box and the modeled explosive stimulant material 202. Regions 616, 626, and 636 illustrate a region generally corresponding to the modeled explosive stimulant material 202 and the combination of the foam packaging material 216 and circuit boards 214 below the modeled explosive stimulant material 202.\n\nFIGS. 7A-7C illustrate basic probability numbers for the analysis window 240 of the radiograph 500 in FIG. 5. FIG. 7A is a top view 710 of the analysis window 240 and FIGS. 7B and 7C are different perspective views (720 and 730, respectively) of the analysis window 240 to better illustrate the various amplitudes of the various voxels within the analysis window 240. The amplitudes of the individual voxels indicate the relative likelihood of there being explosive stimulant at that location determined by basic probabilities according to an embodiment of the present invention. In other words, the taller the voxel, the more likely that there is explosive stimulant at that location. Regions 712, 722, and 732 illustrate a region generally corresponding to the flap 212 of the corrugated cardboard box near the top of analysis window 240 in FIG. 4. Regions 714, 724, and 734 illustrate a region generally corresponding to the foam packaging material 216 between the flap 212 of the corrugated cardboard box and the modeled explosive stimulant material 202. Regions 716, 726, and 736 illustrate a region generally corresponding to the modeled explosive stimulant material 202 and the combination of the foam packaging material 216 and circuit boards 214 below the modeled explosive stimulant material 202.\n\nFIGS. 8A-8C illustrate fused basic probability numbers for the radiographs illustrated in FIGS. 6A-6C and 7A-7C. FIG. 8A is a top view 810 of the analysis window 240 and FIGS. 8B and 8C are different perspective views (820 and 830, respectively) of the analysis window 240 to better illustrate the various amplitudes of the various voxels within the analysis window 240. The amplitudes of the individual voxels indicate the relative likelihood of there being explosive stimulant at that location determined by basic probabilities according to equations 7 and 8. In other words, the taller the voxel, the more likely that there is explosive stimulant at that location. Regions 812, 822, and 832 illustrate a region generally corresponding to the flap 212 of the corrugated cardboard box near the top of analysis window 240 in FIG. 4. Regions 814, 824, and 834 illustrate a region generally corresponding to the foam packaging material 216 between the flap 212 of the corrugated cardboard box and the modeled explosive stimulant material 202. Regions 816, 826, and 836 illustrate a region generally corresponding to the modeled explosive stimulant material 202 and the combination of the foam packaging material 216 and circuit boards 214 below the modeled explosive stimulant material 202.\n\nAlthough the embodiments described above are described in relation to explosive detection, the invention is not so limited. Rather, a person having ordinary skill in the art will understand that the systems and methods described above may be used for detection of other materials such as, for example only, illicit drugs, electronics, or organic materials, among other items.\n\nWhile the present invention has been described herein with respect to certain embodiments, those of ordinary skill in the art will recognize and appreciate that it is not so limited. Rather, many additions, deletions, and modifications to the described embodiments may be made without departing from the scope of the invention as hereinafter claimed, including legal equivalents. In addition, features from one embodiment may be combined with features of another embodiment while still being encompassed within the scope of the invention as contemplated by the inventors."
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https://similargeeks.com/code/python/check-if-number-is-perfect-square-in-python-using-function/
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"Home » Check if Number is Perfect Square in Python using Function\n\n# Check if Number is Perfect Square in Python using Function\n\nLearn about Check if Number is Perfect Square in Python using Function in the below code example. Also refer the comments in the code snippet to get a detailed view about what’s actually happening.\n\nContents\n\n#### Check if Number is Perfect Square in Python using Function\n\n``````# Program to check number is perfect square or not\n\n#importing math-module\nimport math\n\ndef PerfectSquare(x):\nroot = math.sqrt(x)\nif int(root + 0.5) ** 2 == x:\nprint(x, 'is a perfect square')\nelse:\nprint(x, 'is not a perfect square')\nreturn root\n\n# input from user\nnum = int(input('Enter number: '))\n\n# function call\nPerfectSquare(num)``````\n\nOutput:\n\nEnter number: 16\n16 is a perfect square\n\nHope above code works for you and Refer the below Related Codes to gain more insights. Happy coding and come back again."
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https://answers.everydaycalculation.com/multiply-fractions/42-4-times-35-9
|
[
"Solutions by everydaycalculation.com\n\n## Multiply 42/4 with 35/9\n\n1st number: 10 2/4, 2nd number: 3 8/9\n\nThis multiplication involving fractions can also be rephrased as \"What is 42/4 of 3 8/9?\"\n\n42/4 × 35/9 is 245/6.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 42/4 × 35/9 = 42 × 35/4 × 9 = 1470/36\n3. After reducing the fraction, the answer is 245/6\n4. In mixed form: 405/6\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
] |
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https://kmmiles.com/3195-1-miles-in-km
|
[
"kmmiles.com\n\n# 3195.1 miles in km\n\n## Result\n\n3195.1 miles equals 5140.9159 km\n\nYou can also convert 3195.1 km to miles.\n\n## Conversion formula\n\nMultiply the amount of miles by the conversion factor to get the result in km:\n\n3195.1 mi × 1.609 = 5140.9159 km\n\n## How to convert 3195.1 miles to km?\n\nThe conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:\n\n1 mi = 1.609 km\n\nTo convert 3195.1 miles into km we have to multiply 3195.1 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:\n\n1 mi → 1.609 km\n\n3195.1 mi → L(km)\n\nSolve the above proportion to obtain the length L in km:\n\nL(km) = 3195.1 mi × 1.609 km\n\nL(km) = 5140.9159 km\n\nThe final result is:\n\n3195.1 mi → 5140.9159 km\n\nWe conclude that 3195.1 miles is equivalent to 5140.9159 km:\n\n3195.1 miles = 5140.9159 km\n\n## Result approximation\n\nFor practical purposes we can round our final result to an approximate numerical value. In this case three thousand one hundred ninety-five point one miles is approximately five thousand one hundred forty point nine one six km:\n\n3195.1 miles ≅ 5140.916 km\n\n## Conversion table\n\nFor quick reference purposes, below is the miles to kilometers conversion table:\n\nmiles (mi) kilometers (km)\n3196.1 miles 5142.5249 km\n3197.1 miles 5144.1339 km\n3198.1 miles 5145.7429 km\n3199.1 miles 5147.3519 km\n3200.1 miles 5148.9609 km\n3201.1 miles 5150.5699 km\n3202.1 miles 5152.1789 km\n3203.1 miles 5153.7879 km\n3204.1 miles 5155.3969 km\n3205.1 miles 5157.0059 km\n\n## Units definitions\n\nThe units involved in this conversion are miles and kilometers. This is how they are defined:\n\n### Miles\n\nA mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.\n\n### Kilometers\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world."
] |
[
null
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https://crypto.stackexchange.com/questions/2360/x9-62-multiplying-an-elliptic-curve-point-by-a-number
|
[
"# X9.62 Multiplying an elliptic curve point by a number\n\nI'm currently trying to implement ecdsa and the first problem i met --\nmultiply an elliptic curve point by a number.\n\nAs far as i understand X9.62 gives some recommendation for doing it but i haven't managed to find anything.\nIt would be great to see some program like algorithm.\n\nP.S.\nAny help is appreciated. Sorry for my English and thanks.\n\n• Once you can add two points, multiplication by a number follows. The simplest algorithm to multiply by n>0 uses n-1 additions, much like 3*19 is 19+19+19. This can be made O(Log(n)) by methods analog to these for exponentiation.\n– fgrieu\nApr 13, 2012 at 5:13\n• Sorry, NISTs-192..521 with exp. multiplying algorithm sounds not much promising;\n– ted\nApr 13, 2012 at 5:30\n• I think you need to understand the idea about groups, and that \"exponentiation\" (when the group operation is called multiplication) and \"multiplication by number\" (when the group operation is called addition - as is the case for Elliptic Curve groups) are actually exactly the same thing. Apr 13, 2012 at 8:15\n\nWell, you understand that Elliptic Curves define an operation on points we denote as +; that is, if $A$ and $B$ are two (not necessarily distinct) points, then $A+B$ is a third point (which will be distinct unless either $A$ or $B$ are the 'point-at-infinity'). If $A$ and $B$ are the same, the operation is usually called doubling instead of addition.\n\nNow, by \"multiplying an elliptic curve point by a number\" (or point multiplication), what we mean is adding the point to itself the specified number of times. That is:\n\n$\\displaylines{kG =}{\\underbrace{G + G + G + ... + G}}$\n\nwhere exactly k $G$'s are added together.\n\nNow, the naive implementation is just to perform $k-1$ point additions; however, because the integers we're multiplying by are huge, this is not practical.\n\nHowever, point addition is associative (that is, $(A+B)+C = A+(B+C)$, that means that we can use less than $k-1$ additions; for example, to compute $8G$, we can note that:\n\n$2G = G + G$\n\n$4G = G + G + G + G = (G + G) + (G + G) = 2G + 2G$\n\n$8G = 4G + 4G$\n\nand hence we've computed $8G$ using only three additions.\n\nOne straight-forward (decent, but nonoptimal) method to do this (assuming, of course, that you have already implemented the Elliptic Curve addition function) is to use the binary addition method; just note that the Wiki article talks about 'multiplication', while you're doing addition; that is merely a difference in syntax.\n\n• An awesome answer but nothing from X9.62 about point multiplication by number. Still thank you. Cannot upvote for now.\n– ted\nApr 13, 2012 at 5:45\n• Page 91 in ANSI X9.62 defines and describes a way to perform point multiplication (they call it scalar multiplication). This is slightly different from ponchos method, choose whichever one you want. Apr 13, 2012 at 8:03"
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[
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https://www.colorhexa.com/030909
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[
"# #030909 Color Information\n\nIn a RGB color space, hex #030909 is composed of 1.2% red, 3.5% green and 3.5% blue. Whereas in a CMYK color space, it is composed of 66.7% cyan, 0% magenta, 0% yellow and 96.5% black. It has a hue angle of 180 degrees, a saturation of 50% and a lightness of 2.4%. #030909 color hex could be obtained by blending #061212 with #000000. Closest websafe color is: #000000.\n\n• R 1\n• G 4\n• B 4\nRGB color chart\n• C 67\n• M 0\n• Y 0\n• K 96\nCMYK color chart\n\n#030909 color description : Very dark (mostly black) cyan.\n\n# #030909 Color Conversion\n\nThe hexadecimal color #030909 has RGB values of R:3, G:9, B:9 and CMYK values of C:0.67, M:0, Y:0, K:0.96. Its decimal value is 198921.\n\nHex triplet RGB Decimal 030909 `#030909` 3, 9, 9 `rgb(3,9,9)` 1.2, 3.5, 3.5 `rgb(1.2%,3.5%,3.5%)` 67, 0, 0, 96 180°, 50, 2.4 `hsl(180,50%,2.4%)` 180°, 66.7, 3.5 000000 `#000000`\nCIE-LAB 2.118, -1.569, -0.553 0.185, 0.234, 0.294 0.259, 0.329, 0.234 2.118, 1.664, 199.427 2.118, -1.013, -0.219 4.842, -1.671, -0.21 00000011, 00001001, 00001001\n\n# Color Schemes with #030909\n\n• #030909\n``#030909` `rgb(3,9,9)``\n• #090303\n``#090303` `rgb(9,3,3)``\nComplementary Color\n• #030906\n``#030906` `rgb(3,9,6)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #030609\n``#030609` `rgb(3,6,9)``\nAnalogous Color\n• #090603\n``#090603` `rgb(9,6,3)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #090306\n``#090306` `rgb(9,3,6)``\nSplit Complementary Color\n• #090903\n``#090903` `rgb(9,9,3)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #090309\n``#090309` `rgb(9,3,9)``\n• #030903\n``#030903` `rgb(3,9,3)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #090309\n``#090309` `rgb(9,3,9)``\n• #090303\n``#090303` `rgb(9,3,3)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #091c1c\n``#091c1c` `rgb(9,28,28)``\n• #102f2f\n``#102f2f` `rgb(16,47,47)``\n• #164242\n``#164242` `rgb(22,66,66)``\nMonochromatic Color\n\n# Alternatives to #030909\n\nBelow, you can see some colors close to #030909. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #030908\n``#030908` `rgb(3,9,8)``\n• #030908\n``#030908` `rgb(3,9,8)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #030809\n``#030809` `rgb(3,8,9)``\n• #030809\n``#030809` `rgb(3,8,9)``\nSimilar Colors\n\n# #030909 Preview\n\nThis text has a font color of #030909.\n\n``<span style=\"color:#030909;\">Text here</span>``\n#030909 background color\n\nThis paragraph has a background color of #030909.\n\n``<p style=\"background-color:#030909;\">Content here</p>``\n#030909 border color\n\nThis element has a border color of #030909.\n\n``<div style=\"border:1px solid #030909;\">Content here</div>``\nCSS codes\n``.text {color:#030909;}``\n``.background {background-color:#030909;}``\n``.border {border:1px solid #030909;}``\n\n# Shades and Tints of #030909\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #030909 is the darkest color, while #f9fdfd is the lightest one.\n\n• #030909\n``#030909` `rgb(3,9,9)``\n• #081818\n``#081818` `rgb(8,24,24)``\n• #0d2626\n``#0d2626` `rgb(13,38,38)``\n• #123535\n``#123535` `rgb(18,53,53)``\n• #174444\n``#174444` `rgb(23,68,68)``\n• #1c5353\n``#1c5353` `rgb(28,83,83)``\n• #206161\n``#206161` `rgb(32,97,97)``\n• #257070\n``#257070` `rgb(37,112,112)``\n• #2a7f7f\n``#2a7f7f` `rgb(42,127,127)``\n• #2f8d8d\n``#2f8d8d` `rgb(47,141,141)``\n• #349c9c\n``#349c9c` `rgb(52,156,156)``\n• #39abab\n``#39abab` `rgb(57,171,171)``\n• #3ebaba\n``#3ebaba` `rgb(62,186,186)``\n• #49c2c2\n``#49c2c2` `rgb(73,194,194)``\n• #57c7c7\n``#57c7c7` `rgb(87,199,199)``\n• #66cccc\n``#66cccc` `rgb(102,204,204)``\n• #75d1d1\n``#75d1d1` `rgb(117,209,209)``\n• #84d6d6\n``#84d6d6` `rgb(132,214,214)``\n• #92dbdb\n``#92dbdb` `rgb(146,219,219)``\n• #a1e0e0\n``#a1e0e0` `rgb(161,224,224)``\n• #b0e5e5\n``#b0e5e5` `rgb(176,229,229)``\n• #bee9e9\n``#bee9e9` `rgb(190,233,233)``\n• #cdeeee\n``#cdeeee` `rgb(205,238,238)``\n• #dcf3f3\n``#dcf3f3` `rgb(220,243,243)``\n• #ebf8f8\n``#ebf8f8` `rgb(235,248,248)``\n• #f9fdfd\n``#f9fdfd` `rgb(249,253,253)``\nTint Color Variation\n\n# Tones of #030909\n\nA tone is produced by adding gray to any pure hue. In this case, #060606 is the less saturated color, while #000c0c is the most saturated one.\n\n• #060606\n``#060606` `rgb(6,6,6)``\n• #050707\n``#050707` `rgb(5,7,7)``\n• #050707\n``#050707` `rgb(5,7,7)``\n• #040808\n``#040808` `rgb(4,8,8)``\n• #040808\n``#040808` `rgb(4,8,8)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #030909\n``#030909` `rgb(3,9,9)``\n• #020a0a\n``#020a0a` `rgb(2,10,10)``\n• #020a0a\n``#020a0a` `rgb(2,10,10)``\n• #010b0b\n``#010b0b` `rgb(1,11,11)``\n• #010b0b\n``#010b0b` `rgb(1,11,11)``\n• #000c0c\n``#000c0c` `rgb(0,12,12)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #030909 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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[
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http://specialfunctionswiki.org/index.php?title=Bessel_polynomial_generalized_hypergeometric&direction=prev&oldid=2354
|
[
"# Bessel polynomial generalized hypergeometric\n\nTheorem: The following formula holds: $$y_n(x)={}_2F_0 \\left( -n, 1+n;-; -\\dfrac{1}{2}x \\right),$$ where $y_n(x)$ denotes a Bessel polynomial and ${}_2F_0$ denotes the generalized hypergeometric function."
] |
[
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http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff;h=7c6b9557ee34b8106123687679fe43092fc48103;hp=64f42b725746ff9ec1886698f419a173d02272a3;ds=sidebyside
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[
"author jim Mon, 23 Mar 2015 14:17:11 +0000 (10:17 -0400) committer Linux User Mon, 23 Mar 2015 14:17:11 +0000 (10:17 -0400)\n\nindex 7ec7a47..f26f5d8 100644 (file)\n@@ -78,7 +78,7 @@ boxed types. Here, a \"boxed type\" is a type function with one unsaturated\nhole (which may have several occurrences, as in `(α,α) tree`). We can think of the box type\nas a function from a type to a type.\n\nhole (which may have several occurrences, as in `(α,α) tree`). We can think of the box type\nas a function from a type to a type.\n\n-Recall that a monad requires a singleton function <code>mid : P-> <u>P</u></code>, and a\n+Recall that a monad requires a singleton function <code>⇧ (* mid *) : P-> <u>P</u></code>, and a\ncomposition operator like <code>>=> : (P-><u>Q</u>) -> (Q-><u>R</u>) -> (P-><u>R</u>)</code>.\n\nAs we said in the notes, we'll move freely back and forth between using `>=>` and using `<=<` (aka `mcomp`), which\ncomposition operator like <code>>=> : (P-><u>Q</u>) -> (Q-><u>R</u>) -> (P-><u>R</u>)</code>.\n\nAs we said in the notes, we'll move freely back and forth between using `>=>` and using `<=<` (aka `mcomp`), which\n@@ -86,19 +86,19 @@ is just `>=>` with its arguments flipped. `<=<` has the virtue that it correspon\nclosely to the ordinary mathematical symbol `○`. But `>=>` has the virtue\nthat its types flow more naturally from left to right.\n\nclosely to the ordinary mathematical symbol `○`. But `>=>` has the virtue\nthat its types flow more naturally from left to right.\n\n-Anyway, `mid` and (let's say) `<=<` have to obey the Monad Laws:\n+Anyway, `mid`/`⇧` and (let's say) `<=<` have to obey the Monad Laws:\n\n- mid <=< k = k\n- k <=< mid = k\n- j <=< (k <=< l) = (j <=< k) <=< l\n+ ⇧ <=< k == k\n+ k <=< ⇧ == k\n+ j <=< (k <=< l) == (j <=< k) <=< l\n\n-For example, the Identity monad has the identity function `I` for `mid`\n+For example, the Identity monad has the identity function `I` for ``\nand ordinary function composition `○` for `<=<`. It is easy to prove\nthat the laws hold for any terms `j`, `k`, and `l` whose types are\nand ordinary function composition `○` for `<=<`. It is easy to prove\nthat the laws hold for any terms `j`, `k`, and `l` whose types are\n-suitable for `mid` and `<=<`:\n+suitable for `` and `<=<`:\n\n- mid <=< k == I ○ k == \\p. I (k p) ~~> \\p. k p ~~> k\n- k <=< mid == k ○ I == \\p. k (I p) ~~> \\p. k p ~~> k\n+ <=< k == I ○ k == \\p. I (k p) ~~> \\p. k p ~~> k\n+ k <=< == k ○ I == \\p. k (I p) ~~> \\p. k p ~~> k\n\n(j <=< k) <=< l == (\\p. j (k p)) ○ l == \\q. (\\p. j (k p)) (l q) ~~> \\q. j (k (l q))\nj <=< (k <=< l) == j ○ (k ○ l) == j ○ (\\p. k (l p)) == \\q. j ((\\p. k (l p)) q) ~~> \\q. j (k (l q))\n\n(j <=< k) <=< l == (\\p. j (k p)) ○ l == \\q. (\\p. j (k p)) (l q) ~~> \\q. j (k (l q))\nj <=< (k <=< l) == j ○ (k ○ l) == j ○ (\\p. k (l p)) == \\q. j ((\\p. k (l p)) q) ~~> \\q. j (k (l q))\n@@ -119,7 +119,7 @@ More specifically,\nShow that your composition operator obeys the Monad Laws.\n\n2. Do the same with lists. That is, given an arbitrary type\nShow that your composition operator obeys the Monad Laws.\n\n2. Do the same with lists. That is, given an arbitrary type\n-`'a`, let the boxed type be `['a]` or `'a list`, that is, lists of values of type `'a`. The `mid`\n+`'a`, let the boxed type be `['a]` or `'a list`, that is, lists of values of type `'a`. The ``\nis the singleton function `\\p. [p]`, and the composition operator is:\n\nlet (>=>) (j : 'a -> 'b list) (k : 'b -> 'c list) : 'a -> 'c list =\nis the singleton function `\\p. [p]`, and the composition operator is:\n\nlet (>=>) (j : 'a -> 'b list) (k : 'b -> 'c list) : 'a -> 'c list ="
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[
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https://codegolf.stackexchange.com/questions/68748/separate-my-integers
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[
"# Separate my integers\n\n## Introduction\n\nIn the field of mathematics known as topology, there are things called separation axioms. Intuitively, you have a set X and a collection of subsets of X, which we can think of as properties. The system is well separated, if one can distinguish between all items of X based on their properties. The separation axioms formalize this idea. In this challenge, your task is to check three separation axioms, given X and the list of properties.\n\n## Input\n\nYour inputs are an integer n ≥ 2, and a list of lists T of integers. The integers in T are drawn from X = [0, 1, ..., n-1]. The lists in T may be empty and unsorted, but they will not contain duplicates.\n\n## Output\n\nYour output is one of four strings, determined by three separation axioms, each stronger than the last. There are other axioms, but we stick with these for simplicity.\n\n• Suppose that for all distinct x and y in X, there exists a list in T containing exactly one of them. Then X and T satisfy axiom T0.\n• Suppose that for all distinct x and y in X, there exist two lists in T, one of which contains x but not y, and the other contains y but not x. Then X and T satisfy axiom T1.\n• Suppose that the two lists above also contain no common elements. Then X and T satisfy axiom T2.\n\nYour output is one of T2, T1, T0 or TS, depending on which of the above conditions holds (TS means none of them do). Note that T2 is stronger than T1, which is stronger than T0, and you should always output the strongest possible axiom.\n\n## Rules and scoring\n\nYou can write a full program or a function. The lowest byte count wins, and standard loopholes are disallowed.\n\n## Test cases\n\n2 [] -> TS\n2 [[],] -> T0\n2 [,] -> T2\n3 [,[0,1,2],[1,2]] -> TS\n3 [[],,[0,1],] -> T0\n3 [,[0,1],[2,1],[0,1,2]] -> T0\n3 [,[0,1],[2,1],[2,0]] -> T1\n6 [[0,2,4],[0,3,5],[1,2],[3,4,5]] -> TS\n6 [[0,2,4],[0,3,5],[1,2],[2,5],[3,4,5]] -> T0\n6 [[0,2,4],[0,3,5],[1,2],[2,5],[3,1],[3,4,5]] -> T1\n6 [[0,1],[0,2,3],[1,4],[2,4],[2,3,5],[1,3],[4,5]] -> T2\n\n• Is the input n superfluous? In the rest of the challenge, I'm not seeing it used beyond defining what elements can be in T, so is it just a provided shortcut for T.Maximum()? Jan 7 '16 at 14:20\n• @TimmyD, no. See the first test case. 0 [] should give T2. Jan 7 '16 at 14:56\n• @PeterTaylor Aaaahhhhhhhh. Thanks, that helps tremendously. Jan 7 '16 at 15:01\n• Great explanation of what separability means! Jan 7 '16 at 21:15\n• @LuisMendo Weird terminology alert: These are separation axioms, and a topological spaces that satisfies T2 is sometimes called separated, but separability is something else entirely. Jan 8 '16 at 13:36\n\nThe function f does the job.\n\n(#)=elem\nx?y=[1|a<-x,b<-y,not$any(#a)b] f n l|t(++)=\"TS\"|t zip=\"T0\"|t(?)=\"T1\"|1>0=\"T2\"where t p=any null[p(x%y)(y%x)|x<-[0..n-1],y<-[0..x-1]] x%y=[z|z<-l,x#z,not$y#z]\n\n\ntests :\n\nmain=do\nputStrLn $f 2 [] putStrLn$ f 2 [[],]\nputStrLn $f 2 [,] putStrLn$ f 3 [,[0,1,2],[1,2]]\nputStrLn $f 3 [[],,[0,1],] putStrLn$ f 3 [,[0,1],[2,1],[0,1,2]]\nputStrLn $f 3 [,[0,1],[2,1],[2,0]] putStrLn$ f 6 [[0,2,4],[0,3,5],[1,2],[3,4,5]]\nputStrLn $f 6 [[0,2,4],[0,3,5],[1,2],[2,5],[3,4,5]] putStrLn$ f 6 [[0,2,4],[0,3,5],[1,2],[2,5],[3,1],[3,4,5]]\nputStrLn $f 6 [[0,1],[0,2,3],[1,4],[2,4],[2,3,5],[1,3],[4,5]] output: TS T0 T2 TS T0 T0 T1 TS T0 T1 T2 • Giving t a function as input is a clever trick! Jan 7 '16 at 19:44 • In absence of competition, this bounty goes to your answer. Congratulations! Jan 19 '16 at 18:52 • some free bytes - replace f by an operator name, and replace p(x%y)(x%y) by p(x%y)$x%y. and by the way, nice work! Oct 12 '16 at 21:29"
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[
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https://physics.stackexchange.com/questions/593069/why-are-we-allowed-to-make-algebraic-calculation-on-units
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[
"# Why are we allowed to make algebraic calculation on units\n\nIs there math proof that we can cancel out units in Physics? For example:\n\n$$\\require{cancel}distance = \\frac{meters}{\\cancel{second}} * \\cancel{second}$$.\n\nSo we see that seconds cancel out and we left with meters which is correct but how is it possible if it is not actually division (meters per second) it is only our interpretation of speed. Why are we allowed to make algebraic calculations on units? Also, we can use dimensional analysis to check that we do everything correctly but why? Why does it always work?\n\n• See en.wikipedia.org/wiki/Buckingham_%CF%80_theorem and references therein.\n– PPR\nNov 11, 2020 at 13:07\n• \"it is not actually division\" -- who says it isn't? Nov 11, 2020 at 15:04\n• If you want full-on mathematical rigor, I can't really explain it than Terence Tao did. Nov 11, 2020 at 15:19\n• I found this from four years ago: math.stackexchange.com/questions/1980010/… (Why do units from physics behave like numbers?) A lot to digest there. But for your question about dimensional analysis, check out @Jack M's pendulum example. Basically, the result e.g. for period of a pendulum should not depend on the units used to measure the quantities it depends on. Nov 11, 2020 at 15:53\n• My recommendation is to always use dimensional analysis to check answers. If your answer is dimensionally consistent, it has a chance of being correct; if on the other hand your answer is dimensionally inconsistent, it is definitely wrong. Nov 11, 2020 at 16:43\n\nGood question. I don't have any elegant, deep mathematical answer for you. But let's consider an object moving at 3 m/sec and consider a time interval of 5 sec. We can lay out the object's progress along a line consisting of 5 sections, in each of which the object has moved 3 m. So the total distance moved is 3m + 3m +3m +3m +3m = 15m. Knowing multiplication, we can also find the total distance as 3m x 5 = 15m. Notice that in this last calculation time doesn't play a role. I'm simply multiplying a distance by a pure number. So (3 m/sec) x (5 sec) is really 3m x 5. The seconds cancel. That to me would justify treating the sec unit like a number to be canceled.\n\n• Thank you for your answer. it makes sense for this specific example but I'm not sure that it will work for others especially for more difficult ones Nov 13, 2020 at 16:09\n\nThe question seems trivial, but it isn't at all!\n\nSome strange formal properties of algebraic calculations on units (e.g. the elusive rad unit which appears or disappears) should suggest that \"something is wrong\" instead of the more usual W.Allen's \"Whatever works\".\n\nThe problem concerns the logic (syntax and semantics) of the ''mathematical'' language of physical quantities. What does mean \"the product of a force times a distance\" or \"the division of a distance by a time\"? And what is the rigorous logic of the dimensional algebra?\n\nUnits of measure are samples of quantities, but the calculations made with symbols of units are a particular form of dimensional calculus. This (little-shared) statement is not the result of an improvisation: here I am forced to be self-referential by quoting the document:\n\nhttp://pangloss.ilbello.com/Fisica/Metrologia/grf.pdf\n\nUnfortunately the text is not in English language. I emphasize S.3.7 ''Algebra of units of measure'' and the last sentences:\n\nThe usefulness of dimensional monomials consists in making the calculation of the relationships between the various units of measurement of physical quantities algebrically intuitive, conforming it to the ordinary rules of algebra (as illustrated in the examples).\nThe ordinary calculations with units of measure are based on dimensional relations between classes of quantities, they are not univocal relations between units of measure. We must not deceive ourselves to be able to obtain in this way the exact conventional name or even the definition of the calculated unit!\n\nAs Pangloss points out, this is surprisingly non-trivial!\n\nThe sort of manipulations you are talking about are part of a quantity calculus. This is not to be confused with Newton's and Leibniz's Calculus. A calculus is just a method for computation. This is a method of manipulating quantities which yield \"the same\" quantity with different units.\n\nYour method of \"canceling\" the units is the normal way of phrasing one of the rules that can be applied in quantity calculus. And, we show very informally that it works, because nobody has found a way it doesn't work. It's anthropocentric, but its the truth. Like so many systems, we can do the things we can do in that system because we evolved countless systems that didn't work, and this one did.\n\nA major aspect of answering your \"why\" question is the axiomization of unit conversions. This is the process of \"turning the crank\" on these unit manipuations to produce all true unit conversions. In theory, if you could do this, you could start to make statements about how they work and prove something more profound.\n\nAs it turns out, we cant do that. We do not currently believe that units \"in general\" can be axiomized in this way. However, we are confident that the basic scientific units (meters, seconds, etc.) can be treated this way. Perhaps they are a special case, perhaps they are a general rule.\n\nMetrologia keeps their papers behind a paywall, but a particularly interesting paper might be On quantity calculus and units of measurement. This discusses an interesting pattern where our basic scientific units can be treated as quantities in their own right (i.e. \"ft\" corresponds to \"1 ft\"), and thus admit algebraic operations like division. However, it points out that, in general, this is not a valid construction.\n\nThe rules of algebra apply to the manipulation of symbols. The expressions for the units of measure are symbols.\n\n• Only a subset of typical algebraic rules applies to units or physical values, respectively. I cannot arbitrarily add them for instance. Also, how do you define a mathematical symbol other than something to which the rules of algebra apply? And why are units such symbols and not something else? Nov 11, 2020 at 15:24"
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https://www.iso.org/sites/JCGM/VIM/JCGM_200e_FILES/MAIN_JCGM_200e/01_e.html
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[
"# 1 Quantities and units\n\n1.1 (1.1)\nquantity\nproperty of a phenomenon, body, or substance, where the property has a magnitude that can be expressed as a number and a reference\n\nNOTE 1 The generic concept ‘quantity’ can be divided into several levels of specific concepts, as shown in the following table. The left hand side of the table shows specific concepts under ‘quantity’. These are generic concepts for the individual quantities in the right hand column.\n\n length, l radius, r radius of circle A, rA or r(A) wavelength, λ wavelength of the sodium D radiation, λD or λ (D; Na) energy, E kinetic energy, T kinetic energy of particle i in a given system, Ti heat, Q heat of vaporization of sample i of water, Qi electric charge, Q electric charge of the proton, e electric resistance, R electric resistance of resistor i in a given circuit, Ri amount-of-substance concentration of entity B, cB amount-of-substance concentration of ethanol in wine sample i, ci(C2H5OH) number concentration of entity B, CB number concentration of erythrocytes in blood sample i, C(Erys; Bi) Rockwell C hardness (150 kg load), HRC(150 kg) Rockwell C hardness of steel sample i, HRCi (150 kg)\n\nNOTE 2 A reference can be a measurement unit, a measurement procedure, a reference material, or a combination of such.\n\nNOTE 3 Symbols for quantities are given in the ISO 80000 and IEC 80000 series Quantities and units. The symbols for quantities are written in italics. A given symbol can indicate different quantities.\n\nNOTE 4 The preferred IUPAC-IFCC format for designations of quantities in laboratory medicine is “System—Component; kind-of-quantity”.\n\nEXAMPLE “Plasma (Blood)—Sodium ion; amount-of-substance concentration equal to 143 mmol/l in a given person at a given time”.\n\nNOTE 5 A quantity as defined here is a scalar. However, a vector or a tensor, the components of which are quantities, is also considered to be a quantity.\n\nNOTE 6 The concept ‘quantity’ may be generically divided into, e.g. ‘physical quantity’, ‘chemical quantity’, and ‘biological quantity’, or base quantity and derived quantity.\n\n1.2 (1.1, Note 2)\nkind of quantity\nkind\naspect common to mutually comparable quantities\n\nNOTE 1 The division of the concept of ‘quantity’ according to ‘kind of quantity’ is to some extent arbitrary.\n\nEXAMPLE 1 The quantities diameter, circumference, and wavelength are generally considered to be quantities of the same kind, namely of the kind of quantity called length.\n\nEXAMPLE 2 The quantities heat, kinetic energy, and potential energy are generally considered to be quantities of the same kind, namely of the kind of quantity called energy.\n\nNOTE 2 Quantities of the same kind within a given system of quantities have the same quantity dimension. However, quantities of the same dimension are not necessarily of the same kind.\n\nEXAMPLE The quantities moment of force and energy are, by convention, not regarded as being of the same kind, although they have the same dimension. Similarly for heat capacity and entropy, as well as for number of entities, relative permeability, and mass fraction.\n\nNOTE 3 In English, the terms for quantities in the left half of the table in 1.1, Note 1, are often used for the corresponding ‘kinds of quantity’. In French, the term “nature” is only used in expressions such as “grandeurs de même nature” (in English, “quantities of the same kind”).\n\n1.3 (1.2)\nsystem of quantities\nset of quantities together with a set of non-contradictory equations relating those quantities\n\nNOTE Ordinal quantities, such as Rockwell C hardness, are usually not considered to be part of a system of quantities because they are related to other quantities through empirical relations only.\n\n1.4 (1.3)\nbase quantity\nquantity in a conventionally chosen subset of a given system of quantities, where no subset quantity can be expressed in terms of the others\n\nNOTE 1 The subset mentioned in the definition is termed the “set of base quantities”.\n\nEXAMPLE The set of base quantities in the International System of Quantities (ISQ) is given in 1.6.\n\nNOTE 2 Base quantities are referred to as being mutually independent since a base quantity cannot be expressed as a product of powers of the other base quantities.\n\nNOTE 3 ‘Number of entities’ can be regarded as a base quantity in any system of quantities.\n\n1.5 (1.4)\nderived quantity\nquantity, in a system of quantities, defined in terms of the base quantities of that system\n\nEXAMPLE In a system of quantities having the base quantities length and mass, mass density is a derived quantity defined as the quotient of mass and volume (length to the third power).\n\n1.6\nInternational System of Quantities\nISQ\nsystem of quantities based on the seven base quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity\n\nNOTE 1 This system of quantities is published in the ISO 80000 and IEC 80000 series Quantities and units.\n\nNOTE 2 The International System of Units (SI) (see 1.16) is based on the ISQ.\n\n1.7 (1.5)\nquantity dimension\ndimension of a quantity\ndimension\nexpression of the dependence of a quantity on the base quantities of a system of quantities as a product of powers of factors corresponding to the base quantities, omitting any numerical factor\n\nEXAMPLE 1 In the ISQ, the quantity dimension of force is denoted by dim F = LMT−2.\n\nEXAMPLE 2 In the same system of quantities, dim ñB = ML−3 is the quantity dimension of mass concentration of component B, and ML−3 is also the quantity dimension of mass density, ñ, (volumic mass).\n\nEXAMPLE 3 The period T of a pendulum of length l at a place with the local acceleration of free fall g is",
null,
"where",
null,
"Hence dim C(g) = L1/2T.\n\nNOTE 1 A power of a factor is the factor raised to an exponent. Each factor is the dimension of a base quantity.\n\nNOTE 2 The conventional symbolic representation of the dimension of a base quantity is a single upper case letter in roman (upright) sans-serif type. The conventional symbolic representation of the dimension of a derived quantity is the product of powers of the dimensions of the base quantities according to the definition of the derived quantity. The dimension of a quantity Q is denoted by dim Q.\n\nNOTE 3 In deriving the dimension of a quantity, no account is taken of its scalar, vector, or tensor character.\n\nNOTE 4 In a given system of quantities,\n\n• quantities of the same kind have the same quantity dimension,\n• quantities of different quantity dimensions are always of different kinds, and\n• quantities having the same quantity dimension are not necessarily of the same kind.\n\nNOTE 5 Symbols representing the dimensions of the base quantities in the ISQ are:\n\n Base quantity Symbol for dimension length L mass M time T electric current I thermodynamic temperature Θ amount of substance N luminous intensity J\n\nThus, the dimension of a quantity Q is denoted by dim Q = Lα Mβ Tγ Iδ Θε Nζ Jη where the exponents, named dimensional exponents, are positive, negative, or zero.\n\n1.8 (1.6)\nquantity of dimension one\ndimensionless quantity\nquantity for which all the exponents of the factors corresponding to the base quantities in its quantity dimension are zero\n\nNOTE 1 The term “dimensionless quantity” is commonly used and is kept here for historical reasons. It stems from the fact that all exponents are zero in the symbolic representation of the dimension for such quantities. The term “quantity of dimension one” reflects the convention in which the symbolic representation of the dimension for such quantities is the symbol 1 (see ISO 31‑0:1992, 2.2.6).\n\nNOTE 2 The measurement units and values of quantities of dimension one are numbers, but such quantities convey more information than a number.\n\nNOTE 3 Some quantities of dimension one are defined as the ratios of two quantities of the same kind.\n\nEXAMPLES Plane angle, solid angle, refractive index, relative permeability, mass fraction, friction factor, Mach number.\n\nNOTE 4 Numbers of entities are quantities of dimension one.\n\nEXAMPLES Number of turns in a coil, number of molecules in a given sample, degeneracy of the energy levels of a quantum system.\n\n1.9 (1.7)\nmeasurement unit\nunit of measurement\nunit\nreal scalar quantity, defined and adopted by convention, with which any other quantity of the same kind can be compared to express the ratio of the two quantities as a number\n\nNOTE 1 Measurement units are designated by conventionally assigned names and symbols.\n\nNOTE 2 Measurement units of quantities of the same quantity dimension may be designated by the same name and symbol even when the quantities are not of the same kind. For example, joule per kelvin and J ⁄ K are respectively the name and symbol of both a measurement unit of heat capacity and a measurement unit of entropy, which are generally not considered to be quantities of the same kind. However, in some cases special measurement unit names are restricted to be used with quantities of a specific kind only. For example, the measurement unit ‘second to the power minus one’ (1 ⁄ s) is called hertz (Hz) when used for frequencies and becquerel (Bq) when used for activities of radionuclides.\n\nNOTE 3 Measurement units of quantities of dimension one are numbers. In some cases these measurement units are given special names, e.g. radian, steradian, and decibel, or are expressed by quotients such as millimole per mole equal to 10−3 and microgram per kilogram equal to 10−9.\n\nNOTE 4 For a given quantity, the short term “unit” is often combined with the quantity name, such as “mass unit” or “unit of mass”.\n\n1.10 (1.13)\nbase unit\nmeasurement unit that is adopted by convention for a base quantity\n\nNOTE 1 In each coherent system of units, there is only one base unit for each base quantity.\n\nEXAMPLE In the SI, the metre is the base unit of length. In the CGS systems, the centimetre is the base unit of length.\n\nNOTE 2 A base unit may also serve for a derived quantity of the same quantity dimension.\n\nEXAMPLE Rainfall, when defined as areic volume (volume per area), has the metre as a coherent derived unit in the SI.\n\nNOTE 3 For number of entities, the number one, symbol 1, can be regarded as a base unit in any system of units.\n\n1.11 (1.14)\nderived unit\nmeasurement unit for a derived quantity\n\nEXAMPLES The metre per second, symbol m ⁄ s, and the centimetre per second, symbol cm ⁄ s, are derived units of speed in the SI. The kilometre per hour, symbol km ⁄ h, is a measurement unit of speed outside the SI but accepted for use with the SI. The knot, equal to one nautical mile per hour, is a measurement unit of speed outside the SI.\n\n1.12 (1.10)\ncoherent derived unit\nderived unit that, for a given system of quantities and for a chosen set of base units, is a product of powers of base units with no other proportionality factor than one\n\nNOTE 1 A power of a base unit is the base unit raised to an exponent.\n\nNOTE 2 Coherence can be determined only with respect to a particular system of quantities and a given set of base units.\n\nEXAMPLES If the metre, the second, and the mole are base units, the metre per second is the coherent derived unit of velocity when velocity is defined by the quantity equation υ = dr ⁄ dt, and the mole per cubic metre is the coherent derived unit of amount‑of‑substance concentration when amount-of-substance concentration is defined by the quantity equation c = n ⁄ V. The kilometre per hour and the knot, given as examples of derived units in 1.11, are not coherent derived units in such a system of quantities.\n\nNOTE 3 A derived unit can be coherent with respect to one system of quantities but not to another.\n\nEXAMPLE The centimetre per second is the coherent derived unit of speed in a CGS system of units but is not a coherent derived unit in the SI.\n\nNOTE 4 The coherent derived unit for every derived quantity of dimension one in a given system of units is the number one, symbol 1. The name and symbol of the measurement unit one are generally not indicated.\n\n1.13 (1.9)\nsystem of units\nset of base units and derived units, together with their multiples and submultiples, defined in accordance with given rules, for a given system of quantities\n1.14 (1.11)\ncoherent system of units\nsystem of units, based on a given system of quantities, in which the measurement unit for each derived quantity is a coherent derived unit\n\nEXAMPLE Set of coherent SI units and relations between them.\n\nNOTE 1 A system of units can be coherent only with respect to a system of quantities and the adopted base units.\n\nNOTE 2 For a coherent system of units, numerical value equations have the same form, including numerical factors, as the corresponding quantity equations.\n\n1.15 (1.15)\noff‑system measurement unit\noff‑system unit\nmeasurement unit that does not belong to a given system of units\n\nEXAMPLE 1 The electronvolt (about 1.602 18 × 10−19 J) is an off-system measurement unit of energy with respect to the SI.\n\nEXAMPLE 2 Day, hour, minute are off‑system measurement units of time with respect to the SI.\n\n1.16 (1.12)\nInternational System of Units\nSI\nsystem of units, based on the International System of Quantities, their names and symbols, including a series of prefixes and their names and symbols, together with rules for their use, adopted by the General Conference on Weights and Measures (CGPM)\n\nNOTE 1 The SI is founded on the seven base quantities of the ISQ and the names and symbols of the corresponding base units that are contained in the following table.\n\n Base quantity Base unit Name Name Symbol length metre m mass kilogram kg time second s electric current ampere A thermodynamic temperature kelvin K amount of substance mole mol luminous intensity candela cd\n\nNOTE 2 The base units and the coherent derived units of the SI form a coherent set, designated the “set of coherent SI units”.\n\nNOTE 3 For a full description and explanation of the International System of Units, see the current edition of the SI brochure published by the Bureau International des Poids et Mesures (BIPM) and available on the BIPM website.\n\nNOTE 4 In quantity calculus, the quantity ‘number of entities’ is often considered to be a base quantity, with the base unit one, symbol 1.\n\nNOTE 5 The SI prefixes for multiples of units and submultiples of units are:\n\n Factor Prefix Name Symbol 1024 yotta Y 1021 zetta Z 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deca da 10−1 deci d 10−2 centi c 10−3 milli m 10−6 micro µ 10−9 nano n 10−12 pico p 10−15 femto f 10−18 atto a 10−21 zepto z 10−24 yocto y\n1.17 (1.16)\nmultiple of a unit\nmeasurement unit obtained by multiplying a given measurement unit by an integer greater than one\n\nEXAMPLE 1 The kilometre is a decimal multiple of the metre.\n\nEXAMPLE 2 The hour is a non‑decimal multiple of the second.\n\nNOTE 1 SI prefixes for decimal multiples of SI base units and SI derived units are given in Note 5 of 1.16.\n\nNOTE 2 SI prefixes refer strictly to powers of 10, and should not be used for powers of 2. For example, 1 kilobit should not be used to represent 1 024 bits (210 bits), which is 1 kibibit.\n\nPrefixes for binary multiples are:\n\n Factor Prefix Name Symbol (210)8 yobi Yi (210)7 zebi Zi (210)6 exbi Ei (210)5 pebi Pi (210)4 tebi Ti (210)3 gibi Gi (210)2 mebi Mi (210)1 kibi Ki\n\nSource: IEC 80000-13.\n\n1.18 (1.17)\nsubmultiple of a unit\nmeasurement unit obtained by dividing a given measurement unit by an integer greater than one\n\nEXAMPLE 1 The millimetre is a decimal submultiple of the metre.\n\nEXAMPLE 2 For a plane angle, the second is a non-decimal submultiple of the minute.\n\nNOTE SI prefixes for decimal submultiples of SI base units and SI derived units are given in Note 5 of 1.16.\n\n1.19 (1.18)\nquantity value\nvalue of a quantity\nvalue\nnumber and reference together expressing magnitude of a quantity\n\nEXAMPLE 1 Length of a given rod:\n\n5.34 m or 534 cm\n\nEXAMPLE 2 Mass of a given body:\n\n0.152 kg or 152 g\n\nEXAMPLE 3 Curvature of a given arc:\n\n112 m−1\n\nEXAMPLE 4 Celsius temperature of a given sample:\n\n−5 °C\n\nEXAMPLE 5 Electric impedance of a given circuit element at a given frequency, where j is the imaginary unit:\n\n(7 + 3j) Ω\n\nEXAMPLE 6 Refractive index of a given sample of glass:\n\n1.32\n\nEXAMPLE 7 Rockwell C hardness of a given sample (150 kg load):\n\n43.5HRC(150 kg)\n\nEXAMPLE 8 Mass fraction of cadmium in a given sample of copper:\n\n3 µg ⁄ kg or 3 × 10−9\n\nEXAMPLE 9 Molality of Pb2+ in a given sample of water:\n\n1.76 µmol ⁄ kg\n\nEXAMPLE 10 Arbitrary amount-of-substance concentration of lutropin in a given sample of plasma (WHO international standard 80/552):\n\n5.0 International Unit ⁄ l\n\nNOTE 1 According to the type of reference, a quantity value is either\n\nNOTE 2 The number can be complex (see Example 5).\n\nNOTE 3 A quantity value can be presented in more than one way (see Examples 1, 2 and 8).\n\nNOTE 4 In the case of vector or tensor quantities, each component has a quantity value.\n\nEXAMPLE Force acting on a given particle, e.g. in Cartesian components (FxFyFz) = (−31.5; 43.2; 17.0) N.\n\n1.20 (1.21)\nnumerical quantity value\nnumerical value of a quantity\nnumerical value\nnumber in the expression of a quantity value, other than any number serving as the reference\n\nNOTE 1 For quantities of dimension one, the reference is a measurement unit which is a number and this is not considered as a part of the numerical quantity value.\n\nEXAMPLE In an amount‑of‑substance fraction equal to 3 mmol ⁄ mol, the numerical quantity value is 3 and the unit is mmol ⁄ mol. The unit mmol ⁄ mol is numerically equal to 0.001, but this number 0.001 is not part of the numerical quantity value, which remains 3.\n\nNOTE 2 For quantities that have a measurement unit (i.e. those other than ordinal quantities), the numerical value {Q} of a quantity Q is frequently denoted {Q} = Q [Q], where [Q] denotes the measurement unit.\n\nEXAMPLE For a quantity value of 5.7 kg, the numerical quantity value is {m} = (5.7 kg) ⁄ kg = 5.7. The same quantity value can be expressed as 5 700 g in which case the numerical quantity value {m} = (5 700 g) ⁄ g = 5 700.\n\n1.21\nquantity calculus\nset of mathematical rules and operations applied to quantities other than ordinal quantities\n\nNOTE In quantity calculus, quantity equations are preferred to numerical value equations because quantity equations are independent of the choice of measurement units, whereas numerical value equations are not (see ISO 31‑0:1992, 2.2.2).\n\n1.22\nquantity equation\nmathematical relation between quantities in a given system of quantities, independent of measurement units\n\nEXAMPLE 1 Q1 = ζ Q2 Q3 where Q1, Q2 and Q3 denote different quantities, and where ζ is a numerical factor.\n\nEXAMPLE 2 T = (1 ⁄ 2) 2 where T is the kinetic energy and υ the speed of a specified particle of mass m.\n\nEXAMPLE 3 n = It F where n is the amount of substance of a univalent component, I is the electric current and t the duration of the electrolysis, and where F is the Faraday constant.\n\n1.23\nunit equation\nmathematical relation between base units, coherent derived units or other measurement units\n\nEXAMPLE 1 For the quantities in Example 1 of item 1.22, [Q1] = [Q2] [Q3] where [Q1], [Q2] and [Q3] denote the measurement units of Q1, Q2 and Q3, respectively, provided that these measurement units are in a coherent system of units.\n\nEXAMPLE 2 J := kg m2 ⁄ s2, where J, kg, m and s are the symbols for the joule, kilogram, metre and second, respectively. (The symbol := denotes “is by definition equal to” as given in the ISO 80000 and IEC 80000 series.)\n\nEXAMPLE 3 1 km ⁄ h = (1 ⁄ 3.6) m ⁄ s.\n\n1.24\nconversion factor between units\nratio of two measurement units for quantities of the same kind\n\nEXAMPLE km ⁄ m = 1 000 and thus 1 km = 1 000 m.\n\nNOTE The measurement units may belong to different systems of units.\n\nEXAMPLE 1 h ⁄ s = 3 600 and thus 1 h = 3 600 s.\n\nEXAMPLE 2 (km ⁄ h) ⁄ (m ⁄ s) = (1 ⁄ 3.6) and thus 1 km ⁄ h = (1 ⁄ 3.6) m ⁄ s.\n\n1.25\nnumerical value equation\nnumerical quantity value equation\nmathematical relation between numerical quantity values, based on a given quantity equation and specified measurement units\n\nEXAMPLE 1 For the quantities in Example 1 in item 1.22, {Q1} = ζ {Q2} {Q3} where {Q1}, {Q2} and {Q3} denote the numerical values of Q1, Q2 and Q3, respectively, provided that they are expressed in either base units or coherent derived units or both.\n\nEXAMPLE 2 In the quantity equation for kinetic energy of a particle, T = (1 ⁄ 2) 2, if m = 2 kg and υ = 3 m ⁄ s, then {T} = (1 ⁄ 2) × 2 × 32 is a numerical value equation giving the numerical value 9 of T in joules.\n\n1.26\nordinal quantity\nquantity, defined by a conventional measurement procedure, for which a total ordering relation can be established, according to magnitude, with other quantities of the same kind, but for which no algebraic operations among those quantities exist\n\nEXAMPLE 1 Rockwell C hardness.\n\nEXAMPLE 2 Octane number for petroleum fuel.\n\nEXAMPLE 3 Earthquake strength on the Richter scale.\n\nEXAMPLE 4 Subjective level of abdominal pain on a scale from zero to five.\n\nNOTE 1 Ordinal quantities can enter into empirical relations only and have neither measurement units nor quantity dimensions. Differences and ratios of ordinal quantities have no physical meaning.\n\nNOTE 2 Ordinal quantities are arranged according to ordinal quantity‑value scales.\n\n1.27\nquantity‑value scale\nmeasurement scale\nordered set of quantity values of quantities of a given kind of quantity used in ranking, according to magnitude, quantities of that kind\n\nEXAMPLE 1 Celsius temperature scale.\n\nEXAMPLE 2 Time scale.\n\nEXAMPLE 3 Rockwell C hardness scale.\n\n1.28 (1.22)\nordinal quantity‑value scale\nordinal value scale\nquantity‑value scale for ordinal quantities\n\nEXAMPLE 1 Rockwell C hardness scale.\n\nEXAMPLE 2 Scale of octane numbers for petroleum fuel.\n\nNOTE An ordinal quantity‑value scale may be established by measurements according to a measurement procedure.\n\n1.29\nconventional reference scale\nquantity-value scale defined by formal agreement\n1.30\nnominal property\nproperty of a phenomenon, body, or substance, where the property has no magnitude\n\nEXAMPLE 1 Sex of a human being.\n\nEXAMPLE 2 Colour of a paint sample.\n\nEXAMPLE 3 Colour of a spot test in chemistry.\n\nEXAMPLE 4 ISO two-letter country code.\n\nEXAMPLE 5 Sequence of amino acids in a polypeptide.\n\nNOTE 1 A nominal property has a value, which can be expressed in words, by alphanumerical codes, or by other means.\n\nNOTE 2 ‘Nominal property value’ is not to be confused with nominal quantity value."
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"In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular...",
null,
"Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ... - Page xv\nby Robert Potts - 1876 - 403 pages",
null,
"## Euclid's Elements of geometry [book 1-6, 11,12] with explanatory notes ...\n\nEuclides - 1845 - 544 pages\n...triangle, the square of the side subtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle...sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle. Let ABC be any triangle,...",
null,
"## The Elements of Euclid, the parts read in the University of Cambridge [book ...\n\nEuclides - 1846 - 292 pages\n...every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle...line intercepted between the perpendicular let fall upon it from the opposite angle and the acute angle, Let ABC be any triangle, and the angle at B one...",
null,
"## Elements of Geometry: Containing the First Six Books of Euclid, with a ...\n\nEuclid, John Playfair - Euclid's Elements - 1846 - 332 pages\n...every triangle the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle...line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B one...",
null,
"## The Elements of Geometry, Symbolically Arranged\n\nGreat Britain. Admiralty - Geometry - 1846 - 130 pages\n...triangle^ the square of the side sub-tending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle...line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Prop. 46. Prop. 39. Prop. is. P J, 47 Fig. 1....",
null,
"## The Elements of Geometry, Symbolically Arranged\n\nGreat Britain. Admiralty - Geometry - 1846 - 130 pages\n...triangle^ the square of the side sttbtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either oj these sides, and the straight line intercepted between the perpendicular let fall upon it from the...",
null,
"## The English Journal of Education, Volume 1\n\nEducation - 1847 - 508 pages\n...every triangle the square of the side subtending either of the acute angles is less than the squares of the sides containing• that angle by twice the rectangle contained by either of those sides, aud the straight line intercepted between the acute angle and the perpendicular let fall...",
null,
"## Report of the Committee of Council on Education (England and Wales ..., Issue 1\n\n...line. 2. In any triangle the square of the side subtending any acute angle is less than the squares of the sides containing that angle by twice the rectangle...line intercepted between the perpendicular let fall upon it from the opposite angle and the acute angle ; prove only the first case of this proposition....",
null,
"## Minutes of the Committee of Council on Education: with Appendices, Volume 1\n\nGreat Britain. Committee on Education - Education - 1848 - 514 pages\n...line. 2. In any triangle the square of the side subtending any acute angle is less than the squares of the sides containing that angle by twice the rectangle...line intercepted between the perpendicular let fall upon it from the opposite angle and the acute angle ; prove only the first case of this proposition....",
null,
"## The definitions, postulates, axioms, and enunciations of the propositions of ...\n\nEuclides - 1848 - 52 pages\n...every triangle, the square of the side subtending any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle...sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle. PROP. XIV. PROBLEM. BOOK...",
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|
https://dev.to/codr/road-to-genius-advanced-33-eo2
|
[
"",
null,
"Each day I solve several coding challenges and puzzles from Codr's ranked mode. The goal is to reach genius rank, along the way I explain how I solve them. You do not need any programming background to get started, and you will learn a ton of new and interesting things as you go.\n\nfunction LRU(capacity) {\nthis.cache = {};\nthis.capacity = capacity;\nthis.size = 0;\nthis.queue = [];\n}\n;\nLRU.prototype.get = function (key) {\nconst hit = this.cache[key];\nif (hit !== undefined) {\nthis.queue = this.queue.filter(q => 🐼 !== key);\nthis.queue.push(key);\nreturn hit;\n}\nreturn -1;\n};\nLRU.prototype.put = function (key, value) {\nconst hit = this.cache[key];\nthis.cache[key] = value;\nif (!hit) {\nif (this.size === this.capacity) {\nconst key = this.queue.shift();\nthis.cache[key] = undefined;\n} else {\nthis.size = this.size + 1;\n}\nthis.queue.push(😈);\n} else {\nthis.queue = this.queue.filter(q => q !== key);\nthis.queue.push(key);\n}\n};\nlet cache = new LRU(7);\nfor (let i = 0; i < 4; i++)\ncache.put(i, i);\nlet A = cache.queue.length;\n\n// 🐼 = ? (identifier)\n// 😈 = ? (identifier)\n// such that A = 4 (number)\n\n\nIn today's challenge we need to fix two bugs in a relatively large code base. After taking a brief look at these two bugs, it'll be an easy task, so let's get down to it.\n\nThe first bug appears on the following line:\n\nthis.queue = this.queue.filter(q => 🐼 !== key);\n\n\nA filter arrow-function is applied to the queue array. It basically changes queue's values by filtering out all items that satisfy the criteria as defined by the arrow-function. This line of code can be summarized in pseudo-code:\n\nqueue = queue.filter(\nfor each item \"q\" in \"queue\":\nif 🐼 !== key:\nreturn true\nelse:\nreturn false\n)\n\n\nAll that this code does is removing all items from queue that are equal to key; in other words, keeping all items that are not equal to key.\nAs you can see, the bug 🐼 has to be q.\n\nTo fix the 2nd and last bug 😈 we need to analyze a bit more code:\n\nif (!hit) {\nif (this.size === this.capacity) {\nconst key = this.queue.shift();\nthis.cache[key] = undefined;\n} else {\nthis.size = this.size + 1;\n}\nthis.queue.push(😈);\n} else {\nthis.queue = this.queue.filter(q => q !== key);\nthis.queue.push(key);\n}\n\n\nThe bug 😈 must be a variable name, something that's being pushed to the queue array. The else-clause already reveals what this will be: key.",
null,
"If you're interested in the bigger picture, this code is a simple implementation of an LRU cache system (Least Recently Used).\n\nBy solving these challenges you train yourself to be a better programmer. You'll learn newer and better ways of analyzing, debugging and improving code. As a result you'll be more productive and valuable in business. Join me on the Road to Genius and upgrade your programming skills, at https://nevolin.be/codr/\n\n### Discussion",
null,
"",
null,
"",
null,
""
] |
[
null,
"https://res.cloudinary.com/practicaldev/image/fetch/s--Ok1kTGlt--/c_imagga_scale,f_auto,fl_progressive,h_420,q_auto,w_1000/https://res.cloudinary.com/practicaldev/image/fetch/s--I0QvnYp4--/c_imagga_scale%2Cf_auto%2Cfl_progressive%2Ch_420%2Cq_auto%2Cw_1000/https://dev-to-uploads.s3.amazonaws.com/i/bq82jq90brcazxbtad0z.png",
null,
"https://res.cloudinary.com/practicaldev/image/fetch/s--5tyMmRh---/c_limit%2Cf_auto%2Cfl_progressive%2Cq_auto%2Cw_880/https://dev-to-uploads.s3.amazonaws.com/i/314ep2qptjbh8fwqmyra.png",
null,
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null,
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null,
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null
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|
https://get.kintone.help/k/en/user/app_settings/form/autocalc/if_function.html
|
[
"# IF Function: Changing the Value to Display Depending on the Condition\n\nArticle Number:040505\n\nYou can use the IF function in formulas. IF function is a function that changes the value to display depending on the specified conditional expression.\n\n IF(Conditional expression,When TRUE,When FALSE) If the conditional expression is satisfied, returns TRUE. Otherwise, returns FALSE.\n\n## Writing Conditional Expressions",
null,
"By using comparison operators, you can compare field codes and values that you want to use as conditions.\n\n Example: IF(Total>=80,\"OK\",\"NG\") \"Total>=80\" is a conditional expression. In this example, the expression is written in the order of the field code (Total), comparison operators (>=) , and the value (80). You can also specify a field code instead of the value. It displays \"OK\" if the \"Total\" field value is more than or equal to 80. Otherwise, it displays \"NG\".\n• Field Code\nWrite a field code of the field referenced by the formula.\n\n• Comparison Operators\n\n• When comparing numeric type fields (such as Number fields), you can use =, !=, <>, <, >, <=, >= as comparison operators.\n• When you compare string type fields (such as Text field), you can only use =, !=, and <>. You can compare string type fields to check if a string is equal to another string or not. However, you cannot compare them to check which value is bigger or smaller as they are not numeric type fields.\n• For details on the field data types including numeric and string types, refer to Data Types of Values Referenced by Formulas.\nFor details on comparison operators, refer to the description on =, !=, <>, <, >, <=, >= in \"List of Operators and Functions\".\n• Values\nFor numeric values, you can write them as they are. For text strings, you need to enclose them with double quotation marks (\").\n\nFor comparison operators, you can specify field codes and values (such as numeric values and text strings), as well as formulas and functions on both sides. For example, you can specify formulas and functions such as (Numeric_Value_A*0.1) and (SUM(Numeric_Value_A, Numeric_Value_B)), on both sides of comparison operators.\n\n Example: IF(CONTAINS(Workday, \"Sunday\"), \"Working on Sunday\", \"Not Working on Sunday\") \"CONTAINS(Workday, \"Sunday\")\" is a conditional expression. It displays \"Working on Sunday\" if Sunday is selected in the \"Workday\" field. Otherwise, \"Not Working on Sunday\" is displayed.\n\nIf you want to refer to check boxes and multi-choices in a conditional expression with IF function, use CONTAINS function.\nWhen you write \"equal to\" in a conditional expression, \"=\" is used in the case of \"Branching based on the data of a field (1)\". However, for the check boxes and multi-choices, you need to use \",\" instead of \"=\".\nFor details on CONTAINS function, refer to the following pages:\n\n Example: IF(AND(Subject>=80,Practical_Skill>=80),\"OK\",\"NG\") \"AND(Subject>=80,Practical_Skill>=80)\" is a conditional expression. It displays \"OK\" if the values of both \"Subject\" and \"Practical Skill\" fields are more than or equal to 80. Otherwise, it displays \"NG\".\n\nIn the IF function, you can write conditional expressions with multiple fields.\nBy using AND function and OR function, you can concatenate conditional expressions with \",\".\nAvailable comparison operators are the same as the ones that are explained in the \"Branching based on the data of a field (1)\" section.\n\nFor details on AND function and OR function, refer to the \"AND, OR, NOT Functions: Combining Conditions\" page.\n\n## Examples of Formulas Using Conditional Expressions\n\nThis section shows you some examples of formulas using conditional expressions.\n\n Formula for invoice price calculation: IF(Total_Price>=1000, Total_Price*0.85, Total_Price) Displays 15 % discounted price when the total price is more than or equals to 1,000 yen. Otherwise, displays the total price as is.\n Formula for accommodation expense calculation: IF(Breakfast=\"Included\", 9000, 7000) Displays 9,000 yen when breakfast is included. Otherwise, displays 7,000 yen. Whether to include breakfast fee is determined by the selection status of the \"Included\" option in the Drop-down or Radio button field. When you specify a drop-down or a radio button option as a condition, you must enclose the option name with double quotation marks(\"\").\n Formula for budget ratio calculation: IF(Budget>0, (Actual_Amount/Budget)*100, \"\") When the budget is '0', budget ratio becomes empty. When the budget is not '0', calculates the budget ratio. Since a denominator might be '0', IF function should be used to switch conditions.\n Formula for goal achievement: IF(Total>=Target_Value,\"Achieved\",\"Not Achieved\") Displays \"Achieved\" if the total is equal to or greater than the target value. Otherwise, displays \"Not Achieved\". Both \"Total\" and \"Target_Value\" are field codes. Since the result will be returned in text, you must set the formula in the Text field.\n Formula for rating: IF(Score>=80, \"Pass\", \"Fail\") Displays \"Pass\" when the score is more than or equals to 80. Otherwise, displays \"Fail\". Since the result will be returned in text, you must set the formula in the Text field.\n Formula for rating: IF(Score=100,\"Perfect score\",IF(Score>=80,\"Pass\",\"Fail\")) IF function can be nested. Displays \"Perfect score\" when the score is 100, and \"Pass\" when the score is more than or equals to 80. Otherwise, displays \"Fail\". Since the result will be returned in text, you must set the formula in the Text field.\n\n### TIPS: How Empty Fields Are Handled\n\nEven when there is an empty field, that field value is considered as 0 or \"\", and the formula is calculated.\n\n• If you enter \"Calculated field = 0\", the result returns TRUE when the Calculated field is empty, as well as when the Calculated field value is '0'.\n\n• If you specify \"Calculated field = \"\"\", the condition will be used to determine whether the field is empty or not. Returns FALSE when '0' is entered, and TRUE when the field is empty.\n\n• Described below is the way to make the field empty instead of displaying '0'.\nSpecify \"\" as the result.\n\n• Example: IF(Quantity=0, \"\", Unit_Price*Quantity)\nWhen the quantity is 0, the result will be empty. When the quantity is not 0, the result value of multiplication will be displayed.\n\n## Example: Changing the Invoice Price Based on the Total Amount\n\nThis section describes the way to set a formula which changes the invoice price based on the total amount, by using an example of an order management app.\n\nImage of the created app:",
null,
"1. On the screen to set forms, place a field (Calculated field) to show the total amount.\nEnter \"Total Price\" as the field name and \"Total_Price\" as the field code.\nFor details on how to set this field, see the \"SUM Function: Sum Calculation\" page.",
null,
"2. From the parts list shown on the left side, drag and drop a Calculated field.",
null,
"3. Open the field setting screen, and change the field name to \"WSP\".\nKeep the setting screen open.",
null,
"4. Enter a formula that displays 15 % discounted price when the total price is more than or equals to 100,000 yen, and displays the total price as is when the total amount is less than 100,000 yen.\n\n• Formula: IF(Total_Price>=100000, Total_Price*0.85, Total_Price)\nThen, select Number (Example: 1,000).",
null,
"5. Click Save on the calculation setting screen.\n\n6. Click Save Form on the upper left side of the screen, and then click Update App on the upper right side of the screen.",
null,
"Now, the 15 % discounted invoice price is displayed when the total price is more than or equals to 100,000 yen.",
null,
"## Example: Changing the Value to Display Based on the Selected Radio Button\n\nIn a conditional expression in IF function, you can refer to drop-down and radio button fields.\nYou can also change the value to display based on which radio button is selected. For example, you can display 9,000 yen when \"Included\" is selected, and 7,000 yen when \"Not Included\" is selected for the Breakfast option.",
null,
"• Formula example: IF(Breakfast=\"Included\", 9000, 7000)",
null,
""
] |
[
null,
"https://get.kintone.help/k/img-en/if_function_15.png",
null,
"https://get.kintone.help/k/img-en/if_function_01.png",
null,
"https://get.kintone.help/k/img-en/if_function_02.png",
null,
"https://get.kintone.help/k/img-en/if_function_03.png",
null,
"https://get.kintone.help/k/img-en/if_function_01.gif",
null,
"https://get.kintone.help/k/img-en/if_function_04.png",
null,
"https://get.kintone.help/k/img-en/if_function_05.png",
null,
"https://get.kintone.help/k/img-en/if_function_06.png",
null,
"https://get.kintone.help/k/img-en/if_function_11.png",
null,
"https://get.kintone.help/k/img-en/if_function_08.png",
null
] |
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|
https://www.imooc.com/wenda/30/recommend/4
|
[
"### C++",
null,
"C++\nC与C++的区别与联系？",
null,
"C C++",
null,
"C C++ Html/CSS\nC++课中的代码写出来根本运行不了！",
null,
"C++\nint* p=new int什么意思",
null,
"C++",
null,
"C C++",
null,
"C C++",
null,
"C C# C++\nc语言程序设计作业怎么搞",
null,
"C++",
null,
"C++",
null,
"C C# C++\nC语言的for语句为！=的情况",
null,
"C C# C++\nc++的问题",
null,
"C++",
null,
"C++\nC++怎么取对象的指针",
null,
"C++",
null,
"C C++ 数据结构\nerror C2143: syntax error : missing ';' before 'template' 下面是部分代码 谢谢",
null,
"C++",
null,
"C++\n（！find）是什么意思",
null,
"C C++",
null,
"C C++ JAVA\nC语言问题1",
null,
"C++\nC++问题这个算出来是多少",
null,
"C C++",
null,
"C C++",
null,
"C C++",
null,
"C C++ 数据结构",
null,
"C C++",
null,
"C C++",
null,
"C++ JAVA",
null,
"C++"
] |
[
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null
] |
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|
https://www.mathopenref.com/vertical.html
|
[
"# Vertical\n\nA vertical line is one which runs up and down the page.",
null,
"In geometry, a vertical line is one which runs from up and down the page. Its cousin is the horizontal line which runs left to right across the page. A vertical line is perpendicular to a horizontal line.\nSee perpendicular lines.\n\n## Vertical angles\n\nVertical angles are different use of the word 'vertical'. Vertical angles are angles that share a common vertex and need not be drawn one above the other. See Vertical angles.\n\n## Coordinate geometry\n\nA line will be vertical if any two points on the line have the same x-coordinate.\nSee Vertical line (coordinate geometry)."
] |
[
null,
"https://www.mathopenref.com/images/lines/vertical.png",
null
] |
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|
https://encyclopediaofmath.org/wiki/Domain_of_influence
|
[
"# Domain of influence\n\nof a point \\$M\\$ (of a set \\$A\\$ of points \\$M\\$)\n\nThe set \\$B(M)\\$ (respectively, \\$B(A)\\$) of all points at which the solution of a differential equation or of a set of differential equations changes as a result of a perturbation of it at the point \\$M\\$ (or at the set \\$A\\$). In the simplest cases of linear partial differential equations the domain of influence is independent of the solution; for most non-linear problems the domain of influence depends both on the solution itself and on the nature of the perturbations. In such a case infinitely small perturbations are considered. For hyperbolic equations, the domain of influence of the point \\$M\\$ is often the union of the characteristic conoid (cf. Characteristic manifold) passing through the point \\$M\\$ and its interior; for parabolic and elliptic equations, the domain of influence of the point \\$M\\$ is usually the domain of definition of the solution."
] |
[
null
] |
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|
https://doc.cgal.org/Manual/3.1/doc_html/cgal_manual/Point_set_2_ref/Class_Point_set_2.html
|
[
"",
null,
"## CGAL::Point_set_2<Gt,Tds>\n\n### Definition\n\n#include <CGAL/Point_set_2.h>\n\nAn instance PS of the data type Point_set_2<Gt,Tds> is a Delaunay Triangulation of its vertex set. The class Point_set_2<Gt,Tds> is inherited from the CGAL Delaunay triangulation, and provides additional nearest neighbor query operations and range searching operations.\n\nThe Point_set_2<Gt,Tds> class of CGAL depends on template parameters standing for the geometric traits classes used by the point set and by the Delaunay triangulation (Gt) and for the triangulation data structure (Tds).\n\n### Types\n\n typedef Gt::Point_2 Point; the point type typedef Gt::Segment_2 Segment; the segment type typedef Gt::Circle_2 Circle; the circle type typedef Gt::FT Numb_type; the representation field number type.\n\n Point_set_2::Triangulation::Vertex the vertex type of the underlying triangulation. Point_set_2::Triangulation::Edge the edge type of the underlying triangulation. Point_set_2::Triangulation::Vertex_handle handles to vertices.\n\n### Creation\n\n Point_set_2 PS; creates an empty Point_set_2 . template Point_set_2 PS ( InputIterator first, InputIterator last); creates a Point_set_2 PS of the points in the range [first,last).\n\n### Operations\n\nVertex_handle PS.lookup ( Point p)\nif PS contains a Vertex v with |pos(v)| = p the result is a handle to v otherwise the result is NULL.\n\nVertex_handle PS.nearest_neighbor ( Point p)\ncomputes a handle to a vertex v of PS that is closest to p. If PS is empty, NULL is returned.\n\nVertex_handle PS.nearest_neighbor ( Vertex_handle v)\ncomputes a handle to a vertex w of PS that is closest to v. If v is the only vertex in PS , NULL is returned.\n\ntemplate<class OutputIterator>\nOutputIterator\n PS.nearest_neighbors ( Point p, int k, OutputIterator res)\ncomputes the k nearest neighbors of p in PS, and places the handles to the corresponding vertices as a sequence of objects of type Vertex_handle in a container of value type of res which points to the first object in the sequence. The function returns an output iterator pointing to the position beyond the end of the sequence.\n\ntemplate<class OutputIterator>\nOutputIterator\n PS.nearest_neighbors ( Vertex_handle v, int k, OutputIterator res)\ncomputes the k nearest neighbors of v, and places them as a sequence of objects of type Vertex_handle in a container of value type of res which points to the first object in the sequence. The function returns an output iterator pointing to the position beyond the end of the sequence.\n\ntemplate<class OutputIterator>\nOutputIterator PS.range_search ( Circle C, OutputIterator res)\ncomputes handles to all vertices contained in the closure of disk C. The computed vertex handles will be placed as a sequence of objects in a container of value type of res which points to the first object in the sequence. The function returns an output iterator pointing to the position beyond the end of the sequence.\n\ntemplate<class OutputIterator>\nOutputIterator\n PS.range_search ( Point a, Point b, Point c, OutputIterator res)\ncomputes handles to all vertices contained in the closure of the triangle (a,b,c).\n\nPrecondition: a, b, and c must not be collinear. The computed vertex handles will be placed as a sequence of objects in a container of value type of res which points to the first object in the sequence. The function returns an output iterator pointing to the position beyond the end of the sequence.\n\ntemplate<class OutputIterator>\nOutputIterator\n PS.range_search ( Point a1, Point b1, Point c1, Point d1, OutputIterator res)\ncomputes handles to all vertices contained in the closure of the iso-rectangle (a1,b1,c1,d1).\n\nPrecondition: a1 is the upper left point, b1 the lower left, c1 the lower right and d1 the upper right point of the iso rectangle. The computed vertex handles will be placed as a sequence of objects in a container of value type of res which points to the first object in the sequence. The function returns an output iterator pointing to the position beyond the end of the sequence."
] |
[
null,
"https://doc.cgal.org/Manual/3.1/doc_html/cgal_manual/Point_set_2_ref/cc_Class.gif",
null
] |
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|
https://pisquare.osisoft.com/community/all-things-pi/blog/authors/tbernier
|
[
"All Things PI - Ask, Discuss, Connect\n\n1 Post authored by: tbernier",
null,
"",
null,
"AF Analytics –Analyzing an open Batch/Event in Real Time\n\nPosted by tbernier",
null,
"Apr 20, 2017\n\nIn response to user voice :\n\nIntroduction :\n\nEvent frames are really useful to find out when some events happen for a specific time range.\n\nHowever, it´s not possible to run an anaylsis in that event frame while the event frame is running.\n\nHere is a workaround of how you can actually do it. The trick is to run the analysis under certain condition and link this analysis to an event frame.\n\nIntuitively, we try to build an analysis based on an event frame but that is not possible if we want to run calculation during the event frame. We will see instead how to run an event frame based on a analysis.\n\nThat will allow us to do some calculations during an opened event frame.\n\nProblem description:\n\nLet´s say we want to do some calculation over a time range for a periodic signal.\n\nBut the time range is not fixed and the period of the signal can change over time.\n\nIn others words, we don’t know the explicit start and the end times of the calculation.\n\nThe only thing we know is that we want to run the calculation (to find the integral of the ‘sinusoid’ signal in our example) while the signal is increasing and stop the calculation when it’s decreasing.\n\nLet’s plot a graph to summarize what we want to achieve:",
null,
"In our example, we will use instead the Sinusoid signal from the PI Interface for Random, Fractal, Station, Sinusoid Data:",
null,
"As the period of the signal is not always the same we can’t use a fixed time range.\n\nWe will have to retrieve the start and the end times using another tag.\n\nIn this example we have a variable Trigger2 linked to a PI Tag called ST2. (This PI Tag will be needed if we want to create later on an event frame.)\n\nThis variable is equal to 1 when the condition of the calculation is met, otherwise it’s equal to 0.\n\nHere we simply define Trigger2 in that:\n\n• If the previous value in the archive of the tag Sinusoid is lower than the current value of Sinusoid then the value of Trigger2 is equal to 1\n\n• If not; Trigger2 is equal to 0",
null,
"",
null,
"If we were working with a “real “signal, the only difference would have been the variable Trigger2. It would have to take into account the noise present on the signal. We should have ended up with something like that:",
null,
"Then comes the trick of this analysis:\n\nWe create a variable startdate. This variable has to be mapped to a PI Tag (Calcultime in the example) because we will need to store it in the archive. Do not forget in AF to set the Value Type of this tag to “DateTime”.",
null,
"Thanks to the analysis, we will store the actual timestamp of the Sinusoid tag if the trigger is equal to 0. If not, we won´t send any data to the PI Data Archive.\n\nThat way, after the Trigger2 variable jumps from 0 to 1 (in other words, when the Sinusoid signal starts increasing) we won´t update the timestamp in the archive.\n\nThe value stored in the archive is the beginning of our integral calculation.\n\nThen, we can use this timestamp to run our calculation from Calcultime to now ‘*’.",
null,
"Now it’s time to put everything together in a table understand for a better understanding:",
null,
"While Trigger2 is set to 0 (when the Sinusoid value is not increasing) we store the current PI Time of the Sinusoid tag. The result in Calcultime and PI Time column are the same.\n\nWhileTrigger2 is equal to 1, we don´t send the PI Time anymore to the archive so the value of Calcultime is the last value stored in the archive. This value corresponds to the beginning of our integral calculation Tot2.\n\nPlease note that the analysis scheduler is set on Event-Triggered. Indeed, we don’t know how long is the time range of the calculation of the integral so it´s important not to set it to periodic in that case.\n\nNow we are running the integral of the signal from Calcultime to now. We can see the result Tot2 increasing in real time:",
null,
"About the precision of the calculation:\n\nIn our example, we decide to compare the previous event in the archive and the current value of the tag..\n\nThat way, the precision of the integral depends on the configuration of the Sinusoid tag (scan class, exception/compression parameters, etc)\n\nLink an event frame to this calculation:\n\nIt´s possible to link that calculation to an event frame so we can easily find when the calculation was running or not.\n\nTo do that, you should create a PI Tag linked to the variable Trigger 2 in my example. Actually I already did that in the previous part but this should be done if you want to use the event frame.",
null,
"Then we can backfill the data and we can when the calculation was running in the last 2 days :",
null,
"Thanks to the event frame, we can easily find when the calculation was running or not using Coresight for instance :",
null,
"Please note that this method is quite heavy in configuration and required 2 tags at least per Analysis.\n\nThe possibility to run some calculations during an opened event frame will be added to the next release of Analytics (2017 R2)\n\nBy date: By tag:"
] |
[
null,
"https://pisquare.osisoft.com/resources/statics/rolebadges/roleBadge-4-1013-1430934071734.png",
null,
"https://pisquare.osisoft.com/people/tbernier/avatar/46.png",
null,
"https://pisquare.osisoft.com/resources/statics/rolebadges/roleBadge-4-1013-1430934071734.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123350/755-335/pastedImage_5.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123351/741-355/pastedImage_6.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123352/887-113/pastedImage_7.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123353/887-369/pastedImage_0.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123354/889-340/pastedImage_1.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123357/940-47/pastedImage_4.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123358/930-48/pastedImage_5.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123359/671-580/pastedImage_6.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123360/839-532/pastedImage_7.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123362/789-28/pastedImage_8.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123363/845-297/pastedImage_11.png",
null,
"https://pisquare.osisoft.com/servlet/JiveServlet/downloadImage/38-2168-123381/1176-384/pastedImage_1.png",
null
] |
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|
https://www.aijcrnet.com/journal/index/913
|
[
"University Students’ Level of Metacognitive Thinking and their Ability to Solve Problems\nNahil M.Aljaberi, Eman Gheith\n\nAbstract\nThe purpose of this study is to investigate the relationship between the university students’ metacognition thinking and their ability to solve mathematical and scientific problems. 172 university students were involved in this study. The researchers employed two types of instruments: metacognition awareness inventory, and a mathematical & scientific problem solving test; which was constructed by the researchers. After the collection of data, the researchers ran a suitable statistical analysis. The study has concluded that Petra University students have a medium level of metacognitive thinking, and that the variables of sex, faculty, high school stream, and the current year in the university had no effect on their level of metacognitive thinking. The study has also shown that these students suffer from a lack of ability in solving mathematical and scientific problems; no significance correlation between the level of metacognitive thinking in the overall scale and the ability to solve mathematical and scientific problems. However, there was a significant correlation between a few factors of metacognitive thinking and the ability to solve mathematical problems, and these are: Procedural Knowledge, Evaluation, Fault Picking, and Managing Knowledge; as well as a significant correlation between Fault Picking and the ability to solve both mathematical and scientific problems.\n\nFull Text: PDF"
] |
[
null
] |
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|
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/skills-handbook-measurement-conversions-exercises-page-886/4
|
[
"## Geometry: Common Core (15th Edition)\n\n$200\\text{ in} = 16.\\bar{6}\\text{ ft}$\nSince there are $12$ inches in $1$ foot, we divide $200$ by $12$ to convert inches to feet: $200\\text{ in}\\times\\frac{1\\text{ ft}}{12\\text{ in}}=16.\\bar{6}\\text{ ft}$"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8203786,"math_prob":0.99988353,"size":477,"snap":"2021-21-2021-25","text_gpt3_token_len":141,"char_repetition_ratio":0.12684989,"word_repetition_ratio":0.0,"special_character_ratio":0.3186583,"punctuation_ratio":0.08163265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99945575,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T04:25:40Z\",\"WARC-Record-ID\":\"<urn:uuid:4087e13b-3261-41a6-900d-2b9f8c7b1e69>\",\"Content-Length\":\"52943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04414611-d6c8-44e7-a270-ce6f1e63938a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c639c3f-4d91-404b-9e81-681e25eea91d>\",\"WARC-IP-Address\":\"54.164.42.204\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/skills-handbook-measurement-conversions-exercises-page-886/4\",\"WARC-Payload-Digest\":\"sha1:APQDMI5TMDRXMIKFSQHBIPVA5GXGPELN\",\"WARC-Block-Digest\":\"sha1:7YK3KLROPIBHABEQU333OI5PSXIY3IKT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988955.89_warc_CC-MAIN-20210509032519-20210509062519-00056.warc.gz\"}"}
|
https://www.nagwa.com/en/videos/484107202462/
|
[
"# Video: Finding the Magnitude of the Moment of a Force about a Point\n\nIf a force, having a magnitude of 498 N, is 8 cm away from point 𝐴, find the norm of the moment of the force about the point 𝐴, giving your answer in N⋅m.\n\n01:24\n\n### Video Transcript\n\nIf a force, having a magnitude of 498 newtons, is eight centimeters away from point 𝐴, find the norm of the moment of the force about the point 𝐴, giving your answer in newton meters.\n\nWe’ll call our force magnitude, 498 newtons, capital 𝐹. And the distance from point 𝐴 at which the force acts, eight centimeters, we’ll call 𝑑. We want to solve for the norm of the moment of the force about point 𝐴. We’ll call that moment 𝑀.\n\nIn general, if we have a force vector 𝑭 acting at a point separated from the point 𝐴 by a vector 𝒓, then the moment 𝑴 of that force 𝑭 around point 𝐴 is equal to 𝒓 cross 𝑭.\n\nIn our case, because we’re solving for the norm of that moment, we can write a simplified version of the expression: 𝑀 is equal to 𝑑 times 𝐹. Both 𝑑 and 𝐹 are given in our problem statement. So we’re ready to plug in and solve for 𝑀.\n\nWhen we do, we convert our distance into units of meters to be consistent with the required units of 𝑀. This product is equal to 39.84 newton meters. That’s the norm of the moment of the force about the point 𝐴."
] |
[
null
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|
https://www.brainkart.com/article/Backward-Difference-operator(-)_38954/
|
[
"Home | | Business Maths 12th Std | Backward Difference operator(∇)\n\n# Backward Difference operator(∇)\n\nThe operator ∇ is called backward difference operator and pronounced as nepla.\n\nBackward Difference operator ( ∇) :\n\nLet y = f(x) be a given function of x. Let y 0 , y1,..., yn be the values of y at\n\nx= x0 , x1 , x2 ,..., xn respectively. Then\n\ny1 − y0 = ∇y1\n\ny 2 − y1 = ∇y2\n\ny n − yn−1 = ∇yn\n\nare called the first(backward) differences.\n\nThe operator ∇ is called backward difference operator and pronounced as nepla.\n\nSecond(backward) differences: ∇ 2 y n = ∇ y n − ∇yn+1 , n = 1,2,3,…\n\nThird (backward) differences: ∇ 3 y n = ∇ 2 yn − ∇2 yn−1 n = 1,2,3,…\n\nIn general, kth (backward) differences: ∇ k yn = ∇ k −1 yn − ∇k−1 yn−1 n = 1,2,3,…\n\n## Backward difference table:",
null,
"Backward differences can also be defined as follows.\n\n∇ f (x) = f (x) − f (x − h)\n\nFirst differences: ∇ f (x + h) = f (x + h) − f (x)\n\n∇ f (x + 2h) = f (x + 2h) − f (x + h),...,h is the interval of spacing.\n\nSecond differences:\n\n∇ 2 f (x + h) = ∇(∇f (x + h) = ∇( f (x + h) − f (x))\n\n= ∇ f (x + h) − ∇f (x)\n\n∇ 2 f (x + 2h) = ∇ f (x + 2h ) − ∇f (x + h)\n\nThird differences:\n\n∇ 3 f (x + h) = ∇ 2 f (x + h ) − ∇2 f (x)\n\n∇ 3 f (x + 2h) = ∇ 2 f (x + 2h ) − ∇2 f (x + h)\n\nHere we note that, ∇ f (x + h) = f (x + h ) − f (x) = Δf (x)\n\n∇ f (x + 2h) = f (x + 2h ) − f (x + h) = Δf (x + h)\n\n∇ 2 f (x + 2h) = ∇ f (x + 2h ) − ∇f (x + h) = Δf (x + h ) − Δf (x)\n\n= Δ2 f (x)\n\nIn general, ∇n f (x + nh)= Δn f (x)\n\nTags : Finite Differences | Numerical Methods , 12th Business Maths and Statistics : Chapter 5 : Numerical Methods\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n12th Business Maths and Statistics : Chapter 5 : Numerical Methods : Backward Difference operator(∇) | Finite Differences | Numerical Methods"
] |
[
null,
"https://img.brainkart.com/imagebk39/0JGsb2G.jpg",
null
] |
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https://www.centerspace.net/examples/nmath/csharp/one-variable-minimization-example.php
|
[
"# C# One Variable Minimization Example\n\n← All NMath Code Examples\n\n```using System;\n\nusing CenterSpace.NMath.Core;\n\nnamespace CenterSpace.NMath.Examples.CSharp\n{\nclass OneVariableMinimizationExample\n{\n/// <summary>\n/// A .NET example in C# showing how to find a minimum of a univariate function\n/// using a golden minimizer and Brents method.\n/// </summary>\nstatic void Main( string[] args )\n{\n// Create a one-variable function.\nOneVariableFunction function = new Func<double, double>( f );\n\n// Use golden section search to find the minimum of the function\n// in the interval -10000 to 10000.\nint lower = -10000;\nint upper = 10000;\n\n// Create the minimizer with default tolerance and default maximum iterations.\nvar golden = new GoldenMinimizer();\n\n// Perform the minimization.\ndouble min = golden.Minimize( function, lower, upper );\n\nConsole.WriteLine();\n\nConsole.WriteLine( \"GoldenMinimizer found a minimum of \" + function.Evaluate( min ) );\nConsole.WriteLine( \"at \" + min + \" in \" + golden.Iterations + \" iterations.\" );\n\nif ( golden.ToleranceMet )\n{\n// Error is less than desired tolerance\nConsole.Write( \"Error of \" + golden.Error + \" is within tolerance of \" );\nConsole.WriteLine( golden.Tolerance + \".\" );\n}\nelse\n{\n// Minimization ended because maximum iterations was reached\nConsole.WriteLine( \"Ended with maximum iterations.\" );\n}\nConsole.WriteLine();\n\n// Try Brents Method for better efficiency.\nvar brent = new BrentMinimizer();\nmin = brent.Minimize( function, lower, upper );\n\nConsole.WriteLine( \"BrentMinimizer found a minimum of \" + function.Evaluate( min ) );\nConsole.WriteLine( \"at \" + min + \" in \" + brent.Iterations + \" iterations.\" );\nif ( brent.ToleranceMet )\n{\n// Error is less than desired tolerance\nConsole.Write( \"Error of \" + brent.Error + \" is within tolerance of \" );\nConsole.WriteLine( brent.Tolerance + \".\" );\n}\nelse\n{\n// Minimization ended because maximum iterations was reached\nConsole.WriteLine( \"Ended with maximum iterations.\" );\n}\nConsole.WriteLine();\n\n// If the derivative is known, we can utilize a better Brents\n// algorithm.\n\n// Create the derivative function.\nOneVariableFunction derivative = new Func<double, double>( df );\n\n// Instantiate a DBrentMinimizer.\nvar dbrent = new DBrentMinimizer();\nmin = dbrent.Minimize( function, derivative, lower, upper );\n\nConsole.WriteLine( \"DBrentMinimizer found a minimum of \" + function.Evaluate( min ) );\nConsole.WriteLine( \"at \" + min + \" in \" + dbrent.Iterations + \" iterations.\" );\nif ( dbrent.ToleranceMet )\n{\n// Error is less than desired tolerance\nConsole.Write( \"Error of \" + dbrent.Error + \" is within tolerance of \" );\nConsole.WriteLine( dbrent.Tolerance + \".\" );\n}\nelse\n{\n// Minimization ended because maximum iterations was reached\nConsole.WriteLine( \"Ended with maximum iterations.\" );\n}\nConsole.WriteLine();\n\n// Increase the tolerance to 0.1\ndbrent.Tolerance = 0.1;\nmin = dbrent.Minimize( function, derivative, lower, upper );\n\nConsole.WriteLine( \"DBrentMinimizer found a minimum of \" + function.Evaluate( min ) );\nConsole.WriteLine( \"at \" + min + \" in \" + dbrent.Iterations + \" iterations.\" );\nif ( dbrent.ToleranceMet )\n{\n// Error is less than desired tolerance\nConsole.Write( \"Error of \" + dbrent.Error + \" is within tolerance of \" );\nConsole.WriteLine( dbrent.Tolerance + \".\" );\n}\nelse\n{\n// Minimization ended because maximum iterations was reached\nConsole.WriteLine( \"Ended with maximum iterations.\" );\n}\nConsole.WriteLine();\n\n// Decrease tolerance to 1e-13 and increase the interval.\ndbrent.Tolerance = 1e-13;\nmin = dbrent.Minimize( function, derivative, lower * 100, upper * 100 );\n\nConsole.WriteLine( \"DBrentMinimizer found a minimum of \" + function.Evaluate( min ) );\nConsole.WriteLine( \"at \" + min + \" in \" + dbrent.Iterations + \" iterations.\" );\nif ( dbrent.ToleranceMet )\n{\nConsole.Write( \"Error of \" + dbrent.Error + \" is within tolerance of \" );\nConsole.WriteLine( dbrent.Tolerance + \".\" );\n}\nelse\n{\nConsole.WriteLine( \"Ended with maximum iterations.\" );\n}\nConsole.WriteLine();\n\nConsole.WriteLine();\nConsole.WriteLine( \"Press Enter Key\" );\n}\n\nprivate static double df( double x )\n{\nreturn 8 * Math.Cos( x ) + 6 * x - 5;\n}\n\nprivate static double f( double x )\n{\nreturn 8 * Math.Sin( x ) + 3 * x * x - 5 * x;\n}\n\n} // class\n\n} // namepace\n```\n← All NMath Code Examples\nTop"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.5340271,"math_prob":0.9659714,"size":4190,"snap":"2022-40-2023-06","text_gpt3_token_len":989,"char_repetition_ratio":0.21619685,"word_repetition_ratio":0.50308645,"special_character_ratio":0.2940334,"punctuation_ratio":0.24816446,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98938936,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T19:34:16Z\",\"WARC-Record-ID\":\"<urn:uuid:6ed7a0d5-1374-413d-950f-96afd0652cae>\",\"Content-Length\":\"17246\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:511fd171-cfc8-49c4-93ba-b81de5a35a37>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c1ba4c1-2c75-4f64-9cb4-f238079786a7>\",\"WARC-IP-Address\":\"72.47.229.209\",\"WARC-Target-URI\":\"https://www.centerspace.net/examples/nmath/csharp/one-variable-minimization-example.php\",\"WARC-Payload-Digest\":\"sha1:FSSTFMU6LZ6BDBGPSOHDLXB4KXFWKCDK\",\"WARC-Block-Digest\":\"sha1:GJI3PWWZGS43WPOMSZI435U6PZXDE6CH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030333455.97_warc_CC-MAIN-20220924182740-20220924212740-00352.warc.gz\"}"}
|
https://dsp.stackexchange.com/questions/19062/pole-zero-placement-for-filter
|
[
"# Pole/Zero Placement for Filter\n\nI'm attempting to design a FIR high pass filter than keeps signals above 200Hz and rejects signals below 60Hz with a sampling frequency of 500 samples/sec. This is my first time attempting this and I'm a little confused.\n\nI started with looking at pole/zero placement. I'm not exactly sure how to go about this but this is what I did. I want a zero on the unit circle at the 60Hz point to minimize that frequency. To find this spot on the axis I did the following:\n\n$$\\frac{60}{\\frac{500}{2}}\\pi=0.24\\pi$$\n\nSo I believe I need a zero at 0.24pi on the unit circle on the z-plane. Does this sound right? How would I then transform it in a difference equation in the sequence domain?\n\n• well, if it's FIR, all of your poles will be placed at the origin: $z=0$. – robert bristow-johnson Nov 8 '14 at 22:13\n• Is this a toy problem that you want to (have to) solve by hand, or do you want a 'real' design (for which you would normally use some software)? – Matt L. Nov 8 '14 at 23:13\n• By hand preferably. I'm working on the basics right now – codedude Nov 9 '14 at 3:37\n\nThis is how you can try to design a short and simple FIR filter by hand. Note that this method is just supposed to be enlightening, a really useful filter can be designed using some software. If you want a zero at 60 Hz you indeed need to place it at an angle of $0.24\\pi$ on the unit circle of the $z$-plane. The general relation between angle and frequency is\n\n$$\\phi=2\\pi\\frac{f}{f_s}$$\n\nwhere $f$ is the desired frequency, and $f_s$ is the sampling frequency. Note that if the filter coefficients are real-valued, zeros (and poles) always occur in complex conjugate pairs. So if you have a zero at $z=z_0$ you must also have a zero at $z=z_0^*$. This just reflects the fact that the spectrum of a real-valued filter is symmetric: $H(\\omega)=H^*(-\\omega)$. Since you want to design a causal FIR filter, all poles are at the origin of the $z$-plane, and you only have control over the zeros.\n\nSo let's design an FIR filter of order $3$ (i.e. length $4$) with a zero at $z_0=e^{j0.24\\pi}$, and - because it's a high pass filter - a zero at $z=0$. The zero at $z=0$ contributes a factor\n\n$$H_1(z)=1-z^{-1}$$\n\nto the total transfer function, and the zero at $z_0=e^{j0.24\\pi}$ (and its complex conjugate counterpart at $z_0^*=e^{-j0.24\\pi}$) contributes a factor\n\n$$H_2(z)=(1-z_0z^{-1})(1-z_0^*z^{-1})=1-(z_0+z_0^*)z^{-1}+|z_0|^2z^{-2}=\\\\ =1-2\\Re\\{z_0\\}z^{-1}+z^{-2}=1-2\\cos(0.24\\pi)z^{-1}+z^{-2}$$\n\nbecause $z_0+z_0^*=2\\Re\\{z_0\\}=2\\cos(0.24\\pi)$, and $|z_0|=1$ (the zero is on the unit circle). The total transfer function is\n\n$$H(z)=H_1(z)H_2(z)=(1-z^{-1})(1-2\\cos(0.24\\pi)z^{-1}+z^{-2})=\\\\ =1-[1+\\cos(0.24\\pi)]z^{-1}+[1+\\cos(0.24\\pi)]z^{-2}-z^{-3}\\tag{1}$$\n\nFrom (1) you can read the filter coefficients:\n\n$$h_0=1,\\quad h_1=-[1+\\cos(0.24\\pi)]=-2.4579,\\quad h_2=-h_1,\\quad h_3=-h_0\\tag{2}$$\n\nNote that we just imposed the locations of the zeros but we didn't apply any normalization. A useful normalization for a highpass filter is to require $H(-1)=1$, i.e. the response at the Nyquist frequency (half the sampling frequency) is $1$. Since\n\n$$H(-1)=\\sum_n(-1)^nh[n]=h-h+h-h=2h-2h$$\n\nwe can define new normalized filter coefficients by\n\n$$\\hat{h}[n]=\\frac{h[n]}{2(h-h)},\\quad n=0,1,2,3$$\n\nThe magnitude of the resulting frequency response of this filter looks like this (the phase is linear due to the symmetry of the coefficients):",
null,
"You see that the three constraints (zeros at DC and at 60 Hz, and unity response at Nyquist) are satisfied. The transition from stopband to passband is of course not very steep but that's as good as it gets with an FIR filter of order 3."
] |
[
null,
"https://i.stack.imgur.com/2rNXp.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8613839,"math_prob":0.99777627,"size":2580,"snap":"2019-51-2020-05","text_gpt3_token_len":901,"char_repetition_ratio":0.123059005,"word_repetition_ratio":0.005479452,"special_character_ratio":0.36782947,"punctuation_ratio":0.08448276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998969,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T23:12:44Z\",\"WARC-Record-ID\":\"<urn:uuid:814068f6-e33a-4b9f-8990-0a5fcd024485>\",\"Content-Length\":\"137114\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4dc97fb7-76d5-427d-baba-a1f663899884>\",\"WARC-Concurrent-To\":\"<urn:uuid:8988e236-91bb-4b8a-b585-24534995d2ce>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/19062/pole-zero-placement-for-filter\",\"WARC-Payload-Digest\":\"sha1:UFDKUM7WA75FQPBJVDEVMAGIFKQDE4M2\",\"WARC-Block-Digest\":\"sha1:MIAZWQPALILMIINMA2Y2E2UIJ3LJ3B6W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250614086.44_warc_CC-MAIN-20200123221108-20200124010108-00095.warc.gz\"}"}
|
https://www.goldbook.iupac.org/terms/view/M04035
|
[
"## most probable distribution (in macromolecular assemblies)\n\nhttps://doi.org/10.1351/goldbook.M04035\nA discrete distribution with the differential @M03716@ of the form: $f_{\\text{w}}(x) = a^{2}\\ x\\ (1- a)^{x- 1}$ where $$x$$ is a parameter characterizing the @C00956@, such as @R05271@ or @D01569@ and $$a$$ is a positive adjustable parameter. For large values of $$x$$, the most probable distribution converges to the particular case of the @S05502@ with $$b=1$$. In the literature, this distribution is sometimes referred to as the Flory distribution or the Schulz–Flory distribution.\nSource:\nPurple Book, 1st ed., p. 56 [Terms] [Book]"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.78709954,"math_prob":0.99723905,"size":527,"snap":"2020-24-2020-29","text_gpt3_token_len":157,"char_repetition_ratio":0.14913958,"word_repetition_ratio":0.0,"special_character_ratio":0.3434535,"punctuation_ratio":0.11340206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9939318,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-25T20:15:06Z\",\"WARC-Record-ID\":\"<urn:uuid:f787fdbd-549a-493b-8658-1a079f228938>\",\"Content-Length\":\"25648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:706178a3-a6be-4d42-b55a-01d6f5136ea0>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5b3cf11-5fe5-48db-b3ce-ca1041fff60c>\",\"WARC-IP-Address\":\"216.37.42.18\",\"WARC-Target-URI\":\"https://www.goldbook.iupac.org/terms/view/M04035\",\"WARC-Payload-Digest\":\"sha1:N5CDVLDO675QEUDOGB4YQWTFAZJPHCWW\",\"WARC-Block-Digest\":\"sha1:BRCK3AC6MVS24QOD75DE4GN2WRTSYU7E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347389355.2_warc_CC-MAIN-20200525192537-20200525222537-00378.warc.gz\"}"}
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https://www.doorsteptutor.com/Exams/ISS/Paper-1/Questions/Part-25.html
|
[
"# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 135 - 138 of 164\n\nAccess detailed explanations (illustrated with images and videos) to 164 questions. Access all new questions- tracking exam pattern and syllabus. View the complete topic-wise distribution of questions. Unlimited Access, Unlimited Time, on Unlimited Devices!\n\nView Sample Explanation or View Features.\n\nRs. 550.00 -OR-\n\nHow to register? Already Subscribed?\n\n## Question 135\n\nEdit\n\nAppeared in Year: 2011\n\n### Describe in Detail\n\nEssay▾\n\nLet X have the continuous c. d. f. F (x) . Define U = F (x) . Show that both - log U arid -log (1 - U) are exponential random variables:\n\n### Explanation\n\nLet U = F (x) , then the distribution function of G of U is given by\n\nSince F is non-increasing and its continuous.\n\nG (u) = F (F -1 (u) ) implies G (u) = u\n\nThen the p. d. f is\n\nSince F is a distribution function takes value in range [0,1] . Hence\n\nThis is a uniform distribution on [0,1] .\n\nLet Y =-log U. Then\n\nThe probability density function\n\nY =-logU has expon…\n\n… (18 more words) …\n\n## Question 136\n\nAppeared in Year: 2015\n\n### Describe in Detail\n\nEssay▾\n\nLet X1 , X2 , … , X12 be a random sample from a normal N (µ1 , σ1 2) and Y1 , Y2 , … Y10 be another random sample from normal N (µ2 , σ2 2) , independently to each other. Carry out an appropriate test for testing\n\nat 5 % level of significance. It is given that .\n\n### Explanation\n\nLet X1 , X2 , … , X12 be a random sample from a normal N (µ1 , σ1 2) and Y1 , Y2 , … Y10 be another random sample from normal N (µ2 , σ2 2) , independently to each other. The test of hypothesis is\n\nThe sample size of X is n = 12 and Y is m = 10\n\nThe test is depend upon the population variance is known and unknown. Here the variance is known, so the tes…\n\n… (40 more words) …\n\n## Question 137\n\nEdit\n\nAppeared in Year: 2012\n\n### Describe in Detail\n\nEssay▾\n\nYou arc given the following information:\n\n(i) In random testing, you test positive for a disease.\n\n(ii) In 5 % of cases, the test shows positive even when the subject does not have the disease.\n\n(iii) In the population at large, one person in 1000 has the disease. What is the conditional probability that you have the disease given that you have been tested positive, assuming that if someone has the disease, he will test positive with probability 1?\n\n### Explanation\n\nLet X denotes the test is positive and Y denotes the person has disease. Given that 5 % of cases, the test shows positive even when the subject does not have the disease that is\n\nwhere denotes the person has no disease. Also given the test positive with probability 1\n\nThen the conditional probability that you have the disease given that you have been …\n\n… (18 more words) …\n\n## Question 138\n\nEdit\n\nAppeared in Year: 2013\n\n### Describe in Detail\n\nEssay▾\n\nCompute the factorial moments µ(r) and the cumulants kr , r = 1,2, … , of Poisson distribution with parameter m.\n\n### Explanation\n\nLet X follows Poisson distribution with parameter m. The density function is\n\nThe r th factorial moment of Poisson distribution is\n\nThe cumulants of Poisson distribution is\n\nwe known that the moment generating function of X is\n\nSo, the cumulants is\n\nK1 = coefficient of t in = m\n\nK2 = coefficient of in = m\n\nSimilarly, for others K՚s values.\n\n… (2 more words) …"
] |
[
null
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|
https://brainmass.com/physics/velocity/average-power-output-group-velocity-542212
|
[
"Explore BrainMass\n\n# Average Power Output & Group Velocity\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\n(b) A tidal basin in Canada has an area of 240 km^2 with a mean tidal range of 12.4 m. Estimate the average power output assuming that the tidal period is 12.5 hours explaining your calculation.\n\n(c) An ocean wave, with amplitude a = 1m and wavelength...(Please see the attached file for the complete question.)\n\n© BrainMass Inc. brainmass.com March 5, 2021, 12:51 am ad1c9bdddf\nhttps://brainmass.com/physics/velocity/average-power-output-group-velocity-542212\n\n#### Solution Preview\n\n(b) Area of the basin, A= 240km^2 == 2.4*10^8 m^2\nmean tidal range, h = 12.4 m\navaerage tidal period, t = 12.5 hr == 12.5*3600 s\nDensity of water, d = 1*10^3 kg/m^3\ng = 9.8 m/s^2\n\nPotential energy due to, tidal:\nU = m*g*h/2\n= (d*A*h)*g*h/2\n= ...\n\n#### Solution Summary\n\nIn first problem, average power which can be produced by using tidal basin water. In second problem group velocity of a given wave is estimated.\n\n\\$2.49"
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[
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https://www.cnblogs.com/zcybky/p/10022913.html
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[
"",
null,
"# 第三次作业\n\n#include<stdio.h>\nint main()\n{\nint i,sum;\nsum=0;\nfor(i=1;i<=100;i++){\nsum=sum+i;\n}\nprintf(\"sum = %d\",sum);\nreturn 0;\n}",
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"个人总结和遇到的问题\n\n#include<stdio.h>\nint main()\n{\nint a=0,i=2000;\nscanf(\"%d\",&a);\nif(a<i){\nprintf(\"Invalid year!\");\n}\nelse{\nwhile(i<=a){\nif(a>2100){\nprintf(\"Invalid year!\");\nbreak;\n}\nelse if(a>=2004){\ni++;\nif(i%100!=0){\nif(i%4==0){\nprintf(\"%d\\n\",i);\n}\n}\nelse if(i%400==0){\nprintf(\"%d\\n\",i);\n}\n}\nelse{\nprintf(\"None\");\nbreak;\n}\n}}\nreturn 0;\n}",
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"个人总结和遇到的问题\n\n#include<stdio.h>\nint main()\n{\nint a,b,i,n,count=0,sum=0;\nint flag=1;\nscanf(\"%d %d\",&a,&b);\nfor(i=a;i<=b;i++){\nfor(n=2;n<i;n++){\nflag=1;\nif(i%n==0){\nflag=0;\nbreak;\n}\n}\nif(flag==1&&i!=1){\ncount++;\nsum=sum+i;}\n}\nprintf(\"%d %d\",count,sum);\nreturn 0;\n}",
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"posted @ 2018-11-27 19:35 张春宇 阅读(201) 评论(12编辑 收藏 举报"
] |
[
null,
"https://img2023.cnblogs.com/blog/35695/202309/35695-20230925204855354-1690761565.png",
null,
"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181126195043586-1962108835.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202222222255-1736599338.png",
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"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181126195643157-599432867.png",
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null,
"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202222316876-2096657005.png",
null,
"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181126195735702-768042804.png",
null,
"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181126205912530-948268959.png",
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"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181126210028419-842582937.png",
null,
"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202223633532-1868248338.png",
null,
"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202223555972-1006170593.png",
null,
"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202224025175-1678694168.png",
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"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181126210136525-230932113.png",
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"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181127190718202-448212727.png",
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"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181127190859342-286841622.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202224638919-1289011006.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202224734029-1526776285.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202224848970-324871872.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202225057690-1406026256.png",
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"https://img2018.cnblogs.com/blog/1498067/201811/1498067-20181127191013822-682811175.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202225157871-1107268344.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202223115598-206890266.png",
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"https://img2018.cnblogs.com/blog/1498067/201812/1498067-20181202223121323-818914954.png",
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|
https://www.coursehero.com/file/p6d1nhl/SW-Oregon-243121550-255121550-52121550-Central-Oregon-243108550-255108550/
|
[
"Sw oregon 243121550 255121550 52121550 central oregon\n\n• 76\n• 99% (158) 156 out of 158 people found this document helpful\n\nThis preview shows page 8 - 11 out of 76 pages.\n\nSW Oregon =243*(121/550) =255*(121/550) =52*(121/550)\nCentral Oregon =243*(108/550) =255*(108/550) =52*(108/550) Eastern Oregon =243*(111/550) =255*(111/550) =52*(111/550) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to use all 12 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 10 of 20 0.0/ 1.0 Points Click to see additional instructions A local gym is looking in to purchasing more exercise equipment and runs a survey to find out the preference in exercise equipment amongst their members. They categorize the members based on how frequently they use the gym each month – the results are below. Run an independence test at the 0.01 level of significance. Free Weights Weight Machines Endurance Machines Aerobics Equipment 0-10 Uses 12 17 25 13 11-30 Uses 20 18 9 9 31+ Uses 26 12 11 9 Enter the P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal.\n0-10 Uses =58*(67/181) =47*(67/181) =45*(67/181) =31*(67/181) 11-30 Uses =58*(56/181) =47*(56/181) =45*(56/181) =31*(56/181) 31+ Uses =58*(58/181) =47*(58/181) =45*(58/181) =31*(58/181) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.0144 Question 11 of 20 0.0/ 1.0 Points Click to see additional instructions A restaurant chain that has 3 locations in Portland is trying to determine which of their 3 locations they should keep open on New Year’s Eve. They survey a random sample of customers at each location and ask each whether or not they plan on going out to eat on New Year’s Eve. The results are below. Run a test for independence to decide if the proportion of customers that will go out to eat on New Year’s Eve is dependent on location. Use α=0.05 Enter the P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. .\n•",
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"•",
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"•",
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""
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null,
"https://www.coursehero.com/assets/img/doc-landing/start-quote.svg",
null,
"https://www.coursehero.com/assets/img/doc-landing/start-quote.svg",
null,
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null
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|
http://extraconversion.com/force/kiloponds
|
[
"# Kilopond\n\nThe kilopond is a unit of force equal to 1000 ponds. Plural name is kiloponds.\n\n## Sample Force Conversion\n\n 10 kip to kilonewtons, the result is 44.482216 kilonewtons 10 kip to newtons, the result is 44482.216 newtons 10 newton to joules per meter, the result is 10 joules per meter 10 kip to pound-forces, the result is 10000 pound-forces 10 kilonewton to ton-forces (metric), the result is 1.019716212978 ton-forces (metric) 10 kilonewton to kips, the result is 2.24808943871 kips 10 ton-force (metric) to kilonewtons, the result is 98.0665 kilonewtons 10 kilonewton to joules per meter, the result is 10000 joules per meter 10 kip to ton-forces (long), the result is 4.464285714286 ton-forces (long) 10 ton-force (metric) to kips, the result is 22.046226294122 kips\n\n## Table Conversion and Table Chart\n\n### Kilopond to Other Units\n\nKilopond can convert to other units, such as centinewton, decinewton, dekanewton, dyne, exanewton, giganewton, gram-force, hectonewton, joule/meter, kilogram-force, kilonewton, kip, meganewton, micronewton, millinewton, nanonewton, newton, ounce-force, petanewton, piconewton, pond, pound-force, poundal, sthene, teranewton, ton-force (long), ton-force (metric), ton-force (short) and more. You can find the reference table below:\n\nUnit to Other Unit Force Conversion Force Table\n1 kp = 980.665 cNkiloponds to centinewtonskiloponds to centinewtons table\n1 kp = 98.0665 dNkiloponds to decinewtonskiloponds to decinewtons table\n1 kp = 0.9807 daNkiloponds to dekanewtonskiloponds to dekanewtons table\n1 kp = 980665 dynkiloponds to dyneskiloponds to dynes table\n1 kp = 9.80665E-18 ENkiloponds to exanewtonskiloponds to exanewtons table\n1 kp = 9.807E-9 GNkiloponds to giganewtonskiloponds to giganewtons table\n1 kp = 1000 gfkiloponds to gram-forceskiloponds to gram-forces table\n1 kp = 0.0981 hNkiloponds to hectonewtonskiloponds to hectonewtons table\n1 kp = 9.80665 j/mkiloponds to joules per meterkiloponds to joules per meter table\n1 kp = 1 kgfkiloponds to kilogram-forceskiloponds to kilogram-forces table\n1 kp = 0.0098 kNkiloponds to kilonewtonskiloponds to kilonewtons table\n1 kp = 0.0022 kipkiloponds to kipskiloponds to kips table\n1 kp = 9.80665E-6 MNkiloponds to meganewtonskiloponds to meganewtons table\n1 kp = 9806650 µNkiloponds to micronewtonskiloponds to micronewtons table\n1 kp = 9806.65 mNkiloponds to millinewtonskiloponds to millinewtons table\n1 kp = 9806650000 nNkiloponds to nanonewtonskiloponds to nanonewtons table\n1 kp = 9.80665 N kiloponds to newtonskiloponds to newtons table\n1 kp = 35.273962070595 ozfkiloponds to ounce-forceskiloponds to ounce-forces table\n1 kp = 9.80665E-15 PNkiloponds to petanewtonskiloponds to petanewtons table\n1 kp = 9806650000000 pNkiloponds to piconewtonskiloponds to piconewtons table\n1 kp = 1000 p kiloponds to pondskiloponds to ponds table\n1 kp = 2.204622629412 lbfkiloponds to pound-forceskiloponds to pound-forces table\n1 kp = 70.931637529072 pdlkiloponds to poundalskiloponds to poundals table\n1 kp = 0.0098 snkiloponds to stheneskiloponds to sthenes table\n1 kp = 1.0E-11 TNkiloponds to teranewtonskiloponds to teranewtons table\n1 kp = 0.001 tnf (UK)kiloponds to ton-forces (long)kiloponds to ton-forces (long) table\n1 kp = 0.001 tfkiloponds to ton-forces (metric)kiloponds to ton-forces (metric) table\n1 kp = 0.0011 tf (short)kiloponds to ton-forces (short)kiloponds to ton-forces (short) table\n\n### Other Units to Kilopond\n\nKilopond also canvert from other unit such as centinewton, decinewton, dekanewton, dyne, exanewton, giganewton, gram-force, hectonewton, joule/meter, kilogram-force, kilonewton, kip, meganewton, micronewton, millinewton, nanonewton, newton, ounce-force, petanewton, piconewton, pond, pound-force, poundal, sthene, teranewton, ton-force (long), ton-force (metric), ton-force (short) and more. You can find the reference table below:\n\nUnit to Other Unit Force Conversion Force Table\n1 cN = 0.001 kpcentinewtons to kilopondscentinewtons to kiloponds table\n1 dN = 0.0102 kpdecinewtons to kilopondsdecinewtons to kiloponds table\n1 daN = 1.019716212978 kpdekanewtons to kilopondsdekanewtons to kiloponds table\n1 dyn = 1.019716E-6 kpdynes to kilopondsdynes to kiloponds table\n1 EN = 1.0197162129779E+17 kpexanewtons to kilopondsexanewtons to kiloponds table\n1 GN = 101971621.29779 kpgiganewtons to kilopondsgiganewtons to kiloponds table\n1 gf = 0.001 kpgram-forces to kilopondsgram-forces to kiloponds table\n1 hN = 10.197162129779 kphectonewtons to kilopondshectonewtons to kiloponds table\n1 j/m = 0.102 kpjoules per meter to kilopondsjoules per meter to kiloponds table\n1 kgf = 1 kpkilogram-forces to kilopondskilogram-forces to kiloponds table\n1 kN = 101.97162129779 kpkilonewtons to kilopondskilonewtons to kiloponds table\n1 kip = 453.59236844386 kpkips to kilopondskips to kiloponds table\n1 MN = 101971.62129779 kpmeganewtons to kilopondsmeganewtons to kiloponds table\n1 µN = 1.01972E-7 kpmicronewtons to kilopondsmicronewtons to kiloponds table\n1 mN = 0.0001 kpmillinewtons to kilopondsmillinewtons to kiloponds table\n1 nN = 1.02E-10 kpnanonewtons to kilopondsnanonewtons to kiloponds table\n1 N = 0.102 kpnewtons to kilopondsnewtons to kiloponds table\n1 ozf = 0.0283 kpounce-forces to kilopondsounce-forces to kiloponds table\n1 PN = 1.0197162129779E+14 kppetanewtons to kilopondspetanewtons to kiloponds table\n1 pN = 1.0197162129779E-13 kppiconewtons to kilopondspiconewtons to kiloponds table\n1 p = 0.001 kpponds to kilopondsponds to kiloponds table\n1 lbf = 0.4536 kppound-forces to kilopondspound-forces to kiloponds table\n1 pdl = 0.0141 kppoundals to kilopondspoundals to kiloponds table\n1 sn = 101.97162129779 kpsthenes to kilopondssthenes to kiloponds table\n1 TN = 101971621297.79 kpteranewtons to kilopondsteranewtons to kiloponds table\n1 tnf (UK) = 1016.0469053143 kpton-forces (long) to kilopondston-forces (long) to kiloponds table\n1 tf = 1000 kpton-forces (metric) to kilopondston-forces (metric) to kiloponds table\n1 tf (short) = 907.18473688772 kpton-forces (short) to kilopondston-forces (short) to kiloponds table\n\n## Centinewton [ cN ]\n\nCentinewton is a unit of force equal to 1/100 newton and its symbol (cN). It comes from a combination of metrix-prefix \"centi\" with the SI derived unit of force \"newton\".\n\n## Decinewton [ dN ]\n\nDecinewton is a unit of force equal to 1/10 newton and its symbol (dN). It comes from a combination of metrix-prefix \"deci\" with the SI derived unit of force \"newton\".\n\n## Dekanewton [ daN ]\n\nDekanewton is a unit of force equal to 10 newton and its symbol (daN). It comes from a combination of metrix-prefix \"deka\" with the SI derived unit of force \"newton\".\n\n## Dyne [ dyn ]\n\nThe dyne (symbol \"dyn\" ) is a unit of force specified in the centimetre-gram-second (CGS) system of units.\n\n## Exanewton [ EN ]\n\nExanewton is a unit of force equal to 1000000000000000000 (1E+18) newtons and its symbol (EN). It comes from a combination of metrix-prefix \"exa\" with the SI derived unit of force \"newton\".\n\n## Giganewton [ GN ]\n\nGiganewton is a unit of force equal to 1 000 000 000 (1E+9) newtons and its symbol (GN). It comes from a combination of metrix-prefix \"giga\" with the SI derived unit of force \"newton\".\n\n## Gram-force [ gf ]\n\nThe gram-force is a unit of force, 1 gram-force = 0.00980665 newton.\n\n## Hectonewton [ hN ]\n\nHectonewton is a unit of force equal to 100 newtons and its symbol (hN). It comes from a combination of metrix-prefix \"heto\" with the SI derived unit of force \"newton\".\n\n## Joule/meter [ j/m ]\n\nThe joule per meter is a unit of force and equivalent to newton (1 joule per meter = 1 newton).\n\n## Kilogram-force [ kgf ]\n\nThe kilogram-force(kgf) is a gravitational metric unit of force. 1 kilogram-force = 9.80665 newtons.\n\n## Kilonewton [ kN ]\n\nKilonewton is a unit of force equal to 1 000 newtons and its symbol (kN). It comes from a combination of metrix-prefix \"kilo\" with the SI derived unit of force \"newton\".\n\n## Kilopond [ kp ]\n\nThe kilopond is a unit of force equal to 1000 ponds.\n\n## Kip [ kip ]\n\nA kip is a non-SI unit of force equal to 1,000 pounds-force, used primarily by American architects and engineers to measure engineering loads.\n\n## Meganewton [ MN ]\n\nMeganewton is a unit of force equal to 1 000 000 newtons and its symbol (MN). It comes from a combination of metrix-prefix \"mega\" with the SI derived unit of force \"newton\".\n\n## Micronewton [ µN ]\n\nMicronewton is a unit of force equal to 1/1 000 000 newton and its symbol (µN). It comes from a combination of metrix-prefix \"giga\" with the SI derived unit of force \"newton\".\n\n## Millinewton [ mN ]\n\nMillinewton is a unit of force equal to 1/1 000 newton and its symbol (mN). It comes from a combination of metrix-prefix \"milli\" with the SI derived unit of force \"newton\".\n\n## Nanonewton [ nN ]\n\nNanonewton is a unit of force equal to 1/1 000 000 000 newton and its symbol (nN). It comes from a combination of metrix-prefix \"nano\" with the SI derived unit of force \"newton\".\n\n## Newton [ N ]\n\nNewton is the SI derived unit of force and it is equivalent to joule per meter (1 newton = 1 joule/meter).\n\n## Ounce-force [ ozf ]\n\nOunce-force is a unit of force and 1 ounce-force is about 0.278013851 newton.\n\n## Petanewton [ PN ]\n\nPetanewton is a unit of force equal to 1 000 000 000 000 000 newtons and its symbol (PN). It's a combination of metrix-prefix \"peta\" with the SI derived unit of force \"newton\".\n\n## Piconewton [ pN ]\n\nPiconewton is a unit of force equal to 1/1 000 000 000 000 newton and its symbol (pN). It comes from a combination of metrix-prefix \"pico\" with the SI derived unit of force \"newton\".\n\n## Pond [ p ]\n\nThe pond is a unit of force and 1 pond is equal to 0.00980665 newton.\n\n## Pound-force [ lbf ]\n\nThe pound-force is one of force units measurement and equal to the gravitational force exerted on a mass of one avoirdupois pound on the surface of Earth.\n\n## Poundal [ pdl ]\n\nThe poundal is a unit of force that is part of the foot-pound-second system of units with symbol \"lpd\" of system Imperial .\n\n## Sthene [ sn ]\n\nThe sthène is the unit of force in the metre-tonne-second system of units (mts).\n\n## Teranewton [ TN ]\n\nTeranewton is a unit of force equal to 1 000 000 000 000 newtons and its symbol (TN). It comes from a combination of metrix-prefix \"tera\" with the SI derived unit of force \"newton\".\n\n## Ton-force (long) [ tnf (UK) ]\n\nThe ton-forces (long])is a unit of force. 1 newton = 0.000100361 ton-force (long).\n\n## Ton-force (metric) [ tf ]\n\nThe ton-forces (metric) is a unit of force. 1 newton = 0.000101972 fon-force (metric).\n\n## Ton-force (short) [ tf (short) ]\n\nThe ton-forces (short) is a unit of force. 1 ton-force (short) = 8896.4432 newtons."
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https://en.m.wikipedia.org/wiki/Double_coset
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"# Double coset\n\nIn group theory, a field of mathematics, a double coset is a collection of group elements which are equivalent under the symmetries coming from two subgroups. More precisely, let G be a group, and let H and K be subgroups. Let H act on G by left multiplication while K acts on G by right multiplication. For each x in G, the (H, K)-double coset of x is the set\n\n$HxK=\\{hxk\\colon h\\in H,k\\in K\\}.$",
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"When H = K, this is called the H-double coset of x. Equivalently, HxK is the equivalence class of x under the equivalence relation\n\nx ~ y if and only if there exist h in H and k in K such that hxk = y.\n\nThe set of all double cosets is denoted\n\n$H\\backslash G/K.$",
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"## Properties\n\nSuppose that G is a group with subgroups H and K acting by left and right multiplication, respectively. The (H, K)-double cosets of G may be equivalently described as orbits for the product group H × K acting on G by (h, k)⋅x = hxk−1. Many of the basic properties of double cosets follow immediately from the fact that they are orbits. However, because G is a group and H and K are subgroups acting by multiplication, double cosets are more structured than orbits of arbitrary group actions, and they have additional properties that are false for more general actions.\n\n• Two double cosets HxK and HyK are either disjoint or identical.\n• G is the disjoint union of its double cosets.\n• There is a one-to-one correspondence between the two double coset spaces H \\ G / K and K \\ G / H given by identifying HxK with Kx−1H.\n• If H = {1}, then H \\ G / K = G / K. If K = {1}, then H \\ G / K = H \\ G.\n• A double coset HxK is a union of right cosets of H and left cosets of K, specifically,\n{\\begin{aligned}HxK&=\\bigcup _{k\\in K}Hxk=\\coprod _{Hxk\\in H\\backslash HxK}Hxk,\\\\HxK&=\\bigcup _{h\\in H}hxK=\\coprod _{hxK\\in HxK/K}hxK.\\end{aligned}}\n• The set of (H, K)-double cosets is in bijection with the orbits H \\ (G / K), and also with the orbits (H \\ G) / K under the mappings $HgK\\to H(gK)$ and $HgK\\to (Hg)K$ respectively.\n• If H is normal, then H \\ G is a group, and the right action of K on this group factors through the right action of H \\ HK. It follows that H \\ G / K = HK \\ G. Similarly, if K is normal, then H \\ G / K = G / HK.\n• If H is a normal subgroup of G, then the H-double cosets are in one-to-one correspondence with the left (and right) H-cosets.\n• Consider HxK as the union of a K-orbit of right H-cosets. The stabilizer of the right H-coset HxkH \\ HxK with respect to the right action of K is K ∩ (xk)−1Hxk. Similarly, the stabilizer of the left K-coset hxKHxK / K with respect to the left action of H is HhxK(hx)−1.\n• It follows that the number of right cosets of H contained in HxK is the index [K : Kx−1Hx] and the number of left cosets of K contained in HxK is the index [H : HxKx−1]. Therefore\n{\\begin{aligned}|HxK|&=[H:H\\cap xKx^{-1}]|K|=|H|[K:K\\cap x^{-1}Hx],\\\\\\left[G:H\\right]&=\\sum _{HxK\\in H\\backslash G/K}[K:K\\cap x^{-1}Hx],\\\\\\left[G:K\\right]&=\\sum _{HxK\\in H\\backslash G/K}[H:H\\cap xKx^{-1}].\\end{aligned}}\n• If G, H, and K are finite, then it also follows that\n{\\begin{aligned}|HxK|&={\\frac {|H||K|}{|H\\cap xKx^{-1}|}}={\\frac {|H||K|}{|K\\cap x^{-1}Hx|}},\\\\\\left[G:H\\right]&=\\sum _{HxK\\in H\\backslash G/K}{\\frac {|K|}{|K\\cap x^{-1}Hx|}},\\\\\\left[G:K\\right]&=\\sum _{HxK\\in H\\backslash G/K}{\\frac {|H|}{|H\\cap xKx^{-1}|}}.\\end{aligned}}\n• Fix xG, and let (H × K)x denote the double stabilizer {(h, k) : hxk = x}. Then the double stabilizer is a subgroup of H × K.\n• Because G is a group, for each hH there is precisely one gG such that hxg = x, namely g = x−1h−1x; however, g may not be in K. Similarly, for each kK there is precisely one g′ ∈ G such that gxk = x, but g may not be in H. The double stabilizer therefore has the descriptions\n$(H\\times K)_{x}=\\{(h,x^{-1}h^{-1}x)\\colon h\\in H\\}\\cap H\\times K=\\{(xk^{-1}x^{-1},k)\\colon k\\in K\\}\\cap H\\times K.$\n• (Orbit–stabilizer theorem) There is a bijection between HxK and (H × K) / (H × K)x under which hxk corresponds to (h, k−1)(H × K)x. It follows that if G, H, and K are finite, then\n$|HxK|=[H\\times K:(H\\times K)_{x}]=|H\\times K|/|(H\\times K)_{x}|.$\n• (Cauchy–Frobenius lemma) Let G(h, k) denote the elements fixed by the action of (h, k). Then\n$|H\\backslash G/K|={\\frac {1}{|H||K|}}\\sum _{(h,k)\\in H\\times K}|G^{(h,k)}|.$\n• In particular, if G, H, and K are finite, then the number of double cosets equals the average number of points fixed per pair of group elements.\n\nThere is an equivalent description of double cosets in terms of single cosets. Let H and K both act by right multiplication on G. Then G acts by left multiplication on the product of coset spaces G / H × G / K. The orbits of this action are in one-to-one correspondence with H \\ G / K. This correspondence identifies (xH, yK) with the double coset Hx−1yK. Briefly, this is because every G-orbit admits representatives of the form (H, xK), and the representative x is determined only up to left multiplication by an element of H. Similarly, G acts by right multiplication on H \\ G × K \\ G, and the orbits of this action are in one-to-one correspondence with the double cosets H \\ G / K. Conceptually, this identifies the double coset space H \\ G / K with the space of relative configurations of an H-coset and a K-coset. Additionally, this construction generalizes to the case of any number of subgroups. Given subgroups H1, ..., Hn, the space of (H1, ..., Hn)-multicosets is the set of G-orbits of G / H1 × ... × G / Hn.\n\nThe analog of Lagrange's theorem for double cosets is false. This means that the size of a double coset need not divide the order of G. For example, let G = S3 be the symmetric group on three letters, and let H and K be the cyclic subgroups generated by the transpositions (1 2) and (1 3), respectively. If e denotes the identity permutation, then\n\n$HeK=HK=\\{e,(12),(13),(132)\\}.$\n\nThis has four elements, and four does not divide six, the order of S3. It is also false that different double cosets have the same size. Continuing the same example,\n\n$H(23)K=\\{(23),(123)\\},$\n\nwhich has two elements, not four.\n\nHowever, suppose that H is normal. As noted earlier, in this case the double coset space equals the right coset space HK \\ G. Similarly, if K is normal, then H \\ G / K is the left coset space G / HK. Standard results about left and right coset spaces then imply the following facts.\n\n• |HxK| = |HK| for all xG. That is, all double cosets have the same cardinality.\n• If G is finite, then |G| = |HK| ⋅ |H \\ G / K|. In particular, |HK| and |H \\ G / K| divide |G|.\n\n## Examples\n\n• Let G = Sn be the symmetric group, considered as permutations of the set {1, ..., n}. Consider the subgroup H = Sn − 1 which stabilizes n. Then Sn − 1 \\ Sn / Sn − 1 consists of two double cosets. One of these is H = Sn − 1. If γ is a permutation which does not fix n, then the other coset is represented by Sn − 1 γ Sn − 1.\n• Let G be the group GLn(R), and let B be the subgroup of upper triangular matrices. The double coset space B \\ G / B is the Bruhat decomposition of G. Each double coset has a representative BwB, where w is a permutation matrix. For instance, if n = 2, then\n$B\\backslash \\operatorname {GL} _{2}(\\mathbf {R} )/B=\\left\\{B{\\begin{pmatrix}1&0\\\\0&1\\end{pmatrix}}B,B{\\begin{pmatrix}0&1\\\\1&0\\end{pmatrix}}B\\right\\}.$\n\n## Products in the free abelian group on the set of double cosets\n\nSuppose that G is a group and that H, K, and L are subgroups. Under certain finiteness conditions, there is a product on the free abelian group generated by the (H, K)- and (K, L)-double cosets with values in the free abelian group generated by the (H, L)-double cosets. This means there is a bilinear function\n\n$\\mathbf {Z} [H\\backslash G/K]\\times \\mathbf {Z} [K\\backslash G/L]\\to \\mathbf {Z} [H\\backslash G/L].$\n\nAssume for simplicity that G is finite. To define the product, reinterpret these free abelian groups in terms of the group algebra of G as follows. Every element of Z[H \\ G / K] has the form\n\n$\\sum _{HxK\\in H\\backslash G/K}f_{HxK}\\cdot [HxK],$\n\nwhere { fHxK } is a set of integers indexed by the elements of H \\ G / K. This element may be interpreted as a Z-valued function on H \\ G / K, specifically, HxKfHxK. This function may be pulled back along the projection GH \\ G / K which sends x to the double coset HxK. This results in a function xfHxK. By the way in which this function was constructed, it is left invariant under H and right invariant under K. The corresponding element of the group algebra Z[G] is\n\n$\\sum _{x\\in G}f_{HxK}\\cdot [x],$\n\nand this element is invariant under left multiplication by H and right multiplication by K. Conceptually, this element is obtained by replacing HxK by the elements it contains, and the finiteness of G ensures that the sum is still finite. Conversely, every element of Z[G] which is left invariant under H and right invariant under K is the pullback of a function on Z[H \\ G / K]. Parallel statements are true for Z[K \\ G / L] and Z[H \\ G / L].\n\nWhen elements of Z[H \\ G / K], Z[K \\ G / L], and Z[H \\ G / L] are interpreted as invariant elements of Z[G], then the product whose existence was asserted above is precisely the multiplication in Z[G]. Indeed, it is trivial to check that the product of a left-H-invariant element and a right-L-invariant element continues to be left-H-invariant and right-L-invariant. The bilinearity of the product follows immediately from the bilinearity of multiplication in Z[G]. It also follows that if M is a fourth subgroup of G, then the product of (H, K)-, (K, L)-, and (L, M)-double cosets is associative. Because the product in Z[G] corresponds to convolution of functions on G, this product is sometimes called the convolution product.\n\nThere are numerous explicit formulas for the convolution product depending upon the way in which the factors are represented. Fix an (H, K)-double coset HxK and a (K, L)-double coset KyL. These may be written as a union of right H-cosets and a union of left L-cosets, respectively:\n\n$HxK=\\bigcup _{Ha_{i}\\in H\\backslash HxK}Ha_{i},$\n$KyL=\\bigcup _{b_{j}L\\in KxL/L}b_{j}L.$\n\nThen the convolution product is\n\n$[HxK]\\cdot [KyL]=\\sum _{Ha_{i}\\in H\\backslash HxK}\\sum _{b_{j}L\\in KyL/L}[Ha_{i}b_{j}L].$\n\nAn equivalent way of phrasing this is to define homomorphisms\n\n$\\mathbf {Z} [H\\backslash G/K]\\to \\mathbf {Z} [H\\backslash G],$\n$\\mathbf {Z} [H\\backslash G/K]\\to \\mathbf {Z} [G/K],$\n\nby pullback along the projections from H \\ G and G / K to H \\ G / K. These send\n\n$[HxK]\\mapsto \\sum _{Ha_{i}\\in H\\backslash HxK}[Ha_{i}],$\n$[KyL]\\mapsto \\sum _{b_{j}L\\in KyL/L}[b_{j}L].$\n\nThat is, these send a double coset to the single cosets contained in it, analogous to the pullback along GH \\ G / K used above. There is a product\n\n$\\mathbf {Z} [H\\backslash G]\\times \\mathbf {Z} [G/L]\\to \\mathbf {Z} [H\\backslash G/L],$\n\ndefined on generators as [Ha] ⋅ [bL] = [HabL] and extended by bilinearity. Combining this with the above homomorphisms obtains the convolution product.\n\nThe above expression has the additional advantage that it applies even when G is not finite. Assume instead that every (H, K)-double coset is a finite union of right H-cosets and every (K, L)-double coset is a finite union of left L-cosets. Then the above expressions for the product in terms of right H-cosets and left L-cosets are finite sums. Consequently, the product is well-defined in this case, even when G is infinite. The product remains bilinear and associative in this situation.\n\nThe classes [HaibjL] occurring in the above sum may not be distinct. Because of this, the product is not generally equal to the sum of all [HzL], where HzL is a double coset contained in HxKyL = { hxkyl : hH, kK, lL }. To get an expression as a sum over distinct generators, define:\n\n$c_{x,y}(z)=|\\{(i,j)\\colon a_{i}b_{j}\\in HzL\\}|.$\n\nClearly cx, y(z) is zero unless zHxKyL. The product simplifies to\n\n$[HxK]\\cdot [KyL]=\\sum _{HzL\\in H\\backslash G/L}c_{x,y}(z)[HzL].$\n\nThis formula leads to the following formula for a product of general elements:\n\n$\\left(\\sum _{HxK\\in H\\backslash G/K}f_{HxK}[HxK]\\right)\\left(\\sum _{KyL\\in K\\backslash G/L}g_{KyL}[KyL]\\right)=\\sum _{HzL\\in H\\backslash G/L}\\left(\\sum _{HxK\\in H\\backslash G/K}\\sum _{KyL\\in K\\backslash G/L}c_{x,y}(z)f_{HxK}g_{KyL}\\right)[HzL].$\n\nBy decomposing each double coset into single cosets, the product of generators may also be written as a sum over the cosets of H or the cosets of L. Define\n\n{\\begin{aligned}c'_{x,y}(u)&=|\\{(i,j)\\colon a_{i}b_{j}\\in Hu\\}|,\\\\c''_{x,y}(v)&=|\\{(i,j)\\colon a_{i}b_{j}\\in vL\\}|.\\end{aligned}}\n\nThen\n\n$[HxK]\\cdot [KyL]=\\sum _{Hu\\in H\\backslash G}c'_{x,y}(u)[HuL]=\\sum _{vL\\in G/L}c''_{x,y}(v)[HvL],$\n\nbut the sums may no longer be over distinct generators.\n\nAn important special case is when H = K = L. In this case, the product is a bilinear function\n\n$\\mathbf {Z} [H\\backslash G/H]\\times \\mathbf {Z} [H\\backslash G/H]\\to \\mathbf {Z} [H\\backslash G/H].$\n\nThis product turns Z[H \\ G / H] into an associative ring whose identity element is the class of the trivial double coset [H]. In general, this ring is non-commutative. For example, if H = {1}, then the ring is the group algebra Z[G], and a group algebra is a commutative ring if and only if the underlying group is abelian.\n\nIf H is normal, so that the H-double cosets are the same as the elements of the quotient group G / H, then the product on Z[H \\ G / H] is the product in the group algebra Z[G / H]. In particular, it is the usual convolution of functions on G / H. In this case, the ring is commutative if and only if G / H is abelian, or equivalently, if and only if H contains the commutator subgroup of G.\n\nIf H is not normal, then Z[H \\ G / H] may be commutative even if G is non-abelian. A classical example is the product of two Hecke operators. This is the product in the Hecke algebra, which is commutative even though the group G is the modular group, which is non-abelian, and the subgroup is an arithmetic subgroup and in particular does not contain the commutator subgroup. Commutativity of the convolution product is closely tied to Gelfand pairs.\n\nWhen the group G is a topological group, it is possible to weaken the assumption that the number of left and right cosets in each double coset is finite. The group algebra Z[G] is replaced by an algebra of functions such as L2(G) or C(G), and the sums are replaced by integrals. The product still corresponds to convolution. For instance, this happens for the Hecke algebra of a locally compact group.\n\n## Applications\n\nWhen a group $G$ has a transitive group action on a set $S$ , computing certain double coset decompositions of $G$ reveals extra information about structure of the action of $G$ on $S$ . Specifically, if $H$ is the stabilizer subgroup of some element $s\\in S$ , then $G$ decomposes as exactly two double cosets of $(H,H)$ if and only if $G$ acts transitively on the set of distinct pairs of $S$ . See 2-transitive groups for more information about this action.\n\nDouble cosets are important in connection with representation theory, when a representation of H is used to construct an induced representation of G, which is then restricted to K. The corresponding double coset structure carries information about how the resulting representation decomposes. In the case of finite groups, this is Mackey's decomposition theorem.\n\nThey are also important in functional analysis, where in some important cases functions left-invariant and right-invariant by a subgroup K can form a commutative ring under convolution: see Gelfand pair.\n\nIn geometry, a Clifford–Klein form is a double coset space Γ\\G/H, where G is a reductive Lie group, H is a closed subgroup, and Γ is a discrete subgroup (of G) that acts properly discontinuously on the homogeneous space G/H.\n\nIn number theory, the Hecke algebra corresponding to a congruence subgroup Γ of the modular group is spanned by elements of the double coset space $\\Gamma \\backslash \\mathrm {GL} _{2}^{+}(\\mathbb {Q} )/\\Gamma$ ; the algebra structure is that acquired from the multiplication of double cosets described above. Of particular importance are the Hecke operators $T_{m}$ corresponding to the double cosets $\\Gamma _{0}(N)g\\Gamma _{0}(N)$ or $\\Gamma _{1}(N)g\\Gamma _{1}(N)$ , where $g=\\left({\\begin{smallmatrix}1&0\\\\0&m\\end{smallmatrix}}\\right)$ (these have different properties depending on whether m and N are coprime or not), and the diamond operators $\\langle d\\rangle$ given by the double cosets $\\Gamma _{1}(N)\\left({\\begin{smallmatrix}a&b\\\\c&d\\end{smallmatrix}}\\right)\\Gamma _{1}(N)$ where $d\\in (\\mathbb {Z} /N\\mathbb {Z} )^{\\times }$ and we require $\\left({\\begin{smallmatrix}a&b\\\\c&d\\end{smallmatrix}}\\right)\\in \\Gamma _{0}(N)$ (the choice of a, b, c does not affect the answer)."
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/33e8db93a4785947cf61105e66234ab3f20c663f",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/6b0fc27841bd56c039c6be42950781a094b68ad5",
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https://www.esaral.com/q/simplify-each-of-the-following-and-express-it-in-the-form-a-ib-72131
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[
"",
null,
"# Simplify each of the following and express it in the form a + ib :\n\nQuestion:\n\nSimplify each of the following and express it in the form a + ib :\n\n$(-5+6 i)-(-2+i)$\n\nSolution:\n\nGiven: $(-5+6 i)-(-2+i)$\n\nFirstly, we open the brackets\n\n$-5+6 i+2-i$\n\n$=-3+5 i$",
null,
""
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[
null,
"https://www.facebook.com/tr",
null,
"https://www.esaral.com/media/uploads/2022/02/04/image78029.png",
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https://www.wias-berlin.de/publications/wias-publ/run.jsp?lang=0&template=abstract&type=Preprint&year=&number=2487
|
[
"WIAS Preprint No. 2487, (2018)\n\n# Sasa--Satsuma hierarchy of integrable evolution equations\n\nAuthors\n\n• Bandelow, Uwe\nORCID: 0000-0003-3677-2347\n• Amiranashvili, Shalva\nORCID: 0000-0002-8132-882X\n• Pickartz, Sabrina\n• Akhmediev, Nail\n\n2010 Mathematics Subject Classification\n\n• 37K10 37K40 35Q55\n\n2008 Physics and Astronomy Classification Scheme\n\n• 02.30.Ik 05.45.Yv 42.65.Tg 47.35.Fg\n\nKeywords\n\n• Sasa--Satsuma equation, Lax pair, integrable hierarchy\n\nDOI\n\n10.20347/WIAS.PREPRINT.2487\n\nAbstract\n\nWe present the infinite hierarchy of Sasa-Satsuma evolution equations. The corresponding Lax pairs are given, thus proving its integrability. The lowest order member of this hierarchy is the nonlinear Schrödinger equation, while the next one is the Sasa-Satsuma equation that includes third-order terms. Up to sixth- order terms of the hierarchy are given in explicit form, while the provided recurrence relation allows one to explicitly write all higher-order terms. The whole hierarchy can be combined into a single general equation. Each term in this equation contains a real independent coefficient that provides the possibility of adapting the equation to practical needs. A few examples of exact solutions of this general equation with an infinite number of terms are also given explicitly.\n\nAppeared in"
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[
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https://www.lmfdb.org/EllipticCurve/Q/44880/cc/
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[
"# Properties\n\n Label 44880.cc Number of curves $2$ Conductor $44880$ CM no Rank $0$ Graph",
null,
"# Related objects\n\nShow commands for: SageMath\nsage: E = EllipticCurve(\"cc1\")\n\nsage: E.isogeny_class()\n\n## Elliptic curves in class 44880.cc\n\nsage: E.isogeny_class().curves\n\nLMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality\n44880.cc1 44880cc2 [0, 1, 0, -2296, -27820] 49152\n44880.cc2 44880cc1 [0, 1, 0, 424, -2796] 24576 $$\\Gamma_0(N)$$-optimal\n\n## Rank\n\nsage: E.rank()\n\nThe elliptic curves in class 44880.cc have rank $$0$$.\n\n## Complex multiplication\n\nThe elliptic curves in class 44880.cc do not have complex multiplication.\n\n## Modular form 44880.2.a.cc\n\nsage: E.q_eigenform(10)\n\n$$q + q^{3} - q^{5} + 2q^{7} + q^{9} - q^{11} - q^{15} + q^{17} + O(q^{20})$$\n\n## Isogeny matrix\n\nsage: E.isogeny_class().matrix()\n\nThe $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.\n\n$$\\left(\\begin{array}{rr} 1 & 2 \\\\ 2 & 1 \\end{array}\\right)$$\n\n## Isogeny graph\n\nsage: E.isogeny_graph().plot(edge_labels=True)\n\nThe vertices are labelled with LMFDB labels.",
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[
null,
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efviRUo9CmbHH0LP4JhDoZMOGD4/GDn6LQak1PvNcDBs2vNRjUIbsMfQsAhA6WW1tbfzqNy/F1pbWUo/ypra2tMaLLzVHbW1tqUehDNlj6FkEIHSyUaNGRWVlZSxdubrUo7ypb61cHZWVlXHOOeeUehTKkD2GnkUAQierqamJiy66KBbVP3BIv7btUNq7d28srn8gJkyYEDU1NaUehzJkj6FnEYDQBaZOnRqbX9waq9Y1lXqUP2rVuqbY/OLWmDp1aqlHoYzZY%2Bg5vAsYukBRFFE7dGj0O7IiVt06v6w%2BmqIoijj/6pnx8p6IxqamspqN8mKPoedwBRC6QEVFRcyeMyceXr8xlq0qr9dQLV35SDy8fmPMmj3bgyZvyh5Dz%2BEKIHShiRMnxoP33xdPLFsUA2v6lXqc2NLcGmdMnBIfHTMmli5dWupx6CbsMXR/AhC6UGtra5w%2BeHCMOPmEqP/qrJJeqSiKIsZeNzvW/2xzPPnUU9GvX%2BkfyOke7DF0f54Chi7Ur1%2B/WLR4cdz32NqY%2B41lJZ1lzpKlcf%2BP1sXi22/3oMlBscfQ/QlA6GLjxo2LefPmxZw7lsZNy79bkhluWv7dmLtkWcyfPz/Gjh1bkhno3uwxdG9HlHoAyGjGjBmxY8eOuK6uLl7ZuTNmXT6pS55GK4oi5ixZGnOXLIsZM2bEDTfc0OnnpOeyx9B9CUAogYqKipg3b1707t07Zs6cGRuffT4WXX9Vp76gfktza0z5yoK4/0froq6uzoMmfzJ7DN2XN4FAiTU0NMTkz342drW1xa3TJ8ek0ece0qsoRVHE0pWPxNU3L46q6upYfPvtni7jkLPH0L0IQCgDra2tMW3atFi%2BfHmcN3xIXPPJ8TH6fcPisMPe%2Bst09%2B7dG6vWNcUt374nHl6/MSZOnBgLFiyIvn37HsLJ4ffsMXQfAhDKSENDQ8yZPTs2btoUg44dGJPHXRiXjh51UE%2BpbW1pjW%2BtXB2L6x%2BIzS9ujSFnnRWz58yJMWPGdOLk8Hv2GMqfAIQyUxRFrFu3LhYuXBh33313tLe3x3FHD4jaU06MoaecFGeccHy8rVd1VB5xRLTv2RM7drbFEz//RWx49rlofOa5ePGl5qiqqooJEybE1KlTY8SIEb4ZgS5nj6G8CUAoYy0tLbFmzZpobGyMpqbGaGpqim3btr/huKOO6hO1tbUxbNjwqK2tjXPOOSdqampKMDG8kT2G8iMAoRspiiKam5ujra0tdu3aFVVVVVFdXR39%2B/d3dYRuwx5D6QlAAIBkfBMIAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJPM/dLv2FooeUHIAAAAASUVORK5CYII%3D",
null,
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efviRUo9CmbHH0LP4JhDoZMOGD4/GDn6LQak1PvNcDBs2vNRjUIbsMfQsAhA6WW1tbfzqNy/F1pbWUo/ypra2tMaLLzVHbW1tqUehDNlj6FkEIHSyUaNGRWVlZSxdubrUo7ypb61cHZWVlXHOOeeUehTKkD2GnkUAQierqamJiy66KBbVP3BIv7btUNq7d28srn8gJkyYEDU1NaUehzJkj6FnEYDQBaZOnRqbX9waq9Y1lXqUP2rVuqbY/OLWmDp1aqlHoYzZY%2Bg5vAsYukBRFFE7dGj0O7IiVt06v6w%2BmqIoijj/6pnx8p6IxqamspqN8mKPoedwBRC6QEVFRcyeMyceXr8xlq0qr9dQLV35SDy8fmPMmj3bgyZvyh5Dz%2BEKIHShiRMnxoP33xdPLFsUA2v6lXqc2NLcGmdMnBIfHTMmli5dWupx6CbsMXR/AhC6UGtra5w%2BeHCMOPmEqP/qrJJeqSiKIsZeNzvW/2xzPPnUU9GvX%2BkfyOke7DF0f54Chi7Ur1%2B/WLR4cdz32NqY%2B41lJZ1lzpKlcf%2BP1sXi22/3oMlBscfQ/QlA6GLjxo2LefPmxZw7lsZNy79bkhluWv7dmLtkWcyfPz/Gjh1bkhno3uwxdG9HlHoAyGjGjBmxY8eOuK6uLl7ZuTNmXT6pS55GK4oi5ixZGnOXLIsZM2bEDTfc0OnnpOeyx9B9CUAogYqKipg3b1707t07Zs6cGRuffT4WXX9Vp76gfktza0z5yoK4/0froq6uzoMmfzJ7DN2XN4FAiTU0NMTkz342drW1xa3TJ8ek0ece0qsoRVHE0pWPxNU3L46q6upYfPvtni7jkLPH0L0IQCgDra2tMW3atFi%2BfHmcN3xIXPPJ8TH6fcPisMPe%2Bst09%2B7dG6vWNcUt374nHl6/MSZOnBgLFiyIvn37HsLJ4ffsMXQfAhDKSENDQ8yZPTs2btoUg44dGJPHXRiXjh51UE%2BpbW1pjW%2BtXB2L6x%2BIzS9ujSFnnRWz58yJMWPGdOLk8Hv2GMqfAIQyUxRFrFu3LhYuXBh33313tLe3x3FHD4jaU06MoaecFGeccHy8rVd1VB5xRLTv2RM7drbFEz//RWx49rlofOa5ePGl5qiqqooJEybE1KlTY8SIEb4ZgS5nj6G8CUAoYy0tLbFmzZpobGyMpqbGaGpqim3btr/huKOO6hO1tbUxbNjwqK2tjXPOOSdqampKMDG8kT2G8iMAoRspiiKam5ujra0tdu3aFVVVVVFdXR39%2B/d3dYRuwx5D6QlAAIBkfBMIAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJCMAAQCSEYAAAMkIQACAZAQgAEAyAhAAIBkBCACQjAAEAEhGAAIAJPM/dLv2FooeUHIAAAAASUVORK5CYII%3D",
null
] |
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|
https://mathematica.stackexchange.com/questions/60523/error-using-nsolve-with-system-of-two-equations-use-reduce
|
[
"# Error using Nsolve with system of two equations: “Use Reduce”\n\nHaving some odd behavior with some mathematica code I am running. I am solving a simple system of two equations but mathematica gives an error and asks to use Reduce? I looked up in the manual but have no clue what to do to rememdy the situation.\n\ndelay = 4*10^-9;\ngamma = (25*10^3)/0.511;\nbeta = Sqrt[1 - gamma^-2];\nc = 3*10^8;\n\nrho = lb/theta;\ndrift = 4 lb;\n\ntime = (4 rho*theta + 2 drift - 4 lb - 2 drift*Cos[theta])/(beta*c);\ndisp = lb^2/rho + lb/rho*drift*Cos[theta];\n\nNSolve[{time== delay, disp == 0.6},{theta,lb}]\n\n• The expressions for timeDiff and dispersion are lacking – Dr. belisarius Sep 24 '14 at 2:55\n• oops youre right...fixed!! – user1886681 Sep 24 '14 at 3:22\n• NSolve does not handle situations of infinitely many solutions. Could restrict a variable to get to a finite case, e.g. NSolve[{time == delay, disp == 0.6, -30 <= theta <= 30}, {theta, lb}]. – Daniel Lichtblau Sep 24 '14 at 14:53\n\n## 2 Answers\n\ndelay = 4*10^-9;\ngamma = (25*10^3)/0.511;\nbeta = Sqrt[1 - gamma^-2];\nc = 3*10^8;\nrho = lb/theta;\ndrift = 4 lb;\n\ntime = (4 rho*theta + 2 drift - 4 lb - 2 drift*Cos[theta])/(beta*c);\n\ndisp = lb^2/rho + lb/rho*drift*Cos[theta];\n\nContourPlot[{time == delay, disp == .6},\n{theta, -12, 5}, {lb, 0, .3},\nFrameLabel -> (Style[#, 14, Bold] & /@\n{theta, lb})]",
null,
"The intersection points are the solution points and provide the initial estimates for FindRoot\n\nsol = FindRoot[{time == delay, disp == .6},\n{{theta, #[]}, {lb, #[]}}] & /@\n{{-11, .1}, {-8, .1}, {-4, .1}, {-3, .1},\n{2, .2}, {5, .15}}\n\n\n{{theta -> -10.6056, lb -> 0.108685}, {theta -> -8.28907, lb -> 0.105523}, {theta -> -3.97865, lb -> 0.089839}, {theta -> -2.80309, lb -> 0.0771901}, {theta -> 1.26384, lb -> 0.214949}, {theta -> 4.68236, lb -> 0.145628}}\n\n{time == delay, disp == .6} /. sol\n\n\n{{True, True}, {True, True}, {True, True}, {True, True}, {True,\nTrue}, {True, True}}\n\n• I tried using Ersek RootSearch, which is supposed to be able to find roots in given range, but on V 10.01 something changed. Now I get No more memory available. Mathematica kernel has shut down. Try quitting other applications and then retry. This package looks like it is no longer maintained. library.wolfram.com/infocenter/MathSource/4482 It worked ok on version 9.01 – Nasser Sep 24 '14 at 6:43\n• @Nasser I think one can build one by finding the intersections on the interpolation of grids = Cases[ Normal@Graphics@ Cases[ContourPlot[ ...], GraphicsComplex[__], Infinity], Line[__], Infinity] /. Line[a__] -> a; – Dr. belisarius Sep 24 '14 at 17:47\n\nUpdated answer: Thanks to hint by Daniel below, one can give NSolve region to search on.\n\nNSolve[{time == delay, disp == 0.6 && (-4 Pi < theta < 4 Pi) && (0 < lb < 10)},\n{theta, lb}]\n\n\ngives\n\n{{theta -> -10.6056, lb -> 0.108685}, {theta -> -8.28907, lb -> 0.105523},\n{theta -> -3.97865, lb -> 0.089839}, {theta -> -2.80309, lb -> 0.0771901},\n{theta -> 1.26384, lb -> 0.214949}, {theta -> 4.68236, lb -> 0.145628},\n{theta -> 7.96477, lb -> 0.135066}, {theta -> 10.8505, lb -> 0.131059}}\n\n\nOriginal answer\n\nI do not know why NSolve can't solve this since Maple fsolve can. But you can use FindRoot when all else fails. (the problem with FindRoot is that one need to have some idea where to look for)\n\nClear[lb, theta];\ndelay = 4*10^-9;\ngamma = (25*10^3)/0.511;\nbeta = Sqrt[1 - gamma^-2];\nc = 3*10^8;\nrho = lb/theta;\ndrift = 4 lb;\ntime = (4 rho*theta + 2 drift - 4 lb - 2 drift*Cos[theta])/(beta*c);\ndisp = lb^2/rho + lb/rho*drift*Cos[theta];\neqs = {time == delay, disp == 0.6};\nNSolve[eqs, {theta, lb}, Reals]",
null,
"roots = FindRoot[eqs, {{theta, -8}, {lb, .1}}]\neqs /. roots",
null,
"Maple solution:",
null,
"Using Mathematica 10.01 on windows.\n\n• NSolve can handle it but requires a restricted domain. – Daniel Lichtblau Sep 24 '14 at 16:07\n• @DanielLichtblau thanks! I actually did not notice that NSolve accepts ranges, now it works much better. Updated answer. – Nasser Sep 24 '14 at 16:26\n• Hey Guys I thought I would try to make things a little easier and give \"lb=3;\" a try but I get the same error even with lb defined? – user1886681 Sep 24 '14 at 18:56"
] |
[
null,
"https://i.stack.imgur.com/yrDqL.jpg",
null,
"https://i.stack.imgur.com/x94sV.png",
null,
"https://i.stack.imgur.com/I27xW.png",
null,
"https://i.stack.imgur.com/Gms8J.png",
null
] |
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|
https://math.stackexchange.com/questions/3253007/every-relation-in-a-special-structure-m-which-is-first-order-definable-with-pa
|
[
"every relation in a special structure $M$ which is first-order definable with parameters has cardinality $< \\omega$ or $= |M|$.\n\nLet $$M$$ be a special structure. I'm trying to proof that every relation in $$M$$ which is first-order definable with parameters has cardinality $$< \\omega$$ or $$= \\lambda$$.\n\nBy a first-order definable with parameters relation I mean a set $$A = \\left \\{ \\psi(a_1, \\cdots, a_m, b_1, \\cdots, b_n): a_1, \\cdots, a_m \\in M,\\, b_1, \\cdots, b_n \\in X, \\, \\psi(\\overline a, \\overline y) \\text{ formula} \\right \\}$$ for fixed $$m$$ and $$X \\subset M$$.\n\nI don't know even how to start this question. If I suppose that a set $$A$$ as above has cardinality $$\\geq \\omega$$, we proof that $$|A| = |M|$$, but how can I proceed?\n\n• What is a special structure? – Asaf Karagila Jun 6 at 12:57\n• @asaf The definition I know is that a structure $A$ of infinite cardinality $\\kappa$ is special if it is the union (colimit) of an elementary chain $(A_i)_{i < \\kappa}$ where each $A_i$ is $|i|^+$-saturated. – Mark Kamsma Jun 6 at 13:08\n• @Mark: I thought you're on a plane. – Asaf Karagila Jun 6 at 13:10\n• @asaf there is lots of downtime around a flight. For example, now I actually found the time to write an answer. – Mark Kamsma Jun 6 at 14:44\n• @Mark: I've never been more ashamed in my life as I am now, of the fact that we are in the same logic group. – Asaf Karagila Jun 6 at 14:47\n\nIn this answer I will not distinguish between elements/variables and finite tuples of elements/variables (so you may read every lowercase letter as if it denotes a finite tuple).\n\nFirst I should say that your definition of \"definable\" it's not quite right. The way you defined it, every set would be definable. The correct definition is that $$A$$ is definable (in $$M$$) if there is a formula $$\\phi(x,y)$$ and $$b \\in M$$ such that $$A = \\{a \\in M : M \\models \\phi(a,b)\\}.$$ We call $$b$$ the parameters.\n\nLet me also recall the definition of a special structure here (to introduce some notation at the same time). A structure $$M$$ of infinite cardinality $$\\kappa$$ is special if it is the union $$\\bigcup_{i < \\kappa} M_i$$ of a chain $$(M_i)_{i < \\kappa}$$ such that for every $$i < \\kappa$$ the structure $$M_i$$ is $$|i|^+$$-saturated.\n\nNow that is out of the way, we can actually have a look at your question. Let $$A \\subseteq M$$ be infinite and $$|A| < |M|$$. Suppose that $$A$$ is definable by some $$\\phi(x, y)$$ with parameters $$b$$. We will aim for a contradiction. Since $$b$$ is just a finite tuple of parameters, there will be some $$\\lambda < \\kappa$$ such that $$b \\in M_\\lambda$$. We may assume that $$|A| \\leq \\lambda$$ (otherwise replace $$\\lambda$$ by $$|A|$$). Now define $$\\Sigma(x) = \\{x \\neq a : a \\in A \\cap M_\\lambda\\} \\cup \\{\\phi(x, b)\\},$$ then clearly $$\\Sigma(x)$$ is finitely satisfiable. So we can extend $$\\Sigma(x)$$ to a complete type $$p(x)$$ over $$(A \\cap M_\\lambda) \\cup \\{b\\}$$. Since $$|(A \\cap M_\\lambda) \\cup \\{b\\}| < \\lambda^+$$ and $$M_\\lambda$$ is $$\\lambda^+$$-saturated we then must be able to find a realisation $$c \\in M_\\lambda$$ of $$p(x)$$ and thus of $$\\Sigma(x)$$. But that means $$M_\\lambda \\models \\phi(c, b)$$, so $$c \\in A \\cap M_\\lambda$$ while the first part of $$\\Sigma(x)$$ says that $$c$$ must be different from every element in $$A \\cap M_\\lambda$$. We thus reach our desired contradiction, and we can conclude that of $$A$$ is infinite and $$|A| < |M|$$, then $$A$$ cannot be definable in $$M$$.\n\n• Not every special structure is saturated, though... – Alex Kruckman Jun 6 at 15:33\n• Right - the point is that special models (of singular cardinality) provably exist in ZFC for all theories, while saturated models do not. – Alex Kruckman Jun 6 at 15:41\n• @user242964 yes, I defined $\\kappa$ as $|M|$ in the paragraph before. For your other questions. (1) We use here that $A$ is infinite, so every finite part of $\\Sigma(x)$ will say at most \"$x$ is in $A$ but unequal to a finite part of it\". So then we can definitely find some realisation of that. (2) If something is finitely satisfiable it must have a realisation in some elementary extension (indeed by compactness), then taking the type of that realisation is a completion. – Mark Kamsma Jun 7 at 7:07\n• @user242964 For (1) it may also be good to note why $A \\cap M_\\lambda$ is indeed infinite as well. This is because it is defined by $\\phi(x, b)$ and the sentence \"there are at least $n$ different $x$ such that $\\phi(x, b)$\" is true in $M$, and thus in $M_\\lambda$, for all $n \\in \\mathbb{N}$. – Mark Kamsma Jun 7 at 7:10\n• @Kat That is not needed, because if something is $\\lambda$-saturated then it realises any $n$-type over less than $\\lambda$ parameters. If your definition only mentions 1-types, you can still prove this statement from that definition (so it is equivalent). That may be a nice exercise (hint: induction). – Mark Kamsma Jun 9 at 22:08"
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.86534214,"math_prob":0.99990594,"size":1984,"snap":"2019-43-2019-47","text_gpt3_token_len":603,"char_repetition_ratio":0.13131313,"word_repetition_ratio":0.0,"special_character_ratio":0.3200605,"punctuation_ratio":0.08080808,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999785,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T18:33:58Z\",\"WARC-Record-ID\":\"<urn:uuid:fc173b70-afe5-40d0-a052-43fd36c65bd4>\",\"Content-Length\":\"154240\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63c344a9-5023-475e-a3f2-ca298f0513a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6d88117-0922-4204-a1b2-f715f3f57229>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3253007/every-relation-in-a-special-structure-m-which-is-first-order-definable-with-pa\",\"WARC-Payload-Digest\":\"sha1:AD5K5LV53AG5L5CK7PRSCBTQ2CVLIHUK\",\"WARC-Block-Digest\":\"sha1:Q6KTORBZFUYJ74ZBNC2IXZKFVM37OLZI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987835748.66_warc_CC-MAIN-20191023173708-20191023201208-00529.warc.gz\"}"}
|
http://www.brockswood.herts.sch.uk/2019/01/09/rowan-class-maths-learning-a-great-fact-about-angles-in-a-triangle/
|
[
"# Rowan class maths learning – a great fact about angles in a triangle\n\nDid you know why angles in a triangle total 180°? Today we learnt the reason: If you put all the angles of a triangle together with their vertices at the centre point, they make a straight line. We know that all angles on a straight line equal 180° so the angles in a triangle equal 180°.",
null,
""
] |
[
null,
"https://brockswood872.edublogs.org/files/2019/01/L2-Why-angles-in-a-traingle-180-2nd1s38-1fh3o9g.jpg",
null
] |
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|
https://isabelle.in.tum.de/repos/isabelle/file/07eb0ad9bb93/src/HOL/Tools/Sledgehammer/sledgehammer_run.ML
|
[
"src/HOL/Tools/Sledgehammer/sledgehammer_run.ML\n author blanchet Wed, 08 Jun 2011 16:20:18 +0200 changeset 43290 07eb0ad9bb93 parent 43261 a4aeb26a6362 child 43292 ff3d49e77359 permissions -rw-r--r--\nrenamed option to avoid talking about seconds, since this is now the default Isabelle unit\n```\n(* Title: HOL/Tools/Sledgehammer/sledgehammer_run.ML\nAuthor: Fabian Immler, TU Muenchen\nAuthor: Makarius\nAuthor: Jasmin Blanchette, TU Muenchen\n\nSledgehammer's heart.\n*)\n\nsignature SLEDGEHAMMER_RUN =\nsig\ntype minimize_command = ATP_Reconstruct.minimize_command\ntype relevance_override = Sledgehammer_Filter.relevance_override\ntype mode = Sledgehammer_Provers.mode\ntype params = Sledgehammer_Provers.params\ntype prover = Sledgehammer_Provers.prover\n\nval someN : string\nval noneN : string\nval timeoutN : string\nval unknownN : string\nval auto_minimize_min_facts : int Config.T\nval auto_minimize_max_time : real Config.T\nval get_minimizing_prover : Proof.context -> mode -> string -> prover\nval run_sledgehammer :\nparams -> mode -> int -> relevance_override -> (string -> minimize_command)\n-> Proof.state -> bool * (string * Proof.state)\nend;\n\nstructure Sledgehammer_Run : SLEDGEHAMMER_RUN =\nstruct\n\nopen ATP_Util\nopen ATP_Translate\nopen ATP_Reconstruct\nopen Sledgehammer_Util\nopen Sledgehammer_Filter\nopen Sledgehammer_Provers\nopen Sledgehammer_Minimize\n\nval someN = \"some\"\nval noneN = \"none\"\nval timeoutN = \"timeout\"\nval unknownN = \"unknown\"\n\nval ordered_outcome_codes = [someN, unknownN, timeoutN, noneN]\n\nfun max_outcome_code codes =\nNONE\n|> fold (fn candidate =>\nfn accum as SOME _ => accum\n| NONE => if member (op =) codes candidate then SOME candidate\nelse NONE)\nordered_outcome_codes\n|> the_default unknownN\n\nfun prover_description ctxt ({verbose, blocking, ...} : params) name num_facts i\nn goal =\n(name,\n(if verbose then\n\" with \" ^ string_of_int num_facts ^ \" fact\" ^ plural_s num_facts\nelse\n\"\") ^\n\" on \" ^ (if n = 1 then \"goal\" else \"subgoal \" ^ string_of_int i) ^\n(if blocking then\n\".\"\nelse\n\"\\n\" ^ Syntax.string_of_term ctxt (Thm.term_of (Thm.cprem_of goal i))))\n\nval auto_minimize_min_facts =\nAttrib.setup_config_int @{binding sledgehammer_auto_minimize_min_facts}\n(fn generic => Config.get_generic generic binary_min_facts)\nval auto_minimize_max_time =\nAttrib.setup_config_real @{binding sledgehammer_auto_minimize_max_time}\n(K 5.0)\n\nfun minimize ctxt mode name (params as {debug, verbose, isar_proof, ...})\n({state, subgoal, subgoal_count, facts, ...} : prover_problem)\n(result as {outcome, used_facts, run_time_in_msecs, preplay,\nmessage, message_tail} : prover_result) =\nif is_some outcome then\nresult\nelse\nlet\nval num_facts = length used_facts\nval ((minimize, minimize_name), preplay) =\nif mode = Normal then\nif num_facts >= Config.get ctxt auto_minimize_min_facts then\n((true, name), preplay)\nelse\nlet\nfun can_min_fast_enough msecs =\n0.001 * Real.fromInt ((num_facts + 2) * msecs)\n<= Config.get ctxt auto_minimize_max_time\nval prover_fast_enough =\nrun_time_in_msecs |> Option.map can_min_fast_enough\n|> the_default false\nin\nif isar_proof then\n((prover_fast_enough, name), preplay)\nelse\nlet val preplay = preplay () in\n(case preplay of\nPlayed (reconstructor, timeout) =>\nif can_min_fast_enough (Time.toMilliseconds timeout) then\n(true, prover_name_for_reconstructor reconstructor\n|> the_default name)\nelse\n(prover_fast_enough, name)\n| _ => (prover_fast_enough, name),\nK preplay)\nend\nend\nelse\n((false, name), preplay)\nval (used_facts, (preplay, message, _)) =\nif minimize then\nminimize_facts minimize_name params (not verbose) subgoal\nsubgoal_count state\n(filter_used_facts used_facts\n(map (apsnd single o untranslated_fact) facts))\n|>> Option.map (map fst)\nelse\n(SOME used_facts, (preplay, message, \"\"))\nin\ncase used_facts of\nSOME used_facts =>\n(if debug andalso not (null used_facts) then\nfacts ~~ (0 upto length facts - 1)\n|> map (fn (fact, j) => fact |> untranslated_fact |> apsnd (K j))\n|> filter_used_facts used_facts\n|> map (fn ((name, _), j) => name ^ \"@\" ^ string_of_int j)\n|> commas\n|> enclose (\"Fact\" ^ plural_s (length facts) ^ \" in \" ^ quote name ^\n\" proof (of \" ^ string_of_int (length facts) ^ \"): \") \".\"\n|> Output.urgent_message\nelse\n();\n{outcome = NONE, used_facts = used_facts,\nrun_time_in_msecs = run_time_in_msecs, preplay = preplay,\nmessage = message, message_tail = message_tail})\n| NONE => result\nend\n\nfun get_minimizing_prover ctxt mode name params minimize_command problem =\nget_prover ctxt mode name params minimize_command problem\n|> minimize ctxt mode name params problem\n\nfun launch_prover (params as {debug, verbose, blocking, max_relevant, slicing,\ntimeout, expect, ...})\nmode minimize_command only\n{state, goal, subgoal, subgoal_count, facts, smt_filter} name =\nlet\nval ctxt = Proof.context_of state\nval hard_timeout = Time.+ (timeout, timeout)\nval birth_time = Time.now ()\nval death_time = Time.+ (birth_time, hard_timeout)\nval max_relevant =\nmax_relevant\n|> the_default (default_max_relevant_for_prover ctxt slicing name)\nval num_facts = length facts |> not only ? Integer.min max_relevant\nfun desc () =\nprover_description ctxt params name num_facts subgoal subgoal_count goal\nval problem =\n{state = state, goal = goal, subgoal = subgoal,\nsubgoal_count = subgoal_count, facts = facts |> take num_facts,\nsmt_filter = smt_filter}\nfun really_go () =\nproblem\n|> get_minimizing_prover ctxt mode name params minimize_command\n|> (fn {outcome, preplay, message, message_tail, ...} =>\n(if outcome = SOME ATP_Proof.TimedOut then timeoutN\nelse if is_some outcome then noneN\nelse someN, fn () => message (preplay ()) ^ message_tail))\nfun go () =\nlet\nval (outcome_code, message) =\nif debug then\nreally_go ()\nelse\n(really_go ()\nhandle ERROR msg => (unknownN, fn () => \"Error: \" ^ msg ^ \"\\n\")\n| exn =>\nif Exn.is_interrupt exn then\nreraise exn\nelse\n(unknownN, fn () => \"Internal error:\\n\" ^\nML_Compiler.exn_message exn ^ \"\\n\"))\nval _ =\n(* The \"expect\" argument is deliberately ignored if the prover is\nmissing so that the \"Metis_Examples\" can be processed on any\nmachine. *)\nif expect = \"\" orelse outcome_code = expect orelse\nnot (is_prover_installed ctxt name) then\n()\nelse if blocking then\nerror (\"Unexpected outcome: \" ^ quote outcome_code ^ \".\")\nelse\nwarning (\"Unexpected outcome: \" ^ quote outcome_code ^ \".\");\nin (outcome_code, message) end\nin\nif mode = Auto_Try then\nlet val (outcome_code, message) = TimeLimit.timeLimit timeout go () in\n(outcome_code,\nstate\n|> outcome_code = someN\n? Proof.goal_message (fn () =>\n[Pretty.str \"\",\nPretty.mark Markup.hilite (Pretty.str (message ()))]\n|> Pretty.chunks))\nend\nelse if blocking then\nlet\nval (outcome_code, message) = TimeLimit.timeLimit hard_timeout go ()\nin\n(if outcome_code = someN orelse mode = Normal then\nquote name ^ \": \" ^ message ()\nelse\n\"\")\n|> Async_Manager.break_into_chunks\n|> List.app Output.urgent_message;\n(outcome_code, state)\nend\nelse\n(Async_Manager.launch das_tool birth_time death_time (desc ())\n((fn (outcome_code, message) =>\n(verbose orelse outcome_code = someN,\nmessage ())) o go);\n(unknownN, state))\nend\n\nfun class_of_smt_solver ctxt name =\nctxt |> select_smt_solver name\n|> SMT_Config.solver_class_of |> SMT_Utils.string_of_class\n\n(* Makes backtraces more transparent and might well be more efficient as\nwell. *)\nfun smart_par_list_map _ [] = []\n| smart_par_list_map f [x] = [f x]\n| smart_par_list_map f xs = Par_List.map f xs\n\nfun dest_SMT_Weighted_Fact (SMT_Weighted_Fact p) = p\n| dest_SMT_Weighted_Fact _ = raise Fail \"dest_SMT_Weighted_Fact\"\n\nval auto_try_max_relevant_divisor = 2 (* FUDGE *)\n\nfun run_sledgehammer (params as {debug, verbose, blocking, provers,\nrelevance_thresholds, max_relevant, slicing,\ntimeout, ...})\nmode i (relevance_override as {only, ...}) minimize_command state =\nif null provers then\nerror \"No prover is set.\"\nelse case subgoal_count state of\n0 => (Output.urgent_message \"No subgoal!\"; (false, (noneN, state)))\n| n =>\nlet\nval _ = Proof.assert_backward state\nval print = if mode = Normal then Output.urgent_message else K ()\nval state =\nstate |> Proof.map_context (Config.put SMT_Config.verbose debug)\nval ctxt = Proof.context_of state\nval {facts = chained_ths, goal, ...} = Proof.goal state\nval chained_ths = chained_ths |> normalize_chained_theorems\nval (_, hyp_ts, concl_t) = strip_subgoal ctxt goal i\nval _ = () |> not blocking ? kill_provers\nval _ = case find_first (not o is_prover_supported ctxt) provers of\nSOME name => error (\"No such prover: \" ^ name ^ \".\")\n| NONE => ()\nval _ = print \"Sledgehammering...\"\nval (smts, (ueq_atps, full_atps)) =\nprovers |> List.partition (is_smt_prover ctxt)\n||> List.partition (is_unit_equational_atp ctxt)\nfun launch_provers state get_facts translate maybe_smt_filter provers =\nlet\nval facts = get_facts ()\nval num_facts = length facts\nval facts = facts ~~ (0 upto num_facts - 1)\n|> map (translate num_facts)\nval problem =\n{state = state, goal = goal, subgoal = i, subgoal_count = n,\nfacts = facts,\nsmt_filter = maybe_smt_filter\n(fn () => map_filter (try dest_SMT_Weighted_Fact) facts) i}\nval launch = launch_prover params mode minimize_command only\nin\nif mode = Auto_Try orelse mode = Try then\n(unknownN, state)\n|> fold (fn prover => fn accum as (outcome_code, _) =>\nif outcome_code = someN then accum\nelse launch problem prover)\nprovers\nelse\nprovers\n|> (if blocking then smart_par_list_map else map)\n(launch problem #> fst)\n|> max_outcome_code |> rpair state\nend\nfun get_facts label is_appropriate_prop relevance_fudge provers =\nlet\nval max_max_relevant =\ncase max_relevant of\nSOME n => n\n| NONE =>\n0 |> fold (Integer.max\no default_max_relevant_for_prover ctxt slicing)\nprovers\n|> mode = Auto_Try\n? (fn n => n div auto_try_max_relevant_divisor)\nval is_built_in_const =\nis_built_in_const_for_prover ctxt (hd provers)\nin\nrelevant_facts ctxt relevance_thresholds max_max_relevant\nis_appropriate_prop is_built_in_const relevance_fudge\nrelevance_override chained_ths hyp_ts concl_t\n|> tap (fn facts =>\nif debug then\nlabel ^ plural_s (length provers) ^ \": \" ^\n(if null facts then\n\"Found no relevant facts.\"\nelse\n\"Including (up to) \" ^ string_of_int (length facts) ^\n\" relevant fact\" ^ plural_s (length facts) ^ \":\\n\" ^\n(facts |> map (fst o fst) |> space_implode \" \") ^ \".\")\n|> print\nelse\n())\nend\nfun launch_atps label is_appropriate_prop atps accum =\nif null atps then\naccum\nelse if not (is_appropriate_prop concl_t) then\n(if verbose orelse length atps = length provers then\n\"Goal outside the scope of \" ^\nspace_implode \" \" (serial_commas \"and\" (map quote atps)) ^ \".\"\n|> Output.urgent_message\nelse\n();\naccum)\nelse\nlaunch_provers state\n(get_facts label is_appropriate_prop atp_relevance_fudge o K atps)\n(K (Untranslated_Fact o fst)) (K (K NONE)) atps\nfun launch_smts accum =\nif null smts then\naccum\nelse\nlet\nval facts = get_facts \"SMT solver\" (K true) smt_relevance_fudge smts\nval weight = SMT_Weighted_Fact oo weight_smt_fact ctxt\nfun smt_filter facts =\n(if debug then curry (op o) SOME\nelse TimeLimit.timeLimit timeout o try)\n(SMT_Solver.smt_filter_preprocess state (facts ()))\nin\nsmts |> map (`(class_of_smt_solver ctxt))\n|> AList.group (op =)\n|> map (snd #> launch_provers state (K facts) weight smt_filter\n#> fst)\n|> max_outcome_code |> rpair state\nend\nval launch_full_atps = launch_atps \"ATP\" (K true) full_atps\nval launch_ueq_atps =\nlaunch_atps \"Unit equational provers\" is_unit_equality ueq_atps\nfun launch_atps_and_smt_solvers () =\n[launch_full_atps, launch_smts, launch_ueq_atps]\n|> smart_par_list_map (fn f => ignore (f (unknownN, state)))\nhandle ERROR msg => (print (\"Error: \" ^ msg); error msg)\nfun maybe f (accum as (outcome_code, _)) =\naccum |> (mode = Normal orelse outcome_code <> someN) ? f\nin\n(unknownN, state)\n|> (if blocking then\nlaunch_full_atps\n#> mode <> Auto_Try ? (maybe launch_ueq_atps #> maybe launch_smts)\nelse\n(fn p => Future.fork (tap launch_atps_and_smt_solvers) |> K p))\nhandle TimeLimit.TimeOut =>\n(print \"Sledgehammer ran out of time.\"; (unknownN, state))\nend\n|> `(fn (outcome_code, _) => outcome_code = someN)\n\nend;\n```"
] |
[
null
] |
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|
https://www.northernarchitecture.us/resisting-system/seismic-design-example-static-procedure-ibc-2003-asce-702-nfpa-5000.html
|
[
"## Seismic Design Example Static Procedure IBC 2003 ASCE 702 NFPA 5000\n\nGiven. A 12-story building located in downtown, Los Angeles, California. The building properties summarized in Fig. 2.46 are the same as those used in the 1997 UBC example, Section 2.13.19.1.\n\nOccupancy group = II (Table 2.13 ASCE Table 1.1)\n\nSDC = D (Tables 2.17 and 2.18; ASCE Tables 9.4.2.1a and b)\n\nSite class as determined by project geotechnical engineer = D",
null,
"Figure 2.46. Design example; ASCE 7-02 (IBC-03), static force procedure.\n\nBuilding's lateral load system = SMRF with EBF\n\n(Table 2.21 ASCE Table 9.5.2.2) Elastic fundamental period, TB from computer analysis = 2 secs Total seismic weight, W = 22,680 kips Building height hn, above shear base = 160 ft Mapped MCE, 5% damped, Spectral acceleration at short periods, SS = 1.5 Mapped MCE, 5% damped,\n\nSpectral response acceleration at a period of 1 second, S1 = 0.6.\n\nRequired. Using the equivalent lateral force procedure of ASCE 7-02 determines the following:\n\nSeismic base shear, V Vertical distribution of base shear V Seismic story shear Overturning moment Diaphragm design forces Allowable story drifts\n\nSolution. Seismic design coefficients\n\nAcceleration-based site coefficient (at 0.2 sec period), Fa = 1\n\nVelocity-based site coefficient (at 1.0 sec period), Fv = 1.5\n\nMCE spectral response acceleration at short periods for site class effects, SMS = Fa Ss = 1 x 1.5 = 1.5 ASCE Eq. (9.4.1.2.4.1)\n\nMCE spectral response acceleration at a 1-sec period adjusted for site class effects, SM1 = Fv x S1 = 1.5 x 0.6 = 0.9 ASCE Eq. (9.4.1.2.4)\n\nDesign, 5% damped, spectral response acceleration at short periods, SDS = 2/3Sms = 2/3 x 1.5 = 1.0 ASCE Eq. (9.4.1.2.5.2)\n\nDesign, 5% damped, spectral response acceleration at a 1-sec period, SD1 = 2/3Sms = 2/3 x 0.9 = 0.6 ASCE Eq. (9.4.1.2.5.2)\n\nOccupancy importance factor, I = 1 (Table 2.13; ASCE Table 9.1.4)\n\nResponse modification coefficient, R = 8.0 (Table 2.21; ASCE Table 9.5.2.2)\n\nSystem overstrength factor, Qo = 2.5 (Table 2.21; ASCE Table 9.5.2.2)\n\nCoefficient for upper limit on calculated period, Cu = 1.4\n\nTABLE 2.23 Coefficient for Upper Limit on Calculated Period\n\nDesign spectral response acceleration at 1 Second, SD1 Coefficient Cu\n\nTABLE 2.23 Coefficient for Upper Limit on Calculated Period\n\n > 0.4",
null,
""
] |
[
null,
"https://www.northernarchitecture.us/resisting-system/images/3164_141_104-static-seismic-force.jpg",
null,
"https://www.northernarchitecture.us/images/downloads/eJw9yksKgCAQANDbuFT7f0A6Skw65FA6UobXjzZt3ur5nNOsVKGDCm5aDzIws7QclOMSTwa3Boiw46VujE4mnxawmTiaRDY_F4o_kjNdX_Uj2GabWqyFN3WnRfl8AfKhJF4.jpg",
null
] |
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|
https://www.csse.canterbury.ac.nz/greg.ewing/essays/CRC-Reverse-Engineering.html
|
[
"# Reverse-Engineering a CRC Algorithm\n\nBy Gregory Ewing\[email protected]\nhttp://www.cosc.canterbury.ac.nz/greg.ewing\nMarch 2010\n\nI had occasion to want to write a program to make some changes to some files that were protected by a CRC. The program that created the files was closed-source and too big for disassembly to be practical, so I wondered whether it would be possible to work out the polynomial and other parameters being used to generate the CRC from examining examples of the files.\n\nAfter a rather long and interesting journey, I did find a way. I'm writing this essay to document the techniques that I developed, in case anyone else finds them useful.\n\n## CRC Parameters\n\nFirst, a bit of background. I will be discussing CRC algorithms in terms of what is known as the \"Rocksoft model\", described in \"A Painless Guide to CRC Error Detection Algorithms\" by Ross N. Williams. If you're not familiar with the way CRC generation algorithms work, it would be a good idea to go and read that first before coming back here.\n\nThe relevant parameters can be summarised as follows. Here I'm using the names that the Python program pycrc uses for them.\n• Width - the number of bits in the final CRC result.\n• Poly - the polynomial being used to generate the CRC, expressed as a bit string. (I will omit the leading 1 bit that is always understood to be present.)\n• ReflectIn (boolean) - Whether to reverse the input bytes before applying the algorithm to them. (This is equivalent to processing the input bytes LSB-first instead of MSB-first.)\n• XorIn - the initial value of the CRC register.\n• ReflectOut (boolean) - whether to reverse the bits of the CRC before presenting it as the final result. (This is equivalent to mirroring the algorithm so that the register is shifted right instead of left.)\n• XorOut - a value to be exclusive-ored with the final CRC value.\nThere are a number of equivalent ways in which a CRC algorithm can be implemented. I will be discussing things in terms of a so-called \"direct\" bit-by-bit implementation, in which incoming bits are xored with the leading bit shifted out of the CRC register, rather than being shifted through the register.\n\n## The Data Files\n\nThe files concerned were user-defined data tables for a proprietary application having a database component. Examination of the files revealed that they consisted of three parts: a 48-byte header, a description of the schema, and the data records. The schema section appeared to be padded with zeroes to bring the header plus schema up to a multiple of 512 bytes. The test files I was working with were newly-created tables containing no records, so the end of the schema was also the end of the file.\n\nThere were two bytes in the header that changed in an apparently random fashion, which I surmised was a 16-bit CRC. I was assuming it was a CRC rather than some other kind of checksum, because the application would report a \"CRC mismatch\" when attempting to open a modified file.\n\nSince this was on Windows, it also seemed a fair bet that the CRC would be stored little-endian, and this later proved to be correct.\n\nAnother thing I needed to know was how much of the file the CRC covered. Most of the header appeared to be unused, and I found that I could change the last byte of the header without the application complaining. However, if I changed the last byte in the schema, it complained of a CRC mismatch. I concluded that the CRC covered everything in the schema area, including the pad bytes. (This turned out to be almost correct, but not quite.)\n\n## Initial Attempts\n\nFrom some googling and newsgroup enquiries, it appeared at first that there would be no easy way of finding out the CRC parameters, and I would have to rely on trial and error.\n\nI found a program, pycrc, that among other things allows you to calculate a CRC for supplied data given the above parameters. The first thing I did was to try all the standard 16-bit CRC models that pycrc knows about, none of which worked.\n\nNext I considered trying a brute-force search of all possible 16-bit polynomials. This seemed feasible, although not quite as straightforward as it might appear at first, because of the XorIn and XorOut parameters. Without knowing them, I wouldn't be able to tell when I'd found the right polynomial, and trying all possible combinations of Poly+XorIn+XorOut would obviously not be practical.\n\nFortunately, there are some properties of CRC algorithms that provide a way of getting around this difficulty.\n\n## Some CRC Theory\n\nBecause the whole CRC algorithm is based on exclusive-or operations, CRCs obey a kind of superposition principle. You can think of the CRC as being made up of the exclusive-or of a set of component CRCs, each of which depends on just one bit in the message.\n\nWe can also treat the initial value of the CRC register (the XorIn parameter) as generating a component of its own that gets xored into the final CRC. So we can express the CRC of a message in terms of three components:\n\nC = TnI + D + F\n\nwhere + represents exclusive-or, and\n\nI = the initial register contents (XorIn)\nF = the final xor value (XorOut)\nn = the length of the message\nD = the homogeneous CRC of the data in the message\nTi x = the result of applying i shift-xor steps to register contents x with zero data\n\nHere I'm borrowing a term from linear algebra and using \"homogeneous CRC\" to mean a CRC calculated using XorIn = XorOut = 0.\n\nNow suppose we take two messages of the same length and exclusive-or them together. The exclusive-or of their CRCs is given by\n\n C1 + C2 = (TnI + D1 + F) + (TnI + D2 + F) = D1 + D2\n\nThe I and F terms have cancelled out, leaving us with just the homogeneous CRC of the exclusive-ored data. So we can ignore the XorIn and XorOut parameters initially, as long as we confine ourselves to studying this kind of message, which I will call a difference message.\n\n## Applying some Brute Force\n\nI constructed a difference message, and tried all possible 16-bit polys on it (there are actually only 2**15 of them, because it doesn't make sense to use a poly whose LSB isn't 1), and all four combinations of the ReflectIn and ReflectOut parameters. I got some \"hits\" -- polys that produced the expected CRC.\n\nEncouraged by this, I constructed another message with a different length and did the same thing again. I got some more hits. But unfortunately, none of them matched any of the ones I got the first time.\n\nAt this point, I was stuck. I didn't have any more parameters to vary, so I was beginning to doubt that I was dealing with a standard CRC algorithm at all. If I wasn't, it seemed unlikely that I would ever be able to figure it out.\n\nThen I got into a conversation with Patrick Maupin, who suggested a test that might help to clarify whether it was a true CRC or not. Due to the superposition principle, if changing a message by xoring it with a bit pattern B1 causes its CRC to change by C1, and another bit pattern B2 causes the CRC to change by C2, then xoring the message with (B1 xor B2) should change the CRC by (C1 xor C2). If that doesn't happen, the algorithm can't be an ordinary CRC algorithm.\n\nI constructed some suitable test messages, and found that this property did seem to hold. So there was hope that it could be a standard CRC algorithm, or something very similar to it.\n\nI also began to wonder what I could learn by studying what happens to the CRC when individual bits of the file are changed. At first I thought that the CRC corresponding to a single 1 bit might be a rotation of the polynomial, but it turns out to be more complicated than that. However, after some more thinking along those lines, I came up with a possible plan of attack.\n\n## Deducing the Polynomial\n\nConsider what the CRC algorithm does when applied to a message containing a single 1 bit with the rest zeroes. We are considering difference messages here, so the register starts off with a value of zero. As long as the input bits are zero, the register remains zero.\n\nWhen the 1 bit arrives, the polynomial gets loaded into the register. After that, it gets transformed by the shift-xor operation once for each remaining 0 bit in the message, with the leading bit of the register determining whether to xor the polynomial in or not.\n\nSo the final CRC value will be equal to the polynomial transformed by n shift-xor cycles, where n depends on the position of the 1 bit in the message.\n\nNow consider two CRC values obtained from two 1-bit messages, where the 1 bits are in adjacent positions. The resulting CRCs will differ by just one shift-xor cycle. To be precise, if C1 corresponds to the message with a 1 in position i, and C2 corresponds to the message with a 1 in position i+1, then C1 is derived from applying one shift-xor cycle to C2. (If this seems backwards, it's because the further the 1 bit is from the end of the message, the more shift-xor cycles get applied to the CRC.)\n\nThere are two possibilities. If the leading bit of C2 (the one about to be shifted out) is 0, then C1 will be equal to C2 shifted by one place. If it is 1, then C2 will be equal to C1 shifted one place and xored with the polynomial.\n\nSo all we need to do is find a C1 and C2 such that the leading bit of C2 is 1, shift C2 by one place, and xor it with C1. The result will be equal to the polynomial!\n\n## Putting Theory into Practice\n\nAll this seemed almost too magical to be true, so I had to try it out. Although I didn't have complete control over the contents of the files, I was able to construct a set of 1-bit difference messages for several adjacent bits, spanning two consecutive bytes somewhere in the middle of the data.\n\nThe byte values I came up with, and their corresponding CRC values (after byte swapping to correct for little-endianness) were as follows:\n\nBytes CRC\n----- ----\n02 00 763c\n04 00 ec78\n08 00 98f3\n10 00 71e5\n20 00 e3ca\n40 00 8797\n80 00 4f2d\n00 01 9e5a\n00 02 7cb7\n00 04 f96e\n00 08 b2df\n00 10 25bd\n\nWorking upwards through the list of CRCs, it is apparent that whenever the LSB of a CRC is 0, the preceding one is obtained by simply right-shifting it, which is consistent with the theory outlined above.\n\nNow let's see if we can extract the polynomial. Applying the shift-xor operation to adjacent CRC values gives the following results, shown in blue:\n\n02 00 763c\n0000\n04 00 ec78\na001\n08 00 98f3\na001\n10 00 71e5\n0000\n20 00 e3ca\na001\n40 00 8797\na001\n80 00 4f2d\n0000\n00 01 9e5a\na001\n00 02 7cb7\n0000\n00 04 f96e\na001\n00 08 b2df\na001\n00 10 25bd\n\nFrom this it is clear that the polynomial is 0xa001.\n\nA few other things are also apparent. The shifting direction indicates that the ReflectOut parameter should be True, since shifting to the right is equivalent to using the canonical left-shifting version of the algorithm with the polynomial 0x8005 and then reflecting the resulting CRC. It is notable that 0x8005 is one of the standard 16-bit polynomials -- the one that pycrc calls \"crc-16\".\n\nThe fact that the 1 bits proceed from right to left within each byte as we go down the sequence indicates that bits are being processed LSB-first. In terms of the model parameters, this means ReflectIn = True.\n\nI had now established all except two of the CRC algorithm parameters. I was making considerable progress!\n\n## Clearing Up some Puzzles\n\nAt this point there were some puzzling things that I needed to sort out. Having just discovered that one of the standard polynomials was apparently being used after all, I couldn't understand why I hadn't found it during my initial experiments with pycrc.\n\nThe first thing I did was go back and try pycrc again, in case I had made a mistake of some kind the first time around. Armed with knowledge of the polynomial, I should have been able to run pycrc over one of my difference messages and obtain the correct CRC for it. But, it still didn't work.\n\nI was convinced by now that I was dealing with a very standard CRC algorithm, so this failure was perplexing. Thinking about what could cause it, it occurred to me that my assumption about what region of the file was covered by the CRC might be wrong. Perhaps not all of the pad bytes were included, in which case I was going too far and ending up with the wrong CRC.\n\nTrying to think how I could tell how far I should be going, I came up with the following idea. Start by initialising the register with the polynomial -- this corresponds to the state just after encountering the 1 in a 1-bit difference message. Then run the algorithm and count the number of steps required before the known CRC value is reached. Assuming it was eventually reached, that would tell me how many 0 bits following the 1 were included in the CRC.\n\nThe test file I used had a 1 in the byte at offset 0xab. Running the algorithm told me that a further 0x358 zero bytes were needed to reach the correct CRC, meaning that the last byte was at offset 0x403... which was 4 bytes past the end of the file!\n\nSo an extra 4 bytes from somewhere were being included in the CRC. To check this, I added another 4 zero bytes to the end of my difference file and ran pycrc over it. This time, I got the correct checksum.\n\n## In Search of the Extra Bytes\n\nWhere were the extra bytes coming from? One possibility was that they were in the header somewhere. Another was that they were just padding.\n\nPreviously I had assumed that the range of checked bytes was contiguous, but now I retracted that assumption and went back to investigate the header area more thoroughly. Poking around in it, I found that apart from the CRC itself, there were three other bytes that would provoke a CRC mismatch if I changed them.\n\nOne of them had a constant value of 0x04 in all the files I looked at. The other two bytes appeared to contain the length of the schema portion of the file, excluding the header.\n\nMy first thought was that perhaps these were three of the bytes making up the extra data, or that the extra data was derived from them in some way. However, I realised that there was another likely explanation for the CRC depending on the length bytes. If the application was using these bytes to tell how much schema data to read, then changing them would cause it to calculate a CRC for the wrong amount of data. So the contents of the extra bytes needn't have anything to do with the length bytes at all.\n\nThere was another curious thing about the extra bytes. The fact that I was able to successfully calculate CRCs for my difference files meant that the unseen contents of these bytes must have been the same for both of the files that went into each difference file. Otherwise, they would have differed by more than one bit, and my technique for deducing the polynomial wouldn't have worked.\n\nThis suggested that the extra bytes might be constant, or at least they might depend only on the length of the file and not the data in it. However, it might only have been a coincidence, because I had only been working with files having very small differences, and I might just not have happened to change anything that would trigger a difference in the extra bytes.\n\nIt would be fortuitous if the extra bytes were constant, because since they appear at the end of the data being checked, their effect on the CRC depends only on their contents and not on the length of the file. If they were constant, their actual contents wouldn't matter, because I would be able to assume they were zero and compensate for their effect by making an adjustment to the XorOut parameter.\n\nIn any case, I had gone about as far as I could go using difference files, and it was time to start working with the original files.\n\n## Guessing Parameters\n\nWhen an XorIn or XorOut parameter is employed, the value almost always used is 0xffff, so the first thing I tried was calculating CRCs for some real files using both 0 and 0xffff for XorIn, and assuming zeroes for the extra bytes. This gave me the XorOut value that I would need to use in order to get a CRC matching the official one in the file.\n\nThe XorOut values that I obtained this way were not 0 or 0xffff, but this was to be expected, because the actual contents of the extra bytes may not have been zero. However, all of the files of a given length seemed to require the same XorOut value, providing more support for the idea that the extra bytes are the same for all files of a particular length.\n\nTo further check this, I created some more test files with very different contents. The hypothesis still seemed to hold up -- the XorOut values required for a particular length was always the same.\n\nI needed different XorOut values for different lengths, however. This could have been because the extra bytes varied somehow according to the length of the file, but it could also have been because a non-standard XorIn was being used. This would get transformed in different ways as it passed through different numbers of shift-xor steps and change the resulting CRC.\n\nI decided to stick with the hypothesis that the extra bytes were constant, and try to find an XorIn value consistent with it. This meant finding a single pair of XorIn/XorOut values that would work across files of different sizes.\n\nNow, I could have used brute force again here. There are only 2**16 possibilities, because choosing an XorIn for some file uniquely determines the XorOut needed, and then it's just a matter of trying the same combination on another file and seeing if it works.\n\nHowever, I wanted to see if I could come up with a more direct approach, partly for the challenge, and partly because it would give me a technique that could be applied to larger CRCs for which exhaustive searching would not be feasible.\n\n## Deducing XorIn\n\nTime for some more theory. Earlier we considered exclusive-oring together two messages of the same length. Now let's look at what happens if we exclusive-or two messages of different lengths together.\n\nLet n1 and n2 be the lengths of the messages. The exclusive-or of their CRCs will be\n\n C1 + C2 = (Tn1I + D1 + F) + (Tn2I + D2 + F) = Tn1I + Tn2I + (D1 + D2)\n\nNotice that F has cancelled out again, but not I. We can further rearrange the equation to give\n\n Tn1I + Tn2I = C1 + C2 + D1 + D2\n\nor\n\n (Tn1+ Tn2) I = K\n\nwhere KC1 + C2 + D1 + D2. We now have an equation in which I is the only unknown quantity.\n\nAdding Tn1 and Tn2 together may look a bit strange, but it makes perfectly good sense when you realise that T is a linear operator, and can be manipulated using the techniques of linear algebra.\n\nIn particular, we can represent T using a matrix, and calculate Tx using matrix multiplication with mod-2 arithmetic.\n\nTo calculate T, we decompose x into a sum of basis elements bi, each of which corresponds to one bit: b1 = 0x0001, b2 = 0x0002, etc. Then feed each bi through one step of the shift-xor operation. The result becomes column i of the matrix T.\n\nHaving got T, we can calculate M = Tn1Tn2. All that remains then is to solve the matrix equation M I = K.\n\n## Solving the Equation\n\nSolving a matrix equation like this is a standard linear algebra problem, so it seemed like it should be possible to use an algorithm such as Gauss-Jordan elimination, modified to use mod-2 arithmetic.\n\nWhen I tried to do this, the elimination algorithm got stuck, unable to find a pivot element for the last row. Examining the matrix revealed that it had ended up with an all-zero row. When this happens, it is an indication that the equation has more than one solution.\n\nThe final state of the matrix looked like this:\n\n0: 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0\n1: 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1\n2: 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1\n3: 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0\n4: 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0\n5: 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0\n6: 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1\n7: 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0\n8: 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0\n9: 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0\n10: 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0\n11: 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0\n12: 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1\n13: 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0\n14: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1\n15: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n\nThis corresponds to the following set of equations for the bits of I:\n\n b0 + b15 = 0 b8 = 0 b1 = 1 b9 = 0 b2 = 1 b10 = 0 b3 = 0 b11 = 0 b4 = 0 b12 = 1 b5 = 0 b13 = 0 b6 = 1 b14 + b15 = 1 b7 = 0\n\nBy inspection, there are two bit patterns that satisfy these constraints, 0x5046 and 0x9047. Trying them on the relevant files revealed the corresponding XorOut values required:\n\n XorIn XorOut 5046 3e64 5047 fe65\n\nI tried these values on the rest of my test files, and both combinations worked for all of them, regardless of the file size.\n\n An AsideIt's interesting to note that the difference between these two XorIn values (0xc001) is also the difference between the corresponding XorOut values.It turns out that this is not a coincidence, because 0xc001 is a fixed point of the shift-xor operation: (0xc001 >> 1) ^ 0xa001 == 0xc001. So if you change the initial register value by 0xc001, you change its value on all subsequent steps by 0xc001 also.In terms of linear algebra, the value e = 0xc001 is an eigenvector of the matrix T with eigenvalue 1, i.e. T e = e. This also means that Tn e = e for any n. So (Tn1 + Tn2) (x + e) = (Tn1 + Tn2) x + e + e = (Tn1 + Tn2) x.Consequently, if X is a solution of (Tn1 + Tn2) x = K, then X + e must also be a solution.\n\n## Success\n\nEncouraged by this, I tried one of these pairs on a collection of about 50 real files collected from the wild. It worked for all of them.\n\nI never found out exactly what was going on with the extra bytes, or why such strange-looking XorIn/XorOut values were needed, but I had found an algorithm which seemed to be good enough, and I was satisfied with that.\n\n## Conclusion\n\nAs well as solving my immediate problem, I had also developed a general method for determining all of the Rocksoft parameters for a CRC algorithm, given a small collection of specially-chosen example files. My case was complicated by the presence of unknown data, but if the data being checked is fully known, the method requires no searching or guesswork and can be used for CRCs of any size.\n\nSo if I ever need to figure out a 64-bit CRC, I'll be able to do it without needing a Blue Gene. And now that you've read this, you will too!"
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http://steamsia.com/qiche/1710.html
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[
"# 500)this.width=500 align=center hspace=10 vspa=1rel=nofollow",
null,
"车市导购：男人一旦有了家庭，在买车的问题上，想法都会变得很成熟，毕竟家庭永远是第一位的。所以，合适的B级车也是以这位网友为例的众多消费者的很好选择。\n\n推荐理由：看到凯美瑞的第一眼，消费者们大多都会明白，年轻动感的外观一定很受年轻人的青睐。不过不仅仅是受到年轻人的好评，因为结实耐用的品质，凯美瑞也受到众多消费者的欢迎。而在内饰设计上，中控多功能触控屏、7英寸彩色TFT多功能信息显示屏及10英寸彩色HUD平视显示系统在相互配合间增强了车内的科技氛围。同时，凯美瑞提供传动燃油版和混动版车型供消费者选择，丰富的车型配置也将进一步满足消费者的多元化需求。\n\n总结：上述推荐的3款车型基本上都可以满足这位网友的购车需求。因此如果有和这位网友志同道合的朋友就不要再徘徊了，选择一款就抓紧出手吧！\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/\n\n推荐理由：雅阁的年轻化设计也令它充满动感时尚的气息，并且溜背的车型造型也很符合当下流行趋势。当然在内饰设计上，雅阁在采用最新家族式设计的同时，一体化的中控布局很简约干练，8英寸的悬浮式中控大屏页面清晰，也提升了车内的科技视觉。此外，雅阁同样提供传统燃油版和混动版本车型。所以，在你的购车备选清单中，应该出现雅阁这款车。\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/推荐理由：提到B级家用轿车，很多消费者都会立刻想到迈腾，同时在国内B级车销量排行榜上，迈腾也是时长占据榜首的位置。很沉稳的造型设计让其显得低调谦逊。在内饰设计上，迈腾则秉承着大众经典的中控设计，其中12.3 英寸的数字液晶仪表，集驾驶、导航和辅助功能等信息于一体，中控嵌入式8英寸大屏搭配第二代MIB信息娱乐系统。动力总成上，迈腾分别搭载1.4T和2.0T高低功率版本发动机，匹配的均为7速DSG变速箱。因此，想必是各方面都表现的很均衡，让其很受消费者关注和青睐。\n\n网友提问：今年已经30+，上有老下有小，为了满足自己和家人的日常出行，想买一台驾乘舒适且油耗不高的B级车，购车预算在23万左右，求合适的车型推荐。\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>\n\n500)this.width=500 align=center hspace=10 vspace=10 rel=nofollow/>"
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https://greedhead.net/how-do-you-calculate-work-fund/
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[
"# How do you calculate work Fund?\n\n## How do you calculate work Fund?\n\nWorking capital is calculated by using the current ratio, which is current assets divided by current liabilities. A ratio above 1 means current assets exceed liabilities, and, generally, the higher the ratio, the better.\n\n## How is DWC calculated?\n\nDays Working Capital Formula and Calculation Multiply the average working capital by 365 or days in the year. Divide the result by the sales or revenue for the period, which is found on the income statement. You can also take the average sales over multiple periods as well.\n\n## How do you calculate working capital on a balance sheet?\n\nWorking Capital = Current Assets – Current Liabilities The working capital formula tells us the short-term liquid assets available after short-term liabilities have been paid off.\n\n## How do you calculate net working capital?\n\nNet Working Capital Formula\n\n1. Net Working Capital = Current Assets – Current Liabilities.\n2. Net Working Capital = Current Assets (less cash) – Current Liabilities (less debt)\n3. NWC = Accounts Receivable + Inventory – Accounts Payable.\n\n## How do you calculate work?\n\nWork can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.\n\n## How do you interpret the working capital cycle?\n\nThe working capital cycle (WCC) is the amount of time it takes to turn the net current assets and current liabilities into cash. The longer the cycle is, the longer a business is tying up capital in its working capital without earning a return on it.\n\n## Where is working capital in financial statements?\n\nWorking capital is a measure of the company’s liquidity, taken by subtracting a company’s current liabilities from its current assets. It’s possible for you to calculate your working capital from information included in your company’s balance sheet.\n\n## How do you solve working capital problems?\n\n11 Best Way to Manage and Improve Working Capital\n\n1. 1.1 1. Incentivize Receivables.\n2. 1.2 2. Meet Debt Obligations.\n3. 1.3 3. Choose Vendors Who Offer Discounts.\n4. 1.4 4. Analyze Fixed and Variable Costs.\n5. 1.5 5. Examine Interest Payments.\n6. 1.6 6. Manage Inventory.\n7. 1.7 7. Automate Accounts Receivable and Payment Monitoring.\n8. 1.8 8.\n\n## What is the formula to calculate time?\n\nTo solve for time use the formula for time, t = d/s which means time equals distance divided by speed.\n\n## What is net worth example?\n\nYour net worth can tell you many things. If the figure is negative, it means you owe more than you own. For example, if your assets equal \\$200,000 and your liabilities are \\$100,000, you will have a positive net worth of \\$100,000 (\\$200,000 – \\$100,000 = \\$100,000).\n\n## What is normal operating cycle?\n\nDictionary of Business Terms for: normal operating cycle. normal operating cycle. the period of time required to convert cash into raw materials, raw materials into inventory finished goods, finished good inventory into sales and accounts receivable, and accounts receivable into cash.\n\n## Why is the working capital cycle important?\n\nThe working capital cycle is a measure of how quickly a business can turn its current assets into cash. Understanding how it works can help small business owners like you manage their company’s cash flow, improve efficiency, and make money faster.\n\n## How do you prepare a working capital statement?\n\nRules for preparing the statement of changes in working capital: (i) Increase in a current asset, results in increase (+) in “working capital”. (ii) Decrease in a current asset; result in decreases (-) in “working capital”. (iii) Increase in a current liability, results in decrease (-) in “working capital”.\n\nHow to Calculate Working Capital. Working capital is calculated by using the current ratio, which is current assets divided by current liabilities. A ratio above 1 means current assets exceed liabilities, and, generally, the higher the ratio, the better.\n\nWorking Capital = Current Assets – Current Liabilities Both current assets and liabilities can be found directly on your company’s balance sheet.\n\n## How do you calculate working capital requirement?\n\nWorking Capital = Current Assets – Current Liabilities\n\n1. Cash in hand.\n2. Cash equivalent.\n3. Company inventory.\n4. Accounts receivable.\n5. Pre-paid liabilities.\n\n## How is working capital calculated UK?\n\nThis is the working capital calculation: Working capital = current assets – current liabilities. The principal works on the basis that you have cash coming in via payments from customers for example (current assets), and money going out on things like IT support and telecoms etc (current liabilities).\n\n## How do banks calculate average working funds?\n\n\\$ Working funds to be reckoned as average of total assets (excluding accumulated losses, if any) as reported to Reserve Bank of India in Form X under Section 27 of the Banking Regulation Act, 1949, during the 12 months of the financial year.\n\n## What is the cost of sales formula?\n\nThe cost of sales is calculated as beginning inventory + purchases – ending inventory.\n\n## What is NWC formula?\n\nFormula: Net Working Capital = Current Assets (less cash) – Current Liabilities (less debt) or, NWC = Accounts Receivable + Inventory – Accounts Payable.\n\n## What are the 4 main components of working capital?\n\nThe elements of working capital are money coming in, money going out, and the management of inventory. Companies must also prepare reliable cash forecasts and maintain accurate data on transactions and bank balances.\n\n## What is the best way to fund working capital?\n\n6 Ways to Get Working Capital Financing\n\n1. Where to Get the Capital You Need.\n6. Invoice factoring.\n7. Invoice financing.\n\n## What is total fund employed?\n\nCapital employed, also known as funds employed, is the total amount of capital used for the acquisition of profits by a firm or project. Capital employed can also refer to the value of all the assets used by a company to generate earnings.\n\n## How to work out the cost of a fund?\n\nH ere’s a quick way of working out how much your entire portfolio costs to run at the fund level. Simply take each fund in your portfolio and… Multiply the fund’s Ongoing Charge Figure (OCF) by the percentage of your portfolio that’s allocated to the fund. This gives you the weighted OCF of each fund in your portfolio.\n\n## How to calculate mutual fund return on investment?\n\nExample: Mutual Fund Return on Investment Calculation. If the current NAV is 15 and the previous NAV was 13.5, the return would be (15 – 13.5) x 100/13.5 = 150/13.5 = 11.11% over the time period. If the time period is in months say 3 months or in years say 2, or in days say 100, in that case, the above formula can be used as\n\n## Which is the correct way to calculate working capital?\n\nHow to Calculate Working Capital Working capital is calculated by using the current ratio, which is current assets divided by current liabilities. A ratio above 1 means current assets exceed liabilities, and, generally, the higher the ratio, the better.\n\n## How do you calculate the OCF of a fund?\n\nSimply take each fund in your portfolio and…. Multiply the fund’s Ongoing Charge Figure (OCF) by the percentage of your portfolio that’s allocated to the fund. This gives you the weighted OCF of each fund in your portfolio."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9169943,"math_prob":0.94869643,"size":7694,"snap":"2021-43-2021-49","text_gpt3_token_len":1640,"char_repetition_ratio":0.18582575,"word_repetition_ratio":0.16074188,"special_character_ratio":0.22381076,"punctuation_ratio":0.12310866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98970425,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T11:04:34Z\",\"WARC-Record-ID\":\"<urn:uuid:073cff3f-8904-4d06-bddb-7b0c3407ac1a>\",\"Content-Length\":\"108494\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2a5788a-6f1a-47b9-a706-3719eb5a81f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a36df15-3282-4e94-91cd-3cf63ffafcd2>\",\"WARC-IP-Address\":\"104.21.78.80\",\"WARC-Target-URI\":\"https://greedhead.net/how-do-you-calculate-work-fund/\",\"WARC-Payload-Digest\":\"sha1:4TRRXFPIOJW6COO5YNCKER7GGTFACKCF\",\"WARC-Block-Digest\":\"sha1:3JTR5LHEXRGUDRBXFZOQSDJ4SPPR5EO3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363465.47_warc_CC-MAIN-20211208083545-20211208113545-00433.warc.gz\"}"}
|
https://www.snapxam.com/problems/81321937/integral-of-1-1-1sin2-dxd
|
[
"👉 Try now NerdPal! Our new math app on iOS and Android\n\n# Solve the trigonometric integral $\\int\\frac{1}{1- \\sin\\left(2\\right)}ddx$\n\n## Step-by-step Solution\n\nGo!\nMath mode\nText mode\nGo!\n1\n2\n3\n4\n5\n6\n7\n8\n9\n0\na\nb\nc\nd\nf\ng\nm\nn\nu\nv\nw\nx\ny\nz\n.\n(◻)\n+\n-\n×\n◻/◻\n/\n÷\n2\n\ne\nπ\nln\nlog\nlog\nlim\nd/dx\nDx\n|◻|\nθ\n=\n>\n<\n>=\n<=\nsin\ncos\ntan\ncot\nsec\ncsc\n\nasin\nacos\natan\nacot\nasec\nacsc\n\nsinh\ncosh\ntanh\ncoth\nsech\ncsch\n\nasinh\nacosh\natanh\nacoth\nasech\nacsch\n\n### Videos\n\n$\\frac{441}{40}dx+C_0$\nGot another answer? Verify it here!\n\n## Step-by-step Solution \n\nProblem to solve:\n\n$\\int\\frac{1}{1- \\sin\\left(2\\right)}ddx$\n\nSpecify the solving method\n\n1\n\nSimplify the expression inside the integral\n\n$\\int\\frac{441}{40} ddx$\n\nLearn how to solve trigonometric integrals problems step by step online.\n\n$\\int\\frac{441}{40} ddx$\n\nLearn how to solve trigonometric integrals problems step by step online. Solve the trigonometric integral int(1/(1-sin(2))d)dx. Simplify the expression inside the integral. The integral of a constant is equal to the constant times the integral's variable. Simplify the expression inside the integral. As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C.\n\n$\\frac{441}{40}dx+C_0$\n\n## Explore different ways to solve this problem\n\nSolving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more\n\nSolve int(1/(1-sin(2))d)dx using basic integralsSolve int(1/(1-sin(2))d)dx using u-substitutionSolve int(1/(1-sin(2))d)dx using integration by partsSolve int(1/(1-sin(2))d)dx using tabular integrationSolve int(1/(1-sin(2))d)dx using weierstrass substitution\n\nSnapXam A2\n\nGo!\n1\n2\n3\n4\n5\n6\n7\n8\n9\n0\na\nb\nc\nd\nf\ng\nm\nn\nu\nv\nw\nx\ny\nz\n.\n(◻)\n+\n-\n×\n◻/◻\n/\n÷\n2\n\ne\nπ\nln\nlog\nlog\nlim\nd/dx\nDx\n|◻|\nθ\n=\n>\n<\n>=\n<=\nsin\ncos\ntan\ncot\nsec\ncsc\n\nasin\nacos\natan\nacot\nasec\nacsc\n\nsinh\ncosh\ntanh\ncoth\nsech\ncsch\n\nasinh\nacosh\natanh\nacoth\nasech\nacsch\n\n### Main topic:\n\nTrigonometric Integrals\n\n~ 0.02 s\n\n### Join 500k+ students in problem solving.\n\n##### Without automatic renewal.\nCreate an Account"
] |
[
null
] |
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|
https://www.jiskha.com/questions/1799101/which-describes-the-collection-of-all-points-in-the-same-plane-that-are-8-and-1-2-inches
|
[
"# Math\n\nWhich describes the collection of all points in the same plane that are\n8 and 1/2\ninches from Point T?\n\ncircle T with radius 8 and 1/ 2 inches\n\ncircle T with diameter 8 and 1/2 inches\n\ncircle T with radius 17 inches\n\na sphere with center T and radius 8 and 1/2 inches\n\n1. 👍 1\n2. 👎 0\n3. 👁 1,022\n1. last choice\n\n1. 👍 3\n2. 👎 2\n2. First choice,\nIt said in a plane, the last choice would involve 3D\n\n1. 👍 1\n2. 👎 0\n👨🏫\nReiny\n3. Go with Reiny. I missed the term \"sphere\"\n\n1. 👍 0\n2. 👎 0\n4. Tysm :3\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### (Pre-Algebra) A\n\nLesson 7: Coordinate Plane Essential Algebra Readiness(Pre-Algebra) A Unit 4: Real Numbers and the Coordinate Plane Has anyone done the Coordinate Plane Practice AND assessment? 1. Which point is located at (-3, -2)? (1 point)\n\n2. ### Algebra\n\nWhich best describes your ability to work with transformations in the coordinate plane? A. I can graph translations, reflections, and rotations in the coordinate plane. I can use math vocabulary to describe these transformations.\n\n3. ### Physics\n\n1. Two points in a plane have polar coordinates (2.5 m, 30 degrees) and (3.8 m, 120 deg). Determine (a) the cartesian coordinates of both points and (b) the distance between the points. 2. Vector A has magnitude 29 units and\n\n4. ### geometry\n\nDecide which one of the following statements is false. a. any three points lie on a distinct line. b. three noncollinear points determine a plane. c. a line contains at least two points. d. through any two distinct points there\n\n1. ### Physics\n\nAn air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector A , which has a magnitude of 220 km and points in a direction 32 degrees north of\n\n2. ### Chemistry\n\n2. Identify the oxidation numbers for each element in the following equations. Then determine whether each equation describes a redox reaction. Justify your answer in terms of electron transfer. Describe the oxidation and\n\n3. ### Biology\n\nIn conventional human radiological imaging (e.g., MRI, PET, CT) of the head, the axial plane is synonymous with the horizontal plane. Which of the following statements about this plane is most accurate? The axial plane is an axis\n\n4. ### Math\n\nWhich accurately describes a circle with radius 12 centimeters and center Q? A.the collection of all points located 6 centimeters from point Q B.the collection of all points on the same plane located 6 centimeters from point Q\n\n1. ### Calculus\n\nFind a nonzero vector orthogonal to the plane through the points P, Q, and R.? Consider the points below. P(0,-3,0), Q(4,1,-3), R(5,3,1)\n\n2. ### geometry\n\nWhich of the following statements best describe Euclid's axioms about lines? A. Two distinct lines intersect in an inifinite number of points. B. A line contains a finite number of points. C. Lines contain exactly two points. D.\n\n3. ### Science\n\nA scientist from the United States conducts an experiment using miles per hour to measure speed. Why might this be confusing for scientists in other countries? A) Scientists in other countries don't usually perform experiments\n\n4. ### Geometry\n\nA sphere consists of all the points that are 5 units from its center (2, 3,−6). Write an equation that describes this sphere. Does the sphere intersect the xy-plane? Explain."
] |
[
null
] |
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|
https://byjus.com/ncert-solutions-for-class-9-maths-chapter-5-introduction-to-euclids-geometry-ex-5-1/
|
[
"",
null,
"# NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid's Geometry Exercise 5.1\n\n## NCERT Solutions For Class 9 Maths Ex 5.1 PDF Free Download\n\nUnderstanding Euclidian Geometry is easy if students refer NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Excercise 5.1 and 5.2. These exercises are designed to help you grasp the conceptual basis and solve numerical problems.\n\nTraditionally, Euclidian Geometry is an elementary concept but it has significance in various other fields of study. Therefore, it is crucial to have a good grasp of this concept. Students will find that this NCERT solution is one of the best resources to learn. This is due to the fact that the solutions are designed by a team of dedicated teachers with many years of experience.\n\nConsequently, NCERT Solutions is one of the best guides that you might come across for your studies. Important topics are presented in a simple, structured format. Moreover, complex jargons are broken down or explained in the content. Furthermore, content is refreshed regularly as per the prescribed CBSE syllabus.\n\n### Access Answers of Maths NCERT Class 9 Chapter 5 – Introduction to Euclid’s Geometry Exercise 5.1",
null,
"",
null,
"",
null,
"",
null,
"## NCERT Solutions for Class 9 Maths Chapter 5 – Introduction to Euclid’s Geometry Exercise 5.1\n\nEuclid’s Geometry is a system introduced by Euclid, an Alexandrian-Greek Mathematician way back in 300 BC.\n\n2,000 years later, Euclid’s contributions still remain valid and it has been adopted in various applications, ranging from engineering to theoretical physics. Euclidian Geometry also has significant academic implications in disciplines of science and mathematics.\n\nFind out how Euclidean Geometry works and explore the 2000-year-old theorems. Browse other important NCERT Solutions For Class 9 Maths to help you practice various other mathematical concepts.\n\n### Key Features of NCERT Solutions for Class 9 Maths Chapter 5 – Introduction to Euclid’s Geometry Exercise 5.1 Page no 85\n\n1. Content presented in a well-structured format\n2. Chapter highlights and key pointers are provided\n3. Jargons are explained effectively in the content\n4. Well-qualified teachers create accurate solutions with explanations\n5. A thorough analysis of tough questions from each chapter\n\nExplore: NCERT Solutions Class 9\n\n## Frequently Asked Questions on NCERT Solution Class 9 Maths Ex 5.1\n\n### What is parallel lines?\n\nParallel lines are those lines which never intersect each other and are always at a constant distance perpendicular to each other. Parallel lines can be two or more lines.\n\n### Define perpendicular lines?\n\nPerpendicular lines- Perpendicular lines are those lines which intersect each other in a plane at right angles then the lines are said to be perpendicular to each other.\n\n### what is line segment?\n\nWhen a line cannot be extended any further because of its two end points then the line is known as a line segment. A line segment has 2 end points.\n\n### what is Radius of circle?\n\nA radius of a circle is the line from any point on the circumference of the circle to the center of the circle.\n\n### Define Square?\n\nA quadrilateral in which all the four sides are said to be equal and each of its internal angle is right angles is called square"
] |
[
null,
"https://www.facebook.com/tr",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
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"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
null,
"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs=",
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http://www.everyscience.com/Chemistry/General/Symmetry/h.1143.php
|
[
"# Normal Modes of Vibration\n\nVibrations in molecules can be excited by the application of radiation at the same frequency as the vibration itself. This gives rise to spectroscopic methods of studying molecular structure, which depend upon being able to predict or at least rationalise the vibrations that are observed. Symmetry considerations play an important role in this.\n\nNote that there is only one possible mode of vibration for a diatomic molecule, stretching (or compression) of the bond. No bending modes exist, as neither end of the molecule has a fixed position that is defined by being attached to another atom.\n\nFor polyatomic molecules there are multiple vibrational modes, because all of the bonds may stretch or compress, and bond angles may also change, allowing the existence of bending modes.\n\nFor a non-linear molecule containing N atoms, there are 3N – 6 independent vibrational modes. If the molecule is linear, there are 3N – 5 modes of vibration:\n\nEach atom may be displaced in any one of three directions (x, y, z), giving 3N possible displacements. Three coordinates are required to specify the location of the centre of mass of the molecule (fixing the location of the molecule in space). Further, three angles are needed to specify the orientation of a non-linear molecule in space. These six pieces of information restrict the possible vibrational modes to 3N – 6.\n\n(Alternatively, the situation can be viewed as follows. There are 3N possible vibrational modes. However, if all the atoms are vibrating so that they move in the positive x direction, then this is actually a translation of the whole molecule through space in the x direction, with no change in the bond lengths or angles of the molecule. This does not count as a vibration. Similar remarks can be made about translations in the y and z directions. This reduces the number of vibrational modes by three. Similarly, it is possible for the atoms to vibrate together in such a way that the molecule is actually rotating rather than vibrating. Rotation can occur about any of the three Cartesian axes, so a further three of the possible vibrations are counted as rotations not vibrations. Thus of the 3N possible vibrations, 3 are translations, 3 are rotations and 3N – 6 are vibrations.)\n\nFor a linear molecule, only two angles are required to specify the orientation in space (the infinite rotational symmetry about the molecular axis means that all possible orientations about this axis are equivalent), hence there are 3N – 5 vibrations. (In terms of the alternative view of the situation, no vibrational motion of the atoms in a linear molecule can lead to rotation about the axis of the molecule, so one of the rotational modes is not counted.)\n\nIt is necessary for more advanced calculations to find a way of describing the modes (in terms of the different atomic motions that they represent). It proves simplest to do this in terms of the normal modes of a molecule. Normal modes are the independent, synchronous motions of atoms or groups of atoms that may be excited without exciting other normal modes. For example, the symmetric and antisymmetric stretches of CO2 , in which the oxygen atoms are displaced while the carbon atom remains stationary, are two of the normal modes of the molecule:\n\nHowever, stretches of the two individual C=O bonds are not normal modes: When one C=O bond vibration is excited, the motion of the carbon atom causes the other C=O bond to stretch or compress, exciting the other vibrational mode.\n\nNormal modes may be treated as independent harmonic oscillators (assuming that the anharmonicity of the vibrations is neglected). Note that the effective mass of a vibrational mode is a measure of the mass moved in the course of the vibration, and it is in general a highly complicated function of the masses of the atoms involved. It should further be noted that a case such as carbon dioxide, in which all the normal modes are purely stretching or bending is rare. More usually, a normal mode is a motion formed from simultaneous bending and stretching.\n\nBack to Top"
] |
[
null
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https://rdrr.io/cran/pysd2r/f/vignettes/pysd2r.Rmd
|
[
"# Introduction\n\nThe goal of this package is to allow R users run system dynamics models using the pysd, developed by James Houghton. The pysd project \"is a simple library for running System Dynamics models in python, with the purpose of improving integration of Big Data and Machine Learning into the SD workflow.\"\n\npysd2r has been tested with python3, and the following command was used to install pysd from source.\n\npython3 setup.py install\n\nGiven R's facility for also providing big data and machine learning support, this package opens up the functionality of pysd for R users, and provides an interface to the basic set of methods provided by pysd, including the functions:\n\n• model.run()\n\nThe API provide by pysd2r includes the following functions (for list of parameters type ?function_name in R) which call the mapping functions in pysd.\n\n• get_python_info() - Returns the python version currently used by reticulate\n• pysd_connect() - Creates an object to facilitate interaction with pysd\n• run_model() - Runs a simulation model\n• set_components() - Changes a model parameter\n• get_timestep() - Gets the time step (DT) from the model\n• get_initial_time() - Gets the initial time from the model\n• get_final_time() - Gets the final time from the model\n• set_time_values() - Sets the initial time, final time, and timestep\n• print() - Implementation of generic print function of ipysd S3 object\n• get_doc() - Gets the model variables and returns as a tibble\n\nThe following example shows how pysd2r can be used to run a simulation model (Population.mdl which is a one-stock model of population growth)\n\n```library(pysd2r) # load pysd2r\n```\n\nFirst, a connection is made to pysd\n\n```py <- pysd_connect()\n```\n\nNext, the vensim file is opened.\n\n```target <- system.file(\"models/vensim\", \"Population.mdl\", package = \"pysd2r\")\n```\n\nThe returning object (ipysd, and S3 class) can be inspected. This is a list with two elements. The first is the reference to pysd, the second is a referece to the translated python model\n\n```str(py)\n```\n\nWith this reference, the simulation can be run by calling the run_model() function.\n\n```results <- run_model(py)\n```\n\nThe results from the tibble can be shown.\n\n```results\n```\n\nThese results can also be processed using ggplot2.\n\n```ggplot(data=results)+\ngeom_point(aes(x=TIME,y=Population),colour=\"blue\")\n```\n\n## Try the pysd2r package in your browser\n\nAny scripts or data that you put into this service are public.\n\npysd2r documentation built on May 2, 2019, 3:40 p.m."
] |
[
null
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https://www.iue.tuwien.ac.at/phd/hollauer/node44.html
|
[
"",
null,
"",
null,
"",
null,
"",
null,
"Next: 9. Intrinsic Stress Effects Up: 8. Thermo-Mechanical Stress in Previous: 8.1 Simulation Procedure\n\nSubsections\n\n# 8.2 Demonstrative Example\n\nA three-dimensional interconnect layout with (3.0",
null,
"4.2)",
null,
"m floor space, as displayed in Fig. 8.2, is investigated by the two previously defined models. In this structure the bottom layer material is silicon (Si). Above the silicon layer there is a silicon dioxide (SiO",
null,
") layer, where two copper (Cu) lines are embedded. Between the copper lines and the silicon dioxide is a very thin titanium nitride (TiN) passiviation layer. This passiviation layer prevents the diffusion of copper into the silicon dioxide during the manufacturing process .\n\nAs shown in Fig. 8.2, the silicon dioxide layer and the copper lines are covered by a silicon nitride (Si",
null,
"N",
null,
") layer which separates the next upper located SiO",
null,
"layer from the lower one. In the second upper SiO",
null,
"layer three copper lines are embedded. These three copper lines are transverse compared to the two subjacent copper lines. This upper SiO",
null,
"layer is also covered with silicon nitride. On the top of the layout is a third SiO",
null,
"layer.\n\nIn Fig. 8.3 a cut through the interconnect structure given in Fig. 8.2 is presented. As evident from Fig. 8.3 an upper transverse copper line is connected with a lower copper line by a so-called via. The other two copper lines shown in this figure are connected in the same way. The third transverse upper copper line (see Fig. 8.2) does not have an interconnection to another line.",
null,
"",
null,
"## 8.2.1 Simulation Results\n\nFrom the simulation aspect the temperature and stress distribution in the given interconnect structure at two different points of time are of interest. For the simulation a potential difference of 7mV between point A and B in the first interconnect, as marked in Fig. 8.2, is assumed. The other interconnects are assumed to be inactive.\n\n### 8.2.1.1 Temperature Distribution\n\nIf STAP with its electro-thermal model is applied to this interconnect structure, the obtained output is the temperature distribution in the structure. In this analysis it is assumed that the bottom of the silicon layer is connected with a cooling element which holds the temperature at 320K. For the simulation the electric and thermal conductivities given in Table 8.1 are used. Because copper is a metal and an excellent conductor, it has the best thermal and electric conductivity regarding feasible materials for interconnect metals.\n\n Cu Si SiO",
null,
"Si",
null,
"N",
null,
"TiN",
null,
"[S/m] 5.26 x 10",
null,
"0.0 0.0 0.0 1.66 x 10",
null,
"",
null,
"[W/mK] 400.0 1.35 1.39 12.07 48.25\n\nDue to the Joule self-heating in wires with a current flow, the hottest regions are around the active copper line. This is the reason, why after a time of 30",
null,
"s the highest temperature (351K) is in the inner layers which are surrounded by the active copper line, as shown in Fig. 8.4. The relatively high thermal conductivity of copper causes that the temperature values in the copper lines are rather uniform, and so they are not included in Fig. 8.4. In Fig. 8.5 it is demonstrated that after a longer operating time of 100",
null,
"s the self-heating has increased the temperature to 376K.\n\nIn Fig. 8.7 the maximal temperature versus operating time in the interconnect structure is plotted. It can be seen that after approximately 450",
null,
"s the self heating effect reaches a stable steady state. Fig. 8.6 shows the temperature distribution in the steady state where the maximal temperature has reached a steady value of 396K.",
null,
"",
null,
"",
null,
"",
null,
"### 8.2.1.2 Pressure Distribution\n\nFor the pressure calculation, the influence of the temperature on the mechanical parameters can be neglected, because the temperatures are so low that they do not change the material condition perceivably . Also the influence of pressure on the electric and thermal parameters can be neglected for such pressure values . Therefore, a coupling of the electro-thermal and thermo-mechanical system is not necessary and the decoupled alternating solving of the temperature and pressure distribution is acceptable here. With the obtained temperature distribution the mechanical problem can be set up as described in Section 8.1.2. As applied mechanical boundary conditions the bottom surface is fixed and the other surfaces are free. For the simulation the Young modulus E, Poisson ratio",
null,
", and the thermal expansion factor",
null,
"given in Table 8.2 are used.\n\n Cu Si SiO",
null,
"Si",
null,
"N",
null,
"TiN E [GPa] 115 180 73 380 600",
null,
"[-] 0.34 0.22 0.17 0.27 0.25",
null,
"[10",
null,
"/K] 17.7 2.7 0.55 3.3 9.4\n\nThe distribution of pressure is an important quantity for electromigration, because failure risks are increasing with larger pressure. Therefore, the copper lines and their vias are the most interesting regions. The thermal expansion coefficient of copper is the largest and it is enormous compared with the main embedding material silicon dioxide (see Table 8.2). This larger coefficient of the copper lines demands more volume expansion than the other surrounding materials in the heated structure. This means that the copper lines with their vias can not expand as desired and compressive stress is built up.\n\nFig. 8.8 shows the simulation results of the pressure distribution in the copper lines and their vias at time 30",
null,
"s. It can be seen that the via is a high pressure region. The first reason is that the via has less chance to expand in vertical direction because of the over- and underlying copper lines which also have the same demand to extend. The other explanation is the confinement of the via by the passivation layer (see Fig. 8.3) which was made of titanium nitride (TiN). The passivation layer is thin, but the stiffness (Young modulus) of TiN is more than five times larger than the copper one and so this layer is able to prevent the volume increase. The proof is that the largest pressure (188MPa) develops at the bottom of the vias, because it is confined with the passivation layer. The pressure in the bottom region is larger than on the top, where is no limiting TiN-layer.\n\nAfter a longer operating time of 100",
null,
"s the higher temperature in the structure (see Fig. 8.5), causes that the maximum pressure in the via is increased to 354MPa, as displayed in Fig. 8.9. The pressure distribution in the copper lines and vias is nearly the same as at time 30",
null,
"s. As illustrated in Fig. 8.10 the pressure reaches a maximum of 479MPa in the steady state.",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Next: 9. Intrinsic Stress Effects Up: 8. Thermo-Mechanical Stress in Previous: 8.1 Simulation Procedure\n\nCh. Hollauer: Modeling of Thermal Oxidation and Stress Effects"
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https://www.colorhexa.com/818383
|
[
"# #818383 Color Information\n\nIn a RGB color space, hex #818383 is composed of 50.6% red, 51.4% green and 51.4% blue. Whereas in a CMYK color space, it is composed of 1.5% cyan, 0% magenta, 0% yellow and 48.6% black. It has a hue angle of 180 degrees, a saturation of 0.8% and a lightness of 51%. #818383 color hex could be obtained by blending #ffffff with #030707. Closest websafe color is: #999999.\n\n• R 51\n• G 51\n• B 51\nRGB color chart\n• C 2\n• M 0\n• Y 0\n• K 49\nCMYK color chart\n\n#818383 color description : Dark grayish cyan.\n\n# #818383 Color Conversion\n\nThe hexadecimal color #818383 has RGB values of R:129, G:131, B:131 and CMYK values of C:0.02, M:0, Y:0, K:0.49. Its decimal value is 8487811.\n\nHex triplet RGB Decimal 818383 `#818383` 129, 131, 131 `rgb(129,131,131)` 50.6, 51.4, 51.4 `rgb(50.6%,51.4%,51.4%)` 2, 0, 0, 49 180°, 0.8, 51 `hsl(180,0.8%,51%)` 180°, 1.5, 51.4 999999 `#999999`\nCIE-LAB 54.594, -0.743, -0.266 21.266, 22.538, 24.702 0.31, 0.329, 22.538 54.594, 0.789, 199.679 54.594, -1.131, -0.25 47.475, -3.124, 2.383 10000001, 10000011, 10000011\n\n# Color Schemes with #818383\n\n• #818383\n``#818383` `rgb(129,131,131)``\n• #838181\n``#838181` `rgb(131,129,129)``\nComplementary Color\n• #818382\n``#818382` `rgb(129,131,130)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818283\n``#818283` `rgb(129,130,131)``\nAnalogous Color\n• #838281\n``#838281` `rgb(131,130,129)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #838182\n``#838182` `rgb(131,129,130)``\nSplit Complementary Color\n• #838381\n``#838381` `rgb(131,131,129)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #838183\n``#838183` `rgb(131,129,131)``\n• #818381\n``#818381` `rgb(129,131,129)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #838183\n``#838183` `rgb(131,129,131)``\n• #838181\n``#838181` `rgb(131,129,129)``\n• #5b5c5c\n``#5b5c5c` `rgb(91,92,92)``\n• #686969\n``#686969` `rgb(104,105,105)``\n• #747676\n``#747676` `rgb(116,118,118)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #8e9090\n``#8e9090` `rgb(142,144,144)``\n• #9b9c9c\n``#9b9c9c` `rgb(155,156,156)``\n• #a8a9a9\n``#a8a9a9` `rgb(168,169,169)``\nMonochromatic Color\n\n# Alternatives to #818383\n\nBelow, you can see some colors close to #818383. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #818383\n``#818383` `rgb(129,131,131)``\nSimilar Colors\n\n# #818383 Preview\n\nThis text has a font color of #818383.\n\n``<span style=\"color:#818383;\">Text here</span>``\n#818383 background color\n\nThis paragraph has a background color of #818383.\n\n``<p style=\"background-color:#818383;\">Content here</p>``\n#818383 border color\n\nThis element has a border color of #818383.\n\n``<div style=\"border:1px solid #818383;\">Content here</div>``\nCSS codes\n``.text {color:#818383;}``\n``.background {background-color:#818383;}``\n``.border {border:1px solid #818383;}``\n\n# Shades and Tints of #818383\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #020303 is the darkest color, while #f8f8f8 is the lightest one.\n\n• #020303\n``#020303` `rgb(2,3,3)``\n• #0c0c0c\n``#0c0c0c` `rgb(12,12,12)``\n• #161616\n``#161616` `rgb(22,22,22)``\n• #202020\n``#202020` `rgb(32,32,32)``\n• #292a2a\n``#292a2a` `rgb(41,42,42)``\n• #333434\n``#333434` `rgb(51,52,52)``\n• #3d3e3e\n``#3d3e3e` `rgb(61,62,62)``\n• #474848\n``#474848` `rgb(71,72,72)``\n• #505252\n``#505252` `rgb(80,82,82)``\n• #5a5b5b\n``#5a5b5b` `rgb(90,91,91)``\n• #646565\n``#646565` `rgb(100,101,101)``\n• #6e6f6f\n``#6e6f6f` `rgb(110,111,111)``\n• #777979\n``#777979` `rgb(119,121,121)``\n• #818383\n``#818383` `rgb(129,131,131)``\n• #8b8d8d\n``#8b8d8d` `rgb(139,141,141)``\n• #959696\n``#959696` `rgb(149,150,150)``\n• #9fa0a0\n``#9fa0a0` `rgb(159,160,160)``\n• #a9aaaa\n``#a9aaaa` `rgb(169,170,170)``\n• #b2b4b4\n``#b2b4b4` `rgb(178,180,180)``\n• #bcbdbd\n``#bcbdbd` `rgb(188,189,189)``\n• #c6c7c7\n``#c6c7c7` `rgb(198,199,199)``\n• #d0d1d1\n``#d0d1d1` `rgb(208,209,209)``\n``#dadbdb` `rgb(218,219,219)``\n• #e4e4e4\n``#e4e4e4` `rgb(228,228,228)``\n• #eeeeee\n``#eeeeee` `rgb(238,238,238)``\n• #f8f8f8\n``#f8f8f8` `rgb(248,248,248)``\nTint Color Variation\n\n# Tones of #818383\n\nA tone is produced by adding gray to any pure hue. In this case, #818383 is the less saturated color, while #0ef6f6 is the most saturated one.\n\n• #818383\n``#818383` `rgb(129,131,131)``\n• #778d8d\n``#778d8d` `rgb(119,141,141)``\n• #6e9696\n``#6e9696` `rgb(110,150,150)``\n• #64a0a0\n``#64a0a0` `rgb(100,160,160)``\n• #5ba9a9\n``#5ba9a9` `rgb(91,169,169)``\n• #51b3b3\n``#51b3b3` `rgb(81,179,179)``\n• #47bdbd\n``#47bdbd` `rgb(71,189,189)``\n• #3ec6c6\n``#3ec6c6` `rgb(62,198,198)``\n• #34d0d0\n``#34d0d0` `rgb(52,208,208)``\n``#2adada` `rgb(42,218,218)``\n• #21e3e3\n``#21e3e3` `rgb(33,227,227)``\n• #17eded\n``#17eded` `rgb(23,237,237)``\n• #0ef6f6\n``#0ef6f6` `rgb(14,246,246)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #818383 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.5241183,"math_prob":0.81884754,"size":3693,"snap":"2021-04-2021-17","text_gpt3_token_len":1623,"char_repetition_ratio":0.12577935,"word_repetition_ratio":0.0073937154,"special_character_ratio":0.5670187,"punctuation_ratio":0.23250565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931381,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T15:23:23Z\",\"WARC-Record-ID\":\"<urn:uuid:b668e6bb-4061-41b5-928f-04716f73e90b>\",\"Content-Length\":\"36323\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6c9d8c4c-77fe-4a84-a7d7-a3d1c026bb08>\",\"WARC-Concurrent-To\":\"<urn:uuid:10773623-5d0c-4a5b-bb37-98960dbdb487>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/818383\",\"WARC-Payload-Digest\":\"sha1:ZFLTWZ2P7PWUMFUCP65JTSTYO74PNDKM\",\"WARC-Block-Digest\":\"sha1:7ECEUYXSIB6NJUA5ZGDVHPNB5LIHWEHH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703519395.23_warc_CC-MAIN-20210119135001-20210119165001-00713.warc.gz\"}"}
|
http://www.learncheme.com/student-resources/interactive-module-flash-separations/interactive-module-flash-separations-intro
|
[
"Interactive Self-Study Module: Flash Separations\nOverview\n\nThis module uses screencasts and interactive simulations to explain what happens when a feed enters a flash drum and exits as two streams in vapor-liquid equilibrium. In this module, the outlet conditions are calculated, assuming the temperature and pressure is known, by using Raoult's law (ideal solution and ideal gas) and mass balances. Example problems are provided to allow the user to test themselves. We suggest using the learning resources in the following order:\n\n1. Attempt to answer the multiple choice ConcepTests before watching the screencasts or working with the simulations.\n2. Watch the screencast that describe the phase diagrams and answer the questions within the screencasts.\n3. Use the two interactive simulations to further understand the behavior of the phase diagrams.\n4. Try to solve the two example problem screencasts before watching the solution in the screencast.\nMotivation:\n\nMixtures are flashed to effect partial separations so that one or more components is enriched in the vapor phase (relative to the feed). This is referred to as flash distillation. Only binary mixtures are considered in this module to simplify the explanations, but multi-component mixtures are used in real systems.\n\nThis module is intended for Material and Energy Balances, Thermodynamics, and Separations courses.\n\nBefore studying this module, you should:\n\n• Understand how to apply Raoult's Law.\n• Be able to apply the Antoine equation to determine saturation pressure of a single component at a given temperature.\n• Be able to apply mass balances to mixtures.\n• Be able to do energy balances.\n\nAfter studying this module, you should be able to:\n1. Use mass balances and phase equilibrium equations to determine outlet compositions and flow rates for a flash drum if the outlet temperature is known.\n2. Use mass balances, phase equilibrium equations, and an energy balance to determine outlet compositions and flow rates for an adiabatic flash drum."
] |
[
null
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https://www.education.com/worksheets/fourth-grade/?q=word+problems
|
[
"# Search Printable 4th Grade Worksheets\n\n667 filtered results\n667 filtered results\nword problems\nSort by\nSpring Word Problems\nWorksheet\nSpring Word Problems\nSpring into the season with these math word problems!\nMath\nWorksheet\nSummer Word Problems\nWorksheet\nSummer Word Problems\nGet your mind going with these super summer math word problems!\nMath\nWorksheet\nSports Word Problems\nWorksheet\nSports Word Problems\nGet your head in the game with these sports math word problems!\nMath\nWorksheet\nNinja Word Problems\nWorksheet\nNinja Word Problems\nLearn the way of the math ninja with multiplication word problems! Your master mathematician will review times tables as she works through each word problem.\nMath\nWorksheet\nWorksheet\nDo some on-the-road math with our road trip themed word problems. Exercise addition, division, and logical reasoning skills with these road trip word problems.\nMath\nWorksheet\nDivision Word Problems\nWorksheet\nDivision Word Problems\nFourth graders will gain additional practice in strengthening their math skills with this worksheet featuring division word problems.\nMath\nWorksheet\nGlossary: Decimal Word Problems\nWorksheet\nGlossary: Decimal Word Problems\nUse this glossary with the EL Support Lesson: Decimal Word Problems.\nMath\nWorksheet\nWild Word Problems: Multiplication\nWorksheet\nWild Word Problems: Multiplication\nThese multi-digit multiplication word problems draw inspiration from the animal kingdom.\nMath\nWorksheet\nSimple Division Word Problems\nWorksheet\nSimple Division Word Problems\nChildren will solve seven simple division word problems in this appealing math worksheet.\nMath\nWorksheet\nWinter Fractions Word Problems\nWorksheet\nWinter Fractions Word Problems\nGet into the spirit of winter with these word problems related to fractions.\nWorksheet\nAutumn Word Problems\nWorksheet\nAutumn Word Problems\nWelcome the season with autumn-themed word problems. Your student will have the opportunity to practice addition, subtraction, multiplication, and division.\nMath\nWorksheet\nElevations: Word Problems\nWorksheet\nElevations: Word Problems\nStudents practice addition and subtraction with these elevation word problems. Students will reference back to the elevations chart in order to solve them.\nMath\nWorksheet\nWinter Word Problems\nWorksheet\nWinter Word Problems\nThese multi-step winter word problems give kids a way to flex the math muscles they've been building throughout each year of school!\nMath\nWorksheet\nElevations: Multi-Step Word Problems\nWorksheet\nElevations: Multi-Step Word Problems\nStudents will solve these easy multi-step word problems to help them out. They'll use the chart provided to solve the addition and subtraction word problems.\nMath\nWorksheet\nMixed Word Problems\nWorksheet\nMixed Word Problems\nYour fourth grader will hardly realize they are learning as they complete some entertaining mixed word problems to help out with birthday party logistics.\nMath\nWorksheet\nElevations: Challenging Word Problems\nWorksheet\nElevations: Challenging Word Problems\nYour child will hone his math skills with these 3rd grade multiple step word problems.\nMath\nWorksheet\nBeachy Word Problems\nWorksheet\nBeachy Word Problems\nTake math everywhere you go, even to the beach! Students will use subtraction, addition, and time solving skills to complete the beach-themed word problems.\nMath\nWorksheet\nMath Skills: Word Problems\nWorksheet\nMath Skills: Word Problems\nTest your fourth grader's command of general math skills by challenging him to complete a handful of word problems.\nMath\nWorksheet\nPolygon Perimeter Word Problems\nWorksheet\nPolygon Perimeter Word Problems\nIn this worksheet, students will follow four steps to find the perimeter of polygons in word problems.\nMath\nWorksheet\nVocabulary Cards: Decimal Word Problems\nWorksheet\nVocabulary Cards: Decimal Word Problems\nUse these vocabulary cards with the EL Support Lesson: Decimal Word Problems.\nMath\nWorksheet\nGlossary: Deciphering Division Word Problems\nWorksheet\nGlossary: Deciphering Division Word Problems\nUse this glossary with the EL Support Lesson: Deciphering Division Word Problems.\nMath\nWorksheet\nNinja Word Problems #2\nWorksheet\nNinja Word Problems #2\nChallenge your master multiplier with these ninja word problems. He'll use his comprehension skills to translate each problem into a numerical equation.\nMath\nWorksheet\nFunction Tables & Word Problems\nWorksheet\nFunction Tables & Word Problems\nChallenge your students to solve these word problems using function tables. Students will solve for the rule, then complete the function table.\nMath\nWorksheet\nClassroom Math: Multiplication Word Problems\nWorksheet\nClassroom Math: Multiplication Word Problems\nYour fourth grader will be challenged to complete five word problems using multiplication with two-digit numbers."
] |
[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.78483856,"math_prob":0.5751356,"size":7456,"snap":"2023-14-2023-23","text_gpt3_token_len":1744,"char_repetition_ratio":0.28247452,"word_repetition_ratio":0.2146739,"special_character_ratio":0.17596567,"punctuation_ratio":0.058277026,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9945655,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T15:26:45Z\",\"WARC-Record-ID\":\"<urn:uuid:2f30055b-b704-47a6-aed7-a50f9e62f126>\",\"Content-Length\":\"265604\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:01377e75-9990-4573-b5fb-228b4f671b35>\",\"WARC-Concurrent-To\":\"<urn:uuid:84bd6279-8803-4d3e-b8be-52212f118215>\",\"WARC-IP-Address\":\"104.18.12.139\",\"WARC-Target-URI\":\"https://www.education.com/worksheets/fourth-grade/?q=word+problems\",\"WARC-Payload-Digest\":\"sha1:6AIMDWUNSQ2CSPV3H3QTTHVNXD3A2N2L\",\"WARC-Block-Digest\":\"sha1:JXN5SG2MUVZWCO6VSGRCNSZLPHH3KUUV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943484.34_warc_CC-MAIN-20230320144934-20230320174934-00582.warc.gz\"}"}
|
https://saloontokens.info/407-sentimeter-ko-metre-mein-badlo.php
|
[
"# 407 Sentimeter Ko Metre Mein Badlo",
null,
"## What is 407 Sentimeter in Meters?\n\n407 Sentimeter is a unit of measurement that is equivalent to 1.33 meters in the metric system. It is a unit of length that is used in many countries, which have adopted the metric system of measurement. In some countries, this unit is also known as a centimeter. This unit of measurement is commonly used in the fields of engineering, construction, science and architecture. In the metric system, the prefix \"centi\" is used to denote one-hundredth of a meter. Therefore, one centimeter is equal to 0.01 meter. Thus, 407 Sentimeter is equal to 4.07 meters. This is the equivalent of 4 meters and 7 centimeters.\n\n## How to Convert 407 Sentimeter to Meters?\n\nTo convert 407 Sentimeter to Meters, simply multiply the value in Sentimeter by 0.01. This will give you the equivalent value in Meter. For example, if you want to convert 407 Sentimeter to Meters, you will multiply 407 by 0.01, which gives you 4.07 Meters. Alternatively, you can also use a conversion calculator or chart to convert between different units of measurement. While using a calculator, you would need to enter the value of 407 Sentimeter and select the unit of measurement you want to convert it to. Once you have done that, the calculator will display the equivalent value in Meters.\n\n## Why is it Important to Convert 407 Sentimeter to Meters?\n\nIt is important to convert units of measurement to ensure accuracy and consistency in the results. When using different units of measurement, it is important to convert them to the same unit so that the results are more meaningful. This is especially important in fields such as engineering, construction, architecture, science, and other applications where accurate measurements are essential. By converting 407 Sentimeter to Meters, it is easier to compare different measurements and ensure that the results are accurate. This is because Meters is the most common unit of measurement for length in the metric system, and it is easier to compare different measurements when they are all expressed in the same unit.\n\n## What are Some Other Ways to Convert Sentimeter to Meters?\n\nApart from using a calculator or conversion chart, there are other ways to convert Sentimeter to Meters. One way is to use an online conversion tool. There are several websites that offer online conversion tools that can easily convert between different units of measurement. Another way to convert Sentimeter to Meters is to use a formula. To do this, you will need to know the conversion factor for the particular unit of measurement. For example, if you want to convert 407 Sentimeter to Meters, you will need to know that 1 Sentimeter is equal to 0.01 Meters. Then you will need to multiply 407 by 0.01 to get the equivalent value in Meters.\n\n## What are the Advantages of Converting 407 Sentimeter to Meters?\n\nConverting 407 Sentimeter to Meters has several advantages. Firstly, it allows you to compare different measurements more accurately, as all measurements are expressed in the same unit. This is especially important in fields such as engineering or construction, where accurate measurements are essential. Secondly, it is easier to understand and interpret results when measurements are expressed in the same unit, as it eliminates the need to convert between different units. Finally, converting between different units of measurement is also important for data analysis and other scientific applications, as it allows for more accurate comparisons and results.\n\n## Conclusion\n\nConverting 407 Sentimeter to Meters is an important task that ensures accuracy and consistency in the results. There are several ways to convert between different units of measurement, such as using a calculator or conversion chart, an online conversion tool, or a formula. Converting units of measurement also has several advantages, such as allowing for more accurate comparisons and results. Therefore, it is important to convert 407 Sentimeter to Meters whenever needed."
] |
[
null,
"https://i.ytimg.com/vi/A2Qi73FwRhE/maxresdefault.jpg",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.9303359,"math_prob":0.9675013,"size":3939,"snap":"2023-14-2023-23","text_gpt3_token_len":808,"char_repetition_ratio":0.18653113,"word_repetition_ratio":0.11496063,"special_character_ratio":0.20233561,"punctuation_ratio":0.114754096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98127055,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-24T00:31:44Z\",\"WARC-Record-ID\":\"<urn:uuid:2750f1cb-08e8-429c-a663-2462061b171b>\",\"Content-Length\":\"54887\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3287bbd-4543-43e7-9377-87af62bb0bdf>\",\"WARC-Concurrent-To\":\"<urn:uuid:17bf1106-7f31-483c-a3cc-c069383e5261>\",\"WARC-IP-Address\":\"172.67.201.18\",\"WARC-Target-URI\":\"https://saloontokens.info/407-sentimeter-ko-metre-mein-badlo.php\",\"WARC-Payload-Digest\":\"sha1:JB2BILHZ5DKOGXKCN3EDZSZCR7QVBFSY\",\"WARC-Block-Digest\":\"sha1:7IRJD677NN5A6KKKZ46Q5Z5SIMCZSFSH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945218.30_warc_CC-MAIN-20230323225049-20230324015049-00339.warc.gz\"}"}
|
https://www.drugfuture.com/Pharmacopoeia/USP32/pub/data/v32270/usp32nf27s0_m2062.html
|
[
"Aluminum Dichlorohydrate Solution\n» Aluminum Dichlorohydrate Solution consists of complex basic aluminum chloride that is polymeric and encompasses a range of aluminum-to-chloride atomic ratios between 0.90:1 and 1.25:1. The following solvents may be used: water, propylene glycol, dipropylene glycol, or alcohol. It contains the equivalent of not less than 90.0 percent and not more than 110.0 percent of the labeled concentration of anhydrous aluminum dichlorohydrate.\nPackaging and storage— Preserve in well-closed containers.\nLabeling— Label Solution to state the solvent used and the claimed concentration of anhydrous aluminum dichlorohydrate contained therein.\nIdentification—\nTest specimen— Add about 10 mL of isopropyl alcohol to 2 g of Solution, mix, and filter. Evaporate the filtrate to about 1 mL on a steam bath. Deposit this solution on a silver chloride disk.\nStandard specimen: a similar preparation of propylene glycol.\nTest specimen— Add about 10 mL of isopropyl alcohol to 2 g of Solution, mix, and filter. Evaporate the filtrate to about 1 mL on a steam bath. Deposit this solution on a silver chloride disk.\nStandard specimen: a similar preparation of dipropylene glycol.\nD: Identification of alcohol (where stated on the label)—Mix 5 drops of Solution in a small beaker with 1 mL of potassium permanganate solution (1 in 100) and 5 drops of 2 N sulfuric acid, and cover the beaker immediately with filter paper moistened with a freshly prepared solution of 0.1 g of sodium nitroferricyanide and 0.25 g of piperazine in 5 mL of water: an intense blue color is produced on the filter paper, the color becoming paler after a few minutes.\nbetween 3.0 and 5.0, in a solution prepared by diluting 3 g of the Solution with water to obtain 10 mL.\nPrepare the Test Preparation using an accurately weighed quantity of the Solution. The limit is 2 µg per g.\nPrepare the Test Preparation using an accurately weighed quantity of the Solution. The limit is 10 µg per g.\nLimit of iron— Using Aluminum Dichlorohydrate Solution instead of Aluminum Chlorohydrate Solution, proceed as directed in the test for the Limit of iron under Aluminum Chlorohydrate Solution. The limit is 75 µg per g.\nContent of aluminum— Using Aluminum Dichlorohydrate Solution instead of Aluminum Chlorohydrate Solution, proceed as directed in the test for the Content of aluminum under Aluminum Chlorohydrate Solution. Use the result to calculate the Aluminum/chloride atomic ratio.\nContent of chloride— Using Aluminum Dichlorohydrate Solution instead of Aluminum Chlorohydrate Solution, proceed as directed in the test for the Content of chloride under Aluminum Chlorohydrate Solution. Use the result to calculate the Aluminum/chloride atomic ratio.\nAluminum/chloride atomic ratio— Divide the percentage of aluminum found in the test for Content of aluminum by the percentage of chloride found in the test for Content of chloride, and multiply by 35.453/26.98, in which 35.453 and 26.98 are the atomic weights of chlorine and aluminum respectively: the ratio is between 0.90:1 and 1.25:1.\nAssay— Calculate the percentage of anhydrous aluminum dichlorohydrate in the Solution by the formula:\nAl({26.98x + [17.01(3x",
null,
"1)] + 35.453} / 26.98x)\nin which Al is the percentage of aluminum found in the test for Content of aluminum, x is the aluminum/chloride atomic ratio found in the test for aluminum/chloride atomic ratio, 26.98 is the atomic weight of aluminum, 17.01 is the molecular weight of the hydroxide anion (OH), and 35.453 is the atomic weight of chlorine (Cl).\nAuxiliary Information— Please check for your question in the FAQs before contacting USP.\n Topic/Question Contact Expert Committee Monograph Feiwen Mao, M.S. Scientist1-301-816-8320 (MDOOD05) Monograph Development-Ophthalmics Oncologics and Dermatologicals\nUSP32–NF27 Page 1481"
] |
[
null,
"https://www.drugfuture.com/Pharmacopoeia/USP32/pub/data/images/chars/minus.gif",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8160122,"math_prob":0.88765836,"size":4026,"snap":"2019-26-2019-30","text_gpt3_token_len":1009,"char_repetition_ratio":0.18547986,"word_repetition_ratio":0.30126184,"special_character_ratio":0.2225534,"punctuation_ratio":0.11375661,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.954797,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T22:47:13Z\",\"WARC-Record-ID\":\"<urn:uuid:634496c6-d4a8-4f14-85c5-9cc584ec9cbc>\",\"Content-Length\":\"13082\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef2998b6-54e4-48a1-b439-321a85ad2d04>\",\"WARC-Concurrent-To\":\"<urn:uuid:93f5f602-51b7-4da8-882f-2fc22a4cc033>\",\"WARC-IP-Address\":\"122.114.55.184\",\"WARC-Target-URI\":\"https://www.drugfuture.com/Pharmacopoeia/USP32/pub/data/v32270/usp32nf27s0_m2062.html\",\"WARC-Payload-Digest\":\"sha1:HB743KWX2ZQHCNKKPV2CXLAMSRY3YSBV\",\"WARC-Block-Digest\":\"sha1:IW3WT4L6OYIYUPJGM2NJE5Z4V6VREGRH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998581.65_warc_CC-MAIN-20190617223249-20190618005249-00294.warc.gz\"}"}
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https://tex.stackexchange.com/questions/146782/full-screen-beamer-problems
|
[
"# Full screen Beamer problems\n\nIs it possible to do full screen which covers the whole screen? When I use full screen option for my pdf file created from beamer presentation, I see some space is being wasted on right and left side of the screen.. But as we know power point in full screen mode covers the entire screen without leaving any black spaces ...Any help would be really appreciated ..\n\n\\documentclass{beamer}\n\n\\usetheme{CambridgeUS}\n\\usecolortheme {orchid}\n\n\\setbeamercolor*{palette tertiary}{use=structure,fg=white,bg=blue}\n\n\\mode<presentation>\n\n\\usepackage{graphicx}\n\\usepackage{comment}\n\\usepackage{hyperref}\n\\usepackage {xcolor}\n\\usepackage{graphicx,epsfig,setspace,subfig,url,amsmath}\n\\usepackage{indentfirst}\n\\usepackage{listings}\n\\usepackage{array}\n\\usepackage{multirow}\n\\usepackage{multicol}\n\\usepackage{enumerate}\n\\usepackage[english]{babel}\n\\usepackage{booktabs}\n\\usepackage{etex}\n\\usepackage[all]{xypic}\n\\usepackage{color}\n\\usepackage[english]{babel}\n\\usepackage[utf8]{inputenc}\n\\usepackage{times}\n\\usepackage[T1]{fontenc}\n\n\\title[Random Sampling ] % (optional, use only with long paper titles)\n{Random Sampling\\\\ Case Study}\n\n\\pgfdeclareimage[height=1 cm, interpolate=false]{university-logo}{university-logo-filename}\n\\logo{\\pgfuseimage{university-logo}}\n\n\\setbeamercolor*{title}{bg=blue,fg=white}\n\n\\begin{document}\n\n\\section{Introduction}\n\n% FRAME 1\n\\begin{frame}{Introduction}\nRandom sampling is done to gather data about the population in order to make an inference that can be generalized to the population.\\\\\n\\vspace{12pt}\nSteps in sampling\n\\begin{itemize}\n\\item Define population $(N)$ to be sampled\n\\item Determine sample size $(n)$\n\\item Control for bias and error\n\\item Select sample\n\\end{itemize}\n\\end{frame}\n\n% FRAME 2\n\\begin{frame}{Introduction}\n\\textbf{Define population to be sampled}: - Identify the group of interest and its characteristics to which the findings of the study will be generalized.\\\\\n\\vspace{12pt}\n\\textbf{Determine sample size $(n)$}:- The size of the sample influences both the representativeness of the sample and the statistical analysis of the data.\\\\\n\\vspace{12pt}\n\\textbf{Control for bias and error}: - Be aware of the sources of sampling bias and identify how to avoid it and decide whether the bias is so severe that the results of the study will be seriously affected.\\\\\n\\vspace{12pt}\n\\textbf{Select sample}:- A process by which the researcher attempts to ensure that the sample is representative of the population from which it is to be selected.\n\\end{frame}\n\n\\end{document}\n\n• Are you testing this on your monitor or using a projector? If the former, this is normal if your monitor has a different aspect ratio from a standard LCD projector. (If it is with the projector itself, then someonr's answer is probably right.) – cfr Apr 21 '15 at 22:50\n• Off-topic but you really should clean up your preamble and the packages you are using. times is deprecated. You load several packages at least twice and you load packages which you either shouldn't load (e.g. epsfig), or which at the least you don't need to load (e.g. color, hyperref). These are just examples - not exhaustive lists. – cfr Apr 21 '15 at 22:56\n\nCould it be that the aspect ratio of the projector isn't matching the presentation? You could try changing to 16:9 by using \\documentclass[aspectratio=169]{beamer}. The right ratio depends on your projector. Btw. please try to use a MWE.\n\n\\documentclass[aspectratio=169]{beamer}\n\\usetheme{CambridgeUS}\n\\usecolortheme {orchid}\n\n\\setbeamercolor*{palette tertiary}{use=structure,fg=white,bg=blue}\n\n\\mode<presentation>\n\n\\usepackage{graphicx}\n\\usepackage{comment}\n\\usepackage{hyperref}\n\\usepackage {xcolor}\n\\usepackage{graphicx,epsfig,setspace,subfig,url,amsmath}\n\\usepackage{indentfirst}\n\\usepackage{listings}\n\\usepackage{array}\n\\usepackage{multirow}\n\\usepackage{multicol}\n\\usepackage{enumerate}\n\\usepackage[english]{babel}\n\\usepackage{booktabs}\n\\usepackage{etex}\n\\usepackage[all]{xypic}\n\\usepackage{color}\n\\usepackage[english]{babel}\n\\usepackage[utf8]{inputenc}\n\\usepackage{times}\n\\usepackage[T1]{fontenc}\n\n\\title[Random Sampling ] % (optional, use only with long paper titles)\n{Random Sampling\\\\ Case Study}\n\n\\pgfdeclareimage[height=1 cm, interpolate=false]{university-logo}{university-logo-filename}\n\\logo{\\pgfuseimage{university-logo}}\n\n\\setbeamercolor*{title}{bg=blue,fg=white}\n\n\\begin{document}\n\n\\section{Introduction}\n\n% FRAME 1\n\\begin{frame}{Introduction}\nRandom sampling is done to gather data about the population in order to make an inference that can be generalized to the population.\\\\\n\\vspace{12pt}\nSteps in sampling\n\\begin{itemize}\n\\item Define population $(N)$ to be sampled\n\\item Determine sample size $(n)$\n\\item Control for bias and error\n\\item Select sample\n\\end{itemize}\n\\end{frame}\n\n% FRAME 2\n\\begin{frame}{Introduction}\n\\textbf{Define population to be sampled}: - Identify the group of interest and its characteristics to which the findings of the study will be generalized.\\\\\n\\vspace{12pt}\n\\textbf{Determine sample size $(n)$}:- The size of the sample influences both the representativeness of the sample and the statistical analysis of the data.\\\\\n\\vspace{12pt}\n\\textbf{Control for bias and error}: - Be aware of the sources of sampling bias and identify how to avoid it and decide whether the bias is so severe that the results of the study will be seriously affected.\\\\\n\\vspace{12pt}\n\\textbf{Select sample}:- A process by which the researcher attempts to ensure that the sample is representative of the population from which it is to be selected.\n\\end{frame}\n\n\\end{document}"
] |
[
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https://answers.everydaycalculation.com/compare-fractions/72-21-and-90-10
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[
"Solutions by everydaycalculation.com\n\n## Compare 72/21 and 90/10\n\n1st number: 3 9/21, 2nd number: 9 0/10\n\n72/21 is smaller than 90/10\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 21 and 10 is 210\n2. For the 1st fraction, since 21 × 10 = 210,\n72/21 = 72 × 10/21 × 10 = 720/210\n3. Likewise, for the 2nd fraction, since 10 × 21 = 210,\n90/10 = 90 × 21/10 × 21 = 1890/210\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 720/210 < 1890/210 or 72/21 < 90/10\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
] |
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https://www.numwords.com/words-to-number/en/2180
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[
"NumWords.com\n\nHow to write Two thousand one hundred eighty in numbers in English?\n\nWe can write Two thousand one hundred eighty equal to 2180 in numbers in English\n\n< Two thousand one hundred seventy-nine :||: Two thousand one hundred eighty-one >\n\nFour thousand three hundred sixty = 4360 = 2180 × 2\nSix thousand five hundred forty = 6540 = 2180 × 3\nEight thousand seven hundred twenty = 8720 = 2180 × 4\nTen thousand nine hundred = 10900 = 2180 × 5\nThirteen thousand eighty = 13080 = 2180 × 6\nFifteen thousand two hundred sixty = 15260 = 2180 × 7\nSeventeen thousand four hundred forty = 17440 = 2180 × 8\nNineteen thousand six hundred twenty = 19620 = 2180 × 9\nTwenty-one thousand eight hundred = 21800 = 2180 × 10\nTwenty-three thousand nine hundred eighty = 23980 = 2180 × 11\nTwenty-six thousand one hundred sixty = 26160 = 2180 × 12\nTwenty-eight thousand three hundred forty = 28340 = 2180 × 13\nThirty thousand five hundred twenty = 30520 = 2180 × 14\nThirty-two thousand seven hundred = 32700 = 2180 × 15\nThirty-four thousand eight hundred eighty = 34880 = 2180 × 16\nThirty-seven thousand sixty = 37060 = 2180 × 17\nThirty-nine thousand two hundred forty = 39240 = 2180 × 18\nForty-one thousand four hundred twenty = 41420 = 2180 × 19\nForty-three thousand six hundred = 43600 = 2180 × 20\n\nSitemap"
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[
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{"ft_lang_label":"__label__en","ft_lang_prob":0.75753474,"math_prob":0.9990551,"size":1311,"snap":"2019-26-2019-30","text_gpt3_token_len":397,"char_repetition_ratio":0.33052793,"word_repetition_ratio":0.016528925,"special_character_ratio":0.40427154,"punctuation_ratio":0.024509804,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987382,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T13:45:23Z\",\"WARC-Record-ID\":\"<urn:uuid:d4a11bab-6114-4d00-8858-b8c1939f1d14>\",\"Content-Length\":\"3402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4f2ec08-0283-403c-bbab-4ce07edc0b4f>\",\"WARC-Concurrent-To\":\"<urn:uuid:43d70575-7e19-4757-9dda-ceb8931102ce>\",\"WARC-IP-Address\":\"68.66.224.32\",\"WARC-Target-URI\":\"https://www.numwords.com/words-to-number/en/2180\",\"WARC-Payload-Digest\":\"sha1:W4MHAWLDB7BRL3BS7LUPRWM6VQJHOD4M\",\"WARC-Block-Digest\":\"sha1:DNOZRUGEDAOUCHZB6YDQUM6HHSAHCN7J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998475.92_warc_CC-MAIN-20190617123027-20190617145027-00419.warc.gz\"}"}
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https://mooc.es/course/introduction-to-game-theory-and-mechanism-design/
|
[
"",
null,
"# Introduction to Game Theory and Mechanism Design\n\nPor: Swayam . en: , ,\n\nWeek 1: Introduction, the game of chess, proof of chess theorem, normal form games Week 2:Dominance, Nash equilibrium, Maxmin strategies, elimination of dominated strategies, preservation of pure Nash equilibrium (PSNE), matrix games, relation between maxmin and PSNE in matrix games Week 3:Mixed strategies, mixed strategy Nash equilibrium (MSNE), finding MSNE, MSNE characterization theorem, algorithm to find MSNE Week 4:Correlated equilibrium (CE), computing CE, extensive form games, subgame perfection, limitations of subgame perfect Nash equilibrium Week 5:Imperfect information extensive form games (IIEFG), strategies in IIEFGs, equivalence of strategies in IIEFGs, perfect recall Week 6:Equilibrium in IIEFG, game theory in practice: P2P file sharing, Bayesian games, strategy and utility in Bayesian games, equilibrium in Bayesian games Week 7:Introduction to mechanism design, revelation principle, introduction and proof of Arrow’s impossibility result, introduction to social choice setup Week 1:Introduction and proof of Gibbard-Satterthwaite theorem, domain restriction, median voter theorem Week 9:Task sharing domain, uniform rule, mechanism design with transfers, examples of quasi-linear preferences, Pareto optimality and Groves payments Week 10:Introduction to VCG mechanism, VCG in Combinatorial allocations, applications to Internet advertising, slot allocation and payments in position auctions, pros and cons of VCG mechanism Week 11:Affine maximizers, single object allocation, Myerson’s lemma, optimal mechanism design Week 12:Single and multi-agent optimal mechanism design, examples of optimal mechanisms\n\n### Plataforma",
null,
""
] |
[
null,
"https://www.facebook.com/tr",
null,
"https://mooc.es/wp-content/uploads/2019/01/Swayam.png",
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.7843776,"math_prob":0.8224979,"size":1809,"snap":"2022-40-2023-06","text_gpt3_token_len":399,"char_repetition_ratio":0.14903048,"word_repetition_ratio":0.0,"special_character_ratio":0.18794914,"punctuation_ratio":0.19292605,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9855763,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T02:23:20Z\",\"WARC-Record-ID\":\"<urn:uuid:d43381f1-92ff-47a6-8859-72af146e602e>\",\"Content-Length\":\"65087\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b6998184-7499-43b1-be2f-42d823dbc583>\",\"WARC-Concurrent-To\":\"<urn:uuid:93fd18a3-374a-4a32-bee9-0742f50a452c>\",\"WARC-IP-Address\":\"80.85.86.73\",\"WARC-Target-URI\":\"https://mooc.es/course/introduction-to-game-theory-and-mechanism-design/\",\"WARC-Payload-Digest\":\"sha1:65TEO3Z73PT5SN4HQKLIZEB2WZGJH77T\",\"WARC-Block-Digest\":\"sha1:OIWVMMVC5YHIH23JMXK2NUJ2RY4IEXOA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335059.43_warc_CC-MAIN-20220928020513-20220928050513-00544.warc.gz\"}"}
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https://bestphonefaqs.com/pi-phon/
|
[
"# What Is Pi Phon? (pi phon)\n\nWhat Is Pi Phon?\n\nIf you don’t know what Pi Phon is, you’re not alone. This little-known app is taking the world by storm, and for good reason.\n\n## What is the definition of pi phon\n\nPronunciation is the act or manner of pronouncing something. The term can refer to the actual sound produced, as in the case of articulatory phonetics, or to the manner of production, as in the case of phonotactics. It can also refer to how a word or phrase is usually spoken, or how it ought to be spoken according to some prescriptive rule.\n\n## What are the units of pi phon\n\nThe units of pi phon are the phones that make up the sound pattern of the language. They include the sounds of speech, such as the vowels and consonants, as well as the stress and intonation patterns. The phones are the smallest units of sound that can be distinguished from each other.\n\n## How is pi phon used in mathematical equations\n\nPi is a mathematical constant that is equal to the ratio of a circle’s circumference to its diameter. The symbol for pi is the Greek letter π. The value of pi is approximately 3.14159. Pi is an irrational number, which means that it cannot be expressed as a rational number (a number that can be expressed as a fraction). Pi is also a transcendental number, which means that it cannot be expressed as the root of an algebraic equation with integer coefficients. Pi is one of the most important numbers in mathematics. It appears in many formulas in all areas of mathematics, including geometry, trigonometry, and calculus.\n\n## What is the origin of the term pi phon\n\nThe term pi phon originates from the Greek letter π, which is the symbol for the ratio of a circle’s circumference to its diameter. The symbol is pronounced “pie” in English, so the term pi phon is simply a way of referring to the mathematical constant known as pi.\n\n## How did the concept of pi phon come about\n\nThe term “pi phi” was first coined by mathematicians in the 18th century, when they realized that the ratio of a circle’s circumference to its diameter was always the same, no matter what size the circle was. This constant ratio is now known as pi, and is represented by the Greek letter π.\n\nThe term “phi” comes from the fact that the circumference of a circle is also equal to π times the diameter of the circle. Thus, the ratio of a circle’s circumference to its diameter is always equal to π, no matter what size the circle is.\n\nThe value of π is approximately 3.14, but it is an irrational number, meaning that it cannot be expressed as a simple fraction. This means that it is impossible to find a rational number that is equal to π. However, mathematicians have been able to approximate its value with increasing accuracy over the years.\n\n## What is the significance of pi phon in mathematics\n\nThe mathematical constant pi, denoted by the Greek letter π, is one of the most important numbers in all of mathematics. It is the ratio of a circle’s circumference to its diameter, and it also appears in many other formulas in geometry and trigonometry. Pi is an irrational number, which means that it cannot be expressed as a rational fraction (that is, as a ratio of two integers). It is also a transcendental number, which means that it is not the root of any non-zero polynomial equation with integer coefficients.\n\nPi has been known to mathematicians for thousands of years, and it has been studied extensively. It is a fascinating number with many interesting properties. One of the most remarkable things about pi is that it is infinitely long and contains an infinite number of digits. However, despite its infinite length, it is possible to calculate pi to any desired degree of accuracy. In fact, mathematicians have calculated pi to over one trillion decimal places!\n\nThe significance of pi phon in mathematics is that it represents an important and fundamental constant in mathematics that appears in numerous formulas and equations. It is a key ingredient in many mathematical calculations and solutions, making it an essential tool for mathematicians.\n\n## What are some of the applications of pi phon\n\nPi is one of the most important mathematical constants. It is the ratio of a circle’s circumference to its diameter. Pi is also an irrational number, meaning that it cannot be expressed as a fraction p/q for any integers p and q. Despite its simplicity, pi is an incredibly versatile number with a vast array of applications in mathematics, science, and engineering.\n\nIn math, pi is used to calculate the area of a circle. This formula can be applied in physics to determine the strength of a wave or the motion of particles. In engineering, pi is used to design everything from aircraft wings to car engine cylinders.\n\nIn addition to its mathematical uses, pi also pops up in many other places in our everyday lives. For example, the value of pi appears in the measurements of musical instruments and in the decoding of radio signals. It also shows up in art, literature, and even nature.\n\nSo, next time you see pi, remember that this simple number has a lot of hidden power!\n\n## What is the history of pi phon\n\nThe history of pi is fascinating and dates back to Ancient Greece. The first recorded value of pi was by the Greek mathematician Archimedes, who approximated pi to be 3.14. However, the first accurate value of pi was not calculated until almost 2000 years later by the Chinese mathematician Zu Chongzhi, who calculated it to be 3.14159. Since then, mathematicians have continued to improve upon the accuracy of pi, with the most recent calculation being done by the Japanese mathematician Shigeru Kondo, who calculated pi to be 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609…\n\nThis seemingly never-ending number has been studied for centuries and has interesting properties that continue to be discovered. Pi is an irrational number, which means that it cannot be expressed as a rational number (a number that can be expressed as a fraction). This means that it is impossible to write down the exact value of pi, which is why it is represented by the symbol “π”. Pi is also a transcendental number, which means that it is not the root of any algebraic equation with integer coefficients. These properties make pi unique and interesting to mathematicians and laypeople alike.\n\n## What is the importance of pi phon in mathematics\n\nThe importance of pi phon in mathematics cannot be understated. Pi is one of the most important numbers in all of mathematics, and it appears in many formulas throughout the field. One of the most famous examples of where pi appears is in the formula for the circumference of a circle. This formula states that the circumference of a circle is equal to 2πr, where r is the radius of the circle. Thus, without pi, we would not be able to accurately calculate the circumference of a circle.\n\nPi also appears in many other formulas in mathematics, such as the quadratic equation and in solutions to differential equations. It is a truly essential number in the field of mathematics and its importance should not be underestimated.\n\n## What are some interesting facts about pi phon\n\nPi is the name given to the ratio of the circumference of a circle to its diameter. It is also the name given to the sixteenth letter of the Greek alphabet. The value of pi has been known for almost 4000 years and is a very important number in mathematics. Pi is an irrational number, which means that it cannot be expressed as a rational number (a number that can be written as a fraction). This means that it is impossible to find the exact value of pi and that it will continue to infinite decimal places without repeating."
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[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.96037126,"math_prob":0.9746633,"size":7957,"snap":"2023-14-2023-23","text_gpt3_token_len":1693,"char_repetition_ratio":0.15126367,"word_repetition_ratio":0.17571644,"special_character_ratio":0.24255373,"punctuation_ratio":0.089864865,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982663,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-23T11:32:26Z\",\"WARC-Record-ID\":\"<urn:uuid:aa3b5b4d-ed34-42f6-9325-98cf9efdf08e>\",\"Content-Length\":\"60462\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cd4e4394-7417-4419-b912-c3f3cfd590f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:48b49057-f291-4016-a5b9-e9e2044f3d69>\",\"WARC-IP-Address\":\"172.67.149.193\",\"WARC-Target-URI\":\"https://bestphonefaqs.com/pi-phon/\",\"WARC-Payload-Digest\":\"sha1:KQZCMRD3JDKIYSL7WFA6RGZKXHEWZSGJ\",\"WARC-Block-Digest\":\"sha1:KLDKHL53VVEIRPNNLKU6JHB4D2HGOWDM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945144.17_warc_CC-MAIN-20230323100829-20230323130829-00233.warc.gz\"}"}
|
https://oscarforner.com/blog/thread-future-and-promise/
|
[
"Oscar Forner\n\n## Thread, Future, and Promise\n\n### Date: 11/09/2016\n\nIn the C++11 standard several concurrency related classes were added. I will talk about thread, future and promise. Yes, I know there are others more useful than these three. However, I think that it is really important to know well the bases to be able to use the more complex ones properly.\n\nAs usual, I am working from an Arch Linux computer. Therefore, I can install Clang and the tools from the repository (clang). For other distributions you can find the information in the documentation.\n\nAs always, all the code used in this post is available in this repo.\n\nThe videos are made with asciinema, that means you can copy from the video.\n\n### Thread\n\nThread is the basic element of concurrency in the C++11 standard and they are intended to map one-to-one the operating system’s threads.\n\n#### Simple\n\nIn this example we use a thread to perform the task done by the function accum_up_to.\n\n``````#include <iostream>\n#include <functional>\n#include <thread>\n\nvoid accum_up_to(int value, int & result) {\nint accum = 0;\nfor (int i = 0; i <= value; ++i) {\naccum += i;\n}\nresult = accum;\n}\n\nint main()\n{\nint result;\nstd::thread t1(accum_up_to, 50, std::ref(result));\n\nt1.join();\n\nstd::cout << \"Calculated a value of: \" << result << std::endl;\n\nreturn 0;\n}\n``````\n\nSome parts to note in the code are the following:\n\n• The function that is executing the thread has a return type of void. Therefore, to get a result from that function it is required to provide a parameter to hold it.\n• Before being able to use the result stored in the parameters passed we have to make sure the thread has finished. To do so we use the join method of the thread to synchronize it.\n\n#### Two threads\n\nIn this example we use two threads that perform the same task done by the function accum_up_to.\n\n``````#include <iostream>\n#include <functional>\n#include <thread>\n\nvoid accum_up_to(int value, int & result) {\nint accum = 0;\nfor (int i = 0; i <= value; ++i) {\naccum += i;\n}\nresult = accum;\n}\n\nint main()\n{\nint result1, result2;\nstd::thread t1(accum_up_to, 50, std::ref(result1));\nstd::thread t2(accum_up_to, 40, std::ref(result2));\n\nt1.join();\nt2.join();\n\nstd::cout << \"Calculated a value of: \" << result1 << std::endl;\nstd::cout << \"Calculated a value of: \" << result2 << std::endl;\n\nreturn 0;\n}\n``````\n\nAs you can see above once the code is able to hold a thread, it is quite easy to make it hold two. However, now we have two result variables and two calls to the method join. As you can see it is not really easy to keep track once you have several threads.\n\n### Future\n\nFuture provides a way to access the result of asynchronous operations. You can imagine we use this to avoid the tedious part of getting the results from the threads.\n\n#### Simple\n\nIn this example we use a future to perform the task done by the function accum_up_to.\n\n``````#include <iostream>\n#include <future>\n\nint accum_up_to(int value) {\nint accum = 0;\nfor (int i = 0; i <= value; ++i) {\naccum += i;\n}\nreturn accum;\n}\n\nint main()\n{\nstd::future<int> f = std::async(std::launch::async, accum_up_to, 50);\n\nstd::cout << \"Calculated a value of: \" << f.get() << std::endl;\n\nreturn 0;\n}\n``````\n\nSome changes compared to the thread version:\n\n• The function returns the value directly and it can be accesed by the method get. Moreover, the method get synchronizes the future.\n• Future is gotten from the async function. This is really similar to how the threads are created.\n\n#### Two futures\n\nIn this example we use two futures that perform the same task done by the function accum_up_to.\n\n``````#include <iostream>\n#include <future>\n\nint accum_up_to(int value) {\nint accum = 0;\nfor (int i = 0; i <= value; ++i) {\naccum += i;\n}\nreturn accum;\n}\n\nint main()\n{\nstd::future<int> f1 = std::async(std::launch::async, accum_up_to, 50);\nstd::future<int> f2 = std::async(std::launch::async, accum_up_to, 40);\n\nstd::cout << \"Calculated a value of: \" << f1.get() << std::endl;\nstd::cout << \"Calculated a value of: \" << f2.get() << std::endl;\n\nreturn 0;\n}\n``````\n\nThis time the execution of two tasks in separate threads is way cleaner.\n\n### Promise\n\nPromise is where a task can deposit its result to be retrieved through a future. In other words, promise is a way to have a single point of synchronization between two threads without being the end of one of them.\n\n#### Simple\n\nIn this example we use a promise to send some information (from the user) into a future to perform the task done by the function accum_up_to.\n\n``````#include <iostream>\n#include <functional>\n#include <thread>\n#include <future>\n\nvoid accum_up_to(std::future<int>& f) {\nstd::cout << \"Started accum_up_to method\" << std::endl;\nint accum = 0;\nint value = f.get();\nfor (int i = 0; i <= value; ++i) {\naccum += i;\n}\nstd::cout << \"Calculated a value of: \" << accum << std::endl;\n}\n\nint main()\n{\nint limit;\nstd::promise<int> p;\nstd::future<int> f = p.get_future();\nstd::thread t(accum_up_to, std::ref(f));\n\nstd::cout << \"Introduce the limit:\" << std::endl;\nstd::cin >> limit;\n\np.set_value(limit);\nt.join();\n\nreturn 0;\n}\n``````\n\nSome key points from the example:\n\n• The promise provides you a future and that future is used by the accum_up_to function.\n• Once the user gives a number, the promise sends it to the future, as it is waiting for the number where the get method is called.\n• Finally, the method join from the thread is called in order to wait for it to finish the execution of the function accum_up_to.\n##### Execution\n\nIn this execution something called data race is shown. That can be seen because the execution of the same program can generate different outputs (even if the result value is always the same).\n\n### Conclusion\n\nThese are really the basics of concurrency that are provided in the C++11 standard. I know there are other important topics from concurrency such as mutex or condition variable and I will talk about them in the future because they deserve their own post. Futhermore, I know I have not talked about data races, dead locks and all the other problems you face when doing concurrency, again if people are interested in this topic I can write about it too."
] |
[
null
] |
{"ft_lang_label":"__label__en","ft_lang_prob":0.8516361,"math_prob":0.9498141,"size":5965,"snap":"2019-35-2019-39","text_gpt3_token_len":1485,"char_repetition_ratio":0.15098138,"word_repetition_ratio":0.23902439,"special_character_ratio":0.27694887,"punctuation_ratio":0.18860354,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99034095,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T18:19:13Z\",\"WARC-Record-ID\":\"<urn:uuid:fb0e0662-f65c-4a34-92a3-0a91a41766c2>\",\"Content-Length\":\"11622\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e060670c-8dbd-4f94-b984-cd41ed0709b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:68b3cbf7-ee8d-4f8d-94e1-5b1985db4847>\",\"WARC-IP-Address\":\"104.248.120.187\",\"WARC-Target-URI\":\"https://oscarforner.com/blog/thread-future-and-promise/\",\"WARC-Payload-Digest\":\"sha1:DXQJGOFH3P63T4LNPG4RCPKEYUTHNDKW\",\"WARC-Block-Digest\":\"sha1:THCCQCPQOZY2UEMDWUM7SPU37XYLVHXQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027330786.8_warc_CC-MAIN-20190825173827-20190825195827-00468.warc.gz\"}"}
|
https://electronics.stackexchange.com/questions/393906/1-byte-register-broken-into-2-nibble-outputs-not-working-vhdl-modelsim
|
[
"# 1 Byte Register broken into 2 Nibble outputs not working VHDL/ModelSim\n\nI have made a 1 byte instruction register in VHDL. Instead of having a 1 byte output, I have created an upper nibble output and a lower nibble output. The lower nibble output is special because it uses a tri state buffer. It has this because it connects to the main bus. The upper nibble does not as it feeds directly to the control circuit and does not need high impedence. For some reason, I can't get the lower nibble to output anything. The upper nibble seems to work fine.\n\nLIBRARY IEEE;\nUSE IEEE.STD_LOGIC_1164.ALL;\nUSE IEEE.NUMERIC_STD.ALL; --This is required when doing additions to STD_LOGIC_VECTORs\n\nENTITY INST_REG_SAP_1 IS\n\n--LiN is the load signal. It is active low\n--Clk is the clock for the register\n--Din is the register input bus\n--EiN is the enable signal for the low nibble.\n--HighNibble is the high nibble output\n--LowNibble is the low nibble output\n\nPORT(\nLiN, Clk, EiN: IN STD_LOGIC;\nDin: IN STD_LOGIC_VECTOR(7 DOWNTO 0);\nHighNibble: OUT STD_LOGIC_VECTOR(3 DOWNTO 0);\nLowNibble: OUT STD_LOGIC_VECTOR(3 DOWNTO 0));\n\nEND INST_REG_SAP_1;\n\nARCHITECTURE Behavioral OF INST_REG_SAP_1 IS\n\nBEGIN\nPROCESS(LiN, Clk, EiN)\nVARIABLE tempLowNibble: unsigned(3 DOWNTO 0);\n\nBEGIN\n\nIF(rising_edge(Clk)) THEN\n\nIF(LiN='0') THEN\nHighNibble<=Din(7 DOWNTO 4);\ntempLowNibble:=unsigned(Din(3 DOWNTO 0));\nEND IF;\n\nEND IF;\n\nIF(EiN='1') THEN\nLowNibble<=(OTHERS=>'Z');\nELSIF(EiN='0') THEN\nLowNibble<=Std_logic_vector(tempLowNibble);\nEND IF;\n\nEND PROCESS;\n\nEND Behavioral;\n\n\nHere is the testbench\n\nlibrary ieee;\nuse ieee.std_logic_1164.all;\nuse ieee.numeric_std.all;\n\nentity TB_INST_REG_SAP_1 is\nend TB_INST_REG_SAP_1;\n\narchitecture test of TB_INST_REG_SAP_1 is\n\n--create time constant\nconstant CLOCK_PERIOD: time:=2 us;\n\n--create signals for every port\nsignal Clk: std_logic;\nsignal LiN: std_logic;\nsignal EiN: std_logic;\nsignal Din: std_logic_vector(7 DOWNTO 0);\nsignal HighNibble: std_logic_vector(7 DOWNTO 4);\nsignal LowNibble: std_logic_vector(3 DOWNTO 0);\n\nbegin\n\ndut: entity work.INST_REG_SAP_1\nport map(Clk=>Clk,\nLiN=>LiN,\nEiN=>EiN,\nDin=>Din,\nHighNibble=>HighNibble);\n\n--simulate the clock\nClkSimulation: process\nBEGIN\n\nFOR count IN 1 TO 32 LOOP\nClk<= '0';\nwait for CLOCK_PERIOD/2;\nClk<='1';\nwait for CLOCK_PERIOD/2;\nEND LOOP;\n\nEND PROCESS ClkSimulation;\n\n--simulate the load (LiN)\nBEGIN\n\nLiN<='1';\nwait for 4 us;\nLiN<='0';\nwait for 4 us;\nLiN<='1';\nwait for 4 us;\nLiN<='0';\nwait;\n\n--simulate the Low Nibble Enable (EiN)\nenableSimulation: process\nBEGIN\n\nEiN<='0';\nwait for 20 us;\nEiN<='1';\nwait;\n\nEND PROCESS enableSimulation;\n\n--simulate the Din\nDinSimulation: process\nvariable temp: unsigned(7 DOWNTO 0);\nBEGIN\nDin<=\"10000000\";\ntemp:=\"10000000\";\nwait for 4 us;\n\nFOR count IN 1 TO 8 LOOP\ntemp:=temp+1;\nDin<=Std_logic_vector(temp);\nwait for 4 us;\nEND LOOP;\nwait;\n\nEND PROCESS DinSimulation;\n\nend architecture test;\n\n\nHere is the output. Notice no output from the lower nibble.",
null,
"• You have the ability to probe down into DUT in your simulation during debugging, where you'd find the port itself has something besides 'U's.\n– user8352\nSep 3, 2018 at 4:33\n\nIn your simulation, you forgot to map LowNibble.\n\nThis:\n\nport map(Clk=>Clk,\nLiN=>LiN,\nEiN=>EiN,\nDin=>Din,\nHighNibble=>HighNibble);\n\n\nshould be\n\nport map(Clk=>Clk,\nLiN=>LiN,\nEiN=>EiN,\nDin=>Din,\nHighNibble=>HighNibble,\nLowNibble=>LowNibble);"
] |
[
null,
"https://i.stack.imgur.com/4JaYI.jpg",
null
] |
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|
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