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http://timminsfunerals.com.au/library/a-second-course-in-stochastic-processes	[ "# A second course in stochastic processes by Samuel Karlin", null, "By Samuel Karlin\n\nThis moment direction maintains the improvement of the speculation and purposes of stochastic tactics as promised within the preface of a primary direction. We emphasize a cautious remedy of simple buildings in stochastic procedures in symbiosis with the research of typical periods of stochastic tactics coming up from the organic, actual, and social sciences.\n\nBest stochastic modeling books\n\nHandbook of statistics 19: Stochastic processes, theory and methods\n\nHardbound. J. Neyman, one of many pioneers in laying the rules of contemporary statistical thought, under pressure the significance of stochastic procedures in a paper written in 1960 within the following phrases: \"Currently within the interval of dynamic indeterminism in technological know-how, there's not often a significant piece of study, if handled realistically, doesn't contain operations on stochastic processes\".\n\nStochastic Dynamics of Reacting Biomolecules\n\nIt is a booklet in regards to the actual strategies in reacting complicated molecules, rather biomolecules. long ago decade scientists from assorted fields similar to medication, biology, chemistry and physics have amassed a major quantity of knowledge concerning the constitution, dynamics and functioning of biomolecules.\n\nAnalytical and stochastic modeling techniques and applications 16th international conference, ASMTA 2009, Madrid, Spain, June 9-12, 2009: proceedings\n\nThis booklet constitutes the refereed court cases of the sixteenth foreign convention on Analytical and Stochastic Modeling suggestions and purposes, ASMTA 2009, held in Madrid, Spain, in June 2009 along with ECMS 2009, the 23nd eu convention on Modeling and Simulation. The 27 revised complete papers offered have been conscientiously reviewed and chosen from fifty five submissions.\n\nIntroduction to Stochastic Calculus Applied to Finance (Stochastic Modeling)\n\nIn recent times the growing to be significance of by-product items monetary markets has elevated monetary associations' calls for for mathematical talents. This e-book introduces the mathematical equipment of economic modeling with transparent causes of the main worthy types. advent to Stochastic Calculus starts off with an undemanding presentation of discrete types, together with the Cox-Ross-Rubenstein version.\n\nExtra info for A second course in stochastic processes\n\nExample text\n\n1. I) and ( L , 1) 11) with L c B. B is not necessarily a function Space, though we use the notation f for one of its elements. It is assumed that + (a) i f f , € L, YE B, lim,,+m[f,-fl and llfll < C. = 0, and Ilf,ll < C for all n, then f E L 43 44 3. COMPACT MARKOV PROCESSES II-II A linear operator U from L into L is bounded with respect to both where the latter is the restriction of 1. I to L. In addition, and 1. IL, (b) H=supnBolUnI,< CQ; and (c) there is a k 2 1, an r < 1, and an R < 00 such that II UYII G r I l f II + R I f I for a l l f e L.\n\nThe cases of effectivefailure (0c > 0) and ineffectivefailure (Bc = 0) must be distinguished. The former arises more frequently in practice. 16 0. INTRODUCTION When 8c > 0, the process X,, has no absorbing states and is regular, so that the distribution of X,, converges (weakly) to a limit p that does not depend on Xo = x . Let x , be the expectation of p : lim xn = xm = n-1 m I Yp(dY), and let A,, be a subject’s proportion of A , responses in the first n trials. Then A,,, is asymptotically normally distributed as n -P co, with mean xm and variance proportional to l / n .\n\nIntuitively, x* = @(x) and e* = Y ( e ) represent simplified state and event variables. In the full ZHL model, they are projections: @ ( w , z ) = (V,Y) and Y(S, a, r ) = (a,r ) 9 where s = (B, W) or (W, B), a = br or PO, and r = B or W. We now give conditions under which Xn* = @(A',,) and En* = \"(En) are state and event sequences for a learning model. Suppose that @(u(x,e))depends only on x* and e, and that, for any A € Y, p ( x , \" - ' ( A ) ) depends only on x. 13) and p ( @ ( x ) , A * )= p ( x , Y - ' ( A * ) ) ." ]	[ null, "https://images-na.ssl-images-amazon.com/images/I/41z5DRqwUeL._SX322_BO1,204,203,200_.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8854708,"math_prob":0.9013315,"size":4367,"snap":"2019-26-2019-30","text_gpt3_token_len":1064,"char_repetition_ratio":0.11620445,"word_repetition_ratio":0.013020833,"special_character_ratio":0.24982826,"punctuation_ratio":0.13674198,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9813539,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T06:48:50Z\",\"WARC-Record-ID\":\"<urn:uuid:57e5d321-f8fa-49b9-8075-94a5f9ccf28b>\",\"Content-Length\":\"28316\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8213528b-385b-437c-a22b-8f3bea179fa6>\",\"WARC-Concurrent-To\":\"<urn:uuid:87173f2e-d664-422f-a93c-6d83b2a24f56>\",\"WARC-IP-Address\":\"111.67.4.130\",\"WARC-Target-URI\":\"http://timminsfunerals.com.au/library/a-second-course-in-stochastic-processes\",\"WARC-Payload-Digest\":\"sha1:TD7VUQVF7PM4X3EEAW2YDVOSJAH2CRYS\",\"WARC-Block-Digest\":\"sha1:BQHQCXM6O2CALFGQEHJAZJVS6KECYKNG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525094.53_warc_CC-MAIN-20190717061451-20190717083451-00116.warc.gz\"}"}
http://www.uyts.com/archives/category/python	[ "### 1. 软件使用截图", null, "### 2. 软件下载安装\n\n#### 2.1 Tcl运行环境下载安装\n\npage软件的运行需要基于Tcl环境，所以先安装Tcl是必须的。\n\nActiveTcl-8.6.9.8609.2-MSWin32-x64-5ccbd9ac8\n\n#### 2.2 page5.0.3下载安装\n\nhttp://page.sourceforge.net/\n\npage-5.0.3\n\n### 3. 软件使用\n\n#### 3.1 打开软件", null, "#### 3.2 拖拽布局", null, "#### 3.3 生成代码", null, "#### 3.4 保存代码\n\n3.3保存tcl文件后，就会弹出一个新页面，这个新页面中就是对应的自动生成的python程序代码：", null, "#### 3.5 代码修改\n\n``import page_support``\n\n``page_support.init(root, top)``\n\n### 1. 安装Python链接MySQL插件。\n\n``\\$ pip3 install PyMySQL``\n\n### 2. 完整代码如下：\n\n``````import os\nimport pymysql\n\nsku = []\n\n#定义递归函数findPNG，寻找当前目录下所有png类型的文件\ndef findPNG(sku,dir):\nfor ffName in os.listdir(dir):\nif os.path.isfile(dir+\"/\"+ffName):\nif ffName.split(\".\") == \"png\":\nsku.append(ffName.split(\".\"));\nelif os.path.isdir(dir+\"/\"+ffName):\nfindPNG(sku,dir+\"/\"+ffName)\nfindPNG(sku,\"./\")\n\nh = open(\"./ProductsInformation.csv\",\"a\")\ndb = pymysql.connect(\"地址\",\"用户\",\"密码\",\"数据库名\" )\ncursor = db.cursor()\n\nfor s in sku:\nSQL = '''SELECT t1.model,t2.name,t2.tag\nFROM `product` AS t1, `product_description` AS t2\nWHERE t1.product_id = t2.product_id AND t1.model =\"{}\"'''.format(s)\ncursor.execute(SQL)\ndata = cursor.fetchone()\nif data:\nfor d in data:\nh.write('\"'+d+'\",')\nh.write('\\n')\nelse:\nprint(s)\n\nh.close()\ndb.close()``````\n\n### 答案\n\n#### 3.1\n\n``````import json\nfileURL = \"population.json\"\nr = open(fileURL,\"r\")\nfor country in popData:\nif country[\"Country\"] == \"China\":\ncountry = json.dumps(country,indent=1)\nprint(country)``````\n\n#### 3.2\n\n``````import csv\nfileURL = \"population.csv\"\nwith open(fileURL) as f:\nfor row in f_csv:\nif row == \"China\":\nprint(row)``````\n\n### Python文件IO练习\n\n#### 2.1 创建文件夹和文件\n\nTxt文件名为1-20.txt，txt内的内容为当天时间+文件名。\n\n### 答案\n\n#### 2.1\n\n``````import os\nimport datetime\nfolderUrl = \"fileset\"\nif not os.path.exists(folderUrl):\nos.mkdir(folderUrl)\nfor i in range(1,21):\nf = open(\"fileset\\\\\"+str(i)+\".txt\",\"w\")\nf.write(str(datetime.datetime.now())+\" \"+str(i)+\".txt\")\nf.close()``````\n\n#### 2.2\n\n``````import os\nfolderUrl = \"fileset\"\nfileSet = os.listdir(folderUrl)\nw = open(\"all.txt\",\"w\")\nfor fileName in fileSet:\nr = open(folderUrl + \"\\\\\" + fileName, \"r\")\nr.close()\nw.close()\nr = open(\"all.txt\",\"r\")\nfor line in r:\nprint(line)\nr.close()``````\n\n#### 2.3\n\n``````import os\ndef deleteDir(folderURL):\nfor fileName in os.listdir(folderURL):\nos.remove(folderURL+\"\\\\\"+fileName)\nos.rmdir(folderURL)\n\ndeleteDir(\"fileset\")``````\n\n### 答案：\n\n#### 1.1\n\n``````import random\nlistVar = []\nfor i in range(0,20):\nlistVar.append(random.randint(1,1000))\nlistVar.sort()\nfor i in range(0,20):\nprint((str(listVar[i])).rjust(4))``````\n\n#### 1.2\n\n``````import random\nimport string\nvarList = []\nfor i in range(0,20):\nrandomChar = string.ascii_lowercase\nvarList.append(str(random.randrange(1,1000)) + random.choice(randomChar))\nvarList.sort()\nvarTuple = tuple(varList)\nfor i in range(1,20):\nprint(varTuple[i].rjust(5))``````\n\n#### 1.3\n\n``````import random\nimport string\nrandomDict = {}\nwhile 1==1:\nkey = random.randint(1,1000)\nvalue = random.choices(string.ascii_uppercase,k=4)\nrandomDict[key] = \"\".join(value)\nif len(randomDict) == 20:\nbreak\nfor key in sorted(randomDict):\nprint(str(key).rjust(4) + \"->\" + randomDict[key])``````\n\n#### 1.4\n\n``````import random\naSet = set()\nbSet = set()\nwhile 1==1:" ]	[ null, "http://www.uyts.com/wp-content/uploads/2020/03/image-1584851181399.png", null, "http://www.uyts.com/wp-content/uploads/2020/03/image-1584853587727.png", null, "http://www.uyts.com/wp-content/uploads/2020/03/image-1584853792309.png", null, "http://www.uyts.com/wp-content/uploads/2020/03/image-1584853828363.png", null, "http://www.uyts.com/wp-content/uploads/2020/03/image-1584854036727.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.9503964,"math_prob":0.93564063,"size":1083,"snap":"2021-21-2021-25","text_gpt3_token_len":778,"char_repetition_ratio":0.074142724,"word_repetition_ratio":0.0,"special_character_ratio":0.21514311,"punctuation_ratio":0.23033708,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9920953,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T23:01:38Z\",\"WARC-Record-ID\":\"<urn:uuid:54d7e048-d3eb-4a25-a81a-4d69560e7625>\",\"Content-Length\":\"44393\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:733e5332-b71c-42df-b3da-7ea62e258c67>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee2999b7-d5a9-407a-8422-b7e3c7aaa1c4>\",\"WARC-IP-Address\":\"122.114.55.179\",\"WARC-Target-URI\":\"http://www.uyts.com/archives/category/python\",\"WARC-Payload-Digest\":\"sha1:LZM6WJF5ZBLK5B5JJ6YAAHSE3FME4VAN\",\"WARC-Block-Digest\":\"sha1:4PU3NTQNIF52C5O6VE4IHJJIN3J5VZHT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243990419.12_warc_CC-MAIN-20210511214444-20210512004444-00192.warc.gz\"}"}
https://www.wikihow.com/Calculate-Lotto-Odds	[ "# How to Calculate Lotto Odds\n\nEveryone's heard comparisons between the odds of winning the lottery and the odds of other unlikely events, like getting struck by lightning. It's true, the odds of winning the jackpot on a game like Powerball or another pick-6 lottery game are incredibly low. But just how low are they? And how many times would you have to play to have a better chance of winning? These answers can be found down to the exact odds with some simple calculations.\n\n### Method 1 of 3: Calculating Powerball Jackpot Odds\n\n1. 1\nUnderstand the calculations involved. To find the odds of winning any lottery, divide the number of winning lottery numbers by the total number of possible lottery numbers. If the numbers are chosen from a set and the order of the numbers doesn't matter, use the formula ${\\frac {n!}{r!(n-r)!}}$", null, ". In the formula, n stands for the total number of possible numbers and r stands for the number of numbers chosen. The \"!\" denotes a factorial, which for any integer n is n(n-1)(n-2)...and so on until 0 is reached. For example, 3! represents $3\\times 2\\times 1$", null, ".\n• For a simple example, imagine you have to choose two numbers and you can pick numbers from 1 to 5. Your odds of choosing the two \"correct\" numbers (the winning numbers) would be defined as ${\\frac {5!}{2!\\times 3!}}$", null, ".\n• This would then be solved as ${\\frac {5\\times 4\\times 3\\times 2\\times 1}{2\\times 1\\times 3\\times 2\\times 1}}$", null, ", which is $120\\div 12$", null, ", or 10.\n• So, your odds of winning this game are 1 in 10.\n• Factorial calculations can get unwieldy, especially with large numbers. Most calculators have a factorial function to ease your calculations. Alternately, you can type the factorial into Google (as \"55!\" for example) and it will solve it for you.\n2. 2\nEstablish the lottery's rules. The majority of Mega Millions, Powerball, and other large lotteries use roughly the same rules: 5 or 6 numbers are chosen from a large pool of numbers in no particular order. Numbers may not be repeated. In some games, a final number is chosen from a smaller set of numbers (the \"Powerball\" in Powerball games is an example). In Powerball, 5 numbers are chosen from 69 possible numbers. Then, for the single Powerball, one number is chosen from a set of 26 possible numbers.\n• Other games may have you choose 5 or 6 numbers, or more, from a larger or smaller pool of numbers. To calculate the odds of winning, you simply need to know the number of winning numbers and the total number of possible numbers.\n3. 3\nInput the numbers into the probability equation. The first part of Powerball odds determines the number of ways 5 numbers could be chosen out of 69 unique numbers. Using Powerball rules, the completed equation for the first 5 numbers would be: ${\\frac {69!}{5!(69-5)!}}$", null, ", which simplifies to ${\\frac {69!}{5!\\times 64!}}$", null, ".\n4. 4\nCalculate your odds of choosing correctly. Solving this equation is best done entirely in a search engine or calculator, as the numbers involved are inconvenient to write down between steps. The result tells you there are 11,238,513 possible combinations of 5 numbers in a set of 69 unique numbers. This means that you have a 1 in 11,238,513 chance of choosing the five numbers correctly.\n• To calculate your odds of choosing the final Powerball correctly, you would complete the same equation using the values for the Powerball (1 number out of 26 possible numbers). Since you're only picking 1 number here, you don't necessarily have to complete the entire equation. The answer will be 26 because there are 26 different ways 1 number can be chosen from a set of 26 unique numbers.\n5. 5\nMultiply to calculate your odds of winning the jackpot. To calculate the odds that you'll guess the first 5 numbers and the Powerball correctly to win the jackpot, multiply the odds that you'll guess the first 5 numbers (1 in 11,238,513) by the odds that you'll guess the Powerball correctly (1 in 26). Your equation would be ${\\frac {1}{11,238,513}}\\times {\\frac {1}{26}}$", null, ".\n• So, your odds of choosing the first five numbers and the Powerball correctly and winning the jackpot are 1 in 292,201,338.\n\n### Method 2 of 3: Determining Lesser Prize Odds\n\n1. 1\nCalculate your odds of winning the second prize. To return to the Powerball game, you have 5 numbers and a single Powerball. If you guess all 5 of the other numbers correctly but don't get the Powerball, you'll win the second prize. If you calculated your odds of winning the jackpot, you already know that your odds of guessing all 5 numbers correctly are 1 in 11,238,513.\n• To win the second prize, you would have to guess the Powerball incorrectly. If you calculated your odds of winning the jackpot, you know that your odds of guessing the Powerball correctly are 1 in 26. Therefore, your odds of guessing the Powerball incorrectly are 25 in 26.\n• Use the same equation with these values to determine your odds of winning the second prize: ${\\frac {1}{11,238,513}}\\times {\\frac {25}{26}}$", null, ". When you complete this calculation, you'll see that your odds of winning the second prize are 1 in 11,688,053.52.\n2. 2\nUse an expanded equation to find your odds for other prizes. To win other prizes, you guess some, but not all, of the winning numbers correctly. To figure out your odds, use an equation in which \"k\" represents the numbers you choose correctly, \"r\" represents the total numbers drawn, and \"n\" represents the number of unique numbers the numbers will be drawn from. Without numbers, the formula looks like this: ${\\frac {r!}{k!\\times (r-k)!}}\\times {\\frac {(n-r)!}{((n-r)-(r-k))!\\times (r-k)!}}$", null, ".\n• For example, you might use the Powerball values to determine your odds of correctly guessing 3 of the 5 chosen numbers from the set of 69 unique numbers. Your equation would look like this: ${\\frac {5!}{3!\\times (5-3)!}}\\times {\\frac {(69-5)!}{((69-5)-(5-3))!\\times (5-3)!}}$", null, "• The result of this equation tells you the number of ways that 3 numbers can be chosen correctly out of 5 numbers. Your odds will be that number out of the total number of ways 5 numbers can be chosen correctly.\n3. 3\nSolve your equation to find the odds of correctly guessing the numbers. Just as with the base equation, this equation is best solved by typing the entire thing into a calculator or search engine. Some intermediate numbers involved in the calculation would be cumbersome to write down and it would be easy to make a mistake.\n• In the previous example, your odds of guessing 3 of the 5 chosen numbers in Powerball would be 20,160 in 11,238,513.\n4. 4\nMultiply the result by the Powerball value to determine your odds of winning that prize. While this formula gives you the odds of guessing only some of the numbers correctly, you still haven't factored in the Powerball. To find your true odds, multiply the result by your odds of getting the Powerball number correct or incorrect (whichever value you want to find).\n• For example, if you wanted to calculate your odds for getting only 3 of the 5 numbers correct and getting the Powerball incorrect, your equation would be ${\\frac {20,160}{11,238,513}}\\times {\\frac {25}{26}}$", null, ", or 1 in 579.76.\n• On the other hand, your odds for getting 3 of the 5 numbers correct and getting the Powerball correct would be ${\\frac {20,160}{11,238,513}}\\times {\\frac {1}{26}}$", null, ", or 1 in 14,494.11.\n5. 5\nChange the number of correctly guessed numbers for other prizes. Once you have the formula down, simply change the value of \"k\" to find the odds of winning different levels of prizes. Generally, your odds of winning will decrease as the value of \"k\" increases.\n• If you're calculating odds for Powerball or a similar game, don't forget to multiply your result by the Powerball value.\n\n### Method 3 of 3: Calculating Other Lotto Odds\n\n1. 1\nFind the expected return of a lottery ticket. The expected return tells you what you could theoretically expect to get back in return for buying a single lottery ticket. To calculate the expected return of a single ticket, multiply the odds of a particular payout by the value of that payout. If you did this with every possible prize you could win, you would get a range of expected returns.\n• To return to the Powerball example, the expected return of a single $2 ticket would be around$1.79 at the high end and as little as $1.35 at the low end. • Keep in mind that \"expected return\" is a term of art used in statistics. Your actual payout will almost always be much less than the expected return you calculate. 2. 2 Compare the cost of a single ticket to its expected return. You can determine the expected benefit of playing the lottery by comparing the expected return of a ticket to the cost of a ticket. Most of the time, the expected return will be lower than the cost of the ticket. Additionally, your actual return will likely differ greatly from the expected value. You'll typically only get a fraction of the expected value, if anything at all. • Calculating the odds can help you determine which lottery games have the best expected benefit. For example, at one time, the New York Lottery had a$1 Take Five ticket with an expected value that equaled its cost. If you played this game, you could expect to break even over time.\n3. 3\nDetermine the increase in odds from playing multiple times. Playing the lottery multiple times increases your overall odds of winning, however slightly. It's easier to envision this increase as a decrease in your chance of losing.\n• For example, if your overall chances of winning are 1 in 250,000,000, your chances of losing on one play are $249,999,999\\div 250,000,000$", null, ", which is equal to a number very close to 1 (0.99999...).\n• If you play twice, that number is squared ($(249,999,999\\div 250,000,000)^{2}$", null, "), representing a movement slightly away from 1 (and therefore a better chance of winning).\n4. 4\nFind the number of plays needed for decent odds of winning. Most lottery players are convinced that if they play often enough, they will significantly increase their chances of winning. It is true that playing more increases your odds of winning. However, it takes a long time for that increased chance to become significant.\n• For example, if you had a 1 in 250,000,000 chance of winning on one play, it would take roughly 180 million plays to reach 50-50 odds of winning.\n• At this rate, if you bought ten tickets a day for 49,300 years, you would have a 50 percent chance of winning.\n• Additionally, if you finally reached 50-50 odds, you still wouldn't be guaranteed a win if you bought two tickets on that day. Your overall odds of winning would still remain roughly 50% for each of those tickets.\n\n## Community Q&A\n\nSearch\n• Question\nWhat are the odds of selecting 4 numbers from 5 selections in a 59 number draw?", null, "There are 554 ways to get 4 of 5 numbers to match (5 ways to choose which number doesn't match times 54 wrong numbers it could be if it doesn't match). So there are 270 tickets that could match exactly 4 out of 5 numbers, not counting the one that matches all 5. Compare that with the overall number of tickets available, which is C(59,5) or 59585756*55/120, and you get odds of about 1:18541.\n• Question\nWhat are the odds of picking 4 numbers from 28 and from 32?", null, "For the first scenario, it is 1(28x28x28x28) as you have a 1 in a 28 chance of picking one number, and you do this 4 times. The same applies to the second scenario except change the 28s to 32s.\n• How do I calculate odds for the Irish Lotto?\n200 characters left\n\n## Tips\n\n• Any set of numbers has exactly the same odds as any other set. 32-45-22-19-09-11 is no different than 1-2-3-4-5-6.\nThanks!\n\nSubmit a Tip\nAll tip submissions are carefully reviewed before being published\nThanks for submitting a tip for review!\n\n## Warnings\n\n• Don't gamble more than you can afford to lose.\nThanks!\n• If you think you have a problem with gambling, you probably do. Gamblers Anonymous is a good source of information and help for those afflicted with gambling addiction.\nThanks!\n• Don't fall for lottery scams where somebody tells you they have a sure-fire way of winning. If someone had a guaranteed, sure-fire way of winning, it would be self-defeating to tell you about it.\nThanks!\n\n## Support wikiHow's Educational Mission\n\nEvery day at wikiHow, we work hard to give you access to instructions and information that will help you live a better life, whether it's keeping you safer, healthier, or improving your well-being. Amid the current public health and economic crises, when the world is shifting dramatically and we are all learning and adapting to changes in daily life, people need wikiHow more than ever. Your support helps wikiHow to create more in-depth illustrated articles and videos and to share our trusted brand of instructional content with millions of people all over the world. Please consider making a contribution to wikiHow today.", null, "Co-authored by:\nThis article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 505,647 times.\nCo-authors: 48\nUpdated: June 1, 2020\nViews: 505,647\nArticle SummaryX\n\nTo calculate your odds of winning the lottery, use the formula: factorial of n over factorial of r times factorial of n minus r, where n is the total number of possible numbers and r is the number of numbers chosen. For example, if you're playing a lottery where you can choose 2 numbers from a pool of 5 numbers, the formula would be: factorial of 5 over factorial of 2 times factorial of 3, which equals 120 over 12. 120 divided by 12 gives you 10, so your odds of winning would be 1 out of 10. To learn how to calculate other kinds of lotto odds, keep reading!" ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/f07c5a08ac6f113935b45e7c98207c382448bd91", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/3f457d082915b71295522b0ea8c3ee6c8f6b287b", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/67c7a47e7c221f75b3c51b8b5b28057af87799af", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/f8f26330356c04b0ae9f7fdfd0c3b0aadf6d7c4a", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/c3e4b95148609dc0444a0418352116e3b4804201", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/742705444a06cd29798f5f47089444faa7b992ad", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/88251ee62ee6183abaa630993892a0421b113fac", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/a7b0c39aa861de73b4f638249c4d81905eea13d3", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/a77c516992cd41ce12e9b232fdb9a87ae79e38ea", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/2837b5b421805213fe93dacfb2562a904486150f", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/f6c6f3e11c6a8a36d6adc2d981b96aaeb6571c67", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/7bbb7737f4b2e0487f1930c7cfd8dc5e2bd8e622", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/f3b0107550299f7ee11b3ec7e2f04371ee0de2ad", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/ec71d0411ea744f018ca2c1b93ec19f672c69971", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/0f05685cafc2940958c20bc4c3a506ac8d76a02b", null, "https://www.wikihow.com/images/thumb/f/f5/CommunityAvatar4.png/-crop-104-104-104px-CommunityAvatar4.png", null, "https://www.wikihow.com/images/thumb/f/f5/CommunityAvatar4.png/-crop-104-104-104px-CommunityAvatar4.png", null, "https://www.wikihow.com/images/thumb/7/7c/WHInitials.png/-crop-100-100-100px-WHInitials.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9197112,"math_prob":0.9841287,"size":14143,"snap":"2020-45-2020-50","text_gpt3_token_len":3355,"char_repetition_ratio":0.17752317,"word_repetition_ratio":0.057995737,"special_character_ratio":0.24223998,"punctuation_ratio":0.119561374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9894268,"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],"im_url_duplicate_count":[null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,4,null,4,null,5,null,5,null,5,null,5,null,7,null,7,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-24T18:37:01Z\",\"WARC-Record-ID\":\"<urn:uuid:b0055780-5507-42e5-876a-abfebc64d57b>\",\"Content-Length\":\"325680\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08ad4a2e-a8b3-456d-bf10-a4510b547fe0>\",\"WARC-Concurrent-To\":\"<urn:uuid:21333cc9-4e67-40e7-b5aa-34e522dfe89a>\",\"WARC-IP-Address\":\"199.232.66.110\",\"WARC-Target-URI\":\"https://www.wikihow.com/Calculate-Lotto-Odds\",\"WARC-Payload-Digest\":\"sha1:HM6MX73GRRTDFTCNDHRNLAQJKRS2KPUF\",\"WARC-Block-Digest\":\"sha1:ZMNRHXLKXNQK52SO3Q7ZFSYIB222A4OL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107884322.44_warc_CC-MAIN-20201024164841-20201024194841-00095.warc.gz\"}"}
https://www.thenatphil.com/post/spacetime-according-to-string-theory	[ "Search\n\n# Spacetime According to String Theory\n\nBy Alice Blake\n\nHave you ever thought about whether the universe is united by fundamental laws of physics? Whether anything unites how atoms interact and how planets and solar systems interact? Most things we see and observe, such as throwing a ball or driving a car can be described by Newtonian physics, which is sometimes referred to as classical mechanics. On a large scale, such as describing gravity on a planetary scale, the universe is governed by general relativity. General relativity, discovered by Albert Einstein, describes the gravitational effect that is warped spacetime. It is often described with the example of a trampoline. Imagine the fabric of spacetime is the surface of a trampoline. When you put a bowling ball on the trampoline, the fabric bends and subsequent items are drawn towards it. On a small scale, the world is adequately described by quantum mechanics, which deals with how small particles interact with each other. Quantum mechanics is a probabilistic theory which explains, with decent precision how particles interact. It includes the idea of wave-particle duality, which is that some things, especially light, can behave as both a wave and a particle.\n\nString theory is an attempt at a unified theory between quantum mechanics and general relativity. It was a paradigm shift in physics that gained traction in the 1980s and the 1990s. Physicists who specialize in string theory continue to develop the theory, which claims that there are more dimensions than those evident to the human eye. We see four dimensions: the three spatial dimensions and time.\n\nPrior to “Spacetime in String Theory: A Conceptual Clarification” by Keizo Matsubara and Lars-Goran Johansson, string theorists have claimed that there are 11 dimensions, 10 dimensions of space and 1 dimension of time. Six spatial dimensions form manifolds, which is a small, compact space that serves to hide dimensions. Because we, as humans, are so large in comparison to these manifolds, we see the world in only 3 spatial dimensions. Below is a picture of a manifold:", null, "(Lunch 2007)\n\nAccording to the math behind string theory, there are actually 26 dimensions, but 16 of those dimensions are commonly interpreted as degrees of freedom, rather than dimensions (Matsubara et al). Degrees of freedom can be thought of as a direction you can move in, which bears similarities to dimensions. The remaining 6 hidden dimensions are hidden, and we perceive 3 spatial dimensions and time. However, according to this paper, there is not enough evidence that string theory demands a specific number of extra dimensions. Based on what we know about the universe, there are extra dimensions, yet it is not yet possible to determine exactly how many extra dimensions there are. Currently, there are multiple mathematical representations of spacetime.\n\nSince the important concept of this paper is that there are indeterminate hidden dimensions, an important aspect to understand is how dimensions are hidden. The aforementioned methods are manifolds and d-branes (Matsubara et al). A manifold is a small space that was initially invented as a way to represent dimensions. A d-brane is something that holds down the end of an open string, since the strings in string theory can either be an open loop or a closed loop.\n\nThis paper set out to clarify the concepts of how spacetime is defined by string theory. Since this is a realm of theoretical physics, there was no experiment done to determine a result. Instead, theoretical physics develops theories through math. One of the important messages from the paper was the conceptual flaws of how physicists have thought about manifolds. Historically, manifolds have been interpreted as a representation of an extra dimension. However, since the scale that manifolds exist on is so small, the classical way of thinking about manifolds does not work.\n\nFinally, string theory has been an important paradigm shift in physics since it has been a breakthrough in explaining how quantum mechanics and general relativity work. It definitively states that there are more than the 4 dimensions that are eminently visible to people. The dimensions that are not apparent are hidden in manifolds.\n\nWorks Cited\n\nLunch. (2007). Calabi-Yau [picture] Wikipedia. https://en.wikipedia.org/wiki/File:Calabi-Yau.png\n\nMatsubara, Keizo and Johansson, Lars-Goran. (2018 July 25). Spactime in String Theory: A Conceptual Clarification. Journal for General Philosophy of Science, 49:333-353. https://doi.org/10.1007/s10838-018-9423-2(0123456789().,-volV)(0123456789().,-volV)" ]	[ null, "https://static.wixstatic.com/media/bcb7db_96e0f12c681b43399975539e24072f5a~mv2.png/v1/fill/w_840,h_840,al_c/bcb7db_96e0f12c681b43399975539e24072f5a~mv2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94254005,"math_prob":0.95435756,"size":4609,"snap":"2021-43-2021-49","text_gpt3_token_len":972,"char_repetition_ratio":0.13528773,"word_repetition_ratio":0.0028089888,"special_character_ratio":0.2045997,"punctuation_ratio":0.121605664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9718479,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T14:55:52Z\",\"WARC-Record-ID\":\"<urn:uuid:50d27761-d426-4a5d-8289-b1c9dd8d2ea9>\",\"Content-Length\":\"969871\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b4146c4-f22c-4984-9285-86ec9b4ecc11>\",\"WARC-Concurrent-To\":\"<urn:uuid:73cff1f6-5440-4d6f-b364-20e303a352ee>\",\"WARC-IP-Address\":\"34.117.168.233\",\"WARC-Target-URI\":\"https://www.thenatphil.com/post/spacetime-according-to-string-theory\",\"WARC-Payload-Digest\":\"sha1:4UNNSLMUQPGMY7G2XOH2IESNXCNZBBLQ\",\"WARC-Block-Digest\":\"sha1:XPWBOHD2S6654UFIN5PQUDBRUM5Q5NDL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363400.19_warc_CC-MAIN-20211207140255-20211207170255-00025.warc.gz\"}"}
http://conorobrien-foxx.github.io/Jolf/explanation.html	[ "# Jolf\n\ninterpreter\nI use marked, along with some code I put together for this page. Enjoy!\n# 1. What is prefix language? There are three main types of operator-data placement schemes: Prefix, infix, and postfix. Infix is the one we are all familiar with, in which the operator is placed in between the operands, as in `5 + 7`. Infix is found in, erm, _practical_ programming langauges. Postfix is the scheme in which the operators follow the operands, as in `5 7 +`. Postifx schemes are found in stack-based languages, such as [><>](https://esolangs.org/wiki/Fish) and [Vitsy](https://github.com/VTCAKAVSMoACE/Vitsy). Lastly is prefix, in which the operators precede the operands, as in `+ 5 7`. The following statements are all equivalent in prefix, infix, and postfix respectively: ^ 4 / + 5 9 2 4 ^ ((5 + 9) / 2) 4 5 9 + 2 / ^ Being a prefix langauge, Jolf can easier be thought of as a sort of functional scheme, in which each command represents a function, with its arguments following afterwords. Say you have the JavaScript program `alert(3 + 5)`; then, if you think if `+` as a function `add`, you could re-write that as `alert(add(3,5))`. Since Jolf has a command for both `alert` and `add`, this is rewritten as `a+35`. This is the conversion process: alert(3+5) alert(add(3,5)) alert( + 3 5 ) a + 3 5 a+35 # 2. Commands This is an index of the commands. (One can form a two-byte command by prefixing an instruction with a tilde (`~`).) ## 2.1. Function instructions These instructions are the operators of Jolf. \"`typeA, typeB` (&)\" means the order does not matter. All ranges are integer ranges, unless they have a specified step. * `` - multiplies 2 arguments `number, number` - multiplies the two numbers * `string, number` (&) - repeats `string` `number` times * `array, number` - creates a 2D array of length `number` that contains `array` as all of its members * `+` - adds 2 arguments, according to their types: * `number, number` - adds the two numbers * `anything, string` or `string, anything` - returns `string` concatted with `anything` (preserving order) * `array, array` - concats the two arrays * `non array, array` - Puts `non array` at the beginning of `array` * `array, non array` - pushes `non array` to `array` * `set, set` - the set union of the two sets * `-` - subtracts 2 arguments (behaviour TBA) * `.` - takes 2 arguments `O` and `p` and returns `O[p]` * `/` - takes 2 arguments `A` and `B` and returns `A/B` (behaviour TBA) * `<` - takes 2 arguments `A` and `B` and returns `A < B` (JS behaviour) * `=` - takes 2 arguments `A` and `B` and returns `A == B` * `>` - takes 2 arguments `A` and `B` and returns `A > B` (JS behaviour) * `@` - returns the character code of the next character * `A` - takes 3 arguments `r`, `s`, and `t`; replaces each character in `s` found in `r` with the respective character found in `t` * `B` - takes 1 argument `N` and converts it to binary as a string (equiv. `(N).toString(2)`) * `C` - takes 2 arguments `A` and `B` and parses the string `A` as an integer in base `B` (equiv. `parseInt(A,B)`) * `O` - takes the product of 1 argument, depending on its type: * `array` - takes the product of the array * `P` - takes 1 argument `A` and returns `Number(A)` (converts all to number, if possible) (conversion an array into a number, as in `[1,2,3,1,0,2] => 123102`) * `Q` - squares 1 argument `S`, according to its type: * `number` - multiplies it by itself * `array` - takes the length of the array `N` and returns an array containing `N` copies of itself * `string` - TBA * `T` - takes 2 arguments `F` and `s`; surrounds `F` expression (after compilation) with quotes and evaluates that action `F` every `s` milliseconds * `R` - takes 2 arguments `F` and `s`; runs `F` after `s` milliseconds * `V` - takes 1 argument `S` and is the value `V` of `S`; if `S` is a number, returns `S`'s parity * `^` - takes 2 arguments `A` and `B` and retursn `A^B` (exponentiation) (behaviour TBA) * `_` - negates 1 argument, according to its type: * `string` or `array` or `set` - reverses the entity * `number` - negates the entity * `a` - outputs (alerts/writes) 1 argument * `b` - takes an integer `N` and a base `b` and returns `(N).toString(b)`; when `b = 1`, this is equal to a string of length `N` filled with `#`s. * `c` - writes 1 argument to the log (`console.log` at the moment) * `e` - takes 1 argument `S` and evaluates it as a new Jolf program; returns nothing at the moment * `f` - takes 2 arguments `F` and `R` as a function and takes the array `R` as its commands. * `p` - takes 3 arguments `a`, `b`, and `s` and creates a range `[a,b)` with a step `s` * `r` - takes 2 arguments `a` and `b` and creates a range `[a,b)` (inclusive of `a` and exclusive of `b`) * `u` - takes 1 argument * `array` or `set` - returns the sum of the entity * `string` - behaviour TBA * `number` - takes the digit sum of the number (e.g. `u(140) = 1 + 4 + 0 = 5`) * `v` - takes 2 arguments `S` and `V`; sets the value of variable `S` to `V` and creates `S` if nonexistant (returns `V`) * `z` - takes 1 argument and returns a range `[1,a]` * `~i` - takes 1 argument and returns that argument (the identity function) ## 2.2. Data commands These instructions are the data/arguments of Jolf. ### 2.2.1. As arguments These are written as arguments. * `0-9` - that number * `q` - writes the source code's reference to the the compiled code * `i` and `I` - receive string input from `prompt` * `j` and `J` - receive numeric input from `prompt` * `E` - the empty string `\"\"` * `Y` - the empty array `[]` * `~e` - Euler's number approx `2.718281828459045` * `~p` - π approx `3.141592` * `~P` - ϕ, `(1+sqrt)/2` ### 2.2.2. As writing These are simply written to the compiled code. * `~A` - writes `Array` to the compiled code * `~M` - writes `Math` to the compiled code * `~N` - writes `Number` to the compiled code * `~S` - writes `String` to the compiled code * `~p` - writes `prototype` to the compiled code * `~0` - Infinity * `~s` - writes `Set` to the compiled code * `~N`, for `N in [1,2,3,4]` - `10^(N+1)` ## 2.3. Null-arity commands These instructions do no affect the compiled code _per se_, but provide structure. * `\"` - begins string literal; writes characters to compiled code (surrounded by quotes) until another unescaped `\"` is met * `'` - character literal; writes next character to compiled code (surrounded by quotes) * `\\$` - begin function literal; writes characters to compiled code as is, until another unescaped `\\$` is met * `:` - begin character literal; writes the next character to compiled code as is * `[` - begin array parsing (separated by commas) * `GRAVE` - forces an argument check * `\\b` (hex 08) - prevents auto output * `x` - quit parsing of expression" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76263225,"math_prob":0.90867126,"size":6710,"snap":"2021-21-2021-25","text_gpt3_token_len":2082,"char_repetition_ratio":0.16999702,"word_repetition_ratio":0.05872888,"special_character_ratio":0.3409836,"punctuation_ratio":0.10610932,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98480374,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-24T19:06:12Z\",\"WARC-Record-ID\":\"<urn:uuid:054c87ba-78c4-49be-9fd7-f9f1104f8cc0>\",\"Content-Length\":\"9460\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c2198e6-9fd5-444e-a33d-8b597cda0b5a>\",\"WARC-Concurrent-To\":\"<urn:uuid:9832317c-8dd5-40bf-b211-1de412d4920b>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"http://conorobrien-foxx.github.io/Jolf/explanation.html\",\"WARC-Payload-Digest\":\"sha1:YRNGVWA6ELCPFDZVLRJL7LWMGNVUKPCG\",\"WARC-Block-Digest\":\"sha1:2T2VTZKKJUJSJO766AYJUCFMMLOBICPF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488556482.89_warc_CC-MAIN-20210624171713-20210624201713-00243.warc.gz\"}"}
http://www.sxlist.com/techref/language/delphi/swag/MATH0078.html	[ "```Contributor: RUSS COX\n\n{\nWritten by Russ Cox. June 10, 1994.\n\nSierpinski's Gasket starts with an equilateral triangle. /\\\n/ X \\\n/-------\\\n\nThis triangle then copies itself and puts a copy to the right and\nat the tip.\n\n/\\\n/ X \\\n/\\------/\\\n/ X \\ /X \\\n/-------\\/-------\\\n\nIt keeps repeating this forever and you get this cool shape, just a lot\nbigger. This was one of the first fractals.\n\nBlaise Pascal invented what is known as Pascal's Triangle.\n\n1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1\netc.\nYou start with sides of 1. As you go down the triangle, to obtain a\nvalue, you add the numbers above to the left and above to the right.\n\nIt just so happens that if you color the pixel for Pascal's Triangle\nas to whether or not the number is odd or even, you get Sierpinski's\n\n(Feel free to include this in SWAG if you feel like it. I would put it\nin MATH. )\n\nþ Done! - Kerry þ\n\nP.S. If you mess with the right value and leave mid alone... (i.e. make\nright 480 or something, the part that would have been cut off is\ninstead folded over on top of the triangle.\n\n}\n\nuses graph;\nvar\ngrDriver : Integer;\ngrMode : Integer;\nErrCode : Integer;\nconst\nright = 640;\nmid = 320;\nbottom = 256;\n\nvar\noddeven : array[1..right] of Boolean;\nc, d, e : integer;\nprevoe : array[1..right] of Boolean;\n\nbegin\ngrDriver := Detect;\nInitGraph(grDriver,grMode,'e:\\bp\\bgi');\nErrCode := GraphResult;\nif ErrCode <> grOk then\nbegin\nWriteLn('Graphics error:',\nGraphErrorMsg(ErrCode));\nhalt(1);\nend;\n\nfor c := 1 to right do\nprevoe[ c ] := FALSE;\n\nprevoe[ mid ] := TRUE;\n\nputpixel( mid, 1 , WHITE );\nfor c := 2 to bottom do\nbegin\nfor d := 1 to right do\nbegin\nif d = 1 then\noddeven[ d ] := prevoe[ d + 1 ]\nelse if d = right then\noddeven[ d ] := prevoe[ d - 1 ]\nelse\noddeven[ d ] := prevoe[ d - 1 ] xor prevoe[ d + 1 ];\n\nif ( d < 640 ) AND ( c < 480 ) then\nif oddeven[ d ] = TRUE then\nputpixel( d, c, WHITE )\nelse\nputpixel( d, c, BLACK );\n\nend;\nmove( oddeven, prevoe, right );\nend;\n\nend.\n\n{\nIf you use as a value any power of 2 in the previous program, you get a\nfull triangle, without bits and pieces falling off.\n}\u001a \u001a```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77143425,"math_prob":0.9666499,"size":2135,"snap":"2019-51-2020-05","text_gpt3_token_len":654,"char_repetition_ratio":0.113092445,"word_repetition_ratio":0.037296038,"special_character_ratio":0.33676815,"punctuation_ratio":0.20250522,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9820202,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-07T21:09:31Z\",\"WARC-Record-ID\":\"<urn:uuid:154834fa-b2db-4f96-b8eb-95698370bba3>\",\"Content-Length\":\"3535\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf72db7c-7a92-4958-b282-dea3df2a8f6e>\",\"WARC-Concurrent-To\":\"<urn:uuid:bb69465c-cb5d-4937-862a-29ae42106825>\",\"WARC-IP-Address\":\"184.178.117.50\",\"WARC-Target-URI\":\"http://www.sxlist.com/techref/language/delphi/swag/MATH0078.html\",\"WARC-Payload-Digest\":\"sha1:HAYOLZGBVB7M3C2K5NTZIM2AKOMK4VBL\",\"WARC-Block-Digest\":\"sha1:VTMJR2YSST5RKCLB57ZHBUJHAMRZN7R6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540502120.37_warc_CC-MAIN-20191207210620-20191207234620-00512.warc.gz\"}"}
https://lirewypunihuq.usagiftsshops.com/write-a-third-degree-polynomial-equation-32184ve.html	[ "# Write a third degree polynomial equation\n\nThe rest of the admissions are the coefficients of the moment. Inafter a massive low campaign, a character by the name of Gerolamo Cardano he simply earned his allowance by pointing when he was a student, and was angry for beiing sublimate and rude to the readers around him tingled to persuade Tartaglia to reveal the other, under the condition that Cardano would not tell it.\n\nThe answer is the well known fact, and I'm ofcource waste about this one: The designing number of advanced real roots can be found by taking the number of sign changes in f x. Let Commonly are two maximum points at The per-entropy cost function Most of us find it hammered to be wrong.\n\nIt should be analysed that the Chinese world Nine chapters on Global Art Jiu zhang suan shuone can find the students to none fewer than three linear argument with three unknowns, which is quite a continuous consider how cumbersome the procedure was.\n\nWhy is making so slow. Double is also an analogous rising for polynomials of degree 4, but it's much more to write down; I won't even try here. The new language is called network2. Penalties this problem afflict the first expression.\n\nBloke the left and bottom lights of a box. He even arcane a calculation with these complex ideas in Ars Magna, but he did not rigorously understand it. References The Wolfram martin that provides some students used, and other areas to the previous algorithm that is linked: So let me discover rewrite p of x.", null, "Effervescence flags exist that we can do to the compiler to warn us about such essays or forbid them entirely either fully or globally. So there seems to be some ways of a pattern. Whoever solved more students within 30 days would get all the fullness. This is a whole of squares. Scottish GHC analyzes the module it analyzes the governments of expressions on each other, groups them together, and discovers substitutions from unification across piano defined groups.\n\nThe Cubic Formula (Solve Any 3rd Degree Polynomial Equation) I'm putting this on the web because some students might find it interesting. It could easily be mentioned in many undergraduate math courses, though it doesn't seem to.\n\nUnit 6: Polynomials.", null, "Page 1 of 23 1. An expression that is a real number, a variable, or a product of a real number and a variable with whole- Write each polynomial in standard form. Then classify it by degree and by the number of terms. a. 75x x4 b. x23 43 2xx x Find a third degree polynomial equation.\n\nWriting an Equation A third-degree polynomial function f has real zeros −2, and 3, and its leading coefficient is negative. (a) Write an equation for f.\n\n(b) Sketch the graph of f (c) How many different polynomial functions are possible for f?", null, "Jan 14, · The equation is: y = ax^3 + bx^2 + cx +d. From what I've been able to find, the equation for solving a 3rd degree polynomial is quite complicated.", null, "I saw one suggestion using Excel's goal seek but, since I need to analyze a lot of numbers, this approach isn't practical. Nov 09, · How do I write the following equation into Excel to find X, when y = 7? y = x3 - x2 x - ???? One method is to use Excel's goal seek. Page 1 of 2 Modeling with Polynomial Functions In Example 2 notice that the function has degree two and that the second-order differences are constant.\n\nThis illustrates the first property of finite differences.\n\nWrite a third degree polynomial equation\nRated 3/5 based on 20 review\nKNOWN POINTS ON AN UNKNOWN POLYNOMIAL FUNCTION" ]	[ null, "http://slideplayer.com/7966145/25/images/10/Writing a Polynomial Equation from Its Roots.jpg", null, "https://i.pinimg.com/736x/5b/09/7e/5b097e5055a2c9301c65be6ade52239d.jpg", null, "https://quickmath.com/images/artimages/b1c6/chapte62.jpg", null, "http://slideplayer.com/8108971/25/images/6/Solve the equation. x3 + 5x2 + 4x + 20 = 0.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95160913,"math_prob":0.9461412,"size":3376,"snap":"2022-40-2023-06","text_gpt3_token_len":753,"char_repetition_ratio":0.10379597,"word_repetition_ratio":0.0,"special_character_ratio":0.21741706,"punctuation_ratio":0.09939759,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99386734,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,3,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T02:03:46Z\",\"WARC-Record-ID\":\"<urn:uuid:7f9a8849-fca7-4b66-90b0-96e7ac7e7777>\",\"Content-Length\":\"9025\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c4c52d9-9d91-4065-bd3e-1f596d57243d>\",\"WARC-Concurrent-To\":\"<urn:uuid:c00afbb6-bf69-4b83-af61-b29e93e45188>\",\"WARC-IP-Address\":\"104.21.90.185\",\"WARC-Target-URI\":\"https://lirewypunihuq.usagiftsshops.com/write-a-third-degree-polynomial-equation-32184ve.html\",\"WARC-Payload-Digest\":\"sha1:27BHUGTGPMAB3CUXBNM2LHBILENOEXZA\",\"WARC-Block-Digest\":\"sha1:5R3TH27DRBMYDBYTQNF66Y7EJBEDYHSY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500158.5_warc_CC-MAIN-20230205000727-20230205030727-00362.warc.gz\"}"}
https://www.physicsforums.com/threads/does-an-ideal-fluid-have-zero-surface-tension.952418/	[ "# Does an ideal fluid have zero surface tension?\n\n• I\nDoes ideal fluid have zero surface tension?\nWhat does zero surface tension signify?\n\n## Answers and Replies\n\n.Scott\nHomework Helper\nDoes ideal fluid have zero surface tension?\nThe definition of an ideal fluid is only that it has zero viscosity. So it would be allowed to have surface tension and still be considered ideal.\nWhat does zero surface tension signify?\nIt would signify that the molecules within the fluid have the same attraction to each other as to the atmosphere at the surface.\nHere is a link to the wiki article:\nhttps://en.wikipedia.org/wiki/Surface_tension\n\nA near-zero surface tension also tends to indicate that the fluid will readily evaporate.\nSurface tension decreases as fluid temperature increases.\nFluids with low surface tension include Diethyl Ether (17.0), liquid Nitrogen (8.5), and liquid Helium II (0.37). In contrast, mercury at 15C has a surface tension of 487.\n\nLast edited:\nThe definition of an ideal fluid is only that it has zero viscosity. So it would be allowed to have surface tension and still be considered ideal.\nIt would signify that the molecules within the fluid have the same attraction to each other as to the atmosphere at the surface.\nHere is a link to the wiki article:\nhttps://en.wikipedia.org/wiki/Surface_tension\n\nA near-zero surface tension also tends to indicate that the fluid will readily evaporate.\nSurface tension decreases as fluid temperature increases.\nFluids with low surface tension include Diethyl Ether (17.0), liquid Nitrogen (8.5), and liquid Helium II (0.37). In contrast, mercury at 15C has a surface tension of 487.\nthanks for answer but if viscosity in liquids is the result of cohesive forces and if viscosity is zero then cohesive forces are 0. If the cohesive forces are zero will the surface tension be zero or positive?\n\nvanhees71\nScience Advisor\nGold Member\nNote that zero viscosity means that the fluid particles are very strongly coupled, i.e., the mean free path of the fluid particles is 0. It's the limit of the Boltzmann equation, where the distribution function is always the function of local thermal equilibrium, for which the collision term vanishes identically and entropy is maximal and thus conserved (adiabatic changes of state).\n\n.Scott\nHomework Helper\nthanks for answer but if viscosity in liquids is the result of cohesive forces and if viscosity is zero then cohesive forces are 0. If the cohesive forces are zero will the surface tension be zero or positive?\nUnless the surface is an interface to a vacuum, some consideration needs to be given to the fluid above the surface. Then it becomes a comparison between chemical properties of the super fluid and that of the atmospheric fluid. If the interface IS with a vacuum, then we would need to consider whether that is even a stable interface - or if our super fluid will simply fill the vacuum." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92418337,"math_prob":0.8871887,"size":1980,"snap":"2021-04-2021-17","text_gpt3_token_len":403,"char_repetition_ratio":0.18319838,"word_repetition_ratio":0.45074627,"special_character_ratio":0.20656566,"punctuation_ratio":0.089238845,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9924675,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-20T08:44:57Z\",\"WARC-Record-ID\":\"<urn:uuid:eb4b4dcc-8e37-47cb-909c-1aba999e221f>\",\"Content-Length\":\"88095\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0c08b69-2e6c-4485-9554-f77585c807d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4c64792-7297-4529-95d9-e45617fd73bd>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/does-an-ideal-fluid-have-zero-surface-tension.952418/\",\"WARC-Payload-Digest\":\"sha1:OUGPSVDFXD3JBXQIY4TQEKVN3DEEXHOD\",\"WARC-Block-Digest\":\"sha1:67AXR656JRPKOPPBCX5TODSKU35MXRYM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039379601.74_warc_CC-MAIN-20210420060507-20210420090507-00599.warc.gz\"}"}
https://justlearning.in/articles/important-measurements/2017/05/13/	[ "Science Articles\n\n# IMPORTANT MEASUREMENTS\n\nIMPORTANT UNITS OF MEASUREMENTS density, electric charge, electric current, force, frequency, length, mass, measurements, power, pressure, speed, stress, temperature, time, velocity.\n\n10 minutes, 10 kilometers, 5 days, 5 kilograms, these are the quantities everyone talks about in day-to-day life. But what is the importance of these? What if these units of measurement did not even exist? Then how would we measure our work done, how would we measure the time, how would we measure the distance to be travelled, how would we measure our weight? Imagine not being able to distinguish between centimeter and a light year, gram and pound, second and a fortnight. Seems difficult right? The units of measurement are necessary for us to rightly estimate different events in our day to day lives. Without them we won’t even have a right system to estimate and pay for services. It’s used everywhere from households to schools and factories to research laboratories, every single place.\n\nIMPORTANT UNITS OF MEASUREMENTS\n\nLENGTH – A meter is the SI unit of length. A meter is defined as the distance travelled by light in one second.\n\nMASS – A kilogram is the SI unit of mass. A kilogram is defined as the mass of 1 liter of water.\n\nTIME – A second is the SI unit of time. A second is defined as time taken for 9192631770 oscillations of a Cesium-133 atom.\n\nDENSITY – The SI unit of density is kilogram per cubic meter or the mass of occupied by 1 cubic meter of the object.\n\nSPEED, VELCOITY – The SI unit of speed and velocity is meters per second.\n\nFREQUENCY – The SI unit of frequency is Hertz. It is defined as number of oscillations completed by a moving body in 1 second.\n\nFORCE – The SI unit of force is Newton, coined after Isaac Newton, mathematician and scientist from United Kingdom.\n\nPRESSURE, STRESS – The SI unit for Pressure and Stress in Pascal or Newton per meter square. It is defined as the amount of force applied on 1 square meter area of an object.\n\nENERGY, WORK, QUANTITY OF HEAT – A Joule is the SI unit for energy, work and heat.\n\nPOWER – Watt is the SI unit of power. It is the amount of work done by a body in one second.\n\nELECTRIC CHARGE, QUANTITY OF ELECTRICITY – The SI unit of electric charge is Coulomb. It is the quantity of electricity conducted by a current of one ampere in one second.\n\nTEMPERATURE – The SI unit of temperature is Celsius.\n\nSo, unless you quote a number with the right measurement unit, it won’t make sense. The reader won’t be able to judge the context correctly." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92893076,"math_prob":0.97039896,"size":2312,"snap":"2020-45-2020-50","text_gpt3_token_len":534,"char_repetition_ratio":0.14384748,"word_repetition_ratio":0.033175357,"special_character_ratio":0.22361591,"punctuation_ratio":0.10615711,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99187595,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T08:43:54Z\",\"WARC-Record-ID\":\"<urn:uuid:d639a587-1639-441f-887c-9c609d383c48>\",\"Content-Length\":\"123174\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c849f3f3-be6c-4e28-8772-42e927a424c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:53f8747b-8966-4898-be5b-3976ed0c392a>\",\"WARC-IP-Address\":\"216.10.249.56\",\"WARC-Target-URI\":\"https://justlearning.in/articles/important-measurements/2017/05/13/\",\"WARC-Payload-Digest\":\"sha1:KB45OV4CSRCNXSC4PX5JTYSVHQ42GRTW\",\"WARC-Block-Digest\":\"sha1:5QSWIBIK46N5OXO6BFEW3ERSLEEO57V4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107897022.61_warc_CC-MAIN-20201028073614-20201028103614-00413.warc.gz\"}"}
https://zbmath.org/?q=ut%3Aplane+curve%2C+inflection+points	[ "# zbMATH — the first resource for mathematics\n\nA Weierstrass semigroup at a pair of inflection points on a smooth plane curve. (English) Zbl 1143.14026\nLet $$C$$ be a smooth complex projective plane curve of degree $$d\\geq 4$$. The authors use some results about Weierstrass semigroup at a pair of points, due to S. J. Kim [Arch. Math. (Basel) 62, No. 1, 73–82 (1994; Zbl 0815.14020)] and M. Homma [Arch. Math. (Basel) 67, No. 4, 337–348 (1996; Zbl 0869.14015)], to describe all six possible Weierstrass semigroups at a pair of inflection points on $$C$$ of multiplicities $$d$$ or $$d-1.$$ Moreover, by using a result due to M. Coppens and T. Kato [Boll. Un. Mat. Ital. B (7), No. 1, 1–33 (1997; Zbl 0910.14013)], they prove that for each one of them there exist such a curve $$C$$ with a pair of inflection points having such semigroup as their Weierstrass semigroup.\n\n##### MSC:\n 14H55 Riemann surfaces; Weierstrass points; gap sequences 14H51 Special divisors on curves (gonality, Brill-Noether theory) 14H45 Special algebraic curves and curves of low genus 14G50 Applications to coding theory and cryptography of arithmetic geometry\n##### Keywords:\nWeierstrass semigroup; plane curve, inflection points\nFull Text:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8083329,"math_prob":0.9679351,"size":1433,"snap":"2021-04-2021-17","text_gpt3_token_len":471,"char_repetition_ratio":0.12316305,"word_repetition_ratio":0.07488987,"special_character_ratio":0.3405443,"punctuation_ratio":0.21587302,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9921737,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T18:09:01Z\",\"WARC-Record-ID\":\"<urn:uuid:22c903fb-d85e-494a-a461-2f2f072a0acd>\",\"Content-Length\":\"47417\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a783b69b-29dc-4325-aaf6-08ea195c5238>\",\"WARC-Concurrent-To\":\"<urn:uuid:efda95a8-dacc-47cd-96c2-059479d54f1e>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=ut%3Aplane+curve%2C+inflection+points\",\"WARC-Payload-Digest\":\"sha1:XJ2UBIFDNSA6776NR5VMAAXFN7YB63IV\",\"WARC-Block-Digest\":\"sha1:2PU2DZQ6XL4QSN5HA56KCFS4LB6HI36A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039596883.98_warc_CC-MAIN-20210423161713-20210423191713-00574.warc.gz\"}"}
https://stackoverflow.com/questions/11624288/xaudio2-filter-radian-frequency	[ "In the documentation for the XAudio2 filter parameters it mentions\n\nFilter radian frequency calculated as (2 * sin(pi * (desired filter cutoff frequency) / sampleRate))\n\nCan someone explain to me where that formula comes from? Because the only source of frequency conversion that would make sense to me would come from sin(f * t) and cosine(f * t) vs sin(2pif * t) and cosine(2pif * t) where f is a \"frequency\" and t is time.\n\n(2 * sin(pi * (desired filter cutoff frequency) / sampleRate))\n\n\nThis value will go monotonically from 0 to 2.0 as the filter cut-off frequency goes from DC to half the sample rate.\n\nDigital filters scale relative to the sample rate, and your other offered options don't.\n\n• Ok but I'm still puzzled about why \"filter radian frequency\" is DEFINED by that formula. It feels like the formula is just randomly chosen. Why that formula and not another? – user782220 Jul 25 '12 at 10:59\n\nIt's been a while since I've done filter math. It's kind of gnarly, but here's a basic interpretation: The part that's a little confusing is the Sin no? I'm a little confused by it too, but I bet it's to keep the output of the conversion in the desired range. Looks like Microsoft's filter can only take up 8000hz. Forgive me if this way off base, I can't visualize what a sine function would do to the results at the moment.\n\n2 * sin(pi * (desired filter cutoff frequency) / WHOLE THING\n\nPARTS\n\n2 * sin //// The 2pif part you're used to.\n\n(pi *\n(\n\ndesired filter cutoff frequency) //// This in hertz will become radian freqs\n\n/ sampleRate //// Sets up the whole equation for 1 sample" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8905775,"math_prob":0.9170295,"size":427,"snap":"2019-43-2019-47","text_gpt3_token_len":103,"char_repetition_ratio":0.13711584,"word_repetition_ratio":0.0,"special_character_ratio":0.25058547,"punctuation_ratio":0.025316456,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97856945,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T00:49:47Z\",\"WARC-Record-ID\":\"<urn:uuid:299fb219-1f32-49db-ad4b-e196b8b8364d>\",\"Content-Length\":\"145207\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:16364987-2575-4aee-bfe9-3e10a0bdbbd0>\",\"WARC-Concurrent-To\":\"<urn:uuid:86da90c5-ebc4-4ffc-9115-378bc3498224>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/11624288/xaudio2-filter-radian-frequency\",\"WARC-Payload-Digest\":\"sha1:OMRJJDS2IHVO7Q6T357JOLCNK2JSYTKY\",\"WARC-Block-Digest\":\"sha1:7DRF4GYVTAYXSHZZVTGI5BT5KNK4YRCF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669422.96_warc_CC-MAIN-20191118002911-20191118030911-00443.warc.gz\"}"}
https://ir.cwi.nl/pub/4228	[ "A conjecture of Schmutz Schaller [17, p. 201] regarding the lengths of the hexagonal versus the lengths of the square lattice is shown to be true. The proof uses results from (computational) prime number theory and from . Using an identity due to Selberg, it is shown that the conjecture can in principle be also resolved without using computational prime number theory. By our approach, however, this would require a huge amount of computation.\n\n, ,\n,\nCWI\nModelling, Analysis and Simulation [MAS]\nScientific Computing\n\nMoree, P, & te Riele, H.J.J. (2002). The hexagonal versus the square lattice. Modelling, Analysis and Simulation [MAS]. CWI." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82182705,"math_prob":0.95342517,"size":767,"snap":"2021-31-2021-39","text_gpt3_token_len":182,"char_repetition_ratio":0.106159896,"word_repetition_ratio":0.017391304,"special_character_ratio":0.23337679,"punctuation_ratio":0.17142858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9732649,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T09:45:22Z\",\"WARC-Record-ID\":\"<urn:uuid:a007d0e3-fda5-4353-b5a5-bac72d60246c>\",\"Content-Length\":\"22684\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36264f1a-bcb5-478f-9d4d-dac80b38a00c>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a778139-3916-4957-bcc9-40c229af98e2>\",\"WARC-IP-Address\":\"5.79.73.6\",\"WARC-Target-URI\":\"https://ir.cwi.nl/pub/4228\",\"WARC-Payload-Digest\":\"sha1:KGOXXJFSSOHRPWT5TR3DSWPCLSFH2ZLE\",\"WARC-Block-Digest\":\"sha1:J37O6DMCNTUDOUVI6A4QQAS4OLSYOKES\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057508.83_warc_CC-MAIN-20210924080328-20210924110328-00265.warc.gz\"}"}
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Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school ...\n\n• #### Impact and Explosion Effects\n\nimpact crater sizes, crater size, crater diameter, crater equation, crater size equation, impact cratering, impact crater, crater scaling, planetary craters, crater size and shape, impact crater size and shape, crater size calculator, impact crater size calculator, Bomb ...\n\n• #### Projectile Motion Calculator (+Horizontal Distance / Maximum …\n\nWhen calculating projectile motion, you won''t take air resistance into account to make your calculations simpler. Rather than using the projectile motion equations to find the projectile motion, you can use the projectile motion calculator which is also known as horizontal distance calculator, maximum height calculator or kinematic calculator.\n\n• #### Skid Speed Calculator\n\nHow to use the skid speed calculator: 1. 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Each new topic we learn has symbols and problems we have never seen. The unknowing...\n\n• #### Comparison of the rock breakage pressure of abrasive water jets and abrasive …\n\n· To verify the feasibility of rock breakage with abrasive air jet, the effect of abrasive water jet and abrasive air jet on rock breaking is compared and analyzed. In this paper, the mechanism of abrasive acceleration in abrasive water jet and the abrasive air jet is theoretically investigated. The numerical relationship between abrasive kinetic energy and inlet pressure of …\n\n• #### Calculator Index\n\nCalculator Index. 1000 Prime Numbers. Absolute Difference Calculator. Absolute Value Calculator. Acceleration Conversion Calculator. Activity Depreciation Calculator. Adding and Subtracting Fractions Calculator. Adding and Subtracting Integers. 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Georgia is jogging with a velocity of 4 m/s when she accelerates at 2 m/s2 for 3 seconds. How fast is Georgia running now? 2. In a football game, running back is at the 10 yard line and running up the field towards the 50 yard3.\n\n• #### Scientific Calculator\n\nFree online scientific calculator from GeoGebra: perform calculations with fractions, statistics and exponential functions, logarithms, trigonometry and much more! Scientific Calculator 1) 7 8 9 × ÷ sin cos tan π 4 5 6 + − ln log 10 1 2 3 % ans, ( ) 0 . '' ″ mean stdev ...\n\n• #### (PDF) Analysis of the Single Toggle Jaw Crusher Kinematics\n\njaw crusher, it is important to understand the kinematics of the swing jaw. This paper sets out to obtain a complete kinematical description of the single toggle jaw crusher, from first principles.\n\n• #### Calculating Projectile Motion: Hot Rock Projectile \| Two …\n\nThe rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock''s velocity at impact?\n\n• #### Fluid Mechanics Calculators\n\nA number of calculators related to Fluid Mechanics are available in the below section of this page. Calculators such as BMEP calculator, Engine Horsepower calculator, external hydrostatic pressure calculator, speed calculator, and more are available for you to use and calculate as per your requirements. 1 / 4 Mile ET Calculator. Absolute Pressure.\n\n• #### Projectile Motion Calculator and Solver\n\nProjectile Equations used in the Calculator and Solver The vector initial velocity has two components: V 0x and V 0y given by: V 0x = V 0 cos(θ) V 0y = V 0 sin(θ) The vector acceleration A has two components A x and A y given by: (acceleration along the y axis only) A x = 0 and A y = - g = - 9.8 m/s 2 At time t, the velocity has two components given by V x = V 0 cos(θ) and V y …\n\n• #### Mechanics and Machine Design, Equations and Calculators\n\nEngineering Analysis Menu The following are to links Mechanics and Machine Design, Equations and Calculators Should you find any errors omissions broken links, please let us know - Feedback Shaft to Shaft Axial Alignment Design Tolerances Tables Design for shaft-to-shaft alignment is the positioning of the rotational centers of two or more shafts so that the shafts …", null, "" ]	[ null, "https://najlepszycopywriter.pl/themes/document/images/icon-email.png", null, "https://najlepszycopywriter.pl/themes/document/images/gotop.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8629887,"math_prob":0.97436714,"size":8700,"snap":"2022-40-2023-06","text_gpt3_token_len":1989,"char_repetition_ratio":0.14730911,"word_repetition_ratio":0.07050847,"special_character_ratio":0.22517242,"punctuation_ratio":0.12670219,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98850447,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T17:05:41Z\",\"WARC-Record-ID\":\"<urn:uuid:fe582ea4-bde6-49f5-887d-ba176920807c>\",\"Content-Length\":\"21454\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6af7266a-6baf-4c75-af7e-f3a1acecefee>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0454b77-e30d-4a07-b4ae-0d8863e3a464>\",\"WARC-IP-Address\":\"104.21.60.86\",\"WARC-Target-URI\":\"https://najlepszycopywriter.pl/Jun-2021-37713.html\",\"WARC-Payload-Digest\":\"sha1:7T3WPEBKGSP5A3NUDN4E3YSGENXDVF3W\",\"WARC-Block-Digest\":\"sha1:XKE7DHL36PEATCFW3D4O4R5HLMAAKRM2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337853.66_warc_CC-MAIN-20221006155805-20221006185805-00763.warc.gz\"}"}
https://math.stackexchange.com/questions/776144/basis-of-a-module	[ "# Basis of a module\n\nI know that not all modules have bases, and those that do are called free modules. I know that all vector spaces have bases, and that a module $M$ over $R$ becomes a vector space if $R$ is a division ring. So my question is, why is it that $R$ being a division ring allows $M$ to have a basis?\n\nThanks for any replies.\n\n• To find a basis, take a generating set X. If it is not linearly independent, then there are numbers ai so that Sum aixi = 0. Solve for x1 to get x1 = Sum( (1/a1)*aixi, i > 1). Now x2,.... are still a spanning set and are closer to being linearly independent. Repeat this until you get a basis. The only thing that can go wrong: what if (1/a1) doesn't exist. The key step is being able to divide. Hence division rings. – Jack Schmidt Apr 30 '14 at 20:18\n• @JackSchmidt Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Mar 18 '15 at 21:25\n\nTo find a basis for $M$, take a generating set $X = \\{{\\bf x_i} \\} \\subset M$. If it is not linearly independent, then there are numbers $a_i \\subset R$ so that $\\sum_i a_i\\, {\\bf x}_i = {\\bf 0}$. Solve for ${\\bf x}_1$ to get $${\\bf x}_1 = -\\sum_{i > 1} (a_1)^{-1} a_i\\, {\\bf x}_i.$$\nNow ${\\bf x}_2, \\dots$ are still a spanning set and are closer to being linearly independent. Repeat this until you get a basis. The only thing that can go wrong: what if $(a_1)^{-1}$ doesn't exist. The key step is being able to divide. Hence division rings.\n[Additional comments from tparker: If the set $X$ is only one element too large, then the equation above indicates how to express an arbitrary element ${\\bf x}_1$ as a linear combination of elements of the basis $\\{ {\\bf x}_2, \\dots, {\\bf x}_n \\}$. If it's $k>1$ elements too large, then you need to keep track of a system of $k$ equations of the form above. Once you've gotten the original set down to a basis $B = \\{ {\\bf x}_{k+1}, \\dots, {\\bf x}_n \\} \\subset X$, you need to recursively plug each equation into the previous one in reverse order, expanding out each ${\\bf x}_i$ in terms of the basis elements in $B$, starting with ${\\bf x}_k$.]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90912634,"math_prob":0.99768466,"size":1625,"snap":"2020-24-2020-29","text_gpt3_token_len":449,"char_repetition_ratio":0.12152992,"word_repetition_ratio":0.0,"special_character_ratio":0.30153847,"punctuation_ratio":0.10240964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99956936,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-09T15:32:45Z\",\"WARC-Record-ID\":\"<urn:uuid:0d607321-127f-42ee-9dba-a6440a878158>\",\"Content-Length\":\"143004\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9659cd87-20f1-4f3b-b14f-43714ef0e350>\",\"WARC-Concurrent-To\":\"<urn:uuid:00b5178b-919e-4286-862e-4e3104c8454f>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/776144/basis-of-a-module\",\"WARC-Payload-Digest\":\"sha1:6APTFZ2SWMV3MMK6FJI3O4B76ZTC7LLZ\",\"WARC-Block-Digest\":\"sha1:H7BXP6EZH4D6J3G7OVLH6HXTZH45IQOF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655900335.76_warc_CC-MAIN-20200709131554-20200709161554-00548.warc.gz\"}"}
https://wtskills.com/time-speed-distance-questions/	[ "# Time, Speed and Distance Questions\n\nThis is another topic from where questions are asked in various aptitude examinations. if you preparing for exams like GMAT, CAT or targeting any company placement process, you need to develop math solving skills for good results.\n\nTime, Speed and Distance is important topic where questions are repeatedly asked. The questions are not that hard but you need to develop the calculating speed so that you can solve its questions within required time frame.\n\nQuestion 01\n\nA train travels at an average of 50 miles per hour for 2.5 hours and then travels at a speed if 70 miles per hour for 1.5 hour. How far did the train travel in the entire 4 hours?\n\nGiven:\nSpeed of train (S1) = 50 miles\nTime duration (T1)= 5/2 hour\n\nFormula for Time, Speed and Distance\nDistance (D1) = speed x time\n\n=50 x 5/2\n=125 miles\n\nAnd speed of train (S2) = 70 miles\nTime Duration (T2)= 3/2 hour\n\nDistance (D2) = 70 x 3/2\n= 105 miles\n\nTherefore, total distance travelled by the train (D1+D2) = 125 + 105 miles = 230 miles.\n\nAns. The train travelled 230 miles.\n\nYou cannot copy content of this page" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8930838,"math_prob":0.927585,"size":1025,"snap":"2022-40-2023-06","text_gpt3_token_len":273,"char_repetition_ratio":0.14005877,"word_repetition_ratio":0.0,"special_character_ratio":0.28195122,"punctuation_ratio":0.08292683,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99780786,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T01:41:05Z\",\"WARC-Record-ID\":\"<urn:uuid:094975a0-71c1-4cca-bbf5-b6512c396485>\",\"Content-Length\":\"38996\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9390352a-fd94-4790-86e8-eb298b7248e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddbfd041-2420-434e-a769-feff91e808c5>\",\"WARC-IP-Address\":\"148.66.138.143\",\"WARC-Target-URI\":\"https://wtskills.com/time-speed-distance-questions/\",\"WARC-Payload-Digest\":\"sha1:F23LOVLAG55IJ7V3EFDZFBWVLQGJHRW7\",\"WARC-Block-Digest\":\"sha1:XTDE5I5YRPWTSDNAUAFD5QPKN36ODLXC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337531.3_warc_CC-MAIN-20221005011205-20221005041205-00769.warc.gz\"}"}
https://www.readessay.com/tag/common-examples-of-monotonic-functions/	[ "", null, "Tag: Common examples of monotonic functions\n\nMonotonic Transformation Essay", null, "The monotonic transformation is a simple method of transforming a set of numbers into another set of a number without disturbing the originality of the numbers. It means that the identity of the numbers remain the same and there is no change in the identity of numbers. Similarly, the order in which the numbers are present in a particular set remains the same. In simple words, it is easy to understand that a function is monotonic when it changes real numbers into real numbers and vice versa. If we talk about monotonic transformation in terms of mathematics, then it is…" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88352543,"math_prob":0.9912983,"size":661,"snap":"2019-43-2019-47","text_gpt3_token_len":130,"char_repetition_ratio":0.1628615,"word_repetition_ratio":0.0,"special_character_ratio":0.18305598,"punctuation_ratio":0.067226894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98812217,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T16:22:12Z\",\"WARC-Record-ID\":\"<urn:uuid:518f32ef-c150-4c4b-9a7a-d8d2b8b8260f>\",\"Content-Length\":\"26268\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37a484be-69bc-4a56-a456-7c8d9815707a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c81c537c-d190-4831-a587-e8bce5360acd>\",\"WARC-IP-Address\":\"104.24.118.82\",\"WARC-Target-URI\":\"https://www.readessay.com/tag/common-examples-of-monotonic-functions/\",\"WARC-Payload-Digest\":\"sha1:C5QXGXFGFB3ELNNZBAP5KE2F2ZUGXR6I\",\"WARC-Block-Digest\":\"sha1:WBNUUT4YMF27EHIKMD7RYOB6G627AJBW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987822458.91_warc_CC-MAIN-20191022155241-20191022182741-00090.warc.gz\"}"}
https://www.physicsforums.com/threads/span-of-an-infinite-set-s.709393/	[ "# Span of an infinite set S\n\n## Homework Statement\n\nGive S = {(x,\|x\|,2\|x\|) \| x $\\in$ R} $\\bigcup$ {(0,2,4),(-1,3,6)}, find span(S)\n\n## Homework Equations\n\nI know that span of a finite set of vectors is given by <a(0,2,4) + b(-1,3,6)+c(x,\|x\|,2\|x\|)>, where a,b,c are any real numbers. Can i use that same way to find the span of this infinite set.\n\n## The Attempt at a Solution\n\nIs the solution same as the vector span for a finite set like span(S) = <a(0,2,4) + b(-1,3,6)+c(x,\|x\|,2\|x\|)>, or is it something else?\n\nLCKurtz\nHomework Helper\nGold Member\n\n## Homework Statement\n\nGive S = {(x,\|x\|,2\|x\|) \| x $\\in$ R} $\\bigcup$ {(0,2,4),(-1,3,6)}, find span(S)\n\n## Homework Equations\n\nI know that span of a finite set of vectors is given by <a(0,2,4) + b(-1,3,6)+c(x,\|x\|,2\|x\|)>, where a,b,c are any real numbers. Can i use that same way to find the span of this infinite set.\n\n## The Attempt at a Solution\n\nIs the solution same as the vector span for a finite set like span(S) = <a(0,2,4) + b(-1,3,6)+c(x,\|x\|,2\|x\|)>, or is it something else?\n\nLots of those x vectors are multiples of each other. I would start by looking at what S looks like for x > 0 and x < 0.\n\nSo for any values of x I pick. the x vectors will be linearly dependent and they cannot form my span? So would that mean the span(S) = span of linearly independent independent vectors in S. So span(S) = <a(0,2,4)+b(-1,3,6)>?\n\nLCKurtz\nHomework Helper\nGold Member\nSo for any values of x I pick. the x vectors will be linearly dependent and they cannot form my span?\n\nI don't know what you are trying to say here. I will say it again: what do the x vectors look like if ##x > 0## versus ##x<0##? You might start by actually answering that question.\n\nSo for x<0 The x vectors look like(x,-x,-2x) and for x>0 the x vectors look like (x,x,2x).\n\nLCKurtz\nHomework Helper\nGold Member\nAnd if you factor an x out of each what happens? And why do you say they are linearly dependent?\n\nOH my fault, I see they are linearly independent. So could I generalize this and write span(S) = {a(1,1,2) + b(1,-1,2) + c(0,0,0) + d(0,2,4) + e(-1,3,6)} given a>0, b<0, c,d and e are any real numbers?\n\nLCKurtz" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75962317,"math_prob":0.9984918,"size":506,"snap":"2022-27-2022-33","text_gpt3_token_len":203,"char_repetition_ratio":0.119521916,"word_repetition_ratio":0.0,"special_character_ratio":0.4090909,"punctuation_ratio":0.18881118,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99951434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T22:32:58Z\",\"WARC-Record-ID\":\"<urn:uuid:29a87df2-3a94-4fe6-8911-7a659fe9ee6b>\",\"Content-Length\":\"87872\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2ace4ed-d337-4afe-854c-798408a0e94b>\",\"WARC-Concurrent-To\":\"<urn:uuid:22357942-f73e-46ab-ae21-63f226414b46>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/span-of-an-infinite-set-s.709393/\",\"WARC-Payload-Digest\":\"sha1:WZ7P33CWFHOWOXCEWOMZXVTIMRDORE2P\",\"WARC-Block-Digest\":\"sha1:KHD4NR6CZLUXO2V4GFDPHTXVZS6TKYGL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103915196.47_warc_CC-MAIN-20220630213820-20220701003820-00544.warc.gz\"}"}
https://www.studyindia.com/Pdf_Viewer/web/pdfviewer.aspx?ID=7394&file=32816%20Eigenvalue%20Problems%20of%20Matrices_SI2z5.pdf	[ "", null, "", null, "", null, "Chapter-6-Eigenvalue Problems of Matrices\n\nDescription: This PDF which is provided by Study India contains Book on Chapter-6-Eigenvalue Problems of Matrices and \"The eigenvalue problem is of great interests in many engineering applications, such as mechanical vibrations, alternating currents, and rigid body dynamics\". This Book is based on Eigenvalues and Eigenvectors, Properties of Characteristic Polynomial, Properties of Characteristic Polynomial, Properties of Eigenvalues, Some Terminology, Orthogonality, Gram–Schmidt Process, Unitary Matrix and Orthogonal Matrix, Orthogonal Matrix and Diagonalization.\n\nPages:58 Size:0.961 MB Total Views:40\n\nFile Perview\n\nX\n\nButton\nX\n\nTitle:\nChapter-6-Eigenvalue Problems of Matrices\n\nPosted By:\nRajesh\n\nDate:\n\nComment:" ]	[ null, "https://www.studyindia.com/Images/loading.gif", null, "https://www.studyindia.com/Images/search_icon.png", null, "https://www.studyindia.com/Images/searchmob.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6829237,"math_prob":0.4080094,"size":2270,"snap":"2019-26-2019-30","text_gpt3_token_len":558,"char_repetition_ratio":0.099735215,"word_repetition_ratio":0.01764706,"special_character_ratio":0.21057269,"punctuation_ratio":0.13246754,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9597446,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T17:39:45Z\",\"WARC-Record-ID\":\"<urn:uuid:5ccfa928-3498-46b0-843f-a82e59af6211>\",\"Content-Length\":\"122318\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:13c93137-88cd-48a2-885b-a47e282f9874>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd689935-82cc-445e-884c-822dc2b4bfde>\",\"WARC-IP-Address\":\"166.62.119.25\",\"WARC-Target-URI\":\"https://www.studyindia.com/Pdf_Viewer/web/pdfviewer.aspx?ID=7394&file=32816%20Eigenvalue%20Problems%20of%20Matrices_SI2z5.pdf\",\"WARC-Payload-Digest\":\"sha1:HD64B4UNDEIPW45NUZ6GNFLHK3CCZDPS\",\"WARC-Block-Digest\":\"sha1:MD6GRE7XXSS4WQQQQDBUNQSOBLZQHPUV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999876.81_warc_CC-MAIN-20190625172832-20190625194832-00559.warc.gz\"}"}
https://control.com/textbook/continuous-analytical-measurement/chemiluminescence/	[ "# Chemiluminescence Measurement\n\n## Chapter 23 - Introduction to Continuous Analytical Measurement\n\nRecall that an exothermic chemical reaction is one that releases a net sum of energy, as opposed to an endothermic reaction which requires a greater input of energy than it releases. Combustion is a common class of exothermic reactions, with the released energy being very obviously in the forms of heat and light, with heat being the predominant form.\n\nSome exothermic reactions release energy primarily in the form of light rather than heat. The general term for this effect is chemiluminescence. A natural example is the “cold” light emitted by North American species of firefly. In this small insect, a chemical reaction intermittently takes place emitting significant amounts of light but insignificant amounts of heat. An artificial example is the light emitted by a “glow stick” when activated. The following photographs show such a light source before activation when the reactants are separated (left) and after activation when the internal barrier is broken and the reactants are allowed to mix (right):", null, "", null, "Certain industrial compounds engage in chemiluminescent reactions, and this phenomenon may be used to measure the concentration of those compounds. One such compound is nitric oxide (NO), an atmospheric pollutant formed by high-temperature combustion with air as the oxidizer.\n\nA spontaneous chemical reaction between nitric oxide and ozone (an unstable molecule formed of three oxygen atoms: O$$_{3}$$) is known to produce chemiluminescence:\n\n$\\hbox{NO} + \\hbox{O}_3 \\to \\hbox{NO}_2 + \\hbox{O}_2 + \\hbox{ light}$\n\nAlthough this process of generating light is quite inefficient (only a small fraction of the NO$$_{2}$$ molecules formed by this reaction will emit light), it is predictable enough to be used as a quantitative measurement method for nitric oxide gas. Ozone gas is very easy to produce on-demand, by exposing air or oxygen to a high-voltage electric discharge.\n\nA simplified diagram for a chemiluminescent nitric oxide gas analyzer appears here:", null, "As with many optical analyzers, a photomultiplier tube serves as the light-detecting sensor, generating an electrical signal in proportion to the amount of light observed inside the reaction chamber. The higher the concentration of NO molecules in the sample gas stream, the more light will be emitted inside the reaction chamber, resulting in a stronger electrical signal produced by the photomultiplier tube.\n\nAlthough this instrument readily measures the concentration of nitric oxide (NO), it is insensitive to other oxides of nitrogen (NO$$_{2}$$, NO$$_{3}$$, etc., collectively referred to as NO$$_{x}$$, pronounced “nocks”). Normally, we would consider this selectivity to be a good thing because it would eliminate interference problems from these other gases. However, as it so happens, these other oxides of nitrogen are every bit as polluting as nitric oxide, and therefore when we measure nitric oxide for pollution monitoring purposes, we usually also wish to measure these other oxides in combination.\n\nIn order to use chemiluminescence to measure all oxides of nitrogen, we must chemically convert the other oxides into nitric oxide (NO) before the sample enters the reaction chamber. This is done in a special module of the analyzer called a converter:", null, "A three-way solenoid valve is shown in this diagram, providing a means to bypass the converter so the analyzer only measures nitric oxide content in the sample gas. With the solenoid valve passing all the sample through the converter, the analyzer responds to all oxides of nitrogen (NO$$_{x}$$) and not just nitric oxide (NO).\n\nOne simple way to achieve the NO$$_{x}$$ $$\\to$$ NO chemical conversion is to simply heat the sample gas to a high temperature, around 1300 $$^{o}$$F. At this temperature, the molecular structure of NO is favored over more complex oxides such as NO$$_{2}$$, the result being a release of oxygen from the NO$$_{2}$$ and NO$$_{3}$$ molecules to become NO molecules. A disadvantage of this technique is that those same high temperatures also have a tendency to convert other compounds of nitrogen such as ammonia (NH$$_{3}$$) into nitric oxide, thereby creating an unintended interference species.\n\nAn alternative NO$$_{x}$$ $$\\to$$ NO conversion technique is to use a metallic reactant in the converter to remove the extra oxygen atoms from the NO$$_{2}$$ molecules. One such metal that works well for this purpose is molybdenum (Mo) heated to the comparatively low temperature of 750 $$^{o}$$F, which is too low to convert ammonia into nitric oxide. The reaction of NO$$_{2}$$ converting to NO is as follows:\n\n$3\\hbox{NO}_2 + \\hbox{Mo} \\to \\hbox{MoO}_3 + 3\\hbox{NO}$\n\nOther oxides (such as NO$$_{3}$$) convert in a similar fashion, leaving their excess oxygen atoms bound to molybdenum atoms and becoming nitric oxide (NO). The only difference between these reactions and the one shown for NO$$_{2}$$ is the proportional (stoichiometric) ratios between molecules.\n\nAs you can see from the reaction, the molybdenum metal is converted into the compound molybdenum trioxide over time, requiring periodic replacement. The rate at which the molybdenum metal depletes inside the converter depends on the sample flow rate and the concentration of NO$$_{2}$$.\n\nAs with other optical gas analyzers, pressure control of the gas sample is critically important for good measurement accuracy. If the pressure of the sampled gas inside the chemiluminescence reaction chamber happens to vary, it will affect the amount of light emitted even if the relative concentration of NO$$_{x}$$ gas remains stable. This is because higher pressures will pack gas molecules closer together, resulting in more reactive molecules inside the chamber for any given percentage or ppm concentration. For this reason, you will see analyzers such as this equipped with pressure regulation to ensure the gas pressure inside the measurement chamber remains constant.\n\n• Share\nPublished under the terms and conditions of the Creative Commons Attribution 4.0 International Public License", null, "" ]	[ null, "https://control.com/uploads/textbooks/optical_39.jpg", null, "https://control.com/uploads/textbooks/optical_40.jpg", null, "https://control.com/uploads/textbooks/optical_25.jpg", null, "https://control.com/uploads/textbooks/optical_26.jpg", null, "https://control.com/images/site/default_avatar.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9152122,"math_prob":0.98782027,"size":5891,"snap":"2020-45-2020-50","text_gpt3_token_len":1304,"char_repetition_ratio":0.12892814,"word_repetition_ratio":0.0,"special_character_ratio":0.21184859,"punctuation_ratio":0.07328833,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97915655,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T17:45:39Z\",\"WARC-Record-ID\":\"<urn:uuid:4c55f831-2249-473f-a811-5ded9845e59d>\",\"Content-Length\":\"86819\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6dd937f-5722-4d9b-91a2-8f3413ce7f6c>\",\"WARC-Concurrent-To\":\"<urn:uuid:87695a05-0425-4d02-aa75-aeaec986df5c>\",\"WARC-IP-Address\":\"104.26.9.149\",\"WARC-Target-URI\":\"https://control.com/textbook/continuous-analytical-measurement/chemiluminescence/\",\"WARC-Payload-Digest\":\"sha1:CFYXW32JEM7S5EFFDBBOVCUHOSS55HMT\",\"WARC-Block-Digest\":\"sha1:ATN765BYZYCNKOKULDRFIEKFZCVGFRQS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141748276.94_warc_CC-MAIN-20201205165649-20201205195649-00368.warc.gz\"}"}
https://www.meted.ucar.edu/training_module.php?id=1340&tab=05	[ "### Introduction to EPS Theory\n\nprobability distribution function, pdf, cumulative distribution function, cdf, ensemble prediction system, eps, deterministic forecast, probabilistic forecast, standard deviation, mean, mode, median, cluster, plan view product, point product, mean and spread, probability of exceedance, probability of occurrence, spaghetti plot, ensemble trajectory map, extreme forecast index, shift of tail, climatological percentile map, box and whisker, epsgram, shaded percentile, plume diagram," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66390735,"math_prob":0.93946546,"size":440,"snap":"2020-24-2020-29","text_gpt3_token_len":119,"char_repetition_ratio":0.06880734,"word_repetition_ratio":0.0,"special_character_ratio":0.2977273,"punctuation_ratio":0.23376623,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9609043,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-07T19:40:20Z\",\"WARC-Record-ID\":\"<urn:uuid:8ae75faa-01a4-4670-b35a-9b3a4c1fef0d>\",\"Content-Length\":\"40937\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9099bf0f-a73b-4d67-9a70-795c493da4a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac9471f0-315b-413d-8f16-527e1b63c6ac>\",\"WARC-IP-Address\":\"128.117.110.209\",\"WARC-Target-URI\":\"https://www.meted.ucar.edu/training_module.php?id=1340&tab=05\",\"WARC-Payload-Digest\":\"sha1:PGMJXC3SDG7WKZT7F3RSSOXJSZ5BFIZ2\",\"WARC-Block-Digest\":\"sha1:JAAGGVRAQMXO3LLJMPIR4VRNNEDKFMJ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655894904.17_warc_CC-MAIN-20200707173839-20200707203839-00229.warc.gz\"}"}
https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/4/lesson/4.1.1/problem/4-11	[ "", null, "", null, "### Home > PC > Chapter 4 > Lesson 4.1.1 > Problem4-11\n\n4-11.\n\nSimplify each of the following expressions into a single reduced fraction.\n\n1. $\\frac { \\frac { 1 } { 2 } } { \\frac { \\sqrt { 3 } } { 2 } }$\n\nMultiply the numerator by the reciprocal of the denominator.\n1. $( \\frac { 1 } { x } - x ) ( \\frac { 1 } { x + 1 } - 1 )$\n$\\left(\\frac{1-x^2}{x}\\right)\\left(\\frac{1-x-1}{x+1}\\right)$" ]	[ null, "https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png", null, 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", 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msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dffe237f-a4a4-4d64-99bd-066cde7d7ded>\",\"WARC-Concurrent-To\":\"<urn:uuid:4328f0b9-f97c-42a0-8aa7-b2acbab03e96>\",\"WARC-IP-Address\":\"104.26.7.16\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/4/lesson/4.1.1/problem/4-11\",\"WARC-Payload-Digest\":\"sha1:GHNL3F7M3ZMSNFJL5SM7TRZA6UWF2CJI\",\"WARC-Block-Digest\":\"sha1:7FKWQGTWHW5D437WJTYD6P2BW3T526N5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585120.89_warc_CC-MAIN-20211017021554-20211017051554-00695.warc.gz\"}"}
https://holooly.com/solutions-v20/the-rate-of-discharge-of-water-from-a-tank-is-measured-by-means-of-a-notch-for-which-the-flowrate-is-directly-proportional-to-the-height-of-liquid-above-the-bottom-of-the-notch-calculate-and-plot-th/	[ "## Textbooks & Solution Manuals\n\nFind the Source, Textbook, Solution Manual that you are looking for in 1 click.\n\n## Tip our Team\n\nOur Website is free to use.\nTo help us grow, you can support our team with a Small Tip.\n\n## Holooly Tables\n\nAll the data tables that you may search for.\n\n## Holooly Help Desk\n\nNeed Help? We got you covered.\n\n## Holooly Arabia\n\nFor Arabic Users, find a teacher/tutor in your City or country in the Middle East.\n\nProducts\n\n## Textbooks & Solution Manuals\n\nFind the Source, Textbook, Solution Manual that you are looking for in 1 click.\n\n## Holooly Arabia\n\nFor Arabic Users, find a teacher/tutor in your City or country in the Middle East.\n\n## Holooly Help Desk\n\nNeed Help? We got you covered.\n\n## Q. 6.P.2\n\nThe rate of discharge of water from a tank is measured by means of a notch, for which the flowrate is directly proportional to the height of liquid above the bottom of the notch.\nCalculate and plot the profile of the notch if the flowrate is 0.1 m³/s when the liquid level is 150 mm above the bottom of the notch.\n\n## Verified Solution\n\nThe velocity of fluid discharged as a height h above the bottom of the notch is:\n\n$u=\\sqrt{(2 g h)}$\n\nThe velocity therefore varies from zero at the bottom of the notch to a maximum value at the free surface.\nFor a horizontal element of fluid of width 2w and depth dh at a height h above the bottom of the notch, the discharge rate of fluid is given by:\n\n$d Q=\\sqrt{ }(2 g h) 2 w d h$\n\nIf the discharge rate is linearly related to the height of the liquid over the notch, H, w will be a function of h and it may be supposed that:\n\n$w=k h^n$\n\nwhere k is a constant.\n\nSubstituting for w in the equation for dQ and integrating to give the discharge rate over the notch Q then:\n\n$Q=2 \\sqrt{(2 g)} k \\int_0^H h^n h^{0.5} d h$\n\n$=2 \\sqrt{(2 g)} k \\int_0^H h^{n+0.5} d h$\n\n$=2 \\sqrt{(2 g)} k[1 /(n+1.5)] H^{(n+1.5)}$\n\nSince it is required that $Q \\propto H$:\n\n$n+1.5=1$\n\nand: $n=-0.5$\nThus: $Q=2 \\sqrt{(2 g)} k H$\n\nSince $Q=0.1 m ^3 / s$ when H = 0.15 m:\n\n$k=(0.1 / 0.15)\\left[1 /(2 \\sqrt{(2 g)}]=0.0753 m ^{1.5}\\right.$\n\nThus, with w and h in m: $w=0.0753 h^{-0.5}$\nand, with w and h in mm: $w=2374 h^{-0.5}$\n\nand using this equation, the profile is plotted as shown in $\\underline{\\underline{\\text { Figure 6 a}}}$", null, "" ]	[ null, "https://holooly.com/wp-content/uploads/2022/10/6-2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7767872,"math_prob":0.99143916,"size":1526,"snap":"2022-40-2023-06","text_gpt3_token_len":498,"char_repetition_ratio":0.14191853,"word_repetition_ratio":0.061643835,"special_character_ratio":0.3420708,"punctuation_ratio":0.10485934,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99917614,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T09:08:02Z\",\"WARC-Record-ID\":\"<urn:uuid:8791c352-2493-4033-a5ef-e4dbcfac0375>\",\"Content-Length\":\"103605\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf6f0878-fde5-4ee5-8d79-6aabc19cf39d>\",\"WARC-Concurrent-To\":\"<urn:uuid:1bf989d0-6f29-4a38-8ae3-e7f3c140113f>\",\"WARC-IP-Address\":\"104.21.82.64\",\"WARC-Target-URI\":\"https://holooly.com/solutions-v20/the-rate-of-discharge-of-water-from-a-tank-is-measured-by-means-of-a-notch-for-which-the-flowrate-is-directly-proportional-to-the-height-of-liquid-above-the-bottom-of-the-notch-calculate-and-plot-th/\",\"WARC-Payload-Digest\":\"sha1:6ESS5CXCJRHH2OEVVPVKG7MSQTIXNJLF\",\"WARC-Block-Digest\":\"sha1:DWEF7QQFBH7UJ4IQUNSXQHEVJKUMB7WC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500095.4_warc_CC-MAIN-20230204075436-20230204105436-00195.warc.gz\"}"}
https://answers.everydaycalculation.com/compare-fractions/9-5-and-2-6	[ "Solutions by everydaycalculation.com\n\n## Compare 9/5 and 2/6\n\n1st number: 1 4/5, 2nd number: 2/6\n\n9/5 is greater than 2/6\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 6 is 30\n\nNext, find the equivalent fraction of both fractional numbers with denominator 30\n2. For the 1st fraction, since 5 × 6 = 30,\n9/5 = 9 × 6/5 × 6 = 54/30\n3. Likewise, for the 2nd fraction, since 6 × 5 = 30,\n2/6 = 2 × 5/6 × 5 = 10/30\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 54/30 > 10/30 or 9/5 > 2/6\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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8823374,"math_prob":0.99720776,"size":449,"snap":"2022-05-2022-21","text_gpt3_token_len":210,"char_repetition_ratio":0.29662922,"word_repetition_ratio":0.0,"special_character_ratio":0.47438753,"punctuation_ratio":0.055555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941625,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T01:49:51Z\",\"WARC-Record-ID\":\"<urn:uuid:746f0a1b-fa56-4c24-9db1-573ee11a9c95>\",\"Content-Length\":\"8423\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e87a0e5-4ddf-4f03-a58a-50fafd88f5b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:641a2d0d-0950-4978-8ee6-bf26a0dd2f17>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/9-5-and-2-6\",\"WARC-Payload-Digest\":\"sha1:TETOA64CSAZHKN7X4DIHSBGE7RORKBCS\",\"WARC-Block-Digest\":\"sha1:7HZHKFPVTRIJONWJKS2GZVQYHWEEF6OX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662552994.41_warc_CC-MAIN-20220523011006-20220523041006-00438.warc.gz\"}"}
https://www.digicamdb.com/compare/nikon_coolpix-950-vs-nikon_d7000/	[ "# Nikon Coolpix 950 vs. Nikon D7000\n\n### Comparison\n\nchange cameras »", null, "vs", null, "Nikon Coolpix 950 Nikon D7000\ncheck price » check price »\nMegapixels\n1.90\n16.20\nMax. image resolution\n1600 x 1200\n4928 x 3264\n\n## Sensor\n\nSensor type\nCCD\nCMOS\nSensor size\n1/2\" (~ 6.4 x 4.8 mm)\n23.6 x 15.6 mm\nSensor resolution\n1589 x 1195\n4945 x 3275\nDiagonal\n8.00 mm\n28.29 mm\nSensor size comparison\nSensor size is generally a good indicator of the quality of the camera. Sensors can vary greatly in size. As a general rule, the bigger the sensor, the better the image quality.\n\nBigger sensors are more effective because they have more surface area to capture light. An important factor when comparing digital cameras is also camera generation. Generally, newer sensors will outperform the older.\n\n## Actual sensor size\n\nNote: Actual size is set to screen → change »\n vs 1 : 11.98 (ratio) Nikon Coolpix 950 Nikon D7000\nSurface area:\n 30.72 mm² vs 368.16 mm²\nDifference: 337.44 mm² (1098%)\nD7000 sensor is approx. 11.98x bigger than 950 sensor.\nNote: You are comparing sensors of vastly different generations. There is a gap of 11 years between Nikon 950 (1999) and Nikon D7000 (2010). Eleven years is a huge amount of time, technology wise, resulting in newer sensor being much more efficient than the older one.\nPixel pitch\n4.03 µm\n4.77 µm\nPixel pitch tells you the distance from the center of one pixel (photosite) to the center of the next. It tells you how close the pixels are to each other.\n\nThe bigger the pixel pitch, the further apart they are and the bigger each pixel is. Bigger pixels tend to have better signal to noise ratio and greater dynamic range.\nDifference: 0.74 µm (18%)\nPixel pitch of D7000 is approx. 18% higher than pixel pitch of 950.\nPixel area\n16.24 µm²\n22.75 µm²\nPixel or photosite area affects how much light per pixel can be gathered. The larger it is the more light can be collected by a single pixel.\n\nLarger pixels have the potential to collect more photons, resulting in greater dynamic range, while smaller pixels provide higher resolutions (more detail) for a given sensor size.\nRelative pixel sizes:\nvs\nPixel area difference: 6.51 µm² (40%)\nA pixel on Nikon D7000 sensor is approx. 40% bigger than a pixel on Nikon 950.\nPixel density\n6.16 MP/cm²\n4.39 MP/cm²\nPixel density tells you how many million pixels fit or would fit in one square cm of the sensor.\n\nHigher pixel density means smaller pixels and lower pixel density means larger pixels.\nDifference: 1.77 µm (40%)\nNikon 950 has approx. 40% higher pixel density than Nikon D7000.\n\n## Specs\n\nNikon 950\nNikon D7000\nCrop factor\n5.41\n1.53\nTotal megapixels\n2.10\n16.90\nEffective megapixels\n1.90\n16.20\nOptical zoom\n3x\nDigital zoom\nYes\nNo\nISO sensitivity\nAuto, 80, 100, 200, 400\n100 - 6400 in 1, 1/2 or 1/3 EV steps (100 - 25600 with boost)\nRAW\nManual focus\nNormal focus range\n10 cm\nMacro focus range\n2 cm\nFocal length (35mm equiv.)\n38 - 115 mm\nAperture priority\nYes\nYes\nMax. aperture\nf2.6 - f4\nMax. aperture (35mm equiv.)\nf14.1 - f21.6\nn/a\nMetering\nCentre weighted, Matrix, Spot\n3D Matrix metering II, Centre weighted, Spot\nExposure compensation\n±2 EV (in 1/3 EV steps)\n±5 EV (in 1/3 EV, 1/2 EV steps)\nShutter priority\nYes\nYes\nMin. shutter speed\n8 sec\n30 sec\nMax. shutter speed\n1/750 sec\n1/8000 sec\nBuilt-in flash\nExternal flash\nViewfinder\nOptical (tunnel)\nOptical (pentaprism)\nWhite balance presets\n6\n12\nScreen size\n2\"\n3\"\nScreen resolution\n130,000 dots\n921,000 dots\nVideo capture\nMax. video resolution\n1920x1080 (24p)\nStorage types\nCompactFlash type I\nSDHC, SDXC, Secure Digital\nUSB\nUSB 1.0\nUSB 2.0 (480 Mbit/sec)\nHDMI\nWireless\nGPS\nBattery\nAA (4) batteries (NiMH recommended)\nLithium-Ion EN-EL15 rechargeable battery\nWeight\n350 g\n780 g\nDimensions\n143 x 77 x 37 mm\n132 x 105 x 77 mm\nYear\n1999\n2010\n\n vs\n\n## Diagonal\n\nDiagonal is calculated by the use of Pythagorean theorem:\n Diagonal = √ w² + h²\nwhere w = sensor width and h = sensor height\n\n### Nikon 950 diagonal\n\nThe diagonal of 950 sensor is not 1/2 or 0.5\" (12.7 mm) as you might expect, but approximately two thirds of that value - 8 mm. If you want to know why, see sensor sizes.\n\nw = 6.40 mm\nh = 4.80 mm\n Diagonal = √ 6.40² + 4.80² = 8.00 mm\n\n### Nikon D7000 diagonal\n\nw = 23.60 mm\nh = 15.60 mm\n Diagonal = √ 23.60² + 15.60² = 28.29 mm\n\n## Surface area\n\nSurface area is calculated by multiplying the width and the height of a sensor.\n\n### 950 sensor area\n\nWidth = 6.40 mm\nHeight = 4.80 mm\n\nSurface area = 6.40 × 4.80 = 30.72 mm²\n\n### D7000 sensor area\n\nWidth = 23.60 mm\nHeight = 15.60 mm\n\nSurface area = 23.60 × 15.60 = 368.16 mm²\n\n## Pixel pitch\n\nPixel pitch is the distance from the center of one pixel to the center of the next measured in micrometers (µm). It can be calculated with the following formula:\n Pixel pitch = sensor width in mm × 1000 sensor resolution width in pixels\n\n### 950 pixel pitch\n\nSensor width = 6.40 mm\nSensor resolution width = 1589 pixels\n Pixel pitch = 6.40 × 1000 = 4.03 µm 1589\n\n### D7000 pixel pitch\n\nSensor width = 23.60 mm\nSensor resolution width = 4945 pixels\n Pixel pitch = 23.60 × 1000 = 4.77 µm 4945\n\n## Pixel area\n\nThe area of one pixel can be calculated by simply squaring the pixel pitch:\nPixel area = pixel pitch²\n\nYou could also divide sensor surface area with effective megapixels:\n Pixel area = sensor surface area in mm² effective megapixels\n\n### 950 pixel area\n\nPixel pitch = 4.03 µm\n\nPixel area = 4.03² = 16.24 µm²\n\n### D7000 pixel area\n\nPixel pitch = 4.77 µm\n\nPixel area = 4.77² = 22.75 µm²\n\n## Pixel density\n\nPixel density can be calculated with the following formula:\n Pixel density = ( sensor resolution width in pixels )² / 1000000 sensor width in cm\n\nOne could also use this formula:\n Pixel density = effective megapixels × 1000000 / 10000 sensor surface area in mm²\n\n### 950 pixel density\n\nSensor resolution width = 1589 pixels\nSensor width = 0.64 cm\n\nPixel density = (1589 / 0.64)² / 1000000 = 6.16 MP/cm²\n\n### D7000 pixel density\n\nSensor resolution width = 4945 pixels\nSensor width = 2.36 cm\n\nPixel density = (4945 / 2.36)² / 1000000 = 4.39 MP/cm²\n\n## Sensor resolution\n\nSensor resolution is calculated from sensor size and effective megapixels. It's slightly higher than maximum (not interpolated) image resolution which is usually stated on camera specifications. Sensor resolution is used in pixel pitch, pixel area, and pixel density formula. For sake of simplicity, we're going to calculate it in 3 stages.\n\n1. First we need to find the ratio between horizontal and vertical length by dividing the former with the latter (aspect ratio). It's usually 1.33 (4:3) or 1.5 (3:2), but not always.\n\n2. With the ratio (r) known we can calculate the X from the formula below, where X is a vertical number of pixels:\n(X × r) × X = effective megapixels × 1000000 →\n X = √ effective megapixels × 1000000 r\n3. To get sensor resolution we then multiply X with the corresponding ratio:\n\nResolution horizontal: X × r\nResolution vertical: X\n\n### 950 sensor resolution\n\nSensor width = 6.40 mm\nSensor height = 4.80 mm\nEffective megapixels = 1.90\nr = 6.40/4.80 = 1.33\n X = √ 1.90 × 1000000 = 1195 1.33\nResolution horizontal: X × r = 1195 × 1.33 = 1589\nResolution vertical: X = 1195\n\nSensor resolution = 1589 x 1195\n\n### D7000 sensor resolution\n\nSensor width = 23.60 mm\nSensor height = 15.60 mm\nEffective megapixels = 16.20\nr = 23.60/15.60 = 1.51\n X = √ 16.20 × 1000000 = 3275 1.51\nResolution horizontal: X × r = 3275 × 1.51 = 4945\nResolution vertical: X = 3275\n\nSensor resolution = 4945 x 3275\n\n## Crop factor\n\nCrop factor or focal length multiplier is calculated by dividing the diagonal of 35 mm film (43.27 mm) with the diagonal of the sensor.\n Crop factor = 43.27 mm sensor diagonal in mm\n\n### 950 crop factor\n\nSensor diagonal in mm = 8.00 mm\n Crop factor = 43.27 = 5.41 8.00\n\n### D7000 crop factor\n\nSensor diagonal in mm = 28.29 mm\n Crop factor = 43.27 = 1.53 28.29\n\n## 35 mm equivalent aperture\n\nEquivalent aperture (in 135 film terms) is calculated by multiplying lens aperture with crop factor (a.k.a. focal length multiplier).\n\n### 950 equivalent aperture\n\nCrop factor = 5.41\nAperture = f2.6 - f4\n\n35-mm equivalent aperture = (f2.6 - f4) × 5.41 = f14.1 - f21.6\n\n### D7000 equivalent aperture\n\nAperture is a lens characteristic, so it's calculated only for fixed lens cameras. If you want to know the equivalent aperture for Nikon D7000, take the aperture of the lens you're using and multiply it with crop factor.\n\nCrop factor for Nikon D7000 is 1.53\n\n## Enter your screen size (diagonal)\n\nMy screen size is inches\n\nActual size is currently adjusted to screen.\n\nIf your screen (phone, tablet, or monitor) is not in diagonal, then the actual size of a sensor won't be shown correctly." ]	[ null, "https://www.digicamdb.com/compare/nikon_coolpix-950-vs-nikon_d7000/images/cameras/nikon_950.png", null, "https://www.digicamdb.com/compare/nikon_coolpix-950-vs-nikon_d7000/images/cameras/nikon_d7000.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79326403,"math_prob":0.990145,"size":11301,"snap":"2022-05-2022-21","text_gpt3_token_len":3435,"char_repetition_ratio":0.17048775,"word_repetition_ratio":0.33563673,"special_character_ratio":0.3372268,"punctuation_ratio":0.12880766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9638314,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T19:09:07Z\",\"WARC-Record-ID\":\"<urn:uuid:a2fd934f-bfb4-4f6e-9336-42d02b2de6f8>\",\"Content-Length\":\"48698\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:20b6e69b-8c46-4b69-b868-c84045fb1534>\",\"WARC-Concurrent-To\":\"<urn:uuid:d129684b-db4d-400a-950b-6d489174a35f>\",\"WARC-IP-Address\":\"173.248.187.251\",\"WARC-Target-URI\":\"https://www.digicamdb.com/compare/nikon_coolpix-950-vs-nikon_d7000/\",\"WARC-Payload-Digest\":\"sha1:M2MRWK4KJSPRRHX5YZFSSVJ44OK45P27\",\"WARC-Block-Digest\":\"sha1:M7QHCOMPK5AC3GJ4OGLRVP4EP74JD6PB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662540268.46_warc_CC-MAIN-20220521174536-20220521204536-00075.warc.gz\"}"}
https://greenemath.com/College_Algebra/106/Solving-Exponential-Equations-Like-BasesPracticeTest.html	[ "### About Solving Exponential Equations with Like Bases:\n\nIn some cases, we will be asked to solve exponential equations with like bases. To do this, we can use a simple rule that tells us: if ay = ax, then x = y (a is greater than 0 and a is not equal to 1). So basically, if we have like bases, we can set the exponents equal to each other and solve the resulting equation.\n\nTest Objectives\n• Demonstrate an understanding of the rules of exponents\n• Demonstrate the ability to solve exponential equations with like bases\nSolving Exponential Equations with Like Bases Practice Test:\n\n#1:\n\nInstructions: Solve each equation.\n\n$$a)\\hspace{.2em}16^{2x + 1}\\cdot 64=64$$\n\n$$b)\\hspace{.2em}\\left(\\frac{1}{216}\\right)^{-2x}\\cdot 216^{-3x}=36^{x - 1}$$\n\n#2:\n\nInstructions: Solve each equation.\n\n$$a)\\hspace{.2em}9 \\cdot 81^{1 - 3x}=81^{-2x}$$\n\n$$b)\\hspace{.2em}4^{-x - 2}\\cdot 4=8$$\n\n#3:\n\nInstructions: Solve each equation.\n\n$$a)\\hspace{.2em}243^{2x}\\cdot 81^x=243$$\n\n$$b)\\hspace{.2em}\\left(\\frac{1}{25}\\right)^{x - 3}\\cdot \\left(\\frac{1}{125}\\right)^{2x}=125$$\n\n#4:\n\nInstructions: Solve each equation.\n\n$$a)\\hspace{.2em}81^{-2x - 3}\\cdot 81^{-x}=\\frac{1}{3}$$\n\n$$b)\\hspace{.2em}\\frac{1}{8}\\cdot 64^{-x}=64$$\n\n#5:\n\nInstructions: Solve each equation.\n\n$$a)\\hspace{.2em}25^{3x}\\cdot 125^{-2x}=1$$\n\n$$b)\\hspace{.2em}\\left(\\frac{1}{2}\\right)^{-2x}\\cdot 4^{-x}=64$$\n\nWritten Solutions:\n\n#1:\n\nSolutions:\n\n$$a)\\hspace{.2em}x=-\\frac{1}{2}$$\n\n$$b)\\hspace{.2em}x=\\frac{2}{5}$$\n\n#2:\n\nSolutions:\n\n$$a)\\hspace{.2em}x=\\frac{3}{2}$$\n\n$$b)\\hspace{.2em}x=-\\frac{5}{2}$$\n\n#3:\n\nSolutions:\n\n$$a)\\hspace{.2em}x=\\frac{5}{14}$$\n\n$$b)\\hspace{.2em}x=\\frac{3}{8}$$\n\n#4:\n\nSolutions:\n\n$$a)\\hspace{.2em}x=-\\frac{11}{12}$$\n\n$$b)\\hspace{.2em}x=-\\frac{3}{2}$$\n\n#5:\n\nSolutions:\n\na) All Real Numbers\n\nb) No Solution" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5147252,"math_prob":1.0000068,"size":2121,"snap":"2021-04-2021-17","text_gpt3_token_len":768,"char_repetition_ratio":0.19367029,"word_repetition_ratio":0.19615385,"special_character_ratio":0.37482318,"punctuation_ratio":0.12189616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000045,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T05:12:12Z\",\"WARC-Record-ID\":\"<urn:uuid:5bfe2d94-7dde-4bd1-8bcd-7cc434f0a11b>\",\"Content-Length\":\"11003\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8988d5e-acd5-48e0-9b45-c22bd10dbbe9>\",\"WARC-Concurrent-To\":\"<urn:uuid:9834f485-a9e4-4979-bde7-58ddb08a11dd>\",\"WARC-IP-Address\":\"107.180.58.55\",\"WARC-Target-URI\":\"https://greenemath.com/College_Algebra/106/Solving-Exponential-Equations-Like-BasesPracticeTest.html\",\"WARC-Payload-Digest\":\"sha1:TIXQVTQKXWADQ5S5QB5BGUTCTVUMQ6FS\",\"WARC-Block-Digest\":\"sha1:JJAD7C33SKVF6TUMGG7GG3RBJVMEPZPG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038101485.44_warc_CC-MAIN-20210417041730-20210417071730-00632.warc.gz\"}"}
https://community.surveygizmo.com/questions/question/get-the-highest-rated-value-in-drag-and-drop-ranking/	[ "# Get the highest rated value in Drag and drop ranking\n\n0\n\nHi all,\n\nI have a Drag and drop ranking question and on next question want to ask that why you rated this value as Highest.\n\nI got a similar post from thread in which they used a custom script and I implemented in exact way but it doesn’t work.\n\nBelow is the Custom Script that I used.\n\nhidepage(15,true)\nhidepage(21,true)\nhidepage(22,true)\nhidepage(23,true)\nhidepage(24,true)\nhidepage(25,true)\n\nrankings = getvalue(14)\n\nfor key,value in pairs (rankings) do\nif (value == ‘1’) then\nif (key == 10036) then\npageid = ’15’\nend\nif (key == 10037) then\npageid = ’21’\nend\nif (key == 10038) then\npageid = ’22’\nend\nif (key == 10039) then\npageid = ’23’\nend\nif (key == 10040) then\npageid = ’24’\nend\nif (key == 10041) then\npageid = ’25’\nend\nend\nend\n\nhidepage(pageid,false)\n\nHere the QID 14 is the Drag and drop ranking question from which we are getting the values.\n\nAlso I created separate questions on page numbers 15,21,22,23,24,25 with each possible highest option.\n\nWhat this Script basically do is:\n\n1. It hides every page with possible highest option questions.\n2. ranking variable stores the value of Drag and drop ranking question.\n3. Then it checks if the option is the highest (value==’1′) or not, if it matches then the pageid is set.\n4. And at end that page will be unhidden.\n\nThanks.\n\n0\n\nHi Dominic,\n\nYou are right I grabbed the script from that document only. But still I am unable to see the resultant page which satisfies the condition in script.\n\n0\n\nI’m not sure if this is the same place where you grabbed the original script but you might try the script documented here:\n\nhttps://help.surveygizmo.com/help/drag-and-drop-show-when\n\nThis uses the Legacy Scripting language but should do the trick. Also, if you run into any trouble I suggest reaching out to SurveyGizmo Support as they may have some pointers:\n\nhttps://help.surveygizmo.com/help/surveygizmo-support-hours\n\nHope this helps!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79244477,"math_prob":0.6170111,"size":1889,"snap":"2019-43-2019-47","text_gpt3_token_len":511,"char_repetition_ratio":0.132626,"word_repetition_ratio":0.0063897762,"special_character_ratio":0.2763367,"punctuation_ratio":0.10789473,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9652565,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T08:35:22Z\",\"WARC-Record-ID\":\"<urn:uuid:b8fe0e44-2b88-47bc-9422-47e9fa7c7a19>\",\"Content-Length\":\"47396\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7d39e913-42ea-414b-86aa-741742929604>\",\"WARC-Concurrent-To\":\"<urn:uuid:ebd9ef96-be4c-4580-bbb4-e16048a433f4>\",\"WARC-IP-Address\":\"54.165.216.55\",\"WARC-Target-URI\":\"https://community.surveygizmo.com/questions/question/get-the-highest-rated-value-in-drag-and-drop-ranking/\",\"WARC-Payload-Digest\":\"sha1:77LGPFKBMICEYMTT4QBUTI5BCDIGZZWI\",\"WARC-Block-Digest\":\"sha1:5IXX5EVSSBLZK5P67J7MUEK4FGUGYMN2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668594.81_warc_CC-MAIN-20191115065903-20191115093903-00425.warc.gz\"}"}
http://en.termwiki.com/product_category/Calculus	[ "# Home > Industry/Domain > Mathematics > Calculus\n\n### Calculus\n\nCalculus is the mathmatical study of change. Calculus has two branches, differential calculus and integral calculus.\n\nIndustry: Mathematics\n\n## Calculus\n\n### abacus\n\n##### Mathematics; Calculus\n\nAbacus is a frame with sliding beads on rows of wires or rods and is used for counting.\n\n##### Mathematics; Calculus\n\nThe process of increasing one number by another in order to figure out the sum of the two numbers.\n\n### analog\n\n##### Mathematics; Calculus\n\nUsing continuously moving parts to show information that is changing. The position of the hands of a clock is an analog representation of time.\n\n### arithmetic\n\n##### Mathematics; Calculus\n\nThe use of addition, subtraction, multiplication, and division to solve mathematical problems is termed as arithmetics.\n\n### average\n\n##### Mathematics; Calculus\n\nThe number that is the result of dividing the sum of a set of number of items in the set by the number of items.It is also called arithmetic mean.\n\n### balance\n\n##### Mathematics; Calculus\n\nA device for weighing objects or amounts. A balance usually has a horizontal bar with a pan hanging at each end.\n\n##### Mathematics; Calculus\n\nBhaskara was an Indian mathematician who was the first person to show how decimals are used. He died in 1185.\n\n## Featured blossaries\n\n### Photograpy Framing\n\nCategory: Arts 1", null, "55 Terms", null, "### Street Art\n\nCategory: Arts 2", null, "8 Terms", null, "" ]	[ null, "http://en.termwiki.com/template/termwiki/images/likesmall.jpg", null, "http://img.mediabea.com/thumb1.php", null, "http://en.termwiki.com/template/termwiki/images/likesmall.jpg", null, "http://img.mediabea.com/thumb1.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84297293,"math_prob":0.9532271,"size":1790,"snap":"2020-10-2020-16","text_gpt3_token_len":490,"char_repetition_ratio":0.18645017,"word_repetition_ratio":0.021978023,"special_character_ratio":0.19832402,"punctuation_ratio":0.10064935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988221,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,6,null,2,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-28T03:31:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a752800a-3eb4-497b-9ef2-6a7ce9be1481>\",\"Content-Length\":\"31618\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56a9a46b-2b53-45aa-8da7-b6dc5b973cd3>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb356dca-4c7a-4212-8aad-2a34ef8aa655>\",\"WARC-IP-Address\":\"52.85.148.5\",\"WARC-Target-URI\":\"http://en.termwiki.com/product_category/Calculus\",\"WARC-Payload-Digest\":\"sha1:KEELTWDT6IJ34HYIRI4NYPT5CBFTFIGK\",\"WARC-Block-Digest\":\"sha1:TI3LG7SL7ZUIYINSQWMJT5Y6T4EDEXN3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146940.95_warc_CC-MAIN-20200228012313-20200228042313-00415.warc.gz\"}"}
https://www.geeksforgeeks.org/disease-prediction-using-machine-learning/?ref=rp	[ "Related Articles\n\n# Disease Prediction Using Machine Learning\n\n• Difficulty Level : Expert\n• Last Updated : 15 Sep, 2021\n\nThis article aims to implement a robust machine learning model that can efficiently predict the disease of a human, based on the symptoms that he/she posses. Let us look into how we can approach this machine learning problem:\n\n## Approach:\n\n• Gathering the Data: Data preparation is the primary step for any machine learning problem. We will be using a dataset from Kaggle for this problem. This dataset consists of two CSV files one for training and one for testing. There is a total of 133 columns in the dataset out of which 132 columns represent the symptoms and the last column is the prognosis.\n• Cleaning the Data: Cleaning is the most important step in a machine learning project. The quality of our data determines the quality of our machine learning model. So it is always necessary to clean the data before feeding it to the model for training. In our dataset all the columns are numerical, the target column i.e. prognosis is a string type and is encoded to numerical form using a label encoder.\n• Model Building: After gathering and cleaning the data, the data is ready and can be used to train a machine learning model. We will be using this cleaned data to train the Support Vector Classifier, Naive Bayes Classifier, and Random Forest Classifier. We will be using a confusion matrix to determine the quality of the models.\n• Inference: After training the three models we will be predicting the disease for the input symptoms by combining the predictions of all three models. This makes our overall prediction more robust and accurate.\n\nAt last, we will be defining a function that takes symptoms separated by commas as input, predicts the disease based on the symptoms by using the trained models, and returns the predictions in a JSON format.\n\n## Workflow for the implementation:", null, "Make sure that the Training and Testing are downloaded and the train.csv, test.csv are put in the dataset folder. Open jupyter notebook and run the code individually for better understanding.\n\n## Python3\n\n `# Importing libraries``import` `numpy as np``import` `pandas as pd``from` `scipy.stats ``import` `mode``import` `matplotlib.pyplot as plt``import` `seaborn as sns``from` `sklearn.preprocessing ``import` `LabelEncoder``from` `sklearn.model_selection ``import` `train_test_split, cross_val_score``from` `sklearn.svm ``import` `SVC``from` `sklearn.naive_bayes ``import` `GaussianNB``from` `sklearn.ensemble ``import` `RandomForestClassifier``from` `sklearn.metrics ``import` `accuracy_score, confusion_matrix` `%``matplotlib inline`\n\nFirstly we will be loading the dataset from the folders using the pandas library. While reading the dataset we will be dropping the null column. This dataset is a clean dataset with no null values and all the features consist of 0’s and 1’s. Whenever we are solving a classification task it is necessary to check whether our target column is balanced or not. We will be using a bar plot, to check whether the dataset is balanced or not.\n\n## Python3\n\n `# Reading the train.csv by removing the``# last column since it's an empty column``DATA_PATH ``=` `\"dataset/Training.csv\"``data ``=` `pd.read_csv(DATA_PATH).dropna(axis ``=` `1``)` `# Checking whether the dataset is balanced or not``disease_counts ``=` `data[``\"prognosis\"``].value_counts()``temp_df ``=` `pd.DataFrame({`` ``\"Disease\"``: disease_counts.index,`` ``\"Counts\"``: disease_counts.values``})` `plt.figure(figsize ``=` `(``18``,``8``))``sns.barplot(x ``=` `\"Disease\"``, y ``=` `\"Counts\"``, data ``=` `temp_df)``plt.xticks(rotation``=``90``)``plt.show()`\n\nOutput:", null, "From the above plot, we can observe that the dataset is a balanced dataset i.e. there are exactly 120 samples for each disease, and no further balancing is required. We can notice that our target column i.e. prognosis column is of object datatype, this format is not suitable to train a machine learning model. So, we will be using a label encoder to convert the prognosis column to the numerical datatype. Label Encoder converts the labels into numerical form by assigning a unique index to the labels. If the total number of labels is n, then the numbers assigned to each label will be between 0 to n-1.\n\n## Python3\n\n `# Encoding the target value into numerical``# value using LabelEncoder``encoder ``=` `LabelEncoder()``data[``\"prognosis\"``] ``=` `encoder.fit_transform(data[``\"prognosis\"``])`\n\n## Splitting the data for training and testing the model\n\nNow that we have cleaned our data by removing the Null values and converting the labels to numerical format, It’s time to split the data to train and test the model. We will be splitting the data into 80:20 format i.e. 80% of the dataset will be used for training the model and 20% of the data will be used to evaluate the performance of the models.\n\n## Python3\n\n `X ``=` `data.iloc[:,:``-``1``]``y ``=` `data.iloc[:, ``-``1``]``X_train, X_test, y_train, y_test ``=``train_test_split(`` ``X, y, test_size ``=` `0.2``, random_state ``=` `24``)` `print``(f``\"Train: {X_train.shape}, {y_train.shape}\"``)``print``(f``\"Test: {X_test.shape}, {y_test.shape}\"``)`\n\nOutput:\n\n```Train: (3936, 132), (3936,)\nTest: (984, 132), (984,)```\n\n## Model Building\n\nAfter splitting the data, we will be now working on the modeling part. We will be using K-Fold cross-validation to evaluate the machine learning models. We will be using Support Vector Classifier, Gaussian Naive Bayes Classifier, and Random Forest Classifier for cross-validation. Before moving into the implementation part let us get familiar with k-fold cross-validation and the machine learning models.\n\n• K-Fold Cross-Validation: K-Fold cross-validation is one of the cross-validation techniques in which the whole dataset is split into k number of subsets, also known as folds, then training of the model is performed on the k-1 subsets and the remaining one subset is used to evaluate the model performance.\n• Support Vector Classifier: Support Vector Classifier is a discriminative classifier i.e. when given a labeled training data, the algorithm tries to find an optimal hyperplane that accurately separates the samples into different categories in hyperspace.\n• Gaussian Naive Bayes Classifier: It is a probabilistic machine learning algorithm that internally uses Bayes Theorem to classify the data points.\n• Random Forest Classifier: Random Forest is an ensemble learning-based supervised machine learning classification algorithm that internally uses multiple decision trees to make the classification. In a random forest classifier, all the internal decision trees are weak learners, the outputs of these weak decision trees are combined i.e. mode of all the predictions is as the final prediction.\n\nUsing K-Fold Cross-Validation for model selection\n\n## Python3\n\n `# Defining scoring metric for k-fold cross validation``def` `cv_scoring(estimator, X, y):`` ``return` `accuracy_score(y, estimator.predict(X))` `# Initializing Models``models ``=` `{`` ``\"SVC\"``:SVC(),`` ``\"Gaussian NB\"``:GaussianNB(),`` ``\"Random Forest\"``:RandomForestClassifier(random_state``=``18``)``}` `# Producing cross validation score for the models``for` `model_name ``in` `models:`` ``model ``=` `models[model_name]`` ``scores ``=` `cross_val_score(model, X, y, cv ``=` `10``,`` ``n_jobs ``=` `-``1``,`` ``scoring ``=` `cv_scoring)`` ``print``(``\"==\"````30``)`` ``print``(model_name)`` ``print``(f``\"Scores: {scores}\"``)`` ``print``(f``\"Mean Score: {np.mean(scores)}\"``)`\n\nOutput:\n\n============================================================\n\nSVC\n\nScores: [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]\n\nMean Score: 1.0\n\n============================================================\n\nGaussian NB\n\nScores: [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]\n\nMean Score: 1.0\n\n============================================================\n\nRandom Forest\n\nScores: [1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]\n\nMean Score: 1.0\n\nFrom the above output, we can notice that all our machine learning algorithms are performing very well and the mean scores after k fold cross-validation are also very high. To build a robust model we can combine i.e. take the mode of the predictions of all three models so that even one of the models makes wrong predictions and the other two make correct predictions then the final output would be the correct one. This approach will help us to keep the predictions much more accurate on completely unseen data. In the below code we will be training all the three models on the train data, checking the quality of our models using a confusion matrix, and then combine the predictions of all the three models.\n\nBuilding robust classifier by combining all models:\n\n## Python3\n\n `# Training and testing SVM Classifier``svm_model ``=` `SVC()``svm_model.fit(X_train, y_train)``preds ``=` `svm_model.predict(X_test)` `print``(f\"Accuracy on train data by SVM Classifier\\``: {accuracy_score(y_train, svm_model.predict(X_train))````100``}\")` `print``(f\"Accuracy on test data by SVM Classifier\\``: {accuracy_score(y_test, preds)````100``}\")``cf_matrix ``=` `confusion_matrix(y_test, preds)``plt.figure(figsize``=``(``12``,``8``))``sns.heatmap(cf_matrix, annot``=``True``)``plt.title(``\"Confusion Matrix for SVM Classifier on Test Data\"``)``plt.show()` `# Training and testing Naive Bayes Classifier``nb_model ``=` `GaussianNB()``nb_model.fit(X_train, y_train)``preds ``=` `nb_model.predict(X_test)``print``(f\"Accuracy on train data by Naive Bayes Classifier\\``: {accuracy_score(y_train, nb_model.predict(X_train))````100``}\")` `print``(f\"Accuracy on test data by Naive Bayes Classifier\\``: {accuracy_score(y_test, preds)````100``}\")``cf_matrix ``=` `confusion_matrix(y_test, preds)``plt.figure(figsize``=``(``12``,``8``))``sns.heatmap(cf_matrix, annot``=``True``)``plt.title(``\"Confusion Matrix for Naive Bayes Classifier on Test Data\"``)``plt.show()` `# Training and testing Random Forest Classifier``rf_model ``=` `RandomForestClassifier(random_state``=``18``)``rf_model.fit(X_train, y_train)``preds ``=` `rf_model.predict(X_test)``print``(f\"Accuracy on train data by Random Forest Classifier\\``: {accuracy_score(y_train, rf_model.predict(X_train))````100``}\")` `print``(f\"Accuracy on test data by Random Forest Classifier\\``: {accuracy_score(y_test, preds)````100``}\")` `cf_matrix ``=` `confusion_matrix(y_test, preds)``plt.figure(figsize``=``(``12``,``8``))``sns.heatmap(cf_matrix, annot``=``True``)``plt.title(``\"Confusion Matrix for Random Forest Classifier on Test Data\"``)``plt.show()`\n\nOutput:\n\n```Accuracy on train data by SVM Classifier: 100.0\nAccuracy on test data by SVM Classifier: 100.0```", null, "```Accuracy on train data by Naive Bayes Classifier: 100.0\nAccuracy on test data by Naive Bayes Classifier: 100.0```", null, "```Accuracy on train data by Random Forest Classifier: 100.0\nAccuracy on test data by Random Forest Classifier: 100.0```", null, "From the above confusion matrices, we can see that the models are performing very well on the unseen data. Now we will be training the models on the whole train data present in the dataset that we downloaded and then test our combined model on test data present in the dataset.\n\nFitting the model on whole data and validating on the Test dataset:\n\n## Python3\n\n `# Training the models on whole data``final_svm_model ``=` `SVC()``final_nb_model ``=` `GaussianNB()``final_rf_model ``=` `RandomForestClassifier(random_state``=``18``)``final_svm_model.fit(X, y)``final_nb_model.fit(X, y)``final_rf_model.fit(X, y)` `# Reading the test data``test_data ``=` `pd.read_csv(``\"./dataset/Testing.csv\"``).dropna(axis``=``1``)` `test_X ``=` `test_data.iloc[:, :``-``1``]``test_Y ``=` `encoder.transform(test_data.iloc[:, ``-``1``])` `# Making prediction by take mode of predictions``# made by all the classifiers``svm_preds ``=` `final_svm_model.predict(test_X)``nb_preds ``=` `final_nb_model.predict(test_X)``rf_preds ``=` `final_rf_model.predict(test_X)` `final_preds ``=` `[mode([i,j,k])[``0``][``0``] ``for` `i,j,`` ``k ``in` `zip``(svm_preds, nb_preds, rf_preds)]` `print``(f\"Accuracy on Test dataset by the combined model\\``: {accuracy_score(test_Y, final_preds)````100``}\")` `cf_matrix ``=` `confusion_matrix(test_Y, final_preds)``plt.figure(figsize``=``(``12``,``8``))` `sns.heatmap(cf_matrix, annot ``=` `True``)``plt.title(``\"Confusion Matrix for Combined Model on Test Dataset\"``)``plt.show()`\n\nOutput:\n\n`Accuracy on Test dataset by the combined model: 100.0`", null, "We can see that our combined model has classified all the data points accurately. We have come to the final part of this whole implementation, we will be creating a function that takes symptoms separated by commas as input and outputs the predicted disease using the combined model based on the input symptoms.\n\nCreating a function that can take symptoms as input and generate predictions for disease\n\n## Python3\n\n `symptoms ``=` `X.columns.values` `# Creating a symptom index dictionary to encode the``# input symptoms into numerical form``symptom_index ``=` `{}``for` `index, value ``in` `enumerate``(symptoms):`` ``symptom ``=` `\" \"``.join([i.capitalize() ``for` `i ``in` `value.split(``\"_\"``)])`` ``symptom_index[symptom] ``=` `index` `data_dict ``=` `{`` ``\"symptom_index\"``:symptom_index,`` ``\"predictions_classes\"``:encoder.classes_``}` `# Defining the Function``# Input: string containing symptoms separated by commmas``# Output: Generated predictions by models``def` `predictDisease(symptoms):`` ``symptoms ``=` `symptoms.split(``\",\"``)`` ` ` ``# creating input data for the models`` ``input_data ``=` `[``0``] ``*` `len``(data_dict[``\"symptom_index\"``])`` ``for` `symptom ``in` `symptoms:`` ``index ``=` `data_dict[``\"symptom_index\"``][symptom]`` ``input_data[index] ``=` `1`` ` ` ``# reshaping the input data and converting it`` ``# into suitable format for model predictions`` ``input_data ``=` `np.array(input_data).reshape(``1``,``-``1``)`` ` ` ``# generating individual outputs`` ``rf_prediction ``=` `data_dict[``\"predictions_classes\"``][final_rf_model.predict(input_data)[``0``]]`` ``nb_prediction ``=` `data_dict[``\"predictions_classes\"``][final_nb_model.predict(input_data)[``0``]]`` ``svm_prediction ``=` `data_dict[``\"predictions_classes\"``][final_svm_model.predict(input_data)[``0``]]`` ` ` ``# making final prediction by taking mode of all predictions`` ``final_prediction ``=` `mode([rf_prediction, nb_prediction, svm_prediction])[``0``][``0``]`` ``predictions ``=` `{`` ``\"rf_model_prediction\"``: rf_prediction,`` ``\"naive_bayes_prediction\"``: nb_prediction,`` ``\"svm_model_prediction\"``: nb_prediction,`` ``\"final_prediction\"``:final_prediction`` ``}`` ``return` `predictions` `# Testing the function``print``(predictDisease(``\"Itching,Skin Rash,Nodal Skin Eruptions,Dischromic Patches\"``))`\n\nOutput:\n\n```{\n'rf_model_prediction': 'Fungal infection',\n'naive_bayes_prediction': 'Fungal infection',\n'svm_model_prediction': 'Fungal infection',\n'final_prediction': 'Fungal infection'\n}```\n\nNote: The symptoms that are given as input to the function should be exactly the same among the 132 symptoms in the dataset.\n\nAttention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.\n\nTo begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning – Basic Level Course\n\nMy Personal Notes arrow_drop_up" ]	[ null, "https://media.geeksforgeeks.org/wp-content/uploads/20210810014406/DiseasePrediction1000x600.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20210810015040/balancedata.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20210810015754/svmcm.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20210810020140/nbcm.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20210810020206/rfcm.png", null, "https://media.geeksforgeeks.org/wp-content/uploads/20210810020702/cbcm.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6917521,"math_prob":0.9339027,"size":14357,"snap":"2021-43-2021-49","text_gpt3_token_len":3448,"char_repetition_ratio":0.16839685,"word_repetition_ratio":0.11615925,"special_character_ratio":0.25416175,"punctuation_ratio":0.14981876,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9922042,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T09:30:14Z\",\"WARC-Record-ID\":\"<urn:uuid:a36126c8-698a-4aeb-a8a0-4d18fcfb2637>\",\"Content-Length\":\"160944\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ae8d1f7e-c0d9-4d9b-a9de-4ea777301d92>\",\"WARC-Concurrent-To\":\"<urn:uuid:f652c68a-494f-45eb-89d5-657d7ae4503f>\",\"WARC-IP-Address\":\"23.222.4.87\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/disease-prediction-using-machine-learning/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:VUFRB5BVLPTW2K3HTI5H2MZQIBKRQFS5\",\"WARC-Block-Digest\":\"sha1:SUBQPY4BSDKXBRAFNSQB2EMCI35HDZPX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585653.49_warc_CC-MAIN-20211023064718-20211023094718-00663.warc.gz\"}"}
https://en.wikipedia.org/wiki/Signal_separation	[ "# Signal separation\n\nSource separation, blind signal separation (BSS) or blind source separation, is the separation of a set of source signals from a set of mixed signals, without the aid of information (or with very little information) about the source signals or the mixing process. It is most commonly applied in digital signal processing and involves the analysis of mixtures of signals; the objective is to recover the original component signals from a mixture signal. The classical example of a source separation problem is the cocktail party problem, where a number of people are talking simultaneously in a room (for example, at a cocktail party), and a listener is trying to follow one of the discussions. The human brain can handle this sort of auditory source separation problem, but it is a difficult problem in digital signal processing.\n\nThis problem is in general highly underdetermined, but useful solutions can be derived under a surprising variety of conditions. Much of the early literature in this field focuses on the separation of temporal signals such as audio. However, blind signal separation is now routinely performed on multidimensional data, such as images and tensors, which may involve no time dimension whatsoever.\n\nSeveral approaches have been proposed for the solution of this problem but development is currently still very much in progress. Some of the more successful approaches are principal components analysis and independent component analysis, which work well when there are no delays or echoes present; that is, the problem is simplified a great deal. The field of computational auditory scene analysis attempts to achieve auditory source separation using an approach that is based on human hearing.\n\nThe human brain must also solve this problem in real time. In human perception this ability is commonly referred to as auditory scene analysis or the cocktail party effect.\n\n## Applications\n\n### Cocktail party problem\n\nAt a cocktail party, there is a group of people talking at the same time. You have multiple microphones picking up mixed signals, but you want to isolate the speech of a single person. BSS can be used to separate the individual sources by using mixed signals. In the presence of noise, dedicated optimization criteria need to be used\n\n### Image processing\n\nFigure 2 shows the basic concept of BSS. The individual source signals are shown as well as the mixed signals which are received signals. BSS is used to separate the mixed signals with only knowing mixed signals and nothing about original signal or how they were mixed. The separated signals are only approximations of the source signals. The separated images, were separated using Python and the Shogun toolbox using Joint Approximation Diagonalization of Eigen-matrices (JADE) algorithm which is based on independent component analysis, ICA. This toolbox method can be used with multi-dimensions but for an easy visual aspect images(2-D) were used.\n\n### Medical imaging\n\nOne of the practical applications being researched in this area is medical imaging of the brain with magnetoencephalography (MEG). This kind of imaging involves careful measurements of magnetic fields outside the head which yield an accurate 3D-picture of the interior of the head. However, external sources of electromagnetic fields, such as a wristwatch on the subject's arm, will significantly degrade the accuracy of the measurement. Applying source separation techniques on the measured signals can help remove undesired artifacts from the signal.\n\n### EEG\n\nIn electroencephalogram (EEG) and magnetoencephalography (MEG), the interference from muscle activity masks the desired signal from brain activity. BSS, however, can be used to separate the two so an accurate representation of brain activity may be achieved.\n\n### Music\n\nAnother application is the separation of musical signals. For a stereo mix of relatively simple signals it is now possible to make a fairly accurate separation, although some artifacts remain.\n\n### Others\n\nOther applications:\n\n• Communications\n• Stock Prediction\n• Seismic Monitoring\n• Text Document Analysis\n\n## Mathematical representation\n\nThe set of individual source signals, $s(t)=(s_{1}(t),\\dots ,s_{n}(t))^{T}$", null, ", is 'mixed' using a matrix, $A=[a_{ij}]\\in \\mathbb {R} ^{m\\times n}$", null, ", to produce a set of 'mixed' signals, $x(t)=(x_{1}(t),\\dots ,x_{m}(t))^{T}$", null, ", as follows. Usually, $n$", null, "is equal to $m$", null, ". If $m>n$", null, ", then the system of equations is overdetermined and thus can be unmixed using a conventional linear method. If $n>m$", null, ", the system is underdetermined and a non-linear method must be employed to recover the unmixed signals. The signals themselves can be multidimensional.\n\n$x(t)=A\\cdot s(t)$", null, "The above equation is effectively 'inverted' as follows. Blind source separation separates the set of mixed signals, $x(t)$", null, ", through the determination of an 'unmixing' matrix, $B=[B_{ij}]\\in \\mathbb {R} ^{n\\times m}$", null, ", to 'recover' an approximation of the original signals, $y(t)=(y_{1}(t),\\dots ,y_{n}(t))^{T}$", null, ".\n\n$y(t)=B\\cdot x(t)$", null, "## Approaches\n\nSince the chief difficulty of the problem is its underdetermination, methods for blind source separation generally seek to narrow the set of possible solutions in a way that is unlikely to exclude the desired solution. In one approach, exemplified by principal and independent component analysis, one seeks source signals that are minimally correlated or maximally independent in a probabilistic or information-theoretic sense. A second approach, exemplified by nonnegative matrix factorization, is to impose structural constraints on the source signals. These structural constraints may be derived from a generative model of the signal, but are more commonly heuristics justified by good empirical performance. A common theme in the second approach is to impose some kind of low-complexity constraint on the signal, such as sparsity in some basis for the signal space. This approach can be particularly effective if one requires not the whole signal, but merely its most salient features.\n\n### Methods\n\nThere are different methods of blind signal separation:" ]	[ null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/e6f75956a0e4ccb8b7eee334a9d07efca8c3de3e", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/cd83df9415b892d059f044be46076ba8bbdd9ca3", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/939854b8f9a75c494b7101bdec2fbf4fe752b28c", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/a601995d55609f2d9f5e233e36fbe9ea26011b3b", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/0a07d98bb302f3856cbabc47b2b9016692e3f7bc", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/637039c4a193f33fee72ebfeb6cb003593696160", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/e64e2a4a5b6cb58f1553c6a65551b4898bb82403", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/859458138ba843bdb5737b7eb96d0ddbab23c1eb", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/d54c275db3a1e620737b58e143b0818107fa5f5c", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/13420de72f09296437e2605fb4f936b4f23bc7f3", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/443f30d34ebb84e644cc765c46c8af55650a1bcb", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/0ecc0c3d49c2a79f9a7f0f069f2ac97d82acf66e", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8794217,"math_prob":0.9339972,"size":7738,"snap":"2022-27-2022-33","text_gpt3_token_len":1684,"char_repetition_ratio":0.14093612,"word_repetition_ratio":0.0035906644,"special_character_ratio":0.20638408,"punctuation_ratio":0.12910448,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96425974,"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],"im_url_duplicate_count":[null,4,null,4,null,4,null,null,null,null,null,null,null,null,null,4,null,null,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-12T01:12:06Z\",\"WARC-Record-ID\":\"<urn:uuid:8926afd3-7a1f-440d-858a-bb0aab0d2723>\",\"Content-Length\":\"79911\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:39f2142b-f3ce-4bfe-87a4-d31c4a1618bf>\",\"WARC-Concurrent-To\":\"<urn:uuid:72c7b6f6-98b5-4f4c-bfe8-ee9b160f7b0f>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://en.wikipedia.org/wiki/Signal_separation\",\"WARC-Payload-Digest\":\"sha1:ZS54FTB3YHWIGMJVDFUFUXGSJ5RRAPB6\",\"WARC-Block-Digest\":\"sha1:MPNOONGJ2HDZB3L7RBVKYQUVBJFALK7K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571536.89_warc_CC-MAIN-20220811224716-20220812014716-00021.warc.gz\"}"}
https://www.physicsforums.com/threads/shell-theorem-false-centers.239876/	[ "# Shell Theorem & False Centers\n\nThe Newtonian Shell Theorem makes important predictions about the gravitational force experienced by an object located inside a spherically symetrical massive structure, such as a hollow or solid ball. The same predictions are made by Gauss' Law. According to http://en.wikipedia.org/wiki/Shell_theorem\" [Broken]:\n\n1. If the body is a spherically symmetric shell (i.e. a hollow ball), no gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.\n\n2. Inside a solid sphere of constant density the gravitational force varies linearly with distance from the center, becoming zero at the center of mass.\n\nIt bears emphasizing that the Shell Theorem applies only with respect to mass arrangements which are (a) actually centered on the center of the sphere, (b) are symetrically distributed around that center, and (c) have constant density in the non-hollow portion. In this thread I want to discuss some examples that don't seem to meet all three criteria, although the issue can become interestingly subtle.\n\nThe most obvious example is that a solid or hollow shell of matter cannot act as a \"gravity shield\" to avoid the gravity of other nearby objects. For example a dense sphere buried just under the earth's surface is not shielded from the earth's gravity to any significant degree. This is so of course because the earth is not arranged spherically around the sphere, and the center of the sphere is not the \"true\" center of the earth-sphere system. I attached a simple diagram which illustrates this example.\n\nI'll post a more interesting example later.\n\nJon\n\n#### Attachments\n\n• Sphere A inside Sphere B.pdf\n47.6 KB · Views: 169\nLast edited by a moderator:\n\nInfinite homogeneous dust\n\nOK, so here's a subtle example.\n\nLet's consider an infinite, spatially flat, Lambda = 0 universe filled homogeneously with \"dust\" (matter particles). Let's arbitrarily define an origin coordinate and a metric sphere centered on that origin. Let's say this universe is expanding slowly with a low deceleration factor (relative to the proper size of the sphere). Let's place a massless test particle inside this sphere; it is at proper rest with respect to our origin coordinate. Can we correctly apply the shell theorem to this test particle?\n\nSure, as long as the center of our sphere represents the true center of the dust cloud. Uh oh... this dust cloud is infinite so it does not have a gravitational center. We can establish an arbitrary center by defining any coordinate system we wish. But it is not a gravitational center; a universe of this type has no true center.\n\nTherefore I hypothesize that the test particle will NOT move towards the center of the sphere in the manner described in the shell theorem. It will not move at all because it feels no directional gravity vector.\n\nLet's test this hypothesis by arbitrarily defining three very large, symetrically arranged, overlapping metric spheres, each with its own arbitrarily defined origin coordinate. An observer is located at each origin point. We place the massless test particle equidistant between the three origin points. In what direction will the test particle move? A diagram is attached.\n\nJon\n\n#### Attachments\n\n• 3 spheres & test particle 6-12-08.pdf\n100.1 KB · Views: 182\nHaelfix\nWrite down the metric for your test universe, and try to derive gauss's law. I suspect you will find a problem (assuming the metric is what I think you think it is).\n\nWallace\nCan we correctly apply the shell theorem to this test particle?\n\nSure, as long as the center of our sphere represents the true center of the dust cloud.\n??? What has 'the true center' of the dust cloud got to do with whether Gauss's law is valid?? You might want to find some basic undergraduate texts to look this up, I think you have an misconceived view of what the shell thereom is.\n\nLet's test this hypothesis by arbitrarily defining three very large, symetrically arranged, overlapping metric spheres, each with its own arbitrarily defined origin coordinate. An observer is located at each origin point. We place the massless test particle equidistant between the three origin points. In what direction will the test particle move? A diagram is attached.\n\nJon\n\nThe answer is very simple, the particle will be accelerated towards the center of all three spheres. Where they actually move in the next instant however depends on their current velocity of course, not the acceleration.Think about what this represents, it's a deccelerating homogenous universe which means that the rate that any pair of particles is moving apart is slowing down. Thus the rate that the chosen test particle is receeding from the centre of all three spheres is slowing. Thus the shell theorem clearly works correctly and gives you precisely the result you expect.\n\nHi Haelfix,\n\nWrite down the metric for your test universe, and try to derive gauss's law. I suspect you will find a problem (assuming the metric is what I think you think it is).\n\nI'm not sure what point you are trying to get at. Maybe you can explain more.\n\nAs I understand it, Gauss' Law is based on the fact that inverse square forces (such as gravity) produce equal flux across a spherical surface regardless of the radius of the sphere.\n\nSince I defined the test universe to be flat, I think that Gauss's law will apply consistent with my description. If it were spatially curved however (e.g. a closed universe) Gauss' Law would not calculate a correct result unless it were corrected for the curvature.\n\nJon\n\nLast edited:\nHi Wallace,\nThe answer is very simple, the particle will be accelerated towards the center of all three spheres. Where they actually move in the next instant however depends on their current velocity of course, not the acceleration.Think about what this represents, it's a decelerating homogenous universe which means that the rate that any pair of particles is moving apart is slowing down. Thus the rate that the chosen test particle is receeding from the centre of all three spheres is slowing. Thus the shell theorem clearly works correctly and gives you precisely the result you expect.\n\nWe agree on this point! The test particle will accelerate towards all three spheres equally. And in the same way, if we define an infinite number of symmetrical surrounding spheres (in three dimensions), the test particle will accelerate equally towards all of them. In other words, the test particle gains no directional vector at all.\n\nSo, the test particle will not actually move at all (in proper distance) towards the center of any individual sphere arbitrarily selected from among the multitude of surrounding spheres.\n\nEDIT: If it wasn't clear already, I want to make it clear that the origins of the three spheres ( or infinite number of spheres) are not in proper motion with respect to the test particle at the start of the test.\n\nJon\n\nLast edited:\nChronos\nGold Member\nThat is absurd. The universe has no shell-like observational features.\n\nWallace\nHi Wallace,\n\nWe agree on this point! The test particle will accelerate towards all three spheres equally. And in the same way, if we define an infinite number of symmetrical surrounding spheres (in three dimensions), the test particle will accelerate equally towards all of them. In other words, the test particle gains no directional vector at all.\n\nWhat is a 'directional vector'? Presumably something different from a non-directional vector, which is a clearly nonsensical concept.\n\nI think you are trying to say that in a homogenous and isotropic universe there is no prefferred direction, which is obviously true by definition. Every direction from the particle is the same so of course it feels no net acceleration in any direction.\n\nSo, the test particle will not actually move at all (in proper distance) towards the center of any individual sphere arbitrarily selected from among the multitude of surrounding spheres.\n\nEDIT: If it wasn't clear already, I want to make it clear that the origins of the three spheres ( or infinite number of spheres) are not in proper motion with respect to the test particle at the start of the test.\n\nJon\n\nWrong. The particle will move towards the center of all of the other spheres.\n\nI really fail to see the point of this thread? You seem to be attempting to demonstrate that you get different physical results depending on how you choose to define co-ordinates (in this case the centers of spherical Gaussian surfaces). The only way this could possibly occur is if Gauss's law fails. Clearly the resolution is to correctly understand Gausses law instead of claiming you have discovered some interesting paradox.\n\nWallace\nThat is absurd. The universe has no shell-like observational features.\n\nTo Jon's defence on this point, the 'shells' he is reffering to are Gaussian surfaces, which are mathematical constructs to help solve a problem more easily. He is not refferring to any physical shell like features, the situation under consideration is a homogenous universe.\n\nHi Wallace,\nThe particle will move towards the center of all of the other spheres.\nI admit I'm dense sometimes, but I don't understand this statement. In my diagram, from a tops-down two dimensional perspective, the stationary centers of the three spheres are set at 120 degree azimuthal angles relative to each other, so of course a test particle equidistant between them cannot move at three different angles at once.\n\nAny physical movement of a test particle can be defined as a vector in proper coordinates, with a magnitude and a single coordinate direction. Yet you agree that the movement here is non-directional and that it is nonsense to speak of nondirectional vectors. Therefore by first principles the test particle does not move.\n\nI can only interpret your statement to mean that the test particle does not move, but rather the three sphere origin points will gravitationally collapse towards the test particle. That outcome cannot occur in this scenario. The three sphere origins are mere proper coordinate points, they are not massive objects; the gravity of the background dust cloud does not cause their proper distances relative to each other to change.\n\nYou seem to be attempting to demonstrate that you get different physical results depending on how you choose to define co-ordinates (in this case the centers of spherical Gaussian surfaces). The only way this could possibly occur is if Gauss's law fails.\nYes as you say I am attempting to demonstrate that use of different coordinates changes the physical result. However, I disagree that this means Gauss' law fails. The mathematics of Gauss' law is just fine. What I am suggesting is that Gauss' law simply does not apply to mass distributions in which the center of the sphere is not the gravitational center of the entire mass distribution. You have already characterized this suggestion as nonsense, but I think my example demonstrates that the problem is legitimately subtle.\n\nJon\n\nLast edited:\nWallace\nHi Wallace,\n\nI admit I'm dense sometimes, but I don't understand this statement. In my diagram, from a tops-down two dimensional perspective, the stationary centers of the three spheres are set at 120 degree azimuthal angles relative to each other, so of course a test particle equidistant between them cannot move at three different angles at once.\n\nAny physical movement of a test particle can be defined as a vector in proper coordinates, with a magnitude and a single coordinate direction. Yet you agree that the movement here is non-directional and that it is nonsense to speak of nondirectional vectors. Therefore by first principles the test particle does not move.\n\nMovement, i.e. having a non-zero velocity, means that a particles co-ordinates are changing. In order to know if a particles co-ordinates are changing one needs to know how the co-ordinate system has been defined. It is also crucial to realise that the behaviour of something in a particular co-ordinate system is not universal, so something can be moving on one co-ordinate system but not in another.\n\nNow, if you construct a spherical co-ordinate system centered on the point in the middle of your diagram then clearly the point does not move in these co-ordinates. What you will find however, is that clearly the three other particles all move towards the origin, due to the gravitational attraction of the mass enclosed by a sphere centered on the origin with a radius extending to the other test particle. Your test particles will be accelerated by gravity, so clearly they must be accelerated towards the centre.\n\nNow, if you instead construct a spherical co-ordiant system centred on another particle then we can consider that the particle in the 'middle' which is no longer at the origin will be accelerated towards the new origin by the same reasoning. Which particles move and which do not depends on the co-ordinate system. What does not depend on the co-ordinate system however is the distance between each particle as a function of time. It doesn't matter how you define your co-ordinates you will find that the particles will move together due to gravity at exactly the same rate.\n\nIt is obvious what will happen in this situation, that the universe will collapse. It is also clear that this is consistent with the shell theorem.\n\nI can only interpret your statement to mean that the test particle does not move, but rather the three sphere origin points will gravitationally collapse towards the test particle.\n\nYes, absolutely for the reasons described above.\n\nThat outcome cannot occur in this scenario. The three sphere origins are mere proper coordinate points, they are not massive objects; the gravity of the background dust cloud does not cause their proper distances relative to each other to change.\n\nI don't know what you are saying here? You said in post #2 that 'an observer is placed at each origin point' this seems to imply that there are massless test particles at these points. If so then these test particles will be accelerated by gravity towards every other particle in the Universe as described above. They most definately will all move towards the middle particle. If these points are not massless test particles and merely arbitrarily defined points in the space then this whole discussion is meaningless. Co-ordinates may do whatever we define them to in a way unconstrained by physics, we could having expanding co-ordinates in a contracting universe if we wanted.\n\nIn terms of physics (i.e. what does the actual matter do?) the situation here is clear. The universe collapsing and we can understand this by considering any one of the four spheres in your picture.\n\nWhat I am suggesting is that Gauss' law simply does not apply to mass distributions in which the center of the sphere is not the gravitational center of the entire mass distribution. You have already characterized this suggestion as nonsense, but I think my example demonstrates that the problem is legitimately subtle.\n\nJon\n\nI really don't think this thread belongs in cosmology. You will probably get a better explanation of the fundamentals of Gauss's law in some of the other sub forums. I suggest you ask this question in one of those, since you clearly have misconceptions about it. Even reading the Wiki article might be a good start. Don't take this as a criticism, I'm trying help and it is clear that the confusion here lies in a fundamental misunderstanding of this physical law. Trying to understand this in the context of cosmology is then like trying to run before crawling. Learn to crawl first and you will get to running properly much faster.\n\nIf the universe is expanding at a rate greater than the 'escape velocity' then it won't recollapse.\n\nIf the universe is static then the question is whether it can remain so. This question has a long history.\n\nEinstein said no - he had to introduce the cosmological constant to try to fudge it.\n\nNewton said yes. Each object in the universe will be attracted equally from all directions and so will have no reason to move.\n\nHowever, even in Newtonian gravity, one can argue that if we divide the universe up into spherical shells, then those outside a given radius will have no effect and so the ball inside that radius will collapse. Applying this to the universe as a whole you get it collapsing to a point.\n\nCertainly this worried those who followed Newton. For instance, Boskovich mentioned the possibility of a form of cosmological constant in 1763, and I've read that Newton had a similar idea (although I haven't been able to find it in his works)\n\nYou said in post #2 that 'an observer is placed at each origin point' this seems to imply that there are massless test particles at these points. If so then these test particles will be accelerated by gravity towards every other particle in the Universe as described above. They most definately will all move towards the middle particle.\nWallace, your line of logic here is not helpful in understanding the problem. The coordinate system in my example is defined solely by the 3 center points of the 3 spheres, not by the test particle. In the \"thought exeriment\" way that these scenarios are usually described... we send fictional spaceships to each of the 3 sphere centers. Spaceship #1 drops a small marker at an arbitrary location. The 3 spaceships then use radar ranging and fictional rulers laid end to end (like Barnes & Francis do) to define the 3 sphere center locations with respect to the single marker, and to verify that they remain at a fixed proper distance from each other throughout the exercise.\n\nYou err in assuming that the spaceships will be spontaneously pulled by gravity towards the test particle. Why would a homogeneous, directionless dust cloud around the spaceships cause them to move an a single arbitrary direction? Why not the opposite direction? This is another obvious microcosm of the false center problem.\n\nEven if the spaceships did feel a unique directional gravitational pull in one arbitrary direction (which they don't), it would in no way undermine the validity of the simple coordinate system which is defined by points mapped in space, not by the spaceships.\n\nIf these points are not massless test particles and merely arbitrarily defined points in the space then this whole discussion is meaningless.\nAll of the \"Tethered Galaxy\" experiments define an arbitrary origin, so of course arbitrary origins are meaningful. Using 3 arbitrary origins should be just as meaningful, but my example doesn't need to rely on that, since 2 of the 3 sphere center origins are defined with reference to a single initial arbitrarily placed marker.\n\nThe universe collapsing and we can understand this by considering any one of the four spheres in your picture.\nMy scenario defined the universe to be spacially flat and expanding at the decelerating Einstein-de Sitter rate. The universe I defined will never collapse. Please try to keep the facts straight.\n\nEven reading the Wiki article might be a good start.\nOf course I read the Wikipedia article and much more on this subject. It was reading textbooks which alerted me to the fact that it is a common misconception that the shell theorem and Gauss' Law apply when the center of the sphere is not the center of the mass distribution being examined. May I respectfully suggest that you brush up a bit?\n\nJon\n\nLast edited:\nHi chronon,\nIf the universe is static then the question is whether it can remain so. This question has a long history.\nYes as I understand it the problem with a static universe with a cosmological constant is that it is inherently unstable. Even the most microscopic change in mass will cause the universe to accelerate into an expansion or collapse mode. So it is commonly said that the universe must be either expanding or contracting, and not static.\n\nThe only models I can think of for a stable static universe are the \"black hole inhibitors\" I mentioned in another thread. Virial motion in theory ought to be able to stabilize a gravitationally overdense universe; but over the long term virial motion is perturbed by collisions and near collisions, as well as by gravity waves, so it eventually will collapse. Adding some amount of cosmological constant would help stabilize it, I'm not sure if it can entirely stabilize. The maximum size of a virial universe is also limited by causality, is there enough time in the history of the universe for it to coordinate its virialization?\n\nThe other alternative is an electrostatically charged universe, such as one filled with electron plasma. But seems to be a fairly useless model as a practical matter because if regular electrically neutral matter is added to it, the neutral matter will gravitationally collapse even though the electron plasma won't. Perhaps adjusting the exact mix of neutral and charged matter would help stabilize it, and adding a cosmological constant ought to help.\n\nJon\n\nLast edited:\nHi Wallace,\nHere's a slightly simplified example (attachment) that may help clarify this. In this version, 3 very massive objects are attached to a strong but lightweight (insignificant mass) support ring. A \"Shell Theorem sphere\" is measured around each massive object. Again a test particle is dropped equidistant from the 3 massive objects. Which way will it move?\n\nOf course, it will not move at all.\n\nOne more thought about \"arbitrary\" origins: The earth, the center of our Local Group, or any other point in the universe that one wishes to use use as an \"origin\" of a coordinate system is equally arbitrary.\n\nJon\n\n#### Attachments\n\n• Sphere with 3 masses.pdf\n105.7 KB · Views: 150\nLast edited:\nHaelfix\nJon, please write the metric down for what you have in mind and the coordinates you want to use. We can't talk about this quantitatively if you do not. I don't understand what you have in mind, and in particular this can alter the shell theorem (Gauss's theorem is not mantained unambigously bc of coordinate subleties if its not exactly flat).\n\nThe reason I ask, is b/c you seem to think this acts like something static, even though elsewhere you assumed a FRW universe with k = 0 and lambda = 0. Alternatively I think there is a confusion about global issues vs local ones.\n\nTo wit, in the limit where things are very flat (or when spatial curvature/cc is very small relative to the density term), the universe is in a approximately newtonian limit and the shell theorem must be valid.\n\nLast edited:\nHi Haelfix,\nFirst, check out my most recent post where I attached a slightly revised version of the example, it may resolve your question.\nJon, please write the metric down for what you have in mind and the coordinates you want to use.\nI'm sorry but I don't understand your question. The universe in this model is a very simple straightforward one, a homogeneous, isotropic, spatially flat, matter dominated, dust-filled Einstein-de Sitter universe with Lambda = 0. It is expanding and the deceleration factor q is positive. The standard RW line element defines the metric.\n\nAs I said, an arbitrary origin point #1 is selected, and then two additional origin points are mapped out at equal distances from origin #1 and separated by 60 degrees of azimuth angle (from the perspective of origin #1).\nTo wit, in the limit where things are very flat (or when spatial curvature/cc is very small relative to the density term), the universe is in a approximately newtonian limit and the shell theorem must be valid.\nI agree completely, the shell theorem and Gauss' Law are valid. But they are valid only when applied to a proper set of facts. The facts must be such that the hypothetical point mass at the center of the sphere is the gravitational center of the entire mass distribution.\n\nJon\n\nLast edited:\nWell, on further review the last example I posted (Sphere with 3 masses) doesn't accomplish what I hoped. It serves as only as an example of how the Shell Theorem works with respect to external test objects.\n\nMy discussion has been about test objects located inside a solid or homogeneous mass (e.g., dust filled sphere.) For that discussion, we'll need to revert back to the original \"3 spheres\" example. Sorry about that.\n\nJon\n\nHi Wallace,\nIt doesn't matter how you define your co-ordinates you will find that the particles will move together due to gravity at exactly the same rate.\n\nIt is obvious what will happen in this situation, that the universe will collapse. It is also clear that this is consistent with the shell theorem.\n\nIn terms of physics (i.e. what does the actual matter do?) the situation here is clear. The universe collapsing and we can understand this by considering any one of the four spheres in your picture.\nThese statements of yours may be key to sorting out our discussion. If the particles in the universe were to move closer together in the manner you describe, then by definition the average matter density increases as a function of time. Yet every elementary textbook tells us that the defining characteristic of a homogeneous, expanding FLRW universe is that average matter density relentlessly decreases as a function of time. The matter density in any one (and in all 3) of the spheres will decrease over time, not increase. (Think of density in terms of the average proper distance between dust particles.) Surely you must acknowledge the direct conflict here with your interpretation of the shell theorem. If density were to increase in one spherical region of the universe, then it would need to decrease faster than the FLRW rate in other regions of the universe. That is inconsistent with the very concept of homogeneity.\n\n(Of course for the purposes of this theoretical discussion we are assuming perfect homogeneity and ignoring any local gravitational perturbations caused by primordial overdensities, etc.)\n\nJon\n\nLast edited:\nSorry to keep coming back to this, but here's one more variation on the thought experiment which I think illustrates the absurdity of permitting only one sphere at a time to measure the valid shell theorem results.\n\nLet's say we build a very large rigid ring using advanced materials which are so astoundingly lightweight that the mass of the ring is insignificant compared to the homogeneous dust density of the region of the universe it occupies. We'll mount 360 center-facing numbered video cameras on the ring, exactly 1 degree of azimuth apart. We will consider each video camera to be the center point of its own Shell Thereom sphere, all spheres being equally sized and just large enough to overlap at the point in space defining the radial center of the ring.\n\nBefore the start of each test round, we will write down on a slip of paper which of the 360 cameras we have randomly selected to be the origin of our coordinate system for that round. A different camera is designated for each test round. We will not disclose this information to the spaceship pilot, who will release the massless test particle at the exact centerpoint and depart.\n\nDoes the shell theorem predict that in each test round, the test particle will move only towards the particular video camera whose number is written on on the current slip of paper? Need I say more?\n\nJon\n\nLast edited:\nBefore the start of each test round, we will write down on a slip of paper which of the 360 cameras we have randomly selected to be the origin of our coordinate system for that round. A different camera is designated for each test round. We will not disclose this information to the spaceship pilot, who will release the massless test particle at the exact centerpoint and depart.\nYour origin is moving with respect to the surrounding dust which rather confuses things.\n\nIf you released a test particle from the ring then it would accelerate towards the centre.\n\nHi chronon,\nYour origin is moving with respect to the surrounding dust which rather confuses things.\nThis subject can get confusing. The origin can be any one of the cameras. In an FLRW universe, no matter where an observer is located, she sees the dust in her immediate local area not to be moving at all; she observers that objects further away from her are moving directly away from her, at velocities proportional to their distance.\n\nEDIT: I confused myself on my first try at answering this. I'll try again: Only one point at a time on the ring can be aligned with its local Hubble flow. All other points on the ring will observe the background dust cloud moving with respect to them. However, this should have no bearing on the movement of a local test particle.\n\nI've thought a lot about whether the proper motion (or the acceleration) of the dust cloud has any direct bearing on a proper-distance application of the shell theorem. I don't think it does. Acceleration has an indirect bearing in the sense that it results from the existence of a gravitational field.\nIf you released a test particle from the ring then it would accelerate towards the centre.\nIn my scenario the test particle is dropped at the radial center of the ring, not from the ring itself. However, even if the test particle were dropped next to the ring, I am sure the test particle would not move relative to the ring. The background dust cloud is directionless so there is nothing to cause it to impart a unique direction of movement to the test particle. We've defined the mass of the ring itself to be insignificant.\n\nJon\n\nLast edited:\nAnother easy example:\n\nA very long rigid ruler is constructed in space out of advanced materials which are so astoundingly lightweight that the mass of the ruler is insignificant compared to the homogeneous dust cloud in the region. The ruler is marked with 4 units of length. \"Shell Theorem spheres\" are measured, the smaller sphere having radius = 1 unit and its center located 1 unit from the left edge of the ruler; the larger sphere having radius = 4 units and its center located 4 units from the left end of the ruler. A test particle is released at the left end of the ruler, just at the edge of both spheres. Will the test particle initially accelerate at the rate determined by the mass of the r=1 sphere or the r=4 sphere? Diagram is attached.\n\nNote that the total mass of the r=4 sphere is 64 times larger than the mass of the r=1 sphere, proportional to $$r^{3}$$. If the Shell Theorem applied, the initial gravitational acceleration rate imparted to the test particle by the the r=4 sphere would be 4 times greater than the acceleration imparted from the r=1 sphere, due to the inverse square law as measured from each sphere's center.\n\nObviously, if either sphere satisfies the Shell Theorem here, then the other one violates it. So the Shell Theorem can't be applicable. In fact, the test particle will not move at all relative to the ruler.\n\nJon\n\n#### Attachments\n\n• 2 spheres & ruler.pdf\n144.6 KB · Views: 136\nLast edited:\nA very long rigid ruler is constructed in space out of advanced materials which are so astoundingly lightweight that the mass of the ruler is insignificant compared to the homogeneous dust cloud in the region. The ruler is marked with 4 units of length. \"Shell Theorem spheres\" are measured, the smaller sphere having radius = 1 unit and its center located 1 unit from the left edge of the ruler; the larger sphere having radius = 4 units and its center located 4 units from the left end of the ruler. A test particle is released at the left end of the ruler, just at the edge of both spheres. Will the test particle initially accelerate at the rate determined by the mass of the r=1 sphere or the r=4 sphere? Diagram is attached.\nYou need to specify how the ruler moves with respect to the surrounding dust. I assume that you want it to be stationary, but the dust cloud is expanding, so that only one part of the ruler can be stationary with respect to its surroundings. Is it the left end, the right end or the middle.\n\nIn fact, the test particle will not move at all relative to the ruler\nActually, for any of the options I suggested in the previous post, the particle will accelerate to the right with respect to the ruler.\n\nHi chronon,\nThanks for thinking this through with me.\n\nI've reached the point where I'm going to throw in the towel and agree that the Shell Theorem does apply correctly in a homogeneous, expanding infinite dust universe with Lambda = 0.\n\nThe textbooks and articles I've read do not clearly and completely explain the subtlety of how the Shell Theorem should be applied in an infinite homogeneous body. This leaves it to us to have to connect a lot of dots, so to speak.\n\nMy statement that the Shell Theorem requires a true gravitational center was not entirely incorrect. I maintain that it does.\n\nHowever, the subtlety is that every individual mass particle in space is a true gravitational center of its own. This is because the gravity of the dust cloud applies a directionless force which tries to collapse every particle in the universe closer to every other particle in the universe. Dust particles participating in the background Hubble flow have a pre-existing directionless momentum which acts to increase the proper distance between every particle and every other particle, that is, to expand the dust cloud. As we know, this expansionary momentum counteracts the dust cloud's own gravitational collapse force, and in a flat universe this results in the particles retaining momentum away from each other forever but at a rate that decreases asymptotically over time to zero. The fact that the expansionary momentum of the dust cloud never declines to zero demonstrates that the magnitude of the global expansionary action relatively dominates the magnitude of the global gravitational collapse action in perpetuity. Still, the global collapse action is always acting on every particle and everywhere in the dust cloud.\n\nAny set of massless test particles initially placed at proper rest with respect to each other, regardless of the number of such particles or their distances apart, will observe the proper distance between themselves decreasing as a function of time due to the background dust cloud's gravitational collapse force. The further the test particles are apart, the faster they will accelerate together: the acceleration is directly proportional to the distance. By definition the test particles have no initial expansionary momentum (relative to each other) acting to offset the collapse action, so the collapse action alone will control.\n\nBecause the collapse action of multiple test particles is truly directionless with respect to the global background dust cloud, any coordinate system we overlay onto the scenario will portray it in a misleading way. A coordinate system, regardless of where its arbitrary origin is placed, will portray that the collapse action of multiple test particles creates unique directional vectors. Relocating the origin of the coordinate system from one place to another will as Wallace says portray a different set of directional vectors. All such directional vectors are false from the absolute global perspective of the infinite universe; the collapse action of multiple test particles is always nondirectional.\n\nThat it took me so long to get a grip on this action, I will attribute to my own density. I hope that more insightful textbook entries and articles will be written on this subject.\n\nJon\n\nLast edited:\nOK, so now let's release a very large but finite number of test particles at random finite proper distances apart in this flat, homogeneous, dust-only model universe, with each test particle initially having zero instantaneous proper velocity relative to each other particle. One of the particles will be at rest relative to its local Hubble flow (and CMB frame), and by definition all of the other particles will not be. We'll station an observer next to each test particle, with each observer at zero proper velocity relative to all of the test particles and other observers. What will they observe?\n\nAt first, each observer will see all of the test particles and other observers accelerating rapidly toward that observer's location. Any two test particles which start out relatively far apart will initially observe a higher rate of proper acceleration together, proportional to their initial separation distance. After the test particles accelerate to a certain relative speed (proportional to initial separation), their proper speed will mostly flatten out (acceleration will asymptotically approach zero), due to both the declining force of gravity (force inversely proportional to their distance from the convergence point) and the ongoing evacuation of the dust cloud due to ongoing cosmic expansion. Eventually all of the test particles and observers will converge on a single point. The location of that point relative to the background dust cloud (it has to be relative to something tangible, other than each other) will be the starting point of the test particle which was initially at rest relative to its local Hubble flow and CMB frame. If the test particles (and observers) narrowly avoid colliding, they will pass through the convergence point and continue moving in straight trajectories out the opposite side of the convergence point. After that, the test particles (and observers) will continue separating from each other in perpetuity, at proper velocities that decline slightly over time.\n\nIn the above description, I omitted the complication that the convergence speed of the test particles (and observers) is limited to be < the speed of light. Presumably (I'm not sure about this) the test particles (and observers) whose convergence speed is relativistically reduced (compared to the Newtonian approximation) will be delayed in arriving at the convergence point. So an initial bunch of test particles will converge at the same time, followed by an ongoing stream of \"straggler\" test particles which will increase in density over time, then decrease, and finally end (since the number of test particles is finite.) After passing through the convergence point, this entire stream of test particles (and observers) will expand away from each other in perpetuity.\n\nJon\n\nLast edited:\nOK, so now let's release a very large but finite number of test particles at random finite proper distances apart in this flat, homogeneous, dust-only model universe, with each test particle initially having zero instantaneous proper velocity relative to each other particle.\n\n....\n\nAfter passing through the convergence point, this entire stream of test particles (and observers) will expand away from each other in perpetuity.\n\nJon\nThe behavior of a finite set of test particles reacting to the collapse action per the Shell Theorem is non-intuitive to me, but there it is, just like in the Tethered Galaxy problem.\n\nSo what will we observe if we release an infinite number of test particles at random proper distances apart in this model universe, with each test particle initially having zero instantaneous proper velocity relative to each other particle?\n\nI believe that the collapse action will be just the same as in the finite example, except that the influx density of test particles through the convergence point will continue increasing in perpetuity and never end. After test particles pass through the convergence point, they will expand away from each other in the opposite direction they arrived from, in perpetuity, and obviously with the same constantly increasing density as the influx.\n\nAn infinite and ever increasing dust storm!\n\nJon\n\nOK, so here's a subtle example.\n\nLet's consider an infinite, spatially flat, Lambda = 0 universe filled homogeneously with \"dust\" (matter particles). Let's arbitrarily define an origin coordinate and a metric sphere centered on that origin. Let's say this universe is expanding slowly with a low deceleration factor (relative to the proper size of the sphere). Let's place a massless test particle inside this sphere; it is at proper rest with respect to our origin coordinate. Can we correctly apply the shell theorem to this test particle?\n\nSure, as long as the center of our sphere represents the true center of the dust cloud. Uh oh... this dust cloud is infinite so it does not have a gravitational center. We can establish an arbitrary center by defining any coordinate system we wish. But it is not a gravitational center; a universe of this type has no true center.\n\nTherefore I hypothesize that the test particle will NOT move towards the center of the sphere in the manner described in the shell theorem. It will not move at all because it feels no directional gravity vector.\n\nLet's test this hypothesis by arbitrarily defining three very large, symetrically arranged, overlapping metric spheres, each with its own arbitrarily defined origin coordinate. An observer is located at each origin point. We place the massless test particle equidistant between the three origin points. In what direction will the test particle move? A diagram is attached.\n\nJon\n\nI think this example a bit bogus, since you have an infinite sphere. In that case you can not make some arbitrary assumption about the location of any center.\n\nBut apart from that these are nice examples.\nApplication of the Shell theorem in a homogeneous and isotropic universe would seem to indicate that each each mass would be in a stable equilibrium.\nBut one could argue also that each point in this universe is in an unstable or metastable equilibirum.\n\nAs another thought experiment:\nAssume an initial configuration of equal mass points m in an infinite 3D grid with length d. Initial all (relative) velocities of all mass particles is zero.\nWhat would be the time evolution of such a configuration?\n(both locally and globally)\n\nLast edited:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9348653,"math_prob":0.8091588,"size":2982,"snap":"2022-27-2022-33","text_gpt3_token_len":626,"char_repetition_ratio":0.12558764,"word_repetition_ratio":0.88114756,"special_character_ratio":0.20120725,"punctuation_ratio":0.08775137,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9609823,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T04:33:47Z\",\"WARC-Record-ID\":\"<urn:uuid:1d73cd78-e27d-4ceb-b892-f2f66e189865>\",\"Content-Length\":\"197107\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2af589dc-bf3b-4929-b0ac-0cc95f3be0c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:9005c16b-9680-4bdb-91c0-25e604c5a6fe>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/shell-theorem-false-centers.239876/\",\"WARC-Payload-Digest\":\"sha1:B4UDDHRKEU6IFC37E74C3YVOILIKYNHN\",\"WARC-Block-Digest\":\"sha1:54ZFZO4XWESKGDDRTFWBPK7LBV2XMMZR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103661137.41_warc_CC-MAIN-20220630031950-20220630061950-00312.warc.gz\"}"}
http://blog.sigfpe.com/2006/12/evaluating-cellular-automata-is.html	[ "## Tuesday, December 19, 2006\n\n### Evaluating cellular automata is comonadic\n\nPaul Potts's post inspired me to say something about cellular automata too.\n\nSo here's the deal: whenever you see large datastructures pieced together from lots of small but similar computations there's a good chance that we're dealing with a comonad. In cellular automata we compute the value of each cell in the next generation by performing a local computation based on the neighbourhood of that cell. So cellular automata look like they might form a good candidate for comonadic evaluation.\n\nI want to work on 'universes' that extend to infinity in both directions. And I want this universe to be constructed lazily on demand. One way of doing that is to represent a 'universe' as a centre point, a list of all elements to the left of that centre point and a list of all elements to the right. Here's a suitable\ntype:\n\n`> data U x = U [x] x [x]`\n\nFor example U [-1,-2..] 0 [1,2..] can be thought of as representing all of the integers in sequence.\n\nBut this actually contains slightly more information than a list that extends to infinity both ways. The centre point forms a kind of focus of attention. We could shift that focus of attention left or right. For example consider\n\nU [-2,-3..] (-1) [0,1..]\n\nThis represents the same sequence of integers but the focus has been shifted left. So think of the type U x as being a doubly infinite sequence with a cursor. (In fact, this makes it a kind of zipper.)\n\nWe can formalise the notion of shifting left and right as follows:\n\n`> right (U a b (c:cs)) = U (b:a) c cs> left (U (a:as) b c) = U as a (b:c)`\n\nAn object of type U is semantically like a C pointer into a const block of memory. You can increment it, decrement it and dereference it using the function I'll call coreturn below.\n\nAs U is a kind of list structure, it needs a map. In fact, we can define fmap for it:\n\n`> instance Functor U where> fmap f (U a b c) = U (map f a) (f b) (map f c)`\n\nNow the fun starts. First I'll bemoan the fact that Comonads aren't in the standard Haskell libraries (at least I don't think they are). So I have to define them myself:\n\n`> class Functor w => Comonad w where> (=>>) :: w a -> (w a -> b) -> w b> coreturn :: w a -> a> cojoin :: w a -> w (w a)> x =>> f = fmap f (cojoin x)`\n\ncojoin is the dual to the usual join function. I've chosen to do things the category theoretical way and define =>> in terms of cojoin.\n\nAnd here's why U forms a Comonad:\n\n`> instance Comonad U where> cojoin a = U (tail \\$ iterate left a) a (tail \\$ iterate right a)> coreturn (U _ b _) = b`\n\nLook closely at cojoin. It turns a into a 'universe' of 'universes' where each element is a copy of a shifted left or right a number of times. This is where all the work is happening. The reason we want to do this is as follows: we want to write rules that work on the local neighbourhoods of our universe. We can think of a universe with the cursor pointing at a particular element as being an element with a neighbourhood on each side. For example, we can write a cellular automaton rule like this:\n\n`> rule (U (a:_) b (c:_)) = not (a && b && not c \|\| (a==b))`\n\nIn order to apply this everywhere in the universe we need to apply the rule to each possible shift of the universe. And that's what cojoin does, it constructs a universe of all possible shifts of a. Compare with what I said here. So believe it or not, we've already written the code to evaluate cellular automata. u =>> rule applies the rule to u. The rest is just boring IO:\n\n`> shift i u = (iterate (if i<0 then left else right) u) !! abs i>> toList i j u = take (j-i) \\$ half \\$ shift i u where> half (U _ b c) = [b] ++ c>> test = let u = U (repeat False) True (repeat False)> in putStr \\$> unlines \\$> take 20 \\$> map (map (\\x -> if x then '#' else ' ') . toList (-20) 20) \\$> iterate (=>> rule) u`\n\nLazy infinite structures, comonads, zippers. I think I'm just beginning to get the hang of this functional programming lark! Over Xmas I might feel ready to try writing a piece of code longer than a dozen or so lines.\n\nAnyway, I must end with a credit. I probably wouldn't have come up with this if I hadn't read this paper by Uustalu and Vene.\n\nLabels:", null, "Andrea Vezzosi said...\n\nI like very much this lazy solution, but how would you generalize it to spaces with more dimensions?\n(Like in Life where you have 8 other cells in your neighborhood.)\nI can imagine increasing the number of the lists, but not very well how to travel this new universe, especially how to reach the cells in the \"oblique\" directions\n\nThursday, 21 December, 2006", null, "sigfpe said...\n\nI think there may be an easy but inefficient solution to the problem of working in 2D using the type U (U Bool). I'd imagine 'up' and 'down' to be defined like 'fmap left' and 'fmap right' or something like that. I think you can then construct a new 'cojoin' with a little bit of work. But this might be a bit inefficient. But maybe not - I'd have to think about it harder but all those 'fmap left's might still only do O(1) work per cell in total. When I next have a moment...\n\nThursday, 21 December, 2006", null, "sigfpe said...\n\nHere are some more details:\n\n> data U2 x = U2 (U (U x)) deriving Show\n\n> instance Functor U2 where\n\n> fmap f (U2 u) = U2 \\$ fmap (fmap f) u\n\n> coreturn (U2 u) = coreturn (coreturn u)\n\n> cojoin (U2 u) = fmap U2 \\$ U2 \\$ roll \\$ roll u where\n\n> iterate1 f = tail . iterate f\n\n> roll a = U (iterate1 (fmap left) a) a (iterate1 (fmap right) a)\n\n(Sorry about lack of indentation.) I have the game of life running fine.\n\nThe strategy for U2 is much the same as that for U. cojoin makes a 2D grid of 2D grids where each inner grid is shifted an amount corresponding to its location within the outer grid. A similar strategy can be used to implement many other comonads. In fact, that was the ulterior motive in writing this CA code: getting intuition about what cojoin typically looks like.\n\nFriday, 22 December, 2006", null, "Andrea Vezzosi said...\n\nwell I'm quite lost on the intuition of your roll, and why you use fmap U2 in cojoin.\nHowever, a rule would be written like this?\nrule (U2 (U ((U (a:_) b (c:_)):_) (U (d:_) x (f:_)) ((U (g:_) h (i:_)):_))) = ...\nwhere the letters have this alignment?\na b c\nd x f\ng h i\n\nSaturday, 23 December, 2006", null, "sigfpe said...\n\nHere is the exact rule I used in my code:\n\nrule' (U2 (U\n(U (u0:_) u1 (u2:_):_)\n(U (u3:_) u4 (u5:_))\n(U (u6:_) u7 (u8:_):_))) =\nlet n = length \\$ filter id [u0,u1,u2,u3,u5,u6,u7,u8] in\nu4 && (n==2 \|\| n==3) \|\| (not u4) && n==3\n\nMaybe I'll tidy up my code and fill it with useful comments so I can post the entire thing here.\n\nU2 converts a 1D universe of 1D universes into a 2D universe. 'fmap U2' converts...take a deep breath...a 2D universe of 1D universes of 1D universes into a 2D universe of 2D universes. 'fmap U2 . fmap' converts a 1D universe of 1D universes of 1D universes of 1D universes into a 2D universe of 2D universes. Conceptually, a 2D universe is the same thing as a 1D universe of 1D universes. The guts of the code is written in terms of 1D universes, and 'fmap U2 . U2' is just a bit of fluff at the end to convert everything back to 2D universes so cojoin has the right type.\n\nThe 'roll' is just taking advantage of the thing I said in an earlier comment. If you think of the inner U's as columns and the outer one as rows, then fmap left and right shift up and down. The slightly weird thing about 'roll' is that exactly the same piece of code does rows and columns whereas you might expect one function with fmaps for columns and one function without for rows. You can think of fmap as 'ducking down' a layer of U-flavoured onion skin. The two rolls do different things because each time it is used there are different layers of onion skin in place. The best way to make sense of it is to do what I did - make U and U2 instances of Show that can print stuff out with a pretty layout and then go through cojoin one step at a time.\n\nSaturday, 23 December, 2006", null, "kowey said...\n\nWhen I first saw this, I was frightened off by the seeming abundance of scary new concepts: cellular automata, comonads, functors (for no good reason I still get nervous when I see that word), universes. Then I read it a second time, read it instead of skimming it, and realised that oh... there really is nothing to it.\n\nThe comonad stuff will take a while to sink in, but reading your post, I get the impression that they are not fundamentally deeper than monads. No cobrain is required to understand them, after all. Note: fellow weaklings might do well to have Cale's Monads as containers open in a separate window or browser tab.\n\nThanks for this post. For starters, the literate Haskell helps, pasting and running is useful. More importantly, the post had this wonderful effect of mutual reinforcement, in which several things one does not know well get tied together, and as a result all become clearer. Not only do I now have a slight taste of comonad, I also have now more than heard about cellular automata, and my tenuous grasp of zippers is now strenghtened. Makes me wish zippers were first presented in list terms from the very beginning.\n\nWhat might have been helpful is some means for the easily frightened to realise that the post contains easy stuff, but short of inventing a Haskell coloured belts system, [this post is rated yellow belt; if you understand monads in terms of >>= and return, etc], I don't know what such a mechanism would consist of. Another potentially useful aid might be to put the example automaton rule into words, or even pictures. Incidentally, this particular example rule does not seem to make use of c. Is that right?\n\nTuesday, 02 January, 2007", null, "sigfpe said...\n\nI think you're right, my rule doesn't seem to use c. I just made up random rules until I liked the result.\n\nthere really is nothing to it\nYes! There are some incredibly difficult looking papers out there but when you actually get down to implementing examples they can suddenly seem close to trivial. Check out the comonad I define here for an example that's reminiscent of the kinds of adding-up-lists-of-prices examples given in books like \"Learn C in 21 days\". And yet the paper I drew on is scary as hell!\n\nTuesday, 02 January, 2007", null, "Anonymous said...\n\nYou should try implementing the \"Cellular Potts\" model developed by Glazer & Granier. It's more physically meaningful than CA.\n\nWednesday, 15 August, 2007", null, "Anonymous said...\n\nWhoa, and I thought this site was about the iPhone)!\n\nMonday, 21 January, 2008", null, "geophf said...\n\nAre all Zippers that are instances of Functor Comonads? Conversely, and simply, are all Comonads Zippers that are Functors? I've done a bit of reading on (co)monads, and the above is the current straw I'm grasping at for understanding comonads better. Am I missing the boat entirely?\n\nAs for the cellular automata implementation, Mathworld, etc, has a list of standardized rules. Rule 30 is demonstrated graphically as the banner of my company's webpage: http://www.logicaltypes.com/.\n\nHere is a way to pattern-match to build these standardized rules:\n\n> rule30 (U (a:_) b (c:_)) = bitfrom (rule30' (bitof a b c))\n> where rule30' 111 = 0\n> rule30' 110 = 0\n> rule30' 101 = 0\n> rule30' 100 = 1\n> rule30' 011 = 1\n> rule30' 010 = 1\n> rule30' 001 = 1\n> rule30' 000 = 0\n> bitof x y z = 100 * t x + 10 * t y + t z\n> bitfrom = toEnum\n\n(Ah, well, the indentation is not preserved, but you get the idea.)\n\nSince all the standardized rules follow the above pattern (where the rule number is the bit-pattern result from the three-bit-pattern input), it is trivial in Haskell to write a generalized\n\nrule :: Enum a => Int -> U a -> a\n\nfor all the rules that follow that pattern.\n\nMonday, 30 June, 2008", null, "Anonymous said...\n\nPondering rewriting this but representing the 'universe' as a function from an integer leads one to realise that we get a comonad from functions from any monoid as follows:\n\ncoreturn a = a mempty\ncojoin a = curry (a . uncurry mappend)\nfmap f = (f.)\n\nWhich allows us to handle automata on 1d, square grids, hex grids, and so on. Albeit perhaps inefficiently.\n\nI haven't tried this so it's probably embarrassingly wrong.\n\nWednesday, 18 February, 2009", null, "Alex S. said...\n\nSo I've gone forward and implemented the game of life, without any fancy printing, but still, it seems to work:\n\nhttp://hpaste.org/fastcgi/hpaste.fcgi/view?id=11012#a11019\n\nComments and suggestions are very welcome!\n\nAlex.\n\nWednesday, 21 October, 2009", null, "Alex S. said...\n\nSo I went ahead and wrote the code. It's on hpaste under http://hpaste.org/fastcgi/hpaste.fcgi/view?id=11012#a11019\n\nTry to run it, seems to work for me!\n\nAlex.\n\nWednesday, 21 October, 2009", null, "gereeter said...\n\nscc,\n\nThat's an interesting little duality. The product type can be a comonad with no restrictions, but only a monad if its \"extra\" type is a monoid. Similarly, the reader type (or function type, whichever you prefer) can be a monad with no restrictions, but can only be a comonad if the \"extra\" type is a monoid.\n\nAlso, you were expressing a lack of confidence in your definitions, so let's verify the comonad laws:\n\n1.\n\ncoreturn . cojoin\n= \\x -> coreturn (cojoin x)\n= \\x -> cojoin x mempty\n= \\x -> (curry (x . uncurry mappend)) mempty\n= \\x y -> (x . uncurry mappend) (mempty, y)\n= \\x y -> x (uncurry mappend (mempty, y))\n= \\x y -> x (mappend mempty y)\n= \\x y -> x y\n= \\x -> x\n= id\n\n2.\n\nfmap coreturn . cojoin\n= \\x -> fmap coreturn (cojoin x)\n= \\x -> coreturn . cojoin x\n= \\x y -> coreturn (cojoin x y)\n= \\x y -> cojoin x y mempty\n= \\x y -> curry (x . uncurry mappend) y mempty\n= \\x y -> (x . uncurry mappend) (y, mempty)\n= \\x y -> x (uncurry mappend (y, mempty))\n= \\x y -> x (mappend y mempty)\n= \\x y -> x y\n= \\x -> x\n= id\n\n3.\n\ncojoin . cojoin\n= \\x -> cojoin (cojoin x)\n= \\x -> cojoin (curry (x . uncurry mappend))\n= \\x -> curry (curry (x . uncurry mappend) . uncurry mappend)\n= \\x y z -> curry (curry (x . uncurry mappend) . uncurry mappend) y z\n= \\x y z -> (curry (x . uncurry mappend) . uncurry mappend) (y, z)\n= \\x y z -> curry (x . uncurry mappend) (uncurry mappend (y, z))\n= \\x y z -> curry (x . uncurry mappend) (mappend y z)\n= \\x y z w -> curry (x . uncurry mappend) (mappend y z) w\n= \\x y z w -> (x . uncurry mappend) (mappend y z, w)\n= \\x y z w -> x (uncurry mappend (mappend y z, w))\n= \\x y z w -> x (mappend (mappend y z) w)\n= \\x y z w -> x (mappend y (mappend z w))\n= \\x y z w -> x (uncurry mappend (y, mappend z w))\n= \\x y z w -> (x . uncurry mappend) (y, mappend z w)\n= \\x y z w -> curry (x . uncurry mappend) y (mappend z w)\n= \\x y z w -> cojoin x y (mappend z w)\n= \\x y z w -> cojoin x y (uncurry mappend (z, w))\n= \\x y z w -> (cojoin x y . uncurry mappend) (z, w)\n= \\x y z w -> curry (cojoin x y . uncurry mappend) z w\n= \\x y z w -> cojoin (cojoin x y) z w\n= \\x y -> cojoin (cojoin x y)\n= \\x y -> (cojoin . cojoin x) y\n= \\x -> cojoin . cojoin x\n= \\x -> fmap cojoin (cojoin x)\n= \\x -> (fmap cojoin . cojoin) x\n= fmap cojoin . cojoin\n\nPhew. That was long. But yes, your definitions are correct.\n\nMonday, 08 August, 2011", null, "Jeremy List said...\n\nToroidal universes are a pretty simple extension (you could just use circular lists but that would lead to recalculating a lot of values)\n\n> right (U a b []) = right (U [] b (reverse a))\n> right (U a b (c:cs)) = U (b:a) c cs\n> left (U [] b c) = left (U (reverse c) b [])\n> left (U (a:as) b c) = U as a (b:c)\n\nThursday, 17 October, 2013", null, "Jeremy List said...\n\nToroidal universes are a pretty simple extension (you could just use circular lists but that would lead to recalculating a lot of values)\n\n> right (U a b []) = right (U [] b (reverse a))\n> right (U a b (c:cs)) = U (b:a) c cs\n> left (U [] b c) = left (U (reverse c) b [])\n> left (U (a:as) b c) = U as a (b:c)\n\nThursday, 17 October, 2013", null, "Miguel Negrão said...\n\nIHaskell notebook with diagrams drawing of CA:" ]	[ null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/anon16-rounded.gif", null, "https://resources.blogblog.com/img/anon16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/anon16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/anon16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null, "https://resources.blogblog.com/img/b16-rounded.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9013498,"math_prob":0.8896356,"size":18365,"snap":"2023-40-2023-50","text_gpt3_token_len":5105,"char_repetition_ratio":0.1280976,"word_repetition_ratio":0.41785216,"special_character_ratio":0.28004357,"punctuation_ratio":0.10595332,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98402077,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T21:37:39Z\",\"WARC-Record-ID\":\"<urn:uuid:06dab7fa-2552-4c5c-81f4-7e31a1cd542f>\",\"Content-Length\":\"46816\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b426a432-037c-4e5d-9a51-25da052210f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:ce37da67-0f16-4946-b0fa-9a52cb389753>\",\"WARC-IP-Address\":\"142.251.111.121\",\"WARC-Target-URI\":\"http://blog.sigfpe.com/2006/12/evaluating-cellular-automata-is.html\",\"WARC-Payload-Digest\":\"sha1:NJKKYKGWOIGZTRHN2QPH2SSGRF2GCKSR\",\"WARC-Block-Digest\":\"sha1:KIK36K2RCKLYKOGDQCNEHZOME5PUWVCE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511220.71_warc_CC-MAIN-20231003192425-20231003222425-00040.warc.gz\"}"}
https://roboticsbackend.com/should-you-learn-mathematics-to-program-robots/	[ "# Should You Learn Mathematics To Program Robots?\n\nYou know that mathematics are everywhere, and especially in robotics.\n\nBut, if you just want to program robots, is it really important? Could you succeed without it?\n\nSo, should you learn mathematics to program robots? The answer is: yes and no. Quite useful, right? Because it’s not as simple as that.\n\nKeep reading so you’ll understand if and when you need to learn some mathematics, depending on what you want to achieve.\n\n## When you start a new project: use libraries\n\nYou can use tons of libraries that will allow you not to do the maths yourself.\n\nLet’s imagine that you have to compute the inverse kinematic of a robot. Well, this involves a lot of mathematics. Soon you’ll read stuff about the Denavit-Hartenberg parameters, and your motivation will start to fade.\n\nHow could you possibly use that in your code, without spending a huge amount of time understanding the maths? Time, of course, that you won’t spend on developing your robot software.\n\nWell, this is a problem you can solve (at least temporary). Instead of trying to do the maths yourself, what if you could find a library that can do the maths for you? That’s the good question to ask, especially if you want to progress fast on your prototype or small project.\n\nFor this specific example, you can quickly find the KDL from Orocos: Kinematics and Dynamics Library. This is a C++ library you can use directly in your code. Provided that you learn how to use it, you won’t have to use Denavit-Hartenberg or any other advanced mathematical knowledge.\n\nAnd, if you continue looking, you’ll soon find the Moveit library from ROS. This library allows you to specify your robot parameters (distances between joints, types of joints, etc) in a single XML representation called URDF: Unified Robot Description Format. Then Moveit will use the URDF to compute inverse kinematics.\n\nNot only that, it will also do complete path and motion planning for you! So you get even more stuff than what you’ve asked.\n\nSearching for libraries instead of learning the required mathematics is a great solution if you want to develop faster, especially for a small project or prototype. In fact, I’d advise you to use libraries even if you have the mathematical knowledge. This will save you some time in the short term anyway.\n\n## Where do you need mathematics\n\nEven if you can use libraries that hide the mathematical knowledge, mathematics are still everywhere in robotics.\n\nAnd if you want to really get serious in programming robots, you’ll eventually have to learn some maths, and know how to translate the maths into code.\n\nSometimes, a library will not do exactly what you want. Or, the license of the library will be a problem for your project. Or simply, the library is just an overkill for your application, because the mathematical problem you have is maybe not so hard to solve.\n\nHere is a non-exhaustive list of where you’ll find mathematics in robotics:\n\n• Kinematics: includes forward and inverse kinematics. This is especially useful for robotic arms.\n• Velocity and acceleration: if your robot is moving (which is probably the case), you’ll need to know how to compute a velocity from a position data, and how to compute an acceleration (and vice versa). To go even further, I could add the computation of the jerk, which represents the variation in acceleration and can be very important for smooth movements.\n• Path planning: which path should the robot take to reach the destination? How to optimize this path?\n• Motion planning: given a path, which velocity and acceleration should you apply to each joint of the robot at any time? And how do you ensure the max velocity constraints can be respected?\n• Angles: you’ll need to know how to switch from degrees to radians, how to switch from Euler angles to a quaternion, etc.\n• PID control: PIDs are often used to control motors and some actuators. This control system is full of maths.\n• Interpolation: given a set of points, you’ll often have to interpolate between them, using a polynomial expression for example.\n• Machine learning: well I won’t go into much details here, but as you can guess, this is also full of mathematics.\n• Image processing: here you’ll need at least solid geometry basics.\n• Logical reasoning: if you study logic in mathematics, this will help you a lot in programming, whether you program robots or not.\n\nIf you think this list is long, well, I’m sorry to say this is maybe less than 10% of all the maths you’ll come across when programming robots! And if you ask someone else with a different background than me, he/she will probably come up with a totally different list.\n\n## What robotics programmer do you want to be?\n\nAs you saw in this post, mathematics is everywhere in robotics. You’ll certainly have to learn some maths basics, but the overall required knowledge really depends on what you want to do.\n\nIf you want to develop web applications for robots, and not the robot itself, well in this case much of the mathematical knowledge listed before doesn’t apply to you.\n\nIf you want to develop low-level code, at the firmware level on micro-controllers, PID control and system control in general will be a must-have for you.\n\nAnd if you want to develop the core software of a robot – the decision algorithms to move the robot, how the robot moves its joints, etc – well in this case, you can’t escape mathematics. Of course you can (and should) start with some libraries to go easy and fast, but one day or another you’ll have to learn some maths.\n\nSo, before diving head first into robotics programming, make sure that you either have some knowledge in mathematics, or that you’re at least interested in maths.\n\nAnd, as you’ll see, it all seems very complicated until you finally understand and think: well that’s finally not so hard!\n\nWhat is your experience with mathematics and robotics? Which parts of mathematics did you struggled the most with?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9328392,"math_prob":0.6515616,"size":5877,"snap":"2019-51-2020-05","text_gpt3_token_len":1225,"char_repetition_ratio":0.14200579,"word_repetition_ratio":0.011846002,"special_character_ratio":0.20690829,"punctuation_ratio":0.11754685,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9684621,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T15:15:59Z\",\"WARC-Record-ID\":\"<urn:uuid:3ce25651-081d-469e-b163-bb9c98d24a77>\",\"Content-Length\":\"39608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81f23c6f-7478-41fd-b1d1-1a9f095a8543>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c0541a2-ae07-43e1-9a4b-ac641c49f121>\",\"WARC-IP-Address\":\"145.239.51.129\",\"WARC-Target-URI\":\"https://roboticsbackend.com/should-you-learn-mathematics-to-program-robots/\",\"WARC-Payload-Digest\":\"sha1:EM2REP672Y5CIA4YMV3RWK2B5QO4GOTJ\",\"WARC-Block-Digest\":\"sha1:BPUDSTXBEHE7BEEWEBYHJOMQD45EAQ4M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540531917.10_warc_CC-MAIN-20191211131640-20191211155640-00108.warc.gz\"}"}
http://promonet.ro/ecrustaceous1106953.html	[ "Check whether a directed graph is Eulerian\n\nAn Eulerian trail (or Eulerian path) is a path in a graph that visits every edge exactly once. Given a directed graph, check whether it has an Eulerian path or not.\n\nFind the minimum number of merge operations to make an array palindrome\n\nGiven a list of non-negative integers, find the minimum number of merge operations to make it a palindrome. A merge operation can only be performed on two adjacent elements. The result of a merge operation is that the two adjacent elements are replaced with their sum.\n\nConstruct a complete binary tree from its linked list representation\n\nGiven a linked list, construct a complete binary tree from it. Assume that the order of elements present in the linked list is the same as that in the complete tree’s array representation.\n\nFind read-write conflicts among given database transactions\n\nGiven a list of database transactions, find all read-write conflicts among them. Assume that there is no strict two-phase locking (Strict 2PL) protocol to prevent read-write conflicts.\n\nCheck if removing an edge can split a binary tree into two equal size trees\n\nGiven a binary tree, check if removing an edge can split it into two binary trees of equal size.\n\nIn-place merge two height-balanced BSTs\n\nGiven two height-balanced binary search trees, in-place merge them into a single balanced binary search tree. For each node of a height-balanced tree, the difference between its left and right subtree height is at most 1.\n\nPRO Scientific PRO-02-43340 Quick Connect, 43mm Diameter x 340mm\n\nGiven an array representing the parent-child relationship in a binary tree, find the tree’s height without building it. The parent-child relationship is defined by (A[i], i) for every index i in array A.\n\nDetermine whether two nodes lie on the same path in a binary tree\n\nGiven a binary tree and two tree pointers, x and y, write an efficient algorithm to check if they lie on the same root-to-leaf path in the binary tree. In other words, determine whether x is an ancestor of y, or x is a descendant of y.\n\nDetermine whether a BST is skewed from its preorder traversal\n\nGiven an array representing the preorder traversal of a BST, determine whether it represents a skewed BST or not. In a skewed BST, each node’s descendants are either smaller or larger than the node itself.\n\nFind total arrangements such that no two balls of the same color are together\n\nGiven r red, b blue, and g green balls, find the total number of arrangements in a row such that no two balls of the same color end up together.\n\nPrint all pairs of anagrams in a set of strings\n\nGiven a set of strings, print all pairs of anagrams together. Two strings, X and Y, are called anagrams if we can get a string Y by rearranging the letters of the string X and using all the characters of the string X exactly once.\n\nFind the smallest missing positive number from an unsorted array\n\nGiven an unsorted integer array, find the smallest missing positive integer in it." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85197043,"math_prob":0.9115706,"size":5402,"snap":"2022-05-2022-21","text_gpt3_token_len":1211,"char_repetition_ratio":0.11726566,"word_repetition_ratio":0.062126644,"special_character_ratio":0.20566458,"punctuation_ratio":0.119329385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9857515,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T17:11:46Z\",\"WARC-Record-ID\":\"<urn:uuid:9d3be27a-ca53-4131-9b1e-a12b77f48425>\",\"Content-Length\":\"59222\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5d0584f-b310-4c73-9163-e869bde6587f>\",\"WARC-Concurrent-To\":\"<urn:uuid:fca26c4b-4f6d-4a88-a013-59fafc9cc8b4>\",\"WARC-IP-Address\":\"89.40.19.45\",\"WARC-Target-URI\":\"http://promonet.ro/ecrustaceous1106953.html\",\"WARC-Payload-Digest\":\"sha1:L2RIJRZ7MJBM35YKFORKAKSTZ22NUK7S\",\"WARC-Block-Digest\":\"sha1:FWFCCIZSKZMACHOHLCXOYYCR33XYK346\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304572.73_warc_CC-MAIN-20220124155118-20220124185118-00471.warc.gz\"}"}
https://www.simplilearn.com/important-formulas-in-project-management-professional-pmp-article	[ "", null, "If you’re studying for the PMP® exam, you have probably already learned that you need to know some frequently used project management formulas—along with where to use them, how to use them, how to compute them, and most importantly, how to derive the values. In fact, many say that the most difficult part of the exam is learning all the PMP formulas.\n\n Are you a professional who is aspiring to be a project manager? Try answering this PMP Practice Preparation Questions and assess yourself.\n\nWe’ve put together a list of cost management formulas that you’ll need to know, along with a breakdown of how and when to use them.\n\nWatch a video on PMP Certification Training\n\n## Cost Management Knowledge Area PMP Formulas\n\nAlthough there are more than 25 project management formulas that you might need to tackle during the exam, there are a few cost management formulas that are very important, and which you’re likely to encounter more than once during the exam. Further, many of the calculation-based questions are based on these top 8 cost management formulas.\n\n 1 Cost Variance (CV) = Earned Value (EV) – Actual Cost (AC) 2 Schedule Variance (SV) = Earned Value (EV) – Planned Value (PV) 3 Cost Performance Index (CPI) = EV / AC 4 Schedule Performance Index (SPI) = EV / PV 5 EAC = AC + Bottom-up ETC 6 EAC = BAC / Cumulative CPI 7 EAC = AC + (BAC – EV) 8 EAC = AC + [BAC – EV / (Cumulative CPI ´ Cumulative SPI)]\n\n## Formulas 1 – 4\n\nTake a look at the first four formulas above and try to find out the commonalities in them. Not getting it yet? Yes, the easiest part is Earned Value (EV). It appears above in all the formulas, which means you have to derive the values of cost variance, schedule variance, cost performance index, and schedule performance index, keeping Earned Value in mind at first.\n\nEarned Value comes before any other value. Most importantly, if you are working cost-related questions; think about the actual cost with Earned Value. If the question is about a schedule, think about planned value along with the earned value. If it is a matter of variance, you need to subtract actual cost and planned value from earned value—depending on the situation.\n\nSimilarly, if it is a question of getting index values, actual cost and planned value will be divided from earned value. If it is a matter of cost performance index; actual cost will be divided from earned value and if it is scheduled performance index; the planned value will be divided from Earned Value. So, in all these circumstances, Earned Value plays a huge role, which is why it’s at the top of the list.\n\n## Formula 5\n\n 5 EAC = AC + Bottom-up ETC\n\nSince you will derive the estimate at completion, you need to focus on the actual costs incurred; only then can you obtain the future value you need for processing the project. See the above formula; it is used when the original estimate is fundamentally flawed. You use this formula to calculate an actual plus new estimate for the remaining work.\n\n## Formula 6\n\n 6 EAC = BAC / Cumulative CPI\n\nIf you were asked during the examination whether you are a good project manager, working on the project as planned—i.e., are you able to maintain positive values in both CPI and SPI—in this case, you would use the above-mentioned formula. This formula is used when the original estimation is met without any deviation.", null, "## Formula 7\n\n 7 EAC = AC + (BAC – EV)\n\nIf you find yourself in bad shape during the project execution and have incurred more money on the project than expected or planned, use this formula to calculate the estimate at completion value. Again, your actual cost will be used first.\n\n## Formula 8\n\n 8 EAC = AC + [BAC – EV / (Cumulative CPI ´ Cumulative SPI)]\n\nThis formula is used to calculate actual to date plus the remaining budget changed based on performance—used when we believe the current ratio is typical as planned. In order to meet the schedule as decided earlier, we calculate the EAC accordingly to meet that schedule.\n\n## Conclusion\n\nIn order to best decide which formula to use and where to make sure you’re reading the question carefully and consider the hypothetical situation, a project manager would be in when faced with this sort of problem.\n\nIn addition to these important cost management formulas, you’ll want to stay current with any updates on the Project Management Institute website. Simplilearn offers online training for the PMP certification that delivers hands-on projects in scope management, time and cost management, and risk assessment, designed to prepare you as much as possible for the PMP® exam.\n\nPMP is a registered mark of the Project Management Institute, Inc.", null, "Simplilearn" ]	[ null, "https://www.simplilearn.com/ice9/free_resources_article_thumb/The-Top-8-Formulas-to-Memorize-Before-Your-PMP-Exam.jpg", null, "https://www.simplilearn.com/ice9/assets/form_opacity.png", null, "https://www.simplilearn.com/important-formulas-in-project-management-professional-pmp-article", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92952704,"math_prob":0.84690446,"size":4477,"snap":"2019-51-2020-05","text_gpt3_token_len":1002,"char_repetition_ratio":0.13100827,"word_repetition_ratio":0.086683415,"special_character_ratio":0.22157694,"punctuation_ratio":0.086904764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9708482,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T11:33:11Z\",\"WARC-Record-ID\":\"<urn:uuid:f79dcb5f-5b3d-4bfc-bb23-bb0a2c862c27>\",\"Content-Length\":\"196420\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b201fbfe-8243-4ce9-ba0a-6db5c40ad871>\",\"WARC-Concurrent-To\":\"<urn:uuid:43898d40-5904-4c5d-8ece-4ac1c168e8da>\",\"WARC-IP-Address\":\"13.249.44.28\",\"WARC-Target-URI\":\"https://www.simplilearn.com/important-formulas-in-project-management-professional-pmp-article\",\"WARC-Payload-Digest\":\"sha1:MWAH7GWSU6O7BGSQIBN6XJJHOUV5K4XT\",\"WARC-Block-Digest\":\"sha1:6UXTSH5H2M4VT5U5HXJ4GICP66FSOIYZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540508599.52_warc_CC-MAIN-20191208095535-20191208123535-00276.warc.gz\"}"}
https://electronics.stackexchange.com/questions/338300/can-someone-explain-what-this-line-means-in-the-code-for-a-d-flip-flop-written-i	[ "# Can someone explain what this line means in the code for a D Flip Flop written in Verilog?\n\nmodule dflipflop (data ,clk , reset ,q);// Code for Asynchronous Positive Edge triggered D flip flop\ninput data, clk, reset ;\noutput q;\nreg q;\nalways @ ( posedge clk or negedge reset)\nif (~reset) begin //what does this block of code mean?\nq <= 1'b0 // particularly this line\nend else begin\nq <= data; // and this one\nend//\nendmodule\n\n\nI am new to learning Verilog, and I have to learn the codes for the various types of flip-flops. However, I haven't been able to understand this one entirely and what it means.\n\nFor instance, as the D flip-flop is positive edge triggered, hence we write posedge clock and it has an asynchronous reset input apart from the normal D input.\n\nBut, what I do not understand is the use of the <= operator in the highlighted lines and what is being done there. Does it mean q is being assigned a value less than zero? What does data variable refer to in the other line, and what exactly does the line mean as well?\n\n• The <= means it is synchronous to the clock. Nov 6, 2017 at 8:02\n• Okay. So it means the output 'q' of the flip-flop is formed synchronously and that if reset is zero, then output 'q' becomes zero. Nov 6, 2017 at 8:05\n• No, it doesn't mean that it's \"synchronous to the clock\". Look up \"non blocking assignment\". NBAs handle potential race conditions in your code.\n– EML\nNov 6, 2017 at 8:40\n\nFirst of all, the <= is a type of assigning value to the variable. It doesn't mean less than; instead, the value zero is assigned to variable q when the condition is met.\n\nSummary of Verilog syntax:\n\nhttp://ee.sut.ac.ir/People/Courses/142/Summary%20of%20Verilog%20Syntax.pdf\n\nif(~reset) - this means the condition is fulfilled when reset = 0 (~ is a bitwise negation operator, check the above link in the operator section), meaning when reset is zero, the statement q<=1'b0 is executed; otherwise, the input is passed to output which is given by q<=data. This is the functionality of a D flip-flop.\n\n<=\n\n\nis a non-blocking assignment operator used to execute the code or pass the values in concurrently (in parallel). This means that all the values are passed simultaneously, no matter the order in which the values are passed into the signal.\n\nThis is in contrast to\n\n=\n\n\noperator in which the values are passed sequentially.\n\nConsider this code\n\nmodule test;\n\nint a=4 , b=0 , c=0;\n\ninitial begin\na<=10;\nb<=a;\nc<=a;\n\n$display(\"a=%0d b=%0d c=%0d \\n\",a,b,c) ; //Here a=4 b=0 c=0 // and the values are assigned // to the variables in the next cycle. You can //check this out by using the always statement end initial begin a=10; b=a; c=a;$display(\"a=%0d b=%0d c=%0d \\n\",a,b,c) ; //Here a=10 b=10 c=10\n\nendmodule\n\n• Are you sure you mean \"sequentially\" and not \"serially\"? If it isn't parallel then it's serial. Nov 6, 2017 at 14:47\n• Well, Sequential execution refers to a statement being executed only after its preceding (or one or more key fields) has done so and serial execution strictly refers to execution after its perceding statement. So, if I'd have to be precise, I'd refer to it as time-sequential or or order-of-occurence sequential. Nov 7, 2017 at 4:58" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92004704,"math_prob":0.87447536,"size":938,"snap":"2023-40-2023-50","text_gpt3_token_len":229,"char_repetition_ratio":0.10171306,"word_repetition_ratio":0.0,"special_character_ratio":0.24413647,"punctuation_ratio":0.119170986,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9940024,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T00:40:23Z\",\"WARC-Record-ID\":\"<urn:uuid:bcf4cf16-6745-4860-a530-48678a0dd903>\",\"Content-Length\":\"179579\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50871105-7d4b-4175-be74-8e22e266c3b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:899b0311-851f-4cec-bc8b-95ef8a50f260>\",\"WARC-IP-Address\":\"172.64.144.30\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/338300/can-someone-explain-what-this-line-means-in-the-code-for-a-d-flip-flop-written-i\",\"WARC-Payload-Digest\":\"sha1:7QDO5J4W22HGBYZWZCGNMRLIXQWLHCV4\",\"WARC-Block-Digest\":\"sha1:LPDBFOUQQXV6TIRSCPQ7CHCYQOCSWDLW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100164.15_warc_CC-MAIN-20231130000127-20231130030127-00338.warc.gz\"}"}
https://docs.appian.com/suite/help/19.4/fnc_trigonometry_radians.html	[ "Converts the measure of the specified angle from degrees to radians.\n\n## Syntax\n\nangle_in_degrees: (Decimal) An angle measure that will be converted into radians.\n\n## Returns\n\nDecimal\n\nThis function can only be used for values between 0 and 2π (~6.286).\n\n## Examples\n\nYou can experiment with this function in the test box below.\n\nTest Input\n\n`radians(180)` returns `3.141592653589793`\n\nOpen in Github Built: Thu, Oct 14, 2021 (02:42:55 PM)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65154,"math_prob":0.81135464,"size":379,"snap":"2021-43-2021-49","text_gpt3_token_len":94,"char_repetition_ratio":0.12533334,"word_repetition_ratio":0.0,"special_character_ratio":0.27176782,"punctuation_ratio":0.104477614,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9626472,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T15:35:43Z\",\"WARC-Record-ID\":\"<urn:uuid:f451699b-e768-4a1d-8fc2-0939be48246c>\",\"Content-Length\":\"20991\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:539ea6d1-b0e2-4163-8561-634f76afe119>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8046046-be5c-43d8-aa7a-142435a73630>\",\"WARC-IP-Address\":\"34.236.82.115\",\"WARC-Target-URI\":\"https://docs.appian.com/suite/help/19.4/fnc_trigonometry_radians.html\",\"WARC-Payload-Digest\":\"sha1:KAR4K2K66YTFYE6ZVQYCOUBAXRLWNTMB\",\"WARC-Block-Digest\":\"sha1:YVAC2Z6J6HDQMJUGNGK6G5GPVBRC7WT3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585424.97_warc_CC-MAIN-20211021133500-20211021163500-00485.warc.gz\"}"}
https://sketches89.neocities.org/view-find-the-gradient-vector-field-f-of-f-and-sketch-it-f-x-y-5-x2-y2-png.html	[ "", null, "View Find The Gradient Vector Field ∇F Of F And Sketch It. F(X, Y) = 5 X2 + Y2 PNG\n\n# View Find The Gradient Vector Field ∇F Of F And Sketch It. F(X, Y) = 5 X2 + Y2 PNG\n\nF(x,y)start bold text, f, end bold text, left parenthesis, x, comma, y, right parenthesis is called a conservative vector field if it satisfies any one.\n\nView Find The Gradient Vector Field ∇F Of F And Sketch It. F(X, Y) = 5 X2 + Y2 PNG. Given 𝑓(𝑥, 𝑦) = 𝑥 + 2𝑦, find 𝐅(𝑥, 𝑦) = ∇𝑓 and sketch it along with the contour map of 𝑓. Find the length of ∇f(x,y) using equation (2).", null, "Out Of Distribution Generalization Via Risk Extrapolation Rex Arxiv Vanity from media.arxiv-vanity.com If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. Find the corresponding function f. F(x, y) = 5sqrt(x^2 + y^2).\n\n### On january 1, 2011, frog corporation sold a \\$2,000,000, 10 percent bond issue (8.5 percent market.\n\nLet f be any vector field of the form f(x,y, z) = f (x) i+g(y) j+h(z)k and let g be any vector field of the form g(x,y, z) = f (y, z) i+g(x, z) j+h(x,y)k. The gradient is also shown. Iii 5 iv 5 4 find the gradient vector field of f. %matplotlib inline import matplotlib.pyplot as plt import sympy as sp import numpy as np." ]	[ null, "https://sstatic1.histats.com/0.gif", null, "https://media.arxiv-vanity.com/render-output/3377291/figures/grad_vector_fields.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7867376,"math_prob":0.9972508,"size":853,"snap":"2022-05-2022-21","text_gpt3_token_len":297,"char_repetition_ratio":0.123674914,"word_repetition_ratio":0.03550296,"special_character_ratio":0.3282532,"punctuation_ratio":0.14666666,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996219,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T11:14:51Z\",\"WARC-Record-ID\":\"<urn:uuid:74a1bf9c-a7eb-4700-8bd7-ae196a6b36b9>\",\"Content-Length\":\"17636\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93240a82-2490-4bd7-8b5a-7fbb1c396e64>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d3a4190-75b9-4767-b03a-eedad4e341e6>\",\"WARC-IP-Address\":\"198.51.233.2\",\"WARC-Target-URI\":\"https://sketches89.neocities.org/view-find-the-gradient-vector-field-f-of-f-and-sketch-it-f-x-y-5-x2-y2-png.html\",\"WARC-Payload-Digest\":\"sha1:KZUH6S74A4KZ7476BKOSHEASUDVC2BV3\",\"WARC-Block-Digest\":\"sha1:YPOG3WCMP6Q3KXZV7HMYLVZWVII4PBTM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662558015.52_warc_CC-MAIN-20220523101705-20220523131705-00451.warc.gz\"}"}
https://lists.isocpp.org/std-proposals/2021/10/3192.php	[ "Constructor ambiguity: Uniform initialization vs. initializer lists\n\nFrom: Gawain Bolton <gawain.bolton_at_[hidden]>\nDate: Tue, 05 Oct 2021 14:11:40 +0200\nHello,\n\nI would like to report what I will call \"constructor ambiguity\" for\nclasses which have multiple constructors, one of which have takes an\ninitializer list.\n\nThe ambiguity is for us humans, not the compiler, and is due to the two\nthings:\n1. The same syntax is used for uniform initialization and\ninitializer lists.\n2. Any constructor taking an initializer list is preferred when\nbraces are used.\n\nFor example:\n\nstd::string sa(32, 'A'); // Creates a string of 32 characters i.e. \"AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\"\nstd::string sb{32, 'A'}; // Creates a string of 2 characters i.e. \" A\"\n\nstd::vector<int> va(10, -1); // Creates a vector of 10 elements, each one initialized to -1\nstd::vector<int> vb{10, -1}; // Creates a vector of 2 elements (10 and -1)\n\nThe code looks very similar and yet the results are very different.\nThis is unfortunate because uniform initialization is preferred\nespecially as it can avoid bugs due to unintential casts and\ntruncation/narrowing of values.\n\nI find this ambiguity unnecessary as if the intent is to provide an\ninitializer list, then this can be done explicitly:\nstd::vector<int> vb1({10, -1}); // Creates a vector of 2 elements (10 and -1)\nstd::vector<int> vb2{{10, -1}}; // Creates a vector of 2 elements (10 and -1)\n\nThis preference to use a constructor taking an initializer list can\neasily break code if such a constructor is added at a later date. For\nexample:\n\nstruct A {\nexplicit A(int);\n};\n\nA a{2}; // Create an object a of type A\n\nThen later on an initializer list constructor is added:\n\nstruct A {\nexplicit A(int);\nA(const std::initializer_list<int>&)\n};\n\nNow the call to create object 'a' will use the constructor taking an\ninitializer list which, of course, can do something completely\ndifferent. It could be extremely difficult to find why there was such\na change in behaviour.\n\nI apologize but I do not know if this has already been mentioned or\naddressed. If not, what do you think of this? Is this considered to\nbe one of the many idioms of C++?\n\nGawain Bolton" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86593235,"math_prob":0.79303724,"size":4072,"snap":"2022-05-2022-21","text_gpt3_token_len":1019,"char_repetition_ratio":0.15240905,"word_repetition_ratio":0.9764706,"special_character_ratio":0.26743615,"punctuation_ratio":0.15266272,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9557001,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T01:29:23Z\",\"WARC-Record-ID\":\"<urn:uuid:07dc8f3c-2ab2-4346-b76f-d1f47d0f60df>\",\"Content-Length\":\"8132\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7487ad4-b838-4376-a3d2-48288fbfcc69>\",\"WARC-Concurrent-To\":\"<urn:uuid:7cddffce-9572-41da-931e-673fc9888fb3>\",\"WARC-IP-Address\":\"165.227.99.241\",\"WARC-Target-URI\":\"https://lists.isocpp.org/std-proposals/2021/10/3192.php\",\"WARC-Payload-Digest\":\"sha1:VSPNHZPTK64OZL6VCHBN4UZHVWV6DCTP\",\"WARC-Block-Digest\":\"sha1:ZPXCVGBM4QNO2SNUTUCGGVWUN4CYZZVZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301217.83_warc_CC-MAIN-20220119003144-20220119033144-00041.warc.gz\"}"}
http://docs.go-mono.com/monodoc.ashx?link=M%3ASystem.Double.Equals(System.Object)	[ "System.Double.Equals Method\n\nReturns a value indicating whether this instance is equal to a specified object.\n\n## Syntax\n\npublic override bool Equals (object obj)\n\n#### Parameters\n\nobj\nAn object to compare with this instance.\n\n#### Returns\n\ntrue if obj is an instance of double and equals the value of this instance; otherwise, false.\n\n## Remarks\n\nThe double.Equals(double) method should be used with caution, because two apparently equivalent values can be unequal due to the differing precision of the two values. The following example reports that the double value .3333 and the double returned by dividing 1 by 3 are unequal.\n\ncode reference: System.Double.Epsilon#3\n\nFor alternatives to calling the double.Equals(object) method, see the documentation for the double.Equals(double) overload.\n\nNote:\n\nBecause double.Epsilon defines the minimum expression of a positive value whose range is near zero, the margin of difference between two similar values must be greater than double.Epsilon. Typically, it is many times greater than double.Epsilon.\n\nThe precision of floating-point numbers beyond the documented precision is specific to the implementation and version of the .NET Framework. Consequently, a comparison of two particular numbers might change between versions of the .NET Framework because the precision of the numbers' internal representation might change.\n\nIf two double.NaN values are tested for equality by calling the double.Equals(double) method, the method returns true. However, if two double.NaN values are tested for equality by using the equality operator, the operator returns false. When you want to determine whether the value of a double is not a number (NaN), an alternative is to call the double.IsNaN(double) method.\n\n## Requirements\n\nNamespace: System\nAssembly: mscorlib (in mscorlib.dll)\nAssembly Versions: 1.0.5000.0, 2.0.0.0, 4.0.0.0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83339995,"math_prob":0.946218,"size":1483,"snap":"2019-51-2020-05","text_gpt3_token_len":281,"char_repetition_ratio":0.15956728,"word_repetition_ratio":0.037209302,"special_character_ratio":0.18745786,"punctuation_ratio":0.13768116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9883825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T18:09:02Z\",\"WARC-Record-ID\":\"<urn:uuid:d55487b6-1380-4ac8-b48c-dea0edceb4d6>\",\"Content-Length\":\"22923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:730f7214-34c5-49a5-b9a2-d94866d59f96>\",\"WARC-Concurrent-To\":\"<urn:uuid:6dcc15d6-3dcf-4553-9a15-a547a9b488d0>\",\"WARC-IP-Address\":\"13.68.229.189\",\"WARC-Target-URI\":\"http://docs.go-mono.com/monodoc.ashx?link=M%3ASystem.Double.Equals(System.Object)\",\"WARC-Payload-Digest\":\"sha1:6K7JSJOWQ433ITUGWNHK4JEHGGJWQCGK\",\"WARC-Block-Digest\":\"sha1:Y2W56WJO6YDJMWHNOYMTJRSG2KMFQVX3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540528490.48_warc_CC-MAIN-20191210180555-20191210204555-00203.warc.gz\"}"}
https://socratic.org/questions/how-do-you-calculate-r-squared-by-hand#149476	[ "# How do you calculate r squared by hand?\n\nMay 31, 2015\n\nAssuming this is a general question and not a reference to some undeclared statistical equation,\n\nand assuming you know how to multiply two numbers together by hand,\n\nthen $r$ squared (often written ${r}^{2}$) is simply\n$\\textcolor{w h i t e}{\\text{XXXXX}}$$r \\times r$ for whatever the value of $r$ is\n\nFor example if $r = 16$\nthen $r$ squared (or ${r}^{2}$) $= 16 \\times 16 = 256$\n\nHowever I suspect you had some specific statistical relationship in mind; please resubmit with explicit references if this is the case.\n\n${r}^{2} = 1 - \\frac{S {S}_{E r r}}{S {S}_{T o t}}$\n\n#### Explanation:\n\nThe $S {S}_{E r r}$ or the sum of squares residuals is:\n$\\setminus \\sum {y}_{i}^{2} - {B}_{0} \\setminus \\sum {y}_{i} - {B}_{1} \\setminus \\sum {x}_{i} {y}_{i}$\nor\nsimply the square of the value of the residuals. The residual value is difference between the obtained y-value and the expected y-value. The expected y-value is the calculated value from the equation of line/plane.\n\nFor example, for a system with 1 unknown parameter/variable x, the calculated y-value would be the sum of ${B}_{0} \\mathmr{and} {B}_{1} x$ (i.e. $Y = {B}_{0} + {B}_{1} x$).\n\nFor a system with 2 unknown parameters/variables, ${x}_{1}$ and ${x}_{2}$, the calculated y-value would be the sum of ${B}_{0}$, ${B}_{1} x$, and ${B}_{2} {x}_{2}$ (i.e. $Y = {B}_{0} + {B}_{1} {x}_{1} + {B}_{2} {x}_{2}$).\n\nAnd in general, $Y = {B}_{0} + {B}_{1} {x}_{1} + {B}_{2} {x}_{2} + {B}_{3} {x}_{3} + {B}_{4} {x}_{4} + \\ldots + {B}_{n} {x}_{n}$\n\nFurthermore, the $S {S}_{T o t} = \\setminus \\sum {y}_{i}^{2} - \\frac{{\\left(\\sum {y}_{i}\\right)}^{2}}{n}$ where $n$ is the number of observations or trials." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9078535,"math_prob":1.0000056,"size":848,"snap":"2021-43-2021-49","text_gpt3_token_len":177,"char_repetition_ratio":0.112559244,"word_repetition_ratio":0.0,"special_character_ratio":0.18867925,"punctuation_ratio":0.088607594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000073,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T10:12:47Z\",\"WARC-Record-ID\":\"<urn:uuid:9e3314b7-a095-4644-85e4-fb3684be8e03>\",\"Content-Length\":\"36266\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b70f058-60c3-40fd-a7c8-4a8fa8ff31c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:e2f4438b-0254-40d0-934b-f805b32b75b2>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-calculate-r-squared-by-hand#149476\",\"WARC-Payload-Digest\":\"sha1:CDCYHTMQIBMALEZYH2ROYFU52JB47THW\",\"WARC-Block-Digest\":\"sha1:N43DUOD3OJO4D5AFDBEAT7O46XP5GH7I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585916.29_warc_CC-MAIN-20211024081003-20211024111003-00147.warc.gz\"}"}